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a ENNGG> orncite\ ffirs Pozar September 26, 2011 18:2 This page is intentionally left blank ffirs Pozar September 26, 2011 18:2 Microwave Engineering ffirs Pozar September 26, 2011 18:2 This page is intentionally left blank ffirs Pozar September 26, 2011 18:2 Microwave Engineering Fourth Edition David M. Pozar University of Massachusetts at Amherst John Wiley & Sons, Inc. ffirs Pozar September 30, 2011 8:23 Vice President & Executive Publisher Don Fowley Associate Publisher Dan Sayre Content Manager Lucille Buonocore Senior Production Editor Anna Melhorn Marketing Manager Christopher Ruel Creative Director Harry Nolan Senior Designer Jim O’Shea Production Management Services Sherrill Redd of AptaraEditorial Assistant Charlotte Cerf Lead Product Designer Tom Kulesa Cover Designer Jim O’Shea This book was set in Times Roman 10/12 by Aptara R/circlecopyrt, Inc. and printed and bound by Hamilton Printing. The cover was printed by Hamilton Printing. Copyright C/circlecopyrt2012, 2005, 1998 by John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning orotherwise, except as permitted under Sections 107 or 108 of the 1976 United States CopyrightAct, without either the prior written permission of the Publisher, or authorization throughpayment of the appropriate per-copy fee to the Copyright Clearance Center, Inc. 222 Rosewood Drive, Danvers, MA 01923, website www.copyright.com. Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley &Sons, Inc., 111 River Street, Hoboken, NJ 07030-5774, (201)748-6011, fax (201)748-6008,website http://www.wiley.com/go/permissions. Founded in 1807, John Wiley & Sons, Inc. has been a valued source of knowledge and understanding for more than 200 years, helping people around the world meet their needs andfulfill their aspirations. Our company is built on a foundation of principles that includeresponsibility to the communities we serve and where we live and work. In 2008, welaunched a Corporate Citizenship Initiative, a global effort to address the environmental,social, economic, and ethical challenges we face in our business. Among the issues we areaddressing are carbon impact, paper specifications and procurement, ethical conduct withinour business and among our vendors, and community and charitable support. For moreinformation, please visit our website: www.wiley.com/go/citizenship. Evaluation copies are provided to qualified academics and professionals for review purposes only, for use in their courses during the next academic year. These copies are licensed and may not be sold or transferred to a third party. Upon completion of the review period, pleasereturn the evaluation copy to Wiley. Return instructions and a free of charge return shippinglabel are available at www.wiley.com/go/returnlabel. Outside of the United States, pleasecontact your local representative. Library of Congress Cataloging-in-Publication Data Pozar, David M. Microwave engineering/David M. Pozar.—4th ed. p. cm. Includes bibliographical references and index.ISBN 978-0-470-63155-3 (hardback : acid free paper)1. Microwaves. 2. Microwave devices. 3. Microwave circuits. I. Title. TK7876.P69 2011 621.381’3—dc23 2011033196 Printed in the United States of America 1 0987654321 fpref Pozar September 9, 2011 21:39 Preface The continuing popularity of Microwave Engineering is gratifying. I have received many letters and emails from students and teachers from around the world with positive com- ments and suggestions. I think one reason for its success is the emphasis on the funda- mentals of electromagnetics, wave propagation, network analysis, and design principlesas applied to modern RF and microwave engineering. As I have stated in earlier editions, I have tried to avoid the handbook approach in which a large amount of information is presented with little or no explanation or context, but a considerable amount of materialin this book is related to the design of specific microwave circuits and components, for both practical and motivational value. I have tried to base the analysis and logic behind these designs on first principles, so the reader can see and understand the process of ap-plying fundamental concepts to arrive at useful results. The engineer who has a firm grasp of the basic concepts and principles of microwave engineering and knows how these can be applied toward practical problems is the engineer who is the most likely to be rewardedwith a creative and productive career. For this new edition I again solicited detailed feedback from teachers and readers for their thoughts about how the book should be revised. The most common requests werefor more material on active circuits, noise, nonlinear effects, and wireless systems. This edition, therefore, now has separate chapters on noise and nonlinear distortion, and ac- tive devices. In Chapter 10, the coverage of noise has been expanded, along with more material on intermodulation distortion and related nonlinear effects. For Chapter 11, on active devices, I have added updated material on bipolar junction and field effect transis-tors, including data for a number of commercial devices (Schottky and PIN diodes, and Si, GaAs, GaN, and SiGe transistors), and these sections have been reorganized and rewritten. Chapters 12 and 13 treat active circuit design, and discussions of differential amplifiers,inductive degeneration for nMOS amplifiers, and differential FET and Gilbert cell mix- ers have been added. In Chapter 14, on RF and microwave systems, I have updated and added new material on wireless communications systems, including link budget, link mar-gin, digital modulation methods, and bit error rates. The section on radiation hazards has been updated and rewritten. Other new material includes a section on transients on trans- mission lines (material that was originally in the first edition, cut from later editions, andnow brought back by popular demand), the theory of power waves, a discussion of higher order modes and frequency effects for microstrip line, and a discussion of how to deter- mine unloaded Qfrom resonator measurements. This edition also has numerous new or revised problems and examples, including several questions of the “open-ended” variety. Material that has been cut from this edition includes the quasi-static numerical analysis of microstrip line and some material related to microwave tubes. Finally, working from the original source files, I have made hundreds of corrections and rewrites of the original text. v fpref Pozar September 9, 2011 21:39 vi Preface Today, microwave and RF technology is more pervasive than ever. This is especially true in the commercial sector, where modern applications include cellular telephones, smartphones, 3G and WiFi wireless networking, millimeter wave collision sensors for ve- hicles, direct broadcast satellites for radio, television, and networking, global positioningsystems, radio frequency identification tagging, ultra wideband radio and radar systems, and microwave remote sensing systems for the environment. Defense systems continue to rely heavily on microwave technology for passive and active sensing, communications, andweapons control systems. There should be no shortage of challenging problems in RF and microwave engineering in the foreseeable future, and there will be a clear need for engi- neers having both an understanding of the fundamentals of microwave engineering and the creativity to apply this knowledge to problems of practical interest. Modern RF and microwave engineering predominantly involves distributed circuit analysis and design, in contrast to the waveguide and field theory orientation of earlier generations. The majority of microwave engineers today design planar components and in- tegrated circuits without direct recourse to electromagnetic analysis. Microwave computer-aided design (CAD) software and network analyzers are the essential tools of today’s microwave engineer, and microwave engineering education must respond to this shift in emphasis to network analysis, planar circuits and components, and active circuit design.Microwave engineering will always involve electromagnetics (many of the more sophisti- cated microwave CAD packages implement rigorous field theory solutions), and students will still benefit from an exposure to subjects such as waveguide modes and couplingthrough apertures, but the change in emphasis to microwave circuit analysis and design is clear. This text is written for a two-semester course in RF and microwave engineering for seniors or first-year graduate students. It is possible to use Microwave Engineering with or without an electromagnetics emphasis. Many instructors today prefer to focus on circuit analysis and design, and there is more than enough material in Chapters 2, 4–8, and 10–14 for such a program with minimal or no field theory requirement. Some instructors may wish to begin their course with Chapter 14 on systems in order to provide some motivationalcontext for the study of microwave circuit theory and components. This can be done, but some basic material on noise from Chapter 10 may be required. Two important items that should be included in a successful course on microwave engineering are the use of CAD simulation software and a microwave laboratory experi- ence. Providing students with access to CAD software allows them to verify results of the design-oriented problems in the text, giving immediate feedback that builds confidence andmakes the effort more rewarding. Because the drudgery of repetitive calculation is elimi- nated, students can easily try alternative approaches and explore problems in more detail. The effect of line losses, for example, is explored in several examples and problems; thiswould be effectively impossible without the use of modern CAD tools. In addition, class- room exposure to CAD tools provides useful experience upon graduation. Most of the commercially available microwave CAD tools are very expensive, but several manufactur-ers provide academic discounts or free “student versions” of their products. Feedback from reviewers was almost unanimous, however, that the text should not emphasize a particular software product in the text or in supplementary materials. A hands-on microwave instructional laboratory is expensive to equip but provides the best way for students to develop an intuition and physical feeling for microwave phenom-ena. A laboratory with the first semester of the course might cover the measurement of microwave power, frequency, standing wave ratio, impedance, and scattering parameters, as well as the characterization of basic microwave components such as tuners, couplers,resonators, loads, circulators, and filters. Important practical knowledge about connectors, waveguides, and microwave test equipment will be acquired in this way. A more advanced fpref Pozar October 5, 2011 10:43 Preface vii laboratory session can consider topics such as noise figure, intermodulation distortion, and mixing. Naturally, the type of experiments that can be offered is heavily dependent on the test equipment that is available. Additional resources for students and instructors are available on the Wiley website. These include PowerPoint slides, a suggested laboratory manual, and an online solution manual for all problems in the text (available to qualified instructors, who may apply for access at the website http://he-cda.wiley.com/wileycda/). ACKNOWLEDGMENTS It is a pleasure to acknowledge the many students, readers, and teachers who have used the first three editions of Microwave Engineering , and have written with comments, praise, and suggestions. I would also like to thank my colleagues in the microwave engineering group at the University of Massachusetts at Amherst for their support and collegiality over many years. In addition I would like to thank Bob Jackson (University of Massachusetts) for suggestions on MOSFET amplifiers and related material; Juraj Bartolic (University of Zagreb) for the simplified derivation of the µ-parameter stability criteria; and Jussi Rahola (Nokia Research Center) for his discussions of power waves. I am also grateful to the following people for providing new photographs for this edition: Kent Whitney and Chris Koh of Millitech Inc., Tom Linnenbrink and Chris Hay of Hittite Microwave Corp., PhilBeucler and Lamberto Raffaelli of LNX Corp., Michael Adlerstein of Raytheon Company, Bill Wallace of Agilent Technologies Inc., Jim Mead of ProSensing Inc., Bob Jackson and B. Hou of the University of Massachusetts, J. Wendler of M/A-COM Inc., MohamedAbouzahra of Lincoln Laboratory, and Dev Gupta, Abbie Mathew, and Salvador Rivera of Newlans Inc. I would also like to thank Sherrill Redd, Philip Koplin, and the staff at Aptara, Inc. for their professional efforts during production of this book. Also, thanks toBen for help with PhotoShop. David M. Pozar Amherst fpref Pozar September 9, 2011 21:39 This page is intentionally left blank ftoc Pozar September 9, 2011 21:38 Contents 1ELECTROMAGNETICTHEORY 1 1.1 Introduction to Microwave Engineering 1 Applications of Microwave Engineering 2 A Short History of Microwave Engineering 4 1.2 Maxwell’s Equations 61.3 Fields in Media and Boundary Conditions 10 Fields at a General Material Interface 12 Fields at a Dielectric Interface 14 Fields at the Interface with a Perfect Conductor (Electric Wall) 14 The Magnetic Wall Boundary Condition 15 The Radiation Condition 15 1.4 The Wave Equation and Basic Plane Wave Solutions 15 The Helmholtz Equation 15 Plane Waves in a Lossless Medium 16 Plane Waves in a General Lossy Medium 17 Plane Waves in a Good Conductor 19 1.5 General Plane Wave Solutions 20 Circularly Polarized Plane Waves 24 1.6 Energy and Power 25 Power Absorbed by a Good Conductor 27 1.7 Plane Wave Reflection from a Media Interface 28 General Medium 28 Lossless Medium 30 Good Conductor 31 Perfect Conductor 32The Surface Impedance Concept 33 1.8 Oblique Incidence at a Dielectric Interface 35 Parallel Polarization 36 Perpendicular Polarization 37 Total Reflection and Surface Waves 38 1.9 Some Useful Theorems 40 The Reciprocity Theorem 40 Image Theory 42 ix ftoc Pozar September 9, 2011 21:38 xContents 2TRANSMISSIONLINETHEORY 48 2.1 The Lumped-Element Circuit Model for a Transmission Line 48 Wave Propagation on a Transmission Line 50 The Lossless Line 51 2.2 Field Analysis of Transmission Lines 51 Transmission Line Parameters 51 The Telegrapher Equations Derived from Field Analysis of a Coaxial Line 54 Propagation Constant, Impedance, and Power Flow for the Lossless Coaxial Line 56 2.3 The Terminated Lossless Transmission Line 56 Special Cases of Lossless Terminated Lines 59 2.4 The Smith Chart 63 The Combined Impedance–Admittance Smith Chart 67 The Slotted Line 68 2.5 The Quarter-Wave Transformer 72 The Impedance Viewpoint 72 The Multiple-Reflection Viewpoint 74 2.6 Generator and Load Mismatches 76 Load Matched to Line 77 Generator Matched to Loaded Line 77 Conjugate Matching 77 2.7 Lossy Transmission Lines 78 The Low-Loss Line 79 The Distortionless Line 80 The Terminated Lossy Line 81The Perturbation Method for Calculating Attenuation 82 The Wheeler Incremental Inductance Rule 83 2.8 Transients on Transmission Lines 85 Reflection of Pulses from a Terminated Transmission Line 86 Bounce Diagrams for Transient Propagation 87 3TRANSMISSIONLINESANDWAVEGUIDES 95 3.1 General Solutions for TEM, TE, and TM Waves 96 TEM Waves 98 TE Waves 100 TM Waves 100 Attenuation Due to Dielectric Loss 101 3.2 Parallel Plate Waveguide 102 TEM Modes 103 TM Modes 104 TE Modes 107 3.3 Rectangular Waveguide 110 TE Modes 110 TM Modes 115 TEm0Modes of a Partially Loaded Waveguide 119 3.4 Circular Waveguide 121 TE Modes 122 TM Modes 125 3.5 Coaxial Line 130 TEM Modes 130 Higher Order Modes 131 ftoc Pozar September 9, 2011 21:38 Contents xi 3.6 Surface Waves on a Grounded Dielectric Sheet 135 TM Modes 135 TE Modes 137 3.7 Stripline 141 Formulas for Propagation Constant, Characteristic Impedance, and Attenuation 141 An Approximate Electrostatic Solution 144 3.8 Microstrip Line 147 Formulas for Effective Dielectric Constant, Characteristic Impedance, and Attenuation 148Frequency-Dependent Effects and Higher Order Modes 150 3.9 The Transverse Resonance Technique 153 TE 0nModes of a Partially Loaded Rectangular Waveguide 153 3.10 Wave Velocities and Dispersion 154 Group Velocity 155 3.11 Summary of Transmission Lines and Waveguides 157 Other Types of Lines and Guides 158 4MICROWAVENETWORKANALYSIS 165 4.1 Impedance and Equivalent Voltages and Currents 166 Equivalent V oltages and Currents 166 The Concept of Impedance 170 Even and Odd Properties of Z(ω)and/Gamma1(ω) 173 4.2 Impedance and Admittance Matrices 174 Reciprocal Networks 175 Lossless Networks 177 4.3 The Scattering Matrix 178 Reciprocal Networks and Lossless Networks 181 A Shift in Reference Planes 184Power Waves and Generalized Scattering Parameters 185 4.4 The Transmission (ABCD) Matrix 188 Relation to Impedance Matrix 191 Equivalent Circuits for Two-Port Networks 191 4.5 Signal Flow Graphs 194 Decomposition of Signal Flow Graphs 195 Application to Thru-Reflect-Line Network Analyzer Calibration 197 4.6 Discontinuities and Modal Analysis 203 Modal Analysis of an H-Plane Step in Rectangular Waveguide 203 4.7 Excitation of Waveguides—Electric and Magnetic Currents 210 Current Sheets That Excite Only One Waveguide Mode 210 Mode Excitation from an Arbitrary Electric or Magnetic Current Source 212 4.8 Excitation of Waveguides—Aperture Coupling 215 Coupling Through an Aperture in a Transverse Waveguide Wall 218 Coupling Through an Aperture in the Broad Wall of a Waveguide 220 ftoc Pozar September 9, 2011 21:38 xii Contents 5IMPEDANCEMATCHINGANDTUNING 228 5.1 Matching with Lumped Elements ( LNetworks) 229 Analytic Solutions 230 Smith Chart Solutions 231 5.2 Single-Stub Tuning 234 Shunt Stubs 235 Series Stubs 238 5.3 Double-Stub Tuning 241 Smith Chart Solution 242 Analytic Solution 245 5.4 The Quarter-Wave Transformer 2465.5 The Theory of Small Reflections 250 Single-Section Transformer 250 Multisection Transformer 251 5.6 Binomial Multisection Matching Transformers 2525.7 Chebyshev Multisection Matching Transformers 256 Chebyshev Polynomials 257 Design of Chebyshev Transformers 258 5.8 Tapered Lines 261 Exponential Taper 262 Triangular Taper 263 Klopfenstein Taper 264 5.9 The Bode–Fano Criterion 266 6MICROWAVERESONATORS 272 6.1 Series and Parallel Resonant Circuits 272 Series Resonant Circuit 272 Parallel Resonant Circuit 275 Loaded and Unloaded Q277 6.2 Transmission Line Resonators 278 Short-Circuited λ/2 Line 278 Short-Circuited λ/4 Line 281 Open-Circuited λ/2 Line 282 6.3 Rectangular Waveguide Cavity Resonators 284 Resonant Frequencies 284 Unloaded Qof the TE 10/lscriptMode 286 6.4 Circular Waveguide Cavity Resonators 288 Resonant Frequencies 289 Unloaded Qof the TE nm/lscriptMode 291 6.5 Dielectric Resonators 293 Resonant Frequencies of TE 01δMode 294 6.6 Excitation of Resonators 297 The Coupling Coefficient and Critical Coupling 298 A Gap-Coupled Microstrip Resonator 299An Aperture-Coupled Cavity 302 Determining Unloaded Qfrom Two-Port Measurements 305 6.7 Cavity Perturbations 306 Material Perturbations 306 Shape Perturbations 309 ftoc Pozar September 9, 2011 21:38 Contents xiii 7POWERDIVIDERSANDDIRECTIONALCOUPLERS 317 7.1 Basic Properties of Dividers and Couplers 317 Three-Port Networks (T-Junctions) 318 Four-Port Networks (Directional Couplers) 320 7.2 The T-Junction Power Divider 324 Lossless Divider 324 Resistive Divider 326 7.3 The Wilkinson Power Divider 328 Even-Odd Mode Analysis 328 Unequal Power Division and N-Way Wilkinson Dividers 332 7.4 Waveguide Directional Couplers 333 Bethe Hole Coupler 334 Design of Multihole Couplers 338 7.5 The Quadrature (90◦) Hybrid 343 Even-Odd Mode Analysis 344 7.6 Coupled Line Directional Couplers 347 Coupled Line Theory 347 Design of Coupled Line Couplers 351 Design of Multisection Coupled Line Couplers 356 7.7 The Lange Coupler 3597.8 The 180 ◦Hybrid 362 Even-Odd Mode Analysis of the Ring Hybrid 364 Even-Odd Mode Analysis of the Tapered Coupled Line Hybrid 367 Waveguide Magic-T 371 7.9 Other Couplers 372 8MICROWAVEFILTERS 380 8.1 Periodic Structures 381 Analysis of Infinite Periodic Structures 382 Terminated Periodic Structures 384 k-βDiagrams and Wave Velocities 385 8.2 Filter Design by the Image Parameter Method 388 Image Impedances and Transfer Functions for Two-Port Networks 388 Constant-k Filter Sections 390 m-Derived Filter Sections 393 Composite Filters 396 8.3 Filter Design by the Insertion Loss Method 399 Characterization by Power Loss Ratio 399 Maximally Flat Low-Pass Filter Prototype 402 Equal-Ripple Low-Pass Filter Prototype 404Linear Phase Low-Pass Filter Prototypes 406 8.4 Filter Transformations 408 Impedance and Frequency Scaling 408 Bandpass and Bandstop Transformations 411 ftoc Pozar September 9, 2011 21:38 xiv Contents 8.5 Filter Implementation 415 Richards’ Transformation 416 Kuroda’s Identities 416 Impedance and Admittance Inverters 421 8.6 Stepped-Impedance Low-Pass Filters 422 Approximate Equivalent Circuits for Short Transmission Line Sections 422 8.7 Coupled Line Filters 426 Filter Properties of a Coupled Line Section 426 Design of Coupled Line Bandpass Filters 430 8.8 Filters Using Coupled Resonators 437 Bandstop and Bandpass Filters Using Quarter-Wave Resonators 437 Bandpass Filters Using Capacitively Coupled Series Resonators 441 Bandpass Filters Using Capacitively Coupled Shunt Resonators 443 9THEORYANDDESIGNOFFERRIMAGNETICCOMPONENTS 451 9.1 Basic Properties of Ferrimagnetic Materials 452 The Permeability Tensor 452 Circularly Polarized Fields 458 Effect of Loss 460 Demagnetization Factors 462 9.2 Plane Wave Propagation in a Ferrite Medium 465 Propagation in Direction of Bias (Faraday Rotation) 465 Propagation Transverse to Bias (Birefringence) 469 9.3 Propagation in a Ferrite-Loaded Rectangular Waveguide 471 TEm0Modes of Waveguide with a Single Ferrite Slab 471 TEm0Modes of Waveguide with Two Symmetrical Ferrite Slabs 474 9.4 Ferrite Isolators 475 Resonance Isolators 476 The Field Displacement Isolator 479 9.5 Ferrite Phase Shifters 482 Nonreciprocal Latching Phase Shifter 482 Other Types of Ferrite Phase Shifters 485 The Gyrator 486 9.6 Ferrite Circulators 487 Properties of a Mismatched Circulator 488 Junction Circulator 488 10NOISEANDNONLINEARDISTORTION 496 10.1 Noise in Microwave Circuits 496 Dynamic Range and Sources of Noise 497 Noise Power and Equivalent Noise Temperature 498 Measurement of Noise Temperature 501 10.2 Noise Figure 502 Definition of Noise Figure 502 Noise Figure of a Cascaded System 504 Noise Figure of a Passive Two-Port Network 506Noise Figure of a Mismatched Lossy Line 508 Noise Figure of a Mismatched Amplifier 510 ftoc Pozar September 9, 2011 21:38 Contents xv 10.3 Nonlinear Distortion 511 Gain Compression 512 Harmonic and Intermodulation Distortion 513 Third-Order Intercept Point 515 Intercept Point of a Cascaded System 516 Passive Intermodulation 519 10.4 Dynamic Range 519 Linear and Spurious Free Dynamic Range 519 11ACTIVERFANDMICROWAVEDEVICES 524 11.1 Diodes and Diode Circuits 525 Schottky Diodes and Detectors 525 PIN Diodes and Control Circuits 530 Varactor Diodes 537 Other Diodes 538 Power Combining 539 11.2 Bipolar Junction Transistors 540 Bipolar Junction Transistor 540 Heterojunction Bipolar Transistor 542 11.3 Field Effect Transistors 543 Metal Semiconductor Field Effect Transistor 544 Metal Oxide Semiconductor Field Effect Transistor 546 High Electron Mobility Transistor 546 11.4 Microwave Integrated Circuits 547 Hybrid Microwave Integrated Circuits 548 Monolithic Microwave Integrated Circuits 548 11.5 Microwave Tubes 552 12MICROWAVEAMPLIFIERDESIGN 558 12.1 Two-Port Power Gains 558 Definitions of Two-Port Power Gains 559 Further Discussion of Two-Port Power Gains 562 12.2 Stability 564 Stability Circles 564 Tests for Unconditional Stability 567 12.3 Single-Stage Transistor Amplifier Design 571 Design for Maximum Gain (Conjugate Matching) 571 Constant-Gain Circles and Design for Specified Gain 575Low-Noise Amplifier Design 580 Low-Noise MOSFET Amplifier 582 12.4 Broadband Transistor Amplifier Design 585 Balanced Amplifiers 586 Distributed Amplifiers 588 Differential Amplifiers 593 12.5 Power Amplifiers 596 Characteristics of Power Amplifiers and Amplifier Classes 597 Large-Signal Characterization of Transistors 598Design of Class A Power Amplifiers 599 ftoc Pozar September 9, 2011 21:38 xvi Contents 13OSCILLATORSANDMIXERS 604 13.1 RF Oscillators 605 General Analysis 606 Oscillators Using a Common Emitter BJT 607 Oscillators Using a Common Gate FET 609 Practical Considerations 610 Crystal Oscillators 612 13.2 Microwave Oscillators 613 Transistor Oscillators 615 Dielectric Resonator Oscillators 617 13.3 Oscillator Phase Noise 622 Representation of Phase Noise 623 Leeson’s Model for Oscillator Phase Noise 624 13.4 Frequency Multipliers 627 Reactive Diode Multipliers (Manley–Rowe Relations) 628 Resistive Diode Multipliers 631 Transistor Multipliers 633 13.5 Mixers 637 Mixer Characteristics 637 Single-Ended Diode Mixer 642 Single-Ended FET Mixer 643 Balanced Mixer 646 Image Reject Mixer 649Differential FET Mixer and Gilbert Cell Mixer 650 Other Mixers 652 14INTRODUCTIONTOMICROWAVESYSTEMS 658 14.1 System Aspects of Antennas 658 Fields and Power Radiated by an Antenna 660 Antenna Pattern Characteristics 662Antenna Gain and Efficiency 664 Aperture Efficiency and Effective Area 665 Background and Brightness Temperature 666Antenna Noise Temperature and G/T 669 14.2 Wireless Communications 671 The Friis Formula 673 Link Budget and Link Margin 674 Radio Receiver Architectures 676 Noise Characterization of a Receiver 679 Digital Modulation and Bit Error Rate 681 Wireless Communication Systems 684 14.3 Radar Systems 690 The Radar Equation 691 Pulse Radar 693 Doppler Radar 694 Radar Cross Section 695 14.4 Radiometer Systems 696 Theory and Applications of Radiometry 697 Total Power Radiometer 699 The Dicke Radiometer 700 14.5 Microwave Propagation 701 Atmospheric Effects 701 Ground Effects 703 Plasma Effects 704 ftoc Pozar September 30, 2011 11:51 Contents xvii 14.6 Other Applications and Topics 705 Microwave Heating 705 Power Transfer 705 Biological Effects and Safety 706 APPENDICES 712 A Prefixes 713 B Vector Analysis 713C Bessel Functions 715D Other Mathematical Results 718E Physical Constants 718F Conductivities for Some Materials 719G Dielectric Constants and Loss Tangents for Some Materials 719H Properties of Some Microwave Ferrite Materials 720I Standard Rectangular Waveguide Data 720J Standard Coaxial Cable Data 721 ANSWERSTOSELECTEDPROBLEMS 722 INDEX 725 ftoc Pozar September 9, 2011 21:38 This page is intentionally left blank c01ElectromagneticTheory Pozar July 28, 2011 8:7 Chapter One Electromagnetic Theory We begin our study of microwave engineering with a brief overview of the history and major applications of microwave technology, followed by a review of some of the fundamental topics in electromagnetic theory that we will need throughout the book. Further discussion of these topics may be found in references [1–8]. 1.1INTRODUCTIONTOMICROWAVEENGINEERING The field of radio frequency (RF) and microwave engineering generally covers the behavior of alternating current signals with frequencies in the range of 100 MHz (1 MHz = 106Hz) to 1000 GHz (1 GHz = 109Hz). RF frequencies range from very high frequency (VHF) (30–300 MHz) to ultra high frequency (UHF) (300–3000 MHz), while the term microwave is typically used for frequencies between 3 and 300 GHz, with a corresponding electrical wavelength between λ=c/f=10 cm and λ=1 mm, respectively. Signals with wave- lengths on the order of millimeters are often referred to as millimeter waves . Figure 1.1 shows the location of the RF and microwave frequency bands in the electromagnetic spec- trum. Because of the high frequencies (and short wavelengths), standard circuit theoryoften cannot be used directly to solve microwave network problems. In a sense, standard circuit theory is an approximation, or special case, of the broader theory of electromag- netics as described by Maxwell’s equations. This is due to the fact that, in general, the lumped circuit element approximations of circuit theory may not be valid at high RF and microwave frequencies. Microwave components often act as distributed elements, where the phase of the voltage or current changes significantly over the physical extent of the de- vice because the device dimensions are on the order of the electrical wavelength. At much lower frequencies the wavelength is large enough that there is insignificant phase variationacross the dimensions of a component. The other extreme of frequency can be identified as optical engineering, in which the wavelength is much shorter than the dimensions of the component. In this case Maxwell’s equations can be simplified to the geometrical optics regime, and optical systems can be designed with the theory of geometrical optics. Such 1 c01ElectromagneticTheory Pozar July 28, 2011 8:7 2Chapter 1: Electromagnetic Theory 3 × 1053 × 1063 × 1073 × 1083 × 1093 × 10103 × 10113 × 10123 × 10133 × 1014Frequency (Hz)Long wave radio AM broadcast radio Shortwave radio VHF TV FM broadcast radio Infrared Visible lightFar InfraredMicrowaves 10310210 1 10–110–210–310–410–510–6 Wavelength (m) Typical Frequencies AM broadcast band Short wave radio bandFM broadcast bandVHF TV (2–4)VHF TV (5–6)UHF TV (7–13)UHF TV (14–83)US cellular telephone European GSM cellularGPSMicrowave ovens US DBSUS ISM bands US UWB radio535–1605 kHz 3–30 MHz88–108 MHz54–72 MHz76–88 MHz174–216 MHz470–890 MHz824–849 MHz869–894 MHz880–915 MHz925–960 MHz1575.42 MHz1227.60 MHz2.45 GHz11.7–12.5 GHz902–928 MHz2.400–2.484 GHz5.725–5.850 GHz3.1–10.6 GHzApproximate Band Designations Medium frequency High frequency (HF)Very high frequency (VHF)Ultra high frequency (UHF)L bandS bandC bandX bandKu bandK bandKa bandU bandV bandE bandW bandF band300 kHz–3 MHz 3 MHz–30 MHz30 MHz–300 MHz300 MHz–3 GHz1–2 GHz2–4 GHz4–8 GHz8–12 GHz12–18 GHz18–26 GHz26–40 GHz40–60 GHz50–75 GHz60–90 GHz75–110 GHz90–140 GHz FIGURE 1.1 The electromagnetic spectrum. techniques are sometimes applicable to millimeter wave systems, where they are referred to as quasi-optical. In RF and microwave engineering, then, one must often work with Maxwell’s equa- tions and their solutions. It is in the nature of these equations that mathematical complexity arises since Maxwell’s equations involve vector differential or integral operations on vec- tor field quantities, and these fields are functions of spatial coordinates. One of the goals of this book is to try to reduce the complexity of a field theory solution to a result thatcan be expressed in terms of simpler circuit theory, perhaps extended to include distributed elements (such as transmission lines) and concepts (such as reflection coefficients and scat- tering parameters). A field theory solution generally provides a complete description of theelectromagnetic field at every point in space, which is usually much more information than we need for most practical purposes. We are typically more interested in terminal quanti- ties such as power, impedance, voltage, and current, which can often be expressed in termsof these extended circuit theory concepts. It is this complexity that adds to the challenge, as well as the rewards, of microwave engineering. ApplicationsofMicrowaveEngineering Just as the high frequencies and short wavelengths of microwave energy make for diffi- culties in the analysis and design of microwave devices and systems, these same aspects c01ElectromagneticTheory Pozar July 28, 2011 8:7 1.1 Introduction to Microwave Engineering 3 provide unique opportunities for the application of microwave systems. The following con- siderations can be useful in practice: rAntenna gain is proportional to the electrical size of the antenna. At higher frequen-cies, more antenna gain can be obtained for a given physical antenna size, and this has important consequences when implementing microwave systems. rMore bandwidth (directly related to data rate) can be realized at higher frequencies. A 1% bandwidth at 600 MHz is 6 MHz, which (with binary phase shift keying modulation) can provide a data rate of about 6 Mbps (megabits per second), while at 60 GHz a 1% bandwidth is 600 MHz, allowing a 600 Mbps data rate.rMicrowave signals travel by line of sight and are not bent by the ionosphere as are lower frequency signals. Satellite and terrestrial communication links with very high capacities are therefore possible, with frequency reuse at minimally distant locations.rThe effective reflection area (radar cross section) of a radar target is usually propor- tional to the target’s electrical size. This fact, coupled with the frequency character- istics of antenna gain, generally makes microwave frequencies preferred for radar systems.rVarious molecular, atomic, and nuclear resonances occur at microwave frequencies,creating a variety of unique applications in the areas of basic science, remote sens- ing, medical diagnostics and treatment, and heating methods. The majority of today’s applications of RF and microwave technology are to wire- less networking and communications systems, wireless security systems, radar systems, environmental remote sensing, and medical systems. As the frequency allocations listed in Figure 1.1 show, RF and microwave communications systems are pervasive, especiallytoday when wireless connectivity promises to provide voice and data access to “anyone, anywhere, at any time.” Modern wireless telephony is based on the concept of cellular frequency reuse, a tech- nique first proposed by Bell Labs in 1947 but not practically implemented until the 1970s. By this time advances in miniaturization, as well as increasing demand for wireless com-munications, drove the introduction of several early cellular telephone systems in Europe, the United States, and Japan. The Nordic Mobile Telephone (NMT) system was deployed in 1981 in the Nordic countries, the Advanced Mobile Phone System (AMPS) was intro- duced in the United States in 1983 by AT&T, and NTT in Japan introduced its first mobile phone service in 1988. All of these early systems used analog FM modulation, with their allocated frequency bands divided into several hundred narrow band voice channels. Theseearly systems are usually referred to now as first-generation cellular systems, or 1G. Second-generation (2G) cellular systems achieved improved performance by using various digital modulation schemes, with systems such as GSM, CDMA, DAMPS, PCS,and PHS being some of the major standards introduced in the 1990s in the United States, Europe, and Japan. These systems can handle digitized voice, as well as some limited data, with data rates typically in the 8 to 14 kbps range. In recent years there has been a widevariety of new and modified standards to transition to handheld services that include voice, texting, data networking, positioning, and Internet access. These standards are variously known as 2.5G, 3G, 3.5G, 3.75G, and 4G, with current plans to provide data rates up to at least 100 Mbps. The number of subscribers to wireless services seems to be keeping pace with the growing power and access provided by modern handheld wireless devices; as of2010 there were more than five billion cell phone users worldwide. Satellite systems also depend on RF and microwave technology, and satellites have been developed to provide cellular (voice), video, and data connections worldwide. Two largesatellite constellations, Iridium and Globalstar, were deployed in the late 1990s to provide worldwide telephony service. Unfortunately, these systems suffered from both technical c01ElectromagneticTheory Pozar July 28, 2011 8:7 4Chapter 1: Electromagnetic Theory drawbacks and weak business models and have led to multibillion dollar financial failures. However, smaller satellite systems, such as the Global Positioning Satellite (GPS) system and the Direct Broadcast Satellite (DBS) system, have been extremely successful. Wireless local area networks (WLANs) provide high-speed networking between com- puters over short distances, and the demand for this capability is expected to remain strong. One of the newer examples of wireless communications technology is ultra wide band (UWB) radio, where the broadcast signal occupies a very wide frequency band but with avery low power level (typically below the ambient radio noise level) to avoid interference with other systems. Radar systems find application in military, commercial, and scientific fields. Radar is used for detecting and locating air, ground, and seagoing targets, as well as for missile guidance and fire control. In the commercial sector, radar technology is used for air trafficcontrol, motion detectors (door openers and security alarms), vehicle collision avoidance, and distance measurement. Scientific applications of radar include weather prediction, re- mote sensing of the atmosphere, the oceans, and the ground, as well as medical diagnosticsand therapy. Microwave radiometry, which is the passive sensing of microwave energy emitted by an object, is used for remote sensing of the atmosphere and the earth, as well as in medical diagnostics and imaging for security applications. AShortHistoryofMicrowaveEngineering Microwave engineering is often considered a fairly mature discipline because the funda- mental concepts were developed more than 50 years ago, and probably because radar, the first major application of microwave technology, was intensively developed as far back asWorld War II. However, recent years have brought substantial and continuing developments in high-frequency solid-state devices, microwave integrated circuits, and computer-aided design techniques, and the ever-widening applications of RF and microwave technology to wireless communications, networking, sensing, and security have kept the field active and vibrant. The foundations of modern electromagnetic theory were formulated in 1873 by James Clerk Maxwell, who hypothesized, solely from mathematical considerations, electromag- netic wave propagation and the idea that light was a form of electromagnetic energy.Maxwell’s formulation was cast in its modern form by Oliver Heaviside during the period from 1885 to 1887. Heaviside was a reclusive genius whose efforts removed many of the mathematical complexities of Maxwell’s theory, introduced vector notation, and provideda foundation for practical applications of guided waves and transmission lines. Heinrich Hertz, a German professor of physics and a gifted experimentalist who understood the the- ory published by Maxwell, carried out a set of experiments during the period 1887–1891that validated Maxwell’s theory of electromagnetic waves. Figure 1.2 is a photograph of the original equipment used by Hertz in his experiments. It is interesting to observe that this is an instance of a discovery occurring after a prediction has been made on theoreticalgrounds—a characteristic of many of the major discoveries throughout the history of sci- ence. All of the practical applications of electromagnetic theory—radio, television, radar, cellular telephones, and wireless networking—owe their existence to the theoretical work of Maxwell. Because of the lack of reliable microwave sources and other components, the rapid growth of radio technology in the early 1900s occurred primarily in the HF to VHF range. It was not until the 1940s and the advent of radar development during World War II that microwave theory and technology received substantial interest. In the United States, theRadiation Laboratory was established at the Massachusetts Institute of Technology to de- velop radar theory and practice. A number of talented scientists, including N. Marcuvitz, c01ElectromagneticTheory Pozar July 28, 2011 8:7 1.1 Introduction to Microwave Engineering 5 FIGURE 1.2 Original apparatus used by Hertz for his electromagnetics experiments. (1) 50 MHz transmitter spark gap and loaded dipole antenna. (2) Wire grid for polarization ex- periments. (3) Vacuum apparatus for cathode ray experiments. (4) Hot-wire gal-vanometer. (5) Reiss or Knochenhauer spirals. (6) Rolled-paper galvanometer. (7) Metal sphere probe. (8) Reiss spark micrometer. (9) Coaxial line. (10–12) Equip- ment to demonstrate dielectric polarization effects. (13) Mercury induction coilinterrupter. (14) Meidinger cell. (15) Bell jar. (16) Induction coil. (17) Bunsen cells. (18) Large-area conductor for charge storage. (19) Circular loop receiving antenna. (20) Eight-sided receiver detector. (21) Rotating mirror and mercury inter-rupter. (22) Square loop receiving antenna. (23) Equipment for refraction and dielec- tric constant measurement. (24) Two square loop receiving antennas. (25) Square loop receiving antenna. (26) Transmitter dipole. (27) Induction coil. (28) Coaxialline. (29) High-voltage discharger. (30) Cylindrical parabolic reflector/receiver. (31) Cylindrical parabolic reflector/transmitter. (32) Circular loop receiving antenna. (33) Planar reflector. (34, 35) Battery of accumulators. Photographed on October1, 1913, at the Bavarian Academy of Science, Munich, Germany, with Hertz’s as- sistant, Julius Amman. Photograph and identification courtesy of J. H. Bryant. I. I. Rabi, J. S. Schwinger, H. A. Bethe, E. M. Purcell, C. G. Montgomery, and R. H. Dicke, among others, gathered for a very intensive period of development in the microwave field. Their work included the theoretical and experimental treatment of waveguide components, microwave antennas, small-aperture coupling theory, and the beginnings of microwave net- work theory. Many of these researchers were physicists who returned to physics research after the war, but their microwave work is summarized in the classic 28-volume RadiationLaboratory Series of books that still finds application today. Communications systems using microwave technology began to be developed soon after the birth of radar, benefiting from much of the work that was originally done forradar systems. The advantages offered by microwave systems, including wide bandwidths and line-of-sight propagation, have proved to be critical for both terrestrial and satellite c01ElectromagneticTheory Pozar July 28, 2011 8:7 6Chapter 1: Electromagnetic Theory communications systems and have thus provided an impetus for the continuing develop- ment of low-cost miniaturized microwave components. We refer the interested reader to references [1] and [2] for further historical perspectives on the fields of wireless commu- nications and microwave engineering. 1.2MAXWELL’SEQUATIONS Electric and magnetic phenomena at the macroscopic level are described by Maxwell’s equations, as published by Maxwell in 1873. This work summarized the state of electro-magnetic science at that time and hypothesized from theoretical considerations the exis- tence of the electrical displacement current, which led to the experimental discovery by Hertz of electromagnetic wave propagation. Maxwell’s work was based on a large body ofempirical and theoretical knowledge developed by Gauss, Ampere, Faraday, and others. A first course in electromagnetics usually follows this historical (or deductive) approach, and it is assumed that the reader has had such a course as a prerequisite to the present material. Several references are available [3–7] that provide a good treatment of electromagnetic theory at the undergraduate or graduate level. This chapter will outline the fundamental concepts of electromagnetic theory that we will require later in the book. Maxwell’s equations will be presented, and boundary condi- tions and the effect of dielectric and magnetic materials will be discussed. Wave phenom- ena are of essential importance in microwave engineering, and thus much of the chapter is spent on topics related to plane waves. Plane waves are the simplest form of electromag- netic waves and so serve to illustrate a number of basic properties associated with wavepropagation. Although it is assumed that the reader has studied plane waves before, the present material should help to reinforce the basic principles in the reader’s mind and per- haps to introduce some concepts that the reader has not seen previously. This material willalso serve as a useful reference for later chapters. With an awareness of the historical perspective, it is usually advantageous from a pedagogical point of view to present electromagnetic theory from the “inductive,” or ax-iomatic, approach by beginning with Maxwell’s equations. The general form of time- varying Maxwell equations, then, can be written in “point,” or differential, form as ∇ׯE=−∂¯B ∂t−¯M, (1.1a) ∇ׯH=∂¯D ∂t+¯J, (1.1b) ∇·¯D=ρ, (1.1c) ∇·¯B=0. (1.1d) The MKS system of units is used throughout this book. The script quantities represent time-varying vector fields and are real functions of spatial coordinates x,y,z, and the time variable t. These quantities are defined as follows: ¯Eis the electric field, in volts per meter (V/m).1 ¯His the magnetic field, in amperes per meter (A/m). 1As recommended by the IEEE Standard Definitions of Terms for Radio Wave Propagation, IEEE Standard 211-1997, the terms “electric field” and “magnetic field” are used in place of the older terminology of “electricfield intensity” and “magnetic field intensity.” c01ElectromagneticTheory Pozar July 28, 2011 8:7 1.2 Maxwell’s Equations 7 ¯Dis the electric flux density, in coulombs per meter squared (Coul/m2). ¯Bis the magnetic flux density, in webers per meter squared (Wb/m2). ¯Mis the (fictitious) magnetic current density, in volts per meter (V/m2). ¯Jis the electric current density, in amperes per meter squared (A/m2). ρis the electric charge density, in coulombs per meter cubed (Coul/m3). The sources of the electromagnetic field are the currents ¯Mand¯Jand the electric charge density ρ. The magnetic current ¯Mis a fictitious source in the sense that it is only a mathematical convenience: the real source of a magnetic current is always a loop of electric current or some similar type of magnetic dipole, as opposed to the flow of an actual magnetic charge (magnetic monopole charges are not known to exist). The magneticcurrent is included here for completeness, as we will have occasion to use it in Chapter 4 when dealing with apertures. Since electric current is really the flow of charge, it can be said that the electric charge density ρis the ultimate source of the electromagnetic field. In free-space, the following simple relations hold between the electric and magnetic field intensities and flux densities: ¯B=µ 0¯H, (1.2a) ¯D=/epsilon10¯E, (1.2b) where µ0=4π×10−7henry/m is the permeability of free-space, and /epsilon10=8.854×10−12 farad/m is the permittivity of free-space. We will see in the next section how media other than free-space affect these constitutive relations. Equations (1.1a)–(1.1d) are linear but are not independent of each other. For instance, consider the divergence of (1.1a). Since the divergence of the curl of any vector is zero [vector identity (B.12), from Appendix B], we have ∇·∇× ¯E=0=−∂ ∂t(∇·¯B)−∇· ¯M. Since there is no free magnetic charge, ∇·¯M=0, which leads to ∇·¯B=0, or (1.1d). Thecontinuity equation can be similarly derived by taking the divergence of (1.1b), giving ∇·¯J+∂ρ ∂t=0,( 1.3) where (1.1c) was used. This equation states that charge is conserved, or that current is continuous, since ∇·¯Jrepresents the outflow of current at a point, and ∂ρ/∂ trepresents the charge buildup with time at the same point. It is this result that led Maxwell to the conclusion that the displacement current density ∂¯D/∂twas necessary in (1.1b), which can be seen by taking the divergence of this equation. The above differential equations can be converted to integral form through the use of various vector integral theorems. Thus, applying the divergence theorem (B.15) to (1.1c)and (1.1d) yields /contintegraldisplay S¯D·d¯s=/integraldisplay Vρdv=Q, (1.4) /contintegraldisplay S¯B·d¯s=0, (1.5) c01ElectromagneticTheory Pozar July 28, 2011 8:7 8Chapter 1: Electromagnetic Theory C S dl n B ˆ FIGURE 1.3 The closed contour Cand surface Sassociated with Faraday’s law. where Qin (1.4) represents the total charge contained in the closed volume V(enclosed by a closed surface S). Applying Stokes’ theorem (B.16) to (1.1a) gives /contintegraldisplay C¯E·d¯l=−∂ ∂t/integraldisplay S¯B·d¯s−/integraldisplay S¯M·d¯s,( 1.6) which, without the ¯Mterm, is the usual form of Faraday’s law and forms the basis for Kirchhoff’s voltage law. In (1.6), Crepresents a closed contour around the surface S,a s s h o w ni nF i g u r e1 . 3 .Ampere’s law can be derived by applying Stokes’ theorem to (1.1b): /contintegraldisplay C¯H·d¯l=∂ ∂t/integraldisplay S¯D·d¯s+/integraldisplay S¯J·d¯s=∂ ∂t/integraldisplay S¯D·d¯s+I,( 1.7) where I=/integraltext S¯J·d¯sis the total electric current flow through the surface S. Equations (1.4)–(1.7) constitute the integral forms of Maxwell’s equations. The above equations are valid for arbitrary time dependence, but most of our work will be involved with fields having a sinusoidal, or harmonic, time dependence, with steady- state conditions assumed. In this case phasor notation is very convenient, and so all field quantities will be assumed to be complex vectors with an implied ejωttime dependence and written with roman (rather than script) letters. Thus, a sinusoidal electric field polarized in the ˆxdirection of the form ¯E(x,y,z,t)=ˆxA(x,y,z)cos(ωt+φ),( 1.8) where Ais the (real) amplitude, ωis the radian frequency, and φis the phase reference of the wave at t=0, has the phasor for ¯E(x,y,z)=ˆxA(x,y,z)ejφ.( 1.9) We will assume cosine-based phasors in this book, so the conversion from phasor quanti- ties to real time-varying quantities is accomplished by multiplying the phasor by ejωtand taking the real part: ¯E(x,y,z,t)=Re{¯E(x,y,z)ejωt},( 1.10) as substituting (1.9) into (1.10) to obtain (1.8) demonstrates. When working in phasor notation, it is customary to suppress the factor ejωtthat is common to all terms. When dealing with power and energy we will often be interested in the time average of a quadratic quantity. This can be found very easily for time harmonic fields. For example,the average of the square of the magnitude of an electric field, given as ¯E=ˆxE 1cos(ωt+φ1)+ˆyE2cos(ωt+φ2)+ˆzE2cos(ω t+φ3), (1.11) has the phasor form ¯E=ˆxE1ejφ1+ˆyE2ejφ2+ˆzE3ejφ3,( 1.12) c01ElectromagneticTheory Pozar July 28, 2011 8:7 1.2 Maxwell’s Equations 9 can be calculated as |¯E|2 avg=1 T/integraldisplayT 0¯E·¯Edt =1 T/integraldisplayT 0/bracketleftbig E2 1cos2(ωt+φ1)+E2 2cos2(ωt+φ2)+E2 3cos2(ωt+φ3)/bracketrightbig dt =1 2/parenleftbig E2 1+E2 2+E2 3/parenrightbig =1 2|¯E|2=1 2¯E·¯E∗. (1.13) Then the root-mean-square (rms) value is |¯E|rms=|¯E|/√ 2. y yz y x (a)J(x, y, z) A/m2 z y xM(x, y, z) V/m2 z y x (b)Js(x, y) A/m J(x, y, z) = Js(x, y) /H9254(z – zo) A/m2M(x, y, z) = Ms(x, y) /H9254(z – zo) V/m2z y x Ms(x, y) V/m z x (c)xIo(x) A (xo, yo, zo)( xo, yo, zo)J(x, y, z) = xIo(x) /H9254(y – yo) /H9254(z – zo) A/m2M(x, y, z) = xVo(x) /H9254(y – yo) /H9254(z – zo) V/m2z xxVo(x) V z y x (d)J(x, y, z) = xIl/H9254(x – xo) /H9254(y – yo) /H9254(z – zo) A/m2M(x, y, z) = xVl/H9254(x – xo) /H9254(y – yo) /H9254(z – zo) V/m2z y xIl A-m Vl V-mˆˆ ˆ ˆ ˆˆ FIGURE 1.4 Arbitrary volume, surface, and line currents. (a) Arbitrary electric and magnetic vol- ume current densities. (b) Arbitrary electric and magnetic surface current densitiesin the z=z 0plane. (c) Arbitrary electric and magnetic line currents. (d) Infinitesi- mal electric and magnetic dipoles parallel to the x-axis. c01ElectromagneticTheory Pozar July 28, 2011 8:7 10 Chapter 1: Electromagnetic Theory Assuming an ejωttime dependence, we can replace the time derivatives in (1.1a)– (1.1d) with jω. Maxwell’s equations in phasor form then become ∇ׯE=− jω¯B−¯M, (1.14a) ∇ׯH=jω¯D+¯J, (1.14b) ∇·¯D=ρ, (1.14c) ∇·¯B=0. (1.14d) The Fourier transform can be used to convert a solution to Maxwell’s equations for an arbitrary frequency ωinto a solution for arbitrary time dependence. The electric and magnetic current sources, ¯Jand¯M, in (1.14) are volume current densities with units A/m2and V/m2. In many cases, however, the actual currents will be in the form of a current sheet, a line current, or an infinitesimal dipole current. These special types of current distributions can always be written as volume current densities through the use of delta functions. Figure 1.4 shows examples of this procedure for electric and magnetic currents. 1.3FIELDSINMEDIAANDBOUNDARYCONDITIONS In the preceding section it was assumed that the electric and magnetic fields were in free- space, with no material bodies present. In practice, material bodies are often present; thiscomplicates the analysis but also allows the useful application of material properties to microwave components. When electromagnetic fields exist in material media, the field vectors are related to each other by the constitutive relations. For a dielectric material, an applied electric field ¯Ecauses the polarization of the atoms or molecules of the material to create electric dipole moments that augment the total displacement flux, ¯D. This additional polarization vector is called ¯P e,t h e electric polarization , where ¯D=/epsilon10¯E+¯Pe.( 1.15) In a linear medium the electric polarization is linearly related to the applied electric field as ¯Pe=/epsilon10χe¯E,( 1.16) where χe, which may be complex, is called the electric susceptibility . Then, ¯D=/epsilon10¯E+¯Pe=/epsilon10(1+χe)¯E=/epsilon1¯E,( 1.17) where /epsilon1=/epsilon1/prime−j/epsilon1/prime/prime=/epsilon10(1+χe)( 1.18) is the complex permittivity of the medium. The imaginary part of /epsilon1accounts for loss in the medium (heat) due to damping of the vibrating dipole moments. (Free-space, having a real/epsilon1, is lossless.) Due to energy conservation, as we will see in Section 1.6, the imaginary part of /epsilon1must be negative (/epsilon1/prime/primepositive). The loss of a dielectric material may also be considered as an equivalent conductor loss. In a material with conductivity σ, a conduction current density will exist: ¯J=σ¯E,( 1.19) c01ElectromagneticTheory Pozar July 28, 2011 8:7 1.3 Fields in Media and Boundary Conditions 11 which is Ohm’s law from an electromagnetic field point of view. Maxwell’s curl equation for¯Hin (1.14b) then becomes ∇ׯH=jω¯D+¯J =jω/epsilon1¯E+σ¯E =jω/epsilon1/prime¯E+(ω/epsilon1/prime/prime+σ)¯E =jω/parenleftBig /epsilon1/prime−j/epsilon1/prime/prime−jσ ω/parenrightBig ¯E, (1.20) where it is seen that loss due to dielectric damping ( ω/epsilon1/prime/prime)is indistinguishable from conduc- tivity loss (σ) .T h et e r m ω/epsilon1/prime/prime+σcan then be considered as the total effective conductivity. A related quantity of interest is the loss tangent, defined as tanδ=ω/epsilon1/prime/prime+σ ω/epsilon1/prime,( 1.21) which is seen to be the ratio of the real to the imaginary part of the total displacement current. Microwave materials are usually characterized by specifying the real relative per- mittivity (the dielectric constant ),2/epsilon1r, with /epsilon1/prime=/epsilon1r/epsilon10, and the loss tangent at a certain fre- quency. These properties are listed in Appendix G for several types of materials. It is usefulto note that, after a problem has been solved assuming a lossless dielectric, loss can eas- ily be introduced by replacing the real /epsilon1with a complex /epsilon1=/epsilon1 /prime−j/epsilon1/prime/prime=/epsilon1/prime(1−jtanδ)= /epsilon10/epsilon1r(1−jtanδ). In the preceding discussion it was assumed that ¯Pewas a vector in the same direction as¯E. Such materials are called isotropic materials, but not all materials have this property. Some materials are anisotropic and are characterized by a more complicated relation be- tween ¯Peand¯E,o r¯Dand¯E. The most general linear relation between these vectors takes the form of a tensor of rank two (a dyad), which can be written in matrix form as /bracketleftBiggDx Dy Dz/bracketrightBigg =/bracketleftBigg/epsilon1xx/epsilon1xy/epsilon1xz /epsilon1yx/epsilon1yy/epsilon1yz /epsilon1zx/epsilon1zy/epsilon1zz/bracketrightBigg/bracketleftBiggEx Ey Ez/bracketrightBigg =[/epsilon1]/bracketleftBiggEx Ey Ez/bracketrightBigg .( 1.22) It is thus seen that a given vector component of ¯Egives rise, in general, to three components of¯D. Crystal structures and ionized gases are examples of anisotropic dielectrics. For a linear isotropic material, the matrix of (1.22) reduces to a diagonal matrix with elements /epsilon1. An analogous situation occurs for magnetic materials. An applied magnetic field may align magnetic dipole moments in a magnetic material to produce a magnetic polarization (or magnetization) ¯Pm. Then, ¯B=µ0(¯H+¯Pm). (1.23) For a linear magnetic material, ¯Pmis linearly related to ¯Has ¯Pm=χm¯H,( 1.24) where χmis a complex magnetic susceptibility . From (1.23) and (1.24), ¯B=µ0(1+χm)¯H=µ¯H,( 1.25) 2The IEEE Standard Definitions of Terms for Radio Wave Propagation, IEEE Standard 211-1997, suggests that the term “relative permittivity” be used instead of “dielectric constant.” The IEEE Standard Definitions of Terms for Antennas, IEEE Standard 145-1993 , however, still recognizes “dielectric constant.” Since this term is commonly used in microwave engineering work, it will occasionally be used in this book. c01ElectromagneticTheory Pozar July 28, 2011 8:7 12 Chapter 1: Electromagnetic Theory where µ=µ0(1+χm)=µ/prime−jµ/prime/primeis the complex permeability of the medium. Again, the imaginary part of χmorµaccounts for loss due to damping forces; there is no magnetic conductivity because there is no real magnetic current. As in the electric case, magneticmaterials may be anisotropic, in which case a tensor permeability can be written as /bracketleftBiggB x By Bz/bracketrightBigg =/bracketleftBiggµxxµxyµxz µyxµyyµyz µzxµzyµzz/bracketrightBigg/bracketleftBiggHx Hy Hz/bracketrightBigg =[µ]/bracketleftBiggHx Hy Hz/bracketrightBigg .( 1.26) An important example of anisotropic magnetic materials in microwave engineering is the class of ferrimagnetic materials known as ferrites; these materials and their applications will be discussed further in Chapter 9. If linear media are assumed ( /epsilon1, µ not depending on ¯Eor¯H), then Maxwell’s equa- tions can be written in phasor form as ∇ׯE=− jωµ¯H−¯M, (1.27a) ∇ׯH=jω/epsilon1¯E+¯J, (1.27b) ∇·¯D=ρ, (1.27c) ∇·¯B=0. (1.27d) The constitutive relations are ¯D=/epsilon1¯E, (1.28a) ¯B=µ¯H, (1.28b) where /epsilon1andµmay be complex and may be tensors. Note that relations like (1.28a) and (1.28b) generally cannot be written in time domain form, even for linear media, because ofthe possible phase shift between ¯Dand¯E,o r¯Band¯H. The phasor representation accounts for this phase shift by the complex form of /epsilon1andµ. Maxwell’s equations (1.27a)–(1.27d) in differential form require known boundary val- ues for a complete and unique solution. A general method used throughout this book is to solve the source-free Maxwell equations in a certain region to obtain solutions with un- known coefficients and then apply boundary conditions to solve for these coefficients. Anumber of specific cases of boundary conditions arise, as discussed in what follows. FieldsataGeneralMaterialInterface Consider a plane interface between two media, as shown in Figure 1.5. Maxwell’s equa- tions in integral form can be used to deduce conditions involving the normal and tangential Bn2 Bn1Ht1Et2Dn2 Dn1 Et1Ht2Medium 2: /H92802, /H92622 Medium 1: /H92801, /H92621nJsMs /H9267sˆ FIGURE 1.5 Fields, currents, and surface charge at a general interface between two media. c01ElectromagneticTheory Pozar July 28, 2011 8:7 1.3 Fields in Media and Boundary Conditions 13 Dn2 Dn1Medium 2 Medium 1n /H9267s∆S shˆ FIGURE 1.6 Closed surface Sfor equation (1.29). fields at this interface. The time-harmonic version of (1.4), where Sis the closed “pillbox”- shaped surface shown in Figure 1.6, can be written as /contintegraldisplay S¯D·d¯s=/integraldisplay Vρdv. (1.29) In the limit as h→0, the contribution of Dtanthrough the sidewalls goes to zero, so (1.29) reduces to /Delta1SD2n−/Delta1SD1n=/Delta1Sρs, or D2n−D1n=ρs,( 1.30) where ρsis the surface charge density on the interface. In vector form, we can write ˆn·(¯D2−¯D1)=ρs.( 1.31) A similar argument for ¯Bleads to the result that ˆn·¯B2=ˆn·¯B1,( 1.32) because there is no free magnetic charge. For the tangential components of the electric field we use the phasor form of (1.6), /contintegraldisplay C¯E·d¯l=− jω/integraldisplay S¯B·d¯s−/integraldisplay S¯M·d¯s,( 1.33) in connection with the closed contour Cshown in Figure 1.7. In the limit as h→0, the surface integral of ¯Bvanishes (because S=h/Delta1/lscriptvanishes). The contribution from the surface integral of ¯M, however, may be nonzero if a magnetic surface current density ¯Ms exists on the surface. The Dirac delta function can then be used to write ¯M=¯Msδ(h), (1.34) where his a coordinate measured normal from the interface. Equation (1.33) then gives /Delta1/lscriptEt1−/Delta1/lscriptEt2=−/Delta1/lscriptMs, Medium 2 Medium 1n Et2 Msn Et1S Ch ∆lˆ FIGURE 1.7 Closed contour Cfor equation (1.33). c01ElectromagneticTheory Pozar July 28, 2011 8:7 14 Chapter 1: Electromagnetic Theory or Et1−Et2=− Ms,( 1.35) which can be generalized in vector form as (¯E2−¯E1)׈n=¯Ms.( 1.36) A similar argument for the magnetic field leads to ˆn×(¯H2−¯H1)=¯Js,( 1.37) where ¯Jsis an electric surface current density that may exist at the interface. Equations (1.31), (1.32), (1.36), and (1.37) are the most general expressions for the boundary condi- tions at an arbitrary interface of materials and/or surface currents. FieldsataDielectricInterface At an interface between two lossless dielectric materials, no charge or surface current den- sities will ordinarily exist. Equations (1.31), (1.32), (1.36), and (1.37) then reduce to ˆn·¯D1=ˆn·¯D2, (1.38a) ˆn·¯B1=ˆn·¯B2, (1.38b) ˆnׯE1=ˆnׯE2, (1.38c) ˆnׯH1=ˆnׯH2. (1.38d) In words, these equations state that the normal components of ¯Dand¯Bare continuous across the interface, and the tangential components of ¯Eand¯Hare continuous across the interface. Because Maxwell’s equations are not all linearly independent, the six boundary conditions contained in the above equations are not all linearly independent. Thus, theenforcement of (1.38c) and (1.38d) for the four tangential field components, for example, will automatically force the satisfaction of the equations for the continuity of the normal components. FieldsattheInterfacewithaPerfectConductor(ElectricWall) Many problems in microwave engineering involve boundaries with good conductors (e.g., metals), which can often be assumed as lossless ( σ→∞). In this case of a perfect con- ductor, all field components must be zero inside the conducting region. This result can be seen by considering a conductor with finite conductivity ( σ<∞)and noting that the skin depth (the depth to which most of the microwave power penetrates) goes to zero as σ→∞ . (Such an analysis will be performed in Section 1.7.) If we also assume here that ¯M s=0, which would be the case if the perfect conductor filled all the space on one side of the boundary, then (1.31), (1.32), (1.36), and (1.37) reduce to the following: ˆn·¯D=ρs, (1.39a) ˆn·¯B=0, (1.39b) ˆnׯE=0, (1.39c) ˆnׯH=¯Js, (1.39d) where ρsand¯Jsare the electric surface charge density and current density, respectively, on the interface, and ˆnis the normal unit vector pointing out of the perfect conductor. Such c01ElectromagneticTheory Pozar July 28, 2011 8:7 1.4 The Wave Equation and Basic Plane Wave Solutions 15 a boundary is also known as an electric wall because the tangential components of ¯Eare “shorted out,” as seen from (1.39c), and must vanish at the surface of the conductor. TheMagneticWallBoundaryCondition Dual to the preceding boundary condition is the magnetic wall boundary condition, where the tangential components of ¯Hmust vanish. Such a boundary does not really exist in practice but may be approximated by a corrugated surface or in certain planar transmission line problems. In addition, the idealization that ˆnׯH=0 at an interface is often a con- venient simplification, as we will see in later chapters. We will also see that the magnetic wall boundary condition is analogous to the relations between the voltage and current at the end of an open-circuited transmission line, while the electric wall boundary conditionis analogous to the voltage and current at the end of a short-circuited transmission line. The magnetic wall condition, then, provides a degree of completeness in our formulation of boundary conditions and is a useful approximation in several cases of practical interest. The fields at a magnetic wall satisfy the following conditions: ˆn·¯D=0, (1.40a) ˆn·¯B=0, (1.40b) ˆnׯE=−¯M s, (1.40c) ˆnׯH=0, (1.40d) where ˆnis the normal unit vector pointing out of the magnetic wall region. TheRadiationCondition When dealing with problems that have one or more infinite boundaries, such as plane waves in an infinite medium, or infinitely long transmission lines, a condition on the fieldsat infinity must be enforced. This boundary condition is known as the radiation condition and is essentially a statement of energy conservation. It states that, at an infinite distance from a source, the fields must either be vanishingly small (i.e., zero) or propagating in anoutward direction. This result can easily be seen by allowing the infinite medium to contain a small loss factor (as any physical medium would have). Incoming waves (from infinity) of finite amplitude would then require an infinite source at infinity and so are disallowed. 1.4THEWAVEEQUATIONANDBASICPLANEWAVESOLUTIONS TheHelmholtzEquation In a source-free, linear, isotropic, homogeneous region, Maxwell’s curl equations in phasor form are ∇ׯE=− jωµ¯H, (1.41a) ∇ׯH=jω/epsilon1¯E, (1.41b) and constitute two equations for the two unknowns, ¯Eand¯H. As such, they can be solved for either ¯Eor¯H. Taking the curl of (1.41a) and using (1.41b) gives ∇×∇× ¯E=− jωµ∇ׯH=ω2µ/epsilon1¯E, c01ElectromagneticTheory Pozar July 28, 2011 8:7 16 Chapter 1: Electromagnetic Theory which is an equation for ¯E. This result can be simplified through the use of vector identity (B.14), ∇×∇× ¯A=∇(∇·¯A)−∇2¯A, which is valid for the rectangular components of an arbitrary vector ¯A. Then, ∇2¯E+ω2µ/epsilon1¯E=0,( 1.42) because ∇·¯E=0 in a source-free region. Equation (1.42) is the wave equation,o r Helmholtz equation,f o r ¯E. An identical equation for ¯Hcan be derived in the same manner: ∇2¯H+ω2µ/epsilon1¯H=0.( 1.43) A constant k=ω√µ/epsilon1is defined and called the propagation constant (also known as the phase constant,o r wave number ), of the medium; its units are 1/m. As a way of introducing wave behavior, we will next study the solutions to the above wave equations in their simplest forms, first for a lossless medium and then for a lossy(conducting) medium. PlaneWavesinaLosslessMedium In a lossless medium, /epsilon1andµare real numbers, and so kis real. A basic plane wave solution to the above wave equations can be found by considering an electric field with only an ˆx component and uniform (no variation) in the xandydirections. Then, ∂/∂x=∂/∂y=0, and the Helmholtz equation of (1.42) reduces to ∂ 2Ex ∂z2+k2Ex=0.( 1.44) The two independent solutions to this equation are easily seen, by substitution, to be of the form Ex(z)=E+e−jkz+E−ejkz,( 1.45) where E+andE−are arbitrary amplitude constants. The above solution is for the time harmonic case at frequency ω. In the time domain, this result is written as Ex(z,t)=E+cos(ω t−kz)+E−cos(ω t+kz), (1.46) where we have assumed that E+and E−are real constants. Consider the first term in (1.46). This term represents a wave traveling in the +zdirection because, to maintain a fixed point on the wave ( ωt−kz=constant), one must move in the +zdirection as time increases. Similarly, the second term in (1.46) represents a wave traveling in the negative z direction—hence the notation E+andE−for these wave amplitudes. The velocity of the wave in this sense is called the phase velocity because it is the velocity at which a fixed phase point on the wave travels, and it is given by vp=dz dt=d dt/parenleftbiggωt−constant k/parenrightbigg =ω k=1√µ/epsilon1(1.47) In free-space, we have vp=1/√µ0/epsilon10=c=2.998 ×108m/sec, which is the speed of light. The wavelength ,λ, is defined as the distance between two successive maxima (or minima, or any other reference points) on the wave at a fixed instant of time. Thus, (ωt−kz)−[ωt−k(z+λ)]= 2π, c01ElectromagneticTheory Pozar July 28, 2011 8:7 1.4 The Wave Equation and Basic Plane Wave Solutions 17 so λ=2π k=2πv p ω=vp f.( 1.48) A complete specification of the plane wave electromagnetic field should include the magnetic field. In general, whenever ¯Eor¯His known, the other field vector can be readily found by using one of Maxwell’s curl equations. Thus, applying (1.41a) to the electric field of (1.45) gives Hx=Hz=0, and Hy=j ωµ∂Ex ∂z=1 η(E+e−jkz−E−ejkz), (1.49) where η=ωµ/ k=√µ//epsilon1 is known as the intrinsic impedance of the medium. The ratio of the ¯Eand¯Hfield components is seen to have units of impedance, known as the wave impedance; for planes waves the wave impedance is equal to the intrinsic impedance of the medium. In free-space the intrinsic impedance is η0=√µ0//epsilon10=377/Omega1. Note that the ¯E and¯Hvectors are orthogonal to each other and orthogonal to the direction of propagation (±ˆz); this is a characteristic of transverse electromagnetic (TEM) waves. EXAMPLE 1.1 BASIC PLANE WA VE PARAMETERS A plane wave propagating in a lossless dielectric medium has an electric field given as Ex=E0cos(ωt−βz)with a frequency of 5.0 GHz and a wavelength in the material of 3.0 cm. Determine the propagation constant, the phase velocity, the relative permittivity of the medium, and the wave impedance. Solution From (1.48) the propagation constant is k=2π λ=2π 0.03=209.4m−1, and from (1.47) the phase velocity is vp=ω k=2πf k=λf=(0.03)(5×109)=1.5×108m/sec. This is slower than the speed of light by a factor of 2.0. The relative permittivity of the medium can be found from (1.47) as /epsilon1r=/parenleftbiggc vp/parenrightbigg2 =/parenleftbigg3.0×108 1.5×108/parenrightbigg2 =4.0 The wave impedance is η=η0/√/epsilon1r=377√ 4.0=188.5 /Omega1 ■ PlaneWavesinaGeneralLossyMedium Now consider the effect of a lossy medium. If the medium is conductive, with a conductiv- ityσ, Maxwell’s curl equations can be written, from (1.41a) and (1.20) as ∇ׯE=− jωµ¯H, (1.50a) ∇ׯH=jω/epsilon1¯E+σ¯E. (1.50b) c01ElectromagneticTheory Pozar July 28, 2011 8:7 18 Chapter 1: Electromagnetic Theory The resulting wave equation for ¯Ethen becomes ∇2¯E+ω2µ/epsilon1/parenleftBig 1−jσ ω/epsilon1/parenrightBig ¯E=0,( 1.51) where we see a similarity with (1.42), the wave equation for ¯Ein the lossless case. The difference is that the quantity k2=ω2µ/epsilon1of (1.42) is replaced by ω2µ/epsilon1[1−j(σ/ω/epsilon1)] in (1.51). We then define a complex propagation constant for the medium as γ=α+jβ=jω√µ/epsilon1/radicalbigg 1−jσ ω/epsilon1(1.52) where αis the attenuation constant andβis the phase constant. If we again assume an electric field with only an ˆxcomponent and uniform in xand y, the wave equation of (1.51) reduces to ∂2Ex ∂z2−γ2Ex=0,( 1.53) which has solutions Ex(z)=E+e−γz+E−eγz.( 1.54) The positive traveling wave then has a propagation factor of the form e−γz=e−αze−jβz, which in the time domain is of the form e−αzcos(ω t−βz). We see that this represents a wave traveling in the +zdirection with a phase velocity vp=ω/β, a wavelength λ=2π/β , and an exponential damping factor. The rate of decay with distance is given by the attenuation constant, α. The negative traveling wave term of (1.54) is similarly damped along the −zaxis. If the loss is removed, σ=0, and we have γ=jkandα=0,β=k. As discussed in Section 1.3, loss can also be treated through the use of a complex permittivity. From (1.52) and (1.20) with σ=0b u t/epsilon1=/epsilon1/prime−j/epsilon1/prime/primecomplex, we have that γ=jω√µ/epsilon1=jk=jω/radicalbig µ/epsilon1/prime(1−jtanδ), (1.55) where tan δ=/epsilon1/prime/prime//epsilon1/primeis the loss tangent of the material. The associated magnetic field can be calculated as Hy=j ωµ∂Ex ∂z=−jγ ωµ(E+e−γz−E−eγz). (1.56) The intrinsic impedance of the conducting medium is now complex, η=jωµ γ,( 1.57) but is still identified as the wave impedance, which expresses the ratio of electric to mag- netic field components. This allows (1.56) to be rewritten as Hy=1 η(E+e−γz−E−eγz). (1.58) Note that although ηof (1.57) is, in general, complex, it reduces to the lossless case of η=√µ//epsilon1 whenγ=jk=jω√µ/epsilon1. c01ElectromagneticTheory Pozar July 28, 2011 8:7 1.4 The Wave Equation and Basic Plane Wave Solutions 19 PlaneWavesinaGoodConductor Many problems of practical interest involve loss or attenuation due to good (but not perfect) conductors. A good conductor is a special case of the preceding analysis, where the con- ductive current is much greater than the displacement current, which means that σ/greatermuchω/epsilon1. Most metals can be categorized as good conductors. In terms of a complex /epsilon1, rather than conductivity, this condition is equivalent to /epsilon1/prime/prime/greatermuch/epsilon1/prime. The propagation constant of (1.52) can then be adequately approximated by ignoring the displacement current term, to give γ=α+jβ/similarequaljω√µ/epsilon1/radicalbiggσ jω/epsilon1=(1+j)/radicalbiggωµσ 2.( 1.59) Theskin depth, or characteristic depth of penetration, is defined as δs=1 α=/radicalBigg 2 ωµσ.( 1.60) Thus the amplitude of the fields in the conductor will decay by an amount 1/ e, or 36.8%, after traveling a distance of one skin depth, because e−αz=e−αδ s=e−1.A tm i c r o w a v e frequencies, for a good conductor, this distance is very small. The practical importance ofthis result is that only a thin plating of a good conductor (e.g., silver or gold) is necessary for low-loss microwave components. EXAMPLE 1.2 SKIN DEPTH AT MICROWA VE FREQUENCIES Compute the skin depth of aluminum, copper, gold, and silver at a frequency of 10 GHz. Solution The conductivities for these metals are listed in Appendix F. Equation (1.60) givesthe skin depths as δ s=/radicalBigg 2 ωµσ=/radicalBigg 1 πfµ0σ=/radicalBigg 1 π(1010)(4π×10−7)/radicalbigg 1 σ =5.03×10−3/radicalbigg 1 σ. For aluminum: δs=5.03×10−3/radicalbigg 1 3.816×107=8.14×10−7m. For copper: δs=5.03×10−3/radicalbigg 1 5.813×107=6.60×10−7m. For gold: δs=5.03×10−3/radicalbigg 1 4.098×107=7.86×10−7m. For silver: δs=5.03×10−3/radicalbigg 1 6.173×107=6.40×10−7m. These results show that most of the current flow in a good conductor occurs in an extremely thin region near the surface of the conductor. ■ c01ElectromagneticTheory Pozar July 28, 2011 8:7 20 Chapter 1: Electromagnetic Theory TABLE 1.1 Summary of Results for Plane Wave Propagation in Various Media Type of Medium Lossless General Good Conductor Quantity (/epsilon1/prime/prime=σ=0) Lossy (/epsilon1/prime/prime/greatermuch/epsilon1/primeorσ/greatermuchω/epsilon1/prime) Complex propagation γ=jω√µ/epsilon1 γ =jω√µ/epsilon1 γ =(1+j)√ωµσ/ 2 constant =jω/radicalbig µ/epsilon1/prime/radicalbigg 1−jσ ω/epsilon1/prime Phase constant β=k=ω√µ/epsilon1 β =Im{γ} β=Im{γ}=√ωµσ/ 2 (wave number) Attenuation constant α=0 α=Re{γ} α=Re{γ}=√ωµσ/ 2 Impedance η=√µ//epsilon1=ωµ/ k η=jωµ/γ η =(1+j)√ωµ/2σ Skin depth δs=∞ δs=1/α δs=√2/ωµσ Wavelength λ=2π/β λ=2π/β λ=2π/β Phase velocity vp=ω/β vp=ω/β vp=ω/β The intrinsic impedance inside a good conductor can be obtained from (1.57) and (1.59). The result is η=jωµ γ/similarequal(1+j)/radicalbiggωµ 2σ=(1+j)1 σδs.( 1.61) Notice that the phase angle of this impedance is 45◦, a characteristic of good conductors. The phase angle of the impedance for a lossless material is 0◦, and the phase angle of the impedance of an arbitrary lossy medium is somewhere between 0◦and 45◦. Table 1.1 summarizes the results for plane wave propagation in lossless and lossy homogeneous media. 1.5GENERALPLANEWAVESOLUTIONS Some specific features of plane waves were discussed in Section 1.4, but we will now look at plane waves from a more general point of view and solve the wave equation by the method of separation of variables. This technique will find application in succeedingchapters. We will also discuss circularly polarized plane waves, which will be important for the discussion of ferrites in Chapter 9. In free-space, the Helmholtz equation for ¯Ecan be written as ∇ 2¯E+k2 0¯E=∂2¯E ∂x2+∂2¯E ∂y2+∂2¯E ∂z2+k2 0¯E=0,( 1.62) and this vector wave equation holds for each rectangular component of ¯E: ∂2Ei ∂x2+∂2Ei ∂y2+∂2Ei ∂z2+k2 0Ei=0,( 1.63) where the index i=x,y,orz. This equation can be solved by the method of separation of variables , a standard technique for treating such partial differential equations. The method begins by assuming that the solution to (1.63) for, say, Ex, can be written as a product of three functions for each of the three coordinates: Ex(x,y,z)=f(x)g(y)h(z). (1.64) c01ElectromagneticTheory Pozar July 28, 2011 8:7 1.5 General Plane Wave Solutions 21 Substituting this form into (1.63) and dividing by fghgives f/prime/prime f+g/prime/prime g+h/prime/prime h+k2 0=0,( 1.65) where the double primes denote the second derivative. The key step in the argument is to recognize that each of the terms in (1.65) must be equal to a constant because they areindependent of each other. That is, f /prime/prime/fis only a function of x, and the remaining terms in (1.65) do not depend on x,s of/prime/prime/fmust be a constant, and similarly for the other terms in (1.65). Thus, we define three separation constants, kx,ky, and kz, such that f/prime/prime/f=−k2 x;g/prime/prime/g=−k2 y;h/prime/prime/h=−k2 z; or d2f dx2+k2 xf=0;d2g dy2+k2 yg=0;d2h dz2+k2 zh=0.( 1.66) Combining (1.65) and (1.66) shows that k2 x+k2 y+k2 z=k2 0.( 1.67) The partial differential equation of (1.63) has now been reduced to three separate ordinary differential equations in (1.66). Solutions to these equations have the forms e±jkxx,e±jkyy, ande±jkzz, respectively. As we saw in the previous section, the terms with +signs result in waves traveling in the negative x,y,o rzdirection, while the terms with −signs result in waves traveling in the positive direction. Both solutions are possible and are valid; the amount to which these various terms are excited is dependent on the source of the fields and the boundary conditions. For our present discussion we will select a plane wave traveling in the positive direction for each coordinate and write the complete solution for Exas Ex(x,y,z)=Ae−j(kxx+kyy+kzz),( 1.68) where Ais an arbitrary amplitude constant. Now define a wave number vector ¯kas ¯k=kxˆx+kyˆy+kzˆz=k0ˆn.( 1.69) Then from (1.67), |¯k|=k 0, and so ˆnis a unit vector in the direction of propagation. Also define a position vector as ¯r=xˆx+yˆy+zˆz; (1.70) then (1.68) can be written as Ex(x,y,z)=Ae−j¯k·¯r.( 1.71) Solutions to (1.63) for EyandEzare, of course, similar in form to Exof (1.71), but with different amplitude constants: Ey(x,y,z)=Be−j¯k·¯r, (1.72) Ez(x,y,z)=Ce−j¯k·¯r. (1.73) The x,y,andzdependences of the three components of ¯Ein (1.71)–(1.73) must be the same (same kx,ky,kz), because the divergence condition that ∇·¯E=∂Ex ∂x+∂Ey ∂y+∂Ez ∂z=0 c01ElectromagneticTheory Pozar July 28, 2011 8:7 22 Chapter 1: Electromagnetic Theory must also be applied in order to satisfy Maxwell’s equations, and this implies that Ex,Ey, andEzmust each have the same variation in x,y, and z. (Note that the solutions in the preceding section automatically satisfied the divergence condition because Exwas the only component of ¯E, and Exdid not vary with x.) This condition also imposes a constraint on the amplitudes A,B, and Cbecause if ¯E0=Aˆx+Bˆy+Cˆz, we have ¯E=¯E0e−j¯k·¯r, and ∇·¯E=∇· (¯E0e−j¯k·¯r)=¯E0·∇e−j¯k·¯r=− j¯k·¯E0e−j¯k·¯r=0, where vector identity (B.7) was used. Thus, we must have ¯k·¯E0=0,( 1.74) which means that the electric field amplitude vector ¯E0must be perpendicular to the direc- tion of propagation, ¯k. This condition is a general result for plane waves and implies that only two of the three amplitude constants, A,B,andC, can be chosen independently. The magnetic field can be found from Maxwell’s equation, ∇ׯE=− jωµ0¯H,( 1.75) to give ¯H=j ωµ0∇ׯE=j ωµ0∇×(¯E0e−j¯k·¯r) =−j ωµ0¯E0×∇e−j¯k·¯r =−j ωµ0¯E0×(−j¯k)e−j¯k·¯r =k0 ωµ0ˆnׯE0e−j¯k·¯r =1 η0ˆnׯE0e−j¯k·¯r =1 η0ˆnׯE, (1.76) where vector identity (B.9) was used in obtaining the second line. This result shows that the magnetic field vector ¯Hlies in a plane normal to ¯k, the direction of propagation, and that¯His perpendicular to ¯E. See Figure 1.8 for an illustration of these vector relations. The quantity η0=√µ0//epsilon10=377/Omega1in (1.76) is the intrinsic impedance of free-space. The time domain expression for the electric field can be found as ¯E(x,y,z,t)=Re/braceleftbig¯E(x,y,z)ejωt/bracerightbig =Re/braceleftbig¯E0e−j¯k·¯rejωt/bracerightbig =¯E0cos(¯k·¯r−ωt), (1.77) c01ElectromagneticTheory Pozar July 28, 2011 8:7 1.5 General Plane Wave Solutions 23 Enz y xHˆ FIGURE 1.8 Orientation of the ¯E,¯H,and¯k=k0ˆnvectors for a general plane wave. assuming that the amplitude constants A,B, and Ccontained in ¯E0are real. If these constants are not real, their phases should be included inside the cosine term of (1.77). Itis easy to show that the wavelength and phase velocity for this solution are the same as obtained in Section 1.4. EXAMPLE 1.3 CURRENT SHEETS AS SOURCES OF PLANE WA VES An infinite sheet of surface current can be considered as a source for plane waves. If an electric surface current density ¯Js=J0ˆxexists on the z=0 plane in free- space, find the resulting fields by assuming plane waves on either side of the current sheet and enforcing boundary conditions. Solution Since the source does not vary with xory, the fields will not vary with xorybut will propagate away from the source in the ±zdirection. The boundary conditions to be satisfied at z=0a r e ˆn×(¯E2−¯E1)=ˆz×(¯E2−¯E1)=0, ˆn×(¯H2−¯H1)=ˆz×(¯H2−¯H1)=J0ˆx, where ¯E1,¯H1are the fields for z<0, and ¯E2,¯H2are the fields for z>0. To satisfy the second condition, ¯Hmust have a ˆycomponent. Then for ¯Eto be or- thogonal to ¯Handˆz,¯Emust have an ˆxcomponent. Thus the fields will have the following form: forz<0, ¯E1=ˆxAη0ejk0z, ¯H1=−ˆ yA ejk0z, forz>0, ¯E2=ˆxBη0e−jk0z, ¯H2=ˆyBe−jk0z, where AandBare arbitrary amplitude constants. The first boundary condition, that Exis continuous at z=0, yields A=B, while the boundary condition for ¯Hyields the equation −B−A=J0. Solving for A,Bgives A=B=− J0/2, which completes the solution. ■ c01ElectromagneticTheory Pozar July 28, 2011 8:7 24 Chapter 1: Electromagnetic Theory CircularlyPolarizedPlaneWaves The plane waves discussed previously all had their electric field vector pointing in a fixed direction and so are called linearly polarized waves. In general, the polarization of a plane wave refers to the orientation of the electric field vector, which may be in a fixed directionor may change with time. Consider the superposition of an ˆxlinearly polarized wave with amplitude E 1and a ˆy linearly polarized wave with amplitude E2, both traveling in the positive ˆzdirection. The total electric field can be written as ¯E=(E1ˆx+E2ˆy)e−jk0z.( 1.78) A number of possibilities now arise. If E1/negationslash=0 and E2=0, we have a plane wave linearly polarized in the ˆxdirection. Similarly, if E1=0 and E2/negationslash=0, we have a plane wave linearly polarized in the ˆydirection. If E1andE2are both real and nonzero, we have a plane wave linearly polarized at the angle φ=tan−1E2 E1. For example, if E1=E2=E0,w eh a v e ¯E=E0(ˆx+ˆy)e−jk0z, which represents an electric field vector at a 45◦angle from the x-axis. Now consider the case in which E1=jE2=E0, where E0is real, so that ¯E=E0(ˆx−jˆy)e−jk0z.( 1.79) The time domain form of this field is ¯E(z,t)=E0[ˆxcos(ω t−k0z)+ˆycos(ωt−k0z−π/2)].( 1.80) This expression shows that the electric field vector changes with time or, equivalently, with distance along the z-axis. To see this, pick a fixed position, say z=0. Equation (1.80) then reduces to ¯E(0,t)=E0[ˆxcosωt+ˆysinωt],( 1.81) so asωtincreases from zero, the electric field vector rotates counterclockwise from the x-axis. The resulting angle from the x-axis of the electric field vector at time t,a tz=0, is then φ=tan−1/parenleftbiggsinωt cosωt/parenrightbigg =ωt, which shows that the polarization rotates at the uniform angular velocity ω. Since the fingers of the right hand point in the direction of rotation of the electric field vector when the thumb points in the direction of propagation, this type of wave is referred to as a right- hand circularly polarized (RHCP) wave. Similarly, a field of the form ¯E=E0(ˆx+jˆy)e−jk0z(1.82) constitutes a left-hand circularly polarized (LHCP) wave, where the electric field vector rotates in the opposite direction. See Figure 1.9 for a sketch of the polarization vectors for RHCP and LHCP plane waves. The magnetic field associated with a circularly polarized wave may be found from Maxwell’s equations or by using the wave impedance applied to each component of the c01ElectromagneticTheory Pozar July 28, 2011 8:7 1.6 Energy and Power 25 xy (0, t) zPropagation (a)xy (0, t) zPropagation (b)ee FIGURE 1.9 Electric field polarization for (a) RHCP and (b) LHCP plane waves. electric field. For example, applying (1.76) to the electric field of a RHCP wave as given in (1.79) yields ¯H=E0 η0ˆz×(ˆx−jˆy)e−jk0z=E0 η0(ˆy+jˆx)e−jk0z=jE0 η0(ˆx−jˆy)e−jk0z, which is also seen to represent a vector rotating in the RHCP sense. 1.6ENERGYANDPOWER In general, a source of electromagnetic energy sets up fields that store electric and magnetic energy and carry power that may be transmitted or dissipated as loss. In the sinusoidal steady-state case, the time-average stored electric energy in a volume Vis given by We=1 4Re/integraldisplay V¯E·¯D∗dv, (1.83) which in the case of simple lossless isotropic, homogeneous, linear media, where /epsilon1is a real scalar constant, reduces to We=/epsilon1 4/integraldisplay V¯E·¯E∗dv. (1.84) Similarly, the time-average magnetic energy stored in the volume Vis Wm=1 4Re/integraldisplay V¯H·¯B∗dv, (1.85) which becomes Wm=µ 4/integraldisplay V¯H·¯H∗dv, (1.86) for a real, constant, scalar µ. We can now derive Poynting’s theorem, which leads to energy conservation for elec- tromagnetic fields and sources. If we have an electric source current ¯Jsand a conduction current σ¯Eas defined in (1.19), then the total electric current density is ¯J=¯Js+σ¯E. Multiplying (1.27a) by ¯H∗and multiplying the conjugate of (1.27b) by ¯Eyields ¯H∗·(∇× ¯E)=− jωµ|¯H|2−¯H∗·¯Ms, ¯E·(∇× ¯H∗)=¯E·¯J∗−jω/epsilon1∗|¯E|2=¯E·¯J∗ s+σ|¯E|2−jω/epsilon1∗|¯E|2, c01ElectromagneticTheory Pozar July 28, 2011 8:7 26 Chapter 1: Electromagnetic Theory FIGURE 1.10 A volume V, enclosed by the closed surface S, containing fields ¯E,¯H,and current sources ¯Js,¯Ms. where ¯Msis the magnetic source current. Using these two results in vector identity (B.8) gives ∇·(¯EׯH∗)=¯H∗·(∇× ¯E)−¯E·(∇× ¯H∗) =−σ|¯E|2+jω(/epsilon1∗|¯E|2−µ|¯H|2)−(¯E·¯J∗ s+¯H∗·¯Ms). Now integrate over a volume Vand use the divergence theorem: /integraldisplay V∇·(¯EׯH∗)dv=/contintegraldisplay S¯EׯH∗·d¯s =−σ/integraldisplay V|¯E|2dv+jω/integraldisplay V(/epsilon1∗|¯E|2−µ|¯H|2)dv−/integraldisplay V(¯E·¯J∗ s+¯H∗·¯Ms)dv,(1.87) where Sis a closed surface enclosing the volume V, as shown in Figure 1.10. Allowing /epsilon1=/epsilon1/prime−j/epsilon1/prime/primeandµ=µ/prime−jµ/prime/primeto be complex to allow for loss, and rewriting (1.87), gives −1 2/integraldisplay V(¯E·¯J∗ s+¯H∗·¯Ms)dv=1 2/contintegraldisplay S¯EׯH∗·d¯s+σ 2/integraldisplay V|¯E|2dv +ω 2/integraldisplay V(/epsilon1/prime/prime|¯E|2+µ/prime/prime|¯H|2)dv+jω 2/integraldisplay V(µ/prime|¯H|2−/epsilon1/prime|¯E|2)dv. (1.88) This result is known as Poynting’s theorem, after the physicist J. H. Poynting (1852–1914), and is basically a power balance equation. Thus, the integral on the left-hand side repre- sents the complex power Psdelivered by the sources ¯Jsand¯Msinside S: Ps=−1 2/integraldisplay V(¯E·¯J∗ s+¯H∗·¯Ms)dv. (1.89) The first integral on the right-hand side of (1.88) represents complex power flow out of the closed surface S. If we define a quantity ¯S, called the Poynting vector,a s ¯S=¯EׯH∗,( 1.90) then this power can be expressed as Po=1 2/contintegraldisplay S¯EׯH∗·d¯s=1 2/contintegraldisplay S¯S·d¯s.( 1.91) The surface Sin (1.91) must be a closed surface for this interpretation to be valid. The real parts of PsandPoin (1.89) and (1.91) represent time-average powers. The second and third integrals in (1.88) are real quantities representing the time- average power dissipated in the volume Vdue to conductivity, dielectric, and magnetic losses. If we define this power as P/lscriptwe have P/lscript=σ 2/integraldisplay V|¯E|2dv+ω 2/integraldisplay V(/epsilon1/prime/prime|¯E|2+µ/prime/prime|¯H|2)dv, (1.92) c01ElectromagneticTheory Pozar July 28, 2011 8:7 1.6 Energy and Power 27 which is sometimes referred to as Joule’s law. The last integral in (1.88) can be seen to be related to the stored electric and magnetic energies, as defined in (1.84) and (1.86). With the above definitions, Poynting’s theorem can be rewritten as Ps=Po+P/lscript+2jω(Wm−We). (1.93) In words, this complex power balance equation states that the power delivered by the sources ( Ps)is equal to the sum of the power transmitted through the surface ( Po), the power lost to heat in the volume ( P/lscript), and 2ω times the net reactive energy stored in the volume. PowerAbsorbedbyaGoodConductor Practical transmission lines involve imperfect conductors, leading to attenuation and power losses, as well as the generation of noise. To calculate loss and attenuation due to an im- perfect conductor we must find the power dissipated in the conductor. We will show that this can be accomplished using only the fields at the surface of the conductor, which is a very helpful simplification when calculating attenuation. Consider the geometry of Figure 1.11, which shows the interface between a lossless medium and a good conductor. A field is incident from z<0, and the field penetrates into the conducting region, z>0. The real average power entering the conductor volume de- fined by the cross-sectional area S0at the interface and the surface Sis given from (1.91) as Pavg=1 2Re/integraldisplay S0+S¯EׯH∗·ˆnd s,( 1.94) where ˆnis a unit normal vector pointing into the closed surface S0+S, and ¯E,¯Hare the fields over this surface. The contribution to the integral in (1.94) from the surface Scan be made zero by proper selection of this surface. For example, if the field is a normallyincident plane wave, the Poynting vector ¯S=¯EׯH ∗will be in the ˆzdirection, and so tangential to the top, bottom, front, and back of S, if these walls are made parallel to the z-axis. If the wave is obliquely incident, these walls can be slanted to obtain the same result. If the conductor is good, the decay of the fields away from the interface at z=0 will be very rapid, so the right-hand end of Scan be made far enough away from z=0 such that there is negligible contribution to the integral from this part of the surface S.T h e n n Szx S0Pn = z/H9262, /H9280/H9268 >> /H9275/H9280 ˆˆ ˆˆ FIGURE 1.11 An interface between a lossless medium and a good conductor with a closed sur- face S0+Sfor computing the power dissipated in the conductor. c01ElectromagneticTheory Pozar July 28, 2011 8:7 28 Chapter 1: Electromagnetic Theory time-average power entering the conductor through S0can then be written as Pavg=1 2Re/integraldisplay S0¯EׯH∗·ˆzd s.( 1.95) From vector identity (B.3) we have ˆz·(¯EׯH∗)=(ˆzׯE)·¯H∗=η¯H·¯H∗,( 1.96) since ¯H=ˆnׯE/η, as generalized from (1.76) for conductive media, where ηis the in- trinsic impedance (complex) of the conductor. Equation (1.95) can then be written as Pavg=Rs 2/integraldisplay S0|¯H|2ds,( 1.97) where Rs=Re{η}=Re/braceleftbigg (1+j)/radicalbiggωµ 2σ/bracerightbigg =/radicalbiggωµ 2σ=1 σδs(1.98) is defined as the surface resistance of the conductor. The magnetic field ¯Hin (1.97) is tangential to the conductor surface and needs only to be evaluated at the surface of the con-ductor; since H tis continuous at z=0, it does not matter whether this field is evaluated just outside the conductor or just inside the conductor. In the next section we will show how (1.97) can be evaluated in terms of a surface current density flowing on the surface of the conductor, where the conductor can be approximated as perfect. 1.7PLANEWAVEREFLECTIONFROMAMEDIAINTERFACE A number of problems to be considered in later chapters involve the behavior of electro- magnetic fields at the interface of various types of media, including lossless media, lossy media, a good conductor, or a perfect conductor, and so it is beneficial at this time to study the reflection of a plane wave normally incident from free-space onto a half-space of anarbitrary material. The geometry is shown in Figure 1.12, where the material half-space z>0 is characterized by the parameters /epsilon1, µ, andσ. GeneralMedium With no loss of generality we can assume that the incident plane wave has an electric field vector oriented along the x-axis and is propagating along the positive z-axis. The incident x /H92800, /H92620 /H9280, /H9262, /H9268 zEi ErEt FIGURE 1.12 Plane wave reflection from an arbitrary medium; normal incidence. c01ElectromagneticTheory Pozar July 28, 2011 8:7 1.7 Plane Wave Reflection from a Media Interface 29 fields can then be written, for z<0, as ¯Ei=ˆxE0e−jk0z, (1.99a) ¯Hi=ˆy1 η0E0e−jk0z, (1.99b) where η0is the impedance of free-space and E0is an arbitrary amplitude. Also in the region z<0, a reflected wave may exist with the form ¯Er=ˆx/Gamma1E0e+jk0z, (1.100a) ¯Hr=−ˆ y/Gamma1 η0E0e+jk0z, (1.100b) where /Gamma1is the unknown reflection coefficient of the reflected electric field. Note that in (1.100), the sign in the exponential terms has been chosen as positive, to represent waves traveling in the −ˆzdirection of propagation, as derived in (1.46). This is also consis- tent with the Poynting vector ¯Sr=¯ErׯH∗ r=− |/Gamma1|2|E0|2ˆz/η0, which shows power to be traveling in the −ˆzdirection for the reflected wave. As shown in Section 1.4, from equations (1.54) and (1.58), the transmitted fields for z>0 in the lossy medium can be written as ¯Et=ˆxTE 0e−γz, (1.101a) ¯Ht=ˆyTE 0 ηe−γz, (1.101b) where Tis the transmission coefficient of the transmitted electric field and ηis the intrinsic impedance (complex) of the lossy medium in the region z>0. From (1.57) and (1.52) the intrinsic impedance is η=jωµ γ,( 1.102) and the propagation constant is γ=α+jβ=jω√µ/epsilon1/radicalbig 1−jσ/ω/epsilon1. (1.103) We now have a boundary value problem where the general form of the fields are known via (1.99)–(1.101) on either side of the material discontinuity at z=0. The two unknown constants /Gamma1andTare found by applying boundary conditions for ExandHyatz=0. Since these tangential field components must be continuous at z=0, we arrive at the fol- lowing two equations: 1+/Gamma1=T, (1.104a) 1−/Gamma1 η0=T η. (1.104b) Solving these equations for the reflection and transmission coefficients gives /Gamma1=η−η0 η+η0, (1.105a) T=1+/Gamma1=2η η+η0. (1.105b) This is a general solution for reflection and transmission of a normally incident wave at the interface of an arbitrary material, where ηis the intrinsic impedance of the material. We now consider three special cases of this result. c01ElectromagneticTheory Pozar July 28, 2011 8:7 30 Chapter 1: Electromagnetic Theory LosslessMedium If the region for z>0 is a lossless dielectric, then σ=0, and µand/epsilon1are real quantities. The propagation constant in this case is purely imaginary and can be written as γ=jβ=jω√µ/epsilon1=jk0√µr/epsilon1r,( 1.106) where k0=ω√µ0/epsilon10is the propagation constant (wave number) of a plane wave in free- space. The wavelength in the dielectric is λ=2π β=2π ω√µ/epsilon1=λ0√µr/epsilon1r,( 1.107) the phase velocity is vp=ω β=1√µ/epsilon1=c√µr/epsilon1r,( 1.108) (slower than the speed of light in free-space) and the intrinsic impedance of the dielectric is η=jωµ γ=/radicalbiggµ /epsilon1=η0/radicalbiggµr /epsilon1r.( 1.109) For this lossless case, ηis real, so both /Gamma1andTfrom (1.105) are real, and ¯Eand¯Hare in phase with each other in both regions. Power conservation for the incident, reflected, and transmitted waves can be demon- strated by computing the Poynting vectors in the two regions. Thus, for z<0, the complex Poynting vector is found from the total fields in this region as ¯S−=¯EׯH∗=(¯Ei+¯Er)×(¯Hi+¯Hr)∗ =ˆz|E0|21 η0(e−jk0z+/Gamma1ejk0z)(e−jk0z−/Gamma1ejk0z)∗ =ˆz|E0|21 η0(1−|/Gamma1|2+/Gamma1e2jk0z−/Gamma1∗e−2jk0z) =ˆz|E0|21 η0(1−|/Gamma1|2+2j/Gamma1sin 2k 0z), (1.110a) since/Gamma1is real. For z>0 the complex Poynting vector is ¯S+=¯EtׯH∗ t=ˆz|E0|2|T|2 η, which can be rewritten, using (1.105), as ¯S+=ˆz|E0|2 4η (η+η0)2=ˆz|E0|21 η0(1−|/Gamma1|2). (1.110b) Now observe that at z=0,¯S−=¯S+, so that complex power flow is conserved across the interface. Next consider the time-average power flow in the two regions. For z<0t h e time-average power flow through a 1 m2cross section is P−=1 2Re/braceleftbig¯S−·ˆz/bracerightbig =1 2|E0|21 η0(1−|/Gamma1|2). (1.111a) c01ElectromagneticTheory Pozar July 28, 2011 8:7 1.7 Plane Wave Reflection from a Media Interface 31 and for z>0, the time-average power flow through a 1 m2cross section is P+=1 2Re/braceleftbig¯S+·ˆz/bracerightbig =1 2|E0|21 η0(1−|/Gamma1|2)=P−,( 1.111b) so real power flow is conserved. We now note a subtle point. When computing the complex Poynting vector for z<0i n (1.110a), we used the total ¯Eand¯Hfields. If we compute separately the Poynting vectors for the incident and reflected waves, we obtain ¯Si=¯EiׯH∗ i=ˆz|E0|2 η0, (1.112a) ¯Sr=¯ErׯH∗ r=− ˆz|E0|2|/Gamma1|2 η0, (1.112b) and we see that ¯Si+¯Sr/negationslash=¯S−of (1.110a). The missing cross-product terms account for stored reactive energy in the standing wave in the z<0 region. Thus, the decomposition of a Poynting vector into incident and reflected components is not, in general, meaningful. It is possible to define a time-average Poynting vector as (1/2)Re {¯EׯH∗}, and in this case such a definition applied to the individual incident and reflected components will givethe correct result since P i=(1/2)| ¯E0|2/η0andPr=(−1/2)| E0|2|/Gamma1|2/η0,soPi+Pr= P−.However, this definition will fail to provide meaningful results when the medium for z<0 is lossy. GoodConductor If the region for z>0 is a good (but not perfect) conductor, the propagation constant can be written as discussed in Section 1.4: γ=α+jβ=(1+j)/radicalbiggωµσ 2=(1+j)1 δs.( 1.113) Similarly, the intrinsic impedance of the conductor simplifies to η=(1+j)/radicalbiggωµ 2σ=(1+j)1 σδs.( 1.114) Now the impedance is complex, with a phase angle of 45◦,s o¯Eand¯Hwill be 45◦out of phase, and /Gamma1andTwill be complex. In (1.113) and (1.114), δs=1/αis the skin depth, as defined in (1.60). Forz<0 the complex Poynting vector can be evaluated at z=0t og i v e ¯S−(z=0)=ˆz|E0|21 η0(1−|/Gamma1|2+/Gamma1−/Gamma1∗). (1.115a) Forz>0 the complex Poynting vector is ¯S+=¯EtׯH∗ t=ˆz|E0|2|T|21 η∗e−2α z, and using (1.105) for Tand/Gamma1gives ¯S+=ˆz|E0|2 4η |η+η0|2e−2α z=ˆz|E0|21 η0(1−|/Gamma1|2+/Gamma1−/Gamma1∗)e−2α z.( 1.115b) So at the interface at z=0,¯S−=¯S+, and complex power is conserved. c01ElectromagneticTheory Pozar July 28, 2011 8:7 32 Chapter 1: Electromagnetic Theory Observe that if we were to compute the separate incident and reflected Poynting vec- tors for z<0a s ¯Si=¯EiׯH∗ i=ˆz|E0|2 η0, (1.116a) ¯Sr=¯ErׯH∗ r=− ˆz|E0|2|/Gamma1|2 η0, (1.116b) we would not obtain ¯Si+¯Sr=¯S−of (1.115a), even for z=0. It is possible, however, to consider real power flow in terms of the individual traveling wave components. Thus, the time-average power flows through a 1 m2cross section are P−=1 2Re(¯S−·ˆz)=1 2|E0|21 η0(1−|/Gamma1|2), (1.117a) P+=1 2Re(¯S−·ˆz)=1 2|E0|21 η0(1−|/Gamma1|2)e−2α z, (1.117b) which shows power balance at z=0. In addition, Pi=|E0|2/2η 0andPr=− | E0|2|/Gamma1|2/ 2η0, so that Pi+Pr=P−, showing that the real power flow for z<0 can be decomposed into incident and reflected wave components. Notice that ¯S+, the power density in the lossy conductor, decays exponentially accord- i n gt ot h ee−2α zattenuation factor. This means that power is being dissipated in the lossy material as the wave propagates into the medium in the +zdirection. The power, and also the fields, decay to a negligibly small value within a few skin depths of the material, which for a reasonably good conductor is an extremely small distance at microwave frequencies. The electric volume current density flowing in the conducting region is given as ¯Jt=σ¯Et=ˆxσE0Te−γzA/m2,( 1.118) and so the average power dissipated in (or transmitted into) a 1 m2cross-sectional volume of the conductor can be calculated from the conductor loss term of (1.92) (Joule’s law) as Pt=1 2/integraldisplay V¯Et·¯J∗tdv=1 2/integraldisplay1 x=0/integraldisplay1 y=0/integraldisplay∞ z=0(ˆxE0Te−γz)·(ˆxσE0Te−γz)∗dzdy dx =1 2σ|E0|2|T|2/integraldisplay∞ z=0e−2α zdz=σ|E0|2|T|2 4α. (1.119) Since 1/η =σδs/(1+j)=(σ/2α)( 1−j), the real power entering the conductor through a1m2cross section [as given by (1/2)Re{ ¯S+·ˆz}atz=0] can be expressed using (1.115b) asPt=|E0|2|T|2(σ/4α) , which is in agreement with (1.119). PerfectConductor Now assume that the region z>0 contains a perfect conductor. The above results can be specialized to this case by allowing σ→∞ . Then, from (1.113), α→∞ ; from (1.114), η→0; from (1.60), δs→0; and from (1.105a, b), T→0 and /Gamma1→− 1. The fields for z>0 thus decay infinitely fast and are identically zero in the perfect conductor. The perfect conductor can be thought of as “shorting out” the incident electric field. For z<0, from (1.99) and (1.100), the total ¯Eand¯Hfields are, since /Gamma1=−1, ¯E=¯Ei+¯Er=ˆxE0(e−jk0z−ejk0z)=−ˆx2jE0sink0z, (1.120a) ¯H=¯Hi+¯Hr=ˆy1 η0E0(e−jk0z+ejk0z)=ˆy2 η0E0cosk0z. (1.120b) c01ElectromagneticTheory Pozar July 28, 2011 8:7 1.7 Plane Wave Reflection from a Media Interface 33 Observe that at z=0,¯E=0 and ¯H=ˆy(2/η 0)E0. The Poynting vector for z<0i s ¯S−=¯EׯH∗=− ˆzj4 η0|E0|2sink0zcosk0z,( 1.121) which has a zero real part and thus indicates that no real power is delivered to the perfect conductor. The volume current density of (1.118) for the lossy conductor reduces to an infinitely thin sheet of surface current in the limit of infinite conductivity: ¯Js=ˆnׯH=− ˆz×/parenleftbigg ˆy2 η0E0cosk0z/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle z=0=ˆx2 η0E0A/m.( 1.122) TheSurfaceImpedanceConcept In many problems, particularly those in which the effect of attenuation or conductor loss is needed, the presence of an imperfect conductor must be taken into account. The sur- face impedance concept allows us to do this in an approximate, but very convenient andaccurate, manner. We will develop this method from the theory presented in the previous sections. Consider a good conductor in the region z>0. As we have seen, a plane wave nor- mally incident on this conductor is mostly reflected, and the power that is transmitted into the conductor is dissipated as heat within a very short distance from the surface. There are three ways to compute this power. First, we can use Joule’s law, as in (1.119). For a 1 m 2area of conductor surface, the power transmitted through this surface and dissipated as heat is given by (1.119). Using (1.105b) for T, (1.114) for η, and the fact that α=1/δsgives the following result: σ|T|2 α=σδs4|η|2 |η+η0|2/similarequal8 σδsη2 0,( 1.123) where we have assumed η/lessmuchη0, which is true for a good conductor. Then the power of (1.119) can be written as Pt=σ|E0|2|T|2 4α=2|E0|2 σδsη2 0=2|E0|2Rs η2 0,( 1.124) where Rs=Re{η}=Re/braceleftbigg1+j σδs/bracerightbigg =1 σδs=/radicalbiggωµ 2σ(1.125) is the surface resistance of the metal. Another way to find the power loss is to compute the power flow into the conductor using the Poynting vector since all power entering the conductor at z=0 is dissipated. As in (1.115b), we have Pt=1 2Re/braceleftbig¯S+·ˆz/bracerightbig/vextendsingle/vextendsingle z=0=2|E0|2Re{η} |η+η0|2, which for large conductivity becomes, since η/lessmuchη0, Pt=2|E0|2Rs η2 0,( 1.126) which agrees with (1.124). c01ElectromagneticTheory Pozar July 28, 2011 8:7 34 Chapter 1: Electromagnetic Theory A third method uses an effective surface current density and the surface impedance, without the need for knowing the fields inside the conductor. From (1.118), the volume current density in the conductor is ¯Jt=ˆxσTE0e−γzA/m2,( 1.127) so the total (surface) current flow per unit width in the xdirection is ¯Js=/integraldisplay∞ 0¯Jtdz=ˆxσTE0/integraldisplay∞ 0e−γzdz=ˆxσTE0 γA/m. Approximating σT/γfor large σand using (1.113), (1.105b), and (1.114) gives σT γ=σδs (1+j)2η (η+η0)/similarequalσδs (1+j)2(1+j) σδsη0=2 η0, so ¯Js/similarequalˆx2E0 η0A/m.( 1.128) If the conductivity were infinite, then /Gamma1=−1 and a true surface current density of ¯Js=ˆnׯH|z=0=− ˆz×(¯Hi+¯Hr)|z=0=ˆxE01 η0(1−/Gamma1)=ˆx2E0 η0A/m would flow, which is identical to the total current in (1.128). Now replace the exponentially decaying volume current of (1.127) with a uniform volume current extending a distance of one skin depth. Thus, let ¯Jt=/braceleftbigg¯Js/δs for 0<z<δ s 0f o r z>δ s,(1.129) so that the total current flow is the same. Then Joule’s law gives the power lost: Pt=1 2σ/integraldisplay S/integraldisplayδs z=0|¯Js|2 δ2sdzds =Rs 2/integraldisplay S|¯Js|2ds=2|E0|2Rs η2 0,( 1.130) where/integraltext Sdenotes a surface integral over the conductor surface, in this case chosen as 1 m2. The result of (1.130) agrees with our previous results for Ptin (1.126) and (1.124) and shows that the power loss in a good conductor can be accurately and simply calculated as Pt=Rs 2/integraldisplay S|¯Js|2ds=Rs 2/integraldisplay S|¯Ht|2ds,( 1.131) in terms of the surface resistance Rsand the surface current ¯Js, or tangential magnetic field ¯Ht. It is important to realize that the surface current can be found from ¯Js=ˆnׯH,a si f the metal were a perfect conductor. This method is very general, applying to fields other than plane waves and to conductors of arbitrary shape, as long as bends or corners haveradii on the order of a skin depth or larger. The method is also quite accurate, as the only approximation was that η/lessmuchη 0, which is a good approximation. As an example, copper at 1 GHz has |η|=0.012 /Omega1, which is indeed much less than η0=377/Omega1. EXAMPLE 1.4 PLANE WA VE REFLECTION FROM A CONDUCTOR Consider a plane wave normally incident on a half-space of copper. If f=1 GHz, compute the propagation constant, intrinsic impedance, and skin depth forthe conductor. Also compute the reflection and transmission coefficients. c01ElectromagneticTheory Pozar July 28, 2011 8:7 1.8 Oblique Incidence at a Dielectric Interface 35 Solution For copper, σ=5.813×107S/m, so from (1.60) the skin depth is δs=/radicalBigg 2 ωµσ=2.088×10−6m, and the propagation constant is, from (1.113), γ=1+j δs=(4.789 +j4.789) ×105m−1. The intrinsic impedance is, from (1.114), η=1+j σδs=(8.239 +j8.239) ×10−3/Omega1, which is quite small relative to the impedance of free-space ( η0=377/Omega1).T h e reflection coefficient is, from (1.105a), /Gamma1=η−η0 η+η0=1.0/negationslash179.99◦ (practically that of an ideal short circuit), and the transmission coefficient is T=2η η+η0=6.181 ×10−5/negationslash45◦. ■ 1.8OBLIQUEINCIDENCEATADIELECTRICINTERFACE We continue our discussion of plane waves by considering the problem of a plane wave obliquely incident on a plane interface between two lossless dielectric regions, as shown inFigure 1.13. There are two canonical cases of this problem: the electric field is either in the xzplane (parallel polarization) or normal to the xzplane (perpendicular polarization). An arbitrary incident plane wave, of course, may have a polarization that is neither of these,but it can be expressed as a linear combination of these two individual cases. The general method of solution is similar to the problem of normal incidence: we will write expressions for the incident, reflected, and transmitted fields in each region and matchboundary conditions to find the unknown amplitude coefficients and angles. zx Er, Hr Ei, HiEt, Ht/H9258r /H9258i/H9258t /H92802, /H92622 Region 2/H92801, /H92621 Region 1 FIGURE 1.13 Geometry for a plane wave obliquely incident at the interface between two dielec- tric regions. c01ElectromagneticTheory Pozar July 28, 2011 8:7 36 Chapter 1: Electromagnetic Theory ParallelPolarization In this case the electric field vector lies in the xzplane, and the incident fields can be written as ¯Ei=E0(ˆxcosθi−ˆzsinθi)e−jk1(xsinθi+zcosθi), (1.132a) ¯Hi=E0 η1ˆye−jk1(xsinθi+zcosθi), (1.132b) where k1=ω√µ0/epsilon11andη1=√µ0//epsilon11are the propagation constant and impedance of region 1. The reflected and transmitted fields can be written as ¯Er=E0/Gamma1(ˆxcosθr+ˆzsinθr)e−jk1(xsinθr−zcosθr), (1.133a) ¯Hr=−E0/Gamma1 η1ˆye−jk1(xsinθr−zcosθr), (1.133b) ¯Et=E0T(ˆxcosθt−ˆzsinθt)e−jk2(xsinθt+zcosθt), (1.134a) ¯Ht=E0T η2ˆye−jk2(xsinθt+zcosθt). (1.134b) Here,/Gamma1andTare the reflection and transmission coefficients, and k2andη2are the prop- agation constant and impedance of region 2, defined as k2=ω√µ0/epsilon12,η 2=/radicalbig µ0//epsilon12. At this point we have /Gamma1,T,θr,andθtas unknowns. We can obtain two complex equations for these unknowns by enforcing the continuity ofExandHy, the tangential field components, at the interface between the two regions at z=0. We then obtain cosθie−jk1xsinθi+/Gamma1cosθre−jk1xsinθr=Tcosθte−jk2xsinθt, (1.135a) 1 η1e−jk1xsinθi−/Gamma1 η1e−jk1xsinθr=T η2e−jk2xsinθt. (1.135b) Both sides of (1.135a) and (1.135b) are functions of the coordinate x.I fExandHyare to be continuous at the interface z=0 for all x, then this xvariation must be the same on both sides of the equations, leading to the following condition: k1sinθi=k1sinθr=k2sinθt. This results in the well-known Snell’s laws of reflection and refraction: θi=θr, (1.136a) k1sinθi=k2sinθt. (1.136b) The above argument ensures that the phase terms in (1.135) vary with xat the same rate on both sides of the interface, and so is often called the phase matching condition. Using (1.136) in (1.135) allows us to solve for the reflection and transmission coeffi- cients as /Gamma1=η2cosθt−η1cosθi η2cosθt+η1cosθi, (1.137a) T=2η2cosθi η2cosθt+η1cosθi. (1.137b) c01ElectromagneticTheory Pozar July 28, 2011 8:7 1.8 Oblique Incidence at a Dielectric Interface 37 Observe that for normal incidence θi=0, we have θr=θt=0, so then /Gamma1=η2−η1 η2+η1and T=2η2 η2+η1, which is in agreement with the results of Section 1.7. For this polarization a special angle of incidence, θb, called the Brewster angle,e x - ists where /Gamma1=0. This occurs when the numerator of (1.137a) goes to zero ( θi=θb): η2cosθt=η1cosθb,which can be rewritten using cosθt=/radicalBig 1−sin2θt=/radicalBigg 1−k2 1 k2 2sin2θb, to give sinθb=1√1+/epsilon11//epsilon12.( 1.138) PerpendicularPolarization In this case the electric field vector is perpendicular to the xzplane. The incident field can be written as ¯Ei=E0ˆye−jk1(xsinθi+zcosθi), (1.139a) ¯Hi=E0 η1(−ˆxcosθi+ˆzsinθi)e−jk1(xsinθi+zcosθi), (1.139b) where k1=ω√µ0/epsilon11andη1=√µ0//epsilon11are the propagation constant and impedance for region 1, as before. The reflected and transmitted fields can be expressed as ¯Er=E0/Gamma1ˆye−jk1(xsinθr−zcosθr), (1.140a) ¯Hr=E0/Gamma1 η1(ˆxcosθr+ˆzsinθr)e−jk1(xsinθr−zcosθr), (1.140b) ¯Et=E0Tˆye−jk2(xsinθt+zcosθt), (1.141a) ¯Ht=E0T η2(−ˆxcosθt+ˆzsinθt)e−jk2(xsinθt+zcosθt), (1.141b) with k2=ω√µ0/epsilon12andη2=√µ0//epsilon12being the propagation constant and impedance in region 2. Equating the tangential field components EyandHxatz=0g i v e s e−jk1xsinθi+/Gamma1e−jk1xsinθr=Te−jk2xsinθt, (1.142a) −1 η1cosθie−jk1xsinθi+/Gamma1 η1cosθre−jk2xsinθr=−T η2cosθte−jk2xsinθt. (1.142b) By the same phase matching argument that was used in the parallel case, we obtain Snell’s laws k1sinθi=k1sinθr=k2sinθt identical to (1.136). c01ElectromagneticTheory Pozar July 28, 2011 8:7 38 Chapter 1: Electromagnetic Theory Using (1.136) in (1.142) allows us to solve for the reflection and transmission coeffi- cients as /Gamma1=η2cosθi−η1cosθt η2cosθi+η1cosθt, (1.143a) T=2η2cosθi η2cosθi+η1cosθt. (1.143b) Again, for the normally incident case, these results reduce to those of Section 1.7. For this polarization no Brewster angle exists where /Gamma1=0, as we can see by examin- ing the possibility that the numerator of (1.143a) could be zero: η2cosθi=η1cosθt, and using Snell’s law to give k2 2/parenleftbig η2 2−η2 1/parenrightbig =/parenleftbig k2 2η2 2−k2 1η2 1/parenrightbig sin2θi. This leads to a contradiction since the term in parentheses on the right-hand side is identi- cally zero for dielectric media. Thus, no Brewster angle exists for perpendicular polariza-tion for dielectric media. EXAMPLE 1.5 OBLIQUE REFLECTION FROM A DIELECTRIC INTERFACE Plot the reflection coefficients versus incidence angle for parallel and perpendic- ular polarized plane waves incident from free-space onto a dielectric region with /epsilon1r=2.55. Solution The impedances for the two regions are η1=377/Omega1, η2=η0√/epsilon1r=377√ 2.55=236/Omega1. We then evaluate (1.137a) and (1.143a) versus incidence angle; the results are shown in Figure 1.14. ■ TotalReflectio andSurfaceWaves Snell’s law of (1.136b) can be rewritten as sinθt=/radicalbigg/epsilon11 /epsilon12sinθi.( 1.144) Consider the case (for either parallel or perpendicular polarization) where /epsilon11>/epsilon1 2.A sθi increases, the refraction angle θtwill increase, but at a faster rate than θiincreases. The incidence angle θifor which θt=90◦is called the critical angle, θc, where sinθc=/radicalbigg/epsilon12 /epsilon11.( 1.145) At this angle and beyond, the incident wave will be totally reflected, as the transmitted wave will not propagate into region 2. Let us look at this situation more closely for the case of θi>θ cwith parallel polarization. c01ElectromagneticTheory Pozar July 28, 2011 8:7 1.8 Oblique Incidence at a Dielectric Interface 39 /H9258i/H92800/H9280r = 2.55 0 1 02 03 04 05 06 07 08 09 00.00.20.40.60.81.0|Γ| Incidence angle /H9258iParallel polarizationPerpendicular polarization FIGURE 1.14 Reflection coefficient magnitude for parallel and perpendicular polarizations of a plane wave obliquely incident on a dielectric half-space. When θi>θ c(1.144) shows that sin θt>1, so that cos θt=/radicalbig 1−sin2θtmust be imaginary, and the angle θtloses its physical significance. At this point, it is better to replace the expressions for the transmitted fields in region 2 with the following: ¯Et=E0T/parenleftbigg−jα k2ˆx−β k2ˆz/parenrightbigg e−jβxe−αz, (1.146a) ¯Ht=E0T η2ˆye−jβxe−αz. (1.146b) The form of these fields is derived from (1.134) after noting that −jk2sinθtis still imag- inary for sin θt>1b u t−jk2cosθtis real, so we can replace sin θtbyβ/k2and cos θtby −jα/k2. Substituting (1.146b) into the Helmholtz wave equation for ¯Hgives −β2+α2+k2 2=0.( 1.147) Matching ExandHyof (1.146) with the ˆxandˆycomponents of the incident and reflected fields of (1.132) and (1.133) at z=0g i v e s cosθie−jk1xsinθi+/Gamma1cosθre−jk1xsinθr=−jα k2Te−jβx, (1.148a) 1 η1e−jk1xsinθi−/Gamma1 η1e−jk1xsinθr=T η2e−jβx. (1.148b) To obtain phase matching at the z=0 boundary, we must have k1sinθi=k1sinθr=β, c01ElectromagneticTheory Pozar July 28, 2011 8:7 40 Chapter 1: Electromagnetic Theory which leads again to Snell’s law for reflection, θi=θr, and to β=k1sinθi. Then αis determined from (1.147) as α=/radicalBig β2−k2 2=/radicalBig k2 1sin2θi−k2 2,( 1.149) which is seen to be a positive real number since sin2θi>/epsilon1 2//epsilon11. The reflection and trans- mission coefficients can be obtained from (1.148) as /Gamma1=(−jα/k2)η2−η1cosθi (−jα/k2)η2+η1cosθi, (1.150a) T=2η2cosθi (−jα/k2)η2+η1cosθi. (1.150b) Since/Gamma1is of the form (ja−b)/(ja+b), its magnitude is unity, indicating that all incident power is reflected. The transmitted fields of (1.146) show propagation in the xdirection, along the inter- face, but exponential decay in the zdirection. Such a field is known as a surface wave3 since it is tightly bound to the interface. A surface wave is an example of a nonuniform plane wave, so called because it has an amplitude variation in the zdirection, apart from the propagation factor in the xdirection. Finally, it is of interest to calculate the complex Poynting vector for the surface wave fields of (1.146): ¯St=¯EtׯH∗ t=|E0|2|T|2 η2/parenleftbigg ˆz−jα k2+ˆxβ k2/parenrightbigg e−2α z.( 1.151) This shows that no real power flow occurs in the zdirection. The real power flow in the xdirection is that of the surface wave field, and it decays exponentially with distance into region 2. So even though no real power is transmitted into region 2, a nonzero field does exist there, in order to satisfy the boundary conditions at the interface. 1.9SOMEUSEFULTHEOREMS Finally, we discuss several theorems in electromagnetics that we will find useful for later discussions. TheReciprocityTheorem Reciprocity is a general concept that occurs in many areas of physics and engineering, and the reader may already be familiar with the reciprocity theorem of circuit theory. Herewe will derive the Lorentz reciprocity theorem for electromagnetic fields in two different forms. This theorem will be used later in the book to obtain general properties of network matrices representing microwave circuits and to evaluate the coupling of waveguides fromcurrent probes and loops, as well as the coupling of waveguides through apertures. There are a number of other important uses of this powerful concept. 3Some authors argue that the term “surface wave” should not be used for a field of this type since it exists only when plane wave fields exist in the z<0 region, and so prefer the term “surface wave–like” field, or a “forced surface wave.” c01ElectromagneticTheory Pozar July 28, 2011 8:7 1.9 Some Useful Theorems 41 FIGURE 1.15 Geometry for the Lorentz reciprocity theorem. Consider the two separate sets of sources, ¯J1,¯M1and¯J2,¯M2, which generate the fields ¯E1,¯H1,and¯E2,¯H2, respectively, in the volume Venclosed by the closed surface S,a s shown in Figure 1.15. Maxwell’s equations are satisfied individually for these two sets of sources and fields, so we can write ∇ׯE1=− jωµ¯H1−¯M1, (1.152a) ∇ׯH1=jω/epsilon1¯E1+¯J1, (1.152b) ∇ׯE2=− jωµ¯H2−¯M2, (1.153a) ∇ׯH2=jω/epsilon1¯E2+¯J2. (1.153b) Now consider the quantity ∇·(¯E1ׯH2−¯E2ׯH1), which can be expanded using vector identity (B.8) to give ∇·(¯E1ׯH2−¯E2ׯH1)=¯J1·¯E2−¯J2·¯E1+¯M2·¯H1−¯M1·¯H2.( 1.154) Integrating over the volume Vand applying the divergence theorem (B.15), gives /integraldisplay V∇·(¯E1ׯH2−¯E2ׯH1)dv=/contintegraldisplay S(¯E1ׯH2−¯E2ׯH1)·ds (1.155) =/integraldisplay V(¯E2·¯J1−¯E1·¯J2+¯H1·¯M2−¯H2·¯M1)dv Equation (1.155) represents a general form of the reciprocity theorem, but in practice a number of special situations often occur leading to some simplification. We will consider three cases. S encloses no sources : Then ¯J1=¯J2=¯M1=¯M2=0, and the fields ¯E1,¯H1and¯E2,¯H2 are source-free fields. In this case, the right-hand side of (1.155) vanishes, with the result that /contintegraldisplay S¯E1ׯH2·d¯s=/contintegraldisplay S¯E2ׯH1·d¯s.( 1.156) This result will be used in Chapter 4 when we demonstrate the symmetry of the impedance matrix for a reciprocal microwave network. S bounds a perfect conductor: For example, Smay be the inner surface of a perfectly con- ducting closed cavity. Then the surface integral of (1.155) vanishes since ¯E1ׯH2·ˆn= (ˆnׯE1)·¯H2[by vector identity (B.3)], and ˆnׯE1is zero on the surface of a perfect c01ElectromagneticTheory Pozar July 28, 2011 8:7 42 Chapter 1: Electromagnetic Theory conductor (similarly for ¯E2). The result is /integraldisplay V(¯E1·¯J2−¯H1·¯M2)dv=/integraldisplay V(¯E2·¯J1−¯H2·¯M1)dv. ( 1.157) This result is analogous to the reciprocity theorem of circuit theory. In words, this result states that the system response ¯E1or¯E2is not changed when the source and observation points are interchanged. That is, ¯E2(caused by ¯J2)at¯J1is the same as ¯E1(caused by ¯J1) at¯J2. S is a sphere at infinity: In this case the fields evaluated on Sare very far from the sources and so can be considered locally as plane waves. Then the wave impedance relation ¯H= ˆnׯE/ηapplies to (1.155) to give (¯E1ׯH2−¯E2ׯH1)·ˆn=(ˆnׯE1)·¯H2−(ˆnׯE2)·¯H1 =1 η¯H1·¯H2−1 η¯H2·¯H1=0, so that the result of (1.157) is again obtained. This result can also be obtained for the case of a closed surface Swhere the surface impedance boundary condition applies. ImageTheory In many problems a current source (electric or magnetic) is located in the vicinity of a conducting ground plane. Image theory permits the removal of the ground plane by placinga virtual image source on the other side of the ground plane. The reader should be familiar with this concept from electrostatics, so we will prove the result for an infinite current sheet next to an infinite ground plane and then summarize other possible cases. Consider the surface current density ¯J s=Js0ˆxparallel to a ground plane, as shown in Figure 1.16a. Because the current source is of infinite extent and is uniform in the x,y directions, it will excite plane waves traveling outward from it. The negatively traveling FIGURE 1.16 Illustration of image theory as applied to an electric current source next to a ground plane. (a) An electric surface current density parallel to a ground plane. (b) Theground plane of (a) is replaced with image current at z=−d. c01ElectromagneticTheory Pozar July 28, 2011 8:7 1.9 Some Useful Theorems 43 wave will reflect from the ground plane at z=0 and then travel in the positive direction. Thus, there will be a standing wave field in the region 0 <z<dand a positively traveling wave for z>d. The forms of the fields in these two regions can thus be written as Es x=A(ejk0z−e−jk0z), for 0<z<d, (1.158a) Hs y=−A η0(ejk0z+e−jk0z), for 0<z<d, (1.158b) E+ x=Be−jk0z, forz>d, (1.159a) H+ y=B η0e−jk0z, forz>d, (1.159b) where η0is the impedance of free-space. Note that the standing wave fields of (1.158) have been constructed to satisfy the boundary condition that Ex=0a t z=0. The remaining boundary conditions to satisfy are the continuity of ¯Eatz=dand the discontinuity in the ¯Hfield at z=ddue to the current sheet. From (1.36), since ¯Ms=0, Es x=E+ x|z=d,( 1.160a) while from (1.37) we have ¯Js=ˆz׈y(H+ y−Hs y)|z=d.( 1.160b) Using (1.158) and (1.159) then gives 2jAsink0d=Be−jk0d and Js0=−B η0e−jk0d−2A η0cosk0d, which can be solved for AandB: A=−Js0η0 2e−jk0d, B=− jJs0η0sink0d. So the total fields are Es x=− jJs0η0e−jk0dsink0z, for 0<z<d, (1.161a) Hs y=Js0e−jk0dcosk0z, for 0<z<d, (1.161b) E+ x=− jJs0η0sink0de−jk0z,forz>d, (1.162a) H+ y=− jJs0sink0de−jk0z,forz>d. (1.162b) Now consider the application of image theory to this problem. As shown in Figure 1.16b, the ground plane is removed and an image source of −¯Jsis placed at z=−d.B y superposition, the total fields for z>0 can be found by combining the fields from the two sources individually. These fields can be derived by a procedure similar to that in the above, with the following results: Fields due to source at z=d: Ex=⎧ ⎪⎪⎨ ⎪⎪⎩−Js0η0 2e−jk0(z−d)forz>d −Js0η0 2ejk0(z−d)forz<d,(1.163a) c01ElectromagneticTheory Pozar July 28, 2011 8:7 44 Chapter 1: Electromagnetic Theory Hy=⎧ ⎪⎪⎨ ⎪⎪⎩−Js0 2e−jk0(z−d)forz>d Js0 2ejk0(z−d)forz<d.(1.163b) Fields due to source at z=−d: Ex=⎧ ⎪⎪⎨ ⎪⎪⎩J s0η0 2e−jk0(z+d)forz>−d Js0η0 2ejk0(z+d)forz<−d,(1.164a) Hy=⎧ ⎪⎪⎨ ⎪⎪⎩J s0 2e−jk0(z+d)forz>−d −Js0 2ejk0(z+d)forz<−d.(1.164b) The reader can verify that this solution is identical to that of (1.161) for 0 <z<dand to that of (1.162) for z>d, thus verifying the validity of the image theory solution. Note that image theory only gives the correct fields to the right of the conducting plane. Figure 1.17shows more general image theory results for electric and magnetic dipoles. Original GeometryImage Equivalent (a)≡ (b)≡ (c)≡ (d)≡ FIGURE 1.17 Electric and magnetic current images. (a) An electric current parallel to a ground plane. (b) An electric current normal to a ground plane. (c) A magnetic current parallel to a ground plane. (d) A magnetic current normal to a ground plane. c01ElectromagneticTheory Pozar July 28, 2011 8:7 Problems 45 REFERENCES [1] T. S. Sarkar, R. J. Mailloux, A. A. Oliner, M. Salazar-Palma, and D. Sengupta, History of Wireless , John Wiley & Sons, Hoboken, N.J., 2006. [2] A. A. Oliner, “Historical Perspectives on Microwave Field Theory,” IEEE Transactions on Mi- crowave Theory and Techniques , vol. MTT-32, pp. 1022–1045, September 1984 [this special issue contains other articles on the history of microwave engineering]. [3] F. Ulaby, Fundamentals of Applied Electromagnetics, 6th edition, Prentice-Hall, Upper Saddle River, N.J., 2010. [ 4 ] J .D .K r a u sa n dD .A .F l e i s c h ,Electromagnetics, 5th edition, McGraw-Hill, New York, 1999. [5] S. Ramo, T. R. Whinnery, and T. van Duzer, Fields and Waves in Communication Electronics ,3 r d edition, John Wiley & Sons, New York, 1994. [6] R. E. Collin, Foundations for Microwave Engineering , 2nd edition, Wiley-IEEE Press, Hoboken, N.J., 2001. [7] C. A. Balanis, Advanced Engineering Electromagnetics, John Wiley & Sons, New York, 1989. [8] D. M. Pozar, Microwave and RF Design of Wireless Systems, John Wiley & Sons, Hoboken N.J., 2001. PROBLEMS 1.1Who invented radio? Guglielmo Marconi often receives credit for the invention of modern radio, but there were several important developments by other workers before Marconi. Write a brief sum- mary of the early work in wireless during the period of 1865–1900, particularly the work by MahlonLoomis, Oliver Lodge, Nikola Tesla, and Marconi. Explain the difference between inductive com- munication schemes and wireless methods that involve wave propagation. Can the development of radio be attributed to a single individual? Reference [1] may be a good starting point. 1.2A plane wave traveling along the x-axis in a polystyrene-filled region with /epsilon1 r=2.54 has an elec- tric field given by Ey=E0cos(ωt−kx). The frequency is 2.4 GHz, and E0=5.0 V/m. Find the following: (a) the amplitude and direction of the magnetic field, (b) the phase velocity, (c) the wave-length, and (d) the phase shift between the positions x 1=0.1ma n d x2=0.15 m. 1.3Show that a linearly polarized plane wave of the form ¯E=E0(aˆx+bˆy)e−jk0z,w h e r e aandbare real numbers, can be represented as the sum of an RHCP and an LHCP wave. 1.4Compute the Poynting vector for the general plane wave field of (1.76). 1.5A plane wave is normally incident on a dielectric slab of permittivity /epsilon1rand thickness d,w h e r e d= λ0/(4√/epsilon1r)andλ0is the free-space wavelength of the incident wave, as shown in the accompanying figure. If free-space exists on both sides of the slab, find the reflection coefficient of the wave reflectedfrom the front of the slab. 1 ΓT /H92800 /H92800 /H9280r/H92800 d d z 0 1.6Consider an RHCP plane wave normally incident from free-space (z<0)onto a half-space (z>0) consisting of a good conductor. Let the incident electric field be of the form ¯Ei=E0(ˆx−jˆy)e−jk0z, c01ElectromagneticTheory Pozar July 28, 2011 8:7 46 Chapter 1: Electromagnetic Theory and find the electric and magnetic fields in the region z>0. Compute the Poynting vectors for z<0 andz>0 and show that complex power is conserved. What is the polarization of the reflected wave? 1.7Consider a plane wave propagating in a lossy dielectric medium for z<0, with a perfectly conduct- ing plate at z=0. Assume that the lossy medium is characterized by /epsilon1=(5−j2)/epsilon10,µ=µ0,a n d that the frequency of the plane wave is 1.0 GHz, and let the amplitude of the incident electric field be 4V / ma t z=0. Find the reflected electric field for z<0 and plot the magnitude of the total electric field for −0.5≤z≤0. 1.8A plane wave at 1 GHz is normally incident on a thin copper sheet of thickness t. (a) Compute the transmission losses, in dB, of the wave at the air–copper and the copper–air interfaces. (b) If the sheet is to be used as a shield to reduce the level of the transmitted wave by 150 dB, what is the minimumsheet thickness? 1.9A uniform lossy medium with /epsilon1 r=3.0, tanδ=0.1,andµ=µ0fills the region between z=0a n d z=20 cm, with a ground plane at z=20 cm, as shown in the accompanying figure. An incident plane wave with an electric field ¯Ei=ˆx100e−γzV/m is present at z=0 and propagates in the +zdirection. The frequency is 3.0 GHz. (a)Compute Si, the power density of the incident wave, and Sr, the power density of the reflected wave, at z=0. (b)Compute the input power density, Sin,a tz=0 from the total fields at z=0. Does Sin= Si−Sr? 0 l = 20 cm zEi Er/H9280r = 3.0 tan /H9254 = 0.1 1.10 Assume that an infinite sheet of electric surface current density ¯Js=J0ˆxA/m is placed on the z=0 plane between free-space for z<0 and a dielectric with /epsilon1=/epsilon1r/epsilon10forz>0, as in the accompanying figure. Find the resulting ¯Eand¯Hfields in the two regions. HINT: Assume plane wave solutions propagating away from the current sheet, and match boundary conditions to find the amplitudes, as in Example 1.3. zx /H92800 /H9280r/H92800 0Js = xJ0 A/mˆ 1.11 Redo Problem 1.10, but with an electric surface current density of ¯Js=J0ˆxe−jβxA/m, where β< k0. c01ElectromagneticTheory Pozar July 28, 2011 8:7 Problems 47 1.12 A parallel polarized plane wave is obliquely incident from free-space onto a magnetic material with permittivity /epsilon10and permeability µ0µr. Find the reflection and transmission coefficients. Does a Brewster angle exist for this case where the reflection coefficient vanishes for a particular angle of incidence? 1.13 Repeat Problem 1.12 for the perpendicularly polarized case. 1.14 An artificial anisotropic dielectric material has the tensor permittivity [/epsilon1]given as follows: [/epsilon1]=/epsilon1 0/bracketleftBigg13 j0 −3j20 00 4/bracketrightBigg At a certain point in the material the electric field is known to be ¯E=3ˆx−2ˆy+5ˆz.W h a ti s ¯Dat this point? 1.15 The permittivity tensor for a gyrotropic dielectric material is [/epsilon1]=/epsilon1 0/bracketleftBigg/epsilon1r jκ0 −jκ/epsilon1 r0 00 1/bracketrightBigg . Show that the transformations E+=Ex−jEy, D+=Dx−jDy, E−=Ex+jEy, D−=Dx+jDy, allow the relation between ¯Eand¯Dto be written as /bracketleftBiggD+ D− Dz/bracketrightBigg =[/epsilon1/prime]/bracketleftBiggE+ E− Ez/bracketrightBigg , where [/epsilon1/prime]is now a diagonal matrix. What are the elements of [/epsilon1/prime]? Using this result, derive wave equations for E+andE−and find the resulting propagation constants. 1.16 Show that the reciprocity theorem expressed in (1.157) also applies to a region enclosed by a closed surface S, where a surface impedance boundary condition applies. 1.17 Consider an electric surface current density of ¯Js=ˆyJ0e−βxA/m located on the z=dplane. If a perfectly conducting ground plane is located at z=0, use image theory to find the total fields for z>0. 1.18 Let¯E=Eρˆρ+Eφˆφ+Ezˆzbe an electric field vector in cylindrical coordinates. Demonstrate that it is incorrect to interpret the expression ∇2¯Ein cylindrical coordinates as ˆρ∇2Eρ+ˆφ∇2Eφ+ ˆz∇2Ezby evaluating both sides of the vector identity ∇×∇× ¯E=∇(∇·¯E)−∇2¯Efor the given electric field. c02TransmissionLineTheory Pozar July 26, 2011 17:33 Chapter Two Transmission Line Theory Transmission line theory bridges the gap between field analysis and basic circuit theory and therefore is of significant importance in the analysis of microwave circuits and devices. As we will see, the phenomenon of wave propagation on transmission lines can be approached from an extension of circuit theory or from a specialization of Maxwell’s equations; we shallpresent both viewpoints and show how this wave propagation is described by equations verysimilar to those used in Chapter 1 for plane wave propagation. 2.1THELUMPED-ELEMENTCIRCUITMODEL FORATRANSMISSIONLINE The key difference between circuit theory and transmission line theory is electrical size. Circuit analysis assumes that the physical dimensions of the network are much smaller than the electrical wavelength, while transmission lines may be a considerable fractionof a wavelength, or many wavelengths, in size. Thus a transmission line is a distributed- parameter network, where voltages and currents can vary in magnitude and phase over its length, while ordinary circuit analysis deals with lumped elements, where voltage and current do not vary appreciably over the physical dimension of the elements. As shown in Figure 2.1a, a transmission line is often schematically represented as a two-wire line since transmission lines (for transverse electromagnetic [TEM] wave propa-gation) always have at least two conductors. The piece of line of infinitesimal length /Delta1zof Figure 2.1a can be modeled as a lumped-element circuit, as shown in Figure 2.1b, where R,L,G, and Care per-unit-length quantities defined as follows: R=series resistance per unit length, for both conductors, in /Omega1/m. L=series inductance per unit length, for both conductors, in H/m. G=shunt conductance per unit length, in S/m. C=shunt capacitance per unit length, in F/m. 48 c02TransmissionLineTheory Pozar July 26, 2011 17:33 2.1 The Lumped-Element Circuit Model for a Transmission Line 49 ∆z∆zi(z, t) i(z, t) i(z +∆ z, t)z (a) (b)R∆zL∆z G∆zC ∆z v(z + ∆ z, t)v(z, t)+ + –– v(z, t)+ – FIGURE 2.1 V oltage and current definitions and equivalent circuit for an incremental length of transmission line. (a) V oltage and current definitions. (b) Lumped-element equiva- lent circuit. The series inductance Lrepresents the total self-inductance of the two conductors, and the shunt capacitance Cis due to the close proximity of the two conductors. The series resistance Rrepresents the resistance due to the finite conductivity of the individual conductors, and the shunt conductance Gis due to dielectric loss in the material between the conductors. RandG, therefore, represent loss. A finite length of transmission line can be viewed as a cascade of sections of the form shown in Figure 2.1b. From the circuit of Figure 2.1b, Kirchhoff’s voltage law can be applied to give v(z,t)−R/Delta1zi(z,t)−L/Delta1z∂i(z,t) ∂t−v(z+/Delta1z,t)=0,( 2.1a) and Kirchhoff’s current law leads to i(z,t)−G/Delta1zv(z+/Delta1z,t)−C/Delta1z∂v(z+/Delta1z,t) ∂t−i(z+/Delta1z,t)=0.( 2.1b) Dividing (2.1a) and (2.1b) by /Delta1zand taking the limit as /Delta1z→0 gives the following differential equations: ∂v(z,t) ∂z=− Ri(z,t)−L∂i(z,t) ∂t, (2.2a) ∂i(z,t) ∂z=− Gv(z,t)−C∂v(z,t) ∂t. (2.2b) These are the time domain form of the transmission line equations, also known as the telegrapher equations . For the sinusoidal steady-state condition, with cosine-based phasors, (2.2a) and (2.2b) simplify to dV(z) dz=−(R+jωL)I(z), (2.3a) dI(z) dz=−(G+jωC)V(z). (2.3b) c02TransmissionLineTheory Pozar July 26, 2011 17:33 50 Chapter 2: Transmission Line Theory Note the similarity in the form of (2.3a) and (2.3b) and Maxwell’s curl equations of (1.41a) and (1.41b). WavePropagationonaTransmissionLine The two equations (2.3a) and (2.3b) can be solved simultaneously to give wave equations forV(z)andI(z): d2V(z) dz2−γ2V(z)=0, (2.4a) d2I(z) dz2−γ2I(z)=0, (2.4b) where γ=α+jβ=/radicalbig (R+jωL)(G+jωC)( 2.5) is the complex propagation constant, which is a function of frequency. Traveling wave solutions to (2.4) can be found as V(z)=V+ oe−γz+V− oeγz, (2.6a) I(z)=I+ oe−γz+I− oeγz, (2.6b) where the e−γzterm represents wave propagation in the +zdirection, and the eγzterm represents wave propagation in the −zdirection. Applying (2.3a) to the voltage of (2.6a) gives the current on the line: I(z)=γ R+jωL/parenleftbig V+ oe−γz−V− oeγz/parenrightbig . Comparison with (2.6b) shows that a characteristic impedance ,Z0, can be defined as Z0=R+jωL γ=/radicalBigg R+jωL G+jωC,( 2.7) to relate the voltage and current on the line as follows: V+ o I+o=Z0=−V− o I−o. Then (2.6b) can be rewritten in the following form: I(z)=V+ o Z0e−γz−V− o Z0eγz.( 2.8) Converting back to the time domain, we can express the voltage waveform as v(z,t)=|V+ o|cos(ω t−βz+φ+)e−αz +|V− o|cos(ωt+βz+φ−)eαz, (2.9) where φ±is the phase angle of the complex voltage V± o. Using arguments similar to those in Section 1.4, we find that the wavelength on the line is λ=2π β,( 2.10) c02TransmissionLineTheory Pozar July 26, 2011 17:33 2.2 Field Analysis of Transmission Lines 51 and the phase velocity is vp=ω β=λf.( 2.11) TheLosslessLine The above solution is for a general transmission line, including loss effects, and it was seen that the propagation constant and characteristic impedance were complex. In many practi- cal cases, however, the loss of the line is very small and so can be neglected, resulting in a simplification of the results. Setting R=G=0 in (2.5) gives the propagation constant as γ=α+jβ=jω√ LC, or β=ω√ LC, (2.12a) α=0. (2.12b) As expected for a lossless line, the attenuation constant αis zero. The characteristic impedance of (2.7) reduces to Z0=/radicalbigg L C,( 2.13) which is now a real number. The general solutions for voltage and current on a lossless transmission line can then be written as V(z)=V+ oe−jβz+V− oejβz, (2.14a) I(z)=V+ o Z0e−jβz−V− o Z0ejβz. (2.14b) The wavelength is λ=2π β=2π ω√ LC,( 2.15) and the phase velocity is vp=ω β=1√ LC.( 2.16) 2.2FIELDANALYSISOFTRANSMISSIONLINES In this section we will rederive the time-harmonic form of the telegrapher’s equations start- ing from Maxwell’s equations. We will begin by deriving the transmission line parameters (R,L,G,C)in terms of the electric and magnetic fields of the transmission line and then derive the telegrapher equations using these parameters for the specific case of a coaxialline. TransmissionLineParameters Consider a 1 m length of a uniform transmission line with fields ¯Eand¯H,a ss h o w ni n Figure 2.2, where Sis the cross-sectional surface area of the line. Let the voltage between the conductors be V oe±jβzand the current be Ioe±jβz. The time-average stored magnetic c02TransmissionLineTheory Pozar July 26, 2011 17:33 52 Chapter 2: Transmission Line Theory FIGURE 2.2 Field lines on an arbitrary TEM transmission line. energy for this 1 m length of line can be written, from (1.86), as Wm=µ 4/integraldisplay S¯H·¯H∗ds, while circuit theory gives Wm=L|Io|2/4 in terms of the current on the line. We can thus identify the self-inductance per unit length as L=µ |Io|2/integraldisplay S¯H·¯H∗dsH/m.( 2.17) Similarly, the time-average stored electric energy per unit length can be found from (1.84) as We=/epsilon1 4/integraldisplay S¯E·¯E∗ds, while circuit theory gives We=C|Vo|2/4, resulting in the following expression for the capacitance per unit length: C=/epsilon1 |Vo|2/integraldisplay S¯E·¯E∗dsF/m.( 2.18) From (1.131), the power loss per unit length due to the finite conductivity of the metallic conductors is Pc=Rs 2/integraldisplay C1+C2¯H·¯H∗d/lscript (assuming ¯His tangential to S), while circuit theory gives Pc=R|Io|2/2, so the series resistance Rper unit length of line is R=Rs |Io|2/integraldisplay C1+C2¯H·¯H∗dl/Omega1/m.( 2.19) In (2.19), Rs=1/σδ sis the surface resistance of the conductors, and C1+C2represent integration paths over the conductor boundaries. From (1.92), the time-average power dis- sipated per unit length in a lossy dielectric is Pd=ω/epsilon1/prime/prime 2/integraldisplay S¯E·¯E∗ds, where /epsilon1/prime/primeis the imaginary part of the complex permittivity /epsilon1=/epsilon1/prime−j/epsilon1/prime/prime=/epsilon1/prime(1−jtanδ). Circuit theory gives Pd=G|Vo|2/2, so the shunt conductance per unit length can be written as G=ω/epsilon1/prime/prime |Vo|2/integraldisplay S¯E·¯E∗dsS/m.( 2.20) c02TransmissionLineTheory Pozar July 26, 2011 17:33 2.2 Field Analysis of Transmission Lines 53 Rsxy a b/H9278/H9267/H9262, /H9280 FIGURE 2.3 Geometry of a coaxial line with surface resistance Rson the inner and outer conductors. EXAMPLE 2.1 TRANSMISSION LINE PARAMETERS OF A COAXIAL LINE The fields of a traveling TEM wave inside the coaxial line of Figure 2.3 can be expressed as ¯E=Voˆρ ρlnb/ae−γz, ¯H=Ioˆφ 2πρe−γz, where γis the propagation constant of the line. The conductors are assumed to have a surface resistivity Rs, and the material filling the space between the con- ductors is assumed to have a complex permittivity /epsilon1=/epsilon1/prime−j/epsilon1/prime/primeand a permeabil- ityµ=µ0µr. Determine the transmission line parameters. Solution From (2.17)–(2.20) and the given fields the parameters of the coaxial line can be calculated as L=µ (2π)2/integraldisplay2π φ=0/integraldisplayb ρ=a1 ρ2ρdρdφ=µ 2πlnb/aH/m, C=/epsilon1/prime (lnb/a)2/integraldisplay2π φ=0/integraldisplayb ρ=a1 ρ2ρdρdφ=2π/epsilon1/prime lnb/aF/m, R=Rs (2π)2/braceleftBigg/integraldisplay2π φ=01 a2adφ+/integraldisplay2π φ=01 b2bdφ/bracerightBigg =Rs 2π/parenleftbigg1 a+1 b/parenrightbigg /Omega1/m, G=ω/epsilon1/prime/prime (lnb/a)2/integraldisplay2π φ=0/integraldisplayb ρ=a1 ρ2ρdρdφ=2πω/epsilon1/prime/prime lnb/aS/m. ■ Table 2.1 summarizes the parameters for coaxial, two-wire, and parallel plate lines. As we will see in the next chapter, the propagation constant, characteristic impedance, andattenuation of most transmission lines are usually derived directly from a field theory so- lution; the approach here of first finding the equivalent circuit parameters (L,C,R,G)is useful only for relatively simple lines. Nevertheless, it provides a helpful intuitive conceptfor understanding the properties of a transmission line and relates a transmission line to its equivalent circuit model. c02TransmissionLineTheory Pozar July 26, 2011 17:33 54 Chapter 2: Transmission Line Theory TABLE 2.1 Transmission Line Parameters for Some Common Lines COAX TWO-WIRE PARALLEL PLATE a ba aDw d Lµ 2πlnb aµ πcosh−1/parenleftbiggD 2a/parenrightbiggµd w C2π/epsilon1/prime lnb/aπ/epsilon1/prime cosh−1(D/2a)/epsilon1/primew d RRs 2π/parenleftbigg1 a+1 b/parenrightbiggRs πa2Rs w G2πω/epsilon1/prime/prime lnb/aπω/epsilon1/prime/prime cosh−1(D/2a)ω/epsilon1/prime/primew d TheTelegrapherEquationsDerivedfromFieldAnalysis ofaCoaxialLine We now show that the telegrapher equations of (2.3), derived using circuit theory, can also be obtained from Maxwell’s equations. We will consider the specific geometry of thecoaxial line of Figure 2.3. Although we will treat TEM wave propagation more generally in the next chapter, the present discussion should provide some insight into the relationship of circuit and field quantities. A TEM wave on the coaxial line of Figure 2.3 will be characterized by E z=Hz=0; furthermore, due to azimuthal symmetry, the fields will have no φvariation, so ∂/∂φ=0. The fields inside the coaxial line will satisfy Maxwell’s curl equations, ∇ׯE=− jωµ¯H, (2.21a) ∇ׯH=jω/epsilon1¯E, (2.21b) where /epsilon1=/epsilon1/prime−j/epsilon1/prime/primemay be complex to allow for a lossy dielectric filling. Conductor loss will be ignored here. A rigorous field analysis of conductor loss can be carried out but at this point would tend to obscure our purpose; the interested reader is referred to references [1] and [2]. Expanding (2.21a) and (2.21b) gives the following two vector equations: −ˆρ∂Eφ ∂z+ˆφ∂Eρ ∂z+ˆz1 ρ∂ ∂ρ(ρEφ)=− jωµ(ˆρHρ+ˆφHφ), (2.22a) −ˆρ∂Hφ ∂z+ˆφ∂Hρ ∂z+ˆz1 ρ∂ ∂ρ(ρHφ)=jω/epsilon1(ˆρEρ+ˆφEφ). (2.22b) Since the ˆzcomponents of these two equations must vanish, it is seen that EφandHφmust have the forms Eφ=f(z) ρ, (2.23a) Hφ=g(z) ρ. (2.23b) c02TransmissionLineTheory Pozar July 26, 2011 17:33 2.2 Field Analysis of Transmission Lines 55 To satisfy the boundary condition that Eφ=0a tρ =a,b,w em u s th a v e Eφ=0e v e r y - where, due to the form of Eφin (2.23a). Then from the ˆρcomponent of (2.22a), it is seen thatHρ=0. With these results, (2.22) can be reduced to ∂Eρ ∂z=− jωµHφ, (2.24a) ∂Hφ ∂z=− jω/epsilon1Eρ. (2.24b) From the form of Hφin (2.23b) and (2.24a), Eρmust be of the form Eρ=h(z) ρ.( 2.25) Using (2.23b) and (2.25) in (2.24) gives ∂h(z) ∂z=− jωµg(z), (2.26a) ∂g(z) ∂z=− jω/epsilon1h(z). (2.26b) The voltage between the two conductors can be evaluated as V(z)=/integraldisplayb ρ=aEρ(ρ,z)dρ=h(z)/integraldisplayb ρ=adρ ρ=h(z)lnb a,( 2.27a) and the total current on the inner conductor at ρ=acan be evaluated using (2.23b) as I(z)=/integraldisplay2π φ=0Hφ(a,z)adφ=2πg(z). (2.27b) Then h(z)andg(z)can be eliminated from (2.26) by using (2.27) to give ∂V(z) ∂z=− jωµlnb/a 2πI(z), ∂I(z) ∂z=− jω(/epsilon1/prime−j/epsilon1/prime/prime)2πV(z) lnb/a. Finally, using the results for L,G, and Cfor a coaxial line as derived earlier, we obtain the telegrapher equations as ∂V(z) ∂z=− jωLI(z), (2.28a) ∂I(z) ∂z=−(G+jωC)V(z). (2.28b) This result excludes R, the series resistance, since the conductors were assumed to have perfect conductivity. A similar analysis can be carried out for other simple transmission lines. c02TransmissionLineTheory Pozar July 26, 2011 17:33 56 Chapter 2: Transmission Line Theory PropagationConstant,Impedance,andPowerFlow fortheLosslessCoaxialLine Equations (2.24a) and (2.24b) for EρandHφcan be simultaneously solved to yield a wave equation for Eρ(orHφ): ∂2Eρ ∂z2+ω2µ/epsilon1Eρ=0,( 2.29) from which it is seen that the propagation constant is γ2=−ω2µ/epsilon1, which, for lossless media, reduces to β=ω√µ/epsilon1=ω√ LC,( 2.30) where the last result is from (2.12). Observe that this propagation constant is of the same form as that for plane waves in a lossless dielectric medium. This is a general result forTEM transmission lines. The wave impedance for the coaxial line is defined as Z w=Eρ/Hφ, which can be calculated from (2.24a), assuming an e−jβzdependence, to give Zw=Eρ Hφ=ωµ β=/radicalbig µ//epsilon1=η. (2.31) This wave impedance is seen to be identical to the intrinsic impedance of the medium, η, and is a general result for TEM transmission lines. The characteristic impedance of the coaxial line is defined as Z0=Vo Io=Eρlnb/a 2πHφ=ηlnb/a 2π=/radicalbiggµ /epsilon1lnb/a 2π,( 2.32) where the forms for Eρand Hφfrom Example 2.1 have been used. The characteristic impedance is geometry dependent and will be different for other transmission line config-urations. Finally, the power flow (in the zdirection) on the coaxial line may be computed from the Poynting vector as P=1 2/integraldisplay s¯EׯH∗·d¯s=1 2/integraldisplay2π φ=0/integraldisplayb ρ=aVoI∗ o 2πρ2lnb/aρdρdφ=1 2VoI∗ o,( 2.33) a result that is in clear agreement with circuit theory. This shows that the flow of power in a transmission line takes place entirely via the electric and magnetic fields between the two conductors; power is not transmitted through the conductors themselves. As we willsee later, for the case of finite conductivity, power may enter the conductors, but this power is then lost as heat and is not delivered to the load. 2.3THETERMINATEDLOSSLESSTRANSMISSIONLINE Figure 2.4 shows a lossless transmission line terminated in an arbitrary load impedance ZL. This problem will illustrate wave reflection on transmission lines, a fundamental propertyof distributed systems. Assume that an incident wave of the form V + oe−jβzis generated from a source at z<0. We have seen that the ratio of voltage to current for such a traveling wave is Z0,t h e characteristic impedance of the line. However, when the line is terminated in an arbitrary load ZL/negationslash=Z0, the ratio of voltage to current at the load must be ZL. Thus, a reflected wave c02TransmissionLineTheory Pozar July 26, 2011 17:33 2.3 The Terminated Lossless Transmission Line 57 z l0V(z), I (z) Z0, /H9252 VL IL+ –ZL FIGURE 2.4 A transmission line terminated in a load impedance ZL. must be excited with the appropriate amplitude to satisfy this condition. The total voltage on the line can then be written as in (2.14a), as a sum of incident and reflected waves: V(z)=V+ oe−jβz+V− oejβz.( 2.34a) Similarly, the total current on the line is described by (2.14b): I(z)=V+ o Z0e−jβz−V− o Z0ejβz.( 2.34b) The total voltage and current at the load are related by the load impedance, so at z=0w e must have ZL=V(0) I(0)=V+ o+V− o V+o−V−oZ0. Solving for V− ogives V− o=ZL−Z0 ZL+Z0V+ o. The amplitude of the reflected voltage wave normalized to the amplitude of the incident voltage wave is defined as the voltage reflection coefficient ,/Gamma1: /Gamma1=V− o V+o=ZL−Z0 ZL+Z0.( 2.35) The total voltage and current waves on the line can then be written as V(z)=V+ o/parenleftbig e−jβz+/Gamma1ejβz/parenrightbig , (2.36a) I(z)=V+ o Z0/parenleftbig e−jβz−/Gamma1ejβz/parenrightbig . (2.36b) From these equations it is seen that the voltage and current on the line consist of a super- position of an incident and a reflected wave; such waves are called standing waves.O n l y when/Gamma1=0 is there no reflected wave. To obtain /Gamma1=0, the load impedance ZLmust be equal to the characteristic impedance Z0of the transmission line, as seen from (2.35). Such a load is said to be matched to the line since there is no reflection of the incident wave. Now consider the time-average power flow along the line at the point z: Pavg=1 2Re/braceleftbig V(z)I(z)∗/bracerightbig =1 2|V+ o|2 Z0Re/braceleftbig 1−/Gamma1∗e−2jβz+/Gamma1e2jβz−|/Gamma1|2/bracerightbig , where (2.36) has been used. The middle two terms in the brackets are of the form A−A∗= 2jIm{A}and so are purely imaginary. This simplifies the result to Pavg=1 2|V+ o|2 Z0/parenleftbig 1−|/Gamma1|2/parenrightbig ,( 2.37) c02TransmissionLineTheory Pozar July 26, 2011 17:33 58 Chapter 2: Transmission Line Theory which shows that the average power flow is constant at any point on the line and that the total power delivered to the load (Pavg)is equal to the incident power (|V+ o|2/2Z0)minus the reflected power (|Vo|2|/Gamma1|2/2Z0).I f/Gamma1 =0, maximum power is delivered to the load, while no power is delivered for |/Gamma1|=1. The above discussion assumes that the generator is matched, so that there is no re-reflection of the reflected wave from z<0. When the load is mismatched, not all of the available power from the generator is delivered to the load. This “loss” is called return loss (RL), and is defined (in dB) as RL=−20 log |/Gamma1|dB,( 2.38) so that a matched load (/Gamma1=0)has a return loss of ∞dB (no reflected power), while a total reflection (|/Gamma1|=1)has a return loss of 0 dB (all incident power is reflected). Note that return loss is a nonnegative number for reflection from a passive network. If the load is matched to the line, /Gamma1=0 and the magnitude of the voltage on the line is |V(z)|=| V+ o|, which is a constant. Such a line is sometimes said to be flat. When the load is mismatched, however, the presence of a reflected wave leads to standing waves, and the magnitude of the voltage on the line is not constant. Thus, from (2.36a), |V(z)|=| V+ o||1+/Gamma1e2jβz|=| V+ o||1+/Gamma1e−2jβ/lscript| =|V+ o||1+|/Gamma1|ej(θ−2β/lscript)|,(2.39) where /lscript=−zis the positive distance measured from the load at z=0, and θis the phase of the reflection coefficient (/Gamma1=|/Gamma1|ejθ). This result shows that the voltage magnitude oscillates with position zalong the line. The maximum value occurs when the phase term ej(θ−2β/lscript)=1 and is given by Vmax=|V+ o|(1+|/Gamma1|). (2.40a) The minimum value occurs when the phase term ej(θ−2β/lscript)=−1 and is given by Vmin=|V+ o|(1−|/Gamma1|). (2.40b) As|/Gamma1|increases, the ratio of VmaxtoVminincreases, so a measure of the mismatch of a line, called the standing wave ratio (SWR), can be defined as SWR=Vmax Vmin=1+|/Gamma1| 1−|/Gamma1|.( 2.41) This quantity is also known as the voltage standing wave ratio and is sometimes identified as VSWR. From (2.41) it is seen that SWR is a real number such that 1 ≤SWR≤∞ , where SWR =1 implies a matched load. From (2.39), it is seen that the distance between two successive voltage maxima (or minima) is /lscript=2π/2β=πλ/2π=λ/2, while the distance between a maximum and a minimum is /lscript=π/2β=λ/4, where λis the wavelength on the transmission line. The reflection coefficient of (2.35) was defined as the ratio of the reflected to the incident voltage wave amplitudes at the load (/lscript=0), but this quantity can be generalized to any point /lscriptalong the line as follows. From (2.34a), with z=−/lscript, the ratio of the reflected component to the incident component is /Gamma1(/lscript)=V− oe−jβ/lscript V+oejβ/lscript=/Gamma1(0)e−2jβ/lscript,( 2.42) where /Gamma1(0)is the reflection coefficient at z=0, as given by (2.35). This result is useful when transforming the effect of a load mismatch down the line. c02TransmissionLineTheory Pozar July 26, 2011 17:33 2.3 The Terminated Lossless Transmission Line 59 We have seen that the real power flow on the line is a constant (for a lossless line) but that the voltage amplitude, at least for a mismatched line, is oscillatory with position on the line. The perceptive reader may therefore have concluded that the impedance seen looking into the line must vary with position, and this is indeed the case. At a distance /lscript=−zfrom the load, the input impedance seen looking toward the load is Zin=V(−/lscript) I(−/lscript)=V+ o/parenleftbig ejβ/lscript+/Gamma1e−jβ/lscript/parenrightbig V+o/parenleftbig ejβ/lscript−/Gamma1e−jβ/lscript/parenrightbigZ0=1+/Gamma1e−2jβ/lscript 1−/Gamma1e−2jβ/lscriptZ0,( 2.43) where (2.36a,b) have been used for V(z)andI(z). A more usable form may be obtained by using (2.35) for /Gamma1in (2.43): Zin=Z0(ZL+Z0)ejβ/lscript+(ZL−Z0)e−jβ/lscript (ZL+Z0)ejβ/lscript−(ZL−Z0)e−jβ/lscript =Z0ZLcosβ/lscript+jZ0sinβ/lscript Z0cosβ/lscript+jZLsinβ/lscript =Z0ZL+jZ0tanβ/lscript Z0+jZLtanβ/lscript. (2.44) This is an important result giving the input impedance of a length of transmission line with an arbitrary load impedance. We will refer to this result as the transmission line impedance equation; some special cases will be considered next. SpecialCasesofLosslessTerminatedLines A number of special cases of lossless terminated transmission lines will frequently appear in our work, so it is appropriate to consider the properties of such cases here. Consider first the transmission line circuit shown in Figure 2.5, where a line is termi- nated in a short circuit, ZL=0. From (2.35) it is seen that the reflection coefficient for a short circuit load is /Gamma1=−1; it then follows from (2.41) that the standing wave ratio is infinite. From (2.36) the voltage and current on the line are V(z)=V+ o/parenleftbig e−jβz−ejβz/parenrightbig =−2jV+ osinβz, (2.45a) I(z)=V+ o Z0/parenleftbig e−jβz+ejβz/parenrightbig =2V+ o Z0cosβz, (2.45b) which shows that V=0 at the load (as expected, for a short circuit), while the current is a maximum there. From (2.44), or the ratio V(−/lscript)/ I(−/lscript), the input impedance is Zin=jZ0tanβ/lscript, (2.45c) which is seen to be purely imaginary for any length /lscriptand to take on all values between +j∞and−j∞. For example, when /lscript=0w eh a v e Zin=0, but for /lscript=λ/4w eh a v e Zin=∞ (open circuit). Equation (2.45c) also shows that the impedance is periodic in /lscript, z –l 0V(z), I (z) Z0, /H9252 VL = 0 ZL = 0IL+ – FIGURE 2.5 A transmission line terminated in a short circuit. c02TransmissionLineTheory Pozar July 26, 2011 17:33 60 Chapter 2: Transmission Line Theory zV(z) 2jVo /H9261 4–/H9261 3/H9261 2––/H9261 4–1 –1 zI(z)Z0 /H9261 4–/H9261 3/H9261 2––/H9261 4–1 –1 zXin Z0 /H9261 2––/H9261 /H9261 4–(a) (b) (c)3/H9261 4–+ 2Vo+ FIGURE 2.6 (a) V oltage, (b) current, and (c) impedance (Rin=0o r∞) variation along a short- circuited transmission line. repeating for multiples of λ/2. The voltage, current, and input reactance for the short- circuited line are plotted in Figure 2.6. Next consider the open-circuited line shown in Figure 2.7, where ZL=∞ . Dividing the numerator and denominator of (2.35) by ZLand allowing ZL→∞ shows that the reflection coefficient for this case is /Gamma1=1, and the standing wave ratio is again infinite. From (2.36) the voltage and current on the line are V(z)=V+ o/parenleftbig e−jβz+ejβz/parenrightbig =2V+ ocosβz, (2.46a) I(z)=V+ o Z0/parenleftbig e−jβz−ejβz/parenrightbig =−2jV+ o Z0sinβz, (2.46b) z –l 0V(z), I(z) Z0, /H9252 VL ZL = ∞IL = 0 + – FIGURE 2.7 A transmission line terminated in an open circuit. c02TransmissionLineTheory Pozar July 26, 2011 17:33 2.3 The Terminated Lossless Transmission Line 61 zI(z)Z0 –2jVo /H9261 4–/H9261 3/H9261 2––/H9261 4–1 –1zV(z) 2Vo /H9261 4–/H9261 3/H9261 2––/H9261 4–1 –1 zXin Z0 /H9261 4–(a) (b) (c)3/H9261 4–/H9261 2––/H9261+ + FIGURE 2.8 (a) V oltage, (b) current, and (c) impedance (Rin=0o r∞) variation along an open- circuited transmission line. which shows that now I=0 at the load, as expected for an open circuit, while the voltage is a maximum. The input impedance is Zin=− jZ0cotβ/lscript, (2.46c) which is also purely imaginary for any length, /lscript. The voltage, current, and input reactance of the open-circuited line are plotted in Figure 2.8. Now consider terminated transmission lines with some special lengths. If /lscript=λ/2, (2.44) shows that Zin=ZL,( 2.47) meaning that a half-wavelength line (or any multiple of λ/2)does not alter or transform the load impedance, regardless of its characteristic impedance. If the line is a quarter-wavelength long or, more generally, /lscript=λ/4+nλ/2, for n= 1,2,3,..., (2.44) shows that the input impedance is given by Zin=Z2 0 ZL.( 2.48) Such a line is known as a quarter-wave transformer because it has the effect of transform- ing the load impedance in an inverse manner, depending on the characteristic impedance of the line. We will study this case more thoroughly in Section 2.5. c02TransmissionLineTheory Pozar July 26, 2011 17:33 62 Chapter 2: Transmission Line Theory 1Γ T z 0Z1 Z0 FIGURE 2.9 Reflection and transmission at the junction of two transmission lines with different characteristic impedances. Next consider a transmission line of characteristic impedance Z0feeding a line of dif- ferent characteristic impedance, Z1, as shown in Figure 2.9. If the load line is infinitely long, or if it is terminated in its own characteristic impedance, so that there are no reflec-tions from its far end, then the input impedance seen by the feed line is Z 1, so that the reflection coefficient /Gamma1is /Gamma1=Z1−Z0 Z1+Z0.( 2.49) Not all of the incident wave is reflected; some is transmitted onto the second line with a voltage amplitude given by a transmission coefficient. From (2.36a) the voltage for z<0i s V(z)=V+ o/parenleftbig e−jβz+/Gamma1ejβz/parenrightbig ,z<0,( 2.50a) where V+ ois the amplitude of the incident voltage wave on the feed line. The voltage wave forz>0, in the absence of reflections, is outgoing only and can be written as V(z)=V+ oTe−jβzforz>0.( 2.50b) Equating these voltages at z=0g i v e st h etransmission coefficient, T,a s T=1+/Gamma1=1+Z1−Z0 Z1+Z0=2Z1 Z1+Z0.( 2.51) The transmission coefficient between two points in a circuit is often expressed in dB as the insertion loss,I L , IL=−20 log |T|dB.( 2.52) POINT OF INTEREST: Decibels and Nepers Often the ratio of two power levels P1andP2in a microwave system is expressed in decibels (dB) as 10 logP1 P2dB. Thus, a power ratio of 2 is equivalent to 3 dB, while a power ratio of 0.1 is equivalent to −10 dB. Using power ratios in dB makes it easy to calculate power loss or gain through a series of components since multiplicative loss or gain factors can be accounted for by adding the loss or gain in dB for each stage. For example, a signal passing through a 6 dB attenuator followed bya 23 dB amplifier will have an overall gain of 23 −6=17 dB. c02TransmissionLineTheory Pozar July 26, 2011 17:33 2.4 The Smith Chart 63 Decibels are used only to represent power ratios, but if P1=V2 1/R1andP2=V2 2/R2, then the resulting power ratio in terms of voltage ratios is 10 logV2 1R2 V2 2R1=20 logV1 V2/radicalBigg R2 R1dB, where R1,R2are the load resistances and V1,V2are the voltages appearing across these loads. If the load resistances are equal, then this formula simplifies to 20 logV1 V2dB. The ratio of voltages across equal load resistances can also be expressed in terms of nepers (Np) as lnV1 V2Np. The corresponding expression in terms of powers is 1 2lnP1 P2Np, since voltage is proportional to the square root of power. Transmission line attenuation is some- times expressed in nepers. Since 1 Np corresponds to a power ratio of e2, the conversion between nepers and decibels is 1N p=10 log e2=8.686 dB. Absolute power can also be expressed in decibel notation if a reference power level is assumed. If we let P2=1 mW, then the power P1c a nb ee x p r e s s e di nd B ma s 10 logP1 1m WdBm Thus a power of 1 mW is equivalent to 0 dBm, while a power of 1 W is equivalent to 30 dBm, and so on. 2.4THESMITHCHART The Smith chart, shown in Figure 2.10, is a graphical aid that can be very useful for solving transmission line problems. Although there are a number of other impedance and reflec- tion coefficient charts that can be used for such problems [3], the Smith chart is probably the best known and most widely used. It was developed in 1939 by P. Smith at the BellTelephone Laboratories [4]. The reader might feel that, in this day of personal computers and computer-aided design (CAD) tools, graphical solutions have no place in modern engi- neering. The Smith chart, however, is more than just a graphical technique. Besides beingan integral part of much of the current CAD software and test equipment for microwave design, the Smith chart provides a useful way of visualizing transmission line phenomenon without the need for detailed numerical calculations. A microwave engineer can develop a good intuition about transmission line and impedance-matching problems by learning to think in terms of the Smith chart. At first glance the Smith chart may seem intimidating, but the key to its understanding is to realize that it is based on a polar plot of the voltage reflection coefficient, /Gamma1.L e tt h e reflection coefficient be expressed in magnitude and phase (polar) form as /Gamma1=|/Gamma1|e jθ. Then the magnitude |/Gamma1|is plotted as a radius (|/Gamma1|≤1) from the center of the chart, and the angle θ(−180◦≤θ≤180◦)is measured counterclockwise from the right-hand side of c02TransmissionLineTheory Pozar July 26, 2011 17:33 64 Chapter 2: Transmission Line Theory j , 20-2030-3040-4050-5060-6070-7080-8090-90100-100110-110120-120130-130140-140150-150160-160170-1701800.040.040.050.050.060.060.070.070.080.080.090.090.10.10.110.110.120.120.130.130.140.140.150.150.160.160.170.170.180.180.190.190.20.20.210.210.220.220.230.230.240.24 0.250.250.260.260.270.270.280.280.290.290.30.30.310.310.320.320.330.330.340.340.350.350.360.360.370.370.380.380.390.390.40.40.410.410.420.420.430.430.440.440.450.450.460.460.470.470.480.480.490.49 0.00.0AN GLE OFREFLECTION COEFFICIENTINDEGREES—>WAVELENGTHSTOWARDGENERATOR—><— W AVELENGTH STOW ARD LOAD <—II T NDUCTIVEREACTANCECOMPONEN(+X/Zo) ORCAPACTIVESUSCEPTANCE(+jB/Yo)CAPACITIVEREACTANCECOMPONENT(-jX/Zo),ORINDUCTIVESUSCEPTANCE(-jB/Yo) 0.1 0.1 0.1 0.2 0.2 0.2 0.3 0.3 0.3 0.4 0.4 0.4 0.5 0.5 0.5 0.6 0.6 0.6 0.7 0.7 0.7 0.8 0.8 0.8 0.9 0.9 0.9 1.0 1.0 1.01.21.21.21.41.41.41.61.61.61.81.81.82.02.02.03.03.03.04.04.04.05.05.05.01010102020205050500.20.2 0.2 0.20.40.4 0.4 0.40.60.6 0.6 0.60.80.8 0.8 0.81.01.0 1.0 1.0RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) ± FIGURE 2.10 The Smith chart. the horizontal diameter. Any passively realizable (|/Gamma1|≤1) reflection coefficient can then be plotted as a unique point on the Smith chart. The real utility of the Smith chart, however, lies in the fact that it can be used to convert from reflection coefficients to normalized impedances (or admittances) and vice versa by using the impedance (or admittance) circles printed on the chart. When dealingwith impedances on a Smith chart, normalized quantities are generally used, which we will denote by lowercase letters. The normalization constant is usually the characteristic impedance of the transmission line. Thus, z=Z/Z 0represents the normalized version of the impedance Z. If a lossless line of characteristic impedance Z0is terminated with a load impedance ZL, the reflection coefficient at the load can be written from (2.35) as /Gamma1=zL−1 zL+1=|/Gamma1|ejθ,( 2.53) where zL=ZL/Z0is the normalized load impedance. This relation can be solved for zL in terms of /Gamma1to give [or, from (2.43) with /lscript=0] zL=1+|/Gamma1|ejθ 1−|/Gamma1|ejθ.( 2.54) c02TransmissionLineTheory Pozar July 26, 2011 17:33 2.4 The Smith Chart 65 This complex equation can be reduced to two real equations by writing /Gamma1andzLin terms of their real and imaginary parts, /Gamma1=/Gamma1r+j/Gamma1i, and zL=rL+jxL,g i v i n g rL+jxL=(1+/Gamma1r)+j/Gamma1i (1−/Gamma1r)−j/Gamma1i. The real and imaginary parts of this equation can be separated by multiplying the numerator and denominator by the complex conjugate of the denominator to give rL=1−/Gamma12 r−/Gamma12 i (1−/Gamma1r)2+/Gamma12 i, (2.55a) xL=2/Gamma1i (1−/Gamma1r)2+/Gamma12 i. (2.55b) Rearranging (2.55) gives /parenleftbigg /Gamma1r−rL 1+rL/parenrightbigg2 +/Gamma12 i=/parenleftbigg1 1+rL/parenrightbigg2 , (2.56a) (/Gamma1r−1)2+/parenleftbigg /Gamma1i−1 xL/parenrightbigg2 =/parenleftbigg1 xL/parenrightbigg2 , (2.56b) which are seen to represent two families of circles in the /Gamma1r,/Gamma1iplane. Resistance circles are defined by (2.56a) and reactance circles are defined by (2.56b). For example, the rL=1 circle has its center at /Gamma1r=0.5,/Gamma1 i=0, and has a radius of 0.5, and so it passes through the center of the Smith chart. All of the resistance circles of (2.56a) have centers on the horizontal /Gamma1i=0 axis and pass through the /Gamma1=1 point on the right-hand side of the chart. The centers of all of the reactance circles of (2.56b) lie on the vertical /Gamma1r=1 line (off the chart), and these circles also pass through the /Gamma1=1 point. The resistance and reactance circles are orthogonal. The Smith chart can also be used to graphically solve the transmission line impedance equation of (2.44) since this can be written in terms of the generalized reflection coefficient as Zin=Z01+/Gamma1e−2jβ/lscript 1−/Gamma1e−2jβ/lscript,( 2.57) where /Gamma1is the reflection coefficient at the load and /lscriptis the (positive) length of transmission line. We then see that (2.57) is of the same form as (2.54), differing only by the phase angles of the /Gamma1terms. Thus, if we have plotted the reflection coefficient |/Gamma1|ejθat the load, the normalized input impedance seen looking into a length /lscriptof transmission line terminated with zLcan be found by rotating the point clockwise by an amount 2 β/lscript(subtracting 2β/lscript fromθ)around the center of the chart. The radius stays the same since the magnitude of /Gamma1 does not change with position along the line (assuming a lossless line). To facilitate such rotations, the Smith chart has scales around its periphery calibrated in electrical wavelengths, toward and away from the “generator” (which simply means the direction away from the load). These scales are relative, so only the difference in wave- lengths between two points on the Smith chart is meaningful. The scales cover a range of0 to 0.5 wavelength, which reflects the fact that the Smith chart automatically includes the periodicity of transmission line phenomenon. Thus, a line of length λ/2 (or any multiple) requires a rotation of 2β/lscript =2πaround the center of the chart, bringing the point back to its original position, showing that the input impedance of a load seen through a λ/2 line is unchanged. c02TransmissionLineTheory Pozar July 26, 2011 17:33 66 Chapter 2: Transmission Line Theory We will now illustrate the use of the Smith chart for a variety of typical transmission line problems through examples. EXAMPLE 2.2 BASIC SMITH CHART OPERATIONS A load impedance of 40 +j70/Omega1terminates a 100 /Omega1transmission line that is 0.3λ long. Find the reflection coefficient at the load, the reflection coefficient at the input to the line, the input impedance, the standing wave ratio on the line, andthe return loss. Solution The normalized load impedance is z L=ZL Z0=0.4+j0.7, which can be plotted on the Smith chart as shown in Figure 2.11. By using a drawing compass and the voltage coefficient scale printed below the chart, one canread off the reflection coefficient magnitude at the load as |/Gamma1|=0.59. This same compass setting can then be applied to the standing wave ratio (SWR) scale to read SWR =3.87 and to the return loss (RL) (in dB) scale to read RL =4.6d B . ZL Zin FIGURE 2.11 Smith chart for Example 2.2. c02TransmissionLineTheory Pozar July 26, 2011 17:33 2.4 The Smith Chart 67 Now draw a radial line through the load impedance point and read the angle of the reflection coefficient at the load from the outer scale of the chart as 104◦. Now draw an SWR circle through the load impedance point. Reading the reference position of the load on the wavelengths-toward-generator (WTG) scalegives a value of 0.106λ. Moving down the line 0.3λ toward the generator brings us to 0.406λ on the WTG scale. Drawing a radial line at this position gives the normalized input impedance at the intersection with SWR circle of z in=0.365− j0.611. Then the input impedance of the line is Zin=Z0zin=36.5−j61.1/Omega1. The reflection coefficient at the input still has a magnitude of |/Gamma1|=0.59; the phase is read from the radial line at the phase scale as 248◦. ■ TheCombinedImpedance–AdmittanceSmithChart The Smith chart can be used for normalized admittance in the same way that it is used for normalized impedances, and it can be used to convert between impedance and admittance. The latter technique is based on the fact that, in normalized form, the input impedance of a load zLconnected to a λ/4 line is, from (2.44), zin=1/zL, which has the effect of converting a normalized impedance to a normalized admittance. Since a complete revolution around the Smith chart corresponds to a line length of λ/2, aλ/4 transformation is equivalent to a 180◦rotation; this is also equivalent to imag- ing a given impedance (or admittance) point across the center of the chart to obtain the corresponding admittance (or impedance) point. Thus, a Smith chart can be used for both impedance and admittance calculations dur- ing the solution of a given problem. At different stages of the solution, then, the chart maybe either an impedance Smith chart or an admittance Smith chart. This procedure can be made less confusing by using a Smith chart that has a superposition of the scales for a regular Smith chart and the scales of a Smith chart that has been rotated by180 ◦,a ss h o w n in Figure 2.12. Such a chart is referred to as an impedance and admittance Smith chart and usually has different-colored scales for impedance and admittance. EXAMPLE 2.3 SMITH CHART OPERATIONS USING ADMITTANCES A load of ZL=100+j50/Omega1terminates a 50 /Omega1line. What are the load admit- tance and input admittance if the line is 0.15λ long? Solution The normalized load impedance is zL=2+j1. A standard Smith chart can be used for this problem by initially considering it as an impedance chart and plotting zLand the SWR circle. Conversion to admittance can be accomplished with a λ/4r o t a t i o no f zL(easily obtained by drawing a straight line through zLand the center of the chart to intersect the other side of the SWR circle). The chart can now be considered as an admittance chart, and the input admittance can be found by rotating 0.15λ from yL. Alternatively, we can use the combined zychart of Figure 2.12, where conver- sion between impedance and admittance is accomplished merely by reading the c02TransmissionLineTheory Pozar July 26, 2011 17:33 68 Chapter 2: Transmission Line Theory j , 20-2030-3040-4050-5060-6070-7080-8090-90100-100110-110120-120130-130140-140150-150160-160170-1701800.040.040.050.050.060.060.070.070.080.080.090.090.10.10.110.110.120.120.130.130.140.140.150.150.160.160.170.170.180.180.190.190.20.20.210.210.220.220.230.230.240.24 0.250.250.260.260.270.270.280.280.290.290.30.30.310.310.320.320.330.330.340.340.350.350.360.360.370.370.380.380.390.390.40.40.410.410.420.420.430.430.440.440.450.450.460.460.470.470.480.480.490.49 0.00.0AN GLE OFREFLECTION COEFFICIENTINDEGREES—>WAVELENGTHSTOWARDGENERATOR—><— W AVELENGTH STOW ARD LOAD <—II T NDUCTIVEREACTANCECOMPONEN(+X/Zo) ORCAPACTIVESUSCEPTANCE(+jB/Yo)CAPACITIVEREACTANCECOMPONENT(-jX/Zo),ORINDUCTIVESUSCEPTANCE(-jB/Yo) 0.1 0.1 0.1 0.2 0.2 0.2 0.3 0.3 0.3 0.4 0.4 0.4 0.5 0.5 0.5 0.6 0.6 0.6 0.7 0.7 0.7 0.8 0.8 0.8 0.9 0.9 0.9 1.0 1.0 1.01.21.21.21.41.41.41.61.61.61.81.81.82.02.02.03.03.03.04.04.04.05.05.05.01010102020205050500.20.2 0.2 0.20.40.4 0.4 0.40.60.6 0.6 0.60.80.8 0.8 0.81.01.0 1.0 1.0RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) ± 0.15/H9261 TGzL (on impedance scales) yL (on admittance scales) yL (on impedance scales) y 0.10.20.30.40.50.60.70.80.91.01.21.41.61.82.03.04.05.0102000.2 0.10.30.40.50.60.70.80.9 1.01.21.6 1.41.82.03.04.05.0102030 30 20 10 5.0 4.0 3.0 2.0 1.8 1.6 1.4 1.21.0 0.90.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 FIGURE 2.12 ZYSmith chart with solution for Example 2.3. appropriate scales. Plotting zLon the impedance scales and reading the admittance scales at this same point gives yL=0.40−j0.20. The actual load admittance is then YL=yLY0=yL Z0=0.0080 −j0.0040 S. Then, on the WTG scale, the load admittance is seen to have a reference position of 0.214λ. Moving 0.15λ past this point brings us to 0.364λ . A radial line at this point on the WTG scale intersects the SWR circle at an admittance of y= 0.61+j0.66. The actual input admittance is then Y=0.0122 +j0.0132 S. ■ TheSlottedLine A slotted line is a transmission line configuration (usually a waveguide or coaxial line) that allows the sampling of the electric field amplitude of a standing wave on a terminated line.With this device the SWR and the distance of the first voltage minimum from the load can be measured, and from these data the load impedance can be determined. Note that be- cause the load impedance is, in general, a complex number (with two degrees of freedom),two distinct quantities must be measured with the slotted line to uniquely determine this impedance. A typical waveguide slotted line is shown in Figure 2.13. c02TransmissionLineTheory Pozar July 26, 2011 17:33 2.4 The Smith Chart 69 FIGURE 2.13 AnX-band waveguide slotted line. Although slotted lines used to be the principal way of measuring an unknown impedance at microwave frequencies, they have largely been superseded by the modern vector network analyzer in terms of accuracy, versatility, and convenience. The slotted line is still of some use, however, in certain applications such as high millimeter wave frequencies or whereit is desired to avoid connector mismatches by connecting the unknown load directly to the slotted line, thus avoiding the use of imperfect transitions. Another reason for studying the slotted line is that it provides an unexcelled tool for learning the basic concepts of standing waves and mismatched transmission lines. We will derive expressions for finding the unknown load impedance from slotted line measurements and also show how the Smithchart can be used for the same purpose. Assume that, for a certain terminated line, we have measured the SWR on the line and/lscript min, the distance from the load to the first voltage minimum on the line. The load impedance ZLcan then be determined as follows. From (2.41) the magnitude of the reflec- tion coefficient on the line is found from the standing wave ratio as |/Gamma1|=SWR−1 SWR+1.( 2.58) From Section 2.3, we know that a voltage minimum occurs when ej(θ−2β/lscript)=−1, where θis the phase angle of the reflection coefficient, /Gamma1=|/Gamma1|ejθ. The phase of the reflection coefficient is then θ=π+2β/lscript min,( 2.59) where /lscriptminis the distance from the load to the first voltage minimum. Actually, since the voltage minima repeat every λ/2, where λis the wavelength on the line, any multiple of λ/2 can be added to /lscriptminwithout changing the result in (2.59) because this just amounts to adding 2β nλ/2=2πntoθ, which will not change /Gamma1. Thus, the two quantities SWR and/lscriptmincan be used to find the complex reflection coefficient /Gamma1at the load. It is then c02TransmissionLineTheory Pozar July 26, 2011 17:33 70 Chapter 2: Transmission Line Theory straightforward to use (2.43) with /lscript=0 to find the load impedance from /Gamma1: ZL=Z01+/Gamma1 1−/Gamma1.( 2.60) The use of the Smith chart in solving this problem is best illustrated by an example. EXAMPLE 2.4 IMPEDANCE MEASUREMENT WITH A SLOTTED LINE The following two-step procedure has been carried out with a 50 /Omega1coaxial slotted line to determine an unknown load impedance: 1. A short circuit is placed at the load plane, resulting in a standing wave on the line with infinite SWR and sharply defined voltage minima, as shown in Figure 2.14a. On the arbitrarily positioned scale on the slotted line, voltage minima are recorded at z=0.2c m ,2.2c m ,4.2c m . 2. The short circuit is removed and replaced with the unknown load. The standing wave ratio is measured as SWR = 1.5, and voltage minima, which are not as sharply defined as those in step 1, are recorded at z=0.72 cm ,2.72 cm ,4.72 cm , as shown in Figure 2.14b. Find the load impedance. Solution Knowing that voltage minima repeat every λ/2, we have from the data of step 1 thatλ=4.0 cm. In addition, because the reflection coefficient and input impedance also repeat every λ/2, we can consider the load terminals to be effectively located at any of the voltage minima locations listed in step 1. Thus, if we say the load is at 4.2 cm, then the data from step 2 show that the next voltage minimum away from the load occurs at 2.72 cm, giving /lscriptmin=4.2−2.72=1.48 cm =0.37λ. |V| |V|Short circuit Unknown loadVmax Vmin(a) (b)0123 45 0123 45 FIGURE 2.14 V oltage standing wave patterns for Example 2.4. (a) Standing wave for short-circuit load. (b) Standing wave for unknown load. c02TransmissionLineTheory Pozar July 26, 2011 17:33 2.4 The Smith Chart 71 Applying (2.58)–(2.60) to these data gives |/Gamma1|=1.5−1 1.5+1=0.2, θ=π+4π 4.0(1.48) =86.4◦, so /Gamma1=0.2ej86.4◦=0.0126 +j0.1996. The load impedance is then ZL=50/parenleftbigg1+/Gamma1 1−/Gamma1/parenrightbigg =47.3+j19.7/Omega1. For the Smith chart version of the solution, we begin by drawing the SWR circle for SWR =1.5, as shown in Figure 2.15; the unknown normalized load impedance must lie on this circle. The reference that we have is that the load is 0.37λ away from the first voltage minimum. On the Smith chart the position of a voltage minimum corresponds to the minimum impedance point (minimumvoltage, maximum current), which is the horizontal axis (zero reactance) to the j , 20-2030-3040-4050-5060-6070-7080-8090-90100-100110-110120-120130-130140-140150-150160-160170-1701800.040.040.050.050.060.060.070.070.080.080.090.090.10.10.110.110.120.120.130.130.140.140.150.150.160.160.170.170.180.180.190.190.20.20.210.210.220.220.230.230.240.24 0.250.250.260.260.270.270.280.280.290.290.30.30.310.310.320.320.330.330.340.340.350.350.360.360.370.370.380.380.390.390.40.40.410.410.420.420.430.430.440.440.450.450.460.460.470.470.480.480.490.49 0.00.0AN GLE OFREFLECTION COEFFICIENTINDEGREES—>WAVELENGTHSTOWARDGENERATOR—><— W AVELENGTH STOW ARD LOAD <—II T NDUCTIVEREACTANCECOMPONEN(+X/Zo) ORCAPACTIVESUSCEPTANCE(+jB/Yo)CAPACITIVEREACTANCECOMPONENT(-jX/Zo),ORINDUCTIVESUSCEPTANCE(-jB/Yo) 0.1 0.1 0.1 0.2 0.2 0.2 0.3 0.3 0.3 0.4 0.4 0.4 0.5 0.5 0.5 0.6 0.6 0.6 0.7 0.7 0.7 0.8 0.8 0.8 0.9 0.9 0.9 1.0 1.0 1.01.21.21.21.41.41.41.61.61.61.81.81.82.02.02.03.03.03.04.04.04.05.05.05.01010102020205050500.20.2 0.2 0.20.40.4 0.4 0.40.60.6 0.6 0.60.80.8 0.8 0.81.01.0 1.0 1.0RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) ± SWRcircleVoltage max.Voltage min.zL 0.37 /H9261 TL FIGURE 2.15 Smith chart for Example 2.4. c02TransmissionLineTheory Pozar July 26, 2011 17:33 72 Chapter 2: Transmission Line Theory left of the origin. Thus, we begin at the voltage minimum point and move 0 .37λ toward the load (counterclockwise), to the normalized load impedance point, zL=0.95+j0.4, as shown in Figure 2.15. The actual load impedance is then ZL=47.5+j20/Omega1, in close agreement with the above result using equations. Note that, in principle, voltage maxima locations could be used as well as voltage minima positions, but voltage minima are more sharply defined than volt- age maxima and so usually result in greater accuracy. ■ 2.5THEQUARTER-WAVETRANSFORMER The quarter-wave transformer is a useful and practical circuit for impedance matching and also provides a simple transmission line circuit that further illustrates the properties of standing waves on a mismatched line. Although we will study the design and performance of quarter-wave matching transformers more extensively in Chapter 5, the main purposehere is the application of the previously developed transmission line theory to a basic trans- mission line circuit. We will first approach the problem from the impedance viewpoint and then show how this result can also be interpreted in terms of an infinite set of multiple reflections on the matching section. TheImpedanceViewpoint Figure 2.16 shows a circuit employing a quarter-wave transformer. The load resistance R L and the feedline characteristic impedance Z0are both real and assumed to be known. These two components are connected with a lossless piece of transmission line of (unknown)characteristic impedance Z 1and length λ/4. It is desired to match the load to the Z0line by using the λ/4 section of line and so make /Gamma1=0 looking into the λ/4 matching section. From (2.44) the input impedance Zincan be found as Zin=Z1RL+jZ1tanβ/lscript Z1+jRLtanβ/lscript.( 2.61) To evaluate this for β/lscript=(2π/λ)(λ/4) =π/2, we can divide the numerator and denomi- nator by tan β/lscriptand take the limit as β/lscript→π/2 to get Zin=Z2 1 RL.( 2.62) In order for /Gamma1=0, we must have Zin=Z0, which yields the characteristic impedance Z1 as Z1=/radicalbig Z0RL,( 2.63) /H9261/4 Z0 Z1 ZinRLΓ FIGURE 2.16 The quarter-wave matching transformer. c02TransmissionLineTheory Pozar July 26, 2011 17:33 2.5 The Quarter-Wave Transformer 73 which is the geometric mean of the load and source impedances. Then there will be no standing waves on the feedline (SWR =1), although there will be standing waves on the λ/4 matching section. In addition, the above condition applies only when the length of the matching section is λ/4 or an odd multiple of λ/4, long, so that a perfect match may be achieved at one frequency, but impedance mismatch will occur at other frequencies. EXAMPLE 2.5 FREQUENCY RESPONSE OF A QUARTER-WA VE TRANSFORMER Consider a load resistance RL=100/Omega1to be matched to a 50 /Omega1line with a quarter-wave transformer. Find the characteristic impedance of the matching sec- tion and plot the magnitude of the reflection coefficient versus normalized fre-quency, f/f o, where fois the frequency at which the line is λ/4 long. Solution From (2.63), the necessary characteristic impedance is Z1=/radicalbig (50)(100) =70.71 /Omega1. The reflection coefficient magnitude is given as |/Gamma1|=/vextendsingle/vextendsingle/vextendsingle/vextendsingleZ in−Z0 Zin+Z0/vextendsingle/vextendsingle/vextendsingle/vextendsingle, where the input impedance Z inis a function of frequency as given by (2.44). The frequency dependence in (2.44) comes from the β/lscriptterm, which can be written in terms of f/foas β/lscript=/parenleftbigg2π λ/parenrightbigg/parenleftbiggλ0 4/parenrightbigg =/parenleftbigg2πf vp/parenrightbigg/parenleftbiggvp 4fo/parenrightbigg =πf 2fo, where it is seen that β/lscript=π/2f o r f=fo, as expected. For higher frequen- cies the matching section looks electrically longer, and for lower frequencies itlooks shorter. The magnitude of the reflection coefficient is plotted versus f/f oin Figure 2.17. ■ 0.0 1.0 2.0 3.0 4.00.00.10.20.3|Γ| f/fo FIGURE 2.17 Reflection coefficient versus normalized frequency for the quarter-wave trans- former of Example 2.5. c02TransmissionLineTheory Pozar July 26, 2011 17:33 74 Chapter 2: Transmission Line Theory This method of impedance matching is limited to real load impedances, although a complex load impedance can easily be made real, at a single frequency, by transformation through an appropriate length of line. The above analysis shows how useful the impedance concept can be when solving transmission line problems, and this method is probably the preferred method in practice. It may aid our understanding of the quarter-wave transformer (and other transmission line circuits), however, if we now look at it from the viewpoint of multiple reflections. TheMultiple-Reflectio Viewpoint Figure 2.18 shows the quarter-wave transformer circuit with reflection and transmission coefficients defined as follows: /Gamma1=overall, or total, reflection coefficient of a wave incident on the λ/4 transformer (same as /Gamma1in Example 2.5). /Gamma11=partial reflection coefficient of a wave incident on a load Z1,f r o mt h e Z0line. /Gamma12=partial reflection coefficient of a wave incident on a load Z0,f r o mt h e Z1line. /Gamma13=partial reflection coefficient of a wave incident on a load RL,f r o mt h e Z1line. T1=partial transmission coefficient of a wave from the Z0line into the Z1line. T2=partial transmission coefficient of a wave from the Z1line into the Z0line. These coefficients can be expressed as /Gamma11=Z1−Z0 Z1+Z0, (2.64a) /Gamma12=Z0−Z1 Z0+Z1=−/Gamma11, (2.64b) /Gamma13=RL−Z1 RL+Z1, (2.64c) /H9261/4Z0 Z1 RL 11 Γ1Γ1Γ Γ3Γ3 Γ2 Γ3 Γ3Γ2 Γ2–T1T2Γ3 T1T2Γ32Γ2T2T1T2 T1 T2 FIGURE 2.18 Multiple reflection analysis of the quarter-wave transformer. c02TransmissionLineTheory Pozar July 26, 2011 17:33 2.5 The Quarter-Wave Transformer 75 T1=2Z1 Z1+Z0, (2.64d) T2=2Z0 Z1+Z0. (2.64e) Now think of the quarter-wave transformer of Figure 2.18 in the time domain, and imagine aw a v et r a v e l i n gd o w nt h e Z0feedline toward the transformer. When the wave first hits the junction with the Z1line, it sees only an impedance Z1since it has not yet traveled to the load RLand cannot see that effect. Part of the wave is reflected with a coefficient /Gamma11, and part is transmitted onto the Z1line with a coefficient T1. The transmitted wave then travels λ/4 to the load, is reflected with a coefficient /Gamma13, and travels another λ/4 back to the junction with the Z0line. Part of this wave is transmitted through (to the left) to the Z0line, with coefficient T2, and part is reflected back toward the load with coefficient /Gamma12. Clearly, this process continues with an infinite number of bouncing waves, and the totalreflection coefficient, /Gamma1, is the sum of all of these partial reflections. Since each round trip path up and down the λ/4 transformer section results in a 180 ◦phase shift, the total reflection coefficient can be expressed as /Gamma1=/Gamma11−T1T2/Gamma13+T1T2/Gamma12/Gamma12 3−T1T2/Gamma12 2/Gamma13 3+··· =/Gamma11−T1T2/Gamma13∞/summationdisplay n=0(−/Gamma12/Gamma13)n. (2.65) Since |/Gamma13|<1 and |/Gamma12|<1, the infinite series in (2.65) can be summed using the geometric series result that ∞/summationdisplay n=0xn=1 1−x, for|x|<1, to give /Gamma1=/Gamma11−T1T2/Gamma13 1+/Gamma12/Gamma13=/Gamma11+/Gamma11/Gamma12/Gamma13−T1T2/Gamma13 1+/Gamma12/Gamma13.( 2.66) The numerator of this expression can be simplified using (2.64) to give /Gamma11−/Gamma13/parenleftbig /Gamma12 1+T1T2/parenrightbig =/Gamma11−/Gamma13/bracketleftbigg(Z1−Z0)2 (Z1+Z0)2+4Z1Z0 (Z1+Z0)2/bracketrightbigg =/Gamma11−/Gamma13=(Z1−Z0)(RL+Z1)−(RL−Z1)(Z1+Z0) (Z1+Z0)(RL+Z1) =2/parenleftbig Z2 1−Z0RL/parenrightbig (Z1+Z0)(RL+Z1), which is seen to vanish if we choose Z1=√Z0RL, as in (2.63). Then /Gamma1of (2.66) is zero, and the line is matched. This analysis shows that the matching property of the quarter-wave transformer comes about by properly selecting the characteristic impedance and length ofthe matching section so that the superposition of all of the partial reflections adds to zero. Under steady-state conditions, an infinite sum of waves traveling in the same direction with the same phase velocity can be combined into a single traveling wave. Thus, the infiniteset of waves traveling in the forward and reverse directions on the matching section can be reduced to two waves traveling in opposite directions. See Problem 2.25. c02TransmissionLineTheory Pozar July 26, 2011 17:33 76 Chapter 2: Transmission Line Theory 2.6GENERATORANDLOADMISMATCHES In Section 2.3 we treated the terminated (mismatched) transmission line assuming that the generator was matched, so that no reflections occurred at the generator. In general, however, both generator and load may present mismatched impedances to the transmissionline. We will study this case and also see that the condition for maximum power transfer from the generator to the load may, in some situations, involve a standing wave on the line. Figure 2.19 shows a transmission line circuit with arbitrary generator and load impedances Z gandZ/lscript, which may be complex. The transmission line is assumed to be lossless, with a length /lscriptand characteristic impedance Z0. This circuit is general enough to model most passive and active networks that occur in practice. Because both the generator and load are mismatched, multiple reflections can occur on the line, as in the problem of the quarter-wave transformer. The present circuit could thus be analyzed using an infinite series to represent the multiple bounces, as in Section 2.5,but we will use the easier and more useful method of impedance transformation. The input impedance looking into the terminated transmission line from the generator end is, from (2.43) and (2.44), Z in=Z01+/Gamma1/lscripte−2jβ/lscript 1−/Gamma1/lscripte−2jβ/lscript=Z0Z/lscript+jZ0tanβ/lscript Z0+jZ/lscripttanβ/lscript,( 2.67) where /Gamma1/lscriptis the reflection coefficient of the load: /Gamma1/lscript=Z/lscript−Z0 Z/lscript+Z0.( 2.68) The voltage on the line can be written as V(z)=V+ o/parenleftbig e−jβz+/Gamma1/lscriptejβz/parenrightbig ,( 2.69) and we can find V+ ofrom the voltage at the generator end of the line, where z=−/lscript: V(−/lscript)=VgZin Zin+Zg=V+ o/parenleftbig ejβ/lscript+/Gamma1/lscripte−jβ/lscript/parenrightbig , so that V+ o=VgZin Zin+Zg1/parenleftbig ejβ/lscript+/Gamma1/lscripte−jβ/lscript/parenrightbig.( 2.70) This can be rewritten, using (2.67), as V+ o=VgZ0 Z0+Zge−jβ/lscript /parenleftbig 1−/Gamma1/lscript/Gamma1ge−2jβ/lscript/parenrightbig,( 2.71) z –l 0Z0, /H9252 ZlZg Vg Zin VinIinΓ Γl + – FIGURE 2.19 Transmission line circuit for mismatched load and generator. c02TransmissionLineTheory Pozar July 26, 2011 17:33 2.6 Generator and Load Mismatches 77 where /Gamma1gis the reflection coefficient seen looking into the generator: /Gamma1g=Zg−Z0 Zg+Z0.( 2.72) The standing wave ratio on the line is then SWR=1+|/Gamma1/lscript| 1−|/Gamma1/lscript|.( 2.73) The power delivered to the load is P=1 2Re{VinI∗ in}=1 2|Vin|2Re/braceleftbigg1 Zin/bracerightbigg =1 2|Vg|2/vextendsingle/vextendsingle/vextendsingle/vextendsingleZin Zin+Zg/vextendsingle/vextendsingle/vextendsingle/vextendsingle2 Re/braceleftbigg1 Zin/bracerightbigg .( 2.74) Now let Zin=Rin+jXinandZg=Rg+jXg; then (2.74) can be reduced to P=1 2|Vg|2 Rin (Rin+Rg)2+(Xin+Xg)2.( 2.75) We now assume that the generator impedance, Zg, is fixed, and consider three cases of load impedance. LoadMatchedtoLine In this case we have Zl=Z0,s o/Gamma1/lscript=0, and SWR = 1, from (2.68) and (2.73). Then the input impedance is Zin=Z0, and the power delivered to the load is, from (2.75), P=1 2|Vg|2 Z0 (Z0+Rg)2+X2g.( 2.76) GeneratorMatchedtoLoadedLine In this case the load impedance Z/lscriptand/or the transmission line parameters β/lscript,Z0are chosen to make the input impedance Zin=Zg, so that the generator is matched to the load presented by the terminated transmission line. Then the overall reflection coefficient, /Gamma1,i s zero: /Gamma1=Zin−Zg Zin+Zg=0.( 2.77) There may, however, be a standing wave on the line since /Gamma1/lscriptmay not be zero. The power delivered to the load is P=1 2|Vg|2Rg 4/parenleftbig R2g+X2g/parenrightbig.( 2.78) Observe that even though the loaded line is matched to the generator, the power deliv- ered to the load may be less than that of (2.76), where the loaded line was not necessarily matched to the generator. Thus, we are led to the question of what is the optimum load impedance, or equivalently, what is the optimum input impedance, to achieve maximumpower transfer to the load for a given generator impedance. ConjugateMatching Assuming that the generator series impedance Z gis fixed, we may vary the input impedance Zinuntil we achieve the maximum power delivered to the load. Knowing Zin,i ti st h e n easy to find the corresponding load impedance Z/lscriptvia an impedance transformation along c02TransmissionLineTheory Pozar July 26, 2011 17:33 78 Chapter 2: Transmission Line Theory the line. To maximize P, we differentiate with respect to the real and imaginary parts of Zin. Using (2.75) gives ∂P ∂Rin=0→1 (Rin+Rg)2+(Xin+Xg)2+−2Rin(Rin+Rg) [(Rin+Rg)2+(Xin+Xg)2]2=0, or R2 g−R2 in+(Xin+Xg)2=0,( 2.79a) and ∂P ∂Xin=0→−2Rin(Xin+Xg) [(Rin+Rg)2+(Xin+Xg)2]2=0, or Xin(Xin+Xg)=0.( 2.79b) Solving (2.79a) and (2.79b) simultaneously for RinandXingives Rin=Rg,Xin=− Xg, or Zin=Z∗ g.( 2.80) This condition is known as conjugate matching, and it results in maximum power transfer to the load for a fixed generator impedance. The power delivered is, from (2.75) and (2.80), P=1 2|Vg|21 4Rg,( 2.81) which is seen to be greater than or equal to the powers of (2.76) or (2.78). This is also the maximum available power from the generator. Note that the reflection coefficients /Gamma1/lscript,/Gamma1g, and/Gamma1may be nonzero. Physically, this means that in some cases the power in the multiple reflections on a mismatched line may add in phase to deliver more power to the load than would be delivered if the line were flat (no reflections). If the generator impedance is real (Xg=0), then the last two cases reduce to the same result, which is that maximum power is delivered to the load when the loaded line is matched to the generator ( Rin=Rg, with Xin=Xg=0). Finally, note that neither matching for zero reflection (Z/lscript=Z0)nor conjugate match- ing(Zin=Z∗ g)necessarily yields a system with the best efficiency. For example, if Zg= Z/lscript=Z0then both load and generator are matched (no reflections), but only half the power produced by the generator is delivered to the load (the other half is lost in Zg), for a trans- mission efficiency of 50%. This efficiency can only be improved by making Zgas small as possible. 2.7LOSSYTRANSMISSIONLINES In practice, transmission lines have losses due to finite conductivity and/or lossy dielectric, but these losses are usually small. In many practical problems loss may be neglected, but at other times the effect of loss may be very important, as when dealing with the attenuation of a transmission line, noise introduced by a lossy line, or the Qof a resonator, for example. In this section we will study the effects of loss on transmission line behavior and show how the attenuation constant can be calculated. c02TransmissionLineTheory Pozar July 26, 2011 17:33 2.7 Lossy Transmission Lines 79 TheLow-LossLine In most practical microwave and RF transmission lines the loss is small—if this were not the case, the line would be of little practical value. When the loss is small, some approxima- tions can be made to simplify the expressions for the general transmission line parametersofγ=α+jβandZ 0. The general expression for the complex propagation constant is, from (2.5), γ=/radicalbig (R+jωL)(G+jωC), (2.82) which can be rearranged as γ=/radicalBigg (jωL)(jωC)/parenleftbigg 1+R jωL/parenrightbigg/parenleftbigg 1+G jωC/parenrightbigg =jω√ LC/radicalBigg 1−j/parenleftbiggR ωL+G ωC/parenrightbigg −RG ω2LC. (2.83) For a low-loss line both conductor and dielectric loss will be small, and we can assume thatR/lessmuchωLandG/lessmuchωC. Then, RG/lessmuchω2LC, and (2.83) reduces to γ/similarequaljω√ LC/radicalBigg 1−j/parenleftbiggR ωL+G ωC/parenrightbigg .( 2.84) If we were to ignore the (R/ωL+G/ωC)term we would obtain the result that γwas purely imaginary (no loss), so we will instead use the first two terms of the Taylor series expansion for√ 1+x/similarequal1+x/2+··· to give the first higher order real term for γ: γ/similarequaljω√ LC/bracketleftbigg 1−j 2/parenleftbiggR ωL+G ωC/parenrightbigg/bracketrightbigg , so that α/similarequal1 2/parenleftBigg R/radicalbigg C L+G/radicalbigg L C/parenrightBigg =1 2/parenleftbiggR Z0+GZ 0/parenrightbigg , (2.85a) β/similarequalω√ LC, (2.85b) where Z0=√L/Cis the characteristic impedance of the line in the absence of loss. Note from (2.85b) that the propagation constant βis identical to that of the lossless case of (2.12). By the same order of approximation, the characteristic impedance Z0can be approximated as a real quantity: Z0=/radicalBigg R+jωL G+jωC/similarequal/radicalbigg L C.( 2.86) Equations (2.85)–(2.86) are known as the high-frequency, low-loss approximations for transmission lines, and they are important because they show that the propagation constantand characteristic impedance for a low-loss line can be closely approximated by consider- ing the line as lossless. c02TransmissionLineTheory Pozar July 26, 2011 17:33 80 Chapter 2: Transmission Line Theory EXAMPLE 2.6 ATTENUATION CONSTANT OF THE COAXIAL LINE In Example 2.1 the L,C,R, and Gparameters were derived for a lossy coaxial line. Assuming the loss is small, derive the attenuation constant from (2.85a) with the results from Example 2.1. Solution From (2.85a), α=1 2/parenleftBigg R/radicalbigg C L+G/radicalbigg L C/parenrightBigg . Using the results for RandGderived in Example 2.1 gives α=1 2/bracketleftbiggRs ηlnb/a/parenleftbigg1 a+1 b/parenrightbigg +ω/epsilon1/prime/primeη/bracketrightbigg , where η=/radicalbig µ//epsilon1/primeis the intrinsic impedance of the dielectric material filling the coaxial line. In addition, β=ω√ LC=ω√µ/epsilon1/primeand Z0=√L/C=(η/2π) lnb/a. ■ This method for the calculation of attenuation requires that the line parameters L,C,R, andGbe known. These can sometimes be derived using the formulas of (2.17) −(2.20), but a more direct and versatile procedure is to use the perturbation method, to be discussed shortly. TheDistortionlessLine As can be seen from the exact equations (2.82)–(2.83) for the propagation constant of a lossy line, the phase term βis generally a complicated function of frequency ωwhen loss is present. In particular, we note that βis generally not exactly a linear function of frequency, as in (2.85b), unless the line is lossless. If βis not a linear function of frequency (of the formβ=aω), then the phase velocity vp=ω/β will vary with frequency. The implication of this is that the various frequency components of a wideband signal will travel with different phase velocities and so arrive at the receiver end of the transmission line at slightly different times. This will lead to dispersion, a distortion of the signal, and is generally an undesirable effect. Granted, as we have argued, the departure of βfrom a linear function may be quite small, but the effect can be significant if the line is very long. This effect leads to the concept of group velocity, which we will address in detail in Section 3.10. There is a special case, however, of a lossy line that has a linear phase factor as a function of frequency. Such a line is called a distortionless line, and it is characterized by line parameters that satisfy the relation R L=G C.( 2.87) From (2.83) the exact complex propagation constant, under the condition specified by (2.87), reduces to γ=jω√ LC/radicalBigg 1−2jR ωL−R2 ω2L2 =jω√ LC/parenleftbigg 1−jR ωL/parenrightbigg =R/radicalbigg C L+jω√ LC=α+jβ, (2.88) c02TransmissionLineTheory Pozar July 26, 2011 17:33 2.7 Lossy Transmission Lines 81 z –l 0V(z), I (z) Z0, /H9251, /H9252 ZL Zin FIGURE 2.20 A lossy transmission line terminated in the impedance ZL. which shows that β=ω√ LCis now a linear function of frequency. Equation (2.88) also shows that the attenuation constant, α=R√C/L, does not depend on frequency, so that all frequency components of a signal will be attenuated by the same amount (actually, Ris usually a weak function of frequency). Thus, the distortionless line is not loss free but is capable of passing a pulse or modulation envelope without distortion. To obtain a transmission line with parameters that satisfy (2.87) often requires that Lbe increased by adding series loading coils spaced periodically along the line. The above theory for the distortionless line was first developed by Oliver Heavi- side (1850–1925), who solved many problems in transmission line theory and reworkedMaxwell’s original theory of electromagnetism into the modern version that we are famil- iar with today [5]. TheTerminatedLossyLine Figure 2.20 shows a length /lscriptof a lossy transmission line terminated in a load impedance Z L. Thus, γ=α+jβis complex, but we assume the loss is small, so that Z0is approxi- mately real, as in (2.86). In (2.36), expressions for the voltage and current wave on a lossless line are given. The analogous expressions for the lossy case are V(z)=V+ o/parenleftbig e−γz+/Gamma1eγz/parenrightbig , (2.89a) I(z)=V+ o Z0/parenleftbig e−γz−/Gamma1eγz/parenrightbig , (2.89b) where /Gamma1is the reflection coefficient of the load, as given in (2.35), and V+ ois the incident voltage amplitude referenced at z=0. From (2.42) the reflection coefficient at a distance /lscriptfrom the load is /Gamma1(/lscript)=/Gamma1e−2jβ/lscripte−2α/lscript=/Gamma1e−2γ/lscript.( 2.90) The input impedance Zinat a distance /lscriptfrom the load is then Zin=V(−/lscript) I(−/lscript)=Z0ZL+Z0tanhγ/lscript Z0+ZLtanhγ/lscript.( 2.91) We can compute the power delivered to the input of the terminated line at z=−/lscriptas Pin=1 2Re/braceleftbig V(−/lscript) I∗(−/lscript)/bracerightbig =|V+ o|2 2Z0/parenleftbig e2α/lscript−|/Gamma1|2e−2α/lscript/parenrightbig =|V+ o|2 2Z0/parenleftbig 1−|/Gamma1(/lscript)|2/parenrightbig e2α/lscript, (2.92) c02TransmissionLineTheory Pozar July 26, 2011 17:33 82 Chapter 2: Transmission Line Theory where (2.89) has been used for V(−/lscript) andI(−/lscript). The power actually delivered to the load is PL=1 2Re{V(0)I∗(0)}=|V+ o|2 2Z0(1−|/Gamma1|2). (2.93) The difference in these powers corresponds to the power lost in the line: Ploss=Pin−PL=|V+ o|2 2Z0/bracketleftbig/parenleftbig e2α/lscript−1/parenrightbig +|/Gamma1|2/parenleftbig 1−e−2α/lscript/parenrightbig/bracketrightbig .( 2.94) The first term in (2.94) accounts for the power loss of the incident wave, while the second term accounts for the power loss of the reflected wave; note that both terms increase as α increases. ThePerturbationMethodforCalculatingAttenuation Here we derive a useful and standard technique for finding the attenuation constant of a low-loss line. The method avoids the use of the transmission line parameters L,C,R, and Gand instead relies on the fields of the lossless line, with the assumption that the fields of the lossy line are not greatly different from the fields of the lossless line—hence the term,perturbation method . We have seen that the power flow along a lossy transmission line, in the absence of reflections, is of the form P(z)=P oe−2α z,( 2.95) where Pois the power at the z=0 plane and αis the attenuation constant we wish to determine. Now define the power loss per unit length along the line as P/lscript=−∂P ∂z=2αPoe−2α z=2αP(z), where the negative sign on the derivative was introduced so that P/lscriptwould be a positive quantity. From this, the attenuation constant can be determined as α=P/lscript(z) 2P(z)=P/lscript(z=0) 2Po.( 2.96) This equation states that αcan be determined from Po, the power on the line, and P/lscript,t h e power loss per unit length of line. It is important to realize that P/lscriptcan be computed from the fields of the lossless line and can account for both conductor loss [using (1.131)] and dielectric loss [using (1.92)]. EXAMPLE 2.7 USING THE PERTURBATION METHOD TO FIND THE ATTENUATION CONSTANT Use the perturbation method to find the attenuation constant of a coaxial line having a lossy dielectric and lossy conductors. Solution From Example 2.1 and (2.32), the fields of the lossless coaxial line are, for a< ρ< b, ¯E=Voˆρ ρlnb/ae−jβz, ¯H=Voˆφ 2πρZ0e−jβz, c02TransmissionLineTheory Pozar July 26, 2011 17:33 2.7 Lossy Transmission Lines 83 where Z0=(η/2π) lnb/ais the characteristic impedance of the coaxial line and Vois the voltage across the line at z=0. The first step is to find Po, the power flowing on the lossless line: Po=1 2Re/integraldisplay S¯EׯH∗·d¯s=|Vo|2 2Z0/integraldisplayb ρ=a/integraldisplay2π φ=0ρdρdφ 2πρ2lnb/a=|Vo|2 2Z0, as expected from basic circuit theory. The loss per unit length, P/lscript, comes from conductor loss (P/lscriptc)and dielectric loss(P/lscriptd). From (1.131), the conductor loss in a 1 m length of line can be found as P/lscriptc=Rs 2/integraldisplay S|¯Ht|2ds=Rs 2/integraldisplay1 z=0/braceleftbigg/integraldisplay2π φ=0|Hφ(ρ=a)|2adφ +/integraldisplay2π φ=0|Hφ(ρ=b)|2bdφ/bracerightBigg dz =Rs|Vo|2 4πZ2 0/parenleftbigg1 a+1 b/parenrightbigg . The dielectric loss in a 1 m length of line is, from (1.92), P/lscriptd=ω/epsilon1/prime/prime 2/integraldisplay V|¯E|2ds=ω/epsilon1/prime/prime 2/integraldisplayb ρ=a/integraldisplay2π φ=0/integraldisplay1 z=0|Eρ|2ρdρdφdz=πω/epsilon1/prime/prime lnb/a|Vo|2, where /epsilon1/prime/primeis the imaginary part of the complex permittivity, /epsilon1=/epsilon1/prime−j/epsilon1/prime/prime. Finally, applying (2.96) gives α=P/lscriptc+P/lscriptd 2Po=Rs 4πZ0/parenleftbigg1 a+1 b/parenrightbigg +πω/epsilon1/prime/primeZ0 lnb/a =Rs 2ηlnb/a/parenleftbigg1 a+1 b/parenrightbigg +ω/epsilon1/prime/primeη 2, where η=/radicalbig µ//epsilon1/prime. This result is seen to agree with that of Example 2.6. ■ TheWheelerIncrementalInductanceRule Another useful technique for the practical evaluation of attenuation due to conductor loss for TEM or quasi-TEM lines is the Wheeler incremental inductance rule [6]. This method is based on the similarity of the equations for the inductance per unit length and resistance per unit length of a transmission line, as given by (2.17) and (2.19), respectively. In other words, the conductor loss of a line is due to current flow inside the conductor, which, aswas shown in Section 1.7, is directly related to the tangential magnetic field at the surface of the conductor and thus to the inductance of the line. From (1.131), the power loss into a cross section Sof a good (but not perfect) conduc- tor is P /lscript=Rs 2/integraldisplay S|¯Js|2ds=Rs 2/integraldisplay S|¯Ht|2dsW/m2,( 2.97) so the power loss per unit length of a uniform transmission line is P/lscript=Rs 2/integraldisplay C|¯Ht|2d/lscriptW/m,( 2.98) c02TransmissionLineTheory Pozar July 26, 2011 17:33 84 Chapter 2: Transmission Line Theory where the line integral of (2.98) is over the cross-sectional contours of both conductors. From (2.17), the inductance per unit length of the line is L=µ |I|2/integraldisplay S|¯H|2ds,( 2.99) which is computed assuming the conductors are lossless. When the conductors have a small loss, the ¯Hfield in the conductor is no longer zero, and this field contributes a small additional “incremental” inductance, /Delta1L, to that of (2.99). As discussed in Chapter 1, the fields inside the conductor decay exponentially, so that the integration into the conductordimension can be evaluated as /Delta1L=µ 0δs 2|I|2/integraldisplay C|¯Ht|2d/lscript, (2.100) since/integraltext∞ 0e−2z/δsdz=δs/2. (The skin depth is δs=√2/ωµσ .) Then P/lscriptfrom (2.98) can be written in terms of /Delta1Las P/lscript=Rs|I|2/Delta1L µ0δs=|I|2/Delta1L σµ0δ2s=|I|2ω/Delta1L 2W/m,( 2.101) since Rs=√ωµ0/2σ=1/σδ s. Then from (2.96) the attenuation due to conductor loss can be evaluated as αc=P/lscript 2Po=ω/Delta1L 2Z0,( 2.102) since Po, the total power flow down the line, is Po=|I|2Z0/2. In (2.102), /Delta1Lis evaluated as the change in inductance when all conductor walls recede by an amount δs/2. Equation (2.102) can also be written in terms of the change in characteristic impedance since Z0=/radicalbigg L C=L√ LC=Lvp,( 2.103) so that αc=β/Delta1Z0 2Z0,( 2.104) where /Delta1Z0is the change in characteristic impedance when all conductor walls recede by an amount δs/2. Yet another form of the incremental inductance rule can be obtained by using the first two terms of a Taylor series expansion for Z0. Thus, Z0/parenleftbiggδs 2/parenrightbigg /similarequalZ0+δs 2dZ0 d/lscript,( 2.105) so that /Delta1Z0=Z0/parenleftbiggδs 2/parenrightbigg −Z0=δs 2dZ0 d/lscript, where Z0(δs/2)refers to the characteristic impedance of the line when the walls recede by δs/2, and /lscriptrefers to a distance into the conductors. Then (2.104) can be written as αc=βδs 4Z0dZ0 d/lscript=Rs 2Z0ηdZ0 d/lscript,( 2.106) c02TransmissionLineTheory Pozar July 26, 2011 17:33 2.8 Transients on Transmission Lines 85 where η=√µ0//epsilon1is the intrinsic impedance of the dielectric and Rsis the surface resistiv- ity of the conductor. Equation (2.106) is one of the most practical forms of the incremental inductance rule because the characteristic impedance is known for a wide variety of trans- mission lines. EXAMPLE 2.8 USING THE WHEELER INCREMENTAL INDUCTANCE RULE TO FIND THE ATTENUATION CONSTANT Calculate the attenuation due to conductor loss of a coaxial line using the Wheeler incremental inductance rule. Solution From (2.32) the characteristic impedance of the coaxial line is Z0=η 2πlnb a. From the incremental inductance rule of the form given in (2.106), the attenuation due to conductor loss is αc=Rs 2Z0ηdZ0 d/lscript=Rs 4πZ0/parenleftbiggdlnb/a db−dlnb/a da/parenrightbigg =Rs 4πZ0/parenleftbigg1 b+1 a/parenrightbigg , which is seen to be in agreement with the result of Example 2.7. The negative sign on the second differentiation in this equation is because the derivative for the inner conductor is in the −ρdirection (receding wall). ■ Regardless of how attenuation is calculated, measured attenuation values for practical transmission lines are usually higher. One reason for this discrepancy is the fact that real- istic transmission lines have metallic surfaces with a certain amount of roughness, whichincreases loss, while our theoretical calculations assume perfectly smooth conductors. A quasi-empirical formula that can be used to approximately account for surface roughness for a transmission line is [7] α /prime c=αc/bracketleftBigg 1+2 πtan−11.4/parenleftbigg/Delta1 δs/parenrightbigg2/bracketrightBigg ,( 2.107) where αcis the attenuation due to perfectly smooth conductors, α/prime cis the attenuation cor- rected for surface roughness, /Delta1is the rms surface roughness, and δsis the skin depth of the conductors. 2.8TRANSIENTSONTRANSMISSIONLINES So far we have concentrated on the behavior of transmission lines at a single frequency, and in many cases of practical interest this viewpoint is entirely satisfactory. In some situations, however, where short pulses or very wideband signals are propagating on a transmissionline, it is useful to consider wave propagation from a transient, or time domain, point of view. c02TransmissionLineTheory Pozar July 26, 2011 17:33 86 Chapter 2: Transmission Line Theory In this section we will discuss the reflection of transient pulses from terminated trans- mission lines, including the special cases of a matched line, a short-circuited line, and an open-circuited line. We will conclude with a description of bounce diagrams, which can be used to describe multiple reflections of pulses on transmission lines. Reflectio ofPulsesfromaTerminatedTransmissionLine A transient transmission line circuit is shown in Figure 2.21a, where a DC source is switched on at t=0. We first consider the case in which the line has a characteristic impedance of Z0, the source impedance is Z0, and the load impedance is Z0.I ti s assumed that the voltage on the line is initially zero: v(z,t)=0 for all z,f o r t<0. We want to determine the voltage response on the transmission line as a function of time and position. Because of the finite transit time of the line, its input impedance will appear to be equal to the characteristic impedance of the line for t<2/lscript/v p, where vpis the phase velocity of the line. In other words, the line looks infinitely long until the pulse has time to reach theload and (possibly) reflect back to the input. Therefore, when the switch closes at t=0, the circuit appears as a voltage divider consisting of the source impedance and the input impedance, both being Z 0. The initial voltage on the line is thus V0/2, and this voltage waveform propagates toward the load with a velocity vp. The leading edge of the pulse will be at position zon the line at time t=z/vp, as shown in Figure 2.21b. The pulse reaches the load at time t=/lscript/v p. Since the load is matched to the line, there is no reflection of the pulse from the load. The circuit is now in a steady-state condition, and voltage on the line is constant: v(z,t)=V0/2 for all t>/lscript / v p, as shown in Figure 2.21c. This is, of course, the DC value that we would expect for a voltage divider consisting of equal source and input impedances. Next consider the transmission line circuit of Figure 2.22a, where the line is now ter- minated with a short circuit. Initially, the input impedance of the line again appears as Z0, and the initial incident pulse again has an amplitude of V0/2, as shown in Figure 2.22b. z l 0Z0Z0Z0 V0+ –t = 0v(z, t) z l 00v 2V0 zl 00 z = v ptv 2V0(a) (b) (c) FIGURE 2.21 Transient response of a transmission line terminated with a matched load. (a) Transmission line circuit with a step function voltage source. (b) Responsefor 0<t< /lscript/v p. (c) Response for /lscript/vp<t<2/lscript/v p; there is no reflection from the load. c02TransmissionLineTheory Pozar September 12, 2011 18:26 2.8 Transients on Transmission Lines 87 z l 0Z0s.c.Z0 V0+ –t = 0v(z, t) z l 00 z = v ptv 2V0 z0v 2V0 2 V0–(a) (b) (c)l FIGURE 2.22 Transient response of a transmission line terminated with a short circuit. (a) Transmission line circuit with a step function voltage source. (b) Response for0<t< /lscript/v p. (c) Response for /lscript/vp<t<2/lscript/v p; the incident pulse is reflected with/Gamma1=−1. The short-circuit load has a reflection coefficient of /Gamma1=−1, which has the effect of invert- ing the reflected pulse as it travels back toward the source. The superposition of the forward and reverse traveling pulses leads to cancellation, as shown in Figure 2.22c, for the periodwhere /lscript/v p<t<2/lscript/v p. When the return pulse reaches the source, at t=2/lscript/v p, it will not be re-reflected because the source is matched to the line. The circuit is then in steady state, with zero voltage everywhere on the line. Again, this is consistent with DC circuitanalysis, as the shorted line has zero electrical length at DC and thus appears as a short at its input, leading to a terminal voltage of zero. The voltage waveform at a fixed point zon the line will consist of a rectangular pulse of amplitude V 0/2 existing only over the time period z/vp<t<(2/lscript−z)/vp. This effect can be used in practice to generate pulses of very short duration. Finally, consider the effect of a transmission line with an open-circuit termination, as shown in Figure 2.23a. As in previous cases, the input impedance of the line initially appears as Z0, and the initial incident pulse has an amplitude of V0/2, as shown in Figure 2.23b. The open-circuit load has a reflection coefficient of /Gamma1=1, which reflects the incident waveform with the same polarity toward the source. The amplitudes of the for- ward and reverse pulses add to create a wave with an amplitude of V0, as shown in Figure 2.23c. At t=2/lscript/v pthe return pulse reaches the source, but it is not re-reflected since the source is matched to the line. The circuit is then in steady state, with a constant voltage ofV0on the line. By DC analysis, the open-circuited line presents an open circuit at its terminals, leading to a terminal voltage equal to the source voltage. BounceDiagramsforTransientPropagation The plots in Figures 2.21–2.23 show the voltage of a propagating pulse versus position along the transmission line but do not directly show the time variable, nor do they show very clearly the contribution of reflections on the waveform (especially when multiple c02TransmissionLineTheory Pozar September 29, 2011 16:41 88 Chapter 2: Transmission Line Theory z l 0Z0o.c.Z0 V0+ –t = 0v(z, t) zl 00 z = v ptv 2V0 z l0 0v V0 2V0 z = v pt(a) (b) (c) FIGURE 2.23 Transient response of a transmission line terminated with an open circuit. (a) Transmission line circuit with a step function voltage source. (b) Response for 0<t< /lscript/v p. (c) Response for /lscript/vp<t<2/lscript/v p; the incident pulse is reflected with/Gamma1=1. reflections are present). An alternative way of viewing the progress of a pulse propagating in time and position along a transmission line is with a bounce diagram. As an example, Figure 2.24 shows the bounce diagram for the transient circuit of Figure 2.23a. The horizontal axis represents position on the line, while the vertical axis represents time. The ray representing the incident wave begins at t=z=0 and travels to the right (increasing z)and up (for increasing t). This ray is labeled with the amplitude of the incident wave, V0/2. At t=/lscript/v pthe incident wave reaches the open-circuit load and is reflected to produce a wave of amplitude V0/2 traveling back to the source. The ray for this reflected wave thus moves to the left and up, until it reaches the source at z=0 and t=2/lscript/v p, at which point steady state is reached. The total voltage at any position zand time tcan be easily found by drawing a vertical line through the point zand extending up from t=0t ot . The total voltage is found by adding the voltages of each forward or reverse traveling wave component, as represented by the rays that intersect this vertical line. The next example shows how a bounce diagram can be applied to circuits that have multiple reflections. z l 00 zt p2l l 2V02V0 p FIGURE 2.24 Bounce diagram for the transient circuit of Figure 2.23a. c02TransmissionLineTheory Pozar September 29, 2011 16:41 2.8 Transients on Transmission Lines 89 z l 0Z0 = 100 Ω12 V+ –t = 0ΓL Γg50 Ω 200 Ω FIGURE 2.25 Circuit for Example 2.9. EXAMPLE 2.9 BOUNCE DIAGRAM FOR A TRANSIENT CIRCUIT WITH MULTIPLE REFLECTIONS Draw the bounce diagram for the transient circuit of Figure 2.25, including the first three reflections. Solution The amplitude of the incident wave is given by a voltage divider as v+=12100 50+100=8.0V The incident ray can be plotted as a line from the origin to the point z=/lscriptand t=/lscript/v p. The reflection coefficients at the generator and load are /Gamma1g=50−100 50+100=−1/3 and /Gamma1L=200−100 200+100=1/3, so the amplitude of the wave reflected from the load is 8/3 V . When this wave reaches the source, it will be reflected to form a wave of amplitude –8/9 V . The next reflection from the load will have an amplitude of –8/27 V . These four wavesare shown in the bounce diagram of Figure 2.26. ■ z l 00t 2l4l 3l lV 8 27– V8 9– V8 3 8Vp p p p FIGURE 2.26 Bounce diagram for Example 2.9. c02TransmissionLineTheory Pozar July 26, 2011 17:33 90 Chapter 2: Transmission Line Theory REFERENCES [1] S. Ramo, J. R. Winnery, and T. Van Duzer, Fields and Waves in Communication Electronics,3 r d edition, John Wiley & Sons, New York, 1994. [2] J. A. Stratton, Electromagnetic Theory , McGraw-Hill, New York, 1941. [3] H. A. Wheeler, “Reflection Charts Relating to Impedance Matching,” IEEE Transactions on Mi- crowave Theory and Techniques, vol. MTT-32, pp. 1008–1021, September 1984. [4] P. H. Smith, “Transmission Line Calculator,” Electronics , vol. 12, No. 1, pp. 29–31, January 1939. [ 5 ] P .J .N a h i n ,Oliver Heaviside: Sage in Solitude, IEEE Press, New York, 1988.[6] H. A. Wheeler, “Formulas for the Skin Effect,” Proceedings of the IRE, vol. 30, pp. 412–424, September 1942. [7] T. C. Edwards, Foundations for Microstrip Circuit Design, John Wiley & Sons, New York, 1987. PROBLEMS 2.1 A7 5/Omega1coaxial line has a current i(t,z)=1.8c o s (3.77×109t−18.13 z)mA. Determine (a) the frequency, (b) the phase velocity, (c) the wavelength, (d) the relative permittivity of the line, (e) the phasor form of the current, and (f) the time domain voltage on the line. 2.2 A transmission line has the following per-unit-length parameters: L=0.5µH/m, C=200 pF/m, R=4.0/Omega1/m, and G=0.02 S/m. Calculate the propagation constant and characteristic impedance of this line at 800 MHz. If the line is 30 cm long, what is the attenuation in dB? Recalculate thesequantities in the absence of loss ( R=G=0). 2.3 RG-402U semirigid coaxial cable has an inner conductor diameter of 0.91 mm and a dielectric diam- eter (equal to the inner diameter of the outer conductor) of 3.02 mm. Both conductors are copper, andthe dielectric material is Teflon. Compute the R,L,G,a n d Cparameters of this line at 1 GHz, and use these results to find the characteristic impedance and attenuation of the line at 1 GHz. Compare your results to the manufacturer’s specifications of 50 /Omega1and 0.43 dB/m, and discuss reasons for the difference. 2.4 Compute and plot the attenuation of the coaxial line of Problem 2.3, in dB/m, over a frequency range of 1 MHz to 100 GHz. Use log-log graph paper. 2.5 For the parallel plate line shown in the accompanying figure, derive the R,L,G,a n d Cparameters. Assume W/greatermuchd. y x z/H9280r/H9268 dW 2.6 For the parallel plate line of Problem 2.5, derive the telegrapher equations using the field theory approach. 2.7 Show that the T-model of a transmission line shown in the accompanying figure also yields the telegrapher equations derived in Section 2.1. c02TransmissionLineTheory Pozar July 26, 2011 17:33 Problems 91 i(z, t) R∆z 2i(z + ∆ z, t) v(z, t) G∆z C∆z ∆zv(z + ∆ z, t)R∆z 2L∆z 2L∆z 2 + –+ – 2.8 A lossless transmission line of electrical length /lscript=0.3λ is terminated with a complex load impedance as shown in the accompanying figure. Find the reflection coefficient at the load, the SWR on the line, the reflection coefficient at the input of the line, and the input impedance to the line. Z0 = 75 Ω ZL Zinl = 0.3/H9261 ZL = 30 /H11002 j 20 Ω 2.9 A7 5/Omega1 coaxial transmission line has a length of 2.0 cm and is terminated with a load impedance of 37.5 +j75/Omega1. If the relative permittivity of the line is 2.56 and the frequency is 3.0 GHz, find the input impedance to the line, the reflection coefficient at the load, the reflection coefficient at the input, and the SWR on the line. 2.10 A terminated transmission line with Z0=60/Omega1has a reflection coefficient at the load of /Gamma1=0.4/negationslash60◦. (a) What is the load impedance? (b) What is the reflection coefficient 0.3 λaway from the load? (c) What is the input impedance at this point? 2.11 A 100 /Omega1transmission line has an effective dielectric constant of 1.65. Find the shortest open-circuited length of this line that appears at its input as a capacitor of 5 pF at 2.5 GHz. Repeat for an inductance of 5 nH. 2.12 A lossless transmission line is terminated with a 100 /Omega1load. If the SWR on the line is 1.5, find the two possible values for the characteristic impedance of the line. 2.13 LetZscbe the input impedance of a length of coaxial line when one end is short-circuited, and let Zocbe the input impedance of the line when one end is open-circuited. Derive an expression for the characteristic impedance of the cable in terms of ZscandZoc. 2.14 A radio transmitter is connected to an antenna having an impedance 80 +j40/Omega1with a 50 /Omega1coaxial cable. If the 50 /Omega1transmitter can deliver 30 W when connected to a 50 /Omega1load, how much power is delivered to the antenna? 2.15 Calculate standing wave ratio, reflection coefficient magnitude, and return loss values to complete the entries in the following table: SWR |/Gamma1| RL (dB) 1.00 0.00 ∞ 1.01 — — — 0.01 — 1.05 — — — — 30.0 1.10 — — 1.20 — — — 0.10 — 1.50 — — — — 10.0 2.00 — — 2.50 — — c02TransmissionLineTheory Pozar July 26, 2011 17:33 92 Chapter 2: Transmission Line Theory 2.16 The transmission line circuit in the accompanying figure has Vg=15 V rms, Zg=75/Omega1,Z0=75/Omega1, ZL=60−j40/Omega1,a n d /lscript=0.7λ. Compute the power delivered to the load using three different techniques: (a)Find/Gamma1and compute PL=/parenleftbiggVg 2/parenrightbigg21 Z0(1−|/Gamma1|2); (b)find Zinand compute PL=/vextendsingle/vextendsingle/vextendsingle/vextendsingleVg Zg+Zin/vextendsingle/vextendsingle/vextendsingle/vextendsingle2 Re{Zin}; (c)findVLand compute PL=/vextendsingle/vextendsingle/vextendsingle/vextendsingleV L ZL/vextendsingle/vextendsingle/vextendsingle/vextendsingle2 Re{ZL}. Discuss the rationale for each of these methods. Which of these methods can be used if the line is not lossless? 2.17 For a purely reactive load impedance of the form ZL=jX, show that the reflection coefficient magnitude |/Gamma1|is always unity. Assume that the characteristic impedance Z0is real. 2.18 Consider the transmission line circuit shown in the accompanying figure. Compute the incident power, the reflected power, and the power transmitted into the infinite 75 /Omega1line. Show that power conservation is satisfied. Z0 = 50 Ω Z1 = 75 Ω50 Ω /H9261/2 10 V Pinc PrefPtrans 2.19 A generator is connected to a transmission line as shown in the accompanying figure. Find the voltage as a function of zalong the transmission line. Plot the magnitude of this voltage for −/lscript≤z≤0. z –l 0Z0 = 100 Ω100 Ωl = 1.5λ ZL = 80 – j 40 Ω 10 V 2.20 Use the Smith chart to find the following quantities for the transmission line circuit shown in the accompanying figure: (a)The SWR on the line. (b)The reflection coefficient at the load. (c)The load admittance. (d)The input impedance of the line. (e)The distance from the load to the first voltage minimum. c02TransmissionLineTheory Pozar July 26, 2011 17:33 Problems 93 (f)The distance from the load to the first voltage maximum. l = 0.4/H9261 Zin Z0 = 50 Ω ZL = 60 + j 50 Ω 2.21 Use the Smith chart to find the shortest lengths of a short-circuited 75 /Omega1line to give the following input impedance: (a)Zin=0. (b) Zin=∞. (c)Zin=j75/Omega1. (d) Zin=− j50/Omega1. (e)Zin=j10/Omega1. 2.22 Repeat Problem 2.21 for an open-circuited length of 75 /Omega1line. 2.23 A slotted-line experiment is performed with the following results: distance between successive min- ima=2.1 cm; distance of first voltage minimum from load =0.9 cm; SWR of load =2.5. If Z0=50/Omega1, find the load impedance. 2.24 Design a quarter-wave matching transformer to match a 40 /Omega1load to a 75 /Omega1line. Plot the SWR for 0.5≤f/fo≤2.0, where fois the frequency at which the line is λ/4 long. 2.25 Consider the quarter-wave matching transformer circuit shown in the accompanying figure. Derive expressions for V+andV−, the respective amplitudes of the forward and reverse traveling waves on the quarter-wave line section, in terms of Vi, the incident voltage amplitude. z –l 0/H9261/4 Z0 Z0RL RLViV+ V– 2.26 Derive equation (2.71) from (2.70). 2.27 In Example 2.7, the attenuation of a coaxial line due to finite conductivity is αc=Rs 2ηlnb/a/parenleftbigg1 a+1 b/parenrightbigg . Show that αcis minimized for conductor radii such that xlnx=1+x,w h e r e x=b/a. Solve this equation for x, and show that the corresponding characteristic impedance for /epsilon1r=1i s7 7 /Omega1. 2.28 Compute and plot the factor by which attenuation is increased due to surface roughness, for rms roughness ranging from 0 to 0.01 mm. Assume copper conductors at 10 GHz. 2.29 A5 0/Omega1 transmission line is matched to a 10 V source and feeds a load ZL=100/Omega1. If the line is 2.3λ long and has an attenuation constant α=0.5d B / λ, find the powers that are delivered by the source, lost in the line, and delivered to the load. 2.30 Consider a nonreciprocal transmission line having different propagation constants, β+andβ−,f o r propagation in the forward and reverse directions, with corresponding characteristic impedances Z+ 0 andZ− 0. (An example of such a line could be a microstrip transmission line on a magnetized ferrite c02TransmissionLineTheory Pozar July 26, 2011 17:33 94 Chapter 2: Transmission Line Theory substrate.) If the line is terminated as shown in the accompanying figure, derive expressions for the reflection coefficient and impedance seen at the input of the line. 2.31 Plot the bounce diagram for the transient circuit shown in the accompanying figure. Include at least three reflections. What is the total voltage at the midpoint of the line ( z=l/2), at time t=3/lscript/v p? z l 0Z0 = 50 Ω100 Ω 10 V+ –t = 025 Ω c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 Chapter Three Transmission Lines and Waveguides One of the early milestones in microwave engineering was the development of waveguide and other transmission lines for the low-loss transmission of power at high frequencies. Al-though Heaviside considered the possibility of propagation of electromagnetic waves inside a closed hollow tube in 1893, he rejected the idea because he believed that two conductors were necessary for the transfer of electromagnetic energy [1]. In 1897, Lord Rayleigh (JohnWilliam Strutt) mathematically proved that wave propagation in waveguides was possible forboth circular and rectangular cross sections [2]. Rayleigh also noted the infinite set of wave- guide modes of the TE and TM type that were possible and the existence of a cutoff frequency, but no experimental verification was made at the time. The waveguide was then essentially for-gotten until it was rediscovered independently in 1936 by two researchers [3]. After preliminaryexperiments in 1932, George C. Southworth of the AT&T Company in New York presented a paper on the waveguide in 1936. At the same meeting, W. L. Barrow of MIT presented a paper on the circular waveguide, with experimental confirmation of propagation. Early RF and microwave systems relied on waveguides, two-wire lines, and coaxial lines for transmission. Waveguides have the advantage of high power-handling capability and low loss but are bulky and expensive, especially at low frequencies. Two-wire lines are inexpensive but lack shielding. Coaxial lines are shielded but are a difficult medium in which to fabricatecomplex microwave components. Planar transmission lines provide an alternative, in the formof stripline, microstrip lines, slotlines, coplanar waveguides, and several other types of related geometries. Such transmission lines are compact, low in cost, and capable of being easily inte- grated with active circuit devices, such as diodes and transistors, to form microwave integratedcircuits. The first planar transmission line may have been a flat-strip coaxial line, similar toa stripline, used in a production power divider network in World War II [4], but planar lines did not see intensive development until the 1950s. Microstrip lines were developed at ITT laboratories [5] and were competitors of stripline. The first microstrip lines used a relativelythick dielectric substrate, which accentuated the non-TEM mode behavior and frequency dis-persion of the line. This characteristic made it less desirable than stripline until the 1960s, when much thinner substrates began to be used. This reduced the frequency dependence of the line, and now microstrip lines are often the preferred medium for microwave integratedcircuits. 95 c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 96 Chapter 3: Transmission Lines and Waveguides In this chapter we will study the properties of several types of transmission lines and waveguides that are in common use. As we know from Chapter 2, a transmission line is char- acterized by a propagation constant, an attenuation constant, and a characteristic impedance.These quantities will be derived by field theory analysis for the various lines and waveguidestreated here. We begin with a discussion of the different types of wave propagation and modes that can exist on general transmission lines and waveguides. Transmission lines that consist of two ormore conductors may support transverse electromagnetic (TEM) waves, characterized by the lack of longitudinal field components. Such lines have a uniquely defined voltage, current, and characteristic impedance. Waveguides, often consisting of a single conductor, support trans- verse electric (TE) and/or transverse magnetic (TM) waves, characterized by the presence of longitudinal magnetic or electric field components. As we will see in Chapter 4, a unique def-inition of characteristic impedance is not possible for such waves, although definitions can be chosen so that the characteristic impedance concept can be extended to waveguides with meaningful results. 3.1GENERALSOLUTIONSFORTEM,TE,ANDTMWAVES In this section we will find general solutions to Maxwell’s equations for the specific cases of TEM, TE, and TM wave propagation in cylindrical transmission lines or waveguides. The geometry of an arbitrary transmission line or waveguide is shown in Figure 3.1 and is characterized by conductor boundaries that are parallel to the z-axis. These structures are assumed to be uniform in shape and dimension in the zdirection and infinitely long. The conductors will initially be assumed to be perfectly conducting, but attenuation can befound by the perturbation method discussed in Chapter 2. y x zy x z (a) (b) FIGURE 3.1 (a) General two-conductor transmission line and (b) closed waveguide. c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 3.1 General Solutions for TEM, TE, and TM Waves 97 We assume time-harmonic fields with an ejωtdependence and wave propagation along thez-axis. The electric and magnetic fields can then be written as ¯E(x,y,z)=[ ¯e(x,y)+ˆzez(x,y)]e−jβz, (3.1a) ¯H(x,y,z)=[¯h(x,y)+ˆzhz(x,y)]e−jβz, (3.1b) where ¯e(x,y)and¯h(x,y)represent the transverse (ˆx,ˆy)electric and magnetic field com- ponents, and ezandhzare the longitudinal electric and magnetic field components. In (3.1) the wave is propagating in the +zdirection; −zpropagation can be obtained by replacing βwith−β. In addition, if conductor or dielectric loss is present, the propagation constant will be complex; jβshould then be replaced with γ=α+jβ. Assuming that the transmission line or waveguide region is source free, we can write Maxwell’s equations as ∇ׯE=− jωµ¯H, (3.2a) ∇ׯH=jω/epsilon1¯E. (3.2b) With an e−jβzzdependence, the three components of each of these vector equations can be reduced to the following: ∂Ez ∂y+jβEy=− jωµHx, (3.3a) −jβEx−∂Ez ∂x=− jωµHy, (3.3b) ∂Ey ∂x−∂Ex ∂y=− jωµHz, (3.3c) ∂Hz ∂y+jβHy=jω/epsilon1Ex, (3.4a) −jβHx−∂Hz ∂x=jω/epsilon1Ey, (3.4b) ∂Hy ∂x−∂Hx ∂y=jω/epsilon1Ez. (3.4c) These six equations can be solved for the four transverse field components in terms of Ez andHz[e.g., Hxcan be derived by eliminating Eyfrom (3.3a) and (3.4b)] as follows: Hx=j k2c/parenleftbigg ω/epsilon1∂Ez ∂y−β∂Hz ∂x/parenrightbigg , (3.5a) Hy=−j k2c/parenleftbigg ω/epsilon1∂Ez ∂x+β∂Hz ∂y/parenrightbigg , (3.5b) Ex=−j k2c/parenleftbigg β∂Ez ∂x+ωµ∂Hz ∂y/parenrightbigg , (3.5c) Ey=j k2c/parenleftbigg −β∂Ez ∂y+ωµ∂Hz ∂x/parenrightbigg , (3.5d) where k2 c=k2−β2(3.6) c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 98 Chapter 3: Transmission Lines and Waveguides is defined as the cutoff wave number ; the reason for this terminology will become clear later. As in previous chapters, k=ω√µ/epsilon1=2π/λ (3.7) is the wave number of the material filling the transmission line or waveguide region. If dielectric loss is present, /epsilon1can be made complex by using /epsilon1=/epsilon1o/epsilon1r(1−jtanδ), where tanδis the loss tangent of the material. Equations (3.5a)–(3.5d) are general results that can be applied to a variety of wave- guiding systems. We will now specialize these results to specific wave types. TEMWaves Transverse electromagnetic (TEM) waves are characterized by Ez=Hz=0. Observe from (3.5) that if Ez=Hz=0, then the transverse fields are also all zero, unless k2 c= 0(k2=β2), in which case we have an indeterminate result. However, we can return to (3.3)–(3.4) and apply the condition that Ez=Hz=0. Then from (3.3a) and (3.4b), we can eliminate Hxto obtain β2Ey=ω2µ/epsilon1Ey, or β=ω√µ/epsilon1=k,( 3.8) as noted earlier. [This result can also be obtained from (3.3b) and (3.4a).] The cutoff wave number, kc=/radicalbig k2−β2, is thus zero for TEM waves. The Helmholtz wave equation for Exis, from (1.42), /parenleftBigg ∂2 ∂x2+∂2 ∂y2+∂2 ∂z2+k2/parenrightBigg Ex=0,( 3.9) but for e−jβzdependence, (∂2/∂z2)Ex=−β2Ex=−k2Ex, so (3.9) reduces to /parenleftBigg ∂2 ∂x2+∂2 ∂y2/parenrightBigg Ex=0.( 3.10) A similar result also applies to Ey, so using the form of ¯Eassumed in (3.1a), we can write ∇2 t¯e(x,y)=0,( 3.11) where ∇2 t=∂2/∂x2+∂2/∂y2is the Laplacian operator in the two transverse dimensions. The result of (3.11) shows that the transverse electric fields, ¯e(x,y), of a TEM wave satisfy Laplace’s equation. It is easy to show in the same way that the transverse magnetic fields also satisfy Laplace’s equation: ∇2 t¯h(x,y)=0.( 3.12) The transverse fields of a TEM wave are thus the same as the static fields that can exist between the conductors. In the electrostatic case, we know that the electric field can beexpressed as the gradient of a scalar potential, /Phi1(x,y): ¯e(x,y)=− ∇ t/Phi1(x,y), (3.13) c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 3.1 General Solutions for TEM, TE, and TM Waves 99 where ∇t=ˆx(∂/∂x)+ˆy(∂/∂y)is the transverse gradient operator in two dimensions. For the relation in (3.13) to be valid, the curl of ¯emust vanish, and this is indeed the case here since ∇tׯe=− jωµhzˆz=0. Using the fact that ∇·¯D=/epsilon1∇t·¯e=0 with (3.13) shows that /Phi1(x,y)also satisfies Laplace’s equation, ∇2 t/Phi1(x,y)=0,( 3.14) as expected from electrostatics. The voltage between two conductors can be found as V12=/Phi11−/Phi12=/integraldisplay2 1¯E·d¯/lscript, (3.15) where /Phi11and/Phi12represent the potential at conductors 1 and 2, respectively. The current flow on a given conductor can be found from Ampere’s law as I=/contintegraldisplay C¯H·d¯/lscript, (3.16) where Cis the cross-sectional contour of the conductor. TEM waves can exist when two or more conductors are present. Plane waves are also examples of TEM waves since there are no field components in the direction of propaga- tion; in this case the transmission line conductors may be considered to be two infinitely large plates separated to infinity. The above results show that a closed conductor (such as a rectangular waveguide) cannot support TEM waves since the corresponding static potential in such a region would be zero (or possibly a constant), leading to ¯e=0. The wave impedance of a TEM mode can be found as the ratio of the transverse electric and magnetic fields: ZTEM=Ex Hy=ωµ β=/radicalbiggµ /epsilon1=η, (3.17a) where (3.4a) was used. The other pair of transverse field components, from (3.3a), gives ZTEM=−Ey Hx=/radicalbiggµ /epsilon1=η. (3.17b) Combining the results of (3.17a) and (3.17b) gives a general expression for the transverse fields as ¯h(x,y)=1 ZTEMˆzׯe(x,y). (3.18) Note that the wave impedance is the same as that for a plane wave in a lossless medium, as derived in Chapter 1; the reader should not confuse this impedance with the character- istic impedance, Z0, of a transmission line. The latter relates traveling voltage and current and is a function of the line geometry as well as the material filling the line, while thewave impedance relates transverse field components and is dependent only on the material constants. From (2.32), the characteristic impedance of the TEM line is Z 0=V/I, where VandIare the amplitudes of the incident voltage and current waves. The procedure for analyzing a TEM line can be summarized as follows: 1. Solve Laplace’s equation, (3.14), for /Phi1(x,y). The solution will contain several unknown constants. c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 100 Chapter 3: Transmission Lines and Waveguides 2. Find these constants by applying the boundary conditions for the known voltages on the conductors. 3. Compute ¯eand¯Efrom (3.13) and (3.1a). Compute ¯hand¯Hfrom (3.18) and (3.1b). 4. Compute Vfrom (3.15) and Ifrom (3.16). 5. The propagation constant is given by (3.8), and the characteristic impedance is given byZ0=V/I. TEWaves Transverse electric (TE) waves, (also referred to as H-waves) are characterized by Ez=0 andHz/negationslash=0. Equations (3.5) then reduce to Hx=−jβ k2c∂Hz ∂x, (3.19a) Hy=−jβ k2c∂Hz ∂y, (3.19b) Ex=−jωµ k2c∂Hz ∂y, (3.19c) Ey=jωµ k2c∂Hz ∂x. (3.19d) In this case kc/negationslash=0, and the propagation constant β=/radicalbig k2−k2cis generally a function of frequency and the geometry of the line or guide. To apply (3.19), one must first find Hz from the Helmholtz wave equation, /parenleftBigg ∂2 ∂x2+∂2 ∂y2+∂2 ∂z2+k2/parenrightBigg Hz=0,( 3.20) which, since Hz(x,y,z)=hz(x,y)e−jβz, can be reduced to a two-dimensional wave equa- tion for hz: /parenleftBigg ∂2 ∂x2+∂2 ∂y2+k2 c/parenrightBigg hz=0,( 3.21) since k2 c=k2−β2. This equation must be solved subject to the boundary conditions of the specific guide geometry. The TE wave impedance can be found as ZTE=Ex Hy=−Ey Hx=ωµ β=kη β,( 3.22) which is seen to be frequency dependent. TE waves can be supported inside closed con- ductors, as well as between two or more conductors. TMWaves Transverse magnetic (TM) waves (also referred to as E-waves) are characterized by Ez/negationslash=0 and Hz=0. Equations (3.5) then reduce to Hx=jω/epsilon1 k2c∂Ez ∂y, (3.23a) Hy=−jω/epsilon1 k2c∂Ez ∂x, (3.23b) c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 3.1 General Solutions for TEM, TE, and TM Waves 101 Ex=−jβ k2c∂Ez ∂x, (3.23c) Ey=−jβ k2c∂Ez ∂y. (3.23d) As in the TE case, kc/negationslash=0, and the propagation constant β=/radicalbig k2−k2cis a function of frequency and the geometry of the line or guide. Ezis found from the Helmholtz wave equation, /parenleftBigg ∂2 ∂x2+∂2 ∂y2+∂2 ∂z2+k2/parenrightBigg Ez=0,( 3.24) which, since Ez(x,y,z)=ez(x,y)e−jβz, can be reduced to a two-dimensional wave equa- tion for ez: /parenleftBigg ∂2 ∂x2+∂2 ∂y2+k2 c/parenrightBigg ez=0,( 3.25) since k2 c=k2−β2. This equation must be solved subject to the boundary conditions of the specific guide geometry. The TM wave impedance can be found as ZTM=Ex Hy=−Ey Hx=β ω/epsilon1=βη k,( 3.26) which is frequency dependent. As for TE waves, TM waves can be supported inside closed conductors, as well as between two or more conductors. The procedure for analyzing TE and TM waveguides can be summarized as follows: 1. Solve the reduced Helmholtz equation, (3.21) or (3.25), for hzorez. The solution will contain several unknown constants and the unknown cutoff wave number, kc. 2. Use (3.19) or (3.23) to find the transverse fields from hzorez. 3. Apply the boundary conditions to the appropriate field components to find the unknown constants and kc. 4. The propagation constant is given by (3.6) and the wave impedance by (3.22) or (3.26). AttenuationDuetoDielectricLoss Attenuation in a transmission line or waveguide can be caused by either dielectric loss or conductor loss. If αdis the attenuation constant due to dielectric loss and αcis the attenu- ation constant due to conductor loss, then the total attenuation constant is α=αd+αc. Attenuation caused by conductor loss can be calculated using the perturbation method of Section 2.7; this loss depends on the field distribution in the guide and so must be evaluated separately for each type of transmission line or waveguide. However, if the lineor guide is completely filled with a homogeneous dielectric, the attenuation due to a lossy dielectric material can be calculated from the propagation constant, and this result will apply to any guide or line with a homogeneous dielectric filling. Thus, use of the complex permittivity allows the complex propagation constant to be written as γ=α d+jβ=/radicalBig k2c−k2 =/radicalBig k2c−ω2µ0/epsilon10/epsilon1r(1−jtanδ). (3.27) c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 102 Chapter 3: Transmission Lines and Waveguides In practice, most dielectric materials have small losses (tan δ/lessmuch1), and so this expression can be simplified by using the first two terms of the Taylor expansion, /radicalbig a2+x2/similarequala+1 2/parenleftBigg x2 a/parenrightBigg ,forx/lessmucha. Then (3.27) reduces to γ=/radicalBig k2c−k2+jk2tanδ /similarequal/radicalBig k2c−k2+jk2tanδ 2/radicalbig k2c−k2 =k2tanδ 2β+jβ, (3.28) since/radicalbig k2c−k2=jβ. In these results, k=ω√µ0/epsilon10/epsilon1ris the (real) wave number in the absence of loss. Equation (3.28) shows that when the loss is small the phase constant βis unchanged, while the attenuation constant due to dielectric loss is given by αd=k2tanδ 2βNp/m (TE or TM waves).( 3.29) This result applies to any TE or TM wave, as long as the guide is completely filled with the dielectric material. It can also be used for TEM lines, where kc=0, by letting β=k: αd=ktanδ 2Np/m (TEM waves) .( 3.30) 3.2PARALLELPLATEWAVEGUIDE The parallel plate waveguide is the simplest type of guide that can support TM and TE modes; it can also support a TEM mode since it is formed from two flat conducting plates, or strips, as shown in Figure 3.2. Although it is an idealization, understanding the parallelplate guide can be useful because its operation is similar to that of many other waveguides. The parallel plate guide can also be useful for modeling the propagation of higher order modes in stripline. y x z/H9280, /H9262 d W FIGURE 3.2 Geometry of a parallel plate waveguide. c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 3.2 Parallel Plate Waveguide 103 In the geometry of the parallel plate waveguide of Figure 3.2, the strip width, W,i s assumed to be much greater than the separation, d, so that fringing fields and any xvari- ation can be ignored. A material with permittivity /epsilon1and permeability µis assumed to fill the region between the two plates. We will derive solutions for TEM, TM, and TE waves. TEMModes As discussed in Section 3.1, the TEM mode solution can be obtained by solving Laplace’s equation, (3.14), for the electrostatic potential /Phi1(x,y)between the two plates. Thus, ∇2 t/Phi1(x,y)=0, for 0≤x≤W,0≤y≤d.( 3.31) If we assume that the bottom plate is at ground (zero) potential and the top plate at a potential of Vo, then the boundary conditions for /Phi1(x,y)are /Phi1(x,0)=0, (3.32a) /Phi1(x,d)=Vo. (3.32b) Because there is no variation in x, the general solution to (3.31) for /Phi1(x,y)is /Phi1(x,y)=A+By, and the constants A,Bcan be evaluated from the boundary conditions of (3.32) to give the final solution as /Phi1(x,y)=Voy/d.( 3.33) The transverse electric field is, from (3.13), ¯e(x,y)=− ∇ t/Phi1(x,y)=−ˆ yVo d,( 3.34) so that the total electric field is ¯E(x,y,z)=¯e(x,y)e−jkz=−ˆ yVo de−jkz,( 3.35) where k=ω√µ/epsilon1is the propagation constant of the TEM wave, as in (3.8). The magnetic field, from (3.18), is ¯H(x,y,z)=¯h(x,y)e−jkz=1 ηˆzׯE(x,y,z)=ˆxVo ηde−jkz,( 3.36) where η=√µ//epsilon1 is the intrinsic impedance of the medium between the parallel plates. Note that Ez=Hz=0 and that the fields are similar in form to a plane wave in a homo- geneous region. The voltage of the top plate with respect to the bottom plate can be calculated from (3.15) and (3.35) as V=−/integraldisplayd y=0Eydy=Voe−jkz,( 3.37) c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 104 Chapter 3: Transmission Lines and Waveguides as expected. The total current on the top plate can be found from Ampere’s law or the surface current density: I=/integraldisplayW x=0¯Js·ˆzdx=/integraldisplayW x=0(−ˆyׯH)·ˆzdx=/integraldisplayW x=0Hxdx=WV o ηde−jkz.(3.38) Then the characteristic impedance is Z0=V I=ηd W,( 3.39) which is seen to be a constant dependent only on the geometry and material parameters of the guide. The phase velocity is also a constant: vp=ω β=1√µ/epsilon1,( 3.40) which is the speed of light in the material medium. Attenuation due to dielectric loss is given by (3.30). The formula for conductor atten- uation will be derived in the next subsection as a special case of TM mode attenuation. TMModes As discussed in Section 3.1, TM waves are characterized by Hz=0 and a nonzero Ezfield that satisfies the reduced wave equation of (3.25), with ∂/∂x=0: /parenleftBigg ∂2 ∂y2+k2 c/parenrightBigg ez(x,y)=0,( 3.41) where kc=/radicalbig k2−β2is the cutoff wave number, and Ez(x,y,z)=ez(x,y)e−jβz.T h e general solution to (3.41) is of the form ez(x,y)=Asinkcy+Bcoskcy,( 3.42) subject to the boundary conditions that ez(x,y)=0, aty=0,d.( 3.43) This implies that B=0 and kcd=nπforn=0,1,2,3..., or kc=nπ d,n=0,1,2,3,.... (3.44) Thus the cutoff wave number, kc, is constrained to discrete values as given by (3.44); this implies that the propagation constant, β, is given by β=/radicalBig k2−k2c=/radicalBig k2−(nπ/d)2.( 3.45) The solution for ez(x,y)is then ez(x,y)=Ansinnπy d,( 3.46) and thus, Ez(x,y,z)=Ansinnπy de−jβz.( 3.47) c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 3.2 Parallel Plate Waveguide 105 The transverse field components can be found, using (3.23), to be Hx=jω/epsilon1 kcAncosnπy de−jβz, (3.48a) Ey=−jβ kcAncosnπy de−jβz, (3.48b) Ex=Hy=0. (3.48c) Observe that for n=0,β=k=ω√µ/epsilon1, and that Ez=0. The EyandHxfields are then constant in y, so that the TM 0mode is actually identical to the TEM mode. For n>0, however, the situation is different. Each value of ncorresponds to a different TM mode, denoted as the TM nmode, and each mode has its own propagation constant given by (3.45) and field expressions given by (3.48). From (3.45) it can be seen that βis real only when k>kc. Because k=ω√µ/epsilon1is pro- portional to frequency, the TM nmodes (for n>0)exhibit a cutoff phenomenon, whereby no propagation will occur until the frequency is such that k>kc.T h e cutoff frequency of the TM nmode can be found as fc=kc 2π√µ/epsilon1=n 2d√µ/epsilon1.( 3.49) Thus, the TM mode (for n>0) that propagates at the lowest frequency is the TM 1mode, with a cutoff frequency of fc=1/2d√µ/epsilon1;t h eT M 2mode has a cutoff frequency equal to twice this value, and so on. At frequencies below the cutoff frequency of a given mode,the propagation constant is purely imaginary, corresponding to a rapid exponential decay of the fields. Such modes are referred to as cutoff modes,o revanescent modes. Because of the cutoff frequency, below which propagation cannot occur, waveguide mode propagation is analogous to a high-pass filter response. The wave impedance of a TM mode, from (3.26), is a function of frequency: Z TM=−Ey Hx=β ω/epsilon1=βη k,( 3.50) which we see is pure real when f>fcbut pure imaginary when f<fc. The phase velocity is also a function of frequency: vp=ω β,( 3.51) and is seen to be greater than 1/√µ/epsilon1=ω/k, the speed of light in the medium, since β< k. The guide wavelength is defined as λg=2π β,( 3.52) and is the distance between equiphase planes along the z-axis. Note that λg>λ= 2π/k, the wavelength of a plane wave in the material. The phase velocity and guide wavelength are defined only for a propagating mode, for which βis real. One may also define a cutoff wavelength for the TM nmode as λc=2d n.( 3.53) c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 106 Chapter 3: Transmission Lines and Waveguides It is instructive to compute the Poynting vector to see how power propagates in the TM nmode. From (1.91), the time-average power passing a transverse cross section of the parallel plate guide is Po=1 2Re/integraldisplayW x=0/integraldisplayd y=0¯EׯH∗·ˆzdydx =−1 2Re/integraldisplayW x=0/integraldisplayd y=0EyH∗ xdydx =WRe(β)ω/epsilon1 2k2c|An|2/integraldisplayd y=0cos2nπy ddy=⎧ ⎪⎪⎪⎪⎨ ⎪⎪⎪⎪⎩WRe(β)ω/epsilon1 d 4k2c|An|2forn>0 WRe(β)ω/epsilon1 d 2k2c|An|2forn=0 (3.54) where (3.48a, b) were used for Ey,Hx. Thus, Pois positive and nonzero when βis real, which occurs when f>fc. When the mode is below cutoff, βis imaginary, and then Po=0. TM (or TE) waveguide mode propagation has an interesting interpretation when viewed as a pair of bouncing plane waves. For example, consider the dominant TM 1mode, which has propagation constant β1=/radicalBig k2−(π/d)2,( 3.55) andEzfield Ez=A1sinπy de−jβ1z, which can be rewritten as Ez=A1 2j/bracketleftBig ej(πy/d−β1z)−e−j(πy/d+β1z)/bracketrightBig .( 3.56) This result is in the form of two plane waves traveling obliquely in the −y,+zand+y,+z directions, respectively, as shown in Figure 3.3. By comparison with the phase factor of(1.132), the angle θthat each plane wave makes with the z-axis satisfies the relations ksinθ=π d, (3.57a) kcosθ=β1, (3.57b) so that (π/d)2+β2 1=k2, as in (3.55). For f>fc,βis real and less than k1,s oθ is some angle between 0◦and 90◦, and the mode can be thought of as two plane waves alternately bouncing off of the top and bottom plates. y zd/H9258 /H9258 0 FIGURE 3.3 Bouncing plane wave interpretation of the TM 1parallel plate waveguide mode. c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 3.2 Parallel Plate Waveguide 107 The phase velocity of each plane wave along its direction of propagation ( θdirection) isω/k=1/√µ/epsilon1, which is the speed of light in the material filling the guide. However, the phase velocity of the plane waves in the zdirection is ω/β1=1/√µ/epsilon1cosθ, which is greater than the speed of light in the material. (This situation is analogous to ocean waves hittinga shoreline: the intersection point of the shore and an obliquely incident wave crest moves faster than the wave crest itself.) The superposition of the two plane wave fields is such that complete cancellation occurs at y=0 and y=d, to satisfy the boundary condition that E z=0 at these planes. As fdecreases to fc,β1approaches zero, so that, by (3.57b), θ approaches 90◦. The two plane waves are then bouncing up and down, with no motion in the+zdirection, and no real power flow occurs in the zdirection. Attenuation due to dielectric loss can be found from (3.29). Conductor loss can be treated using the perturbation method. Thus, αc=P/lscript 2Po,( 3.58) where Pois the power flow down the guide in the absence of conductor loss, as given by (3.54). P/lscriptis the power dissipated per unit length in the two lossy conductors and can be found from (2.97) as P/lscript=2/parenleftbiggRs 2/parenrightbigg/integraldisplayW x=0|¯Js|2dx=ω2/epsilon12RsW k2c|An|2,( 3.59) where Rsis the surface resistivity of the conductors. Using (3.54) and (3.59) in (3.58) gives the attenuation due to conductor loss as αc=2ω/epsilon1Rs βd=2kR s βηdNp/m, forn>0.( 3.60) As discussed previously, the TEM mode is identical to the TM 0mode for the parallel plate waveguide, so the above attenuation results for the TM nmode can be used to obtain the TEM mode attenuation by letting n=0. For this case, the n=0 result of (3.54) must be used in (3.58), to obtain αc=Rs ηdNp/m.( 3.61) TEModes TE modes, characterized by Ez=0, can also propagate in a parallel plate waveguide. From (3.21), with ∂/∂x=0,Hzmust satisfy the reduced wave equation, /parenleftBigg ∂2 ∂y2+k2 c/parenrightBigg hz(x,y)=0,( 3.62) where kc=/radicalbig k2−β2is the cutoff wave number and Hz(x,y,z)=hz(x,y)e−jβz.T h e general solution to (3.62) is hz(x,y)=Asinkcy+Bcoskcy.( 3.63) The boundary conditions are that Ex=0a ty=0,d;Ezis identically zero for TE modes. From (3.19c) we have Ex=−jωµ kc(Acoskcy−Bsinkcy)e−jβz,( 3.64) c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 108 Chapter 3: Transmission Lines and Waveguides and applying the boundary conditions shows that A=0 and kc=nπ d,n=1,2,3..., (3.65) as for the TM case. The final solution for Hzis then Hz(x,y)=Bncosnπy de−jβz.( 3.66) The transverse fields can be computed from (3.19) as Ex=jωµ kcBnsinnπy de−jβz, (3.67a) Hy=jβ kcBnsinnπy de−jβz, (3.67b) Ey=Hx=0. (3.67c) The propagation constant of the TE nmode is given as β=/radicalbigg k2−/parenleftBignπ d/parenrightBig2 ,( 3.68) which is the same as the propagation constant of the TM nmode. The cutoff frequency of the TE nmode is fc=n 2d√µ/epsilon1,( 3.69) which is also identical to that of the TM nmode. The wave impedance of the TE nmode is, from (3.22), ZTE=Ex Hy=ωµ β=kη β,( 3.70) which is seen to be real for propagating modes and imaginary for nonpropagating, or cutoff, modes. The phase velocity, guide wavelength, and cutoff wavelength are similar to the results obtained for the TM modes. The power flow down the guide for a TE nmode can be calculated as Po=1 2Re/integraldisplayW x=0/integraldisplayd y=0¯EׯH∗·ˆzdydx =1 2Re/integraldisplayW x=0/integraldisplayd y=0ExH∗ ydydx =ωµdW 4k2c|Bn|2Re(β), forn>0, (3.71) which is zero if the operating frequency is below the cutoff frequency (β imaginary). Note that if n=0, then Ex=Hy=0 from (3.67), and thus Po=0, implying that there is no TE 0mode. Attenuation can be calculated in the same way as for the TM modes. The attenuation due to dielectric loss is given by (3.29). It is left as a problem to show that the attenuationdue to conductor loss for TE modes is given by α c=2k2 cRs ωµβ d=2k2 cRs kβηdNp/m.( 3.72) c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 3.2 Parallel Plate Waveguide 109 0123 0123456789 1 0TE1TM1 TEM k kckd /H9266=cutoff/H9251c/H9257d Rs FIGURE 3.4 Attenuation due to conductor loss for the TEM, TM 1,a n dT E 1modes of a parallel plate waveguide. Figure 3.4 shows attenuation versus frequency due to conductor loss for the TEM, TM 1, and TE 1modes. Observe that αc→∞ as cutoff is approached for the TM and TE modes. Table 3.1 summarizes a number of useful results for parallel plate waveguide modes. Field lines for the TEM, TM 1, and TE 1modes are shown in Figure 3.5. TABLE 3.1 Summary of Results for Parallel Plate Waveguide Quantity TEM Mode TMnMode TEnMode k ω√µ/epsilon1 ω√µ/epsilon1 ω√µ/epsilon1 kc 0 nπ/dn π/d β k=ω√µ/epsilon1/radicalBig k2−k2c/radicalBig k2−k2c λc ∞ 2π/kc=2d/n 2π/kc=2d/n λg 2π/k 2π/β 2π/β vp ω/k=1/√µ/epsilon1 ω/β ω/β αd (ktanδ)/2 (k2tanδ)/2β( k2tanδ)/2β αc Rs/ηd 2kR s/βηd 2k2cRs/kβηd Ez 0 Asin(nπy/d)e−jβz0 Hz 00 Bcos(nπy/d)e−jβz Ex 00 (jωµ/ kc)Bsin(nπy/d)e−jβz Ey (−Vo/d)e−jβz(−jβ/kc)Acos(nπy/d)e−jβz0 Hx (Vo/ηd)e−jβz(jω/epsilon1/kc)Acos(nπy/d)e−jβz0 Hy 00 (jβ/kc)Bnsin(nπy/d)e−jβz ZZ TEM=ηd/WZ TM=βη/kZ TE=kη/β c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 110 Chapter 3: Transmission Lines and Waveguides E H(b) (c)(a) FIGURE 3.5 Field lines for the (a) TEM, (b) TM 1,a n d( c )T E 1modes of a parallel plate wave- guide. There is no variation across the width of the waveguide. 3.3RECTANGULARWAVEGUIDE Rectangular waveguides were one of the earliest types of transmission lines used to transport microwave signals, and they are still used for many applications. A large variety of components such as couplers, detectors, isolators, attenuators, and slotted lines are commercially availablefor various standard waveguide bands from 1 to 220 GHz. Figure 3.6 shows some of the standard rectangular waveguide components that are available. Because of the trend toward miniaturization and integration, most modern microwave circuitry is fabricated using planartransmission lines such as microstrips and stripline rather than waveguides. There is, however, still a need for waveguides in many cases, including high-power systems, millimeter wave applications, satellite systems, and some precision test applications. The hollow rectangular waveguide can propagate TM and TE modes but not TEM waves since only one conductor is present. We will see that the TM and TE modes of a rectangular waveguide have cutoff frequencies below which propagation is not possible, similar to the TM and TE modes of the parallel plate guide. TEModes The geometry of a rectangular waveguide is shown in Figure 3.7, where it is assumed that the guide is filled with a material of permittivity /epsilon1and permeability µ. It is standard convention to have the longest side of the waveguide along the x-axis, so that a>b. TE waveguide modes are characterized by fields with E z=0, while Hzmust satisfy the reduced wave equation of (3.21): /parenleftBigg ∂2 ∂x2+∂2 ∂y2+k2 c/parenrightBigg hz(x,y)=0,( 3.73) with Hz(x,y,z)=hz(x,y)e−jβz; here kc=/radicalbig k2−β2is the cutoff wave number. The partial differential equation (3.73) can be solved by the method of separation of variables by letting hz(x,y)=X(x)Y(y)( 3.74) c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 3.3 Rectangular Waveguide 111 FIGURE 3.6 Photograph of Ka-band (WR-28) rectangular waveguide components. Clockwise from top: a variable attenuator, an E-H (magic) tee junction, a directional coupler, an adaptor to ridge waveguide, an E-plane swept bend, an adjustable short, and asliding matched load. and substituting into (3.73) to obtain 1 Xd2X dx2+1 Yd2Y dy2+k2 c=0.( 3.75) Then, by the usual separation-of-variables argument (see Section 1.5), each of the terms in (3.75) must be equal to a constant, so we define separation constants kxandkysuch that d2X dx2+k2 xX=0, (3.76a) d2Y dy2+k2 yY=0, (3.76b) y x z/H9262, /H9280b 0 a FIGURE 3.7 Geometry of a rectangular waveguide. c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 112 Chapter 3: Transmission Lines and Waveguides and k2 x+k2 y=k2 c.( 3.77) The general solution for hzcan then be written as hz(x,y)=(Acoskxx+Bsinkxx)(Ccoskyy+Dsinkyy). (3.78) To evaluate the constants in (3.78) we must apply the boundary conditions on the electric field components tangential to the waveguide walls. That is, ex(x,y)=0, aty=0,b, (3.79a) ey(x,y)=0, atx=0,a. (3.79b) We therefore cannot use hzof (3.78) directly but must first use (3.19c) and (3.19d) to find exandeyfrom hz: ex=−jωµ k2cky(Acoskxx+Bsinkxx)(−Csinkyy+Dcoskyy), (3.80a) ey=jωµ k2ckx(−Asinkxx+Bcoskxx)(Ccoskyy+Dsinkyy). (3.80b) Then from (3.79a) and (3.80a) we see that D=0, and ky=nπ/bforn=0,1,2.... From (3.79b) and (3.80b) we have that B=0 and kx=mπ/aform=0,1,2....T h e final solution for Hzis then Hz(x,y,z)=Amncosmπx acosnπy be−jβz,( 3.81) where Amnis an arbitrary amplitude constant composed of the remaining constants A andCof (3.78). The transverse field components of the TE mnmode can be found using (3.19) and (3.81): Ex=jωµnπ k2cbAmncosmπx asinnπy be−jβz, (3.82a) Ey=−jωµmπ k2caAmnsinmπx acosnπy be−jβz, (3.82b) Hx=jβmπ k2caAmnsinmπx acosnπy be−jβz, (3.82c) Hy=jβnπ k2cbAmncosmπx asinnπy be−jβz. (3.82d) The propagation constant is β=/radicalBig k2−k2c=/radicalbigg k2−/parenleftBigmπ a/parenrightBig2 −/parenleftBignπ b/parenrightBig2 ,( 3.83) which is seen to be real, corresponding to a propagating mode, when k>kc=/radicalbigg/parenleftBigmπ a/parenrightBig2 +/parenleftBignπ b/parenrightBig2 . c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 3.3 Rectangular Waveguide 113 Each mode (each combination of mandn)has a cutoff frequency fcmngiven by fcmn=kc 2π√µ/epsilon1=1 2π√µ/epsilon1/radicalbigg/parenleftBigmπ a/parenrightBig2 +/parenleftBignπ b/parenrightBig2 .( 3.84) The mode with the lowest cutoff frequency is called the dominant mode; because we have assumed a>b, the lowest cutoff frequency occurs for the TE 10(m=1,n=0)mode: fc10=1 2a√µ/epsilon1.( 3.85) Thus the TE 10mode is the dominant TE mode and, as we will see, the overall dominant mode of the rectangular waveguide. Observe that the field expressions for ¯Eand¯Hin (3.82) are all zero if both m=n=0; there is no TE 00mode. At a given operating frequency fonly those modes having f>fcwill propagate; modes with f<fcwill lead to an imaginary β(or real α), meaning that all field compo- nents will decay exponentially away from the source of excitation. Such modes are referred to as cutoff modes,o r evanescent modes. If more than one mode is propagating, the wave- guide is said to be overmoded. From (3.22) the wave impedance that relates the transverse electric and magnetic fields is ZTE=Ex Hy=−Ey Hx=kη β,( 3.86) where η=√µ//epsilon1 is the intrinsic impedance of the material filling the waveguide. Note thatZTEis real whenβ is real (a propagating mode) but is imaginary when βis imaginary (a cutoff mode). The guide wavelength is defined as the distance between two equal-phase planes along the waveguide and is equal to λg=2π β>2π k=λ, (3.87) which is thus greater than λ, the wavelength of a plane wave in the medium filling the guide. The phase velocity is vp=ω β>ω k=1/√µ/epsilon1, (3.88) which is greater than 1/√µ/epsilon1, the speed of light (plane wave) in the medium. In the vast majority of waveguide applications the operating frequency and guide dimensions are chosen so that only the dominant TE 10mode will propagate. Because of the practical importance of the TE 10mode, we will list the field components and derive the attenuation due to conductor loss for this case. Specializing (3.81) and (3.82) to the m=1,n=0 case gives the following results for the TE 10mode fields: Hz=A10cosπx ae−jβz, (3.89a) Ey=−jωµa πA10sinπx ae−jβz, (3.89b) Hx=jβa πA10sinπx ae−jβz, (3.89c) Ex=Ez=Hy=0. (3.89d) c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 114 Chapter 3: Transmission Lines and Waveguides The cutoff wave number and propagation constant for the TE 10mode are, respectively, kc=π/a, (3.90) β=/radicalbig k2−(π/a)2. (3.91) The power flow down the guide for the TE 10mode can be calculated as P10=1 2Re/integraldisplaya x=0/integraldisplayb y=0¯EׯH∗·ˆzdydx =1 2Re/integraldisplaya x=0/integraldisplayb y=0EyH∗ xdydx =ωµa2 2π2Re(β)|A10|2/integraldisplaya x=0/integraldisplayb y=0sin2πx adydx =ωµa3|A10|2b 4π2Re(β). (3.92) Note that this result gives nonzero real power only when βis real, corresponding to a propagating mode. Attenuation in a rectangular waveguide may occur due to dielectric loss or conductor loss. Dielectric loss can be treated by making /epsilon1complex and using the general result given in (3.29). Conductor loss is best treated using the perturbation method. The power lost per unit length due to finite wall conductivity is, from (1.131), P/lscript=Rs 2/integraldisplay C|¯Js|2d/lscript, (3.93) where Rsis the wall surface resistance, and the integration contour Cencloses the inside perimeter of the guide walls. There are surface currents on all four walls, but from sym- metry the currents on the top and bottom walls are identical, as are the currents on the left and right side walls. So we can compute the power lost in the walls at x=0 and y=0 and double their sum to obtain the total power loss. The surface current on the x=0( l e f t ) wall is ¯Js=ˆnׯH|x=0=ˆx׈zHz|x=0=−ˆ yHz|x=0=−ˆ yA10e−jβz,( 3.94a) and the surface current on the y=0 (bottom) wall is ¯Js=ˆnׯH|y=0=ˆy×(ˆxHx|y=0+ˆzHz|y=0) =− ˆ zjβa πA10sinπx ae−jβz+ˆxA10cosπx ae−jβz. (3.94b) Substituting (3.94) into (3.93) gives P/lscript=Rs/integraldisplayb y=0|Jsy|2dy+Rs/integraldisplaya x=0/bracketleftBig |Jsx|2+|Jsz|2/bracketrightBig dx =Rs|A10|2/parenleftBigg b+a 2+β2a3 2π2/parenrightBigg . (3.95) c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 3.3 Rectangular Waveguide 115 The attenuation due to conductor loss for the TE 10mode is then αc=P/lscript 2P10=2π2Rs(b+a/2+β2a3/2π2) ωµa3bβ =Rs a3bβkη(2bπ2+a3k2)Np/m. (3.96) TMModes TM modes are characterized by fields with Hz=0, while Ezmust satisfy the reduced wave equation (3.25): /parenleftBigg ∂2 ∂x2+∂2 ∂y2+k2 c/parenrightBigg ez(x,y)=0,( 3.97) with Ez(x,y,z)=ez(x,y)e−jβzandk2 c=k2−β2. Equation (3.97) can be solved by the separation-of-variables procedure that was used for TE modes. The general solution is ez(x,y)=(Acoskxx+Bsinkxx)(Ccoskyy+Dsinkyy). (3.98) The boundary conditions can be applied directly to ez: ez(x,y)=0, atx=0,a, (3.99a) ez(x,y)=0, aty=0,b. (3.99b) We will see that satisfaction of these conditions on ezwill lead to satisfaction of the bound- ary conditions by exandey. Applying (3.99a) to (3.98) shows that A=0 and kx=mπ/aform=1,2,3.... Similarly, applying (3.99b) to (3.98) shows that C=0 and ky=nπ/bforn=1,2,3.... The solution for Ezthen reduces to Ez(x,y,z)=Bmnsinmπx asinnπy be−jβz,( 3.100) where Bmnis an arbitrary amplitude constant. The transverse field components for the TM mnmode can be computed from (3.23) and (3.100) as Ex=−jβmπ ak2cBmncosmπx asinnπy be−jβz, (3.101a) Ey=−jβnπ bk2cBmnsinmπx acosnπy be−jβz, (3.101b) Hx=jω/epsilon1nπ bk2cBmnsinmπx acosnπy be−jβz, (3.101c) Hy=−jω/epsilon1mπ ak2cBmncosmπx asinnπy be−jβz. (3.101d) As for the TE modes, the propagation constant is β=/radicalBig k2−k2c=/radicalbigg k2−/parenleftBigmπ a/parenrightBig2 −/parenleftBignπ b/parenrightBig2 (3.102) c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 116 Chapter 3: Transmission Lines and Waveguides 00.10.20.30.40.5 0 8 10 12 14 16 18TE10 cutoff TM11 cutoffTE20 cutoff Frequency (GHz)/H9251c(dB/m)TM11 b = aTE10 b = a/2 TE10 b = a FIGURE 3.8 Attenuation of various modes in a rectangular brass waveguide with a=2.0 cm. and is real for propagating modes and imaginary for cutoff modes. The cutoff frequencies for the TM mnmodes are also the same as those of the TE mnmodes, as given in (3.84). The guide wavelength and phase velocity for TM modes are also the same as those for TE modes. Observe that the field expressions for ¯Eand¯Hin (3.101) are identically zero if either mornis zero. Thus there is no TM 00,T M 01,o rT M 10mode, and the lowest order TM mode to propagate (lowest fc)is the TM 11mode, having a cutoff frequency of fc11=1 2π√µ/epsilon1/radicalbigg/parenleftBigπ a/parenrightBig2 +/parenleftBigπ b/parenrightBig2 ,( 3.103) which is seen to be larger than fc10, the cutoff frequency of the TE 10mode. The wave impedance relating the transverse electric and magnetic fields for TM modes is, from (3.26), ZTM=Ex Hy=−Ey Hx=βη k.( 3.104) Attenuation due to dielectric loss is computed in the same way as for TE modes, with the same result. The calculation of attenuation due to conductor loss is left as a problem;Figure 3.8 shows attenuation versus frequency for some TE and TM modes in a rectangular waveguide. Table 3.2 summarizes results for TE and TM wave propagation in rectangular waveguides, and Figure 3.9 shows the field lines for several of the lowest order TE and TMmodes. EXAMPLE 3.1 CHARACTERISTICS OF A RECTANGULAR WA VEGUIDE Consider a length of Teflon-filled, copper K-band rectangular waveguide having dimensions a=1.07 cm and b=0.43 cm. Find the cutoff frequencies of the first five propagating modes. If the operating frequency is 15 GHz, find the attenuation due to dielectric and conductor losses. c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 3.3 Rectangular Waveguide 117 TABLE 3.2 Summary of Results for Rectangular Waveguide Quantity TEmnMode TMmnMode k ω√µ/epsilon1 ω√µ/epsilon1 kc/radicalbig (mπ/a)2+(nπ/b)2/radicalbig (mπ/a)2+(nπ/b)2 β/radicalBig k2−k2c/radicalBig k2−k2c λc2π kc2π kc λg2π β2π β vpω βω β αdk2tanδ 2βk2tanδ 2β Ez 0 Bsinmπx asinnπy be−jβz Hz Acosmπx acosnπy be−jβz0 Exjωµnπ k2cbAcosmπx asinnπy be−jβz −jβmπ k2caBcosmπx asinnπy be−jβz Ey−jωµmπ k2caAsinmπx acosnπy be−jβz −jβnπ k2cbBsinmπx acosnπy be−jβz Hxjβmπ k2caAsinmπx acosnπy be−jβz jω/epsilon1nπ k2cbBsinmπx acosnπy be−jβz Hyjβnπ k2cbAcosmπx asinnπy be−jβz −jω/epsilon1mπ k2caBcosmπx asinnπy be−jβz ZZ TE=kη βZTM=βη k Solution From Appendix G, for Teflon, /epsilon1r=2.08 and tan δ=0.0004. From (3.84) the cutoff frequencies are given by fcmn=c 2π√/epsilon1r/radicalbigg/parenleftBigmπ a/parenrightBig2 +/parenleftBignπ b/parenrightBig2 . Computing fcfor the first few values of mandngives the following results: Mode mn f c(GHz) TE 1 0 9.72 TE 2 0 19.44 TE 0 1 24.19 TE, TM 1 1 26.07TE, TM 2 1 31.03 c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 118 Chapter 3: Transmission Lines and Waveguides xyz1 23 xyz2 13 xyz xyz2 13 2 13xyz1 23 2 13TE10 TE11 TE21 TE20 TM11 TM211 2 31 2 31 2 3 1 2 31 2 31 2 3 FIGURE 3.9 Field lines for some of the lower order modes of a rectangular waveguide. Reprinted with permission from S. Ramo, J. R. Whinnery, and T. Van Duzer, Fields and Waves in Communication Electronics. Copyright c/circlecopyrt1965 by John Wiley & Sons, Inc. Table 8.02. c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 3.3 Rectangular Waveguide 119 Thus the TE 10,T E 20,T E 01,T E 11, and TM 11modes will be the first five modes to propagate. At 15 GHz, k=453.1m−1, and the propagation constant for the TE 10mode is β=/radicalBigg/parenleftbigg2πf√/epsilon1r c/parenrightbigg2 −/parenleftBigπ a/parenrightBig2 =/radicalbigg k2−/parenleftBigπ a/parenrightBig2 =345.1m−1. From (3.29), the attenuation due to dielectric loss is αd=k2tanδ 2β=0.119 Np/m =1.03 dB/m. The surface resistivity of the copper walls is (σ=5.8×107S/m) Rs=/radicalbiggωµ0 2σ=0.032 /Omega1, and the attenuation due to conductor loss, from (3.96), is αc=Rs a3bβkη(2bπ2+a3k2)=0.050 Np/m =0.434 dB/m.■ TEm0ModesofaPartiallyLoadedWaveguide The above results apply to an empty waveguide as well as one filled with a homogeneous dielectric or magnetic material, but in some cases of practical interest (such as impedance matching or phase-shifting sections) a waveguide is used with a partial dielectric filling.In this case an additional set of boundary conditions are introduced at the material inter- face, necessitating a new analysis. To illustrate the technique we will consider the TE m0 modes of a rectangular waveguide that is partially filled with a dielectric slab, as shown in Figure 3.10. The analysis still follows the basic procedure outlined at the end of Section 3.1. Since the geometry is uniform in the ydirection and n=0, the TE m0modes have no ydependence. Then the wave equation of (3.21) for hzcan be written separately for the dielectric and air regions as /parenleftbigg∂2 ∂x2+k2 d/parenrightbigg hz=0, for 0≤x≤t, (3.105a) /parenleftbigg∂2 ∂x2+k2 a/parenrightbigg hz=0, fort≤x≤a, (3.105b) y x/H9280r/H92800 /H92800b 0a t FIGURE 3.10 Geometry of a partially loaded rectangular waveguide. c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 120 Chapter 3: Transmission Lines and Waveguides where kdandkaare the cutoff wave numbers for the dielectric and air regions, defined as follows: β=/radicalBig /epsilon1rk2 0−k2 d, (3.106a) β=/radicalBig k2 0−k2a. (3.106b) These relations incorporate the fact that the propagation constant, β, must be the same in both regions to ensure phase matching (see Section 1.8) of the fields along the interface at x= t. The solutions to (3.105) can be written as hz=/braceleftbiggAcoskdx+Bsinkdx for 0≤x≤t Ccoska(a−x)+Dsinka(a−x) fort≤x≤a,(3.107) where the form of the solution for t<x<awas chosen to simplify the evaluation of boundary conditions at x=a. We need ˆyandˆzelectric and magnetic field components to apply the boundary condi- tions at x=0,t,anda.Ez=0 for TE modes, and Hy=0 since ∂/∂y=0.Eyis found from (3.19d) as ey=⎧ ⎪⎪⎨ ⎪⎪⎩jωµ0 kd(−Asinkdx+Bcoskdx) for 0≤x≤t jωµ0 ka[Csinka(a−x)−Dcoska(a−x)] fort≤x≤a.(3.108) To satisfy the boundary conditions that Ey=0a t x= 0 and x=arequires that B= D=0. We next enforce continuity of tangential fields ( Ey,Hz)a tx=t. Equations (3.107) and (3.108) then give the following: −A kdsinkdt=C kasinka(a−t), Acoskdt=Ccoska(a−t). Because this is a homogeneous set of equations, the determinant must vanish in order to have a nontrivial solution. Thus, katankdt+kdtanka(a−t)=0.( 3.109) Using (3.106) allows kaandkdto be expressed in terms of β, so (3.109) can be solved numerically for β. There is an infinite number of solutions to (3.109), corresponding to the propagation constants of the TE m0modes. This technique can be applied to many other waveguide geometries involving dielec- tric or magnetic material inhomogeneities, such as the surface waveguide of Section 3.6 or the ferrite-loaded waveguide of Section 9.3. In some cases, however, it will be impossibleto satisfy all the necessary boundary conditions with only TE- or TM-type modes, and a hybrid combination of both types of modes may be required. POINT OF INTEREST: Waveguide Flanges There are two commonly used waveguide flanges: the cover flange and the choke flange. As shown in the accompanying figure, two waveguides with cover-type flanges can be bolted to- gether to form a contacting joint. To avoid reflections and resistive loss at this joint it is neces-sary that the contacting surfaces be smooth, clean, and square because RF currents must flow across this discontinuity. In high-power applications voltage breakdown may occur at an imper- fect junction. Otherwise, the simplicity of the cover-to-cover connection makes it preferable forgeneral use. The SWR from such a joint is typically less than 1.03. c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 3.4 Circular Waveguide 121 An alternative waveguide connection uses a cover flange against a choke flange, as shown in the figure. The choke flange is machined to form an effective radial transmission line in the narrow gap between the two flanges; this line is approximately λg/4 in length between the guide and the point of contact for the two flanges. Another λg/4 line is formed by a circular axial groove in the choke flange. Then the short circuit at the right-hand end of this groove is trans- formed into an open circuit at the contact point of the flanges. Any resistance in this contact is in series with an infinite (or very high) impedance and thus has little effect. This high impedanceis transformed back into a short circuit (or very low impedance) at the edges of the waveguides to provide an effective low-resistance path for current flow across the joint. Because there is a negligible voltage drop across the ohmic contact between the flanges, voltage breakdown isavoided. Thus, the cover-to-choke connection can be useful for high-power applications. The SWR for this joint is typically less than 1.05 but is more frequency dependent than that of the cover-to-cover joint. Contact Contact Cover-to-cover connectionCover-to-choke connection/H9261g/4/H9261g/4 Reference: C. G. Montgomery, R. H. Dicke, and E. M. Purcell, Principles of Microwave Circuits , McGraw-Hill, New York, 1948. 3.4CIRCULARWAVEGUIDE A hollow, round metal pipe also supports TE and TM waveguide modes. Figure 3.11 shows the geometry of such a circular waveguide, with inner radius a. Because cylindrical geom- etry is involved, it is appropriate to employ cylindrical coordinates. As in the rectangular y x za/H9267 /H9278 FIGURE 3.11 Geometry of a circular waveguide. c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 122 Chapter 3: Transmission Lines and Waveguides coordinate case, the transverse fields in cylindrical coordinates can be derived from Ezor Hzfield components for TM and TE modes, respectively. Paralleling the development of Section 3.1, we can derive the cylindrical components of the transverse fields from the longitudinal components as Eρ=−j k2c/parenleftbigg β∂Ez ∂ρ+ωµ ρ∂Hz ∂φ/parenrightbigg , (3.110a) Eφ=−j k2c/parenleftbiggβ ρ∂Ez ∂φ−ωµ∂Hz ∂ρ/parenrightbigg , (3.110b) Hρ=j k2c/parenleftbiggω/epsilon1 ρ∂Ez ∂φ−β∂Hz ∂ρ/parenrightbigg , (3.110c) Hφ=−j k2c/parenleftbigg ω/epsilon1∂Ez ∂ρ+β ρ∂Hz ∂φ/parenrightbigg , (3.110d) where k2 c=k2−β2, and e−jβzpropagation has been assumed. For e+jβzpropagation, replace βwith−βin all expressions. TEModes For TE modes, Ez=0, and Hzis a solution to the wave equation, ∇2Hz+k2Hz=0.( 3.111) IfHz(ρ,φ, z)=hz(ρ,φ)e−jβz, (3.111) can be expressed in cylindrical coordinates as /parenleftBigg ∂2 ∂ρ2+1 ρ∂ ∂ρ+1 ρ2∂2 ∂φ2+k2 c/parenrightBigg hz(ρ,φ) =0.( 3.112) As before, we apply the method of separation of variables. Thus, let hz(ρ,φ) =R(ρ)P(φ), (3.113) and substitute into (3.112) to obtain 1 Rd2R dρ2+1 ρRdR dρ+1 ρ2Pd2P dφ2+k2 c=0, or ρ2 Rd2R dρ2+ρ RdR dρ+ρ2k2 c=−1 Pd2P dφ2.( 3.114) The left side of this equation depends only on ρ(notφ), while the right side depends only onφ. Thus, each side must be equal to a constant, which we will call k2 φ. Then, −1 Pd2P dφ2=k2 φ, or d2P dφ2+k2 φP=0.( 3.115) c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 3.4 Circular Waveguide 123 In addition, ρ2d2R dρ2+ρdR dρ+/parenleftBig ρ2k2 c−k2 φ/parenrightBig R=0.( 3.116) The general solution to (3.115) is P(φ)=Asinkφφ+Bcoskφφ. (3.117) Because the solution to hzmust be periodic in φ[i.e., hz(ρ, φ) =hz(ρ,φ±2mπ)],kφ must be an integer, n. Thus (3.117) becomes P(φ)=Asinnφ+Bcosnφ, (3.118) and (3.116) becomes ρ2d2R dρ2+ρdR dρ+/parenleftBig ρ2k2 c−n2/parenrightBig R=0,( 3.119) which is recognized as Bessel’s differential equation. The solution is R(ρ)=CJn(kcρ)+DYn(kcρ), (3.120) where Jn(x)andYn(x)are the Bessel functions of first and second kinds, respectively. Because Yn(kcρ)becomes infinite at ρ=0, this term is physically unacceptable for a circular waveguide, so D=0. The solution for hzcan then be simplified to hz(ρ,φ) =(Asinnφ+Bcosnφ)Jn(kcρ), (3.121) where the constant Cof (3.120) has been absorbed into the constants AandBof (3.121). We must still determine the cutoff wave number kc, which we can do by enforcing the boundary condition that Etan=0 on the waveguide wall. Because Ez=0, we must have that Eφ(ρ,φ) =0a t ρ=a.( 3.122) From (3.110b), we find Eφfrom Hzas Eφ(ρ,φ, z)=jωµ kc(Asinnφ+Bcosnφ)J/prime n(kcρ)e−jβz,( 3.123) where the notation J/prime n(kcρ)refers to the derivative of Jnwith respect to its argument. For Eφto vanish at ρ=a,w em u s th a v e J/prime n(kca)=0.( 3.124) If the roots of J/prime n(x)are defined as p/prime nm, so that J/prime n(p/prime nm)=0, where p/prime nmis the mth root of J/prime n, then kcmust have the value kcnm=p/prime nm a.( 3.125) Values of p/prime nmare given in mathematical tables; the first few values are listed in Table 3.3. TABLE 3.3 Values of p/primenmfor TE Modes of a Circular Waveguide np/prime n1p/prime n2p/prime n3 0 3.832 7.016 10.174 1 1.841 5.331 8.536 2 3.054 6.706 9.970 c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 124 Chapter 3: Transmission Lines and Waveguides The TE nmmodes are thus defined by the cutoff wave number kcnm=p/prime nm/a, where nrefers to the number of circumferential (φ)variations and mrefers to the number of radial (ρ) variations. The propagation constant of the TE nmmode is βnm=/radicalBig k2−k2c=/radicalBigg k2−/parenleftbiggp/primenm a/parenrightbigg2 ,( 3.126) with a cutoff frequency of fcnm=kc 2π√µ/epsilon1=p/prime nm 2πa√µ/epsilon1.( 3.127) The first TE mode to propagate is the mode with the smallest p/prime nm, which from Table 3.3 is seen to be the TE 11mode. This mode is therefore the dominant circular waveguide mode and the one most frequently used. Because m≥1, there is no TE 10mode, but there is a TE01mode. The transverse field components are, from (3.110) and (3.121), Eρ=−jωµn k2cρ(Acosnφ−Bsinnφ)Jn(kcρ)e−jβz, (3.128a) Eφ=jωµ kc(Asinnφ+Bcosnφ)J/prime n(kcρ)e−jβz, (3.128b) Hρ=−jβ kc(Asinnφ+Bcosnφ)J/prime n(kcρ)e−jβz, (3.128c) Hφ=−jβn k2cρ(Acosnφ−Bsinnφ)Jn(kcρ)e−jβz. (3.128d) The wave impedance is ZTE=Eρ Hφ=−Eφ Hρ=ηk β.( 3.129) In the above solutions there are two remaining arbitrary amplitude constants, AandB. These constants control the amplitude of the sin nφand cos nφterms, which are indepen- dent. That is, because of the azimuthal symmetry of the circular waveguide, both the sin nφ and cos nφterms represent valid solutions, and both may be present in a specific problem. The actual amplitudes of these terms will depend on the excitation of the waveguide. From a different viewpoint, the coordinate system can be rotated about the z-axis to obtain an hz with either A=0o r B=0. Now consider the dominant TE 11mode with an excitation such that B=0. The fields can be written as Hz=AsinφJ1(kcρ)e−jβz, (3.130a) Eρ=−jωµ k2cρAcosφJ1(kcρ)e−jβz, (3.130b) Eφ=jωµ kcAsinφJ/prime 1(kcρ)e−jβz, (3.130c) Hρ=−jβ kcAsinφJ/prime 1(kcρ)e−jβz, (3.130d) Hφ=−jβ k2cρAcosφJ1(kcρ)e−jβz, (3.130e) Ez=0. (3.130f) c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 3.4 Circular Waveguide 125 The power flow down the guide can be computed as Po=1 2Re/integraldisplaya ρ=0/integraldisplay2π φ=0¯EׯH∗·ˆzρdφdρ =1 2Re/integraldisplaya ρ=0/integraldisplay2π φ=0/parenleftBig EρH∗ φ−EφH∗ ρ/parenrightBig ρdφdρ =ωµ|A|2Re(β) 2k4c/integraldisplaya ρ=0/integraldisplay2π φ=0/bracketleftbigg1 ρ2cos2φJ2 1(kcρ)+k2 csin2φJ/prime2 1(kcρ)/bracketrightbigg ρdφdρ =πωµ|A|2Re(β) 2k4c/integraldisplaya ρ=0/bracketleftbigg1 ρJ2 1(kcρ)+ρk2 cJ/prime2 1(kcρ)/bracketrightbigg dρ =πωµ|A|2Re(β) 4k4c/parenleftBig p/prime2 11−1/parenrightBig J2 1(kca), (3.131) which is seen to be nonzero only when βis real, corresponding to a propagating mode. (The required integral for this result is given in Appendix C.) Attenuation due to dielectric loss is given by (3.29). The attenuation due to a lossy waveguide conductor can be found by computing the power loss per unit length of guide: P/lscript=Rs 2/integraldisplay2π φ=0|¯Js|2adφ =Rs 2/integraldisplay2π φ=0/parenleftBig |Hφ|2+|Hz|2/parenrightBig adφ =|A|2Rs 2/integraldisplay2π φ=0/parenleftBigg β2 k4ca2cos2φ+sin2φ/parenrightBigg J2 1(kca)adφ =π|A|2Rsa 2/parenleftBigg 1+β2 k4ca2/parenrightBigg J2 1(kca). (3.132) The attenuation constant is then αc=P/lscript 2Po=Rs/parenleftbig k4 ca2+β2/parenrightbig ηkβa(p/prime2 11−1) =Rs akηβ/parenleftBigg k2 c+k2 p/prime2 11−1/parenrightBigg Np/m. (3.133) TMModes For the TM modes of the circular waveguide, we must solve for Ezfrom the wave equation in cylindrical coordinates: /parenleftBigg ∂2 ∂ρ2+1 ρ∂ ∂ρ+1 ρ2∂2 ∂φ2+k2 c/parenrightBigg ez=0,( 3.134) where Ez(ρ,φ, z)=ez(ρ,φ)e−jβz, and k2 c=k2−β2. Because this equation is identical to (3.107), the general solutions are the same. Thus, from (3.121), ez(ρ,φ) =(Asinnφ+Bcosnφ)Jn(kcρ). (3.135) c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 126 Chapter 3: Transmission Lines and Waveguides TABLE 3.4 Values of pnmfor TM Modes of a Circular Waveguide np n1 pn2 pn3 0 2.405 5.520 8.654 1 3.832 7.016 10.174 2 5.135 8.417 11.620 The difference between the TE solution and the present solution is that the boundary con- ditions can now be applied directly to ezof (3.135) since Ez(ρ,φ) =0a t ρ=a.( 3.136) Thus, we must have Jn(kca)=0,( 3.137) or kc=pnm/a,( 3.138) where pnmis the mth root of Jn(x), that is, Jn(pnm)=0. Values of pnmare given in mathematical tables; the first few values are listed in Table 3.4. The propagation constant of the TM nmmode is βnm=/radicalBig k2−k2c=/radicalBig k2−(pnm/a)2,( 3.139) and the cutoff frequency is fcnm=kc 2π√µ/epsilon1=pnm 2πa√µ/epsilon1.( 3.140) Thus, the first TM mode to propagate is the TM 01mode, with p01=2.405. Because this is greater than p/prime 11=1.841 for the lowest order TE 11mode, the TE 11mode is the dominant mode of the circular waveguide. As with the TE modes, m≥1, so there is no TM 10mode. From (3.110), the transverse fields can be derived as Eρ=−jβ kc(Asinnφ+Bcosnφ)J/prime n(kcρ)e−jβz, (3.141a) Eφ=−jβn k2cρ(Acosnφ−Bsinnφ)Jn(kcρ)e−jβz, (3.141b) Hρ=jω/epsilon1n k2cρ(Acosnφ−Bsinnφ)Jn(kcρ)e−jβz, (3.141c) Hφ=−jω/epsilon1 kc(Asinnφ+Bcosnφ)J/prime n(kcρ)e−jβz. (3.141d) The wave impedance is ZTM=Eρ Hφ=−Eφ Hρ=ηβ k.( 3.142) Calculation of the attenuation for TM modes is left as a problem. Figure 3.12 shows the attenuation due to conductor loss versus frequency for various modes of a circular wave- guide. Observe that the attenuation of the TE 01mode decreases to a very small value with c03TransmissionLinesandWaveguides Pozar September 12, 2011 18:18 3.4 Circular Waveguide 127 00.010.020.030.040.050.060.07 1 3 5 7 9 1 11 31 5TE11 cutoffTM01 cutoffTE01 cutoff Frequency (GHz)/H9251c(dB/m)TM01TE01 TE11 FIGURE 3.12 Attenuation of various modes in a circular copper waveguide with a=2.54 cm. fcTE11 TE21 TM01 TM11 TM21TM02TE01TE31TE41TE12 23 1 0 fc(TE 11) FIGURE 3.13 Cutoff frequencies of the first few TE and TM modes of a circular waveguide relative to the cutoff frequency of the dominant TE 11mode. increasing frequency. This property makes the TE 01mode of interest for low-loss trans- mission over long distances. Unfortunately, this mode is not the dominant mode of thecircular waveguide, so in practice power can be lost from the TE 01mode to lower order propagating modes. Figure 3.13 shows the relative cutoff frequencies of the TE and TM modes, and Table 3.5 summarizes results for wave propagation in circular waveguide. Field lines for some of the lowest order TE and TM modes are shown in Figure 3.14. EXAMPLE 3.2 CHARACTERISTICS OF A CIRCULAR WA VEGUIDE Find the cutoff frequencies of the first two propagating modes of a Teflon-filled circular waveguide with a=0.5 cm. If the interior of the guide is gold plated, calculate the overall loss in dB for a 30 cm length operating at 14 GHz. Solution From Figure 3.13, the first two propagating modes of a circular waveguide are the TE11and TM 01modes. The cutoff frequencies can be found using (3.127) and (3.140): TE11: fc=p/prime 11c 2πa√/epsilon1r=1.841(3 ×108) 2π(0.005)√ 2.08=12.19 GHz, TM 01: fc=p01c 2πa√/epsilon1r=2.405(3 ×108) 2π(0.005)√ 2.08=15.92 GHz. c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 128 Chapter 3: Transmission Lines and Waveguides TABLE 3.5 Summary of Results for Circular Waveguide Quantity TEnmMode TMnmMode k ω√µ/epsilon1 ω√µ/epsilon1 kcp/primenm apnm a β/radicalBig k2−k2c/radicalBig k2−k2c λc2π kc2π kc λg2π β2π β vpω βω β αdk2tanδ 2βk2tanδ 2β Ez 0 (Asinnφ+Bcosnφ)Jn(kcρ)e−jβz Hz (Asinnφ+Bcosnφ)Jn(kcρ)e−jβz0 Eρ−jωµn k2cρ(Acosnφ−Bsinnφ)Jn(kcρ)e−jβz−jβ kc(Asinnφ+Bcosnφ)J/prime n(kcρ)e−jβz Eφjωµ kc(Asinnφ+Bcosnφ)J/prime n(kcρ)e−jβz−jβn k2cρ(Acosnφ−Bsinnφ)Jn(kcρ)e−jβz Hρ−jβ kc(Asinnφ+Bcosnφ)J/prime n(kcρ)e−jβz jω/epsilon1n k2cρ(Acosnφ−Bsinnφ)Jn(kcρ)e−jβz Hφ−jβn k2cρ(Acosnφ−Bsinnφ)Jn(kcρ)e−jβz−jω/epsilon1 kc(Asinnφ+Bcosnφ)J/prime n(kcρ)e−jβz ZZ TE=kη βZTM=βη k So only the TE 11mode is propagating at 14 GHz. The wave number is k=2πf√/epsilon1r c=2π(14×109)√ 2.08 3×108=422.9m−1, and the propagation constant of the TE 11mode is β=/radicalBigg k2−/parenleftbiggp/prime 11 a/parenrightbigg2 =/radicalBigg (422.9)2−/parenleftbigg1.841 0.005/parenrightbigg2 =208.0m−1. The attenuation due to dielectric loss is calculated from (3.29) as αd=k2tanδ 2β=(422.9)2(0.0004) 2(208.0)=0.172 Np/m =1.49 dB/m. The conductivity of gold is σ=4.1×107S/m, so the surface resistance is Rs=/radicalbiggωµ0 2σ=0.0367 /Omega1. c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 3.4 Circular Waveguide 129TM01 TM02 TM11 TE01 TE11 Distributions below along this planeDistributions below along this plane FIGURE 3.14 Field lines for some of the lower order modes of a circular waveguide. Reprinted with permission from S. Ramo, J. R. Whinnery, and T. Van Duzer, Fields and Waves in Communication Electronics , Copyright c/circlecopyrt 1965 by John Wiley & Sons, Inc. Table 8.04. c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 130 Chapter 3: Transmission Lines and Waveguides Then from (3.133) the attenuation due to conductor loss is αc=Rs akηβ/parenleftBigg k2 c+k2 p/prime2 11−1/parenrightBigg =0.0672 Np/m =0.583 dB/m. The total attenuation is α=αd+αc=2.07 dB/m, and the loss in the 30 cm length of guide is attenuation (dB) =α(dB/m) ×L(m)=(2.07)(0.3) =0.62 dB . ■ 3.5COAXIALLINE TEMModes Although we have already discussed TEM mode propagation on a coaxial line in Chapter 2, we will briefly reconsider it here in the context of the general framework that is being used in this chapter. The coaxial transmission line geometry is shown in Figure 3.15, where the inner con- ductor is at a potential of Vovolts and the outer conductor is at zero volts. From Section 3.1 we know that the fields can be derived from a scalar potential function, /Phi1(ρ,φ) , which is a solution to Laplace’s equation (3.14). In cylindrical coordinates Laplace’s equation takesthe form 1 ρ∂ ∂ρ/parenleftbigg ρ∂/Phi1(ρ,φ) ∂ρ/parenrightbigg +1 ρ2∂2/Phi1(ρ,φ) ∂φ2=0.( 3.143) This equation must be solved for /Phi1(ρ,φ) subject to the boundary conditions /Phi1(a,φ)=Vo, (3.144a) /Phi1(b,φ)=0. (3.144b) By the method of separation of variables, let /Phi1(ρ,φ) be expressed in product form as /Phi1(ρ,φ) =R(ρ)P(φ). (3.145) y x za b/H9267 /H9278V = 0 V = Vo FIGURE 3.15 Coaxial line geometry. c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 3.5 Coaxial Line 131 Substituting (3.145) into (3.143) and dividing by RPgives ρ R∂ ∂ρ/parenleftbigg ρdR dρ/parenrightbigg +1 Pd2P dφ2=0.( 3.146) By the usual separation-of-variables argument, the two terms in (3.146) must be equal to constants, so that ρ R∂ ∂ρ/parenleftbigg ρdR dρ/parenrightbigg =−k2 ρ, (3.147) 1 Pd2P dφ2=−k2 φ, (3.148) k2 ρ+k2 φ=0. (3.149) The general solution to (3.148) is P(φ)=Acosnφ+Bsinnφ, (3.150) where kφ=nmust be an integer since increasing φby a multiple of 2π should not change the result. Now, because the boundary conditions of (3.144) do not vary with φ, the poten- tial/Phi1(ρ, φ) should not vary with φ. Thus, nmust be zero. By (3.149), this implies that kρ must also be zero, so that the equation for R(ρ)in (3.147) reduces to ∂ ∂ρ/parenleftbigg ρdR dρ/parenrightbigg =0. The solution for R(ρ)is then R(ρ)=Clnρ+D, and so /Phi1(ρ,φ) =Clnρ+D.( 3.151) Applying the boundary conditions of (3.144) gives two equations for the constants C andD: /Phi1(a,φ)=Vo=Clna+D, (3.152a) /Phi1(b,φ)=0=Clnb+D. (3.152b) After solving for CandD, we can write the final solution for /Phi1(ρ,φ) as /Phi1(ρ,φ) =Volnb/ρ lnb/a.( 3.153) The¯Eand¯Hfields can now be found using (3.13) and (3.18), and the voltage, current, and characteristic impedance can be determined as in Chapter 2. Attenuation due to dielectric or conductor loss has already been treated in Chapter 2. HigherOrderModes The coaxial line, like the parallel plate waveguide, can also support TE and TM waveguide modes in addition to the TEM mode. In practice, these modes are usually cut off (evanes- cent), and so have only a reactive effect near discontinuities or sources, where they maybe excited. It is important in practice, however, to be aware of the cutoff frequency of the lowest order waveguide-type modes to avoid the propagation of these modes. Undesirable c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 132 Chapter 3: Transmission Lines and Waveguides effects can occur if two or more modes with different propagation constants are propagat- ing at the same time. Avoiding propagation of higher order modes sets an upper limit on the size of a coaxial cable or, equivalently, an upper limit on the frequency of operation for a given cable. This also affects the power handling capacity of a coaxial line (see the Pointof Interest on power capacity of transmission lines). We will derive the solution for the TE modes of the coaxial line; the TE 11mode is the dominant waveguide mode of the coaxial line and so is of primary importance. For TE modes, Ez=0, and Hzsatisfies the wave equation of (3.112): /parenleftBigg ∂2 ∂ρ2+1 ρ∂ ∂ρ+1 ρ2∂2 ∂φ2+k2 c/parenrightBigg hz(ρ,φ) =0,( 3.154) where Hz(ρ,φ, z)=hz(ρ,φ)e−jβz, and k2 c=k2−β2.The general solution to this equa- tion, as derived in Section 3.4, is given by the product of (3.118) and (3.120): hz(ρ,φ) =(Asinnφ+Bcosnφ)(CJn(kcρ)+DYn(kcρ)). (3.155) In this case, a≤ρ≤b, so we have no reason to discard the Ynterm. The boundary condi- tions are Eφ(ρ,φ, z)=0f o rρ =a,b.( 3.156) Using (3.110b) to find Eφfrom Hzgives Eφ=jωµ kc(Asinnφ+Bcosnφ)[CJ/prime n(kcρ)+DY/prime n(kcρ)]e−jβz.( 3.157) Applying (3.156) to (3.157) gives two equations: CJ/prime n(kca)+DY/prime n(kca)=0, (3.158a) CJ/prime n(kcb)+DY/prime n(kcb)=0. (3.158b) Because this is a homogeneous set of equations, the only nontrivial (C/negationslash=0,D/negationslash=0)solu- tion occurs when the determinant is zero. Thus we must have J/prime n(kca)Y/prime n(kcb)=J/prime n(kcb)Y/prime n(kca). (3.159) This is a characteristic (or eigenvalue) equation for kc. The values of kcthat satisfy (3.159) then define the TE nmmodes of the coaxial line. Equation (3.159) is a transcendental equation, which must be solved numerically for kc. Figure 3.16 shows the result of such a solution for n=1 for various b/aratios. An approximate solution that is often used in practice is kc=2 a+b. Once kcis known, the propagation constant or cutoff frequency can be determined. Solutions for the TM modes can be found in a similar manner; the required determinantal equation is the same as (3.159), except for the derivatives. Field lines for the TEM andTE 11modes of the coaxial line are shown in Figure 3.17. EXAMPLE 3.3 HIGHER ORDER MODE OF A COAXIAL LINE Consider a RG-401U semirigid coaxial cable, with inner and outer conductordiameters of 0.0645 in. and 0.215 in., and a Teflon dielectric with /epsilon1 r=2.2. What is the highest usable frequency before the TE 11waveguide mode starts to propagate? c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 3.5 Coaxial Line 133 123456789 1 0 1 1 1 200.10.20.30.40.50.60.70.80.91.0 b/akcaa b FIGURE 3.16 Normalized cutoff frequency of the dominant TE 11waveguide mode for a coaxial line. Solution We have b a=2b 2a=0.215 0.0645=3.33. From Figure 3.16 this value of b/agives kca=0.45 [the approximate result is kca=2/(1+b/a)=0.462]. Thus, kc=549.4m−1, and the cutoff frequency of theTE11mode is fc=ckc 2π√/epsilon1r=17.7 GHz. In practice, a 5% safety margin is usually recommended, so fmax=(0.95)(17.7 GHz )=16.8 GHz. ■ (a) (b) FIGURE 3.17 Field lines for the (a) TEM and (b) TE 11modes of a coaxial line. c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 134 Chapter 3: Transmission Lines and Waveguides POINT OF INTEREST: Coaxial Connectors Most coaxial cables and connectors in common use have a 50 /Omega1characteristic impedance, with an exception being the 75 /Omega1cable used in television systems. The reasoning behind these choices is that an air-filled coaxial line has minimum attenuation for a characteristic impedance of about 77 /Omega1(Problem 2.27), while maximum power capacity occurs for a characteristic impedance of about 30 /Omega1(Problem 3.28). A 50 /Omega1characteristic impedance thus represents a compromise between minimum attenuation and maximum power capacity. Other requirements for coaxial connectors include low SWR, higher-order-mode–free operation at a high frequency, high repeatability after a connect–disconnect cycle, and mechanical strength. Connectors are used in pairs, with a male end and a female end (or plug and jack). The accompanying photo shows several types of commonly used coaxial connectors and adapters. From top left: Type-N,TNC, SMA, APC-7, and 2.4 mm. Type-N: This connector was developed in 1942 and is named after its inventor, P. Neil, of Bell Labs. The outer diameter of the female end is about 0.625 in. The recommended upper frequency limit ranges from 11 to 18 GHz, depending on cable size. This rugged but large connector is often found on older equipment. TNC: This is a threaded version of the very common BNC connector. Its use is limited to frequencies below 1 GHz. SMA: The need for smaller and lighter connectors led to the development of this connector in the 1960s. The outer diameter of the female end is about 0.25 in. It can be used up to frequencies in the range of 18–25 GHz and is probably the most commonly used microwave connector today. APC-7: This is a precision connector (Amphenol Precision Connector) that can repeatedly achieve SWR less than 1.04 at frequencies up to 18 GHz. The connectors are “sexless,” with butt contact between both inner conductors and outer conductors. This connector is most commonlyused for measurement and instrumentation applications. c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 3.6 Surface Waves on a Grounded Dielectric Sheet 135 2.4 mm: The need for connectors at millimeter wave frequencies led to the development of several variations of the SMA connector. One of the most common is the 2.4 mm connector, which is useful to about 50 GHz. The size of this connector is similar to that of the SMA connector. 3.6SURFACEWAVESONAGROUNDEDDIELECTRICSHEET We briefly discussed surface waves in Chapter 1 in connection with the field of a plane wave totally reflected from a dielectric interface, but surface waves can exist in a variety of geometries involving dielectric interfaces. Here we consider the TM and TE surface waves that can be excited along a grounded dielectric sheet. Other geometries that can be used as surface waveguides include an ungrounded dielectric sheet, a dielectric rod, a corrugated conductor, and a dielectric-coated conducting rod. Surface waves are typified by a field that decays exponentially away from the dielectric surface, with most of the field contained in or near the dielectric. At higher frequencies the field generally becomes more tightly bound to the dielectric, making such waveguides practical. Because of the presence of the dielectric, the phase velocity of a surface wave is less than the velocity of light in a vacuum. Another reason for studying surface waves is that they may be excited on some types of planar transmission lines, such as microstrip line and slotline. TMModes Figure 3.18 shows the geometry of a grounded dielectric slab waveguide. The dielectric sheet, of thickness dand relative permittivity /epsilon1r, is assumed to be of infinite extent in the y andzdirections. We will assume propagation in the +zdirection with an e−jβzpropagation factor and no variation in the ydirection (∂/∂y=0). Because there are two distinct regions, with and without a dielectric, we must sepa- rately consider the field in these regions and then match tangential fields across the inter- face. Ezmust satisfy the wave equation of (3.25) in each region: /parenleftBigg ∂2 ∂x2+/epsilon1rk2 0−β2/parenrightBigg ez(x,y)=0, for 0≤x≤d, (3.160a) /parenleftBigg ∂2 ∂x2+k2 0−β2/parenrightBigg ez(x,y)=0, ford≤x<∞, (3.160b) where Ez(x,y,z)=ez(x,y)e−jβz. We define the cutoff wave numbers for the two regions as k2 c=/epsilon1rk2 0−β2, (3.161a) h2=β2−k2 0, (3.161b) x zDielectric Ground planed/H92800 /H9280r/H92800 FIGURE 3.18 Geometry of a grounded dielectric sheet. c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 136 Chapter 3: Transmission Lines and Waveguides where the sign on h2has been selected in anticipation of an exponentially decaying result forx>d. Observe that the same propagation constant, β, has been used for both regions. This must be the case to achieve phase matching of the tangential fields at the x=dinter- face for all values of z. The general solutions to (3.160) are ez(x,y)=Asinkcx+Bcoskcx, for 0≤x≤d (3.162a) ez(x,y)=Cehx+De−hx, ford≤x<∞ (3.162b) Note that these solutions are valid for kcandheither real or imaginary; it will turn out that both kcandhare real because of the choice of definitions in (3.161). The boundary conditions that must be satisfied are Ez(x,y,z)=0, atx=0, (3.163a) Ez(x,y,z)<∞, asx→∞, (3.163b) Ez(x,y,z)continuous at x=d, (3.163c) Hy(x,y,z)continuous at x=d. (3.163d) From (3.23), Hx=Ey=Hz=0. Condition (3.163a) implies that B=0 in (3.162a). Con- dition (3.163b) is a result of a requirement for finite fields (and energy) infinitely far awayfrom a source and implies that C=0. The continuity of E zleads to Asinkcd=De−hd,( 3.164a) while (3.23b) must be used to apply continuity to Hy, to obtain /epsilon1rA kccoskcd=D he−hd.( 3.164b) For a nontrivial solution, the determinant of the two equations of (3.164) must vanish, leading to kctankcd=/epsilon1rh.( 3.165) Eliminating βfrom (3.161a) and (3.161b) gives k2 c+h2=(/epsilon1r−1)k2 0.( 3.166) Equations (3.165) and (3.166) constitute a set of simultaneous transcendental equations that must be solved for the propagation constants kcandh,g i v e n koand/epsilon1r. These equa- tions are easily solved numerically, but Figure 3.19 shows a graphical representation of the solutions. Multiplying both sides of (3.166) by d2gives (kcd)2+(hd)2=(/epsilon1r−1)(k 0d)2, which is the equation of a circle in the kcd,hdplane, as shown in Figure 3.19. The radius of the circle is√/epsilon1r−1k0d, which is proportional to the electrical thickness of the dielectric sheet. Multiplying (3.165) by dgives kcdtankcd=/epsilon1rhd, which is also plotted in Figure 3.19. The intersection of these curves implies a solution to both (3.165) and (3.166). Observe that kcmay be positive or negative; from (3.162a) this is seen to merely change the sign of the constant A.A s√/epsilon1r−1k0dbecomes larger, the circle may intersect more than one branch of the tangent function, implying that more than c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 3.6 Surface Waves on a Grounded Dielectric Sheet 137 hd /H9266 2/H9266/H9266/H9266 2– – kcdValid solutions FIGURE 3.19 Graphical solution of the transcendental equation for the cutoff frequency of a TM surface wave mode of the grounded dielectric sheet. one TM mode can propagate. Solutions for negative h, however, must be excluded since we assumed hwas positive real when applying boundary condition (3.163b). For any nonzero-thickness sheet with a relative permittivity greater than unity, there is at least one propagating TM mode, which we will call the TM 0mode. This is the dominant mode of the dielectric slab waveguide, and it has a zero cutoff frequency. (Although for k0=0,kc=h=0, and all fields vanish.) From Figure 3.19 it can be seen that the next TM mode, the TM 1mode, will not begin to propagate until the radius of the circle becomes greater than π. The cutoff frequency of the TM nmode can then be derived as fc=nc 2d√/epsilon1r−1,n=0,1,2,.... (3.167) Once kcandhhave been found for a particular surface wave mode, the field expres- sions can be found as Ez(x,y,z)=/braceleftBigg Asinkcxe−jβzfor 0≤x≤d Asinkcde−h(x−d)e−jβzford≤x<∞,(3.168a) Ex(x,y,z)=⎧ ⎪⎪⎨ ⎪⎪⎩−jβ kcAcoskcxe−jβzfor 0≤x≤d −jβ hAsinkcde−h(x−d)e−jβzford≤x<∞,(3.168b) Hy(x,y,z)=⎧ ⎪⎪⎨ ⎪⎪⎩−jω/epsilon1 0/epsilon1r kcAcoskcxe−jβzfor 0≤x≤d −jω/epsilon10 hAsinkcde−h(x−d)e−jβzford≤x<∞.(3.168c) TEModes TE modes can also be supported by the grounded dielectric sheet. The Hzfield satisfies the wave equations /parenleftBigg ∂2 ∂x2+k2 c/parenrightBigg hz(x,y)=0, for 0≤x≤d, (3.169a) /parenleftBigg ∂2 ∂x2−h2/parenrightBigg hz(x,y)=0, ford≤x<∞, (3.169b) c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 138 Chapter 3: Transmission Lines and Waveguides with Hz(x,y,z)=hz(x,y)e−jβzandk2 candh2defined in (3.161a) and (3.161b). As for the TM modes, the general solutions to (3.169) are hz(x,y)=Asinkcx+Bcoskcx, (3.170a) hz(x,y)=Cehx+De−hx. (3.170b) To satisfy the radiation condition, C=0. Using (3.19d) to find Eyfrom Hzleads to A=0 forEy=0a tx=0 and to the equation −B kcsinkcd=D he−hd(3.171a) for continuity of Eyatx=d. Continuity of Hzatx=dgives Bcoskcd=De−hd.( 3.171b) Simultaneously solving (3.171a) and (3.171b) leads to the determinantal equation −kccotkcd=h.( 3.172) From (3.161a) and (3.161b) we also have that k2 c+h2=(/epsilon1r−1)k2 0.( 3.173) Equations (3.172) and (3.173) must be solved simultaneously for the variables kcand h. Equation (3.173) again represents circles in the kcd,hdplane, while (3.172) can be rewritten as −kcdcotkcd=hd, and plotted as a family of curves in the kcd,hdplane, as shown in Figure 3.20. Because negative values of hmust be excluded, we see from Figure 3.20 that the first TE mode does not start to propagate until the radius of the circle,√/epsilon1r−1k0d, becomes greater than π/2. The cutoff frequency of the TE nmodes can then be found as fc=(2n−1)c 4d√/epsilon1r−1forn=1,2,3,.... (3.174) hd /H9266 2 –/H9266/H9266 2 /H9266– kcd Invalid solutions FIGURE 3.20 Graphical solution of the transcendental equation for the cutoff frequency of a TE surface wave mode. The figure depicts a mode below cutoff. c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 3.6 Surface Waves on a Grounded Dielectric Sheet 139 Comparing with (3.167) shows that the order of propagation for the TM nand TE nmodes is TM 0,T E 1,T M 1,T E 2,T M 2,.... After finding the constants kcandh, the field expressions can be derived as Hz(x,y,z)=/braceleftBigg Bcoskcxe−jβzfor 0≤x≤d Bcoskcde−h(x−d)e−jβzford≤x<∞,(3.175a) Hx(x,y,z)=⎧ ⎪⎪⎨ ⎪⎪⎩jβ kcBsinkcxe−jβzfor 0≤x≤d −jβ hBcoskcde−h(x−d)e−jβzford≤x<∞,(3.175b) Ey(x,y,z)=⎧ ⎪⎪⎨ ⎪⎪⎩−jωµ 0 kcBsinkcxe−jβzfor 0≤x≤d jωµ0 hBcoskcde−h(x−d)e−jβzford≤x<∞.(3.175c) EXAMPLE 3.4 SURFACE WA VE PROPAGATION CONSTANTS Calculate and plot the propagation constants of the first three propagating surface wave modes of a grounded dielectric sheet with /epsilon1r=2.55, for d/λ0=0t o1 . 2 . Solution The first three propagating surface wave modes are the TM 0,TE1, and TM 1 modes. The cutoff frequencies for these modes can be found from (3.167) and (3.174) as TM 0:fc=0/equal1⇒d λ0=0, TE1:fc=c 4d√/epsilon1r−1/equal1⇒d λ0=1 (4√/epsilon1r−1), TM 1:fc=c 2d√/epsilon1r−1/equal1⇒d λ0=1 (2√/epsilon1r−1). The propagation constants can be found from the numerical solution of (3.165) and (3.166) for the TM modes and (3.172) and (3.173) for the TE modes. This can be done with a relatively simple root-finding algorithm (see the Point of Interest on root-finding algorithms); the results are shown in Figure 3.21. ■ POINT OF INTEREST: Root-Finding Algorithms In several examples throughout this book we will need to numerically find the root of a tran- scendental equation, so it may be useful to review two relatively simple but effective algorithms for doing this. Both methods can be easily programmed. In the interval-halving method the root of f(x)=0 is first bracketed between the values x1andx2. These values can often be estimated from the problem under consideration. If a single root lies between x1andx2,t h e n f(x1)f(x2)<0. An estimate, x3, of the root is made by halving the interval between x1andx2. Thus, x3=x1+x2 2. c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 140 Chapter 3: Transmission Lines and Waveguides 0 .2 .4 .6 .8 1.0 1.21.01.11.21.31.41.51.6/H9252/ko/H9280r = 2.55/H9280r TM0 TE1 TM1 TE2 TM2 TE3 d//H9261o FIGURE 3.21 Surface wave propagation constants for a grounded dielectric slab with /epsilon1r=2.55. Iff(x1)f(x3)<0, then the root must lie in the interval x1<x<x3;i ff(x3)f(x2)<0, the root must be in the interval x3<x<x2. A new estimate, x4, can be made by halving the appropriate interval, and this process is repeated until the location of the root has been deter- mined with the desired accuracy. The accompanying figure illustrates this algorithm for severaliterations. TheNewton–Raphson method begins with an estimate, x 1, of the root of f(x)=0. Then a new estimate, x2, is obtained from the formula x2=x1−f(x1) f/prime(x1), where f/prime(x1)is the derivative of f(x)atx1. This result is easily derived from a two-term Taylor series expansion of f(x)near x=x1:f(x)=f(x1)+(x−x1)f/prime(x1). It can also be interpreted geometrically as fitting a straight line at x=x1with the same slope as f(x)at this point; this line then intercepts the x-axis at x=x2, as shown in the figure. Reapplying the above formula gives improved estimates of the root. Convergence is generally much faster than with the interval-halving method, but a disadvantage is that the derivative of f(x)is required; this can often be computed numerically. The Newton–Raphson technique can easily be applied to the case where the root is complex (a situation that occurs, for example, when finding the propagation constant of a line or guide with loss). x x1 x3x5x4 x2f(x) Interval halvingx x1x3x2f(x) Newton–Raphson Reference: R. W. Hornbeck, Numerical Methods, Quantum Publishers, New York, 1975. c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 3.7 Stripline 141 y xb /H9280r zW Ground planeGround plane E H (a) (b) FIGURE 3.22 Stripline transmission line. (a) Geometry. (b) Electric and magnetic field lines. 3.7STRIPLINE Stripline is a planar type of transmission line that lends itself well to microwave integrated circuitry, miniaturization, and photolithographic fabrication. The geometry of stripline isshown in Figure 3.22a. A thin conducting strip of width Wis centered between two wide conducting ground planes of separation b, and the region between the ground planes is filled with a dielectric material. In practice stripline is usually constructed by etching thecenter conductor on a grounded dielectric substrate of thickness b/2 and then covering with another grounded substrate. Variations of the basic geometry of Figure 3.22a include stripline with differing dielectric substrate thicknesses ( asymmetric stripline) or different dielectric constants (inhomogeneous stripline ). Air dielectric is sometimes used when it is necessary to minimize loss. An example of a stripline circuit is shown in Figure 3.23. Because stripline has two conductors and a homogeneous dielectric, it supports a TEM wave, and this is the usual mode of operation. Like parallel plate guide and coaxial line, however, stripline can also support higher order waveguide modes. These can usually beavoided in practice by restricting both the ground plane spacing and the sidewall width to less than λ d/2. Shorting vias between the ground planes are often used to enforce this condition relative to the sidewall width. Shorting vias should also be used to eliminatehigher order modes that can be generated when an asymmetry is introduced between the ground planes (e.g., when a surface-mounted coaxial transition is used). Intuitively, one can think of stripline as a sort of “flattened-out” coax—both have a center conductor completely enclosed by an outer conductor and are uniformly filled with a dielectric medium. A sketch of the field lines for stripline is shown in Figure 3.22b. The geometry of stripline does not lend itself to the simple analyses that were used for previously treated transmission lines and waveguides. Because we will be concerned primarily with the TEM mode of stripline, an electrostatic analysis is sufficient to give the propagation constant and characteristic impedance. An exact solution of Laplace’s equa-tion is possible by a conformal mapping approach [6], but the procedure and results are cumbersome. Instead, we will present closed-form expressions that give good approxima- tions to the exact results and then discuss an approximate numerical technique for solving Laplace’s equation for a geometry similar to stripline. FormulasforPropagationConstant,CharacteristicImpedance, andAttenuation From Section 3.1 we know that the phase velocity of a TEM mode is given by v p=1/√µ0/epsilon10/epsilon1r=c/√/epsilon1r,( 3.176) c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 142 Chapter 3: Transmission Lines and Waveguides FIGURE 3.23 Photograph of a stripline circuit assembly (cover removed), showing four quadra- ture hybrids, open-circuit tuning stubs, and coaxial transitions. and thus the propagation constant of stripline is β=ω vp=ω√µ0/epsilon10/epsilon1r=√/epsilon1rk0.( 3.177) In (3.176), c=3×108m/sec is the speed of light in free-space. Using (2.13) and (2.16) allows us to write the characteristic impedance of a transmission line as Z0=/radicalbigg L C=√ LC C=1 vpC,( 3.178) where LandCare the inductance and capacitance per unit length of the line. Thus, we can find Z0if we know C. As mentioned previously, Laplace’s equation can be solved by conformal mapping to find the capacitance per unit length of stripline, but the resulting solution involves complicated special functions [6], so for practical computations simpleformulas have been developed by curve fitting to the exact solution [6, 7]. The resulting formula for characteristic impedance is Z 0=30π√/epsilon1rb We+0.441b,( 3.179a) c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 3.7 Stripline 143 where Weis the effective width of the center conductor given by We b=W b−⎧ ⎪⎪⎨ ⎪⎪⎩0f orW b>0.35 (0.35−W/b)2forW b<0.35.(3.179b) These formulas assume a strip with zero thickness and are quoted as being accurate to about 1% of the exact results. It is seen from (3.179) that the characteristic impedance decreases as the strip width Wincreases. When designing stripline circuits one usually needs to find the strip width, given the characteristic impedance (and height band relative permittivity /epsilon1r), which requires the inverse of the formulas in (3.179). Such formulas have been derived as W b=/braceleftBiggx for√/epsilon1rZ0<120/Omega1 0.85−√ 0.6−x for√/epsilon1rZ0>120/Omega1,(3.180a) where x=30π√/epsilon1rZ0−0.441.( 3.180b) Since stripline is a TEM line, the attenuation due to dielectric loss is of the same form as that for other TEM lines and is given in (3.30). The attenuation due to conductor loss can be found by the perturbation method or Wheeler’s incremental inductance rule. Anapproximate result is α c=⎧ ⎪⎪⎪⎨ ⎪⎪⎪⎩2.7×10 −3Rs/epsilon1rZ0 30π( b−t)A for√/epsilon1rZ0<120/Omega1 0.16Rs Z0bB for√/epsilon1rZ0>120/Omega1Np/m,( 3.181) with A=1+2W b−t+1 πb+t b−tln/parenleftbigg2b−t t/parenrightbigg , B=1+b (0.5W+0.7t)/parenleftbigg 0.5+0.414t W+1 2πln4πW t/parenrightbigg , where tis the thickness of the strip. EXAMPLE 3.5 STRIPLINE DESIGN Find the width for a 50 /Omega1copper stripline conductor with b= 0.32 cm and /epsilon1r=2.20. If the dielectric loss tangent is 0.001 and the operating frequency is 10 GHz, calculate the attenuation in dB/λ. Assume a conductor thickness oft=0.01 mm. Solution Because√ /epsilon1rZ0=√ 2.2(50)=74.2<120 and x=30π/(√/epsilon1rZ0)−0.441 = 0.830, (3.180) gives the strip width as W=bx=(0.32)(0.830) =0.266 cm. c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 144 Chapter 3: Transmission Lines and Waveguides At 10 GHz, the wave number is k=2πf√/epsilon1r c=310.6m−1. From (3.30) the dielectric attenuation is αd=ktanδ 2=(310.6)(0.001) 2=0.155 Np/m . The surface resistance of copper at 10 GHz is Rs=0.026 /Omega1. Then from (3.181) the conductor attenuation is αc=2.7×10−3Rs/epsilon1rZ0A 30π( b−t)=0.122 Np/m , since A=4.74. The total attenuation constant is α=αd+αc=0.277 Np/m . In dB, α(dB)=20 log eα=2.41 dB/m. At 10 GHz, the wavelength on the stripline is λ=c√/epsilon1rf=2.02 cm , so in terms of wavelength the attenuation is α(dB)=(2.41)(0.0202) =0.049 dB/λ. ■ AnApproximateElectrostaticSolution Many practical problems in microwave engineering are very complicated and do not lend themselves to straightforward analytic solutions but require some sort of numerical approach. Thus it is useful for the student to become aware of such techniques; we will introduce such methods when appropriate throughout this book, beginning with a numeri-cal solution for the characteristic impedance of stripline. We know that the fields of the TEM mode on stripline must satisfy Laplace’s equation, (3.11), in the region between the two parallel plates. The idealized stripline geometry ofFigure 3.22a extends to ±∞, which makes the analysis more difficult. Because we suspect, from the field line drawing of Figure 3.22b, that the field lines do not extend very far away from the center conductor, we can simplify the geometry by truncating the plates beyondsome distance, say |x|>a/2, and placing metal walls on the sides. Thus, the geometry we will analyze is shown in Figure 3.24, where a/greatermuchb, so that the fields around the center εr xy Wb a 2a2– 0/H9280r FIGURE 3.24 Geometry of enclosed stripline. c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 3.7 Stripline 145 conductor are not perturbed by the sidewalls. We then have a closed finite region in which the potential /Phi1(x,y)satisfies Laplace’s equation, ∇2 t/Phi1(x,y)=0f o r |x|≤a/2,0≤y≤b,( 3.182) with the boundary conditions /Phi1(x,y)=0, atx=±a/2, (3.183a) /Phi1(x,y)=0, aty=0,b. (3.183b) Laplace’s equation can be solved by the method of separation of variables. Because the center conductor at y=b/2 will contain a surface charge density, the potential /Phi1(x,y) will have a slope discontinuity there because ¯D=−/epsilon10/epsilon1r∇t/Phi1is discontinuous at y=b/2. Therefore, separate solutions for /Phi1(x,y)must be found for 0 <y<b/2 and b/2<y<b. The general solutions for /Phi1(x,y)in these two regions can be written as /Phi1(x,y)=⎧ ⎪⎪⎪⎪⎪⎨ ⎪⎪⎪⎪⎪⎩∞/summationtext n=1 oddAncosnπx asinhnπy afor 0≤y≤b/2 ∞/summationtext n=1 oddBncosnπx asinhnπ a(b−y) forb/2≤y≤b.(3.184) Only the odd-n terms are needed in (3.184) because the solution is an even function of x. The reader can verify by substitution that (3.184) satisfies Laplace’s equation in the tworegions and satisfies the boundary conditions of (3.183). The potential must be continuous at y=b/2, which from (3.184) leads to A n=Bn.( 3.185) The remaining set of unknown coefficients, An, can be found by solving for the charge density on the center strip. Because Ey=−∂/Phi1/∂y,w eh a v e Ey=⎧ ⎪⎪⎪⎪⎪⎨ ⎪⎪⎪⎪⎪⎩− ∞/summationtext n=1 oddAn/parenleftBignπ a/parenrightBig cosnπx acoshnπy afor 0≤y≤b/2 ∞/summationtext n=1 oddAn/parenleftBignπ a/parenrightBig cosnπx acoshnπ a(b−y) forb/2≤y≤b.(3.186) The surface charge density on the strip at y=b/2 is ρs=Dy(x,y=b/2+)−Dy(x,y=b/2−) =/epsilon10/epsilon1r[Ey(x,y=b/2+)−Ey(x,y=b/2−)] =2/epsilon10/epsilon1r∞/summationdisplay n=1 oddAn/parenleftBignπ a/parenrightBig cosnπx acoshnπb 2a, (3.187) which is seen to be a Fourier series in xfor the surface charge density, ρs, on the strip at y=b/2. If we know the surface charge density we could easily find the unknown con- stants, An, and then the capacitance. We do not know the exact surface charge density, but we can make a good guess by approximating it as a constant over the width of the strip, ρs(x)=/braceleftbigg1f o r |x|<W/2 0f o r |x|>W/2.(3.188) c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 146 Chapter 3: Transmission Lines and Waveguides Equating this to (3.187) and using the orthogonality properties of the cos (nπx/a)functions gives the constants Anas An=2asin(nπW/2a) (nπ)2/epsilon10/epsilon1rcosh(n πb/2a).( 3.189) The voltage of the strip conductor relative to the bottom conductor can be found by inte- grating the vertical electric field from y=0t ob /2. Because the solution is approximate, this voltage is not constant over the width of the strip but varies with position, x. Rather than choosing the voltage at an arbitrary position, we can obtain an improved result by averaging the voltage over the width of the strip: Vavg=1 WW/2/integraldisplay −W/2/integraldisplayb/2 0Ey(x,y)dydx =∞/summationdisplay n=1 oddAn/parenleftbigg2a nπW/parenrightbigg sinnπW 2asinhnπb 2a.(3.190) The total charge per unit length on the center conductor is Q=/integraldisplayW/2 −W/2ρs(x)dx=WCoul/m,( 3.191) so the capacitance per unit length of the stripline is C=Q Vavg=W ∞/summationtext n=1 oddAn/parenleftbigg2a nπW/parenrightbigg sinnπW 2asinhnπb 2aF/m.( 3.192) Finally, the characteristic impedance is given by Z0=/radicalbigg L C=√ LC C=1 vpC=√/epsilon1r cC, where c=3×108m/sec. EXAMPLE 3.6 NUMERICAL CALCULATION OF STRIPLINE IMPEDANCE Evaluate the above expressions for a stripline having /epsilon1r=2.55 and a=100b to find the characteristic impedance for W/b=0.25 to 5.0. Compare with the results from (3.179). Solution A computer program was written to evaluate (3.192). The series was truncated after 500 terms, and the results for Z0are as follows. Z0,/Omega1 Numerical, Formula, Commercial W/b Eq. (3.192) Eq. (3.179) CAD 0.25 90.9 86.6 85.3 0.50 66.4 62.7 61.7 1.0 43.6 41.0 40.2 2.0 25.5 24.2 24.4 5.0 11.1 10.8 11.9 c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 3.8 Microstrip Line 147 We see that the results are in reasonable agreement with the closed-form equa- tions of (3.179) and the results from a commercial CAD package, particularly for wider strips where the charge density is closer to uniform. Better results could be obtained if more sophisticated estimates were used for the charge density. ■ 3.8MICROSTRIPLINE Microstrip line is one of the most popular types of planar transmission lines primarily because it can be fabricated by photolithographic processes and is easily miniaturized and integrated with both passive and active microwave devices. The geometry of a microstrip line is shown in Figure 3.25a. A conductor of width Wis printed on a thin, grounded dielectric substrate of thickness dand relative permittivity /epsilon1r; a sketch of the field lines is shown in Figure 3.25b. If the dielectric substrate were not present ( /epsilon1r=1), we would have a two-wire line consisting of a flat strip conductor over a ground plane, embedded in a homogeneous medium (air). This would constitute a simple TEM transmission line with phase veloc-ityv p=cand propagation constant β=k0. The presence of the dielectric, particularly the fact that the dielectric does not fill the region above the strip (y>d), complicates the behavior and analysis of microstrip line. Unlike stripline, where all the fields are contained within a homogeneous dielectric region, microstrip has some (usually most) of its field lines in the dielectric region between the strip conductor and the ground plane and some fraction in the air region above the substrate. Forthis reason microstrip line cannot support a pure TEM wave since the phase velocity of TEM fields in the dielectric region would be c/√ /epsilon1r, while the phase velocity of TEM fields in the air region would be c, so a phase-matching condition at the dielectric–air interface would be impossible to enforce. In actuality, the exact fields of a microstrip line constitute a hybrid TM-TE wave and require more advanced analysis techniques than we are prepared to deal with here. In mostpractical applications, however, the dielectric substrate is electrically very thin (d/lessmuchλ), and so the fields are quasi-TEM. In other words, the fields are essentially the same as those of the static (DC) case. Thus, good approximations for the phase velocity, propagation con-stant, and characteristic impedance can be obtained from static, or quasi-static , solutions. Then the phase velocity and propagation constant can be expressed as v p=c√/epsilon1e, (3.193) β=k0√/epsilon1e, (3.194) y xd /H9280r zW Ground planeE H (a) (b) FIGURE 3.25 Microstrip transmission line. (a) Geometry. (b) Electric and magnetic field lines. c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 148 Chapter 3: Transmission Lines and Waveguides where /epsilon1eis the effective dielectric constant of the microstrip line. Because some of the field lines are in the dielectric region and some are in air, the effective dielectric constant satisfies the relation 1</epsilon1 e</epsilon1 r and depends on the substrate dielectric constant, the substrate thickness, the conductor width, and the frequency. We will present approximate design formulas for the effective dielectric constant, charac- teristic impedance, and attenuation of microstrip line; these results are curve-fit approximations to rigorous quasi-static solutions [8, 9]. Then we will discuss additional aspects of microstriplines, including frequency-dependent effects, higher order modes, and parasitic effects. FormulasforEffectiveDielectricConstant,Characteristic Impedance,andAttenuation The effective dielectric constant of a microstrip line is given approximately by /epsilon1 e=/epsilon1r+1 2+/epsilon1r−1 21√ 1+12d/W.( 3.195) The effective dielectric constant can be interpreted as the dielectric constant of a homo- geneous medium that equivalently replaces the air and dielectric regions of the microstripline, as shown in Figure 3.26. The phase velocity and propagation constant are then given by (3.193) and (3.194). Given the dimensions of the microstrip line, the characteristic impedance can be cal- culated as Z 0=⎧ ⎪⎪⎪⎨ ⎪⎪⎪⎩60 √/epsilon1eln/parenleftbigg8d W+W 4d/parenrightbigg forW/d≤1 120π√/epsilon1e[W/d+1.393 +0.667 ln (W/d+1.444)]forW/d≥1.(3.196) For a given characteristic impedance Z0and dielectric constant /epsilon1r,t h e W/dratio can be found as W d=⎧ ⎪⎪⎪⎨ ⎪⎪⎪⎩8eA e2A−2forW/d<2 2 π/bracketleftbigg B−1−ln(2B−1)+/epsilon1r−1 2/epsilon1r/braceleftbigg ln(B−1)+0.39−0.61 /epsilon1r/bracerightbigg/bracketrightbigg forW/d>2, (3.197) /H9280rW dW/H9280e d (a) (b) FIGURE 3.26 Equivalent geometry of a quasi-TEM microstrip line. (a) Original geometry. (b) Equivalent geometry, where the dielectric substrate of relative permittivity /epsilon1r is replaced with a homogeneous medium of effective relative permittivity /epsilon1e. c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 3.8 Microstrip Line 149 where A=Z0 60/radicalbigg /epsilon1r+1 2+/epsilon1r−1 /epsilon1r+1/parenleftbigg 0.23+0.11 /epsilon1r/parenrightbigg B=377π 2Z0√/epsilon1r. Considering a microstrip line as a quasi-TEM line, we can determine the attenuation due to dielectric loss as αd=k0/epsilon1r(/epsilon1e−1)tanδ 2√/epsilon1e(/epsilon1r−1)Np/m,( 3.198) where tan δis the loss tangent of the dielectric. This result is derived from (3.30) by multi- plying by a “filling factor,” /epsilon1r(/epsilon1e−1) /epsilon1e(/epsilon1r−1), which accounts for the fact that the fields around the microstrip line are partly in air (loss- less) and partly in the dielectric (lossy). The attenuation due to conductor loss is givenapproximately by [8] α c=Rs Z0WNp/m,( 3.199) where Rs=√ωµ0/2σis the surface resistivity of the conductor. For most microstrip sub- strates, conductor loss is more significant than dielectric loss; exceptions may occur, how- ever, with some semiconductor substrates. EXAMPLE 3.7 MICROSTRIP LINE DESIGN Design a microstrip line on a 0.5 mm alumina substrate ( /epsilon1r=9.9,tanδ=0.001) for a 50 /Omega1characteristic impedance. Find the length of this line required to produce a phase delay of 270◦at 10 GHz, and compute the total loss on this line, assuming copper conductors. Compare the results obtained from the approx- imate formulas of (3.195)–(3.199) with those from a microwave CAD package. Solution First find W/dforZ0=50/Omega1, and initially guess that W/d<2. From (3.197), A=2.142, W/d=0.9654. So the condition that W/d<2 is satisfied; otherwise we would use the expression forW/d>2. Then the required line width is W=0.9654d =0.483 mm. From (3.195) the effective dielectric constant is /epsilon1e=6.665. The line length, /lscript,f o ra 270◦phase shift is found as φ=270◦=β/lscript=√/epsilon1ek0/lscript, k0=2πf c=209.4m−1, /lscript=270◦(π/180◦)√/epsilon1ek0=8.72 mm. c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 150 Chapter 3: Transmission Lines and Waveguides Attenuation due to dielectric loss is found from (3.198) as αd=0.255 Np/m = 0.022 dB/cm. The surface resistivity for copper at 10 GHz is 0.026 /Omega1, and the attenuation due to conductor loss is, from (3.199), αc=0.0108 Np/cm =0.094 dB/cm. The total loss on the line is then 0.101 dB. A commercial microwave CAD package gives the following results: W= 0.478 mm, /epsilon1e=6.83,/lscript=8.61 mm, αd=0.022 dB/cm, and αc=0.054 dB/cm. The approximate formulas give results that are within a few percent of the CADdata for linewidth, effective dielectric constant, line length, and dielectric attenu- ation. The greatest discrepancy occurs for the attenuation constant for conductor loss. ■ Frequency-DependentEffectsandHigherOrderModes The results for the parameters of microstrip line presented in the previous section were based on the quasi-static approximation and are strictly valid only at DC (or very low frequencies). At higher frequencies a number of effects can occur that lead to variations from the quasi-static results for effective dielectric constant, characteristic impedance, and attenuation of microstrip line. In addition, new effects can arise, such as higher order modes and parasitic reactances. Because microstrip line is not a true TEM line, its propagation constant is not a linear function of frequency, meaning that the effective dielectric constant varies with frequency. The electromagnetic field that exists on microstrip line involves a hybrid coupling of TMand TE modes, complicated by the boundary condition imposed by the air and dielectric substrate interface. In addition, the current on the strip conductor is not uniform across the width of the strip, and this distribution varies with frequency. The thickness of the stripconductor also has an effect on the current distribution and hence affects the line parameters (especially the conductor loss). The variation with frequency of the parameters of a transmission line is important for several reasons. First, if the variation is significant it becomes important to know and use the parameters at the particular frequency of interest to avoid errors in design or analysis. Typically, for microstrip line, the frequency variation of the effective dielectric constant is more significant than the variation of characteristic impedance, both in terms of relative change and the relative effect on performance. A change in the effective dielectric con-stant may have a substantial effect on the phase delay through a long section of line, while a small change in characteristic impedance has the primary effect of introducing a small impedance mismatch. Second, a variation in line parameters with frequency means thatdifferent frequency components of a broadband signal will propagate differently. A varia- tion in phase velocity, for example, means that different frequency components will arrive at the output of the line at different times, leading to signal dispersion and distortion of the input signal. Third, because of the complexity of modeling these effects, approximate formulas are generally useful only for a limited range of frequency and line parameters, and numerical computer models are usually more accurate and useful. There are a number of approximate formulas, developed from numerical computer solutions and/or experimental data, that have been suggested for predicting the frequency variation of microstrip line parameters [8, 9]. A popular frequency-dependent model forthe effective dielectric constant has a form similar to the following formula [8]: /epsilon1 e(f)=/epsilon1r−/epsilon1r−/epsilon1e(0) 1+G(f),( 3.200) where /epsilon1e(f)represents the frequency-dependent effective dielectric constant, /epsilon1ris the rel- ative permittivity of the substrate, and /epsilon1e(0)is the effective dielectric constant of the line at c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 3.8 Microstrip Line 151 DC, as given by (3.195). The function G(f)can take various forms, but one suggested in reference [8] is that G(f)=g/parenleftbig f/fp/parenrightbig2, with g=0.6+0.009 Z0and fp=Z0/8πd(Z0is in ohms, fis in GHz, and dis in cm). It can be seen from the form of (3.200) that /epsilon1e(f) reduces to the DC value /epsilon1e(0)when f=0 and increases toward /epsilon1ras frequency increases. Approximate formulas like the above were primarily developed in the years before computer-aided design tools for RF and microwave engineering became commonly avail- able (see the Point of Interest on computer-aided design in Chapter 4). Such tools usuallygive accurate results for a wide range of line parameters and today are usually preferred over closed-form approximations. Another potential difficulty with microstrip line is that it may support several types of higher order modes, particularly at higher frequencies. Some of these are directly related to the TM and TE surface waves modes that were discussed in Section 3.6, while others are related to waveguide-type modes in the cross section of the line. The TM 0surface wave mode for a grounded dielectric substrate has a zero cutoff frequency, as we know from (3.167). Because some of the field lines of this mode arealigned with the field lines of the quasi-TEM mode of a microstrip line, it is possible for coupling to occur from the desired microstrip mode to a surface wave, leading to excess power loss and possibly undesired coupling to adjacent microstrip elements. Because thefields of the TM 0surface wave are zero at DC, there is little coupling to the quasi-TEM microstrip mode until a critical frequency is reached. Studies have shown that this threshold frequency is greater than zero and less than the cutoff frequency of theTM 1surface wave mode. A commonly used approximation is [8] fT1/similarequalc 2πd/radicalBigg 2 /epsilon1r−1tan−1/epsilon1r.( 3.201) For/epsilon1rranging from 1 to 10, (3.201) gives a frequency that is 35% to 66% of fc1, the cutoff frequency of the TM 1surface wave mode. When a microstrip circuit has transverse discontinuities (such as bends, junctions, or even step changes in width), the transverse currents on the conductors that are generated may allow coupling to TE surface wave modes. Most practical microstrip circuits involvesuch discontinuities, so this type of coupling is often important. The minimum threshold frequency where such coupling becomes important is given by the cutoff of the TE 1surface wave, from (3.174): fT2/similarequalc 4d√/epsilon1r−1.( 3.202) For wide microstrip lines, it is possible to excite a transverse resonance along the xaxis of the microstrip line below the strip in the dielectric region because the sides below thestrip conductor appear approximately as magnetic walls. This condition occurs when the width is about λ/2 in the dielectric, but because of field fringing the effective width of the strip is somewhat larger than the physical width. A rough approximation for the effective width is W+d/2, so the approximate threshold frequency for transverse resonance is f T3/similarequalc√/epsilon1r(2W+d).( 3.203) It is rare that a microstrip line is wide enough to approach this limit in practice. Finally, a parallel plate–type waveguide mode may propagate when the vertical spac- ing between the strip conductor and ground plane approaches λ/2 in the dielectric. Thus, an c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 152 Chapter 3: Transmission Lines and Waveguides approximation for the threshold frequency for this mode (valid for wide microstrip lines) can be given as fT4/similarequalc 2d√/epsilon1r.( 3.204) Thinner microstrip lines will have more fringing field that effectively lengthens the path between the strip and ground plane, thus reducing the threshold frequency by as much as 50%. The net effect of the threshold frequencies given in (3.201)–(3.204) is to impose an upper frequency limit of operation for a given microstrip geometry. This limit is a function of the substrate thickness, dielectric constant, and strip width. EXAMPLE 3.8 FREQUENCY DEPENDENCE OF EFFECTIVE DIELECTRIC CONSTANT Use the approximate formula of (3.200) to plot the change in effective dielectric constant over frequency for a 25 /Omega1microstrip line on a substrate having a rela- tive permittivity of 10.0 and a thickness of 0.65 mm. Compare the approximate data with results from a CAD model for frequencies up to 20 GHz. Compare thecalculated phase delay at 10 GHz through a 1.093 cm length of line when using /epsilon1 e(0)versus /epsilon1e(10 GHz ). Solution The required linewidth for a 25 /Omega1impedance is w=2.00 mm. The effective dielectric constant for this line at low frequencies can be found from (3.195) to be/epsilon1e(0)=7.53. A short computer program was used to calculate the effective dielectric constant as a function of frequency using (3.200), and the result is shown in Figure 3.27. Comparison with a commercial microwave CAD package shows that the approximate model is reasonably accurate up to about 10 GHz butgives an overestimate at higher frequencies. 0 5 10 15 201.07.08.0 CAD Eq. (3.200)9.0Effective Dielectric Constant (/H9280e) Frequency (GHz) FIGURE 3.27 Effective dielectric constant versus frequency for the microstrip line of Example 3.8, comparing the approximate model of (3.200) with data from a computer-aided design package. c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 3.9 The Transverse Resonance Technique 153 Using an effective dielectric constant of /epsilon1e(0)=7.53, we find the phase delay through a 1.093 cm length of line to be φ0=√/epsilon1e(0)k0/lscript=360◦.T h e effective dielectric constant at 10 GHz is 8.120 (CAD), with a corresponding phase delay of φ10=√/epsilon1e(10 GHz )k0/lscript=374◦—an error of about 14◦. ■ 3.9THETRANSVERSERESONANCETECHNIQUE According to the general solutions of Maxwell’s equations for TE or TM waves given in Section 3.1, a uniform waveguide structure always has a propagation constant of the form β=/radicalBig k2−k2c=/radicalBig k2−k2x−k2y,( 3.205) where kc=/radicalBig k2x+k2yis the cutoff wave number of the guide and, for a given mode, is a fixed function of the cross-sectional geometry of the guide. Thus, if we know kcwe can determine the propagation constant of the guide. In previous sections we determined kc by solving the wave equation in the guide, subject to the appropriate boundary conditions. Although this technique is very powerful and general, it can be complicated for complexwaveguides, especially if dielectric layers are present. In addition, the wave equation solu- tion gives a complete field description inside the waveguide, which is often more informa- tion than we really need if we are only interested in the propagation constant of the guide. Thetransverse resonance technique employs a transmission line model of the transverse cross section of the waveguide and gives a much simpler and more direct solution for thecutoff frequency. This is another example where circuit and transmission line theory offers a simplified alternative to a field theory solution. The transverse resonance procedure is based on the fact that in a waveguide at cutoff, the fields form standing waves in the transverse plane of the guide, as can be inferred from the “bouncing plane wave” interpretation of waveguide modes discussed in Section 3.2. This situation can be modeled with an equivalent transmission line circuit operating atresonance. One of the conditions of such a resonant line is the fact that, at any point on the line, the sum of the input impedances seen looking to either side must be zero. That is, Z r in(x)+Z/lscript in(x)=0 for all x,( 3.206) where Zr in(x)and Z/lscript in(x)are the input impedances seen looking to the right and left, respectively, at any point xon the resonant line. The transverse resonance technique only gives results for the cutoff frequency of the guide. If fields or attenuation due to conductor loss are needed, the complete field theorysolution will be required. The procedure will now be illustrated with an example. TE 0nModesofaPartiallyLoadedRectangularWaveguide The transverse resonance technique is particularly useful when the guide contains dielec- tric layers because the boundary conditions at the dielectric interfaces, which require thesolution of simultaneous algebraic equations in the field theory approach, can be easily handled as junctions of different transmission lines. As an example, consider a rectangu- lar waveguide partially filled with dielectric, as shown in Figure 3.28. To find the cutofffrequencies for the TE 0nmodes, the equivalent transverse resonance circuit shown in the figure can be used. The line for 0 <y<trepresents the dielectric-filled part of the guide c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 154 Chapter 3: Transmission Lines and Waveguides /H92800y y xb at 0tb 0kya, Za kyd, Zd(Air) (Dielectric) /H9280r/H92800 FIGURE 3.28 A rectangular waveguide partially filled with dielectric and the transverse reso- nance equivalent circuit. and has a transverse propagation constant kydand a characteristic impedance for TE modes given by Zd=kη kyd=k0η0 kyd,( 3.207a) where k0=ω√µ0/epsilon10andη0=√µ0//epsilon10.F o r t<y<b, the guide is air filled and has a transverse propagation constant kyaand an equivalent characteristic impedance given by Za=k0η0 kya.( 3.207b) Applying condition (3.206) yields kyatankydt+kydtankya(b−t)=0.( 3.208) This equation contains two unknowns, kyaandkyd. An additional equation is obtained from the fact that the longitudinal propagation constant, β, must be the same in both regions for phase matching of the tangential fields at the dielectric interface. Thus, with kx=0, β=/radicalBig /epsilon1rk2 0−k2 yd=/radicalBig k2 0−k2ya, or /epsilon1rk2 0−k2 yd=k2 0−k2 ya.( 3.209) Equations (3.208) and (3.209) can be solved (numerically or graphically) to obtain kyd andkya. There will be an infinite number of solutions, corresponding to the ndependence (number of variations in y)o ft h eT E 0nmode. 3.10WAVEVELOCITIESANDDISPERSION We have so far encountered two types of velocities related to the propagation of electro- magnetic waves: rThe speed of light in a medium (1/√µ/epsilon1 )rThe phase velocity (vp=ω/β) The speed of light in a medium is the velocity at which a plane wave would propagate in that medium, while the phase velocity is the speed at which a constant phase point travels.For a TEM plane wave, these two velocities are identical, but for other types of guided wave propagation the phase velocity may be greater or less than the speed of light. c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 3.10 Wave Velocities and Dispersion 155 If the phase velocity and attenuation of a line or guide are constants that do not change with frequency, then the phase of a signal that contains more than one frequency component will not be distorted. If the phase velocity is different for different frequencies, then the individual frequency components will not maintain their original phase relationships asthey propagate down the transmission line or waveguide, and signal distortion will occur. Such an effect is called dispersion since different phase velocities allow the “faster” waves to lead in phase relative to the “slower” waves, and the original phase relationships willgradually be dispersed as the signal propagates down the line. In such a case, there is no single phase velocity that can be attributed to the signal as a whole. However, if the bandwidth of the signal is relatively small or if the dispersion is not too severe, a group velocity can be defined in a meaningful way. This velocity can be used to describe the speed at which the signal propagates. GroupVelocity As discussed earlier, the physical interpretation of group velocity is the velocity at which a narrowband signal propagates. We will derive the relation of group velocity to the propa-gation constant by considering a signal f(t)in the time domain. The Fourier transform of this signal is defined as F(ω)=/integraldisplay ∞ −∞f(t)e−jωtdt,( 3.210a) and the inverse transform is f(t)=1 2π/integraldisplay∞ −∞F(ω)ejωtdω. (3.210b) Now consider the transmission line or waveguide on which the signal f(t)is propa- gating as a linear system, with a transfer function Z(ω)that relates the output, Fo(ω),o f the line to the input, F(ω), of the line, as shown in Figure 3.29. Thus, Fo(ω)=Z(ω)F(ω). (3.211) For a lossless matched transmission line or waveguide, the transfer function Z(ω)can be expressed as Z(ω)=Ae−jβz=|Z(ω)|e−jψ,( 3.212) where Ais a constant and βis the propagation constant of the line or guide. The time domain representation of the output signal, fo(t), can then be written as fo(t)=1 2π/integraldisplay∞ −∞F(ω)|Z(ω)|ej(ωt−ψ)dω. (3.213) F(/H9275) Z(/H9275) Fo(/H9275) Fo(/H9275) = Z (/H9275)F(/H9275) FIGURE 3.29 A transmission line or waveguide represented as a linear system with transfer function Z(ω). c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 156 Chapter 3: Transmission Lines and Waveguides If|Z(ω)|= Ais a constant and the phase ψofZ(ω)is a linear function of ω,s a yψ=aω, the output can be expressed as fo(t)=1 2π/integraldisplay∞ −∞AF(ω)ejω(t−a)dω=Af(t−a), (3.214) which is seen to be a replica of f(t), except for an amplitude factor Aand time shift a. Thus, a transfer function of the form Z(ω)=Ae−jωadoes not distort the input signal. A lossless TEM wave has a propagation constant β=ω/c, which is of this form, so a TEM line is dispersionless and does not lead to signal distortion. If the TEM line is lossy, however, the attenuation may be a function of frequency, which could lead to signal distortion. Now consider a narrowband input signal of the form s(t)=f(t)cosω0t=Re/braceleftBig f(t)ejωot/bracerightBig ,( 3.215) which represents an amplitude-modulated carrier wave of frequency ωo. Assume that the highest frequency component of f(t)isωm, where ωm/lessmuchωo. The Fourier transform, S(ω), ofs(t),i s S(ω)=/integraldisplay∞ −∞f(t)e−jωotejωtdt=F(ω−ωo), (3.216) where we have used the complex form of the input signal as expressed in (3.215). We will need to take the real part of the output inverse transform to obtain the time domain output signal. The spectra of F(ω)andS(ω)are depicted in Figure 3.30. The output signal spectrum is So(ω)=AF(ω−ωo)e−jβz,( 3.217) and in the time domain, so(t)=1 2πRe/integraldisplay∞ −∞So(ω)ejωtdω =1 2πRe/integraldisplayωo+ωm ωo−ωmAF(ω−ωo)ej(ωt−βz)dω.(3.218) In general, the propagation constant βmay be a complicated function of ω. However, ifF(ω)is narrowband (ωm/lessmuchωo), then βcan often be linearized by using a Taylor series expansion about ωo: β(ω)=β(ω o)+dβ dω/vextendsingle/vextendsingle/vextendsingle/vextendsingle ω=ωo(ω−ωo)+1 2d2β dω2/vextendsingle/vextendsingle/vextendsingle/vextendsingle ω=ωo(ω−ωo)2+···.( 3.219) –/H9275m /H9275m /H9275 0F(/H9275) (a)–/H9275o /H9275o /H9275 0S(/H9275) (b) FIGURE 3.30 Fourier spectra of the signals (a) f(t)and (b) s(t). c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 3.11 Summary of Transmission Lines and Waveguides 157 Retaining the first two terms gives β(ω)/similarequalβo+β/prime o(ω−ωo), (3.220) where βo=β(ω o), β/prime o=dβ dω/vextendsingle/vextendsingle/vextendsingle/vextendsingle ω=ωo. After a change of variables to y=ω−ωo, the expression for so(t)becomes so(t)=A 2πRe/braceleftbigg ej(ωot−βoz)/integraldisplayωm −ωmF(y)ej(t−β/prime oz)ydy/bracerightbigg =ARe/braceleftBig f(t−β/prime oz)ej(ωot−βoz)/bracerightBig =Af(t−β/prime oz)cos(ωot−βoz), (3.221) which is a time-shifted replica of the original modulation envelope, f(t), of (3.215). The velocity of this envelope is the group velocity, vg: vg=1 β/primeo=/parenleftbiggdβ dω/parenrightbigg−1/vextendsingle/vextendsingle/vextendsingle/vextendsingle ω=ωo.( 3.222) EXAMPLE 3.9 WA VEGUIDE WA VE VELOCITIES Calculate the group velocity for a waveguide mode propagating in an air-filled guide. Compare this velocity to the phase velocity and speed of light. Solution The propagation constant for a mode in an air-filled waveguide is β=/radicalBig k2 0−k2c=/radicalBig (ω/c)2−k2c. Taking the derivative with respect to frequency gives dβ dω=ω/c2 /radicalbig (ω/c)2−k2c=ko cβ, so from (3.234) the group velocity is vg=/parenleftbiggdβ dω/parenrightbigg−1 =cβ k0. The phase velocity is vp=ω/β=ck0/β. Since β< k0, we have that vg< c<v p, which indicates that the phase velocity of a waveguide mode may be greater than the speed of light, but the group velocity (the velocity of a narrow-band signal) will be less than the speed of light. ■ 3.11SUMMARYOFTRANSMISSIONLINESANDWAVEGUIDES We have discussed a variety of transmission lines and waveguides in this chapter, and here we will summarize some of the basic properties of these transmission media and their relative advantages in a broader context. c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 158 Chapter 3: Transmission Lines and Waveguides TABLE 3.6 Comparison of Common Transmission Lines and Waveguides Characteristic Coax Waveguide Stripline Microstrip Modes: Preferred TEM TE10 TEM Quasi-TEM Other TM,TE TM,TE TM,TE Hybrid TM,TE Dispersion None Medium None Low Bandwidth High Low High High Loss Medium Low High High Power capacity Medium High Low Low Physical size Large Large Medium Small Ease of fabrication Medium Medium Easy Easy Integration with Hard Hard Fair Easy We made a distinction between TEM, TM, and TE waves and saw that transmission lines and waveguides can be categorized according to which type of waves they can sup- port. We saw that TEM waves are nondispersive, with no cutoff frequency, while TM andTE waves exhibit dispersion and generally have nonzero cutoff frequencies. Other electri- cal considerations include bandwidth, attenuation, and power-handling capacity. Mechan- ical factors are also very important, however, and include such considerations as physical size (volume and weight), ease of fabrication (cost), and the ability to be integrated with other devices (active or passive). Table 3.6 compares several types of transmission mediawith regard to these considerations; this table only gives general guidelines, as specific cases may give better or worse results than those indicated. OtherTypesofLinesandGuides Although we have discussed the most common types of waveguides and transmission lines, there are many other guides and lines (and many variations) that we are not able to presentin detail. A few of the more popular types are briefly mentioned here. Ridge waveguide: The practical bandwidth of rectangular waveguide is slightly less than an octave (a 2:1 frequency range). This is because the TE 20mode begins to propagate at a frequency equal to twice the cutoff frequency of the TE 10mode. The ridge waveguide, shown in Figure 3.31, consists of a rectangular waveguide loaded with conducting ridges on the top and/or bottom walls. This loading tends to lower the cutoff frequency of the dominant mode, leading to increased bandwidth and better (more constant) impedancecharacteristics. Ridge waveguides are often used for impedance matching purposes, where FIGURE 3.31 Cross section of a ridge waveguide. c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 3.11 Summary of Transmission Lines and Waveguides 159 /H9280r1/H9280r2 FIGURE 3.32 Dielectric waveguide geometry. the ridge may be tapered along the length of the guide. The presence of the ridge, however, reduces the power-handling capacity of the waveguide. Dielectric waveguide : As we have seen from our study of surface waves, metallic con- ductors are not necessary to confine and support a propagating electromagnetic field. The dielectric waveguide shown in Figure 3.32 is another example of such a guide, where /epsilon1r2, the dielectric constant of the ridge, is usually greater than /epsilon1r1, the dielectric constant of the substrate. The fields are thus mostly confined to the ridge and the surrounding area. This type of guide supports TM and TE modes, and is convenient for miniaturization andintegration with active devices. Its small size makes it useful for millimeter wave to optical frequencies, although it can be very lossy at bends or junctions in the ridge line. Many variations in this basic geometry are possible. Slotline: Slotline is another one of the many possible types of planar transmission lines. The geometry of a slotline is shown in Figure 3.33. It consists of a thin slot in the groundplane on one side of a dielectric substrate. Thus, like microstrip line, the two conductors of slotline lead to a quasi-TEM type of mode. The width of the slot controls the characteristic impedance of the line. Coplanar waveguide: The coplanar waveguide, shown in Figure 3.34, is similar to the slot- line, and can be viewed as a slotline with a third conductor centered in the slot region.Because of the presence of this additional conductor, this type of line can support even or odd quasi-TEM modes, depending on whether the electric fields in the two slots are in the opposite direction or the same direction. Coplanar waveguides are particularly useful for fabricating active circuitry due to the presence of the center conductor and the close proximity of the ground planes. Covered microstrip: Many variations of the basic microstrip line geometry are possible, but one of the more common is the covered microstrip, shown in Figure 3.35. The metallic cover plate is often used for electrical shielding and physical protection of the microstrip circuitry and is usually situated several substrate thicknesses away from the circuit. Its presence, however, can perturb the operation of the circuit enough so that its effect mustbe taken into account during design. /H9280r FIGURE 3.33 Geometry of a printed slotline. c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 160 Chapter 3: Transmission Lines and Waveguides /H9280r FIGURE 3.34 Coplanar waveguide geometry. POINT OF INTEREST: Power Capacity of Transmission Lines The power-handling capacity of an air-filled transmission line or waveguide is usually limited by voltage breakdown, which occurs at a field strength of about Ed=3×106V/m for room temperature air at sea level pressure. Thermal effects may also serve to limit the power capacity of some types of lines. In an air-filled coaxial line the electric field varies as Eρ=Vo/(ρlnb/a), which has a maximum at ρ=a(at the inner conductor). Thus the maximum voltage before breakdown is Vmax=Edalnb a(peak-to-peak), and the maximum power capacity is then Pmax=V2max 2Z0=πa2E2 d η0lnb a. As might be expected, this result shows that power capacity can be increased by using a larger coaxial cable (larger a,bwith fixed b/afor the same characteristic impedance). However, prop- agation of higher order modes limits the maximum operating frequency for a given cable size. Thus, there is an upper limit on the power capacity of a coaxial line for a given maximum operating frequency, fmax, which can be shown to be given by Pmax=0.025 η0/parenleftbiggcEd fmax/parenrightbigg2 =5.8×1012/parenleftbiggEd fmax/parenrightbigg2 . As an example, at 10 GHz the maximum peak power capacity of any coaxial line with no higher order modes is about 520 kW. In an air-filled rectangular waveguide the electric field varies as Ey=Eosin(π x/a),w h i c h has a maximum value of Eoatx=a/2 (the middle of the guide). Thus the maximum power capacity before breakdown is Pmax=abE2o 4Zw=abE2 d 4Zw, which shows that power capacity increases with guide size. For most standard waveguides, b/similarequal2a. To avoid propagation of the TE 20mode we must have a<c/fmax,w h e r e fmaxis the /H9280r FIGURE 3.35 Covered microstrip line. c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 Problems 161 maximum operating frequency. Then the maximum power capacity of the guide can be shown to be Pmax=0.11 η0/parenleftbiggcEd fmax/parenrightbigg2 =2.6×1013/parenleftbiggEd fmax/parenrightbigg2 . As an example, at 10 GHz the maximum peak power capacity of a rectangular waveguide oper- a t i n gi nt h eT E 10mode is about 2300 kW, which is considerably higher than the power capacity of a coaxial cable at the same frequency. Because arcing and voltage breakdown are high-speed transient effects, these voltage and power limits are peak values; average power capacity is lower. In addition, it is good engineering practice to provide a safety factor of at least two, so the maximum powers that can be safelytransmitted should be limited to about half of the above values. If there are reflections on the line or guide, the power capacity is further reduced. In the worst case, a reflection coefficient magnitude of unity will double the maximum voltage on the line, so the power capacity will be reduced by a factor of four. The power capacity of a line can be increased by pressurizing the line with air or an inert gas or by using a dielectric. The dielectric strength (E d)of most dielectric materials is greater than that of air, but the power capacity may be further limited by the heating of the dielectric due to ohmic loss. Reference: P. A. Rizzi, Microwave Engineering—Passive Circuits , Prentice-Hall, Englewood Cliffs, N.J., 1988. REFERENCES [1] O. Heaviside, Electromagnetic Theory, V ol. 1, 1893. Reprinted by Dover, New York, 1950. [2] Lord Rayleigh, “On the Passage of Electric Waves through Tubes,” Philosophical Magazine , vol. 43, pp. 125–132, 1897. Reprinted in Collected Papers, Cambridge University Press, Cambridge, 1903. [3] K. S. Packard, “The Origin of Waveguides: A Case of Multiple Rediscovery,” IEEE Transactions on Microwave Theory and Techniques, vol. MTT-32, pp. 961–969, September 1984. [4] R. M. Barrett, “Microwave Printed Circuits—An Historical Perspective,” IEEE Transactions on Microwave Theory and Techniques, vol. MTT-32, pp. 983–990, September 1984. [5] D. D. Grieg and H. F. Englemann, “Microstrip—A New Transmission Technique for the Kilomega- cycle Range,” Proceedings of the IRE , vol. 40, pp. 1644–1650, December 1952. [6] H. Howe, Jr., Stripline Circuit Design, Artech House, Dedham, Mass., 1974. [7] I. J. Bahl and R. Garg, “A Designer’s Guide to Stripline Circuits,” Microwaves , January 1978, pp. 90–96. [8] I. J. Bahl and D. K. Trivedi, “A Designer’s Guide to Microstrip Line,” Microwaves , May 1977, pp. 174–182. [9] K. C. Gupta, R. Garg, and I. J. Bahl, Microstrip Lines and Slotlines , Artech House, Dedham, Mass., 1979. PROBLEMS 3.1Devise at least two variations of the basic coaxial transmission line geometry of Section 3.5, and discuss the advantages and disadvantages of your proposed lines in terms of size, loss, cost, higher order modes, dispersion, or other considerations. Repeat this exercise for the microstrip line geometry of Section 3.8. 3.2Derive equations (3.5a)–(3.5d) from equations (3.3) and (3.4). 3.3Calculate the attenuation due to conductor loss for the TE nmode of a parallel plate waveguide. 3.4Consider a section of air-filled K-band waveguide. From the dimensions given in Appendix I, determine the cutoff frequencies of the first two propagating modes. From the recommendedoperating range given in Appendix I for this guide, determine the percentage reduction in bandwidth c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 162 Chapter 3: Transmission Lines and Waveguides that this operating range represents, relative to the theoretical bandwidth for a single propagating mode. 3.5A 10 cm length of a K-band copper waveguide is filled with a dielectric material with /epsilon1r=2.55 and tanδ=0.0015. If the operating frequency is 15 GHz, find the total loss through the guide and the phase delay from the input to the output of the guide. 3.6An attenuator can be made using a section of waveguide operating below cutoff, as shown in the accompanying figure. If a=2.286 cm and the operating frequency is 12 GHz, determine the required length of the below-cutoff section of waveguide to achieve an attenuation of 100 dB between the input and output guides. Ignore the effect of reflections at the step discontinuities. a al a/2 Propagating wavePropagating waveEvanescent waves 3.7Find expressions for the electric surface current density on the walls of a rectangular waveguide for aT E 10mode. Why can a narrow slot be cut along the centerline of the broad wall of a rectangular waveguide without perturbing the operation of the guide? (Such a slot is often used in a slotted line for a probe to sample the standing wave field inside the guide.) 3.8Derive the expression for the attenuation of the TM mnmode of a rectangular waveguide due to imperfectly conducting walls. 3.9For the partially loaded rectangular waveguide shown in the accompanying figure, solve (3.109) withβ=0 to find the cutoff frequency of the TE 10mode. Assume a=2.286 cm, t=a/2, and /epsilon1r=2.25. a 2y xb 0a/H9280r = 2.25 /H9280r = 1 3.10 Consider the partially filled parallel plate waveguide shown in the accompanying figure. Derive the solution (fields and cutoff frequency) for the lowest order TE mode of this structure. Assume themetal plates are infinitely wide. Can a TEM wave propagate on this structure? xy d /H92800 /H92800 /H9280r/H92800 W 3.11 Derive equations (3.110a)–(3.110d) for the transverse field components in terms of longitudinal fields, in cylindrical coordinates. c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 Problems 163 3.12 Derive the expression for the attenuation of the TM nmmode in a circular waveguide with finite conductivity. 3.13 A circular copper waveguide has a radius of 0.4 cm and is filled with a dielectric material having /epsilon1r=1.5 and tan δ=0.0002. Identify the first four propagating modes and their cutoff frequencies. For the dominant mode, calculate the total attenuation at 20 GHz. 3.14 Derive the ¯Eand¯Hfields of a coaxial line from the expression for the potential given in (3.153). Also find expressions for the voltage and current on the line and the characteristic impedance. 3.15 Derive a transcendental equation for the cutoff frequency of the TM modes of a coaxial waveguide. Using tables, obtain an approximate value of kcafor the TM 01mode if b/a=2. 3.16 Derive an expression for the attenuation of a TE surface wave on a grounded dielectric substrate when the ground plane has finite conductivity. 3.17 Consider the grounded magnetic substrate shown in the accompanying figure. Derive a solution for the TM surface waves that can propagate on this structure. y zxd /H92800, /H92620/H9262r/H92800, /H92620 3.18 Consider the partially filled coaxial line shown in the accompanying figure. Can a TEM wave propa- gate on this line? Derive the solution for the TM 0m(no azimuthal variation) modes of this geometry. /H92800/H92800/H9280ry xa b c 3.19 A copper stripline transmission line is to be designed for a 100 /Omega1characteristic impedance. The ground plane separation is 1.02 mm and the dielectric constant is 2.20, with tan δ=0.001. At 5 GHz, find the guide wavelength on the line and the total attenuation. 3.20 A copper microstrip transmission line is to be designed for a 100 /Omega1characteristic impedance. The substrate is 0.51 mm thick, with /epsilon1r=2.20 and tan δ=0.001. At 5 GHz, find the guide wavelength on the line and the total attenuation. Compare these results with those for the similar stripline case ofthe preceding problem. 3.21 A 100 /Omega1microstrip line is printed on a substrate of thickness 0.0762 cm with a dielectric constant of 2.2. Ignoring losses and fringing fields, find the shortest length of this line that appears at its input asa capacitor of 5 pF at 2.5 GHz. Repeat for an inductance of 5 nH. Using a microwave CAD package with a physical model for the microstrip line, compute the actual input impedance seen when losses are included (assume copper conductors and tan δ=0.001). 3.22 A microwave antenna feed network operating at 5 GHz requires a 50 /Omega1printed transmission line that is 16λlong. Possible choices are (1) copper microstrip, with d=0.16 cm, /epsilon1 r=2.20, and tan δ= 0.001, or (2) copper stripline, with b=0.32 cm, /epsilon1r=2.20, t=0.01 mm, and tan δ=0.001. Which line should be used if attenuation is to be minimized? c03TransmissionLinesandWaveguides Pozar July 29, 2011 20:41 164 Chapter 3: Transmission Lines and Waveguides 3.23 Consider the TE modes of an arbitrary uniform waveguiding structure in which the transverse fields are related to Hzas in (3.19). If Hzis of the form Hz(x,y,z)=hz(x,y)e−jβz,w h e r e hz(x,y)is a real function, compute the Poynting vector and show that real power flow occurs only in the z direction. Assume that βis real, corresponding to a propagating mode. 3.24 A piece of rectangular waveguide is air filled for z<0 and dielectric filled for z>0. Assume that both regions can support only the dominant TE 10mode and that a TE 10mode is incident on the inter- face from z<0. Using a field analysis, write general expressions for the transverse field components of the incident, reflected, and transmitted waves in the two regions and enforce the boundary con- ditions at the dielectric interface to find the reflection and transmission coefficients. Compare these results to those obtained with an impedance approach, using ZTEfor each region. 3.25 Use the transverse resonance technique to derive a transcendental equation for the propagation con- stant of the TM modes of a rectangular waveguide that is air filled for 0 <x<dand dielectric filled ford<x<a. 3.26 Apply the transverse resonance technique to find the propagation constants for the TE surface waves that can be supported by the structure of Problem 3.17. 3.27 An X-band waveguide filled with Rexolite is operating at 9.0 GHz. Calculate the speed of light in this material and the phase and group velocities in the waveguide. 3.28 As discussed in the Point of Interest on the power-handling capacity of transmission lines, the maxi- mum power capacity of a coaxial line is limited by voltage breakdown and is given by Pmax=πa2E2 d η0lnb a, where Edis the field strength at breakdown. Find the value of b/athat maximizes the maximum power capacity and show that the corresponding characteristic impedance is about 30 /Omega1. 3.29 A microstrip circuit is fabricated on an alumina substrate having a dielectric constant of 9.9, a thick- ness of 2.0 mm, and a 50 /Omega1linewidth of 1.93 mm. Find the threshold frequencies of the four higher order modes discussed in Section 3.8, and recommend the maximum operating frequency for thismicrostrip circuit. c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 Chapter Four Microwave Network Analysis Circuits operating at low frequencies, for which the circuit dimensions are small relative to the wavelength, can be treated as an interconnection of lumped passive or active components with unique voltages and currents defined at any point in the circuit. In this situation the circuit dimensions are small enough such that there is negligible phase delay from one point in the cir-cuit to another. In addition, the fields can be considered as TEM fields supported by two or moreconductors. This leads to a quasi-static type of solution to Maxwell’s equations and to the well- known Kirchhoff voltage and current laws and impedance concepts of circuit theory [1]. As the reader is aware, there is a powerful and useful set of techniques for analyzing low-frequencycircuits. In general, these techniques cannot be directly applied to microwave circuits, but itis the purpose of the present chapter to show how basic circuit and network concepts can be extended to handle many microwave analysis and design problems of practical interest. The main reason for doing this is that it is usually much easier to apply the simple and intuitive ideas of circuit analysis to a microwave problem than it is to solve Maxwell’s equa-tions for the same problem. In a way, field analysis gives us much more information about the particular problem under consideration than we really want or need. That is, because the solution to Maxwell’s equations for a given problem is complete, it gives the electric and mag-netic fields at all points in space. However, usually we are only interested in the voltage orcurrent at a set of terminals, the power flow through a device, or some other type of “terminal” quantity, as opposed to a minute description of the fields at all points in space. Another reason for using circuit or network analysis is that it is then very easy to modify the original prob-lem, or combine several elements together and find the response, without having to reanalyzein detail the behavior of each element in combination with its neighbors. A field analysis us- ing Maxwell’s equations for such problems would be hopelessly difficult. There are situations, however, in which such circuit analysis techniques are an oversimplification and may lead toerroneous results. In such cases one must resort to a field analysis approach, using Maxwell’sequations. Fortunately, there are a number of commercially available computer-aided design packages that can model RF and microwave problems using both field theory analysis and net- work analysis. It is part of the education of a microwave engineer to be able to determine whennetwork analysis concepts apply and when they should be cast aside in favor of more rigorousanalysis. 165 c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 166 Chapter 4: Microwave Network Analysis The basic procedure for microwave network analysis is as follows. We first treat a set of basic, canonical problems rigorously, using field analysis and Maxwell’s equations (as we have done in Chapters 2 and 3, for a variety of transmission line and waveguide problems). Whenso doing, we try to obtain quantities that can be directly related to a circuit or transmissionline parameter. For example, when we treated various transmission lines and waveguides in Chapter 3 we derived the propagation constant and characteristic impedance of the line. This allowed the transmission line or waveguide to be treated as an idealized distributed componentcharacterized by its length, propagation constant, and characteristic impedance. At this point,we can interconnect various components and use network and/or transmission line theory to analyze the behavior of the entire system of components, including effects such as multiple reflections, loss, impedance transformations, and transitions from one type of transmissionmedium to another (e.g., coax to microstrip). As we will see, a transition between differenttransmission lines, or a discontinuity on a transmission line, generally cannot be treated as a simple junction between two transmission lines, but typically includes some type of equivalent circuit to account for reactances associated with the transition or discontinuity. Microwave network theory was originally developed in the service of radar system and component development at the MIT Radiation Lab in the 1940s. This work was continued at the Polytechnic Institute of Brooklyn and other locations by researchers such as E. Weber,N. Marcuvitz, A. A. Oliner, L. B. Felsen, A. Hessel, and many others [2]. 4.1IMPEDANCEANDEQUIVALENTVOLTAGESANDCURRENTS EquivalentVoltagesandCurrents At microwave frequencies the measurement of voltage or current is difficult (or impossi- ble), unless a clearly defined terminal pair is available. Such a terminal pair may be present in the case of TEM-type lines (such as coaxial cable, microstrip line, or stripline), but doesnot strictly exist for non-TEM lines (such as rectangular, circular, or surface waveguides). Figure 4.1 shows the electric and magnetic field lines for an arbitrary two-conductor TEM transmission line. As in Chapter 3, the voltage, V, of the + conductor relative to the −conductor can be found as V=/integraldisplay − +¯E·d¯/lscript, (4.1) where the integration path begins on the + conductor and ends on the −conductor. It is important to realize that, because of the electrostatic nature of the transverse fields between the two conductors, the voltage defined in (4.1) is unique and does not depend on the shapeof the integration path. The total current flowing on the +conductor can be determined from an application of Ampere’s law as I=/contintegraldisplay C+¯H·d¯/lscript, (4.2) c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 4.1 Impedance and Equivalent Voltages and Currents 167 + – E H FIGURE 4.1 Electric and magnetic field lines for an arbitrary two-conductor TEM line. where the integration contour is any closed path enclosing the + conductor (but not the −conductor). A characteristic impedance Z0can then be defined for traveling waves as Z0=V I.( 4.3) At this point, after having defined and determined a voltage, current, and characteristic impedance (and assuming we know the propagation constant for the line), we can proceedto apply the circuit theory for transmission lines developed in Chapter 2 to characterize this line as a circuit element. The situation is more difficult for waveguides. To see why, we will look at the case of a rectangular waveguide, as shown in Figure 4.2. For the dominant TE 10mode, the transverse fields can be written, from Table 3.2, as Ey(x,y,z)=jωµa πAsinπx ae−jβz=Aey(x,y)e−jβz, (4.4a) Hx(x,y,z)=jβa πAsinπx ae−jβz=Ahx(x,y)e−jβz. (4.4b) by x a 0 FIGURE 4.2 Electric field lines for the TE 10mode of a rectangular waveguide. c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 168 Chapter 4: Microwave Network Analysis Applying (4.1) to the electric field of (4.4a) gives V=−jωµa πAsinπx ae−jβz/integraldisplay ydy.( 4.5) Thus it is seen that this voltage depends on the position, x, as well as the length of the integration contour along the ydirection. For example, integrating from y=0t o bfor x=a/2 gives a voltage that is quite different from that obtained by integrating from y=0 tobforx=0. What, then, is the correct voltage? The answer is that there is no “correct” voltage in the sense of being unique or pertinent for all applications. A similar problem arises with current, and also impedance. We will now show how we can define equivalent voltages, currents, and impedances that can be useful for non-TEM lines. There are many ways to define equivalent voltage, current, and impedance for wave- guides since these quantities are not unique for non-TEM lines, but the following consid- erations usually lead to the most useful results [1, 3, 4]: rV oltage and current are defined only for a particular waveguide mode, and are defined so that the voltage is proportional to the transverse electric field and thecurrent is proportional to the transverse magnetic field. rIn order to be useful in a manner similar to voltages and currents of circuit theory,the equivalent voltages and currents should be defined so that their product gives the power flow of the waveguide mode. rThe ratio of the voltage to the current for a single traveling wave should be equal to the characteristic impedance of the line. This impedance may be chosen arbitrarily, but is usually selected as equal to the wave impedance of the line, or else normalized to unity. For an arbitrary waveguide mode with both positively and negatively traveling waves, the transverse fields can be written as ¯Et(x,y,z)=¯e(x,y)(A+e−jβz+A−ejβz)=¯e(x,y) C1/parenleftbig V+e−jβz+V−ejβz/parenrightbig ,(4.6a) ¯Ht(x,y,z)=¯h(x,y)/parenleftbig A+e−jβz−A−ejβz/parenrightbig =¯h(x,y) C2/parenleftbig I+e−jβz−I−ejβz/parenrightbig ,(4.6b) where ¯eand¯hare the transverse field variations of the mode, and A+,A−are the field amplitudes of the traveling waves. Because ¯Etand¯Htare related by the wave impedance, Zw, according to (3.22) or (3.26), we also have that ¯h(x,y)=ˆzׯe(x,y) Zw.( 4.7) Equation (4.6) also defines equivalent voltage and current waves as V(z)=V+e−jβz+V−ejβz, (4.8a) I(z)=I+e−jβz−I−ejβz, (4.8b) with V+/I+=V−/I−=Z0. This definition embodies the idea of making the equivalent voltage and current proportional to the transverse electric and magnetic fields, respectively. The proportionality constants for this relationship are C1=V+/A+=V−/A−andC2= I+/A+=I−/A−, and can be determined from the remaining two conditions for power and impedance. c04MicrowaveNetworkAnalysis Pozar September 12, 2011 17:20 4.1 Impedance and Equivalent Voltages and Currents 169 The complex power flow for the incident wave is given by P+=1 2|A+|2/integraldisplay S¯eׯh∗·ˆzds=V+I+∗ 2C1C∗ 2/integraldisplay S¯eׯh∗·ˆzds.(4.9) Because we want this power to be equal to (1/2) V+I+∗, we have the result that C1C∗ 2=/integraldisplay S¯eׯh∗·ˆzds,( 4.10) where the surface integration is over the cross section of the waveguide. The characteristic impedance is Z0=V+ I+=V− I−=C1 C2,( 4.11) since V+=C1AandI+=C2A, from (4.6a) and (4.6b). If it is desired to have Z0=Zw, the wave impedance ( ZTEorZTM)of the mode, then C1 C2=Zw(ZTEorZTM). (4.12a) Alternatively, it may be desirable to normalize the characteristic impedance to unity (Z0=1), in which case we have C1 C2=1.( 4.12b) For a given waveguide mode, (4.10) and (4.12) can be solved for the constants C1and C2, and equivalent voltages and currents defined. Higher order modes can be treated in the same way, so that a general field in a waveguide can be expressed in the following form: ¯Et(x,y,z)=N/summationdisplay n=1/parenleftbiggV+ n C1ne−jβnz+V− n C1nejβnz/parenrightbigg ¯en(x,y), (4.13a) ¯Ht(x,y,z)=N/summationdisplay n=1/parenleftbiggI+ n C2ne−jβnz−I− n C2nejβnz/parenrightbigg ¯hn(x,y), (4.13b) where V± nandI± nare the equivalent voltages and currents for the nth mode, and C1nand C2nare the proportionality constants for each mode. EXAMPLE 4.1 EQUIV ALENT VOLTAGE AND CURRENT FOR A RECTANGULAR WA VEGUIDE Find the equivalent voltages and currents for a TE 10mode in a rectangular wave- guide. Solution The transverse field components and power flow of the TE 10rectangular wave- guide mode and the equivalent transmission line model of this mode can be written as follows: c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 170 Chapter 4: Microwave Network Analysis Waveguide Fields Transmission Line Model Ey=/parenleftBig A+e−jβz+A−ejβz/parenrightBig sinπx aV(z)=V+e−jβz+V−ejβz Hx=−1 ZTE/parenleftBig A+e−jβz−A−ejβz/parenrightBig sinπx aI(z)=I+e−jβz−I−ejβz =1 Z0/parenleftBig V+e−jβz−V−ejβz/parenrightBig P+=−1 2/integraldisplay SEyH∗ xdxdy =ab 4ZTE|A+|2P+=1 2V+I+∗ We now find the constants C1=V+/A+=V−/A−andC2=I+/A+=I−/A− that relate the equivalent voltages V±and currents I±to the field amplitudes, A±. Equating incident powers gives ab/vextendsingle/vextendsingleA+/vextendsingle/vextendsingle2 4ZTE=1 2V+I+∗=1 2/vextendsingle/vextendsingleA+/vextendsingle/vextendsingle2C1C∗ 2. If we choose Z0=ZTE, then we also have that V+ I+=C1 C2=ZTE. Solving for C1,C2gives C1=/radicalbigg ab 2, C2=1 ZTE/radicalbigg ab 2, which completes the transmission line equivalence for the TE 10mode. ■ TheConceptofImpedance We have used the idea of impedance in several different ways, so it may be useful at this point to summarize this important concept. The term impedance w a sfi r s tu s e db yO l i v e r Heaviside in the nineteenth century to describe the complex ratio V/Iin AC circuits con- sisting of resistors, inductors, and capacitors; the impedance concept quickly became indis- pensable in the analysis of AC circuits. It was then applied to transmission lines, in terms of lumped-element equivalent circuits and the distributed series impedance and shunt ad- mittance of the line. In the 1930s, S. A. Schelkunoff recognized that the impedance conceptcould be extended to electromagnetic fields in a systematic way, and noted that impedance should be regarded as characteristic of the type of field, as well as of the medium [2]. In addition, in relation to the analogy between transmission lines and plane wave propa-gation, impedance may even be dependent on direction. The concept of impedance, then, forms an important link between field theory and transmission line or circuit theory. c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 4.1 Impedance and Equivalent Voltages and Currents 171 We summarize the various types of impedance we have used so far, and their notation: rη=√µ//epsilon1=intrinsic impedance of the medium. This impedance is dependent only on the material parameters of the medium, and is equal to the wave impedance forplane waves. rZw=Et/Ht=1/Yw=wave impedance. This impedance is a characteristic of the particular type of wave. TEM, TM, and TE waves each have different wave impedances (ZTEM,ZTM,ZTE), which may depend on the type of line or guide, the material, and the operating frequency.rZ0=1/Y0=V+/I+=characteristic impedance. Characteristic impedance is the ratio of voltage to current for a traveling wave on a transmission line. Because volt- age and current are uniquely defined for TEM waves, the characteristic impedanceof a TEM wave is unique. TE and TM waves, however, do not have a uniquely defined voltage and current, so the characteristic impedance for such waves may be defined in different ways. EXAMPLE 4.2 APPLICATION OF WA VEGUIDE IMPEDANCE Consider a rectangular waveguide with a=2.286 cm and b=1.016 cm (X-band guide), air filled for z<0 and Rexolite filled (/epsilon1 r=2.54) forz>0, as shown in Figure 4.3. If the operating frequency is 10 GHz, use an equivalent transmission line model to compute the reflection coefficient of a TE 10wave incident on the interface from z<0. Solution The waveguide propagation constants in the air ( z<0) and the dielectric ( z>0) regions are βa=/radicalbigg k2 0−/parenleftBigπ a/parenrightBig2 =158.0m−1, βd=/radicalbigg /epsilon1rk2 0−/parenleftBigπ a/parenrightBig2 =304.1m−1, where k0=209.4m−1. The reader may verify that the TE 10mode is the only propagating mode in ei- ther waveguide region. We can set up an equivalent transmission line for the TE 10 mode in each waveguide, and treat the problem as the reflection of an incidentvoltage wave at the junction of two infinite transmission lines. z = 0z zΓ/H92800/H9280r /H92800TE10 Z0a Z0d FIGURE 4.3 Geometry of a partially filled waveguide and its transmission line equivalent for Example 4.2. c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 172 Chapter 4: Microwave Network Analysis By Example 4.1 and Table 3.2, the equivalent characteristic impedances for the two lines are Z0a=k0η0 βa=(209.4)(377) 158.0=500.0 /Omega1, Z0d=kη βd=k0η0 βd=(209.4)(377) 304.1=259.6 /Omega1. The reflection coefficient seen looking into the dielectric filled region is then /Gamma1=Z0d−Z0a Z0d+Z0a=−0.316. With this result, expressions for the incident, reflected, and transmitted waves can be written in terms of fields, or in terms of equivalent voltages and currents. ■ We now consider the arbitrary one-port network shown in Figure 4.4 and derive a general relation between its impedance properties and electromagnetic energy stored in, and the power dissipated by, the network. The complex power delivered to this network isgiven by (1.91): P=1 2/contintegraldisplay S¯EׯH∗·d¯s=P/lscript+2jω(Wm−We), (4.14) where P/lscriptis real and represents the average power dissipated by the network, and Wm andWerepresent the stored magnetic and electric energy, respectively. Note that the unit normal vector in Figure 4.4 is pointing into the volume. If we define real transverse modal fields ¯eand¯hover the terminal plane of the network such that ¯Et(x,y,z)=V(z)¯e(x,y)e−jβz, (4.15a) ¯Ht(x,y,z)=I(z)¯h(x,y)e−jβz, (4.15b) with a normalization such that /integraldisplay S¯eׯh·d¯s=1, then we can express (4.14) in terms of the terminal voltage and current: P=1 2/integraldisplay SVI∗¯eׯh·d¯s=1 2VI∗.( 4.16) ZinI E, HnS V+ –One-port network ˆ FIGURE 4.4 An arbitrary one-port network. c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 4.1 Impedance and Equivalent Voltages and Currents 173 Then the input impedance is Zin=R+jX=V I=VI∗ |I|2=P 1 2|I|2=P/lscript+2jω(Wm−We) 1 2|I|2.( 4.17) Thus we see that the real part, R, of the input impedance is related to the dissipated power, while the imaginary part, X, is related to the net energy stored in the network. If the net- work is lossless, then P/lscript=0 and R=0. Then Zinis purely imaginary, with a reactance X=4ω(Wm−We) |I|2,( 4.18) which is positive for an inductive load ( Wm>We), and negative for a capacitive load (Wm<We). EvenandOddPropertiesof Z(ω)and/Gamma1(ω) Consider the driving point impedance, Z(ω), at the input port of an electrical network. The voltage and current at this port are related as V(ω)=Z(ω)I(ω). For an arbitrary frequency dependence, we can find the time-domain voltage by taking the inverse Fourier transformofV(ω): v(t)=1 2π/integraldisplay∞ −∞V(ω)ejωtdω. (4.19) Because v(t)must be real, we have that v(t)=v∗(t),o r /integraldisplay∞ −∞V(ω)ejωtdω=/integraldisplay∞ −∞V∗(ω)e−jωtdω=/integraldisplay∞ −∞V∗(−ω) ejωtdω, where the last term was obtained by a change of variable from ωto−ω. This shows that V(ω)must satisfy the relation V(−ω)=V∗(ω), (4.20) which means that Re {V(ω)} is even in ω, while Im {V(ω)} is odd in ω. Similar results hold forI(ω), and for Z(ω)since V∗(−ω)=Z∗(−ω) I∗(−ω)=Z∗(−ω) I(ω)=V(ω)=Z(ω)I(ω). Thus, if Z(ω)=R(ω)+jX(ω), then R(ω)is even in ωand X(ω)is odd in ω. These results can also be inferred from (4.17). Now consider the reflection coefficient at the input port: /Gamma1(ω)=Z(ω)−Z0 Z(ω)+Z0=R(ω)−Z0+jX(ω) R(ω)+Z0+jX(ω).( 4.21) Then /Gamma1(−ω)=R(ω)−Z0−jX(ω) R(ω)+Z0−jX(ω)=/Gamma1∗(ω), (4.22) which shows that the real and imaginary parts of /Gamma1(ω) are even and odd, respectively, inω. Finally, the magnitude of the reflection coefficient is |/Gamma1(ω)|2=/Gamma1(ω)/Gamma1∗(ω)=/Gamma1(ω)/Gamma1( −ω)=|/Gamma1(−ω)|2,( 4.23) which shows that |/Gamma1(ω)|2and|/Gamma1(ω)|are even functions of ω. This result implies that only even series of the form a+bω2+cω4+··· can be used to represent |/Gamma1(ω)|or|/Gamma1(ω)|2. c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 174 Chapter 4: Microwave Network Analysis 4.2IMPEDANCEANDADMITTANCEMATRICES In the previous section we have seen how equivalent voltages and currents can be defined for TEM and non-TEM waves. Once such voltages and currents have been defined at vari- ous points in a microwave network, we can use the impedance and/or admittance matricesof circuit theory to relate these terminal or port quantities to each other, and thus to essen- tially arrive at a matrix description of the network. This type of representation lends itself to the development of equivalent circuits of arbitrary networks, which will be quite usefulwhen we discuss the design of passive components such as couplers and filters. (The term port was introduced by H. A. Wheeler in the 1950s to replace the less descriptive and more cumbersome phrase “two-terminal pair” [2, 3].) We begin by considering an arbitrary N-port microwave network, as depicted in Figure 4.5. The ports in Figure 4.5 may be any type of transmission line or transmission line equivalent of a single propagating waveguide mode. If one of the physical ports of thenetwork is a waveguide supporting more than one propagating mode, additional electri- cal ports can be added to account for these modes. At a specific point on the nth port, a terminal plane, t n, is defined along with equivalent voltages and currents for the incident (V+ n,I+ n)and reflected (V− n,I− n)waves. The terminal planes are important in providing a phase reference for the voltage and current phasors. Now, at the nth terminal plane, the total voltage and current are given by Vn=V+ n+V− n, (4.24a ) In=I+ n−I− n, (4.24b ) as seen from (4.8) when z=0. The impedance matrix [ Z] of the microwave network then relates these voltages and currents:  V 1 V2 ... VN = Z 11 Z12··· Z1N Z21... ...... Z N1··· ··· ZNN  I 1 I2 ... IN , tNt4t3 St2 t1 VN, IN++VN, – IN––V4, – I4– –V4, I4++V3, I3++ V3,– I3––V2, – I2–– V2, I2++ V1, I1++ V1, – I1–– FIGURE 4.5 An arbitrary N-port microwave network. c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 4.2 Impedance and Admittance Matrices 175 or in matrix form as [V]=[ Z][I].( 4.25) Similarly, we can define an admittance matrix [Y] as  I 1 I2 ... IN = Y 11 Y12··· Y1N Y21... ...... Y N1··· ··· YNN  V 1 V2 ... VN , or in matrix form as [I]=[ Y][V].( 4.26) Of course, the [Z]and[Y]matrices are the inverses of each other: [Y]=[ Z]−1.( 4.27) Note that both the [Z]and[Y]matrices relate the total port voltages and currents. From (4.25), we see that Zijcan be found as Zij=Vi Ij/vextendsingle/vextendsingle/vextendsingle/vextendsingle Ik=0f o r k/negationslash=j.( 4.28) In words, (4.28) states that Zijcan be found by driving port jwith the current Ij, open- circuiting all other ports (so Ik=0f o r k/negationslash=j), and measuring the open-circuit voltage at port i. Thus, Ziiis the input impedance seen looking into port iwhen all other ports are open-circuited, and Zijis the transfer impedance between ports iand jwhen all other ports are open-circuited. Similarly, from (4.26), Yijcan be found as Yij=Ii Vj/vextendsingle/vextendsingle/vextendsingle/vextendsingle Vk=0f o r k/negationslash=j,( 4.29) which states that Yijcan be determined by driving port jwith the voltage Vj, short- circuiting all other ports (so Vk=0f o r k/negationslash=j), and measuring the short-circuit current at port i. In general, each ZijorYijelement may be complex. For an arbitrary N-port network, the impedance and admittance matrices are N×Nin size, so there are 2 N2independent quantities or degrees of freedom. In practice, however, many networks are either recipro- cal or lossless, or both. If the network is reciprocal (not containing any active devices or nonreciprocal media, such as ferrites or plasmas), we will show that the impedance andadmittance matrices are symmetric, so that Z ij=Zji, and Yij=Yji. If the network is lossless, we can show that all the ZijorYijelements are purely imaginary. Either of these special cases serves to reduce the number of independent quantities or degrees of freedomthat an N-port network may have. We now derive the above characteristics for reciprocal and lossless networks. ReciprocalNetworks Consider the arbitrary network of Figure 4.5 to be reciprocal (no active devices, ferrites, or plasmas), with short circuits placed at all terminal planes except those of ports 1 and 2. Let ¯E a,¯Haand¯Eb,¯Hbbe the fields anywhere in the network due to two independent sources, c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 176 Chapter 4: Microwave Network Analysis aandb, located somewhere in the network. Then the reciprocity theorem of (1.156) states that/contintegraldisplay S¯EaׯHb·d¯s=/contintegraldisplay S¯EbׯHa·d¯s,( 4.30) where Sis the closed surface along the boundaries of the network and through the terminal planes of the ports. If the boundary walls of the network and transmission lines are metal, then¯Etan=0 on these walls (assuming perfect conductors). If the network or the transmis- sion lines are open structures, like microstrip line or slotline, the boundaries of the network can be taken arbitrarily far from the lines so that ¯Etanis negligible. Then the only nonzero contribution to the integrals of (4.30) come from the cross-sectional areas of ports 1 and 2. From Section 4.1, the fields due to sources aandbcan be evaluated at the terminal planes t1andt2as ¯E1a=V1a¯e1,¯H1a=I1a¯h1, (4.31a) ¯E1b=V1b¯e1,¯H1b=I1b¯h1, (4.31b) ¯E2a=V2a¯e2,¯H2a=I2a¯h2, (4.31c) ¯E2b=V2b¯e2,¯H2b=I2b¯h2, (4.31d) where ¯e1,¯h1and¯e2,¯h2are the transverse modal fields of ports 1 and 2, respectively, and the Vs and Is are the equivalent total voltages and currents. (For instance, ¯E1bis the transverse electric field at terminal plane t1of port 1 due to source b.) Substituting the fields of (4.31) into (4.30) gives (V1aI1b−V1bI1a)/integraldisplay S1¯e1ׯh1·d¯s+(V2aI2b−V2bI2a)/integraldisplay S2¯e2ׯh2·d¯s=0,( 4.32) where S1andS2are the cross-sectional areas at the terminal planes of ports 1 and 2. As in Section 4.1, the equivalent voltages and currents have been defined so that the power through a given port can be expressed as VI∗/2; then, comparing (4.31) to (4.6) implies that C1=C2=1 for each port, so that /integraldisplay S1¯e1ׯh1·d¯s=/integraldisplay S2¯e2ׯh2·d¯s=1.( 4.33) This reduces (4.32) to V1aI1b−V1bI1a+V2aI2b−V2bI2a=0.( 4.34) Now use the 2 ×2 admittance matrix of the (effectively) two-port network to eliminate the Is: I1=Y11V1+Y12V2, I2=Y21V1+Y22V2. Substitution into (4.34) gives (V1aV2b−V1bV2a)(Y12−Y21)=0.( 4.35) Because the sources aandbare independent, the voltages V1a,V1b,V2a, and V2bcan take on arbitrary values. So in order for (4.35) to be satisfied for any choice of sources, we must have Y12=Y21, and since the choice of which ports are labeled as 1 and 2 is arbitrary, we have the general result that Yij=Yji.( 4.36) Then if [ Y] is a symmetric matrix, its inverse, [ Z], is also symmetric. c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 4.2 Impedance and Admittance Matrices 177 LosslessNetworks Now consider a reciprocal lossless N-port junction; we will show that the elements of the impedance and admittance matrices must be pure imaginary. If the network is lossless, then the net real power delivered to the network must be zero. Thus, Re{ Pavg}=0, where Pavg=1 2[V]t[I]∗=1 2([Z][I])t[I]∗=1 2[I]t[Z][I]∗ =1 2(I1Z11I∗ 1+I1Z12I∗ 2+I2Z21I∗ 1+···) =1 2N/summationdisplay n=1N/summationdisplay m=1ImZmnI∗ n. (4.37) We have used the result from matrix algebra that ([A][B])t=[B]t[A]t. Because the Inare independent, we must have the real part of each self term ( InZnnI∗ n)equal to zero, since we could set all port currents equal to zero except for the nth current. So, Re{InZnnI∗ n}=| In|2Re{Znn}=0, or Re{Znn}=0. (4.38) Now let all port currents be zero except for ImandIn. Then (4.37) reduces to Re/braceleftbig (InI∗ m+ImI∗ n)Zmn/bracerightbig =0, since Zmn=Znm. However, ( InI∗ m+ImI∗ n)is a purely real quantity that is, in general, nonzero. Thus we must have that Re{Zmn}=0.( 4.39) Then (4.38) and (4.39) imply that Re {Zmn}=0 for any m,n. The reader can verify that this also leads to an imaginary [Y ] matrix. EXAMPLE 4.3 EV ALUATION OF IMPEDANCE PARAMETERS Find the Zparameters of the two-port T-network shown in Figure 4.6. Solution From (4.28), Z11can be found as the input impedance of port 1 when port 2 is open-circuited: Z11=V1 I1/vextendsingle/vextendsingle/vextendsingle/vextendsingle I2=0=ZA+ZC. + –+ –Port 1Port 2V1 V2ZA ZCZB FIGURE 4.6 A two-port T-network. c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 178 Chapter 4: Microwave Network Analysis The transfer impedance Z12can be found measuring the open-circuit voltage at port 1 when a current I2is applied at port 2. By voltage division, Z12=V1 I2/vextendsingle/vextendsingle/vextendsingle/vextendsingle I1=0=V2 I2ZC ZB+ZC=ZC. The reader can verify that Z21=Z12, indicating that the circuit is reciprocal. Finally, Z22is found as Z22=V2 I2/vextendsingle/vextendsingle/vextendsingle/vextendsingle I1=0=ZB+ZC. ■ 4.3THESCATTERINGMATRIX We have already discussed the difficulty in defining voltages and currents for non-TEM lines. In addition, a practical problem exists when trying to measure voltages and currents at microwave frequencies because direct measurements usually involve the magnitude (inferred from power) and phase of a wave traveling in a given direction or of a standing wave. Thus, equivalent voltages and currents, and the related impedance and admittance matrices, become somewhat of an abstraction when dealing with high-frequency networks.A representation more in accord with direct measurements, and with the ideas of incident, reflected, and transmitted waves, is given by the scattering matrix. Like the impedance or admittance matrix for an N-port network, the scattering matrix provides a complete description of the network as seen at its Nports. While the impedance and admittance matrices relate the total voltages and currents at the ports, the scattering matrix relates the voltage waves incident on the ports to those reflected from the ports.For some components and circuits, the scattering parameters can be calculated using net- work analysis techniques. Otherwise, the scattering parameters can be measured directly with a vector network analyzer; a photograph of a modern network analyzer is shown inFigure 4.7. Once the scattering parameters of the network are known, conversion to other matrix parameters can be performed, if needed. Consider the N-port network shown in Figure 4.5, where V + nis the amplitude of the voltage wave incident on port nandV− nis the amplitude of the voltage wave reflected from port n. The scattering matrix, or [S]matrix, is defined in relation to these incident and reflected voltage waves as  V − 1 V− 2 ... V− N = S 11 S12··· S1N S21... SN1··· SNN ...  V + 1 V+ 2 ... V+ N , or [V −]=[ S][V+].( 4.40) A specific element of the scattering matrix can be determined as Sij=V− i V+ j/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle V+ k=0f o r k/negationslash=j.( 4.41) In words, (4.41) says that Sijis found by driving port jwith an incident wave of voltage V+ jand measuring the reflected wave amplitude V− icoming out of port i. The incident c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 4.3 The Scattering Matrix 179 FIGURE 4.7 Photograph of the Agilent N5247A Programmable Network Analyzer. This instru- ment is used to measure the scattering parameters of RF and microwave networks from 10 MHz to 67 GHz. The instrument is programmable, performs error correc- tion, and has a wide variety of display formats and data conversions. Courtesy of Agilent Technologies. waves on all ports except the jth port are set to zero, which means that all ports should be terminated in matched loads to avoid reflections. Thus, Siiis the reflection coefficient seen looking into port iwhen all other ports are terminated in matched loads, and Sijis the transmission coefficient from port jto port iwhen all other ports are terminated in matched loads. EXAMPLE 4.4 EV ALUATION OF SCATTERING PARAMETERS Find the scattering parameters of the 3 dB attenuator circuit shown in Figure 4.8. Solution From (4.41), S11can be found as the reflection coefficient seen at port 1 when port 2 is terminated in a matched load ( Z0=50/Omega1): S11=V− 1 V+ 1/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle V+ 2=0=/Gamma1(1)|V+ 2=0=Z(1) in−Z0 Z(1) in+Z0/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle Z0on port 2, 8.56 Ω 8.56 Ω 141.8 ΩPort 2Port 1 FIGURE 4.8 A matched 3 dB attenuator with a 50 /Omega1characteristic impedance (Example 4.4). c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 180 Chapter 4: Microwave Network Analysis butZ(1) in=8.56+[141.8(8.56 +50)]/(141.8 +8.56+50)=50/Omega1,s o S11=0. Because of the symmetry of the circuit, S22=0. We can find S21by applying an incident wave at port 1, V+ 1, and measuring the outcoming wave at port 2, V− 2. This is equivalent to the transmission coeffi- cient from port 1 to port 2: S21=V− 2 V+ 1/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle V+ 2=0. From the fact that S11=S22=0, we know that V− 1=0 when port 2 is terminated inZ0=50/Omega1, and that V+ 2=0. In this case we have that V+ 1=V1andV− 2= V2. By applying a voltage V1at port 1 and using voltage division twice we find V− 2=V2as the voltage across the 50 /Omega1load resistor at port 2: V− 2=V2=V1/parenleftbigg41.44 41.44+8.56/parenrightbigg/parenleftbigg50 50+8.56/parenrightbigg =0.707 V1, where 41.44 =141.8(58.56)/(141.8 +58.56) is the resistance of the parallel com- bination of the 50 /Omega1load and the 8.56 /Omega1resistor with the 141.8 /Omega1resistor. Thus, S12=S21=0.707. If the input power is |V+ 1|2/2Z0, then the output power is |V− 2|2/2Z0= |S21V+ 1|2/2Z0=|S21|2/2Z0|V+ 1|2=|V+ 1|2/4Z0, which is one-half ( −3d B )o f the input power. ■ We now show how the scattering matrix can be determined from the [Z](or[Y]) matrix and vice versa. First, we must assume that the characteristic impedances, Z0n,o f all the ports are identical. (This restriction will be removed when we discuss generalizedscattering parameters.) Then, for convenience, we can set Z 0n=1. From (4.24) the total voltage and current at the nth port can be written as Vn=V+ n+V− n, (4.42a) In=I+ n−I− n=V+ n−V− n. (4.42b) Using the definition of [ Z] from (4.25) with (4.42) gives [Z][I]=[ Z][V+]−[ Z][V−]=[ V]=[ V+]+[ V−], which can be rewritten as ([Z]+[ U])[V−]=([Z]−[U])[V+],( 4.43) where [U]is the unit, or identity, matrix defined as [U]= 10 ··· 0 01... ...... 0 ··· 1 . c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 4.3 The Scattering Matrix 181 Comparing (4.43) to (4.40) suggests that [S]=([Z]+[U])−1([Z]−[U]),( 4.44) giving the scattering matrix in terms of the impedance matrix. Note that for a one-port network (4.44) reduces to S11=z11−1 z11+1, in agreement with the result for the reflection coefficient seen looking into a load with a normalized input impedance of z11. To find [Z]in terms of [S], rewrite (4.44) as [Z][S]+[U][S]=[ Z]−[U], and solve for[Z]to give [Z]=([U]+[S])([U]−[S])−1.( 4.45) ReciprocalNetworksandLosslessNetworks As we discussed in Section 4.2, the impedance and admittance matrices are symmetric for reciprocal networks, and are purely imaginary for lossless networks. The scattering matrices for these particular types of networks also have special properties. We will show that the scattering matrix for a reciprocal network is symmetric, and that the scattering matrix for a lossless network is unitary. By adding (4.42a) and (4.42b) we obtain V+ n=1 2(Vn+In), or [V+]=1 2([Z]+[U])[I].( 4.46a) By subtracting (4.42a) and (4.42b) we obtain V− n=1 2(Vn−In), or [V−]=1 2([Z]−[U])[I].( 4.46b) Eliminating [I]from (4.46a) and (4.46b) gives [V−]=([ Z]−[U])([Z]+[U])−1[V+], so that [S]=([Z]−[U])([Z]+[U])−1.( 4.47) Taking the transpose of (4.47) gives [S]t={([Z]+[U])−1}t([Z]−[U])t. Now[U]is diagonal, so [U]t=[U], and if the network is reciprocal, [Z]is symmetric. so that[Z]t=[Z]. The above equation then reduces to [S]t=([Z]+[U])−1([Z]−[U]), c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 182 Chapter 4: Microwave Network Analysis which is equivalent to (4.44). We have thus shown that [S]=[S]t,( 4.48) so the scattering matrix is symmetric for reciprocal networks. If the network is lossless, no real power can be delivered to the network. Thus, if the characteristic impedances of all the ports are identical and assumed to be unity, the average power delivered to the network is Pavg=1 2Re{[V]t[I]∗}=1 2Re{([V+]t+[V−]t)([V+]∗−[V−]∗)} =1 2Re{[V+]t[V+]∗−[V+]t[V−]∗+[V−]t[V+]∗−[V−]t[V−]∗} =1 2[V+]t[V+]∗−1 2[V−]t[V−]∗=0, (4.49) since the terms −[V+]t[V−]∗+[V−]t[V+]∗are of the form A−A∗, and so are purely imaginary. Of the remaining terms in (4.49), (1/2)[ V+]t[V+]∗represents the total inci- dent power, while (1/2)[ V−]t[V−]∗represents the total reflected power. So, for a lossless junction, we have the intuitive result that the incident and reflected powers are equal: [V+]t[V+]∗=[V−]t[V−]∗.( 4.50) Using [V−]=[ S][V+]in (4.50) gives [V+]t[V+]∗=[V+]t[S]t[S]∗[V+]∗, so that, for nonzero [V+], [S]t[S]∗=[U], (4.51) or[S]∗={ [S]t}−1. A matrix that satisfies the condition of (4.51) is called a unitary matrix . The matrix equation of (4.51) can be written in summation form as N/summationdisplay k=1SkiS∗ kj=δij,for all i,j,( 4.52) where δij=1i fi=j, andδij=0i fi/negationslash=j, is the Kronecker delta symbol. Thus, if i=j, (4.52) reduces to N/summationdisplay k=1SkiS∗ ki=1,( 4.53a) while if i/negationslash=j, (4.52) reduces to N/summationdisplay k=1SkiS∗ kj=0,fori/negationslash=j.( 4.53b) In words, (4.53a) states that the dot product of any column of [S]with the conjugate of that same column gives unity, while (4.53b) states that the dot product of any column with the c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 4.3 The Scattering Matrix 183 conjugate of a different column gives zero (the columns are orthonormal). From (4.51) we also have that [S][S]∗t=[U], so the same statements can be made about the rows of the scattering matrix. EXAMPLE 4.5 APPLICATION OF SCATTERING PARAMETERS A two-port network is known to have the following scattering matrix: [S]=/bracketleftbigg 0.15/negationslash0◦0.85/negationslash−45◦ 0.85/negationslash45◦0.2/negationslash0◦/bracketrightbigg Determine if the network is reciprocal and lossless. If port 2 is terminated with a matched load, what is the return loss seen at port 1? If port 2 is terminated with a short circuit, what is the return loss seen at port 1? Solution Because [ S] is not symmetric, the network is not reciprocal. To be lossless, the scattering parameters must satisfy (4.53). Taking the first column [ i=1 in (4.53a)] gives |S11|2+|S21|2=(0.15)2+(0.85)2=0.745 /negationslash=1, so the network is not lossless. When port 2 is terminated with a matched load, the reflection coefficient seen at port 1 is /Gamma1=S11=0.15. So the return loss is RL=−20 log |/Gamma1|=− 20 log(0.15) =16.5d B . When port 2 is terminated with a short circuit, the reflection coefficient seen at port 1 can be found as follows. From the definition of the scattering matrix andthe fact that V + 2=−V− 2(for a short circuit at port 2), we can write V− 1=S11V+ 1+S12V+ 2=S11V+ 1−S12V− 2, V− 2=S21V+ 1+S22V+ 2=S21V+ 1−S22V− 2. The second equation gives V− 2=S21 1+S22V+ 1. Dividing the first equation by V+ 1and using the above result gives the reflection coefficient seen at port 1 as /Gamma1=V− 1 V+ 1=S11−S12V− 2 V+ 1=S11−S12S21 1+S22 =0.15−(0.85 /negationslash−45◦)(0.85 /negationslash45◦) 1+0.2=−0.452. So the return loss is RL =−20 log |/Gamma1|=− 20 log(0.452) =6.9d B . ■ An important point to understand about scattering parameters is that the reflection coefficient looking into port nis not equal to Snnunless all other ports are matched (this c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 184 Chapter 4: Microwave Network Analysis is illustrated in the above example). Similarly, the transmission coefficient from port mto port nis not equal to Snmunless all other ports are matched. The scattering parameters of a network are properties only of the network itself (assuming the network is linear), and are defined under the condition that all ports are matched. Changing the terminationsor excitations of a network does not change its scattering parameters, but may change the reflection coefficient seen at a given port, or the transmission coefficient between two ports. AShiftinReferencePlanes Because scattering parameters relate amplitudes (magnitude and phase) of traveling waves incident on and reflected from a microwave network, phase reference planes must be speci- fied for each port of the network. We now show how scattering parameters are transformed when the reference planes are moved from their original locations. Consider the N-port microwave network shown in Figure 4.9, where the original ter- minal planes are assumed to be located at z n=0f o rt h en th port, where znis an arbitrary coordinate measured along the transmission line feeding the nth port. The scattering matrix for the network with this set of terminal planes is denoted by [S]. Now consider a new set of reference planes defined at zn=/lscriptnfor the nth port, and let the new scattering matrix be denoted as [S/prime]. Then in terms of the incident and reflected port voltages we have that [V−]=[ S][V+], (4.54a) [V/prime−]=[ S/prime][V/prime+], (4.54b) where the unprimed quantities are referenced to the original terminal planes at zn=0, and the primed quantities are referenced to the new terminal planes at zn=/lscriptn. From the theory of traveling waves on lossless transmission lines we can relate the new wave amplitudes to the original ones as V/prime+ n=V+ nejθn, (4.55a) V/prime− n=V− ne−jθn, (4.55b) zn = ln zn = 0z1 = l1 z1 = 0 V'n–V'n+V'1–V'1+ Vn–Vn+V1–V1+ Port 1 N-port network [S], [S'] Port n FIGURE 4.9 Shifting reference planes for an N-port network. c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 4.3 The Scattering Matrix 185 where θn=βn/lscriptnis the electrical length of the outward shift of the reference plane of portn. Writing (4.55) in matrix form and substituting into (4.54a) gives  ejθ1 0 ejθ2 ... 0 ejθN [V/prime−]=[ S] e−jθ1 0 e−jθ2 ... 0 e−jθN [V/prime+]. Multiplying by the inverse of the first matrix on the left gives [V/prime−]= e−jθ1 0 e−jθ2 ... 0 e−jθN [S] e−jθ1 0 e−jθ2 ... 0 e−jθN [V/prime+]. Comparing with (4.54b) shows that [S/prime]= e−jθ1 0 e−jθ2 ... 0 e−jθN [S] e−jθ1 0 e−jθ2 ... 0 e−jθN ,( 4.56) which is the desired result. Note that S/prime nn=e−2jθnSnn, meaning that the phase of Snnis shifted by twice the electrical length of the shift in terminal plane nbecause the wave travels twice over this length upon incidence and reflection. This result is consistent with(2.42), which gives the change in the reflection coefficient on a transmission line due to a shift in the reference plane. PowerWavesandGeneralizedScatteringParameters We previously expressed the total voltage and current on a transmission line in terms of the incident and reflected voltage wave amplitudes, as in (2.34) or (4.42): V=V + 0+V− 0, (4.57a) I=1 Z0/parenleftbig V+ 0−V− 0/parenrightbig , (4.57b) with Z0being the characteristic impedance of the line. Inverting (4.57) gives the incident and reflected voltage wave amplitudes in terms of the total voltage and current: V+ 0=V+Z0I 2, (4.58a) V− 0=V−Z0I 2. (4.58b) The average power delivered to a load can be expressed as PL=1 2Re/braceleftbig VI∗/bracerightbig =1 2Z0Re/braceleftBig/vextendsingle/vextendsingleV+ 0/vextendsingle/vextendsingle2−V+ 0V−∗ 0+V+∗ 0V− 0−/vextendsingle/vextendsingleV− 0/vextendsingle/vextendsingle2/bracerightBig =1 2Z0/parenleftBig/vextendsingle/vextendsingleV+ 0/vextendsingle/vextendsingle2−/vextendsingle/vextendsingleV− 0/vextendsingle/vextendsingle2/parenrightBig , (4.59) where the last step follows because the quantity V+∗ 0V− 0−V+ 0V−∗ 0is pure imaginary. This is a physically satisfying result since it expresses the net power delivered to the load as the difference between the incident and reflected powers. Unfortunately, this result is only c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 186 Chapter 4: Microwave Network Analysis Zg ZL V0I V+ – FIGURE 4.10 A generator with impedance Zgconnected to a load impedance ZL. valid when the characteristic impedance is real; it does not apply when Z0is complex, as in the case of a lossy line. In addition, these results are not useful when no transmission line is present between the generator and load, as in the circuit shown in Figure 4.10. In the circuit of Figure 4.10 there is no defined characteristic impedance, nor is there a voltage reflection coefficient, or incident and reflected voltage or current waves. It is possi- ble, however, to define a new set of waves, called power waves , which have useful proper- ties when dealing with power transfer between a generator and a load, and can be applied to circuits like that of Figure 4.10, as well as to problems with lossless or lossy transmission lines. We will also see how power waves lead to a generalization of scattering parameters. The incident and reflected power wave amplitudes aandbare defined as the following linear transformations of the total voltage and current: a=V+ZRI 2√RR, (4.60a) b=V−Z∗ RI 2√RR, (4.60b) where ZR=RR+jXRis known as the reference impedance, and may be complex. Note that the power wave amplitudes of (4.60) are similar in form to the voltage waves of (4.58), but do not have units of power, voltage, or current. Inverting (4.60) gives the total voltage and current in terms of the power wave ampli- tudes: V=Z∗ Ra+ZRb√RR, (4.61a) I=a−b√RR. (4.61b) Then the power delivered to the load can be expressed as PL=1 2Re/braceleftbig VI∗/bracerightbig =1 2RRRe/braceleftBig Z∗ R|a|2−Z∗ Rab∗+ZRa∗b−ZR|b|2/bracerightBig =1 2|a|2−1 2|b|2, (4.62) since the quantity ZRa∗b−Z∗ Rab∗is pure imaginary. Once again we have the satisfying result that the load power is the difference between the powers of the incident and reflectedpower waves. It is important to note that this result is valid for any reference impedance Z R. The reflection coefficient, /Gamma1p, for the reflected power wave can be found by using (4.60) and the fact that V=ZLIat the load: /Gamma1p=b a=V−Z∗ RI V+ZRI=ZL−Z∗ R ZL+ZR.( 4.63) Observe that this reflection coefficient reduces to our usual voltage reflection coefficient of (2.35) when ZR=Z0is a real characteristic impedance. Equation (4.63) suggests that c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 4.3 The Scattering Matrix 187 choosing the reference impedance as the conjugate of the load impedance [5], ZR=Z∗ L,( 4.64) will have the useful effect of making the reflected power wave amplitude go to zero.1 From basic circuit theory, the voltage, current, and load power for the circuit of Figure 4.10 are V=V0ZL ZL+Zg,I=V0 ZL+Zg,PL=V2 0 2RL/vextendsingle/vextendsingleZL+Zg/vextendsingle/vextendsingle2,( 4.65a, b,c) where ZL=RL+jXL. Then the power wave amplitudes can be found from (4.60), with ZR=Z∗ L,a s a=V+ZRI 2√RR=V0ZL ZL+Zg+Z∗ L ZL+Zg 2√RR=V0√RL ZL+Zg, (4.66a) b=V−Z∗ RI 2√RR=V0ZL ZL+Zg−ZL ZL+Zg 2√RR=0. (4.66b) From (4.62) the power delivered to the load is PL=1 2|a|2=V2 0 2RL/vextendsingle/vextendsingleZL+Zg/vextendsingle/vextendsingle2, in agreement with (4.65c). When the load is conjugately matched to the generator, so that Zg=Z∗ L,w eh a v e PL=V2 0/8RL. Note that selecting the reference impedance as ZR=Z∗ Lresults in the condition that b=0 (and /Gamma1p=0), but this does not necessarily mean that the load is conjugately matched to the generator, nor that maximum power is delivered to the load. The incident power wave amplitude of (4.66a) depends on ZLandZg, and is maximum only when Zg=Z∗ L. To define the scattering matrix for power waves for an N-port network, we assume the reference impedance for port iisZRi. Then, analogous to (4.60), we define the power wave amplitude vectors in terms of the total voltage and current vectors: [a]=[F]([V]+[ZR][I]), (4.67a) [b]=[F]/parenleftbig [V]−[ZR]∗[I]/parenrightbig , (4.67b) where [F]is a diagonal matrix with elements 1 /2√Re{ZRi}and[ZR]is a diagonal matrix with elements ZRi. By the impedance matrix relation that [V]=[Z][I], (4.67) can be written as [b]=[F]/parenleftbig [Z]−[ZR]∗/parenrightbig ([Z]+[ZR])−1[F]−1[a]. Because the scattering matrix for power waves,/bracketleftbig Sp/bracketrightbig , should relate [b]to[a],w eh a v e /bracketleftbig Sp/bracketrightbig =[F]/parenleftbig [Z]−[ZR]∗/parenrightbig ([Z]+[ZR])−1[F]−1.( 4.68) 1Some authors choose the reference impedance equal to the generator impedance. This has the same effect as (4.64) when the generator and load are conjugately matched, but the choice of (4.64) leads to a zero reflectedwave even when the conjugate matching condition is not satisfied, and so can be more useful in general. c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 188 Chapter 4: Microwave Network Analysis The ordinary scattering matrix for a network can first be converted to an impedance matrix, using a relation similar to (4.45), then converted to the generalized power wave scattering matrix using (4.68). The generalized scattering matrix has the useful property that the diagonal elements can be made to be zero by proper selection of the reference impedances. POINT OF INTEREST: The Vector Network Analyzer The scattering parameters of passive and active networks can be measured with a vector network analyzer , which is a two-channel (or four-channel) microwave receiver designed to process the magnitude and phase of the transmitted and reflected waves from the network. A simplified block diagram of a network analyzer is shown in the accompanying figure. In operation, the RF source is usually set to sweep over a specified bandwidth. A four-port reflectometer samples theincident, reflected, and transmitted RF waves; a switch allows the network to be driven from either port 1 or port 2. Four dual-conversion channels convert these signals to 100-kHz IF fre- quencies, which are then detected and converted to digital form. An internal computer is used tocalculate and display the magnitude and phase of the scattering parameters or other quantities that can be derived from these data, such as SWR, return loss, group delay, impedance, etc. An important feature of the network analyzer is the substantial improvement in accuracy made pos-sible with error-correcting software. Errors caused by directional coupler mismatch, imperfect directivity, loss, and variations in the frequency response of the analyzer system are accounted for by using a 12-term error model and a calibration procedure. Another useful feature is theability to determine the time-domain response of the network by calculating the inverse Fourier transform of the frequency-domain data. (S11, S12) (S22, S12)(S21, S22)(S11, S21) Port 2 RF sourcePort 1Device under test FWD REV RF source and test set IF processing Digital processingHarmonic generatorRef.Ref. TestTest20 MHz 1ST IF100 kHz 2ND IF 19.9 MHz Phase lockPanel controlDisplayComputer processing and error correctionRef. det.Sample and holdIF amp and input selector A/D conv.Test det.X Y X Y 4.4THETRANSMISSION (ABCD)MATRIX The Z,Y, and Sparameter representations can be used to characterize a microwave net- work with an arbitrary number of ports, but in practice many microwave networks consist of a cascade connection of two or more two-port networks. In this case it is convenient c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 4.4 The Transmission (ABCD) Matrix189 I1I1 I2 I2 I3 V1V1 + –+ –V2+ – V2+ –V3+ –A1 C1B1 D1A2 C2B2 D2Port 1Port 2A CB D (a) (b) FIGURE 4.11 (a) A two-port network; (b) a cascade connection of two-port networks. to define a 2 ×2transmission,o rABCD, matrix , for each two-port network. We will see that the ABCD matrix of the cascade connection of two or more two-port networks can be easily found by multiplying the ABCD matrices of the individual two-ports. TheABCD matrix is defined for a two-port network in terms of the total voltages and currents as shown in Figure 4.11a and the following: V1=AV2+BI2, I1=CV2+DI2, or in matrix form as /bracketleftbigg V1 I1/bracketrightbigg =/bracketleftbigg AB CD/bracketrightbigg/bracketleftbigg V2 I2/bracketrightbigg .( 4.69) It is important to note from Figure 4.11a that a change in the sign convention of I2 has been made from our previous definitions, which had I2as the current flowing into port 2. The convention that I2flows outof port 2 will be used when dealing with ABCD matrices so that in a cascade network I2will be the same current that flows into the adjacent network, as shown in Figure 4.11b. Then the left-hand side of (4.69) represents the voltage and current at port 1 of the network, while the column on the right-hand side of (4.69) represents the voltage and current at port 2. In the cascade connection of two two-port networks shown in Figure 4.11b we have that /bracketleftbigg V1 I1/bracketrightbigg =/bracketleftbigg A1B1 C1D1/bracketrightbigg/bracketleftbigg V2 I2/bracketrightbigg , (4.70ab )/bracketleftbigg V2 I2/bracketrightbigg =/bracketleftbigg A2B2 C2D2/bracketrightbigg/bracketleftbigg V3 I3/bracketrightbigg . Substituting (4.70b) into (4.70a) gives /bracketleftbigg V1 I1/bracketrightbigg =/bracketleftbigg A1B1 C1C1/bracketrightbigg/bracketleftbigg A2B2 C2D2/bracketrightbigg/bracketleftbigg V3 I3/bracketrightbigg ,( 4.71) which shows that the ABCD matrix of the cascade connection of the two networks is equal to the product of the ABCD matrices representing the individual two-ports. Note that the c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 190 Chapter 4: Microwave Network Analysis TABLE 4.1 ABCD Parameters of Some Useful Two-Port Circuits Circuit ABCD Parameters ZA=1 C=0B=Z D=1 YA=1 C=YB=0 D=1 Z0, lA=cosβ/lscript C=jY0sinβ/lscriptB=jZ0sinβ/lscript D=cosβ/lscript N : 1 A=N C=0B=0 D=1 N Y1 Y2Y3 A=1+Y2 Y3 C=Y1+Y2+Y1Y2 Y3B=1 Y3 D=1+Y1 Y3 Z1 Z2 Z3A=1+Z1 Z3 C=1 Z3B=Z1+Z2+Z1Z2 Z3 D=1+Z2 Z3 order of multiplication of the matrix must be the same as the order in which the networks are arranged since matrix multiplication is not, in general, commutative. The usefulness of the ABCD matrix representation lies in the fact that a library of ABCD matrices for elementary two-port networks can be built up, and applied in building- block fashion to more complicated microwave networks that consist of cascades of thesesimpler two-ports. Table 4.1 lists a number of useful two-port networks and their ABCD matrices. EXAMPLE 4.6 EV ALUATION OF ABCD PARAMETERS Find the ABCD parameters of a two-port network consisting of a series impedance Zbetween ports 1 and 2 (the first entry in Table 4.1). Solution From the defining relations of (4.69), we have that A=V1 V2/vextendsingle/vextendsingle/vextendsingle/vextendsingle I2=0, c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 4.4 The Transmission (ABCD) Matrix191 which indicates that Ais found by applying a voltage V1at port 1, and measuring the open-circuit voltage V2at port 2. Thus, A=1. Similarly, B=V1 I2/vextendsingle/vextendsingle/vextendsingle/vextendsingle V2=0=V1 V1/Z=Z, C=I1 V2/vextendsingle/vextendsingle/vextendsingle/vextendsingle I2=0=0, D=I1 I2/vextendsingle/vextendsingle/vextendsingle/vextendsingle V2=0=I1 I1=1.■ RelationtoImpedanceMatrix The impedance parameters of a network can be easily converted to ABCD parameters. Thus, from the definition of the ABCD parameters in (4.69), and from the defining relations for the Zparameters of (4.25) for a two-port network with I2to be consistent with the sign convention used with ABCD parameters, V1=I1Z11−I2Z12, (4.72a) V2=I1Z21−I2Z22, (4.72b) we have that A=V1 V2/vextendsingle/vextendsingle/vextendsingle/vextendsingle I2=0=I1Z11 I1Z21=Z11/Z21, (4.73a) B=V1 I2/vextendsingle/vextendsingle/vextendsingle/vextendsingle V2=0=I1Z11−I2Z12 I2/vextendsingle/vextendsingle/vextendsingle/vextendsingle V2=0=Z11I1 I2/vextendsingle/vextendsingle/vextendsingle/vextendsingle V2=0−Z12 =Z11I1Z22 I1Z21−Z12=Z11Z22−Z12Z21 Z21, (4.73b) C=I1 V2/vextendsingle/vextendsingle/vextendsingle/vextendsingle I2=0=I1 I1Z21=1/Z21, (4.73c) D=I1 I2/vextendsingle/vextendsingle/vextendsingle/vextendsingle V2=0=I2Z22/Z21 I2=Z22/Z21. (4.73d) If the network is reciprocal, then Z12=Z21and (4.73) can be used to show that AD− BC=1. EquivalentCircuitsforTwo-PortNetworks The special case of a two-port microwave network occurs so frequently in practice that it deserves further attention. Here we will discuss the use of equivalent circuits to represent an arbitrary two-port network. Useful conversions between two-port network parameters are given in Table 4.2. Figure 4.12a shows a transition between a coaxial line and a microstrip line, and is an example of a two-port network. Terminal planes can be defined at arbitrary points onthe two transmission lines; a convenient choice might be as shown in the figure. However, because of the physical discontinuity in the transition from a coaxial line to a microstrip line, electric and/or magnetic energy can be stored in the vicinity of the junction, leadingto reactive effects. Characterization of such effects can be obtained by measurement or by numerical analysis (such analysis may be quite complicated), and represented by the c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 192 Chapter 4: Microwave Network AnalysisTABLE 4.2 Conversions Between Two-Port Network Parameters S Z Y ABCD S11 S11(Z11−Z0)(Z22+Z0)−Z12Z21 /Delta1Z(Y0−Y11)(Y0+Y22)+Y12Y21 /Delta1YA+B/Z0−CZ0−D A+B/Z0+CZ0+D S12 S122Z12Z0 /Delta1Z−2Y12Y0 /Delta1Y2(AD−BC) A+B/Z0+CZ0+D S21 S212Z21Z0 /Delta1Z−2Y21Y0 /Delta1Y2 A+B/Z0+CZ0+D S22 S22(Z11+Z0)(Z22−Z0)−Z12Z21 /Delta1Z(Y0+Y11)(Y0−Y22)+Y12Y21 /Delta1Y−A+B/Z0−CZ0+D A+B/Z0+CZ0+D Z11 Z0(1+S11)(1−S22)+S12S21 (1−S11)(1−S22)−S12S21Z11Y22 |Y|A C Z12 Z02S12 (1−S11)(1−S22)−S12S21Z12−Y12 |Y|AD−BC C Z21 Z02S21 (1−S11)(1−S22)−S12S21Z21−Y21 |Y|1 C Z22 Z0(1−S11)(1+S22)+S12S21 (1−S11)(1−S22)−S12S21Z22Y11 |Y|D C Y11 Y0(1−S11)(1+S22)+S12S21 (1+S11)(1+S22)−S12S21Z22 |Z|Y11D B Y12 Y0−2S12 (1+S11)(1+S22)−S12S21−Z12 |Z|Y12BC−AD B Y21 Y0−2S21 (1+S11)(1+S22)−S12S21−Z21 |Z|Y21−1 B Y22 Y0(1+S11)(1−S22)+S12S21 (1+S11)(1+S22)−S12S21Z11 |Z|Y22A B A(1+S11)(1−S22)+S12S21 2S21Z11 Z21−Y22 Y21A BZ 0(1+S11)(1+S22)−S12S21 2S21|Z| Z21−1 Y21B C1 Z0(1−S11)(1−S22)−S12S21 2S211 Z21−|Y| Y21C D(1−S11)(1+S22)+S12S21 2S21Z22 Z21−Y11 Y21D |Z|=Z11Z22−Z12Z21;| Y|=Y 11Y22−Y12Y21; /Delta1Y=(Y11+Y0)(Y22+Y0)−Y12Y21; /Delta1Z=(Z11+Z0)(Z22+Z0)−Z12Z21; Y0=1/Z0. c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 4.4 The Transmission (ABCD) Matrix193 Z0c Z0m[S] Microstrip lineCoaxial lineZ0m Z0ct2 t1Microstripline Coaxial line/H9280 r (a) (b) (c)C2 C1 Z0c Z0mL FIGURE 4.12 A coax-to-microstrip transition and equivalent circuit representations. (a) Geom- etry of the transition. (b) Representation of the transition by a “black box.” (c) A possible equivalent circuit for the transition [6]. two-port “black box” shown in Figure 4.12b. The properties of the transition can then be expressed in terms of the network parameters ( Z,Y,S,o rABCD) of the two-port network. This type of treatment can be applied to a variety of two-port junctions, such as transitions from one type of transmission line to another, transmission line discontinuities such asstep changes in width or bends, etc. When modeling a microwave junction in this way, it is often useful to replace the two-port “black box” with an equivalent circuit containing a few idealized components, as shown in Figure 4.12c. This is particularly useful if thecomponent values can be related to some physical features of the actual junction. There is an unlimited number of ways in which such equivalent circuits can be defined; we will discuss some of the most common and useful types below. As we have seen, an arbitrary two-port network can be described in terms of impedance parameters as V 1=Z11I1+Z12I2, V2=Z21I1+Z22I2,(4.74a) or in terms of admittance parameters as I1=Y11V1+Y12V2, I2=Y21V1+Y22V2.(4.74b) If the network is reciprocal, then Z12=Z21andY12=Y21. These representations lead naturally to the T and πequivalent circuits shown in Figures 4.13a and 4.13b. The relations in Table 4.2 can be used to relate the component values to other network parameters. Other equivalent circuits can also be used to represent a two-port network. If the network is reciprocal, there are six degrees of freedom (the real and imaginary parts of three matrix elements), so the equivalent circuit should have six independent parameters. c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 194 Chapter 4: Microwave Network Analysis FIGURE 4.13 Equivalent circuits for a reciprocal two-port network. (a) T equivalent. (b) πequi- valent. A nonreciprocal network cannot be represented by a passive equivalent circuit using recip- rocal elements. If the network is lossless, which is a good approximation for many practical two- port junctions, some simplifications can be made in the equivalent circuit. As was shownin Section 4.2, the impedance or admittance matrix elements are purely imaginary for a lossless network. This reduces the degrees of freedom for such a network to three, and implies that the T and πequivalent circuits of Figure 4.13 can be constructed from purely reactive elements. 4.5SIGNALFLOWGRAPHS We have seen how transmitted and reflected waves can be represented by scattering parameters, and how the interconnection of sources, networks, and loads can be treated with various matrix representations. In this section we discuss the signal flow graph, which is an additional technique that is very useful for the analysis of microwave networks in terms of transmitted and reflected waves. We first discuss the features and the construction of the flow graph itself, and then present a technique for the reduction, or solution, of theflow graph. The primary components of a signal flow graph are nodes and branches: rNodes: Each port iof a microwave network has two nodes, aiandbi. Node ai is identified with a wave entering port i, while node biis identified with a wave reflected from port i. The voltage at a node is equal to the sum of all signals entering that node.rBranches: A branch is a directed path between two nodes representing signal flowfrom one node to another. Every branch has an associated scattering parameter or reflection coefficient. At this point it is useful to consider the flow graph of an arbitrary two-port network, as shown in Figure 4.14. Figure 4.14a shows a two-port network with incident and reflected waves at each port, and Figure 4.14b shows the corresponding signal flow graph represen- tation. The flow graph gives an intuitive graphical illustration of the network behavior. For example, a wave of amplitude a 1incident at port 1 is split, with part going through S11and out port 1 as a reflected wave, and part transmitted through S21to node b2. c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 4.5 Signal Flow Graphs 195 Port 2Port 1[S] (a) (b)S22S21 S12S11a1 a2 b2 b1 a1 b2 b1 a2 FIGURE 4.14 The signal flow graph representation of a two-port network. (a) Definition of inci- dent and reflected waves. (b) Signal flow graph. At node b2, the wave goes out port 2; if a load with nonzero reflection coefficient is con- nected at port 2, this wave will be at least partly reflected and reenter the two-port network at node a2. Part of this wave can be reflected back out port 2 via S22, and part can be transmitted out port 1 through S12. Two other special networks—a one-port network and a voltage source—are shown in Figure 4.15, along with their signal flow graph representations. Once a microwave network has been represented in signal flow graph form, it is a relatively easy matter to solve for theratio of any combination of wave amplitudes. We will discuss how this can be done using four basic decomposition rules, but the same results can also be obtained using Mason’s rule from control system theory. DecompositionofSignalFlowGraphs A signal flow graph can be reduced to a single branch between two nodes using the fol- lowing four basic decomposition rules to obtain any desired wave amplitude ratio. rRule 1 (Series Rule). Two branches, whose common node has only one incoming and one outgoing wave (branches in series), may be combined to form a single branch whose coefficient is the product of the coefficients of the original branches. (a) (b)Vi Vs Γs ΓsΓlΓl abb b aa ba Zs FIGURE 4.15 The signal flow graph representations of a one-port network and a source. (a) A one-port network and its flow graph. (b) A source and its flow graph. c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 196 Chapter 4: Microwave Network Analysis V1 V3 V2 V1 V2S21 S21 S32S42 V4 S21 S42 S32 V3V4 V'V3 V2 V1S21 S32S22S21 S32 V3 V2 V1S22S21 S32 2 (d)(c)(b)(a) 1 –V2 V1V2 V1Sa + SbSa SbV3 V1S21S32 V3 V2 V1 FIGURE 4.16 Decomposition rules. (a) Series rule. (b) Parallel rule. (c) Self-loop rule. (d) Split- ting rule. Figure 4.16a shows the flow graphs for this rule. Its derivation follows from the basic relation V3=S32V2=S32S21V1.( 4.75) rRule 2 (Parallel Rule). Two branches from one common node to another common node (branches in parallel) may be combined into a single branch whose coefficient is the sum of the coefficients of the original branches. Figure 4.16b shows the flowgraphs for this rule. The derivation follows from the obvious relation V 2=SaV1+SbV1=(Sa+Sb)V1.( 4.76) rRule 3 (Self-Loop Rule). When a node has a self-loop (a branch that begins and ends on the same node) of coefficient S, the self-loop can be eliminated by multiplying coefficients of the branches feeding that node by 1/(1 −S). Figure 4.16c shows the flow graphs for this rule, which can be derived as follows. From the original networkwe have V 2=S21V1+S22V2, (4.77a) V3=S32V2. (4.77b) Eliminating V2gives V3=S32S21 1−S22V1,( 4.78) which is seen to be the transfer function for the reduced graph of Figure 4.16c. c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 4.5 Signal Flow Graphs 197 FIGURE 4.17 A terminated two-port network. rRule 4 (Splitting Rule). A node may be split into two separate nodes as long as the resulting flow graph contains, once and only once, each combination of separate (not self-loops) input and output branches that connect to the original node. Thisrule is illustrated in Figure 4.16d and follows from the observation that V 4=S42V2=S21S42V1 (4.79) in both the original flow graph and the flow graph with the split node. We now illustrate the use of each of these rules with an example. EXAMPLE 4.7 APPLICATION OF SIGNAL FLOW GRAPH Use signal flow graphs to derive expressions for /Gamma1inand/Gamma1outfor the microwave network shown in Figure 4.17. Solution The signal flow graph for the circuit of Figure 4.17 is shown in Figure 4.18. In terms of node voltages, /Gamma1inis given by the ratio b1/a1. The first two steps of the required decomposition of the flow graph are shown in Figures 4.19a and 4.19b,from which the desired result follows by inspection: /Gamma1 in=b1 a1=S11+S12S21/Gamma1/lscript 1−S22/Gamma1/lscript. Next,/Gamma1outis given by the ratio b2/a2. The first two steps for this decomposition are shown in Figures 4.19c and 4.19d. The desired result is /Gamma1out=b2 a2=S22+S12S21/Gamma1s 1−S11/Gamma1s ■ ApplicationtoThru-Reflect-Lin NetworkAnalyzerCalibration As a further application of signal flow graphs we consider the calibration of a network analyzer using the Thru-Reflect-Line (TRL) technique [7]. The general problem is shown in Figure 4.20, where it is intended to measure the scattering parameters of a two-port device Vs S11 S22 S12S21 b1 a2a1 b2 Γ/lscript Γs1 FIGURE 4.18 Signal flow graph for the two-port network with general source and load impe- dances of Figure 4.17. c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 198 Chapter 4: Microwave Network Analysis S21 S21S22 S22S12 S12S11a1 a2 a2b2 b2b1 b1Γ/lscript S11Γs Γs ΓsΓ/lscript a1 S22 S12 a2b2 b1a1(a)S12S11a1 a2b2 b1Γ/lscript (c)S21 1 – S22Γ/lscript S21 1 – S11Γs (d)(b) FIGURE 4.19 Decompositions of the flow graph of Figure 4.18 to find /Gamma1in=b1/a1and/Gamma1out= b2/a2. (a) Using Rule 4 on node a2. (b) Using Rule 3 for the self-loop at node b2. (c) Using Rule 4 on node b1. (d) Using Rule 3 for the self-loop at node a1. at the indicated reference planes. As discussed in the previous Point of Interest, a network analyzer measures scattering parameters as ratios of complex voltage amplitudes. The pri-mary reference plane for such measurements is generally at some point within the analyzer itself, so the measurement will include losses and phase delays caused by the effects of the connectors, cables, and transitions that must be used to connect the device under test (DUT)to the analyzer. In the block diagram of Figure 4.20 these effects are lumped together in a two-port error box placed at each port between the actual measurement reference plane and the desired reference plane for the two-port DUT. A calibration procedure is used to char- acterize the error boxes before measurement of the DUT; then the actual error-corrected scattering parameters of the DUT can be calculated from the measured data. Measurementof a one-port network can be considered as a reduced version of the two-port case. The simplest way to calibrate a network analyzer is to use three or more known loads, such as shorts, opens, and matched loads. The problem with this approach is that suchstandards are always imperfect to some degree, and therefore introduce errors into the measurement. These errors become increasingly significant at higher frequencies and as the quality of the measurement system improves. The TRL calibration scheme does not FIGURE 4.20 Block diagram of a network analyzer measurement of a two-port device. c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 4.5 Signal Flow Graphs 199 rely on known standard loads, but uses three simple connections to allow the error boxes to be characterized completely. These three connections are shown in Figure 4.21. The Thru connection is made by directly connecting port 1 to port 2 at the desired reference planes. The Reflect connection uses a load having a large reflection coefficient, /Gamma1L, such as a nominal open or short. It is not necessary to know the exact value of /Gamma1L,a st h i sw i l l be determined by the TRL calibration procedure. The Line connection involves connecting ports 1 and 3 together through a length of matched transmission line. It is not necessary toknow the length of the line, and it is not required that the line be lossless; these parameters will be determined by the TRL procedure. We can use signal flow graphs to derive the set of equations necessary to find the scat- tering parameters for the error boxes in the TRL calibration procedure. With reference to Figure 4.20, we will apply the Thru, Reflect, and Line connections at the reference plane for the DUT, and measure the scattering parameters for these three cases at the measurement planes. For simplicity, we assume the same characteristic impedance for ports 1 and 2, and that the error boxes are reciprocal and identical for both ports. The error boxes are charac-terized by a scattering matrix [ S] and, alternatively, by an ABCD matrix. Thus S 21=S12 for both error boxes. Also note that ports 1 and 2 of the error boxes are in opposite posi- tions since they are symmetrically connected, as shown in the figure. To avoid confusion innotation we will denote the measured scattering parameters for the Thru ,Reflect, and Line connections as the [ T], [R], and [ L] matrices, respectively. Figure 4.21a shows the arrangement for the Thru connection and the corresponding signal flow graph. Observe that we have made use of the fact that S 21=S12and that the error boxes are identical and symmetrically arranged. The signal flow graph can be easily reduced using the decomposition rules to give the measured scattering parameters at themeasurement planes in terms of the scattering parameters of the error boxes as T 11=b1 a1/vextendsingle/vextendsingle/vextendsingle/vextendsingle a2=0=S11+S22S2 12 1−S2 22(4.80a) T12=b1 a2/vextendsingle/vextendsingle/vextendsingle/vextendsingle a1=0=S2 12 1−S2 22(4.80b ) By symmetry we have T22=T11, and by reciprocity we have T21=T12. FIGURE 4.21a Block diagram and signal flow graph for the Thru connection. c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 200 Chapter 4: Microwave Network Analysis FIGURE 4.21b Block diagram and signal flow graph for the Reflect connection. TheReflect connection is shown in Figure 4.21b, with the corresponding signal flow graph. Note that this arrangement effectively decouples the two measurement ports, so R12=R21=0. The signal flow graph can be easily reduced to show that R11=b1 a1/vextendsingle/vextendsingle/vextendsingle/vextendsingle a2=0=S11+S2 12/Gamma1L 1−S22/Gamma1L.( 4.81) By symmetry we have R22=R11. The Line connection is shown in Figure 4.21c, with its corresponding signal flow graph. A reduction similar to that used for the Thru case gives L11=b1 a1/vextendsingle/vextendsingle/vextendsingle/vextendsingle a2=0=S11+S22S2 12e−2γ/lscript 1−S2 22e−2γ/lscript, (4.82a) L12=b1 a2/vextendsingle/vextendsingle/vextendsingle/vextendsingle a1=0=S2 12e−γ/lscript 1−S2 22e−2γ/lscript. (4.82b) Reference plane for DUT (c)S12 S11 S22S22 S11S12 S12 S12 e–/H9253le–/H9253lZ0, e–/H9253l 2 b2 a2a1 b1[S][L] lError boxError box A B C D[ ][S] D B C A[ ]1a1 b1b2 a2 FIGURE 4.21c Block diagram and signal flow graph for the Line connection. c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 4.5 Signal Flow Graphs 201 By symmetry and reciprocity we have L22=L11andL21=L12. We now have five equations (4.80)–(4.82) for the five unknowns S11,S12,S22,/Gamma1L, andeγ/lscript; the solution is straightforward but lengthy. Because (4.81) is the only equation that contains /Gamma1L, we can first solve the four equations in (4.80) and (4.82) for the other four unknowns. Equation (4.80b) can be used to eliminate S12from (4.80a) and (4.82), and then S11can be eliminated from (4.80a) and (4.82a). This leaves two equations for S22 andeγ/lscript: L12e2γ/lscript−L12S2 22=T12eγ/lscript−T12S2 22eγ/lscript, (4.83a) e2γ/lscript(T11−S22T12)−T11S2 22=L11/parenleftbig e2γ/lscript−S2 22/parenrightbig −S22T12. (4.83b) Equation (4.83a) can be solved for S22and substituted into (4.83b) to give a quadratic equation for eγ/lscript. Application of the quadratic formula then gives the solution for eγ/lscriptin terms of the measured TRL scattering parameters as eγ/lscript=L2 12+T2 12−(T11−L11)2±/radicalBig/bracketleftbig L2 12+T2 12−(T11−L11)2/bracketrightbig2−4L2 12T2 12 2L12T12.(4.84) The choice of sign can be determined by the requirement that the real and imaginary parts ofγbe positive, or by knowing the phase of /Gamma1L[as determined from (4.83)] to within 180◦. Now multiply (4.80b) by S22and subtract from (4.80a) to get T11=S11+S22T12,( 4.85a) and similarly multiply (4.82b) by S22e−γ/lscriptand subtract from (4.82a) to get L11=S11+S22L12e−γ/lscript.( 4.85b) Eliminating S11from these two equations gives S22in terms of e−γ/lscriptas S22=T11−L11 T12−L12e−γ/lscript.( 4.86) Solving (4.85a) for S11gives S11=T11−S22T12,( 4.87) and solving (4.80b) for S12gives S2 12=T12(1−S2 22). (4.88) Finally, (4.81) can be solved for /Gamma1Lto give /Gamma1L=R11−S11 S2 12+S22(R11−S11).( 4.89) Equations (4.84) and (4.86)–(4.89) give the scattering parameters for the error boxes, as well as the unknown reflection coefficient /Gamma1L(to within the sign), and the propagation factor e−γ/lscript. This completes the calibration procedure for the TRL method. The scattering parameters of the DUT can now be measured at the measurement refer- ence planes shown in Figure 4.20, and corrected using the above TRL error box parameters c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 202 Chapter 4: Microwave Network Analysis to give the scattering parameters at the reference planes of the DUT. Because we are work- ing with a cascade of three two-port networks, it is convenient to use ABCD parameters. Thus, we convert the error box scattering parameters to the corresponding ABCD param- eters, and convert the measured scattering parameters of the cascade to the corresponding AmBmCmDmparameters. If we use A/primeB/primeC/primeD/primeto denote the parameters for the DUT, then we have /bracketleftbigg AmBm CmDm/bracketrightbigg =/bracketleftbigg AB CD/bracketrightbigg/bracketleftbigg A/primeB/prime C/primeD/prime/bracketrightbigg/bracketleftbigg DB CA/bracketrightbigg , where the change in the elements of the last matrix account for the reversal of ports for the error box at port 2 of the DUT (see Problem 4.25). Then the ABCD parameters for the DUT can be determined as /bracketleftbigg A/primeB/prime C/primeD/prime/bracketrightbigg =/bracketleftbigg AB CD/bracketrightbigg−1/bracketleftbigg AmBm CmDm/bracketrightbigg/bracketleftbigg DB CA/bracketrightbigg−1 .( 4.90) POINT OF INTEREST: Computer-Aided Design for Microwave Circuits Computer-aided design (CAD) software packages have become essential tools for the analysis, design, and optimization of RF and microwave circuits and systems. Several microwave CAD products are commercially available, including Microwave Office (Applied Wave Research), ADS (Agilent Technologies), Microwave Studio (CST), Designer (Ansoft), and many others.RF and microwave CAD packages can be divided into two types: those that use “physics-based” solutions, where Maxwell’s equations are numerically solved for physical geometries such as printed circuit geometries or waveguides, and “circuit-based” solutions, which use equivalentcircuits for various elements, including distributed elements, discontinuities, coupled lines, and active devices. Some packages combine these two approaches. Both linear and nonlinear mod- eling, as well as circuit optimization, are generally possible. Although such computer programs can be fast, powerful, and accurate, they cannot serve as a substitute for engineering experience and a good understanding of microwave principles. A typical design process usually begins with specifications or design goals for the circuit or system. Based on previous designs and his or her experience, an engineer can develop an initial design, including specific components and a circuit layout. CAD can then be used to model andanalyze the design, using data for each of the components and including effects such as loss and discontinuities. The software can be used to optimize the design by adjusting some of the circuit parameters to achieve the best performance. If the specifications are not met, the design mayhave to be revised. CAD tools can also be used to study the effects of component tolerances and errors to improve circuit reliability and robustness. When the design meets the specifications, an engineering prototype can be built and tested. If the measured results satisfy the specifications,the design process is completed. Otherwise the design will need to be revised and the procedure repeated. Without CAD tools the design process would require the construction and measurement of laboratory prototypes at each iteration, which is expensive and time consuming. Thus, CAD can greatly decrease the time and cost of a design while enhancing its quality. The simulation and optimization process is especially important for monolithic microwave integrated circuitsbecause these circuits cannot easily be tuned or trimmed after fabrication. CAD techniques are not without limitations, however. Of primary importance is the fact that any computer model is only an approximation to a “real-world” physical circuit and cannotcompletely account for the inevitable differences due to component and fabrication tolerances, surface roughness, spurious coupling, higher order modes, junction discontinuities, thermal effects, and a number of other practical issues that can occur with a physical circuit or device. c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 4.6 Discontinuities and Modal Analysis 203 4.6DISCONTINUITIESANDMODALANALYSIS By either necessity or design, microwave circuits and networks often consist of transmis- sion lines with various types of discontinuities. In some cases discontinuities are an un- avoidable result of mechanical or electrical transitions from one medium to another (e.g.,a junction between two waveguides, or a coax-to-microstrip transition), and the discon- tinuity effect is unwanted but may be significant enough to warrant characterization. In other cases discontinuities may be deliberately introduced into the circuit to perform a cer-tain electrical function (e.g., reactive diaphragms in waveguide, or stubs on a microstrip line for matching or filter circuits). In any event, a transmission line discontinuity can be represented as an equivalent circuit at some point on the transmission line. Depend-ing on the type of discontinuity, the equivalent circuit may be a simple shunt or series element across the line or, in the more general case, a T- or π-equivalent circuit may be required. The component values of an equivalent circuit depend on the parameters of theline and the discontinuity, as well as on the frequency of operation. In some cases the equivalent circuit involves a shift in the phase reference planes on the transmission lines. Once the equivalent circuit of a given discontinuity is known, its effect can be incorporatedinto the analysis or design of the network using the theory developed previously in this chapter. The purpose of the present section is to discuss how equivalent circuits are obtained for transmission line discontinuities; we will see that one approach is to start with a field theory solution to a canonical discontinuity problem and develop a circuit model withcomponent values. This is thus another example of our objective of replacing complicated field analyses with circuit concepts. In other cases, it may be easier to measure the network parameters of an isolated discontinuity. Figures 4.22 and 4.23 show some common transmission line discontinuities and their equivalent circuits. As shown in Figures 4.22a–4.22c, thin metallic diaphragms (or “irises”) can be placed in the cross section of a waveguide to yield equivalent shunt inductance,capacitance, or a resonant combination. Similar effects occur with step changes in the height or width of the waveguide, as shown in Figures 4.22d and 4.22e. Similar disconti- nuities can also be made in circular waveguide. The classic reference for waveguide dis-continuities and their equivalent circuits is the Waveguide Handbook [8]. Some typical microstrip discontinuities and transitions are shown in Figure 4.23; sim- ilar geometries exist for stripline and other printed transmission lines such as slotline, cov-ered microstrip, coplanar waveguide, etc. Although approximate equivalent circuits have been developed for some printed transmission line discontinuities [9], many do not lend themselves to easy or accurate modeling, and must be treated by numerical analysis. Mod- ern CAD tools are usually capable of accurately modeling such problems. ModalAnalysisofan H-PlaneStepinRectangularWaveguide The field analysis of most transmission line discontinuity problems is difficult, and beyond the scope of this book. The technique of waveguide modal analysis, however, is relatively straightforward and similar in principle to the reflection/transmission problems that werediscussed in Chapters 1 and 2. In addition, modal analysis is a rigorous and versatile tech- nique that can be applied to a number of waveguide and coax discontinuity problems, and lends itself well to computer implementation. We will illustrate the technique by applyingit to the problem of finding the equivalent circuit of an H-plane step (change in width) in a rectangular waveguide. c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 204 Chapter 4: Microwave Network Analysis Symmetrical inductive diaphragmAsymmetrical inductive diaphragm Symmetrical capacitive diaphragmAsymmetrical capacitive diaphragm Rectangular resonant irisCircular resonant iris(a) (b) (c) (d) (e)Change in height Change in widthE E Equivalent circuitZ01 Z02Equivalent circuitZ01 Z02Equivalent circuitEquivalent circuitEquivalent circuit FIGURE 4.22 Rectangular waveguide discontinuities. The geometry of the H-plane waveguide step is shown in Figure 4.24. It is assumed that only the dominant TE 10mode is propagating in guide 1 (z<0)and is incident on the junction from z<0. It is also assumed that no modes are propagating in guide 2, although the analysis to follow is still valid if propagation can occur in guide 2. From Section 3.3, the transverse components of the incident TE 10mode can be written, for z<0, as Ei y=sinπx ae−jβa 1z, (4.91a) Hi x=−1 Za 1sinπx ae−jβa 1z, (4.91b) where βa n=/radicalbigg k2 0−/parenleftBignπ a/parenrightBig2 (4.92) c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 4.6 Discontinuities and Modal Analysis 205 Z0cZ0mZ01Z01 Z02 Z02 Z03Z0 Z0 Z01 Z01L1 LL2 L3Z02 Z03CZ02Z0 Cp Cp CL LCg Z0 Z0CZ0 (a) (b) (c) (d) (e)C1 C2Z0c Z0m FIGURE 4.23 Some common microstrip discontinuities. (a) Open-ended microstrip. (b) Gap in microstrip. (c) Change in width. (d) T-junction. (e) Coax-to-microstrip junction. is the propagation constant of the TE n0mode in guide 1 (of width a), and Za n=k0η0 βan(4.93) is the wave impedance of the TE n0mode in guide 1. Because of the discontinuity at z=0 there will be reflected and transmitted waves in both guides, consisting of infinite sets ofTE n0modes in guides 1 and 2. Only the TE 10mode will propagate in guide 1, but higher order modes are also important in this problem because they account for stored energy, c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 206 Chapter 4: Microwave Network Analysis yy x xz cab 0Guide 2Guide 1 FIGURE 4.24 Geometry of an H-plane step (change in width) in a rectangular waveguide. localized near z=0. Because there is no yvariation introduced by this discontinuity, TE nm modes for m/negationslash=0 are not excited, nor are any TM modes. A more general discontinuity, however, may excite such modes. The reflected modes in guide 1 may be written, for z<0, as Er y=∞/summationdisplay n=1Ansinnπx aejβanz, (4.94a) Hr x=∞/summationdisplay n=1An Zansinnπx aejβa nz, (4.94b) where Anis the unknown amplitude coefficient of the reflected TE n0mode in guide 1. The reflection coefficient of the incident TE 10mode is then A1. Similarly, the transmitted modes into guide 2 can be written, for z>0, as Et y=∞/summationdisplay n=1Bnsinnπx ce−jβcnz, (4.95a) Ht x=−∞/summationdisplay n=1Bn Zcnsinnπx ce−jβc nz, (4.95b) where the propagation constant in guide 2 is βc n=/radicalbigg k2 0−/parenleftBignπ c/parenrightBig2 ,( 4.96) and the wave impedance in guide 2 is Zc n=k0η0 βcn.( 4.97) c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 4.6 Discontinuities and Modal Analysis 207 Atz=0, the transverse fields ( Ey,Hx)must be continuous for 0 <x<c; in addi- tion, Eymust be zero for c<x<abecause of the step. Enforcing these boundary condi- tions leads to the following equations: Ey=sinπx a+∞/summationdisplay n=1Ansinnπx a=  ∞/summationdisplay n=1Bnsinnπx cfor 0<x<c, 0f orc<x<a,(4.98a) Hx=−1 Za 1sinπx a+∞/summationdisplay n=1An Zansinnπx a=−∞/summationdisplay n=1Bn Zcnsinnπx cfor 0<x<c.(4.98b) Equations (4.98a) and (4.98b) constitute a doubly infinite set of linear equations for the modal coefficients AnandBn. We will first eliminate the Bnand then truncate the resulting equation to a finite number of terms and solve for the An. Multiplying (4.98a) by sin(m πx/a), integrating from x=0t o a, and using the or- thogonality relations from Appendix D yields a 2δm1+a 2Am=∞/summationdisplay n=1BnImn=∞/summationdisplay k=1BkImk,( 4.99) where Imn=/integraldisplayc x=0sinmπx asinnπx cdx (4.100) is an integral that can be easily evaluated, and δmn=/braceleftbigg1i f m=n 0i f m/negationslash=n(4.101) is the Kronecker delta symbol. Now solve (4.98b) for Bkby multiplying (4.98b) by sin(kπx/c)and integrating from x=0t oc.After using orthogonality relations, we ob- tain −1 Za 1Ik1+∞/summationdisplay n=1An ZanIkn=−cB k 2Zc k.( 4.102) Substituting Bkfrom (4.102) into (4.99) gives an infinite set of linear equations for the An, where m=1,2,... , a 2Am+∞/summationdisplay n=1∞/summationdisplay k=12Zc kImkIknAn cZan=∞/summationdisplay k=12Zc kImkIk1 cZa 1−a 2δm1.( 4.103) For numerical calculation we can truncate these summations to Nterms, which will result inNlinear equations for the first Ncoefficients, An. For example, let N=1. Then (4.103) reduces to a 2A1+2Zc 1I2 11 cZa 1A1=2Zc 1I2 11 cZa 1−a 2.( 4.104) Solving for A1(the reflection coefficient of the incident TE 10mode) gives A1=Z/lscript−Za 1 Z/lscript+Za 1forN=1,( 4.105) c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 208 Chapter 4: Microwave Network Analysis where Z/lscript=4Zc 1I2 11/ac, which looks like an effective load impedance to guide 1. Accuracy is improved by using larger values of Nand leads to a set of equations that can be written in matrix form as [Q][A]=[ P],( 4.106) where [Q]is a square N×Nmatrix of coefficients, Qmn=a 2δmn+N/summationdisplay k=12Zc kImkIkn cZan,( 4.107) [P]is an N×1 column vector of coefficients given by Pm=N/summationdisplay k=12Zc kImkIk1 cZa 1−a 2δm1,( 4.108) and[A]is an N×1 column vector of the coefficients An.A f t e rt h e Anare found, the Bn can be calculated from (4.102), if desired. Equations (4.106)–(4.108) lend themselves well to computer implementation, and Figure 4.25 shows the results of such a calculation for various matrix sizes. If the width cof guide 2 is such that all modes are cut off (evanescent), then no real power can be transmitted into guide 2, and all the incident power is reflected back into guide 1. The evanescent fields on both sides of the discontinuity store reactive power, however, which implies that the step discontinuity and guide 2 beyond the discontinuitylook like a reactance (in this case an inductive reactance) to an incident TE 10mode in guide 1. Thus the equivalent circuit of the H-plane step looks like a shunt inductor at the z=0 plane of guide 1, as shown in Figure 4.22e. The equivalent reactance can be found from the reflection coefficient A1[after solving (4.106)] as X=− jZa 11+A1 1−A1.( 4.109) Figure 4.25 shows the normalized equivalent inductance versus the ratio of the guide widths c/afor a free-space wavelength λ=1.4aand for N=1,2, and 10 equations. The 000.20.40.60.81.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7N = 1 2 10 c/aX/H9261g 2aZ1aaaCalculated data from Marcuvitz [8].Modal analysis using N equations. /H9261 = 1.4a Z1 = ko/H9257o//H92521 FIGURE 4.25 Equivalent inductance of an H-plane asymmetric step. c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 4.6 Discontinuities and Modal Analysis 209 modal analysis results are compared to data from reference [8]. Note that the solution con- verges very quickly (because of the fast exponential decay of the higher order evanescent modes), and that the result using just two modes is very close to the data of reference [8]. The fact that the H-plane step appears inductive is a result of the actual value of the reflection coefficient, A1, but we can verify the inductive nature of the discontinuity by computing the complex power flow into the evanescent modes on either side of the discon- tinuity. For example, the complex power flow into guide 2 can be found as P=/integraldisplayc x=0/integraldisplayb y=0¯EׯH∗/vextendsingle/vextendsingle/vextendsingle/vextendsingle z=0+·ˆzdxdy =−b/integraldisplayc x=0EyH∗ xdx =−b/integraldisplayc x=0/bracketleftBigg∞/summationdisplay n=1Bnsinnπx c/bracketrightBigg/bracketleftBigg −∞/summationdisplay m=1B∗ m Zc∗msinmπx c/bracketrightBigg dx =bc 2∞/summationdisplay n=1|Bn|2 Zc∗n =jbc 2k0η0∞/summationdisplay n=1|Bn|2|βc n|, (4.110) where the orthogonality property of the sine functions was used, as well as (4.95)–(4.97). Equation (4.110) shows that the complex power flow into guide 2 is positive imaginary,implying stored magnetic energy and an inductive reactance. A similar result can be de- rived for the evanescent modes in guide 1; this is left as a problem. POINT OF INTEREST: Microstrip Discontinuity Compensation Because a microstrip circuit is easy to fabricate and allows the convenient integration of pas- sive and active components, many types of microwave circuits and subsystems are made in microstrip form. One problem with microstrip circuits (and other planar circuits) is that the inevitable discontinuities at bends, step changes in widths, and junctions can cause degrada- tion in circuit performance. This is because such discontinuities introduce parasitic reactancesthat can lead to phase and amplitude errors, input and output mismatch, and possibly spurious coupling or radiation. One approach for eliminating such effects is to construct an equivalent circuit for the discontinuity (perhaps by measurement), including it in the design of the circuit,and compensating for its effect by adjusting other circuit parameters (such as line lengths and characteristic impedances, or tuning stubs). Another approach is to minimize the effect of a discontinuity by compensating the discontinuity directly, often by chamfering or mitering theconductor. Consider the case of a bend in a microstrip line. The straightforward right-angle bend shown below has a parasitic discontinuity capacitance caused by the increased conductor areaat the corner of the bend. This effect could be eliminated by making a smooth, “swept” bend with a radius r≥3W, but this takes up more space. Alternatively, the right-angle bend can be compensated by mitering the corner, which has the effect of reducing the excess capacitance atthe bend. As shown later, this technique can be applied to bends of arbitrary angle. The optimum value of the miter length, a, depends on the characteristic impedance and the bend angle, but a value of a=1.8Wis often used in practice. The technique of mitering can also be used to compensate step and T-junction discontinuities, as shown on the next page. c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 210 Chapter 4: Microwave Network Analysis W Right-angle bendWa Mitered bends Mitered T-junction Mitered step Wa W Swept bendr 3 W Reference: T. C. Edwards, Foundations for Microwave Circuit Design, John Wiley & Sons, New York, 1981. 4.7EXCITATIONOFWAVEGUIDES—ELECTRIC ANDMAGNETICCURRENTS So far we have considered the propagation, reflection, and transmission of guided waves in the absence of sources, but obviously the waveguide or transmission line must be coupled to a generator or some other source of power. For TEM or quasi-TEM lines, there is usuallyonly one propagating mode that can be excited by a given source, although there may be reactance (stored energy) associated with a given feed. In the waveguide case, it may be possible for several propagating modes to be excited, along with evanescent modes thatstore energy. In this section we will develop a formalism for determining the excitation of a given waveguide mode due to an arbitrary electric or magnetic current source. This theory can then be used to find the excitation and input impedance of probe and loop feedsand, in the next section, to determine the excitation of waveguides by apertures. CurrentSheetsThatExciteOnlyOneWaveguideMode Consider an infinitely long rectangular waveguide with a transverse sheet of electric surface current density at z=0, as shown in Figure 4.26. First assume that this current has ˆxand ˆycomponents given as ¯J TE s(x,y)=−ˆx2A+ mnnπ bcosmπx asinnπy b+ˆy2A+ mnmπ asinmπx acosnπy b.(4.111) We will show that such a current excites a single TE mnwaveguide mode traveling away from the current source in both the +zand−zdirections. ba Js or Mszxy FIGURE 4.26 An infinitely long rectangular waveguide with surface current densities at z=0. c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 4.7 Excitation of Waveguides—Electric and Magnetic Currents 211 From Table 3.2, the transverse fields for positive and negative traveling TE mnwave- guide modes can be written as E± x=ZTE/parenleftBignπ b/parenrightBig A± mncosmπx asinnπy be∓jβz, (4.112a) E± y=− ZTE/parenleftBigmπ a/parenrightBig A± mnsinmπx acosnπy be∓jβz, (4.112b) H± x=±/parenleftBigmπ a/parenrightBig A± mnsinmπx acosnπy be∓jβz, (4.112c) H± y=±/parenleftBignπ b/parenrightBig A± mncosmπx asinnπy be∓jβz, (4.112d) where the ±notation refers to waves traveling in the +zdirection or −zdirection with amplitude coefficients A+ mnandA− mn, respectively. From (1.36) and (1.37), the following boundary conditions must be satisfied at z=0: (¯E+−¯E−)׈z=0, (4.113a) ˆz×(¯H+−¯H−)=¯Js. (4.113b) Equation (4.112a) states that the transverse components of the electric field must be con- tinuous at z=0, which when applied to (4.112a) and (4.112b), gives A+ mn=A− mn.( 4.114) Equation (4.113b) states that the discontinuity in the transverse magnetic field is equal to the electric surface current density. Thus, the surface current density at z=0m u s tb e ¯Js=ˆy/parenleftbig H+ x−H− x/parenrightbig −ˆx/parenleftbig H+ y−H− y/parenrightbig =−ˆ x2A+ mnnπ bcosmπx asinnπy b+ˆy2A+ mnmπ asinmπx acosnπy b,(4.115) where (4.114) was used. This current is seen to be the same as the current of (4.111), which shows, by the uniqueness theorem, that such a current will excite only the TE mnmode propagating in each direction, since Maxwell’s equations and all boundary conditions are satisfied. The analogous electric current that excites only the TM mnmode can be shown to be ¯JTM s(x,y)=ˆx2B+ mnmπ acosmπx asinnπy b+ˆy2B+ mnnπ bsinmπx acosnπy b.(4.116) It is left as a problem to verify that this current excites TM mnmodes that satisfy the appro- priate boundary conditions. Similar results can be derived for magnetic surface current sheets. From (1.36) and (1.37) the appropriate boundary conditions are (¯E+−¯E−)׈z=¯Ms, (4.117a) ˆz×(¯H+−¯H−)=0. (4.117b) For a magnetic current sheet at z=0, the TE mnwaveguide mode fields of (4.112) must now have continuous HxandHyfield components, due to (4.117b). This results in the condition that A+ mn=− A− mn.( 4.118) c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 212 Chapter 4: Microwave Network Analysis Then applying (4.117a) gives the source current as ¯MTE s=−ˆx2ZTEA+ mnmπ asinmπx acosnπy b−ˆy2ZTEA+ mnnπ bcosmπx asinnπy b. (4.119) The corresponding magnetic surface current that excites only the TM mnmode can be shown to be ¯MTM s=−ˆx2B+ mnnπ bsinmπx acosnπy b+ˆy2B+ mnmπ acosmπx asinnπy b.( 4.120) These results show that a single waveguide mode can be selectively excited, to the exclu- sion of all other modes, by either an electric or magnetic current sheet of the appropriate form. In practice, however, such currents are difficult to generate and are usually onlyapproximated with one or two probes or loops. In this case many modes may be excited, but usually most of these modes are evanescent. ModeExcitationfromanArbitraryElectric orMagneticCurrentSource We now consider the excitation of waveguide modes by an arbitrary electric or magnetic current source [4]. With reference to Figure 4.27, first consider an electric current source ¯Jlocated between two transverse planes at z 1andz2, which generates the fields ¯E+,¯H+ traveling in the +zdirection, and the fields ¯E−,¯H−traveling in the −zdirection. These fields can be expressed in terms of the waveguide modes as follows: ¯E+=/summationdisplay nA+ n¯E+ n=/summationdisplay nA+ n(¯en+ˆzezn)e−jβnz,z>z2, (4.121a) ¯H+=/summationdisplay nA+ n¯H+ n=/summationdisplay nA+ n(¯hn+ˆzhzn)e−jβnz,z>z2, (4.121b) ¯E−=/summationdisplay nA− n¯E− n=/summationdisplay nA− n(¯en−ˆzezn)ejβnz,z<z1, (4.121c) ¯H−=/summationdisplay nA− n¯H− n=/summationdisplay nA− n(−¯hn+ˆzhzn)ejβnz,z<z1, (4.121d) where the single index nis used to represent any possible TE or TM mode. For a given current ¯J, we can determine the unknown amplitude A+ nby using the Lorentz reciprocity theorem of (1.155) with ¯M1=¯M2=0 (since here we are only considering an electric current source), /contintegraldisplay S(¯E1ׯH2−¯E2ׯH1)·d¯s=/integraldisplay V(¯E2·¯J1−¯E1·¯J2)dv, E–, H–E+, H+ J or M z1 z2V z FIGURE 4.27 An arbitrary electric or magnetic current source in an infinitely long waveguide. c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 4.7 Excitation of Waveguides—Electric and Magnetic Currents 213 where Sis a closed surface enclosing the volume V, and ¯Ei,¯Hiare the fields due to the current source ¯Ji(fori=1o r2 ) . To apply the reciprocity theorem to the present problem we let the volume Vbe the region between the waveguide walls and the transverse cross-section planes at z1andz2. Then let ¯E1=¯E±and¯H1=¯H±, depending on whether z≥z2orz≤z1, and let ¯E2,¯H2 be the nth waveguide mode traveling in the negative zdirection: ¯E2=¯E− n=(¯en−ˆzezn)ejβnz, ¯H2=¯H− n=(−¯hn+ˆzhzn)ejβnz. Substitution into the above form of the reciprocity theorem gives, with ¯J1=¯Jand¯J2=0, /contintegraldisplay S(¯E±×¯H− n−¯E− nׯH±)·d¯s=/integraldisplay V¯E− n·¯Jdv. ( 4.122) The portion of the surface integral over the waveguide walls vanishes because the tan- gential electric field is zero there; that is, ¯EׯH·ˆz=¯H·(ˆzׯE)=0 on the waveguide walls. This reduces the integration to the guide cross section, S0, at the planes z1andz2.I n addition, the waveguide modes are orthogonal over the guide cross section: /integraldisplay S0¯E± mׯH± n·d¯s=/integraldisplay S0(¯em±ˆzezn)×(±¯hn+ˆzhzn)·ˆzds =±/integraldisplay S0¯emׯhn·ˆzds=0,form/negationslash=n. (4.123) Using (4.121) and (4.123) then reduces (4.122) to A+ n/integraldisplay z2(¯E+ nׯH− n−¯E− nׯH+ n)·d¯s+A− n/integraldisplay z1(¯E− nׯH− n−¯E− nׯH− n)·d¯s =/integraldisplay V¯E− n·¯Jdv. Because the second integral vanishes, this further reduces to A+ n/integraldisplay z2[(¯en+ˆzezn)×(−¯hn+ˆzhzn)−(¯en−ˆzezn)×(¯hn+ˆzhzn)]·ˆzds =−2A+ n/integraldisplay z2¯enׯhn·ˆzds=/integraldisplay V¯E− n·¯Jdv, or A+ n=−1 Pn/integraldisplay V¯E− n·¯Jdv=−1 Pn/integraldisplay V(¯en−ˆzezn)·¯Jejβnzdv, ( 4.124) where Pn=2/integraldisplay S0¯enׯhn·ˆzds (4.125) is a normalization constant proportional to the power flow of the nth mode. By repeating the above procedure with ¯E2=¯E+ nand¯H2=¯H+ n, we can derive the amplitude of the negatively traveling waves as A− n=−1 Pn/integraldisplay V¯E+ n·¯Jdv=−1 Pn/integraldisplay V(¯en+ˆzezn)·¯Je−jβnzdv. ( 4.126) c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 214 Chapter 4: Microwave Network Analysis These results are quite general, being applicable to any type of waveguide (includ- ing planar lines such as stripline and microstrip), where modal fields can be defined. Ex- ample 4.8 applies this theory to the problem of a probe-fed rectangular waveguide. EXAMPLE 4.8 PROBE-FED RECTANGULAR WA VEGUIDE For the probe-fed rectangular waveguide shown in Figure 4.28, determine the amplitudes of the forward and backward traveling TE 10modes, and the input resistance seen by the probe. Assume that the TE 10mode is the only propagating mode. Solution If the current probe is assumed to have an infinitesimal diameter, the source vol- ume current density ¯Jcan be written as ¯J(x,y,z)=I0δ/parenleftBig x−a 2/parenrightBig δ(z)ˆyfor 0≤y≤b. From Chapter 3 the TE 10modal fields can be written as ¯e1=ˆysinπx a, ¯h1=−ˆx Z1sinπx a, where Z1=k0η0/β1is the TE 10wave impedance. From (4.125) the normaliza- tion constant P1is P1=2 Z1/integraldisplaya x=0/integraldisplayb y=0sin2πx adxdy =ab Z1. Then from (4.124) the amplitude A+ 1is A+ 1=−1 P1/integraldisplay Vsinπx aejβ1zI0δ/parenleftBig x−a 2/parenrightBig δ(z)dxdydz =−I0b P1=−Z1I0 a. Similarly, A− 1=−Z1I0 a. I0 x aby FIGURE 4.28 A uniform current probe in a rectangular waveguide. c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 4.8 Excitation of Waveguides—Aperture Coupling 215 If the TE 10mode is the only propagating mode in the waveguide, then this mode carries all of the average power, which can be calculated for real Z1as P=1 2/integraldisplay S0¯E+ׯH+∗·d¯s+1 2/integraldisplay S0¯E−ׯH−∗·d¯s =/integraldisplay S0¯E+ׯH+∗·d¯s =/integraldisplaya x=0/integraldisplayb y=0|A+ 1|2 Z1sin2πx adxdy =ab|A+ 1|2 2Z1. If the input resistance seen looking into the probe is Rin, and the terminal current isI0, then P=I2 0Rin/2, so that the input resistance is Rin=2P I2 0=ab|A+ 1|2 I2 0Z1=bZ1 a, which is real for real Z1(corresponding to a propagating TE 10mode). ■ A similar derivation can be carried out for a magnetic current source ¯M(e.g., a small loop). This source will also generate positively and negatively traveling waves, which can be expressed as a superposition of waveguide modes, as in (4.121). For ¯J1=¯J2=0, the reciprocity theorem of (1.155) reduces to /contintegraldisplay S(¯E1ׯH2−¯E2ׯH1)·d¯s=/integraldisplay V(¯H1·¯M2−¯H2·¯M1)dv. ( 4.127) By following the same procedure as for the electric current case, we can derive the excita- tion coefficients of the nth waveguide mode as A+ n=1 Pn/integraldisplay V¯H− n·¯Mdv=1 Pn/integraldisplay V(−¯hn+ˆzhzn)·¯Mejβnzdv, (4.128) A− n=1 Pn/integraldisplay V¯H+ n·¯Mdv=1 Pn/integraldisplay V(¯hn+ˆzhzn)·¯Me−jβnzdv, (4.129) where Pnis defined in (4.125). 4.8EXCITATIONOFWAVEGUIDES—APERTURECOUPLING Besides the probe and loop feeds of the previous section, waveguides and other transmis- sion lines can also be coupled through small apertures. One common application of such coupling is in directional couplers and power dividers, where power from one guide is coupled to another guide through small apertures in a common wall. Figure 4.29 shows a variety of waveguide and other transmission line configurations in which aperture cou-pling can be employed. We will first develop an intuitive explanation for the fact that a small aperture can be represented as an infinitesimal electric and/or magnetic dipole, then we will use the results of Section 4.7 to find the fields generated by these equivalent cur-rents. Our analysis will be somewhat phenomenological [4, 10]; a more advanced theory of aperture coupling based on the equivalence theorem can be found in reference [11]. c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 216 Chapter 4: Microwave Network Analysis Coupling aperture Coupling aperture Ground planeWaveguide StriplineMicrostrip 1 Microstrip 2Waveguide 1Feed waveguideCavity Waveguide 2 (a) (b) (d) (c)/H9280r /H9280r/H9280r FIGURE 4.29 Various waveguide and other transmission line configurations using aperture cou- pling. (a) Coupling between two waveguides via an aperture in the common broad wall. (b) Coupling to a waveguide cavity via an aperture in a transverse wall. (c) Coupling between two microstrip lines via an aperture in the common groundplane. (d) Coupling from a waveguide to a stripline via an aperture. Consider Figure 4.30a, which shows the normal electric field lines near a conducting wall (the tangential electric field is zero near the wall). If a small aperture is cut into theconductor, the electric field lines will fringe through and around the aperture as shown in Figure 4.30b. Now consider Figure 4.30c, which shows the fringing field lines around two infinitesimal electric polarization currents, ¯P e, normal to a conducting wall (without (a) (d)Hn(c) (b)E Pe Pm (e) (f)n ˆˆ FIGURE 4.30 Illustrating the development of equivalent electric and magnetic polarization cur- rents at an aperture in a conducting wall. (a) Normal electric field at a conductingwall. (b) Electric field lines around an aperture in a conducting wall. (c) Elec- tric field lines around electric polarization currents normal to a conducting wall. (d) Magnetic field lines near a conducting wall. (e) Magnetic field lines near anaperture in a conducting wall. (f) Magnetic field lines near magnetic polarization currents parallel to a conducting wall. c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 4.8 Excitation of Waveguides—Aperture Coupling 217 an aperture). The similarity of the field lines of Figures 4.30c and 4.30b suggests that an aperture excited by a normal electric field can be represented by two oppositely directed infinitesimal electric polarization currents, ¯Pe, normal to the closed conducting wall. The strength of this polarization current is proportional to the normal electric field; thus, ¯Pe=/epsilon10αeˆnEnδ(x−x0)δ(y−y0)δ(z−z0), (4.130) where the proportionality constant αeis defined as the electric polarizability of the aper- ture, and ( x0,y0,z0)are the coordinates of the center of the aperture. Similarly, Figure 4.30e shows the fringing of tangential magnetic field lines (the nor- mal magnetic field is zero at the conductor) near a small aperture. Because these field lines are similar to those produced by two magnetic polarization currents located parallel to the conducting wall (as shown in Figure 4.30f), we can conclude that the aperture can be replaced by two oppositely directed infinitesimal polarization currents, ¯Pm, where ¯Pm=−αm¯Htδ(x−x0)δ(y−y0)δ(z−z0). (4.131) In (4.131), αmis defined as the magnetic polarizability of the aperture. The electric and magnetic polarizabilities are constants that depend on the size and shape of the aperture and have been derived for a variety of simple shapes [3, 10, 11]. The polarizabilities for circular and rectangular apertures, which are probably the most commonly used shapes, are given in Table 4.3. We now show that the electric and magnetic polarization currents, ¯Peand¯Pm, can be related to electric and magnetic current sources, ¯Jand¯M, respectively. From Maxwell’s equations (1.27a) and (1.27b) we have /triangleinvׯE=− jωµ¯H−¯M, (4.132a) /triangleinvׯH=jω/epsilon1¯E+¯J. (4.132b ) Then using (1.15) and (1.23), which define ¯Peand¯Pm, we obtain /triangleinvׯE=− jωµ0¯H−jωµ0¯Pm−¯M, (4.133a) /triangleinvׯH=jω/epsilon10¯E+jω¯Pe+¯J. (4.133b) Thus, since ¯Mhas the same role in these equations as jωµ0¯Pm, and ¯Jhas the same role asjω¯Pe, we can define equivalent currents as ¯J=jω¯Pe, (4.134a) ¯M=jωµ0¯Pm. (4.134b) These results allow us to use the formulas of (4.124), (4.126), (4.128), and (4.129) to compute the fields from these currents. TABLE 4.3 Electric and Magnetic Polarizations Aperture Shape αe αm Round hole2r3 0 34r3 0 3 Rectangular slotπ/lscriptd2 16π/lscriptd2 16(¯Hacross slot) c04MicrowaveNetworkAnalysis Pozar September 12, 2011 17:20 218 Chapter 4: Microwave Network Analysis The above theory is approximate because of various assumptions involved in the evaluation of the polarizabilities, but generally it gives reasonable results for apertures that are small (where the term small implies small relative to an electrical wavelength), and not located too close to edges or corners of the guide. In addition, it is important to realize that the equivalent dipolesgiven by (4.130) and (4.131) radiate in the presence of the conducting wall to give the fields transmitted through the aperture. The fields on the input side of the conducting wall are also affected by the presence of the aperture, and this effect is accounted for by the equivalent dipoleson the incident side of the conductor (which are the negative of those on the output side). In this way, continuity of tangential fields is preserved across the aperture. In both cases, the presence of the (closed) conducting wall can be accounted for by using image theory to remove the wall and double the strength of the dipoles. These details will be clarified by applying this theory to apertures in transverse and broad walls of waveguides. CouplingThroughanApertureinaTransverseWaveguideWall Consider a small circular aperture centered in the transverse wall of a waveguide, as shown in Figure 4.31a. Assume that only the TE 10mode propagates in the guide, and is incident on the transverse wall from z<0. Then, if the aperture is assumed to be closed, as in Figure 4.31b, the standing wave fields in the region z<0 can be written as Ey=A/parenleftbig e−jβz−ejβz/parenrightbig sinπx a, (4.135a) Hx=−A Z10/parenleftbig e−jβz+ejβz/parenrightbig sinπx a, (4.135b) where βandZ10are the propagation constant and wave impedance of the TE 10mode. From (4.130) and (4.131) we can determine the equivalent electric and magnetic polarizationcurrents from the above fields as ¯P e=ˆz/epsilon10αeEzδ/parenleftBig x−a 2/parenrightBig δ/parenleftbigg y−b 2/parenrightbigg δ(z)=0, (4.136a) ¯Pm=−ˆ xαmHxδ/parenleftBig x−a 2/parenrightBig δ/parenleftbigg y−b 2/parenrightbigg δ(z) =ˆx2Aαm Z10δ/parenleftBig x−a 2/parenrightBig δ/parenleftbigg y−b 2/parenrightbigg δ(z), (4.136b) since Ez=0 for a TE mode. Now, by (4.134b), the magnetic polarization current ¯Pmis equivalent to the magnetic current density ¯M=jωµ0¯Pm=ˆx2jωµ0Aαm Z10δ/parenleftBig x−a 2/parenrightBig δ/parenleftbigg y−b 2/parenrightbigg δ(z). (4.137) As shown in Figure 4.31d, the fields scattered by the aperture are considered as being produced by the equivalent currents ¯Pmand−¯Pmon either side of the closed wall. The presence of the conducting wall is easily accounted for using image theory, which hasthe effect of doubling the dipole strengths and removing the wall, as depicted in Figure 4.31e (for z<0)and Figure 4.31f (for z>0). Thus the coefficients of the transmitted and reflected waves caused by the equivalent aperture currents can be found by using (4.137)in (4.128) and (4.129) to give A + 10=−1 P10/integraldisplay ¯h10·(2jωµ0¯Pm)dv=4jAωµ0αm abZ 10=4jAβαm ab, (4.138a) A− 10=−1 P10/integraldisplay ¯h10·(−2jωµ0¯Pm)dv=4jAωµ0αm abZ 10=4jAβαm ab,(4.138b) c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 4.8 Excitation of Waveguides—Aperture Coupling 219 (e)(d)(c)(b)(a) –2Pm (f)yy 2PmPm –Pm y z zz E–, H–E– H– E+, H+E+, H+y zzz x ay b b/2 a/20 E + E– H + H–E, H E+, H+yy 2r0 FIGURE 4.31 Applying small-hole coupling theory and image theory to the problem of an aper- ture in the transverse wall of a waveguide. (a) Geometry of a circular aperture in the transverse wall of a waveguide. (b) Fields with aperture closed. (c) Fields with aperture open. (d) Fields with aperture closed and replaced with equivalent dipoles. (e) Fields radiated by equivalent dipoles for z<0; wall removed by image theory. (f) Fields radiated by equivalent dipoles for z>0; wall removed by image theory. since ¯h10=(−ˆx/Z10)sin(πx/a), and P10=ab/Z10. The magnetic polarizability αmis given in Table 4.3. The complete fields can now be written as Ey=/bracketleftbig Ae−jβz+(A− 10−A)ejβz/bracketrightbig sinπx a, forz<0, (4.139a) Hx=1 Z10[−Ae−jβz+(A− 10−A)ejβz]sinπx a,forz<0, (4.139b) and Ey=A+ 10e−jβzsinπx a, forz>0, (4.140a) Hx=−A+ 10 Z10e−jβzsinπx a, forz>0. (4.140b) c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 220 Chapter 4: Microwave Network Analysis jB Z10 Z10 z = 0 z FIGURE 4.32 Equivalent circuit of the aperture in a transverse waveguide wall. Then the reflection and transmission coefficients can be found as /Gamma1=A− 10−A A=4jβαm ab−1, (4.141a) T=A+ 10 A=4jβαm ab, (4.141b ) since Z10=k0η0/β. Note that |/Gamma1|>1; this physically unrealizable result (for a passive network) is an artifact of the approximations used in the above theory. An equivalent circuit for this problem can be obtained by comparing the reflection coefficient of (4.141a) with that of the transmission line with a normalized shunt susceptance, jB, shown in Figure 4.32. The reflection coefficient seen looking into this line is /Gamma1=1−yin 1+yin=1−(1+jB) 1+(1+jB)=−jB 2+jB. If the shunt susceptance is very large (low impedance), /Gamma1can be approximated as /Gamma1=−1 1+(2/jB)/similarequal−1−j2 B. Comparison with (4.141a) suggests that the aperture is equivalent to a normalized inductive susceptance, B=−ab 2βα m. CouplingThroughanApertureintheBroadWallofaWaveguide Another common configuration for aperture coupling is shown in Figure 4.33, where two parallel waveguides share a common broad wall and are coupled with a small centered aperture. We will assume a TE 10mode incident from z<0 in the lower guide (guide 1), 4 3 2 1 x z az y b2b a/20 FIGURE 4.33 Two parallel waveguides coupled through an aperture in a common broad wall. c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 4.8 Excitation of Waveguides—Aperture Coupling 221 and compute the fields coupled to the upper guide. The incident fields can be written as Ey=Asinπx ae−jβz, (4.142a) Hx=−A Z10sinπx ae−jβz. (4.142b) The excitation field at the center of the aperture at ( x=a/2,y=b,z=0)is Ey=A, (4.143a) Hx=−A Z10. (4.143b) (If the aperture were not centered at x=a/2, the Hzfield would be nonzero and would have to be included.) From (4.130), (4.131), and (4.134), the equivalent electric and magnetic dipoles for coupling to the fields in the upper guide are Jy=jω/epsilon10αeAδ/parenleftBig x−a 2/parenrightBig δ(y−b)δ(z), (4.144a) Mx=jωµ0αmA Z10δ/parenleftBig x−a 2/parenrightBig δ(y−b)δ(z). (4.144b) Note that in this case we have excited both an electric and a magnetic dipole. Let the fields in the upper guide be expressed as E− y=A−sinπx ae+jβzforz<0, (4.145a) H− x=A− Z10sinπx ae+jβzforz<0, (4.145b) E+ y=A+sinπx ae−jβzforz>0, (4.146a) H+ x=−A+ Z10sinπx ae−jβzforz>0, (4.146b) where A+,A−are the unknown amplitudes of the forward and backward traveling waves in the upper guide, respectively. By superposition, the total fields in the upper guide due to the electric and magnetic currents of (4.144) can be found from (4.124) and (4.128) for the forward wave as A+=−1 P10/integraldisplay V/parenleftbig E− yJy−H− xMx/parenrightbig dv=−jωA P10/parenleftBigg /epsilon10αe−µ0αm Z2 10/parenrightBigg ,( 4.147a) and from (4.126) and (4.129) for the backward wave as A−=−1 P10/integraldisplay V/parenleftbig E+ yJy−H+ xMx/parenrightbig dv=−jωA P10/parenleftBigg /epsilon10αe+µ0αm Z2 10/parenrightBigg ,( 4.147b) where P10=ab/Z10. Note that the electric dipole excites the same fields in both direc- tions, but the magnetic dipole excites oppositely polarized fields in the forward and back- ward directions. c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 222 Chapter 4: Microwave Network Analysis REFERENCES [1] S. Ramo, T. R. Whinnery, and T. van Duzer, Fields and Waves in Communication Electronics, John Wiley & Sons, New York, 1965. [2] A. A. Oliner, “Historical Perspectives on Microwave Field Theory,” IEEE Transactions on Microwave Theory and Techniques, vol. MTT-32, pp. 1022–1045, September 1984. [3] C. G. Montgomery, R. H. Dicke, and E. M. Purcell, eds., Principles of Microwave Circuits ,M I T Radiation Laboratory Series, V ol. 8, McGraw-Hill, New York, 1948. [4] R. E. Collin, Foundations for Microwave Engineering , 2nd edition, McGraw-Hill, New York, 1992. [5] J. Rahola, “Power Waves and Conjugate Matching,” IEEE Transactions on Circuits and Systems, vol. 55, pp. 92–96, January 2008. [6] J. S. Wright, O. P. Jain, W. J. Chudobiak, and V . Makios, “Equivalent Circuits of Microstrip Impedance Discontinuities and Launchers,” IEEE Transactions on Microwave Theory and Tech- niques, vol. MTT-22, pp. 48–52, January 1974. [7] G. F. Engen and C. A. Hoer, “Thru-Reflect-Line: An Improved Technique for Calibrating the Dual Six-Port Automatic Network Analyzer,” IEEE Transactions on Microwave Theory and Techniques , vol. MTT-27, pp. 987–998, December 1979. [8] N. Marcuvitz, ed., Waveguide Handbook , MIT Radiation Laboratory Series, V ol. 10, McGraw-Hill, New York, 1948. [9] K. C. Gupta, R. Garg, and I. J. Bahl, Microstrip Lines and Slotlines , Artech House, Dedham, Mass., 1979. [10] G. Matthaei, L. Young, and E. M. T. Jones, Microwave Filters, Impedance-Matching Networks, and Coupling Structures, Artech House, Dedham, Mass., 1980, Chapter 5. [11] R. E. Collin, Field Theory of Guided Waves , McGraw-Hill, New York, 1960. PROBLEMS 4.1 Consider the reflection of a TE 10mode, incident from z<0, at a step change in the height of a rectangular waveguide, as shown below. Show that if the method of Example 4.2 is used, the result /Gamma1=0 is obtained. Do you think this is the correct solution? Why? (This problem shows that the one-mode impedance viewpoint does not always provide a correct analysis.) y x za b z = 0b/2 4.2 Consider a series RLC circuit with a current I. Calculate the power lost and the stored electric and magnetic energies, and show that the input impedance can be expressed as in (4.17). 4.3 Show that the input impedance Zof a parallel RLC circuit satisfies the condition that Z(−ω)= Z∗(ω). 4.4 A two-port network is driven at both ports such that the port voltages and currents have the following values ( Z0=50/Omega1): V1=10/negationslash90◦,I1=0.2/negationslash90◦, V2=8/negationslash0◦, I2=0.16/negationslash−90◦. Determine the input impedance seen at each port, and find the incident and reflected voltages at each port. c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 Problems 223 4.5 Show that the admittance matrix of a lossless N-port network has purely imaginary elements. 4.6 Does a nonreciprocal lossless network always have a purely imaginary impedance matrix? 4.7 Derive the [Z]and[Y]matrices for the two-port networks shown in the figure below. (a) (b)Port 1Port 2Port 1Port 2ZA YBYA YA ZAZB 4.8 Consider a two-port network, and let Z(1) SC,Z(2) SC,Z(1) OC,andZ(2) OCbe the input impedance seen when port 2 is short-circuited, when port 1 is short-circuited, when port 2 is open-circuited, and when port 1 is open-circuited, respectively. Show that the impedance matrix elements are given by Z11=Z(1) OC,Z22=Z(2) OC,Z2 12=Z2 21=/parenleftBig Z(1) OC−Z(1) SC/parenrightBig Z(2) OC. 4.9 Find the impedance parameters of a section of transmission line with length /lscript, characteristic impedance Z0, and propagation constant β. 4.10 Show that the admittance matrix of the two parallel-connected two-port πnetworks shown below can be found by adding the admittance matrices of the individual two-ports. Apply this result to find the admittance matrix of the bridged-T circuit shown. What is the corresponding result for the impedance matrix of two series-connected T-networks? 4.11 Find the scattering parameters for the series and shunt loads shown below. Show that S12=1−S11 for the series case, and that S12=1+S11for the shunt case. Assume a characteristic impedance Z0. Port 1Port 2Z Port 1Port 2Z 4.12 Consider two two-port networks with individual scattering matrices [SA]and[SB]. Show that the overall S21parameter of the cascade of these networks is given by S21=SA 21SB 21 1−SA 22SB 11. 4.13 Consider a lossless two-port network. (a) If the network is reciprocal, show that |S21|2=1−|S11|2. (b) If the network is nonreciprocal, show that it is impossible to have unidirectional transmission,where S 12=0a n d S21/negationslash=0. c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 224 Chapter 4: Microwave Network Analysis 4.14 A four-port network has the scattering matrix shown as follows. (a) Is this network lossless? (b) Is this network reciprocal? (c) What is the return loss at port 1 when all other ports are terminated with matched loads? (d) What is the insertion loss and phase delay between ports 2 and 4 when all other ports are terminated with matched loads? (e) What is the reflection coefficient seen at port 1 if a shortcircuit is placed at the terminal plane of port 3 and all other ports are terminated with matched loads? [S]= 0.178 /negationslash90◦0.6/negationslash45◦0.4/negationslash45◦0 0.6/negationslash45◦00 0 .3/negationslash−45◦ 0.4/negationslash45◦00 0 .5/negationslash−45◦ 00 .3/negationslash−45◦0.5/negationslash−45◦0 . 4.15 Show that it is impossible to construct a three-port network that is lossless, reciprocal, and matched at all ports. Is it possible to construct a nonreciprocal three-port network that is lossless and matched at all ports? 4.16 Prove the following decoupling theorem: For any lossless reciprocal three-port network, one port (say port 3) can be terminated in a reactance so that the other two ports (say ports 1 and 2) are decoupled (no power flow from port 1 to port 2, or from port 2 to port 1). 4.17 A certain three-port network is lossless and reciprocal, and has S13=S23andS11=S22. Show that if port 2 is terminated with a matched load, then port 1 can be matched by placing an appropriate reactance at port 3. 4.18 A four-port network has the scattering matrix shown as follows. If ports 3 and 4 are connected with a lossless matched transmission line with an electrical length of 45◦, find the resulting insertion loss and phase delay between ports 1 and 2. [S]= 0.2/negationslash50◦00 0 .4/negationslash−45◦ 00 .6/negationslash45◦0.7/negationslash−45◦0 00 .7/negationslash−45◦0.6/negationslash45◦0 0.4/negationslash−45◦00 0 .5/negationslash45◦ . 4.19 When normalized to a single characteristic impedance Z0, a certain two-port network has scatter- ing parameters Sij. Find the generalized scattering parameters, Sp ij, in terms of the real reference impedances, R01andR02, at ports 1 and 2, respectively. 4.20 At reference plane A, for the circuit shown below, choose an appropriate reference impedance, find the power wave amplitudes, and compute the power delivered to the load. Repeat this procedure forreference plane B. Assume the transmission line is lossless. Z0 = 70.7 Ω 30 V AB100 Ω100 Ω l = /H9261/H114084 4.21 TheABCD parameters of the first entry in Table 4.1 were derived in Example 4.6. Verify the ABCD parameters for the second, third, and fourth entries. 4.22 Derive expressions that give the impedance parameters in terms of the ABCD parameters. 4.23 Find the ABCD matrix for the circuit shown below by direct calculation using the definition of the ABCD matrix, and compare with the ABCD matrix of the appropriate cascade of canonical circuits from Table 4.1. c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 Problems 225 4.24 UseABCD matrices to find the voltage VLacross the load resistor in the circuit shown below. VL Z0 = 50 Ω50 Ω 90° 30°ZL = 25 Ω/H11001 /H110021 : 2 V 4.25 A reciprocal two-port network with its ABCD matrix is shown below at left. Prove that the network with ports 1 and 2 in reversed positions has the ABCD matrix shown below at right. Choose a simple asymmetrical network to demonstrate this result. Port 1Port 2 D CB A21Port 1Port 2 A CB D12 4.26 Derive the expressions for Sparameters in terms of the ABCD parameters, as given in Table 4.2. 4.27 As shown in the figure below, a variable attenuator can be implemented using a four-port 90◦hybrid coupler by terminating ports 2 and 3 with equal but adjustable loads. (a) Using the given scattering matrix for the coupler, show that the transmission coefficient between the input (port 1) and the output (port 4) is given as T=j/Gamma1,w h e r e /Gamma1is the reflection coefficient of the mismatch at ports 2 and 3. Also show that the input port is matched for all values of /Gamma1. (b) Plot the attenuation, in dB, from the input to the output as a function of ZL/Z0,f o r0 ≤ZL/Z0≤10 (let ZLbe real). Port 1 Port 4Port 2 Port 3ZL ZLΓΓ [S][S] = /H110021–290° HybridIn Out√–0 j 10j 0011 00 j0 1 j 0 4.28 Use signal flow graphs to find the power ratios P2/P1and P3/P1for the mismatched three-port network shown in the accompanying figure. Port 2 Port 1Port 3P1P2 P3Γ3Γ2 [S] =0 S12 0 S12 0 S23 0 S23 0 4.29 TheABCD parameters are useful for treating cascades of two-port networks in terms of the total port voltages and currents, but it is also possible to use incident and reflected voltages to treat cascades. One way of doing this is with the transfer,o r T-,parameters, defined as follows: /bracketleftbigga1 b1/bracketrightbigg =/bracketleftbiggT11 T12 T21 T22/bracketrightbigg/bracketleftbiggb2 a2/bracketrightbigg , c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 226 Chapter 4: Microwave Network Analysis where a1,b1anda2,b2are the incident and reflected voltages at ports 1 and 2, respectively. Derive theT-parameters in terms of the scattering parameters of a two-port network. Show how the T-parameters can be used for a cascade of two two-port networks. 4.30 The end of an open-circuited microstrip line has fringing fields that can be modeled as a shunt capac- itor, Cf, at the end of the line, as shown below. This capacitance can be replaced with an additional length, /Delta1, of microstrip line. Derive an expression for the length extension in terms of the fringing capacitance. Evaluate the length extension for a 50 /Omega1open-circuited microstrip line on a substrate with d=0.158 cm and /epsilon1r=2.2(w=0.487 cm, /epsilon1e=1.894), if the fringing capacitance is known to beCf=0.075 pF. Compare your result with the approximation given by Hammerstad and Bekkadal: /Delta1=0.412d/parenleftbigg/epsilon1e+0.3 /epsilon1e−0.258/parenrightbigg/parenleftbiggw+0.262d w+0.813d/parenrightbigg . Cf Z0 Z0 O.C./H9004 4.31 For the H-plane step analysis of Section 4.6, compute the complex power flow in the reflected modes in guide 1, and show that the reactive power is inductive. 4.32 Derive the modal analysis equations for the symmetric H-plane step shown below. (HINT: Because of symmetry, only the TE n0modes for nodd will be excited.) zxy 0y b a c x 4.33 Find the transverse ¯Eand¯Hfields excited by the current of (4.116) by postulating traveling TM mn modes on either side of the source at z=0 and applying the appropriate boundary conditions. 4.34 An infinitely long rectangular waveguide is fed with a probe of length d as shown below. The current on this probe can be approximated as I(y)=I0sink(d−y)/sinkd.I ft h eT E 10mode is the only propagating mode in the waveguide, compute the input resistance seen at the probe terminals. y b ada/2 x c04MicrowaveNetworkAnalysis Pozar July 30, 2011 12:0 Problems 227 4.35 Consider the infinitely long waveguide fed with two probes driven 180◦out of phase, as shown below. What are the resulting excitation coefficients for the TE 10and TE 20modes? What other modes can be excited by this feeding arrangement? y b aIIa/4 a/4 x 4.36 Consider a small current loop on the sidewall of a rectangular waveguide, as shown below. Find the TE10fields excited by this loop if the loop is of radius r0. y b I0r0 ax 4.37 A rectangular waveguide is shorted at z=0 and has an electric current sheet, Jsy, located at z=d, where Jsy=2πA asinπx a (see the accompanying figure). Find expressions for the fields generated by this current by assuming standing wave fields for 0 <z<d, and traveling wave fields for z>d, and applying boundary conditions at z=0a n d z=d. Now solve the problem using image theory, by placing a current sheet−Jsyatz=−d, and removing the shorting wall at z=0. Use the results of Section 4.7 and superposition to find the fields radiated by these two currents, which should be the same as the first results for z>0. y 0Jsy a dz c05ImpedanceMatchingandTuning Pozar July 29, 2011 20:34 Chapter Five Impedance Matching and Tuning This chapter marks a turning point, in that we now begin to apply the theory and tech- niques of previous chapters to practical problems in microwave engineering. We start with the topic of impedance matching, which is often an important part of a larger design process for a microwave component or system. The basic idea of impedance matching is illustrated inFigure 5.1, which shows an impedance matching network placed between a load impedanceand a transmission line. The matching network is ideally lossless, to avoid unnecessary loss of power, and is usually designed so that the impedance seen looking into the matching network isZ 0. Then reflections will be eliminated on the transmission line to the left of the matching network, although there will usually be multiple reflections between the matching network andthe load. This procedure is sometimes referred to as tuning. Impedance matching or tuning is important for the following reasons: rMaximum power is delivered when the load is matched to the line (assuming the gener- ator is matched), and power loss in the feed line is minimized. rImpedance matching sensitive receiver components (antenna, low-noise amplifier, etc.) may improve the signal-to-noise ratio of the system. rImpedance matching in a power distribution network (such as an antenna array feed network) may reduce amplitude and phase errors. As long as the load impedance, ZL, has a positive real part, a matching network can always be found. Many choices are available, however, and we will discuss the design and performanceof several types of practical matching networks. Factors that may be important in the selectionof a particular matching network include the following: rComplexity —As with most engineering solutions, the simplest design that satisfies the required specifications is generally preferable. A simpler matching network is usuallycheaper, smaller, more reliable, and less lossy than a more complex design. rBandwidth —Any type of matching network can ideally give a perfect match (zero reflection) at a single frequency. In many applications, however, it is desirable to matcha load over a band of frequencies. There are several ways of doing this, with, of course,a corresponding increase in complexity. 228 c05ImpedanceMatchingandTuning Pozar July 29, 2011 20:34 5.1 Matching with Lumped Elements ( LNetworks) 229 Z0Matching networkLoad ZL FIGURE 5.1 A lossless network matching an arbitrary load impedance to a transmission line. rImplementation—Depending on the type of transmission line or waveguide being used, one type of matching network may be preferable to another. For example, tuningstubs are much easier to implement in waveguide than are multisection quarter-wavetransformers. rAdjustability —In some applications the matching network may require adjustment to match a variable load impedance. Some types of matching networks are more amenablethan others in this regard. 5.1MATCHINGWITHLUMPEDELEMENTS( LNETWORKS) Probably the simplest type of matching network is the L-section, which uses two reac- tive elements to match an arbitrary load impedance to a transmission line. There are twopossible configurations for this network, as shown in Figure 5.2. If the normalized load impedance, z L=ZL/Z0, is inside the 1 +jxcircle on the Smith chart, then the circuit of Figure 5.2a should be used. If the normalized load impedance is outside the 1 +jxcir- cle on the Smith chart, the circuit of Figure 5.2b should be used. The 1 +jxcircle is the resistance circle on the impedance Smith chart for which r=1. In either of the configurations of Figure 5.2, the reactive elements may be either induc- tors or capacitors, depending on the load impedance. Thus, there are eight distinct possibil- ities for the matching circuit for various load impedances. If the frequency is low enough and/or the circuit size is small enough, actual lumped-element capacitors and inductors can be used. This may be feasible for frequencies up to about 1 GHz or so, although modern microwave integrated circuits may be small enough such that lumped elements can be used at higher frequencies as well. There is, however, a large range of frequencies and circuit sizes where lumped elements may not be realizable. This is a limitation of the L-section Z0jX ZL (a) (b)jBjX ZL jB FIGURE 5.2 L-section matching networks. (a) Network for zLinside the 1 +jxcircle. (b) Net- work for zLoutside the 1 +jxcircle. c05ImpedanceMatchingandTuning Pozar July 29, 2011 20:34 230 Chapter 5: Impedance Matching and Tuning matching technique. We will first derive analytic expressions for the matching network elements of the two cases in Figure 5.2, and then illustrate an alternative design procedure using the Smith chart. AnalyticSolutions Although we will discuss a simple graphical solution using the Smith chart, it is also useful to have simple expressions for the L-section matching network components. These expres- sions can be used in a computer-aided design program for L-section matching, or when it is necessary to have more accuracy than the Smith chart can provide. Consider first the circuit of Figure 5.2a, and let ZL=RL+jXL. We stated that this circuit would be used when zL=ZL/Z0is inside the 1 +jxcircle on the Smith chart, which implies that RL>Z0for this case. The impedance seen looking into the matching network, followed by the load impedance, must be equal to Z0for an impedance-matched condition: Z0=jX+1 jB+1/(RL+jXL).( 5.1) Rearranging and separating into real and imaginary parts gives two equations for the two unknowns, XandB: B(XR L−XLZ0)=RL−Z0, (5.2a) X(1−BX L)=BZ0RL−XL. (5.2b) Solving (5.2a) for Xand substituting into (5.2b) gives a quadratic equation for B.T h e solution is B=XL±√RL/Z0/radicalBig R2 L+X2 L−Z0RL R2 L+X2 L.( 5.3a) Note that since RL>Z0, the argument of the second square root is always positive. Then the series reactance can be found as X=1 B+XLZ0 RL−Z0 BRL.( 5.3b) Equation (5.3a) indicates that two solutions are possible for BandX. Both of these solutions are physically realizable since both positive and negative values of BandXare possible (positive Ximplies an inductor and negative Ximplies a capacitor, while positive Bimplies a capacitor and negative Bimplies an inductor). One solution, however, may result in significantly smaller values for the reactive components, or may be the preferred solution if the bandwidth of the match is better, or if the SWR on the line between the matching network and the load is smaller. Next consider the circuit of Figure 5.2b. This circuit is used when zLis outside the 1+jxcircle on the Smith chart, which implies that RL<Z0. The admittance seen look- ing into the matching network, followed by the load impedance, must be equal to 1 /Z0for an impedance-matched condition: 1 Z0=jB+1 RL+j(X+XL).( 5.4) c05ImpedanceMatchingandTuning Pozar July 29, 2011 20:34 5.1 Matching with Lumped Elements ( LNetworks) 231 Rearranging and separating into real and imaginary parts gives two equations for the two unknowns, XandB: BZ0(X+XL)=Z0−RL, (5.5a) (X+XL)=BZ0RL. (5.5b) Solving for XandBgives X=±/radicalbig RL(Z0−RL)−XL, (5.6a) B=±√(Z0−RL)/RL Z0. (5.6b) Because RL<Z0, the arguments of the square roots are always positive. Again, note that two solutions are possible. In order to match an arbitrary complex load to a line of characteristic impedance Z0, the real part of the input impedance to the matching network must be Z0, while the imag- inary part must be zero. This implies that a general matching network must have at leasttwo degrees of freedom; in the L-section matching circuit these two degrees of freedom are provided by the values of the two reactive components. SmithChartSolutions Instead of the above formulas, the Smith chart can be used to quickly and accurately design L-section matching networks. The procedure is best illustrated by an example. EXAMPLE 5.1 L-SECTION IMPEDANCE MATCHING Design an L-section matching network to match a series RCload with an impedance ZL=200−j100/Omega1to a 100 /Omega1line at a frequency of 500 MHz. Solution The normalized load impedance is zL=2−j1, which is plotted on the Smith chart of Figure 5.3a. This point is inside the 1 +jxcircle, so we use the match- ing circuit of Figure 5.2a. Because the first element from the load is a shunt sus- ceptance, it makes sense to convert to admittance by drawing the SWR circlethrough the load, and a straight line from the load through the center of the chart, as shown in Figure 5.3a. After we add the shunt susceptance and convert back to impedance, we want to be on the 1 +jxcircle so that we can add a series reactance to cancel jxand match the load. This means that the shunt suscep- tance must move us from y Lto the 1 +jxcircle on the admittance Smith chart. Thus, we construct the rotated 1 +jxcircle as shown in Figure 5.3a (center at r=0.333). (A combined ZYchart may be convenient to use here, if it is not too confusing.) Then we see that adding a susceptance of jb=j0.3 will move us along a constant-conductance circle to y=0.4+j0.5 (this choice is the short- est distance from yLto the shifted 1 +jxcircle). Converting back to impedance leaves us at z=1−j1.2, indicating that a series reactance of x=j1.2 will bring us to the center of the chart. For comparison, the formulas (5.3a) and (5.3b) give the solution as b=0.29, x=1.22. c05ImpedanceMatchingandTuning Pozar July 29, 2011 20:34 232 Chapter 5: Impedance Matching and Tuning This matching circuit consists of a shunt capacitor and a series inductor, as shown in Figure 5.3b. For a matching frequency of 500 MHz, the capacitor has a value of C=b 2πfZ0=0.92 pF, and the inductor has a value of L=xZ0 2πf=38.8n H . It is also interesting to look at the second solution to this matching problem. If instead of adding a shunt susceptance of b=0.3, we use a shunt susceptance of b=−0.7, we will move to a point on the lower half of the shifted 1 +jxcircle, toy=0.4−j0.5. Then converting to impedance and adding a series reactance of x=−1.2 leads to a match as well. Formulas (5.3a) and (5.3b) give this solution as b=−0.69, x=−1.22. This matching circuit is also shown in Figure 5.3b, and is seen to have the positions of the inductor and capacitor reversed from the first matching network. At a frequency of f=500 MHz, the capacitor has a value of C=−1 2πfxZ 0=2.61 pF, j , 20-2030-3040-4050-5060-6070-7080-8090-90100-100110-110120-120130-130140-140150-150160-160170-1701800.040.040.050.050.060.060.070.070.080.080.090.090.10.10.110.110.120.120.130.130.140.140.150.150.160.160.170.170.180.180.190.190.20.20.210.210.220.220.230.230.240.24 0.250.250.260.260.270.270.280.280.290.290.30.30.310.310.320.320.330.330.340.340.350.350.360.360.370.370.380.380.390.390.40.40.410.410.420.420.430.430.440.440.450.450.460.460.470.470.480.480.490.49 0.00.0AN GLE OFREFLECTION COEFFICIENTINDEGREES—>WAVELENGTHSTOWARDGENERATOR—><— W AVELENGTH STOW ARD LOAD <—II T NDUCTIVEREACTANCECOMPONEN(+X/Zo) ORCAPACTIVESUSCEPTANCE(+jB/Yo)CAPACITIVEREACTANCECOMPONENT(-jX/Zo),ORINDUCTIVESUSCEPTANCE(-jB/Yo) 0.1 0.1 0.1 0.2 0.2 0.2 0.3 0.3 0.3 0.4 0.4 0.4 0.5 0.5 0.5 0.6 0.6 0.6 0.7 0.7 0.7 0.8 0.8 0.8 0.9 0.9 0.9 1.0 1.0 1.01.21.21.21.41.41.41.61.61.61.81.81.82.02.02.03.03.03.04.04.04.05.05.05.01010102020205050500.20.2 0.2 0.20.40.4 0.4 0.40.60.6 0.6 0.60.80.8 0.8 0.81.01.0 1.0 1.0RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) ± zLyL +j1.2 (a)+j0.3Rotated 1 + jx circle on admittance chart FIGURE 5.3 Solution to Example 5.1. (a) Smith chart for the L-section matching networks. c05ImpedanceMatchingandTuning Pozar July 29, 2011 20:34 5.1 Matching with Lumped Elements ( LNetworks) 233 Z0 = 100 Ω ZL = 200 – j100 Ω 0.92 pF38.8 nH Solution 1 (b) (c)Solution 2 Solution 1 f(GHz)Z0 = 100 Ω ZL = 200 – j100 Ω 46.1 nH2.61pF Solution 2 000.250.50.751 0.25 0.5 0.75 1⎪Γ⎪ FIGURE 5.3 Continued. (b) The two possible L-section matching circuits. (c) Reflection coeffi- cient magnitudes versus frequency for the matching circuits of (b). while the inductor has a value of L=−Z0 2πfb=46.1n H . Figure 5.3c shows the reflection coefficient magnitude versus frequency for these two matching networks, assuming that the load impedance of ZL=200−j100/Omega1 at 500 MHz consists of a 200 /Omega1resistor and a 3.18 pF capacitor in series. There is not a substantial difference in bandwidth for these two solutions. ■ POINT OF INTEREST: Lumped Elements for Microwave Integrated Circuits Lumped R,L,a n d Celements can be practically realized at microwave frequencies if the length, /lscript, of the component is very small relative to the operating wavelength. Over a limited range of values, such components can be used in hybrid and monolithic microwave integratedcircuits at frequencies up to 60 GHz, or higher, if the condition that /lscript<λ / 10 is satisfied. Usually, however, the characteristics of such an element are far from ideal, requiring that un- desirable effects such as parasitic capacitance and/or inductance, spurious resonances, fringingfields, loss, and perturbations caused by a ground plane be incorporated in the design via a CAD model (see the Point of Interest concerning CAD). c05ImpedanceMatchingandTuning Pozar July 29, 2011 20:34 234 Chapter 5: Impedance Matching and Tuning Lossy filmLossy film Planar resistor Interdigital gap capacitorChip resistor Dielectric Metal-insulator- metal capacitorLoop inductor Spiral inductor Chip capacitorεr εrAir bridge Resistors are fabricated with thin films of lossy material such as nichrome, tantalum nitride, or doped semiconductor material. In monolithic circuits such films can be deposited or grown,whereas chip resistors made from a lossy film deposited on a ceramic chip can be bonded or soldered in a hybrid circuit. Low resistances are hard to obtain. Small values of inductance can be realized with a short length or loop of transmission line, and larger values (up to about 10 nH) can be obtained with a spiral inductor, as shown in the following figures. Larger inductance values generally incur more loss and more shunt capacitance; this leads to a resonance that limits the maximum operating frequency. Capacitors can be fabricated in several ways. A short transmission line stub can provide a shunt capacitance in the range of 0–0.1 pF. A single gap, or an interdigital set of gaps, ina transmission line can provide a series capacitance up to about 0.5 pF. Greater values (up to about 25 pF) can be obtained using a metal-insulator-metal sandwich in either monolithic or chip (hybrid) form. 5.2SINGLE-STUBTUNING Another popular matching technique uses a single open-circuited or short-circuited length of transmission line (a stub) connected either in parallel or in series with the transmission feed line at a certain distance from the load, as shown in Figure 5.4. Such a single-stub tuning circuit is often very convenient because the stub can be fabricated as part of the transmission line media of the circuit, and lumped elements are avoided. Shunt stubs arepreferred for microstrip line or stripline, while series stubs are preferred for slotline or coplanar waveguide. In single-stub tuning the two adjustable parameters are the distance, d, from the load to the stub position, and the value of susceptance or reactance provided by the stub. For the shunt-stub case, the basic idea is to select dso that the admittance, Y, seen looking into the line at distance dfrom the load is of the form Y 0+jB. Then the stub susceptance is chosen as −jB, resulting in a matched condition. For the series-stub case, the distance dis selected so that the impedance, Z, seen looking into the line at a distance dfrom the load is of the form Z0+jX. Then the stub reactance is chosen as −jX, resulting in a matched condition. As discussed in Chapter 2, the proper length of an open or shorted transmission line section can provide any desired value of reactance or susceptance. For a given suscep- tance or reactance, the difference in lengths of an open- or short-circuited stub is λ/4. c05ImpedanceMatchingandTuning Pozar July 29, 2011 20:34 5.2 Single-Stub Tuning 235 Y0 Y0d Y =Y0 Open or shorted stub1 ZYL Z0 Z0Z0d lZ =(a) (b)Open or shorted stub1 YZLl FIGURE 5.4 Single-stub tuning circuits. (a) Shunt stub. (b) Series stub. For transmission line media such as microstrip or stripline, open-circuited stubs are easier to fabricate since a via hole through the substrate to the ground plane is not needed. For lines like coax or waveguide, however, short-circuited stubs are usually preferred because the cross-sectional area of such an open-circuited line may be large enough (electrically)to radiate, in which case the stub is no longer purely reactive. We will discuss both Smith chart and analytic solutions for shunt- and series-stub tun- ing. The Smith chart solutions are fast, intuitive, and usually accurate enough in practice.The analytic expressions are more precise, and are useful for computer analysis. ShuntStubs The single-stub shunt tuning circuit is shown in Figure 5.4a. We will first discuss an exam- ple illustrating the Smith chart solution and then derive formulas for dand/lscript. EXAMPLE 5.2 SINGLE-STUB SHUNT TUNING For a load impedance ZL=60−j80/Omega1, design two single-stub (short circuit) shunt tuning networks to match this load to a 50 /Omega1line. Assuming that the load is matched at 2 GHz and that the load consists of a resistor and capacitor in series, plot the reflection coefficient magnitude from 1 to 3 GHz for each solution. Solution The first step is to plot the normalized load impedance zL=1.2−j1.6, construct the appropriate SWR circle, and convert to the load admittance, yL, as shown on c05ImpedanceMatchingandTuning Pozar September 13, 2011 15:31 236 Chapter 5: Impedance Matching and Tuning the Smith chart in Figure 5.5a. For the remaining steps we consider the Smith chart as an admittance chart. Notice that the SWR circle intersects the 1 +jb circle at two points, denoted as y1andy2in Figure 5.5a. Thus the distance dfrom the load to the stub is given by either of these two intersections. Reading the WTGscale, we obtain d 1=0.176−0.065 =0.110λ, d2=0.325−0.065 =0.260λ. Actually, there is an infinite number of distances daround the SWR circle that intersect the 1 +jbcircle. Usually it is desired to keep the matching stub as close as possible to the load to improve the bandwidth of the match and to reduce losses caused by a possibly large standing wave ratio on the line between the stuband the load. At the two intersection points, the normalized admittances are y 1=1.00+j1.47, y2=1.00−j1.47. y2yL zLd2 d1 y1 (a) FIGURE 5.5 Solution to Example 5.2. (a) Smith chart for the shunt-stub tuners. c05ImpedanceMatchingandTuning Pozar July 29, 2011 20:34 5.2 Single-Stub Tuning 237 f(GHz)1.000.40.60.81.0 1.5 2.0 2.5 3.0Solution #2 Solution #1Solution #250 Ω 0.405/H92610.260/H9261 50 Ω 50 Ω Solution #150 Ω 0.095/H926160 Ω 0.995 pF0.110/H9261 50 Ω 50 Ω (b) (c)⎪Γ⎪60 Ω 0.995 pF 0.2 FIGURE 5.5 Continued. (b) The two shunt-stub tuning solutions. (c) Reflection coefficient mag- nitudes versus frequency for the tuning circuits of (b). Thus, the first tuning solution requires a stub with a susceptance of −j1.47. The length of a short-circuited stub that gives this susceptance can be found on the Smith chart by starting at y=∞ (the short circuit) and moving along the outer edge of the chart ( g=0)toward the generator to the −j1.47 point. The stub length is then /lscript1=0.095λ. Similarly, the required short-circuit stub length for the second solution is /lscript2=0.405λ. This completes the two tuner designs. To analyze the frequency dependence of these two designs, we need to know the load impedance as a function of frequency. The series-RC load impedance isZL=60−j80/Omega1at 2 GHz, so R=60/Omega1andC=0.995 pF. The two tun- ing circuits are shown in Figure 5.5b. Figure 5.5c shows the calculated reflection coefficient magnitudes for these two solutions. Observe that solution 1 has a sig- nificantly better bandwidth than solution 2; this is because both dand/lscriptare shorter for solution 1, which reduces the frequency variation of the match. ■ c05ImpedanceMatchingandTuning Pozar July 29, 2011 20:34 238 Chapter 5: Impedance Matching and Tuning To derive formulas for dand/lscript, let the load impedance be written as ZL=1/YL= RL+jXL. Then the impedance Zdown a length dof line from the load is Z=Z0(RL+jXL)+jZ0t Z0+j(RL+jXL)t,( 5.7) where t=tanβd. The admittance at this point is Y=G+jB=1 Z, where G=RL(1+t2) R2 L+(XL+Z0t)2, (5.8a) B=R2 Lt−(Z0−XLt)(XL+Z0t) Z0/bracketleftbig R2 L+(XL+Z0t)2/bracketrightbig. (5.8b) Now d(which implies t)is chosen so that G=Y0=1/Z0. From (5.8a), this results in a quadratic equation for t: Z0(RL−Z0)t2−2XLZ0t+/parenleftbig RLZ0−R2 L−X2 L/parenrightbig =0. Solving for tgives t=XL±/radicalBig RL/bracketleftbig (Z0−RL)2+X2 L/bracketrightbig /Z0 RL−Z0forRL/negationslash=Z0.( 5.9) IfRL=Z0, then t=− XL/2Z0. Thus, the two principal solutions for dare d λ=⎧ ⎪⎪⎨ ⎪⎪⎩1 2πtan−1t fort≥0 1 2π(π+tan−1t) fort<0.(5.10) To find the required stub lengths, first use tin (5.8b) to find the stub susceptance, Bs=− B. Then, for an open-circuited stub, /lscripto λ=1 2πtan−1/parenleftbiggBs Y0/parenrightbigg =−1 2πtan−1/parenleftbiggB Y0/parenrightbigg ,( 5.11a) and for a short-circuited stub, /lscripts λ=−1 2πtan−1/parenleftbiggY0 Bs/parenrightbigg =1 2πtan−1/parenleftbiggY0 B/parenrightbigg .( 5.11b) If the length given by (5.11a) or (5.11b) is negative, λ/2 can be added to give a positive result. SeriesStubs The series-stub tuning circuit is shown in Figure 5.4b. We will illustrate the Smith chart solution by an example, and then derive expressions for dand/lscript. EXAMPLE 5.3 SINGLE-STUB SERIES TUNING Match a load impedance of ZL=100+j8 0t oa5 0/Omega1 line using a single series open-circuit stub. Assuming that the load is matched at 2 GHz and that the load c05ImpedanceMatchingandTuning Pozar July 29, 2011 20:34 5.2 Single-Stub Tuning 239 consists of a resistor and inductor in series, plot the reflection coefficient magni- tude from 1 to 3 GHz. Solution First plot the normalized load impedance, zL=2+j1.6, and draw the SWR circle. For the series-stub design the chart is an impedance chart. Note that the SWR circle intersects the 1 +jxcircle at two points, denoted as z1andz2in Figure 5.6a. The shortest distance, d1, from the load to the stub is, from the WTG scale, d1=0.328 −0.208 =0.120λ, and the second distance is d2=(0.5−0.208) +0.172 =0.463λ. As in the shunt-stub case, additional rotations around the SWR circle lead to ad- ditional solutions, but these are usually not of practical interest. j , 20-2030-3040-4050-5060-6070-7080-8090-90100-100110-110120-120130-130140-140150-150160-160170-1701800.040.040.050.050.060.060.070.070.080.080.090.090.10.10.110.110.120.120.130.130.140.140.150.150.160.160.170.170.180.180.190.190.20.20.210.210.220.220.230.230.240.24 0.250.250.260.260.270.270.280.280.290.290.30.30.310.310.320.320.330.330.340.340.350.350.360.360.370.370.380.380.390.390.40.40.410.410.420.420.430.430.440.440.450.450.460.460.470.470.480.480.490.49 0.00.0AN GLE OFREFLECTION COEFFICIENTINDEGREES—>WAVELENGTHSTOWARDGENERATOR—><— W AVELENGTH STOW ARD LOAD <—II T NDUCTIVEREACTANCECOMPONEN(+X/Zo) ORCAPACTIVESUSCEPTANCE(+jB/Yo)CAPACITIVEREACTANCECOMPONENT(-jX/Zo),ORINDUCTIVESUSCEPTANCE(-jB/Yo) 0.1 0.1 0.1 0.2 0.2 0.2 0.3 0.3 0.3 0.4 0.4 0.4 0.5 0.5 0.5 0.6 0.6 0.6 0.7 0.7 0.7 0.8 0.8 0.8 0.9 0.9 0.9 1.0 1.0 1.01.21.21.21.41.41.41.61.61.61.81.81.82.02.02.03.03.03.04.04.04.05.05.05.01010102020205050500.20.2 0.2 0.20.40.4 0.4 0.40.60.6 0.6 0.60.80.8 0.8 0.81.01.0 1.0 1.0RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) ±z2 d2 l2 (a)d1 z1zLl1 FIGURE 5.6 Solution to Example 5.3. (a) Smith chart for the series-stub tuners. c05ImpedanceMatchingandTuning Pozar July 29, 2011 20:34 240 Chapter 5: Impedance Matching and Tuning f(GHz)1.000.250.50.751.0 1.5 2.0 2.5 3.0Solution 2Solution 150 Ω100 Ω 6.37 nH0.397/H9261 0.120/H926150 Ω50 Ω Solution 1 (c)⎪Γ⎪50 Ω100 Ω 6.37 nH0.103/H9261 0.463/H926150 Ω50 Ω Solution 2 (b) FIGURE 5.6 Continued. (b) The two series-stub tuning solutions. (c) Reflection coefficient mag- nitudes versus frequency for the tuning circuits of (b). The normalized impedances at the two intersection points are z1=1−j1.33, z2=1+j1.33. Thus, the first solution requires a stub with a reactance of j1.33. The length of an open-circuited stub that gives this reactance can be found on the Smith chart by starting at z=∞ (open circuit), and moving along the outer edge of the chart (r=0)toward the generator to the j1.33 point. This gives a stub length of /lscript1=0.397λ. Similarly, the required open-circuited stub length for the second solution is /lscript2=0.103λ. This completes the tuner designs. If the load is a series resistor and inductor with ZL=100+j80/Omega1at 2 GHz, then R=100/Omega1and L= 6.37 nH. The two matching circuits are shown in Figure 5.6b. Figure 5.6c shows the calculated reflection coefficient magnitudes versus frequency for the two solutions. ■ c05ImpedanceMatchingandTuning Pozar July 29, 2011 20:34 5.3 Double-Stub Tuning 241 To derive formulas for dand/lscriptfor the series-stub tuner, let the load admittance be written as YL=1/ZL=GL+jBL. Then the admittance Ydown a length dof line from the load is Y=Y0(GL+jBL)+jtY0 Y0+jt(GL+jBL),( 5.12) where t=tanβdandY0=1/Z0. The impedance at this point is Z=R+jX=1 Y, where R=GL(1+t2) G2 L+(BL+Y0t)2, (5.13a) X=G2 Lt−(Y0−tBL)(BL+tY0) Y0/bracketleftbig G2 L+(BL+Y0t)2/bracketrightbig. (5.13b) Now d(which implies t)is chosen so that R=Z0=1/Y0. From (5.13a), this results in a quadratic equation for t: Y0(GL−Y0)t2−2BLY0t+/parenleftbig GLY0−G2 L−B2 L/parenrightbig =0. Solving for tgives t=BL±/radicalBig GL/bracketleftbig (Y0−GL)2+B2 L/bracketrightbig /Y0 GL−Y0forGL/negationslash=Y0.( 5.14) IfGL=Y0, then t=− BL/2Y 0. Then the two principal solutions for dare d/λ=⎧ ⎪⎪⎨ ⎪⎪⎩1 2πtan−1t fort≥0 1 2π(π+tan−1t)fort<0.(5.15) The required stub lengths are determined by first using tin (5.13b) to find the reactance X. This reactance is the negative of the necessary stub reactance, Xs. Thus, for a short- circuited stub, /lscripts λ=1 2πtan−1/parenleftbiggXs Z0/parenrightbigg =−1 2πtan−1/parenleftbiggX Z0/parenrightbigg ,( 5.16a) and for an open-circuited stub, /lscripto λ=−1 2πtan−1/parenleftbiggZ0 Xs/parenrightbigg =1 2πtan−1/parenleftbiggZ0 X/parenrightbigg . (5.16b) If the length given by (5.16a) or (5.16b) is negative, λ/2 can be added to give a positive result. 5.3DOUBLE-STUBTUNING The single-stub tuner of the previous section is able to match any load impedance (having a positive real part) to a transmission line, but suffers from the disadvantage of requiringa variable length of line between the load and the stub. This may not be a problem for a fixed matching circuit, but would probably pose some difficulty if an adjustable tuner was c05ImpedanceMatchingandTuning Pozar July 29, 2011 20:34 242 Chapter 5: Impedance Matching and Tuning Y0 Y0 jB2 jB1 Y0d Open or shortOpen or shortl1 l2 (a) (b)Y'L Y0 jB2 jB1 Y0d Open or shortOpen or shortl1 l2YL FIGURE 5.7 Double-stub tuning. (a) Original circuit with the load an arbitrary distance from the first stub. (b) Equivalent circuit with the load transformed to the first stub. desired. In this case, the double-stub tuner , which uses two tuning stubs in fixed positions, can be used. Such tuners are often fabricated in coaxial line with adjustable stubs connected in shunt to the main coaxial line. We will see, however, that a double-stub tuner cannot match all load impedances. The double-stub tuner circuit is shown in Figure 5.7a, where the load may be an ar- bitrary distance from the first stub. Although this is more representative of a practical sit- uation, the circuit of Figure 5.7b, where the load Y/prime Lhas been transformed back to the position of the first stub, is easier to deal with and does not lose any generality. The shunt stubs shown in Figure 5.7 can be conveniently implemented for some types of transmission lines, while series stubs are more appropriate for other types of lines. In either case, thestubs can be open-circuited or short-circuited. SmithChartSolution The Smith chart of Figure 5.8 illustrates the basic operation of the double-stub tuner. As in the case of the single-stub tuner, two solutions are possible. The susceptance of the firststub, b 1(orb/prime 1, for the second solution), moves the load admittance to y1(ory/prime 1). These points lie on the rotated 1 +jbcircle; the amount of rotation is dwavelengths toward the load, where dis the electrical distance between the two stubs. Then transforming y1(or y/prime 1)toward the generator through a length dof line leaves us at the point y2(ory/prime 2), which must be on the 1 +jbcircle. The second stub then adds a susceptance b2(orb/prime 2), which brings us to the center of the chart and completes the match. Notice from Figure 5.8 that if the load admittance, yL, were inside the shaded region of the g0+jbcircle, no value of stub susceptance b1could ever bring the load point to intersect the rotated 1 +jbcircle. This shaded region thus forms a forbidden range of load admittances that cannot be matched with this particular double-stub tuner. A simple way c05ImpedanceMatchingandTuning Pozar July 29, 2011 20:34 5.3 Double-Stub Tuning 243 FIGURE 5.8 Smith chart diagram for the operation of a double-stub tuner. of reducing the forbidden range is to reduce the distance dbetween the stubs. This has the effect of swinging the rotated 1 +jbcircle back toward the y=∞ point, but dmust be kept large enough for the practical purpose of fabricating the two separate stubs. Inaddition, stub spacings near 0 or λ/2 lead to matching networks that are very frequency sensitive. In practice, stub spacings are usually chosen as λ/8o r3λ/8. If the length of line between the load and the first stub can be adjusted, then the load admittance y Lcan always be moved out of the forbidden region. EXAMPLE 5.4 DOUBLE-STUB TUNING Design a double-stub shunt tuner to match a load impedance ZL=60−j80/Omega1 t oa5 0/Omega1 line. The stubs are to be open-circuited stubs and are spaced λ/8 apart. Assuming that this load consists of a series resistor and capacitor and that the match frequency is 2 GHz, plot the reflection coefficient magnitude versus fre-quency from 1 to 3 GHz. Solution The normalized load admittance is y L=0.3+j0.4, which is plotted on the Smith chart of Figure 5.9a. Next we construct the rotated 1 +jbconductance circle by moving every point on the g=1c i r c l e λ/8 toward the load. We then find the susceptance of the first stub, which can be one of two possible values: b1=1.314 or b/prime 1=−0.114. We now transform through the λ/8 section of line by rotating along a constant- radius (SWR) circle λ/8 toward the generator. This brings the two solutions to the c05ImpedanceMatchingandTuning Pozar July 29, 2011 20:34 244 Chapter 5: Impedance Matching and Tuning following points: y2=1−j3.38 or y/prime 2=1+j1.38. Then the susceptance of the second stub should be b2=3.38 or b/prime 2=−1.38. The lengths of the open-circuited stubs are then found as /lscript1=0.146λ,/lscript 2=0.204λ or/lscript/prime 1=0.482λ,/lscript/prime 2=0.350λ. This completes both solutions for the double-stub tuner design. Atf=2 GHz the resistor-capacitor load of ZL=60−j80/Omega1implies that R=60/Omega1and C=0.995 pF. The two tuning circuits are then as shown in Figure 5.9b, and the reflection coefficient magnitudes are plotted versus frequency in Figure 5.9c. Note that the first solution has a much narrower bandwidth than the second (primed) solution due to the fact that both stubs for the first solution are somewhat longer (and closer to λ/2)than the stubs of the second solution. ■ j , 20-2030-3040-4050-5060-6070-7080-8090-90100-100110-110120-120130-130140-140150-150160-160170-1701800.040.040.050.050.060.060.070.070.080.080.090.090.10.10.110.110.120.120.130.130.140.140.150.150.160.160.170.170.180.180.190.190.20.20.210.210.220.220.230.230.240.24 0.250.250.260.260.270.270.280.280.290.290.30.30.310.310.320.320.330.330.340.340.350.350.360.360.370.370.380.380.390.390.40.40.410.410.420.420.430.430.440.440.450.450.460.460.470.470.480.480.490.49 0.00.0AN GLE OFREFLECTION COEFFICIENTINDEGREES—>WAVELENGTHSTOWARDGENERATOR—><— W AVELENGTH STOW ARD LOAD <—II T NDUCTIVEREACTANCECOMPONEN(+X/Zo) ORCAPACTIVESUSCEPTANCE(+jB/Yo)CAPACITIVEREACTANCECOMPONENT(-jX/Zo),ORINDUCTIVESUSCEPTANCE(-jB/Yo) 0.1 0.1 0.1 0.2 0.2 0.2 0.3 0.3 0.3 0.4 0.4 0.4 0.5 0.5 0.5 0.6 0.6 0.6 0.7 0.7 0.7 0.8 0.8 0.8 0.9 0.9 0.9 1.0 1.0 1.01.21.21.21.41.41.41.61.61.61.81.81.82.02.02.03.03.03.04.04.04.05.05.05.01010102020205050500.20.2 0.2 0.20.40.4 0.4 0.40.60.6 0.6 0.60.80.8 0.8 0.81.01.0 1.0 1.0RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) ±y1'y1Rotated 1 + jb circle (a)b2b2b1 ' 'b1 y2yLy2' FIGURE 5.9 Solution to Example 5.4. (a) Smith chart for the double-stub tuners. c05ImpedanceMatchingandTuning Pozar July 29, 2011 20:34 5.3 Double-Stub Tuning 245 f(GHz)1.000.40.60.81.0 1.5 2.0 2.5 3.0Solution #2Solution #1Solution 250 Ω 0.350/H9261/H9261/8 50 Ω (b) (c)⎪Γ⎪0.482/H9261 Solution 150 Ω 0.204/H9261/H9261/8 50 Ω 0.146/H926160 Ω 0.995 pF60 Ω 0.995 pF 0.2 FIGURE 5.9 Continued. (b) The two double-stub tuning solutions. (c) Reflection coefficient mag- nitudes versus frequency for the tuning circuits of (b). AnalyticSolution The admittance just to the left of the first stub in Figure 5.7b is Y1=GL+j(BL+B1), (5.17) where YL=GL+jBLis the load admittance, and B1is the susceptance of the first stub. After transforming through a length dof transmission line, we find that the admittance just to the right of the second stub is Y2=Y0GL+j(BL+B1+Y0t) Y0+jt(GL+jBL+jB1),( 5.18) where t=tanβdandY0=1/Z0. At this point the real part of Y2must equal Y0, which leads to the equation G2 L−GLY01+t2 t2+(Y0−BLt−B1t)2 t2=0.( 5.19) Solving for GLgives GL=Y01+t2 2t2/bracketleftBigg 1±/radicalBigg 1−4t2(Y0−BLt−B1t)2 Y2 0(1+t2)2/bracketrightBigg .( 5.20) c05ImpedanceMatchingandTuning Pozar July 29, 2011 20:34 246 Chapter 5: Impedance Matching and Tuning Because GLis real, the quantity within the square root must be nonnegative, and so 0≤4t2(Y0−BLt−B1t)2 Y2 0(1+t2)2≤1. This implies that 0≤GL≤Y01+t2 t2=Y0 sin2βd,( 5.21) which gives the range on GLthat can be matched for a given stub spacing d.A f t e rd has been set, the first stub susceptance can be determined from (5.19) as B1=− BL+Y0±/radicalBig (1+t2)GLY0−G2 Lt2 t.( 5.22) Then the second stub susceptance can be found from the negative of the imaginary part of (5.18) to be B2=±Y0/radicalBig Y0GL(1+t2)−G2 Lt2+GLY0 GLt.( 5.23) The upper and lower signs in (5.22) and (5.23) correspond to the same solutions. The open-circuited stub length is found as /lscripto λ=1 2πtan−1/parenleftbiggB Y0/parenrightbigg ,( 5.24a) and the short-circuited stub length is found as /lscripts λ=−1 2πtan−1/parenleftbiggY0 B/parenrightbigg ,( 5.24b) where B=B1orB2. 5.4THEQUARTER-WAVETRANSFORMER As introduced in Section 2.5, the quarter-wave transformer is a simple and useful circuit for matching a real load impedance to a transmission line. An additional feature of the quarter-wave transformer is that it can be extended to multisection designs in a methodical manner to provide broader bandwidth. If only a narrow band impedance match is required,a single-section transformer may suffice. However, as we will see in the next few sec- tions, multisection quarter-wave transformer designs can be synthesized to yield optimum matching characteristics over a desired frequency band. We will see in Chapter 8 that suchnetworks are closely related to bandpass filters. One drawback of the quarter-wave transformer is that it can only match a real load impedance. A complex load impedance can always be transformed into a real impedance, however, by using an appropriate length of transmission line between the load and the transformer, or an appropriate series or shunt reactive element. These techniques will usu-ally alter the frequency dependence of the load, and this often has the effect of reducing the bandwidth of the match. In Section 2.5 we analyzed the operation of a quarter-wave transformer from both an impedance viewpoint and a multiple reflection viewpoint. Here we will concentrate on the bandwidth performance of the transformer as a function of the load mismatch; this c05ImpedanceMatchingandTuning Pozar July 29, 2011 20:34 5.4 The Quarter-Wave Transformer 247 l Z0 Z1 ZL (real) FIGURE 5.10 A single-section quarter-wave matching transformer. /lscript=λ0/4 at the design fre- quency f0. discussion will also serve as a prelude to the more general case of multisection transformers in the sections to follow. The single-section quarter-wave matching transformer circuit is shown in Figure 5.10, with the characteristic impedance of the matching section given as Z1=/radicalbig Z0ZL.( 5.25) At the design frequency, f0, the electrical length of the matching section is λ0/4, but at other frequencies the length is different, so a perfect match is no longer obtained. We will derive an approximate expression for the resulting impedance mismatch versus frequency. The input impedance seen looking into the matching section is Zin=Z1ZL+jZ1t Z1+jZLt,( 5.26) where t=tanβ/lscript=tanθ, andβ/lscript=θ=π/2 at the design frequency f0. The resulting re- flection coefficient is /Gamma1=Zin−Z0 Zin+Z0=Z1(ZL−Z0)+jt/parenleftbig Z2 1−Z0ZL/parenrightbig Z1(ZL+Z0)+jt/parenleftbig Z2 1+Z0ZL/parenrightbig.( 5.27) Because Z2 1=Z0ZL, this reduces to /Gamma1=ZL−Z0 ZL+Z0+j2t√Z0ZL.( 5.28) The reflection coefficient magnitude is |/Gamma1|=|ZL−Z0| /bracketleftbig (ZL+Z0)2+4t2Z0ZL/bracketrightbig1/2 =1 /braceleftbig (ZL+Z0)2/(ZL−Z0)2+[4t2Z0ZL/(ZL−Z0)2]/bracerightbig1/2 =1 /braceleftbig 1+[4Z0ZL/(ZL−Z0)2]+[ 4Z0ZLt2/(ZL−Z0)2]/bracerightbig1/2 =1 /braceleftbig 1+[4Z0ZL/(ZL−Z0)2]sec2θ/bracerightbig1/2, (5.29) since 1 +t2=1+tan2θ=sec2θ. If we assume that the operating frequency is near the design frequency f0, then /lscript/similarequal λ0/4 and θ/similarequalπ/2. Then sec2θ/greatermuch1, and (5.29) simplifies to |/Gamma1|/similarequal|ZL−Z0| 2√Z0ZL|cosθ|forθnearπ/2.( 5.30) c05ImpedanceMatchingandTuning Pozar July 29, 2011 20:34 248 Chapter 5: Impedance Matching and Tuning /H9266/H9266 – /H9266 2⎪Γ⎪ Γm /H9258m /H9258 = /H9252l /H9258m0∆/H9258 FIGURE 5.11 Approximate behavior of the reflection coefficient magnitude for a single-section quarter-wave transformer operating near its design frequency. This result gives the approximate mismatch of the quarter-wave transformer near the design frequency, as sketched in Figure 5.11. If we set a maximum value, /Gamma1m, for an acceptable reflection coefficient magnitude, then the bandwidth of the matching transformer can be defined as /Delta1θ=2/parenleftBigπ 2−θm/parenrightBig ,( 5.31) since the response of (5.29) is symmetric about θ=π/2, and /Gamma1=/Gamma1matθ=θmand at θ=π−θm. Equating /Gamma1mto the exact expression for the reflection coefficient magnitude in (5.29) allows us to solve for θm: 1 /Gamma12m=1+/parenleftbigg2√Z0ZL ZL−Z0secθm/parenrightbigg2 , or cosθm=/Gamma1m/radicalbig 1−/Gamma12m2√Z0ZL |ZL−Z0|.( 5.32) If we assume TEM lines, then θ=β/lscript=2πf vpvp 4f0=πf 2f0, and so the frequency of the lower band edge at θ=θmis fm=2θmf0 π, and the fractional bandwidth is, using (5.32), /Delta1f f0=2(f0−fm) f0=2−2fm f0=2−4θm π =2−4 πcos−1/bracketleftBigg /Gamma1m/radicalbig 1−/Gamma12m2√Z0ZL |ZL−Z0|/bracketrightBigg . (5.33) Fractional bandwidth is usually expressed as a percentage, 100/Delta1 f/f0%.Note that the bandwidth of the transformer increases as ZLbecomes closer to Z0(a less mismatched load). The above results are strictly valid only for TEM lines. When non-TEM lines (such as waveguides) are used, the propagation constant is no longer a linear function of frequency, and the wave impedance will be frequency dependent. These factors serve to complicate c05ImpedanceMatchingandTuning Pozar July 29, 2011 20:34 5.4 The Quarter-Wave Transformer 249 f/f001200.250.50.751.0 ⎪Γ⎪ ZL/Z0 = 2, 0.5ZL/Z0 = 4, 0.25ZL/Z0 = 10, 0.1 FIGURE 5.12 Reflection coefficient magnitude versus frequency for a single-section quarter- wave matching transformer with various load mismatches. the general behavior of quarter-wave transformers for non-TEM lines, but in practice the bandwidth of the transformer is often small enough that these complications do not sub- stantially affect the result. Another factor ignored in the above analysis is the effect of reactances associated with discontinuities when there is a step change in the dimensions ofa transmission line. This can often be compensated by making a small adjustment in the length of the matching section. Figure 5.12 shows a plot of the reflection coefficient magnitude versus normalized frequency for various mismatched loads. Note the trend of increased bandwidth for smaller load mismatches. EXAMPLE 5.5 QUARTER-WA VE TRANSFORMER BANDWIDTH Design a single-section quarter-wave matching transformer to match a 10 /Omega1load to a 50 /Omega1transmission line at f0=3 GHz. Determine the percent bandwidth for which the SWR ≤1.5. Solution From (5.25), the characteristic impedance of the matching section is Z1=/radicalbig Z0ZL=/radicalbig (50)(10) =22.36 /Omega1, and the length of the matching section is λ/4 at 3 GHz (the physical length de- pends on the dielectric constant of the line). An SWR of 1.5 corresponds to a reflection coefficient magnitude of /Gamma1m=SWR−1 SWR+1=1.5−1 1.5+1=0.2. The fractional bandwidth is computed from (5.33) as /Delta1f f0=2−4 πcos−1/bracketleftBigg /Gamma1m/radicalbig 1−/Gamma12m2√Z0ZL |ZL−Z0|/bracketrightBigg =2−4 πcos−1/bracketleftBigg 0.2/radicalbig 1−(0.2)22√(50)(10) |10−50|/bracketrightBigg =0.29, or 29%. ■ c05ImpedanceMatchingandTuning Pozar July 29, 2011 20:34 250 Chapter 5: Impedance Matching and Tuning 5.5THETHEORYOFSMALLREFLECTIONS The quarter-wave transformer provides a simple means of matching any real load imped- ance to any transmission line impedance. For applications requiring more bandwidth than a single quarter-wave section can provide, multisection transformers can be used. The design of such transformers is the subject of the next two sections, but prior to that material we need to derive some approximate results for the total reflection coefficient caused by the partial reflections from several small discontinuities. This topic is generally referred to asthetheory of small reflections [1]. Single-SectionTransformer We will derive an approximate expression for the overall reflection coefficient, /Gamma1,f o r the single-section matching transformer shown in Figure 5.13. The partial reflection and transmission coefficients are /Gamma1 1=Z2−Z1 Z2+Z1, (5.34) /Gamma12=−/Gamma11, (5.35) /Gamma13=ZL−Z2 ZL+Z2, (5.36) T21=1+/Gamma11=2Z2 Z1+Z2, (5.37) T12=1+/Gamma12=2Z1 Z1+Z2. (5.38) We can compute the total reflection, /Gamma1, seen by the feed line using either the impedance method, or the multiple reflection method, as discussed in Section 2.5. For our present T21 T12 T12ZLZ2 Z1T21 T12 Γ3 Γ2 Γ1Γ/H9252l = θ Γ3 Γ3 Γ3Γ11 e–j/H9258 e–j/H9258 e–j/H9258e–j/H9258 e–j/H9258 FIGURE 5.13 Partial reflections and transmissions on a single-section matching transformer. c05ImpedanceMatchingandTuning Pozar July 29, 2011 20:34 5.5 The Theory of Small Reflections 251 purpose the latter technique is preferred, so we express the total reflection as an infinite sum of partial reflections and transmissions as follows: /Gamma1=/Gamma11+T12T21/Gamma13e−2jθ+T12T21/Gamma12 3/Gamma12e−4jθ+··· =/Gamma11+T12T21/Gamma13e−2jθ∞/summationdisplay n=0/Gamma1n 2/Gamma1n 3e−2jnθ. (5.39) The summation of the geometric series ∞/summationdisplay n=0xn=1 1−xfor|x|<1 allows us to express (5.39) in closed form as /Gamma1=/Gamma11+T12T21/Gamma13e−2jθ 1−/Gamma12/Gamma13e−2jθ.( 5.40) From (5.35), (5.37), and (5.38), we use /Gamma12=−/Gamma11,T21=1+/Gamma11, and T12=1−/Gamma11in (5.40) to give /Gamma1=/Gamma11+/Gamma13e−2jθ 1+/Gamma11/Gamma13e−2jθ.( 5.41) If the discontinuities between the impedances Z1,Z2andZ2,ZLare small, then |/Gamma11/Gamma13|=1, so we can approximate (5.41) as /Gamma1/similarequal/Gamma11+/Gamma13e−2jθ.( 5.42) This result expresses the intuitive idea that the total reflection is dominated by the reflection from the initial discontinuity between Z1and Z2(/Gamma11), and the first reflection from the discontinuity between Z2andZL(/Gamma13e−2jθ).T h e e−2jθterm accounts for the phase delay when the incident wave travels up and down the line. The accuracy of this approximationis illustrated in Problem 5.14. MultisectionTransformer Now consider the multisection transformer shown in Figure 5.14, which consists of N equal-length ( commensurate ) sections of transmission lines. We will derive an approximate expression for the total reflection coefficient /Gamma1. Partial reflection coefficients can be defined at each junction, as follows: /Gamma1 0=Z1−Z0 Z1+Z0, (5.43a) /Gamma1n=Zn+1−Zn Zn+1+Zn, (5.43b) /Gamma1N=ZL−ZN ZL+ZN. (5.43c) Z0Γ Γ0 Γ1 Γ2 ΓNZ1 Z2 ZN ZL/H9258 /H9258 /H9258 FIGURE 5.14 Partial reflection coefficients for a multisection matching transformer. c05ImpedanceMatchingandTuning Pozar July 29, 2011 20:34 252 Chapter 5: Impedance Matching and Tuning We also assume that all Znincrease or decrease monotonically across the transformer and that ZLis real. This implies that all /Gamma1nwill be real and of the same sign (/Gamma1n>0 ifZL>Z0;/Gamma1n<0i fZL<Z0). Using the results of the previous section allows us to approximate the overall reflection coefficient as /Gamma1(θ)=/Gamma10+/Gamma11e−2jθ+/Gamma12e−4jθ+···+ /Gamma1Ne−2jNθ.( 5.44) Further assume that the transformer can be made symmetrical, so that /Gamma10=/Gamma1N,/Gamma11= /Gamma1N−1,/Gamma12=/Gamma1N−2, and so on. (Note that this does notimply that the Znare symmetrical.) Then (5.44) can be written as /Gamma1(θ)=e−jNθ/braceleftBig /Gamma10[ejNθ+e−jNθ]+/Gamma11[ej(N−2)θ+e−j(N−2)θ]+···/bracerightBig .( 5.45) IfNis odd, the last term is /Gamma1(N−1)/2(ejθ+e−jθ), while if Nis even, the last term is /Gamma1N/2. Equation (5.45) is seen to be of the form of a finite Fourier cosine series in θ, which can be written as /Gamma1(θ)=2e−jNθ/bracketleftbigg /Gamma10cosNθ+/Gamma11cos(N−2)θ+···+ /Gamma1ncos(N−2n)θ + ···+1 2/Gamma1N/2/bracketrightbigg forNeven, (5.46a ) /Gamma1(θ)=2e−jNθ[/Gamma10cosNθ+/Gamma11cos(N−2)θ+···+ /Gamma1ncos(N−2n)θ +···+ /Gamma1(N−1)/2 cosθ]forNodd. (5.46b) The importance of these results lies in the fact that we can synthesize any desired reflection coefficient response as a function of frequency (θ ) by properly choosing the /Gamma1n and using enough sections ( N). This should be clear from the realization that a Fourier se- ries can approximate an arbitrary smooth function if enough terms are used. In the next two sections we will show how to use this theory to design multisection transformers for two of the most commonly used passband responses: the binomial (maximally flat) response, and the Chebyshev (equal-ripple) response. 5.6BINOMIALMULTISECTIONMATCHINGTRANSFORMERS The passband response (the frequency band where a good impedance match is achieved) of a binomial matching transformer is optimum in the sense that, for a given number of sections, the response is as flat as possible near the design frequency. This type of response, which is also known as maximally flat, is determined for an N-section transformer by setting the first N−1 derivatives of |/Gamma1(θ)|to zero at the center frequency, f0. Such a response can be obtained with a reflection coefficient of the following form: /Gamma1(θ)=A(1+e−2jθ)N.( 5.47) Then the reflection coefficient magnitude is |/Gamma1(θ)|=| A||e−jθ|N|ejθ+e−jθ|N =2N|A||cosθ|N(5.48) Note that |/Gamma1(θ)|=0f o rθ =π/2, and that dn|/Gamma1(θ)|/dθn=0a tθ=π/2f o rn =1,2,..., N−1. (θ=π/2 corresponds to the center frequency, f0,f o rw h i c h/lscript =λ/4 and θ= β/lscript=π/2.) c05ImpedanceMatchingandTuning Pozar July 29, 2011 20:34 5.6 Binomial Multisection Matching Transformers 253 We can determine the constant Aby letting f→0. Then θ=β/lscript=0, and (5.47) reduces to /Gamma1(0)=2NA=ZL−Z0 ZL+Z0, since for f=0 all sections are of zero electrical length. The constant Acan then be written as A=2−NZL−Z0 ZL+Z0.( 5.49) Next we expand /Gamma1(θ) in (5.47) according to the binomial expansion: /Gamma1(θ)=A(1+e−2jθ)N=AN/summationdisplay n=0CN ne−2jnθ,( 5.50) where CN n=N! (N−n)!n!(5.51) are the binomial coefficients. Note that CN n=CN N−n,CN 0=1, and CN 1=N=CN N−1.T h e key step is now to equate the desired passband response, given by (5.50), to the actual response as given (approximately) by (5.44): /Gamma1(θ)=AN/summationdisplay n=0CN ne−2jnθ=/Gamma10+/Gamma11e−2jθ+/Gamma12e−4jθ+···+ /Gamma1Ne−2jNθ. This shows that the /Gamma1nmust be chosen as /Gamma1n=ACN n.( 5.52) where Ais given by (5.49) and CN nis a binomial coefficient. At this point, the characteristic impedances, Zn, can be found via (5.43), but a simpler solution can be obtained using the following approximation [1]. Because we assumed that the/Gamma1nare small, we can write /Gamma1n=Zn+1−Zn Zn+1+Zn/similarequal1 2lnZn+1 Zn, since ln x/similarequal2(x−1)/(x+1)forxclose to unity. Then, using (5.52) and (5.49) gives lnZn+1 Zn/similarequal2/Gamma1n=2ACN n=2(2−N)ZL−Z0 ZL+Z0CN n/similarequal2−NCN nlnZL Z0,( 5.53) which can be used to find Zn+1, starting with n=0. This technique has the advantage of ensuring self-consistency, in that ZN+1computed from (5.53) will be equal to ZL,a si t should. Exact design results, including the effect of multiple reflections in each section, can be found by using the transmission line equations for each section and numerically solv-ing for the characteristic impedances [2]. The results of such calculations are listed in Table 5.1, which gives the exact line impedances for N=2-, 3-, 4-, 5-, and 6-section c05ImpedanceMatchingandTuning Pozar July 29, 2011 20:34 254 Chapter 5: Impedance Matching and Tuning TABLE 5.1 Binomial Transformer Design N=2 N=3 N=4 ZL/Z0 Z1/Z0 Z2/Z0 Z1/Z0 Z2/Z0 Z3/Z0 Z1/Z0 Z2/Z0 Z3/Z0 Z4/Z0 1.0 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.5 1.1067 1.3554 1.0520 1.2247 1.4259 1.0257 1.1351 1.3215 1.46242.0 1.1892 1.6818 1.0907 1.4142 1.8337 1.0444 1.2421 1.6102 1.9150 3.0 1.3161 2.2795 1.1479 1.7321 2.6135 1.0718 1.4105 2.1269 2.7990 4.0 1.4142 2.8285 1.1907 2.0000 3.3594 1.0919 1.5442 2.5903 3.66336.0 1.5651 3.8336 1.2544 2.4495 4.7832 1.1215 1.7553 3.4182 5.35008.0 1.6818 4.7568 1.3022 2.8284 6.1434 1.1436 1.9232 4.1597 6.9955 10.0 1.7783 5.6233 1.3409 3.1623 7.4577 1.1613 2.0651 4.8424 8.6110 N=5 N=6 ZL/Z0 Z1/Z0 Z2/Z0 Z3/Z0 Z4/Z0 Z5/Z0 Z1/Z0 Z2/Z0 Z3/Z0 Z4/Z0 Z5/Z0 Z6/Z0 1.0 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.5 1.0128 1.0790 1.2247 1.3902 1.4810 1.0064 1.0454 1.1496 1.3048 1.4349 1.49052.0 1.0220 1.1391 1.4142 1.7558 1.9569 1.0110 1.0790 1.2693 1.5757 1.8536 1.97823.0 1.0354 1.2300 1.7321 2.4390 2.8974 1.0176 1.1288 1.4599 2.0549 2.6577 2.94814.0 1.0452 1.2995 2.0000 3.0781 3.8270 1.0225 1.1661 1.6129 2.4800 3.4302 3.91206.0 1.0596 1.4055 2.4495 4.2689 5.6625 1.0296 1.2219 1.8573 3.2305 4.9104 5.82758.0 1.0703 1.4870 2.8284 5.3800 7.4745 1.0349 1.2640 2.0539 3.8950 6.3291 7.7302 10.0 1.0789 1.5541 3.1623 6.4346 9.2687 1.0392 1.2982 2.2215 4.5015 7.7030 9.6228 c05ImpedanceMatchingandTuning Pozar July 29, 2011 20:34 5.6 Binomial Multisection Matching Transformers 255 binomial matching transformers for various ratios of load impedance, ZL, to feed line impedance, Z0. The table gives results only for ZL/Z0>1; if ZL/Z0<1, the results for Z0/ZLshould be used but with Z1starting at the load end. This is because the solution is symmetric about ZL/Z0=1; the same transformer that matches ZLtoZ0can be reversed and used to match Z0toZL. More extensive tables can be found in reference [2]. The bandwidth of the binomial transformer can be evaluated as follows. As in Section 5.4, let /Gamma1mbe the maximum value of reflection coefficient that can be tolerated over the passband. Then from (5.48), /Gamma1m=2N|A|cosNθm, where θm<π/ 2 is the lower edge of the passband, as shown in Figure 5.11. Thus, θm=cos−1/bracketleftBigg 1 2/parenleftbigg/Gamma1m |A|/parenrightbigg1/N/bracketrightBigg ,( 5.54) and using (5.33) gives the fractional bandwidth as /Delta1f f0=2(f0−fm) f0=2−4θm π =2−4 πcos−1/bracketleftBigg 1 2/parenleftbigg/Gamma1m |A|/parenrightbigg1/N/bracketrightBigg . (5.55) EXAMPLE 5.6 BINOMIAL TRANSFORMER DESIGN Design a three-section binomial transformer to match a 50 /Omega1load to a 100 /Omega1 line and calculate the bandwidth for /Gamma1m=0.05. Plot the reflection coefficient magnitude versus normalized frequency for the exact designs using 1, 2, 3, 4, and 5 sections. Solution ForN=3,ZL=50/Omega1,andZ0=100/Omega1we have, from (5.49) and (5.53), A=2−NZL−Z0 ZL+Z0/similarequal1 2N+1lnZL Z0=−0.0433. From (5.55) the bandwidth is /Delta1f f0=2−4 πcos−1/bracketleftBigg 1 2/parenleftbigg/Gamma1m |A|/parenrightbigg1/N/bracketrightBigg =2−4 πcos−1/bracketleftBigg 1 2/parenleftbigg0.05 0.0433/parenrightbigg1/3/bracketrightBigg =0.70, or 70%. The necessary binomial coefficients are C3 0=3! 3!0!=1, C3 1=3! 2!1!=3, C3 2=3! 1!2!=3. c05ImpedanceMatchingandTuning Pozar July 29, 2011 20:34 256 Chapter 5: Impedance Matching and Tuning f/f01/300.10.20.3 1 5/3⎪Γ⎪N = 1 2 3 4 5 FIGURE 5.15 Reflection coefficient magnitude versus frequency for multisection binomial matching transformers of Example 5.6. ZL=50/Omega1andZ0=100/Omega1. Using (5.53) gives the required characteristic impedances as n=0:lnZ1=lnZ0+2−NC3 0lnZL Z0 =ln 100 +2−3(1)ln50 100=4.518, Z1=91.7/Omega1; n=1:lnZ2=lnZ1+2−NC3 1lnZL Z0 =ln 91.7 +2−3(3)ln50 100=4.26, Z2=70.7/Omega1; n=2:lnZ3=lnZ2+2−NC3 2lnZL Z0 =ln 70.7 +2−3(3)ln50 100=4.00, Z3=54.5/Omega1. To use the data in Table 5.1 we reverse the source and load impedances and consider the problem of matching a 100 /Omega1load to a 50 /Omega1line. Then ZL/Z0=2.0, and we obtain the exact characteristic impedances as Z1=91.7/Omega1,Z2=70.7/Omega1, andZ3=54.5/Omega1,which agree with the approximate results to three significant digits. Figure 5.15 shows the reflection coefficient magnitude versus frequency for exact designs using N=1, 2, 3, 4, and 5 sections. Observe that greater bandwidth is obtained for transformers using more sections. ■ 5.7CHEBYSHEVMULTISECTIONMATCHINGTRANSFORMERS In contrast with the binomial transformer, the multisection Chebyshev matching trans- former optimizes bandwidth at the expense of passband ripple. Compromising on the flat- ness of the passband response leads to a bandwidth that is substantially better than that of the binomial transformer for a given number of sections. The Chebyshev transformer is c05ImpedanceMatchingandTuning Pozar July 29, 2011 20:34 5.7 Chebyshev Multisection Matching Transformers 257 designed by equating /Gamma1(θ) to a Chebyshev polynomial, which has the optimum character- istics needed for this type of transformer. We will first discuss the properties of Chebyshev polynomials and then derive a design procedure for Chebyshev matching transformers us- ing the small-reflection theory of Section 5.5. ChebyshevPolynomials Thenth-order Chebyshev polynomial is a polynomial of degree n, denoted by Tn(x).T h e first four Chebyshev polynomials are T1(x)=x, (5.56a) T2(x)=2x2−1, (5.56b) T3(x)=4x3−3x, (5.56c) T4(x)=8x4−8x2+1. (5.56d) Higher order polynomials can be found using the following recurrence formula: Tn(x)=2xTn−1(x)−Tn−2(x). (5.57) The first four Chebyshev polynomials are plotted in Figure 5.16, from which the fol- lowing very useful properties of Chebyshev polynomials can be noted: rFor−1≤x≤1,|Tn(x)|≤ 1. In this range the Chebyshev polynomials oscillate between ±1. This is the equal-ripple property, and this region will be mapped to the passband of the matching transformer.rFor|x|>1,|Tn(x)|>1. This region will map to the frequency range outside the passband.rFor|x|>1, the |Tn(x)|increases faster with xasnincreases. 2 –2 –4 –646Tn(x) 0.5 –1.5 –1.0 1.54 3 2 n = 1 –0.5 x 1.0 FIGURE 5.16 The first four Chebyshev polynomials, Tn(x). c05ImpedanceMatchingandTuning Pozar July 29, 2011 20:34 258 Chapter 5: Impedance Matching and Tuning Now let x=cosθfor|x|<1. Then it can be shown that the Chebyshev polynomials can be expressed as Tn(cosθ)=cosnθ, or more generally as Tn(x)=cos(ncos−1x) for|x|<1, (5.58a) Tn(x)=cosh(ncosh−1x)forx>1. (5.58b) We desire equal ripple for the passband response of the transformer, so it is necessary to mapθmtox=1 and π−θmtox=−1, where θmandπ−θmare the lower and upper edges of the passband, respectively, as shown in Figure 5.11. This can be accomplished by replacing cos θin (5.58a) with cos θ/cosθm: Tn/parenleftbiggcosθ cosθm/parenrightbigg =Tn(secθmcosθ)=cosn/bracketleftbigg cos−1/parenleftbiggcosθ cosθm/parenrightbigg/bracketrightbigg .( 5.59) Then|secθmcosθ|≤1f o r θm<θ<π −θm,so|Tn(secθmcosθ)|≤1 over this same range. Because cosnθcan be expanded into a sum of terms of the form cos(n −2m)θ,t h e Chebyshev polynomials of (5.56) can be rewritten in the following useful form: T1(secθmcosθ)=secθmcosθ, (5.60a) T2(secθmcosθ)=sec2θm(1+cos 2θ) −1, (5.60b) T3(secθmcosθ)=sec3θm(cos 3θ +3 cosθ)−3s e cθmcosθ, (5.60c) T4(secθmcosθ)=sec4θm(cos 4θ +4 cos 2θ +3) −4s e c2θm(cos 2θ +1)+1. (5.60d) These results can be used to design matching transformers with up to four sections, and will also be used in later chapters for the design of directional couplers and filters. DesignofChebyshevTransformers We can now synthesize a Chebyshev equal-ripple passband by making /Gamma1(θ) proportional toTN(secθmcosθ), where Nis the number of sections in the transformer. Thus, using (5.46), we have /Gamma1(θ)=2e−jNθ[/Gamma10cosNθ+/Gamma11cos(N−2)θ+···+ /Gamma1ncos(N−2n)θ+···] =Ae−jNθTN(secθmcosθ), (5.61) where the last term in the series of (5.61) is (1/2)/Gamma1 N/2forNeven and /Gamma1(N−1)/2 cosθfor Nodd. As in the binomial transformer case, we can find the constant Aby letting θ=0, corresponding to zero frequency. Thus, /Gamma1(0)=ZL−Z0 ZL+Z0=ATN(secθm), so we have A=ZL−Z0 ZL+Z01 TN(secθm).( 5.62) If the maximum allowable reflection coefficient magnitude in the passband is /Gamma1m, then from (5.61) /Gamma1m=|A|since the maximum value of Tn(secθmcosθ)in the passband is unity. c05ImpedanceMatchingandTuning Pozar July 29, 2011 20:34 5.7 Chebyshev Multisection Matching Transformers 259 Then (5.62) gives TN(secθm)=1 /Gamma1m/vextendsingle/vextendsingle/vextendsingle/vextendsingleZL−Z0 ZL+Z0/vextendsingle/vextendsingle/vextendsingle/vextendsingle, which, after using (5.58b) and the approximations introduced in Section 5.6, allows us to determine θmas secθm=cosh/bracketleftbigg1 Ncosh−1/parenleftbigg1 /Gamma1m/vextendsingle/vextendsingle/vextendsingle/vextendsingleZ L−Z0 ZL+Z0/vextendsingle/vextendsingle/vextendsingle/vextendsingle/parenrightbigg/bracketrightbigg /similarequalcosh/bracketleftbigg1 Ncosh−1/parenleftbigg/vextendsingle/vextendsingle/vextendsingle/vextendsinglelnZ L/Z0 2/Gamma1m/vextendsingle/vextendsingle/vextendsingle/vextendsingle/parenrightbigg/bracketrightbigg . (5.63) Onceθ mis known, the fractional bandwidth can be calculated from (5.33) as /Delta1f f0=2−4θm π.( 5.64) From (5.61), the /Gamma1ncan be determined using the results of (5.60) to expand TN(secθmcosθ) and equating similar terms of the form cos (N−2n)θ. The characteristic impedances Zn can be found from (5.43), although, as in the case of the binomial transformer, accuracy can be improved and self-consistency can be achieved by using the approximation that /Gamma1n/similarequal1 2lnZn+1 Zn. This procedure will be illustrated in Example 5.7. The above results are approximate because of the reliance on small-reflection theory but are general enough to design transformers with an arbitrary ripple level, /Gamma1m. Table 5.2 gives exact results [2] for a few specific values of /Gamma1mforN=2, 3, and 4 sections; more extensive tables can be found in reference [2]. EXAMPLE 5.7 CHEBYSHEV TRANSFORMER DESIGN Design a three-section Chebyshev transformer to match a 100 /Omega1load to a 50 /Omega1 line with /Gamma1m=0.05, using the above theory. Plot the reflection coefficient mag- nitude versus normalized frequency for exact designs using 1, 2, 3, and 4 sections. Solution From (5.61) with N=3, /Gamma1(θ)=2e−j3θ(/Gamma10cos 3θ +/Gamma11cosθ)=Ae−j3θT3(secθmcosθ). Then A=/Gamma1m=0.05, and from (5.63), secθm=cosh/bracketleftbigg1 Ncosh−1/parenleftbigglnZL/Z0 2/Gamma1m/parenrightbigg/bracketrightbigg =cosh/bracketleftbigg1 3cosh−1/parenleftbiggln(100/50) 2(0.05)/parenrightbigg/bracketrightbigg =1.408, soθm=44.7◦. Using (5.60c) for T3gives 2(/Gamma10cos 3θ +/Gamma11cosθ)=Asec3θm(cos 3θ +3 cosθ)−3Asecθmcosθ. c05ImpedanceMatchingandTuning Pozar July 29, 2011 20:34 260 Chapter 5: Impedance Matching and Tuning TABLE 5.2 Chebyshev Transformer Design N=2 N=3 /Gamma1m=0.05 /Gamma1m=0.20 /Gamma1m=0.05 /Gamma1m=0.20 ZL/Z0 Z1/Z0 Z2/Z0 Z1/Z0 Z2/Z0 Z1/Z0 Z2/Z0 Z3/Z0 Z1/Z0 Z2/Z0 Z3/Z0 1.0 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.5 1.1347 1.3219 1.2247 1.2247 1.1029 1.2247 1.3601 1.2247 1.2247 1.22472.0 1.2193 1.6402 1.3161 1.5197 1.1475 1.4142 1.7429 1.2855 1.4142 1.55583.0 1.3494 2.2232 1.4565 2.0598 1.2171 1.7321 2.4649 1.3743 1.7321 2.18294.0 1.4500 2.7585 1.5651 2.5558 1.2662 2.0000 3.1591 1.4333 2.0000 2.79086.0 1.6047 3.7389 1.7321 3.4641 1.3383 2.4495 4.4833 1.5193 2.4495 3.94928.0 1.7244 4.6393 1.8612 4.2983 1.3944 2.8284 5.7372 1.5766 2.8284 5.0742 10.0 1.8233 5.4845 1.9680 5.0813 1.4385 3.1623 6.9517 1.6415 3.1623 6.0920 N=4 /Gamma1m=0.05 /Gamma1m=0.20 ZL/Z0 Z1/Z0 Z2/Z0 Z3/Z0 Z4/Z0 Z1/Z0 Z2/Z0 Z3/Z0 Z4/Z0 1.0 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.5 1.0892 1.1742 1.2775 1.3772 1.2247 1.2247 1.2247 1.22472.0 1.1201 1.2979 1.5409 1.7855 1.2727 1.3634 1.4669 1.57153.0 1.1586 1.4876 2.0167 2.5893 1.4879 1.5819 1.8965 2.01634.0 1.1906 1.6414 2.4369 3.3597 1.3692 1.7490 2.2870 2.92146.0 1.2290 1.8773 3.1961 4.8820 1.4415 2.0231 2.9657 4.16238.0 1.2583 2.0657 3.8728 6.3578 1.4914 2.2428 3.5670 5.3641 10.0 1.2832 2.2268 4.4907 7.7930 1.5163 2.4210 4.1305 6.5950 Equating similar terms in cos nθgives the following results: cos 3θ :2/Gamma10=Asec3θm, /Gamma10=0.0698; cosθ:2/Gamma11=3A(sec3θm−secθm), /Gamma11=0.1037. From symmetry we also have that /Gamma13=/Gamma10=0.0698, /Gamma12=/Gamma11=0.1037. Then the characteristic impedances are: n=0: lnZ1=lnZ0+2/Gamma10 =ln 50+2(0.0698) =4.051 Z1=57.5/Omega1; n=1: lnZ2=lnZ1+2/Gamma11 =ln 57.5 +2(0.1037) =4.259 Z2=70.7/Omega1; n=2: lnZ3=lnZ2+2/Gamma12 =ln 70.7 +2(0.1037) =4.466 Z3=87.0/Omega1. c05ImpedanceMatchingandTuning Pozar July 29, 2011 20:34 5.8 Tapered Lines 261 f/f01/300.10.20.3 1 5/3⎪Γ⎪N = 1 2 3 4 FIGURE 5.17 Reflection coefficient magnitude versus frequency for the multisection matching transformers of Example 5.7. These values can be compared to the exact values from Table 5.2 of Z1=57.37 /Omega1, Z2=70.71 /Omega1, and Z3=87.15 /Omega1. The bandwidth, from (5.64), is /Delta1f f0=2−4θm π=2−4/parenleftbigg44.7◦ 180◦/parenrightbigg =1.01, or 101%. This is significantly greater than the bandwidth of the binomial trans- former of Example 5.6 (70%), which involved the same impedance mismatch.The trade-off, of course, is a nonzero ripple in the passband of the Chebyshev transformer. Figure 5.17 shows reflection coefficient magnitudes versus frequency for the exact designs from Table 5.2 for N=1, 2, 3, and 4 sections. ■ 5.8TAPEREDLINES In the preceding sections we discussed how an arbitrary real load impedance could be matched to a line over a desired bandwidth by using multisection matching transformers.As the number Nof discrete transformer sections increases, the step changes in charac- teristic impedance between the sections become smaller, and the transformer geometry approaches a continuously tapered line. In practice, of course, a matching transformermust be of finite length—often no more than a few sections long. This suggests that, instead of discrete sections, the transformer can be continuously tapered, as shown in Figure 5.18a. Different passband characteristics can be obtained by using different types of taper. In this section we will derive an approximate theory, again based on the theory of small reflections, to predict the reflection coefficient response as a function of the impedance taper versus position, Z(z). We will apply these results to a few common types of imped- ance tapers. Consider the continuously tapered line of Figure 5.18a as being made up of a num- ber of incremental sections of length /Delta1z, with an impedance change /Delta1Z(z)from one c05ImpedanceMatchingandTuning Pozar July 29, 2011 20:34 262 Chapter 5: Impedance Matching and Tuning z z zz + ∆z0 (a) (b)LZL Z(z) Z0 ZZ + ∆ Z∆Γ FIGURE 5.18 A tapered transmission line matching section and the model for an incremen- tal length of tapered line. (a) The tapered transmission line matching section. (b) Model for an incremental step change in impedance of the tapered line. section to the next, as shown in Figure 5.18b. The incremental reflection coefficient from the impedance step at zis given by /Delta1/Gamma1=(Z+/Delta1Z)−Z (Z+/Delta1Z)+Z/similarequal/Delta1Z 2Z.( 5.65) In the limit as /Delta1z→0 we have an exact differential: d/Gamma1=dZ 2Z=1 2d(lnZ/Z0) dzdz,( 5.66) since d(lnf(z)) dz=1 fdf(z) dz. By using the theory of small reflections, we can find the total reflection coefficient at z=0 by summing all the partial reflections with their appropriate phase shifts: /Gamma1(θ)=1 2/integraldisplayL z=0e−2jβzd dzln/parenleftbiggZ Z0/parenrightbigg dz,( 5.67) where θ=2β/lscript.I fZ(z)is known, /Gamma1(θ) can be found as a function of frequency. Alter- natively, if /Gamma1(θ) is specified, then in principle Z(z)can be found by inversion. This latter procedure is difficult, and is generally avoided in practice; the reader is referred to refer- ences [1] and [4] for further discussion of this topic. Here we will consider three specialcases of Z(z)impedance tapers, and evaluate the resulting responses. ExponentialTaper Consider first an exponential taper , where Z(z)=Z 0eazfor 0<z<L,( 5.68) c05ImpedanceMatchingandTuning Pozar July 29, 2011 20:34 5.8 Tapered Lines 263 Z(z) ZL Z0 z L 0 (a) /H9252L 0 /H9266 2/H9266 3/H9266 4/H9266 5/H9266 (b)⎪Γ⎪ FIGURE 5.19 A matching section with an exponential impedance taper. (a) Variation of imped- ance. (b) Resulting reflection coefficient magnitude response. as indicated in Figure 5.19a. At z=0,Z(0)=Z0, as desired. At z=Lwe wish to have Z(L)=ZL=Z0eaL, which determines the constant aas a=1 Lln/parenleftbiggZL Z0/parenrightbigg .( 5.69) We find /Gamma1(θ) by using (5.68) and (5.69) in (5.67): /Gamma1=1 2/integraldisplayL 0e−2jβzd dz(lneaz)dz =lnZL/Z0 2L/integraldisplayL 0e−2jβzdz =lnZL/Z0 2e−jβLsinβL βL. (5.70) Observe that this derivation assumes that β, the propagation constant of the tapered line, is not a function of z—an assumption generally valid only for TEM lines. The magnitude of the reflection coefficient in (5.70) is sketched in Figure 5.19b; note that the peaks in |/Gamma1|decrease with increasing length, as one might expect, and that the length should be greater than λ/2(βL>π ) to minimize the mismatch at low frequencies. TriangularTaper Next consider a triangular taper fordln(Z/Z0)/dz, that is, Z(z)=/braceleftBigg Z0e2(z/L)2lnZL/Z0 for 0≤z≤L/2 Z0e(4z/L−2z2/L2−1)lnZL/Z0forL/2≤z≤L,(5.71) c05ImpedanceMatchingandTuning Pozar July 29, 2011 20:34 264 Chapter 5: Impedance Matching and Tuning Z(z) ZL Z0Z0 ZL z L L 0 ⎪Γ⎪(a) /H9252L 003/H9266 2/H9266 /H9266 4/H9266 5/H9266 6/H9266 (b)2 FIGURE 5.20 A matching section with a triangular taper for d(lnZ/Z0)/dz.(a) Variation of impedance. (b) Resulting reflection coefficient magnitude response. so that the derivative is triangular in form: d(lnZ/Z0) dz=/braceleftBigg 4z/L2lnZL/Z0 for 0≤z≤L/2 (4/L−4z/L2)lnZL/Z0forL/2≤z≤L.(5.72) Z(z)is plotted in Figure 5.20a. Evaluating /Gamma1from (5.67) gives /Gamma1(θ)=1 2e−jβLln/parenleftbiggZL Z0/parenrightbigg/bracketleftbiggsin(βL/2) βL/2/bracketrightbigg2 .( 5.73) The magnitude of this result is sketched in Figure 5.20b. Note that, for βL>2π,t h e peaks of the triangular taper are lower than the corresponding peaks of the exponential case. However, the first null for the triangular taper occurs at βL=2π, whereas for the exponential taper it occurs at βL=π. KlopfensteinTaper Considering the fact that there is an infinite number of possibilities for choosing an impedance matching taper, it is logical to ask if there is a design that is “best.” For a given taper length (greater than some critical value), the Klopfenstein impedance taper [4, 5] has been shown to be optimum in the sense that the reflection coefficient is minimum over thepassband. Alternatively, for a maximum reflection coefficient specification in the passband, the Klopfenstein taper yields the shortest matching section. The Klopfenstein taper is derived from a stepped Chebyshev transformer as the num- ber of sections increases to infinity, and is analogous to the Taylor distribution of antenna array theory. We will not present the details of this derivation, which can be found in c05ImpedanceMatchingandTuning Pozar July 29, 2011 20:34 5.8 Tapered Lines 265 references [1] and [4]; only the necessary results for the design of Klopfenstein tapers are given in what follows. The logarithm of the characteristic impedance variation for the Klopfenstein taper is given by lnZ(z)=1 2lnZ0ZL+/Gamma10 cosh AA2φ(2z/L−1,A)for 0≤z≤L,( 5.74) where the function φ(x,A)is defined as φ(x,A)=−φ(−x,A)=/integraldisplayx 0I1(A/radicalbig 1−y2) A/radicalbig 1−y2dy for|x|≤1,( 5.75) where I1(x)is the modified Bessel function. The function of (5.75) has the following spe- cial values: φ(0,A)=0 φ(x,0)=x 2 φ(1,A)=cosh A−1 A2, but otherwise (5.75) must be calculated numerically. A simple and efficient method for doing this is available [6]. The resulting reflection coefficient is given by /Gamma1(θ)=/Gamma10e−jβLcos/radicalbig (βL)2−A2 cosh AforβL>A.( 5.76) IfβL<A, the cos/radicalbig (βL)2−A2term becomes cosh/radicalbig A2−(βL)2. In (5.74) and (5.76), /Gamma10is the reflection coefficient at zero frequency, given as /Gamma10=ZL−Z0 ZL+Z0/similarequal1 2ln/parenleftbiggZL Z0/parenrightbigg .( 5.77) The passband is defined as βL≥A, and so the maximum ripple in the passband is /Gamma1m=/Gamma10 cosh A(5.78) because /Gamma1(θ) oscillates between ±/Gamma10/cosh AforβL>A. It is interesting to note that the impedance taper of (5.74) has steps at z=0 and L(the ends of the tapered section) and so does not smoothly join the source and load impedances. A typical Klopfenstein impedance taper and its response are given in the fol- lowing example. EXAMPLE 5.8 DESIGN OF TAPERED MATCHING SECTIONS Design a triangular taper, an exponential taper, and a Klopfenstein taper (with /Gamma1m=0.02) to match a 50 /Omega1load to a 100 /Omega1line. Plot the impedance variations and resulting reflection coefficient magnitudes versus βL. Solution Triangular taper: From (5.71) the impedance variation is Z(z)=Z0/braceleftBigg e2(z/L)2lnZL/Z0 for 0≤z≤L/2 e(4z/L−2z2/L2−1)lnZL/Z0forL/2≤z≤L, c05ImpedanceMatchingandTuning Pozar July 29, 2011 20:34 266 Chapter 5: Impedance Matching and Tuning FIGURE 5.21 Solution to Example 5.8. (a) Impedance variations for the triangular, exponential, and Klopfenstein tapers. (b) Resulting reflection coefficient magnitude versus fre- quency for the tapers of (a). with Z0=100/Omega1andZL=50/Omega1. The resulting reflection coefficient response is given by (5.73): |/Gamma1(θ)|=1 2ln/parenleftbiggZL Z0/parenrightbigg/bracketleftbiggsin(βL/2) βL/2/bracketrightbigg2 . Exponential taper: From (5.68) the impedance variation is Z(z)=Z0eazfor 0<z<L, with a=(1/L)lnZL/Z0=0.693/ L.The reflection coefficient response is, from (5.70), |/Gamma1(θ)|=1 2ln/parenleftbiggZL Z0/parenrightbiggsinβL βL. Klopfenstein taper: Using (5.77) gives /Gamma10as /Gamma10=1 2ln/parenleftbiggZL Z0/parenrightbigg =0.346, c05ImpedanceMatchingandTuning Pozar July 29, 2011 20:34 5.9 The Bode–Fano Criterion 267 and (5.78) gives Aas A=cosh−1/parenleftbigg/Gamma10 /Gamma1m/parenrightbigg =cosh−1/parenleftbigg0.346 0.02/parenrightbigg =3.543. The impedance taper must be numerically evaluated from (5.74). The reflection coefficient magnitude is given by (5.76): |/Gamma1(θ)|=/Gamma10cos/radicalbig (βL)2−A2 cosh A. The passband for the Klopfenstein taper is defined as βL>A=3.543 =1.13π . Figure 5.21 shows the impedance variations (vs. z/L), and the resulting re- flection coefficient magnitude (vs. βL)for the three types of tapers. The Klopfen- stein taper gives the desired response of |/Gamma1|≤/Gamma1 m=0.02 for βL≥1.13π , which is smaller than the corresponding lengths of either the triangular or the expo-nential taper transformer. Also note that, like the stepped-Chebyshev matching transformer, the response of the Klopfenstein taper has equal-ripple lobes versus frequency in its passband. ■ 5.9THEBODE–FANOCRITERION In this chapter we discussed several techniques for matching an arbitrary load at a single frequency, using lumped elements, tuning stubs, and single-section quarter-wave trans-formers. We presented multisection matching transformers and tapered lines as a means of obtaining broader bandwidths with various passband characteristics. We close our study of impedance matching with a somewhat qualitative discussion of the theoretical limits thatconstrain the performance of an impedance matching network. We limit our discussion to the circuit of Figure 5.1, where a lossless network is used to match an arbitrary complex load, generally over a nonzero bandwidth. From a very general perspective, we might raise the following questions in regard to this problem: rCan we achieve a perfect match (zero reflection) over a specified bandwidth?rIf not, how well can we do? What is the trade-off between /Gamma1m, the maximum allow- able reflection in the passband, and the bandwidth?rHow complex must the matching network be for a given specification? These questions can be answered by the Bode–Fano criterion [7, 8] which gives, for certain canonical types of load impedances, a theoretical limit on the minimum reflec- tion coefficient magnitude that can be obtained with an arbitrary matching network. TheBode–Fano criterion thus represents an optimum result that can be ideally achieved, even though such a result may only be approximated in practice. Such optimal results are always important, however, because they specify an upper limit of performance, and so provide a benchmark against which a practical design can be compared. Figure 5.22a shows a lossless network used to match a parallel RCload impedance. The Bode–Fano criterion states that/integraldisplay ∞ 0ln1 |/Gamma1(ω)|dω≤π RC,( 5.79) where /Gamma1(ω) is the reflection coefficient seen looking into the arbitrary lossless match- ing network. The derivation of this result is beyond the scope of this text (the interested reader is referred to references [7] and [8]); our goal here is to discuss the implications of this result. c05ImpedanceMatchingandTuning Pozar July 29, 2011 20:34 268 Chapter 5: Impedance Matching and Tuning Circuit Bode-Fano limit (a)R CΓ(/H9275)Lossless matching network (b)R CΓ(/H9275)Lossless matching network (c) (d)R LΓ(/H9275)Lossless matching network R LΓ(/H9275)Lossless matching network0ln d/H9275/H9266<RC1 ⎪Γ(/H9275)⎪ ln d/H9275<1 ⎪Γ(/H9275)⎪ ln d/H9275/H9266R<L1 ⎪Γ(/H9275)⎪/H9266L R/H9266RC∫ 0∫ 0∫/H11009/H11009/H11009 /H927521 ln d/H9275<1 ⎪Γ(/H9275)⎪0∫/H11009 /H927521 FIGURE 5.22 The Bode–Fano limits for RCandRLloads matched with passive and lossless networks ( ω0is the center frequency of the matching bandwidth). (a) Parallel RC. (b) Series RC. (c) Parallel RL.( d )S e r i e s RL. Assume that we desire to synthesize a matching network with a reflection coefficient response like that shown in Figure 5.23a. Applying (5.79) to this function gives /integraldisplay∞ 0ln1 |/Gamma1|dω=/integraldisplay /Delta1ωln1 /Gamma1mdω=/Delta1ωln1 /Gamma1m≤π RC,( 5.80) which leads to the following conclusions: rFor a given load (a fixed RCproduct), a broader bandwidth (/Delta1ω ) can be achieved only at the expense of a higher reflection coefficient in the passband ( /Gamma1m).rThe passband reflection coefficient, /Gamma1m, cannot be zero unless /Delta1ω=0. Thus a perfect match can be achieved only at a finite number of discrete frequencies, as illustrated in Figure 5.23b.rAsRand/or Cincreases, the quality of the match ( /Delta1ωand/or 1//Gamma1 m) must decrease. Thus, higher- Qcircuits are intrinsically harder to match than are lower- Qcircuits (we will discuss Qin Chapter 6). Because ln (1/|/Gamma1|)is proportional to the return loss (in dB) at the input of the matching network, (5.79) can be interpreted as requiring that the area between the return loss curve c05ImpedanceMatchingandTuning Pozar July 29, 2011 20:34 References 269 ⎪Γ⎪ /H9275 ∆/H927501 Γm ⎪Γ⎪ /H9275Not realizable⎪Γ⎪ /H9275Realizable(a) (b) FIGURE 5.23 Illustrating the Bode–Fano criterion. (a) A possible reflection coefficient response. (b) Nonrealizable and realizable reflection coefficient responses. and the |/Gamma1|=1(RL=0 dB) axis must be less than or equal to a particular constant. Optimization then implies that the return loss curve be adjusted so that |/Gamma1|=/Gamma1mover the passband and |/Gamma1|=1 elsewhere, as in Figure 5.23a. In this way, no area under the return loss curve is wasted outside the passband, or lost in regions within the passband for which |/Gamma1|</Gamma1 m. The square-shaped response of Figure 5.23a is therefore the optimum response, but cannot be realized in practice because it would require an infinite number of elements in the matching network. It can be approximated, however, with a reasonably small number of elements, as described in reference [8]. Finally, note that the Chebyshevmatching transformer can be considered as a close approximation to the ideal passband of Figure 5.23a when the ripple of the Chebyshev response is made equal to /Gamma1 m. Figure 5.22 lists the Bode–Fano limits for other types of RCandRLloads. REFERENCES [1] R. E. Collin, Foundations for Microwave Engineering , 2nd edition, McGraw-Hill, New York, 1992. [2] G. L. Matthaei, L. Young, and E. M. T. Jones, Microwave Filters, Impedance-Matching Networks, and Coupling Structures, Artech House Books, Dedham, Mass. 1980. [3] P. Bhartia and I. J. Bahl, Millimeter Wave Engineering and Applications , Wiley Interscience, New York, 1984. [4] R. E. Collin, “The Optimum Tapered Transmission Line Matching Section,” Proceedings of the IRE, vol. 44, pp. 539–548, April 1956. [5] R. W. Klopfenstein, “A Transmission Line Taper of Improved Design,” Proceedings of the IRE, vol. 44, pp. 31–15, January 1956. [6] M. A. Grossberg, “Extremely Rapid Computation of the Klopfenstein Impedance Taper,” Proceed- ings of the IEEE, vol. 56, pp. 1629–1630, September 1968. [7] H. W. Bode, Network Analysis and Feedback Amplifier Design, Van Nostrand, New York, 1945. [8] R. M. Fano, “Theoretical Limitations on the Broad-Band Matching of Arbitrary Impedances,” Journal of the Franklin Institute, vol. 249, pp. 57–83, January 1950, and pp. 139–154, February 1950. c05ImpedanceMatchingandTuning Pozar July 29, 2011 20:34 270 Chapter 5: Impedance Matching and Tuning PROBLEMS 5.1 Design two lossless L-section matching circuits to match each of the following loads to a 100 /Omega1 generator at 3 GHz. (a) ZL=150−j200/Omega1and (b) ZL=20−j90/Omega1. 5.2 We have seen that the matching of an arbitrary load impedance requires a network with at least two degrees of freedom. Determine the types of load impedances/admittances that can be matched with the two single-element networks shown below. Z0 ZL (a)jX Z0 YL (b)jB 5.3 A load impedance ZL=100+j80/Omega1is to be matched to a 75 /Omega1line using a single shunt-stub tuner. Find two designs using open-circuited stubs. 5.4 Repeat Problem 5.3 using short-circuited stubs. 5.5 A load impedance ZL=90+j60/Omega1is to be matched to a 75 /Omega1line using a single series-stub tuner. Find two designs using open-circuited stubs. 5.6 Repeat Problem 5.5 using short-circuited stubs. 5.7 In the circuit shown below a load ZL=200+j100/Omega1is to be matched to a 40 /Omega1line, using a length /lscriptof lossless transmission line of characteristic impedance Z1. Find /lscriptandZ1. Determine, in general, what type of load impedances can be matched using such a circuit. l Z0 = 40 Ω Z1ZLZL = 200 + j100 Ω 5.8 An open-circuit tuning stub is to be made from a lossy transmission line with an attenuation con- stantα. What is the maximum value of normalized reactance that can be obtained with this stub? What is the maximum value of normalized reactance that can be obtained with a shorted stub of thesame type of transmission line? Assume α/lscriptis small. 5.9 Design a double-stub tuner using open-circuited stubs with a λ/8 spacing to match a load admittance Y L=(0.4+j1.2)Y0. 5.10 Repeat Problem 5.9 using a double-stub tuner with short-circuited stubs and a 3 λ/8 spacing. 5.11 Derive the design equations for a double-stub tuner using two series stubs spaced a distance dapart. Assume the load impedance is ZL=RL+jXL. 5.12 Consider matching a load ZL=200/Omega1to a 100 /Omega1line, using single shunt-stub, single series stub, and double shunt-stub tuners, with short-circuited stubs. Which tuner will give the best bandwidth? Justify your answer by calculating the reflection coefficient for all six solutions at 1 .1f0,w h e r e f0is the match frequency, or use CAD to plot the reflection coefficient versus frequency. 5.13 Design a single-section quarter-wave matching transformer to match a 350 /Omega1load to a 100 /Omega1line. What is the percent bandwidth of this transformer, for SWR ≤2? If the design frequency is 4 GHz, sketch the layout of a microstrip circuit, including dimensions, to implement this matching trans-former. Assume the substrate is 0.159 cm thick, with a relative permittivity of 2.2. 5.14 Consider the quarter-wave transformer of Figure 5.13 with Z 1=100/Omega1,Z2=150/Omega1,a n d ZL= 225/Omega1.Evaluate the worst-case percent error in computing |/Gamma1|from the approximate expression (5.42), compared to the exact result. c05ImpedanceMatchingandTuning Pozar July 29, 2011 20:34 Problems 271 5.15 A waveguide load with an equivalent TE 10wave impedance of 377 /Omega1must be matched to an air-filled X-band rectangular guide at 10 GHz. A quarter-wave matching transformer is to be used, and is to consist of a section of guide filled with dielectric. Find the required dielectric constant and physical length of the matching section. What restrictions on the load impedance apply to this technique? 5.16 A four-section binomial matching transformer is to be used to match a 12.5 /Omega1load to a 50 /Omega1 line at a center frequency of 1 GHz. (a) Design the matching transformer, and compute the band- width for /Gamma1m=0.05. Use CAD to plot the input reflection coefficient versus frequency. (b) Lay out the microstrip implementation of this circuit on an FR4 substrate having /epsilon1r=4.2,d=0.158 cm, and tan δ=0.02, with copper conductors 0.5 mil thick. Use CAD to plot the insertion loss versus frequency. 5.17 Derive the exact characteristic impedance for a two-section binomial matching transformer for a normalized load impedance ZL/Z0=1.5. Check your results with Table 5.1. 5.18 Calculate and plot the percent bandwidth for an N=1-, 2-, and 4-section binomial matching trans- former versus ZL/Z0=1.5t o6f o r/Gamma1 m=0.2. 5.19 Design a four-section Chebyshev matching transformer to match a 50 /Omega1line to a 30 /Omega1load. The maximum permissible SWR over the passband is 1.25. What is the resulting bandwidth? Use the approximate theory developed in the text, as opposed to the tables. Use CAD to plot the input SWRversus frequency. 5.20 Derive the exact characteristic impedances for a two-section Chebyshev matching transformer for a normalized load impedance Z L/Z0=1.5. Check your results with Table 5.2 for /Gamma1m=0.05. 5.21 Al o a do f ZL/Z0=1.5 is to be matched to a feed line using a multisection transformer, and it is desired to have a passband response with |/Gamma1(θ)|=A(0.1+cos2θ)for 0≤θ≤π. Use the approx- imate theory for multisection transformers to design a two-section transformer. 5.22 A tapered matching section has dln(Z/Z0)/dz=Asinπz/L. Find the constant Aso that Z(0)= Z0andZ(L)=ZL.Compute /Gamma1, and plot |/Gamma1|versus βL. 5.23 Design an exponentially tapered matching transformer to match a 100 /Omega1load to a 50 /Omega1line. Plot |/Gamma1| versus βL, and find the length of the matching section (at the center frequency) required to obtain |/Gamma1|≤0.05 over a 100% bandwidth. How many sections would be required if a Chebyshev matching transformer were used to achieve the same specifications? 5.24 An ultra wideband (UWB) radio transmitter, operating from 3.1 to 10.6 GHz, drives a parallel RC load with R=75/Omega1andC=0.6 pF. What is the best return loss that can be obtained with an optimum matching network? 5.25 Consider a series RLload with R=80/Omega1andL=5 nH. Design a lumped-element L-section match- ing network to match this load to a 50 /Omega1line at 2 GHz. Plot |/Gamma1|versus frequency for this network to determine the bandwidth for which |/Gamma1|≤/Gamma1m=0.1. Compare this with the maximum possible bandwidth for this load, as given by the Bode–Fano criterion. (Assume a square reflection coefficientresponse like that of Figure 5.23a.) c06MicrowaveResonators Pozar August 5, 2011 18:28 Chapter Six Microwave Resonators Microwave resonators are used in a variety of applications, including filters, oscillators, frequency meters, and tuned amplifiers. Because the operation of microwave resonators is very similar to that of lumped-element resonators of circuit theory, we will begin by reviewing the basic characteristics of series and parallel RLC resonant circuits. We will then discuss various implementations of resonators at microwave frequencies using distributed elements such astransmission lines, rectangular and circular waveguides, and dielectric cavities. We will also discuss the excitation of resonators using apertures and current sheets. 6.1SERIESANDPARALLELRESONANTCIRCUITS At frequencies near resonance, a microwave resonator can usually be modeled by either a series or parallel RLC lumped-element equivalent circuit, and so we will now review some of the basic properties of these circuits. SeriesResonantCircuit As e r i e sRLC resonant circuit is shown in Figure 6.1a. The input impedance is Zin=R+jωL−j1 ωC,( 6.1) and the complex power delivered to the resonator is Pin=1 2VI∗=1 2Zin|I|2=1 2Zin/vextendsingle/vextendsingle/vextendsingle/vextendsingleV Zin/vextendsingle/vextendsingle/vextendsingle/vextendsingle2 =1 2|I|2/parenleftbigg R+jωL−j1 ωC/parenrightbigg . (6.2) 272 c06MicrowaveResonators Pozar August 5, 2011 18:28 6.1 Series and Parallel Resonant Circuits 273 ⎪Zin(/H9275)⎪ RR /H9275//H92750010.707ZinI V+ –CL R (a) (b)BW FIGURE 6.1 A series RLC resonator and its response. (a) A series RLC resonator circuit. (b) Input impedance magnitude versus frequency. The power dissipated by the resistor Ris Ploss=1 2|I|2R, (6.3a) the average magnetic energy stored in the inductor Lis Wm=1 4|I|2L, (6.3b) and the average electric energy stored in the capacitor Cis We=1 4|Vc|2C=1 4|I|21 ω2C, (6.3c) where Vcis the voltage across the capacitor. Then the complex power of (6.2) can be rewritten as Pin=Ploss+2jω(Wm−We), (6.4) and the input impedance of (6.1) can be rewritten as Zin=2Pin |I|2=Ploss+2jω(Wm−We) 1 2|I|2.( 6.5) Resonance occurs when the average stored magnetic and electric energies are equal, or Wm=We. Then from (6.5) and (6.3a), the input impedance at resonance is Zin=Ploss 1 2|I|2=R, c06MicrowaveResonators Pozar August 5, 2011 18:28 274 Chapter 6: Microwave Resonators which is purely real. From (6.3b,c), Wm=Weimplies that the resonant frequency, ω0, can be defined as ω0=1√ LC.( 6.6) Another important parameter of a resonant circuit is its Q,o rquality factor, which is defined as Q=ωaverage energy stored energy loss/second =ωWm+We Ploss. (6.7) Thus Qis a measure of the loss of a resonant circuit—lower loss implies a higher Q. Resonator losses may be due to conductor loss, dielectric loss, or radiation loss, and are represented by the resistance, R, of the equivalent circuit. An external connecting network may introduce additional loss. Each of these loss mechanisms will have the effect of low- ering the Q.T h e Qof the resonator itself, disregarding external loading effects, is called theunloaded Q , denoted as Q0. For the series resonant circuit of Figure 6.1a, the unloaded Qcan be evaluated from (6.7), using (6.3) and the fact that Wm=Weat resonance, to give Q0=ω02Wm Ploss=ω0L R=1 ω0RC,( 6.8) which shows that Qincreases as Rdecreases. Next, consider the behavior of the input impedance of this resonator near its resonant frequency [1]. Let ω=ω0+/Delta1ω, where /Delta1ωis small. The input impedance can then be rewritten from (6.1) as Zin=R+jωL/parenleftbigg 1−1 ω2LC/parenrightbigg =R+jωL/parenleftBigg ω2−ω2 0 ω2/parenrightBigg , since ω2 0=1/LC.N o w ω2−ω2 0=(ω−ω0)(ω+ω0)=/Delta1ω(2ω −/Delta1ω)/similarequal2ω/Delta1ω for small/Delta1ω. Thus, Zin/similarequalR+j2L/Delta1ω /similarequalR+j2RQ 0/Delta1ω ω0. (6.9) This form will be useful for identifying equivalent circuits with distributed element resonators. Alternatively, a resonator with loss can be modeled as a lossless resonator whose res- onant frequency, ω0, has been replaced with a complex effective resonant frequency: ω0←ω0/parenleftbigg 1+j 2Q0/parenrightbigg .( 6.10) This can be seen by considering the input impedance of a series resonator with no loss, as given by (6.9) with R=0: Zin=j2L(ω−ω0). c06MicrowaveResonators Pozar September 12, 2011 21:3 6.1 Series and Parallel Resonant Circuits 275 Then substituting the complex frequency of (6.10) for ω0gives Zin=j2L/parenleftbigg ω−ω0−jω0 2Q0/parenrightbigg =ω0L Q0+j2L(ω−ω0)=R+j2L/Delta1ω, which is identical to (6.9). This is a useful procedure because for most practical resonators the loss is very small, so the Qcan be found using the perturbation method, beginning with the solution for the lossless case. Then the effect of loss can be added to the input impedance by replacing ω0with the complex resonant frequency given in (6.10). Finally, consider the half-power fractional bandwidth of the resonator. Figure 6.1b shows the variation of the magnitude of the input impedance versus frequency. When thefrequency is such that |Z in|2=2R2, then by (6.2) the average (real) power delivered to the circuit is one-half that delivered at resonance. If BW is the fractional bandwidth, then /Delta1ω/ω 0=BW/2 at the upper band edge. Using (6.9) gives |R+jRQ0(BW)|2=2R2, or BW=1 Q0.( 6.11) ParallelResonantCircuit The parallel RLC resonant circuit, shown in Figure 6.2a, is the dual of the series RLC circuit. The input impedance is Zin=/parenleftbigg1 R+1 jωL+jωC/parenrightbigg−1 ,( 6.12) ⎪Zin(/H9275)⎪ R /H9275//H92750010.707RZinCI L V+ –R (a) (b)BW FIGURE 6.2 A parallel RLC resonator and its response. (a) A parallel RLC circuit. (b) Input impedance magnitude versus frequency. c06MicrowaveResonators Pozar August 5, 2011 18:28 276 Chapter 6: Microwave Resonators and the complex power delivered to the resonator is Pin=1 2VI∗=1 2Zin|I|2=1 2|V|21 Z∗ in =1 2|V|2/parenleftbigg1 R+j ωL−jωC/parenrightbigg . (6.13) The power dissipated by the resistor, R,i s Ploss=1 2|V|2 R, (6.14a) the average electric energy stored in the capacitor, C,i s We=1 4|V|2C, (6.14b) and the average magnetic energy stored in the inductor, L,i s Wm=1 4|IL|2L=1 4|V|21 ω2L, (6.14c) where ILis the current through the inductor. Then the complex power of (6.13) can be rewritten as Pin=Ploss+2jω(Wm−We), (6.15) which is identical to (6.4). Similarly, the input impedance can be expressed as Zin=2Pin |I|2=Ploss+2jω(Wm−We) 1 2|I|2,( 6.16) which is identical to (6.5). As in the series case, resonance occurs when Wm=We. Then from (6.16) and (6.14a) the input impedance at resonance is Zin=Ploss 1 2|I|2=R, which is a purely real impedance. From (6.14b) and (6.14c), Wm=Weimplies that the resonant frequency, ω0, can be defined as ω0=1√ LC,( 6.17) which is identical to the series resonant circuit case. Resonance in the case of a parallel RLC circuit is sometimes referred to as an antiresonance. From the definition of (6.7), and the results in (6.14), the unloaded Qof the parallel resonant circuit can be expressed as Q0=ω02Wm Ploss=R ω0L=ω0RC,( 6.18) since Wm=Weat resonance. This result shows that the Qof the parallel resonant circuit increases as Rincreases. c06MicrowaveResonators Pozar August 5, 2011 18:28 6.1 Series and Parallel Resonant Circuits 277 Near resonance, the input impedance of (6.12) can be simplified using the series ex- pansion result that 1 1+x/similarequal1−x+···. Again letting ω=ω0+/Delta1ω, where /Delta1ωis small, allows (6.12) to be rewritten as [1] Zin/similarequal/parenleftbigg1 R+1−/Delta1ω/ω 0 jω0L+jω0C+j/Delta1ωC/parenrightbigg−1 /similarequal/parenleftBigg 1 R+j/Delta1ω ω2 0L+j/Delta1ωC/parenrightBigg−1 /similarequal/parenleftbigg1 R+2j/Delta1ωC/parenrightbigg−1 /similarequalR 1+2j/Delta1ωRC=R 1+2jQ0/Delta1ω/ω 0, (6.19) sinceω2 0=1/LC. When R=∞ (6.19) reduces to Zin=1 j2C(ω−ω0). As in the series resonator case, the effect of loss can be accounted for by replacing ω0 in this expression with a complex effective resonant frequency: ω0←ω0/parenleftbigg 1+j 2Q0/parenrightbigg .( 6.20) Figure 6.2b shows the behavior of the magnitude of the input impedance versus frequency. The half-power bandwidth edges occur at frequencies (/Delta1ω/ω 0=BW/2) such that |Zin|2=R2 2, which, from (6.19), implies that BW=1 Q0,( 6.21) as in the series resonance case. LoadedandUnloaded Q The unloaded Q,Q0, defined in the preceding sections is a characteristic of the resonator it- self, in the absence of any loading effects caused by external circuitry. In practice, however,a resonator is invariably coupled to other circuitry, which will have the effect of lowering the overall, or loaded Q ,Q L, of the circuit. Figure 6.3 depicts a resonator coupled to an RLResonant circuit Q FIGURE 6.3 A resonant circuit connected to an external load, RL. c06MicrowaveResonators Pozar August 5, 2011 18:28 278 Chapter 6: Microwave Resonators TABLE 6.1 Summary of Results for Series and Parallel Resonators Quantity Series Resonator Parallel Resonator Input impedance/admittance Zin=R+jωL−j1 ωCYin=1 R+jωC−j1 ωL /similarequalR+j2RQ 0/Delta1ω ω0/similarequal1 R+j2Q0/Delta1ω Rω0 Power loss Ploss=1 2|I|2RP loss=1 2|V|2 R Stored magnetic energy Wm=1 4|I|2LW m=1 4|V|21 ω2L Stored electric energy We=1 4|I|21 ω2CWe=1 4|V|2C Resonant frequency ω0=1√ LCω0=1√ LC Unloaded QQ 0=ω0L R=1 ω0RCQ0=ω0RC=R ω0L External QQ e=ω0L RLQe=RL ω0L external load resistor, RL. If the resonator is a series RLC circuit, the load resistor RLadds in series with R, so the effective resistance in (6.8) is R+RL. If the resonator is a parallel RLC circuit, the load resistor RLcombines in parallel with R, so the effective resistance in (6.18) is RRL/(R+RL). If we define an external Q ,Qe,a s Qe=⎧ ⎪⎪⎨ ⎪⎪⎩ω 0L RLfor series circuits RL ω0Lfor parallel circuits,(6.22) then the loaded Qcan be expressed as 1 QL=1 Qe+1 Q0.( 6.23) Table 6.1 summarizes the above results for series and parallel resonant circuits. 6.2TRANSMISSIONLINERESONATORS As we have seen, ideal lumped circuit elements are often unattainable at microwave fre- quencies, so distributed elements are frequently used. In this section we will study the useof transmission line sections with various lengths and terminations (usually open- or short- circuited) to form resonators. Because we are interested in the Qof these resonators, we must consider transmission lines with losses. Short-Circuited λ/2Line A length of lossy transmission line, short circuited at one end, is shown in Figure 6.4. The line has a characteristic impedance, Z 0, propagation constant, β, and attenuation c06MicrowaveResonators Pozar August 5, 2011 18:28 6.2 Transmission Line Resonators 279 V 0 Z0, /H9252, /H9251 Zinn = 2n = 1 FIGURE 6.4 A short-circuited length of lossy transmission line, and the voltage distributions for n=1(/lscript=λ/2)andn=2(/lscript=λ)resonators. constant, α. At the resonant frequency ω=ω0, the length of the line is /lscript=λ/2. From (2.91), the input impedance is Zin=Z0tanh(α+jβ)/lscript. Using an identity for the hyperbolic tangent gives Zin=Z0tanhα/lscript+jtanβ/lscript 1+jtanβ/lscripttanhα/lscript.( 6.24) Observe that Zin=jZ0tanβ/lscriptifα=0 (a lossless line). In practice it is usually desirable to use a low-loss transmission line, so we assume thatα/lscript/lessmuch1, and then tanh α/lscript/similarequalα/lscript. Again let ω=ω0+/Delta1ω, where /Delta1ωis small. Then, assuming a TEM line, we have β/lscript=ω/lscript vp=ω0/lscript vp+/Delta1ω/lscript vp, where vpis the phase velocity of the transmission line. Because /lscript=λ/2=πvp/ω0for ω=ω0,w eh a v e β/lscript=π+/Delta1ωπ ω0, and then tanβ/lscript=tan/parenleftbigg π+/Delta1ωπ ω0/parenrightbigg =tan/Delta1ωπ ω0/similarequal/Delta1ωπ ω0. Using these results in (6.24) gives Zin/similarequalZ0α/lscript+j(/Delta1ωπ/ω 0) 1+j(/Delta1ωπ/ω 0)α/lscript/similarequalZ0/parenleftbigg α/lscript+j/Delta1ωπ ω0/parenrightbigg ,( 6.25) since/Delta1ωα/lscript/ω 0/lessmuch1. Equation (6.25) is of the form Zin=R+2jL/Delta1ω, c06MicrowaveResonators Pozar August 5, 2011 18:28 280 Chapter 6: Microwave Resonators which is the input impedance of a series RLC resonant circuit, as given by (6.9). We can identify the resistance of the equivalent circuit as R=Z0α/lscript, (6.26a) and the inductance of the equivalent circuit as L=Z0π 2ω0.( 6.26b) The capacitance of the equivalent circuit can be found from (6.6) as C=1 ω2 0L.( 6.26c) The resonator of Figure 6.4 thus resonates for /Delta1ω=0(/lscript=λ/2), and its input impedance at resonance is Zin=R=Z0α/lscript. Resonance also occurs for /lscript=nλ/2,n= 1,2,3,.... The voltage distributions for the n=1 and n=2 resonant modes are shown in Figure 6.4. The unloaded Qof this resonator can be found from (6.8) and (6.26) as Q0=ω0L R=π 2α/lscript=β 2α,( 6.27) sinceβ/lscript=πat the first resonance. This result shows that the Qdecreases as the attenua- tion of the line increases, as expected. EXAMPLE 6.1 QOF HALF-WA VE COAXIAL LINE RESONATORS Aλ/2 resonator is made from a piece of copper coaxial line having an inner conductor radius of 1 mm and an outer conductor radius of 4 mm. If the resonant frequency is 5 GHz, compare the unloaded Qof an air-filled coaxial line resonator to that of a Teflon-filled coaxial line resonator. Solution We first compute the attenuation of the coaxial line, using the results of Examples 2.6 or 2.7. From Appendix F, the conductivity of copper is σ=5.813 ×107S/m. The surface resistivity at 5 GHz is Rs=/radicalbiggωµ0 2σ=1.84×10−2/Omega1. The attenuation due to conductor loss for the air-filled line is αc=Rs 2ηlnb/a/parenleftbigg1 a+1 b/parenrightbigg =1.84×10−2 2(377) ln(0.004 /0.001)/parenleftbigg1 0.001+1 0.004/parenrightbigg =0.022 Np/m. c06MicrowaveResonators Pozar August 5, 2011 18:28 6.2 Transmission Line Resonators 281 For Teflon, /epsilon1r=2.08 and tan δ=0.0004, so the attenuation due to conductor loss for the Teflon-filled line is αc=1.84×10−2√ 2.08 2(377) ln(0.004 /0.001)/parenleftbigg1 0.001+1 0.004/parenrightbigg =0.032 Np/m. The dielectric loss of the air-filled line is zero, but the dielectric loss of the Teflon- filled line is αd=k0√/epsilon1r 2tanδ =(104.7)√ 2.08(0.0004) 2=0.030 Np/m . Finally, from (6.27), the unloaded Qs can be computed as Qair=β 2α=104.7 2(0.022)=2380, QTeflon=β 2α=104.7√ 2.08 2(0.032 +0.030)=1218. Thus it is seen that the Qof the air-filled line is almost twice that of the Teflon- filled line. The Qcan be further increased by using silver-plated conductors. ■ Short-Circuited λ/4Line A parallel type of resonance (antiresonance) can be achieved using a short-circuited trans- mission line of length λ/4. The input impedance of a shorted line of length /lscriptis Zin=Z0tanh(α+jβ)/lscript =Z0tanhα/lscript+jtanβ/lscript 1+jtanβ/lscripttanhα/lscript =Z01−jtanhα/lscriptcotβ/lscript tanhα/lscript−jcotβ/lscript, (6.28) where the last result was obtained by multiplying both numerator and denominator by −jcotβ/lscript. Now assume that /lscript=λ/4a tω =ω0, and let ω=ω0+/Delta1ω. Then, for a TEM line, β/lscript=ω0/lscript vp+/Delta1ω/lscript vp=π 2+π/Delta1ω 2ω0, and so cotβ/lscript=cot/parenleftbiggπ 2+π/Delta1ω 2ω0/parenrightbigg =− tanπ/Delta1ω 2ω0/similarequal−π/Delta1ω 2ω0. Also, as before, tanh α/lscript/similarequalα/lscriptfor small loss. Using these results in (6.28) gives Zin=Z01+jα/lscriptπ/Delta1ω/ 2ω0 α/lscript+jπ/Delta1ω / 2ω0/similarequalZ0 α/lscript+jπ/Delta1ω / 2ω0,( 6.29) sinceα/lscriptπ/Delta1ω/ 2ω0/lessmuch1. This result is of the same form as the impedance of a parallel RLC circuit, as given in (6.19): Zin=1 (1/R)+2j/Delta1ωC. c06MicrowaveResonators Pozar August 5, 2011 18:28 282 Chapter 6: Microwave Resonators We can identify the resistance of the equivalent circuit as R=Z0 α/lscript(6.30a) and the capacitance of the equivalent circuit as C=π 4ω0Z0.( 6.30b) The inductance of the equivalent circuit can be found as L=1 ω2 0C.( 6.30c) The resonator of Figure 6.4 therefore has a parallel-type resonance for /lscript=λ/4, with an input impedance at resonance of Zin=R=Z0/α/lscript. From (6.18) and (6.30) the unloaded Qof this resonator is Q0=ω0RC=π 4α/lscript=β 2α,( 6.31) since/lscript=π/2β at resonance. Open-Circuited λ/2Line A practical resonator that is often used in microstrip circuits consists of an open-circuited length of transmission line, as shown in Figure 6.5. This resonator will behave as a parallel resonant circuit when the length is λ/2, or multiples of λ/2. The input impedance of an open-circuited lossy transmission line of length /lscriptis Zin=Z0coth(α+jβ)/lscript=Z01+jtanβ/lscripttanhα/lscript tanhα/lscript+jtanβ/lscript.( 6.32) As before, assume that /lscript=λ/2a tω=ω0, and let ω=ω0+/Delta1ω. Then, β/lscript=π+π/Delta1ω ω0, V 0 Z0, /H9252, /H9251 Zinn = 2n = 1 FIGURE 6.5 An open-circuited length of lossy transmission line, and the voltage distributions for n=1(/lscript=λ/2)andn=2(/lscript=λ)resonators. c06MicrowaveResonators Pozar August 5, 2011 18:28 6.2 Transmission Line Resonators 283 and so tanβ/lscript=tan/Delta1ωπ ω/similarequal/Delta1ωπ ω0, and tanh α/lscript/similarequalα/lscript. Using these results in (6.32) gives Zin=Z0 α/lscript+j(/Delta1ωπ/ω 0).( 6.33) Comparison with the input impedance of a parallel resonant circuit, as given by (6.19), suggests that the resistance of the equivalent RLC circuit is R=Z0 α/lscript,( 6.34a) and the capacitance of the equivalent circuit is C=π 2ω0Z0.( 6.34b) The inductance of the equivalent circuit is L=1 ω2 0C.( 6.34c) From (6.18) and (6.34) the unloaded Qis Q0=ω0RC=π 2α/lscript=β 2α,( 6.35) since/lscript=π/βat resonance. EXAMPLE 6.2 A HALF-WA VE MICROSTRIP RESONATOR Consider a microstrip resonator constructed from a λ/2 length of 50 /Omega1open- circuited microstrip line. The substrate is Teflon ( /epsilon1r=2.08, tanδ=0.0004), with a thickness of 0.159 cm, and the conductors are copper. Compute the required length of the line for resonance at 5 GHz, and the unloaded Qof the resonator. Ignore fringing fields at the end of the line. Solution From (3.197), the width of a 50 /Omega1microstrip line on this substrate is found to be W=0.508 cm, and the effective permittivity is /epsilon1e=1.80. The resonant length can then be calculated as /lscript=λ 2=vp 2f=c 2f√/epsilon1e=3×108 2(5×109)√ 1.80=2.24 cm . The propagation constant is β=2πf vp=2πf√/epsilon1e c=2π(5×109)√ 1.80 3×108=151.0 rad/m. From (3.199), the attenuation due to conductor loss is αc=Rs Z0W=1.84×10−2 50(0.00508)=0.0724 Np/m, c06MicrowaveResonators Pozar August 5, 2011 18:28 284 Chapter 6: Microwave Resonators where we used Rsfrom Example 6.1. From (3.198), the attenuation due to dielec- tric loss is αd=k0/epsilon1r(/epsilon1e−1)tanδ 2√/epsilon1e(/epsilon1r−1)=(104.7)(2.08)(0.80)(0.0004) 2√ 1.80(1.08)=0.024 Np/m . Then from (6.35) the unloaded Qis Q0=β 2α=151.0 2(0.0724 +0.024)=783. ■ 6.3RECTANGULARWAVEGUIDECAVITYRESONATORS Microwave resonators can also be constructed from closed sections of waveguide. Because radiation loss from an open-ended waveguide can be significant, waveguide resonators are usually short circuited at both ends, thus forming a closed box, or cavity. Electric and magnetic energy is stored within the cavity enclosure, and power is dissipated in the metallic walls of the cavity as well as in the dielectric material that may fill the cavity. Coupling to a cavity resonator may be by a small aperture, or a small probe or loop. We will see that there are many possible resonant modes for a cavity resonator, corresponding to field variations along the three dimensions of the structure. We will first derive the resonant frequencies for a general TE or TM resonant mode of a rectangular cavity, and then derive an expression for the unloaded Qof the TE 10/lscriptmode. A complete treatment of the unloaded Qfor arbitrary TE and TM modes can be made using the same procedure, but is not included here because of its length and complexity. ResonantFrequencies The geometry of a rectangular cavity is shown in Figure 6.6. It consists of a length, d, of rectangular waveguide shorted at both ends ( z=0,d). We will find the resonant x a a xyEy b dd = 2m = 1 zz0 = 1 FIGURE 6.6 A rectangular cavity resonator, and the electric field variations for the TE 101and TE102resonant modes. c06MicrowaveResonators Pozar August 5, 2011 18:28 6.3 Rectangular Waveguide Cavity Resonators 285 frequencies of this cavity under the assumption that the cavity is lossless, then determine the unloaded Qusing the perturbation method outlined in Section 2.7. Although we could begin with the Helmholtz wave equation and the method of separation of variables to solve for the electric and magnetic fields that satisfy the boundary conditions of the cavity, it iseasier to start with the fields of the TE or TM waveguide modes since these already satisfy the necessary boundary conditions on the side walls ( x=0,aandy=0,b)of the cavity. Then it is only necessary to enforce the boundary conditions that E x=Ey=0 on the end walls at z=0,d. From Table 3.2 the transverse electric fields ( Ex,Ey)o ft h eT E mnor TM mnrectangu- lar waveguide mode can be written as ¯Et(x,y,z)=¯e(x,y)/parenleftBig A+e−jβmnz+A−ejβmnz/parenrightBig ,( 6.36) where ¯e(x,y)is the transverse variation of the mode, and A+,A−are arbitrary amplitudes of the forward and backward traveling waves. The propagation constant of the m,nth TE or TM mode is βmn=/radicalbigg k2−/parenleftBigmπ a/parenrightBig2 −/parenleftBignπ b/parenrightBig2 ,( 6.37) where k=ω√µ/epsilon1, andµand/epsilon1are the permeability and permittivity of the material filling the cavity. Applying the condition that ¯Et=0a t z=0 to (6.36) implies that A+=− A−(as we should expect for reflection from a perfectly conducting wall). Then the condition that ¯Et=0a tz=dleads to the equation ¯Et(x,y,d)=− ¯e(x,y)A+2jsinβmnd=0. The only nontrivial ( A+/negationslash=0)solution occurs for βmnd=/lscriptπ, /lscript =1,2,3,..., (6.38) which implies that the cavity must be an integer multiple of a half-guide wavelength long at the resonant frequency. No nontrivial solutions are possible for other lengths, or forfrequencies other than the resonant frequencies. A resonance wave number for the rectangular cavity can be defined as k mn/lscript=/radicalbigg/parenleftBigmπ a/parenrightBig2 +/parenleftBignπ b/parenrightBig2 +/parenleftBig/lscriptπ d/parenrightBig2 .( 6.39) Then we can refer to the TE mn/lscriptor TM mn/lscriptresonant mode of the cavity, where the in- dices m,n,/lscriptindicate the number of variations in the standing wave pattern in the x,y,z directions, respectively. The resonant frequency of the TE mn/lscriptor TM mn/lscriptmode is given by fmn/lscript=ckmn/lscript 2π√µr/epsilon1r=c 2π√µr/epsilon1r/radicalbigg/parenleftBigmπ a/parenrightBig2 +/parenleftBignπ b/parenrightBig2 +/parenleftBig/lscriptπ d/parenrightBig2 .( 6.40) Ifb<a<d, the dominant resonant mode (lowest resonant frequency) will be the TE 101 mode, corresponding to the TE 10dominant waveguide mode in a shorted guide of length λg/2, and is similar to the short-circuited λ/2 transmission line resonator. The dominant TM resonant mode is the TM 110mode. c06MicrowaveResonators Pozar August 5, 2011 18:28 286 Chapter 6: Microwave Resonators Unloaded QoftheTE 10/lscriptMode From Table 3.2, (6.36), and the fact that A−=− A+, the total fields for the TE 10/lscriptresonant mode can be written as Ey=A+sinπx a/parenleftBig e−jβz−ejβz/parenrightBig , (6.41a) Hx=−A+ ZTEsinπx a/parenleftBig e−jβz+ejβz/parenrightBig , (6.41b) Hz=jπA+ kηacosπx a/parenleftBig e−jβz−ejβz/parenrightBig . (6.41c) Letting E0=−2jA+and using (6.38) allows these expressions to be simplified to Ey=E0sinπx asin/lscriptπz d, (6.42a) Hx=−jE0 ZTEsinπx acos/lscriptπz d, (6.42b) Hz=jπE0 kηacosπx asin/lscriptπz d, (6.42c) which clearly show that the fields form standing waves inside the cavity. We can now compute the unloaded Qof this mode by finding the stored electric and magnetic energies, and the power lost in the conducting walls and the dielectric filling. The stored electric energy is, from (1.84), We=/epsilon1 4/integraldisplay VEyE∗ ydv=/epsilon1abd 16E2 0,( 6.43a) while the stored magnetic energy is, from (1.86), Wm=µ 4/integraldisplay V(HxH∗ x+HzH∗ z)dv =µabd 16E2 0/parenleftBigg 1 Z2 TE+π2 k2η2a2/parenrightBigg . (6.43b) Because ZTE=kη/β, with β=β10=/radicalbig k2−(π/a)2, the quantity in parentheses in (6.43b) can be reduced to /parenleftBigg 1 Z2 TE+π2 k2η2a2/parenrightBigg =β2+(π/a)2 k2η2=1 η2=/epsilon1 µ, showing that We=Wmat resonance. The condition of equal stored electric and magnetic energies at resonance also applied to the RLC resonant circuits of Section 6.1. For small losses we can find the power dissipated in the cavity walls using the per- turbation method of Section 2.7. Thus, the power lost in the conducting walls is given by (1.131) as Pc=Rs 2/integraldisplay walls|Ht|2ds,( 6.44) where Rs=√ωµ0/2σ is the surface resistivity of the metallic walls, and Htis the tangential magnetic field at the surface of the walls. Using (6.42b), (6.42c) in (6.44) c06MicrowaveResonators Pozar August 5, 2011 18:28 6.3 Rectangular Waveguide Cavity Resonators 287 gives Pc=Rs 2/braceleftbigg 2/integraldisplayb y=0/integraldisplaya x=0|Hx(z=0)|2dxdy+2/integraldisplayd z=0/integraldisplayb y=0|Hz(x=0)|2dydz +2/integraldisplayd z=0/integraldisplaya x=0/bracketleftBig |Hx(y=0)|2+|Hz(y=0)|2/bracketrightBig dxdz/bracerightbigg =RsE2 0λ2 8η2/parenleftBigg /lscript2ab d2+bd a2+/lscript2a 2d+d 2a/parenrightBigg , (6.45) where use has been made of the symmetry of the cavity in doubling the contributions from the walls at x=0,y=0, and z=0 to account for the contributions from the walls at x=a,y=b, and z=d, respectively. The relations k=2π/λ andZTE=kη/β=2dη//lscriptλ were also used in simplifying (6.45). Then, from (6.7), the unloaded Qof the cavity with lossy conducting walls but lossless dielectric can be found as Qc=2ω0We Pc =k3abdη 4π2Rs1 [(/lscript2ab/d2)+(bd/a2)+(/lscript2a/2d)+(d/2a)] =(kad)3bη 2π2Rs1 (2/lscript2a3b+2bd3+/lscript2a3d+ad3). (6.46) Next we compute the power lost in the dielectric material that may fill the cavity. As discussed in Chapter 1, a lossy dielectric has an effective conductivity σ=ω/epsilon1/prime/prime= ω/epsilon1r/epsilon10tanδ, where /epsilon1=/epsilon1/prime−j/epsilon1/prime/prime=/epsilon1r/epsilon10(1−jtanδ), and tan δis the loss tangent of the material. The power dissipated in the dielectric is, from (1.92), Pd=1 2/integraldisplay V¯J·¯E∗dv=ω/epsilon1/prime/prime 2/integraldisplay V|¯E|2dv=abdω/epsilon1/prime/prime|E0|2 8,( 6.47) where ¯Eis given by (6.42a). Then from (6.7) the unloaded Qof the cavity with a lossy dielectric filling, but with perfectly conducting walls, is Qd=2ωWe Pd=/epsilon1/prime /epsilon1/prime/prime=1 tanδ.( 6.48) The simplicity of this result is due to the fact that the integral in (6.43a) for Wecancels with the identical integral in (6.47) for Pd. This result therefore applies to Qdfor an arbitrary resonant cavity mode. When both wall losses and dielectric losses are present, the totalpower loss is P c+Pd, so (6.7) gives the total unloaded Qas Q0=/parenleftbigg1 Qc+1 Qd/parenrightbigg−1 .( 6.49) EXAMPLE 6.3 DESIGN OF A RECTANGULAR CA VITY RESONATOR A rectangular waveguide cavity is made from a piece of copper WR-187 H-band waveguide, with a=4.755 cm and b=2.215 cm. The cavity is filled with poly- ethylene ( /epsilon1r=2.25, tan δ=0.0004). If resonance is to occur at f=5 GHz, find the required length, d, and the resulting unloaded Qfor the /lscript=1 and /lscript=2 resonant modes. c06MicrowaveResonators Pozar August 5, 2011 18:28 288 Chapter 6: Microwave Resonators Solution The wave number kis k=2πf√/epsilon1r c=157.08 m−1. From (6.40) the required length for resonance can be found as (m=1,n=0) d=/lscriptπ/radicalbig k2−(π/a)2, for/lscript=1, d=π/radicalbig (157.08)2−(π/0.04755)2=2.20 cm, for/lscript=2, d=2(2.20) =4.40 cm. From Example 6.1, the surface resistivity of copper at 5 GHz is Rs=1.84× 10−2/Omega1. The intrinsic impedance is η=377√/epsilon1r=251.3 /Omega1. Then from (6.46) the Qdue to conductor loss only is for/lscript=1, Qc=8,403, for/lscript=2, Qc=11,898 . From (6.48) the Qdue to dielectric loss only is, for both /lscript=1 and/lscript=2, Qd=1 tanδ=1 0.0004=2500. Then total unloaded Qs are, from (6.49) for/lscript=1, Q0=/parenleftbigg1 8403+1 2500/parenrightbigg−1 =1927, for/lscript=2, Q0=/parenleftbigg1 11,898+1 2500/parenrightbigg−1 =2065. Note that the dielectric loss has the dominant effect on the Q; higher Qcould be obtained using an air-filled cavity. These results can be compared to those of Examples 6.1 and 6.2, which used similar types of materials at the same frequency. ■ 6.4CIRCULARWAVEGUIDECAVITYRESONATORS A cylindrical cavity resonator can be constructed from a section of circular waveguide shorted at both ends, similar to rectangular cavities. Because the dominant circular wave- guide mode is the TE 11mode, the dominant cylindrical cavity mode is the TE 111mode. We will derive the resonant frequencies for the TE nm/lscriptand TM nm/lscriptcircular cavity modes, and an expression for the unloaded Qof the TE nm/lscriptmode. Circular cavities are often used for microwave frequency meters. The cavity is con- structed with a movable top wall to allow mechanical tuning of the resonant frequency, and the cavity is loosely coupled to a waveguide through a small aperture. In operation,power will be absorbed by the cavity as it is tuned to the operating frequency of the system; this absorption can be monitored with a power meter elsewhere in the system. The c06MicrowaveResonators Pozar August 5, 2011 18:28 6.4 Circular Waveguide Cavity Resonators 289 FIGURE 6.7 Photograph of a W-band waveguide frequency meter. The knob rotates to change the length of the circular cavity resonator; the scale gives a readout of the frequency. Photograph courtesy of Millitech Inc., Northampton, Mass. mechanical tuning dial is usually directly calibrated in frequency, as in the model shown in Figure 6.7. Because frequency resolution is determined by the Qof the resonator, the TE011mode is often used for frequency meters because its Qis much higher than the Q of the dominant circular cavity mode. This is also the reason for a loose coupling to the cavity. ResonantFrequencies The geometry of a cylindrical cavity is shown in Figure 6.8. As in the case of the rectan- gular cavity, the solution is simplified by beginning with the circular waveguide modes, which already satisfy the necessary boundary conditions on the wall of the circular wave- guide. From Table 3.5, the transverse electric fields ( Eρ,Eφ)of the TE nmor TM nmcircular waveguide mode can be written as ¯Et(ρ,φ, z)=¯e(ρ,φ)/parenleftbig A+e−jβnmz+A−ejβnmz/parenrightbig ,( 6.50) where ¯e(ρ,φ) represents the transverse variation of the mode, and A+andA−are arbitrary amplitudes of the forward and backward traveling waves. The propagation constant of the TEnmmode is, from (3.126), βnm=/radicalBigg k2−/parenleftbiggp/primenm a/parenrightbigg2 ,( 6.51a) z d E/H9267, E/H9278z x = 2a d /H9278 = 1 FIGURE 6.8 A cylindrical resonant cavity, and the electric field distribution for resonant modes with/lscript=1o r/lscript =2. c06MicrowaveResonators Pozar August 5, 2011 18:28 290 Chapter 6: Microwave Resonators while the propagation constant of the TM nmmode is, from (3.139), βnm=/radicalbigg k2−/parenleftBigpnm a/parenrightBig2 ,( 6.51b) where k=ω√µ/epsilon1. In order to have ¯Et=0a t z=0,d, we must choose A+=− A−, and A+sinβnm d=0, or βnmd=/lscriptπ, for/lscript=0,1,2,3,..., (6.52) which implies that the waveguide must be an integer number of half-guide wavelengths long. Thus, the resonant frequency of the TE nm/lscriptmode is fnm/lscript=c 2π√µr/epsilon1r/radicalBigg/parenleftbiggp/primenm a/parenrightbigg2 +/parenleftbigg/lscriptπ d/parenrightbigg2 ,( 6.53a) and the resonant frequency of the TM nm/lscriptmode is fnm/lscript=c 2π√µr/epsilon1r/radicalBigg /parenleftBigpnm a/parenrightBig2 +/parenleftbigg/lscriptπ d/parenrightbigg2 .( 6.53b) Thus the dominant TE mode is the TE 111mode, while the dominant TM mode is the TM 010 mode. Figure 6.9 shows a mode chart for the lower order resonant modes of a cylindrical cavity. Such a chart is useful for the design of circular cavity resonators, as it shows what modes can be excited at a given frequency for a given cavity size. 05 × 10810 × 10815 × 10820 × 108 24 (2a/ d )2(2af)2, (MHz – cm)2 6TM010TM110TE111 TM011 TE211TE112 TM012 TE212 TM112 TE011TM111 FIGURE 6.9 Resonant mode chart for a cylindrical cavity. Adapted from data from R. E. Collin, Foundations for Microwave Engineering, 2nd edition, Wiley–IEEE Press, Hoboken, N.J., 2001. Used with permission. c06MicrowaveResonators Pozar August 5, 2011 18:28 6.4 Circular Waveguide Cavity Resonators 291 Unloaded QoftheTE nm/lscriptMode From Table 3.5, (6.50), and the fact that A+=− A−, the fields of the TE nm/lscriptmode can be written as Hz=H0Jn/parenleftbiggp/prime nmρ a/parenrightbigg cosnφsin/lscriptπz d, (6.54a) Hρ=βaH0 p/primenmJ/prime n/parenleftbiggp/prime nmρ a/parenrightbigg cosnφcos/lscriptπz d, (6.54b) Hφ=−βa2nH0 (p/primenm)2ρJn/parenleftbiggp/prime nmρ a/parenrightbigg sinnφcos/lscriptπz d, (6.54c) Eρ=jkηa2nH0 (p/primenm)2ρJn/parenleftbiggp/prime nmρ a/parenrightbigg sinnφsin/lscriptπz d, (6.54d) Eφ=jkηaH0 p/primenmJ/prime n/parenleftbiggp/prime nmρ a/parenrightbigg cosnφsin/lscriptπz d, (6.54e) Ez=0, (6.54f) where η=√µ//epsilon1andH0=−2jA+. Because the time-average stored electric and magnetic energies are equal, the total stored energy is W=2We=/epsilon1 2/integraldisplayd z=0/integraldisplay2π φ=0/integraldisplaya ρ=0/parenleftBig |Eρ|2+|Eφ|2/parenrightBig ρdρdφdz =/epsilon1k2η2a2πdH2 0 4(p/primenm)2/integraldisplaya ρ=0/bracketleftBigg J/prime2 n/parenleftbiggp/prime nmρ a/parenrightbigg +/parenleftbiggna p/primenmρ/parenrightbigg2 J2 n/parenleftbiggp/prime nmρ a/parenrightbigg/bracketrightBigg ρdρ =/epsilon1k2η2a4H2 0πd 8(p/primenm)2/bracketleftBigg 1−/parenleftbiggn p/primenm/parenrightbigg2/bracketrightBigg J2 n(p/prime nm), (6.55) where the integral identity of Appendix C.17 has been used. The power loss in the con- ducting walls is Pc=Rs 2/integraldisplay S|¯Htan|2ds =Rs 2/braceleftBigg/integraldisplayd z=0/integraldisplay2π φ=0/bracketleftBig |Hφ(ρ=a)|2+|Hz(ρ=a)|2/bracketrightBig adφdz +2/integraldisplay2π φ=0/integraldisplaya ρ=0/bracketleftBig |Hρ(z=0)|2+|Hφ(z=0)|2/bracketrightBig ρdρdφ/bracerightBigg =Rs 2πH2 0J2 n(p/prime nm)⎧ ⎨ ⎩da 2/bracketleftBigg 1+/parenleftbiggβan (p/primenm)2/parenrightbigg2/bracketrightBigg +/parenleftBigg βa2 p/primenm/parenrightBigg2/parenleftBigg 1−n2 (p/primenm)2/parenrightBigg⎫ ⎬ ⎭.(6.56) Then, from (6.8), the unloaded Qof the cavity with imperfectly conducting walls but lossless dielectric is Qc=ω0W Pc=(ka)3ηad 4(p/primenm)2Rs1−/parenleftbiggn p/primenm/parenrightbigg2 ⎧ ⎨ ⎩ad 2⎡ ⎣1+/parenleftBigg βan (p/primenm)2/parenrightBigg2⎤ ⎦+/parenleftbigg βa2 p/primenm/parenrightbigg2/parenleftbigg 1−n2 (p/primenm)2/parenrightbigg⎫ ⎬ ⎭.(6.57) c06MicrowaveResonators Pozar August 5, 2011 18:28 292 Chapter 6: Microwave Resonators 0.5 1.0 1.5 2.0 2.5 3.0 0.00.00.20.40.60.81.0 2a/dQRs//H9266/H9257 = Q/H9254s//H92610 TE111TM010TM111TE011TE012 FIGURE 6.10 Normalized unloaded Qfor various cylindrical cavity modes (air filled). Adapted from data from R. E. Collin, Foundations for Microwave Engineering , 2nd edition, Wiley–IEEE Press, Hoboken, N.J., 2001. Used with permission. From (6.52) and (6.51) we see that β=/lscriptπ/dand(ka)2are constants that do not vary with frequency, for a cavity with fixed dimensions. Thus, the frequency dependence of Qcis given by k/Rs, which varies as 1 /√f; this gives the variation in Qcfor a given resonant mode and cavity shape (fixed n,m,/lscript, and a/d). Figure 6.10 shows the normalized unloaded Qdue to conductor loss for various res- onant modes of a cylindrical cavity. Observe that the TE 011mode has an unloaded Q significantly higher than that of the lower order TE 111,T M 010,o rT M 111mode. To compute the unloaded Qdue to dielectric loss, we must compute the power dissi- pated in the dielectric. Thus, Pd=1 2/integraldisplay V¯J·¯E∗dv=ω/epsilon1/prime/prime 2/integraldisplay V/bracketleftBig |Eρ|2+|Eφ|2/bracketrightBig dv =ω/epsilon1/prime/primek2η2a2H2 0πd 4(p/primenm)2/integraldisplaya ρ=0/bracketleftBigg/parenleftbiggna p/primenmρ/parenrightbigg2 J2 n/parenleftbiggp/prime nmρ a/parenrightbigg +J/prime2 n/parenleftbiggp/prime nmρ a/parenrightbigg/bracketrightBigg ρdρ =ω/epsilon1/prime/primek2η2a4H2 0 8(p/primenm)2/bracketleftBigg 1−/parenleftbiggn p/primenm/parenrightbigg2/bracketrightBigg J2 n(p/prime nm). (6.58) Then (6.8) gives the unloaded Qdue to dielectric loss as Qd=ωW Pd=/epsilon1 /epsilon1/prime/prime=1 tanδ,( 6.59) where tan δis the loss tangent of the dielectric. This is the same as the result for Qdof (6.48) for the rectangular cavity. When both conductor and dielectric losses are present, the total unloaded cavity Qcan be found from (6.49). EXAMPLE 6.4 DESIGN OF A CIRCULAR CA VITY RESONATOR A circular cavity resonator with d=2ais to be designed to resonate at 5.0 GHz in the TE 011mode. If the cavity is made from copper and is Teflon filled ( /epsilon1r= 2.08, tanδ=0.0004), find its dimensions and unloaded Q. c06MicrowaveResonators Pozar August 5, 2011 18:28 6.5 Dielectric Resonators 293 Solution k=2πf011√/epsilon1r c=2π(5×109)√ 2.08 3×108=151.0m−1 From (6.53a) the resonant frequency of the TE 011mode is f011=c 2π√/epsilon1r/radicalBigg/parenleftbiggp/prime 01 a/parenrightbigg2 +/parenleftBigπ d/parenrightBig2 , with p/prime 01=3.832. Then, since d=2a 2πf011√/epsilon1r c=k=/radicalBigg/parenleftbiggp/prime 01 a/parenrightbigg2 +/parenleftBigπ d/parenrightBig2 . Solving for agives a=/radicalBig (p/prime 01)2+(π/2)2 k=/radicalbig (3.832)2+(π/2)2 151.0=2.74 cm, so we have d=5.48 cm. The surface resistivity of copper at 5 GHz is Rs=0.0184 /Omega1. Then from (6.57), with n=0,m=/lscript=1, and d=2a, the unloaded Qdue to conductor losses is Qc=(ka)3ηad 4(p/prime 01)2Rs1 [ad/2+(βa2/p/prime 01)2]=kaη 2Rs=29,390 , where (6.51a) was used to simplify the expression. From (6.59) the unloaded Q due to dielectric loss is Qd=1 tanδ=1 0.0004=2500, and the total unloaded Qof the cavity is Q0=/parenleftbigg1 Qc+1 Qd/parenrightbigg−1 =2300. This result can be compared with the rectangular cavity case of Example 6.3, which had Q0=1927 for the TE 101mode and Q0=2065 for the TE 102mode. If this cavity were air filled, the Qwould increase to 42,400. ■ 6.5DIELECTRICRESONATORS A small disc or cube (or other shape) of dielectric material can also be used as a microwave resonator. The operation of such a dielectric resonator is similar in principle to the rectan- gular or cylindrical cavity resonators previously discussed. Dielectric resonators typically use materials with low loss and a high dielectric constant, ensuring that most of the fields will be contained within the dielectric. Unlike metallic cavities, however, there is somefield fringing or leakage from the sides and ends of a dielectric resonator (which are not metalized), leading to a small radiation loss and consequent lowering of Q. A dielectric resonator is generally smaller in size, cost, and weight than an equivalent metallic cavity,and it can easily be incorporated into microwave integrated circuits and coupled to planar transmission lines. Materials with dielectric constants in the range of 10–100 are generally c06MicrowaveResonators Pozar August 5, 2011 18:28 294 Chapter 6: Microwave Resonators used, with barium tetratitanate and titanium dioxide being typical examples. Conductor losses are absent, but dielectric loss usually increases with dielectric constant; Qso fu p to several thousand can sometimes be achieved, however. By using an adjustable metal plate above the resonator, the resonant frequency can be mechanically tuned. Because ofthese desirable features, dielectric resonators have become key components for integrated microwave filters and oscillators. Below we present an approximate analysis for the resonant frequencies of the TE 01δ mode of a cylindrical dielectric resonator; this mode is the one most commonly used in practice, and is analogous to the TE 011mode of a circular metallic cavity. ResonantFrequenciesofTE 01δMode The geometry of a cylindrical dielectric resonator is shown in Figure 6.11. The basic oper- ation of the TE 01δmode can be explained as follows. The dielectric resonator is considered as a short length, L, of dielectric waveguide open at both ends. The lowest order TE mode of this guide is the TE 01mode, and is the dual of the TM 01mode of a circular metal- lic waveguide. Because of the high permittivity of the resonator, propagation along the z-axis can occur inside the dielectric at the resonant frequency, but the fields will be cut off (evanescent) in the air regions around the dielectric. Thus the Hzfield will look like that sketched in Figure 6.12; higher order resonant modes will have more variations in the zdirection inside the resonator. Because the resonant length for the TE 01δmode is less thanλg/2 (where λgis the guide wavelength of the TE 01dielectric waveguide mode), the symbol δ=2L/λg<1 is used to denote the zvariation of the resonant mode. The equiv- alent circuit of the resonator looks like a length of transmission line terminated in purely reactive loads at both ends. Our analysis follows that of reference [2], and involves the assumption that a magnetic wall boundary condition can be imposed at ρ=a. This approximation is based on the fact that the reflection coefficient of a wave in a high dielectric constant region incident on an air-filled region approaches +1: /Gamma1=η0−η η0+η=√/epsilon1r−1√/epsilon1r+1→1a s /epsilon1r→∞. This reflection coefficient is the same as that obtained at an ideal magnetic wall boundary condition, or a perfect open circuit. We begin by finding the fields of the TE 01dielectric waveguide mode with a magnetic wall boundary condition at ρ=a. For TE modes, Ez=0, and Hzmust satisfy the wave yz xL a /H9280r2 L 2– FIGURE 6.11 Geometry of a cylindrical dielectric resonator. c06MicrowaveResonators Pozar August 5, 2011 18:28 6.5 Dielectric Resonators 295 Ht = 0Hz (/H9267 = 0)z /H92800 /H92800/H9280r > 1 L FIGURE 6.12 Magnetic wall boundary condition approximation and distribution of Hzversus z forρ=0 of the first mode of a cylindrical dielectric resonator. equation (∇2+k2)Hz=0,( 6.60) where k=/braceleftBigg√/epsilon1rk0for|z|<L/2 k0 for|z|>L/2.(6.61) Because ∂/∂φ=0, the transverse fields are given by (3.110) as follows: Eφ=jωµ0 k2c∂Hz ∂ρ, (6.62a) Hρ=−jβ k2c∂Hz ∂ρ, (6.62b) where k2 c=k2−β2. Because Hzmust be finite at ρ=0 and zero at ρ=a(the magnetic wall), we have Hz=H0J0(kcρ)e±jβz,( 6.63) where kc=p01/a, and J0(p01)=0(p01=2.405). Then from (6.62) the transverse fields are Eφ=jωµ0H0 kcJ/prime 0(kcρ)e±jβz, (6.64a) Hρ=∓jβH0 kcJ/prime 0(kcρ)e±jβz. (6.64b) c06MicrowaveResonators Pozar August 5, 2011 18:28 296 Chapter 6: Microwave Resonators In the dielectric region, for |z|<L/2, the propagation constant is real: β=/radicalBig /epsilon1rk2 0−k2c=/radicalbigg /epsilon1rk2 0−/parenleftBigp01 a/parenrightBig2 , (6.65a) and a wave impedance can be defined as Zd=Eφ Hρ=ωµ0 β. (6.65b) In the air region, for |z|>L/2, the propagation constant will be imaginary, so it is conve- nient to write α=/radicalBig k2c−k2 0=/radicalbigg/parenleftBigp01 a/parenrightBig2 −k2 0, (6.66a) and to define a wave impedance in the air region as Za=jωµ0 α, (6.66b) which is seen to be imaginary. From symmetry, the HzandEφfield distributions for the lowest order mode will be even functions about z=0. Then the transverse fields for the TE 01δmode can be written for|z|<L/2a s Eφ=AJ/prime 0(kcρ)cosβz, (6.67a) Hρ=−jA ZdJ/prime 0(kcρ)sinβz, (6.67b) and for |z|>L/2a s Eφ=BJ/prime 0(kcρ)e−α|z|, (6.68a) Hρ=±B ZaJ/prime 0(kcρ)e−α|z|, (6.68b) where Aand Bare unknown amplitude coefficients. In (6.68b), the ±sign is used for z>L/2o r z<−L/2, respectively. Matching tangential fields at z=L/2( o r z=− L/2)leads to the following two equations: AcosβL 2=Be−αL/2, (6.69a) −jA ZdsinβL 2=B Zae−αL/2, (6.69b) which can be reduced to a single transcendental equation: −jZasinβL 2=ZdcosβL 2. Using (6.65b) and (6.66b) allows this to be simplified as tanβL 2=α β,( 6.70) where βis given by (6.65a) and αis given by (6.66a). This equation can be solved numer- ically for k0, which determines the resonant frequency. c06MicrowaveResonators Pozar August 5, 2011 18:28 6.6 Excitation of Resonators 297 This solution is approximate since it ignores fringing fields at the sides of the res- onator, and it yields accuracies only on the order of 10% (usually not accurate enough for practical purposes), but it serves to illustrate the basic behavior of dielectric resonators. More accurate solutions are available in the literature [3]. The unloaded Qof the resonator can be calculated by determining the stored energy (inside and outside the dielectric cylinder), and the power dissipated in the dielectric and possibly lost to radiation. If the latter is small, the unloaded Qcan be approximated as 1/tanδ, as in the case of the metallic cavity resonators. EXAMPLE 6.5 RESONANT FREQUENCY AND QOF A DIELECTRIC RESONATOR Find the resonant frequency and approximate unloaded Qfor the TE 01δmode of a dielectric resonator made from titania, with /epsilon1r=95 and tan δ=0.001. The resonator dimensions are a=0.413 cm and L=0.8255 cm. Solution The transcendental equation of (6.70) must be solved for k0, with βandαgiven by (6.65a) and (6.66a). Thus, tanβL 2=α β, where α=/radicalBig (2.405 /a)2−k2 0, β=/radicalBig /epsilon1rk2 0−(2.405 /a)2, and k0=2πf c. Because αandβmust both be real, the possible frequency range is from f1tof2, where f1=ck0 2π=c(2.405) 2π√/epsilon1ra=2.853 GHz, f2=ck0 2π=c(2.405) 2πa=27.804 GHz. Using the interval-halving method (see the Point of Interest on root-finding algorithms in Chapter 3) to find the root of the above equation gives a resonant frequency of about 3.152 GHz. This compares with a measured value of about3.4 GHz from reference [2], indicating a 10% error. The approximate unloaded Q, due to dielectric loss, is Q d=1 tanδ=1000. ■ 6.6EXCITATIONOFRESONATORS Resonators are not useful unless they are coupled to external circuitry, so we now discuss how resonators can be coupled to transmission lines and waveguides. In practice, the way in which this is done depends on the type of resonator under consideration; some examples c06MicrowaveResonators Pozar August 5, 2011 18:28 298 Chapter 6: Microwave Resonators (a) (b) (c) (d) FIGURE 6.13 Coupling to microwave resonators. (a) A microstrip transmission line resonator gap coupled to a microstrip feedline. (b) A rectangular cavity resonator fed by a coaxial probe. (c) A circular cavity resonator aperture coupled to a rectangularwaveguide. (d) A dielectric resonator coupled to a microstrip line. of resonator coupling techniques are shown in Figure 6.13. We will discuss the operation of some of the more common coupling techniques, notably gap coupling and aperturecoupling. We begin by discussing the coupling coefficient for a resonator connected to a feed line, and the subject of critical coupling. A related topic of practical interest is how the unloaded Qof a resonator can be determined from the two-port response of a resonator coupled to a transmission line. TheCouplingCoefficien andCriticalCoupling The level of coupling required between a resonator and its attached circuitry depends on the application. A waveguide cavity used as a frequency meter, for example, is usually loosely coupled to its feed guide in order to maintain high Qand good accuracy. A resonator used in an oscillator or tuned amplifier, however, may be tightly coupled in order to achieve maximum power transfer. A measure of the level of coupling between a resonator and a feed is given by the coupling coefficient . To obtain maximum power transfer between a resonator and a feed line, the resonator should be matched to the line at the resonant frequency; the resonator is then said to be critically coupled to the feed. We will illustrate these concepts by considering the series resonant circuit shown in Figure 6.14. ZinZ0CL R FIGURE 6.14 A series resonant circuit coupled to a feedline. c06MicrowaveResonators Pozar August 5, 2011 18:28 6.6 Excitation of Resonators 299 From (6.9), the input impedance near resonance of the series resonant circuit of Figure 6.14 is given by Zin=R+j2L/Delta1ω=R+j2RQ 0/Delta1ω ω0,( 6.71) and the unloaded Qis, from (6.8), Q0=ω0L R.( 6.72) At resonance, /Delta1ω=0, so from (6.71) the input impedance is Zin=R. In order to match the resonator to the line we must have R=Z0.( 6.73) In this case the unloaded Qis Q0=ω0L Z0.( 6.74) From (6.22), the external Qis Qe=ω0L Z0=Q0,( 6.75) which shows that the external and unloaded Qs are equal under the condition of critical coupling. The loaded Qis half this value. We can define the coupling coefficient, g,a s g=Q0 Qe,( 6.76) which can be applied to both series (g=Z0/R)and parallel (g=R/Z0)resonant circuits, when connected to a transmission line of characteristic impedance Z0. Three cases can be distinguished: 1.g<1: The resonator is said to be undercoupled to the feedline. 2.g=1: The resonator is critically coupled to the feedline. 3.g>1: The resonator is said to be overcoupled to the feedline. Figure 6.15 shows a Smith chart sketch of the impedance loci for the series resonant circuit, as given by (6.71), for various values of Rcorresponding to the above cases. AGap-CoupledMicrostripResonator Consider a λ/2 open-circuited microstrip resonator proximity coupled to the open end of a microstrip transmission line, as shown in Figure 6.13a. The gap between the resonatorand the microstrip line can be modeled as a series capacitor, so the equivalent circuit can be constructed as shown in Figure 6.16. The normalized input impedance seen by the feedline is z=Z Z0=− j(1/ωC+Z0cotβ/lscript) Z0=− j/parenleftbiggtanβ/lscript+bc bctanβ/lscript/parenrightbigg ,( 6.77) where bc=Z0ωCis the normalized susceptance of the coupling capacitor, C. Resonance occurs with z=0, or when tanβ/lscript+bc=0.( 6.78) c06MicrowaveResonators Pozar August 5, 2011 18:28 300 Chapter 6: Microwave Resonators Overcoupled (R < Z0) Critically coupled (R = Z0) Undercoupled (R > Z0) FIGURE 6.15 Smith chart illustrating coupling to a series RLC circuit. The solutions to this transcendental equation are shown in the graph of Figure 6.17. In practice, bc/lessmuch1, so the first resonant frequency, ω1, will be close to the frequency for which β/lscript=π(the first resonant frequency of the unloaded resonator). The coupling of the resonator to the feedline has the effect of lowering its resonant frequency. We now wish to simplify the driving point impedance of (6.77) to relate this resonator to a series RLC equivalent circuit. This can be accomplished by expanding z(ω)in a Taylor series about the resonant frequency, ω1, and assuming that bcis small. Thus, z(ω)=z(ω1)+(ω−ω1)dz(ω) dω/vextendsingle/vextendsingle/vextendsingle/vextendsingle ω1+···= (ω−ω1)dz(ω) d(β/lscript)d(β/lscript) dω/vextendsingle/vextendsingle/vextendsingle/vextendsingle ω1+···,( 6.79) ZZ0 Z0C Feed line Gap capacitanceOpen-circuit /H9261/2 resonator FIGURE 6.16 Equivalent circuit of the gap-coupled microstrip resonator of Figure 6.13a. c06MicrowaveResonators Pozar August 5, 2011 18:28 6.6 Excitation of Resonators 301 /H9252 = /H9275 /vp –bc = –/H9275CZ02/H9266 /H9266tan /H9252 vp/H92751 vp/H92752 FIGURE 6.17 Solutions to (6.78) for the resonant frequencies of the gap-coupled microstrip resonator. since, from (6.77) and (6.78), z(ω1)=0. Then, dz d(β/lscript)/vextendsingle/vextendsingle/vextendsingle/vextendsingle ω1=jsec2β/lscript tan2β/lscript=j1+tan2β/lscript tan2β/lscript=j1+b2 c b2c/similarequalj b2c, where we have used (6.78) and the assumption that bc/lessmuch1. Assuming a TEM line, we have d(β/lscript)/ dω=/lscript/vp, where vpis the phase velocity of the line. Because /lscript/similarequalπvp/ω1,t h e normalized impedance can be written as z(ω)/similarequalj/lscript(ω−ω1) b2cvp/similarequaljπ(ω−ω1) ω1b2c.( 6.80) So far we have ignored losses, but for a high- Qresonator loss can be included by re- placing the resonant frequency, ω1, with the complex resonant frequency given by ω1(1+ j/2Q0), which follows from (6.10). Applying this procedure to (6.80) gives the input impedance of the gap-coupled lossy resonator as z(ω)=π 2Q0b2c+jπ(ω−ω1) ω1b2c.( 6.81) Note that an uncoupled λ/2 open-circuited transmission line resonator looks like a parallel RLC circuit near resonance, but the present case of a capacitive coupled λ/2 resonator looks like a series RLC circuit near resonance. This is because the series coupling capacitor has the effect of inverting the driving point impedance of the resonator (see the discussion ofimpedance inverters in Section 8.5). At resonance the input resistance is R=Z 0π/2Q0b2 c. For critical coupling we must have R=Z0,o r bc=/radicalbiggπ 2Q0.( 6.82) The coupling coefficient of (6.76) is found to be g=Z0 R=2Q0b2 c π.( 6.83) Ifbc<√π/2Q, then g<1 and the resonator is undercoupled; if bc>√π/2Q, then g>1 and the resonator is overcoupled. c06MicrowaveResonators Pozar August 5, 2011 18:28 302 Chapter 6: Microwave Resonators EXAMPLE 6.6 DESIGN OF A GAP-COUPLED MICROSTRIP RESONATOR A resonator is made from an open-circuited 50 /Omega1microstrip line and is gap cou- pled to a 50 /Omega1feedline, as in Figure 6.13a. The resonator has a length of 2.175 cm, an effective dielectric constant of 1.9, and an attenuation of 0.01 dB/cm near its resonance. Find the value of the coupling capacitor required for critical coupling, and the resulting resonant frequency. Solution The first resonant frequency will occur when the resonator is about /lscript=λg/2 in length. Ignoring fringing fields, we find that the approximate resonant fre-quency is f 0=vp λg=c 2/lscript√/epsilon1e=3×108 2(0.02175)√ 1.9=5.00 GHz. This result does not include the effect of the coupling capacitor. From (6.35) the unloaded Qof this resonator is Q0=β 2α=π λgα=π 2/lscriptα=π(8.7d B / N p ) 2(0.02175 m)(1dB/m)=628. From (6.82) the normalized coupling capacitor susceptance is bc=/radicalbiggπ 2Q0=/radicalbiggπ 2(628)=0.05, so the coupling capacitor has a value of C=bc ωZ0=0.05 2π(5×109)(50)=0.032 pF, which should provide critical coupling of the resonator to the 50 /Omega1feedline. Now that Cis determined, the exact resonant frequency can be found by solv- ing the transcendental equation of (6.78). Because we know from the graphical so- lution of Figure 6.17 that the actual resonant frequency is slightly lower than the unloaded resonant frequency of 5.0 GHz, it is an easy matter to calculate (6.78)for several frequencies in this vicinity, which leads to a value of about 4.918 GHz. This is about 1.6% lower than the unloaded resonant frequency. Figure 6.18 shows a Smith chart plot of the input impedance of the gap-coupled resonator forcoupling capacitor values that lead to undercoupled, critically coupled, and over- coupled resonators. ■ AnAperture-CoupledCavity As a final example of resonator excitation, we consider the aperture coupled waveguide cavity shown in Figure 6.19. As discussed in Section 4.8, a small aperture in the transverse wall of a waveguide acts as a shunt inductance. If we consider the first resonant mode ofthe cavity, which occurs for the cavity length /lscript=λ g/2, then the cavity can be considered as a transmission line resonator shorted at one end. The aperture-coupled cavity can then be modeled by the equivalent circuit shown in Figure 6.20. This circuit is basically thedual of the equivalent circuit of Figure 6.16, for the gap-coupled microstrip resonator, so we will approach the solution in the same manner. c06MicrowaveResonators Pozar August 5, 2011 18:28 6.6 Excitation of Resonators 303 C=0.06 pF C=0.033 pF C=0.02 pF FIGURE 6.18 Smith chart plot of input impedance of the gap-coupled microstrip resonator of Example 6.6 versus frequency for various values of the coupling capacitor. zx ay bShort circui t Cavity WaveguideAperture FIGURE 6.19 A rectangular waveguide aperture coupled to a rectangular cavity. YL Z0 Z0, /H9252 FIGURE 6.20 Equivalent circuit of an aperture-coupled cavity resonator. c06MicrowaveResonators Pozar August 5, 2011 18:28 304 Chapter 6: Microwave Resonators The normalized input admittance seen by the feedline is y=Z0Y=− j/parenleftbiggZ0 XL+cotβ/lscript/parenrightbigg =− j/parenleftbiggtanβ/lscript+xL xLtanβ/lscript/parenrightbigg ,( 6.84) where xL=ωL/Z0is the normalized reactance of the aperture. An antiresonance occurs when the numerator of (6.84) vanishes, or when tanβ/lscript+xL=0,( 6.85) which is similar in form to (6.78), for the case of the gap-coupled microstrip resonator. In practice, xL/lessmuch1, so the first resonant frequency, ω1, will be close to the resonant frequency for which β/lscript=π, similar to the solution illustrated in Figure 6.17. Using the same procedure as in the previous section, we can expand the input admit- tance of (6.84) in a Taylor series about the resonant frequency, ω1, assuming xL/lessmuch1, to obtain y(ω)=y(ω1)+(ω−ω1)dy(ω) dω/vextendsingle/vextendsingle/vextendsingle/vextendsingle ω1+··· = (ω−ω1)dy(ω) d(β/lscript)d(β/lscript) dω/vextendsingle/vextendsingle/vextendsingle/vextendsingle ω1+···, (6.86) since, from (6.84) and (6.85), y(ω1)=0. Then, dy(ω) d(β/lscript)=jsec2β/lscript tan2β/lscript=j1+tan2β/lscript tan2β/lscript=j1+x2 L x2 L/similarequalj x2 L. For the rectangular waveguide, dβ dω=d dω/radicalBig k2 0−k2c=k0 βc, where cis the speed of light. Then the normalized admittance of (6.86) can be reduced to y(ω)/similarequaljk0/lscript x2 Lβc(ω−ω1)/similarequaljπk0 x2 Lβ2c(ω−ω1). (6.87) In (6.87), k0,β,andxLshould be evaluated at the resonant frequency, ω1. Loss can now be included by assuming a high- Qcavity, and replacing ω1in the numer- ator of (6.87) with ω1(1+j/2Q0), to obtain y(ω)/similarequalπk0ω1 2Q0β2cx2 L+jπk0(ω−ω1) β2cx2 L.( 6.88) At resonance the input resistance is R=2Q0β2cx2 LZ0/πk0ω1. To obtain critical cou- pling we must have R=Z0, which yields the required aperture reactance as XL=Z0/radicalBigg πk0ω1 2Q0β2c.( 6.89) From XL, the necessary aperture size can be found. The next resonant mode for the aperture-coupled cavity occurs when the input impe- dance becomes zero, or Y→∞ . From (6.84) it is seen that this occurs at a frequency such that tan β/lscript=0, orβ/lscript=π. In this case the cavity is exactly λg/2 long, so a null in the transverse electric field exists at the aperture plane, and the aperture has no effect. This mode is of little practical interest because of this negligible coupling. The excitation of a cavity resonator by an electric current probe or loop can be ana- lyzed by the method of modal analysis, similar to that discussed in Sections 4.7 and 4.8. c06MicrowaveResonators Pozar September 13, 2011 17:11 6.6 Excitation of Resonators 305 Z0 Z0C L R 1 2 FIGURE 6.21 A two-port network consisting of a series RLC resonator in series with a transmis- sion line. The procedure is complicated, however, by the fact that a complete modal expansion re- quires fields having irrotational (zero curl) components. The interested reader is referred to references [1] and [4]. DeterminingUnloaded QfromTwo-PortMeasurements Direct measurement of the unloaded Qof a resonator is generally not possible because of the loading effect of the measurement system, but it is possible to determine unloaded Q from measurements of the frequency response of the loaded resonator when it is connected to a transmission line. Both one-port (reflection measurement) and two-port (transmission measurement) techniques are possible; we will describe how unloaded Qcan be found from a two-port measurement. Figure 6.21 shows a series RLC resonator inserted in series in a transmission line of characteristic impedance Z0, forming a two-port network. Maximum transmission oc- curs at resonance since the impedance of the series resonator is minimum at resonance. Off resonance, the resonator impedance increases, and the insertion loss increases. The result is that the network of Figure 6.21 has a two-port transmission response (as given by |S21|) of the form shown in Figure 6.22. The loaded Qcan be determined from (6.21) as QL=f0/BW, where f0is the resonant frequency, and BW is the half-power bandwidth (in Hz), where the transmission response is 3 dB lower than at resonance. The unloaded Qcan be expressed in terms of the loaded Qand the coupling coeffi- cient, g. From (6.23), 1 QL=1 Qe+1 Q0=1 Q0/parenleftbigg 1+Q0 Qe/parenrightbigg =1 Q0(1+g), (6.90) –30–25–20–15–10–50 0.985 0.990 0.995 1.000 1.005 1.010 1.015S21 (dB)Q0 = 1000, g = 0.5 Q0 = 1000, g = 4.0 /H9275//H92750 FIGURE 6.22 Frequency response of the transmission characteristics of the resonator network of Figure 6.21 for two values of unloaded Qand coupling coefficient. c06MicrowaveResonators Pozar August 5, 2011 18:28 306 Chapter 6: Microwave Resonators since g=Q0/Qefrom (6.76). Rewriting (6.90) gives Q0=(1+g)QL.( 6.91) Because Q0=ω0L/Rfor the series resonator, and the external QisQe=ω0L/2Z0,a sa result of the line loading at each end of the resonator, the coupling coefficient is g=2Z0 R.( 6.92) At resonance, the impedance of the series RLC resonator reduces to Z=R. The scattering parameter S21for the two-port network of Figure 6.21 can be found in terms of the series resonator impedance using the results of Table 4.2 (or from Problem 4.11). At resonance, S21(ω0)=2Z0 2Z0+Z(ω0)=2Z0 2Z0+R=g 1+g.( 6.93) Solving for ggives g=S21(ω0) 1−S21(ω0).( 6.94) The procedure for finding the unloaded Qfrom measured scattering parameter data (or from data produced by computer modeling) is to first find the coupling coefficient using (6.94), then find the loaded Qfrom the 3 dB bandwidth, and finally, using (6.91), find Q0. Note that S21should be a real number at resonance, assuming phase reference planes at the resonator circuit. If the resonator appears as a parallel RLC circuit, it is easy to show that the result for gin (6.94) should be inverted. 6.7CAVITYPERTURBATIONS In practical applications cavity resonators are often modified by making small changes in their shape, or by introducing small pieces of dielectric or metallic materials. For exam-ple, the resonant frequency of a cavity resonator can be easily tuned with a small screw (dielectric or metallic) that enters the cavity volume, or by changing the size of the cavity with a movable wall. Another application involves the determination of dielectric constantby measuring the shift in resonant frequency when a small dielectric sample is introduced into the cavity. In some cases, the effect of such perturbations on the cavity performance can be cal- culated exactly, but often approximations must be made. One useful technique for doing this is the perturbational method, which assumes that the actual fields of a cavity with a small shape or material perturbation are not greatly different from those of the unperturbed cavity. Thus, this technique is similar in concept to the perturbational method introduced in Section 2.7 for treating loss in good conductors, where it was assumed that there wasnot a significant difference between the fields of a device with good conductors and one with perfect conductors. In this section we derive expressions for the approximate change in resonant frequency when a resonant cavity is perturbed by small changes in the material filling the cavity, or by small changes in its shape. MaterialPerturbations Figure 6.23 shows a cavity perturbed by a change in the permittivity (/Delta1/epsilon1), or permeability (/Delta1µ), of all or part of the material filling the cavity. If ¯E 0,¯H0are the fields of the original c06MicrowaveResonators Pozar August 5, 2011 18:28 6.7 Cavity Perturbations 307 ˆ E0, H0E, Hnˆn (a) (b)/H9280, /H9262 /H92750/H9275V0V0 S0S0/H9280 + ∆/H9280, /H9262 + ∆/H9262 FIGURE 6.23 A resonant cavity perturbed by a change in the permittivity or permeability of the material in the cavity. (a) Original cavity. (b) Perturbed cavity. cavity, and ¯E,¯Hare the fields of the perturbed cavity, then Maxwell’s curl equations can be written for the two cases as ∇ׯE0=− jω0µ¯H0, (6.95a) ∇ׯH0=jω0/epsilon1¯E0, (6.95b) ∇ׯE=− jω(µ+/Delta1µ)¯H, (6.96a) ∇ׯH=jω(/epsilon1+/Delta1/epsilon1)¯E, (6.96b) where ω0is the resonant frequency of the original cavity, and ωis the resonant frequency of the perturbed cavity. Multiply the conjugate of (6.95a) by ¯H, and multiply (6.96b) by ¯E∗ 0, to get ¯H·∇× ¯E∗ 0=jω0µ¯H·¯H∗ 0, ¯E∗ 0·∇× ¯H=jω(/epsilon1+/Delta1/epsilon1)¯E∗ 0·¯E. Subtracting these two equations and using the vector identity (B.8) that ∇·(¯AׯB)= ¯B·∇× ¯A−¯A·∇× ¯Bgives ∇·(¯E∗ 0ׯH)=jω0µ¯H·¯H∗ 0−jω(/epsilon1+/Delta1/epsilon1)¯E∗ 0·¯E. (6.97a) Similarly, multiply the conjugate of (6.95b) by ¯E, and multiply (6.96a) by ¯H∗ 0, to get ¯E·∇× ¯H∗ 0=− jω0/epsilon1¯E∗ 0·¯E, ¯H∗ 0·∇× ¯E=− jω(µ+/Delta1µ)¯H∗ 0·¯H. Subtracting these two equations and using vector identity (B.8) gives ∇·(¯EׯH∗ 0)=− jω(µ+/Delta1µ)¯H∗ 0·¯H+jω0/epsilon1¯E∗ 0·¯E. (6.97b) Now add (6.97a) and (6.97b), integrate over the volume V0, and use the divergence theorem to obtain /integraldisplay V0∇·(¯E∗ 0ׯH+¯EׯH∗ 0)dv=/contintegraldisplay S0(¯E∗ 0ׯH+¯EׯH∗ 0)·d¯s=0 =j/integraldisplay V0{[ω0/epsilon1−ω(/epsilon1+/Delta1/epsilon1)]¯E∗ 0·¯E+[ω0µ−ω(µ+/Delta1µ)] ¯H∗ 0·¯H}dv, (6.98) c06MicrowaveResonators Pozar August 5, 2011 18:28 308 Chapter 6: Microwave Resonators where the surface integral is zero because ˆnׯE=0o n S0. Rewriting gives ω−ω0 ω=−/integraltext V0(/Delta1/epsilon1¯E·¯E∗ 0+/Delta1µ¯H·¯H∗ 0)dv /integraltext V0(/epsilon1¯E·¯E∗ 0+µ¯H·¯H∗ 0)dv.( 6.99) This is an exact equation for the change in resonant frequency due to material pertur- bations, but is not in a very usable form since we generally do not know ¯Eand¯H,t h e exact fields in the perturbed cavity. However, if we assume that /Delta1/epsilon1and/Delta1µare small, we can approximate the perturbed fields ¯E,¯Hby the original fields ¯E0,¯H0, andωin the denominator of (6.99) by ω0, to give the approximate fractional change in resonant frequency as ω−ω0 ω0/similarequal−/integraltext V0(/Delta1/epsilon1|¯E0|2+/Delta1µ|¯H0|2)dv /integraltext V0(/epsilon1|¯E0|2+µ|¯H0|2)dv.( 6.100) This result shows that any increase in /epsilon1orµat any point in the cavity will decrease the resonant frequency. The reader may also observe that the terms in (6.100) can be related to the stored electric and magnetic energies in the original and perturbed cavities, so that the decrease in resonant frequency can be related to the increase in stored energy of theperturbed cavity. EXAMPLE 6.7 MATERIAL PERTURBATION OF A RECTANGULAR CA VITY A rectangular cavity operating in the TE 101mode is perturbed by the insertion of a thin dielectric slab into the bottom of the cavity, as shown in Figure 6.24.Use the perturbational result of (6.100) to derive an expression for the change in resonant frequency. Solution From (6.42a)–(6.42c), the fields for the unperturbed TE 101cavity mode can be written as Ey=Asinπx asinπz d, Hx=−jA ZTEsinπx acosπz d, Hz=jπA kηacosπx asinπz d. In the numerator of (6.100), /Delta1/epsilon1=(/epsilon1r−1)/epsilon10for 0≤y≤tand zero elsewhere. tbax z d/H9280ry FIGURE 6.24 A rectangular cavity perturbed by a thin dielectric slab. c06MicrowaveResonators Pozar August 5, 2011 18:28 6.7 Cavity Perturbations 309 The integral can then be evaluated as /integraldisplay V(/Delta1/epsilon1|¯E0|2+/Delta1µ¯H0|2)dv=(/epsilon1r−1)/epsilon10/integraldisplaya x=0/integraldisplayt y=0/integraldisplayd z=0|Ey|2dzdydx =(/epsilon1r−1)/epsilon10A2atd 4. The denominator of (6.100) is proportional to the total energy in the unperturbed cavity, which was evaluated in (6.43); thus, /integraldisplay V(/epsilon1|¯E0|2+µ|¯H0|2)dv=abd/epsilon10 2A2. Then (6.100) gives the fractional change (decrease) in resonant frequency as ω−ω0 ω0=−(/epsilon1 r−1)t 2b.■ ShapePerturbations Changing the size of a cavity, or inserting a tuning screw, can be considered as a change in the shape of the cavity and, for small changes, can also be treated by the perturbationtechnique. Figure 6.25 shows an arbitrary cavity with a perturbation in its shape; we will derive an expression for the change in resonant frequency. As in the case of material perturbations, let ¯E 0,¯H0,ω0be the fields and resonant frequency of the original cavity and let ¯E,¯H,ωbe the fields and resonant frequency of the perturbed cavity. Then Maxwell’s curl equations can be written for the two cases as ∇ׯE0=− jω0µ¯H0, (6.101a) ∇ׯH0=jω0/epsilon1¯E0, (6.101b) ∇ׯE=− jωµ¯H, (6.102a) ∇ׯH=jω/epsilon1¯E. (6.102b) Multiply the conjugate of (6.101a) by ¯H, and multiply (6.102b) by ¯E∗ 0, to get ¯H·∇× ¯E∗ 0=jω0µ¯H·¯H∗ 0, ¯E∗ 0·∇× ¯H=jω/epsilon1¯E∗ 0·¯E. ˆ E0, H0E, Hnˆn (a) (b)/H92750/H9275V0V∆V ∆S S0S FIGURE 6.25 A resonant cavity perturbed by a change in shape. (a) Original cavity. (b) Perturbed cavity. c06MicrowaveResonators Pozar August 5, 2011 18:28 310 Chapter 6: Microwave Resonators Subtracting these two equations and using vector identity (B.8) gives ∇·(¯E∗ 0ׯH)=jω0µ¯H·¯H∗ 0−jω/epsilon1¯E∗ 0·¯E. (6.103a) Similarly, multiply the conjugate of (6.101b) by ¯Eand (6.102a) by ¯H∗ 0to get ¯E·∇× ¯H∗ 0=− jω0/epsilon1¯E·¯E∗ 0, ¯H∗ 0·∇× ¯E=− jωµ¯H∗ 0·¯H. Subtracting and applying vector identity (B.8) gives ∇·(¯EׯH∗ 0)=− jωµ¯H∗ 0·¯H+jω0/epsilon1¯E·¯E∗ 0 (6.103b) Now add (6.103a) and (6.103b), integrate over the volume V, and use the divergence the- orem to obtain /integraldisplay V∇·(¯EׯH∗ 0+¯E∗ 0ׯH)dv=/contintegraldisplay S(¯EׯH∗ 0+¯E∗ 0ׯH)·d¯s =/contintegraldisplay S¯E∗ 0ׯH·d¯s=− j(ω−ω0)/integraldisplay V(/epsilon1¯E·¯E∗ 0+µ¯H·¯H∗ 0)dv, (6.104) sinceˆnׯE=0o n S. Because the perturbed surface S=S0−/Delta1S, we can write /contintegraldisplay S¯E∗ 0ׯH·d¯s=/contintegraldisplay S0¯E∗ 0ׯH·d¯s−/contintegraldisplay /Delta1S¯E∗ 0ׯH·d¯s=−/contintegraldisplay /Delta1S¯E∗ 0ׯH·ds, because ˆnׯE0=0o n S0. Using this result in (6.104) gives ω−ω0=−j/contintegraltext /Delta1S¯E∗ 0ׯH·d¯s/integraltext V(/epsilon1¯E·¯E∗ 0+µ¯H·¯H∗ 0)dv,( 6.105) which is an exact expression for the new resonant frequency, but not a very usable one since we generally do not initially know ¯E,¯H,o rω. If we assume /Delta1Sis small, and approximate ¯E,¯Hby the unperturbed values of ¯E0,¯H0, then the numerator of (6.105) can be reduced as follows: /contintegraldisplay /Delta1S¯E∗ 0ׯH·d¯s/similarequal/contintegraldisplay /Delta1S¯E∗ 0ׯH0·d¯s=− jω0/integraldisplay /Delta1V(/epsilon1|¯E0|2−µ|¯H0|2)dv, ( 6.106) where the last identity follows from conservation of power, as derived from the conjugate of (1.87) with σ,¯Js, and ¯Msset to zero. Using this result in (6.106) gives an expression for the approximate fractional change in resonant frequency as ω−ω0 ω0/similarequal/integraltext /Delta1V(µ|¯H0|2−/epsilon1|¯E0|2)dv/integraltext V0(µ|¯H0|2+/epsilon1|¯E0|2)dv,( 6.107) where we have also assumed that the denominator of (6.105), which represents the total energy stored in the perturbed cavity, is approximately the same as that for the unperturbedcavity. Equation (6.107) can be written in terms of stored energies as follows: ω−ω 0 ω0=/Delta1Wm−/Delta1We Wm+We,( 6.108) where /Delta1Wmand/Delta1Weare the changes in the stored magnetic energy and electric energy, respectively, after the shape perturbation, and Wm+Weis the total stored energy in the c06MicrowaveResonators Pozar September 12, 2011 21:3 6.7 Cavity Perturbations 311 cavity. These results show that the resonant frequency may either increase or decrease, depending on where the perturbation is located and whether it increases or decreases the cavity volume. EXAMPLE 6.8 SHAPE PERTURBATION OF A RECTANGULAR CA VITY A thin screw of radius r0extends a distance /lscriptthrough the center of the top wall of a rectangular cavity operating in the TE 101mode, as shown in Figure 6.26. If the cavity is air filled, use (6.107) to derive an expression for the change in resonant frequency from the unperturbed cavity. Solution From (6.42a)–(6.42c), the fields for the unperturbed TE 101cavity can be written as Ey=Asinπx asinπz d, Hx=−jA ZTEsinπx acosπz d, Hz=jπA kηacosπx asinπz d. If the screw is thin, we can assume that the fields are constant over the cross section of the screw and can be represented by the fields at x=a/2,z=d/2: Ey/parenleftbigg x=a 2,y,z=d 2/parenrightbigg =A, Hx/parenleftbigg x=a 2,y,z=d 2/parenrightbigg =0, Hz/parenleftbigg x=a 2,y,z=d 2/parenrightbigg =0. Then the numerator of (6.107) can be evaluated as /integraldisplay /Delta1V(µ|¯H0|2−/epsilon1|¯E0|2)dv=−/epsilon10/integraldisplay /Delta1VA2dv=−/epsilon10A2/Delta1V, where /Delta1V=π/lscriptr2 0is the volume of the screw. The denominator of (6.107) is, from (6.43), /integraldisplay V0(µ|¯H0|2+/epsilon1|¯E0|2)dv=abd/epsilon10A2 2=V0/epsilon10A2 2, b 2r0x z d 0y FIGURE 6.26 A rectangular cavity perturbed by a tuning post in the center of the top wall. c06MicrowaveResonators Pozar August 5, 2011 18:28 312 Chapter 6: Microwave Resonators where V0=abd is the volume of the unperturbed cavity. Then (6.107) gives ω−ω0 ω0=−2/lscriptπ r2 0 abd=−2/Delta1 V V0, which indicates a lowering of the resonant frequency. ■ REFERENCES [1] R. E. Collin, Foundations for Microwave Engineering , 2nd edition, Wiley–IEEE Press, Hoboken, N.J., 2001. [2] S. B. Cohn, “Microwave Bandpass Filters Containing High- QDielectric Resonators,” IEEE Trans- actions on Microwave Theory and Techniques, vol. MTT-16, pp. 218–227, April 1968. [3] M. W. Pospieszalski, “Cylindrical Dielectric Resonators and Their Applications in TEM Line Mi- crowave Circuits,” IEEE Transactions on Microwave Theory and Techniques , vol. MTT-27, pp. 233– 238, March 1979. [4] R. E. Collin, Field Theory of Guided Waves , McGraw-Hill, New York, 1960. PROBLEMS 6.1 A series RLC resonator with an external load is shown below. Find the resonant frequency, the un- loaded Q, and the loaded Q. Resonator Load2.5 Ω2.5 Ω 50 nH 0.79 pF 6.2 Derive an expression for the unloaded Qof a transmission line resonator consisting of a short- circuited transmission line 1λ long. 6.3 A transmission line resonator is fabricated from a λ/4 length of open-circuited line. Find the unloaded Qof this resonator if the complex propagation constant of the line is α+jβ. 6.4 Consider the resonator shown below, consisting of a λ/2 length of lossless transmission line shorted at both ends. At an arbitrary point, z, on the line, compute the impedances ZLandZRseen looking to the left and to the right, respectively, and show that ZL=Z∗ R. (This condition holds true for any lossless transmission line resonator and is the basis for the transverse resonance technique discussed in Section 3.9.) 0 z z = /H9261/2ZL ZRZ0, /H9252 6.5 A resonator is constructed from a 3.0 cm length of 100 /Omega1air-filled coaxial line, shorted at one end and terminated with a capacitor at the other end, as shown below. (a) Determine the capacitor value c06MicrowaveResonators Pozar August 5, 2011 18:28 Problems 313 to achieve the lowest order resonance at 6.0 GHz. (b) Now assume that loss is introduced by placing a 10,000 /Omega1resistor in parallel with the capacitor. Calculate the unloaded Q. R = 104 Ω Z0 = 100 Ω C3.0 cm 6.6 A transmission line resonator is made from a length /lscriptof lossless transmission line of characteristic impedance Z0=100/Omega1. If the line is terminated at both ends as shown below, find /lscript/λfor the first resonance, and the unloaded Qof this resonator. 6.7 Write the expressions for the ¯Eand¯Hfields for a short-circuited λ/2 coaxial line resonator, and show that the time-average stored electric and magnetic energies are equal. 6.8 A series RLC resonant circuit is connected to a length of transmission line that is λ/4 long at its resonant frequency, as shown below. Show that, in the vicinity of resonance, the input impedance behaves like that of a parallel RLC circuit. Z0 ZinRL C (f0, Q) /H9261/4 @ f0 6.9 A rectangular cavity resonator is constructed from a 2.0 cm length of aluminum X-band waveguide. The cavity is air filled. Find the resonant frequency and unloaded Qof the TE 101and TE 102resonant modes. 6.10 Derive the unloaded Qfor the TM 111mode of a rectangular cavity, assuming lossy conducting walls and lossless dielectric. 6.11 Consider the rectangular cavity resonator partially filled with dielectric as shown below. Derive a transcendental equation for the resonant frequency of the dominant mode by writing the fields in the air- and dielectric-filled regions in terms of TE 10waveguide modes, and enforcing boundary conditions at z=0,d–t,a n dd . z dtby x a /H92800/H9280r 0 c06MicrowaveResonators Pozar August 5, 2011 18:28 314 Chapter 6: Microwave Resonators 6.12 Determine the resonant frequencies of a rectangular cavity by carrying out a full separation-of- variables solution to the wave equation for Ez(for TM modes) and Hz(for TE modes), subject to the appropriate boundary conditions of the cavity. [Assume a solution of the form X(x)Y(y)Z(z).] 6.13 Find the unloaded Qfor the TM nm0resonant mode of a circular cavity. Consider both conductor and dielectric losses. 6.14 Design a circular cavity resonator to operate in the TE 111mode with maximum unloaded Qat a frequency of 6 GHz. The cavity is gold plated and filled with a dielectric material having /epsilon1r=1.5 and tan δ=0.0005. Find the cavity dimensions and the resulting unloaded Q. 6.15 An air-filled rectangular cavity resonator has its first three resonant modes at the frequencies 5.2, 6.5, and 7.2 GHz. Find the dimensions of the cavity. 6.16 Consider the microstrip ring resonator shown below. If the effective dielectric constant of the microstrip line is /epsilon1e, find an equation for the frequency of the first resonance. Suggest some methods of coupling to this resonator. dW a /H9280r 6.17 A circular microstrip disk resonator is shown below. Solve the wave equation for TM nm0modes for this structure, using the magnetic wall approximation that Hϕ=0a tρ=a. If fringing fields are neglected, show that the resonant frequency of the dominant mode is given by f110=1.841c 2πa√/epsilon1r 6.18 Compute the resonant frequency of a cylindrical dielectric resonator with /epsilon1r=36.2, 2 a=7.99 mm, andL=2.14 mm. 6.19 Extend the analysis of Section 6.5 to derive a transcendental equation for the resonant frequency of the next resonant mode of the cylindrical dielectric resonator. ( Hzodd in z.) 6.20 Consider the rectangular dielectric resonator shown below. Assume a magnetic wall boundary con- dition around the edges of the cavity, and allow evanescent fields in the ±zdirections away from the c06MicrowaveResonators Pozar August 5, 2011 18:28 Problems 315 dielectric, similar to the analysis of Section 6.5. Derive a transcendental equation for the resonant frequency. cb axz y 6.21 A high- Qresonator useful at millimeter wave frequencies is the Fabry-Perot resonator, which con- sists of two parallel metal plates (see figure below). A plane wave traveling at normal incidence between the two plates will exhibit resonance when the plate separation is equal to a multiple ofλ/2. (a) Derive an expression for the resonant frequency of a Fabry-Perot resonator having a plate separation dand mode number /lscript. (b) If the plates have conductivity σ, derive an expression for the unloaded Qof the resonator. (c) Use these results to find the resonant frequency and unloaded Qof a Fabry-Perot resonator having d=4.0 cm, with copper plates, and with a mode number /lscript=25. z dEx–Ex+ 0 6.22 A parallel RLC circuit, with R=1000/Omega1,L=1.26 nH, C=0.804 pF, is coupled with a series capacitor, C0, to a 50-/Omega1 transmission line, as shown below. Determine C0for critical coupling to the line. What is the resonant frequency? Z0C0 RL C 6.23 An aperture-coupled rectangular waveguide cavity has a resonant frequency of 9.0 GHz and an unloaded Qof 11,000. If the waveguide dimensions are a=2.5c ma n db =1.25 cm, find the nor- malized aperture reactance required for critical coupling. 6.24 A microwave resonator is connected as a one-port circuit, and its return loss is measured versus frequency. At resonance the return loss is 14 dB, while at 2.9985 GHz and at 3.0015 GHz the return loss is 11 dB (the half-power points). Determine the unloaded Qof the resonator. Do this for both series and parallel resonators. 6.25 A microwave resonator is measured in a two-port configuration like that shown in Figure 6.21. The minimum insertion loss is measured as 1.94 dB at 3.0000 GHz. The insertion loss is 4.95 dB at2.9925 GHz and at 3.0075 GHz. What is the unloaded Qof the resonator? c06MicrowaveResonators Pozar August 5, 2011 18:28 316 Chapter 6: Microwave Resonators 6.26 A thin slab of magnetic material is inserted next to the z=0 wall of the rectangular cavity shown below. If the cavity is operating in the TE 101mode, derive a perturbational expression for the change in resonant frequency caused by the magnetic material. z d tbyx a/H9262r /H92620 6.27 Derive an expression for the change in resonant frequency for the screw-tuned rectangular cavity of Example 6.8 if the screw is located at x=a/2,z=0, where Hxis maximum and Eyis minimum. c07PowerDividers Pozar August 24, 2011 15:53 Chapter Seven Power Dividers and Directional Couplers Power dividers and directional couplers are passive microwave components used for power division or power combining, as illustrated in Figure 7.1. In power division, an input signal is divided into two (or more) output signals of lesser power, while a power combiner accepts two or more input signals and combines them at an output port. The coupler or divider mayhave three ports, four ports, or more, and may be (ideally) lossless. Three-port networks takethe form of T-junctions and other power dividers, while four-port networks take the form of directional couplers and hybrids. Power dividers usually provide in-phase output signals with an equal power division ratio (3 dB), but unequal power division ratios are also possible. Di-rectional couplers can be designed for arbitrary power division, while hybrid junctions usuallyhave equal power division. Hybrid junctions have either a 90 ◦or a 180◦phase shift between the output ports. A wide variety of waveguide couplers and power dividers were invented and characterized at the MIT Radiation Laboratory in the 1940s. These included E-a n d H-plane waveguide T-junctions, the Bethe hole coupler, multihole directional couplers, the Schwinger coupler, the waveguide magic-T, and various types of couplers using coaxial probes. In the mid-1950s through the 1960s, many of these couplers were reinvented to use stripline or microstrip tech-nology. The increasing use of planar lines also led to the development of new types of couplersand dividers, such as the Wilkinson divider, the branch line hybrid, and the coupled line direc- tional coupler. We will first discuss some of the general properties of three- and four-port networks, and then treat the analysis and design of several of the most common types of power dividers,couplers, and hybrids. 7.1BASICPROPERTIESOFDIVIDERSANDDOUPLERS In this section we will use properties of the scattering matrix developed in Section 4.3 to de- rive some of the basic characteristics of three- and four-port networks. We will also define 317 c07PowerDividers Pozar August 24, 2011 15:53 318 Chapter 7: Power Dividers and Directional Couplers Divider or couplerP1 (a) (b)P2 = /H9251P1 P1 = P2 + P3 P3 = (1 – /H9251) P1Divider or couplerP2 P3 FIGURE 7.1 Power division and combining. (a) Power division. (b) Power combining. isolation, coupling, and directivity, which are important quantities for the characterization of couplers and hybrids. Three-PortNetworks(T-Junctions) The simplest type of power divider is a T-junction , which is a three-port network with two inputs and one output. The scattering matrix of an arbitrary three-port network has nine independent elements: [S]=/bracketleftBiggS11S12S13 S21S22S23 S31S32S33/bracketrightBigg .( 7.1) If the device is passive and contains no anisotropic materials, then it must be reciprocal and its scattering matrix will be symmetric ( Sij=Sji). Usually, to avoid power loss, we would like to have a junction that is lossless and matched at all ports. We can easily show, however, that it is impossible to construct such a three-port lossless reciprocal network thatis matched at all ports. If all ports are matched, then S ii=0, and if the network is reciprocal, the scattering matrix of (7.1) reduces to [S]=/bracketleftBigg0 S12S13 S12 0 S23 S13S23 0/bracketrightBigg .( 7.2) If the network is also lossless, then energy conservation requires that the scattering matrix satisfy the unitary properties of (4.53), which leads to the following conditions [1, 2]: |S12|2+|S13|2=1, (7.3a) |S12|2+|S23|2=1, (7.3b) |S13|2+|S23|2=1, (7.3c) S∗ 13S23=0, (7.3d) S∗ 23S12=0, (7.3e) S∗ 12S13=0. (7.3f) Equations (7.3d)–(7.3f) show that at least two of the three parameters ( S12,S13,S23)m u s t be zero. However, this condition will always be inconsistent with one of equations (7.3a)– (7.3c), implying that a three-port network cannot be simultaneously lossless, reciprocal, and matched at all ports. If any one of these three conditions is relaxed, then a physicallyrealizable device is possible. If the three-port network is nonreciprocal, then S ij/negationslash=Sji, and the conditions of input matching at all ports and energy conservation can be satisfied. Such a device is known as acirculator , and generally relies on an anisotropic material, such as ferrite, to achieve non- reciprocal behavior. Ferrite circulators will be discussed in more detail in Chapter 9, but c07PowerDividers Pozar August 24, 2011 15:53 7.1 Basic Properties of Dividers and Douplers 319 we can demonstrate here that any matched lossless three-port network must be nonrecip- rocal and, thus, a circulator. The scattering matrix of a matched three-port network has the following form: [S]=/bracketleftBigg0 S12S13 S21 0 S23 S31S32 0/bracketrightBigg .( 7.4) If the network is lossless, [S]must be unitary, which implies the following conditions: S∗ 31S32=0, (7.5a) S∗ 21S23=0, (7.5b) S∗ 12S13=0, (7.5c) |S12|2+|S13|2=1, (7.5d) |S21|2+|S23|2=1, (7.5e) |S31|2+|S32|2=1. (7.5f) These equations can be satisfied in one of two ways. Either S12=S23=S31=0,|S21|=| S32|=| S13|=1,( 7.6a) or S21=S32=S13=0,|S12|=| S23|=| S31|=1.( 7.6b) These results shows that Sij/negationslash=Sjifori/negationslash=j, which implies that the device must be non- reciprocal. The scattering matrices for the two solutions of (7.6) are shown in Figure 7.2, together with the symbols for the two possible types of circulators. The only difference between the two cases is in the direction of power flow between the ports: solution (7.6a) corresponds to a circulator that allows power flow only from port 1 to 2, or port 2 to 3, orport 3 to 1, while solution (7.6b) corresponds to a circulator with the opposite direction of power flow. Alternatively, a lossless and reciprocal three-port network can be physically realized if only two of its ports are matched [1]. If ports1and 2 are the matched ports, then the scattering matrix can be written as [S]=/bracketleftBigg0 S 12S13 S12 0 S23 S13S23S33/bracketrightBigg .( 7.7) 12 312 3001 1 [S ] = 0 0 010 (a)010 0 [S ] = 0 1 100 (b) FIGURE 7.2 Two types of circulators and their scattering matrices. (a) Clockwise circulation. (b) Counterclockwise circulation. The phase references for the ports are arbitrary. c07PowerDividers Pozar August 24, 2011 15:53 320 Chapter 7: Power Dividers and Directional Couplers 1 2 3[S] =0 ej/H9258 0ej/H9258 0 00 0 ej/H9278S21 = ej/H9258 S12 = ej/H9258 S33 = ej/H9278 FIGURE 7.3 A reciprocal lossless three-port network matched at ports 1 and 2. To be lossless, the following unitarity conditions must be satisfied: S∗ 13S23=0, (7.8a) S∗ 12S13+S∗ 23S33=0, (7.8b) S∗ 23S12+S∗ 33S13=0, (7.8c) |S12|2+|S13|2=1, (7.8d) |S12|2+|S23|2=1, (7.8e) |S13|2+|S23|2+|S33|2=1. (7.8f) Equations (7.8d) and (7.8e) show that |S13|=| S23|, so (7.8a) leads to the result that S13= S23=0. Then, |S12|=| S33|=1. The scattering matrix and corresponding signal flow graph for this network are shown in Figure 7.3, where it is seen that the network actu- ally degenerates into two separate components—one a matched two-port line and the other a totally mismatched one-port. Finally, if the three-port network is allowed to be lossy, it can be reciprocal and matched at all ports; this is the case of the resistive divider, which will be discussed in Section 7.2. In addition, a lossy three-port network can be made to have isolation between its output ports (e.g., S23=S32=0). Four-PortNetworks(DirectionalCouplers) The scattering matrix of a reciprocal four-port network matched at all ports has the follow- ing form: [S]=⎡ ⎢⎣0 S12S13S14 S12 0 S23S24 S13S23 0 S34 S14S24S34 0⎤ ⎥⎦.( 7.9) If the network is lossless, 10 equations result from the unitarity, or energy conservation, condition [1, 2]. Consider the multiplication of row 1 and row 2, and the multiplication of row 4 and row 3: S∗ 13S23+S∗ 14S24=0, (7.10a) S∗ 14S13+S∗ 24S23=0. (7.10b) c07PowerDividers Pozar August 24, 2011 15:53 7.1 Basic Properties of Dividers and Douplers 321 Multiply (7.10a) by S∗ 24, and (7.10b) by S∗ 13, and subtract to obtain S∗ 14(|S13|2−|S24|2)=0.( 7.11) Similarly, the multiplication of row 1 and row 3, and the multiplication of row 4 and row 2, gives S∗ 12S23+S∗ 14S34=0, (7.12a) S∗ 14S12+S∗ 34S23=0. (7.12b) Multiply (7.12a) by S12, and (7.12b) by S34, and subtract to obtain S23(|S12|2−|S34|2)=0.( 7.13) One way for (7.11) and (7.13) to be satisfied is if S14=S23=0, which results in a direc- tional coupler. Then the self-products of the rows of the unitary scattering matrix of (7.9) yield the following equations: |S12|2+|S13|2=1, (7.14a) |S12|2+|S24|2=1, (7.14b) |S13|2+|S34|2=1, (7.14c) |S24|2+|S34|2=1, (7.14d) which imply that |S13|=| S24|[using (7.14a) and (7.14b)], and that |S12|=| S34|[using (7.14b) and (7.14d)]. Further simplification can be made by choosing the phase references on three of the four ports. Thus, we choose S12=S34=α,S13=βejθ,andS24=βejφ, where αandβ are real, and θandφare phase constants to be determined (one of which we are still free to choose). The dot product of rows 2 and 3 gives S∗ 12S13+S∗ 24S34=0,( 7.15) which yields a relation between the remaining phase constants as θ+φ=π±2nπ. (7.16) If we ignore integer multiples of 2π , there are two particular choices that commonly occur in practice: 1.A Symmetric Coupler :θ=φ=π/2. The phases of the terms having amplitude β are chosen equal. Then the scattering matrix has the following form: [S]=⎡ ⎢⎣0α jβ 0 α 00 jβ jβ 00 α 0 jβα 0⎤ ⎥⎦.( 7.17) 2.An Antisymmetric Coupler :θ=0,φ=π. The phases of the terms having ampli- tudeβare chosen to be 180◦apart. Then the scattering matrix has the following form: [S]=⎡ ⎢⎣0αβ 0 α 00 −β β 00 α 0−βα 0⎤ ⎥⎦.( 7.18) c07PowerDividers Pozar August 24, 2011 15:53 322 Chapter 7: Power Dividers and Directional Couplers 1 2 4 3Input IsolatedThrough Coupled 1 2 4 3Input IsolatedThrough Coupled FIGURE 7.4 Two commonly used symbols for directional couplers, and power flow conventions. Note that these two couplers differ only in the choice of reference planes. In addition, the amplitudes αandβare not independent, as (7.14a) requires that α2+β2=1.( 7.19) Thus, apart from phase references, an ideal four-port directional coupler has only one de- gree of freedom, leading to two possible configurations. Another way for (7.11) and (7.13) to be satisfied is if |S13|=| S24|and|S12|=| S34|. If we choose phase references, however, such that S13=S24=αand S12=S34=jβ [which satisfies (7.16)], then (7.10a) yields α(S23+S∗ 14)=0, and (7.12a) yields β(S∗ 14− S23)=0. These two equations have two possible solutions. First, S14=S23=0, which is the same as the above solution for the directional coupler. The other solution occurs for α=β=0, which implies that S12=S13=S24=S34=0. This is the degenerate case of two decoupled two-port networks (between ports 1 and 4, and ports 2 and 3), which is oftrivial interest and will not be considered further. We are thus left with the conclusion that any reciprocal, lossless, matched four-port network is a directional coupler. The basic operation of a directional coupler can be illustrated with the aid of Figure 7.4, which shows two commonly used symbols for a directional coupler and the port definitions. Power supplied to port 1 is coupled to port 3 (the coupled port) with the coupling factor |S 13|2=β2, while the remainder of the input power is delivered to port 2 (the through port) with the coefficient |S12|2=α2=1−β2. In an ideal directional coupler, no power is delivered to port 4 (the isolated port). The following quantities are commonly used to characterize a directional coupler: Coupling =C=10 logP1 P3=−20 log βdB, (7.20a) Directivity =D=10 logP3 P4=20 logβ |S14|dB, (7.20b) Isolation =I=10 logP1 P4=−20 log |S14|dB, (7.20c) Insertion loss =L=10 logP1 P2=−20 log |S12|dB. (7.20d) The coupling factor indicates the fraction of the input power that is coupled to the out- put port. The directivity is a measure of the coupler’s ability to isolate forward and back- ward waves (or the coupled and uncoupled ports). The isolation is a measure of the power c07PowerDividers Pozar August 24, 2011 15:53 7.1 Basic Properties of Dividers and Douplers 323 delivered to the uncoupled port. These quantities are related as I=D+CdB. (7.21) Theinsertion loss accounts for the input power delivered to the through port, diminished by power delivered to the coupled and isolated ports. The ideal coupler has infinite directivity and isolation (S14=0). Then both αandβcan be determined from the coupling factor, C. Hybrid couplers are special cases of directional couplers, where the coupling factor is 3 dB, which implies that α=β=1/√ 2. There are two types of hybrids. The quadrature hybrid has a 90◦phase shift between ports 2 and 3 ( θ=φ=π/2) when fed at port 1, and is an example of a symmetric coupler. Its scattering matrix has the following form: [S]=1√ 2⎡ ⎢⎣01 j0 100 j j001 0j10⎤ ⎥⎦.( 7.22) Themagic-T hybrid and the rat-race hybrid have a 180◦phase difference between ports 2 and 3 when fed at port 4, and are examples of an antisymmetric coupler. Its scattering matrix has the following form: [S]=1√ 2⎡ ⎢⎣0110 100 −1 1001 0−11 0⎤ ⎥⎦.( 7.23) POINT OF INTEREST: Measuring Coupler Directivity The directivity of a directional coupler is a measure of the coupler’s ability to separate forward and reverse wave components, and applications of directional couplers often require high (35 dB or greater) directivity. Poor directivity will limit the accuracy of a reflectometer, and can cause variations in the coupled power level from a coupler when there is even a small mismatch onthe through line. The directivity of a coupler generally cannot be measured directly because it involves a low-level signal that can be masked by coupled power from a reflected wave on the througharm. For example, if a coupler has C=20 dB and D=35 dB, with a load having a return loss RL=30 dB, the signal level through the directivity path will be D+C=55 dB below the input power, but the reflected power through the coupled arm will only be RL +C=50 dB below the input power. One way to measure coupler directivity uses a sliding matched load, as follows. First, the coupler is connected to a source and a matched load, as shown in the accompanying left-handfigure, and the coupled output power is measured. If we assume an input power P i,t h i sp o w e r will be Pc=C2Pi,w h e r e C=10(−CdB)/20is the numerical voltage coupling factor of the coupler. Next, the position of the coupler is reversed, and the through line is terminated with asliding load, as shown in the right-hand figure. Vi, PiCPc Load Vi, PiCC DV0 Sliding load(Pmax, Pmin) Γ Changing the position of the sliding load introduces a variable phase shift in the signal re- flected from the load and coupled to the output port. The voltage at the output port can be c07PowerDividers Pozar August 24, 2011 15:53 324 Chapter 7: Power Dividers and Directional Couplers written as V0=Vi/parenleftbiggC D+C|/Gamma1|e−jθ/parenrightbigg , where Viis the input voltage, D=10(DdB)/20≥1 is the numerical value of the directivity, |/Gamma1| is the reflection coefficient magnitude of the load, and θis the path length difference between the directivity and reflected signals. Moving the sliding load changes θ, so the two signals will combine to trace out a circular locus, as shown in the following figure. Im V0 0ReV0VmaxCΓVi VminV0 ViC D/H9258 The minimum and maximum output powers are given by Pmin=Pi/parenleftbiggC D−C|/Gamma1|/parenrightbigg2 ,Pmax=Pi/parenleftbiggC D+C|/Gamma1|/parenrightbigg2 . LetMandmbe defined in terms of these powers as follows: M=Pc Pmax=/parenleftbiggD 1+|/Gamma1|D/parenrightbigg2 ,m=Pmax Pmin=/parenleftbigg1+|/Gamma1|D 1−|/Gamma1|D/parenrightbigg2 . These ratios can be accurately measured directly by using a variable attenuator between the source and coupler. The coupler directivity (numerical) can then be found as D=M/parenleftbigg2m m+1/parenrightbigg . This method requires that |/Gamma1|<1/Dor, in dB, RL >D. Reference: M. Sucher and J. Fox, eds., Handbook of Microwave Measurements, 3rd edition, V olume II, Polytech- nic Press, New York, 1963. 7.2THET-JUNCTIONPOWERDIVIDER The T-junction power divider is a simple three-port network that can be used for power division or power combining, and it can be implemented in virtually any type of transmis- sion line medium. Figure 7.5 shows some commonly used T-junctions in waveguide and microstrip line or stripline form. The junctions shown here are, in the absence of transmis- sion line loss, lossless junctions. Thus, as discussed in the preceding section, such junctions cannot be matched simultaneously at all ports. We will analyze the T-junction divider be-low, followed by a discussion of the resistive power divider, which can be matched at all ports but is not lossless. LosslessDivider The lossless T-junction dividers of Figure 7.5 can all be modeled as a junction of three transmission lines, as shown in Figure 7.6 [3]. In general, there may be fringing fields and c07PowerDividers Pozar August 24, 2011 15:53 7.2 The T-Junction Power Divider 325 (a) (c)(b) FIGURE 7.5 Various T-junction power dividers. (a) E-plane waveguide T. (b) H-plane wave- guide T. (c) Microstrip line T-junction divider. higher order modes associated with the discontinuity at such a junction, leading to stored energy that can be accounted for by a lumped susceptance, B. In order for the divider to be matched to the input line of characteristic impedance Z0,w em u s th a v e Yin=jB+1 Z1+1 Z2=1 Z0.( 7.24) If the transmission lines are assumed to be lossless (or of low loss), then the characteristic impedances are real. If we also assume B=0, then (7.24) reduces to 1 Z1+1 Z2=1 Z0.( 7.25) In practice, if Bis not negligible, some type of discontinuity compensation or a reac- tive tuning element can usually be used to cancel this susceptance, at least over a narrow frequency range. YinjB V0 Z0Z1 Z2+ – FIGURE 7.6 Transmission line model of a lossless T-junction divider. c07PowerDividers Pozar August 24, 2011 15:53 326 Chapter 7: Power Dividers and Directional Couplers The output line impedances, Z1and Z2, can be selected to provide various power division ratios. Thus, for a 50 /Omega1input line, a 3 dB (equal split) power divider can be made by using two 100 /Omega1output lines. If necessary, quarter-wave transformers can be used to bring the output line impedances back to the desired levels. If the output lines are matched,then the input line will be matched. There will be no isolation between the two output ports, however, and there will be a mismatch looking into the output ports. EXAMPLE 7.1 THE T-JUNCTION POWER DIVIDER A lossless T-junction power divider has a source impedance of 50 /Omega1. Find the out- put characteristic impedances so that the output powers are in a 2:1 ratio. Computethe reflection coefficients seen looking into the output ports. Solution If the voltage at the junction is V 0, as shown in Figure 7.6, the input power to the matched divider is Pin=1 2V2 0 Z0, while the output powers are P1=1 2V2 0 Z1=1 3Pin, P2=1 2V2 0 Z2=2 3Pin. These results yield the characteristic impedances as Z1=3Z0=150/Omega1, Z2=3Z0 2=75/Omega1. The input impedance to the junction is Zin=75||150 =50/Omega1, so that the input is matched to the 50 /Omega1source. Looking into the 150 /Omega1output line, we see an impedance of 50 ||75=30/Omega1, while at the 75 /Omega1output line we see an impedance of 50 ||150=37.5/Omega1.T h e reflection coefficients seen looking into these ports are /Gamma11=30−150 30+150=−0.666, /Gamma12=37.5−75 37.5+75=−0.333.■ ResistiveDivider If a three-port divider contains lossy components, it can be made to be matched at all ports, although the two output ports may not be isolated [3]. The circuit for such a divider isillustrated in Figure 7.7, using lumped-element resistors. An equal-split ( −3 dB) divider is shown, but unequal power division ratios are also possible. c07PowerDividers Pozar August 24, 2011 15:53 7.2 The T-Junction Power Divider 327 ZinPort 1 Port 3Port 2 Z0/3 V1 Z0Z0 Z0P2 P3P1+ –VZ+ –Z0/3 Z0/3V2+ – V3+ – FIGURE 7.7 An equal-split three-port resistive power divider. The resistive divider of Figure 7.7 can easily be analyzed using circuit theory. Assum- ing that all ports are terminated in the characteristic impedance Z0, the impedance Z, seen looking into the Z0/3 resistor followed by a terminated output line, is Z=Z0 3+Z0=4Z0 3.( 7.26) Then the input impedance of the divider is Zin=Z0 3+2Z0 3=Z0,( 7.27) which shows that the input is matched to the feed line. Because the network is symmetric from all three ports, the output ports are also matched. Thus, S11=S22=S33=0. If the voltage at port 1 is V1, then by voltage division the voltage Vat the center of the junction is V=V12Z0/3 Z0/3+2Z0/3=2 3V1,( 7.28) and the output voltages are, again by voltage division, V2=V3=VZ0 Z0+Z0/3=3 4V=1 2V1.( 7.29) Thus, S21=S31=S23=1/2, so the output powers are 6 dB below the input power level. The network is reciprocal, so the scattering matrix is symmetric, and it can be written as [S]=1 2/bracketleftBigg011 101110/bracketrightBigg .( 7.30) The reader may verify that this is not a unitary matrix. The power delivered to the input of the divider is P in=1 2V2 1 Z0,( 7.31) c07PowerDividers Pozar August 24, 2011 15:53 328 Chapter 7: Power Dividers and Directional Couplers while the output powers are P2=P3=1 2(1/2V1)2 Z0=1 8V2 1 Z0=1 4Pin,( 7.32) which shows that half of the supplied power is dissipated in the resistors. 7.3THEWILKINSONPOWERDIVIDER The lossless T-junction divider suffers from the disadvantage of not being matched at all ports, and it does not have isolation between output ports. The resistive divider can be matched at all ports, but even though it is not lossless, isolation is still not achieved. Fromthe discussion in Section 7.1, however, we know that a lossy three-port network can be made having all ports matched, with isolation between output ports. The Wilkinson power divider [4] is such a network, with the useful property of appearing lossless when the output ports are matched; that is, only reflected power from the output ports is dissipated. The Wilkinson power divider can be made with arbitrary power division, but we will first consider the equal-split (3 dB) case. This divider is often made in microstrip line or stripline form, as depicted in Figure 7.8a; the corresponding transmission line circuit is given in Figure 7.8b. We will analyze this circuit by reducing it to two simpler circuitsdriven by symmetric and antisymmetric sources at the output ports. This “even-odd” mode analysis technique [5] will also be useful for other networks that we will study in later sections. Even-OddModeAnalysis For simplicity, we can normalize all impedances to the characteristic impedance Z 0, and redraw the circuit of Figure 7.8b with voltage generators at the output ports as shown inFigure 7.9. This network has been drawn in a form that is symmetric across the midplane; the two source resistors of normalized value 2 combine in parallel to give a resistor of normalized value 1, representing the impedance of a matched source. The quarter-wavelines have a normalized characteristic impedance Z, and the shunt resistor has a normalized value of r; we shall show that, for the equal-split power divider, these values should be Z=√ 2 and r=2, as given in Figure 7.8. /H9261/4 /H9261/4/H9261 4Z02Z0 2Z0Z0 Z0Z0 2Z0 Z02Z0 2Z0 (a) (b) FIGURE 7.8 The Wilkinson power divider. (a) An equal-split Wilkinson power divider in mi- crostrip line form. (b) Equivalent transmission line circuit. c07PowerDividers Pozar August 24, 2011 15:53 7.3 The Wilkinson Power Divider 329 Port 3/H9261/4 /H9261/4Port 2 1 22 1Vg2 Vg3r/2 r/2 +V3+V2 +V1Z ZPort 1 FIGURE 7.9 The Wilkinson power divider circuit in normalized and symmetric form. Now define two separate modes of excitation for the circuit of Figure 7.9: the even mode, where Vg2=Vg3=2V0, and the odd mode , where Vg2=−Vg3=2V0. Superpo- sition of these two modes effectively produces an excitation of Vg2=4V0andVg3=0, from which we can find the scattering parameters of the network. We now treat these two modes separately. Even mode: For even-mode excitation, Vg2=Vg3=2V0,s oVe 2=Ve 3, and therefore no current flows through the r/2 resistors or the short circuit between the inputs of the two transmission lines at port 1. We can then bisect the network of Figure 7.9 with open circuits at these points to obtain the network of Figure 7.10a (the grounded side of the λ/4 line is not shown). Then, looking into port 2, we see an impedance Ze in=Z2 2,( 7.33) FIGURE 7.10 Bisection of the circuit of Figure 7.9. (a) Even-mode excitation. (b) Odd-mode excitation. c07PowerDividers Pozar August 24, 2011 15:53 330 Chapter 7: Power Dividers and Directional Couplers since the transmission line looks like a quarter-wave transformer. Thus, if Z=√ 2, port 2 will be matched for even-mode excitation; then Ve 2=V0since Ze in=1. The r/2 resistor is superfluous in this case since one end is open-circuited. Next, we find Ve 1from the transmission line equations. If we let x=0 at port 1 and x=−λ/4 at port 2, we can write the voltage on the transmission line section as V(x)=V+(e−jβx+/Gamma1ejβx). Then Ve 2=V(−λ/ 4)=jV+(1−/Gamma1)=V0, (7.34a) Ve 1=V(0)=V+(1+/Gamma1)=jV0/Gamma1+1 /Gamma1−1. (7.34b) The reflection coefficient /Gamma1is that seen at port 1 looking toward the resistor of normalized value 2, so /Gamma1=2−√ 2 2+√ 2, and Ve 1=− jV0√ 2.( 7.35) Odd mode: For odd-mode excitation, Vg2=−Vg3=2V0, and so Vo 2=−Vo 3, and there is a voltage null along the middle of the circuit in Figure 7.9. We can then bisect this circuit by grounding it at two points on its midplane to give the network of Figure 7.10b. Looking into port 2, we see an impedance of r/2 since the parallel-connected transmission line is λ/4 long and shorted at port 1, and so looks like an open circuit at port 2. Thus, port 2 will be matched for odd-mode excitation if we select r=2. Then Vo 2=V0andVo 1=0; for this mode of excitation all power is delivered to the r/2 resistors, with none going to port 1. Finally, we must find the input impedance at port 1 of the Wilkinson divider when ports 2 and 3 are terminated in matched loads. The resulting circuit is shown in Figure 7.11a, where it is seen that this is similar to an even mode of excitation since V2=V3. No current flows through the resistor of normalized value 2, so it can be removed, leav- ing the circuit of Figure 7.11b. We then have the parallel connection of two quarter-wave transformers terminated in loads of unity (normalized). The input impedance is Zin=1 2/parenleftbig√ 2/parenrightbig2=1.( 7.36) In summary, we can establish the following scattering parameters for the Wilkinson divider: S11=0 (Zin=1 at port 1) S22=S33=0 (ports 2 and 3 matched for even and odd modes ) S12=S21=Ve 1+Vo 1 Ve 2+Vo 2=− j/√ 2(symmetry due to reciprocity ) S13=S31=− j/√ 2 (symmetry of ports 2 and 3) S23=S32=0 (due to short or open at bisection) c07PowerDividers Pozar August 24, 2011 15:53 7.3 The Wilkinson Power Divider 331 /H9261/4/H9261/4 11 1 Port 3Port 2 2 Zin 211 1 Port 3Port 2 2 Zin 22 (a) (b)Port 1Port 1 FIGURE 7.11 Analysis of the Wilkinson divider to find S11. (a) The terminated Wilkinson di- vider. (b) Bisection of the circuit in (a). The preceding formula for S12applies because all ports are matched when terminated with matched loads. Note that when the divider is driven at port 1 and the outputs are matched, no power is dissipated in the resistor. Thus the divider is lossless when the outputs are matched; only reflected power from ports 2 or 3 is dissipated in the resistor. Because S23=S32=0, ports 2 and 3 are isolated. EXAMPLE 7.2 DESIGN AND PERFORMANCE OF A WILKINSON DIVIDER Design an equal-split Wilkinson power divider for a 50 /Omega1system impedance at frequency f0, and plot the return loss ( S11), insertion loss ( S21=S31), and isola- tion ( S23=S32) versus frequency from 0 .5f0to 1.5f0. Solution From Figure 7.8 and the above derivation, we have that the quarter-wave trans- mission lines in the divider should have a characteristic impedance of Z=√ 2Z0=70.7/Omega1, and the shunt resistor a value of R=2Z0=100/Omega1. The transmission lines are λ/4 long at the frequency f0. Using a computer-aided design tool for the analysis of microwave circuits, the scattering parameter mag-nitudes were calculated and plotted in Figure 7.12. ■ c07PowerDividers Pozar August 24, 2011 15:53 332 Chapter 7: Power Dividers and Directional Couplers 0.5 f0 f0 1.5 f0–40–30–20 ⎢Sij⎢ dB–100 ⎢S12⎢ ⎢S11⎢⎢S23⎢ FIGURE 7.12 Frequency response of an equal-split Wilkinson power divider. Port 1 is the input port; ports 2 and 3 are the output ports. UnequalPowerDivisionand N-WayWilkinsonDividers Wilkinson-type power dividers can also be made with unequal power splits; a microstrip line version is shown in Figure 7.13. If the power ratio between ports 2 and 3 is K2=P3/P2, then the following design equations apply: Z03=Z0/radicalBigg 1+K2 K3, (7.37a) Z02=K2Z03=Z0/radicalbig K(1+K2), (7.37b) R=Z0/parenleftbigg K+1 K/parenrightbigg . (7.37c) Note that the above results reduce to the equal-split case for K=1. Also observe that the output lines are matched to the impedances R2=Z0KandR3=Z0/K, as opposed to the impedance Z0; matching transformers can be used to transform these output impedances. The Wilkinson divider can also be generalized to an N-way divider or combiner [4], as shown in Figure 7.14. This circuit can be matched at all ports, with isolation between all ports. A disadvantage, however, is the fact that the divider requires crossovers for the re-sistors for N≥3, which makes fabrication difficult in planar form. The Wilkinson divider can also be made with stepped multiple sections, for increased bandwidth. A photograph of a four-way Wilkinson divider network is shown in Figure 7.15. 12 3Z0 Z03Z02 R2 = Z0K R3 = Z0/KR FIGURE 7.13 A Wilkinson power divider in microstrip form having unequal power division. c07PowerDividers Pozar August 24, 2011 15:53 7.4 Waveguide Directional Couplers 333 /H9261/4Z0N Z0N Z0N Z0Z0 Z0 Z0 Z0Z0Z0Z0Z0 Z0N. .. FIGURE 7.14 AnN-way, equal-split Wilkinson power divider. 7.4WAVEGUIDEDIRECTIONALCOUPLERS We now turn our attention to directional couplers, which are four-port devices with the characteristics discussed in Section 7.1. To review the basic operation, consider the direc- tional coupler schematic symbols shown in Figure 7.4. Power incident at port 1 will coupleto port 2 (the through port) and to port 3 (the coupled port), but not to port 4 (the isolated port). Similarly, power incident in port 2 will couple to ports 1 and 4, but not 3. Thus, ports 1 and 4 are decoupled, as are ports 2 and 3. The fraction of power coupled from port 1 to port 3 is given by C, the coupling coefficient, as defined in (7.20a), and the leakage of power from port 1 to port 4 is given by I, the isolation, as defined in (7.20c). Another quantity that characterizes a coupler is the directivity, D=I−C(dB), which is the ratio FIGURE 7.15 Photograph of a four-way corporate power divider network using three microstrip Wilkinson power dividers. Note the isolation chip resistors. Courtesy of M. D. Abouzahra, MIT Lincoln Laboratory, Lexington, Mass. c07PowerDividers Pozar August 24, 2011 15:53 334 Chapter 7: Power Dividers and Directional Couplers of the powers delivered to the coupled port and the isolated port. The ideal coupler is char- acterized solely by the coupling factor, as the isolation and directivity are infinite. The ideal coupler is also lossless and matched at all ports. Directional couplers can be made in many different forms. We will first discuss wave- guide couplers, followed by hybrid junctions. A hybrid junction is a special case of a directional coupler, where the coupling factor is 3 dB (equal split), and the phase relation between the output ports is either 90◦(quadrature hybrid), or 180◦(magic-T or rat-race hybrid). Then we will discuss the implementation of directional couplers in coupled trans- mission line form. BetheHoleCoupler The directional property of all directional couplers is produced through the use of two sep- arate waves or wave components, which add in phase at the coupled port and are canceled at the isolated port. One of the simplest ways of doing this is to couple one waveguideto another through a single small hole in the common broad wall between the two wave- guides. Such a coupler is known as a Bethe hole coupler , two versions of which are shown in Figure 7.16. From the small-aperture coupling theory of Section 4.8, we know that anaperture can be replaced with equivalent sources consisting of electric and magnetic dipole moments [6]. The normal electric dipole moment and the axial magnetic dipole moment radiate with even symmetry in the coupled guide, while the transverse magnetic dipole mo-ment radiates with odd symmetry. Thus, by adjusting the relative amplitudes of these two equivalent sources, we can cancel the radiation in the direction of the isolated port, while enhancing the radiation in the direction of the coupled port. Figure 7.16 shows two ways inwhich these wave amplitudes can be controlled; in the coupler shown in Figure 7.16a, the a34 2(Isolated) (Through) 1 (Input)3 (Coupled) 2 (Through) 4( Isolated)1S b by x z(Coupled) (Input) (a) (b)/H9258 FIGURE 7.16 Two versions of the Bethe hole directional coupler. (a) Parallel waveguides. (b) Skewed waveguides. c07PowerDividers Pozar September 12, 2011 21:26 7.4 Waveguide Directional Couplers 335 two waveguides are parallel and the coupling is controlled by s, the aperture offset from the sidewall of the waveguide. For the coupler of Figure 7.16b, the wave amplitudes are controlled by the angle, θ, between the two waveguides. First consider the configuration of Figure 7.16a, with an incident TE 10mode into port 1. These fields can be written as Ey=Asinπx ae−jβz, (7.38a) Hx=−A Z10sinπx ae−jβz, (7.38b) Hz=jπA βaZ10cosπx ae−jβz, (7.38c) where Z10=k0η0/βis the wave impedance of the TE 10mode. Then, from (4.124) and (4.125), this incident wave generates the following equivalent polarization currents at the aperture at x=s,y=b,z=0: ¯Pe=/epsilon10αeˆyAsinπs aδ(x−s)δ(y−b)δ(z), (7.39a) ¯Pm=−αmA/bracketleftbigg−ˆx Z10sinπs a+ˆzjπ βaZ10cosπs a/bracketrightbigg δ(x−s)δ(y−b)δ(z).(7.39b) Using (4.128a) and (4.128b) to relate ¯Peand¯Pmto the currents ¯Jand¯M, and then using (4.118), (4.120), (4.122), and (4.123), gives the amplitudes of the forward and reverse traveling waves in the top guide as A+ 10=−1 P10/integraldisplay v¯E− 10·¯Jdv+1 P10/integraldisplay v¯H− 10·¯Mdv =−jωA P10/bracketleftBigg /epsilon10αesin2πs a−µ0αm Z2 10/parenleftBigg sin2πs a+π2 β2a2cos2πs a/parenrightBigg/bracketrightBigg ,(7.40a) A− 10=−1 P10/integraldisplay v¯E+ 10·¯Jdv+1 P10/integraldisplay v¯H+ 10·¯Mdv =−jωA P10/bracketleftBigg /epsilon10αesin2πs a+µ0αm Z2 10/parenleftBigg sin2πs a−π2 β2a2cos2πs a/parenrightBigg/bracketrightBigg ,(7.40b) where P10=ab/Z10is the power normalization constant. Note from (7.40a) and (7.40b) that the amplitude of the wave excited toward port 4 ( A+ 10)is generally different from that excited toward port 3 ( A− 10)(because H+ x=− H− x), so we can cancel the power delivered to port 4 by setting A+ 10=0. If we assume that the aperture is round, then Table 4.3 gives the polarizabilities as αe=2r3 0/3 and αm=4r3 0/3, where r0is the radius of the aperture. Then from (7.40a) we obtain the following condition for A+ 10=0: /parenleftBigg 2/epsilon10−4µ0 Z2 10/parenrightBigg sin2πs a−4π2µ0 β2a2Z2 10cos2πs a=0, /parenleftbig k2 0−2β2/parenrightbig sin2πs a=2π2 a2cos2πs a, /parenleftBigg 4π2 a2−k2 0/parenrightBigg sin2πs a=2π2 a2, c07PowerDividers Pozar August 24, 2011 15:53 336 Chapter 7: Power Dividers and Directional Couplers or sinπs a=π/radicalBigg 2 4π2−k2 0a2=λ0/radicalBig 2/parenleftbig λ2 0−a2/parenrightbig.( 7.41) The coupling factor is then given by C=20 log/vextendsingle/vextendsingle/vextendsingle/vextendsingleA A− 10/vextendsingle/vextendsingle/vextendsingle/vextendsingledB (7.42a) and the directivity by D=20 log/vextendsingle/vextendsingle/vextendsingle/vextendsingleA− 10 A+10/vextendsingle/vextendsingle/vextendsingle/vextendsingledB. (7.42b) Thus, a Bethe hole coupler of the type shown in Figure 7.16a can be designed by first using (7.41) to find s, the position of the aperture, and then using (7.42a) to determine the aperture size, r 0, to give the required coupling factor. For the skewed geometry of Figure 7.16b, the aperture may be centered at s=a/2, and the skew angle θadjusted for cancellation at port 4. In this case, the normal electric field does not change with θ, but the transverse magnetic field components are reduced by cosθ. We can account for the skew angle by replacing αmin the previous derivation by αmcosθ. The wave amplitudes of (7.40a) and (7.40b) then become, for s=a/2, A+ 10=−jωA P10/parenleftBigg /epsilon10αe−µ0αm Z2 10cosθ/parenrightBigg , (7.43a) A− 10=−jωA P10/parenleftBigg /epsilon10αe+µ0αm Z2 10cosθ/parenrightBigg . (7.43b) Setting A+ 10=0 results in the following condition for the angle θ: 2/epsilon10−4µ0 Z2 10cosθ=0, or cosθ=k2 0 2β2.( 7.44) The coupling factor then simplifies to C=20 log/vextendsingle/vextendsingle/vextendsingle/vextendsingleA A− 10/vextendsingle/vextendsingle/vextendsingle/vextendsingle=−20 log4k2 0r3 0 3abβdB.( 7.45) The angular geometry of the skewed Bethe hole coupler is often a disadvantage in terms of fabrication and application. In addition, both coupler designs operate properlyonly at the design frequency; deviation from this frequency will alter the coupling level and the directivity, as shown in the following example. EXAMPLE 7.3 BETHE HOLE COUPLER DESIGN AND PERFORMANCE Design a Bethe hole coupler of the type shown in Figure 7.16a for an X-band waveguide operating at 9 GHz, with a coupling of 20 dB. Calculate and plot thecoupling and directivity from 7 to 11 GHz. Assume a round aperture. c07PowerDividers Pozar August 24, 2011 15:53 7.4 Waveguide Directional Couplers 337 Solution For an X-band waveguide at 9 GHz, we have the following constants: a=0.02286 m, b=0.01016 m, λ0=0.0333 m, k0=188.5m−1, β=129.0m−1, Z10=550.9 /Omega1, P10=4.22×10−7m2//Omega1. Equation (7.41) can be used to find the aperture position s: sinπs a=λ0/radicalBig 2/parenleftbig λ2 0−a2/parenrightbig=0.972, s=a πsin−10.972 =0.424 a=9.69 mm. The coupling is 20 dB, so C=20 dB =20 log/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleA A− 10/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle, or /vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleA A− 10/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=10 20/20=10; thus,|A− 10/A|=1/10. Now use (7.40b) to find r0: /vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleA − 10 A/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=1 10=ω P10/bracketleftBigg/parenleftBigg /epsilon10αe+µ0αm Z2 10/parenrightBigg (0.944) −π2µ0αm β2a2Z2 10(0.056)/bracketrightBigg . Because αe=2r3 0/3 and αm=4r3 0/3, we obtain 0.1=1.44×106r3 0, or r0=4.15 mm. This completes the design of the Bethe hole coupler. To compute the coupling and directivity versus frequency, we evaluate (7.42a) and (7.42b), using the ex- pressions for A− 10andA+ 10given in (7.40a) and (7.40b). In these expressions the aperture position and size are fixed at s=9.69 mm and r0=4.15 mm, and the frequency is varied. A short computer program was used to calculate the data shown in Figure 7.17. Observe that the coupling varies by less than 1 dB over theband. The directivity is very large (>60 dB) at the design frequency but decreases to 15–20 dB at the band edges. The directivity is a more sensitive function of frequency because it depends on the cancellation of two wave components. ■ c07PowerDividers Pozar August 24, 2011 15:53 338 Chapter 7: Power Dividers and Directional Couplers 7.0 7.5 8.0 8.5 9.0 Frequency GHz9.5 10.0 10.5 11.060504030C, D dB20100 C D FIGURE 7.17 Coupling and directivity versus frequency for the Bethe hole coupler of Exam- ple 7.3. DesignofMultiholeCouplers As seen from Example 7.3, a single-hole coupler has a relatively narrow bandwidth, at least in terms of its directivity. However, if the coupler is designed with a series of coupling holes, the extra degrees of freedom can be used to increase this bandwidth. The principleof operation and design of such a multihole waveguide coupler is very similar to that of the multisection matching transformer. First let us consider the operation of the two-hole coupler shown in Figure 7.18. Two parallel waveguides sharing a common broad wall are shown, although the same type of structure could be made in microstrip line or stripline form. Two small apertures are spaced λ g/4 apart and couple the two guides. A wave entering at port 1 is mostly transmitted through to port 2, but some power is coupled through the two apertures. If a phase reference is taken at the first aperture, then the phase of the wave incident at the second aperture will be−90◦. Each aperture will radiate a forward wave component and a backward wave component into the upper guide; in general, the forward and backward amplitudes are different. In the direction of port 3, both wave components are in phase because both have traveled λg/4 to the second aperture. However, we obtain a cancellation in the direction of port 4 because the wave coming through the second aperture travels λg/2 further than the wave component coming through the first aperture. Clearly, this cancellation is frequency sensitive, making the directivity a sensitive function of frequency. The coupling is less frequency dependent since the path lengths from port 1 to port 3 are always the same. 4 3 1 2/H9261g/4 0° –90°(Isolated) (Out of phase) (In phase) (Input)(Coupled) (Through) FIGURE 7.18 Basic operation of a two-hole directional coupler. c07PowerDividers Pozar August 24, 2011 15:53 7.4 Waveguide Directional Couplers 339 dB0AB F AAF0A n = 0B1AF1A n = 1B2AF2A n = 2B3AF3A . . .BNAFNA n = N FIGURE 7.19 G e o m e t r yo fa n( N+1)-hole waveguide directional coupler. Thus, in the multihole coupler design, we synthesize the directivity response, as opposed to the coupling response, as a function of frequency. Now consider the general case of the multihole coupler shown in Figure 7.19, where N+1 equally spaced apertures couple two parallel waveguides. The amplitude of the inci- dent wave in the lower left guide is Aand, for small coupling, is essentially the same as the amplitude of the through wave. For instance, a 20 dB coupler has a power coupling factor of 10−20/ 10=0.01, so the power transmitted through waveguide Ais 1−0.01=0.99 of the incident power (1% coupled to the upper guide). The voltage (or field) drop in wave- guide Ais√ 0.99=0.995, or 0.5%. Thus, the assumption that the amplitude of the incident field is identical at each aperture is a good one. Of course, the phase will change from oneaperture to the next. As we saw in the previous section for the Bethe hole coupler, an aperture generally excites forward and backward traveling waves with different amplitudes. Thus, let F ndenote the coupling coefficient of the nth aperture in the forward direction. Bndenote the coupling coefficient of the nth aperture in the backward direction. Then the amplitude of the forward wave can be written as F=Ae−jβNdN/summationdisplay n=0Fn,( 7.46) since all components travel the same path length. The amplitude of the backward wave is B=AN/summationdisplay n=0Bne−2jβnd,( 7.47) since the path length for the nth component is 2 βnd, where dis the spacing between the apertures. In (7.46) and (7.47) the phase reference is taken at the n=0 aperture. From the definitions in (7.20a) and (7.20b) the coupling and directivity can be com- puted as C=−20 log/vextendsingle/vextendsingle/vextendsingle/vextendsingleF A/vextendsingle/vextendsingle/vextendsingle/vextendsingle=−20 log/vextendsingle/vextendsingle/vextendsingle/vextendsingleN/summationdisplay n=0Fn/vextendsingle/vextendsingle/vextendsingle/vextendsingledB, (7.48) D=−20 log/vextendsingle/vextendsingle/vextendsingle/vextendsingleB F/vextendsingle/vextendsingle/vextendsingle/vextendsingle=−20 log/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/summationtext N n=0Bne−2jβnd /summationtextN n=0Fn/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle =−C−20 log/vextendsingle/vextendsingle/vextendsingle/vextendsingleN/summationdisplay n=0Bne−2jβnd/vextendsingle/vextendsingle/vextendsingle/vextendsingledB. (7.49) c07PowerDividers Pozar August 24, 2011 15:53 340 Chapter 7: Power Dividers and Directional Couplers Now assume that the apertures are round holes with identical positions srelative to the edge of the guide, with rnbeing the radius of the nth aperture. Then we know from Section 4.8 and the preceding section that the coupling coefficients will be proportional to the polarizabilities αeandαmof the aperture, and hence proportional to r3 n.S ow ec a n write Fn=Kfr3 n, (7.50a) Bn=Kbr3 n, (7.50b) where KfandKbare constants for the forward and backward coupling coefficients that are the same for all apertures, but are functions of frequency. Then (7.48) and (7.49) reduce to C=−20 log |Kf|−20 logN/summationdisplay n=0r3 ndB, (7.51) D=−C−20 log |Kb|−20 log/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleN/summationdisplay n=0r3 ne−2jβnd/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle =−C−20 log |K b|−20 log SdB. (7.52) In (7.51), the second term is constant with frequency. The first term is not affected by the choice of rn, but is a relatively slowly varying function of frequency. Similarly, in (7.52) the first two terms are slowly varying functions of frequency, representing the directivity of a single aperture, but the last term ( S)is a sensitive function of frequency due to phase cancellation in the summation. Thus we can choose the rnto synthesize a desired frequency response for the directivity, while the coupling should be relatively constant with frequency. Observe that the last term in (7.52), S=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleN/summationdisplay n=0r3 ne−2jβnd/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle,( 7.53) is very similar in form to the expression obtained in Section 5.5 for multisection quarter- wave matching transformers. As in that case, we can develop coupler designs that yieldeither a binomial (maximally flat) or a Chebyshev (equal ripple) response for the directivity. Another interpretation of (7.53) may be recognizable to the student familiar with basic antenna theory, as this expression is identical to the array pattern factor of an ( N+1)- element array with element weights r 3 n. In that case, too, the pattern may be synthesized in terms of binomial or Chebyshev polynomials. Binomial response: As in the case of the multisection quarter-wave matching transformers, we can obtain a binomial, or maximally flat, response for the directivity of the multiholecoupler by making the coupling coefficients proportional to the binomial coefficients. Thus, r 3 n=kCN n,( 7.54) where kis a constant to be determined, and CN nis a binomial coefficient given in (5.51). To find k, we evaluate the coupling using (7.51) to give C=−20 log |Kf|−20 log k−20 logN/summationdisplay n=0CN ndB.( 7.55) c07PowerDividers Pozar August 24, 2011 15:53 7.4 Waveguide Directional Couplers 341 Because we know Kf,N,andC, we can solve for kand then find the required aperture radii from (7.54). The spacing, d, should be λg/4 at the center frequency. Chebyshev response : First assume that Nis even (an odd number of holes), and that the coupler is symmetric, so that r0=rN,r1=rN−1, etc. Then from (7.53) we can write Sas S=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleN/summationdisplay n=0r3 ne−2jnθ/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=2N/2/summationdisplay n=0r3 ncos(N−2n)θ, where θ=βd. To achieve a Chebyshev response we equate this to the Chebyshev polyno- mial of degree N: S=2N/2/summationdisplay n=0r3 ncos(N−2n)θ=k|TN(secθmcosθ)|,( 7.56) where kandθmare constants to be determined. From (7.53) and (7.56), we see that for θ=0,S=/summationtextN n=0r3 n=k|TN(secθm)|. Using this result in (7.51) gives the coupling as C=− 20 log |Kf|−20 log S/vextendsingle/vextendsingle θ=0 =−20 log |Kf|−20 log k−20 log |TN(secθm)|dB. (7.57) From (7.52) the directivity is D=−C−20 log |Kb|−20 log S =20 logKf Kb+20 logTN(secθm) TN(secθmcosθ)dB. (7.58) The term log Kf/Kbis a function of frequency, so Dwill not have an exact Chebyshev response. This error is usually small, however. We can assume that the smallest value of Dwill occur when TN(secθmcosθ)=1, since |TN(secθm)|≥| TN(secθmcosθ)|.S oi f Dminis the specified minimum value of directivity in the passband, then θmcan be found from the relation Dmin=20 log TN(secθm)dB.( 7.59) Alternatively, we could specify the bandwidth, which then dictates θmandDmin. In either case, (7.57) can then be used to find k, and then (7.56) solved for the radii, rn. IfNis odd (an even number of holes), the results for C,D,andDminin (7.57), (7.58), and (7.59) still apply, but instead of (7.56), the following relation is used to find the apertureradii: S=2(N−1)/2/summationdisplay n=0r3 ncos(N−2n)θ=k|TN(secθmcosθ)|.( 7.60) EXAMPLE 7.4 MULTIHOLE WA VEGUIDE COUPLER DESIGN Design a four-hole Chebyshev coupler in an X-band waveguide using round aper- tures located at s=a/4. The center frequency is 9 GHz, the coupling is 20 dB, and the minimum directivity is 40 dB. Plot the coupling and directivity response from 7 to 11 GHz. c07PowerDividers Pozar August 24, 2011 15:53 342 Chapter 7: Power Dividers and Directional Couplers Solution For an X-band waveguide at 9 GHz, we have the following constants: a=0.02286 m, b=0.01016 m, λ0=0.0333 m, k0=188.5m−1, β=129.0m−1, Z10=550.9 /Omega1, P10=4.22×10−7m2//Omega1. From (7.40a) and (7.40b), we obtain for an aperture at s=a/4: |Kf|=2k0 3η0P10/bracketleftBigg sin2πs a−2β2 k2 0/parenleftBigg sin2πs a+π2 β2a2cos2πs a/parenrightBigg/bracketrightBigg =3.953×105, |Kb|=2k0 3η0P10/bracketleftBigg sin2πs a+2β2 k2 0/parenleftBigg sin2πs a−π2 β2a2cos2πs a/parenrightBigg/bracketrightBigg =3.454×105. For a four-hole coupler, N=3, so (7.59) gives 40=20 log T3(secθm)dB, 100=T3(secθm)=cosh/bracketleftbig 3 cosh−1(secθm)/bracketrightbig , secθm=3.01, where (5.58b) was used. Thus θm=70.6◦and 109.4◦at the band edges. Then from (7.57) we can solve for k: C=20=−20 log(3.953 ×105)−20 log k−40 dB, 20 log k=−171.94, k=2.53×10−9. Finally, (7.60) and the expansion from (5.60c) for T3allow us to solve for the radii as follows: S=2/parenleftbig r3 0cos 3θ +r3 1cosθ/parenrightbig =k/bracketleftbig sec3θm(cos 3θ +3 cosθ)−3s e cθmcosθ/bracketrightbig , 2r3 0=ksec3θm⇒ r0=r3=3.26 mm, 2r3 1=3k(sec3θm−secθm)⇒ r1=r2=4.51 mm. The resulting coupling and directivity are plotted in Figure 7.20; note the in- creased directivity bandwidth compared to that of the Bethe hole coupler ofExample 7.3. ■ c07PowerDividers Pozar August 24, 2011 15:53 7.5 The Quadrature (90◦) Hybrid 343 7.0 7.5 8.0C D 8.5 9.0 Frequency GHz9.5 10.0 10.5 11.060504030C, D dB20100 FIGURE 7.20 Coupling and directivity versus frequency for the four-hole coupler of Exam- ple 7.4. 7.5THEQUADRATURE(90◦)HYBRID Quadrature hybrids are 3 dB directional couplers with a 90◦phase difference in the out- puts of the through and coupled arms. This type of hybrid is often made in microstrip line or stripline form as shown in Figure 7.21 and is also known as a branch-line hybrid. Other 3 dB couplers, such as coupled line couplers or Lange couplers, can also be used as quadra- ture couplers; these components will be discussed in later sections. Here we will analyze the operation of the quadrature hybrid using an even-odd mode decomposition techniquesimilar to that used for the Wilkinson power divider. With reference to Figure 7.21, the basic operation of the branch-line coupler is as follows. With all ports matched, power entering port 1 is evenly divided between ports 2and 3, with a 90 ◦phase shift between these outputs. No power is coupled to port 4 (the isolated port). The scattering matrix has the following form: [S]=−1√ 2⎡ ⎢⎣0j10 j001 100 j 01 j0⎤ ⎥⎦.( 7.61) Observe that the branch-line hybrid has a high degree of symmetry, as any port can be used as the input port. The output ports will always be on the opposite side of the junction fromthe input port, and the isolated port will be the remaining port on the same side as the input port. This symmetry is reflected in the scattering matrix, as each row can be obtained as a transposition of the first row. /H9261 4Z0 Z0 Z0Z0/2 Z0/2Z0 Z0 Z0/H9261 41 4(Input) (Isolated)(Output) (Output)2 3 FIGURE 7.21 Geometry of a branch-line coupler. c07PowerDividers Pozar August 24, 2011 15:53 344 Chapter 7: Power Dividers and Directional Couplers 4 31 2A1 = 1 B3 B4B1 B21 1 11 1 111 11/ 2 1/ 2 FIGURE 7.22 Circuit of the branch-line hybrid coupler in normalized form. Even-OddModeAnalysis We first draw the schematic circuit of the branch-line coupler in normalized form, as in Figure 7.22, where it is understood that each line represents a transmission line with in- dicated characteristic impedance normalized to Z0. The common ground return for each transmission line is not shown. We assume that a wave of unit amplitude A1=1 is incident at port 1. The circuit of Figure 7.22 can be decomposed into the superposition of an even-mode excitation and an odd-mode excitation [5], as shown in Figure 7.23. Note that superimpos- ing the two sets of excitations produces the original excitation of Figure 7.22, and since the circuit is linear, the actual response (the scattered waves) can be obtained from the sum ofthe responses to the even and odd excitations. Because of the symmetry or antisymmetry of the excitation, the four-port network can be decomposed into a set of two decoupled two-port networks, as shown in Figure 7.23. Because the amplitudes of the incident waves for these two-ports are ±1/2, the amplitudes 1 2 4 3+1/2+1/2 11 1111 1 11/ 2 1/ 2 Line of symmetry I = 0V = max+1/2+1/2 Te11 11 11 1 11/ 2 1/ 2 Open-circuited stubs (2 separate 2-ports)Γe 1 2 4 3–1/2+1/2 11 1111 1 11/ 2 1/ 2 Line of antisymmetry V = 0I = max–1/2+1/2 To11 11 11 11 1 11/ 2 1/ 2 Short-circuited stubs (2 separate 2-ports)Γo(a) (b) FIGURE 7.23 Decomposition of the branch-line coupler into even- and odd-mode excitations. (a) Even mode (e ). (b) Odd mode (o ). c07PowerDividers Pozar August 24, 2011 15:53 7.5 The Quadrature (90◦) Hybrid 345 of the emerging wave at each port of the branch-line hybrid can be expressed as B1=1 2/Gamma1e+1 2/Gamma1o, (7.62a) B2=1 2Te+1 2To, (7.62b) B3=1 2Te−1 2To, (7.62c) B4=1 2/Gamma1e−1 2/Gamma1o, (7.62d) where /Gamma1e,oandTe,oare the even- and odd-mode reflection and transmission coefficients for the two-port networks of Figure 7.23. First consider the calculation of /Gamma1eandTefor the even-mode two-port circuit. This can best be done by multiplying the ABCD matrices of each cascade component in that circuit, to give /bracketleftbigg AB CD/bracketrightbigg e=/bracketleftbigg 10 j1/bracketrightbigg /bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright ShuntY=j/bracketleftbigg 0 j/√ 2 j√ 20/bracketrightbigg /bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright λ/4 Transmissionline/bracketleftbigg 10 j1/bracketrightbigg /bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright ShuntY=J=1√ 2/bracketleftbigg −1 j j−1/bracketrightbigg , (7.63) where the individual matrices can be found from Table 4.1, and the admittance of the shunt open-circuited λ/8 stubs is Y=jtanβ/lscript=j. Then Table 4.2 can be used to convert from ABCD parameters (defined here with Zo=1)toSparameters, which are equivalent to the reflection and transmission coefficients. Thus, /Gamma1e=A+B−C−D A+B+C+D=(−1+j−j+1)/√ 2 (−1+j+j−1)/√ 2=0, (7.64a) Te=2 A+B+C+D=2 (−1+j+j−1)/√ 2=−1√ 2(1+j). (7.64b) Similarly, for the odd mode we obtain /bracketleftbigg AB CD/bracketrightbigg o=1√ 2/bracketleftbigg 1j j1/bracketrightbigg ,( 7.65) which gives the reflection and transmission coefficients as /Gamma1o=0, (7.66a) To=1√ 2(1−j). (7.66b) Using (7.64) and (7.66) in (7.62) gives the following results: B1=0 (port 1 is matched), (7.67a) B2=−j√ 2(half-power ,−90◦phase shift from port 1 to 2), (7.67b) B3=−1√ 2(half-power ,−180◦phase shift from port 1 to 3), (7.67c) B4=0 (no power to port 4) . (7.67d) These results agree with the first row and column of the scattering matrix given in (7.61); the remaining elements can easily be found by transposition. c07PowerDividers Pozar August 24, 2011 15:53 346 Chapter 7: Power Dividers and Directional Couplers FIGURE 7.24 Photograph of an eight-way microstrip power divider for an array antenna feed net- work at 1.26 GHz. The circuit uses six quadrature hybrids in a Bailey configurationfor unequal power division ratios (see Problem 7.33). Courtesy of ProSensing, Inc., Amherst, Mass. In practice, due to the quarter-wave length requirement, the bandwidth of a branch-line hybrid is limited to 10%–20%. However, as with multisection matching transformers andmultihole directional couplers, the bandwidth of a branch-line hybrid can be increased to a decade or more by using multiple sections in cascade. In addition, the basic design can be modified for unequal power division and/or different characteristic impedances at theoutput ports. Another practical point to be aware of is the fact that discontinuity effects at the junctions of the branch-line coupler may require that the shunt arms be lengthened by 10 ◦–20◦. Figure 7.24 shows a photograph of a circuit using several quadrature hybrids. EXAMPLE 7.5 DESIGN AND PERFORMANCE OF A QUADRATURE HYBRID Design a 50 /Omega1branch-line quadrature hybrid junction, and plot the scattering parameter magnitudes from 0.5 f0to 1.5f0, where f0is the design frequency. Solution After the preceding analysis, the design of a quadrature hybrid is trivial. The linesareλ/4 at the design frequency f 0, and the branch-line impedances are Z0√ 2=50√ 2=35.4/Omega1. The calculated frequency response is plotted in Figure 7.25. Note that we obtain perfect 3 dB power division at ports 2 and 3, and perfect isolation and return lossat ports 4 and 1, respectively, at the design frequency f 0. All of these quantities, however, degrade quickly as the frequency departs from f0. ■ 0.5f0 f0 1.5f0–40–30–20 ⎪Sij⎪–100S13 S14S12S11 FIGURE 7.25 Scattering parameter magnitudes versus frequency for the branch-line coupler of Example 7.5. c07PowerDividers Pozar August 24, 2011 15:53 7.6 Coupled Line Directional Couplers 347 bW S (a)W /H9280r dW (c)/H9280rW SbW S (b)/H9280r FIGURE 7.26 Various coupled transmission line geometries. (a) Coupled stripline (planar, or edge-coupled). (b) Coupled stripline (stacked, or broadside-coupled). (c) Coupledmicrostrip lines. 7.6COUPLEDLINEDIRECTIONALCOUPLERS When two unshielded transmission lines are in close proximity, power can be coupled from one line to the other due to the interaction of the electromagnetic fields. Such lines are re-f e r r e dt oa scoupled transmission lines, and they usually consist of three conductors in close proximity, although more conductors can be used. Figure 7.26 shows several examples of coupled transmission lines. Coupled transmission lines are sometimes assumed to operatein the TEM mode, which is rigorously valid for coaxial line and stripline structures, but only approximately valid for microstrip line, coplanar waveguide, or slotline structures. Coupled transmission lines can support two distinct propagating modes, and this featurecan be used to implement a variety of practical directional couplers, hybrids, and filters. The coupled lines shown in Figure 7.26 are symmetric, meaning that the two conduct- ing strips have the same width and position relative to ground; this simplifies the analysisof their operation. We will first discuss the basic theory of coupled lines and present some design data for coupled stripline and coupled microstrip line. We will then analyze the operation of a single-section coupled line directional coupler and extend these results to multisection coupled line coupler design. CoupledLineTheory The coupled lines of Figure 7.26, or other symmetric three-wire lines, can be represented by the structure and equivalent circuit shown in Figure 7.27. If we assume TEM prop- agation, then the electrical characteristics of the coupled lines can be completely deter-mined from the effective capacitances between the lines and the velocity of propagation on the line. As depicted in Figure 7.27, C 12represents the capacitance between the two strip conductors, and C11andC22represent the capacitance between one strip conductor C11 C22C12 FIGURE 7.27 A three-wire coupled transmission line and its equivalent capacitance network. c07PowerDividers Pozar August 24, 2011 15:53 348 Chapter 7: Power Dividers and Directional Couplers C11 C22 C11 C222C12 2C12 FIGURE 7.28 Even- and odd-mode excitations for a coupled line, and the resulting equivalent capacitance networks. (a) Even-mode excitation. (b) Odd-mode excitation. and ground. Because the strip conductors are identical in size and location relative to the ground conductor, we have C11=C22. Note that the designation of “ground” for the third conductor has no special relevance beyond the fact that it is convenient, since in manyapplications this conductor is the ground plane of a stripline or microstrip circuit. Now consider two special types of excitations for the coupled line: the even mode, where the currents in the strip conductors are equal in amplitude and in the same direction,and the odd mode, where the currents in the strip conductors are equal in amplitude but in opposite directions. The electric field lines for these two cases are sketched in Figure 7.28. Because the line is TEM, the propagation constant and phase velocity are the same forboth of these modes: β=ω/v pandvp=c/√/epsilon1r, where /epsilon1ris the relative permittivity of the TEM line. For the even mode, the electric field has even symmetry about the center line, and no current flows between the two strip conductors. This leads to the equivalent circuit shown, where C12is effectively open-circuited. The resulting capacitance of either line to ground for the even mode is Ce=C11=C22,( 7.68) assuming that the two strip conductors are identical in size and location. Then the charac- teristic impedance for the even mode is Z0e=/radicalBigg Le Ce=√LeCe Ce=1 vpCe,( 7.69) where vp=c/√/epsilon1r=1/√LeCe=1/√LoCois the phase velocity of propagation on the line. For the odd mode, the electric field lines have an odd symmetry about the center line, and a voltage null exists between the two strip conductors. We can imagine this as a ground c07PowerDividers Pozar August 24, 2011 15:53 7.6 Coupled Line Directional Couplers 349 plane through the middle of C12, which leads to the equivalent circuit shown. In this case the effective capacitance between either strip conductor and ground is Co=C11+2C12=C22+2C12,( 7.70) and the characteristic impedance for the odd mode is Z0o=/radicalBigg Lo Co=√LoCo Co=1 vpCo.( 7.71) In words, Z0e(Z0o)is the characteristic impedance of one of the strip conductors relative to ground when the coupled line is operated in the even (odd) mode. An arbitrary excitation of a coupled line can always be treated as a superposition of appropriate amplitudes of even- and odd-mode excitations. This analysis assumes the lines are symmetric, and that fringing capacitances are identical for even and odd modes. If the coupled line supports a pure TEM mode, such as coaxial line, parallel plate guide, or stripline, analytical techniques such as conformal mapping [7] can be used to evaluate the capacitances per unit length of line, and the even- and odd-mode character-istic impedances can then be determined. For quasi-TEM lines, such as microstrip line, these results can be obtained numerically or by approximate quasi-static techniques [8]. In either case, such calculations are generally too involved for our consideration, but manycommercial microwave CAD packages can provide design data for a variety of coupled lines. Here we will present only graphical design data for two cases of coupled lines. For a symmetric coupled stripline of the type shown in Figure 7.26a, the design graph in Figure 7.29 can be used to determine the necessary strip widths and spacing for a given set of characteristic impedances, Z 0eand Z0o, and the dielectric constant. This graph 20 40 60 80 100 /H9280rZ0o120 140 16020406080100120140160180200220/H9280rZ0e0.010.05 0.1 0.2 0.5 1.0 0.2 0.3 0.4 0.6 0.8 1.0 1.5 2.0S/b W/b W SW b /H9280r FIGURE 7.29 Normalized even- and odd-mode characteristic impedance design data for sym- metric edge-coupled striplines. c07PowerDividers Pozar August 24, 2011 15:53 350 Chapter 7: Power Dividers and Directional Couplers 20 40 60 80 Z0oZ0e 100 120204060801001201401601800.05 0.1 0.2 0.4 0.6 1.0 2.0 0.07 0.1 0.15 0.2 0.3 0.4 0.5 0.7 1.0 1.5 2.0S/d W/d WWd S /H9280r = 10 FIGURE 7.30 Even- and odd-mode characteristic impedance design data for symmetric coupled microstrip lines on a substrate with /epsilon1r=10. should cover ranges of parameters for most practical applications, and can be used for any dielectric constant, since the TEM mode of stripline allows scaling by the dielectricconstant. For coupled microstrip lines, the results do not scale with dielectric constant, so design graphs must be made for specific values of dielectric constant. Figure 7.30 shows such adesign graph for symmetric coupled microstrip lines on a substrate with /epsilon1 r=10. Another difficulty with coupled microstrip lines is the fact that the phase velocity is usually different for the two modes of propagation because the two modes operate with different field con-figurations in the vicinity of the air–dielectric interface. This can have a degrading effect on coupler directivity. EXAMPLE 7.6 IMPEDANCE OF A SIMPLE COUPLED LINE For the broadside coupled stripline geometry of Figure 7.26b, assume W/greatermuchS andW/greatermuchb, so that fringing fields can be ignored, and determine the even- and odd-mode characteristic impedances. Solution We first find the equivalent network capacitances, C11andC12(because the line is symmetric, C22=C11). The capacitance per unit length of broadside parallel lines with width Wand separation dis ¯C=/epsilon1W dF/m, where /epsilon1is the substrate permittivity. This formula ignores fringing fields. c07PowerDividers Pozar August 24, 2011 15:53 7.6 Coupled Line Directional Couplers 351 C11is formed by the capacitance of one strip to the ground planes. Thus the capacitance per unit length is ¯C11=2/epsilon1r/epsilon10W b−sF/m. The capacitance per unit length between the strips is ¯C12=/epsilon1r/epsilon10W SF/m. Then from (7.68) and (7.70), the even- and odd-mode capacitances are ¯Ce=¯C11=2/epsilon1r/epsilon10W b−SF/m, ¯Co=¯C11+2¯C12=2/epsilon1r/epsilon10W/parenleftbigg1 b−S+1 S/parenrightbigg F/m. The phase velocity on the line is vp=1/√/epsilon1r/epsilon10µ0=c/√/epsilon1r, so the characteristic impedances are Z0e=1 vp¯Ce=η0b−S 2W√/epsilon1r, Z0o=1 vp¯Co=η01 2W√/epsilon1r[1/(b−S)+1/S].■ DesignofCoupledLineCouplers With the preceding definitions of the even- and odd-mode characteristic impedances, we can apply an even-odd mode analysis to a length of coupled line to arrive at the design equations for a single-section coupled line coupler. Such a line is shown in Figure 7.31. This four-port network is terminated in the impedance Z0at three of its ports, and driven with a generator of voltage 2 V0and internal impedance Z0at port 1. We will show that a coupler can be designed with arbitrary coupling such that the input (port 1) is matched, while port 4 is isolated. Port 2 is the through port, and port 3 is the coupled port. In Figure7.31, a ground conductor is understood to be common to both strip conductors. For this problem we will apply the even-odd mode analysis technique in conjunction with the voltages and currents on the line, as opposed to the reflection and transmission coefficients. So, by superposition, the excitation at port 1 in Figure 7.31 can be treated as the sum of the even- and odd-mode excitations shown in Figure 7.32. From symmetry we can see that I e 1=Ie 3,Ie 4=Ie 2,Ve 1=Ve 3, and Ve 4=Ve 2for the even mode, while Io 1=− Io 3, Io 4=− Io 2,Vo 1=−Vo 3, and Vo 4=−Vo 2for the odd mode. The input impedance at port 1 of the coupler of Figure 7.31 can then be expressed as Zin=V1 I1=Ve 1+Vo 1 Ie 1+Io 1.( 7.72) If we let Ze inbe the input impedance at port 1 for the even mode, and Zo inbe the input impedance for the odd mode, then we have Ze in=Z0eZ0+jZ0etanθ Z0e+jZ0tanθ, (7.73a) Zo in=Z0oZ0+jZ0otanθ Z0o+jZ0tanθ, (7.73b) c07PowerDividers Pozar August 24, 2011 15:53 352 Chapter 7: Power Dividers and Directional Couplers 134 2Coupled Isolated Through Input (a) (b)/H9258 Z0e, Z0oI3 Z0 Z0 +V3 +V4I4 I1 Z0 Z0 +V1 +V2I2 2V0124 3 FIGURE 7.31 A single-section coupled line coupler. (a) Geometry and port designations. (b) The schematic circuit. FIGURE 7.32 Decomposition of the coupled line coupler circuit of Figure 7.31 into even- and odd-mode excitations. (a) Even mode. (b) Odd mode. c07PowerDividers Pozar August 24, 2011 15:53 7.6 Coupled Line Directional Couplers 353 since, for each mode, the line looks like a transmission line of characteristic impedance Z0eorZ0o, terminated in a load impedance, Z0. Then by voltage division, Vo 1=V0Zo in Zo in+Z0, (7.74a) Ve 1=V0Ze in Ze in+Z0, (7.74b) and Io 1=V0 Zo in+Z0, (7.75a) Ie 1=V0 Ze in+Z0. (7.75b) Using these results in (7.72) yields Zin=Zo in/parenleftbig Ze in+Z0/parenrightbig +Ze in/parenleftbig Zo in+Z0/parenrightbig Ze in+Zo in+2Z0=Z0+2/parenleftbig Zo inZe in−Z2 0/parenrightbig Ze in+Zo in+2Z0.( 7.76) Now if we let Z0=/radicalbig Z0eZ0o,( 7.77) then (7.73a) and (7.73b) reduce to Ze in=Z0e√Z0o+j√Z0etanθ√Z0e+j√Z0otanθ, Zo in=Z0o√Z0e+j√Z0otanθ√Z0o+j√Z0etanθ, so that Ze inZo in=Z0eZ0o=Z2 0, and (7.76) reduces to Zin=Z0.( 7.78) Thus, as long as (7.77) is satisfied, port 1 (and, by symmetry, all other ports) will be matched. Now if (7.77) is satisfied, so that Zin=Z0, we have that V1=V0, by voltage division. The voltage at port 3 is V3=Ve 3+Vo 3=Ve 1−Vo 1=V0/bracketleftbiggZe in Ze in+Z0−Zo in Zo in+Z0/bracketrightbigg ,( 7.79) where (7.74) has been used. From (7.73) and (7.77), we can show that Ze in Ze in+Z0=Z0+jZ0etanθ 2Z0+j(Z0e+Z0o)tanθ, Zo in Zo in+Z0=Z0+jZ0otanθ 2Z0+j(Z0e+Z0o)tanθ, so that (7.79) reduces to V3=V0j(Z0e−Z0o)tanθ 2Z0+j(Z0e+Z0o)tanθ.( 7.80) c07PowerDividers Pozar August 24, 2011 15:53 354 Chapter 7: Power Dividers and Directional Couplers Now define the coupling coefficient, C,a s C=Z0e−Z0o Z0e+Z0o,( 7.81) which we will soon see is actually the midband voltage coupling coefficient, V3/V0. Then, /radicalbig 1−C2=2Z0 Z0e+Z0o, so that V3=V0jCtanθ√ 1−C2+jtanθ.( 7.82) Similarly, we can show that V4=Ve 4+Vo 4=Ve 2−Vo 2=0, (7.83) V2=Ve 2+Vo 2=V0√ 1−C2 √ 1−C2cosθ+jsinθ. (7.84) Equations (7.82) and (7.84) can be used to plot the coupled and through port voltages versus frequency, as shown in Figure 7.33. At very low frequencies ( θ/lessmuchπ/2), virtually all power is transmitted through port 2, with none being coupled to port 3. For θ=π/2, the coupling to port 3 is at its first maximum; this is where the coupler is generally operated,for small size and minimum line loss. Otherwise, the response is periodic, with maxima in V 3forθ=π/2,3π/2,... Forθ=π/2, the coupler is λ/4 long, and (7.82) and (7.84) reduce to V3 V0=C, (7.85) V2 V0=− j/radicalbig 1−C2, (7.86) which shows that C<1 is the voltage coupling factor at the design frequency, θ=π/2. Note that these results satisfy power conservation since Pin=(1/2)| V0|2/Z0, while the output powers are P2=(1/2)| V2|2/Z0=(1/2)(1 −C2)|V0|2/Z0,P3=(1/2)| C|2|V0|2/Z0, andP4=0,so that Pin=P2+P3+P4. Also observe that there is a 90◦phase shift be- tween the two output port voltages; thus this coupler can be used as a quadrature hybrid.In addition, as long as (7.77) is satisfied, the coupler will be matched at the input and have perfect isolation, at any frequency. /H9258 0 /H9266 2/H9266 3/H9266 2/H926601 1 – C2 C2 2V22 V0 V32 V0 FIGURE 7.33 Coupled and through port voltages (squared) versus frequency for the coupled line coupler of Figure 7.31. c07PowerDividers Pozar August 24, 2011 15:53 7.6 Coupled Line Directional Couplers 355 Finally, if the characteristic impedance, Z0, and the voltage coupling coefficient, C, are specified, then the following design equations for the required even- and odd-mode characteristic impedances can be easily derived from (7.77) and (7.81): Z0e=Z0/radicalbigg 1+C 1−C, (7.87a) Z0o=Z0/radicalbigg 1−C 1+C. (7.87b) In the above analysis it was assumed that the even and odd modes of the coupled line structure have the same velocities of propagation, so that the line has the same electrical length for both modes. For coupled microstrip lines, or other non-TEM lines, this conditionwill generally not be satisfied exactly, and the coupler will have poor directivity. The fact that coupled microstrip lines have unequal even- and odd-mode phase velocities can be intuitively explained by considering the field line plots of Figure 7.28, which show that theeven mode has less fringing field in the air region than the odd mode. Thus its effective dielectric constant should be higher, indicating a smaller phase velocity for the even mode. Techniques for compensating coupled microstrip lines to achieve equal even- and odd- mode phase velocities include the use of dielectric overlays and anisotropic substrates. This type of coupler is best suited for weak coupling, as a large coupling factor requires lines that are too close together to be practical, or a combination of even- and odd-mode characteristic impedances that is nonrealizable. EXAMPLE 7.7 SINGLE-SECTION COUPLER DESIGN AND PERFORMANCE Design a 20 dB single-section coupled line coupler in stripline with a groundplane spacing of 0.32 cm, a dielectric constant of 2.2, a characteristic impedance of 50/Omega1, and a center frequency of 3 GHz. Plot the coupling and directivity from 1 to 5 GHz. Include the effect of losses by assuming a loss tangent of 0.05 for the dielectric material and copper conductors of 2 mil thickness. Solution The voltage coupling factor is C=10 −20/ 20=0.1. From (7.87), the even- and odd-mode characteristic impedances are Z0e=Z0/radicalbigg 1+C 1−C=55.28 /Omega1, Z0o=Z0/radicalbigg 1−C 1+C=45.23 /Omega1. To use Figure 7.29, we have that √/epsilon1rZ0e=82.0, √/epsilon1rZ0o=67.1, and so W/b=0.809 and S/b=0.306. This gives a conductor width of W= 0.259 cm and a conductor separation of S=0.098 cm (these values were actually found using a commercial microwave CAD package). Figure 7.34 shows the resulting coupling and directivity versus frequency, including the effect of dielectric and conductor losses. Losses have the effect of reducing the directivity, which is typically greater than 70 dB in the absence of loss. ■ c07PowerDividers Pozar August 24, 2011 15:53 356 Chapter 7: Power Dividers and Directional Couplers 1.0 2.0 3.0 Frequency (GHz)4.0 5.040 50603020100Coupling, Directivity (dB)DC FIGURE 7.34 Coupling versus frequency for the single-section coupler of Example 7.7. DesignofMultisectionCoupledLineCouplers As Figure 7.33 shows, the coupling of a single-section coupled line coupler is limited in bandwidth due to the λ/4 length requirement. As in the case of matching transformers and waveguide couplers, bandwidth can be increased by using multiple sections. In fact, there is a very close relationship between multisection coupled line couplers and multisection quarter-wave transformers [9]. Because the phase characteristics are usually better, multisection coupled line couplers are generally made with an odd number of sections, as shown in Figure 7.35. Thus, we will assume that Nis odd. We will also assume that the coupling is weak ( C≥10 dB), and that each section is λ/4 long (θ =π/2)at the center frequency. For a single coupled line section, with C/lessmuch1, (7.82) and (7.84) simplify to V3 V1=jCtanθ√ 1−C2+jtanθ/similarequaljCtanθ 1+jtanθ=jCsinθe−jθ, (7.88a) V2 V1=√ 1−C2 √ 1−C2cosθ+jsinθ/similarequale−jθ. (7.88b) Then for θ=π/2w eh a v et h a t V3/V1=CandV2/V1=− j. The above approximation is equivalent to assuming that no power is lost on the through path from one section tothe next, and is similar to the approximation used for the multisection waveguide coupler analysis. It is a good assumption for small C, even though power conservation is violated. V3 V4 V1 V2C1 C2 CN – 1 CNCoupled Isolated Input Through/H9258/H9258/H9258/H9258 ... FIGURE 7.35 AnN-section coupled line coupler. c07PowerDividers Pozar September 12, 2011 21:26 7.6 Coupled Line Directional Couplers 357 Using these results, we can express the total voltage at the coupled port (port 3) of the cascaded coupler in Figure 7.35 as V3=/parenleftbig jC1sinθe−jθ/parenrightbig V1+/parenleftbig jC2sinθe−jθ/parenrightbig V1e−2jθ +···+/parenleftbig jCNsinθe−jθ/parenrightbig V1e−2j(N−1)θ, (7.89) where Cnis the voltage coupling coefficient of the nth section. If we assume that the coupler is symmetric, so that C1=CN,C2=CN−1, etc., we can simplify (7.89) to V3=jV1sinθe−jθ/bracketleftBig C1/parenleftbig 1+e−2j(N−1)θ/parenrightbig +C2/parenleftbig e−2jθ+e−2j(N−2)θ/parenrightbig +···+ CMe−j(N−1)θ/bracketrightBig =2jV1sinθe−jNθ/bracketleftbigg C1cos(N−1)θ+C2cos(N−3)θ+···+1 2CM/bracketrightbigg , (7.90) where M=(N+1)/2. At the center frequency, we define the voltage coupling factor C0: C0=/vextendsingle/vextendsingle/vextendsingle/vextendsingleV 3 V1/vextendsingle/vextendsingle/vextendsingle/vextendsingle θ=π/ 2.( 7.91) Equation (7.90) is in the form of a Fourier series for the coupling as a function of fre- quency. Thus, we can synthesize a desired coupling response by choosing the coupling co-efficients, C n. Note that in this case we synthesize the coupling response, while in the case of the multihole waveguide coupler we synthesized the directivity response. This is because the path for the uncoupled arm of the multisection coupled line coupler is in the forwarddirection, and so is less dependent on frequency than the coupled arm path, which is in the reverse direction; this is the opposite situation from the multihole waveguide coupler. Multisection couplers of this form can achieve decade bandwidths, but coupling lev- els must be low. Because of the longer electrical length, it is more critical to have equal even- and odd-mode phase velocities than it is for the single-section coupler. This usually means that stripline is the preferred medium for such couplers. Mismatched phase veloci-ties will degrade the coupler directivity, as will junction discontinuities, load mismatches, and fabrication tolerances. A photograph of a coupled line coupler is shown in Figure 7.36. FIGURE 7.36 Photograph of a single-section microstrip coupled line coupler. Courtesy of M. D. Abouzahra, MIT Lincoln Laboratory, Lexington, Mass. c07PowerDividers Pozar August 24, 2011 15:53 358 Chapter 7: Power Dividers and Directional Couplers EXAMPLE 7.8 MULTISECTION COUPLER DESIGN AND PERFORMANCE Design a three-section 20 dB coupled line coupler with a binomial (maximally flat) response, a system impedance of 50 /Omega1, and a center frequency of 3 GHz. Plot the coupling and directivity from 1 to 5 GHz. Solution For a maximally flat response for a three-section ( N=3) coupler, we require that dn dθnC(θ)/vextendsingle/vextendsingle/vextendsingle/vextendsingle θ=π/ 2=0f o r n=1,2. From (7.90), C=/vextendsingle/vextendsingle/vextendsingle/vextendsingleV3 V1/vextendsingle/vextendsingle/vextendsingle/vextendsingle=2s i nθ/parenleftbigg C1cos 2θ +1 2C2/parenrightbigg =C1(sin 3θ −sinθ)+C2sinθ =C1sin 3θ +(C2−C1)sinθ, so dC dθ=[3C1cos 3θ +(C2−C1)cosθ]/vextendsingle/vextendsingle/vextendsingle π/2=0, d2C dθ2=[ − 9C1sin 3θ −(C2−C1)sinθ]/vextendsingle/vextendsingle/vextendsingle π/2=10C1−C2=0. At midband, θ=π/2 and C0=20 dB. Thus, C=10−20/ 20=0.1=C2−2C1. Solving these two equations for C1andC2gives C1=C3=0.0125 , C2=0.125. From (7.87) the even- and odd-mode characteristic impedances for each section are Z1 0e=Z3 0e=50/radicalbigg 1.0125 0.9875=50.63 /Omega1, Z1 0o=Z3 0o=50/radicalbigg 0.9875 1.0125=49.38 /Omega1, Z2 0e=50/radicalbigg 1.125 0.875=56.69 /Omega1, Z2 0o=50/radicalbigg 0.875 1.125=44.10 /Omega1. The coupling and directivity for this coupler are plotted in Figure 7.37. ■ c07PowerDividers Pozar August 24, 2011 15:53 7.7 The Lange Coupler 359 1 2 3 Frequency (GHz)45403020100Coupling (dB)D > 100 dB FIGURE 7.37 Coupling versus frequency for the three-section binomial coupler of Example 7.8. 7.7THELANGECOUPLER Generally the coupling in a coupled line coupler is too loose to achieve coupling factors of 3 or 6 dB. One way to increase the coupling between edge-coupled lines is to use several lines parallel to each other, so that the fringing fields at both edges of a line contribute to the coupling. One of the most practical implementations of this idea is the Lange coupler [10], shown in Figure 7.38a, where four parallel coupled lines are used with interconnections to provide tight coupling. This coupler can easily achieve 3 dB coupling ratios, with an octave or more bandwidth. The design tends to compensate for unequal even- and odd-mode phasevelocities, which also improves the bandwidth. There is a 90 ◦phase difference between the output lines (ports 2 and 3), so the Lange coupler is a type of quadrature hybrid. The main disadvantage of the Lange coupler is probably practical, as the lines are very narrow and close together, and the required bonding wires across the lines increases complexity. This type of coupled line geometry is also referred to as interdigitated ; such structures can also be used for filter circuits. Theunfolded Lange coupler [11], shown in Figure 7.38b, operates in essentially the same way as the original Lange coupler but is easier to model with an equivalent circuit.Such an equivalent circuit consists of a four-wire coupled line structure, as shown in Fig- ure 7.39a. All of the lines have the same width and spacing. If we make the reasonable assumption that each line couples only to its nearest neighbor, and ignore more-distantcouplings, then we effectively have a two-wire coupled line circuit, as shown in Figure 7.39b. Then, if we can derive the even- and odd-mode characteristic impedances, Z e4and Zo4, of the four-wire circuit of Figure 7.39a in terms of Z0eandZ0o, the even- and odd- mode characteristic impedances of any adjacent pair of lines, we can apply the coupled line coupler results of Section 7.6 to analyze the Lange coupler. Figure 7.40a shows the effective capacitances between the conductors of the four-wire coupled line of Figure 7.39a. Unlike the two-line case of Section 7.6, the capacitances of the four lines to ground are different depending on whether the line is on the outside (1 and 4), or on the inside (2 and 3). An approximate relation between these capacitances is [12] Cin=Cex−CexCm Cex+Cm.( 7.92) c07PowerDividers Pozar August 24, 2011 15:53 360 Chapter 7: Power Dividers and Directional Couplers 4 1WIsolated Input2 3 CoupledThrough Z0S Z0 Z0Z0 /H9261/4 2Through Input4 1 3CoupledIsolated(a) (b) FIGURE 7.38 The Lange coupler. (a) Layout in microstrip form. (b) The unfolded Lange coupler. For an even-mode excitation, all four conductors in Figure 7.40a are at the same po- tential, so Cm, has no effect, and the total capacitance of any line to ground is Ce4=Cex+Cin.( 7.93a) For an odd-mode excitation, electric walls effectively exist through the middle of each Cm, so the capacitance of any line to ground is Co4=Cex+Cin+6Cm.( 7.93b) The even- and odd-mode characteristic impedances are then Ze4=1 vpCe4, (7.94a) Zo4=1 vpCo4, (7.94b) where vpis the phase velocity of propagation on the line. Now consider any isolated pair of adjacent conductors in the four-line model; the effective capacitances are as shown in Figure 7.40b. The even- and odd-mode capacitances c07PowerDividers Pozar August 24, 2011 15:53 7.7 The Lange Coupler 361 3 14 290° (a) (b)90°Coupled Isolated Input Through Input Through3 4 1 2Coupled IsolatedZe4, Zo4 Ze4, Zo4 FIGURE 7.39 Equivalent circuits for the unfolded Lange coupler. (a) Four-wire coupled line model. (b) Approximate two-wire coupled line model. are Ce=Cex, (7.95a) Co=Cex+2Cm. (7.95b) Solving (7.95) for CexandCm, and substituting into (7.93) with the aid of (7.92) gives the even-odd mode capacitances of the four-wire line in terms of a two-wire coupled line: Ce4=Ce(3Ce+Co) Ce+Co, (7.96a) Co4=Co(3Co+Ce) Ce+Co. (7.96b) Because characteristic impedances are related to capacitance as Z0=1/v pC, we can rewrite (7.96) to give the even/odd mode characteristic impedances of the Lange coupler in terms of the characteristic impedances of a two-conductor line that is identical to any pair 1 2 3 4Cm Cex Cin Cin CexCm Cm (a)1 2 Cex CexCm (b) FIGURE 7.40 Effective capacitance networks for the unfolded Lange coupler equivalent cir- cuits of Figure 7.39. (a) Effective capacitance network for the four-wire model. (b) Effective capacitance network for the two-wire model. c07PowerDividers Pozar August 24, 2011 15:53 362 Chapter 7: Power Dividers and Directional Couplers of adjacent lines in the coupler: Ze4=Z0o+Z0e 3Z0o+Z0eZ0e, (7.97a) Zo4=Z0o+Z0e 3Z0e+Z0oZ0o, (7.97b) where Z0eand Z0oare the even- and odd-mode characteristic impedances of the two- conductor pair. Now we can apply the results of Section 7.6 to the coupler of Figure 7.39b. From (7.77) the characteristic impedance is Z0=/radicalbig Ze4Zo4=/radicalBigg Z0eZ0o(Z0o+Z0e)2 (3Z0o+Z0e)(3Z0e+Z0o),( 7.98) and the voltage coupling coefficient is, from (7.81), C=Ze4−Zo4 Ze4+Zo4=3/parenleftbig Z2 0e−Z2 0o/parenrightbig 3/parenleftbig Z2 0e+Z2 0o/parenrightbig +2Z0eZ0o,( 7.99) where (7.97) was used. For design purposes it is useful to invert these results to give the necessary even- and odd-mode impedances in terms of a desired characteristic impedance and coupling coefficient: Z0e=4C−3+√ 9−8C2 2C√(1−C)/(1+C)Z0, (7.100a) Z0o=4C+3−√ 9−8C2 2C√(1+C)/(1−C)Z0. (7.100b) These results are approximate because of the simplifications involved with the appli- cation of two-line characteristic impedances to the four-line circuit, and because of the assumption of equal even- and odd-mode phase velocities. In practice, however, these re-sults generally give sufficient accuracy. If necessary, a more complete analysis can be made to directly determine Z e4andZo4for the four-line circuit, as in reference [13]. 7.8THE180◦HYBRID The 180◦hybrid junction is a four-port network with a 180◦phase shift between the two output ports. It can also be operated so that the outputs are in phase. With reference to the 180◦hybrid symbol shown in Figure 7.41, a signal applied to port 1 will be evenly split into two in-phase components at ports 2 and 3, and port 4 will be isolated. If the input is applied to port 4, it will be equally split into two components with a 180◦phase difference at ports 2 and 3, and port 1 will be isolated. When operated as a combiner, with inputsignals applied at ports 2 and 3, the sum of the inputs will be formed at port 1, while the 1 42 3180° hybrid(Σ) (∆) FIGURE 7.41 Symbol for a 180◦hybrid junction. c07PowerDividers Pozar August 24, 2011 15:53 7.8 The 180◦Hybrid 363 difference will be formed at port 4. Hence, ports 1 and 4 are referred to as the sum and difference ports, respectively. The scattering matrix for the ideal 3 dB 180◦hybrid thus has the following form: [S]=−j√ 2⎡ ⎢⎣0110 100 −1 1001 0−11 0⎤ ⎥⎦.( 7.101) The reader may verify that this matrix is unitary and symmetric. The 180◦hybrid can be fabricated in several forms. The ring hybrid, or rat-race, shown in Figures 7.42a and 7.43, can easily be constructed in planar (microstrip or stripline) form, although waveguide versions are also possible. Another type of planar 180◦hybrid uses tapered matching sections and coupled lines, as shown in Figure 7.42b. Yet another type is the hybrid waveguide junction, or magic-T, shown in Figure 7.42c. We will first analyze the ring hybrid, using an even-odd mode analysis similar to that used for the branch-line hybrid, and use a similar technique for the analysis of the tapered line hybrid. Then we will qualitatively discuss the operation of the waveguide magic-T. 2 4 4 2 1331 /H9261/4 /H9261/4 /H9261/43/H9261/4 2Z0(Σ) (∆)Z0Z0Z0Z0 (a) (b) (c)OutputSum input Outpu t Difference input (Σ)(∆) FIGURE 7.42 Three types of hybrid junctions. (a) A ring hybrid, or rat-race , in microstrip line or stripline form. (b) A tapered coupled line hybrid. (c) A waveguide hybrid junction, ormagic-T. c07PowerDividers Pozar August 24, 2011 15:53 364 Chapter 7: Power Dividers and Directional Couplers FIGURE 7.43 Photograph of a microstrip power divider network using three ring hybrids. Courtesy of M. D. Abouzahra, MIT Lincoln Laboratory, Lexington, Mass. Even-OddModeAnalysisoftheRingHybrid First consider a unit amplitude wave incident at port 1 (the sum port) of the ring hybrid of Figure 7.42a. At the ring junction this wave will divide into two components, which both arrive in phase at ports 2 and 3, and 180◦out of phase at port 4. Using the even-odd mode analysis technique [5], we can decompose this case into a superposition of the two simpler circuits and excitations shown in Figure 7.44. The amplitudes of the scattered waves from 41 22 1 3+1/2 +1/211 O.C. O.C. O.C./H9261/4 /H9261/8 3/H9261/8+1/2 22 2Te Γe (a) 41 22 1 3+1/2 –1/211 S.C. S.C. S.C./H9261/4 /H9261/8 3/H9261/8+1/2 22 2To Γo (b) FIGURE 7.44 Even- and odd-mode decomposition of the ring hybrid when port 1 is excited with a unit amplitude incident wave. (a) Even mode. (b) Odd mode. c07PowerDividers Pozar August 24, 2011 15:53 7.8 The 180◦Hybrid 365 the ring hybrid will be B1=1 2/Gamma1e+1 2/Gamma1o, (7.102a) B2=1 2Te+1 2To, (7.102b) B3=1 2/Gamma1e−1 2/Gamma1o, (7.102c) B4=1 2Te−1 2To. (7.102d) We can evaluate the required reflection and transmission coefficients defined in Figure 7.44 using the ABCD matrix for the even- and odd-mode two-port circuits in Figure 7.44. The results are /bracketleftbigg AB CD/bracketrightbigg e=/bracketleftbigg 1 j√ 2 j√ 2−1/bracketrightbigg , (7.103a) /bracketleftbigg AB CD/bracketrightbigg o=/bracketleftbigg −1 j√ 2 j√ 21/bracketrightbigg . (7.103b) Then with the aid of Table 4.2 we have /Gamma1e=−j√ 2, (7.104a) Te=−j√ 2, (7.104b) /Gamma1o=j√ 2, (7.104c) To=−j√ 2. (7.104d) Using these results in (7.102) gives B1=0, (7.105a) B2=−j√ 2, (7.105b) B3=−j√ 2, (7.105c) B4=0, (7.105d) which shows that the input port is matched, port 4 is isolated, and the input power is evenly divided and in phase between ports 2 and 3. These results form the first row and column of the scattering matrix in (7.101). Next consider a unit amplitude wave incident at port 4 (the difference port) of the ring hybrid of Figure 7.42a. The two wave components on the ring will arrive in phase at port 2 and at port 3, with a relative phase difference of 180◦between these two ports. The two wave components will be 180◦out of phase at port 1. This case can be decomposed into a superposition of the two simpler circuits and excitations shown in Figure 7.45. The c07PowerDividers Pozar August 24, 2011 15:53 366 Chapter 7: Power Dividers and Directional Couplers 431 22 1 O.C. O.C. O.C./H9261/4 /H9261/8 3/H9261/822 2Γe (a)1 1+1/2 +1/2Te 431 22 1 S.C. S.C. S.C./H9261/4 /H9261/8 3/H9261/822 2Γo (b)1 1–1/2 –1/2+1/2 +1/2To FIGURE 7.45 Even- and odd-mode decomposition of the ring hybrid when port 4 is excited with a unit amplitude incident wave. (a) Even mode. (b) Odd mode. amplitudes of the scattered waves will be B1=1 2Te−1 2To, (7.106a) B2=1 2/Gamma1e−1 2/Gamma1o, (7.106b) B3=1 2Te+1 2To, (7.106c) B4=1 2/Gamma1e+1 2/Gamma1o. (7.106d) TheABCD matrices for the even- and odd-mode circuits of Figure 7.45 are /bracketleftbigg AB CD/bracketrightbigg e=/bracketleftbigg −1 j√ 2 j√ 21/bracketrightbigg , (7.107a) /bracketleftbigg AB CD/bracketrightbigg o=/bracketleftbigg 1 j√ 2 j√ 2−1/bracketrightbigg . (7.107b) Then, from Table 4.2, the necessary reflection and transmission coefficients are /Gamma1e=j√ 2, (7.108a) Te=−j√ 2, (7.108b) /Gamma1o=−j√ 2, (7.108c) To=−j√ 2. (7.108d) c07PowerDividers Pozar August 24, 2011 15:53 7.8 The 180◦Hybrid 367 Using these results in (7.106) gives B1=0, (7.109a) B2=j√ 2, (7.109b) B3=−j√ 2, (7.109c) B4=0, (7.109d) which shows that the input port is matched, port 1 is isolated, and the input power is evenly divided into ports 2 and 3 with a 180◦phase difference. These results form the fourth row and column of the scattering matrix of (7.101). The remaining elements in this matrix can be found from symmetry considerations. The bandwidth of the ring hybrid is limited by the frequency dependence of the ring lengths, but is generally on the order of 20%–30%. Increased bandwidth can be obtained by using additional sections, or a symmetric ring circuit as suggested in reference [14]. EXAMPLE 7.9 DESIGN AND PERFORMANCE OF A RING HYBRID Design a 180◦ring hybrid for a 50 /Omega1system impedance, and plot the magnitude of the scattering parameters (S1j)from 0.5 f0to 1.5f0, where f0is the design frequency. Solution With reference to Figure 7.42a, the characteristic impedance of the ring transmis-sion line is√ 2Z0=70.7/Omega1, while the feedline impedances are 50 /Omega1. The scattering parameter magnitudes are plotted versus frequency in Figure 7.46. ■ Even-OddModeAnalysisoftheTaperedCoupledLineHybrid The tapered coupled line 180◦hybrid [15], shown in Figure 7.42b, can provide an arbitrary power division ratio with a bandwidth of a decade or more. This hybrid is also referred toas an asymmetric tapered coupled line coupler. 0.5f0 f0 1.5f0–40–30–20–100 ⎪Sij⎪ dB S11S12S13 FIGURE 7.46 Scattering parameter magnitudes versus frequency for the ring hybrid of Exam- ple 7.9. c07PowerDividers Pozar August 24, 2011 15:53 368 Chapter 7: Power Dividers and Directional Couplers z 0 L 2LkZ0 0Z0Z0/k2Z0z 0 L 2L2 1 3 4Output Difference inputZ0e(z), Z0o(z)Z0Z0 Z0Z0Sum input Output Z0e(z) Z0o(z)(a) (b) FIGURE 7.47 (a) Schematic diagram of the tapered coupled line hybrid. (b) The variation of characteristic impedances. The schematic circuit of this coupler is shown in Figure 7.47, with the ports numbered to correspond functionally to the ports of the 180◦hybrids in Figures 7.41 and 7.42. The coupler consists of two coupled lines with tapering characteristic impedances over the length 0 <z<L.A tz=0 the lines are very weakly coupled, so that Z0e(0)=Z0o(0)= Z0, while at z=Lthe coupling is such that Z0e(L)=Z0/kandZ0o(L)=kZ0, where 0≤k≤1 is a coupling factor that will be related to the voltage coupling factor. The even mode of the coupled line thus matches a load impedance of Z0/k(atz=L)toZ0, while the odd mode matches a load of kZ0toZ0; note that Z0e(z)Z0o(z)=Z2 0for all z.T h e Klopfenstein taper is generally used for these tapered matching lines. For L<z<2Lthe lines are uncoupled, and both have a characteristic impedance Z0; these lines are required for phase compensation of the coupled line section. The length of each section, θ=βL, must be the same, and they should be electrically long to provide a good impedance match over the desired bandwidth. First consider an incident voltage wave of amplitude V0applied to port 4, the differ- ence input. This excitation can be reduced to the superposition of an even-mode excitation and an odd-mode excitation, as shown in Figures 7.48a and 7.48b, respectively. At thejunctions of the coupled and uncoupled lines (z=L), the reflection coefficients seen by the even or odd modes of the tapered lines are /Gamma1 /prime e=Z0−Z0/k Z0+Z0/k=k−1 k+1, (7.110a) /Gamma1/prime o=Z0−kZ0 Z0+kZ0=1−k 1+k. (7.110b) Atz=0 these coefficients are transformed to /Gamma1e=k−1 k+1e−2jθ, (7.111a) /Gamma1o=1−k 1+ke−2jθ. (7.111b) c07PowerDividers Pozar August 24, 2011 15:53 7.8 The 180◦Hybrid 369 2 1 3 4Z0e(z)Z0Z0 Z0 Z0 TeV0/2 Z0 V0/2Z0 ΓeΓe' (a) 2 1 3 4Z0o(z)Z0Z0 Z0 Z0 To/H11002V0/2 Z0 /H11001V0/2Z0 ΓoΓo' (b) FIGURE 7.48 Excitation of the tapered coupled line hybrid. (a) Even-mode excitation. (b) Odd- mode excitation. Then by superposition the scattering parameters of ports 2 and 4 are as follows: S44=1 2(/Gamma1e+/Gamma1o)=0, (7.112a) S24=1 2(/Gamma1e−/Gamma1o)=k−1 k+1e−2jθ. (7.112b) By symmetry, we also have that S22=0 and S42=S24. To evaluate the transmission coefficients into ports 1 and 3, we will use the ABCD parameters for the equivalent circuits shown in Figure 7.49, where the tapered matching sections have been assumed to be ideal, and replaced with transformers. The ABCD matrix 3 4 /H9258 /H9258Z0Z0 Z0 Z0 Tekk : 1 (a) 3 4 /H9258 /H9258Z0 kZ0 Z0 Z0 To1 : k (b) FIGURE 7.49 Equivalent circuits for the tapered coupled line hybrid, for transmission from port 4 to port 3. (a) Even-mode case. (b) Odd-mode case. c07PowerDividers Pozar August 24, 2011 15:53 370 Chapter 7: Power Dividers and Directional Couplers of the transmission line–transformer–transmission line cascade can be found by multiply- ing the three individual ABCD matrices for these components, but it is easier to use the fact that the transmission line sections affect only the phase of the transmission coefficients. TheABCD matrix of the transformer is, for the even mode, /bracketleftbigg√ k 0 01/√ k/bracketrightbigg , and for the odd mode is /bracketleftbigg 1/√ k 0 0√ k/bracketrightbigg . Then the even- and odd-mode transmission coefficients are Te=To=2√ k k+1e−2jθ,( 7.113) since T=2/(A+B/Z0+CZ 0+D)=2√ k/(k+1)for both modes; the e−2jθfactor accounts for the phase delay of the two transmission line sections. We can then evaluatethe following scattering parameters: S 34=1 2(Te+To)=2√ k k+1e−2jθ, (7.114a) S14=1 2(Te−To)=0. (7.114b) The voltage coupling factor from port 4 to port 3 is β=|S34|=2√ k k+1,0<β<1,( 7.115a) while the voltage coupling factor from port 4 to port 2 is α=|S24|=−k−1 k+1,0<α<1.( 7.115b) Power conservation is verified by the fact that |S24|2+|S34|2=α2+β2=1. If we now apply even- and odd-mode excitations at ports 1 and 3, so that superpo- sition yields an incident voltage wave at port 1, we can derive the remaining scattering parameters. With a phase reference at the input ports, the even- and odd-mode reflection coefficients at port 1 will be /Gamma1e=1−k 1+ke−2jθ, (7.116a) /Gamma1o=k−1 k+1e−2jθ. (7.116b) Then we can calculate the following scattering parameters: S11=1 2(/Gamma1e+/Gamma1o)=0, (7.117a) S31=1 2(/Gamma1e−/Gamma1o)=1−k 1+ke−2jθ=αe−2jθ. (7.117b) c07PowerDividers Pozar August 24, 2011 15:53 7.8 The 180◦Hybrid 371 From symmetry we have that S33=0,S13=S31, and S14=S32,S12=S34.T h et a - pered coupled line 180◦hybrid thus has the following scattering matrix: [S]=⎡ ⎢⎢⎢⎣0βα 0 β 00 −α α 00 β 0−αβ 0⎤ ⎥⎥⎥⎦e −2jθ.( 7.118) WaveguideMagic-T The waveguide magic-T hybrid junction of Figure 7.42c has terminal properties similar to those of the ring hybrid, and a scattering matrix similar in form to (7.101). A rigorous analysis of this junction is too complicated to present here, but we can explain its operation in a qualitative sense by considering the field lines for excitations at the sum and difference ports. First consider a TE 10mode incident at port 1. The resulting Eyfield lines are illus- trated in Figure 7.50a, where it is seen that there is an odd symmetry about guide 4. Becausethe field lines of a TE 10mode in guide 4 would have even symmetry, there is no coupling between ports 1 and 4. There is identical coupling to ports 2 and 3, however, resulting in an in-phase, equal-split power division. For a TE 10mode incident at port 4, the field lines are as shown in Figure 7.50b. Again ports 1 and 4 are decoupled, due to symmetry (or reciprocity). Ports 2 and 3 are excitedequally by the incident wave, but with a 180 ◦phase difference. In practice, tuning posts or irises are often used for matching; such components must be placed symmetrically to maintain proper operation of the hybrid. 2 34 (a) 2 34 (b) FIGURE 7.50 Electric field lines for a waveguide hybrid junction. (a) Incident wave at port 1. (b) Incident wave at port 4. c07PowerDividers Pozar August 24, 2011 15:53 372 Chapter 7: Power Dividers and Directional Couplers Isolated InputCoupledThrough FIGURE 7.51 The Moreno crossed-guide coupler. 7.9OTHERCOUPLERS Although we have discussed the general properties of couplers and have analyzed and derived design data for several of the most frequently used couplers, there are many other types that we have not treated in detail. In this section we will briefly describe some of these. Moreno crossed-guide coupler: This is a waveguide directional coupler consisting of two waveguides at right angles, with coupling provided by two apertures in the common broad wall of the guides. See Figure 7.51. By proper design [16], the two wave components ex-cited by these apertures can be made to cancel in the back direction. The apertures usually consist of crossed slots, in order to couple tightly to the fields of both guides. Schwinger reversed-phase coupler : This waveguide coupler is designed so that the path lengths for the two coupling apertures are the same for the uncoupled port, so that the directivity is essentially independent of frequency. Cancellation in the isolated port is ac-complished by placing the slots on opposite sides of the centerline of the waveguide walls, as shown in Figure 7.52, which couple to magnetic dipoles with a 180 ◦phase difference. /H9261g/4+M –MIsolated Coupled InputThrough FIGURE 7.52 The Schwinger reversed-phase coupler. c07PowerDividers Pozar August 24, 2011 15:53 7.9 Other Couplers 373 Coupled Through Isolated Input FIGURE 7.53 The Riblet short-slot coupler. Theλg/4 slot spacing leads to in-phase combining at the coupled (backward) port, but this coupling is very frequency sensitive. This is the opposite situation from that of the multihole waveguide coupler discussed in Section 7.4. Riblet short-slot coupler: This coupler, shown in Figure 7.53, consists of two waveguides with a common sidewall. Coupling takes place in the region where part of the common wall has been removed. In this region both the TE 10(even) and the TE 20(odd) modes are excited, and by proper design can be made to cause cancellation at the isolated port and addition at the coupled port. The width of the interaction region must be small enough to prevent propagation of the undesired TE 30mode. This coupler can usually be made smaller than other waveguide couplers. Symmetric tapered coupled line coupler: We saw that a continuously tapered transmission line matching transformer was the logical extension of the multisection matching trans-former. Similarly, the multisection coupled line coupler can be extended to a continuous taper, yielding a coupled line coupler with good bandwidth characteristics. Such a coupler is shown in Figure 7.54. Generally, both the conductor width and separation can be adjustedto provide a synthesized coupling or directivity response. One way to do this involves the computer optimization of a stepped-section approximation to the continuous taper [17]. This coupler provides a 90 ◦phase shift between the outputs. Couplers with apertures in planar lines : Many of the above-mentioned waveguide couplers can also be fabricated with planar lines such as microstrip line, stripline, dielectric image lines, or various combinations of these. Some possibilities are illustrated in Figure 7.55. FIGURE 7.54 A symmetric tapered coupled line coupler. c07PowerDividers Pozar August 24, 2011 15:53 374 Chapter 7: Power Dividers and Directional Couplers (a) (b)Coupling aperture Ground plane Microstrip linesCoupling apertureWaveguideMicrostrip line (c)Coupling apertureDielectric image guideMicrostrip line/H9280r /H9280r /H9280r /H9280r/H9280r FIGURE 7.55 Various aperture coupled planar line couplers. (a) Microstrip-to-microstrip cou- pler. (b) Microstrip-to-waveguide coupler. (c) Microstrip-to-dielectric image line coupler. In principle, the design of such couplers can be carried out using the small-hole coupling theory and analysis techniques used in this chapter. The evaluation of the fields of planarlines, however, is usually much more complicated than for rectangular waveguides. POINT OF INTEREST: The Reflectometer Areflectometer is a circuit that uses a directional coupler to isolate and sample the incident and reflected powers from a load. It is a key component in a scalar or vector network analyzer, as it can be used to measure the reflection coefficient of a one-port network and, in a more general configuration, the scattering parameters of a two-port network. It can also be used as an SWRmeter, or as a power monitor in systems applications. The basic reflectometer circuit shown on the left in the accompanying figure can be used to measure the reflection coefficient magnitude of an unknown load. If we assume a reasonably matched coupler with loose coupling ( C/lessmuch1), so that/radicalbig 1−C2/similarequal1, then the circuit can be represented by the signal flow graph shown on the right in the accompanying figure. In opera- tion, the directional coupler provides a sample, Vi, of the incident wave, and a sample, Vr,o f the reflected wave. A ratio meter with an appropriately calibrated scale can then measure these voltages and provide a reading in terms of reflection coefficient magnitude, or SWR. 1 2143 24 3 ΓLoadVr ViVr Vi 1C D ΓC DCC C, D c07PowerDividers Pozar August 24, 2011 15:53 References 375 Realistic directional couplers, however, have finite directivity, which means that both inci- dent and reflected powers will contribute to both ViandVr,l e a d i n gt oa ne r r o r .I fw ea s s u m e a unit incident wave from the source, inspection of the signal flow graph leads to the following expressions for ViandVr: Vi=C+C D/Gamma1ejθ, Vr=C D+C/Gamma1ejφ, where /Gamma1is the reflection coefficient of the load, D=10(DdB/20)is the numerical directivity of the coupler, and θ,φ are the unknown phase delays through the circuit. Then the maximum and minimum values of the magnitude of Vr/Vican be written as /vextendsingle/vextendsingle/vextendsingle/vextendsingleV r Vi/vextendsingle/vextendsingle/vextendsingle/vextendsingle max min=|/Gamma1|±1 D 1∓|/Gamma1| D. For a coupler with infinite directivity this reduces to the desired result of |/Gamma1|. Otherwise a measurement uncertainty of approximately ±1/Dis introduced. Good accuracy thus requires a coupler with high directivity, preferably greater than 40 dB. REFERENCES [ 1 ] A .E .B a i l e y ,e d . ,Microwave Measurement, Peter Peregrinus, London, 1985. [2] R. E. Collin, Foundations for Microwave Engineering , 2nd edition, Wiley–IEEE Press, Hoboken, N.J., 2001. [3] F. E. Gardiol, Introduction to Microwaves , Artech House, Dedham, Mass., 1984. [4] E. Wilkinson, “An N-Way Hybrid Power Divider,” IRE Transactions on Microwave Theory and Techniques, vol. MTT-8, pp. 116–118, January 1960. [5] J. Reed and G. J. Wheeler, “A Method of Analysis of Symmetrical Four-Port Networks,” IRE Trans- actions on Microwave Theory and Techniques, vol. MTT-4, pp. 246–252, October 1956. [6] C. G. Montgomery, R. H. Dicke, and E. M. Purcell, Principles of Microwave Circuits , MIT Radiation Laboratory Series, vol. 8, McGraw-Hill, New York, 1948. [7] H. Howe, Stripline Circuit Design, Artech House, Dedham, Mass., 1974. [8] K. C. Gupta, R. Garg, and I. J. Bahl, Microstrip Lines and Slot Lines, Artech House, Dedham, Mass., 1979. [9] L. Young, “The Analytical Equivalence of the TEM-Mode Directional Couplers and Transmission- Line Stepped Impedance Filters,” Proceedings of the IEEE, vol. 110, pp. 275–281, February 1963. [10] J. Lange, “Interdigitated Stripline Quadrature Hybrid,” IEEE Transactions on Microwave Theory and Techniques, vol. MTT-17, pp. 1150–1151, December 1969. [11] R. Waugh and D. LaCombe, “Unfolding the Lange Coupler,” IEEE Transactions on Microwave Theory and Techniques, vol. MTT-20, pp. 777–779, November 1972. [12] W. P. Ou, “Design Equations for an Interdigitated Directional Coupler,” IEEE Transactions on Microwave Theory and Techniques, vol. MTT-23, pp. 253–255, February 1973. [13] D. Paolino, “Design More Accurate Interdigitated Couplers,” Microwaves , vol. 15, pp. 34–38, May 1976. [14] J. Hughes and K. Wilson, “High Power Multiple IMPATT Amplifiers,” in: Proceedings of the 4th European Microwave Conference, Montreux, Switzerland, pp. 118–122, 1974. [15] R. H. DuHamel and M. E. Armstrong, “The Tapered-Line Magic-T,” in: Abstracts of 15th Annual Symposium of the USAF Antenna Research and Development Program , Monticello, Ill., October 12–14, 1965. [16] T. N. Anderson, “Directional Coupler Design Nomograms,” Microwave Journal , vol. 2, pp. 34–38, May 1959. c07PowerDividers Pozar August 24, 2011 15:53 376 Chapter 7: Power Dividers and Directional Couplers [17] D. W. Kammler, “The Design of Discrete N-Section and Continuously Tapered Symmetrical TEM Directional Couplers,” IEEE Transactions on Microwave Theory and Techniques , vol. MTT-17, pp. 577–590, August 1969. PROBLEMS 7.1 Write the scattering matrix for a nonideal symmetric hybrid coupler in terms of the coupler param- eters, C,D,I,a n d L, as defined in (7.20). Repeat for a nonideal antisymmetric hybrid coupler. Assume that the couplers are matched at all ports. 7.2 A 20 dBm power source is connected to the input of a directional coupler having a coupling factor of 20 dB, a directivity of 35 dB, and an insertion loss of 0.5 dB. If all ports are matched, find the output powers (in dBm) at the through, coupled, and isolated ports. 7.3 A directional coupler has the scattering matrix given below. Find the return loss, coupling factor, directivity, and insertion loss. Assume that the ports are terminated in matched loads. [S]=⎡ ⎢⎢⎢⎣0.1/negationslash40◦0.944 /negationslash90◦0.178 /negationslash180◦0.0056 /negationslash90◦ 0.944 /negationslash90◦0.1/negationslash40◦0.0056 /negationslash90◦0.178 /negationslash180◦ 0.178 /negationslash180◦0.0056 /negationslash90◦0.1/negationslash40◦0.944 /negationslash90◦ 0.0056 /negationslash90◦0.178 /negationslash180◦0.944 /negationslash90◦0.1/negationslash40◦⎤ ⎥⎥⎥⎦ 7.4 Two identical 90◦couplers with C=8.34 dB are connected as shown below. Find the resulting phase and amplitudes at ports 2/primeand 3/prime, relative to port 1. 4 3 4' 3' 1 2 1' 2' 7.5 Consider the T-junction of three lines with characteristic impedances Z1,Z2,and Z3,as shown below. Demonstrate that it is impossible for all three lines to be matched when looking toward thejunction. Z1Z2 Z3 7.6 Design a lossless T-junction divider with a 30 /Omega1source impedance to give a 3:1 power split. Design quarter-wave matching transformers to convert the impedances of the output lines to 30 /Omega1. Determine the magnitude of the scattering parameters for this circuit, using a 30 /Omega1characteristic impedance. 7.7 Consider the T and πresistive attenuator circuits shown below. If the input and output are matched toZ0, and the ratio of output voltage to input voltage is α, derive the design equations for R1andR2 for each circuit. If Z0=50/Omega1, compute R1andR2for 3, 10, and 20 dB attenuators of each type. c07PowerDividers Pozar August 24, 2011 15:53 Problems 377 7.8 Design a three-port resistive divider for an equal power split and a 100 /Omega1system impedance. If port 3 is matched, calculate the change in output power at port 3 (in dB) when port 2 is connected first to a matched load, and then to a load having a mismatch of /Gamma1=0.3. See the figure below. 1 32 Γ = 0 or 0.3 Pin Po 7.9 Consider the general resistive divider shown below. For an arbitrary power division ratio α=P2/P3, derive expressions for the resistors R1,R2,andR3, and the output characteristic impedances Zo2 andZo3so that all ports are matched, assuming the source impedance is Z0. P1 Z0R1P2 R2Z02 R3 Z03 P3 7.10 Design a Wilkinson power divider with a power division ratio of P3/P2=1/3 and a source imped- ance of 50 /Omega1. 7.11 Derive the design equations in (7.37a)–(7.37c) for the unequal-split Wilkinson divider. 7.12 For the Bethe hole coupler of the type shown in Figure 7.16a, derive a design for sso that port 3 is the isolated port. 7.13 Design a Bethe hole coupler of the type shown in Figure 7.16a for a Ku-band waveguide operating at 11 GHz. The required coupling is 20 dB. 7.14 Design a Bethe hole coupler of the type shown in Figure 7.16b for a Ku-band waveguide operating at 17 GHz. The required coupling is 30 dB. 7.15 Design a five-hole directional coupler in a Ku-band waveguide with a binomial directivity response. The center frequency is 17.5 GHz, and the required coupling is 20 dB. Use round apertures centered across the broad wall of the waveguides. 7.16 Repeat Problem 7.14 for a design with a Chebyshev response, having a minimum directivity of 30 dB. 7.17 Develop the necessary equations required to design a two-hole directional coupler using two wave- guides with apertures in a common sidewall, as shown below. c07PowerDividers Pozar August 24, 2011 15:53 378 Chapter 7: Power Dividers and Directional Couplers db 2r0a a 7.18 Consider the general branch-line coupler shown below, with shunt arm characteristic impedances Zaand series arm characteristic impedances Zb. Using an even-odd mode analysis, derive design equations for a quadrature hybrid coupler with an arbitrary power division ratio of α=P2/P3,a n d with the input port (port 1) matched. Assume all arms are λ/4 long. Is port 4 isolated, in general? 1 2 4 3Z0 P1 P2 P4 P3Zb ZbZ0Za Za Z0Z0 7.19 An edge-coupled stripline with a ground plane spacing of 2.0 mm and a dielectric constant of 4.2 has strip widths of 0.6 mm and a separation of 0.2 mm between the edges of the strips. Use the graph ofFigure 7.29 to find the resulting even- and odd-mode characteristic impedances. If possible, compare your results to those obtained from a microwave CAD tool. 7.20 A coupled microstrip line is to be designed for a substrate having a thickness of 2.0 mm and dielectric constant of 10.0. The required even- and odd-mode characteristic impedances are 133 /Omega1and 71.5 /Omega1, respectively. Use the graph of Figure 7.30 to find the required line widths and separation. If possible, compare your results to those obtained from a microwave CAD tool. 7.21 Repeat the derivation in Section 7.6 for the design equations of a single-section coupled line coupler using reflection and transmission coefficients instead of voltages and currents. 7.22 Design a single-section coupled line coupler with a coupling of 19.1 dB, a system impedance of 60/Omega1, and a center frequency of 8 GHz. If the coupler is to be made in stripline (edge-coupled), with /epsilon1 r=2.2a n d b=0.32 cm, find the necessary strip widths and separation. 7.23 Repeat Problem 7.22 for a coupling factor of 5 dB. Is this a practical design? 7.24 Derive Equations (7.83) and (7.84). 7.25 A 20-dB three-section coupled line coupler is required to have a maximally flat coupling response with a center frequency of 3 GHz and Z0=50/Omega1. (a) Design the coupler and find Z0eandZ0o for each section. Use CAD to plot the resulting coupling (in dB) from 1 to 5 GHz. (b) Lay out the microstrip implementation of the coupler on an FR4 substrate having /epsilon1r=4.2,d=0.158 cm, and tan δ=0.02, with copper conductors 0.5 mil thick. Use CAD to plot the insertion loss versus frequency. 7.26 Repeat Problem 7.25 for a coupler with an equal-ripple coupling response, where the ripple in the coupling is 1 dB over the passband. 7.27 For the Lange coupler, derive the design equations (7.100) for Z0eandZ0ofrom (7.98) and (7.99). 7.28 Design a 3 dB Lange coupler for operation at 5 GHz. If the coupler is to be fabricated in microstrip on an alumina substrate with /epsilon1r=10 and d=1.0 mm, compute Z0eandZ0ofor the two adjacent lines, and find the necessary spacing and widths of the lines. 7.29 An input signal V1is applied to the sum port of a 180◦hybrid, and another signal V4is applied to the difference port. What are the output signals? c07PowerDividers Pozar August 24, 2011 15:53 Problems 379 7.30 Calculate the even- and odd-mode characteristic impedances for a tapered coupled line 180◦hybrid coupler with a 3 dB coupling ratio and a 50 /Omega1characteristic impedance. 7.31 Find the scattering parameters for the four-port Bagley polygon power divider shown below. 3 2 4 12Z0 3Z0 Z0Z0 Z0/H9261/2 /H9261/2 /H9261/4 /H9261/4 7.32 For the symmetric hybrid shown below, calculate the output voltages if port 1 is fed with an incident wave of 1 /negationslash0 V . Assume that the outputs are matched. /H9261/2 /H9261/4 /H9261/4 /H9261/4 /H9261/44 3 2 15Z0Z0 Z0 Z0 Z02Z0 7.33 The Bailey unequal-split power divider uses a 90◦hybrid coupler and a T-junction, as shown below. The power division ratio is controlled by adjusting the feed position, a, along the transmission line of length bthat connects ports 1 and 4 of the hybrid. A quarter-wave transformer of impedance Z0/√ 2 is used to match the input of the divider. (a) For b=λ/4, show that the output power division ratio is given by P3/P2=tan2(πa/2b). (b) Using a branch-line hybrid with Z0=50/Omega1, design a power divider with a division ratio of P3/P2=0.5, and plot the resulting input return loss and transmission coefficients versus frequency. 2 1 3P2 P3Z0Z0Pin abZ0/2 4 c08MicrowaveFilters Pozar August 25, 2011 18:16 Chapter Eight Microwave Filters A filter is a two-port network used to control the frequency response at a certain point in an RF or microwave system by providing transmission at frequencies within the passband of the filter and attenuation in the stopband of the filter. Typical frequency responses include low-pass, high-pass, bandpass, and band-reject characteristics. Applications can be found invirtually any type of RF or microwave communication, radar, or test and measurement system. The development of filter theory and practice began in the years preceding World War II by pioneers such as Mason, Sykes, Darlington, Fano, Lawson, and Richards. The image parame- ter method of filter design was developed in the late 1930s and was useful for low-frequencyfilters in radio and telephony. In the early 1950s a group at Stanford Research Institute, consist-ing of G. Matthaei, L. Young, E. Jones, S. Cohn, and others, became very active in microwave filter and coupler development. A voluminous handbook on filters and couplers resulted from this work and remains a valuable reference [1]. Today, most microwave filter design is donewith sophisticated computer-aided design (CAD) packages based on the insertion loss method.Because of continuing advances in network synthesis with distributed elements, the use of low- temperature superconductors and other new materials, and the incorporation of active devices in filter circuits, microwave filter design remains an active research area. We begin our discussion of filter theory and design with the frequency characteristics of periodic structures, which consist of a transmission line or waveguide periodically loaded with reactive elements. These structures are of interest in themselves because of their application to slow-wave components and traveling-wave amplifier design, and also because they exhibitbasic passband-stopband responses that lead to the image parameter method of filter design. Filters designed using the image parameter method consist of a cascade of simpler two- port filter sections to provide the desired cutoff frequencies and attenuation characteristics but do not allow the specification of a particular frequency response over the complete operatingrange. Thus, although the procedure is relatively simple, the design of filters by the imageparameter method often must be iterated many times to achieve the desired results. A more modern procedure, called the insertion loss method, uses network synthesis tech- niques to design filters with a completely specified frequency response. The design is simplifiedby beginning with low-pass filter prototypes that are normalized in terms of impedance and 380 c08MicrowaveFilters Pozar August 25, 2011 18:16 8.1 Periodic Structures 381 frequency. Transformations are then applied to convert the prototype designs to the desired frequency range and impedance level. Both the image parameter and insertion loss methods of filter design lead to circuits using lumped elements (capacitors and inductors). For microwave applications such designs usuallymust be modified to employ distributed elements consisting of transmission line sections. The Richards transformation and the Kuroda identities provide this step. We will also discuss trans- mission line filters using stepped impedances and coupled lines; filters using coupled resonatorswill also be briefly described. The subject of microwave filters is quite extensive due to the importance of these compo- nents in practical systems and the wide variety of possible implementations. Here we can treat only the basic principles and some of the more common filter designs, and we refer the readerto references such as [1–4] for further discussion. 8.1PERIODICSTRUCTURES An infinite transmission line or waveguide periodically loaded with reactive elements is an example of a periodic structure. As shown in Figure 8.1, periodic structures can take various forms, depending on the transmission line media being used. Often the loading elements are formed as discontinuities in the line itself, but in any case they can be modeled as lumped reactances in shunt (or series) on a transmission line, as shown in Figure 8.2.Periodic structures support slow-wave propagation (slower than the phase velocity of the (a) (b) FIGURE 8.1 Examples of periodic structures. (a) Periodic stubs on a microstrip line. (b) Periodic diaphragms in a waveguide. c08MicrowaveFilters Pozar August 25, 2011 18:16 382 Chapter 8: Microwave Filters dZ0, kjb jb jb jb jb jb Unit cellzVn+1In+1 + –VnIn + – FIGURE 8.2 Equivalent circuit of a periodically loaded transmission line. The unloaded line has characteristic impedance Z0and propagation constant k. unloaded line), and have passband and stopband characteristics similar to those of filters; they find application in traveling-wave tubes, masers, phase shifters, and antennas. AnalysisofInfinit PeriodicStructures We first consider the propagation characteristics of the infinite loaded line shown in Fig- ure 8.2. Each unit cell of this line consists of a length, d, of transmission line with a shunt susceptance across the midpoint of the line; the susceptance, b, is normalized to the char- acteristic impedance, Z0. If we consider the infinite line as being composed of a cascade of identical two-port networks, we can relate the voltages and currents on either side of the nth unit cell using the ABCD matrix: /bracketleftbiggVn In/bracketrightbigg =/bracketleftbiggAB CD/bracketrightbigg/bracketleftbiggVn+1 In+1/bracketrightbigg ,( 8.1) where A,B,C,andDare the matrix parameters for a cascade of a transmission line section of length d/2, a shunt susceptance b, and another transmission line section of length d/2. From Table 4.1 we then have, in normalized form, /bracketleftbigg AB CD/bracketrightbigg =⎡ ⎢⎢⎣cosθ 2jsinθ 2 jsinθ 2cosθ 2⎤ ⎥⎥⎦/bracketleftbigg 10 jb 1/bracketrightbigg⎡ ⎢⎢⎣cosθ 2jsinθ 2 jsinθ 2cosθ 2⎤ ⎥⎥⎦ =⎡ ⎢⎢⎢⎣/parenleftbigg cosθ−b 2sinθ/parenrightbigg j/parenleftbigg sinθ+b 2cosθ−b 2/parenrightbigg j/parenleftbigg sinθ+b 2cosθ+b 2/parenrightbigg/parenleftbigg cosθ−b 2sinθ/parenrightbigg⎤ ⎥⎥⎥⎦, (8.2) where θ=kd, and kis the propagation constant of the unloaded line. The reader can verify thatAD−BC=1, as required for reciprocal networks. For a wave propagating in the +zdirection, we must have V(z)=V(0)e−γz, (8.3a) I(z)=I(0)e−γz, (8.3b) for a phase reference at z=0. Since the structure is infinitely long, the voltage and current at the nth terminals can differ from the voltage and current at the n+1 terminals only by c08MicrowaveFilters Pozar August 25, 2011 18:16 8.1 Periodic Structures 383 the propagation factor, e−γd. Thus, Vn+1=Vne−γd, (8.4a) In+1=Ine−γd. (8.4b) Using this result in (8.1) gives the following: /bracketleftbigg Vn In/bracketrightbigg =/bracketleftbigg AB CD/bracketrightbigg/bracketleftbigg Vn+1 In+1/bracketrightbigg =/bracketleftbigg Vn+1eγd In+1eγd/bracketrightbigg , or /bracketleftbigg A−eγdB CD −eγd/bracketrightbigg/bracketleftbigg Vn+1 In+1/bracketrightbigg =0.( 8.5) For a nontrivial solution, the determinant of the above matrix must vanish: AD+e2γd−(A+D)eγd−BC=0,( 8.6) or, since AD−BC=1, 1+e2γd−(A+D)eγd=0, e−γd+eγd=A+D, coshγd=A+D 2=cosθ−b 2sinθ, (8.7) where (8.2) was used for the values of AandD.N o w ,i fγ =α+jβ, we have that coshγd=coshαdcosβd+jsinhαdsinβd=cosθ−b 2sinθ. ( 8.8) Since the right-hand side of (8.8) is purely real, we must have either α=0o rβ=0. Case 1: α=0,β/negationslash=0. This case corresponds to a nonattenuated propagating wave on the periodic structure, and defines the passband of the structure. Equation (8.8) reduces to cosβd=cosθ−b 2sinθ, (8.9a) which can be solved for βif the magnitude of the right-hand side is less than or equal to unity. Note that there are an infinite number of values of βthat can satisfy (8.9a). Case 2: α/negationslash=0,β=0,π. In this case the wave does not propagate, but is attenuated along the line; this defines the stopband of the structure. Because the line is lossless, power is not dissipated, but is reflected back to the input of the line. The magnitude of (8.8) reduces to coshαd=/vextendsingle/vextendsingle/vextendsingle/vextendsinglecosθ−b 2sinθ/vextendsingle/vextendsingle/vextendsingle/vextendsingle≥1,( 8.9b) which has only one solution ( α> 0) for positively traveling waves; α< 0 applies for nega- tively traveling waves. If cos θ−(b/2)sinθ≤−1, (8.9b) is obtained from (8.8) by letting β=π; then all the lumped loads on the line are λ/2 apart, yielding an input impedance the same as if β=0. Thus, depending on the frequency and normalized susceptance values, the periodically loaded line will exhibit either passbands or stopbands, and so can be considered as a type of filter. It is important to note that the voltage and current waves defined in (8.3) and (8.4) are meaningful only when measured at the terminals of the unit cells, and do not apply to voltages and currents that may exist at points within a unit cell. These waves are similar to the elastic waves (Bloch waves ) that propagate through periodic crystal lattices. c08MicrowaveFilters Pozar August 25, 2011 18:16 384 Chapter 8: Microwave Filters Besides the propagation constant of the waves on the periodically loaded line, we will also be interested in the characteristic impedance for these waves. We can define a characteristic impedance at the unit cell terminals as ZB=Z0Vn+1 In+1,( 8.10) since Vn+1andIn+1in the above derivation were normalized quantities. This impedance is also referred to as the Bloch impedance. From (8.5) we have that (A−eγd)Vn+1+BIn+1=0, so (8.10) yields ZB=−BZ0 A−eγd. From (8.6) we can solve for eγdin terms of AandDas follows: eγd=(A+D)±/radicalbig (A+D)2−4 2. Then the Bloch impedance has two solutions given by Z± B=−2BZ0 A−D∓/radicalbig (A+D)2−4.( 8.11) For symmetric unit cells (as assumed in Figure 8.2) we will always have A=D.I nt h i s case (8.11) reduces to Z± B=±BZ0√ A2−1.( 8.12) The±solutions correspond to the characteristic impedance for positively and negatively traveling waves, respectively. For symmetric networks these impedances are the same ex- cept for the sign; the characteristic impedance for a negatively traveling wave is negative because we have defined Inin Figure 8.2 as always being in the positive direction. From (8.2) we see that Bis always purely imaginary. If α=0,β/negationslash=0 (passband), then (8.7) shows that cosh γd=A≤1 (for symmetric networks), and (8.12) shows that ZBwill be real. If α/negationslash=0,β=0 (stopband), then (8.7) shows that cosh γd=A≥1, and (8.12) shows that ZBis imaginary. This situation is similar to that for the wave impedance of a waveguide, which is real for propagating modes and imaginary for cutoff, or evanescent, modes. TerminatedPeriodicStructures Next consider a truncated periodic structure terminated in a load impedance ZL,a ss h o w n in Figure 8.3. At the terminals of an arbitrary unit cell, the incident and reflected voltages Unit cellUnit cellUnit cellIn Vn+ –IN VN ZL+ – FIGURE 8.3 A periodic structure terminated in a normalized load impedance ZL. c08MicrowaveFilters Pozar August 25, 2011 18:16 8.1 Periodic Structures 385 and currents can be written as (assuming operation in the passband) Vn=V+ 0e−jβnd+V− 0ejβnd(8.13a) In=I+ 0e−jβnd+I− 0ejβnd=V+ 0 Z+ Be−jβnd+V− 0 Z− Bejβnd(8.13b) where we have replaced γzin (8.3) with jβndsince we are interested only in terminal quantities. Now define the following incident and reflected voltages at the nth unit cell: V+ n=V+ 0e−jβnd(8.14a) V− n=V− 0ejβnd(8.14b) Then (8.13) can be written as Vn=V+ n+V− n, (8.15a) In=V+ n Z+ B+V− n Z− B. (8.15b) At the load, where n=N,w eh a v e VN=V+ N+V− N=ZLIN=ZL/parenleftBigg V+ N Z+ B+V− N Z− B/parenrightBigg ,( 8.16) and the reflection coefficient at the load can be found as /Gamma1=V− N V+ N=−ZL/Z+ B−1 ZL/Z− B−1.( 8.17) If the unit cell network is symmetric (A=D), then Z+ B=− Z− B=ZB, which reduces (8.17) to the familiar result /Gamma1=ZL−ZB ZL+ZB.( 8.18) In order to avoid reflections on the terminated periodic structure we must have ZL= ZB, which is real for a lossless structure operating in a passband. If necessary, a quarter- wave transformer can be used between the periodically loaded line and the load. k-βDiagramsandWaveVelocities When studying the passband and stopband characteristics of a periodic structure, it is useful to plot the propagation constant, β, versus the propagation constant of the unloaded line, k(orω). Such a graph is called a k-βdiagram,o rBrillouin diagram, after L. Brillouin, a physicist who studied wave propagation in periodic crystal structures. The k-βdiagram can be plotted from (8.9a), which is the dispersion relation for a general periodic structure. In fact, a k-βdiagram can be used to study the dispersion char- acteristics of many types of microwave components and transmission lines. For instance, consider the dispersion relation for a waveguide mode: β=/radicalBig k2−k2c,ork=/radicalBig β2+k2c,( 8.19) where kcis the cutoff wave number of the mode, kis the free-space wave number, and β is the propagation constant of the mode. Relation (8.19) is plotted in the k-βdiagram of c08MicrowaveFilters Pozar August 25, 2011 18:16 386 Chapter 8: Microwave Filters k /H9252Propagation Operating point Cutoffk = /H9252 /H9252 kc Slope = vg/c Slope = vp/c 0 FIGURE 8.4 k-βdiagram for a waveguide mode. Figure 8.4. For values of k<kcthere is no real solution for β, so the mode is nonpropa- gating. For k>kcthe mode propagates, and kapproaches βfor large values of β(TEM propagation). Thek-βdiagram is also useful for interpreting the various wave velocities associated with a dispersive structure. The phase velocity is vp=ω β=ck β,( 8.20) which is seen to be equal to c(speed of light) times the slope of the line from the origin to the operating point on the k-βdiagram. The group velocity is vg=dω dβ=cdk dβ,( 8.21) which is the slope of the k-βcurve at the operating point. Thus, referring to Figure 8.4, we see that the phase velocity for a propagating waveguide mode is infinite at cutoff and approaches c(from above) as kincreases. The group velocity, however, is zero at cutoff and approaches c(from below) as kincreases. We finish our discussion of periodic structures with a practical example of a capacitively loaded line. EXAMPLE 8.1 ANALYSIS OF A PERIODIC STRUCTURE Consider the periodic capacitively loaded line shown in Figure 8.5 (such a linemay be implemented as in Figure 8.1, with short capacitive stubs). If Z 0=50/Omega1, d=1.0 cm, and C0=2.666 pF, sketch the k-βdiagram and compute the prop- agation constant, phase velocity, and Bloch impedance at f=3.0 GHz. Assume k=k0. dC0 C0 C0 C0 C0Z0, k FIGURE 8.5 A capacitively loaded line. c08MicrowaveFilters Pozar August 25, 2011 18:16 8.1 Periodic Structures 387 Solution We can rewrite the dispersion relation of (8.9a) as cosβd=cosk0d−/parenleftbiggC0Z0c 2d/parenrightbigg k0dsink0d. Then C0Z0c 2d=(2.666 ×10−12)(50)(3 ×108) 2(0.01)=2.0, so we have cosβd=cosk0d−2k0dsink0d. The most straightforward way to proceed at this point is to numerically evaluate the right-hand side of the above equation for a set of values of k0dstarting at zero. When the magnitude of the right-hand side is unity or less, we have a passband and can solve for βd. Otherwise we have a stopband. Calculation shows that the first passband exists for 0 ≤k0d≤0.96. The second passband does not begin until the sin k0dterm changes sign at k0d=π.A sk 0dincreases, an infinite number of passbands are possible, but they become narrower. Figure 8.6 shows the k-β diagram for the first two passbands. At 3.0 GHz, we have k0d=2π(3×109) 3×108(0.01) =0.6283 =36◦, soβd=1.5 and the propagation constant is β=150 rad/m. The phase velocity is vp=k0c β=0.6283 1.5c=0.42c, which is much less than the speed of light, indicating that this is a slow-wave 1234k0d –3 –2 –1 0 1 2 3 /H9252d–/H9266/H9266PassbandPassband FIGURE 8.6 k-βdiagram for Example 8.1. c08MicrowaveFilters Pozar August 25, 2011 18:16 388 Chapter 8: Microwave Filters structure. To evaluate the Bloch impedance, we use (8.2) and (8.12): b 2=ωC0Z0 2=1.256, θ=k0d=36◦, A=cosθ−b 2sinθ=0.0707, B=j/parenleftbigg sinθ+b 2cosθ−b 2/parenrightbigg =j0.3479. Then, ZB=BZ0√ A2−1=(j0.3479)(50) j/radicalbig 1−(0.0707)2=17.4/Omega1. ■ 8.2FILTERDESIGNBYTHEIMAGEPARAMETERMETHOD The image parameter method of filter design involves the specification of passband and stopband characteristics for a cascade of simple two-port networks, and so is related in con-cept to the periodic structures of Section 8.1. The method is relatively simple but has the disadvantage that an arbitrary frequency response cannot be incorporated into the design. This is in contrast to the insertion loss method, which is the subject of the following section.Nevertheless, the image parameter method is useful for simple filters, and it provides a link between infinite periodic structures and practical filter design. The image parameter method also finds application in solid-state traveling-wave amplifier design. ImageImpedancesandTransferFunctionsforTwo-PortNetworks We begin with definitions of the image impedances and voltage transfer function for an arbitrary reciprocal two-port network; these results are required for the analysis and design of filters by the image parameter method. Consider the arbitrary two-port network shown in Figure 8.7, where the network is specified by its ABCD parameters. Note that the reference direction for the current at port 2 has been chosen according to the convention for ABCD parameters. The image impedances, Z i1andZi2, are defined for this network as follows: Zi1=input impedance at port 1 when port 2 is terminated with Zi2 Zi2=input impedance at port 2 when port 1 is terminated with Zi1. Thus both ports are matched when terminated in their image impedances. We can derive expressions for the image impedances in terms of the ABCD parameters of the network. I2 I1 V2Zi2+ –V1Zi1+ –A CB D Zin 1 Zin 2 FIGURE 8.7 A two-port network terminated in its image impedances. c08MicrowaveFilters Pozar August 25, 2011 18:16 8.2 Filter Design by the Image Parameter Method 389 The port voltages and currents are related as V1=AV2+BI2, (8.22a) I1=CV2+DI2. (8.22b) The input impedance at port 1, with port 2 terminated in Zi2,i s Zin1=V1 I1=AV2+BI2 CV2+DI2=AZi2+B CZ i2+D,( 8.23) since V2=Zi2I2. Now solve (8.22) for V2,I2by inverting the ABCD matrix. Since AD−BC=1f o r a reciprocal network, we obtain V2=DV1−BI1, (8.24a) I2=−CV1+AI1. (8.24b) Then the input impedance at port 2, with port 1 terminated in Zi1, can be found as Zin2=−V2 I2=−DV1−BI1 −CV1+AI1=DZ i1+B CZ i1+A,( 8.25) since V1=− Zi1I1(circuit of Figure 8.7). We desire that Zin1=Zi1andZin2=Zi2, so (8.23) and (8.25) give two equations for the image impedances: Zi1(CZ i2+D)=AZi2+B, (8.26a) Zi1D−B=Zi2(A−CZ i1). (8.26b) Solving for Zi1andZi2gives Zi1=/radicalbigg AB CD, (8.27a) Zi2=/radicalbigg BD AC, (8.27b) with Zi2=DZ i1/A. If the network is symmetric, then A=DandZi1=Zi2as expected. Now consider the voltage transfer function for a two-port network terminated in its image impedances. With reference to Figure 8.8 and (8.24a), the output voltage at port 2 can be expressed as V2=DV1−BI1=/parenleftbigg D−B Zi1/parenrightbigg V1 (8.28) I2 I1 V2Zi2+ –V1Zi1+ –A CB D Zi1 Zi22V1 FIGURE 8.8 A two-port network terminated in its image impedances and driven with a voltage generator. c08MicrowaveFilters Pozar August 25, 2011 18:16 390 Chapter 8: Microwave Filters (since we now have V1=I1Zi1), so the voltage ratio is V2 V1=D−B Zi1=D−B/radicalbigg CD AB=/radicalbigg D A(√ AD−√ BC). (8.29a) Similarly, the current ratio is I2 I1=−CV1 I1+A=−CZ i1+A=/radicalbigg A D(√ AD−√ BC). (8.29b) The factor√D/Aoccurs in reciprocal positions in (8.29a) and (8.29b), and so can be interpreted as a transformer turns ratio. Apart from this factor, we can define a propagationfactor for the network as e −γ=√ AD−√ BC,( 8.30) withγ=α+jβas usual. Since eγ=1/(√ AD−√ BC)=(AD−BC)/(√ AD−√ BC)=√ AD+√ BC and cosh γ=(eγ+e−γ)/2, we also have that coshγ=√ AD.( 8.31) Two important types of two-port networks are the T and πcircuits, which can be made in symmetric form. Table 8.1 lists the image impedances and propagation factors, along with other useful parameters, for these two networks. Constant- kFilterSections We can now develop low-pass and high-pass filter sections. First consider the T-network shown in Figure 8.9. Intuitively, we can see that this is a low-pass filter network becausethe series inductors and shunt capacitor tend to block high-frequency signals while passing low-frequency signals. Comparing with the results given in Table 8.1, we have Z 1=jωL andZ2=1/jωC, so the image impedance is ZiT=/radicalbigg L C/radicalBigg 1−ω2LC 4.( 8.32) If we define a cutoff frequency, ωc,a s ωc=2√ LC(8.33) L/2 L/2 CC /2 C/2L (a) (b) FIGURE 8.9 Low-pass constant-k filter sections in T and πforms. (a) T-section. (b) π-section. c08MicrowaveFilters Pozar August 25, 2011 18:16 8.2 Filter Design by the Image Parameter Method 391 and a nominal characteristic impedance, R0,a s R0=/radicalbigg L C=k,( 8.34) where kis a constant, then we can rewrite (8.32) as ZiT=R0/radicalBigg 1−ω2 ω2c.( 8.35) Then ZiT=R0forω=0. The propagation factor, also from Table 8.1, is eγ=1−2ω2 ω2c+2ω ωc/radicalBigg ω2 ω2c−1.( 8.36) Now consider two frequency regions: 1. For ω<ω c: This is the passband of the filter section. Equation (8.35) shows that ZiTis real, and (8.36) shows that γis imaginary, since ω2/ω2 c−1 is negative and |eγ|=1: |eγ|2=/parenleftBigg 1−2ω2 ω2c/parenrightBigg2 +4ω2 ω2c/parenleftBigg 1−ω2 ω2c/parenrightBigg =1. 2. For ω>ω c: This is the stopband of the filter section. Equation (8.35) shows that ZiT is imaginary, and (8.36) shows that eγis real and −1<eγ<0 (as seen from the limits as ω→ωcandω→∞). The attenuation rate for ω/greatermuchωcis 40 dB/decade. Typical phase and attenuation constants are sketched in Figure 8.10. Observe that the attenuation, α, is zero or relatively small near the cutoff frequency, although α→∞ as TABLE 8.1 Image Parameters for T- and π-Networks Z1/2 Z1/2 Z2Z1 2Z2 2Z2 T-Network π-Network ABCD parameters: ABCD parameters: A=1+Z1/2Z2 A=1+Z1/2Z2 B=Z1+Z2 1/4Z2 B=Z1 C=1/Z2 C=1/Z2+Z1/4Z2 2 D=1+Z1/2Z2 D=1+Z1/2Z2 Zparameters: Yparameters: Z11=Z22=Z2+Z1/2 Y11=Y22=1/Z1+1/2Z2 Z12=Z21=Z2 Y12=Y21=1/Z1 Image impedance: Image impedance: ZiT=/radicalBig Z1Z2/radicalBig 1+Z1/4Z2Ziπ=/radicalBig Z1Z2//radicalBig 1+Z1/4Z2=Z1Z2/ZiT Propagation constant: Propagation constant: eγ=1+Z1/2Z2+/radicalBig Z1/Z2+Z2 1/4Z2 2eγ=1+Z1/2Z2+/radicalBig Z1/Z2+Z2 1/4Z2 2 c08MicrowaveFilters Pozar August 25, 2011 18:16 392 Chapter 8: Microwave Filters /H9275 0Passband Stopband/H9251, /H9252 /H9266 /H9252/H9251 /H9275c FIGURE 8.10 Typical passband and stopband characteristics of the low-pass constant- ksections of Figure 8.9. ω→∞ . This type of filter is known as a constant -klow-pass prototype. There are only two parameters to choose ( LandC), which are determined by ωc, the cutoff frequency, andR0, the image impedance at zero frequency. The above results are valid only when the filter section is terminated in its image impedance at both ports. This is a major weakness of the design because the image imped- ance is a function of frequency, and is not likely to match a given source or load impedance. This disadvantage, as well as the fact that the attenuation is rather low near cutoff, can be remedied with the modified m-derived sections to be discussed shortly. For the low-pass π-network of Figure 8.9, we have that Z1=jωLandZ2=1/jωC, so the propagation factor is the same as that for the low-pass T-network. The cutoff fre- quency, ωc, and nominal characteristic impedance, R0, are the same as the correspond- ing quantities for the T-network as given in (8.33) and (8.34). At ω=0w eh a v et h a t ZiT=Ziπ=R0, where Ziπis the image impedance of the low-pass π-network, but ZiT andZiπare generally not equal at other frequencies. High-pass constant-k sections are shown in Figure 8.11; we see that the positions of the inductors and capacitors are reversed from those in the low-pass prototype. The design equations are easily shown to be R0=/radicalbigg L C. (8.37) ωc=1 2√ LC. (8.38) 2C 2C 2L 2L LC (a) (b) FIGURE 8.11 High-pass constant-k filter sections in T and πforms. (a) T-section. (b) π-section. c08MicrowaveFilters Pozar August 25, 2011 18:16 8.2 Filter Design by the Image Parameter Method 393 1 – m2 4m(a)Z1/2 Z1/2 Z2 (c)mZ1/2 mZ1/2 Z2/m Z1(b)'' 'Z1/2 Z1/2 Z2 FIGURE 8.12 Development of an m-derived filter section from a constant- ksection. (a) Constant- ksection. (b) General m-derived section. (c) Final m-derived section. m-DerivedFilterSections We have seen that the constant-k filter section suffers from the disadvantages of a relatively slow attenuation rate past cutoff, and a nonconstant image impedance. The m-derived filter section is a modification of the constant- ksection designed to overcome these problems. As shown in Figure 8.12a, b the impedances Z1andZ2in a constant-k T-section are replaced with Z/prime 1andZ/prime 2, and we let Z/prime 1=mZ 1.( 8.39) Then we choose Z/prime 2to obtain the same value of ZiTas for the constant-k section. Thus, from Table 8.1, ZiT=/radicalBigg Z1Z2+Z2 1 4=/radicalBigg Z/prime 1Z/prime 2+Z/prime2 1 4=/radicalBigg mZ 1Z/prime 2+m2Z2 1 4.( 8.40) Solving for Z/prime 2gives Z/prime 2=Z2 m+Z1 4m−mZ 1 4=Z2 m+(1−m2) 4mZ1.( 8.41) Because the impedances Z1and Z2represent reactive elements, Z/prime 2represents two el- ements in series, as indicated in Figure 8.12c. Note that m=1 reduces to the original constant- ksection. For a low-pass filter, we have Z1=jωLandZ2=1/jωC. Then (8.39) and (8.41) give the m-derived components as Z/prime 1=jωLm, (8.42a) Z/prime 2=1 jωCm+(1−m2) 4mjωL, (8.42b) which results in the circuit of Figure 8.13. Now consider the propagation factor for the c08MicrowaveFilters Pozar August 25, 2011 18:16 394 Chapter 8: Microwave Filters mL/2 mL/2 mC (a) (b)1 – m2 4mL 1 – m24mCL/m2C/m 2C/m FIGURE 8.13 m-Derived filter sections. (a) Low-pass T-section. (b) High-pass T-section. m-derived section. From Table 8.1, eγ=1+Z/prime 1 2Z/prime 2+/radicalBigg Z/prime 1 Z/prime 2/parenleftbigg 1+Z/prime 1 4Z/prime 2/parenrightbigg .( 8.43) For the low-pass m-derived filter, Z/prime 1 Z/prime 2=jωLm (1/jωCm)+jωL(1−m2)/4m=−(2ω m/ωc)2 1−(1−m2)(ω/ω c)2, where ωc=2/√ LCas before. Then, 1+Z/prime 1 4Z/prime 2=1−(ω/ω c)2 1−(1−m2)(ω/ω c)2. If we restrict 0 <m<1, then these results show that eγis real and |eγ|>1f o rω>ω c. Thus the stopband begins at ω=ωc, as for the constant- ksection. However, when ω= ω∞, where ω∞=ωc√ 1−m2,( 8.44) the denominators vanish and eγbecomes infinite, implying infinite attenuation. Physically, this pole in the attenuation characteristic is caused by the resonance of the series LCres- onator in the shunt arm of the T; this is easily verified by showing that the resonant fre-quency of this LCresonator is ω ∞. Note that (8.44) indicates that ω∞>ω c, so infinite attenuation occurs after the cutoff frequency, ωc, as illustrated in Figure 8.14. The position of the pole at ω∞can be controlled with the value of m. /H9275 0/H9251 /H9275c /H9275/H11009m-derived attenuationComposite response Constant-k attenuation FIGURE 8.14 Typical attenuation responses for constant- k,m-derived, and composite filters. c08MicrowaveFilters Pozar August 25, 2011 18:16 8.2 Filter Design by the Image Parameter Method 395 Z2/mZ2/m (1 – m2) 4mZ1(1 – m2) 4mZ1mZ1/2 mZ1/2 mZ1/2 mZ1/2 (a) (b)2Z2 m 2(1 – m2)Z1 4mmZ1 FIGURE 8.15 Development of an m-derived π-section. (a) Infinite cascade of m-derived T-sections. (b) A de-embedded π-equivalent. We now have a very sharp cutoff response, but one problem with the m-derived section is that its attenuation decreases for ω>ω ∞. Since it is often desirable to have infinite attenuation as ω→∞ ,t h em -derived section can be cascaded with a constant- ksection to give the composite attenuation response shown in Figure 8.14. The m-derived T-section was designed so that its image impedance was identical to that of the constant- ksection (independent of m), so we still have the problem of a non- constant image impedance. However, the image impedance of the π-equivalent will depend onm, and this extra degree of freedom can be used to design an optimum matching section. The easiest way to obtain the corresponding π-section is to consider it as a piece of an infinite cascade of m-derived T-sections, as shown in Figure 8.15. Then the image impedance of this network is, using the results of Table 8.1 and (8.35), Ziπ=Z/prime 1Z/prime 2 ZiT=Z1Z2+Z2 1(1−m2)/4 R0/radicalbig 1−(ω/ω c)2.( 8.45) Now Z1Z2=L/C=R2 0andZ2 1=−ω2L2=−4R2 0(ω/ω c)2, so (8.45) reduces to Ziπ=1−(1−m2)(ω/ω c)2 /radicalbig 1−(ω/ω c)2R0.( 8.46) Since this impedance is a function of m, we can choose mto minimize the variation of Ziπ over the passband of the filter. Figure 8.16 shows this variation with frequency for several values of m;av a l u eo f m=0.6 generally gives the best results. This type of m-derived section can then be used at the input and output of the filter to provide a nearly constant impedance match to and from R0. However, the image impedance of the constant- kandm-derived T-sections, ZiT, does not match Ziπ; this problem can be surmounted by bisecting the π-sections, as shown in Figure 8.17. The image impedances c08MicrowaveFilters Pozar August 25, 2011 18:16 396 Chapter 8: Microwave Filters /H9275c /H9275 0Zi/H9266 R0 m = 0.3m = 0.6m = 1 FIGURE 8.16 Variation of Ziπin the passband of a low-pass m-derived section for various values ofm. of this circuit are Zi1=ZiTandZi2=Ziπ, which can be shown by finding its ABCD parameters: A=1+Z/prime 1 4Z/prime 2, (8.47a) B=Z/prime 1 2, (8.47b) C=1 2Z/prime 2, (8.47c) D=1, (8.47d) and then using (8.27) for Zi1andZi2: Zi1=/radicalBigg Z/prime 1Z/prime 2+Z/prime2 1 4=ZiT, (8.48a) Zi2=/radicalBigg Z/prime 1Z/prime 2 1+Z/prime 1/4Z/prime 2=Z/prime 1Z/prime 2 ZiT=Ziπ, (8.48b) where (8.40) has been used for ZiT. CompositeFilters By combining in cascade the constant-k ,m-derived sharp cutoff and the m-derived match- ing sections we can realize a filter with the desired attenuation and matching properties. Z1/2 Zi1 = ZiT Zi2 = Zi/H9266 2Z2' ' FIGURE 8.17 A bisected π- s e c t i o nu s e dt om a t c h ZiπtoZiT. c08MicrowaveFilters Pozar September 12, 2011 21:42 8.2 Filter Design by the Image Parameter Method 397 m = 0.6 1 2/H9266m = 0.6m < 0.6 T1 2/H9266Constant- k T~R0 ~R0Matching sectionMatching sectionHigh-f cutoffSharp cutoff ZiT ZiT ZiT FIGURE 8.18 The final four-stage composite filter. This type of design is called a composite filter, and is shown in Figure 8.18. The sharp- cutoff section, with m<0.6, places an attenuation pole near the cutoff frequency to provide a sharp attenuation response; the constant- ksection provides high attenuation further into the stopband. The bisected-π sections at the ends of the filter match the nominal source and load impedance, R0, to the internal image impedances, ZiT, of the constant- kand m-derived sections. Table 8.2 summarizes the design equations for low- and high-pass TABLE 8.2 Summary of Composite Filter Design Low-Pass High-Pass (1 – m2) 2mL(1 – m2) 2mLmL/2 mL/2 mC 2L/m2C/m 2C/mL/2 L/2 2C 2C mC(1 – m2) 4mLmL/2 mL/2 mC 2 R0 R0– – ZiT(1 – m2)2mC(1 – m2)2mC(1 – m2)4mC 2L/m 2L/m2C/m 2C/m R0 R0– – ZiTC LConstant-k T section Constant-k T section R0 = L /C ωc = 2/ LCL = 2R0/c C = 2/cR0 m-derived T section L, C Same as constant-k section m =1 – (c/∞)2 for sharp-cutoff 0.6 for matching Bisected- matching sectionL, C Same as constant-k section m =1 – (∞/c)2 for sharp-cutoff 0.6 for matching Bisected- matching sectionR0 = L /C c = 1/2 LCL = R0/2c C = 1/2cR0 m-derived T section c08MicrowaveFilters Pozar August 25, 2011 18:16 398 Chapter 8: Microwave Filters composite filters; notice that once the cutoff frequency and impedance are specified, there is only one degree of freedom (the value of mfor the sharp-cutoff section) left to control the filter response. The following example illustrates the design procedure. EXAMPLE 8.2 LOW-PASS COMPOSITE FILTER DESIGN Design a low-pass composite filter with a cutoff frequency of 2 MHz and imped-ance of 75 /Omega1. Place the infinite attenuation pole at 2.05 MHz, and plot the fre- quency response from 0 to 4 MHz. Solution All of the component values can be found from Table 8.2. For the constant- k section L=2R 0 ωc=11.94 µH, C=2 R0ωc=2.122 nF. For the m-derived sharp-cutoff section m=/radicalBigg 1−/parenleftBigfc f∞/parenrightBig2 =0.2195, mL 2=1.310 µH, mC=465.8p F , 1−m2 4mL=12.94 µH. For the m=0.6 matching sections mL 2=3.582 µH, mC 2=636.5p F , 1−m2 2mL=6.368 µH. The completed filter circuit is shown in Figure 8.19; the series pairs of induc- tors between the sections have been combined. Figure 8.20 shows the resultingfrequency response for |S 12|. Note the sharp dip at f=2.05 MHz due to the m=0.2195 section, and the pole at 2.50 MHz, which is due to the m=0.6 matching sections. ■ 3.582 /H9262H 1.310 /H9262H 1.310 /H9262H 3.582 /H9262H 5.97 /H9262H 5.97 /H9262H 6.368 /H9262H 636.5 pF12.94 /H9262H 465.8 pF6.368 /H9262H 636.5 pF2122 pF Matching Constant-km -derived Matching FIGURE 8.19 Low-pass composite filter for Example 8.2. c08MicrowaveFilters Pozar August 25, 2011 18:16 8.3 Filter Design by the Insertion Loss Method 399 01234–60–50–40–30–20–100 Pole due to = 0.6 section Pole due to = 0.2195 section⎜S12⎜ dB Frequency (MHz)m m FIGURE 8.20 Frequency response for the low-pass filter of Example 8.2. 8.3FILTERDESIGNBYTHEINSERTIONLOSSMETHOD A perfect filter would have zero insertion loss in the passband, infinite attenuation in the stopband, and a linear phase response (to avoid signal distortion) in the passband. Of course,such filters do not exist in practice, so compromises must be made; herein lies the art of filter design. The image parameter method of the previous section may yield a usable filter response for some applications, but there is no methodical way of improving the design. The inser- tion loss method, however, allows a high degree of control over the passband and stop- band amplitude and phase characteristics, with a systematic way to synthesize a desiredresponse. The necessary design trade-offs can be evaluated to best meet the application requirements. If, for example, a minimum insertion loss is most important, a binomial re- sponse could be used; a Chebyshev response would satisfy a requirement for the sharpestcutoff. If it is possible to sacrifice the attenuation rate, a better phase response can be obtained by using a linear phase filter design. In addition, in all cases, the insertion loss method allows filter performance to be improved in a straightforward manner, at the ex-pense of a higher order filter. For the filter prototypes to be discussed below, the order of the filter is equal to the number of reactive elements. CharacterizationbyPowerLossRatio In the insertion loss method a filter response is defined by its insertion loss, or power loss ratio, P LR: PLR=Power available from source Power delivered to load=Pinc Pload=1 1−|/Gamma1(ω)|2.( 8.49) Observe that this quantity is the reciprocal of |S12|2if both load and source are matched. The insertion loss (IL) in dB is IL=10 log PLR.( 8.50) c08MicrowaveFilters Pozar August 25, 2011 18:16 400 Chapter 8: Microwave Filters From Section 4.1 we know that |/Gamma1(ω)|2is an even function of ω; therefore it can be expressed as a polynomial in ω2. Thus we can write |/Gamma1(ω)|2=M(ω2) M(ω2)+N(ω2),( 8.51) where Mand Nare real polynomials in ω2. Substituting this form in (8.49) gives the following: PLR=1+M(ω2) N(ω2).( 8.52) For a filter to be physically realizable its power loss ratio must be of the form in (8.52). Notice that specifying the power loss ratio simultaneously constrains the magnitude of the reflection coefficient, |/Gamma1(ω)|. We now discuss some practical filter responses. Maximally flat: This characteristic is also called the binomial orButterworth response, and is optimum in the sense that it provides the flattest possible passband response for a givenfilter complexity, or order. For a low-pass filter, it is specified by P LR=1+k2/parenleftbiggω ωc/parenrightbigg2N ,( 8.53) where Nis the order of the filter and ωcis the cutoff frequency. The passband extends fromω=0t oω=ωc; at the band edge the power loss ratio is 1 +k2. If we choose this as the −3 dB point, as is common, we have k=1, which we will assume from now on. Forω>ω c, the attenuation increases monotonically with frequency, as shown in Figure 8.21. For ω/greatermuchωc,PLR/similarequalk2(ω/ω c)2N, which shows that the insertion loss increases at the rate of 20 NdB/decade. Like the binomial response for multisection quarter-wave matching transformers, the first (2N−1)derivatives of (8.53) are zero at ω=0. Equal ripple: If a Chebyshev polynomial is used to specify the insertion loss of an Nth- order low-pass filter as PLR=1+k2T2 N/parenleftbiggω ωc/parenrightbigg ,( 8.54) then a sharper cutoff will result, although the passband response will have ripples of ampli- tude 1 +k2, as shown in Figure 8.21, since TN(x)oscillates between ±1f o r |x|21. Thus, FIGURE 8.21 Maximally flat and equal-ripple low-pass filter responses ( N=3). c08MicrowaveFilters Pozar August 25, 2011 18:16 8.3 Filter Design by the Insertion Loss Method 401 PLR /H9275//H9275c012Amin Amax FIGURE 8.22 Elliptic function low-pass filter response. k2determines the passband ripple level. For large x,TN(x)/similarequal1 2(2x)N,s of o rω /greatermuchωcthe insertion loss becomes PLR/similarequalk2 4/parenleftbigg2ω ωc/parenrightbigg2N , which also increases at the rate of 20 NdB/decade. However, the insertion loss for the Chebyshev case is (22N)/4 greater than the binomial response at any given frequency where ω/greatermuchωc. Elliptic function : The maximally flat and equal-ripple responses both have monotonically increasing attenuation in the stopband. In many applications it is adequate to specify a min-imum stopband attenuation, in which case a better cutoff rate can be obtained. Such filters are called elliptic function filters [3], and they have equal-ripple responses in the passband as well as in the stopband, as shown in Figure 8.22. The maximum attenuation in the pass-band, A max, can be specified, as well as the minimum attenuation in the stopband, Amin. Elliptic function filters are difficult to synthesize, so we will not consider them further; the interested reader is referred to reference [3]. Linear phase: The above filters specify the amplitude response, but in some applications (such as multiplexing filters for communication systems) it is important to have a linearphase response in the passband to avoid signal distortion. Since a sharp-cutoff response is generally incompatible with a good phase response, the phase response of a filter must be deliberately synthesized, usually resulting in an inferior attenuation characteristic. A linearphase characteristic can be achieved with the following phase response: φ(ω)=Aω/bracketleftBigg 1+p/parenleftbiggω ωc/parenrightbigg2N/bracketrightBigg ,( 8.55) where φ(ω) is the phase of the voltage transfer function of the filter, and pis a constant. A related quantity is the group delay, defined as τd=dφ dω=A/bracketleftBigg 1+p(2N+1)/parenleftbiggω ωc/parenrightbigg2N/bracketrightBigg ,( 8.56) which shows that the group delay for a linear phase filter is a maximally flat function. More general filter specifications can be obtained, but the above cases are the most common. We will next discuss the design of low-pass filter prototypes that are normalized in terms of impedance and frequency; this normalization simplifies the design of filters c08MicrowaveFilters Pozar August 25, 2011 18:16 402 Chapter 8: Microwave Filters Filter specificationsLow-pass prototype designScaling and conversionImplementation FIGURE 8.23 The process of filter design by the insertion loss method. for arbitrary frequency, impedance, and type (low-pass, high-pass, bandpass, or bandstop). The low-pass prototypes are then scaled to the desired frequency and impedance, and the lumped-element components replaced with distributed circuit elements for implementation at microwave frequencies. This design process is illustrated in Figure 8.23. MaximallyFlatLow-PassFilterPrototype Consider the two-element low-pass filter prototype shown in Figure 8.24; we will derive the normalized element values, LandC, for a maximally flat response. We assume a source impedance of 1 /Omega1, and a cutoff frequency ωc=1 rad/sec. From (8.53), the desired power loss ratio will be, for N=2, PLR=1+ω4.( 8.57) The input impedance of this filter is Zin=jωL+R(1−jωRC) 1+ω2R2C2.( 8.58) Because /Gamma1=Zin−1 Zin+1, the power loss ratio can be written as PLR=1 1−|/Gamma1|2=1 1−/bracketleftbig/parenleftbig Zin−1/parenrightbig //parenleftbig Zin+1/parenrightbig/bracketrightbig/bracketleftbig/parenleftbig Z∗ in−1/parenrightbig //parenleftbig Z∗ in+1/parenrightbig/bracketrightbig=|Zin+1|2 2/parenleftbig Zin+Z∗ in/parenrightbig. (8.59) Now Zin+Z∗ in=2R 1+ω2R2C2, |Zin+1|2=/parenleftbiggR 1+ω2R2C2+1/parenrightbigg2 +/parenleftBigg ωL−ωCR2 1+ω2R2C2/parenrightBigg2 , 1 L CR Zin FIGURE 8.24 Low-pass filter prototype, N=2. c08MicrowaveFilters Pozar August 25, 2011 18:16 8.3 Filter Design by the Insertion Loss Method 403 so (8.59) becomes PLR=1+ω2R2C2 4R⎡ ⎣/parenleftbiggR 1+ω2R2C2+1/parenrightbigg2 +/parenleftBigg ωL−ωCR2 1+ω2R2C2/parenrightBigg2⎤ ⎦ =1 4R(R2+2R+1+R2ω2C2+ω2L2+ω4L2C2R2−2ω2LCR2) =1+1 4R[(1−R)2+(R2C2+L2−2LCR2)ω2+L2C2R2ω4]. (8.60) Observe that this expression is a polynomial in ω2. Comparing to the desired response of (8.57) shows that R=1, since PLR=1f o rω=0. In addition, the coefficient of ω2must vanish, so C2+L2−2LC=(C−L)2=0, orL=C. Then, for the coefficient of ω4to be unity, we must have 1 4L2C2=1 4L4=1, or L=C=√ 2. In principle, this procedure can be extended to find the element values for filters with an ar- bitrary number of elements, N, but clearly this is not practical for large N. For a normalized low-pass design, where the source impedance is 1 /Omega1and the cutoff frequency is ωc=1 rad/sec, however, the element values for the ladder-type circuits of Figure 8.25 can be tabu- lated [1]. Table 8.3 gives such element values for maximally flat low-pass filter prototypes forN=1 to 10. (Notice that the values for N=2 agree with the above analytical so- lution.) These data can be used with either of the ladder circuits of Figure 8.25 in the following way. The element values are numbered from g0at the generator impedance to gN+1at the load impedance for a filter having Nreactive elements. The elements alternate C3 = g3 C1 = g1 C2 = g2 G0 = g0 = 1R0 = g0 = 1 L3 = g3 L1 = g1L2 = g2 gN+1 gN+1(a) (b) FIGURE 8.25 Ladder circuits for low-pass filter prototypes and their element definitions. (a) Pro- totype beginning with a shunt element. (b) Prototype beginning with a series element. c08MicrowaveFilters Pozar September 14, 2011 7:27 404 Chapter 8: Microwave Filters TABLE 8.3 Element Values for Maximally Flat Low-Pass Filter Prototypes ( g0=1, ωc=1,N=1 to 10) Ng 1 g2 g3 g4 g5 g6 g7 g8 g9 g10 g11 1 2.0000 1.0000 2 1.4142 1.4142 1.00003 1.0000 2.0000 1.0000 1.00004 0.7654 1.8478 1.8478 0.7654 1.00005 0.6180 1.6180 2.0000 1.6180 0.6180 1.00006 0.5176 1.4142 1.9318 1.9318 1.4142 0.5176 1.00007 0.4450 1.2470 1.8019 2.0000 1.8019 1.2470 0.4450 1.00008 0.3902 1.1111 1.6629 1.9615 1.9615 1.6629 1.1111 0.3902 1.00009 0.3473 1.0000 1.5321 1.8794 2.0000 1.8794 1.5321 1.0000 0.3473 1.0000 10 0.3129 0.9080 1.4142 1.7820 1.9754 1.9754 1.7820 1.4142 0.9080 0.3129 1.0000 Source: Reprinted from G. L. Matthaei, L. Young, and E. M. T. Jones, Microwave Filters, Impedance-Matching Networks, and Coupling Structures , Artech House, Dedham, Mass., 1980, with permission. between series and shunt connections, and gkhas the following definition: g0=/braceleftbigggenerator resistance (network of Figure 8.25a) generator conductance (network of Figure 8.25b) gk (k=1t o N)=/braceleftbigginductance for series inductors capacitance for shunt capacitors gN+1=/braceleftbiggload resistance if gNis a shunt capacitor load conductance if gNis a series inductor Then the circuits of Figure 8.25 can be considered as the dual of each other, and both will give the same filter response. Finally, as a matter of practical design procedure, it will be necessary to determine the size, or order, of the filter. This is usually dictated by a specification on the insertion loss at some frequency in the stopband of the filter. Figure 8.26 shows the attenuation charac- teristics for various Nversus normalized frequency. If a filter with N>10 is required, a good result can usually be obtained by cascading two designs of lower order. Equal-RippleLow-PassFilterPrototype For an equal-ripple low-pass filter with a cutoff frequency ωc=1 rad/sec, the power loss ratio from (8.54) is PLR=1+k2T2 N(ω), (8.61) where 1 +k2is the ripple level in the passband. Since the Chebyshev polynomials have the property that TN(0)=/braceleftbigg 0f o r Nodd, 1f o r Neven, equation (8.61) shows that the filter will have a unity power loss ratio at ω=0f o r Nodd, but a power loss ratio of 1 +k2atω=0f o r Neven. Thus, there are two cases to consider, depending on N. c08MicrowaveFilters Pozar August 25, 2011 18:16 8.3 Filter Design by the Insertion Loss Method 405 0.1 0.2 0.3 0.5 0.7 1.0 2.0 3.0 5.0 7.0 10010203040506070Attenuation (dB) /H9275 /H9275c– 1n = 1n = 4n = 5 n = 2n = 3n = 7 n = 6n = 8n = 9n = 10 FIGURE 8.26 Attenuation versus normalized frequency for maximally flat filter prototypes. Adapted from G. L. Matthaei, L. Young, and E. M. T. Jones, Microwave Filters, Impedance- Matching Networks, and Coupling Structures, Artech House, Dedham, Mass., 1980, withpermission. For the two-element filter of Figure 8.24, the power loss ratio is given in terms of the component values in (8.60). From (5.56b), we see that T2(x)=2x2−1, so equating (8.61) to (8.60) gives 1+k2(4ω4−4ω2+1)=1+1 4R[(1−R)2+(R2C2+L2−2LCR2)ω2+L2C2R2ω4], (8.62) which can be solved for R,L, and Cif the ripple level (as determined by k2)is known. Thus, at ω=0w eh a v et h a t k2=(1−R)2 4R, or R=1+2k2±2k/radicalbig 1+k2 (forNeven). (8.63) Equating coefficients of ω2andω4yields the additional relations 4k2=1 4RL2C2R2, −4k2=1 4R/parenleftbig R2C2+L2−2LCR2/parenrightbig , which can be used to find LandC. Note that (8.63) gives a value for Rthat is not unity, so there will be an impedance mismatch if the load has a unity (normalized) impedance; this c08MicrowaveFilters Pozar September 14, 2011 7:27 406 Chapter 8: Microwave Filters TABLE 8.4 Element Values for Equal-Ripple Low-Pass Filter Prototypes ( g0=1,ω c= 1,N=1 to 10, 0.5 dB and 3.0 dB ripple) 0.5 dB Ripple Ng 1 g2 g3 g4 g5 g6 g7 g8 g9 g10 g11 1 0.6986 1.00002 1.4029 0.7071 1.98413 1.5963 1.0967 1.5963 1.00004 1.6703 1.1926 2.3661 0.8419 1.98415 1.7058 1.2296 2.5408 1.2296 1.7058 1.00006 1.7254 1.2479 2.6064 1.3137 2.4758 0.8696 1.98417 1.7372 1.2583 2.6381 1.3444 2.6381 1.2583 1.7372 1.00008 1.7451 1.2647 2.6564 1.3590 2.6964 1.3389 2.5093 0.8796 1.98419 1.7504 1.2690 2.6678 1.3673 2.7239 1.3673 2.6678 1.2690 1.7504 1.0000 10 1.7543 1.2721 2.6754 1.3725 2.7392 1.3806 2.7231 1.3485 2.5239 0.8842 1.9841 3.0 dB Ripple Ng 1 g2 g3 g4 g5 g6 g7 g8 g9 g10 g11 1 1.9953 1.00002 3.1013 0.5339 5.80953 3.3487 0.7117 3.3487 1.00004 3.4389 0.7483 4.3471 0.5920 5.80955 3.4817 0.7618 4.5381 0.7618 3.4817 1.00006 3.5045 0.7685 4.6061 0.7929 4.4641 0.6033 5.80957 3.5182 0.7723 4.6386 0.8039 4.6386 0.7723 3.5182 1.00008 3.5277 0.7745 4.6575 0.8089 4.6990 0.8018 4.4990 0.6073 5.80959 3.5340 0.7760 4.6692 0.8118 4.7272 0.8118 4.6692 0.7760 3.5340 1.0000 10 3.5384 0.7771 4.6768 0.8136 4.7425 0.8164 4.7260 0.8051 4.5142 0.6091 5.8095 Source: Reprinted from G. L. Matthaei, L. Young, and E. M. T. Jones, Microwave Filters, Impedance-Matching Networks, and Coupling Structures , Artech House, Dedham, Mass.,1980, with permission. can be corrected with a quarter-wave transformer, or by using an additional filter element to make Nodd. For odd N, it can be shown that R=1. (This is because there is a unity power loss ratio at ω=0f o r Nodd.) Tables exist for designing equal-ripple low-pass filters with a normalized source im- pedance and cutoff frequency (ω/prime c=1r a d/sec) [1], and these can be applied to either of the ladder circuits of Figure 8.25. This design data depends on the specified passband ripple level; Table 8.4 lists element values for normalized low-pass filter prototypes having 0.5 or 3.0 dB ripple for N=1 to 10. Notice that the load impedance gN+1/negationslash=1 for even N.I ft h e stopband attenuation is specified, the curves in Figure 8.27 can be used to determine thenecessary value of Nfor these ripple values. LinearPhaseLow-PassFilterPrototypes Filters having a maximally flat time delay, or a linear phase response, can be designed in the same way, but things are somewhat more complicated because the phase of the volt- age transfer function is not as simply expressed as is its amplitude. Design values have c08MicrowaveFilters Pozar August 25, 2011 18:16 8.3 Filter Design by the Insertion Loss Method 407 0.01 0.02 0.03 0.05 0.07 0.10 0.20 0.30 0.50 0.70 1.0 2.0 3.0 5.0 7.0 10.0010203040506070Attenuation (dB) /H9275 /H9275c– 1n = 1n = 2n = 3n = 7 n = 6n = 8n = 9n = 10 (b)0.01 0.02 0.03 0.05 0.07 0.10 0.20 0.30 0.50 0.70 1.0 2.0 3.0 5.0 7.0 10.0010203040506070Attenuation (dB) /H9275 /H9275c– 1n = 2n = 3n = 7 n = 6n = 8n = 9n = 10 (a)n = 5 n = 4 n = 1 n = 5 n = 4 FIGURE 8.27 Attenuation versus normalized frequency for equal-ripple filter prototypes. (a) 0.5 dB ripple level. (b) 3.0 dB ripple level. Adapted from G. L. Matthaei, L. Young, and E. M. T. Jones, Microwave Filters, Impedance- Matching Networks, and Coupling Structures, Artech House, Dedham, Mass., 1980, withpermission. c08MicrowaveFilters Pozar September 14, 2011 7:27 408 Chapter 8: Microwave Filters TABLE 8.5 Element Values for Maximally Flat Time Delay Low-Pass Filter Prototypes (g0=1,ω c=1,N=1t o1 0 ) Ng 1 g2 g3 g4 g5 g6 g7 g8 g9 g10 g11 1 2.0000 1.0000 2 1.5774 0.4226 1.00003 1.2550 0.5528 0.1922 1.00004 1.0598 0.5116 0.3181 0.1104 1.00005 0.9303 0.4577 0.3312 0.2090 0.0718 1.00006 0.8377 0.4116 0.3158 0.2364 0.1480 0.0505 1.00007 0.7677 0.3744 0.2944 0.2378 0.1778 0.1104 0.0375 1.00008 0.7125 0.3446 0.2735 0.2297 0.1867 0.1387 0.0855 0.0289 1.00009 0.6678 0.3203 0.2547 0.2184 0.1859 0.1506 0.1111 0.0682 0.0230 1.0000 10 0.6305 0.3002 0.2384 0.2066 0.1808 0.1539 0.1240 0.0911 0.0557 0.0187 1.0000 Source: Reprinted from G. L. Matthaei, L. Young, and E. M. T. Jones, Microwave Filters, Impedance-Matching Networks, and Coupling Structures , Artech House, Dedham, Mass., 1980, with permission. been derived for such filters [1], however, again for the ladder circuits of Figure 8.25, and they are given in Table 8.5 for a normalized source impedance and cutoff frequency (ω/prime c=1r a d/sec). The resulting normalized group delay in the passband will be τd= 1/ω/prime c=1 sec. 8.4FILTERTRANSFORMATIONS The low-pass filter prototypes of the previous section were normalized designs having a source impedance of Rs=1/Omega1and a cutoff frequency of ωc=1 rad/sec. Here we show how these designs can be scaled in terms of impedance and frequency, and converted to give high-pass, bandpass, or bandstop characteristics. Several examples will be presentedto illustrate the design procedure. ImpedanceandFrequencyScaling Impedance scaling: In the prototype design, the source and load resistances are unity (ex- cept for equal-ripple filters with even N, which have nonunity load resistance). A source resistance of R 0can be obtained by multiplying all the impedances of the prototype design byR0. Thus, if we let primes denote impedance scaled quantities, the new filter component values are given by L/prime=R0L, (8.64a) C/prime=C R0, (8.64b) R/prime s=R0, (8.64c) R/prime L=R0RL, (8.64d) where L,C, and RLare the component values for the original prototype. c08MicrowaveFilters Pozar August 25, 2011 18:16 8.4 Filter Transformations 409 Frequency scaling for low-pass filters: To change the cutoff frequency of a low-pass pro- totype from unity to ωcrequires that we scale the frequency dependence of the filter by the factor 1/ω c, which is accomplished by replacing ωbyω/ω c: ω←ω ωc.( 8.65) Then the new power loss ratio will be P/prime LR(ω)=PLR/parenleftbiggω ωc/parenrightbigg , where ωcis the new cutoff frequency; cutoff occurs when ω/ω c=1, orω=ωc.T h i s transformation can be viewed as a stretching, or expansion, of the original passband, as illustrated in Figure 8.28a, b. The new element values are determined by applying the substitution of (8.65) to the series reactances, jωLk, and shunt susceptances, jωCk, of the prototype filter. Thus, jXk=jω ωcLk=jωL/prime k, jBk=jω ωcCk=jωC/prime k, which shows that the new element values are given by L/prime k=Lk ωc, (8.66a) C/prime k=Ck ωc. (8.66b) When both impedance and frequency scaling are required, the results of (8.64) can be combined with (8.66) to give L/prime k=R0Lk ωc, (8.67a) C/prime k=Ck R0ωc. (8.67b) PLR 0 –1 1 /H9275 (a)PLR 0 –/H9275c /H9275c/H9275 (b)PLR 0 –/H9275c /H9275c/H9275 (c) FIGURE 8.28 Frequency scaling for low-pass filters and transformation to a high-pass response. (a) Low-pass filter prototype response for ωc=1r a d/sec. (b) Frequency scaling for low-pass response. (c) Transformation to high-pass response. c08MicrowaveFilters Pozar August 25, 2011 18:16 410 Chapter 8: Microwave Filters Low-pass to high-pass transformation : The frequency substitution ω←−ωc ω(8.68) can be used to convert a low-pass response to a high-pass response, as shown in Figure 8.28c. This substitution maps ω=0t oω =± ∞ , and vice versa; cutoff occurs whenω=±ωc. The negative sign is needed to convert inductors (and capacitors) to real- izable capacitors (and inductors). Applying (8.68) to the series reactances, jωLk, and the shunt susceptances, jωCk, of the prototype filter gives jXk=− jωc ωLk=1 jωC/prime k, jBk=− jωc ωCk=1 jωL/prime k, which shows that series inductors Lkmust be replaced with capacitors C/prime k, and shunt ca- pacitors Ckmust be replaced with inductors L/prime k. The new component values are given by C/prime k=1 ωcLk, (8.69a) L/prime k=1 ωcCk. (8.69b) Impedance scaling can be included by using (8.64) to give C/prime k=1 R0ωcLk, (8.70a) L/prime k=R0 ωcCk. (8.70b) EXAMPLE 8.3 LOW-PASS FILTER DESIGN COMPARISON Design a maximally flat low-pass filter with a cutoff frequency of 2 GHz, imped- ance of 50 /Omega1, and at least 15 dB insertion loss at 3 GHz. Compute and plot the amplitude response and group delay for f=0 to 4 GHz, and compare with an equal-ripple (3.0 dB ripple) and linear phase filter having the same order. Solution First find the required order of the maximally flat filter to satisfy the insertion loss specification at 3 GHz. We have that |ω/ω c|−1=0.5; from Figure 8.26 we see thatN=5 will be sufficient. Then Table 8.3 gives the prototype element values as g1=0.618, g2=1.618, g3=2.000, g4=1.618, g5=0.618. c08MicrowaveFilters Pozar August 25, 2011 18:16 8.4 Filter Transformations 411 RS = 50 Ω L2' L4' C1' C3' C5' RL = 50 Ω FIGURE 8.29 Low-pass, maximally flat filter circuit for Example 8.3. Then (8.67) can be used to obtain the scaled element values: C/prime 1=0.984 pF, L/prime 2=6.438 nH, C/prime 3=3.183 pF, L/prime 4=6.438 nH, C/prime 5=0.984 pF. The final filter circuit is shown in Figure 8.29; the ladder circuit of Figure 8.25a was used, but that of Figure 8.25b could have been used just as well. The component values for the equal-ripple filter and the linear phase filter, forN=5, can be determined from Tables 8.4 and 8.5. The amplitude and group delay results for these three filters are shown in Figure 8.30. These results clearlyshow the trade-offs involved with the three types of filters. The equal-ripple re- sponse has the sharpest cutoff but the worst group delay characteristics. The max- imally flat response has a flatter attenuation characteristic in the passband but aslightly lower cutoff rate. The linear phase filter has the worst cutoff rate but a very good group delay characteristic. ■ BandpassandBandstopTransformations Low-pass prototype filter designs can also be transformed to have the bandpass or bandstop responses illustrated in Figure 8.31. If ω 1andω2denote the edges of the passband, then a bandpass response can be obtained using the following frequency substitution: ω←ω0 ω2−ω1/parenleftbiggω ω0−ω0 ω/parenrightbigg =1 /Delta1/parenleftbiggω ω0−ω0 ω/parenrightbigg ,( 8.71) where /Delta1=ω2−ω1 ω0(8.72) is the fractional bandwidth of the passband. The center frequency, ω0, could be chosen as the arithmetic mean of ω1andω2, but the equations are simpler if it is chosen as the geometric mean: ω0=√ω1ω2.( 8.73) Then the transformation of (8.71) maps the bandpass characteristics of Figure 8.31b to the c08MicrowaveFilters Pozar August 25, 2011 18:16 412 Chapter 8: Microwave Filters 403020100 0 1.0 2.0 3.0 4.000.250.50.751.01.251.5 Frequency (GHz)0 1.0 2.0 3.0 4.0 Frequency (GHz) (b)(a)Group delay (nsec) Attenuation (dB)Equal-ripple N = 5Linear phase N = 5 Maximally flat N = 5 Equal-ripple Linear phaseMaximally flat FIGURE 8.30 Frequency response of the filter design of Example 8.3. (a) Amplitude response. (b) Group delay response. FIGURE 8.31 Bandpass and bandstop frequency transformations. (a) Low-pass filter prototype response for ωc=1. (b) Transformation to bandpass response. (c) Transformation to bandstop response. c08MicrowaveFilters Pozar August 25, 2011 18:16 8.4 Filter Transformations 413 low-pass response of Figure 8.31a as follows: When ω=ω0,1 /Delta1/parenleftbiggω ω0−ω0 ω/parenrightbigg =0. When ω=ω1,1 /Delta1/parenleftbiggω ω0−ω0 ω/parenrightbigg =1 /Delta1/parenleftBigg ω2 1−ω2 0 ω0ω1/parenrightBigg =−1. When ω=ω2,1 /Delta1/parenleftbiggω ω0−ω0 ω/parenrightbigg =1 /Delta1/parenleftBigg ω2 2−ω2 0 ω0ω2/parenrightBigg =1. The new filter elements are determined by using (8.71) in the expressions for the series reactance and shunt susceptances. Thus, jXk=j /Delta1/parenleftbiggω ω0−ω0 ω/parenrightbigg Lk=jωLk /Delta1ω 0−jω0Lk /Delta1ω=jωL/prime k−j1 ωC/prime k, which shows that a series inductor, Lk, is transformed to a series LCcircuit with element values L/prime k=Lk /Delta1ω 0, (8.74a) C/prime k=/Delta1 ω0Lk. (8.74b) Similarly, jBk=j /Delta1/parenleftbiggω ω0−ω0 ω/parenrightbigg Ck=jωCk /Delta1ω 0−jω0Ck /Delta1ω=jωC/prime k−j1 ωL/prime k, which shows that a shunt capacitor, Ck, is transformed to a shunt LCcircuit with element values L/prime k=/Delta1 ω0Ck, (8.74c) C/prime k=Ck /Delta1ω 0. (8.74d) The low-pass filter elements are thus converted to series resonant circuits (having a low impedance at resonance) in the series arms, and to parallel resonant circuits (having a high impedance at resonance) in the shunt arms. Notice that both series and parallel resonator elements have a resonant frequency of ω0. The inverse transformation can be used to obtain a bandstop response. Thus, ω←−/Delta1/parenleftbiggω ω0−ω0 ω/parenrightbigg−1 ,( 8.75) where /Delta1andω0have the same definitions as in (8.72) and (8.73). Then series inductors of the low-pass prototype are converted to parallel LCcircuits having element values given by L/prime k=/Delta1Lk ω0, (8.76a) C/prime k=1 ω0/Delta1Lk. (8.76b) c08MicrowaveFilters Pozar August 25, 2011 18:16 414 Chapter 8: Microwave Filters TABLE 8.6 Summary of Prototype Filter Transformations/parenleftbigg /Delta1=ω2−ω1 ω0/parenrightbigg Low-pass High-pass Bandpass Bandstop CL 1 cC1 cL ∆ 0CC 0∆1 0C∆ C∆ 0∆ 0LL 0∆1 0L∆L∆ 0 The shunt capacitor of the low-pass prototype is converted to series LCcircuits having element values given by L/prime k=1 ω0/Delta1Ck, (8.76c) C/prime k=/Delta1Ck ω0. (8.76d) The element transformations from a low-pass prototype to a high-pass, bandpass, or bandstop filter are summarized in Table 8.6. These results do not include impedance scal- ing, which can be made using (8.64). EXAMPLE 8.4 BANDPASS FILTER DESIGN Design a bandpass filter having a 0.5 dB equal-ripple response, with N=3. The center frequency is 1 GHz, the bandwidth is 10%, and the impedance is 50/Omega1. 50 Ω L2' C2'L3' C1' L1' C3' 50 Ω FIGURE 8.32 Bandpass filter circuit for Example 8.4. c08MicrowaveFilters Pozar August 25, 2011 18:16 8.5 Filter Implementation 415 0.5 0.75 1.0 1.25 1.550403020100Attenuation (dB) Frequency (GHz) FIGURE 8.33 Amplitude response for the bandpass filter of Example 8.4. Solution From Table 8.4 the element values for the low-pass prototype circuit of Figure8.25b are given as g 1=1.5963 =L1, g2=1.0967 =C2, g3=1.5963 =L3, g4=1.000 =RL. Equations (8.64) and (8.74) give the impedance-scaled and frequency-transformed element values for the circuit of Figure 8.32 as L/prime 1=L1R0 ω0/Delta1=127.0n H , C/prime 1=/Delta1 ω0L1R0=0.199 pF, L/prime 2=/Delta1R0 ω0C2=0.726 nH, C/prime 2=C2 ω0/Delta1R0=34.91 pF, L/prime 3=L3R0 ω0/Delta1=127.0n H , C/prime 3=/Delta1 ω0L3R0=0.199 pF. The resulting amplitude response is shown in Figure 8.33. ■ 8.5FILTERIMPLEMENTATION The lumped-element filter designs discussed in the previous sections generally work well at low frequencies, but two problems arise at higher RF and microwave frequencies. First,lumped-element inductors and capacitors are generally available only for a limited range of values, and can be difficult to implement at microwave frequencies. Distributed elements, c08MicrowaveFilters Pozar August 25, 2011 18:16 416 Chapter 8: Microwave Filters such as open-circuited or short-circuited transmission line stubs, are often used to approx- imate ideal lumped elements. In addition, at microwave frequencies the distances between filter components is not negligible. The first problem is treated with Richards’ transforma- tion, which can be used to convert lumped elements to transmission line sections. Kuroda’s identities can then be used to physically separate filter elements by using transmission line sections. Because such additional transmission line sections do not affect the filter response, this type of design is called redundant filter synthesis. It is possible to design microwave filters that take advantage of these sections to improve the filter response [4]; such nonredundant synthesis does not have a lumped-element counterpart. Richards’Transformation The transformation /Omega1=tanβ/lscript=tan/parenleftbiggω/lscript vp/parenrightbigg (8.77) maps the ωplane to the /Omega1plane, which repeats with a period of ω/lscript/v p=2π. This trans- formation was introduced by P. Richards [6] to synthesize an LCnetwork using open- and short-circuited transmission line stubs. Thus, if we replace the frequency variable ωwith /Omega1, we can write the reactance of an inductor as jXL=j/Omega1L=jLtanβ/lscript, (8.78a) and the susceptance of a capacitor as jBC=j/Omega1C=jCtanβ/lscript. (8.78b) These results indicate that an inductor can be replaced with a short-circuited stub of length β/lscriptand characteristic impedance L,while a capacitor can be replaced with an open-circuited stub of length β/lscriptand characteristic impedance 1/ C. A unity filter impedance is assumed. Cutoff occurs at unity frequency for a low-pass filter prototype; to obtain the same cutoff frequency for the Richards’-transformed filter, (8.77) shows that /Omega1=1=tanβ/lscript, which gives a stub length of /lscript=λ/8, where λis the wavelength of the line at the cutoff frequency, ωc. At the frequency ω0=2ωc, the lines will be λ/4 long, and an attenuation pole will occur. At frequencies away from ωc, the impedances of the stubs will no longer match the original lumped-element impedances, and the filter response will differ from the desired prototype response. In addition, the response will be periodic in frequency, repeating every 4ω c. In principle, then, Richards’ transformation allows the inductors and capacitors of a lumped-element filter to be replaced with short-circuited and open-circuited transmission line stubs, as illustrated in Figure 8.34. Since the electrical lengths of all the stubs are the same (λ/ 8a tωc), these lines are called commensurate lines. Kuroda’sIdentities The four Kuroda identities use redundant transmission line sections to achieve a more practical microwave filter implementation by performing any of the following operations: rPhysically separate transmission line stubsrTransform series stubs into shunt stubs, or vice versarChange impractical characteristic impedances into more realizable values c08MicrowaveFilters Pozar August 25, 2011 18:16 8.5 Filter Implementation 417 jXLjXL Z0 = L/H9261/8 at /H9275c S.C. jBc Z0 = jBc/H9261/8 at /H9275c O.C. 1 CL C(a) (b) FIGURE 8.34 Richards’ transformation. (a) For an inductor to a short-circuited stub. (b) For a capacitor to an open-circuited stub. The additional transmission line sections are called unit elements and are λ/8 long at ωc; the unit elements are thus commensurate with the stubs used to implement the inductors and capacitors of the prototype design. The four Kuroda identities are illustrated in Table 8.7, where each box represents a unit element, or transmission line, of the indicated characteristic impedance and length (λ/8a tωc). The inductors and capacitors represent short-circuit and open-circuit stubs, TABLE 8.7 The Four Kuroda Identities (n2=1+Z2/Z1) Z1 n2 Z1 n2Z2 n2Z2 n2 n2Z2n2Z1 1 : n2 n2 : 11Z21 Z21 n2Z2 n2Z1 Z1Z2Z2Z1 Z1 Z1 1≡ (a) ≡ (b) ≡ (c) ≡ (d) c08MicrowaveFilters Pozar August 25, 2011 18:16 418 Chapter 8: Microwave Filters l ll l O.C. shunt stubUnit elementZ1 Z2≡ Z2/n2Z1/n2 Unit element n2 = 1 + Z2/Z1S.C. series stub FIGURE 8.35 Equivalent circuits illustrating Kuroda identity (a) in Table 8.7. respectively. We will prove the equivalence of the first case, and then show how to use these identities in Example 8.5. The two circuits of identity (a) in Table 8.7 can be redrawn as shown in Figure 8.35; we will show that these two networks are equivalent by showing that their ABCD matrices are identical. From Table 4.1, the ABCD matrix of a length /lscriptof transmission line with characteristic impedance Z1is /bracketleftbigg AB CD/bracketrightbigg =/bracketleftBiggcosβ/lscript jZ1sinβ/lscript j Z1sinβ/lscript cosβ/lscript/bracketrightBigg =1√ 1+/Omega12/bracketleftBigg1 j/Omega1Z1 j/Omega1 Z11/bracketrightBigg ,( 8.79) where /Omega1=tanβ/lscript. The open-circuited shunt stub in the first circuit in Figure 8.35 has an impedance of −jZ2cotβ/lscript=− jZ2//Omega1,s ot h eABCD matrix of the entire circuit is /bracketleftbigg AB CD/bracketrightbigg L=⎡ ⎣10 j/Omega1 Z21⎤ ⎦⎡ ⎣1 j/Omega1Z1 j/Omega1 Z11⎤ ⎦1√ 1+/Omega12 =1√ 1+/Omega12⎡ ⎣1 j/Omega1Z1 j/Omega1/parenleftbigg1 Z1+1 Z2/parenrightbigg 1−/Omega12Z1 Z2⎤ ⎦. (8.80a) The short-circuited series stub in the second circuit in Figure 8.35 has an impedance ofj(Z1/n2)tanβ/lscript=j/Omega1Z1/n2,s ot h eABCD matrix of the entire circuit is /bracketleftbigg AB CD/bracketrightbigg R=⎡ ⎢⎣1 j/Omega1Z2 n2 j/Omega1n2 Z21⎤ ⎥⎦⎡ ⎣1j/Omega1Z1 n2 01⎤ ⎦1√ 1+/Omega12 =1√ 1+/Omega12⎡ ⎢⎣1j/Omega1 n2(Z1+Z2) j/Omega1n2 Z21−/Omega12Z1 Z2⎤ ⎥⎦. (8.80b) c08MicrowaveFilters Pozar August 25, 2011 18:16 8.5 Filter Implementation 419 The results in (8.80a) and (8.80b) are identical if we choose n2=1+Z2/Z1. The other identities in Table 8.7 can be proved in the same way. EXAMPLE 8.5 LOW-PASS FILTER DESIGN USING STUBS Design a low-pass filter for fabrication using microstrip lines. The specificationsinclude a cutoff frequency of 4 GHz, an impedance of 50 /Omega1, and a third-order 3 dB equal-ripple passband response. Solution From Table 8.4 the normalized low-pass prototype element values are g 1=3.3487 =L1, g2=0.7117 =C2, g3=3.3487 =L3, g4=1.0000 =RL, with the lumped-element circuit shown in Figure 8.36a. 11 1 1L1 = 3.3487 Z0 = 3.3487 Z0 = 3.3487 Z0 = 1.405L3 = 3.3487 C2 = 0.7117 (a) (b) (c)l = /H9261/8 at /H9275 = 1l l l 1 1Z0 = 3.3487 Z0 = 3.3487 Z0 = 1.405Z0 = 1 Z0 = 1 l = /H9261/8 at /H9275 = 1l l ll l FIGURE 8.36 Filter design procedure for Example 8.5. (a) Lumped-element low-pass filter pro- totype. (b) Using Richards’ transformations to convert inductors and capacitors to series and shunt stubs. (c) Adding unit elements at the ends of the filter. c08MicrowaveFilters Pozar August 25, 2011 18:16 420 Chapter 8: Microwave Filters Z0 = 217.5 Ω Z0 = 217.5 Ω Z0 = 64.9 ΩZ0 = 70.3 ΩZ0 = 64.9 Ω (e) (f)l = /H9261/8 at 4 GHzl lll l 50 Ω 217.5 Ω 217.5 Ω50 Ω50 Ω50 ΩZ0 = 4.350 Z0 = 4.350 Z0 = 1.299Z0 = 1.405Z0 = 1.299 (d)l = /H9261/8 at /H9275 = 1l lll l1 1 64.9 Ω 70.3 Ω 64.9 Ω FIGURE 8.36 Continued. (d) Applying the second Kuroda identity. (e) After impedance and fre- quency scaling. (f) Microstrip fabrication of the final filter. We now use Richards’ transformations to convert series inductors to series stubs, and shunt capacitors to shunt stubs, as shown in Figure 8.36b. According to (8.78), the characteristic impedance of a series stub (inductor) is L, and the characteristic impedance of a shunt stub (capacitor) is 1 /C. For commensurate line synthesis, all stubs are λ/8 long at ω=ωc. (It is usually most convenient to work with normalized quantities until the last step in the design.) The series stubs of Figure 8.36b would be very difficult to implement in mi- crostrip line form, so we will use one of the Kuroda identities to convert these to shunt stubs. First we add unit elements at either end of the filter, as shownin Figure 8.36c. These redundant elements do not affect filter performance since they are matched to the source and load ( Z 0=1). Then we can apply Kuroda identity (b) from Table 8.7 to both ends of the filter. In both cases we have that n2=1+Z2 Z1=1+1 3.3487=1.299. The result is shown in Figure 8.36d. Finally, we impedance and frequency scale the circuit, which simply involves multiplying the normalized characteristic impedances by 50 /Omega1and choosing the line and stub lengths to be λ/8 at 4 GHz. The final circuit is shown in Figure 8.36e, with a microstrip layout in Figure 8.36f. c08MicrowaveFilters Pozar August 25, 2011 18:16 8.5 Filter Implementation 421 0 5 10 15 2050403020100 Frequency (GHz)Attenuation (dB)Lumped elements Distributed elements FIGURE 8.37 Amplitude responses of lumped-element and distributed-element low-pass filter of Example 8.5. The calculated amplitude response of this filter is plotted in Figure 8.37, along with the response of the lumped-element version. Note that the pass- band characteristics are very similar up to 4 GHz, but the distributed-elementfilter has a sharper cutoff. Also notice that the distributed-element filter has a re- sponse that repeats every 16 GHz, as a result of the periodic nature of Richards’ transformation. ■ Similar procedures can be used for bandstop filters, but the Kuroda identities are not useful for high-pass or bandpass filters. ImpedanceandAdmittanceInverters As we have seen, it is often desirable to use only series, or only shunt, elements when implementing a filter with a particular type of transmission line. The Kuroda identitiescan be used for conversions of this form, but another possibility is to use impedance (K) oradmittance (J)inverters [1, 4, 7]. Such inverters are especially useful for bandpass or bandstop filters with narrow ( <10%) bandwidths. The conceptual operation of impedance and admittance inverters is illustrated in Figure 8.38; since these inverters essentially form the inverse of the load impedance or admittance, they can be used to transform series-connected elements to shunt-connectedelements, or vice versa. This procedure will be illustrated in later sections for bandpass and bandstop filters. In its simplest form, an impedance or admittance inverter can be constructed using a quarter-wave transformer of the appropriate characteristic impedance, as shown in Figure 8.38b. This implementation also allows the ABCD matrix of the inverter to be easily found from the ABCD parameters for a length of transmission line, as given in Table 4.1. Many other types of circuits can also be used as impedance or admittance inverters, with one such alternative being shown in Figure 8.38c. Inverters of this form turn out to be useful formodeling the coupled resonator filters of Section 8.8. The lengths, θ/2, of the transmission line sections are generally required to be negative for this type of inverter, but this poses no problem if these lines can be absorbed into connecting transmission lines on either side. c08MicrowaveFilters Pozar August 25, 2011 18:16 422 Chapter 8: Microwave Filters /H9261/4 /H9258/2 /H9258/2Zin = K2/ZL Z0 = K Z0 Z0 jXK ± 90°ZL K = Z0 tan ⎜/H9258/2 ⎜ X =K 1 – (K/Z0)2 /H9258 = –tan–12X Z0 (c)(b)(a)Impedance inverters /H9261/4 /H9258/2 /H9258/2Yin = J2/YL Y0 = J Y0 Y0jBJ ± 90° J = Y0 tan ⎜/H9258/2 ⎜ B =J 1 – (J/Y0)2 /H9258 = –tan–12B Y0Admittance inverters YL /H11002C K = 1//H9275C /H11002C C /H11002C J = /H9275C/H11002CC (d) FIGURE 8.38 Impedance and admittance inverters. (a) Operation of impedance and admittance inverters. (b) Implementation as quarter-wave transformers. (c) Implementation using transmission lines and reactive elements. (d) Implementation using capacitornetworks. 8.6STEPPED-IMPEDANCELOW-PASSFILTERS A relatively easy way to implement low-pass filters in microstrip or stripline is to use alter- nating sections of very high and very low characteristic impedance lines. Such filters areusually referred to as stepped-impedance ,o rh i - Z,l o w - Zfilters, and are popular because they are easier to design and take up less space than a similar low-pass filter using stubs. Because of the approximations involved, however, their electrical performance is not asgood, so the use of such filters is usually limited to applications where a sharp cutoff is not required (for instance, in rejecting out-of-band mixer products). ApproximateEquivalentCircuitsforShortTransmissionLineSections We begin by finding the approximate equivalent circuits for a short length of transmis- sion line having either a very large or a very small characteristic impedance. The ABCD c08MicrowaveFilters Pozar August 25, 2011 18:16 8.6 Stepped-Impedance Low-Pass Filters 423 X = Z0/H9252l B = Y0/H9252ljBjX 2jX 2 (a) (b) (c) FIGURE 8.39 Approximate equivalent circuits for short sections of transmission lines. (a) T- equivalent circuit for a transmission line section having β/lscript/lessmuchπ/2. (b) Equivalent circuit for small β/lscriptand large Z0. (c) Equivalent circuit for small β/lscriptand small Z0. parameters of a length /lscriptof line having characteristic impedance Z0are given in Table 4.1; the conversion in Table 4.2 can then be used to find the impedance parameters as Z11=Z22=A C=− jZ0cotβ/lscript, (8.81a) Z12=Z21=1 C=− jZ0cscβ/lscript. (8.81b) The series elements of the T-equivalent circuit are Z11−Z12=− jZ0/parenleftbiggcosβ/lscript−1 sinβ/lscript/parenrightbigg =jZ0tan/parenleftbiggβ/lscript 2/parenrightbigg ,( 8.82) while the shunt element of the T-equivalent is Z12.I fβ/lscript < π/ 2, the series elements have a positive reactance (inductors), while the shunt element has a negative reactance (capacitor). We thus have the equivalent circuit shown in Figure 8.39a, where X 2=Z0tan/parenleftbiggβ/lscript 2/parenrightbigg , (8.83a) B=1 Z0sinβ/lscript. (8.83b) Now assume a short length of line (say β/lscript<π/ 4)and a large characteristic impedance. Then (8.83) approximately reduces to X/similarequalZ0β/lscript, (8.84a) B/similarequal0, (8.84b) which implies the equivalent circuit of Figure 8.39b (a series inductor). For a short length of line and a small characteristic impedance, (8.83) approximately reduces to X/similarequal0, (8.85a) B/similarequalY0β/lscript, (8.85b) which implies the equivalent circuit of Figure 8.39c (a shunt capacitor). So the series induc- tors of a low-pass prototype can be replaced with high-impedance line sections ( Z0=Zh), and the shunt capacitors can be replaced with low-impedance line sections ( Z0=Z/lscript). The c08MicrowaveFilters Pozar August 25, 2011 18:16 424 Chapter 8: Microwave Filters ratio Zh/Z/lscriptshould be as large as possible, so the actual values of ZhandZ/lscriptare usually set to the highest and lowest characteristic impedance that can be practically fabricated. The lengths of the lines can then be determined from (8.84) and (8.85); to get the best response near cutoff, these lengths should be evaluated at ω=ωc. Combining the results of (8.84) and (8.85) with the scaling equations of (8.67) allows the electrical lengths of the inductor sections to be calculated as β/lscript=LR0 Zh(inductor) (8.86a) and the electrical length of the capacitor sections as β/lscript=CZ/lscript R0(capacitor) ,( 8.86b) where R0is the filter impedance and LandCare the normalized element values (the gk) of the low-pass prototype. EXAMPLE 8.6 STEPPED-IMPEDANCE FILTER DESIGN Design a stepped-impedance low-pass filter having a maximally flat response and a cutoff frequency of 2.5 GHz. It is desired to have more than 20 dB insertion loss at 4 GHz. The filter impedance is 50 /Omega1; the highest practical line impedance is 120 /Omega1, and the lowest is 20 /Omega1. Consider the effect of losses when this filter is implemented with a microstrip substrate having d=0.158 cm,/epsilon1 r=4.2,tanδ= 0.02, and copper conductors of 0.5 mil thickness. Solution To use Figure 8.26 we calculate ω ωc−1=4.0 2.5−1=0.6; (a) (b) (c)C1 C3 C5L2 Z0l1 l2 l3 l4 l5 l6 Z0 Zl Zh Zl Zh Zl ZhL4 L6 FIGURE 8.40 Filter design for Example 8.6. (a) Low-pass filter prototype circuit. (b) Stepped- impedance implementation. (c) Microstrip layout of final filter. c08MicrowaveFilters Pozar August 25, 2011 18:16 8.6 Stepped-Impedance Low-Pass Filters 425 then the figure indicates N=6 should give the required attenuation at 4.0 GHz. Table 8.3 gives the low-pass prototype values as g1=0.517 =C1, g2=1.414 =L2, g3=1.932 =C3, g4=1.932 =L4, g5=1.414 =C5, g6=0.517 =L6. The low-pass prototype filter is shown in Figure 8.40a. Next, (8.86a) and (8.86b) are used to replace the series inductors and shunt capacitors with sections of low-impedance and high-impedance lines. The re- quired electrical line lengths, β/lscripti, along with the physical microstrip line widths, Wi, and lengths, /lscripti, are given in the table below. Section Zi=Z/lscriptorZh(/Omega1) β/lscript i(deg) Wi(mm) /lscripti(mm) 1 20 11.8 11.3 2.05 2 120 33.8 0.428 6.63 3 20 44.3 11.3 7.69 4 120 46.1 0.428 9.04 5 20 32.4 11.3 5.63 6 120 12.3 0.428 2.41 The final filter circuit is shown in Figure 8.40b, with Z/lscript=20/Omega1andZh= 120/Omega1. Note that β/lscript < 45◦for all but one section. The microstrip layout of the filter is shown in Figure 8.40c. Figure 8.41 shows the calculated amplitude response of the filter, with and without losses. The effect of loss is to increase the passband attenuation to about 0 1.0 2.0 3.0 5.030100Attenuation (dB) Frequency (GHz)Lumped elementHi-Z, Lo-Z20 4.0 FIGURE 8.41 Amplitude response of the stepped-impedance low-pass filter of Example 8.6, with (dotted line) and without (solid line) losses. The response of the corresponding lumped-element filter is also shown. c08MicrowaveFilters Pozar August 25, 2011 18:16 426 Chapter 8: Microwave Filters 1 dB at 2 GHz. The response of the corresponding lumped-element filter is also shown in Figure 8.41. The passband characteristic is similar to that of the stepped impedance filter, but the lumped-element filter gives more attenuation at higher frequencies. This is because the stepped-impedance filter elements depart sig-nificantly from the lumped-element values at higher frequencies. The stepped- impedance filter may have other passbands at higher frequencies, but the response will not be perfectly periodic because the lines are not commensurate. ■ 8.7COUPLEDLINEFILTERS The parallel coupled transmission lines discussed in Section 7.6 (for directional couplers) can be used to construct many types of filters. Fabrication of multisection bandpass or bandstop coupled line filters is particularly easy in microstrip or stripline form for band-widths less than about 20%. Wider bandwidth filters generally require very tightly coupled lines, which are difficult to fabricate. We will first study the filter characteristics of a single quarter-wave coupled line section, and then show how these sections can be used to designa bandpass filter [7]. Other filter designs using coupled lines can be found in reference [1]. FilterPropertiesofaCoupledLineSection A parallel coupled line section is shown in Figure 8.42a, with port voltage and current definitions. We will derive the open-circuit impedance matrix for this four-port network byconsidering the superposition of even- and odd-mode excitations [8], which are shown in Figure 8.42b. Thus, the current sources i 1andi3drive the line in the even mode, while i2 andi4drive the line in the odd mode. By superposition, we see that the total port currents, Ii, can be expressed in terms of the even- and odd-mode currents as I1=i1+i2, (8.87a) I2=i1−i2, (8.87b) I3=i3−i4, (8.87c) I4=i3+i4. (8.87d) First consider the line as being driven in the even mode by the i1current sources. If the other ports are open-circuited, the impedance seen at port 1 or 2 is Ze in=− jZ0ecotβ/lscript. (8.88) The voltage on either conductor can be expressed as v1 a(z)=v1 b(z)=V+ e[e−jβ(z−/lscript)+ejβ(z−/lscript)] =2V+ ecosβ(/lscript−z), (8.89) so the voltage at port 1 or 2 is v1 a(0)=v1 b(0)=2V+ ecosβ/lscript=i1Ze in. This result and (8.88) can be used to rewrite (8.89) in terms of i1as v1 a(z)=v1 b(z)=− jZ0ecosβ(/lscript−z) sinβ/lscripti1.( 8.90) c08MicrowaveFilters Pozar August 25, 2011 18:16 8.7 Coupled Line Filters 427 1 I1 4 I42 I2 3 I3+V2 +V1 +V4+V3 Z0e, Z0o 0 lz (a) 1 i1 4 i3i42 i1 i23 i3 Z0e, Z0ovb va 0 lz (b) 1 I1 42 O.C. 3 I3 Z0e, Z0o (c)O.C. FIGURE 8.42 Definitions pertaining to a coupled line filter section. (a) A parallel coupled line section with port voltage and current definitions. (b) A parallel coupled line sec- tion with even- and odd-mode current sources. (c) A two-port coupled line section having a bandpass response. Similarly, the voltages due to current sources i3driving the line in the even mode are v3 a(z)=v3 b(z)=− jZ0ecosβz sinβ/lscripti3.( 8.91) Now consider the line as being driven in the odd mode by current i2. If the other ports are open-circuited, the impedance seen at port 1 or 2 is Zo in=− jZ0ocotβ/lscript. (8.92) The voltage on either conductor can be expressed as v2 a(z)=−v2 b(z)=V+ 0/bracketleftbig e−jβ(z−/lscript)+ejβ(z−/lscript)/bracketrightbig =2V+ 0cosβ(/lscript−z). (8.93) Then the voltage at port 1 or port 2 is v2 a(0)=−v2 b(0)=2V+ 0cosβ/lscript=i2Zo in. c08MicrowaveFilters Pozar August 25, 2011 18:16 428 Chapter 8: Microwave Filters This result and (8.92) can be used to rewrite (8.93) in terms of i2as v2 a(z)=−v2 b(z)=− jZ0ocosβ(/lscript−z) sinβ/lscripti2.( 8.94) Similarly, the voltages due to current i4driving the line in the odd mode are v4 a(z)=−v4 b(z)=− jZ0ocosβz sinβ/lscripti4.( 8.95) The total voltage at port 1 is V1=v1 a(0)+v2 a(0)+v3 a(0)+v4 a(0) =− j(Z0ei1+Z0oi2)cotθ−j(Z0ei3+Z0oi4)cscθ, (8.96) where the results of (8.90), (8.91), (8.94), and (8.95) were used, and θ=β/lscript.N e x t ,w e solve (8.87) for the ijin terms of the Is: i1=1 2(I1+I2), (8.97a) i2=1 2(I1−I2), (8.97b) i3=1 2(I3+I4), (8.97c) i4=1 2(I4−I3), (8.97d) and use these results in (8.96): V1=−j 2(Z0eI1+Z0eI2+Z0oI1−Z0oI2)cotθ −j 2(Z0eI3+Z0eI4+Z0oI4−Z0oI3)cscθ. (8.98) This result yields the top row of the open-circuit impedance matrix [ Z] that describes the coupled line section. From symmetry, all other matrix elements can be found once the firstrow is known. The matrix elements are then Z 11=Z22=Z33=Z44=−j 2(Z0e+Z0o)cotθ (8.99a) Z12=Z21=Z34=Z43=−j 2(Z0e−Z0o)cotθ (8.99b) Z13=Z31=Z24=Z42=−j 2(Z0e−Z0o)cscθ (8.99c) Z14=Z41=Z23=Z32=−j 2(Z0e+Z0o)cscθ (8.99d) A two-port network can be formed from a coupled line section by terminating two of the four ports with either open or short circuits, or by connecting two ends; there are 10 possible combinations, as illustrated in Table 8.8. As indicated in the table, the various circuits have different frequency responses, including low-pass, bandpass, all pass, and all stop. For bandpass filters, we are most interested in the case shown in Figure 8.42c, as open circuits are easier to fabricate in microstrip than are short circuits. In this case, I2=I4=0, c08MicrowaveFilters Pozar August 25, 2011 18:16 8.7 Coupled Line Filters 429 TABLE 8.8 Ten Canonical Coupled Line Circuits Circuit Image Impedance Response Re(Zi1) 0Zi1 Zi2 Zi1Zi1 Zi1Zi2 Zi1Zi1 Zi1Zi1Zi1 Zi1Zi1Zi2Zi1Zi1 Zi1 Zi1 Zi1Zi123 2Low-pass Re(Zi1) 0 23 2Bandpass Re(Zi1) 0 23 2Bandpass Re(Zi1) 0 23 2BandpassZi1 = Zi1Zi2 =2Z0eZ0o cos (Z0e + Z0o)2 cos 2 – (Z0e – Z0o)2 Z0eZ0o Zi1 =2Z0eZ0o sin (Z0e – Z0o)2 – (Z0e + Z0o)2 cos 2 Zi1 = 2 sin (Z0e – Z0o)2 – (Z0e + Z0o)2 cos 2 Zi1 = Z0eZ0o(Z0e – Z0o)2 – (Z0e + Z0o)2 cos 2 (Z0e + Z0o) sin Z0e Z0o Zi1Zi2 =Zi1 = –j Z0eZ0oZi1Zi2 =Z0eZ0o 2Zi1 =Z0e + Z0o Zi1 = Zi1 = Z0eZ0o Zi1 = j2Z0eZ0o Z0e + Z0o 2Z0eZ0o Z0e + Z0ocot tan Z0eZ0o Zi1 = –j cot All pass All pass All pass All stop All stop All stop c08MicrowaveFilters Pozar August 25, 2011 18:16 430 Chapter 8: Microwave Filters 00 /H92581 /H92582 /H9258 π /H9266 2 23/H9266Z0e – Z0o 2Re(Zi) FIGURE 8.43 The real part of the image impedance of the bandpass network of Figure 8.42c. so the four-port impedance matrix equations reduce to V1=Z11I1+Z13I3, (8.100a) V3=Z31I1+Z33I3, (8.100b) where Zijis given in (8.99). We can analyze the filter characteristics of this circuit by calculating the image imped- ance (which is the same at ports 1 and 3), and the propagation constant. From Table 8.1, the image impedance in terms of the impedance parameters is Zi=/radicalBigg Z2 11−Z11Z2 13 Z33 =1 2/radicalbig (Z0e−Z0o)2csc2θ−(Z0e+Z0o)2cot2θ. (8.101) When the coupled line section is λ/4 long (θ =π/2), the image impedance reduces to Zi=1 2(Z0e−Z0o), (8.102) which is real and positive since Z0e>Z0o. However, when θ→0o rπ, Zi→± j∞, indicating a stopband. The real part of the image impedance is sketched in Figure 8.43, where the cutoff frequencies can be found from (8.101) as cosθ1=− cosθ2=Z0e−Z0o Z0e+Z0o. The propagation constant can also be calculated from the results of Table 8.1 as cosβ=/radicalBigg Z11Z33 Z2 13=Z11 Z13=Z0e+Z0o Z0e−Z0ocosθ, (8.103) which shows βis real for θ1<θ<θ 2=π−θ1, where cos θ1=(Z0e−Z0o)/(Z0e+ Z0o). DesignofCoupledLineBandpassFilters Narrowband bandpass filters can be made with cascaded coupled line sections of the form shown in Figure 8.42c. To derive the design equations for filters of this type, we first show that a single coupled line section can be approximately modeled by the equivalent circuitshown in Figure 8.44. We will do this by calculating the image impedance and propagation constant of the equivalent circuit and showing that they are approximately equal to those c08MicrowaveFilters Pozar August 25, 2011 18:16 8.7 Coupled Line Filters 431 /H9258 /H9258 J –90°Z0 Z0 FIGURE 8.44 Equivalent circuit of the coupled line section of Figure 8.42c. of the coupled line section for θ=π/2, which will correspond to the center frequency of the bandpass response. The ABCD parameters of the equivalent circuit can be computed using the ABCD matrices for transmission lines from Table 4.1: /bracketleftbigg AB CD/bracketrightbigg =⎡ ⎢⎣cosθ jZ0sinθ jsinθ Z0cosθ⎤ ⎥⎦/bracketleftbigg 0−j/J −jJ 0/bracketrightbigg⎡ ⎢⎣cosθ jZ0sinθ jsinθ Z0cosθ⎤ ⎥⎦ =⎡ ⎢⎢⎢⎢⎢⎣/parenleftbigg JZ 0+1 JZ0/parenrightbigg sinθcosθ j/parenleftBigg JZ2 0sin2θ−cos2θ J/parenrightBigg j/parenleftBigg 1 JZ2 0sin2θ−Jcos2θ/parenrightBigg/parenleftbigg JZ0+1 JZ0/parenrightbigg sinθcosθ⎤ ⎥⎥⎥⎥⎥⎦. (8.104) The ABCD parameters of the admittance inverter were obtained by considering it as a quarter-wave length of transmission of characteristic impedance, 1 /J. From (8.27) the image impedance of the equivalent circuit is Z i=/radicalbigg B C=/radicaltp/radicalvertex/radicalvertex/radicalbtJZ2 0sin2θ−(1/J)cos2θ (1/JZ2 0)sin2θ−Jcos2θ,( 8.105) which reduces to the following value at the center frequency, θ=π/2: Zi=JZ2 0.( 8.106) From (8.31) the propagation constant is cosβ=A=/parenleftbigg JZ0+1 JZ0/parenrightbigg sinθcosθ. (8.107) Equating the image impedances in (8.102) and (8.106), and the propagation constants of (8.103) and (8.107), yields the following equations: 1 2(Z0e−Z0o)=JZ2 0, Z0e+Z0o Z0e−Z0o=JZ0+1 JZ0, w h e r ew eh a v ea s s u m e ds i n θ/similarequal1f o rθnearπ/2. These equations can be solved for the even- and odd-mode line impedances to give Z0e=Z0[1+JZ0+(JZ0)2], (8.108a) Z0o=Z0[1−JZ0+(JZ0)2]. (8.108b) Now consider a bandpass filter composed of a cascade of N+1 coupled line sections, as shown in Figure 8.45a. The sections are numbered from left to right, with the load on the c08MicrowaveFilters Pozar August 25, 2011 18:16 432 Chapter 8: Microwave Filters FIGURE 8.45 Development of an equivalent circuit for derivation of design equations for a cou- pled line bandpass filter. (a) Layout of an ( N+1)-section coupled line bandpass filter. (b) Using the equivalent circuit of Figure 8.44 for each coupled line section. (c) Equivalent circuit for transmission lines of length 2 θ. (d) Equivalent circuit of the admittance inverters. (e) Using results of (c) and (d) for the N=2 case. (f) Lumped-element circuit for a bandpass filter for N=2. c08MicrowaveFilters Pozar August 25, 2011 18:16 8.7 Coupled Line Filters 433 right, but the filter can be reversed without affecting the response. Since each coupled line section has an equivalent circuit of the form shown in Figure 8.44, the equivalent circuit of the cascade is as shown in Figure 8.45b. Between any two consecutive inverters we have a transmission line section that is effectively 2 θin length. This line is approximately λ/2 long in the vicinity of the bandpass region of the filter, and has an approximate equivalent circuit that consists of a shunt parallel LCresonator, as in Figure 8.45c. The first step in establishing this equivalence is to find the parameters for the T- equivalent and ideal transformer circuit of Figure 8.45c (an exact equivalent). The ABCD matrix for this circuit can be calculated using the results in Table 4.1 for a T-circuit and an ideal transformer: /bracketleftbigg AB CD/bracketrightbigg =⎡ ⎢⎢⎣Z11 Z12Z2 11−Z2 12 Z12 1 Z12Z11 Z12⎤ ⎥⎥⎦/bracketleftbigg −10 0−1/bracketrightbigg =⎡ ⎢⎢⎣−Z11 Z12Z2 12−Z2 11 Z12 −1 Z12−Z11 Z12⎤ ⎥⎥⎦.(8.109) Equating this result to the ABCD parameters for a transmission line of length 2θ and char- acteristic impedance Z0gives the parameters of the equivalent circuit as Z12=−1 C=jZ0 sin 2θ, (8.110a) Z11=Z22=− Z12A=− jZ0cot 2θ. (8.110b) Then the series arm impedance is Z11−Z12=− jZ0cos 2θ +1 sin 2θ=− jZ0cotθ. (8.111) The 1: −1 transformer provides a 180◦phase shift, which cannot be obtained with the T-network alone; since this does not affect the amplitude response of the filter, it can bediscarded. For θ∼π/2 the series arm impedances of (8.111) are near zero and can also be ignored. The shunt impedance Z 12, however, looks like the impedance of a parallel reso- nant circuit for θ∼π/2. If we let ω=ω0+/Delta1ω, where θ=π/2 at the center frequency ω0, then we have 2θ =β/lscript=ω/lscript/v p=(ω0+/Delta1ω)π/ω 0=π(1+/Delta1ω/ω 0), so (8.110a) can be written for small /Delta1ωas Z12=jZ0 sinπ(1+/Delta1ω/ω 0)/similarequal−jZ0ω0 π(ω−ω0).( 8.112) From Section 6.1 the impedance near resonance of a parallel LCcircuit is Z=−jLω2 0 2(ω−ω0),( 8.113) withω2 0=1/LC. Equating this to (8.112) gives the equivalent inductor and capacitor val- ues as L=2Z0 πω0, (8.114a) C=1 ω2 0L=π 2Z0ω0. (8.114b) The end sections of the circuit of Figure 8.45b require a different treatment. The lines of length θon either end of the filter are matched to Z0and so can be ignored. The end inverters, J1andJN+1, can each be represented as a transformer followed by a λ/4 section c08MicrowaveFilters Pozar August 25, 2011 18:16 434 Chapter 8: Microwave Filters of line, as shown in Figure 8.45d. The ABCD matrix of a transformer with a turns ratio N in cascade with a quarter-wave line is /bracketleftbigg AB CD/bracketrightbigg =/bracketleftBigg1 N0 0N/bracketrightBigg/bracketleftBigg0−jZ0 −j Z00/bracketrightBigg =⎡ ⎣0−jZ0 N−jN Z00⎤ ⎦.( 8.115) Comparing this to the ABCD matrix of an admittance inverter [part of (8.104)] shows that the necessary turns ratio is N=JZ0.T h eλ/ 4 line merely produces a phase shift and so can be ignored. Using these results for the interior and end sections allows the circuit of Figure 8.45b to be transformed into the circuit of Figure 8.45e, which is specialized to the N=2 case. We see that each pair of coupled line sections leads to an equivalent shunt LCresonator, and an admittance inverter occurs between each pair of LCresonators. Next, we show that the admittance inverters have the effect of transforming a shunt LCresonator into as e r i e s LCresonator, leading to the final equivalent circuit of Figure 8.45f (shown for N=2). This will then allow the admittance inverter constants, Jn, to be determined from the element values of a low-pass prototype. We will demonstrate this for the N=2 case. With reference to Figure 8.45e, the admittance just to the right of the J2inverter is jωC2+1 jωL2+Z0J2 3=j/radicalBigg C2 L2/parenleftbiggω ω0−ω0 ω/parenrightbigg +Z0J2 3, since the transformer scales the load admittance by the square of the turns ratio. Then the admittance seen at the input of the filter is Y=1 J2 1Z2 0/braceleftBigg jωC1+1 jωL1+J2 2 j√C2/L2[(ω/ω 0)−(ω0/ω)]+Z0J2 3/bracerightBigg =1 J2 1Z2 0/braceleftBigg j/radicalBigg C1 L1/parenleftbiggω ω0−ω0 ω/parenrightbigg +J2 2 j√C2/L2[(ω/ω 0)−(ω0/ω)]+Z0J2 3/bracerightBigg .(8.116) These results also use the fact, from (8.114), that LnCn=1/ω2 0for all LCresonators. Now the admittance seen looking into the circuit of Figure 8.45f is Y=jωC/prime 1+1 jωL/prime 1+1 jωL/prime 2+1/jωC/prime 2+Z0 =j/radicalBigg C/prime 1 L/prime1/parenleftbiggω ω0−ω0 ω/parenrightbigg +1 j/radicalBig L/prime 2/C/prime 2[(ω/ω 0)−(ω0/ω)]+Z0, (8.117) which is identical in form to (8.116). Thus, the two circuits will be equivalent if the fol- lowing conditions are met: 1 J2 1Z2 0/radicalBigg C1 L1=/radicalBigg C/prime 1 L/prime1, (8.118a) J2 1Z2 0 J2 2/radicalBigg C2 L2=/radicalBigg L/prime 2 C/prime 2, (8.118b) J2 1Z3 0J2 3 J2 2=Z0. (8.118c) c08MicrowaveFilters Pozar August 25, 2011 18:16 8.7 Coupled Line Filters 435 We know LnandCnfrom (8.114); L/prime nandC/prime nare determined from the element values of a lumped-element low-pass prototype that has been impedance scaled and frequency transformed to a bandpass filter. Using the results in Table 8.6 and the impedance scalingformulas of (8.64) allows the L /prime nandC/prime nvalues to be written as L/prime 1=/Delta1Z0 ω0g1, (8.119a) C/prime 1=g1 /Delta1ω 0Z0, (8.119b) L/prime 2=g2Z0 /Delta1ω 0, (8.119c) C/prime 2=/Delta1 ω0g2Z0, (8.119d) where /Delta1=(ω2−ω1)/ω 0is the fractional bandwidth of the filter. Then (8.118) can be solved for the inverter constants with the following results (for N=2): J1Z0=/parenleftbiggC1L/prime 1 L1C/prime 1/parenrightbigg1/4 =/radicalBigg π/Delta1 2g1, (8.120a) J2Z0=J1Z2 0/parenleftbiggC2C/prime 2 L2L/prime2/parenrightbigg1/4 =π/Delta1 2√g1g2, (8.120b) J3Z0=J2 J1=/radicalBigg π/Delta1 2g2. (8.120c) After the Jnare found, Z0eandZ0ofor each coupled line section can be calculated from (8.108). The above results were derived for the special case of N=2 (three coupled line sec- tions), but more general results can be derived for any number of sections, and for the casewhere Z L/negationslash=Z0(orgN+1/negationslash=1, as in the case of an equal-ripple response with Neven). Thus, the design equations for a bandpass filter with N+1 coupled line sections are Z0J1=/radicalBigg π/Delta1 2g1, (8.121a) Z0Jn=π/Delta1 2√gn−1gnforn=2,3,..., N, (8.121b) Z0JN+1=/radicalBigg π/Delta1 2gNgN+1. (8.121c) The even- and odd-mode characteristic impedances for each section are found from (8.108). EXAMPLE 8.7 COUPLED LINE BANDPASS FILTER DESIGN Design a coupled line bandpass filter with N=3 and a 0.5 dB equal-ripple re- sponse. The center frequency is 2.0 GHz, the bandwidth is 10%, and Z0=50/Omega1. What is the attenuation at 1.8 GHz? c08MicrowaveFilters Pozar August 25, 2011 18:16 436 Chapter 8: Microwave Filters Solution The fractional bandwidth is /Delta1=0.1. We can use Figure 8.27a to obtain the at- tenuation at 1.8 GHz, but first we must use (8.71) to convert this frequency to the normalized low-pass form (ω c=1): ω←1 /Delta1/parenleftbiggω ω0−ω0 ω/parenrightbigg =1 0.1/parenleftbigg1.8 2.0−2.0 1.8/parenrightbigg =−2.11. Then the value on the horizontal scale of Figure 8.27a is /vextendsingle/vextendsingle/vextendsingle/vextendsingleω ωc/vextendsingle/vextendsingle/vextendsingle/vextendsingle−1=|− 2.11|− 1=1.11, which indicates an attenuation of about 20 dB for N=3. The low-pass prototype values, gn, are given in Table 8.4; then (8.121) can be used to calculate the admittance inverter constants, Jn. Finally, the even- and odd-mode characteristic impedances can be found from (8.108). These results are summarized in the following table: ng n Z0Jn Z0e(/Omega1) Z0o(/Omega1) 1 1.5963 0.3137 70.61 39.24 2 1.0967 0.1187 56.64 44.773 1.5963 0.1187 56.64 44.774 1.0000 0.3137 70.61 39.24 Note that the filter sections are symmetric about the midpoint. The calculated response of this filter is shown in Figure 8.46; passbands also occur at 6 GHz,10 GHz, etc. ■ Many other types of filters can be constructed using coupled line sections; most of these are of the bandpass or bandstop variety. One particularly compact design is the inter- digitated filter, which can be obtained from a coupled line filter by folding the lines at their midpoints; see references [1] and [3] for details. 1.0 1.5 2.0 2.5 3.050403020100Attenuation (dB) Frequency (GHz) FIGURE 8.46 Amplitude response of the coupled line bandpass filter of Example 8.7. c08MicrowaveFilters Pozar August 25, 2011 18:16 8.8 Filters Using Coupled Resonators 437 8.8FILTERSUSINGCOUPLEDRESONATORS We have seen that bandpass and bandstop filters require elements that behave as series or parallel resonant circuits; the coupled line bandpass filters of the previous section were of this type. Here we will consider several other types of microwave filters that use transmis-sion line or cavity resonators. BandstopandBandpassFiltersUsingQuarter-WaveResonators From Chapter 6 we know that quarter-wave open-circuited or short-circuited transmission line stubs look like series or parallel resonant circuits, respectively. We can therefore usesuch stubs in shunt along a transmission line to implement bandpass or bandstop filters, as shown in Figure 8.47. Quarter-wavelength sections of line between the stubs act as admittance inverters to effectively convert alternate shunt resonators to series resonators.The stubs and the transmission line sections are λ/4 long at the center frequency, ω 0. For narrow bandwidths the response of such a filter using Nstubs is essentially the same as that of a coupled line filter using N+1 sections. The internal impedance of the stub filter is Z0, while in the case of the coupled line filter end sections are required to trans- form the impedance level. This makes the stub filter more compact and easier to design. A disadvantage, however, is that a filter using stub resonators often requires characteristic impedances that are difficult to realize in practice. We first consider a bandstop filter using Nopen-circuited stubs, as shown in Fig- ure 8.47a. The design equations for the required stub characteristic impedances, Z0n, will be derived in terms of the element values of a low-pass prototype through the use of an equivalent circuit. The analysis of the bandpass version, using short-circuited stubs, fol-lows the same procedure, so the design equations for this case are presented without de- tailed derivation. As indicated in Figure 8.48a, an open-circuited stub can be approximated as a se- riesLCresonator when its length is near 90 ◦. The input impedance of an open-circuited /H9258 /H9258 /H9258Z0 Z0 Z0N Z0N – 1/H9258 /H9258 /H9258Z0 Z0 Z02 Z01 (a) /H9258 /H9258 /H9258Z0 Z0 Z0N Z0N – 1/H9258 /H9258 /H9258Z0 Z0 Z02 Z01 (b) FIGURE 8.47 Bandstop and bandpass filters using shunt transmission line resonators (θ =π/2 at the center frequency). (a) Bandstop filter. (b) Bandpass filter. c08MicrowaveFilters Pozar August 25, 2011 18:16 438 Chapter 8: Microwave Filters CNJ = 1/Z0 –90°J = 1/Z0 –90°J = 1/Z0 –90°Z0 Z0LNCnLn L3 C3L2 C2L1 C1 L3' L1' C3' C1' C2'L2'Y Y (c)(b)(a)/H9258 = /H9266/2 at /H9275 = /H92750Z Z0n/H9258 FIGURE 8.48 Equivalent circuit for the bandstop filter of Figure 8.47a. (a) Equivalent circuit of an open-circuited stub for θnearπ/2. (b) Equivalent filter circuit using resonators and admittance inverters. (c) Equivalent lumped-element bandstop filter. transmission line of characteristic impedance Z0nis Z=− jZ0ncotθ, where θ=π/2f o rω =ω0.I fw el e tω =ω0+/Delta1ω, where /Delta1ω/lessmuchω0, then θ=(π/2) (1+/Delta1ω/ω 0), and this impedance can be approximated as Z=jZ0ntanπ/Delta1ω 2ω0/similarequaljZ0nπ(ω−ω0) 2ω0(8.122) for frequencies in the vicinity of the center frequency, ω0. The impedance of a series LC circuit is Z=jωLn+1 jωCn=j/radicalBigg Ln Cn/parenleftbiggω ω0−ω0 ω/parenrightbigg /similarequal2j/radicalBigg Ln Cnω−ω0 ω0/similarequal2jLn(ω−ω0), (8.123) where LnCn=1/ω2 0. Equating (8.122) and (8.123) gives the characteristic impedance of the stub in terms of the resonator parameters: Z0n=4ω0Ln π.( 8.124) c08MicrowaveFilters Pozar August 25, 2011 18:16 8.8 Filters Using Coupled Resonators 439 Then, if we consider the quarter-wave sections of line between the stubs as ideal ad- mittance inverters, the bandstop filter of Figure 8.47a can be represented by the equivalent circuit of Figure 8.48b. Next, the circuit elements of this equivalent circuit can be related to those of the lumped-element bandstop filter prototype of Figure 8.48c. With reference to Figure 8.48b, the admittance Yseen looking toward the L2C2res- onator is Y=1 jωL2+(1/jωC2)+1 Z2 0/parenleftbigg1 jωL1+1/jωC1+1 Z0/parenrightbigg−1 =1 j√L2/C2/bracketleftbig (ω/ω 0)−(ω0/ω)/bracketrightbig +1 Z0/braceleftBigg 1 j√L1/C1/bracketleftbig (ω/ω 0)−(ω0/ω)/bracketrightbig+1 Z0/bracerightBigg . (8.125) The admittance at the corresponding point in the circuit of Figure 8.48c is Y=1 jωL/prime 2+1/jωC/prime 2+/parenleftbigg1 jωC/prime 1+1/jωL/prime 1+Z0/parenrightbigg−1 =1 j/radicalBig L/prime 2/C/prime 2/bracketleftbig (ω/ω 0)−(ω0/ω)/bracketrightbig +⎧ ⎨ ⎩1 j/radicalBig C/prime 1/L/prime 1/bracketleftbig (ω/ω 0)−(ω0/ω)/bracketrightbig+Z0⎫ ⎬ ⎭−1 . (8.126) These two results will be equivalent if the following conditions are satisfied: 1 Z2 0/radicalBigg L1 C1=/radicalBigg C/prime 1 L/prime1, (8.127a) /radicalBigg L2 C2=/radicalBigg L/prime 2 C/prime 2. (8.127b) Since LnCn=L/prime nC/prime n=1/ω2 0, these results can be solved for Ln: L1=Z2 0 ω2 0L/prime1, (8.128a) L2=L/prime 2. (8.128b) Using (8.124) and the impedance-scaled bandstop filter elements from Table 8.6 gives the stub characteristic impedances as Z01=4Z2 0 πω0L/prime 1=4Z0 πg1/Delta1, (8.129a) Z02=4ω0L/prime 2 π=4Z0 πg2/Delta1, (8.129b) where /Delta1=(ω2−ω1)/ω 0is the fractional bandwidth of the filter. It is easy to show that the general result for the characteristic impedances of a bandstop filter is Z0n=4Z0 πgn/Delta1.( 8.130) c08MicrowaveFilters Pozar August 25, 2011 18:16 440 Chapter 8: Microwave Filters For a bandpass filter using short-circuited stub resonators the corresponding result is Z0n=πZ0/Delta1 4gn.( 8.131) These results only apply to filters having input and output impedances of Z0and so cannot be used for equal-ripple designs with Neven. EXAMPLE 8.8 BANDSTOP FILTER DESIGN Design a bandstop filter using three quarter-wave open-circuit stubs. The center frequency is 2.0 GHz, the bandwidth is 15%, and the impedance is 50 /Omega1.U s ea n equal-ripple response, with a 0.5 dB ripple level. Solution The fractional bandwidth is /Delta1=0.15. Table 8.4 gives the low-pass prototype values, gn,f o r N=3. Then the characteristic impedances of the stubs can be found from (8.130). The results are listed in the following table: ng n Z0n(/Omega1) 1 1.5963 265.9 2 1.0967 387.03 1.5963 265.9 The filter circuit is shown in Figure 8.47a, with all stubs and transmission line sections λ/4 long at 2.0 GHz. The calculated attenuation for this filter is shown in Figure 8.49; the ripple in the passbands is somewhat greater than 0.5 dB as a result of the approximations involved in the development of the design equations. ■ The performance of quarter-wave resonator filters can be improved by allowing the characteristic impedances of the interconnecting lines to be variable; then an exact cor-respondence with coupled line bandpass or bandstop filters can be demonstrated. Design details for this case can be found in reference [1]. 1.0 1.5 2.0 2.5 3.050403020100Attenuation (dB) Frequency (GHz) FIGURE 8.49 Amplitude response of the bandstop filter of Example 8.8. c08MicrowaveFilters Pozar August 25, 2011 18:16 8.8 Filters Using Coupled Resonators 441 J1 +90°J2 +90°JN + 1 +90°Z0 Z0Z0 Z0 Z0 Z0Z0 Z0 Z0 Z0 Z0 Z0 Z0Z0B1 /H92581 /H92581 /H92781/H92582 /H92583 /H92582/H9258N /H9258NB2 B3 BN + 1 jB1 jB1jB2 jB2jB3 jBN jBN + 1 jBN + 12/H92781 2/H92782 2/H92782 /H9278 Z0 Z0 Z0 Z0 Z0 Z0 Z0Z0 Z0/H9278 /H9278/H92782/H9278N + 1 2/H9278N + 1 2 (d)(c)(b)(a) FIGURE 8.50 Development of the equivalence of a capacitive-gap coupled resonator bandpass filter to the coupled line bandpass filter of Figure 8.45. (a) The capacitive-gap coupled resonator bandpass filter. (b) Transmission line model. (c) Transmission line model with negative-length sections forming admittance inverters (φ i/2<0). (d) Equivalent circuit using inverters and λ/2 resonators (φ =πatω0). This cir- cuit is identical in form with the coupled line bandpass filter equivalent circuit of Figure 8.45b. BandpassFiltersUsingCapacitivelyCoupledSeriesResonators Another type of bandpass filter that can be conveniently fabricated in microstrip or stripline form is the capacitive-gap coupled resonator filter shown in Figure 8.50. An Nth-order filter of this form will use Nresonant series sections of transmission line with N+1 capacitive gaps between them. These gaps can be approximated as series capacitors; design data relating the capacitance to the gap size and transmission line parameters is given ingraphical form in reference [1]. The filter can then be modeled as shown in Figure 8.50b. The resonators are approximately λ/2 long at the center frequency, ω 0. Next, we redraw the equivalent circuit of Figure 8.50b with negative-length transmis- sion line sections on either side of the series capacitors. The lines of length φwill be λ/2 long at ω0, so the electrical length θiof the ith section in Figures 8.50a, b is θi=π+1 2φi+1 2φi+1 fori=1,2,..., N,( 8.132) withφi<0. The reason for doing this is that the combination of series capacitor and negative-length transmission lines forms the equivalent circuit of an admittance inverter, as seen from Figure 8.38c. In order for this equivalence to be valid, the following relationship c08MicrowaveFilters Pozar August 25, 2011 18:16 442 Chapter 8: Microwave Filters must hold between the electrical length of the lines and the capacitive susceptance: φi=− tan−1(2Z0Bi). (8.133) Then the resulting inverter constant can be related to the capacitive susceptance as Bi=Ji 1−(Z0Ji)2.( 8.134) (These results are given in Figure 8.38, and their derivation is requested in Problem 8.14.) The capacitive-gap coupled filter can then be modeled as shown in Figure 8.50d. Now consider the equivalent circuit shown in Figure 8.45b for a coupled line bandpass filter.Since these two circuits are identical (as φ=2θ=πat the center frequency), we can use the results from the coupled line filter analysis to complete the present problem. Thus, we can use (8.121) to find the admittance inverter constants, J i, from the low-pass proto- type values, gi, and the fractional bandwidth, /Delta1. As in the case of the coupled line filter, there will be N+1 inverter constants for an Nth-order filter. Then (8.134) can be used to find the susceptance, Bi,f o rt h e ith coupling gap. Finally, the electrical length of the resonator sections can be found from (8.132) and (8.133): θi=π−1 2[tan−1(2Z0Bi)+tan−1(2Z0Bi+1)].( 8.135) EXAMPLE 8.9 CAPACITIVELY COUPLED SERIES RESONATOR BANDPASS FILTER DESIGN Design a bandpass filter using capacitive coupled series resonators, with a 0.5 dB equal-ripple passband characteristic. The center frequency is 2.0 GHz, the band- width is 10%, and the impedance is 50 /Omega1. At least 20 dB of attenuation is required at 2.2 GHz. Solution We first determine the order of the filter to satisfy the attenuation specification at2.2 GHz. Using (8.71) to convert to normalized frequency gives ω←1 /Delta1/parenleftbiggω ω0−ω0 ω/parenrightbigg =1 0.1/parenleftbigg2.2 2.0−2.0 2.2/parenrightbigg =1.91. Then, /vextendsingle/vextendsingle/vextendsingle/vextendsingleω ωc/vextendsingle/vextendsingle/vextendsingle/vextendsingle−1=1.91−1.0=0.91. From Figure 8.27a, we see that N=3 should satisfy the attenuation specification at 2.2 GHz. The low-pass prototype values are given in Table 8.4, from which the inverter constants can be calculated using (8.121). Then the coupling susceptances can be found from (8.134), and the coupling capacitor values as Cn=Bn ω0. Finally, the resonator lengths can be calculated from (8.135). The following table summarizes these results. c08MicrowaveFilters Pozar September 12, 2011 21:42 8.8 Filters Using Coupled Resonators 443 1.0 1.5 2.0 2.5 3.050403020100Attenuation (dB) Frequency (GHz) FIGURE 8.51 Amplitude response for the capacitive-gap coupled series resonator bandpass filter of Example 8.9. ng n Z0Jn Bn Cn(pF) θn(deg) 1 1.5963 0.3137 6.96 ×10−30.554 155.8 2 1.0967 0.1187 2.41 ×10−30.192 166.5 3 1.5963 0.1187 2.41 ×10−30.192 155.8 4 1.0000 0.3137 6.96 ×10−30.554 — The calculated amplitude response is plotted in Figure 8.51. The specifica- tions of this filter are the same as the coupled line bandpass filter of Example 8.8,and comparison of the results in Figures 8.51 and 8.46 shows that the responses are identical near the passband region. ■ BandpassFiltersUsingCapacitivelyCoupledShuntResonators A related type of bandpass filter is shown in Figure 8.52, where short-circuited shunt res- onators are capacitively coupled with series capacitors. An Nth-order filter will use N stubs, which are slightly shorter than λ/4 at the filter center frequency. The short-circuited stub resonators can be made from sections of coaxial line using ceramic materials having a very high dielectric constant and low loss, resulting in a very compact design even at UHF frequencies [9]. Such filters are often referred to as ceramic resonator filters and are FIGURE 8.52 A bandpass filter using capacitively coupled shunt stub resonators. c08MicrowaveFilters Pozar August 25, 2011 18:16 444 Chapter 8: Microwave Filters among the most common types of RF bandpass filters used in portable wireless systems. Most cellular telephones, GPS receivers, and other wireless devices employ two or more filters of this type. Operation and design of this filter can be understood by beginning with the general bandpass filter circuit of Figure 8.53a, where shunt LCresonators alternate with admit- tance inverters. As in the case of previous coupled resonator bandpass and bandstop filters, the function of the admittance inverters is to convert alternate shunt resonators to seriesresonators; the extra inverters at the ends serve to scale the impedance level of the filter to a realistic level. Using an analysis similar to that used for the bandstop filter, we can derive the admittance inverter constants as Z 0J01=/radicalBigg π/Delta1 4g1, (8.136a) Z0Jn,n+1=π/Delta1 4√gngn+1, (8.136b) Z0JN,N+1=/radicalBigg π/Delta1 4gNgN+1. (8.136c) Similarly, the coupling capacitor values can be found as C01=J01 ω0/radicalbig 1−(Z0J01)2, (8.137a) Cn,n+1=Jn,n+1 ω0, (8.137b) CN,N+1=JN,N+1 ω0/radicalbig 1−(Z0JN,N+1)2. (8.137c) Note that the end capacitors are treated differently than the internal elements. Next, we replace the admittance inverters of Figure 8.53a with the equivalent π- network of Figure 8.38d, to produce the equivalent lumped-element circuit shown in Figure 8.53b. Note that the shunt capacitors of the admittance inverter circuits are negative, butthese elements combine in parallel with the larger capacitor of the LCresonator to yield a net capacitance value that is positive. The resulting circuit is shown in Figure 8.53c, where the effective resonator capacitor values are given by C /prime n=Cn+/Delta1Cn=Cn−Cn−1,n−Cn,n+1,( 8.138) where /Delta1Cn=−Cn−1,n−Cn,n+1represents the change in the resonator capacitance caused by the parallel addition of the inverter elements. Finally, the shunt LCresonators of Figure 8.53c are replaced with short-circuited trans- mission stubs, as in the circuit of Figure 8.52. Note that the resonant frequency of the stub resonators is no longer ω0, since the resonator capacitor values have been modified by the /Delta1Cn. This implies that the length of the resonator is less than λ/4a tω 0, the filter center frequency. The transformation of the stub length to account for the change in capacitance is illustrated in Figure 8.53d. A short-circuited length of line with a shunt capacitor at its c08MicrowaveFilters Pozar September 12, 2011 21:42 8.8 Filters Using Coupled Resonators 445 FIGURE 8.53 Equivalent circuit for the bandpass filter of Figure 8.52. (a) A general bandpass filter circuit using shunt resonators with admittance inverters. (b) Replacement of admittance inverters with the circuit implementation of Figure 8.38d. (c) After combining shunt capacitorelements. (d) Change in resonant stub length caused by a shunt capacitor. c08MicrowaveFilters Pozar September 12, 2011 21:42 446 Chapter 8: Microwave Filters input has an input admittance of Y=YL+jω0C,( 8.139a) where YL=−jcotβ/lscript Z0. If the capacitor is replaced with a short length, /Delta1/lscript, of transmission line, the input admit- tance would be Y=1 Z0YL+j1 Z0tanβ/Delta1/lscript 1 Z0+jYLtanβ/Delta1/lscript∼=YL+jβ/Delta1/lscript Z0.( 8.139b) The last approximation follows for β/Delta1/lscript/lessmuch1, which is true in practice for filters of this type. Comparing (8.139b) with (8.139a) gives the change in stub length in terms of the capacitor value: /Delta1/lscript=Z0ω0C β=/parenleftbiggZ0ω0C 2π/parenrightbigg λ. (8.140) Note that if C<0, then /Delta1/lscript < 0, indicating a shortening of the stub length. Thus the overall stub length is given by /lscriptn=λ 4+/parenleftbiggZ0ω0/Delta1Cn 2π/parenrightbigg λ, (8.141) where /Delta1Cnis defined in (8.138). The characteristic impedance of the stub resonators is Z0. Dielectric material properties play a critical role in the performance of ceramic res- onator filters. Materials with high dielectric constants are required in order to provide miniaturization at the frequencies typically used for wireless applications. Losses must be low to provide resonators with high Q, leading to low passband insertion loss and maximum attenuation in the stopbands. In addition, the dielectric constant must be stable with changes in temperature to avoid drifting of the filter passband over normal operat- ing conditions. Most materials that are commonly used in dielectric resonator filters areceramics, such as barium tetratitanate, zinc/strontium titanate, and various titanium oxide compounds. For example, a zinc/strontium titanate ceramic material has a dielectric con- stant of 36, a Qof 10,000 at 4 GHz, and a dielectric constant temperature coefficient of −7 ppm/C ◦. EXAMPLE 8.10 CAPACITIVELY COUPLED SHUNT RESONATOR BANDPASS FILTER DESIGN Design a third-order bandpass filter with a 0.5 dB equal-ripple response using capacitively coupled short-circuited shunt stub resonators. The center frequency is2.5 GHz, and the bandwidth is 10%. The impedance is 50 /Omega1. What is the resulting attenuation at 3.0 GHz? Solution We first calculate the attenuation at 3.0 GHz. Using (8.71) to convert 3.0 GHz tonormalized low-pass form gives ω←1 /Delta1/parenleftbiggω ω0−ω0 ω/parenrightbigg =1 0.1/parenleftbigg3.0 2.5−2.5 3.0/parenrightbigg =3.667. c08MicrowaveFilters Pozar August 25, 2011 18:16 8.8 Filters Using Coupled Resonators 447 Then, to use Figure 8.27a, the value on the horizontal axis is /vextendsingle/vextendsingle/vextendsingle/vextendsingleω ωc/vextendsingle/vextendsingle/vextendsingle/vextendsingle−1=|−3.667 |−1=2.667, from which we find the attenuation as 35 dB. Next we calculate the admittance inverter constants and coupling capacitor values using (8.136) and (8.137): ng n Z0Jn−1,n Cn−1,n(pF) 1 1.5963 Z0J01=0.2218 C01=0.2896 2 1.0967 Z0J12=0.0594 C12=0.0756 3 1.5963 Z0J23=0.0594 C23=0.0756 4 1.0000 Z0J34=0.2218 C34=0.2896 Then we use (8.138), (8.140), and (8.141) to find the required resonator lengths: n /Delta1Cn(pF) /Delta1/lscriptn(λ) /lscript(deg) 1 −0.3652 −0.04565 73.6 2 −0.1512 −0.0189 83.2 3 −0.3652 −0.04565 73.6 Note that the resonator lengths are slightly less than 90◦(λ/4). The calculated amplitude response of this design is shown in Figure 8.54. The stopband rolloff at high frequencies is less than at lower frequencies, and the attenuation at 3 GHz is seen to be about 30 dB, while our calculated value for a canonical lumped-elementbandpass filter was 35 dB. ■ FIGURE 8.54 Amplitude response of the capacitively coupled shunt resonator bandpass filter of Example 8.10. c08MicrowaveFilters Pozar August 25, 2011 18:16 448 Chapter 8: Microwave Filters FIGURE 8.55 Photograph of a wideband down converter. Multiple PIN diode switches, filter banks, amplifiers, and mixers are used to cover the range of 10 MHz to 27 GHz. Several types of filters can be seen in this module, including stub filters, coupledline filters, and stepped impedance filters. The microstrip stub filter at left is a low- pass 21-pole filter with a cutoff frequency of 5.4 GHz and more than 75 dB of rejection from 5.8 to 9.0 GHz. Courtesy of LNX Corporation, Salem, N.H. Figure 8.55 shows a wideband receiver downconverter module employing a variety of different filter types. REFERENCES [1] G. L. Matthaei, L. Young, and E. M. T. Jones, Microwave Filters, Impedance-Matching Networks, and Coupling Structures, Artech House, Dedham, Mass., 1980. [2] R. E. Collin, Foundations for Microwave Engineering , 2nd edition, Wiley-IEEE Press, Hoboken, N.J., 2001. [3] J. A. G. Malherbe, Microwave Transmission Line Filters, Artech House, Dedham, Mass., 1979. [4] W. A. Davis, Microwave Semiconductor Circuit Design, Van Nostrand Reinhold, New York, 1984. [5] R. F. Harrington, Time-Harmonic Electromagnetic Fields, McGraw-Hill, New York, 1961. [6] P. I. Richards, “Resistor-Transmission Line Circuits,” Proceedings of the IRE, vol. 36, pp. 217–220, February 1948. [7] S. B. Cohn, “Parallel-Coupled Transmission-Line-Resonator Filters,” IRE Transactions on Micro- wave Theory and Techniques , vol. MTT-6, pp. 223–231, April 1958. [8] E. M. T. Jones and J. T. Bolljahn, “Coupled-Strip-Transmission Line Filters and Directional Cou- plers,” IRE Transactions on Microwave Theory and Techniques , vol. MTT-4, pp. 78–81, April 1956. [9] M. Sagawa, M. Makimoto, and S. Yamashita, “A Design Method of Bandpass Filters Using Dielectric-Filled Coaxial Resonators,” IEEE Transactions on Microwave Theory and Techniques , vol. MTT-33, pp. 152–157, February 1985. PROBLEMS 8.1 Sketch the k-βdiagram for the infinite periodic structure shown below. Assume Z0=75/Omega1,d= 1.0c m ,k =k0,a n d L0=1.25 nH. c08MicrowaveFilters Pozar August 25, 2011 18:16 Problems 449 dL0Z0, k L0 L0 L0 8.2 Verify the expression for the image impedance of a π-network given in Table 8.1. 8.3 Compute the image impedances and propagation factor for the network shown below. L L CPort 1Port 2 8.4 Design a composite low-pass filter by the image parameter method with the following specifications: R0=50/Omega1,fc=50 MHz, and f∞=52 MHz. Use CAD to plot the insertion loss versus frequency. 8.5 Design a composite high-pass filter by the image parameter method with the following specifications: R0=75/Omega1,fc=50 MHz, and f∞=48 MHz. Use CAD to plot the insertion loss versus frequency. 8.6 Solve the design equations in Section 8.3 for the elements of an N=2 equal-ripple filter if the ripple specification is 1.0 dB. 8.7 Design a low-pass, maximally flat lumped-element filter having a passband of 0–2 GHz, and an attenuation of at least 20 dB at 3.4 GHz. The characteristic impedance is 50 /Omega1.U s eC A Dt op l o tt h e insertion loss versus frequency. 8.8 Design a high-pass lumped-element filter with a 3 dB equal-ripple response, a cutoff frequency of 3 GHz, and at least 30 dB insertion loss at 2.0 GHz. The characteristic impedance is 75 /Omega1.U s eC A D to plot the insertion loss versus frequency. 8.9 Design a four-section bandpass lumped-element filter having a maximally flat group delay response. The bandwidth should be 5% with a center frequency of 2 GHz. The impedance is 50 /Omega1.U s eC A D to plot the insertion loss versus frequency. 8.10 Design a three-section bandstop lumped-element filter with a 0.5 dB equal-ripple response, a band- width of 10% centered at 3 GHz, and an impedance of 75 /Omega1. What is the resulting attenuation at 3.1 GHz? Use CAD to plot the insertion loss versus frequency. 8.11 Verify the second Kuroda identity in Table 8.7 by calculating the ABCD matrices for both circuits. 8.12 Design a low-pass, third-order, maximally flat filter using only series stubs. The cutoff frequency is 6 GHz and the impedance is 50 /Omega1. Use CAD to plot the insertion loss versus frequency. 8.13 Design a low-pass, fourth-order, maximally flat filter using only shunt stubs. The cutoff frequency is 8 GHz and the impedance is 50 /Omega1. Use CAD to plot the insertion loss versus frequency. 8.14 Verify the operation of the admittance inverter of Figure 8.38c by calculating its ABCD matrix and comparing it to the ABCD matrix of the admittance inverter made from a quarter-wave line. 8.15 Show that the π-equivalent circuit for a short length of transmission line leads to equivalent cir- cuits identical to those in Figures 8.39b and 8.39c for large and small characteristic impedance, respectively. 8.16 Design a stepped-impedance low-pass filter having a cutoff frequency of 3 GHz and a fifth-order 0.5 dB equal-ripple response. Assume R0=50/Omega1,Z/lscript=15/Omega1,a n d Zh=120/Omega1.( a )F i n dt h er e - quired electrical lengths of the five sections, and use CAD to plot the insertion loss from 0 to6 GHz. (b) Lay out the microstrip implementation of the filter on an FR4 substrate having /epsilon1 r=4.2, c08MicrowaveFilters Pozar August 25, 2011 18:16 450 Chapter 8: Microwave Filters d=0.079 cm, and tan δ=0.02, and with copper conductors 0.5 mil thick. Use CAD to plot the insertion loss versus frequency in the passband of the filter, and compare with the lossless case. 8.17 Design a stepped-impedance low-pass filter with fc=2.0 GHz and R0=50/Omega1, using the exact transmission line equivalent circuit of Figure 8.39a. Assume a maximally flat N=5 response, and solve for the necessary line lengths and impedances if Z/lscript=10/Omega1andZh=150/Omega1. Use CAD to plot the insertion loss versus frequency. 8.18 Design a four-section coupled line bandpass filter with a 0.5 dB equal ripple response. The center frequency is 2.45 GHz, the bandwidth is 10%, and the impedance is 50 /Omega1. (a) Find the required even- and odd-mode impedances of the coupled line sections, and calculate the expected attenuation at 2.1 GHz. Use CAD to plot the insertion loss from 1.55 to 3.35 GHz. (b) Lay out the microstripimplementation of the filter on an FR4 substrate having /epsilon1 r=4.2, d=0.158 cm, and tan δ=0.01, and with copper conductors 0.5 mil thick. Use CAD to plot the insertion loss versus frequency in the passband of the filter, and compare with the lossless case. 8.19 TheSchiffman phase shifter , shown below, can produce a 90◦differential phase shift over a relatively broad frequency range. It consists of a coupled line section of length θ, and a transmission line s e c t i o no fl e n g t h3 θ;θ=π/2 at midband. The characteristic impedance of the transmission line is Z0=√Z0eZ0o,w h e r e Z0eandZ0oare the even- and odd-mode impedances of the coupled line section. Use the analysis of Section 8.7 to find the phase shift through the coupled line section, and then find the differential phase shift between the two outputs. Plot the differential phase shift forθ=0t oπ forZ 0e/Z0o=2.7 and determine the bandwidth for which the phase shift is 90◦±2.5◦. V0 3/H9258/H9258 /H9278Z0e Z0o /H9004 8.20 Design a maximally flat bandstop filter using four open-circuited quarter-wave stub resonators. The center frequency is 3 GHz, the bandwidth is 15%, and the impedance is 40 /Omega1. Use CAD to plot the insertion loss versus frequency. 8.21 Design a bandpass filter using three quarter-wave short-circuited stub resonators. The filter should have a 0.5 dB equal-ripple response, a center frequency of 3 GHz, a 20% bandwidth, and an impedance of 100 /Omega1. (a) Find the required characteristic impedances of the resonators, and use CAD to plot the insertion loss from 1 to 5 GHz. (b) Lay out the microstrip implementation of the filter on an FR4 substrate having /epsilon1r=4.2, d=0.079 cm, and tan δ=0.02, and with copper conductors 0.5 mil thick. Use CAD to plot the insertion loss versus frequency in the passband of the filter, and compare with the lossless case. 8.22 Derive the design equation of (8.131) for bandpass filters using quarter-wave shorted stub resonators. 8.23 Design a bandpass filter using capacitive-gap coupled resonators. The response should be maximally flat, with a center frequency of 4 GHz, a bandwidth of 12%, and at least 12 dB attenuation at 3.6 GHz. The characteristic impedance is 50 /Omega1. Find the electrical line lengths and the coupling capacitor values. Use CAD to plot the insertion loss versus frequency. 8.24 A bandpass filter is to be used in a PCS receiver operating in the 824–849 MHz band, and must provide at least 30 dB isolation at the lowest end of the transmit frequency band (869–894 MHz). Design a 1 dB equal-ripple bandpass filter meeting these specifications using capacitively coupledshort-circuited shunt stub resonators. Assume an impedance of 50 /Omega1. 8.25 Derive the design equations of (8.136) and (8.137) for the capacitively coupled shunt stub resonator bandpass filter. c09FerrimagneticComponents Pozar August 25, 2011 18:36 Chapter Nine Theory and Design of Ferrimagnetic Components The components and networks discussed up to this point have all been reciprocal. That is, the response between any two ports iand jof a component did not depend on the direction of signal flow (thus, Sij=Sji). This will always be the case when the component is passive and consists of only isotropic materials, but if the component contains either active devices oranisotropic material, nonreciprocal behavior can be obtained. In some cases nonreciprocity is auseful property (e.g., circulators, isolators), while in other cases nonreciprocity is an ancillary property (e.g., transistor amplifiers, ferrite phase shifters). In Chapter 1 we discussed materials having electric anisotropy (tensor permittivity), and magnetic anisotropy (tensor permeability). Some of the most practical anisotropic materials formicrowave applications are ferrimagnetic compounds , also known as ferrites, such as yttrium iron garnet (YIG) and materials composed of iron oxides and various other elements such as aluminum, cobalt, manganese, and nickel. In contrast to ferromagnetic materials (e.g., iron,steel), ferrimagnetic compounds have high resistivity and a significant amount of anisotropy atmicrowave frequencies. As we will see, the magnetic anisotropy of a ferrimagnetic material is actually induced by applying a DC magnetic bias field. This field aligns the magnetic dipoles in the ferrite material to produce a net (nonzero) magnetic dipole moment, and causes the mag-netic dipoles to precess at a frequency controlled by the strength of the bias field. A microwavesignal circularly polarized in the same direction as this precession will interact strongly with the dipole moments, while an oppositely polarized field will interact less strongly. Since, for a given direction of rotation, the sense of polarization changes with the direction of propa-gation, a microwave signal will propagate through a magnetically biased ferrite differently indifferent directions. This effect can be utilized to fabricate directional devices such as isola- tors, circulators, and gyrators. Another useful characteristic of ferrimagnetic materials is that the interaction with an applied microwave signal can be controlled by adjusting the strength ofthe bias field. This effect leads to a variety of control devices such as phase shifters, switches,and tunable resonators and filters. It is interesting to compare ferrimagnetic materials to paraelectric materials, which are almost the dual of ferrimagnetic materials. Certain ceramic compounds, such as lithium nio-bate and barium titanate, have the property that their dielectric permittivity can be controlledwith the application of a DC bias electric field. Paraelectric materials can therefore be used for 451 c09FerrimagneticComponents Pozar August 25, 2011 18:36 452 Chapter 9: Theory and Design of Ferrimagnetic Components variable phase shifters and other control components. Unlike ferrimagnetic materials, para- electric materials are isotropic, and therefore paraelectric devices are reciprocal. Paraelectric materials typically have very high dielectric constants and loss tangents when used in bulkform, so modern applications generally use thin films of paraelectric material layered on a sub-strate. An important advantage of paraelectric devices over ferrite devices is that the need for a large and heavy magnet or biasing coil is eliminated. We will begin by considering the microscopic behavior of a ferrimagnetic material and its interaction with a microwave signal to derive the permeability tensor. This macroscopic de-scription of the material can then be used with Maxwell’s equations to analyze wave propaga- tion in an infinite ferrite medium and in a ferrite-loaded waveguide. These canonical problems will illustrate the nonreciprocal propagation properties of ferrimagnetic materials, includingFaraday rotation and birefringence effects, and will be used in later sections when we discussthe operation and design of waveguide phase shifters and isolators. 9.1BASICPROPERTIESOFFERRIMAGNETICMATERIALS In this section we will show how the permeability tensor for a ferrimagnetic material can be deduced from a relatively simple microscopic view of the atom. We will also discusshow loss affects the permeability tensor and the demagnetization field inside a finite-sized piece of ferrite. ThePermeabilityTensor The magnetic properties of a material are due to the existence of magnetic dipole moments, which arise primarily from electron spin. From quantum mechanical considerations [1], the magnetic dipole moment of an electron due to its spin is given by m=qh − 2me=9.27×10−24A-m2,( 9.1) where h−is Planck’s constant divided by 2 π,qis the electron charge, and meis the mass of the electron. An electron in orbit around a nucleus gives rise to an effective current loopand thus an additional magnetic moment, but this effect is generally insignificant compared to the magnetic moment due to spin. The Land ´e g factor is a measure of the relative con- tributions of the orbital moment and the spin moment to the total magnetic moment; g=1 when the moment is due only to orbital motion, and g=2 when the moment is due only to spin. For most microwave ferrite materials, gis in the range 1.98–2.01, so g=2 is a good approximation. In most solids, electron spins occur in pairs with opposite signs, so the overall mag- netic moment is negligible. In a magnetic material, however, a large fraction of the electronspins are unpaired (more left-hand spins than right-hand spins, or vice versa), but are gen- erally oriented in random directions so that the net magnetic moment is still small. An c09FerrimagneticComponents Pozar August 25, 2011 18:36 9.1 Basic Properties of Ferrimagnetic Materials 453 z H0 m s/H9258 Spinning electron FIGURE 9.1 Spin magnetic dipole moment and angular momentum vectors for a spinning electron. external magnetic field, however, can cause the dipole moments to align in the same di- rection to produce a large overall magnetic moment. The existence of exchange forces can keep adjacent electron spins aligned after the external field is removed; the material is thensaid to be permanently magnetized. An electron has a spin angular momentum given in terms of Planck’s constant as [1, 2] s=h − 2.( 9.2) The vector direction of this momentum is opposite the direction of the spin magnetic dipole moment, as indicated in Figure 9.1. The ratio of the spin magnetic moment to the spin an-gular momentum is a constant called the gyromagnetic ratio : γ=m s=q me=1.759×1011C/kg,( 9.3) where (9.1) and (9.2) have been used. Then we can write the following vector relation between the magnetic moment and the angular momentum: ¯m=−γ¯s,( 9.4) where the negative sign is due to the fact that these vectors are oppositely directed. When a magnetic bias field ¯H0=ˆzH0is present, a torque will be exerted on the mag- netic dipole: ¯T=¯mׯB0=µ0¯mׯH0=−µ0γ¯sׯH0.( 9.5) Since torque is equal to the time rate of change of angular momentum, we have d¯s dt=−1 γd¯m dt=¯T=µ0¯mׯH0, c09FerrimagneticComponents Pozar August 25, 2011 18:36 454 Chapter 9: Theory and Design of Ferrimagnetic Components or d¯m dt=−µ0γ¯mׯH0.( 9.6) This is the equation of motion for the magnetic dipole moment, ¯m. We will solve this equation to show that the magnetic dipole precesses around the H0-field vector, similar to a spinning top precessing around a vertical axis. Writing (9.6) in terms of its three vector components gives dmx dt=−µ0γmyH0, (9.7a) dmy dt=µ0γmxH0, (9.7b) dmz dt=0. (9.7c) Now use (9.7a) and (9.7b) to obtain two equations for mxandmy: d2mx dt2+ω2 0mx=0, (9.8a) d2my dt2+ω2 0my=0, (9.8b) where ω0=µ0γH0 (9.9) is called the Larmor,o rprecession, frequency. One solution to (9.8) that is compatible with (9.7a) and (9.7b) is given by mx=Acosω0t, (9.10a) my=Asinω0t. (9.10b) Equation (9.7c) shows that mzis a constant, and (9.1) shows that the magnitude of ¯mis also a constant, so we have the relation that |¯m|2=/parenleftbiggqh− 2me/parenrightbigg2 =m2 x+m2 y+m2 z=A2+m2 z.( 9.11) Thus the precession angle, θ, between ¯mand¯H0(thez-axis) is given by sinθ=/radicalBig m2x+m2y |¯m|=A |¯m|.( 9.12) The projection of ¯mon the xyplane is given by (9.10), which shows that ¯mtraces a circular path in this plane. The position of this projection at time tis given by φ=ω0t,s ot h e angular rate of rotation is dφ/dt=ω0, the precession frequency. In the absence of any damping forces, the actual precession angle will be determined by the initial position of the magnetic dipole, and the dipole will precess about ¯H0at this angle indefinitely (free precession). In reality, however, the existence of damping forces will cause the magnetic dipole moment to spiral in from its initial angle until ¯mis aligned with ¯H0(θ=0). c09FerrimagneticComponents Pozar August 25, 2011 18:36 9.1 Basic Properties of Ferrimagnetic Materials 455 Ms Applied bias field H0Magnetic moment M 0 FIGURE 9.2 Magnetic moment of a ferrimagnetic material versus bias field, H0. Now assume that there are Nunbalanced electron spins (magnetic dipoles) per unit volume, so that the total magnetization is ¯M=N¯m,( 9.13) and the equation of motion in (9.6) becomes d¯M dt=−µ0γ¯MׯH,( 9.14) where ¯His the internal applied field. (Note: In Chapter 1 we used ¯Pmfor magnetization and¯Mfor magnetic currents; here we use ¯Mfor magnetization, as this is common practice in ferrimagnetics work. Since we will not be using magnetic currents in this chapter, thereshould be no confusion.) As the strength of the bias field H 0is increased, more magnetic dipole moments will align with H0until all are aligned, and ¯Mreaches an upper limit. See Figure 9.2. The material is then said to be magnetically saturated , and Msis denoted as thesaturation magnetization. Msis thus a physical property of the ferrite material, and it typically ranges from 4 πMs=300–5000 G. (Appendix H lists the saturation magnetiza- tion and other physical properties of several types of microwave ferrite materials.) Below saturation, ferrite materials can be very lossy at microwave frequencies, and the RF inter- action is reduced. For this reason ferrites are usually operated in the saturated state, andthis assumption is made for the remainder of this chapter. The saturation magnetization of a material is a strong function of temperature, de- creasing as temperature increases. This effect can be understood by noting that the vibra-tional energy of an atom increases with temperature, making it more difficult to align all the magnetic dipoles. At a high enough temperature the thermal energy is greater than the energy supplied by the internal magnetic field, and a zero net magnetization results. Thistemperature is called the Curie temperature, T C. Now consider the interaction of a small AC (microwave) magnetic field with a mag- netically saturated ferrite material. Such a field will cause a forced precession of the dipolemoments around the ¯H 0(ˆz)axis at the frequency of the applied AC field, much like the operation of an AC synchronous motor. The small-signal approximation will apply to all the ferrite components of interest to us, but there are applications where high-power signalscan be used to obtain useful nonlinear effects. If¯His the applied AC field, the total magnetic field is ¯H t=H0ˆz+¯H,( 9.15) c09FerrimagneticComponents Pozar August 25, 2011 18:36 456 Chapter 9: Theory and Design of Ferrimagnetic Components where we assume that |¯H|/lessmuch H0. This field produces a total magnetization in the ferrite material given by ¯Mt=Msˆz+¯M,( 9.16) where Msis the (DC) saturation magnetization and ¯Mis the additional (AC) magnetization (in the xyplane) caused by ¯H. Substituting (9.16) and (9.15) into (9.14) gives the following component equations of motion: dMx dt=−µ0γMy(H0+Hz)+µ0γ(Ms+Mz)Hy, (9.17a) dM y dt=µ0γMx(H0+Hz)−µ0γ(Ms+Mz)Hx, (9.17b) dM z dt=−µ0γMxHy+µ0γMyHx, (9.17c) since dM s/dt=0. Since |¯H|/lessmuch H0,w eh a v e |¯M||¯H|/lessmuch| ¯M|H0and|¯M||¯H|/lessmuch Ms|¯H|, so we can ignore MHproducts. Then (9.17) reduces to dM x dt=−ω0My+ωmHy, (9.18a) dM y dt=ω0Mx−ωmHx, (9.18b) dM z dt=0, (9.18c) where ω0=µ0γH0andωm=µ0γMs. Solving (9.18a) and (9.18b) for MxandMygives the following equations: d2Mx dt2+ω2 0Mx=ωmdHy dt+ω0ωmHx, (9.19a) d2My dt2+ω2 0My=−ωmdHx dt+ω0ωmHy. (9.19b) These are the equations of motion for the forced precession of the magnetic dipoles, as- suming small-signal conditions. It is now an easy step to arrive at the permeability tensor for ferrites; after doing this, we will try to gain some physical insight into the magnetic interaction process by considering circularly polarized AC fields. If the AC ¯Hfield has an ejωttime-harmonic dependence, the AC steady-state form of (9.19) reduces to the following phasor equations: /parenleftbig ω2 0−ω2/parenrightbig Mx=ω0ωmHx+jωωmHy, (9.20a) /parenleftbig ω2 0−ω2/parenrightbig My=− jωωmHx+ω0ωmHy, (9.20b) which shows the linear relationship between ¯Hand¯M. As in (1.24), (9.20) can be written with a tensor susceptibility, [χ], to relate ¯Hand¯M: ¯M=[χ]¯H=/bracketleftBiggχxxχxy0 χyxχyy0 00 0/bracketrightBigg ¯H,( 9.21) c09FerrimagneticComponents Pozar August 25, 2011 18:36 9.1 Basic Properties of Ferrimagnetic Materials 457 where the elements of [χ]are given by χxx=χyy=ω0ωm ω2 0−ω2, (9.22a) χxy=−χyx=jωωm ω2 0−ω2. (9.22b) Theˆzcomponent of ¯Hdoes not affect the magnetic moment of the material under the above assumptions. To relate ¯Band¯H, we have from (1.23) that ¯B=µ0(¯M+¯H)=[µ]¯H,( 9.23) where the tensor permeability [µ]is given by [µ]=µ 0([U]+[χ])=/bracketleftBiggµ jκ 0 −jκµ 0 00 µ0/bracketrightBigg (ˆzbias). (9.24) The elements of the permeability tensor are then µ=µ0(1+χxx)=µ0(1+χyy)=µ0/parenleftBigg 1+ω0ωm ω2 0−ω2/parenrightBigg , (9.25a) κ=− jµ0χxy=jµ0χyx=µ0ωωm ω2 0−ω2. (9.25b) A material having a permeability tensor of this form is called gyrotropic; note that an ˆx(or ˆy) component of ¯Hgives rise to both ˆxandˆycomponents of ¯B,w i t ha9 0◦phase shift between them. If the direction of bias is reversed, both H0andMswill change signs, so ω0andωm will change signs. Equation (9.25) then shows that µwill be unchanged, but κwill change sign. If the bias field is suddenly removed ( H0=0), the ferrite will generally remain mag- netized (0 <|M|<Ms); only by demagnetizing the ferrite (e.g., with a decreasing AC bias field) can M=0 be obtained. Since the results of (9.22) and (9.25) assume a satu- rated ferrite sample, both MsandH0should be set to zero for the unbiased, demagnetized case. Then ω0=ωm=0 and (9.25) show that µ=µ0andκ=0, as expected for a non- magnetic material. The tensor results of (9.24) assume magnetic bias in the ˆzdirection. If the ferrite is biased in a different direction, the permeability tensor will be transformed according to the change in coordinates. Thus, if ¯H0=ˆxH0, the permeability tensor will be [µ]=/bracketleftBiggµ0 00 0µ jκ 0−jκµ/bracketrightBigg (ˆxbias), (9.26) while if ¯H0=ˆyH0the permeability tensor will be [µ]=/bracketleftBiggµ 0−jκ 0µ0 0 jκ 0µ/bracketrightBigg (ˆybias). (9.27) A comment must be made about units. By tradition most practical work in mag- netics is done with CGS units, with magnetization measured in gauss (1 gauss [G] = 10−4weber/m2), and field strength measured in oersteds (4π ×10−3oersted [Oe] =1 A/m). Thus, µ0=1 G/Oe in CGS units, implying that BandHhave the same numerical c09FerrimagneticComponents Pozar September 12, 2011 22:27 458 Chapter 9: Theory and Design of Ferrimagnetic Components values in a nonmagnetic material. Saturation magnetization is usually expressed as 4 πMs gauss; the corresponding MKS value is then µ0Msweber/m2=10−4(4πMsgauss). In CGS units, the Larmor frequency can be expressed as f0=ω0/2π=µ0γH0/2π=(2.8 MHz/Oe) ×(H0oersted), and fm=ωm/2π=µ0γMs/2π=(2.8 MHz/Oe) ×(4πMs gauss). In practice, these units are convenient and easy to use. CircularlyPolarizedFields To get a better physical understanding of the interaction of an AC signal with a saturated ferrimagnetic material we will consider circularly polarized fields. As discussed in Section 1.5, a right-hand circularly polarized (RHCP) field can be expressed in phasor form as ¯H+=H+(ˆx−jˆy), (9.28a ) and in time domain form as ¯H+=Re{¯H+ejωt}=H+(ˆxcosωt+ˆysinωt), (9.28b) where we have assumed the amplitude H+as real. This latter form shows that ¯H+is a vector that rotates with time, such that at time tit is oriented at the angle ωtfrom the x-axis; thus its angular velocity is ω. (Also note that |¯H+|=H+/negationslash=|¯H+|.) Applying the RHCP field of (9.28a) to (9.20) gives the magnetization components as M+ x=ωm ω0−ωH+, M+ y=−jωm ω0−ωH+, so the magnetization vector resulting from ¯H+can be written as ¯M+=M+ xˆx+M+ yˆy=ωm ω0−ωH+(ˆx−jˆy), (9.29) which shows that the magnetization is also RHCP, and so it rotates with angular veloc- ityωin synchronism with the driving field, ¯H+. Since ¯M+and¯H+are vectors in the same direction, we can write ¯B+=µ0(¯M++¯H+)=µ+¯H+, where µ+is the effective permeability for an RHCP wave given by µ+=µ0/parenleftbigg 1+ωm ω0−ω/parenrightbigg =µ+κ. (9.30) The angle, θM, between M+and the z-axis is given by tanθM=|M+| Ms=ωmH+ (ω0−ω)Ms=ω0H+ (ω0−ω)H0,( 9.31) while the angle, θH, between ¯H+and the z-axis, is given by tanθH=|H+| H0=H+ H0.( 9.32) For frequencies such that ω< 2ω0, (9.31) and (9.32) show that θM>θ H, as illustrated in Figure 9.3a. In this case the magnetic dipole is precessing in the same direction as it wouldfreely precess in the absence of ¯H +. Now consider a left-hand circularly polarized field (LHCP), expressed in phasor form as ¯H−=H−(ˆx+jˆy), (9.33a) c09FerrimagneticComponents Pozar August 25, 2011 18:36 9.1 Basic Properties of Ferrimagnetic Materials 459 z H0 xy/H9258H /H9258M/H9258H /H9258MH0 HtHtH+H– M+M–z xy (a) (b) FIGURE 9.3 Forced precession of a magnetic dipole with circularly polarized fields. (a) RHCP, θM>θH.( b )L H C P ,θ M<θH. and in time domain form as ¯H−=Re{¯H−ejωt}=H−(ˆxcosωt−ˆysinωt). (9.33b) Equation (9.33b) shows that ¯H−is a vector rotating in the −ω(left-hand) direction. Ap- plying the LHCP field of (9.33a) to (9.20) gives the magnetization components as M− x=ωm ω0+ωH−, M− y=jωm ω0+ωH−, so the vector magnetization can be written as ¯M−=M− xˆx+M− yˆy=ωm ω0+ωH−(ˆx+jˆy), (9.34) which shows that the magnetization is LHCP, rotating in synchronism with ¯H−. Writing ¯B−=µ0(¯M−+¯H−)=µ−¯H−gives the effective permeability for an LHCP wave as µ−=µ0/parenleftbigg 1+ωm ω0+ω/parenrightbigg =µ−κ. (9.35) The angle, θM, between ¯M−and the z-axis is given by tanθM=|¯M−| Ms=ωmH− (ω0+ω)Ms=ω0H− (ω0+ω)H0,( 9.36) which is seen to be less than θHof (9.32), as shown in Figure 9.3b. In this case the magnetic dipole is precessing in the opposite direction of its free precession. Thus we see that the interaction of a circularly polarized wave with a magnetically biased ferrite depends on the sense of the polarization (RHCP or LHCP). This is because the bias field sets up a preferential precession direction coinciding with the direction of c09FerrimagneticComponents Pozar August 25, 2011 18:36 460 Chapter 9: Theory and Design of Ferrimagnetic Components forced precession for an RHCP wave, but opposite to that of an LHCP wave. As we will see in Section 9.2, this effect leads to nonreciprocal propagation characteristics. EffectofLoss Equations (9.22) and (9.25) show that the elements of the susceptibility or permeability ten- sors become infinite when the frequency, ω, equals the Larmor frequency, ω0. This effect is known as gyromagnetic resonance , and it occurs when the forced precession frequency is equal to the free precession frequency. In the absence of loss the response may be un- bounded, in the same way that the response of an LCresonant circuit will be unbounded when driven with an AC signal having a frequency equal to the resonant frequency of the LCcircuit. All real ferrite materials, however, have various magnetic loss mechanisms that damp out such singularities. As with other resonant systems, loss can be accounted for by making the resonant frequency complex: ω0←ω0+jαω, (9.37) where αis a damping factor. Substituting (9.37) into (9.22) makes the susceptibilities complex: χxx=χ/prime xx−jχ/prime/prime xx (9.38a) χxy=χ/prime/prime xy+jχ/prime xy (9.38b) where the real and imaginary parts are given by χ/prime xx=ω0ωm/parenleftbig ω2 0−ω2/parenrightbig +ω0ωmω2α2 /bracketleftbig ω2 0−ω2(1+α2)/bracketrightbig2+4ω2 0ω2α2, (9.39a) χ/prime/prime xx=αωω m/bracketleftbig ω2 0+ω2(1+α2)/bracketrightbig /bracketleftbig ω2 0−ω2(1+α2)/bracketrightbig2+4ω2 0ω2α2, (9.39b) χ/prime xy=ωωm/bracketleftbig ω2 0−ω2(1+α2)/bracketrightbig /bracketleftbig ω2 0−ω2(1+α2)/bracketrightbig2+4ω2 0ω2α2, (9.39c) χ/prime/prime xy=2ω0ωmω2α /bracketleftbig ω2 0−ω2(1+α2)/bracketrightbig2+4ω2 0ω2α2. (9.39d) Equation (9.37) can also be applied to (9.25) to give a complex µ=µ/prime−jµ/prime/prime, andκ= κ/prime−jκ/prime/prime; this is why (9.38b) appears to define χ/prime xyandχ/prime/prime xybackward, as χxy=jκ/µ 0. For most ferrite materials the loss is small, so α/lessmuch1, and the (1+α2)terms in (9.39) can be approximated as unity. The real and imaginary parts of the susceptibilities of (9.39) are sketched in Figure 9.4 for a typical ferrite. The damping factor, α, is related to the linewidth, /Delta1H, of the susceptibility curve near resonance. Consider the plot of χ/prime/prime xxversus bias field, H0, shown in Figure 9.5. For a fixed frequency ω, resonance occurs when H0=Hr, such that ω0=µ0γHr. The linewidth, /Delta1H, is defined as the width of the curve of χ/prime/prime xxversus H0where χ/prime/prime xxhas decreased to half its peak value. If we assume (1+α2)/similarequal1, (9.39b) shows that the maximum value of χ/prime/prime xx isωm/2αω , and occurs when ω=ω0.N o wl e tω 02be the Larmor frequency for which H0=H2, where χ/prime/prime xxhas decreased to half its maximum value. Then we can solve (9.39b) c09FerrimagneticComponents Pozar August 25, 2011 18:36 9.1 Basic Properties of Ferrimagnetic Materials 461 /H9273xx 0.5 /H92750//H9275' 1.0 1.5 2.0/H9273xx 0.5 /H92750//H9275" 1.0 1.5 2.0 /H9273xy 0.5 /H92750//H9275' 1.0 1.5 2.0/H9273xy 0.5 /H92750//H9275" 1.0 1.5 2.0(a) (b) FIGURE 9.4 Complex susceptibilities for a typical ferrite. (a) Real and imaginary parts of χxx. (b) Real and imaginary parts of χxy. forαin terms of ω02: αωω m/parenleftbig ω2 02+ω2/parenrightbig /parenleftbig ω2 02−ω2/parenrightbig2+4ω2 02ω2α2=ωm 4αω, 4α2ω4=/parenleftBig ω2 02−ω2/parenrightBig2 , ω02=ω√ 1+2α/similarequalω(1+α). Then /Delta1ω 0=2(ω 02−ω0)/similarequal2[ω( 1+α)−ω]=2αω , and using (9.9) gives the line- width as /Delta1H=/Delta1ω 0 µ0γ=2αω µ0γ.( 9.40) 0/H9273xx" /H9273max /H9273max/2 H1HrH2 H0(/H92750//H9275)H FIGURE 9.5 Definition of the linewidth, /Delta1H, of the gyromagnetic resonance. c09FerrimagneticComponents Pozar August 25, 2011 18:36 462 Chapter 9: Theory and Design of Ferrimagnetic Components FerriteAir AirFerriteAir AirHa Ha HaHa MsMsH0 H0 (b) (a) FIGURE 9.6 Internal and external fields for a thin ferrite plate. (a) Normal bias. (b) Tangential bias. Typical linewidths range from less than 100 Oe (for YIG) to 100–500 Oe (for ferrites); single-crystal YIG can have a linewidth as low as 0.3 Oe. Also note that this loss is separate from the dielectric loss that a ferrimagnetic material may have. DemagnetizationFactors The DC bias field, H0, internal to a ferrite sample is generally different from the externally applied field, Ha, because of the boundary conditions at the surface of the ferrite. To illus- trate this effect, consider a thin ferrite plate, as shown in Figure 9.6. When the applied field is normal to the plate, continuity of Bnat the surface of the plate gives Bn=µ0Ha=µ0(Ms+H0), so the internal magnetic bias field is H0=Ha−Ms. This shows that the internal field is less than the applied field by an amount equal to the saturation magnetization. When the applied field is parallel to the ferrite plate, continuity ofHtat the surfaces of the plate gives Ht=Ha=H0. In this case the internal field is not reduced. In general, the internal field (AC or DC), ¯H, is affected by the shape of the ferrite sample and its orientation with respect to the externalfield, ¯H e, and can be expressed as ¯H=¯He−N¯M,( 9.41) where N=Nx,Ny,o rNzis called the demagnetization factor for that direction of the external field. Different shapes have different demagnetization factors, which depend on the direction of the applied field. Table 9.1 lists the demagnetization factors for a fewsimple shapes. The demagnetization factors are defined such that N x+Ny+Nz=1. The demagnetization factors can also be used to relate the internal and external RF fields near the boundary of a ferrite sample. For a z-biased ferrite with transverse RF fields, (9.41) reduces to Hx=Hxe−NxMx, (9.42a) Hy=Hye−NyMy, (9.42b) Hz=Ha−NzMs, (9.42c) c09FerrimagneticComponents Pozar August 25, 2011 18:36 9.1 Basic Properties of Ferrimagnetic Materials 463 TABLE 9.1 Demagnetization Factors for Some Simple Shapes z zyx x xShape Thin disk or plate Thin rod Sphere0 01 1 2 1 31 31 31 20Nx Ny Nz z y where Hxe,Hyeare the RF fields external to the ferrite, and Hais the externally applied bias field. Equation (9.21) relates the internal transverse RF fields and magnetization as Mx=χxxHx+χxyHy, My=χyxHx+χyyHy. Using (9.42a, b) to eliminate HxandHygives Mx=χxxHxe+χxyHye−χxxNxMx−χxyNyMy, My=χyxHxe+χyyHye−χyxNxMx−χyyNyMy. These equations can be solved for Mx,Myto give Mx=χxx(1+χyyNy)−χxyχyxNy DHxe+χxy DHye, (9.43a) My=χyx DHxe+χyy(1+χxxNx)−χyxχxyNx DHye, (9.43b) where D=(1+χxxNx)(1+χyyNy)−χyxχxyNxNy.( 9.44) This result is of the form ¯M=[χe]¯H, where the coefficients of HxeandHyein (9.43) can be defined as “external” susceptibilities since they relate magnetization to the externalRF fields. For an infinite ferrite medium gyromagnetic resonance occurs when the denominator of the susceptibilities of (9.22) vanishes, at the frequency ω r=ω=ω0. However, for a finite-sized ferrite sample the gyromagnetic resonance frequency is altered by the demag- netization factors, and is given by the condition that D=0 in (9.43). Using the expressions c09FerrimagneticComponents Pozar August 25, 2011 18:36 464 Chapter 9: Theory and Design of Ferrimagnetic Components in (9.22) for the susceptibilities in (9.44), and setting the result equal to zero, gives /parenleftBigg 1+ω0ωmNx ω2 0−ω2/parenrightBigg/parenleftBigg 1+ω0ωmNy ω2 0−ω2/parenrightBigg −ω2ω2 m/parenleftbig ω2 0−ω2/parenrightbig2NxNy=0. After some algebraic manipulations this result can be reduced to give the resonance fre- quency, ωr,a s ωr=ω=/radicalBig (ω0+ωmNx)(ω0+ωmNy). (9.45) Since ω0=µ0γH0=µ0γ(Ha−NzMs), and ωm=µ0γMs, (9.45) can be rewritten in terms of the applied bias field strength and saturation magnetization as ωr=µ0γ/radicalBig [Ha+(Nx−Nz)Ms][Ha+(Ny−Nz)Ms].( 9.46) This result is known as Kittel’s equation [4]. POINT OF INTEREST: Permanent Magnets Since ferrite components such as isolators, gyrators, and circulators generally use permanent magnets to supply the required DC bias field, it may be useful to discuss some of the important characteristics of permanent magnets. A permanent magnet is made by placing the magnetic material in a strong magnetic field, and then removing the field to leave the material magnetized in a remanent state. Unless the magnet shape forms a closed path (like a toroid), the demagnetization factors at the magnetends will cause a slightly negative Hfield to be induced in the magnet. Thus the “operating point” of a permanent magnet will be in the second quadrant of the B–Hhysteresis curve for the magnet material. This portion of the curve is called the demagnetization curve. A typical example is shown below. 0Typical permanent magnet (BH)maxB Br –Hc –H The residual magnetization, for H=0, is called the remanence, Br, of the material. This quantity characterizes the strength of the magnet, so generally a magnet material is chosento have a large remanence. Another important parameter is the coercivity, H c, which is the value of the negative Hfield required to reduce the magnetization to zero. A good permanent magnet should have a high coercivity to reduce the effects of vibration, temperature changes,and external fields, which can lead to a loss of magnetization. An overall figure of merit for a permanent magnet is sometimes given as the maximum value of the BHproduct, (BH ) max,o n the demagnetization curve. This quantity is essentially the maximum magnetic energy densitythat can be stored by the magnet, and can be useful in electromechanical applications. The c09FerrimagneticComponents Pozar August 25, 2011 18:36 9.2 Plane Wave Propagation in a Ferrite Medium 465 following table lists the remanence, coercivity, and ( BH)max for some of the most common permanent magnet materials. Br Hc (BH)max Material Composition (Oe) (G) (G-Oe) ×106 ALNICO 5 Al, Ni, Co, Cu 12,000 720 5.0 ALNICO 8 Al, Ni, Co, Cu, Ti 7100 2000 5.5 ALNICO 9 Al, Ni, Co, Cu, Ti 10,400 1600 8.5Remalloy Mo, Co, Fe 10,500 250 1.1 Platinum cobalt Pt, Co 6450 4300 9.5 Ceramic BaO 6Fe2O3 3950 2400 3.5 Cobalt samarium Co, Sm 8400 7000 16.0 9.2PLANEWAVEPROPAGATIONINAFERRITEMEDIUM The previous section gives an explanation of the microscopic phenomena that occur inside a biased ferrite material to produce a tensor permeability of the form given in (9.24) [or in (9.26) or (9.27), depending on the bias direction]. Once we have this macroscopic de-scription of the ferrite material, we can solve Maxwell’s equations for wave propagation in various geometries involving ferrite materials. We begin with plane wave propagation in an infinite ferrite medium, for propagation either in the direction of bias, or transverse tothe bias field. These problems will illustrate the important effects of Faraday rotation and birefringence. PropagationinDirectionofBias(FaradayRotation) Consider an infinite ferrite-filled region with a DC magnetic bias field given by ¯H 0=ˆzH0, and a tensor permittivity [µ] given by (9.24). Maxwell’s equations can be written as ∇ׯE=− jω[µ]¯H, (9.47a) ∇ׯH=jω/epsilon1¯E, (9.47b) ∇·¯D=0, (9.47c) ∇·¯B=0. (9.47d) Now assume plane wave propagation in the zdirection, with ∂/∂x=∂/∂y=0. Then the electric and magnetic fields will have the following form: ¯E=¯E0e−jβz, (9.48a) ¯H=¯H0e−jβz. (9.48b) The two curl equations of (9.47a) and (9.47b) reduce to the following, after using (9.24): jβEy=− jω(µH x+jκHy), (9.49a) −jβEx=− jω(−jκHx+µHy), (9.49b) 0=− jωµ0Hz, (9.49c) jβHy=jω/epsilon1Ex, (9.49d) −jβHx=jω/epsilon1Ey, (9.49e) 0=jω/epsilon1Ez. (9.49f) c09FerrimagneticComponents Pozar August 25, 2011 18:36 466 Chapter 9: Theory and Design of Ferrimagnetic Components Equations (9.49c) and (9.49f) show that Ez=Hz=0, as expected for TEM plane waves. We also have ∇·¯D=∇· ¯B=0 since ∂/∂x=∂/∂y=0. Equations (9.49d) and (9.49e) give relations between the transverse field components as Y=Hy Ex=−Hx Ey=ω/epsilon1 β,( 9.50) where Yis the wave admittance. Using (9.50) in (9.49a) and (9.49b) to eliminate Hxand Hygives the following results: jω2/epsilon1κEx+(β2−ω2µ/epsilon1)Ey=0, (9.51a) (β2−ω2µ/epsilon1)Ex−jω2/epsilon1κEy=0. (9.51b) For a nontrivial solution for ExandEythe determinant of this set of equations must vanish: ω4/epsilon12κ2−(β2−ω2µ/epsilon1)2=0, or β±=ω/radicalbig /epsilon1(µ±κ). (9.52) There are two possible propagation constants, β+andβ−. First consider the fields associated with β+, which can be found by substituting β+ into (9.51a) or (9.51b): jω2/epsilon1κEx+ω2/epsilon1κEy=0, or Ey=− jEx. Then the electric field of (9.48a) must have the following form: ¯E+=E0(ˆx−jˆy)e−jβ+z,( 9.53a) which is seen to be a right-hand circularly polarized plane wave. Using (9.50) gives the associated magnetic field as ¯H+=E0Y+(jˆx+ˆy)e−jβ+z,( 9.53b) where Y+is the wave admittance for this wave: Y+=ω/epsilon1 β+=/radicalbigg/epsilon1 µ+κ.( 9.53c) Similarly, the fields associated with β−are left-hand circularly polarized: ¯E−=E0(ˆx+jˆy)e−jβ−z, (9.54a) ¯H−=E0Y−(−jˆx+ˆy)e−jβ−z, (9.54b) where Y−is the wave admittance for this wave: Y−=ω/epsilon1 β−=/radicalbigg/epsilon1 µ−κ.( 9.54c) Thus we see that RHCP and LHCP plane waves are the source-free modes of the ˆz- biased ferrite medium, and these waves propagate through the ferrite medium with differentpropagation constants. As discussed in the previous section, the physical explanation for this effect is that the magnetic bias field creates a preferred direction for magnetic dipole c09FerrimagneticComponents Pozar August 25, 2011 18:36 9.2 Plane Wave Propagation in a Ferrite Medium 467 precession, and one sense of circular polarization causes precession in this preferred direc- tion, while the other sense of polarization causes precession in the opposite direction. Also note that for an RHCP wave, the ferrite material can be represented with an effective per- meability of µ+κ, while for an LHCP wave the effective permeability is µ−κ. In math- ematical terms, we can state that (µ+κ)and(µ−κ),o rβ+andβ−,a r et h e eigenvalues of the system of equations in (9.51), and that ¯E+and¯E−are the associated eigenvectors . When losses are present, the attenuation constants for RHCP and LHCP waves will also bedifferent. Now consider a linearly polarized electric field at z=0, represented as the sum of an RHCP and an LHCP wave: ¯E| z=0=ˆxE0=E0 2(ˆx−jˆy)+E0 2(ˆx+jˆy). (9.55) The RHCP component will propagate in the zdirection as e−jβ+z, and the LHCP compo- nent will propagate as e−jβ−z, so the total field of (9.55) will propagate as ¯E=E0 2(ˆx−jˆy)e−jβ+z+E0 2(ˆx+jˆy)e−jβ−z =E0 2ˆx(e−jβ+z+e−jβ−z)−jE0 2ˆy(e−jβ+z−e−jβ−z) =E0/bracketleftbigg ˆxcos/parenleftbiggβ+−β− 2/parenrightbigg z−ˆysin/parenleftbiggβ+−β− 2/parenrightbigg z/bracketrightbigg e−j(β++β−)z/2.(9.56) This is still a linearly polarized wave, but one whose direction of polarization rotates as the wave propagates along the z-axis. At a given point along the z-axis the polarization direction measured from the x-axis is given by φ=tan−1Ey Ex=tan−1/bracketleftbigg −tan/parenleftbiggβ+−β− 2/parenrightbigg z/bracketrightbigg =−/parenleftbiggβ+−β− 2/parenrightbigg z.( 9.57) This effect is called Faraday rotation, after Michael Faraday, who first observed this phe- nomenon during his study of the propagation of light through liquids that had magneticproperties. Note that for a fixed position on the z-axis, the polarization angle is fixed, unlike the case for a circularly polarized wave, where the polarization direction rotates with time. Forω<ω 0,µandκare positive and µ>κ . Then β+>β−, and (9.57) shows that φbecomes more negative as zincreases, meaning that the polarization (direction of ¯E) rotates counterclockwise as we look in the +zdirection. Reversing the bias direction (sign ofH0andMs) changes the sign of κ, which changes the direction of rotation to clockwise. Similarly, for +zbias, a wave traveling in the −zdirection will rotate its polarization clockwise as we look in the direction of propagation (− z); if we were looking in the +z direction, however, the direction of rotation would be counterclockwise (same as a wave propagating in the +zdirection). Thus, a wave that travels from z=0t oz=Land back again to z=0 undergoes a total polarization rotation of 2 φ, where φis given in (9.57) with z=L. So, unlike the situation of a screw being driven into a block of wood and then backed out, the polarization does not “unwind” when the direction of propagation is reversed. Faraday rotation is thus seen to be a nonreciprocal effect. EXAMPLE 9.1 PLANE WA VE PROPAGATION IN A FERRITE MEDIUM Consider an infinite ferrite medium with 4π Ms=1800 G, /Delta1H=75 Oe, /epsilon1r= 14, and tan δ=0.001. If the bias field strength is H0=3570 Oe, calculate and plot the phase and attenuation constants for RHCP and LHCP plane waves versus frequency for f=0t o2 0G H z . c09FerrimagneticComponents Pozar August 25, 2011 18:36 468 Chapter 9: Theory and Design of Ferrimagnetic Components Solution The Larmor precession frequency is f0=ω0 2π=(2.8M H z / O e )(3570 Oe) =10.0 GHz , and fm=ωm 2π=(2.8M H z / O e )(1800 G) =5.04 GHz. At each frequency we can compute the complex propagation constant as γ±=α±+jβ±=jω/radicalbig /epsilon1(µ±κ), where /epsilon1=/epsilon10/epsilon1r(1−jtanδ)is the complex permittivity, and µ,κ are given by (9.25). The following substitution for ω0is used to account for ferrimagnetic loss: ω0←ω0+jµ0γ/Delta1H 2, or f0←f0+j(2.8M H z /Oe)(75 Oe) 2=(10.0+j0.105) GHz, which is derived from (9.37) and (9.40). The quantities µ±κcan be simplified to the following, by using (9.25): µ+κ=µ0/parenleftbigg 1+ωm ω0−ω/parenrightbigg , µ−κ=µ0/parenleftbigg 1+ωm ω0+ω/parenrightbigg . The phase and attenuation constants are plotted in Figure 9.7, normalized to the free-space wave number, k0. 0 2 4 6 8 1 01 21 41 61 82 0024681012141618202224 /H9252+/k0/H9251+/k0 /H9252–/k0 100/H9251–/k0f0 + fm f0Stopband for RHCP wave Frequenc y (GHz)Normalized phase and attenuation constants FIGURE 9.7 Normalized phase and attenuation constants for circularly polarized plane waves in the ferrite medium of Example 9.1. c09FerrimagneticComponents Pozar August 25, 2011 18:36 9.2 Plane Wave Propagation in a Ferrite Medium 469 Observe that β+andα+(for an RHCP wave) show a resonance near f= f0=10 GHz; β−andα−(for an LHCP wave) do not, however, because the singularities in µandκcancel in the (µ−κ)term contained in γ−. Also note from Figure 9.7 that a stopband (β +near zero, large α+)exists for RHCP waves for frequencies between f0and f0+fm(between ω0andω0+ωm). For fre- quencies in this range, the above expression for (µ+κ)shows that this quantity is negative, and β+=0 (in the absence of loss), so an RHCP wave incident on such a ferrite medium would be totally reflected. ■ PropagationTransversetoBias(Birefringence) Next consider the case where an infinite ferrite region is biased in the ˆxdirection, transverse to the direction of propagation; the permeability tensor is given in (9.26). For plane wavefields of the form in (9.48), Maxwell’s curl equations reduce to jβE y=− jωµ0Hx, (9.58a) −jβEx=− jω(µH y+jκHz), (9.58b) 0=− jω(−jκHy+µHz), (9.58c) jβHy=jω/epsilon1Ex, (9.58d) −jβHx=jω/epsilon1Ey, (9.58e) 0=jω/epsilon1Ez. (9.58f) Then Ez=0, and ∇·¯D=0 since ∂/∂x=∂/∂y=0. Equations (9.58d) and (9.58e) give an admittance relation between the transverse field components: Y=Hy Ex=−Hx Ey=ω/epsilon1 β.( 9.59) Using (9.59) in (9.58a) and (9.58b) to eliminate HxandHy, and using (9.58c) in (9.58b) to eliminate Hz, gives the following results: β2Ey=ω2µ0/epsilon1Ey, (9.60a) µ(β2−ω2µ/epsilon1)Ex=−ω2/epsilon1κ2Ex. (9.60b) One solution to (9.60) occurs for βo=ω√µ0/epsilon1, (9.61) with Ex=0. Then the complete fields are ¯Eo=ˆyE0e−jβoz, (9.62a) ¯Ho=−ˆ xE0Yoe−jβoz, (9.62b) since (9.59) shows that Hy=0 when Ex=0, and (9.58c) shows that Hz=0 when Hy=0. The admittance is Yo=ω/epsilon1 βo=/radicalbigg/epsilon1 µ0.( 9.63) This wave is called the ordinary wave because it is unaffected by the magnetic properties of the ferrite. This happens whenever the magnetic field components transverse to the bias c09FerrimagneticComponents Pozar August 25, 2011 18:36 470 Chapter 9: Theory and Design of Ferrimagnetic Components direction are zero ( Hy=Hz=0). The wave propagates in either the +zor−zdirection with the same propagation constant, which is independent of H0. Another solution to (9.60) occurs for βe=ω√µe/epsilon1, (9.64) with Ey=0, where µeis an effective permeability given by µe=µ2−κ2 µ.( 9.65) This wave is called the extraordinary wave and is affected by the ferrite magnetization. Note that the effective permeability may be negative for certain values of ω, ω 0. The elec- tric field is ¯Ee=ˆxE0e−jβez.( 9.66a) Since Ey=0, (9.58e) shows that Hx=0.Hycan be found from (9.58d), and Hzfrom (9.58c), giving the complete magnetic field as ¯He=E0Ye/parenleftbigg ˆy+ˆzjκ µ/parenrightbigg e−jβez,( 9.66b) where Ye=ω/epsilon1 βe=/radicalbigg/epsilon1 µe.( 9.67) These fields constitute a linearly polarized wave, but note that the magnetic field has a com- ponent in the direction of propagation. Except for the existence of Hz, the extraordinary wave has electric and magnetic fields that are perpendicular to the corresponding fields ofthe ordinary wave. Thus, a wave polarized in the ydirection will have a propagation con- stantβ o(ordinary wave), but a wave polarized in the xdirection will have a propagation constant βe(extraordinary wave). This effect, where the propagation constant depends on the polarization direction, is called birefringence [2]. Birefringence often occurs in optics 0 200 400 600 800 1000 1200 1400 1600 1800–4–20246 20006 GHz 4/H9266Ms = 3000G 4/H9266Ms = 1700G 11 GHz 6 GHz 11 GHz H0 (OERSTEDS)/H9262e//H9262o FIGURE 9.8 Effective permeability, µe, versus bias field, H0, for various saturation magnetiza- tions and frequencies. c09FerrimagneticComponents Pozar August 25, 2011 18:36 9.3 Propagation in a Ferrite-Loaded Rectangular Waveguide 471 work, where the index of refraction can have different values depending on the polariza- tion. The double image seen through a calcite crystal is an example of this effect. From (9.65) we can see that µe, the effective permeability for the extraordinary wave, can be negative if κ2>µ2. This condition depends on the values of ω,ω 0, andωm,o r f,H0, and Ms, but for a fixed frequency and saturation magnetization there will always be some range of bias field for which µe<0 (ignoring loss). When this occurs βewill become imaginary, as seen from (9.64), which implies that the wave will be cut off, orevanescent. An ˆx-polarized plane wave incident at the interface of such a ferrite region would be totally reflected. The effective permeability is plotted versus bias field strength in Figure 9.8 for several values of frequency and saturation magnetization. 9.3PROPAGATIONINAFERRITE-LOADEDRECTANGULARWAVEGUIDE In the previous section we introduced the effects of a ferrite material on electromagnetic fields by considering the propagation of plane waves in an infinite ferrite medium. In prac-tice, however, most ferrite components use waveguide or other types of transmission lines loaded with ferrite material. Many of these geometries are very difficult to treat without the use of complex numerical methods, but it is possible to analyze some of the simplercases involving ferrite-loaded rectangular waveguides. This will allow us to quantitatively demonstrate the operation and design of several types of practical ferrite components. TE m0ModesofWaveguidewithaSingleFerriteSlab We first consider the geometry shown in Figure 9.9, where a rectangular waveguide is loaded with a vertical slab of ferrite material biased in the ˆydirection. This geometry and its analysis will be used in later sections to treat the operation and design of resonance isolators, field displacement isolators, and remanent (nonreciprocal) phase shifters. In the ferrite slab, Maxwell’s equations can be written as ∇ׯE=− jω[µ]¯H, (9.68a) ∇ׯH=jω/epsilon1¯E, (9.68b) where [µ]is the permeability tensor for ˆybias, as given in (9.27). If we let ¯E(x,y,z)= [¯e(x,y)+ˆzez(x,y)]e−jβzand¯H(x,y,z)=[¯h(x,y)+ˆzhz(x,y)]e−jβz, (9.68) reduces y b xt dH0 ca z0 FIGURE 9.9 Geometry of a rectangular waveguide loaded with a transversely biased ferrite slab. c09FerrimagneticComponents Pozar August 25, 2011 18:36 472 Chapter 9: Theory and Design of Ferrimagnetic Components to ∂ez ∂y+jβey=− jω(µhx−jκhz), (9.69a) −jβex−∂ez ∂x=− jωµ0hy, (9.69b) ∂ey ∂x−∂ex ∂y=− jω(jκhx+µhz), (9.69c) ∂hz ∂y+jβhy=jω/epsilon1ex, (9.69d) −jβhx−∂hz ∂x=jω/epsilon1ey, (9.69e) ∂hy ∂x−∂hx ∂y=jω/epsilon1ez. (9.69f) For TE m0modes, we know that Ez=0 and∂/∂y=0. Then (9.69b) and (9.69d) imply thatex=hy=0 (since β2/negationslash=ω2µ0/epsilon1for a waveguide mode) and so (9.69) reduces to three equations: jβey=− jω(µhx−jκhz), (9.70a) ∂ey ∂x=− jω(jκhx+µhz), (9.70b) jω/epsilon1ey=− jβhx−∂hz ∂x. (9.70c) We can solve (9.70a) and (9.70b) for hxandhzas follows. Multiply (9.70a) by µand (9.70b) by jκ, then add to obtain hx=1 ωµµ e/parenleftbigg −µβ ey−κ∂ey ∂x/parenrightbigg .( 9.71a) Now multiply (9.70a) by jκand (9.71a) by µ, then add to obtain hz=j ωµµ e/parenleftbigg κβey+µ∂ey ∂x/parenrightbigg ,( 9.71b) where µe=(µ2−κ2)/µ. Substituting (9.71) into (9.70c) gives a wave equation for ey: jω/epsilon1ey=−jβ ωµµ e/parenleftbigg −µβ ey−κ∂ey ∂x/parenrightbigg −j ωµµ e/parenleftBigg κβ∂ey ∂x+µ∂2ey ∂x2/parenrightBigg , or /parenleftBigg ∂2 ∂x2+k2 f/parenrightBigg ey=0,( 9.72) where kfis defined as a cutoff wave number for the ferrite: k2 f=ω2µe/epsilon1−β2.( 9.73) We can obtain the corresponding results for the air regions by letting µ=µ0,κ=0, and/epsilon1r=1, to obtain /parenleftBigg ∂2 ∂x2+k2 a/parenrightBigg ey=0,( 9.74) c09FerrimagneticComponents Pozar August 25, 2011 18:36 9.3 Propagation in a Ferrite-Loaded Rectangular Waveguide 473 where kais the cutoff wave number for the air regions: k2 a=k2 0−β2.( 9.75) The magnetic field in the air region is given by hx=−β ωµ0ey=−1 Zwey, (9.76a) hz=j ωµ0∂ey ∂x. (9.76b) The general solutions for eyin the air-ferrite-air regions of the waveguide are then ey=⎧ ⎪⎨ ⎪⎩Asinkax for 0<x<c, Bsinkf(x−c)+Csinkf(c+t−x) forc<x<c+t, Dsinka(a−x) forc+t<x<a,(9.77a) which have been constructed to facilitate the enforcement of boundary conditions at x=0, c,c+t, and a[3]. We will also need hz, which can be found from (9.77a), (9.71b), and (9.76b): hz=⎧ ⎪⎪⎪⎪⎨ ⎪⎪⎪⎪⎩(jk aA/ωµ 0)coskax for 0<x<c, (j/ωµµ e){κβ[Bsinkf(x−c)+Csinkf(c+t−x)] +µkf[Bcoskf(x−c)−Ccoskf(c+t−x)]} forc<x<c+t, (−jkaD/ωµ 0)coska(a−x) forc+t<x<a. (9.77b) Matching eyandhzatx=candx=c+t=a−dgives four equations for the constants A,B,C,D: Asinkac=Csinkft, (9.78a) Bsinkft=Dsinkad, (9.78b) Aka µ0coskac=Bkf µe−C1 µµe(−κβ sinkft+µkfcoskft), (9.78c) B1 µµe(κβsinkft+µkfcoskft)−Ckf µe=− Dka µ0coskad. (9.78d) Solving (9.78a) and (9.78b) for CandD, substituting into (9.78c) and (9.78d), and then eliminating AorBgives the following transcendental equation for the propagation con- stant,β: /parenleftbiggkf µe/parenrightbigg2 +/parenleftbiggκβ µµe/parenrightbigg2 −kacotkac/parenleftbiggkf µ0µecotkft+κβ µ0µµe/parenrightbigg −/parenleftbiggka µ0/parenrightbigg2 ×cotkaccotkad−kacotkad/parenleftbiggkf µ0µecotkft−κβ µ0µµe/parenrightbigg =0. (9.79) After using (9.73) and (9.75) to express the cutoff wave numbers kfandkain terms of β, we can solve (9.79) numerically. The fact that (9.79) contains terms that are odd in κβindicates that the resulting wave propagation will be nonreciprocal since changing the direction of the bias field (which is equivalent to changing the direction of propagation)changes the sign of κ, which leads to a different solution for β. We will identify these two solutions as β +andβ−for positive bias and propagation in the +zdirection (positive κ) c09FerrimagneticComponents Pozar August 25, 2011 18:36 474 Chapter 9: Theory and Design of Ferrimagnetic Components and in the −zdirection (negative κ), respectively. The effects of magnetic loss can easily be included by allowing ωoto be complex, as in (9.37). In later sections we will also need to evaluate the electric field in the guide, as given in (9.77a). If we choose the arbitrary amplitude constant as A, then B,C,andDcan be found in terms of Aby using (9.78a), (9.78b), and (9.78c). Note from (9.75) that if β> ko, then kawill be imaginary. In this case, the sin kaxfunction of (9.77a) becomes jsinh|ka|x, indicating an almost exponential variation in the field distribution. A useful approximate result can be obtained for the differential phase shift, β+−β−, by expanding βin (9.79) in a Taylor series about t=0. This can be accomplished with implicit differentiation after using (9.73) and (9.75) to express kfandkain terms of β[4]. The result is β+−β−/similarequal2kctκ aµsin 2k cc=2kcκ µ/Delta1S Ssin 2k cc,( 9.80) where kc=π/ais the cutoff frequency of the empty guide, and /Delta1S/S=t/ais the filling factor , or ratio of slab cross-sectional area to waveguide cross-sectional area. Thus, this formula can be applied to other geometries, such as waveguides loaded with small ferrite strips or rods, although the appropriate demagnetization factors may be required for someferrite shapes. The result in (9.80) is accurate, however, only for very small ferrite cross sections, typically for /Delta1S/S<0.01. This same technique can be used to obtain an approximate expression for the forward and reverse attenuation constants in terms of the imaginary parts of the susceptibilities defined in (9.39): α ±/similarequal/Delta1S Sβ0/parenleftBig β2 0χ/prime/prime xxsin2kcx+k2 cχ/prime/prime zzcos2kcx∓χ/prime/prime xykcβ0sin 2k cx/parenrightBig ,( 9.81) where β0=/radicalBig k2 0−k2cis the propagation constant of the empty guide. This result will be useful in the design of resonance isolators. Both (9.80) and (9.81) can also be derived using a perturbation method with the empty waveguide fields [4], and so are usually referred to as the perturbation theory results. TEm0ModesofWaveguidewithTwoSymmetricFerriteSlabs A related geometry is a rectangular waveguide loaded with two symmetrically placed fer- rite slabs, as shown in Figure 9.10. With equal but opposite ˆy-directed bias fields on the ferrite slabs, this configuration provides a useful model for the nonreciprocal remanent y b x ttd dH0 cca a/2 zH0Magnetic wall 0 FIGURE 9.10 Geometry of a rectangular waveguide loaded with two symmetric ferrite slabs. c09FerrimagneticComponents Pozar August 25, 2011 18:36 9.4 Ferrite Isolators 475 phase shifter, which will be discussed in Section 9.5. Its analysis is very similar to that of the single-slab geometry. Since the hyandhzfields (including the bias fields) are antisymmetric about the mid- plane of the waveguide at x=a/2, a magnetic wall can be placed at this point. Then we only need to consider the region for 0 <x<a/2. The electric field in this region can be written as ey=⎧ ⎨ ⎩Asinkax, 0<x<c, Bsinkf(x−c)+Csinkf(c+t−x), c<x<c+t, Dcoska(a/2−x), c+t<x<a/2,(9.82a) which is similar in form to (9.77a), except that the expression for c+t<x<a/2w a s constructed to have a maximum at x=a/2 (since hzmust be zero at x=a/2). The cutoff wave numbers kfandkaare defined in (9.73) and (9.75). Using (9.71) and (9.76) gives the hzfield as hz=⎧ ⎪⎪⎪⎨ ⎪⎪⎪⎩(jk aA/ωµ 0)coskax, 0<x<c, (j/ωµµ e){−κβ[Bsinkf(x−c)+Csinkf(c+t−x)] +µk f[Bcoskf(x−c)−Ccoskf(c+t−x)]}, c<x<c+t, (jkaD/ωµ 0)sinka(a/2−x), c+t<x<a/2. (9.82b) Matching eyandhzatx=candx=c+t=a/2−dgives four equations for the con- stants A,B,C,D: Asinkac=Csinkft, (9.83a) Bsinkft=Dcoskad, (9.83b) Aka µ0coskac=Bkf µe−C1 µµe(−κβsinkft+µkfcoskft), (9.83c) B µµe(κβsinkft+µkfcoskft)−Ckf µe=Dka µ0sinkad. (9.83d) Reducing these results gives a transcendental equation for the propagation constant, β: /parenleftbiggkf µe/parenrightbigg2 +/parenleftbiggκβ µµe/parenrightbigg2 −kacotkac/parenleftbiggkf µ0µecotkft+κβ µ0µµe/parenrightbigg +/parenleftbiggka µ0/parenrightbigg2 ×cotkactankad+katankad/parenleftbiggkf µ0µecotkft−κβ µ0µµe/parenrightbigg =0. (9.84) This equation can be solved numerically for β. As in (9.79) for the single-slab case, κandβappear in (9.84) only as κβ,κ2,orβ2, which implies nonreciprocal propagation since changing the sign of κ(or bias fields) necessitates a change in sign for β(propagation direction) for the same root. At first glance it may seem that, for the same waveguide and slab dimensions and parameters, two slabs would give twice the phase shift of one slab,but this is generally untrue because the fields are highly concentrated in the ferrite regions. 9.4FERRITEISOLATORS One of the most useful microwave ferrite components is the isolator , which is a two-port device having unidirectional transmission characteristics. The scattering matrix for an ideal isolator has the form [S]=/bracketleftbigg00 10/bracketrightbigg ,( 9.85) c09FerrimagneticComponents Pozar August 25, 2011 18:36 476 Chapter 9: Theory and Design of Ferrimagnetic Components indicating that both ports are matched, but transmission occurs only in the direction from port 1 to port 2. Since the scattering matrix is not unitary, the isolator must be lossy. And, of course, [S]is not symmetric, since an isolator is a nonreciprocal component. A common application uses an isolator between a high-power source and a load to prevent possible reflections from damaging the source. An isolator can be used in place of a matching or tuning network, but it should be realized that any power reflected from the load will be absorbed by the isolator, as opposed to being reflected back to the load, whichis the case when a matching network is used. Although there are several types of ferrite isolators, we will concentrate on the res- onance isolator and the field displacement isolator. These devices are of practical impor- tance, and can be analyzed and designed using the results for the ferrite slab-loaded wave- guide of the previous section. ResonanceIsolators We have seen that a circularly polarized plane wave rotating in the same direction as the precessing magnetic dipoles of a ferrite medium will have a strong interaction with thematerial, while a circularly polarized wave rotating in the opposite direction will have a weaker interaction. Such a result was illustrated in Example 9.1, where the attenuation of a circularly polarized wave was very large near the gyromagnetic resonance of the ferrite, while the attenuation of a wave propagating in the opposite direction was very small. This effect can be used to construct an isolator; such isolators must operate near gyromagneticresonance and so are called resonance isolators . Resonance isolators usually consist of a ferrite slab or strip mounted at a certain point in a waveguide. We will discuss the two isolator geometries shown in Figure 9.11. Ideally, the RF fields inside the ferrite material should be circularly polarized. In an empty rectangular waveguide the magnetic fields of the TE 10mode can be written as Hx=jβ0 kcAsinkcxe−jβ0z, Hz=Acoskcxe−jβ0z, where kc=π/ais the cutoff wave number and β0=/radicalBig k2 0−k2cis the propagation con- stant of the empty guide. Since a circularly polarized wave must satisfy the condition that y xtH0 ca zy xS ∆sH0 ca z (a)00 (b) FIGURE 9.11 Two resonance isolator geometries. (a) E-plane, full-height slab. (b) H-plane, partial-height slab. c09FerrimagneticComponents Pozar August 25, 2011 18:36 9.4 Ferrite Isolators 477 Hx/Hz=± j, the location xof the point of circular polarization in the empty guide is given by tankcx=±kc β0.( 9.86) Ferrite loading, however, may perturb the fields so that (9.86) may not give the actual optimum position, or it may prevent the internal fields from being circularly polarized for any position. First consider the full-height E-plane slab geometry of Figure 9.11a; we can analyze this case using the exact results from the previous section. Alternatively, we could use the perturbation result of (9.81), but this would require the use of a demagnetization factor for hx, and would be less accurate than the exact results. Thus, for a given set of parameters, (9.79) can be solved numerically for the complex propagation constants of the forward and reverse waves of the ferrite-loaded guide. It is necessary to include the effect of magnetic loss, which can be done by using (9.37) for the complex resonant frequency, ω0,i nt h e expressions for µandκ. The imaginary part of ω0can be related to the linewidth, /Delta1H, of the ferrite through (9.40). Usually the waveguide width, a, frequency, ω, and ferrite parameters, 4π Msand/epsilon1r, will be fixed, and the bias field and slab position and thickness will be determined to give the optimum design. Ideally, the forward attenuation constant (α+)would be zero, with a nonzero attenua- tion constant (α−)in the reverse direction. However, for the E-plane ferrite slab there is no position x=cwhere the fields are perfectly circularly polarized in the ferrite (this is be- cause the demagnetization factor Nx/similarequal1 [4]). Hence, the forward and reverse waves both contain an RHCP component and an LHCP component, so ideal attenuation characteristics cannot be obtained. The optimum design, then, generally minimizes the forward attenua- tion, which determines the slab position. Alternatively, it may be desired to maximize theratio of the reverse to forward attenuations. Since the maximum reverse attenuation gener- ally does not occur at the same slab position as the minimum forward attenuation, such a design will involve a trade-off of the forward loss. For a long, thin slab, the demagnetization factors are approximately those of a thin disk: N x/similarequal1,Ny=Nz=0. It can then be shown via the Kittel equation of (9.45) that the gyromagnetic resonance frequency of the slab is given by ω=/radicalbig ω0(ω0+ωm), (9.87) which determines H0, given the operating frequency and saturation magnetization. This is an approximate result; the transcendental equation of (9.79) accounts for demagnetization exactly, so the actual internal bias field, H0, can be found by numerically solving (9.79) for the attenuation constants for values of H0near the approximate value given by (9.87). Once the slab position, c, and bias field, H0, have been found, the slab length, L, can be chosen to give the desired total reverse attenuation (or isolation) as (α−)L.T h e slab thickness can also be used to adjust this value. Typical numerical results are given inExample 9.2. One advantage of this geometry is that the full-height slab is easy to bias with an external C-shaped permanent magnet, with no demagnetization factor. However, it suffers from several disadvantages: rZero forward attenuation cannot be obtained because the internal magnetic field is not truly circularly polarized.rThe bandwidth of the isolator is relatively narrow, dictated essentially by the line- width, /Delta1H, of the ferrite. c09FerrimagneticComponents Pozar August 25, 2011 18:36 478 Chapter 9: Theory and Design of Ferrimagnetic Components rThe geometry is not well suited for high-power applications because of poor heat transfer from the middle of the slab, and an increase in temperature will cause a change in Ms, which will degrade performance. The first two problems noted above can be remedied to a significant degree with the addi- tion of a dielectric loading slab; see reference [5] for details. EXAMPLE 9.2 FERRITE RESONANCE ISOLATOR DESIGN Design an E-plane resonance isolator in an X-band waveguide to operate at 10 GHz with a minimum forward insertion loss and 30 dB reverse attenuation. Use a 0.5 mm thick ferrite slab with 4π Ms=1700 G ,/Delta1H=200 Oe, and /epsilon1r=13. Determine the bandwidth for which the reverse attenuation is at least 27 dB. Solution The complex roots of (9.79) were found numerically using an interval-halving routine followed by a Newton–Raphson iteration. The approximate bias field, H0, given by (9.87) is 2820 Oe, but numerical results indicate the actual field to be closer to 2840 Oe for resonance at 10 GHz. Figure 9.12a shows the calculated for- ward(α+)and reverse (α−)attenuation constants at 10 GHz versus slab position, and it can be seen that the minimum forward attenuation occurs for c/a=0.125; the reverse attenuation at this point is α−=12.4 dB/cm. Figure 9.12b shows the attenuation constants versus frequency for this slab position. For a total reverse attenuation of 20 dB, the length of the slab must be L=30 dB 12.4d B / c m=2.4c m . 0 0.1 0.2 0.3 0.4 0.502468101214/H9251+, /H9251– dB/cm /H9251+/H9251– /H9251+/H9251– 9.0 9.4 9.8 10.2 10.6 11.002468101214/H9251+, /H9251– dB/cm min. c/ac/a = 0.125 (a)f(GHz) (b)H0 c a f = 10 GHz 4/H9266Ms = 1700 G H0 = 2840 Oe ∆H = 200 Oe a = 2.286 cm t = 0.05 cm /H9280r = 13 FIGURE 9.12 Forward and reverse attenuation constants for the resonance isolator of Example 9.2. (a) Versus slab position. (b) Versus frequency. c09FerrimagneticComponents Pozar August 25, 2011 18:36 9.4 Ferrite Isolators 479 For the total reverse attenuation to be at least 27 dB, we must have α−>27 dB 2.4c m=11.3d B / c m . The bandwidth according to the above definition is, from the data of Figure 9.12b, less than 2%. This result could be improved by using a ferrite with a largerlinewidth, at the expense of a longer or thicker slab and a higher forward attenuation. ■ Next we consider a resonance isolator using the H-plane slab geometry of Figure 9.11b. If the slab is much thinner than it is wide, the demagnetization factors will be ap-proximately N x=Nz=0,Ny=1. This means that a stronger applied bias field will be required to produce the internal field H0in the ydirection. However, the RF magnetic field components, hxandhz, will not be affected by the air–ferrite boundary since Nx=Nz= 0, and perfect circular polarized fields will exist in the ferrite when it is positioned at the circular polarization point of the empty guide, as given by (9.86). Another advantage of this geometry is that it has better thermal properties than the E-plane version since the ferrite slab has a large surface area in contact with the waveguide wall for heat dissipation. Unlike the full-height E-plane slab case, the H-plane geometry of Figure 9.11b cannot be analyzed exactly. However, if the slab occupies only a very small fraction of the total guide cross section ( /Delta1S/S/lessmuch1, where /Delta1SandSare the cross-sectional areas of the slab and waveguide, respectively), the perturbational result for α+in (9.81) can be used with reasonable results. This expression is given in terms of the susceptibilities χxx=χ/prime xx− jχ/prime/prime xx,χzz=χ/prime zz−jχ/prime/prime zz, andχxy=χ/prime/prime xy+jχ/prime xy, as defined for a ˆy-biased ferrite in a man- ner similar to (9.22). For ferrite shapes other than a thin H-plane slab, these susceptibilities would have to be modified with the appropriate demagnetization factors, as in (9.43) [4]. As seen from the susceptibility expressions of (9.22), gyromagnetic resonance for this geometry will occur when ω=ω0, which determines the internal bias field, H0. The cen- ter of the slab is positioned at the circular polarization point of the empty guide, as given by (9.86). This should result in a near-zero forward attenuation constant. The total reverse attenuation, or isolation, can be controlled with either the length, L, of the ferrite slab or its cross section, /Delta1S, since (9.81) shows α±is proportional to /Delta1S/S.I f/Delta1S/Sis too large, however, the purity of circular polarization over the slab cross section will be degraded, andforward loss will increase. One practical alternative is to use a second, identical ferrite slab on the top wall of the guide to double /Delta1S/Swithout significantly degrading polarization purity. TheFieldDisplacementIsolator Another type of isolator uses the fact that the electric field distributions of the forward and reverse waves in a ferrite slab-loaded waveguide can be quite different. As illustratedin Figure 9.13, the electric field for the forward wave can be made to vanish at the side of the ferrite slab at x=c+t, while the electric field of the reverse wave can be quite large at this same plane. If a thin resistive sheet is placed in this position the forward wave will be essentially unaffected, but the reverse wave will be attenuated. Such an isolator is called a field displacement isolator ; high values of isolation with a relatively compact device can be obtained with bandwidths on the order of 10%. Another advantage of the field displacement isolator over the resonance isolator is that a much smaller bias field is required since it operates well below gyromagnetic resonance. The main problem in designing a field displacement isolator is to determine the design parameters that produce field distributions like those shown in Figure 9.13. The general c09FerrimagneticComponents Pozar August 25, 2011 18:36 480 Chapter 9: Theory and Design of Ferrimagnetic Components y xH0 ca z0c + tReverse EyForward Ey Resistive sheet FIGURE 9.13 Geometry and electric fields of a field displacement isolator. form of the electric field is given in (9.77a), from the analysis of the ferrite slab-loaded waveguide. This shows that for the electric field of the forward wave to have a sinusoidal dependence for c+t<x<a, and to vanish at x=c+t, the cutoff wave number k+ a must be real and satisfy the condition that k+ a=π d,( 9.88) where d=a−c−t. In addition, the electric field of the reverse wave should have a hy- perbolic dependence for c+t<x<a, which implies that k− amust be imaginary. Since from (9.75), k2 a=k2 0−β2, the above conditions imply that β+<k0andβ−>k0, where k0=ω√µ0/epsilon10. These conditions on β±depend critically on the slab position, which must be determined by numerically solving (9.79) for the propagation constants. The slab thick- ness also affects this result, but less critically; a typical value is t=a/10. It also turns out that in order to satisfy (9.88), to force Ey=0a t x=c+t,µe= (µ2−κ2)/µmust be negative. This requirement can be intuitively understood by think- ing of the waveguide mode for c+t<x<aas a superposition of two obliquely traveling plane waves. The magnetic field components HxandHzof these waves are both perpendic- ular to the bias field, a situation that is similar to the extraordinary plane waves discussed in Section 9.2, where it was seen that propagation would not occur for µe<0. Applying this cutoff condition to the ferrite-loaded waveguide will allow a null in Eyfor the forward wave to be formed at x=c+t. The condition that µebe negative depends on the frequency, saturation magnetiza- tion, and bias field. Figure 9.8 shows the dependence of µeversus bias field for several frequencies and saturation magnetization. This type of data can be used to select the sat-uration magnetization and bias field to give µ e<0 at the design frequency. Observe that higher frequencies will require a ferrite with higher saturation magnetization and a higher bias field, but µe<0 always occurs before the resonance in µeat√ω0(ω0+ωm). Further design details will be given in the following example. EXAMPLE 9.3 FIELD DISPLACEMENT ISOLATOR DESIGN Design a field displacement isolator in an X-band waveguide to operate at 11 GHz.The ferrite has 4π M s=3000 G and /epsilon1r=13. Ferrite loss can be ignored. c09FerrimagneticComponents Pozar August 25, 2011 18:36 9.4 Ferrite Isolators 481 0 0.02 0.04 0.06 0.08 0.10 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.00.60.81.01.21.41.61.82.0 0.00.10.20.30.40.50.60.70.80.91.0x/a (b)c/a (a)Electric field Ey (arbitrary units)Reverse /H9252–/k0 = 1.607 Forward /H9252+/k0 = 0.724 c/a = 0.028 t/a = 0.109/H9252±/k0/H9252– /H9252+ ka imaginary ka real H0 = 1200 Oe t/a = 0.109ka+d = /H9266 FIGURE 9.14 Propagation constants and electric field distribution for the field displacement iso- lator of Example 9.3. (a) Forward and reverse propagation constants versus slabposition. (b) Electric field amplitudes for the forward and reverse waves. Solution We first determine the internal bias field, H0, such that µe<0. This can be found from Figure 9.8, which shows µe/µoversus H0for 4π Ms=3000 G at 11 GHz. We see that H0=1200 Oe should be sufficient. Also note from this figure that a ferrite with a smaller saturation magnetization would require a much larger bias field. Next we determine the slab position, c/a, by numerically solving (9.79) for the propagation constants, β±, as a function of c/a. The slab thickness was set tot=0.25 cm, which is approximately a/10. Figure 9.14a shows the resulting propagation constants, as well as the locus of points where β+andc/asatisfy the condition of (9.88). The intersection of β+with this locus will ensure that Ey=0 atx=c+tfor the forward wave; this intersection occurs for a slab position ofc/a=0.028. The resulting propagation constants are β+=0.724k 0<k0and β−=1.607k 0>k0. c09FerrimagneticComponents Pozar August 25, 2011 18:36 482 Chapter 9: Theory and Design of Ferrimagnetic Components The electric fields are plotted in Figure 9.14b. Note that the forward wave has a null at the face of the ferrite slab, while the reverse wave has a peak (the relative amplitudes of these fields are arbitrary). A resistive sheet can be placed at this point to attenuate the reverse wave. The actual isolation will depend on theresistivity of this sheet; a value of 75 /Omega1per square is typical. ■ 9.5FERRITEPHASESHIFTERS Another important application of ferrite materials is in phase shifters, which are two-port components that provide variable phase shift by changing the bias field of the ferrite. (Mi-crowave diodes and FETs can also be used to implement phase shifters; see Section 10.3.) Phase shifters find application in test and measurements systems, but the most significant use is in phased array antennas where the antenna beam can be steered in space by elec-tronically controlled phase shifters. Because of this demand, many different types of phase shifters have been developed, both reciprocal (same phase shift in either direction) and nonreciprocal [2, 6]. One of the most useful designs is the latching (or remanent ) nonre- ciprocal phase shifter using a ferrite toroid in a rectangular waveguide; we can analyze this geometry with a reasonable degree of approximation using the double ferrite slab geometry discussed in Section 9.3. Then we will qualitatively discuss the operation of a few othertypes of phase shifters. NonreciprocalLatchingPhaseShifter The geometry of a latching phase shifter is shown in Figure 9.15; it consists of a toroidal ferrite core symmetrically located in the waveguide with a bias wire passing through its center. When the ferrite is magnetized, the magnetization of the sidewalls of the toroidwill be oppositely directed and perpendicular to the plane of circular polarization of the RF fields. Since the sense of circular polarization is also opposite on opposite sides of the waveguide, a strong interaction between the RF fields and the ferrite can be obtained. Of course, the presence of the ferrite perturbs the waveguide fields (the fields tend to concen- trate in the ferrite), so the circular polarization point does not occur at tan k cx=kc/β0,a s it does for an empty guide. In principle, such a geometry can be used to provide a continuously variable (analog) phase shift by varying the bias current. However, a more useful technique employs themagnetic hysteresis of the ferrite to provide a phase shift that can be switched between two values (digital). A typical hysteresis curve is shown in Figure 9.16, showing the variation in magnetization, M, with bias field, H 0. When the ferrite is initially demagnetized and the bias field is off, both MandH0are zero. As the bias field is increased, the magnetization I Bias line Toroidal ferrite FIGURE 9.15 Geometry of a nonreciprocal latching phase shifter using a ferrite toroid. c09FerrimagneticComponents Pozar August 25, 2011 18:36 9.5 Ferrite Phase Shifters 483 M Ms –MsMr –MrH0 FIGURE 9.16 A hysteresis curve for a ferrite toroid. increases along the dashed-line path until the ferrite is magnetically saturated, and M= Ms. If the bias field is now reduced to zero, the magnetization will decrease to a remanent condition (like a permanent magnet), where M=Mr. A bias field in the opposite direction will saturate the ferrite with M=− Ms, whereupon removal of the bias field will leave the ferrite in a remanent state with M=− Mr. Thus we can “latch” the ferrite magnetization in one of two states, where M=± Mr, giving a digital phase shift. The amount of differ- ential phase shift between these two states is controlled by the length of the ferrite toroid. In practice, several sections having individual bias lines and decreasing lengths are used in series to give binary differential phase shifts of 180◦,9 0◦,4 5◦, etc., to as fine a resolution as desired (or can be afforded). An important advantage of the latching mode of operationis that the bias current does not have to be continuously applied, but only pulsed with one polarity or the other to change the polarity of the remanent magnetization; switching speeds can be on the order of a few microseconds. The bias wire can be oriented perpen-dicular to the electric field in the guide, with a negligible perturbing effect. The top and bottom walls of the ferrite toroid have very little magnetic interaction with the RF fields because the magnetization is not perpendicular to the plane of circular polarization, andthe top and bottom magnetizations are oppositely directed. These walls provide mainly a dielectric loading effect, and the essential operating features of the remanent phase shifters can be obtained by considering the simpler dual ferrite slab geometry of Section 9.3. For a given operating frequency and waveguide size, the design of a remanent dual- slab phase shifter mainly involves the determination of the slab thickness, t, the spacing between the slabs, s=2d=a−2c−2t(see Figure 9.10), and the length of the slabs for the desired phase shift. This requires the propagation constants, β ±, for the dual-slab geometry, which can be numerically evaluated from the transcendental equation of (9.84). This equation requires values for µandκ, which can be determined from (9.25) for the remanent state by setting H0=0(ω0=0)andMs=Mr(ωm=µ0γMr): µ=µ0, (9.89a) κ=−µ0ωm ω. (9.89b) The differential phase shift, β+−β−, is linearly proportional to κforκ/µ 0up to about 0.5. Then, since κis proportional to Mr, as seen by (9.89b), it follows that a shorter length of ferrite can be used to provide a given phase shift if a ferrite with a higher remanent magnetization is selected. The insertion loss of the phase shifter increases with length butis a function of the ferrite linewidth, /Delta1H. A figure of merit commonly used to characterize phase shifters is the ratio of phase shift to insertion loss, measured in degrees/dB. c09FerrimagneticComponents Pozar August 25, 2011 18:36 484 Chapter 9: Theory and Design of Ferrimagnetic Components EXAMPLE 9.4 REMANENT PHASE SHIFTER DESIGN Design a two-slab remanent phase shifter at 10 GHz using an X-band waveguide with ferrite having 4π Mr=1786 G and /epsilon1r=13. Assume that the ferrite slabs are spaced 1 mm apart. Determine the slab thickness for maximum differential phase shift, and the lengths of the slabs for 180◦and 90◦phase shifter sections. Solution From (9.89) we have that µ µ0=1, κ µ0=±ωm ω=±(2.8 MHz/Oe)(1786 G) 10,000 MHz=±0.5. Using a numerical root-finding technique, such as interval halving, we can solve (9.84) for the propagation constants β+andβ−by using positive and negative val- ues of κ. Figure 9.17 shows the resulting differential phase shift, (β+−β−)/k0, versus slab thickness, t, for several slab spacings. Observe that the phase shift increases as the spacing, s, between the slabs decreases and as the slab thickness increases, for t/aup to about 0.12. From the curve in Figure 9.17 for s=1 mm, we see that the optimum slab thickness for maximum phase shift is t/a=0.12, or t=2.74 mm, since a= 2.286 cm for an X-band guide. The corresponding normalized differential phaseshift is 0.40, so β +−β−=0.4k 0=0.4/parenleftbigg2.09 rad cm/parenrightbigg =0.836 rad/cm =48◦/cm. 0 0.04 0.08 0.12 0.1600.20.40.60.8 tt asH0 Slab thickness t/aDifferential phase shift (/H9252+ – /H9252–)/k0S = 0 S = 1 mm S = 2 mm S = 3 mm FIGURE 9.17 Differential phase shift for the two-slab remanent phase shifter of Example 9.4. c09FerrimagneticComponents Pozar August 25, 2011 18:36 9.5 Ferrite Phase Shifters 485 The ferrite length required for the 180◦phase shift section is then L=180◦ 48◦/cm=3.75 cm , while the length required for a 90◦section is L=90◦ 48◦/cm=1.88 cm .■ OtherTypesofFerritePhaseShifters Many other types of ferrite phase shifters have been developed, with various combina- tions of rectangular or circular waveguide, transverse or longitudinal biasing, latching orcontinuous phase variation, and reciprocal or nonreciprocal operation. Phase shifters us- ing printed transmission lines have also been proposed. Even though PIN diode and FET circuits offer a less bulky and more integratable alternative to ferrite components, ferritephase shifters often have advantages in terms of cost, power-handling capacity, and power requirements. However, there is still a great need for a low-cost, compact phase shifter, primarily for phased array antenna systems. Several waveguide phase shifter designs are derived from the nonreciprocal Faraday rotation phase shifter shown in Figure 9.18. In operation, a rectangular waveguide TE 10 mode entering at the left is converted to a TE 11circular waveguide mode with a short transition section. Then a quarter-wave dielectric plate, oriented 45◦from the electric field vector, converts the wave to an RHCP wave by providing a 90◦phase difference between the field components that are parallel and perpendicular to the plate. In the ferrite-loaded region the phase delay is β+z, which can be controlled with the bias field strength. A sec- ond quarter-wave plate converts the wave back to a linearly polarized field. The operationis similar for a wave entering at the right, except now the phase delay is β −z; the phase shift is thus nonreciprocal. The ferrite rod is biased longitudinally, in the direction of prop- agation, with a solenoid coil. This type of phase shifter can be made reciprocal by usingnonreciprocal quarter-wave plates to convert a linearly polarized wave to the same sense of circular polarization for either propagation direction. Ferrite rodQuarter-wave plate Quarter-wave plateBias coilH0 FIGURE 9.18 Nonreciprocal Faraday rotation phase shifter. c09FerrimagneticComponents Pozar August 25, 2011 18:36 486 Chapter 9: Theory and Design of Ferrimagnetic Components Ferrite rod Bias coilH0 FIGURE 9.19 Reggia-Spencer reciprocal phase shifter. TheReggia-Spencer phase shifter, shown in Figure 9.19, is a popular reciprocal phase shifter. In either rectangular or circular waveguide form, a longitudinally biased ferrite rod is centered in the guide. When the diameter of the rod is greater than a certain critical size, the fields become tightly bound to the ferrite and are circularly polarized. A large reciprocal phase shift can be obtained over relatively short lengths, although the phase shift is rather frequency sensitive. TheGyrator An important canonical nonreciprocal component is the gyrator , which is a two-port device having a 180◦differential phase shift. The schematic symbol for a gyrator is shown in Figure 9.20, and the scattering matrix for an ideal gyrator is [S]=/bracketleftbigg 01 −10/bracketrightbigg ,( 9.90) which shows that it is lossless, matched, and nonreciprocal. Use of the gyrator as a basic nonreciprocal building block in combination with reciprocal dividers and couplers can leadto useful equivalent circuits for nonreciprocal components such as isolators and circulators. Figure 9.21, for example, shows an equivalent circuit for an isolator using a gyrator and two quadrature hybrids. /H9266 FIGURE 9.20 Symbol for a gyrator, which has a differential phase shift of 180◦. /H9266 Z0Z0 FIGURE 9.21 An isolator constructed with a gyrator and two quadrature hybrids. The forward wave ( →) is passed, while the reverse wave (←) is absorbed in the matched load of the first hybrid. c09FerrimagneticComponents Pozar August 25, 2011 18:36 9.6 Ferrite Circulators 487 The gyrator can be implemented as a phase shifter with a 180◦differential phase shift; bias can be provided with a permanent magnet, making the gyrator a passive device. 9.6FERRITECIRCULATORS As discussed in Section 7.1, a circulator is a three-port microwave device that can be lossless and matched at all ports; by using the unitary properties of the scattering matrixwe were able to show that such a device must be nonreciprocal. The scattering matrix for an ideal circulator thus has the following form: [S]=/bracketleftBigg001 100010/bracketrightBigg ,( 9.91) which shows that power can flow from port 1 to port 2, port 2 to port 3, and port 3 to port 1, but not in the reverse directions. By transposing the port indices, the opposite circularity can be obtained. For a ferrite circulator, this result can be produced by changing the polarity of the magnetic bias field. Most circulators use permanent magnets for the bias field, but if an electromagnet is used the circulator can operate in a latching (remanent) mode as a single-pole double-throw (SPDT) switch. A circulator can also be used as an isolator byterminating one of the ports with a matched load. A photograph of a disassembled stripline circulator is shown in Figure 9.22. FIGURE 9.22 Photograph of a disassembled ferrite junction circulator, showing the stripline con- ductor, the ferrite disks, and the bias magnet. The middle port of the circulator is terminated with a matched load, so this circulator is actually configured as an iso-lator. Note the change in the width of the stripline conductors due to the different dielectric constants of the ferrite and the surrounding plastic material. c09FerrimagneticComponents Pozar August 25, 2011 18:36 488 Chapter 9: Theory and Design of Ferrimagnetic Components We will first discuss the properties of an imperfectly matched circulator in terms of its scattering matrix, and then we will analyze the operation of the stripline junction circulator. The operation of waveguide circulators is similar in principle. PropertiesofaMismatchedCirculator If we assume that a circulator has circular symmetry around its three ports and is lossless, but not perfectly matched, we can write its scattering matrix as [S]=/bracketleftBigg/Gamma1βα α/Gamma1β βα/Gamma1/bracketrightBigg .( 9.92) Since the circulator is assumed lossless, the scattering matrix must be unitary, which im- plies the following two conditions: |/Gamma1|2+|β|2+|α|2=1, (9.93a) /Gamma1β∗+α/Gamma1∗+βα∗=0. (9.93b) If the circulator were matched (/Gamma1=0), then (9.93) shows that either α=0 and |β|=1, orβ=0 and |α|=1; this describes the ideal circulator with its two possible circularity states. Observe that this condition depends only on a lossless and matched device. Now assume small imperfections, such that |/Gamma1|/lessmuch1. To be specific, consider the cir- cularity state where power flows primarily in the 1-2-3 direction, so that |α|is close to unity and|β|is small. Then β/Gamma1∼0, and (9.93b) shows that α/Gamma1∗+βα∗/similarequal0, so|/Gamma1|/similarequal|β|. Then (9.93a) shows that |α|2/similarequal1−2|β|2/similarequal1−2|/Gamma1|2,o r|α |/similarequal1−|/Gamma1|2. Then the scattering matrix of (9.92) can be written as [S]=/bracketleftBigg/Gamma1/Gamma1 1−/Gamma12 1−/Gamma12/Gamma1/Gamma1 /Gamma1 1−/Gamma12/Gamma1/bracketrightBigg ,( 9.94) ignoring phase factors. This result shows that circulator isolation, β/similarequal/Gamma1, and transmission, α/similarequal1−/Gamma12, both deteriorate as the input ports become mismatched. JunctionCirculator The geometry of a stripline junction circulator is shown in Figure 9.23, and in the photo- graph of Figure 9.22. Two ferrite disks fill the spaces between the center metallic disk and the ground planes of the stripline. Three stripline conductors are attached to the peripheryof the center disk at 120 ◦intervals, forming the three ports of the circulator. The DC bias field is applied normal to the ground planes. In operation, the ferrite disks form a dielectric resonator; in the absence of a bias field, this resonator has a single lowest order resonant mode with a cos φ(or sin φ) dependence. When the ferrite is magnetically biased this mode breaks into two resonant modes having slightly different resonant frequencies. The operating frequency of the circulator can then be chosen so that the superposition of these two modes adds at the output port and cancels at the isolated port. We can analyze the junction circulator by treating it as a thin cavity resonator with electric walls on the top and bottom, and an approximate magnetic wall on the side. Then Eρ=Eφ/similarequal0, and ∂/∂z=0, so we have TM modes. Since Ezon either side of the center conducting disk is antisymmetric, we need only consider the solution for one of the ferrite disks [7]. c09FerrimagneticComponents Pozar August 25, 2011 18:36 9.6 Ferrite Circulators 489 2 31Output IsolatedInputay 2/H9274/H9278 x Ground planes Stripline conductorFerrite disks (a) (b)H0 FIGURE 9.23 A stripline junction circulator. (a) Pictorial view. (b) Geometry. We first transform (9.23), ¯B=[µ]¯H, from rectangular to cylindrical coordinates: Bρ=Bxcosφ+Bysinφ =(µHx+jκHy)cosφ+(−jκHx+µHy)sinφ =µHρ+jκHφ, (9.95a) Bφ=− Bxsinφ+Bycosφ =−(µHx+jκHy)sinφ+(−jκHx+µHy)cosφ =− jκHρ+µHφ. (9.95b) So we have that /bracketleftBiggBρ Bφ Bz/bracketrightBigg =[µ]/bracketleftBiggHρ Hφ Hz/bracketrightBigg ,( 9.96) where [µ]is the same matrix as for rectangular coordinates, as given in (9.24). In cylindrical coordinates, with ∂/∂z=0, Maxwell’s curl equations reduce to the following: 1 ρ∂Ez ∂φ=− jω(µH ρ+jκHφ), (9.97a) −∂Ez ∂ρ=− jω(−jκHρ+µHφ), (9.97b) 1 ρ/bracketleftbigg∂(ρHφ) ∂ρ−∂Hρ ∂φ/bracketrightbigg =jω/epsilon1Ez. (9.97c) Solving (9.97a) and (9.97b) for HρandHφin terms of Ezgives Hρ=jY kµ/parenleftbiggµ ρ∂Ez ∂φ+jκ∂Ez ∂ρ/parenrightbigg , (9.98a) Hφ=−jY kµ/parenleftbigg−jκ ρ∂Ez ∂φ+µ∂Ez ∂ρ/parenrightbigg , (9.98b) where k2=ω2/epsilon1(µ2−κ2)/µ=ω2/epsilon1µeis an effective wave number, and Y=√/epsilon1/µ eis an effective admittance. Using (9.98) to eliminate HρandHφin (9.97c) gives a wave equation c09FerrimagneticComponents Pozar August 25, 2011 18:36 490 Chapter 9: Theory and Design of Ferrimagnetic Components forEz: ∂2Ez ∂ρ2+1 ρ∂Ez ∂ρ+1 ρ2∂2Ez ∂φ2+k2Ez=0.( 9.99) This equation is identical in form to the equation for Ezfor the TM mode of a circular waveguide, so the general solution can be written as Ezn=/bracketleftBig A+nejnφ+A−ne−jnφ/bracketrightBig Jn(kρ), (9.100a) where we have excluded the solution with Yn(kρ)because Ezmust be finite at ρ=0. We will also need Hφn, which can be found using (9.98b): Hφn=− jY/braceleftbigg A+nejnφ/bracketleftbigg J/prime n(kρ)+nκ kρµJn(kρ)/bracketrightbigg +A−ne−jnφ/bracketleftbigg J/prime n(kρ)−nκ kρµJn(kρ)/bracketrightbigg/bracerightbigg . (9.100b) The resonant modes can now be found by enforcing the boundary condition that Hφ=0 atρ=a. If the ferrite is not magnetized, then H0=Ms=0 and ω0=ωm=0, so that κ=0 andµ=µe=µ0, and resonance occurs when J/prime n(ka)=0, orka=x0=p/prime 11=1.841. Define this frequency as ω0(not to be confused with ω0= γµ0H0): ω0=x0 a√/epsilon1µe=1.841 a√/epsilon1µ0.( 9.101) When the ferrite is magnetized there are two possible resonant modes for each value of n, as associated with either a ejnφvariation or a e−jnφvariation. The resonance condition for the two n=1 modes is κ µxJ1(x)±J/prime 1(x)=0,( 9.102) where x=ka. This result shows the nonreciprocal property of the circulator, since chang- ing the sign of κ(the polarity of the bias field) in (9.102) leads to the other root and propagation in the opposite direction in φ. If we let x+andx−be the two roots of (9.102), then we can express the resonant frequencies for these two n=1 modes as ω±=x± a√/epsilon1µe.( 9.103) We can develop an approximate result for ω±if we assume that κ/µ is small, so that ω±will be close to ω0of (9.101). Using a Taylor series about x0for the two terms in (9.102) gives the following results, since J/prime 1(x0)=0: J1(x)/similarequalJ1(x0)+(x−x0)J/prime 1(x0)=J1(x0), J/prime 1(x)/similarequalJ/prime 1(x0)+(x−x0)J/prime/prime 1(x0) =−(x−x0)/parenleftBigg 1−1 x2 0/parenrightBigg J1(x0). c09FerrimagneticComponents Pozar August 25, 2011 18:36 9.6 Ferrite Circulators 491 Then (9.102) becomes κ µx0∓(x±−x0)/parenleftBigg 1−1 x2 0/parenrightBigg =0, or x±/similarequalx0/parenleftbigg 1±0.418κ µ/parenrightbigg , (9.104) since x0=1.841. This result gives the resonant frequencies as ω±/similarequalω0/parenleftbigg 1±0.418κ µ/parenrightbigg .( 9.105) Note that ω±approaches ω0asκ→0, and that ω−≤ω0≤ω+. We can use a superposition of these two modes to design a circulator. The amplitudes of these modes give two degrees of freedom that can be used to provide coupling from the input to the output port, and to provide cancellation at the isolated port. It will turn out that ω0will be the operating frequency, which will be between the resonances of the ω±modes. Thus, Hφ/negationslash=0 over the periphery of the ferrite disks since ω/negationslash=ω±. If we select port 1 as the input port, port 2 as the output port, and port 3 as the isolated port, as in Figure 9.23, we can assume the following Ezfield at the ports at ρ=a: Ez(ρ=a,φ )=/braceleftBiggE0, forφ=0 (port 1) , −E0, forφ=120◦(port 2), 0, forφ=240◦(port 3).(9.106a) If the feedlines are narrow, the Ezfield will be relatively constant across their width. The corresponding Hφfield should be Hφ(ρ=a,φ)=/braceleftBiggH0 for−ψ<φ<ψ , H0 for 120◦−ψ<φ< 120◦+ψ, 0 elsewhere .(9.106b) Equating (9.106a) to Ezof (9.100a) gives the mode amplitude constants as A+1=E0(1+j/√ 3) 2J1(ka), (9.107a) A−1=E0(1−j/√ 3) 2J1(ka). (9.107b) Then (9.100a) and (9.100b) can be reduced to give the electric and magnetic fields as Ez1=E0J1(kρ) 2J1(ka)/bracketleftbigg/parenleftbigg 1+j√ 3/parenrightbigg ejφ+/parenleftbigg 1−j√ 3/parenrightbigg e−jφ/bracketrightbigg =E0J1(kρ) J1(ka)/parenleftbigg cosφ−sinφ√ 3/parenrightbigg , (9.108a) Hφ1=−jYE 0 2J1(ka)/braceleftbigg/parenleftbigg 1+j√ 3/parenrightbigg/bracketleftbigg J/prime 1(kρ)+κ kρµJ1(kρ)/bracketrightbigg ejφ +/parenleftbigg 1−j√ 3/parenrightbigg/bracketleftbigg J/prime 1(kρ)−κ kρµJ1(kρ)/bracketrightbigg e−jφ/bracerightbigg . (9.108b) c09FerrimagneticComponents Pozar August 25, 2011 18:36 492 Chapter 9: Theory and Design of Ferrimagnetic Components To approximately equate Hφ1toHφin (9.106b) requires that Hφbe expanded in a Fourier series: Hφ(ρ=a,φ)=∞/summationdisplay n=−∞Cnejnφ=2H0ψ π +H0 π∞/summationdisplay n=1/bracketleftbig (1+e−j2πn/3)ejnφ+(1+ej2πn/3)e−jnφ/bracketrightbig ×sinnψ n. (9.109) Then=1 term of this result is Hφ1(ρ=a,φ)=−j√ 3H0sinψ 2π/bracketleftbigg/parenleftbigg 1+j√ 3/parenrightbigg ejφ−/parenleftbigg 1−j√ 3/parenrightbigg e−jφ/bracketrightbigg , which can be equated to (9.108b) for ρ=a. Equivalence can be obtained if two conditions are met: J/prime 1(ka)=0, and YE0κ kaµ=√ 3H0sinψ π. The first condition is identical to the condition for resonance in the absence of bias, which implies that the operating frequency is ω0, as given by (9.101). For a given operating frequency, (9.101) can then be used to find the disk radius, a. The second condition can be related to the wave impedance at port 1 or 2: Zw=E0 H0=√ 3kaµsinψ πYκ/similarequalµsinψ κY,( 9.110) since√ 3ka/π=√ 3(1.841)/π /similarequal1.0. Thus, Zwcan be controlled for impedance match- ing by adjusting κ/µ via the bias field. We can compute power flows at the three ports as follows: Pin=P1=−ˆρ·¯EׯH∗=EzHφ/vextendsingle/vextendsingle/vextendsingle φ=0=E0H0sinψ π=E2 0κY πµ,(9.111a) Pout=P2=ˆρ·¯EׯH∗=− EzHφ/vextendsingle/vextendsingle/vextendsingle φ=120◦=E0H0sinψ π=E2 0κY πµ,(9.111b) Piso=P3=ˆρ·¯EׯH∗=− EzHφ/vextendsingle/vextendsingle/vextendsingle φ=240◦=0. (9.111c) These results show that power flow occurs from port 1 to port 2, but not from port 1 to port 3. By the azimuthal symmetry of the circulator, this also implies that power can be coupled from port 2 to port 3, or from port 3 to port 1, but not in the reverse directions. The electric field of (9.108a) is sketched in Figure 9.24 along the periphery of the circulator, showing that the amplitudes and phases of the e±jφmodes are such that their superposition gives a null at the isolated port, with equal voltages at the input and output ports. This result ignores the loading effect of the input and output lines, which will distort the field from that shown in Figure 9.24. This design is narrowband, but bandwidth can beimproved using dielectric loading; the analysis then requires consideration of higher order modes. c09FerrimagneticComponents Pozar September 13, 2011 17:26 Problems 493 ⎜Ez⎜ /H9278 –120° 0 120° 240° 360° Input Output Isolated FIGURE 9.24 Magnitude of the electric field around the periphery of the junction circulator. REFERENCES [1] R. F. Soohoo, Microwave Magnetics , Harper and Row, New York, 1985. [ 2 ] A .J .B a d e nF u l l e r , Ferrites at Microwave Frequencies , Peter Peregrinus, London, 1987. [3] R. E. Collin, Field Theory of Guided Waves , McGraw-Hill, New York, 1960. [4] B. Lax and K. J. Button, Microwave Ferrites and Ferrimagnetics, McGraw-Hill, New York, 1962. [5] F. E. Gardiol and A. S. Vander V orst, “Computer Analysis of E-Plane Resonance Isolators,” IEEE Transactions on Microwave Theory and Techniques , vol. MTT-19, pp. 315–322, March 1971. [6] G. P. Rodrigue, “A Generation of Microwave Ferrite Devices,” Proceedings of the IEEE, vol. 76, pp. 121–137, February 1988. [7] C. E. Fay and R. L. Comstock, “Operation of the Ferrite Junction Circulator,” IEEE Transactions on Microwave Theory and Techniques, vol. MTT-13, pp. 15–27, January 1965. PROBLEMS 9.1 A LHCP RF magnetic field of ¯H=/parenleftbig 0.5ˆx+j0.5ˆy/parenrightbig A/m is applied to a calcium vanadium garnet (CVG) ferrite medium having a saturation magnetization of 4 πMs=900 G. Ignoring loss, calculate the resulting magnetic flux density ¯Batf=2 GHz for two cases: (a) no magnetic bias field and ferrite demagnetized ( Ms=H0=0),a n d( b )a z-directed magnetic bias field of 800 Oe. 9.2 Consider the following field transformations from linearly polarized to circular polarized compo- nents: B+=(Bx+jBy)/2, H+=(Hx+jHy)/2, (RHCP) B−=(Bx−jBy)/2, H−=(Hx−jHy)/2, (LHCP) Bz=Bz, Hz=Hz. For a z-biased ferrite medium, show that the relation between ¯Band¯Hcan be expressed in terms of a diagonal tensor permeability as follows: ⎡ ⎣B+ B− Bz⎤ ⎦=⎡ ⎣(µ+κ) 00 0 (µ−κ) 0 00 µ0⎤ ⎦⎡ ⎣H+ H− Hz⎤ ⎦. 9.3 A tunable oscillator uses a YIG sphere with 4 πMs=1780 G and requires an magnetic bias field of 700 Oe internal to the sphere. What is the external applied magnetic field strength that must be applied to produce this internal field? 9.4 A thin ferrite rod with 4π Ms=800 G is magnetically biased along its axis. Find the external bias field strength required to produce a gyromagnetic resonance at 2.52 GHz. c09FerrimagneticComponents Pozar August 25, 2011 18:36 494 Chapter 9: Theory and Design of Ferrimagnetic Components 9.5 An infinite lossless ferrite medium with a saturation magnetization of 4 πMs=1200 G and a dielec- tric constant of 10 is biased to a field strength of 500 Oe. At 8 GHz, calculate the differential phase shift per meter between an RHCP and an LHCP plane wave propagating in the direction of bias. If a linearly polarized wave is propagating in this material, what is the distance it must travel in orderthat its polarization is rotated 90 ◦? 9.6 An infinite lossless ferrite medium with a saturation magnetization of 4 πMs=1780 G and a dielec- tric constant of 13 is biased in the ˆxdirection with a field strength of 2000 Oe. At 5 GHz, two plane waves propagate in the + zdirection, one linearly polarized in xand the other linearly polarized in y. What is the distance these two waves must travel so that the differential phase shift between them is 180◦? 9.7 Consider a circularly polarized plane wave normally incident on an infinite ferrite medium, as shown below. Calculate the reflection and transmission coefficients for an RHCP (/Gamma1+,T+)and an LHCP (/Gamma1−,T−)incident wave. HINT: The transmitted wave will be polarized in the same sense as the incident wave, but the reflected wave will be oppositely polarized. /H92620, /H92800[/H9262r], /H9280r/H92800 zEi ΓEi TEiH0 0 9.8 An infinite lossless ferrite material with 4 πMs=1200 G is biased in the ˆxdirection with ¯H0=H0ˆx. Determine the range of H0, in oersteds, where an extraordinary wave (polarized in ˆx, propagating in ˆz)will be cut off. The frequency is 4 GHz. 9.9 Find the forward and reverse propagation constants for a waveguide half-filled with a transversely biased ferrite. (The geometry of Figure 9.9 with c=0a n d t=a/2.) Assume a=1.0c m , f= 10 GHz, 4π Ms=1700 G, and /epsilon1r=13. Plot versus H0=0 to 1500 Oe. Ignore loss and the fact that the ferrite may not be saturated for small H0. 9.10 Find the forward and reverse propagation constants for a waveguide filled with two pieces of oppo- sitely biased ferrite. (The geometry of Figure 9.10 with c=0a n dt =a/2.) Assume a=1.0c m , f=10 GHz, 4π Ms=1700 G, and /epsilon1r=13. Plot versus H0=0 to 1500 Oe. Ignore loss and the fact that the ferrite may not be saturated for small H0. 9.11 Consider a wide, thin ferrite slab in a rectangular X-band waveguide, as shown in Figure 9.11b. If f=10 GHz, 4π Ms=1700 G, c=a/4, and /Delta1S=2m m2, use the perturbation formula of (9.80) to plot the differential phase shift, (β+−β−)/k0, versus the bias field for H0=0 to 1200 Oe. Ignore loss. 9.12 AnE-plane resonance isolator with the geometry of Figure 9.11a is to be designed to operate at 8 GHz, with a ferrite having a saturation magnetization of 4π Ms=1500 G. (a) What is the ap- proximate bias field, H0, required for resonance? (b) What is the required bias field if the H-plane geometry of Figure 9.11b is used? 9.13 Design a resonance isolator using the H-plane ferrite slab geometry of Figure 9.11b in an X-band waveguide. The isolator should have minimum forward insertion loss, and a reverse attenuation of 30 dB at 10 GHz. Use a ferrite slab having /Delta1S/S=0.01, 4π Ms=1700 G, and /Delta1H=200 Oe. 9.14 Calculate and plot the two normalized positions, x/a, where the magnetic fields of the TE 10mode of an empty rectangular waveguide are circularly polarized, for k0=kcto 2kc. c09FerrimagneticComponents Pozar August 25, 2011 18:36 Problems 495 9.15 A conceptual latching ferrite phase shifter using the birefringence effect is shown below. In state 1, the ferrite is magnetized so that H0=0a n d ¯M=Mrˆx. In state 2, the ferrite is magnetized so that H0=0a n d ¯M=Mrˆy.I ff=10 GHz, /epsilon1r=12, and 4π Mr=1500 G, find the required length L to achieve a differential phase shift of 90◦. Assume the incident plane wave is ˆxpolarized for both states, and ignore reflections. yzxL Mr1 = Mrxˆ Mr2 = MryˆFerrite HE 9.16 Rework Example 9.4 with a slab spacing of s=2 mm and a remanent magnetization of 1000 G. (Assume all other parameters as unchanged and that the differential phase shift is linearly propor- tional to κ.) 9.17 Consider a latching phase shifter constructed with a wide, thin H-plane ferrite slab in an X-band waveguide, as shown in Figure 9.11b. If f=9 GHz, 4π Mr=1200 G, c=a/4, and /Delta1S=2m m2, use the perturbation formula of (9.80) to calculate the required length for a differential phase shiftof 22.5 ◦. 9.18 Design a gyrator using the twin H-plane ferrite slab geometry shown below. The frequency is 9.0 GHz, and the saturation magnetization is 4 πMs=1700 G. The cross-sectional area of each slab is 3.0 mm2and the guide is X-band waveguide. The permanent magnet has a field strength of Ha=4000 Oe. Determine the internal field in the ferrite, H0, and use the perturbation formula of (9.80) to determine the optimum location of the slabs and the length, L, to give the necessary 180◦ differential phase shift. ∆S alb HaPermanent magnet 9.19 Draw an equivalent circuit for a circulator using a gyrator and two couplers. 9.20 A certain lossless circulator has a return loss of 10 dB. What is the isolation? What is the isolation if the return loss is 20 dB? c10NoiseAndNonlinearDistortion Pozar August 26, 2011 15:49 Chapter Ten Noise and Nonlinear Distortion The effect of noise is critical to the performance of most RF and microwave communica- tions, radar, and remote sensing systems because noise ultimately determines the threshold for the minimum signal that can be reliably detected by a receiver. Noise power in a receiver will be introduced from the external environment through the receiving antenna, as well as gen-erated internally by the receiver circuitry. Here we will study the sources of noise in RF andmicrowave systems, and the characterization of components in terms of noise temperature and noise figure, including the effect of impedance mismatch. The additional noise-related topics of transistor amplifier noise figure, oscillator phase noise, and antenna noise temperature willbe discussed in later chapters. We will also discuss the related topics of compression, harmonic distortion, intermodula- tion distortion, and dynamic range. These can have important limiting effects when large signal levels are present in mixers, amplifiers, and other components that use nonlinear devices suchas diodes and transistors. 10.1NOISEINMICROWAVECIRCUITS Noise power is a result of random processes such as the flow of charges or holes in an electron tube or solid-state device, propagation through the ionosphere or other ionized gas, or, most basic of all, the thermal vibrations in any component at a temperature above absolute zero. Noise can be passed into a microwave system from external sources, or gen-erated within the system itself. In either case the noise level of a system sets the lower limit on the strength of a signal that can be detected in the presence of the noise. Thus, it is generally desired to minimize the residual noise level of a radar or communicationsreceiver to achieve the best performance. In some cases, such as radiometers or radio as- tronomy systems, the desired signal is actually the noise power received by an antenna, and it is necessary to distinguish between the received noise power and the undesired noise generated by the receiver system itself. 496 c10NoiseAndNonlinearDistortion Pozar August 26, 2011 15:49 10.1 Noise in Microwave Circuits 497 PoutPout (dBm) Pin (dBm)Pin Compression Noise floorFailureG FIGURE 10.1 Illustrating the dynamic range of a realistic amplifier. DynamicRangeandSourcesofNoise In previous chapters we have implicitly assumed that all components were linear (meaning that the output signal level is directly proportional to the input signal level), and determin- istic (meaning that the output signal is predictable from the input signal). In reality no component can perform in this way over an unlimited range of input/output signal levels. In practice, however, there is usually a range of signal levels over which such assumptionsare approximately valid; this range is called the dynamic range of the component. As an example, consider a realistic microwave transistor amplifier having a power gain G, as shown in Figure 10.1. If the amplifier were ideal, the output power would be related to the input power as P out=GPin,and this relation would hold true for any value of Pin. Thus, if Pin=0, we would have Pout=0, and if Pin=106W and G=10 dB, we would have Pout=107W. Neither of these results would actually occur in practice, however. Because of noise generated by the amplifier itself, some nonzero noise power will always be delivered by the amplifier, even when the input power is zero. At the other extreme,very high input power will cause the amplifier to fail. Thus, the actual relation between the output and input power will be as shown in Figure 10.1. At very low input power levels, the output will be dominated by the noise generated by the amplifier. This level is oftencalled the noise floor of the component or system; typical values may range from −80 to−140 dBm over the bandwidth of the system, with the lowest values being obtained with thermally cooled components. Above the noise floor, the amplifier will have a range of input power for which P out=GPinis closely approximated. This is the usable dynamic range of the component. At the upper end of this range, the output will begin to saturate, meaning that the output power no longer increases linearly as the input power increases.Excessive input power will lead to failure of the amplifier. Noise that is generated internally in a device or component is usually caused by ran- dom motions of charges or charge carriers in devices and materials. Such motions may bedue to any of several mechanisms, leading to various types of noise: rThermal noise is the most basic type of noise, being caused by thermal vibration of bound charges. It is also known as Johnson orNyquist noise.rShot noise is due to random fluctuations of charge carriers in an electron tube or solid-state device.rFlicker noise occurs in solid-state components and vacuum tubes. Flicker noise power varies inversely with frequency, and so is often called 1/ f-noise. c10NoiseAndNonlinearDistortion Pozar August 26, 2011 15:49 498 Chapter 10: Noise and Nonlinear Distortion rPlasma noise is caused by random motion of charges in an ionized gas, such as a plasma, the ionosphere, or sparking electrical contacts.rQuantum noise results from the quantized nature of charge carriers and photons; it is often insignificant relative to other noise sources. External noise may be introduced into a system either by a receiving antenna or by electromagnetic coupling. Some sources of external RF noise include the following: rThermal noise from the groundrCosmic background noise from the skyrNoise from stars (including the sun)rLightningrGas discharge lampsrRadio, TV , and cellular stationsrWireless devicesrMicrowave ovensrDeliberate jamming devices The characterization of noise effects in RF and microwave systems in terms of noise temperature and noise figure will apply to all types of noise, regardless of the source, aslong as the spectrum of the noise is relatively flat over the bandwidth of the system. Noise with a flat frequency spectrum is called white noise . NoisePowerandEquivalentNoiseTemperature Consider a resistor at a physical temperature of Tdegrees kelvin (K), as depicted in Figure 10.2. The electrons in the resistor are in random motion, with a kinetic energy that isproportional to the temperature. These random motions produce small, random voltage fluctuations at the resistor terminals, as illustrated in Figure 10.2. This voltage has a zero average value but a nonzero root mean square (rms) value given by Planck’s blackbodyradiation law, V n=/radicalbigg 4hf B R ehf/kT−1,( 10.1) where h=6.626×10−34J-sec is Planck’s constant. k=1.380×10−23J/K is Boltzmann’s constant. T=the temperature in degrees kelvin (K). B=the bandwidth of the system in Hz. f=the center frequency of the bandwidth in Hz. R=the resistance in /Omega1. T°K R v(t) v(t) t FIGURE 10.2 A random voltage generated by a noisy resistor. c10NoiseAndNonlinearDistortion Pozar August 26, 2011 15:49 10.1 Noise in Microwave Circuits 499 R R VnIdeal bandpass filter B FIGURE 10.3 Equivalent circuit of a noisy resistor delivering maximum power to a load resistor through an ideal bandpass filter. This result comes from quantum mechanical considerations, and is valid for any frequency f. At microwave frequencies the above result can be simplified by making use of the fact thathf/lessmuchkT. (As a worst-case example, let f=100 GHz and T=100 K. Then hf= 6.6×10−23/lessmuchkT=1.4×10−21.) Using the first two terms of a Taylor series expansion for the exponential in (10.1) gives ehf/kt−1/similarequalhf kT, so that (10.1) reduces to Vn=√ 4kTBR .( 10.2) This is the Rayleigh–Jeans approximation, and is the result that is most commonly used in microwave work [1]. For very high frequencies or very low temperatures, however, this approximation may be invalid, in which case (10.1) should be used. The noisy resistor of Figure 10.2 can be replaced with a Thevenin equivalent circuit consisting of a noiseless resistor and a generator with a voltage given by (10.2), as shown in Figure 10.3. Connecting a load resistor Rresults in maximum power transfer from the noisy resistor, with the result that power delivered to the load in a bandwidth Bis Pn=/parenleftbiggVn 2R/parenrightbigg2 R=V2 n 4R=kTB,( 10.3) since Vnis an rms voltage. This important result gives the maximum available noise power from the noisy resistor at temperature T. Note that this noise power is independent of frequency; such a noise source has a power spectral density that is constant with frequency, and is an example of a white noise source. The noise power is directly proportional to thebandwidth, which in practice is usually limited by the passband of the RF or microwave system. Independent white noise sources can be treated as Gaussian-distributed random variables, so the noise powers (variances) of independent noise sources are additive. The following trends can be observed from (10.3): rAsB→0,Pn→0. This means that systems with smaller bandwidths collect less noise power.rAsT→0,Pn→0. This means that cooler devices and components generate less noise power.rAsB→∞,Pn→∞ . This is the so-called ultra violet catastrophe , which does not occur in reality because (10.2)–(10.3) are not valid as f(orB)→∞ ; (10.1) must be used in this case. If an arbitrary source of noise (thermal or nonthermal) is “white,” so that the noise power is not a strong function of frequency over the bandwidth of interest, it can be mod- eled as an equivalent thermal noise source, and characterized with an equivalent noise c10NoiseAndNonlinearDistortion Pozar August 26, 2011 15:49 500 Chapter 10: Noise and Nonlinear Distortion R RNo NoTe Te =kBR RNoArbitrary white noise source FIGURE 10.4 The equivalent noise temperature, Te, of an arbitrary white noise source. temperature. Thus, consider the arbitrary white noise source of Figure 10.4, which has a driving-point impedance of Rand delivers a noise power Noto a load resistor R.T h i s noise source can be replaced by a noisy resistor of value Rat temperature Te, where Teis an equivalent temperature selected so that the same noise power is delivered to the load. That is, Te=No kB.( 10.4) Components and systems can then be characterized by saying that they have an equiva- lent noise temperature Te; this implies some fixed bandwidth B, which is generally the operational bandwidth of the component or system. For example, consider a noisy amplifier with a bandwidth Band gain G.L e tt h ea m - plifier be matched to noiseless source and load resistors, as shown in Figure 10.5. If the source resistor is at a (hypothetical) temperature of Ts=0K, then the input power to the amplifier will be Ni=0, and the output noise power Nowill be due only to the noise gen- erated by the amplifier itself. We can obtain the same load noise power by driving an ideal noiseless amplifier with a resistor at the temperature Te=No GkB,( 10.5) so that the output power in both cases is No=GkT eB. Then Teis the equivalent noise temperature of the amplifier. It is sometimes useful for measurement purposes to have a calibrated noise source. A passive noise source may simply consist of a resistor held at a constant temperature, either R R Noisy amplifier R R Noiseless amplifierTs = 0 KNi = 0 NiNo No = GkTeB NoG GTe (a) (b)Te =GkB FIGURE 10.5 Defining the equivalent noise temperature of a noisy amplifier. (a) Noisy amplifier. (b) Noiseless amplifier. c10NoiseAndNonlinearDistortion Pozar August 26, 2011 15:49 10.1 Noise in Microwave Circuits 501 in a temperature-controlled oven, or in a cryogenic flask. Active noise sources may use a diode, transistor, or tube to provide a calibrated noise power output. Noise generators can be characterized by an equivalent noise temperature, but a more common measure of noise power for such components is the excess noise ratio (ENR), defined as ENR (dB) =10 logNg−No No=10 logTg−T0 T0,( 10.6) where NgandTgare the noise power and equivalent noise temperature of the generator, andNoandT0are the noise power and temperature associated with a room-temperature (T0=290 K) passive source (a matched load). Solid-state noise generators typically have ENRs ranging from 20 to 40 dB. MeasurementofNoiseTemperature In principle, the equivalent noise temperature of a component can be determined by mea- suring the output power when a matched load at 0 K is connected at the input of the com-ponent. In practice, of course, a 0 K source temperature cannot be obtained, so a different method must be used. If two matched loads at significantly different temperatures are avail- able, then the Y-factor method can be applied. This technique is illustrated in Figure 10.6, where the amplifier (or other component) under test is connected to one of two matched loads at different temperatures, and the output power is measured for each case. Let T 1be the temperature of the hot load and T2the temperature of the cold load ( T1>T2), and let P1andP2be the respective powers measured at the amplifier output. The output noise power consists of noise power generated by the amplifier as well as noise power from the source resistor. Thus we have N1=GkT 1B+GkT eB, (10.7a) N2=GkT 2B+GkT eB, (10.7b) which are two equations for the two unknowns, TeandGB(the gain–bandwidth product of the amplifier). Define the Y-factor as Y=N1 N2=T1+Te T2+Te>1,( 10.8) which is determined as the ratio of the output power measurements. Then (10.7) can be solved for the equivalent noise temperature of the device under test as Te=T1−YT2 Y−1,( 10.9) in terms of the load temperatures and the Y-factor. R RT1(hot) T2(cold)G, B, TeN1, N2 FIGURE 10.6 The Y-factor method for measuring the equivalent noise temperature of an amplifier. c10NoiseAndNonlinearDistortion Pozar August 26, 2011 15:49 502 Chapter 10: Noise and Nonlinear Distortion Note that to obtain accurate results from this method, the two source temperatures must not be too close together. If they are, N1will be close to N2,Ywill be close to unity, and the evaluation of (10.9) will involve the subtractions of numbers close to each other, resulting in a loss of accuracy. In practice, one noise source is usually a load resistor atroom temperature (T 0=290 K), while the other noise source is either “hotter” or “colder,” depending on whether Teis greater or less than T0. An active noise generator can be used as a “hot” source, while a “cold” source can be obtained by immersing a load resistor inliquid nitrogen ( T=77 K) or liquid helium ( T=4K ) . EXAMPLE 10.1 NOISE TEMPERATURE MEASUREMENT An X-band amplifier has a gain of 20 dB and a 1 GHz bandwidth. Its equivalent noise temperature is to be measured via the Y-factor method. The following data are obtained: ForT1=290 K, N1=−62.0d B m . ForT2=77 K, N2=−64.7d B m . Determine the equivalent noise temperature of the amplifier. If the amplifier is used with a source having an equivalent noise temperature of Ts=450 K, what is the output noise power from the amplifier, in dBm? Solution From (10.8), the Y-factor in dB is Y=(N1−N2)dB=(−62.0) −(−64.7) =2.7d B , which is a numeric value of Y=1.86. Using (10.9) gives the equivalent noise temperature as Te=T1−YT2 Y−1=290−(1.86)(77) 1.86−1=170 K. If a source with an equivalent noise temperature of Ts=450 K drives the amplifier, the noise power into the amplifier will be kTsB. The total noise power out of the amplifier will be No=GkT sB+GkT eB=100(1.38 ×10−23)(109)(450+170) =8.56×10−10W=−60.7d B m . ■ 10.2NOISEFIGURE Definitio ofNoiseFigure We have seen that a noisy microwave component can be characterized by an equivalent noise temperature. An alternative characterization is the noise figure of the component, which is a measure of the degradation in the signal-to-noise ratio between the input andoutput of the component. The signal-to-noise ratio is the ratio of desired signal power to undesired noise power, and so is dependent on the signal power. When noise and a desired signal are applied to the input of a noiseless network, both noise and signal will be atten-uated or amplified by the same factor, so that the signal-to-noise ratio will be unchanged. However, if the network is noisy, the output noise power will be increased more than the c10NoiseAndNonlinearDistortion Pozar August 26, 2011 15:49 10.2 Noise Figure 503 R R Pi = Si + Ni Po = So + NoNoisy network G, B, TeT0 = 290 K FIGURE 10.7 Determining the noise figure of a noisy network. output signal power, so that the output signal-to-noise ratio will be reduced. The noise figure, F, is a measure of this reduction in signal-to-noise ratio, and is defined as F=Si/Ni So/No≥1,( 10.10) where Si,Niare the input signal and noise powers, and So,Noare the output signal and noise powers. By definition, the input noise power is assumed to be the noise power result- ing from a matched resistor at T0=290 K; that is, Ni=kT0B. Consider Figure 10.7, which shows noise power Niand signal power Sibeing fed into a noisy two-port network. The network is characterized by a gain, G, a bandwidth, B, and an equivalent noise temperature, Te. The input noise power is Ni=kT0B, and the output noise power is a sum of the amplified input noise and the internally generated noise: No=kGB(T0+Te). The output signal power is So=GSi. Using these results in (10.10) gives the noise figure as F=Si kT0BkGB(T0+Te) GSi=1+Te T0≥1.( 10.11) In dB, F=10 log(1+Te/T0)dB≥0. If the network were noiseless, Tewould be zero, giving F=1, or 0 dB. Solving (10.11) for Tegives Te=(F−1)T0.( 10.12) It is important to keep in mind two things concerning the definition of noise figure: noise figure is defined for a matched input source, and for a noise source equivalent to a matched load at temperature T0=290 K. Noise figure and equivalent noise temperatures are inter- changeable characterizations of the noise properties of a component. An important special case occurs in practice for a two-port network consisting of a passive, lossy component, such as an attenuator or lossy transmission line, held at a phys- ical temperature T. Consider such a network with a matched source resistor that is also at temperature T, as shown in Figure 10.8. The power gain, G, of a lossy network is less than unity; the loss factor, L, can be defined as L=1/G>1. Because the entire system is in thermal equilibrium at the temperature T, and has a driving point impedance of R, the out- put noise power must be No=kTB. However, we can also think of this power as coming from the source resistor (attenuated by the lossy line), and from the noise generated by the line itself. Thus we also have that No=kTB=GkTB +GN added,( 10.13) where Nadded is the noise generated by the line, as if it appeared at the input terminals of the line. Solving (10.13) for this power gives Nadded=1−G GkTB=(L−1)kTB .( 10.14) c10NoiseAndNonlinearDistortion Pozar August 26, 2011 15:49 504 Chapter 10: Noise and Nonlinear Distortion RT No = kTBL, T, Zo = RNi = kTB FIGURE 10.8 Determining the noise figure of a lossy line or attenuator with loss Land tempera- tureT. Then (10.4) shows that the lossy line has an equivalent noise temperature (referred to the input) given by Te=1−G GT=(L−1)T.( 10.15) From (10.11) the noise figure is F=1+(L−1)T T0.( 10.16) If the line is at temperature T0, then F=L. For instance, a 6 dB attenuator at room tem- perature has a noise figure of F=6d B . NoiseFigureofaCascadedSystem In a typical microwave system the input signal travels through a cascade of many different components, each of which may degrade the signal-to-noise ratio to some degree. If we know the noise figure (or noise temperature) of the individual stages, we can determine thenoise figure (or noise temperature) of the cascade connection of stages. We will see that the noise performance of the first stage is usually the most critical, an interesting result that is very important in practice. Consider the cascade of two components, having gains G 1,G2, noise figures F1,F2, and equivalent noise temperatures Te1,Te2, as shown in Figure 10.9. We wish to find the overall noise figure and equivalent noise temperature of the cascade, as if it were a singlecomponent. The overall gain of the cascade is G 1G2. Using noise temperatures, we can write the noise power at the output of the first stage as N1=G1kT0B+G1kTe1B,( 10.17) G1 F1 Te1G2 F2 Te2 G1G2 Fcas TecasNi No NoT0 Ni T0N1 (a) (b) FIGURE 10.9 Noise figure and equivalent noise temperature of a cascaded system. (a) Two cas- caded networks. (b) Equivalent network. c10NoiseAndNonlinearDistortion Pozar August 26, 2011 15:49 10.2 Noise Figure 505 since Ni=kT0Bfor noise figure calculations. The noise power at the output of the second stage is No=G2N1+G2kTe2B =G1G2kB/parenleftbigg T0+Te1+1 G1Te2/parenrightbigg . (10.18) For the equivalent system we have No=G1G2kB(Tcas+T0), (10.19) so comparison with (10.18) gives the noise temperature of the cascade system as Tcas=Te1+1 G1Te2.( 10.20) Using (10.12) to convert the temperatures in (10.20) to noise figures yields the noise figure of the cascade system as Fcas=F1+1 G1(F2−1). (10.21) Equations (10.20) and (10.21) show that the noise characteristics of a cascaded system are dominated by the characteristics of the first stage since the effect of the second stage is reduced by the gain of the first (assuming G1>1). Thus, for the best overall system noise performance, the first stage should have a low noise figure and at least moderate gain. Expense and effort should be devoted primarily to the first stage, as opposed to later stages, since later stages have a diminished impact on the overall noise performance. Equations (10.20) and (10.21) can be generalized to an arbitrary number of stages, as follows: Tcas=Te1+Te2 G1+Te3 G1G2+···, (10.22) Fcas=F1+F2−1 G1+F3−1 G1G2+···. (10.23) EXAMPLE 10.2 NOISE ANALYSIS OF A WIRELESS RECEIVER The block diagram of a wireless receiver front-end is shown in Figure 10.10. Compute the overall noise figure of this subsystem. If the input noise power from a feeding antenna is Ni=kTAB, where TA=150 K, find the output noise power in dBm. If we require a minimum signal-to-noise ratio (SNR) of 20 dB at theoutput of the receiver, what is the minimum signal voltage that should be applied Low noise amplifierBandpass filter Mixer Ga = 10 dB Fa = 2 dBLm = 3 dB Fm = 4 dBSi, Ni So, No Lf = 1 dB FIGURE 10.10 Block diagram of a wireless receiver front-end for Example 10.2. c10NoiseAndNonlinearDistortion Pozar August 26, 2011 15:49 506 Chapter 10: Noise and Nonlinear Distortion at the receiver input? Assume the system is at temperature T0, with a characteristic impedance of 50 /Omega1, and an IF bandwidth of 10 MHz. Solution We first perform the required conversions from dB to numerical values: Ga=10 dB =10 Gf=−1.0d B =0.79 Gm=−3.0d B =0.5 Fa=2d B=1.58 Ff=1d B=1.26 Fm=4d B=2.51 Next, use (10.23) to find the overall noise figure of the system: F=Fa+Ff−1 Ga+Fm−1 GaGf=1.58+(1.26−1) 10+(2.51−1) (10)(0.79) =1.80=2.55 dB. The best way to compute the output noise power is to use noise temperatures. From (10.12), the equivalent noise temperature of the overall system is Te=(F−1)T0=(1.80−1)(290) =232 K. The overall gain of the system is G=(10)(0.79)(0.5) =3.95. Then we can find the output noise power as No=k(TA+Te)BG=(1.38×10−23)(150+232)(10 ×106)(3.95) =2.08×10−13W=−96.8d B m . For an output SNR of 20 dB =100, the input signal power must be Si=So G=So NoNo G=1002.08×10−13 3.95=5.27×10−12W=−82.8d B m . For a 50 /Omega1system impedance, this corresponds to an input signal voltage of Vi=/radicalbig ZoSi=/radicalbig (50)(5.27 ×10−12)=1.62×10−5V=16.2µV (rms). Note: It may be tempting to compute the output noise power from the definition of the noise figure, as No=NiF/parenleftbiggSo Si/parenrightbigg =NiFG=kTABFG =(1.38×10−23)(150)(10 ×106)(1.8)(3.95) =1.47×10−13W. This is an incorrect result! The reason for the disparity with the earlier result is that the definition of noise figure assumes an input noise level of kT0B, while this problem involves an input noise of kTAB, with TA=150 K /negationslash=T0.T h i si sa common error, and suggests that when computing absolute noise power it is often safer to use noise temperatures to avoid this confusion. ■ NoiseFigureofaPassiveTwo-PortNetwork We previously derived the noise figure for a matched lossy line or attenuator by using a thermodynamic argument. Here we generalize that technique to evaluate the noise figure of general passive networks (networks that do not contain active devices such as diodes ortransistors, which generate nonthermal noise). In addition, this method will account for the change in noise figure that occurs when a component is impedance mismatched at either its c10NoiseAndNonlinearDistortion Pozar August 26, 2011 15:49 10.2 Noise Figure 507 FIGURE 10.11 A passive two-port network with impedance mismatches. The network is at phys- ical temperature T. input or output port. Generally it is easier and more accurate to find the noise characteristics of an active device, such as a diode or transistor, by direct measurement than by calculation from first principles. Figure 10.11 shows an arbitrary passive two-port network, with a generator at port 1 and a load at port 2. The network is characterized by its scattering matrix, [ S]. In the general case, impedance mismatches may exist at each port, and we define these mismatches interms of the following reflection coefficients: /Gamma1 s=reflection coefficient looking toward generator , /Gamma1in=reflection coefficient looking toward port 1 of network , /Gamma1out=reflection coefficient looking toward port 2 of network , /Gamma1L=reflection coefficient looking toward load. If we assume the network is at temperature T, and that an available input noise power of N1=kTB is applied to the input of the network, we can write the available output noise power at port 2 as N2=G21kTB+G21Nadded,( 10.24) where Nadded is the noise power generated internally by the network (referenced to port 1), andG21is the available power gain of the network from port 1 to port 2. The available power gain can be expressed in terms of the scattering parameters of the network and theport mismatches as (also see Section 12.1), G 21=power available from network power available from source=|S21|2(1−|/Gamma1S|2) |1−S11/Gamma1S|2(1−|/Gamma1out|2).( 10.25) As derived in Example 4.7, the output port mismatch is given by /Gamma1out=S22+S12S21/Gamma1S 1−S11/Gamma1S.( 10.26) Observe that when the network is matched to its external circuitry, so that /Gamma1s=0 and S22=0, we have /Gamma1out=0 and G21=|S21|2, which is the gain of the network when it is matched. Also observe that the available gain of the network does not depend on the loadmismatch, /Gamma1 L. This is because available gain is defined in terms of the maximum power that is available from the network, which occurs when the load impedance is conjugately matched to the output impedance of the network. Since the input noise power is kTB, and the network is passive and at temperature T, the network is in thermodynamic equilibrium, and so the available output noise power must c10NoiseAndNonlinearDistortion Pozar August 26, 2011 15:49 508 Chapter 10: Noise and Nonlinear Distortion beN2=kTB. Then we can solve for Nadded from (10.24) to give Nadded=1−G21 G21kTB.( 10.27) Then the equivalent noise temperature of the network is Te=Nadded kB=1−G21 G21T,( 10.28) and the noise figure of the network is F=1+Te T0=1+1−G21 G21T T0.( 10.29) Note the similarity of (10.27)–(10.29) to the results in (10.14)–(10.16) for the lossy line— the essential difference is that here we are using the available gain of the network, which accounts for impedance mismatches between the network and the external circuit. We can illustrate the use of this result with some applications to problems of practical interest. NoiseFigureofaMismatchedLossyLine Earlier we found the noise figure of a lossy transmission line under the assumption that it was matched to its input and output circuits. Now we consider the case where the line is mismatched to its input circuit. Figure 10.12 shows a transmission line of length /lscriptat temperature T, with a power loss factor L=1/G, and an impedance mismatch between the line and the generator. Thus, Zg/negationslash=Z0, and the reflection coefficient looking toward the generator is /Gamma1s=Zg−Z0 Zg+Z0/negationslash=0. The scattering matrix of the lossy line of characteristic impedance Z0can be written as [S]=/bracketleftbigg 01 10/bracketrightbigge−jβ/lscript √ L,( 10.30) where βis the propagation constant of the line. Using the elements of (10.30) in (10.26) gives the reflection coefficient looking into port 2 of the line as /Gamma1out=S22+S12S21/Gamma1s 1−S11/Gamma1s=/Gamma1s Le−2jβ/lscript.( 10.31) /lscript FIGURE 10.12 A lossy transmission line at temperature Twith an impedance mismatch at its input port. c10NoiseAndNonlinearDistortion Pozar August 26, 2011 15:49 10.2 Noise Figure 509 Then the available gain, from (10.25), is G21=1 L(1−|/Gamma1s|2) 1−|/Gamma1out|2=L(1−|/Gamma1s|2) L2−|/Gamma1s|2.( 10.32) We can verify two limiting cases of (10.32): when L=1w eh a v e G21=1, and when /Gamma1s=0w eh a v e G21=1/L. Using (10.32) in (10.28) gives the equivalent noise tempera- ture of the mismatched lossy line as Te=1−G21 G21T=(L−1)(L+|/Gamma1s|2) L(1−|/Gamma1s|2)T.( 10.33) The corresponding noise figure can then be evaluated using (10.11). Observe that when the line is matched, /Gamma1s=0, and (10.33) reduces to Te=(L−1)T, in agreement with the result for the matched lossy line given by (10.15). If the line is lossless, then L=1, and (10.33) reduces to Te=0 regardless of mismatch, as expected. However, when the line is lossy and mismatched, so that L>1 and |/Gamma1s|>0, then the noise temperature given by (10.33) is greater than Te=(L−1)T, the noise temperature of the matched lossy line. The reason for this increase is that the lossy line actually delivers noise power out of bothits ports, but when the input port is mismatched some of the available noise power at port 1 is reflected from the source back into port 1 and appears at port 2. When the generator is matched to port 1, none of the available power from port 1 is reflected back into the line,so the noise power available at port 2 is a minimum. This result implies that impedance matching is important in minimizing noise temperature and noise figure. EXAMPLE 10.3 APPLICATION TO A WILKINSON POWER DIVIDER Find the noise figure of a Wilkinson power divider when one of the output ports isterminated in a matched load. Assume an insertion loss factor of Lfrom the input to either output port. Solution From Chapter 7 the scattering matrix of a Wilkinson divider is given as [S]=−j √ 2L/bracketleftBigg011 100 100/bracketrightBigg , where the factor L≥1 accounts for the dissipative loss from port 1 to port 2 or 3 (note that dissipative loss is distinct from the −3 dB power division ratio). To evaluate the noise figure of the Wilkinson divider, we first terminate port 3 with a matched load; this converts the three-port device into a two-port device. If we assume a matched source at port 1, we have /Gamma1s=0. Equation (10.26) then gives /Gamma1out=S22=0, and so the available gain can be calculated from (10.25) as G21=|S21|2=1 2L. The equivalent noise temperature of the Wilkinson divider is, from (10.28), Te=1−G21 G21T=(2L−1)T, c10NoiseAndNonlinearDistortion Pozar September 15, 2011 17:13 510 Chapter 10: Noise and Nonlinear Distortion where Tis the physical temperature of the divider. Using (10.11) gives the noise figure as F=1+Te T0=1+(2L−1)T T0. Observe that if the divider is at room temperature, then T=T0and the above reduces to F=2L. If the divider is at room temperature and lossless, this reduces toF=2=3 dB. In this case the source of the noise power is the isolation resistor contained in the Wilkinson divider circuit. Because the network is matched at its input and output, it is easy to obtain these same results using the thermodynamic argument directly. Thus, if we applyan input noise power of kTB to port 1 of the matched divider at temperature T,t h e system will be in thermal equilibrium and the output noise power must be kTB. We can also express the output noise power as the sum of the input power times the gain of the divider, and N added, the noise power added by the divider itself (referenced to the input to the divider): kTB=kTB 2L+Nadded 2L. Solving for Nadded gives Nadded=kTB(2L−1), so the equivalent noise temper- ature is Te=Nadded kB=(2L−1)T, in agreement with the above. ■ NoiseFigureofaMismatchedAmplifie Finally, consider the effect of an input impedance mismatch on the noise figure of an ampli- fier. As shown in Figure 10.13, the amplifier, when matched, has a gain G, a noise figure F, and a bandwidth B. The amplifier output is matched, but there is an impedance mismatch at the input represented by the reflection coefficient, /Gamma1. Our previous results involving the effect of mismatch on noise figure made use of (10.29), but that was derived for a passive network and so cannot be directly used in this case. Instead we will use noise temperatures. Since we are dealing with noise figure, let the input noise power to the amplifier be Ni=kT0B. Then the output noise power from the amplifier (referenced to the input) is given by No=kT0GB/parenleftbig 1−|/Gamma1|2/parenrightbig +kT0(F−1)GB (10.34) where the first term is due to the input noise power, decreased by the reflection at the input, and the second term is the noise power due to the amplifier itself, based on the equivalent noise temperature as given by (10.12). For an applied signal power Si, the output signal So + No Si + Ni Z0Z0 G, F, B/H9003 FIGURE 10.13 A noisy amplifier with an impedance mismatch at its input. c10NoiseAndNonlinearDistortion Pozar August 26, 2011 15:49 10.3 Nonlinear Distortion 511 power is So=G/parenleftbig 1−|/Gamma1|2/parenrightbig Si.( 10.35) The overall noise figure, Fm, of the mismatched amplifier can be found from (10.10) as Fm=SiNo SoNi=1+F−1 1−|/Gamma1|2.( 10.36) Observe from (10.36) the limiting case that Fm=Fwhen|/Gamma1|=0 (no mismatch), and that this is the minimum noise figure that can be achieved since Fmincreases as the mis- match increases. This result demonstrates that good noise figure requires good impedance matching. This problem would be more complicated if a mismatch also existed at the out-put of the amplifier, particularly if the amplifier is not unilateral. 10.3NONLINEARDISTORTION We have seen that thermal noise is generated by any lossy component. Since all realistic components have at least a small loss, the ideal linear component does not exist in practice because all realistic devices are nonlinear at very low signal levels due to noise effects.In addition, practical components may also become nonlinear at high signal levels. In the case of active devices, such as diodes and transistors, this may be due to effects such as gain compression or the generation of spurious frequency components due to device non-linearities, but all devices ultimately fail at very high power levels. In either case, these effects set a minimum and maximum realistic power range, or dynamic range, over which a given component or network will operate as desired. In this section we will study the re-sponse of nonlinear devices in general, and two definitions of dynamic range. These results will be useful for our later discussions of amplifiers (Chapter 12), mixers (Chapter 13), and wireless receivers (Chapter 14). Devices such as diodes and transistors have nonlinear characteristics, and it is this nonlinearity that is of great utility for desirable functions such as amplification, detection,and frequency conversion [2]. Nonlinear device characteristics, however, can also lead to undesirable effects such as gain compression and the generation of spurious frequency components. These effects may lead to increased losses, signal distortion, and possibleinterference with other radio channels or services. Some of the many possible effects of nonlinearity in RF and microwave circuits are listed below [3]: rHarmonic generation (multiples of a fundamental signal)rSaturation (gain reduction in an amplifier)rIntermodulation distortion (products of a two-tone input signal)rCross-modulation (modulation transfer from one signal to another)rAM-PM conversion (amplitude variation causes phase shift)rSpectral regrowth (intermodulation with many closely spaced signals) Figure 10.14 shows a general nonlinear network, having an input voltage viand an output voltage vo. In the most general sense, the output response of a nonlinear circuit can FIGURE 10.14 A general nonlinear device or network. c10NoiseAndNonlinearDistortion Pozar September 15, 2011 17:17 512 Chapter 10: Noise and Nonlinear Distortion be modeled as a Taylor series in terms of the input signal voltage: vo=a0+a1vi+a2v2 i+a3v3 i+···,( 10.37) where the Taylor coefficients are defined as a0=vo(0) (DC output) (10.38a) a1=dvo dvi/vextendsingle/vextendsingle/vextendsingle/vextendsingle vi=0(linear output) (10.38b) a2=d2vo dv2 i/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle vi=0(squared output) (10.38c) and higher order terms. Different functions can be obtained from the nonlinear network depending on the dominance of particular terms in the expansion. The constant term, with coefficient a0, in (10.37) leads to rectification, converting an AC input signal to DC. The linear term, with coefficient a1, models a linear attenuator ( a1<1) or amplifier ( a1>1). The second-order term, with coefficient a2, can be used for mixing and other frequency conversion functions. Practical nonlinear devices usually have a series expansion contain-ing many nonzero terms, and a combination of several of the above effects will occur. We will consider some important special cases below. GainCompression First consider the case where a single-frequency sinusoid is applied to the input of a general nonlinear network, such as an amplifier: v i=V0cosω0t.( 10.39) Equation (10.37) gives the output voltage as vo=a0+a1V0cosω0t+a2V2 0cos2ω0t+a3V3 0cos3ω0t+··· =/parenleftbigg a0+1 2a2V2 0/parenrightbigg +/parenleftbigg a1V0+3 4a3V3 0/parenrightbigg cosω0t+1 2a2V2 0cos 2ω 0t +1 4a3V3 0cos 3ω 0t+···. (10.40) This result leads to the voltage gain of the signal component at frequency ω0: Gv=v(ω0) o v(ω0) i=a1V0+3 4a3V3 0 V0=a1+3 4a3V2 0,( 10.41) where we have retained only terms through the third order. The result of (10.41) shows that the voltage gain is equal to a1, the coefficient of the linear term, as expected, but with an additional term proportional to the square of the inputvoltage amplitude. In most practical amplifiers a 3typically has the opposite sign of a1,s o that the output of the amplifier tends to be reduced from the expected linear dependence for large values of V0. This effect is called gain compression,o r saturation . Physically, this is usually due to the fact that the instantaneous output voltage of an amplifier is limited by the power supply voltage used to bias the active device. c10NoiseAndNonlinearDistortion Pozar September 15, 2011 17:17 10.3 Nonlinear Distortion 513Pout (dBm) Pin (dBm)1 dB compression point, P1dB IP1dBLinear response (slope = 1)OP1dB1 dB FIGURE 10.15 Definition of the 1 dB compression point for a nonlinear amplifier. A typical amplifier response is shown in Figure 10.15. For an ideal linear amplifier a plot of the output power versus input power would be a straight line with a slope of unity, and the power gain of the amplifier given by the ratio of the output power to the input power. The amplifier response of Figure 10.15 tracks the ideal response over a limited range, thenbegins to saturate, resulting in reduced gain. To quantify the linear operating range of the amplifier, we define the 1 dB compression point as the power level for which the output power has decreased by 1 dB from the ideal linear characteristic. This power level is usuallydenoted by P 1dB, and can be stated in terms of either input power ( IP1dB) or output power (OP 1dB). The 1 dB compression point is typically given as the larger of these two options, so for amplifiers P1dBis usually specified as an output power, while for mixers P1dBis usu- ally specified in terms of input power. The relation between a compression point referenced at the input versus the output is given as, in dB, OP1dB=IP1dB+G−1 dB [4, 5]. HarmonicandIntermodulationDistortion Observe from the expansion of (10.40) that a portion of the input signal at frequency ω0is converted to other frequency components. For example, the first term of (10.40) represents a DC voltage, which would be a useful response in a rectifier application. The voltage components at frequencies 2ω 0or 3ω 0can be useful for frequency multiplier circuits. In amplifiers, however, the presence of other frequency components will lead to signal distortion if those components are in the passband of the amplifier. For a single input frequency, or tone,ω0, the output will in general consist of har- monics of the input frequency of the form nω0,f o rn =0,1,2,... . Often these harmonics lie outside the passband of the amplifier and so do not interfere with the desired signal at frequency ω0. The situation is different, however, when the input signal consists of two closely spaced frequencies. Consider a two-tone input voltage, consisting of two closely spaced frequencies ω1 andω2: vi=V0(cosω1t+cosω2t). (10.42) c10NoiseAndNonlinearDistortion Pozar August 26, 2011 15:49 514 Chapter 10: Noise and Nonlinear Distortion From (10.37) the output is vo=a0+a1V0(cosω1t+cosω2t)+a2V2 0(cosω1t+cosω2t)2 +a3V3 0(cosω1t+cosω2t)3+··· =a0+a1V0cosω1t+a1V0cosω2t+1 2a2V2 0(1+cos 2ω 1t)+1 2a2V2 0(1+cos 2ω 2t) +a2V2 0cos(ω1−ω2)t+a2V2 0cos(ω1+ω2)t +a3V3 0/parenleftbigg3 4cosω1t+1 4cos 3ω 1t/parenrightbigg +a3V3 0/parenleftbigg3 4cosω2t+1 4cos 3ω 2t/parenrightbigg +a3V3 0/bracketleftbigg3 2cosω2t+3 4cos(2ω 1−ω2)t+3 4cos(2ω 1+ω2)t/bracketrightbigg +a3V3 0/bracketleftbigg3 2cosω1t+3 4cos(2ω 2−ω1)t+3 4cos(2ω 2+ω1)t/bracketrightbigg +···. (10.43) where standard trigonometric identities have been used to expand the initial expression. We see that the output spectrum consists of harmonics of the form mω1+nω2,( 10.44) with m,n=0,±1,±2,±3,.... These combinations of the two input frequencies are called intermodulation products , and the order of a given product is defined as |m|+|n|. For example, the squared term of (10.43) gives rise to the following four intermodulation products of second order: 2ω1 (second harmonic of ω1)m=2n=0 order =2, 2ω2 (second harmonic of ω2)m=0n=2 order =2, ω1−ω2(difference frequency) m=1n=−1 order =2, ω1+ω2(sum frequency) m=1n=1 order =2. All of these second-order products are undesired in an amplifier, but in a mixer the sum or difference frequencies form the desired outputs. In either case, if ω1andω2are close, all of the second-order products will be far from ω1orω2and can easily be filtered (either passed or rejected) from the output of the component. Note from (10.43) that the ratio ofthe amplitude of the second-order intermodulation product ω 1−ω2(orω1+ω2)t ot h e amplitude of a second harmonic 2 ω1(or 2ω2) is 2.0, so the second-order harmonic power will be 6 dB less than the power in the second-order sum or difference terms. The cubed term of (10.43) leads to six third-order intermodulation products: 3 ω1,3ω2, 2ω1+ω2,2ω2+ω1,2ω1−ω2, and 2ω 2−ω1. The first four of these will again be located far from ω1orω2, and will typically be outside the passband of the component. However, the two difference terms produce products located near the original input signals at ω1and ω2, and so cannot be easily filtered from the passband of an amplifier. Figure 10.16 shows a typical spectrum of the second- and third-order two-tone intermodulation products. For an arbitrary input signal consisting of many frequencies of varying amplitude and phase, the resulting in-band intermodulation products will cause distortion of the output signal.This effect is called third-order intermodulation distortion . It can be seen from (10.43) that the ratio of the amplitude of the third-order intermod- ulation product 2ω 1−ω2(or 2ω 2−ω1) to the amplitude of the third harmonic 3 ω1(or 3ω2) is 3.0, so the third-order harmonic power will be 9.54 dB less than the power in the third-order intermodulation terms. c10NoiseAndNonlinearDistortion Pozar August 26, 2011 15:49 10.3 Nonlinear Distortion 515 FIGURE 10.16 Output spectrum of second- and third-order two-tone intermodulation products, assuming ω1<ω 2. Third-OrderInterceptPoint Equation (10.43) shows that as the input voltage V0increases, the voltage associated with the third-order products increases as V3 0. Since power is proportional to the square of volt- age, we can also say that the output power of third-order products must increase as the cube of the input power. So for small input powers the third-order intermodulation products willbe very small, but will increase quickly as input power increases. We can view this effect graphically by plotting the output power for the first- and third-order products versus input power on log-log scales (or in dB), as shown in Figure 10.17. The output power of the first-order, or linear, product is proportional to the input power, and so the line describing this response has a slope of unity (before the onset of compression). The line describing the response of the third-order products has a slope of 3.(The second-order products would have a slope of 2, but since these products are generally not in the passband of the component, we have not plotted their response in Figure 10.17.) Both the linear and third-order responses will exhibit compression at high input powers, sowe show the extension of their idealized responses with dotted lines. Since these two lines have different slopes, they will intersect, typically at a point above the onset of compres- sion, as shown in the figure. This hypothetical intersection point where the first-order andthird-order powers would be equal is called the third-order intercept point, denoted as IP 3;Pout (dBm) Pin (dBm)CompressionIntercept point, IP3 Linear response (slope = 1) Cubic response (slope = 3) IP1dB1 dBOP1dBOIP3 IIP3 FIGURE 10.17 Third-order intercept diagram for a nonlinear component. c10NoiseAndNonlinearDistortion Pozar September 15, 2011 17:17 516 Chapter 10: Noise and Nonlinear Distortion it may be specified as either an input power level ( IIP3), or an output power level ( OIP 3). The relation between an intercept point referenced at the input versus the output is simply OIP 3=G(IIP3). As with the 1 dB compression point, the reference for IP3is typically chosen to result in the largest value, so IP3is usually referenced at the output for amplifiers and at the input for mixers. As depicted in Figure 10.17, IP3generally occurs at a higher power level than P1dB, the 1 dB compression point. Many practical components follow the approximate rule that IP3is 10–15 dB greater than P1dB, assuming these powers are referenced at the same point. We can express IP3in terms of the Taylor coefficients of the expansion of (10.43) as follows. Define Pω1as the output power of the desired signal at frequency ω1. Then from (10.43) we have Pω1=1 2a2 1V2 0.( 10.45) Similarly, define P2ω1−ω2as the output power of the intermodulation product of frequency 2ω1−ω2. Then from (10.43) we have P2ω1−ω2=1 2/parenleftbigg3 4a3V3 0/parenrightbigg2 =9 32a2 3V6 0.( 10.46) By definition, these two powers are equal at the third-order intercept point. If we define the input signal voltage at the intercept point as VIP, then equating (10.45) and (10.46) gives 1 2a2 1V2 IP=9 32a2 3V6 IP. Solving for VIPyields VIP=/radicalBigg 4a1 3a3.( 10.47) Since OIP 3is equal to the linear response of Pω1at the intercept point, we have from (10.45) and (10.47) that OIP 3=Pω1/vextendsingle/vextendsingle V0=VIP=1 2a2 1V2 IP=2a3 1 3a3,( 10.48) where IP3in this case is referred to the output port. These expressions will be useful in the following sections. InterceptPointofaCascadedSystem As in the case of noise figure, a cascade connection of components usually has the ef- fect of degrading (lowering) the third-order intercept point. Unlike noise powers, however, intermodulation products in a cascaded system are deterministic and may be in phase co- herence, in which case we cannot simply add powers but must deal with voltages [5]. We will first consider the coherent (in-phase) cascade case, then the noncoherent case. With reference to Figure 10.18, G1andOIP/prime 3are the power gain and third-order inter- cept point for the first stage, and G2andOIP/prime/prime 3are the corresponding values for the second stage. Let P/prime ω1be the first-stage output power of the desired signal at frequency ω1, and let P/prime 2ω1−ω2be the first-stage output power at the third-order intermodulation product. From (10.46), P/prime 2ω1−ω2can be rewritten in terms of P/prime ω1andOIP/prime 3, using (10.45) and (10.48), as c10NoiseAndNonlinearDistortion Pozar August 26, 2011 15:49 10.3 Nonlinear Distortion 517 FIGURE 10.18 Third-order intercept point for a cascaded system. (a) Two cascaded networks. (b) Equivalent network. follows: P/prime 2ω1−ω2=9a2 3V6 0 32=1 8a6 1V6 0 4a6 1 9a2 3=(P/prime ω1)3 (OIP/prime 3)2.( 10.49) The first-stage output voltage associated with this power is V/prime 2ω1−ω2=/radicalBig P/prime 2ω1−ω2Z0=/radicalBig/parenleftbig P/primeω1/parenrightbig3Z0 OIP/prime 3,( 10.50) where Z0is the system impedance. For coherent intermodulation products, the total third-order distortion voltage at the output of the second stage is the sum of the above voltage times the voltage gain of the second stage, and the distortion voltage generated by the second stage. This is becausethese voltages are deterministic and phase related, unlike uncorrelated noise powers that arise in cascaded components. Adding these voltages gives the worst-case result for the overall distortion level because there may be phase delays within the stages that couldcause partial cancellation. Thus we can write the worst-case total distortion voltage at the output of the second stage as V /prime/prime 2ω1−ω2=/radicalBig G2/parenleftbig P/primeω1/parenrightbig3Z0 OIP/prime 3+/radicalBig/parenleftbig P/prime/primeω1/parenrightbig3Z0 OIP/prime/prime 3. Since P/prime/prime ω1=G2P/prime ω1,w eh a v e V/prime/prime 2ω1−ω2=/parenleftBigg 1 G2/parenleftbig OIP/prime 3/parenrightbig+1 OIP/prime/prime 3/parenrightBigg/radicalBig/parenleftbig P/prime/primeω1/parenrightbig3Z0.( 10.51) The total output distortion power is P/prime/prime 2ω1−ω2=/parenleftBig V/prime/prime 2ω1−ω2/parenrightBig2 Z0=/parenleftBigg 1 G2/parenleftbig OIP/prime 3/parenrightbig+1 OIP/prime/prime 3/parenrightBigg2/parenleftbig P/prime/prime ω1/parenrightbig3=/parenleftbig P/prime/prime ω1/parenrightbig3 (OIP 3)2.( 10.52) Thus the third-order intercept point of the cascaded system with coherent products is OIP 3=/parenleftBigg 1 G2/parenleftbig OIP/prime 3/parenrightbig+1 OIP/prime/prime 3/parenrightBigg−1 .( 10.53) Note that OIP 3=G2/parenleftbig OIP/prime 3/parenrightbig forOIP/prime/prime 3→∞ , which is the limiting case when the second stage has no third-order distortion. c10NoiseAndNonlinearDistortion Pozar September 15, 2011 17:17 518 Chapter 10: Noise and Nonlinear Distortion FIGURE 10.19 System for Example 10.4. If the intermodulation products from each stage have relatively random phases, which may occur when the intermodulation products are not very close to the fundamental signals, it may be proper to treat the individual contributions as incoherent, allowing us to add powers. It is straightforward to show that the overall intercept point in this case is given by OIP 3=/parenleftBigg 1 G2 2/parenleftbig OIP/prime 3/parenrightbig2+1 /parenleftbig OIP/prime/prime 3/parenrightbig2/parenrightBigg−1/2 .( 10.54) EXAMPLE 10.4 CALCULATION OF CASCADE INTERCEPT POINT A low-noise amplifier and mixer are shown in Figure 10.19. The amplifier has a gain of 20 dB and a third-order intercept point of 22 dBm (referenced at output), and the mixer has a conversion loss of 6 dB and a third-order intercept point of 13 dBm (referenced at input). Find the intercept points of the cascade network forboth a phase coherence assumption and a random-phase (noncoherence) assump- tion. Solution First we transfer the reference of IP 3for the mixer from its input to its output: OIP/prime/prime 3=/parenleftbig IIP/prime/prime 3/parenrightbig G2=13 dBm −6d B=7d B m . Converting the necessary dB values to numerical values yields: OIP/prime 3=22 dBm =158 mW (for amplifier) , OIP/prime/prime 3=7d B m =5m W ( f o rm i x e r ) , G2=−6d B=0.25 (for mixer). Assuming coherence, equation (10.53) gives the intercept point of the cascade as OIP 3=/parenleftBigg 1 G2/parenleftbig OIP/prime 3/parenrightbig+1 OIP/prime/prime 3/parenrightBigg−1 =/parenleftbigg1 (0.25)(158)+1 5/parenrightbigg−1 =4.4m W =6.4d B m , which is seen to be lower than the minimum IP3of the individual components. Equation (10.54) gives the results for the noncoherent case as OIP 3=/parenleftBigg 1 G2 2/parenleftbig OIP/prime 3/parenrightbig2+1 /parenleftbig OIP/prime/prime 3/parenrightbig2/parenrightBigg−1/2 =/parenleftbigg1 (0.25)2(158)2+1 (5)2/parenrightbigg−1/2 =4.96 mW =6.9d B m . As expected, the noncoherent case results in a slightly higher intercept point. ■ c10NoiseAndNonlinearDistortion Pozar September 15, 2011 17:17 10.4 Dynamic Range 519 PassiveIntermodulation The above discussion of intermodulation distortion was in the context of active circuits involving diodes and transistors, but it is also possible for intermodulation products to be generated by passive nonlinear effects in connectors, cables, antennas, or almost anycomponent where there is a metal-to-metal contact. This effect is called passive intermod- ulation (PIM) and, as in the case of intermodulation in amplifiers and mixers, it occurs when signals at two or more closely spaced frequencies mix to produce spurious products. Passive intermodulation can be caused by a number of factors, such as poor mechan- ical contact, oxidation of junctions between ferrous-based metals, contamination of con-ducting surfaces at RF junctions, or the use of nonlinear materials such as carbon fiber composites or ferromagnetic materials. In addition, when high powers are involved, ther- mal effects may contribute to the overall nonlinearity of a junction. It is very difficultto predict PIM levels from first principles, so measurement techniques must usually be used. Because of the third-power dependence of the third-order intermodulation products with input power, passive intermodulation is usually only significant when input signal powers are relatively large. This is frequently the case in cellular telephone base station transmitters, which may operate with powers of 30–40 dBm, with many closely spacedRF channels. It is often desired to maintain the PIM level below −125 dBm, with two 40 dBm transmit signals. This is a very wide dynamic range, and requires careful selec- tion of components used in the high-power portions of the transmitter, including cables,connectors, and antenna components. Because these components are often exposed to the weather, deterioration due to oxidation, vibration, and sunlight must be offset by a careful maintenance program. Communications satellites often face similar problems with passive intermodulation. Passive intermodulation is generally not a problem in receiver systems due to the much lower power levels. 10.4DYNAMICRANGE LinearandSpuriousFreeDynamicRange We can define dynamic range in a general sense as the operating range for which a compo- nent or system has desirable characteristics. For a power amplifier this may be the powerrange that is limited at the low end by noise and at the high end by the compression point. This is essentially the linear operating range for the amplifier, and is called the linear dy- namic range (LDR). For low-noise amplifiers or mixers, operation may be limited by noise at the low end and the maximum power level for which intermodulation distortion be- comes unacceptable. This is effectively the operating range for which spurious responses are minimal, and it is called the spurious-free dynamic range (SFDR). We can find the linear dynamic range LDR as the ratio of P 1dB, the 1 dB compression point, to the noise level of the component, as shown in Figure 10.20. In dB, this can be written in terms of output powers as LDR(dB)=OP1dB−No,( 10.55) forOP1dBandNoexpressed in dBm. Note that some authors prefer to define the linear dynamic range in terms of a minimum detectable power level. This definition is more ap- propriate for a receiver system rather than an individual component, as it depends on factorsexternal to the component itself, such as the type of modulation used, the recommended system SNR, effects of error-correcting coding, and related factors. c10NoiseAndNonlinearDistortion Pozar August 26, 2011 15:49 520 Chapter 10: Noise and Nonlinear Distortion Pout (dBm) Pin (dBm)n = 1n = 3 IP1dB1 dBOP1dBOIP3 IIP3SFDR LDR Noise level No FIGURE 10.20 Illustrating linear dynamic range (LDR) and spurious free dynamic range (SFDR). The spurious free dynamic range is defined as the maximum output signal power for which the power of the third-order intermodulation product is equal to the noise level of the component, divided by the output noise level. This situation is shown in Figure 10.20. IfPω1is the output power of the desired signal at frequency ω1, and P2ω1−ω2is the output power of the third-order intermodulation product, then the spurious free dynamic range can be expressed as SFDR =Pω1 P2ω1−ω2,( 10.56) with P2ω1−ω2taken equal to the noise level of the component. As in (10.49), P2ω1−ω2can be written in terms of OIP 3andPω1as P2ω1−ω2=(Pω1)3 (OIP 3)2.( 10.57) Observe that this result clearly shows that the third-order intermodulation power increases as the cube of the input signal power. Solving (10.57) for Pω1and applying the result to (10.56) gives the spurious free dynamic range in terms of OIP 3andNo, the output noise power of the component: SFDR =Pω1 P2ω1−ω2/vextendsingle/vextendsingle/vextendsingle/vextendsingle P2ω1−ω2=No=/parenleftbiggOIP 3 No/parenrightbigg2/3 .( 10.58) This result can be written in terms of dB as SFDR (dB)=2 3(OIP 3−No), (10.59) forOIP 3andNoexpressed in dBm. Although this result was derived for the 2 ω1−ω2 product, the same result applies for the 2ω 2−ω1product. In a receiver it may be required to have a minimum detectable signal level, or min- imum SNR, in order to achieve a specified performance level. This requires an increasein the input signal level, resulting in a corresponding decrease in dynamic range, since the spurious power level is still equal to the noise power. In this case, the spurious free dynamic range of (10.59) would be modified as [5, 6]: SFDR (dB)=2 3(OIP 3−No)−SNR.( 10.60) c10NoiseAndNonlinearDistortion Pozar September 15, 2011 17:17 Problems 521 EXAMPLE 10.5 DYNAMIC RANGES A receiver has a noise figure of 7 dB, a 1 dB compression point of 25 dBm (ref- erenced to output), a gain of 40 dB, and a third-order intercept point of 35 dBm(referenced to output). If the receiver is fed with an antenna having a noise tem- perature of T A=150 K, and the desired output SNR is 10 dB, find the linear and spurious free dynamic ranges. Assume a receiver bandwidth of 100 MHz. Solution The noise power at the receiver output can be calculated using noise temperatures as No=GkB[TA+(F−1)T0]=104(1.38×10−23)(108)[150+(4.01)(290)] =1.8×10−8W=−47.4d B m . The linear dynamic range is, from (10.55), in dB, LDR=OP1dB−No=25 dBm +47.4d B m =72.4d B . Equation (10.60) gives the spurious free dynamic range as SFDR =2 3(OIP 3−No)−SNR=2 3(35+47.4)−10=44.9d B . Observe that SFDR /lessmuchLDR. ■ REFERENCES [1] F. T. Ulaby, R. K. Moore, and A. K. Fung, Microwave Remote Sensing: Active and Passive, Volume I, Microwave Remote Sensing, Fundamentals and Radiometry. Addison-Wesley, Reading, Mass., 1981. [2] M. E. Hines, “The Virtues of Nonlinearity—Detection, Frequency Conversion, Parametric Ampli- fication and Harmonic Generation,” IEEE Transactions on Microwave Theory and Techniques, vol. MTT-32, pp. 1097–1104, September 1984. [ 3 ] S .A .M a a s ,Nonlinear Microwave and RF Circuits, 2nd ed., Artech House, Norwood, Mass., 2003.[4] K. Chang, RF and Microwave Wireless Systems , John Wiley & Sons, New York, 2000. [5] W. Egan, Practical RF System Design , John Wiley & Sons, Hoboken, N.J., 2003. [6] M. Steer, Microwave and RF Design: A Systems Approach , SciTech, Raleigh, N.C., 2010. PROBLEMS 10.1 The noise figure of a microwave receiver front-end is measured using the Y-factor method. A noise source having an ENR of 22 dB, and a liquid nitrogen cold load (77 K) are used, resulting in a measured Y-factor ratio of 15.83 dB. What is the noise figure of the receiver? 10.2 Assume that measurement error introduces an uncertainty of /Delta1Yinto the measurement of Yin aY- factor measurement. Derive an expression for the normalized error, /Delta1Te/Te, of the equivalent noise temperature in terms of /Delta1Y/Yand the temperatures T1,T2,andTe. Minimize this result with respect toTeto obtain an expression for Tein terms of T1andT2that will result in minimum error. 10.3 A lossy transmission line has a noise figure of F0at temperature T0=290 K. Calculate and plot the noise figure of this line as its physical temperature ranges from T=0 K to 1000 K, for F0=1d B and for F0=3d B . 10.4 An amplifier with a gain of 12 dB, a bandwidth of 150 MHz, and a noise figure of 4 dB feeds a receiver with a noise temperature of 900 K. Find the noise figure of the overall system. c10NoiseAndNonlinearDistortion Pozar August 26, 2011 15:49 522 Chapter 10: Noise and Nonlinear Distortion 10.5 A cellular telephone receiver front-end circuit is shown below. The operating frequency is 1805– 1880 MHz, and the physical temperature of the system is 300 K. A noise source with Ni=−95 dBm is applied to the receiver input. (a) What is the equivalent noise temperature of the source over the operating bandwidth? (b) What is the noise figure (in dB) of the amplifier? (c) What is the noisefigure (in dB) of the cascaded transmission line and amplifier? (d) What is the total noise power output (in dBm) of the receiver over the operating bandwidth? Noise sourceTransmission line Amplifier Ni = /H1100295 dBm G = 12 dB Te = 180 KNo L = 1.5 dB 10.6 Consider the wireless local area network (WLAN) receiver front-end shown below, where the band- width of the bandpass filter is 100 MHz centered at 2.4 GHz. If the system is at room temperature, find the noise figure of the overall system. What is the resulting signal-to-noise ratio at the output if the input signal power level is −90 dBm? Can the components be rearranged to give a better noise figure? IL = 1.5 dB G = 10 dB F = 2 dBG = 20 dB F = 2 dB 10.7 A two-way power divider has one output port terminated in a matched load, as shown below. Find the noise figure of the resulting two-port network if the divider is (a) an equal-split two-way resistive divider, (b) a two-way Wilkinson divider, and (c) a 3 dB quadrature hybrid. Assume the divider in each case is matched, and at room temperature. Power divider Z0 10.8 Show that, for fixed loss L>1, the equivalent noise temperature of a mismatched lossy line given in (10.33) is minimized when |/Gamma1s|=0. 10.9 Consider the mismatched amplifier of Figure 10.13, having a noise figure Fwhen matched at its input. Calculate and plot the resulting noise figure as the input reflection coefficient magnitude, |/Gamma1|, varies from 0 to 0.9 for F=1, 3, and 10 dB. 10.10 A lossy line at temperature Tfeeds an amplifier with noise figure F, as shown below. If an impedance mismatch /Gamma1is present at the input of the amplifier, find the overall noise figure of the system. Z0, L, T /H9003 G, F 10.11 A balanced amplifier circuit is shown below. The two amplifiers are identical, each with power gain Gand noise figure F. The two quadrature hybrids are also identical, with an insertion loss from the input to either output of L>1 (not including the 3 dB power division factor). Derive an expression c10NoiseAndNonlinearDistortion Pozar September 29, 2011 16:54 Problems 523 for the overall noise figure of the balanced amplifier. What does this result reduce to when the hybrids are lossless? Z0 So, NoZ0Ni, Si L L G, FG, F 10.12 Show that the following relations involving the third-order intercept point of a two-port nonlinear network are valid. Piω1andPoω1are the input and output power levels of an applied two-tone signal, andPi 2ω1−ω2andPo 2ω1−ω2are the power levels of the third-order products referenced to the input and output. OIP 3−Poω1 IIP3−Piω1=1,OIP 3−Po 2ω1−ω2 IIP3−Pi 2ω1−ω2=3. 10.13 In practice, the third-order intercept point is extrapolated from measured data taken at input power levels well below IP3. For the spectrum analyzer display shown below, where /Delta1Pis the difference in power between Pω1andP2ω1−ω2, show that the third-order intercept point is given by OIP 3= Pω1+(1/2)/Delta1 P. Calculate the input and output third-order intercept points for the following data: Pω1=5d B m ,P2ω1−ω2=−27 dBm ,Pin=−4d B m . 10.14 A two-tone input with a 6 dB difference in the two signal levels is applied to a nonlinear component. What is the relative power ratio of the resulting two third-order intermodulation products 2 ω1−ω2 and 2ω 2−ω1,i fω1andω2are close together? 10.15 Find the third-order intercept points for the problem of Example 10.4 when the positions of the amplifier and mixer are reversed. 10.16 It is possible to approximately relate the 1 dB compression point to the third-order intercept point. For a single-tone input, use (10.40) to find the amplitudes of the fundamental and third harmonic terms, and assume that a3is of opposite sign to a1.L e t V0be the voltage where the third-order term reduces the first-order power by 1 dB, and solve for |a3/a1|. For a two-tone input, use (10.43) to find the amplitude of the third-order intermodulation product, then use (10.44) to relate OP1dBtoOIP 3. 10.17 An amplifier with a bandwidth of 1 GHz has a gain of 15 dB and a noise temperature of 250 K. If the 1 dB compression point occurs for an output power level of 5 dBm, what is the linear dynamicrange of the amplifier? 10.18 A receiver subsystem has a noise figure of 6 dB, a 1 dB compression point of 21 dBm (referenced to output), a gain of 30 dB, and a third-order intercept point of 33 dBm (referenced to output). Ifthe subsystem is fed with a noise source with N i=−105 dBm and the desired output SNR is 8 dB, find the linear and spurious free dynamic ranges of the subsystem. Assume a system bandwidth of 20 MHz. c11ActiveRFAndMicrowaveDevices Pozar August 26, 2011 17:16 Chapter Eleven Active RF and Microwave Devices Active devices include diodes, transistors, and electron tubes, which can be used for signal detection, mixing, amplification, frequency multiplication, and switching, and as sources of RF and microwave signals. We will discuss some of the basic characteristics of such devices in this chapter. We will avoid detailed discussion of the physics of active devices (see references[1–5] for such material) since for our purposes it will be adequate to work with the termi-nal characteristics of diodes and transistors using equivalent circuits or scattering parameters. These results will be used to study some basic diode detector and control circuits, and in later chapters for the design of amplifier, mixer, and oscillator circuits using diodes and transistors.We will conclude this chapter with an overview of microwave integrated circuits (MICs) and abrief discussion of some microwave tubes. Historically, the development of useful RF and microwave active devices has been a long and slow process. The first detector diode was probably the “cat-whisker” crystal detector usedin early radio work of the nineteenth century. The advent of electron tubes used as detectors andamplifiers later eliminated this component in most radio systems, but crystal diodes were used by Southworth in his 1930s experiments with waveguides since tube detectors could not oper- ate at such high frequencies. Frequency conversion and heterodyning were also first developedfor radio applications in the 1920s. These same techniques were later applied to microwaveradars at the MIT Radiation Laboratory during World War II (using crystal diodes as mixers) [1], but it was not until the 1960s that the subject of microwave semiconductor devices saw significant progress. The invention of the transistor led to advances in the theory of solid-statematerials and devices, as well as the availability of new semiconductor materials. This led tothe development of many new types of diodes and transistors for high-frequency applications. The invention of the gallium arsenide field effect transistor (FET) in the late 1960s [2] was one of the most far-reaching developments in modern microwave engineering. RF and microwavetransistors are critical components in wireless systems, finding application as amplifiers, oscil-lators, switches, phase shifters, mixers, and active filters. Following the lead from integrated circuitry at lower frequencies, monolithic microwave integrated circuits (MMICs) combine transmission lines, active devices, and other compo-nents on a semiconductor substrate. The first single-function MMICs were developed in thelate 1960s, but more sophisticated circuits and subsystems, such as multistage FET amplifiers, 524 c11ActiveRFAndMicrowaveDevices Pozar August 26, 2011 17:16 11.1 Diodes and Diode Circuits 525 transmit/receive radar modules, front ends for wireless products, and many other circuits, are now being fabricated as MMICs [2]. The trend is toward MMICs having higher performance, lower power requirements, greater complexity, and lower cost. 11.1DIODESANDDIODECIRCUITS We begin our discussion of active devices with some of the major types of diodes used in RF and microwave circuits. A diode is a two-terminal semiconductor device having a nonlinear V–Irelationship. This nonlinearity can be exploited for the useful functions of signal detection, demodulation, switching, frequency multiplication, and oscillation [1]. RF and microwave diodes can be packaged as axial or beam lead components or as surface- mountable chips, or be monolithically integrated with other components on a single semi-conductor substrate. We first consider detector diodes and circuits, then discuss PIN diodes and control circuits, varactor diodes, and a summary of other types of diodes. SchottkyDiodesandDetectors The classical pnjunction diode commonly used at low frequencies has a relatively large junction capacitance that makes it unsuitable for high frequency application. The Schottky barrier diode, however, relies on a semiconductor–metal junction that results in a much lower junction capacitance [3, 4], allowing operation at higher frequencies. Commerciallyavailable microwave Schottky diodes generally use n-type gallium arsenide (GaAs) ma- terial, while lower frequency versions may use n-type silicon. Schottky diodes are often biased with a small DC forward current, but can be used without bias. The primary application of Schottky diodes is in frequency conversion of an input signal. Figure 11.1 illustrates the three basic frequency conversion operations of rectifica- tion(conversion to DC), detection (demodulation of an amplitude-modulated signal), and mixing (frequency shifting). A junction diode can be modeled as a nonlinear resistor, with a small-signal V–I relationship expressed as I(V)=I s(eαV−1), (11.1) where α=q/nkT, and qis the charge of an electron, kis Boltzmann’s constant, Tis temperature, nis the ideality factor, and Isis the saturation current [3–5]. Typically, Isis between 10−6and 10−15A, and α=q/nkT is approximately 1/(25 mV) for T=290 K. The ideality factor, n, depends on the structure of the diode, and can vary from about 1.05 for Schottky barrier diodes to about 2.0 for point-contact silicon diodes. Figure 11.2 shows a typical diode V–Icharacteristic for a Schottky diode. Small-signal approximation : Let the diode voltage be expressed as V=V0+v, (11.2) c11ActiveRFAndMicrowaveDevices Pozar August 26, 2011 17:16 526 Chapter 11: Active RF and Microwave Devices t ft f f ftfRF fRFRF DC Modulated RFModulation(a) (b)t fm ffRF – fLOfRF + fLOffRF ffLO (c)IF RF LO FIGURE 11.1 Basic frequency conversion operations of rectification, detection, and mixing. (a) Diode rectifier. (b) Diode detector. (c) Mixer. where V0is a DC bias voltage and vis a small AC signal voltage. Then (11.1) can be expanded in a Taylor series about V0as follows: I(V)=I0+vdI dV/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle V0+1 2v2d2I dV2/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle V0+···,( 11.3) where I0=I(V0)is the DC bias current. The first derivative can be evaluated as dI dV/vextendsingle/vextendsingle/vextendsingle/vextendsingle V0=αIseαV0=α(I0+Is)=Gd=1 Rj,( 11.4) I V+ –VI Is FIGURE 11.2 V–Icharacteristics of a Schottky diode. c11ActiveRFAndMicrowaveDevices Pozar August 26, 2011 17:16 11.1 Diodes and Diode Circuits 527 + –V Cj(V)Rj(V)CpRs Ls FIGURE 11.3 Equivalent AC circuit model for a Schottky diode. which defines Rj, the junction resistance of the diode, and Gd=1/Rj, which is called the dynamic conductance of the diode. The second derivative is d2I dV2/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle V0=dGd dV/vextendsingle/vextendsingle/vextendsingle/vextendsingle V0=α2IseαV0=α2(I0+Is)=αGd=G/prime d.( 11.5) Then (11.3) can be rewritten as the sum of the DC bias current, I0, and an AC current, i: I(V)=I0+i=I0+vGd+v2 2G/prime d+··· (11.6) The three-term approximation for the diode current in (11.6) is called the small-signal approximation, and will be adequate for most of our purposes. The small-signal approximation is based on the DC voltage–current relationship of (11.1), and shows that the equivalent circuit of a diode will involve a nonlinear resistance.In practice, however, the AC characteristics of a diode also involve reactive effects due to the structure and packaging of the diode. A typical equivalent circuit for an RF diode is shown in Figure 11.3. The leads or contacts of the diode package are modeled as a seriesinductance, L s, and shunt capacitance, Cp. The series resistor, Rs, accounts for contact and current-spreading resistance. The junction capacitance, Cj, and the junction resistance, Rj, are bias dependent. Table 11.1 lists some parameters for a few commercially available Schottky diodes. Diode rectifiers and detectors : In a rectifier application, a diode is used to convert a fraction of an RF input signal to DC power. Rectification is a very common function and is used forpower monitors, automatic gain control circuits, and signal strength indicators. If the total diode voltage consists of a DC bias voltage and a small-signal RF voltage, V=V 0+v0cosω0t,( 11.7) then (11.6) shows that the diode current will be I=I0+v0Gdcosω0t+v2 0 2G/prime dcos2ω0t =I0+v2 0 4G/prime d+v0Gdcosω0t+v2 0 4G/prime dcos 2ω 0t, (11.8 ) TABLE 11.1 Parameters for Some Commercial Schottky Diodes Schottky Diode Is(A) Rs(/Omega1) Cj(pF) Ls(nH) Cp(pF) Skyworks SMS1546 3 ×10−74 0.38 1.0 0.07 Skyworks SMS7630 5 ×10−620 0.14 0.05 0.005 Avago HSMS2800 3 ×10−830 1.6 — — Macom MA4E2054 3 ×10−811 0.1 — 0.11 c11ActiveRFAndMicrowaveDevices Pozar August 26, 2011 17:16 528 Chapter 11: Active RF and Microwave Devices where I0is the bias current and v2 0G/prime d/4 is the DC rectified current. The output also con- tains AC signals of frequency ω0, and 2ω 0(as well as higher order harmonics), which are usually filtered out with a simple low-pass filter. A current sensitivity ,βi, can be defined as a measure of the change in the DC output current for a given RF input power. From (11.6) the RF input power is v2 0Gd/2 (using only the first term), while (11.8) shows the change in DC current is v2 0G/prime d/4. The current sensitivity is then βi=/Delta1Idc Pin=G/prime d 2GdA/W.( 11.9) An open-circuit voltage sensitivity, βv, can be defined in terms of the voltage drop across the junction resistance when the diode is open circuited. Thus, βv=βiRj.( 11.10) Typical values for the voltage sensitivity of an RF diode range from 400 to 1500 mV/mW. In a detector application the nonlinearity of a diode is used to demodulate an amplitude- modulated (AM) RF carrier. In this case, the diode voltage can be expressed as v(t)=v0(1+mcosωmt)cosω0t,( 11.11) where ωmis the modulation frequency, ω0is the RF carrier frequency (ω0/greatermuchωm), and m is defined as the modulation index (0≤m≤1). Using (11.11) in (11.6) gives the diode current: i(t)=v0Gd(1+mcosωmt)cosω0t+v2 0 2G/prime d(1+mcosωmt)2cos2ω0t =v0Gd/bracketleftBig cosω0t+m 2cos(ω0+ωm)t+m 2cos(ω 0−ωm)t/bracketrightBig +v2 0 4G/prime d/bracketleftBigg 1+m2 2+2mcosωmt+m2 2cos 2ω mt+cos 2ω 0t +mcos(2ω 0+ωm)t+mcos(2ω 0−ωm)t+m2 2cos 2ω 0t +m2 4cos 2(ω 0+ωm)t+m2 4cos 2(ω 0−ωm)t/bracketrightBigg . (11.12) The frequency spectrum of this output is shown in Figure 11.4. The output current terms that are linear in the diode voltage (terms multiplying v0Gd)have frequencies of ω0and 0 /H9275m 2/H9275m /H92750 – /H9275m 2(/H92750 – /H9275m) 2(/H92750 + /H9275m) /H92752/H92750 – /H9275m 2/H92750 + /H9275m2/H92750/H92750 /H92750 + /H9275mRelative amplitude of detector output FIGURE 11.4 Output spectrum of a detected AM signal. c11ActiveRFAndMicrowaveDevices Pozar August 26, 2011 17:16 11.1 Diodes and Diode Circuits 529 TABLE 11.2 Frequencies and Relative Amplitudes of the Square-Law Output of a Detected AM Signal Frequency Relative Amplitude 01 +m2/2 ωm 2m 2ωm m2/2 2ω0 1+m2/2 2ω0±ωm m 2(ω0±ωm) m2/4 ω0±ωm, while the terms that are proportional to the square of the diode voltage (terms multiplying v2 0G/prime d/2)include the frequencies and relative amplitudes listed in Table 11.2. The desired demodulated output of frequency ωmis easily separated from the un- desired frequency components with a low-pass filter. Observe that the amplitude of this current is mv2 0G/prime d/2, which is proportional to the square of the input signal voltage, and hence the input signal power. This square-law behavior is the usual operating condition for detector diodes, but it can be obtained only over a restricted range of input power. If the input power is too large, small-signal conditions will not apply, and the output will become saturated and approach a linear, and then a constant, iversus Pcharacteristic. At very low signal levels the input signal will be lost in the noise floor of the device. Figure 11.5 shows a typical voutversus Pincharacteristic, where the output voltage can be considered as the voltage drop across a resistor in series with the diode. Square-law operation is particu-larly important for applications where power levels are inferred from detector voltage, as in SWR indicators and signal level indicators. Detectors may be DC biased to an operating point that provides the best sensitivity. POINT OF INTEREST: The Spectrum Analyzer A spectrum analyzer gives a frequency domain representation of a signal, displaying the average power density versus frequency. Thus, its function is dual to that of an oscilloscope, which –40 –30 –20 –10 0 10 20 30 4010 /H9262V100 /H9262V1 mV10 mV100 mV1 V Noise levellog vout Square-law region vout ~ v02 ~ PinSaturation log Pin (dBm) FIGURE 11.5 Square-law region for a typical diode detector. c11ActiveRFAndMicrowaveDevices Pozar August 26, 2011 17:16 530 Chapter 11: Active RF and Microwave Devices displays a time domain representation of a signal. A spectrum analyzer is basically a sensitive receiver that tunes over a specified frequency band and gives a video output that is proportional to the signal power in a narrow bandwidth. Spectrum analyzers are invaluable for measuring modulation products, harmonic and intermodulation distortion, noise, and interference effects. The diagram below shows a simplified block diagram of a spectrum analyzer. A microwave spectrum analyzer can typically cover any frequency band in the range of several hundred mega- hertz to tens of gigahertz. The frequency resolution is set by the IF bandwidth, and is typicallyadjustable from about 100 Hz to 1 MHz. A sweep generator is used to repetitively scan the receiver over the desired frequency band by adjusting the local oscillator frequency, and to pro- vide horizontal deflection of the display. An important part of a modern spectrum analyzer is theYIG-tuned bandpass filter at the input to the mixer. This filter is tuned along with the local oscil- lator, and acts as a preselector to reduce spurious intermodulation products. An IF amplifier with a logarithmic response is generally used to accommodate a wide dynamic range. Modern spec-trum analyzers usually contain a computer to control the system and the measurement process. This improves performance and makes the analyzer more versatile, but can be a disadvantage in that the computer can sometimes remove the user from the physical reality of the measurement. LP filter InputVariable attenuatorYIG-tuned filter Tuning control Sweep generatorYIG oscillatorMixerIF Amp. (log) DetectorVideo amp. HV Display PINDiodesandControlCircuits Switches are used extensively in microwave systems for directing signal or power flow between components. Switches can also be used to construct other types of control circuits, such as phase shifters and attenuators. Mechanical switches can be made in waveguide or coaxial form, and can handle high powers but are bulky and slow. PIN diodes, however, can be used to construct an electronic switching element easily integrated with planar circuitryand capable of high-speed operation. Switching speeds typically range from 1 to 10 µs, although speeds as fast as 20 ns are possible with careful design of the diode driving circuit. PIN diodes can also be used as power limiters, modulators, and variable attenuators. PIN diode characteristics : A PIN diode contains an intrinsic (lightly doped) layer between thepandnsemiconductor layers. When reverse biased, a small series junction capacitance leads to a relatively high diode impedance, while a forward bias current removes the junc- tion capacitance and leaves the diode in a low-impedance state [3, 4]. These characteristics make the PIN diode a useful RF switching element. Equivalent circuits for the forward- and reverse-biased states are shown in Figure 11.6. The parasitic inductance, L i, is typically less than 1 nH. The reverse resistance, Rr, is usually small relative to the series reactance due to the junction capacitance and is often ignored. The forward bias current is typi- cally 10–30 mA, and the reverse bias voltage is typically 10–60 V . The bias voltages mustbe applied to the diode with RF chokes and DC blocks for isolation from the RF signal. Table 11.3 lists parameters for some commercially available PIN diodes. c11ActiveRFAndMicrowaveDevices Pozar August 26, 2011 17:16 11.1 Diodes and Diode Circuits 531 Li Zf Cj Rf (b)Li Zr Rr (a) FIGURE 11.6 Equivalent circuits for the reverse- and forward-biased states of a PIN diode. (a) Reverse bias state. (b) Forward bias state. TABLE 11.3 Parameters for Some Commercial PIN Diodes PIN Diode Rf(/Omega1) Cj(pF) ASI 8001 3.0 0.03 Skyworks DSG9500 4.0 0.025 Infineon BA592 0.36 1.4 Microsemi UM9605 1.5 0.5 Single-pole PIN diode switches: A PIN diode can be used in either a series or a shunt con- figuration to form a single-pole, single-throw RF switch. These circuits are shown in Fig- ure 11.7, along with the required bias networks. In the series configuration of Figure 11.7a, the switch is ON when the diode is forward biased, while in the shunt configuration theswitch is ON when the diode is reverse biased. In both cases, input power is reflected when the switch is in the OFF state. The DC blocking capacitors should have a relatively low impedance at the RF operating frequency, while the RF choke inductors should have arelatively high RF impedance. In some designs, high-impedance quarter-wavelength lines can be used in place of the chokes, to provide RF blocking. An ideal switch would have zero insertion loss in the ON state, and infinite attenuation in the OFF state. Realistic switching elements, of course, result in some insertion loss for the ON state and finite attenuation for the OFF state. Knowing the diode parameters for the equivalent circuits of Figure 11.6 allows the insertion loss for the ON and OFF statesto be calculated for the series and shunt switches. With reference to the simplified switch Z0 Z0 Z0 Z0Bias RF choke DC block DC blockBias RF choke DC block DC blockRF chokeDiodeDiode (a) (b) FIGURE 11.7 Single-pole PIN diode switches. (a) Series configuration. (b) Shunt configuration. c11ActiveRFAndMicrowaveDevices Pozar August 26, 2011 17:16 532 Chapter 11: Active RF and Microwave Devices Z0 Z0 2 V0 Zd VL+ – (b)Z0 Z0 2 V0Zd VL+ – (a) FIGURE 11.8 Simplified equivalent circuits for the series and shunt single-pole PIN diode switches. (a) Series switch. (b) Shunt switch. circuits of Figure 11.8, we can define the insertion loss in terms of the actual load voltage, VL, and V0, which is the load voltage that would appear if the switch (Zd)were absent: IL=−20 log/vextendsingle/vextendsingle/vextendsingle/vextendsingleV L V0/vextendsingle/vextendsingle/vextendsingle/vextendsingle.( 11.13) Simple circuit analysis applied to the two cases of Figure 11.8 gives the following results: IL=−20 log/vextendsingle/vextendsingle/vextendsingle/vextendsingle2Z 0 2Z0+Zd/vextendsingle/vextendsingle/vextendsingle/vextendsingle(series switch), (11.14a) IL=−20 log/vextendsingle/vextendsingle/vextendsingle/vextendsingle2Z d 2Zd+Z0/vextendsingle/vextendsingle/vextendsingle/vextendsingle(shunt switch). (11.14b) In both cases, Z dis the diode impedance for either the reverse or forward bias state. Thus, Zd=/braceleftbiggZr=Rr+j(ωLi−1/ωCj)for reverse bias Zf=Rf+jωLi for forward bias.(11.15) The ON-state or OFF-state insertion loss of a switch can usually be improved by adding an external reactance in series or in parallel with the diode, to compensate for the diodereactance. This technique usually reduces the bandwidth, however. Several single-throw switches can be combined to form a variety of multiple-pole and/or multiple-throw configurations. Figure 11.9 shows series and shunt circuits for a single-pole, double-throw switch; such a switch requires at least two switching elements. In operation, one diode is forward biased in the low-impedance state, with the other diode reverse biased in the high-impedance state. The input signal is switched from one outputto the other by reversing the diode bias states. The quarter-wave lines of the shunt circuit limit the bandwidth of this configuration. A photograph of a PIN SP3T switch is shown in Figure 11.10. EXAMPLE 11.1 SINGLE-POLE PIN DIODE SWITCH A single-pole switch operating at 1.8 GHz is to be constructed using a MicrosemiUM 9605 PIN diode with C j=0.5p Fa n d Rf=1.5/Omega1. What switch circuit (se- ries or shunt) should be used to obtain the greatest ratio of off-to-on attenuation? Assume that Li=0.5nH,Rr=2.0/Omega1, and Z0=50/Omega1, c11ActiveRFAndMicrowaveDevices Pozar August 26, 2011 17:16 11.1 Diodes and Diode Circuits 533 Output 1 Output 2 InputOutput 1 Output 2 Input (b)(a) /H9261/4 /H9261/4 FIGURE 11.9 Circuits for single-pole, double-throw PIN diode switches. (a) Series. (b) Shunt. FIGURE 11.10 Photograph of a SP3T GaAs PIN diode switch, operating from 6 to 27 GHz. The diode chips are 15 mils square. Courtesy of LNX Corporation, Salem, N.H. c11ActiveRFAndMicrowaveDevices Pozar August 26, 2011 17:16 534 Chapter 11: Active RF and Microwave Devices Solution First use (11.15) to compute the diode impedance for the reverse and forward bias states: Zd=/braceleftbiggZr=Rr+j(ωLi−1/ωCj)=2.0−j171.2 /Omega1 Zf=Rf+jωLi =1.5+j5.6/Omega1. Then (11.14) gives the insertion losses for the ON and OFF states of the series and shunt switches as follows: For the series circuit, ILon=−20 log/vextendsingle/vextendsingle/vextendsingle/vextendsingle2Z 0 2Z0+Zf/vextendsingle/vextendsingle/vextendsingle/vextendsingle=0.14 dB , IL off=−20 log/vextendsingle/vextendsingle/vextendsingle/vextendsingle2Z0 2Z0+Zr/vextendsingle/vextendsingle/vextendsingle/vextendsingle=6.0d B . For the shunt circuit, ILon=−20 log/vextendsingle/vextendsingle/vextendsingle/vextendsingle2Z r 2Zr+Z0/vextendsingle/vextendsingle/vextendsingle/vextendsingle=0.11 dB , IL off=−20 log/vextendsingle/vextendsingle/vextendsingle/vextendsingle2Z f 2Zf+Z0/vextendsingle/vextendsingle/vextendsingle/vextendsingle=13.3d B . The shunt configuration has the greatest difference in attenuation between the ON and OFF states and has the lowest ON insertion loss. ■ PIN diode phase shifters: Several types of microwave phase shifters can be constructed with PIN diode switching elements. Compared with ferrite phase shifters, diode phase shifters have the advantages of small size, integrability with planar circuitry, and high speed. The power requirements for diode phase shifters, however, are generally greaterthan those for a latching ferrite phase shifter (Section 9.5) because diodes require continu- ous bias current, while a latching ferrite device requires only a pulsed current to change its magnetic state. There are basically three types of PIN diode phase shifters: switched line, loaded line, and reflection. The switched-line phase shifter, shown in Figure 11.11, is the most straightforward type, using two single-pole, double-throw switches to route the signal flow between one of two transmission lines of different length. The differential phase shift between the two paths is given by /Delta1φ=β(/lscript 2−/lscript1), (11.16) 2 In Out 1 FIGURE 11.11 A switched-line phase shifter. c11ActiveRFAndMicrowaveDevices Pozar August 26, 2011 17:16 11.1 Diodes and Diode Circuits 535 where βis the propagation constant of the line. If the transmission lines are TEM (or quasi- TEM, like microstrip), this phase shift is a linear function of frequency, which implies a true time delay between the input and output ports. This is a useful feature in wideband systems. This type of phase shifter is also inherently reciprocal, and so it can be used forboth receive and transmit functions. The insertion loss of the switched line phase shifter is equal to the loss of the SPDT switches plus line losses. Like many other types of phase shifters, the switched-line phase shifter is usually de- signed for discrete binary phase shifts of /Delta1φ=180 ◦,90◦,45◦, etc. One potential problem with this type of phase shifter is that resonances can occur in the OFF line if its length is near a multiple of λ/2. The resonant frequency will be slightly shifted due to the series junction capacitances of the reversed biased diodes, so the lengths /lscript1and/lscript2should be determined with this effect taken into account. A design that is useful for small amounts of phase shift (generally 45◦, or less) is the loaded-line phase shifter. The basic principle of this type of phase shifter can be illustrated with the circuit of Figure 11.12a, which shows a transmission line loaded with a shuntsusceptance, jB. The reflection and transmission coefficients can be written as /Gamma1=1−(1+jb) 1+(1+jb)=−jb 2+jb, (11.17a) T=1+/Gamma1=2 2+jb, (11.17b) where b=BZ0is the normalized susceptance. Thus the phase shift in the transmitted wave introduced by the load is /Delta1φ=tan−1b 2,( 11.18) which can be made positive or negative, depending on the sign of b. A disadvantage is the insertion loss that is inherently present due to the reflection from the shunt load. In addition,increasing bto obtain a larger /Delta1φentails a greater insertion loss, as seen from (11.17b). The reflections from the shunt susceptance can be reduced by using the circuit of Figure 11.12b, where two shunt loads are separated by a λ/4 length of line. Then the jBZ0 Z0jB Z0 Z0 Z0 Z0 Ze Z0jB(a) (b)/H9261/4 /H9258eΓ T FIGURE 11.12 Loaded-line phase shifters. (a) Basic circuit. (b) Practical loaded-line phase shifter and its equivalent circuit. c11ActiveRFAndMicrowaveDevices Pozar August 26, 2011 17:16 536 Chapter 11: Active RF and Microwave Devices partial reflection from the second load will be 180◦out of phase with the partial reflection from the first load, leading to cancellation. We can analyze this circuit by calculating its ABCD matrix and comparing it to the ABCD matrix of an equivalent line having a length θeand characteristic impedance Ze. Thus, for the loaded line, /bracketleftbigg AB CD/bracketrightbigg =/bracketleftbigg 10 jB 1/bracketrightbigg/bracketleftbigg 0 jZ0 j/Z0 0/bracketrightbigg/bracketleftbigg 10 jB 1/bracketrightbigg =/bracketleftbigg−BZ0 jZ0 j(1/Z0−B2Z0)−BZ0/bracketrightbigg , (11.19a) while the equivalent transmission line has an ABCD matrix given by /bracketleftbigg AB CD/bracketrightbigg =/bracketleftbigg cosθe jZesinθe jsinθe/Ze cosθe/bracketrightbigg .( 11.19b ) Then we have that cosθe=− BZ0=−b, (11.20a) Ze=Z0cosθe=Z0√ 1−b2. (11.20b) For small values of b,θewill be close to π/2, and these results will reduce to θe/similarequalπ 2+b, (11.21a) Ze/similarequalZ0/parenleftbigg 1+b 2/parenrightbigg . (11.21b) The susceptance, B, can be implemented with a lumped inductor or capacitor, or with a stub, and switched between two states with an SPST diode switch. The third type of PIN diode phase shifter is the reflection phase shifter, which uses an SPST switch to control the path length of a reflected signal. Usually a quadrature hybrid is used to provide a two-port circuit, although other types of hybrids, or even a circulator, could be used for this purpose. Figure 11.13 shows a reflection-type phase shifter using a quadrature hybrid. In oper- ation, an input signal divides equally between the two right-hand ports of the hybrid. The diodes are both biased in the same state (forward or reverse biased), so the waves reflectedfrom the two terminations will add in phase at the indicated output port. Turning the diodes on or off changes the total path length for both reflected waves by /Delta1φ, producing a phase shift of /Delta1φat the output. Ideally, the diodes would look like short circuits in their ON state, and open circuits in their OFF state, so that the reflection coefficients at the right side of the hybrid can be written as /Gamma1=e −j(φ+π)for the diodes in their ON state, and ∆/H9278 2/H9278 290° Hybrid In OutΓ Γ FIGURE 11.13 A reflection phase shifter using a quadrature hybrid. c11ActiveRFAndMicrowaveDevices Pozar August 26, 2011 17:16 11.1 Diodes and Diode Circuits 537 /Gamma1=e−j(φ+/Delta1φ)for the diodes in their OFF state. There is an infinite number of choices of line lengths that give the desired /Delta1φ(i.e., the value of φ/2 is a degree of freedom), but it can be shown that bandwidth is optimized if the reflection coefficients for the two states are phase conjugates. Thus, if /Delta1φ=90◦, the best bandwidth will be obtained for φ=45◦. A good input match for the reflection-type phase shifter requires that the diodes be well matched. The insertion loss is limited by the loss of the hybrid, as well as by the forward and reverse resistances of the diodes. Impedance transformation sections can beused to improve performance in this regard. VaractorDiodes We have seen that a PIN diode has a junction capacitance that can be switched on or off with bias voltage. This effect can be enhanced by tailoring the size and doping profile of the intrinsic layer of the diode to provide a desired junction capacitance versus junction volt- age ( Cvs.V) behavior when reverse biased. Such a device is called a varactor diode, and it produces a junction capacitance that varies smoothly with bias voltage, thus providing an electrically adjustable reactive circuit element. One of the most common applications of varactor diodes is to provide electronic frequency tuning of the local oscillator in a mul-tichannel receiver, such as those used in cellular telephones, wireless local area network radios, and television receivers. This is accomplished by using a varactor diode in the res- onant circuit of a transistor oscillator, and controlling the DC reverse bias voltage applied to the diode. The nonlinearity of varactor diodes also makes them useful for frequency multipliers (discussed in Chapter 13). Varactor diodes are generally made from silicon forRF applications, and gallium arsenide for microwave applications. A simplified equivalent circuit for a reverse-biased varactor diode is shown in Figure 11.14. The junction capacitance is dependent on the (negative) junction bias voltage, V, according to C j(V)=C0 (1−V/V0)γ,( 11.22) where C0is the junction capacitance with no bias; V0=0.5 V for silicon diodes, and V0=1.3 V for GaAs diodes. The exponent γdepends on the doping profile of the intrinsic layer of the diode. An ideal hyperabrupt varactor diode has γ=0.5; many practical diodes have an exponent of about γ=0.47, although the value can be as high as 1.5 or 2.0 for some diodes. In the equivalent circuit, Rsis the series junction and contact resistance, typically on the order of a few ohms. A typical GaAs varactor diode may have C0=0.5– 2.0 pF, and a junction capacitance that varies from about 0.1 to 2.0 pF as the bias voltage ranges from −20 to 0 V . Parasitic reactances due to the diode package should be included in a realistic design. + –V Cj(V)Rs FIGURE 11.14 Equivalent circuit of a reverse-biased varactor diode. c11ActiveRFAndMicrowaveDevices Pozar August 26, 2011 17:16 538 Chapter 11: Active RF and Microwave Devices I VNegative differentialresistance FIGURE 11.15 The DC I–Vcharacteristic of a Gunn diode, showing the region of negative dif- ferential resistance. Other negative resistance devices, such as impact avalanche and transit time (IMPATT) and tunnel diodes, have similar I–Vcharacteristics. OtherDiodes Here we briefly summarize the characteristics of several other diode devices that are com- monly used in microwave circuits. Historically, diode devices were developed long before high-frequency three-terminal devices (e.g., junction and field effect transistors), and for many years diodes provided the only means of microwave power generation and ampli- fication without using electron tubes. Today, many of these devices are most useful atmillimeter wave frequencies, since there are now many types of transistors that offer better performance and more design flexibility at RF and microwave frequencies. Further infor- mation on diode devices can be found in the literature. Gunn diodes: The operation of a Gunn diode is based on the transferred electron effect (also known as the Gunn effect), which was discovered by J. B. Gunn in 1963. Practical Gunn diodes typically use GaAs or InP materials in a specially doped bulk form, as op- posed to a traditional pnjunction. The Gunn diode has an I–Vcharacteristic that exhibits a negative differential resistance (negative slope) that can be used to generate RF power directly from a DC source when properly biased. Figure 11.15 shows a DC I–Vcurve that is characteristic of Gunn diodes, where the region of negative differential resistance(negative slope) corresponds to the operating point of the device. Gunn diodes can produce continuous power of up to several hundred milliwatts, at frequencies from 1 to 200 GHz, with efficiencies ranging from 5% to 15%. Oscillator circuits using Gunn diodes require a high- Qresonant circuit or cavity, which is often tuned mechanically. Electronic tuning by bias adjustment is limited to 1% or less, but varactor diodes are sometimes included inthe resonant circuit to provide a greater range of electronic tuning. Gunn diode sources are used extensively in low-cost applications such as traffic radars, motion detectors for door openers and security alarms, and test and measurement systems. FIGURE 11.16 A W-band Gunn diode oscillator. The output power is 16 dBm, and the source is mechanically tunable over a frequency range of 4 GHz. Courtesy of Millitech, Inc., Northampton, Mass. c11ActiveRFAndMicrowaveDevices Pozar September 14, 2011 22:30 11.1 Diodes and Diode Circuits 539 IMPATT diodes:A n impact avalanche and transit time (IMPATT) diode has a physical structure similar to a PIN diode, but is operated with a relatively high voltage (70–100 V) to produce a reverse-biased avalanche breakdown current. It exhibits a negative resistance over a broad frequency band that can extend into the submillimeter range, and it can beused to directly convert DC to RF power. IMPATT sources are generally noisier than Gunn diodes but are capable of higher powers and higher DC-to-RF conversion efficiencies. IMPATTs also have better temperature stability than Gunn diodes. Typical IMPATTs op-erate at frequencies from 10 to 300 GHz, with efficiencies ranging up to 15%. IMPATT diodes are among the few practical solid-state devices that can provide fundamental fre- quency power above 100 GHz. IMPATT devices can also be used for frequency multipli- cation and amplification. Silicon IMPATT diodes can provide CW power ranging from 10 W at 10 GHz to 1 W at 94 GHz, with efficiencies typically below 10%. GaAs IMPATTs can provide CW power ranging from 20 W at 10 GHz to 5 mW at 130 GHz. Pulsed operation generally results in higher powers and higher efficiencies. Because of the low efficiency of thesedevices, thermal considerations are a limiting factor for both CW and pulsed operation. IMPATT oscillators can be mechanically or electrically tuned. A disadvantage of IMPATT oscillators is that their AM noise level is generally higher than that of other sources. Tunnel diodes:T h e tunnel diode, invented by L. Esaki in 1957, is a pnjunction diode with a doping profile that allows electron tunneling through a narrow energy band gap, leadingto negative resistance at high frequencies. Tunnel diodes can be used for oscillators as well as amplifiers. Before high-frequency transistors were available, tunnel diodes provided the only means of high-frequency amplification with a solid-state device. Such an amplifieremploys the diode in a one-port reflection circuit, where the negative RF resistance of the device produces a reflection coefficient with a magnitude greater than unity, and therefore amplification of an incident signal. Such amplifiers have been made obsolete by modernRF and microwave transistors, but tunnel diodes are still used in some applications today. BARITT diodes :Abarrier injection transit time (BARITT ) diode has a structure similar to a junction transistor without a base contact. Like the IMPATT diode, it is a transit time device. It generally has a lower power capability than the IMPATT diode, but the advantage of lower AM noise. This makes it useful for local oscillator applications at frequencies upto 94 GHz. BARITT diodes are also useful for detector and mixer applications. PowerCombining In many applications RF power requirement exceeds the power capacity of a single solid- state source; this is especially common at millimeter wave frequencies. Because of themany advantages offered by solid-state sources compared to electron tubes, substantial effort has been directed toward increasing output power through the use of various power combining techniques. Thus, the outputs of two or more sources are combined, effectivelymultiplying the output power of a single source by the number of individual sources being used. It is important that the individual sources to be combined are coherent and in phase. In principle, an unlimited amount of RF power can be generated in this manner; in practice, however, factors such as high-order modes and combiner losses limit the multiplication factor to about 10–20 dB. Power combining can be done by combining at the device level or at the circuit level. In addition, in some applications power can be combined spatially by using an array of antennas, where each radiating element is fed with a separate source (the sources must havephase coherence, perhaps by using injection-locked oscillators). At the device level, several diode (or transistor) junctions are essentially connected in parallel over an electrically small c11ActiveRFAndMicrowaveDevices Pozar August 26, 2011 17:16 540 Chapter 11: Active RF and Microwave Devices region and used as a single device. This technique is limited to a relatively few device junctions. At the circuit level, the power output from Ndevices can be combined with anN-way combiner. The combining circuit may be an N-way Wilkinson-type network or a similar type of planar combining network. Resonant cavities can also be used for thispurpose; cavity combiners often have the advantage of providing self-locking of individual oscillator phases. These various power combining techniques all have their own advantages and disadvantages in terms of efficiency, bandwidth, isolation between sources, and circuitcomplexity. 11.2BIPOLARJUNCTIONTRANSISTORS Transistors are three-terminal semiconductor devices, and can be categorized as either junction transistors orfield effect transistors [3–6]. Junction transistors include bipolar junction transistors (BJTs) that use a single semiconductor material (usually silicon), and heterojunction bipolar transistors (HBTs) that use compound semiconductors. Both npn andpnpconfigurations are possible, but most RF junction transistors are usually of the npn type due to higher electron mobility at higher frequencies BipolarJunctionTransistor RF bipolar junction transistors (BJTs) are usually made using silicon (Si), and this tran- sistor is one of the oldest and most popular active RF devices in use today because of its low cost and good operating performance in terms of frequency range, power capacity, and noise characteristics. Silicon junction transistors are useful for amplifiers up to the range of 2–10 GHz, and in oscillators up to about 20 GHz. Bipolar transistors typically havevery low 1 /f-noise characteristics, making them well suited for oscillators with low-phase noise. Bipolar junction transistors are sometimes preferred over FETs at frequencies below about 2–4 GHz because of higher gain and lower cost, and the possibility of biasing with a single power supply. Bipolar transistors are subject to shot noise as well as thermal noise effects, so their noise figure is not as good as that of FETs. Figure 11.17 shows the con-struction of a typical silicon bipolar transistor having multiple fingers for the base and emitter electrodes. The BJT is current driven, with the base current modulating the collec- tor current. The upper frequency limit of the bipolar transistor is controlled primarily bythe base length, which is typically on the order of 0.1 µm. A small-signal equivalent circuit model for an RF bipolar transistor is shown in Figure 11.18 for a common emitter configuration. This model, known as the hybrid-π model, is popular because of its similarity to the equivalent circuit of a FET, and because of its utility BaseEmitter (b)˜150 mµE BE B CB p-base n-type collector˜0.1 mµ (a)p n p p n FIGURE 11.17 (a) Cross section of an interdigitated microwave bipolar junction transistor; (b) top view, showing base and emitter contacts. c11ActiveRFAndMicrowaveDevices Pozar August 26, 2011 17:16 11.2 Bipolar Junction Transistors 541 Base EmitterCollectorCc Rb V Rπ πC+ –gmVπ π FIGURE 11.18 Simplified hybrid- πequivalent circuit for a microwave bipolar junction transistor in the common emitter configuration. in circuit analysis. This model does not include parasitic resistances and inductances due to the base and emitter leads. More sophisticated equivalent circuits may be advantageous for computer-aided design and modeling over wide frequency ranges. The Gummel–Poonmodel [6], for example, is used extensively in computer modeling with the SPICE circuit analysis software package, and can include parasitic effects. In many cases the capacitor, C c, between the base and collector in the hybrid- πmodel, has a relatively small value and may be ignored. This has the effect of making S12=0, implying that power only flows in one direction through the device (from port 1 to port 2); such a device is called unilateral. This approximation is often used to simplify analysis. The hybrid-π model is roughly based on the physics of the junction transistor, and can be useful under circumstances where the element values of the model are fairly constant over a range of operating bias conditions, load conditions, and frequency. Otherwise, the el- ement values become frequency, bias, or load dependent, in which case the hybrid- πmodel (or any other equivalent circuit model) becomes much less useful. In this case, it is simplerto treat the transistor as a two-port network, characterized by two-port parameters. In prac- tice, scattering parameters, measured under typical operating conditions, are usually used for this purpose and are supplied by the device manufacturer. Table 11.4 shows scatteringparameters for a typical RF silicon junction transistor in a common emitter configuration. Note that there are relatively large mismatches at the base (port 1) and the collector (port 2), and that the gain (given roughly by |S 21|) drops quickly with an increase in frequency. Also note that |S12|is relatively small (particularly at low frequencies), making the device approximately unilateral. The equivalent circuit of Figure 11.18 can be used to estimate the upper frequency limit, fT, defined as the threshold frequency where the short-circuit current gain of the transistor is unity. If we assume an input current Iinat the base, and ignore the series base resistance, Rb(typically small), and the shunt resistance, Rπ(typically large), then the volt- age across the capacitor CπisVπ=Iin/jωCπ. The output short-circuit current at the col- lector is Iout=gmVπ, so the short-circuit current gain is Gsc I=/vextendsingle/vextendsingle/vextendsingle/vextendsingleIout Iin/vextendsingle/vextendsingle/vextendsingle/vextendsingle=gm ωCπ. TABLE 11.4 Scattering Parameters for an NPN Silicon BJT (NEC NE 58219, Vce=5.0V , Ic=5.0 mA, common emitter) Frequency (GHz) S11 S12 S21 S22 0.1 0.78/negationslash−33◦0.03/negationslash71◦12.7/negationslash155◦0.93/negationslash−17◦ 0.5 0.46/negationslash−113◦0.08/negationslash52◦6.3/negationslash104◦0.53/negationslash−38◦ 1.0 0.38/negationslash−158◦0.11/negationslash54◦3.5/negationslash80◦0.40/negationslash−43◦ 2.0 0.40/negationslash157◦0.19/negationslash56◦1.9/negationslash52◦0.33/negationslash−63◦ 4.0 0.52/negationslash117◦0.38/negationslash45◦1.1/negationslash14◦0.33/negationslash−127◦ c11ActiveRFAndMicrowaveDevices Pozar August 26, 2011 17:16 542 Chapter 11: Active RF and Microwave Devices 00255075Ice (mA) Vce (V)IB = 1.0 mA IB = 0.75 mA IB = 0.50 mA IB = 0.25 mA 5 10 15 20 (a) +VBB +Vcc RB (b) FIGURE 11.19 (a) DC characteristics of an npnBJT; (b) biasing and decoupling circuit for an npnBJT. The current gain is seen to decrease with frequency, and is unity at the threshold frequency, fT=gm 2πCπ.( 11.23) Figure 11.19a shows typical DC operating characteristics for a BJT. The biasing point for the transistor depends on the application and type of device, with low collector currents generally giving the best noise figure, and higher collector currents giving the best power gain. Figure 11.19b shows a typical bias and decoupling circuit for a bipolar transistor in acommon emitter configuration. HeterojunctionBipolarTransistor The operation of a heterojunction bipolar transistor (HBT) is essentially the same as that of a BJT, but an HBT has a base-emitter junction made from a compound semiconduc- tor material such as GaAs, indium phosphide (InP), or silicon germanium (SiGe), oftenin conjunction with thin layers of other materials (e.g., aluminum). This structure offers much improved performance at high frequencies. Some HBTs can operate at frequencies exceeding 100 GHz, and recent developments with HBTs using SiGe have demonstratedthat these devices are useful in low-cost circuits operating at frequencies of 60 GHz or higher. Since the HBT is similar in structure and operation to the BJT, the equivalent circuit model of Figure 11.18 can be used for both transistor types. As with BJTs, equivalent circuit models may have limited applicability when attempting to model HBTs over a rangeof operating conditions, so scattering parameter data, measured for a particular bias point, may be more useful. Table 11.5 gives the scattering parameters at several frequencies for a popular microwave HBT. Observe that |S 21|decreases much less rapidly with frequency when compared with the BJT of Table 11.4. The device also is seen to be approximately unilateral, as |S12|is relatively small. c11ActiveRFAndMicrowaveDevices Pozar September 14, 2011 22:30 11.3 Field Effect Transistors 543 TABLE 11.5 Scattering Parameters for a SiGe HBT (Infineon BFP640F, Vce=2.0V , Ic=1.2 mA, common emitter) Frequency (GHz) S11 S12 S21 S22 1.0 0.91/negationslash−44◦0.06/negationslash68◦3.92/negationslash149◦0.93/negationslash−17◦ 2.0 0.75/negationslash−86◦0.10/negationslash46◦3.39/negationslash120◦0.79/negationslash−31◦ 4.0 0.59 /negationslash−144◦0.11/negationslash29◦2.18/negationslash82◦0.64/negationslash−43◦ 6.0 0.54/negationslash176◦0.11/negationslash34◦1.64/negationslash57◦0.58/negationslash−53◦ High levels of monolithic integration are easy and inexpensive with SiGe HBTs, so this technology is proving to be very useful for low-cost millimeter wave circuits for both defense and commercial applications. 11.3FIELDEFFECTTRANSISTORS In contrast to BJTs, field effect transistors (FETs) are monopolar, with only one carrier type (holes or electrons) providing current flow through the device: n-channel FETs em- ploy electrons, while p-channel devices use holes. In addition, while a BJT is a current- controlled device, an FET is a voltage-controlled device, having a source-to-drain charac- teristic that is similar to that of a voltage-dependent variable resistor. Field effect transistors can take many forms, including the MESFET (metal semicon- ductor FET), the MOSFET (metal oxide semiconductor FET), the HEMT (high electron mobility transistor), and the PHEMT (pseudomorphic HEMT). FET transistor technology has been under continuous development for more than 50 years—the first junction FETs were developed in the 1950s, while the HEMT was proposed in the early 1980s. GaAs MESFETs are among the most commonly used transistors for microwave and millimeterwave applications, being usable at frequencies up to 60 GHz or more. Even higher op- erating frequencies can be obtained with GaAs HEMTs. GaAs MESFETs and HEMTs are especially useful for low-noise amplifiers since these transistors have lower noise fig-ures than any other active devices. Recently developed gallium nitride (GaN) HEMTs are very useful for high power RF and microwave amplifiers. CMOS FETs are increasingly being used for RF integrated circuits, offering high levels of integration at low cost andlow power requirements, for commercial wireless applications. Table 11.6 summarizes the performance characteristics of some of the most popular microwave transistors. TABLE 11.6 Performance Characteristics of Microwave Transistors Device BJT HBT CMOS MESFET HEMT HEMT Semiconductor Si SiGe Si GaAs GaAs GaN Frequency range (GHz) 10 30 20 60 100 10 Typical gain (dB) 10–15 10–15 10–20 5–20 10–20 10–15Noise figure (dB) 2.0 0.6 1.0 1.0 0.5 1.6(frequency, GHz) (2) (8) (4) (10) (12) (6) Power capacity High Medium Low Medium Medium High Cost Low Medium Low Medium High Medium Single-polarity supply Yes Yes Yes No No No c11ActiveRFAndMicrowaveDevices Pozar August 26, 2011 17:16 544 Chapter 11: Active RF and Microwave Devices SG D n Semi-insulatingn +n + FIGURE 11.20 Cross section of an n-channel GaAs MESFET. MetalSemiconductorFieldEffectTransistor One of the most important developments in microwave technology has been the GaAs metal semiconductor field effect transistor (MESFET), as this device permitted the first practical solid-state implementation of amplifiers, oscillators, and mixers at microwave frequencies, leading to key applications in radar, GPS, remote sensing, and wireless com-munications. GaAs MESFETs can be used at frequencies well into the millimeter wave range, with high gain and low noise figure, often making them the device of choice for hybrid and monolithic integrated circuits at frequencies above 10 GHz. Figure 11.20 shows the cross section of a typical n-channel GaAs MESFET. The gate junction is formed as a Schottky barrier. The desirable gain and noise features of this tran- sistor are a result of the higher electron mobility of GaAs compared to silicon, and theabsence of shot noise. The device is biased with a drain-to-source voltage, V ds, and a gate- to-source voltage, Vgs. In operation, electrons are drawn from the source to the drain by the positive Vdssupply voltage. An applied signal voltage on the gate then modulates these majority electron carriers, producing voltage amplification. The maximum frequency of operation is limited by the gate length; present FETs have gate lengths on the order of 0.2–0.6 µm, with corresponding upper frequency limits of 100 to 50 GHz. A small-signal equivalent circuit for a microwave MESFET is shown in Figure 11.21 for a common-source configuration. The components and some typical values for this model are listed below: Ri(series gate resistance) =7/Omega1 Rds(drain-to-source resistance) =400/Omega1 Cgs(gate-to-source capacitance) =0.3p F Cds(drain-to-source capacitance) =0.12 pF Cgd(gate-to-drain capacitance) =0.01 pF gm(transconductance) =40 mS This model does not include package parasitics, which typically introduce small series resistances and inductances at the three terminals due to ohmic contacts and bonding leads.The dependent current generator g mVcdepends on the voltage across the gate-to-source capacitor Cgs, leading to a value of |S21|>1 under normal operating conditions (where Gate SourceDrain Cgd Ri –+Rds Cds CgsgmVcVc FIGURE 11.21 Small-signal equivalent circuit for a microwave FET in the common-source configuration. c11ActiveRFAndMicrowaveDevices Pozar August 26, 2011 17:16 11.3 Field Effect Transistors 545 TABLE 11.7 Scattering Parameters for an n-Channel GaAs MESFET (NEC NE76184A, VDS=3.0V , ID=10.0 mA, common source) Frequency (GHz) S11 S12 S21 S22 1.0 0.97/negationslash−28◦0.04/negationslash72◦3.82/negationslash154◦0.70/negationslash−19◦ 2.0 0.90/negationslash−55◦0.08/negationslash54◦3.56/negationslash129◦0.65/negationslash−37◦ 4.0 0.72/negationslash−103◦0.12/negationslash28◦2.91/negationslash86◦0.53/negationslash−68◦ 8.0 0.52/negationslash179◦0.14/negationslash−1◦2.0/negationslash20◦0.42/negationslash−129◦ 12.0 0.49/negationslash103◦0.17/negationslash−19◦1.5/negationslash−38◦0.44/negationslash170◦ port 1 is at the gate, and port 2 is at the drain). The reverse signal path, given by S12, is due solely to the capacitance Cgd. As seen from the above data, this is typically a very small capacitor, which can often be ignored in practice. In this case, S12=0, and the device is unilateral. The scattering parameters for a typical GaAs MESFET are given in Table 11.7. As we did for the BJT, we can use the equivalent circuit model of Figure 11.21 to determine the upper frequency of operation for a MESFET. For a FET, the short-circuitcurrent gain, G sc I, is defined as the ratio of drain current to gate current when the output is short circuited. For the unilateral case, where Cgd=0, the short circuit current gain is Gsc I=/vextendsingle/vextendsingle/vextendsingle/vextendsingleI d Ig/vextendsingle/vextendsingle/vextendsingle/vextendsingle=/vextendsingle/vextendsingle/vextendsingle/vextendsingleg mVc Ig/vextendsingle/vextendsingle/vextendsingle/vextendsingle=g m ωCgs. The upper frequency threshold, fT, where the short-circuit current gain is unity, is then given by fT=gm 2πCgs,( 11.24) a result that is equivalent to (11.23) for a bipolar junction transistor. For proper operation, the transistor must be biased at an appropriate operating point. This depends on the application (low noise, high gain, high power), the class of the ampli- fier (class A, class AB, class B), and the transistor. Figure 11.22a shows a typical family of DC Idsversus Vdscurves for a GaAs MESFET. For low-noise design, the drain cur- rent is generally chosen to be about 15% of Idss(the saturated drain-to-source current). High-power circuits generally use higher values of drain current. DC bias voltage must beapplied to both the gate and drain, without disturbing the RF signal paths. This can be done 0255075Ids (mA) Vds (V)Vgs = 0V Vgs = –1V Vgs = –2V Vgs = –3V 51 0 (a) (b)+VDD –VG FIGURE 11.22 (a) DC characteristics of an n-channel GaAs MESFET; (b) biasing and decou- pling circuitry. c11ActiveRFAndMicrowaveDevices Pozar August 26, 2011 17:16 546 Chapter 11: Active RF and Microwave Devices SGDinsulator p-type substraten +n + FIGURE 11.23 Cross section of an n-channel MOSFET. with biasing and decoupling circuitry for a dual-polarity supply, as shown in Figure 11.22b. The RF chokes provide a very low DC resistance for biasing, and a very high impedance at RF frequencies to isolate the signal from the bias supply. Similarly, the input and outputdecoupling capacitors block DC from the input and output lines while allowing passage of RF signals. More sophisticated bias circuits can provide compensation for temperature and device variations, and may work with single-polarity power supplies. MetalOxideSemiconductorFieldEffectTransistor The silicon metal oxide semiconductor field effect transistor (MOSFET) is the most com- mon type of FET, being used extensively in analog and digital integrated circuits. Figure 11.23 shows a cross section of an n-channel MOSFET. It consists of a lightly doped p substrate, and differs from a MESFET by having a thin insulating layer (SiO 2) between the gate contact and the channel region. Because the gate is insulated, it does not conduct DC bias current. MOSFETs can be used at frequencies into the UHF range, and can provide powers of several hundred watts when devices are packaged in parallel. Laterally diffused MOSFETs (LDMOS) have direct grounding of the source, and can operate at low microwave frequen-cies with high powers. These devices are commonly used for high-power transmitters for cellular base stations at 900 and 1900 MHz. High-density integrated circuits typically use complementary MOS (CMOS), where both n-channel and p-channel devices are used. This technology is very mature, and has the advantages of low power requirements and low unit cost. Most RF and microwave MOSFETs use n-channel silicon devices, although GaN devices are possible. The small-signal equivalent circuit for a MOSFET is the same as that of the MESFET, given in Figure 11.21. Scattering parameters are available for most nMOS devices intended for high-frequency applications. HighElectronMobilityTransistor The high electron mobility transistor (HEMT) is a heterojunction FET, meaning that it does not use a single semiconductor material, but instead is constructed with several lay- ers of compound semiconductor materials. These may include transitions between gal- lium aluminum arsenide (GaAlAs), GaAs, gallium indium arsenide (GaInAs), and similar compounds. These structures result in high carrier mobility—about twice that found in a standard MESFET. GaAs HEMTs can operate at frequencies above 100 GHz. Figure 11.24 shows the cross section of a HEMT device. It consists of semi-insulating GaAs substrate, followed by an undoped GaAs layer, and then a very thin undoped GaAlAs layer. This is topped with an n-doped GaAlAs layer. To reduce thermal and mechanical stress the layers usually have matched crystal lattices. Several variations on this device are possible, including the use of different compound semiconductors, and the pseudomorphic c11ActiveRFAndMicrowaveDevices Pozar August 26, 2011 17:16 11.4 Microwave Integrated Circuits 547 SG D n-GaAlAs GaAlAs GaAs substrateGaAsn +n + FIGURE 11.24 Cross section of an n-channel HEMT. TABLE 11.8 Scattering Parameters for a GaN HEMT (Cree CGH21120, VDD=328 V , ID=500 mA, common source) Frequency (GHz) S11 S12 S21 S22 0.5 0.96/negationslash180◦0.007 /negationslash−16◦3.67/negationslash68◦0.72/negationslash−174◦ 1.0 0.95/negationslash172◦0.008 /negationslash−35◦2.03/negationslash44◦0.78/negationslash−172◦ 2.0 0.78/negationslash153◦0.014 /negationslash−83◦2.09/negationslash−17◦0.91/negationslash−174◦ 4.0 0.88/negationslash−51◦0.008 /negationslash79◦0.84/negationslash88◦0.88/negationslash171◦ HEMT, which uses a lattice mismatch between the layers. The relatively complicated struc- ture of the HEMT requires sophisticated fabrication techniques, leading to a relatively highcost. The HEMT is also referred to in the literature as a MODFET (modulation-doped FET), a TEGFET (two-dimensional electron gas FET), and an SDFET (selectively doped FET). A relatively new type of HEMT uses GaN and aluminum gallium nitride (AlGaN) on a silicon or SiC substrate. GaN HEMTs operate with drain voltages in the range of 20–40 V , and can deliver powers up to 100 W at frequencies in the low microwave range, makingthese devices popular for high-power transmitters. The equivalent circuit model of Figure 11.21 can also be used for HEMTs, and the DC bias characteristics of a HEMT are similar to those of the MESFET. Table 11.8 gives the scattering parameters for a medium power GaN HEMT. 11.4MICROWAVEINTEGRATEDCIRCUITS The trend of any maturing electronic technology is toward smaller size, lighter weight, lower power requirements, lower cost, and increased complexity. Microwave technology has been moving in this direction for the last 10–30 years with the development of mi- crowave integrated circuits (MICs) [2]. This technology strives to replace bulky and expen-sive waveguide and coaxial components with small and inexpensive planar components, and is analogous to the digital integrated circuitry that has led to the rapid increase in sophistication of computer systems. Microwave integrated circuitry can incorporate trans- mission lines, discrete resistors, capacitors, and inductors, as well as active devices such as diodes and transistors. MIC technology has advanced to the point where complete mi-crowave subsystems, such as receiver front ends and radar transmit/receive modules, can be integrated on a chip that is only a few square millimeters in size. There are two distinct types of microwave integrated circuits. Hybrid MICs have one layer of metallization for conductors and transmission lines, with discrete components (re- sistors, capacitors, integrated circuit chips, transistors, diodes, etc.) bonded to the substrate. c11ActiveRFAndMicrowaveDevices Pozar August 26, 2011 17:16 548 Chapter 11: Active RF and Microwave Devices In a thin-film hybrid MIC, some of the simpler components are deposited on the sub- strate. Hybrid MICs were first developed in the 1960s, and still provide a very flexible and cost-effective means for circuit implementation. Monolithic microwave integrated circuits (MMICs) are a more recent development, where the active and passive circuit elementsare grown on the substrate. The substrate is a semiconductor material, and several layers of metal, dielectric, and resistive films are used. Below we will briefly describe these two types of MICs in terms of the materials and fabrication processes that are required and therelative merits of each type of circuitry. HybridMicrowaveIntegratedCircuits Material selection is an important consideration for a hybrid integrated circuit; character- istics such as electrical conductivity, dielectric constant, loss tangent, thermal transfer, me- chanical strength, and manufacturing compatibility must be evaluated. Generally the sub- strate material is of primary importance. For hybrid circuits, alumina, quartz, and Teflonfiber are commonly used for substrates. Alumina is a rigid, ceramic-like material with a dielectric constant of about 9–10. A high dielectric constant is often desirable for lower frequency circuits because it results in a smaller circuit size. At higher frequencies, how-ever, the substrate thickness must be decreased to prevent radiation loss and other spurious effects; then the transmission lines (typically microstrip, slotline, or coplanar waveguide) can become too narrow to be practical. Quartz has a lower dielectric constant ( ∼4), which, with its rigidity, makes it useful for higher frequency ( >20 GHz) circuits. Teflon and sim- ilar types of soft plastic substrates have dielectric constants ranging from 2 to 10, and can provide a large substrate area at a low cost as long as rigidity and good thermal transferare not required. Transmission line conductors for hybrid integrated circuits are typically copper or gold. Computer-aided design (CAD) tools are used extensively for MIC design, optimiza- tion, layout, and mask generation. Commonly used software packages include CADENCE (Cadence Design Systems), ADS (Agilent Technologies), Microwave Office (Applied WaveResearch), and DESIGNER (Ansoft). The mask itself may be made on Rubylith (a soft Mylar film), usually at a magnified scale (2 ×,5×,1 0×, etc.) for high accuracy. Then an actual-size mask is made on a thin sheet of glass or quartz. The metallized substrate iscoated with photoresist, covered with the mask, and exposed to a light source. The substrate can then be etched to remove the unwanted areas of metal. Plated-through, or via, holes can be made by evaporating a layer of metal inside a hole that has been drilled in the substrate.Finally, the discrete components are soldered or wire bonded to the conductors. This may be done manually, but today the process is usually automated using computer-controlled pick-and-place machines. The fabricated MIC can then be tested. Often provision is madefor variations in component values and other circuit tolerances by providing tuning or trim stubs that can be manually trimmed for each circuit. This increases circuit yield but also increases cost since trimming involves labor at a highly skilled level. A photograph of ahybrid MIC module is shown in Figure 11.25. MonolithicMicrowaveIntegratedCircuits Progress in GaAs and related semiconductor material processing and device development since the late 1970s has led to the feasibility of the MMIC, where all passive and active components required for a given circuit can be grown or implanted in the substrate. Po- tentially, an MMIC can be made at low cost because the labor involved with fabricatinghybrid MICs is reduced. In addition, a single wafer can contain a large number of circuits, all of which can be processed and fabricated simultaneously. c11ActiveRFAndMicrowaveDevices Pozar August 26, 2011 17:16 11.4 Microwave Integrated Circuits 549 FIGURE 11.25 Photograph of one of the 25,344 hybrid integrated T/R modules used in Raytheon’s Ground Based Radar system. This X-band module contains phase shifters, amplifiers, switches, couplers, a ferrite circulator, and associated controland bias circuitry. Courtesy of Raytheon Company, Waltham, Mass. The substrate of an MMIC must be a semiconductor material to accommodate the fabrication of active devices; the type of devices and the frequency range dictate the typeof substrate material. The GaAs MESFET is a very versatile device, finding applications in low-noise amplifiers, high-gain amplifiers, broadband amplifiers, mixers, oscillators, phase shifters, and switches. Thus, GaAs is one of the most common substrates for MMICs, but silicon, silicon-on-sapphire (SoS), silicon carbide (SiC), and InP are also used. Transmission lines and other conductors are usually made with gold metallization. To improve adhesion of the gold to the substrate, a thin layer of chromium or titanium may be deposited first. These metals are relatively lossy, so the gold layer must be made at least several skin depths thick to reduce attenuation. Capacitors and overlaying linesrequire insulating dielectric films, such as SiO, SiO 2,S i2N4, and Ta 2O5. These materials have high dielectric constants and low loss, and are compatible with integrated circuit processing. Resistors require the deposition of lossy films; NiCr, Ta, Ti, and doped GaAsare commonly used. Designing an MMIC requires extensive use of CAD software for circuit design and optimization, as well as for mask generation. Careful consideration must be given to thecircuit design to allow for component variations and tolerances, and the fact that circuit trimming after fabrication will be difficult or impossible (and defeats the goal of low-cost production). Thus, effects such as transmission line discontinuities, bias networks, spuriouscoupling, and package resonances must be taken into account. After the circuit design has been finalized, the masks can be generated. One or more masks are generally required for each processing step. Processing begins by forming an active layer in the semiconductor substrate for the necessary active devices; this can be done by ion implantation or by epitaxial techniques. Then, active areas are isolated byetching or additional implantation, leaving mesas for the active devices. Next, ohmic con- tacts are made to the active device areas by alloying a gold or gold/germanium layer onto the substrate. FET gates are then formed with a titanium/platinum/gold compound de-posited between the source and drain areas. At this time, the active device processing has been essentially completed, and intermediate tests may be made to evaluate the wafer. c11ActiveRFAndMicrowaveDevices Pozar September 14, 2011 22:30 550 Chapter 11: Active RF and Microwave Devices Air bridgeMicrostrip input lineMIM capacitorInductor Via holeGround plane metalizationThin film resistor GaAs FETImplanted resistorGaAs substrateG S D FIGURE 11.26 Layout of a monolithic microwave integrated circuit. If it meets specifications, the next step is to deposit the first layer of metallization for con- tacts, transmission lines, inductors, and other conducting areas. Then, resistors are formed by depositing resistive films, and the dielectric films required for capacitors and overlaysare deposited. A second layer of metallization completes the formation of capacitors and any remaining interconnections. The final processing steps involve the bottom, or back, of the substrate. First it is lapped to the required thickness, and then via holes are formed by etching and plating. Via holes provide ground connections to the circuitry on the top side of the substrate, and a heat dissipation path from the active devices to the ground plane.After processing has been completed, the individual circuits can be cut from the wafer and tested. Figure 11.26 shows the structure of a typical MMIC, and Figure 11.27 shows a photograph of an X-band monolithic integrated GaAs FET amplifier. Monolithic microwave integrated circuits are not without some disadvantages when compared with hybrid MICs or other types of circuitry. First, MMICs tend to waste large areas of relatively expensive semiconductor substrate for components such as transmis-sion lines and hybrids. In addition, the processing steps and required tolerances for an MMIC are very critical, often resulting in low yields. These factors tend to make MMICs expensive, especially when made in small quantities (less than several hundred). MMICsgenerally require a more thorough design procedure to include effects such as component tolerances and discontinuities, and debugging, tuning, or trimming after fabrication is dif- ficult. Because their small size limits heat dissipation, MMICs cannot be used for circuitsrequiring more than moderate power levels. In addition, high- Qresonators and filters are difficult to implement in MMIC form because of the inherent resistive losses in MMIC materials. Besides the obvious features of small size and weight, MMICs have some unique ad- vantages over other types of circuits. Since it is very easy to fabricate additional FETsin an MMIC design, circuit flexibility and performance can often be enhanced with little additional cost. In addition, monolithically integrated devices have much less parasitic re- actance than discrete packaged devices, so MMIC circuits can often be made with broaderbandwidth than hybrid circuits. MMICs generally give very reproducible results, especially for circuits from the same wafer. c11ActiveRFAndMicrowaveDevices Pozar August 26, 2011 17:16 11.4 Microwave Integrated Circuits 551 FIGURE 11.27 Photograph of a typical GaAs X-band MMIC incorporating a pair of two-stage amplifiers. Courtesy of M. Adlerstein, Raytheon Company, Waltham, Mass. POINT OF INTEREST: RF MEMS Switch Technology An exciting new field is the use of micromachining techniques to form suspended or mov- able structures in a silicon substrate that can be used for microwave resonators, antennas, and switches. A micromachined RF switch having a mechanically movable contact is an example of a micro-electro-mechanical system (MEMS), where the unique properties of silicon can be usedto construct extremely small devices that employ miniaturized mechanical components such as levers, gears, motors, and actuators. RF MEMS switches are among the most promising applications of this new technology. A MEMS switch can be made in several different configurations, depending on the signal path (capacitive or direct contact), the actuation mechanism (electrostatic, magnetic, or thermal), the pull-back mechanism (spring or active), and the type of structure (cantilever, bridge, lever arm,or rotary). One popular configuration for microwave switches is shown below, where the capaci- tance of the signal path is switched between a low-capacitance state and a high-capacitance state by moving a flexible conductive membrane through the application of a DC control voltage. Low-capacitance (open circuit)High-capacitance (closed circuit)Dielectric layer MEMS switches have very good loss characteristics, very low power consumption, and wide bandwidth, and (unlike diode or transistor switches) they exhibit virtually no intermodula- tion distortion or other nonlinear effects. The table below compares some of the key parameters of MEMS switches with those of popular solid-state switch technology over the 10–20 GHzband. c11ActiveRFAndMicrowaveDevices Pozar August 26, 2011 17:16 552 Chapter 11: Active RF and Microwave Devices Switch Insertion Switching Switching Technology Loss (dB) Isolation (dB) Power DC V oltage (V) Speed PIN diode 0.1–0.8 25–45 1–5 mW 1–10 1–5 nsFET 0.5–1.0 20–50 1–5 mW 1–10 2–10 ns MEMS 0.1–1.0 25–60 1 µW 10–20 >30µs Probably the most significant drawbacks of RF MEMS switches are the relatively slow switching time and potential lifetime limitations; both of these are a result of the mechanicalnature of the device. One of the most important applications foreseen for MEMS switches is for low-cost switched-line-length phase shifters, which are required in large numbers for phased array antennas. 11.5MICROWAVETUBES Although solid-state sources of RF and microwave power are preferred due to size, weight, power, and cost considerations, electron tubes provided the first practical sources of high- frequency power, and for several decades they were the only sources available at these frequencies. Today, solid-state diode or transistor sources are used in the majority of RF ormicrowave applications, and progress in solid-state technology is steadily improving the power versus frequency performance of solid-state sources. There are, however, still some systems that are best served by electron tubes. These are generally microwave or millimeterwave applications that require very high powers and/or very high frequencies. Radar systems generally require a relatively high-power source, sometimes as high as 1–10 kW, for the transmitter (in addition to one or more low-power sources for local oscil- lator and down conversion functions in the receiver). Radar transmitters are often operated in a pulsed mode, and peak powers that are much greater than the continuous power ratingof a given source can then be attained. Electronic warfare systems use sources with powers in the range of 100 W to 1 kW, with the additional requirement for tunability over a wide bandwidth. In addition, the microwave oven, that most common of all microwave systems,requires a single-frequency, high-power source in the range of 700 W. Usually the most practical way to meet the power requirements of these systems is with electron tubes. As frequency increases into the millimeter and submillimeter ranges it becomes in- creasingly difficult to produce even moderate power with solid-state devices, so tubes become more useful at these frequencies. Generally, the division is between solid-state sources for low to moderate powers at low to moderate frequencies, and tubes for highpowers and/or high frequencies. Figure 11.28 illustrates the power versus frequency per- formance for solid state and tube sources. Solid-state sources have the advantages of small size, ruggedness, low cost, and compatibility with microwave integrated circuits, and sothey are usually preferred whenever they can meet the necessary power and frequency requirements. However, very high power applications are dominated by microwave tubes, and even though the power and frequency performance of solid-state sources is steadily im- proving, it appears that the need for microwave tubes will not be eliminated any time soon. The first practical microwave source was the magnetron tube, developed in England in the 1930s, which later provided the impetus for the development of microwave radar during World War II. Since then, a large variety of microwave tubes have been designed for the generation and amplification of microwave power. Although solid-state devices have beenprogressively filling roles that once could only be met by microwave tubes, tubes are still essential for the generation of very high powers (10 kW and higher), and at the higher c11ActiveRFAndMicrowaveDevices Pozar August 26, 2011 17:16 11.5 Microwave Tubes 553 0.1 1.0 10 100 1000110102 10–1103104105106 Frequency (GHz)Continuous power (W) Solid state sources predominateMicrowave tubes predominate FIGURE 11.28 Power versus frequency performance of solid-state sources and microwave tubes. millimeter wave frequencies (100 GHz and higher). Here we will provide a brief overview of some of the most common microwave tubes and their basic characteristics. Several of these tubes are not actually sources by themselves but are high-power amplifiers. Suchtubes can used in conjunction with lower power sources (such as solid-state sources), and the combination is referred to as a microwave power module (MPM). There is a wide variety of microwave tube geometries, as well as a wide variety of principles on which tube operation is based, but all tubes have several common features. First, all tubes involve the interaction of an electron beam with an electromagnetic field, inside a glass or metal vacuum envelope. Thus, a way must be provided for RF energy to be coupled outside the envelope; this is usually accomplished with transparent windows or coaxial coupling probes or loops. Next, a hot cathode is used to generate a stream of elec-trons by thermionic emission. Cathodes are usually fabricated from a barium oxide–coated metal surface, or an impregnated tungsten surface. The electron stream is then focused into a narrow beam by a focusing anode with a high voltage bias. Alternatively, a solenoidalelectromagnet can be used to focus the electron beam. For pulsed operation, a beam modu- lating electrode is used between the cathode and anode. A positive bias voltage will attract electrons from the cathode and turn the beam on, while a negative bias will turn the beamoff. After the electron beam leaves the region of the tube where the desired interaction with the RF field takes place, a collector element is used to provide a complete current path back to the cathode power supply. The assembly of the cathode, focusing anode, and modulat-ing electrode is called the electron gun. Because of the requirement for a high vacuum and the need to dissipate large amounts of heat, microwave tubes are generally very large and bulky. In addition, tubes often require large, heavy biasing magnets and high voltage powersupplies. Factors to consider when choosing a particular type of tube include power output, frequency, bandwidth, tuning range, and noise. Microwave tubes can be grouped into two categories, depending on the type of electron beam–field interaction. In linear-beam, or “O,” type tubes the electron beam traverses the length of the tube and is parallel to the electric field. In the crossed-field, or “m,” type tube the focusing field is perpendicular to the accelerating electric field. Microwave tubes can also be classified as either oscillators or amplifiers. Theklystron is a linear-beam tube that can be used as either an amplifier or an oscilla- tor. In a klystron amplifier, the electron beam passes through two or more resonant cavities. The first cavity accepts an RF input and modulates the electron beam by bunching it into c11ActiveRFAndMicrowaveDevices Pozar August 26, 2011 17:16 554 Chapter 11: Active RF and Microwave Devices high-density and low-density regions. The bunched beam then travels to the next cavity, which accentuates the bunching effect. At the final cavity the RF power is extracted, at a highly amplified level. Two cavities can produce up to about 20 dB of gain, while using four cavities (about the practical limit) can give 80–90 dB of gain. Klystrons are capable ofpeak powers in the megawatt range, with RF output/DC input power conversion efficien- cies of 30%–50%. The reflex klystron is a single-cavity klystron tube that operates as an oscillator by using a reflector electrode after the cavity to provide positive feedback via the electron beam. It can be tuned by mechanically adjusting the cavity size. The major disadvantage of klystrons is their narrow bandwidth, which is a result of the high- Qcavities required for electron bunching. Klystrons have very low AM and FM noise levels. The narrow bandwidth of the klystron amplifier is overcome in the traveling wave tube (TWT). The TWT is a linear-beam amplifier that uses an electron gun and a focusing mag- net to accelerate a beam of electrons through an interaction region. Usually the interaction region consists of a slow-wave helix structure, with an RF input at the electron gun endand an RF output at the collector end. The helical structure slows down the propagating RF wave so that it travels at the same velocity as the wave and beam travel along the inter- action region, and amplification is achieved. Then the amplified signal is coupled from theend of the helix. The TWT has the highest bandwidth of any amplifier tube, ranging from 30% to 120%; this makes it very useful for electronic warfare systems, which require high power over broad bandwidths. It has a power rating of several hundred watts (typically),but this can be increased to several kilowatts by using an interaction region consisting of a set of coupled cavities; the bandwidth will be reduced, however. The efficiency of the TWT is relatively small, typically ranging from 20% to 40%. A variation of the TWT is the backward wave oscillator (BWO). The difference be- tween a TWT and the BWO is that in a BWO, the RF wave travels along the helix from the collector toward the electron gun. Thus the signal for amplification is provided by the bunched electron beam itself, and oscillation occurs. A very useful feature of the BWO is that its output frequency can be tuned by varying the DC voltage between the cathodeand the helix; tuning ranges of an octave or more can be achieved. The power output of the BWO, however, is relatively low (typically less than 1 W), so these tubes are generally being replaced with solid-state sources. Another type of linear-beam oscillator tube is the extended interaction oscillator (EIO). The EIO is very similar to a klystron, and uses an interaction region consisting of several cavities coupled together, with positive feedback to support oscillation. It has a narrow tun-ing bandwidth and a moderate efficiency, but it can supply high powers at frequencies up to several hundred GHz. Only the gyratron can deliver more power. Crossed-field tubes include the magnetron,t h e crossed-field amplifier , and the gyra- tron. As previously mentioned, the magnetron was the first high-power microwave source. It consists of a cylindrical cathode surrounded by a cylindrical anode with several cavity resonators along the inside of its periphery. A magnetic bias field is applied parallel to thecathode–anode axis. In operation, a cloud of electrons is formed that rotates around the cathode in the interaction region. As with linear-beam devices, electron bunching occurs, and energy is transferred from the electron beam to the RF wave. RF power can be coupled out of the tube with a probe, loop, or aperture window. Magnetrons are capable of very high power outputs, on the order of several kilowatts, and with efficiencies of 80% or more. A significant disadvantage, however, is that they are very noisy and cannot maintain frequency or phase coherence when operated in a pulsed mode. These factors are important for high-performance pulsed radars, where processingtechniques operate on a sequence of returned pulses. (Modern radars of this type today generally use a stable low-noise solid-state source, followed by a TWT for power amplifi- cation.) The main application of magnetrons today is primarily for microwave cooking. c11ActiveRFAndMicrowaveDevices Pozar August 26, 2011 17:16 11.5 Microwave Tubes 555 0.3 1 3 10 30 100 3001 mW1 W1 kW1 MwCW power Frequency (GHz)Magnetron (fixed tuned or coaxial) Solid state limit Helix BWO Reflex klystronEIK VTMCarcinotronGyroton 2-cavity klystron FIGURE 11.29 Power versus frequency performance of microwave oscillator tubes. Thecrossed-field amplifier (CFA) has a geometry similar to that of a TWT, but em- ploys a crossed-field interaction region that is similar to that of the magnetron. The RF input is applied to a slow-wave structure in the interaction region of the CFA, but the elec- tron beam is deflected by a negatively biased electrode to force the beam perpendicular tothe slow-wave structure. In addition, a magnetic bias field is applied perpendicular to this electric field, and perpendicular to the electron beam direction. The magnetic field exerts a force on the electron beam that counteracts the field from the sole. In the absence of an RF input, the electric and magnetic fields are adjusted so that their effects on the elec- tron beam cancel, leaving the beam to travel parallel to the slow-wave structure. Applyingan RF field causes velocity modulation of the beam, and bunching occurs. The beam is also periodically deflected toward the slow-wave circuit, producing an amplified signal. Crossed-field amplifiers have very good efficiencies—up to 80%, but the gain is limitedto 10–15 dB. In addition, the CFA has a noisier output than either a klystron amplifier or TWT. Its bandwidth can be up to 40%. Another crossed-field tube is the gyratron, which can be used as an amplifier or an oscillator. This tube consists of an electron gun with input and output cavities along the axis of the electron beam, similar to a klystron amplifier. However, the gyratron also has a solenoidal bias magnet that provides an axial magnetic field. This field forces the electronsto travel in tight spirals down the length of the tube. The electron velocity is high enough that relativistic effects are important. Bunching occurs, and energy from the transverse component of the electron velocity is coupled to the RF field. A significant feature of the gyratron is that the frequency of operation is determined by the bias field strength and the electron velocity, as opposed to the dimensions of the tube itself. This makes the gyratron especially useful for millimeter wave frequencies; it offers the highest output power (10–100 kW) of any tube in this frequency range. It also has a high efficiency for tubes in the millimeter wave range. The gyratron is a relatively newtype of tube, but it is rapidly replacing tubes such as reflex klystrons and EIOs as sources of millimeter wave power. Figures 11.29 and 11.30 summarize the power versus frequency performance of mi- crowave tube oscillators and amplifiers. c11ActiveRFAndMicrowaveDevices Pozar August 26, 2011 17:16 556 Chapter 11: Active RF and Microwave Devices 0.3 1 3 10 30 100 300100 W1 kW10 kW100 kW1 MW10 MWAverage power Frequency (GHz)CFAKlystron GyrotronCoupled cavity TWT Helix TWT Gridded FIGURE 11.30 Power versus frequency performance of microwave amplifier tubes. REFERENCES [1] M. E. Hines, “The Virtues of Nonlinearity—Detection, Frequency Conversion, Parametric Ampli- fication and Harmonic Generation,” IEEE Transactions on Microwave Theory and Techniques, vol. MTT-32, pp. 1097–1104, September 1984. [2] D. N. McQuiddy, Jr., J. W. Wassel, J. B. Lagrange, and W. R. Wisseman, “Monolithic Microwave Integrated Circuits: An Historical Perspective,” IEEE Transactions on Microwave Theory and Tech- niques, vol. MTT-32, pp. 997–1008, September 1984. [3] S. Yngvesson, Microwave Semiconductor Devices , Kluwer, Norwell, Mass., 1991. [4] R. Ludwig and P. Bretchko, RF Circuit Design: Theory and Applications, Prentice-Hall, Upper Sad- dle River, N.J., 2000. [5] S. A. Maas, Nonlinear Microwave and RF Circuits, 2nd edition, Artech House, Norwood, Mass., 2003. [6] M. Steer, Microwave and RF Design: A Systems Approach , SciTech, Raleigh, N.C., 2010. PROBLEMS 11.1 The Skyworks SMS1546 Schottky diode has the following parameters: Cj=0.38 pF, Rs=4/Omega1, Is=0.3µA, and Lp=Cp/similarequal0. Compute the open-circuit voltage sensitivity at 10 GHz for I0= 0, 20, and 50 µA. Assume α=1/(25 mV), and neglect the effect of bias current on the junction capacitance. 11.2 A single-pole, single-throw switch uses an Infineon BA592 PIN diode in a shunt configuration. The operating frequency is 4 GHz, Z0=50/Omega1, and the diode parameters are Cj=1.4p F , Rr= 0.5/Omega1,Rf=0.36/Omega1,a n d Li=0.5 nH. Find the electrical length of an open-circuited shunt stub placed across the diode to minimize the insertion loss for the ON state of the switch. Calculate the resulting insertion losses for the ON and OFF states. 11.3 A single-pole, single-throw switch is constructed using two identical PIN diodes in the arrangement shown below. In the ON state, the series diode is forward biased and the shunt diode is reversed biased; and vice versa for the OFF state. If f=6 GHz, Z0=50/Omega1,Cj=0.1p F , Rr=0.5/Omega1,Rf= 0.3/Omega1,a n d Li=0.4 nH, determine the insertion losses for the ON and OFF states. c11ActiveRFAndMicrowaveDevices Pozar September 14, 2011 22:30 Problems 557 Z0 Z0 11.4 Consider the loaded-line phase shifter shown below. If Z0=50/Omega1, find the necessary stub lengths for a differential phase shift of 45◦, and calculate the resulting insertion loss for both states of the phase shifter. Assume all lines are lossless and that the diodes can be approximated as ideal shorts or opens. Z0 Z0 Z0/H9258/H9261/4 3Z0/H9258 3Z0 11.5 Use the equivalent circuit of Figure 11.17 to derive the expression for the short-circuit current gain of a bipolar transistor. Assume a unilateral device, where Cc=0. 11.6 Show that the scattering parameters for an FET can be expressed in terms of the parameters of the equivalent circuit of Figure 11.21 as given below. Assume the device is unilateral. S11=Z11−Z0 Z11+Z0,S12=0, S21=2jZ0gm/ωCgs (Z11+Z0)(Z22+Z0),S22=Z22−Z0 Z22+Z0, where Z11=Ri−j/ωCgsandZ22=(1/Rds+jωCds)−1. 11.7 Given the scattering parameters for an FET, derive expressions for the parameters of the equivalent circuit model of Figure 11.21, assuming a unilateral device. Use these results to find the equivalent circuit parameters for the HEMT device whose scattering parameters are given in Table 11.7, at a frequency of 2.0 GHz. Ignore S12. c12MicrowaveAmplifier Pozar September 16, 2011 14:56 Chapter Twelve Microwave Amplifier Design Signal amplification is one of the most basic and prevalent circuit functions in modern RF and microwave systems. Early microwave amplifiers relied on tubes, such as klystrons and traveling-wave tubes, or solid-state reflection amplifiers based on the negative resistance char- acteristics of tunnel or varactor diodes. However, due to the dramatic improvements and inno-vations in solid-state technology that have occurred since the 1970s, most RF and microwaveamplifiers today use transistor devices such as Si BJTs, GaAs or SiGe HBTs, Si MOSFETs, GaAs MESFETs, or GaAs or GaN HEMTs [1–5]. Microwave transistor amplifiers are rugged, low-cost, and reliable and can be easily integrated in both hybrid and monolithic integratedcircuitry. Transistor amplifiers can be used at frequencies in excess of 100 GHz in a wide rangeof applications requiring small size, low noise figure, broad bandwidth, and medium to high power capacity. Although microwave tubes are still useful for very high power and/or very high frequency applications, continuing improvement in the performance of microwave transistorsis steadily reducing the need for microwave tubes. Our discussion of transistor amplifier design will primarily rely on the terminal character- istics of the transistor, as represented by either scattering parameters or one of the equivalent circuit models introduced in the previous chapter. We will begin with some general definitionsof two-port power gains that are useful for amplifier design and then discuss the subject of sta-bility. These results will then be applied to single-stage transistor amplifiers, including designs for maximum gain, specified gain, and low noise figure. Broadband balanced and distributed amplifiers are discussed in Section 12.4. We conclude with a brief treatment of transistor poweramplifiers. 12.1TWO-PORTPOWERGAINS In this section we develop several expressions for the gain and stability of a general two- port amplifier circuit in terms of the scattering parameters of the transistor. These results 558 c12MicrowaveAmplifier Pozar September 16, 2011 14:56 12.1 Two-Port Power Gains 559 Zs Vs ZLV2 (Z0)[S] V1+ –+ – sV+ 2 V+ 1 V– 1 V– 2 /H9003in/H9003out/H9003L/H9003 FIGURE 12.1 A two-port network with arbitrary source and load impedances. will be used in the following sections for amplifier design and in Chapter 13 for oscillator design. Definition ofTwo-PortPowerGains Consider an arbitrary two-port network, characterized by its scattering matrix [ S], con- nected to source and load impedances ZSandZL, respectively, as shown in Figure 12.1. We will derive expressions for three types of power gain in terms of the scattering param-eters of the two-port network and the reflection coefficients, /Gamma1 Sand/Gamma1L, of the source and load. rPower gain =G=PL/Pinis the ratio of power dissipated in the load ZLto the power delivered to the input of the two-port network. This gain is independent of ZS, although the characteristics of some active devices may be dependent on ZS.rAvailable power gain =GA=Pavn/Pavsis the ratio of the power available from the two-port network to the power available from the source. This assumes conjugate matching of both the source and the load, and depends on ZS, but not ZL.rTransducer power gain =GT=PL/Pavsis the ratio of the power delivered to the load to the power available from the source. This depends on both ZSandZL. These definitions differ primarily in the way the source and load are matched to the two- port device; if the input and output are both conjugately matched to the two-port device, then the gain is maximized and G=GA=GT. With reference to Figure 12.1, the reflection coefficient seen looking toward the load is /Gamma1L=ZL−Z0 ZL+Z0,( 12.1a) while the reflection coefficient seen looking toward the source is /Gamma1S=ZS−Z0 ZS+Z0,( 12.1b) where Z0is the characteristic impedance reference for the scattering parameters of the two-port network. In general, the input impedance of the terminated two-port network will be mis- matched with a reflection coefficient given by /Gamma1in, which can be determined using a signal flow graph (see Example 4.7) or by the following analysis. From the definition of the scat- tering parameters, and the fact that V+ 2=/Gamma1LV− 2,w eh a v e V− 1=S11V+ 1+S12V+ 2=S11V+ 1+S12/Gamma1LV− 2, (12.2a) V− 2=S21V+ 1+S22V+ 2=S21V+ 1+S22/Gamma1LV− 2. (12.2b) c12MicrowaveAmplifier Pozar September 16, 2011 14:56 560 Chapter 12: Microwave Amplifier Design Eliminating V− 2from (12.2a) and solving for V− 1/V+ 1gives /Gamma1in=V− 1 V+ 1=S11+S12S21/Gamma1L 1−S22/Gamma1L=Zin−Z0 Zin+Z0,( 12.3a) where Zinis the impedance seen looking into port 1 of the terminated network. Similarly, the reflection coefficient seen looking into port 2 of the network when port 1 is terminatedbyZ Sis /Gamma1out=V− 2 V+ 2=S22+S12S21/Gamma1S 1−S11/Gamma1S.( 12.3b) By voltage division, V1=VSZin ZS+Zin=V+ 1+V− 1=V+ 1(1+/Gamma1in). Using Zin=Z01+/Gamma1in 1−/Gamma1in from (12.3a) and solving for V+ 1in terms of VSgives V+ 1=VS 2(1−/Gamma1S) (1−/Gamma1S/Gamma1in).( 12.4) If peak values are assumed for all voltages, the average power delivered to the network is Pin=1 2Z0/vextendsingle/vextendsingleV+ 1/vextendsingle/vextendsingle2/parenleftbig 1−|/Gamma1in|2/parenrightbig =|VS|2 8Z0|1−/Gamma1S|2 |1−/Gamma1S/Gamma1in|2/parenleftbig 1−|/Gamma1in|2/parenrightbig ,( 12.5) where (12.4) was used. The power delivered to the load is PL=/vextendsingle/vextendsingleV− 2/vextendsingle/vextendsingle2 2Z0/parenleftbig 1−|/Gamma1L|2/parenrightbig .( 12.6) Solving for V− 2from (12.2b), substituting into (12.6), and using (12.4) gives PL=/vextendsingle/vextendsingleV+ 1/vextendsingle/vextendsingle2 2Z0|S21|2/parenleftbig 1−|/Gamma1L|2/parenrightbig |1−S22/Gamma1L|2=|VS|2 8Z0|S21|2/parenleftbig 1−|/Gamma1L|2/parenrightbig|1−/Gamma1S|2 |1−S22/Gamma1L|2|1−/Gamma1S/Gamma1in|2.( 12.7) The power gain can then be expressed as G=PL Pin=|S21|2/parenleftbig 1−|/Gamma1L|2/parenrightbig /parenleftbig 1−|/Gamma1in|2/parenrightbig|1−S22/Gamma1L|2.( 12.8) The power available from the source, Pavs, is the maximum power that can be delivered to the network. This occurs when the input impedance of the terminated network is conju- gately matched to the source impedance, as discussed in Section 2.6. Thus, from (12.5), Pavs=Pin/vextendsingle/vextendsingle/vextendsingle/vextendsingle /Gamma1in=/Gamma1∗ S=|VS|2 8Z0|1−/Gamma1S|2 /parenleftbig 1−|/Gamma1S|2/parenrightbig.( 12.9) c12MicrowaveAmplifier Pozar September 16, 2011 14:56 12.1 Two-Port Power Gains 561 Similarly, the power available from the network, Pavn, is the maximum power that can be delivered to the load. Thus, from (12.7), Pavn=PL/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle /Gamma1L=/Gamma1∗ out=|VS|2 8Z0|S21|2/parenleftbig 1−|/Gamma1out|2/parenrightbig|1−/Gamma1S|2 /vextendsingle/vextendsingle1−S22/Gamma1∗ out/vextendsingle/vextendsingle2|1−/Gamma1S/Gamma1in|2/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle /Gamma1L=/Gamma1∗ out.( 12.10) In (12.10), /Gamma1inmust be evaluated for /Gamma1L=/Gamma1∗ out. From (12.3a), it can be shown that |1−/Gamma1S/Gamma1in|2/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle /Gamma1L=/Gamma1∗ out=|1−S11/Gamma1S|2/parenleftbig 1−|/Gamma1out|2/parenrightbig2 /vextendsingle/vextendsingle1−S22/Gamma1∗ out/vextendsingle/vextendsingle2, which reduces (12.10) to Pavn=|VS|2 8Z0|S21|2|1−/Gamma1S|2 |1−S11/Gamma1S|2/parenleftbig 1−|/Gamma1out|2/parenrightbig.( 12.11) Observe that PavsandPavnhave been expressed in terms of the source voltage, VS, which is independent of the input or load impedances. There would be confusion if these quantitieswere expressed in terms of V + 1since V+ 1is different for each of the calculations of PL, Pavs, and Pavn. Using (12.11) and (12.9), we obtain the available power gain as GA=Pavn Pavs=|S21|2/parenleftbig 1−|/Gamma1S|2/parenrightbig |1−S11/Gamma1S|2/parenleftbig 1−|/Gamma1out|2/parenrightbig.( 12.12) From (12.7) and (12.9), the transducer power gain is GT=PL Pavs=|S21|2/parenleftbig 1−|/Gamma1S|2/parenrightbig/parenleftbig 1−|/Gamma1L|2/parenrightbig |1−/Gamma1S/Gamma1in|2|1−S22/Gamma1L|2.( 12.13) A special case of the transducer power gain occurs when both the input and output are matched for zero reflection (in contrast to conjugate matching). Then /Gamma1L=/Gamma1S=0, and (12.13) reduces to GT=|S21|2.( 12.14) Another special case is the unilateral tran sducer power gain ,GTU, where S12=0( o ri s negligibly small). This nonreciprocal characteristic is approximately true for many transis- tors devices. From (12.3a), /Gamma1in=S11when S12=0, so (12.13) gives the unilateral trans- ducer power gain as GTU=|S21|2/parenleftbig 1−|/Gamma1S|2/parenrightbig/parenleftbig 1−|/Gamma1L|2/parenrightbig |1−S11/Gamma1S|2|1−S22/Gamma1L|2.( 12.15) EXAMPLE 12.1 COMPARISON OF POWER GAIN DEFINITIONS A silicon bipolar junction transistor has the following scattering parameters at 1.0 GHz, with a 50 /Omega1reference impedance: S11=0.38/negationslash−158◦ S12=0.11/negationslash54◦ S21=3.50/negationslash80◦ S22=0.40/negationslash−43◦ c12MicrowaveAmplifier Pozar September 16, 2011 14:56 562 Chapter 12: Microwave Amplifier Design The source impedance is ZS=25/Omega1and the load impedance is ZL=40/Omega1. Compute the power gain, the available power gain, and the transducer power gain. Solution From (12.1a) and (12.1b) the reflection coefficients at the source and load are /Gamma1S=ZS−Z0 ZS+Z0=25−50 25+50=−0.333, /Gamma1L=ZL−Z0 ZL+Z0=40−50 40+50=−0.111. From (12.3a) and (12.3b) the reflection coefficients seen looking at the input and output of the terminated network are /Gamma1in=S11+S12S21/Gamma1L 1−S22/Gamma1L=0.365 /negationslash−152◦, /Gamma1out=S22+S12S21/Gamma1S 1−S11/Gamma1S=0.545 /negationslash−43◦. Then from (12.8) the power gain is G=|S21|2/parenleftbig 1−|/Gamma1L|2/parenrightbig /parenleftbig 1−|/Gamma1in|2/parenrightbig|1−S22/Gamma1L|2=13.1. From (12.12) the available power gain is GA=|S21|2/parenleftbig 1−|/Gamma1S|2/parenrightbig |1−S11/Gamma1S|2/parenleftbig 1−|/Gamma1out|2/parenrightbig=19.8. From (12.13) the transducer power gain is GT=|S21|2/parenleftbig 1−|/Gamma1S|2/parenrightbig/parenleftbig 1−|/Gamma1L|2/parenrightbig |1−/Gamma1S/Gamma1in|2|1−S22/Gamma1L|2=12.6. ■ FurtherDiscussionofTwo-PortPowerGains A single-stage microwave transistor amplifier can be modeled by the circuit of Figure 12.2, where matching networks are used on both sides of the transistor to transform the input and output impedance Z0to the source and load impedances ZSandZL. The most useful gain definition for amplifier design is the transducer power gain of (12.13), which accounts for both source and load mismatch. From (12.13) we can define separate effective gain factors for the input (source) matching network, the transistor itself, and the output (load) Γin Γs Γout ΓLInput matching circuit GsTransistor [S] G0Output matching circuit GLZ0Z0 FIGURE 12.2 The general transistor amplifier circuit. c12MicrowaveAmplifier Pozar September 16, 2011 14:56 12.1 Two-Port Power Gains 563 matching network as follows: GS=1−|/Gamma1S|2 |1−/Gamma1in/Gamma1S|2, (12.16a) G0=|S21|2, (12.16b) GL=1−|/Gamma1L|2 |1−S22/Gamma1L|2. (12.16c) The overall transducer gain is then GT=GSG0GL. The effective gains GSandGLof the matching networks may be greater than unity. This is because the unmatched transistor would incur power loss due to reflections at the input and output of the transistor, and the matching sections can reduce these losses. If the transistor is unilateral, so that S12=0 (or is small enough to be ignored), then (12.3) reduces to /Gamma1in=S11,/Gamma1out=S22, and the unilateral transducer gain reduces toGTU=GSG0GL, where GS=1−|/Gamma1S|2 |1−S11/Gamma1S|2, (12.17a) G0=|S21|2, (12.17b) GL=1−|/Gamma1L|2 |1−S22/Gamma1L|2. (12.17c) The above results have been derived using the scattering parameters of the transistor, but it is possible to obtain alternative expressions for gain in terms of the equivalent circuit parameters of the transistor. As an example, consider the evaluation of the unilateral trans- ducer gain for a conjugately matched FET using the equivalent circuit of Figure 11.21 (with Cgd=0). To conjugately match the transistor we choose source and load impedances as shown in Figure 12.3. Setting the series source inductive reactance X=1/ωCgswill make Zin=Z∗ S, and setting the shunt load inductive susceptance B=−ωCdswill make Zout= Z∗ L; this effectively eliminates the reactive elements from the transistor equivalent circuit. Then by voltage division Vc=VS/2jωRiCgs, and the gain can be easily evaluated as GTU=PL Pavs=1 8|gmVc|2Rds 1 8|VS|2/Ri=g2 mRds 4ω2RiC2gs=Rds 4Ri/parenleftbiggfT f/parenrightbigg2 .( 12.18) where the last step has been written in terms of the cutoff frequency, fT, from (11.24). This shows the interesting result that the gain of a conjugately matched FET amplifier drops off as 1/ f2, or 6 dB per octave. A photograph of a low-noise MMIC amplifier is shown in Figure 12.4. + – SourceDrain Gate ZinRi Ri CgsgmVc VcRds RdsjBCdsVsjX Zout FIGURE 12.3 Unilateral FET equivalent circuit and source and load terminations for the calcula- tion of unilateral transducer power gain. c12MicrowaveAmplifier Pozar September 16, 2011 14:56 564 Chapter 12: Microwave Amplifier Design FIGURE 12.4 Photograph of a low-noise MMIC amplifier that is switchable between 2.4, 3.6, and 5.8 GHz. The amplifier uses pHEMTs in a cascode configuration with source inductance, followed by a common source stage with feedback. Gain is approxi- mately 13 dB in each band. Chip dimensions are 1.85 mm by 1 mm. Courtesy of J. Shatzman and R. W. Jackson of the University of Massachusetts at Amherst and H. Yu of TriQuint, Lowell, Mass. 12.2STABILITY We now discuss the necessary conditions for a transistor amplifier to be stable. In the circuit of Figure 12.2, oscillation is possible if either the input or output port impedancehas a negative real part; this would then imply that |/Gamma1 in|>1o r|/Gamma1out|>1. Because /Gamma1in and/Gamma1outdepend on the source and load matching networks, the stability of the amplifier depends on /Gamma1Sand/Gamma1Las presented by the matching networks. Thus, we define two types of stability: rUnconditional stability : The network is unconditionally stable if |/Gamma1in|<1 and |/Gamma1out|<1 for all passive source and load impedances (i.e., |/Gamma1S|<1 and|/Gamma1L|<1).rConditional stability : The network is conditionally stable if |/Gamma1in|<1 and|/Gamma1out|<1 only for a certain range of passive source and load impedances. This case is also referred to as potentially un stable. Note that the stability condition of an amplifier circuit is usually frequency depen- dent since the input and output matching networks generally depend on frequency. It is therefore possible for an amplifier to be stable at its design frequency but unstable at other frequencies. Careful amplifier design should consider this possibility. We must also point out that the following discussion of stability is limited to two-port amplifier circuits of the type shown in Figure 12.2, and where the scattering parameters of the active device canbe measured without oscillations over the frequency band of interest. The rigorous gen- eral treatment of stability requires that the network scattering parameters (or other network parameters) have no poles in the right-half complex frequency plane, in addition to theconditions that |/Gamma1 in|<1 and|/Gamma1out|<1 [6]. This can be a difficult assessment in practice, but for the special case considered here, where the scattering parameters are known to be pole free (as confirmed by measurability), the following stability conditions are adequate. StabilityCircles Applying the above requirements for unconditional stability to (12.3) gives the following conditions that must be satisfied by /Gamma1Sand/Gamma1Lif the amplifier is to be unconditionally c12MicrowaveAmplifier Pozar September 16, 2011 14:56 12.2 Stability 565 stable: |/Gamma1in|=/vextendsingle/vextendsingle/vextendsingle/vextendsingleS 11+S12S21/Gamma1L 1−S22/Gamma1L/vextendsingle/vextendsingle/vextendsingle/vextendsingle<1, (12.19a) |/Gamma1 out|=/vextendsingle/vextendsingle/vextendsingle/vextendsingleS22+S12S21/Gamma1S 1−S11/Gamma1S/vextendsingle/vextendsingle/vextendsingle/vextendsingle<1. (12.19b) If the device is unilateral (S12=0), these conditions reduce to the simple results that |S11|<1 and |S22|<1 are sufficient for unconditional stability. Otherwise, the inequali- ties of (12.19) define a range of values for /Gamma1Sand/Gamma1Lwhere the amplifier will be stable. Finding this range for /Gamma1Sand/Gamma1Lcan be facilitated by using the Smith chart and plotting the input and output stability circle s. The stability circles are defined as the loci in the /Gamma1L(or/Gamma1S)plane for which |/Gamma1in|=1( o r |/Gamma1out|=1). The stability circles then define the boundaries between stable and potentially unstable regions of /Gamma1Sand/Gamma1L./Gamma1Sand/Gamma1Lmust lie on the Smith chart (|/Gamma1S|<1,|/Gamma1L|<1 for passive matching networks). We can derive the equation for the output stability circle as follows. First use (12.19a) to express the condition that |/Gamma1in|=1a s /vextendsingle/vextendsingle/vextendsingle/vextendsingleS11+S12S21/Gamma1L 1−S22/Gamma1L/vextendsingle/vextendsingle/vextendsingle/vextendsingle=1,( 12.20) or |S11(1−S22/Gamma1L)+S12S21/Gamma1L|=| 1−S22/Gamma1L|. Now define /Delta1as the determinant of the scattering matrix: /Delta1=S11S22−S12S21.( 12.21) Then we can write the above result as |S11−/Delta1/Gamma1 L|=| 1−S22/Gamma1L|.( 12.22) Now square both sides and simplify to obtain |S11|2+|/Delta1|2|/Gamma1L|2−/parenleftbig /Delta1/Gamma1 LS∗ 11+/Delta1∗/Gamma1∗ LS11/parenrightbig =1+|S22|2|/Gamma1L|2−/parenleftbig S∗ 22/Gamma1∗ L+S22/Gamma1L/parenrightbig /parenleftbig |S22|2−|/Delta1|2/parenrightbig /Gamma1L/Gamma1∗ L−/parenleftbig S22−/Delta1S∗ 11/parenrightbig /Gamma1L−/parenleftbig S∗ 22−/Delta1∗S11/parenrightbig /Gamma1∗ L=|S11|2−1 /Gamma1L/Gamma1∗ L−/parenleftbig S22−/Delta1S∗ 11/parenrightbig /Gamma1L+/parenleftbig S∗ 22−/Delta1∗S11/parenrightbig /Gamma1∗ L |S22|2−|/Delta1|2=|S11|2−1 |S22|2−|/Delta1|2. (12.23) Next, complete the square by adding/vextendsingle/vextendsingleS22−/Delta1S∗ 11/vextendsingle/vextendsingle2//parenleftbig |S22|2−|/Delta1|2/parenrightbig2to both sides: /vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/Gamma1 L−/parenleftbig S22−/Delta1S∗ 11/parenrightbig∗ |S22|2−|/Delta1|2/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle2 =/vextendsingle/vextendsingleS2 11/vextendsingle/vextendsingle−1 |S22|2−|/Delta1|2+/vextendsingle/vextendsingleS22−/Delta1S∗ 11/vextendsingle/vextendsingle2 /parenleftbig |S22|2−|/Delta1|2/parenrightbig2, or /vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/Gamma1 L−/parenleftbig S22−/Delta1S∗ 11/parenrightbig∗ |S22|2−|/Delta1|2/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=/vextendsingle/vextendsingle/vextendsingle/vextendsingleS 12S21 |S22|2−|/Delta1|2/vextendsingle/vextendsingle/vextendsingle/vextendsingle.( 12.24) In the complex /Gamma1plane, an equation of the form |/Gamma1−C|=Rrepresents a circle having a center at C(a complex number) and a radius R(a real number). Thus, (12.24) defines the c12MicrowaveAmplifier Pozar September 16, 2011 14:56 566 Chapter 12: Microwave Amplifier Design output stability circle with a center CLand radius RL, where CL=/parenleftbig S22−/Delta1S∗ 11/parenrightbig∗ |S22|2−|/Delta1|2(center), (12.25a) RL=/vextendsingle/vextendsingle/vextendsingle/vextendsingleS12S21 |S22|2−|/Delta1|2/vextendsingle/vextendsingle/vextendsingle/vextendsingle(radius). (12.25b) Similar results can be obtained for the input stability circle by interchanging S11andS22: CS=/parenleftbig S11−/Delta1S∗ 22/parenrightbig∗ |S11|2−|/Delta1|2(center), (12.26a) RS=/vextendsingle/vextendsingle/vextendsingle/vextendsingleS 12S21 |S11|2−|/Delta1|2/vextendsingle/vextendsingle/vextendsingle/vextendsingle(radius). (12.26b) Given the scattering parameters of the transistor, we can plot the input and output stability circles to define where |/Gamma1 in|=1 and |/Gamma1out|=1. On one side of the input stability circle we will have |/Gamma1out|<1, while on the other side we will have |/Gamma1out|>1. Similarly, we will have |/Gamma1in|<1 on one side of the output stability circle, and |/Gamma1in|>1 on the other side. We need to determine which areas on the Smith chart represent the stable region, for which |/Gamma1in|<1 and|/Gamma1out|<1. Consider the output stability circles plotted in the /Gamma1Lplane for |S11|<1 and|S11|> 1, as shown in Figure 12.5. If we set ZL=Z0, then /Gamma1L=0, and (12.19a) shows that |/Gamma1in|=|S11|.N o wi f| S11|<1, then |/Gamma1in|<1, so/Gamma1L=0 must be in a stable region. This means that the center of the Smith chart (/Gamma1 L=0)is in the stable region, so all of the Smith chart (|/Gamma1 L|<1)that is exterior to the stability circle defines the stable range for /Gamma1L.T h i s region is shaded in Figure 12.5a. Alternatively, if we set ZL=Z0but have |S11|>1, then |/Gamma1in|>1f o r/Gamma1 L=0, and the center of the Smith chart must be in an unstable region. In this case the stable region is the inside region of the stability circle that intersects the Smith chart, as illustrated in Figure 12.5b. Similar results apply to the input stability circle. If the device is unconditionally stable, the stability circles must be completely outside (or totally enclose) the Smith chart. We can state this result mathematically as ||CL|−RL|>1f o r |S11|<1, (12.27a) ||CS|−RS|>1f o r |S22|<1. (12.27b) CL RLCL RL |Γin| < 1 (stable)|Γin| < 1 (stable) (a) (b) FIGURE 12.5 Output stability circles for a conditionally stable device. (a) |S11|<1. (b)|S11|>1. c12MicrowaveAmplifier Pozar September 16, 2011 14:56 12.2 Stability 567 If|S11|>1o r| S22|>1, the amplifier cannot be unconditionally stable because we can always have a source or load impedance of Z0leading to /Gamma1S=0o r/Gamma1 L=0, thus causing |/Gamma1in|>1o r|/Gamma1 out|>1. If the device is only conditionally stable, operating points for /Gamma1S and/Gamma1Lmust be chosen in stable regions, and it is good practice to check stability at several frequencies over the range where the device operates. Also note that the scattering param- eters of a transistor depend on the bias conditions, and so stability will also depend on bias conditions. If it is possible to accept a design with less than maximum gain, a transistorcan usually be made to be unconditionally stable by using resistive loading. TestsforUnconditionalStability The stability circles discussed above can be used to determine regions for /Gamma1 Sand/Gamma1Lwhere the amplifier circuit will be conditionally stable, but simpler tests can be used to determine unconditional stability. One of these is the K−/Delta1test, where it can be shown that a device will be unconditionally stable if Rollet’ scondition, defined as K=1−|S11|2−|S22|2+|/Delta1|2 2|S12S21|>1,( 12.28) along with the auxiliary condition that |/Delta1|=| S11S22−S12S21|<1,( 12.29) are simultaneously satisfied. These two conditions are necessary and sufficient for uncon- ditional stability, and are easily evaluated. If the device scattering parameters do not satisfy theK−/Delta1test, the device is not unconditionally stable, and stability circles must be used to determine if there are values of /Gamma1Sand/Gamma1Lfor which the device will be conditionally stable. Also recall that we must have |S11|<1 and|S22|<1 if the device is to be uncon- ditionally stable. While the K−/Delta1test of (12.28)–(12.29) is a mathematically rigorous condition for unconditional stability, it cannot be used to compare the relative stability of two or moredevices because it involves constraints on two separate parameters. Recently, however, a new criterion has been proposed [7] that combines the scattering parameters in a test involving only a single parameter, µ, defined as µ=1−|S 11|2 /vextendsingle/vextendsingleS22−/Delta1S∗ 11/vextendsingle/vextendsingle+|S12S21|>1.( 12.30) Thus, if µ>1, the device is unconditionally stable. In addition, it can be said that larger values of µimply greater stability. We can derive the µ-test of (12.30) by starting with the expression from (12.3b) for /Gamma1out: /Gamma1out=S22+S12S21/Gamma1S 1−S11/Gamma1S=S22−/Delta1/Gamma1S 1−S11/Gamma1S,( 12.31) where /Delta1is the determinant of the scattering matrix defined in (12.21). Unconditional sta- bility implies that |/Gamma1out|<1 for any passive source termination, /Gamma1S. The reflection coeffi- cient for a passive source impedance must lie within the unit circle on a Smith chart, and theouter boundary of this circle can be written as /Gamma1 S=ejφ. The expression given in (12.31) maps this circle into another circle in the /Gamma1outplane. We can show this by substituting /Gamma1S=ejφinto (12.31) and solving for ejφ: ejφ=S22−/Gamma1out /Delta1−S11/Gamma1out. c12MicrowaveAmplifier Pozar September 16, 2011 14:56 568 Chapter 12: Microwave Amplifier Design Taking the magnitude of both sides gives /vextendsingle/vextendsingle/vextendsingle/vextendsingleS 22−/Gamma1out /Delta1−S11/Gamma1out/vextendsingle/vextendsingle/vextendsingle/vextendsingle=1. Squaring both sides and expanding gives |/Gamma1 out|2/parenleftbig 1−|S11|2/parenrightbig +/Gamma1out/parenleftbig /Delta1∗S11−S∗ 22/parenrightbig +/Gamma1∗ out/parenleftbig /Delta1S∗ 11−S22/parenrightbig =|/Delta1|2−|S22|2. N o wd i v i d eb y1 −|S11|2to obtain |/Gamma1out|2+/parenleftbig /Delta1∗S11−S∗ 22/parenrightbig /Gamma1out+/parenleftbig /Delta1S∗ 11−S22/parenrightbig /Gamma1∗ out 1−|S11|2=|/Delta1|2−|S22|2 1−|S11|2. Complete the square by adding/vextendsingle/vextendsingle/Delta1∗S11−S∗ 22/vextendsingle/vextendsingle2 /parenleftbig 1−|S11|2/parenrightbig2to both sides: /vextendsingle/vextendsingle/vextendsingle/vextendsingle/Gamma1 out+/Delta1S∗ 11−S22 1−|S11|2/vextendsingle/vextendsingle/vextendsingle/vextendsingle2 =|/Delta1|2−|S22|2 1−|S11|2+/vextendsingle/vextendsingle/Delta1∗S11−S∗ 22/vextendsingle/vextendsingle2 /parenleftbig 1−|S11|2/parenrightbig2=|S12S21|2 /parenleftbig 1−|S11|2/parenrightbig2.(12.32) This equation is of the form |/Gamma1out−C|=R, representing a circle with center Cand radius Rin the/Gamma1outplane. Thus the center and radius of the mapped |/Gamma1S|=1 circle are given by C=S22−/Delta1S∗ 11 1−|S11|2, (12.33a) R=|S12S21| 1−|S11|2. (12.33b) If points within this circular region are to satisfy |/Gamma1out|<1, then we must have that |C|+R<1.( 12.34) Substituting (12.33) into (12.34) gives /vextendsingle/vextendsingleS22−/Delta1S∗ 11/vextendsingle/vextendsingle+|S12S21|<1−|S11|2, which after rearranging yields the µ-test of (12.30): 1−|S11|2 /vextendsingle/vextendsingleS22−/Delta1S∗ 11/vextendsingle/vextendsingle+|S12S21|>1. The K−/Delta1test of (12.28)–(12.29) can be derived from a similar starting point, or more simply from the µ-test of (12.30). Rearranging (12.30) and squaring gives /vextendsingle/vextendsingleS22−/Delta1S∗ 11/vextendsingle/vextendsingle2</parenleftbig 1−|S11|2−|S12S21|/parenrightbig2.( 12.35) It can be verified by direct expansion that /vextendsingle/vextendsingleS22−/Delta1S∗ 11/vextendsingle/vextendsingle2=|S12S21|2+/parenleftbig 1−|S11|2/parenrightbig/parenleftbig |S22|2−|/Delta1|2/parenrightbig , so (12.35) expands to |S12S21|2+/parenleftbig 1−|S11|2/parenrightbig/parenleftbig |S22|2−|/Delta1|2/parenrightbig </parenleftbig 1−|S11|2/parenrightbig/parenleftbig 1−|S11|2−2|S12S21|/parenrightbig +|S12S21|2. Simplifying gives |S22|2−|/Delta1|2<1−|S11|2−2|S12S21|, c12MicrowaveAmplifier Pozar September 16, 2011 14:56 12.2 Stability 569 which yields Rollet’s condition of (12.28) after rearranging: 1−|S11|2−|S22|2+|/Delta1|2 2|S12S21|=K>1. In addition to (12.28), the K−/Delta1test also requires the auxiliary condition of (12.29) to guarantee unconditional stability. Although we derived Rollet’s condition from the neces- sary and sufficient result of the µ-test, the squaring step used in (12.35) introduces an ambi- guity in the sign of the right-hand side, thus requiring the additional condition. This can bederived by requiring that the right-hand side of (12.35) be positive before squaring. Thus, |S 12S21|<1−|S11|2. Because similar conditions can be derived for the input side of the circuit, we can inter- change S11andS22to obtain the analogous condition that |S12S21|<1−|S22|2. Adding these two inequalities gives 2|S12S21|<2−|S11|2−|S22|2. From the triangle inequality we know that |/Delta1|=| S11S22−S12S21|≤|S11S22|+|S12S21|, so we have that |/Delta1|<|S11||S22|+1−1 2|S11|2−1 2|S22|2<1−1 2/parenleftbig |S11|2−|S22|2/parenrightbig <1, which is identical to (12.29). EXAMPLE 12.2 TRANSISTOR STABILITY The Triquint T1G6000528 GaN HEMT has the following scattering parameters at 1.9 GHz (Z0=50/Omega1): S11=0.869 /negationslash−159◦, S12=0.031 /negationslash−9◦, S21=4.250 /negationslash61◦, S22=0.507 /negationslash−117◦. Determine the stability of this transistor by using the K−/Delta1test and the µ-test, and plot the stability circles on a Smith chart. Solution From (12.28) and (12.29) we compute Kand|/Delta1|as |/Delta1|= |S11S22−S12S21|=0.336, K=1−|S11|2−|S22|2+|/Delta1|2 2|S12S21|=0.383. Thus we have |/Delta1|<1 but not K>1, so the unconditional stability criteria of (12.28)–(12.29) are not satisfied, and the device is potentially unstable. The sta-bility of this device can also be evaluated using the µ-test, for which (12.30) gives µ=0.678, again indicating potential instability. c12MicrowaveAmplifier Pozar September 16, 2011 14:56 570 Chapter 12: Microwave Amplifier Design The centers and radii of the stability circles are given by (12.25) and (12.26): CL=/parenleftbig S22−/Delta1S∗ 11/parenrightbig∗ |S22|2−|/Delta1|2=1.59/negationslash132◦, RL=|S12S21| |S22|2−|/Delta1|2=0.915, CS=/parenleftbig S11−/Delta1S∗ 22/parenrightbig∗ |S11|2−|/Delta1|2=1.09/negationslash162◦, RS=|S12S21| |S11|2−|/Delta1|2=0.205. These data can be used to plot the input and output stability circles, as shown in Figure 12.6. Since |S11|<1 and |S22|<1, the central part of the Smith chart represents the stable operating region for /Gamma1Sand/Gamma1L. The unstable regions are shaded. ■ j , 20-2030-3040-4050-5060-6070-7080-8090-90100-100110-110120-120130-130140-140150-150160-160170-1701800.040.040.050.050.060.060.070.070.080.080.090.090.10.10.110.110.120.120.130.130.140.140.150.150.160.160.170.170.180.180.190.190.20.20.210.210.220.220.230.230.240.24 0.250.250.260.260.270.270.280.280.290.290.30.30.310.310.320.320.330.330.340.340.350.350.360.360.370.370.380.380.390.390.40.40.410.410.420.420.430.430.440.440.450.450.460.460.470.470.480.480.490.49 0.00.0AN GLE OFREFLECTION COEFFICIENTINDEGREES—>WAVELENGTHSTOWARDGENERATOR—><— W AVELENGTH STOW ARD LOAD <—II T NDUCTIVEREACTANCECOMPONEN(+X/Zo) ORCAPACTIVESUSCEPTANCE(+jB/Yo)CAPACITIVEREACTANCECOMPONENT(-jX/Zo),ORINDUCTIVESUSCEPTANCE(-jB/Yo) ± 0.1 0.1 0.1 0.2 0.2 0.2 0.3 0.3 0.3 0.4 0.4 0.4 0.5 0.5 0.5 0.6 0.6 0.6 0.7 0.7 0.7 0.8 0.8 0.8 0.9 0.9 0.9 1.0 1.0 1.01.21.21.21.41.41.41.61.61.61.81.81.82.02.02.03.03.03.04.04.04.05.05.05.01010102020205050500.20.2 0.2 0.20.4 0.4 0.40.60.6 0.6 0.60.80.8 0.8 0.81.01.0 1.0 1.0RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo)0.4RL CL Input stability circleOutput stability circle Unstable regionsCS FIGURE 12.6 Stability circles for Example 12.2. c12MicrowaveAmplifier Pozar September 16, 2011 14:56 12.3 Single-Stage Transistor Amplifier Design 571 12.3SINGLE-STAGETRANSISTORAMPLIFIERDESIGN DesignforMaximumGain(ConjugateMatching) After the stability of the transistor has been determined and the stable regions for /Gamma1Sand /Gamma1Lhave been located on the Smith chart, the input and output matching sections can be designed. Since G0of (12.16b) is fixed for a given transistor, the overall transducer gain of the amplifier will be controlled by the gains, GSandGL, of the matching sections. Maximum gain will be realized when these sections provide a conjugate match between the amplifier source or load impedance and the transistor. Because most transistors exhibita significant impedance mismatch (large |S 11|and|S22|), the resulting frequency response may be narrowband. In the following section we will discuss how to design for less than maximum gain, with a corresponding improvement in bandwidth. Broadband amplifier design will be discussed in Section 12.4. With reference to Figure 12.2 and our discussion in Section 2.6 on conjugate imped- ance matching, we know that maximum power transfer from the input matching network to the transistor will occur when /Gamma1in=/Gamma1∗ S,( 12.36a) and that maximum power transfer from the transistor to the output matching network will occur when /Gamma1out=/Gamma1∗ L.( 12.36b) With the assumption of lossless matching sections, these conditions will maximize the overall transducer gain. From (12.13), this maximum gain will be given by GTmax=1 1−|/Gamma1S|2|S21|21−|/Gamma1L|2 |1−S22/Gamma1L|2.( 12.37) In addition, with conjugate matching and lossless matching sections, the input and output ports of the amplifier will be matched to Z0. In the general case with a bilateral (S12/negationslash=0) transistor, /Gamma1inis affected by /Gamma1outand vice versa, so the input and output sections must be matched simultaneously. Using (12.36) in (12.3) gives the necessary equations: /Gamma1∗ S=S11+S12S21/Gamma1L 1−S22/Gamma1L, (12.38a) /Gamma1∗ L=S22+S12S21/Gamma1S 1−S11/Gamma1S. (12.38b) We can solve for /Gamma1Sby first rewriting these equations as follows: /Gamma1S=S∗ 11+S∗ 12S∗ 21 1//Gamma1∗ L−S∗ 22, /Gamma1∗ L=S22−/Delta1/Gamma1S 1−S11/Gamma1S, where /Delta1=S11S22−S12S21. Substituting the expression for /Gamma1∗ Linto the expression for /Gamma1S and expanding gives /Gamma1S/parenleftbig 1−|S22|2/parenrightbig +/Gamma12 S/parenleftbig /Delta1S∗ 22−S11/parenrightbig =/Gamma1S/parenleftbig /Delta1S∗ 11S∗ 22−|S11|2−/Delta1S∗ 12S∗ 21/parenrightbig +S∗ 11/parenleftbig 1−|S22|2/parenrightbig +S∗ 12S∗ 21S22. c12MicrowaveAmplifier Pozar September 16, 2011 14:56 572 Chapter 12: Microwave Amplifier Design Using the result that /Delta1/parenleftbig S∗ 11S∗ 22−S∗ 12S∗ 21/parenrightbig =|/Delta1|2allows this to be rewritten as a quadratic equation for /Gamma1S: /parenleftbig S11−/Delta1S∗ 22/parenrightbig /Gamma12 S+/parenleftbig |/Delta1|2−|S11|2+|S22|2−1/parenrightbig /Gamma1S+/parenleftbig S∗ 11−/Delta1∗S22/parenrightbig =0.( 12.39) The solution is /Gamma1S=B1±/radicalBig B2 1−4|C1|2 2C1.( 12.40a) Similarly, the solution for /Gamma1Lcan be written as /Gamma1L=B2±/radicalBig B2 2−4|C2|2 2C2.( 12.40b) The variables B1,C1,B2,C2are defined as B1=1+|S11|2−|S22|2−|/Delta1|2, (12.41a) B2=1+|S22|2−|S11|2−|/Delta1|2, (12.41b) C1=S11−/Delta1S∗ 22, (12.41c) C2=S22−/Delta1S∗ 11. (12.41d) Solutions to (12.40) are only possible if the quantity within the square root is positive, and it can be shown that this is equivalent to requiring K>1. Thus, unconditionally stable devices can always be conjugately matched for maximum gain, and potentially unstable devices can be conjugately matched if K>1 and |/Delta1|<1. The results are much simpler for the unilateral case. When S12=0, (12.38) shows that /Gamma1S=S∗ 11and/Gamma1L=S∗ 22, and then maximum transducer gain of (12.37) reduces to GTUmax=1 1−|S11|2|S21|2 1 1−|S22|2.( 12.42) The maximum transducer power gain given by (12.37) occurs when the source and load are conjugately matched to the transistor, as given by the conditions of (12.36). If the transistor is unconditionally stable, so that K>1, the maximum transducer power gain of (12.37) can be simply rewritten as follows: GTmax=|S21| |S12|/parenleftbig K−/radicalbig K2−1/parenrightbig .( 12.43) This result can be obtained by substituting (12.40) and (12.41) for /Gamma1Sand/Gamma1Linto (12.37) and simplifying. The maximum transducer power gain is also sometimes referred to as the matched gain. The maximum gain does not provide a meaningful result if the device is onlyconditionally stable since simultaneous conjugate matching of the source and load is not possible if K<1 (see Problem 12.8). In this case a useful figure of merit is the maximum stable gain, defined as the maximum transducer power gain of (12.43) with K=1. Thus, G msg=|S21| |S12|.( 12.44) The maximum stable gain is easy to compute and offers a convenient way to compare the gain of various devices under stable operating conditions. c12MicrowaveAmplifier Pozar September 16, 2011 14:56 12.3 Single-Stage Transistor Amplifier Design 573 EXAMPLE 12.3 CONJUGATELY MATCHED AMPLIFIER DESIGN Design an amplifier for maximum gain at 4 GHz using single-stub matching sec- tions. Calculate and plot the input return loss and the gain from 3 to 5 GHz. Thetransistor is a GaAs MESFET with the following scattering parameters ( Z 0= 50/Omega1): f(GHz) S11 S12 S21 S22 3.0 0.80 /negationslash−89◦0.03/negationslash56◦2.86/negationslash99◦0.76/negationslash−41◦ 4.0 0.72 /negationslash−116◦0.03/negationslash57◦2.60/negationslash76◦0.73/negationslash−54◦ 5.0 0.66 /negationslash−142◦0.03/negationslash62◦2.39/negationslash54◦0.72/negationslash−68◦ Solution In practice, scattering parameters are usually provided by the manufacturer over a wide frequency range, and it is prudent to check stability over the entire range. Here we have limited the data to three frequencies to illustrate the point with- out undue computational burden. Using (12.28) and (12.29) to calculate Kand /Delta1from the scattering parameters at each frequency in the above table gives the following results: f(GHz) K /Delta1 3.0 0.77 0.592 4.0 1.19 0.4875.0 1.53 0.418 We see that K>1 and|/Delta1|<1 at 4 and 5 GHz, so the transistor is unconditionally stable at these frequencies, but it is only conditionally stable at 3 GHz. We can proceed with the design at 4 GHz, but should check stability at 3 GHz after we find the matching networks (which determine /Gamma1Sand/Gamma1L). For maximum gain, we should design the matching sections for a conjugate match to the transistor. Thus, /Gamma1S=/Gamma1∗ inand/Gamma1L=/Gamma1∗ out, and/Gamma1S,/Gamma1Lcan be deter- mined from (12.40): /Gamma1S=B1±/radicalBig B2 1−4|C1|2 2C1=0.872 /negationslash123◦, /Gamma1L=B2±/radicalBig B2 2−4|C2|2 2C2=0.876 /negationslash61◦. The effective gain factors of (12.16) can be calculated as GS=1 1−|/Gamma1S|2=4.17=6.20 dB, G0=|S21|2=6.76=8.30 dB, GL=1−|/Gamma1L|2 |1−S22/Gamma1L|2=1.67=2.22 dB. c12MicrowaveAmplifier Pozar September 16, 2011 14:56 574 Chapter 12: Microwave Amplifier Design Then the overall transducer gain is GTmax=6.20+8.30+2.22=16.7d B . The matching networks can easily be determined using the Smith chart. For the input matching section, first plot /Gamma1S, as shown in Figure 12.7a. The impedance, ZS, represented by this reflection coefficient is the impedance seen looking into the matching section toward the source impedance, Z0. Thus, the matching sec- tion must transform Z0to the impedance ZS. There are several ways of doing this, but we will use an open-circuited shunt stub followed by a length of line. We convert to the normalized admittance ys, and work backward (toward the load on the Smith chart) to find that a line of length 0 .120λ will bring us to the 1 +jbcir- cle. Then we see that the required stub admittance is +j3.5, for an open-circuited stub length of 0.206λ. A similar procedure gives a line length of 0.206λ and a stub length of 0.206λ for the output matching circuit. The final amplifier circuit is shown in Figure 12.7b. This circuit only shows the RF components; the amplifier will also require bias circuitry. The return loss and gain were calculated using a CAD package, interpolating the necessary j , 20-2030-3040-4050-5060-6070-7080-8090-90100-100110-110120-120130-130140-140150-150160-160170-1701800.040.040.050.050.060.060.070.070.080.080.090.090.10.10.110.110.120.120.130.130.140.140.150.150.160.160.170.170.180.180.190.190.20.20.210.210.220.220.230.230.240.24 0.250.250.260.260.270.270.280.280.290.290.30.30.310.310.320.320.330.330.340.340.350.350.360.360.370.370.380.380.390.390.40.40.410.410.420.420.430.430.440.440.450.450.460.460.470.470.480.480.490.49 0.00.0AN GLE OFREFLECTION COEFFICIENTINDEGREES—>WAVELENGTHSTOWARDGENERATOR—><— W AVELENGTH STOW ARD LOAD <—II T NDUCTIVEREACTANCECOMPONEN(+X/Zo) ORCAPACTIVESUSCEPTANCE(+jB/Yo)CAPACITIVEREACTANCECOMPONENT(-jX/Zo),ORINDUCTIVESUSCEPTANCE(-jB/Yo) 0.1 0.1 0.1 0.2 0.2 0.2 0.3 0.3 0.3 0.4 0.4 0.4 0.5 0.5 0.5 0.6 0.6 0.6 0.7 0.7 0.7 0.8 0.8 0.8 0.9 0.9 0.9 1.0 1.0 1.01.21.21.21.41.41.41.61.61.61.81.81.82.02.02.03.03.03.04.04.04.05.05.05.01010102020205050500.20.2 0.2 0.20.40.4 0.4 0.40.60.6 0.6 0.60.80.8 0.8 0.81.01.0 1.0 1.0RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) ±Length of open-circuited stub 0.206/H9261 Length of series line 0.120/H9261 (a)ΓS Stub Line yS FIGURE 12.7 Circuit design and frequency response for the transistor amplifier of Example 12.3. (a) Smith chart for the design of the input matching network. c12MicrowaveAmplifier Pozar September 16, 2011 14:56 12.3 Single-Stage Transistor Amplifier Design 575 3.0 3.5 4.0 4.5 5.0–20–100102050 Ω0.120 /H9261 0.206 /H9261 0.206 /H92610.206 /H9261 50 Ω50 Ω 50 Ω50 Ω50 Ω GT –RLGT, –RL (dB) Frequency (GHz) (c)(b) FIGURE 12.7 Continued. (b) RF circuit. (c) Frequency response. scattering parameters from the data given above. The results are plotted in Figure 12.7c, and show the expected gain of 16.7 dB at 4 GHz, with a very good returnloss. The bandwidth where the gain drops by 1 dB is about 2.5%. With regard to the potential instability at 3 GHz, we leave it to the reader to show that the designed matching sections present source and load impedances that lie within the stable regions of the appropriate stability circles. Note that the matching sections are frequency dependent, so the impedances and reflec-tion coefficients are different at 3 GHz than their design values at 4 GHz. The fact that CAD simulation did not show any indication of instability over the fre- quency range of 3–5 GHz is evidence that the circuit is stable over this frequencyrange. ■ Constant-GainCirclesandDesignforSpecifie Gain In many cases it is preferable to design for less than the maximum obtainable gain, to improve bandwidth or to obtain a specific value of amplifier gain. This can be done bydesigning the input and output matching sections to have less than maximum gains; in other words, mismatches are purposely introduced to reduce the overall gain. The de- sign procedure is facilitated by plotting constant-gain circle son the Smith chart to rep- resent loci of /Gamma1 Sand/Gamma1Lthat give fixed values of gain ( GSandGL). To simplify our discussion, we will only treat the case of a unilateral device; the more general case of a c12MicrowaveAmplifier Pozar September 16, 2011 14:56 576 Chapter 12: Microwave Amplifier Design bilateral device must sometimes be considered in practice, and is discussed in detail in references [1–2]. For many transistors |S12|is small enough to be ignored, and the device can be as- sumed to be unilateral. This greatly simplifies the design procedure. The error in the trans-ducer gain caused by approximating |S 12|as zero is given by the ratio GT/GTU. It can be shown that this ratio is bounded by 1 (1+U)2<GT GTU<1 (1−U)2,( 12.45) where Uis defined as the unilateral figure of merit, U=|S12||S21||S11||S22|/parenleftbig 1−|S11|2/parenrightbig/parenleftbig 1−|S22|2/parenrightbig.( 12.46) Usually an error of a few tenths of a dB or less justifies the unilateral assumption. The expression for GSand GLfor the unilateral case are given by (12.17a) and (12.17c): GS=1−|/Gamma1S|2 |1−S11/Gamma1S|2, GL=1−|/Gamma1L|2 |1−S22/Gamma1L|2. These gains are maximized when /Gamma1S=S∗ 11and/Gamma1L=S∗ 22, resulting in the maximum val- ues given by GSmax=1 1−|S11|2, (12.47a) GLmax=1 1−|S22|2. (12.47b) Define normalized gain factors gSandgLas gS=GS GSmax=1−|/Gamma1S|2 |1−S11/Gamma1S|2/parenleftbig 1−|S11|2/parenrightbig , (12.48a) gL=GL GLmax=1−|/Gamma1L|2 |1−S22/Gamma1L|2/parenleftbig 1−|S22|2/parenrightbig . (12.48b) Then we have that 0 ≤gS≤1 and 0 ≤gL≤1. For fixed values of gSandgL, (12.48) represents circles in the /Gamma1Sor/Gamma1Lplane. To show this, consider (12.48a), which can be expanded to give gS|1−S11/Gamma1S|2=/parenleftbig 1−|/Gamma1S|2/parenrightbig/parenleftbig 1−|S11|2/parenrightbig , /parenleftbig gS|S11|2+1−|S11|2/parenrightbig |/Gamma1S|2−gS/parenleftbig S11/Gamma1S+S∗ 11/Gamma1∗ S/parenrightbig =1−|S11|2−gS, /Gamma1S/Gamma1∗ S−gS/parenleftbig S11/Gamma1S+S∗ 11/Gamma1∗ S/parenrightbig 1−(1−gS)|S11|2=1−|S11|2−gS 1−(1−gS)|S11|2. (12.49) c12MicrowaveAmplifier Pozar September 16, 2011 14:56 12.3 Single-Stage Transistor Amplifier Design 577 Now add/parenleftbig g2 S|S11|2/parenrightbig //bracketleftbig 1−(1−gS)|S11|2/bracketrightbig2to both sides to complete the square: /vextendsingle/vextendsingle/vextendsingle/vextendsingle/Gamma1 S−gSS∗ 11 1−(1−gS)|S11|2/vextendsingle/vextendsingle/vextendsingle/vextendsingle2 =/parenleftbig 1−|S11|2−gS/parenrightbig/bracketleftbig 1−(1−gS)|S11|2/bracketrightbig +g2 S|S11|2 /bracketleftbig 1−(1−gS)|S11|2/bracketrightbig2. Simplifying gives /vextendsingle/vextendsingle/vextendsingle/vextendsingle/Gamma1 S−gSS∗ 11 1−(1−gS)|S11|2/vextendsingle/vextendsingle/vextendsingle/vextendsingle=√ 1−gS/parenleftbig 1−|S11|2/parenrightbig 1−(1−gS)|S11|2,( 12.50) which is the equation of a circle with its center and radius given by CS=gSS∗ 11 1−(1−gS)|S11|2, (12.51a) RS=√1−gS/parenleftbig 1−|S11|2/parenrightbig 1−(1−gS)|S11|2. (12.51b) The results for the constant gain circles of the output section can be shown to be CL=gLS∗ 22 1−(1−gL)|S22|2, (12.52a) RL=√1−gL/parenleftbig 1−|S22|2/parenrightbig 1−(1−gL)|S22|2. (12.52b) The centers of each family of circles lie along straight lines given by the angle of S∗ 11orS∗ 22. Note that when gS(orgL)=1 (maximum gain), the radius RS(orRL)=0, and the center reduces to S∗ 11(orS∗ 22), as expected. In addition, it can be shown that the 0 dB gain circles (GS=1o rGL=1)will always pass through the center of the Smith chart. These results can be used to plot a family of circles of constant gain for the input and output sections. Then/Gamma1Sand/Gamma1Lcan be chosen along these circles to provide the desired gains. The choices for/Gamma1Sand/Gamma1Lare not unique, but it makes sense to choose points close to the center of the Smith chart to minimize mismatch, and thus maximize bandwidth. Alternatively, as we will see in the next section, the input network mismatch can be chosen to provide a low-noise design. EXAMPLE 12.4 AMPLIFIER DESIGN FOR SPECIFIED GAIN Design an amplifier to have a gain of 11 dB at 4.0 GHz. Plot constant-gain circles forGS=2 and 3 dB, and GL=0 and 1 dB. Calculate and plot the input return loss and overall amplifier gain from 3 to 5 GHz. The transistor has the following scattering parameters (Z0=50/Omega1): f(GHz) S11 S12 S21 S22 30 .80/negationslash−90◦02 .8/negationslash100◦0.66/negationslash−50◦ 40 .75/negationslash−120◦02 .5/negationslash80◦0.60/negationslash−70◦ 50 .71/negationslash−140◦02 .3/negationslash60◦0.58/negationslash−85◦ Solution Since S12=0 and|S11|<1 and|S22|<1, the transistor is unilateral and uncon- ditionally stable at each frequency in the above table. From (12.47) we calculate c12MicrowaveAmplifier Pozar September 16, 2011 14:56 578 Chapter 12: Microwave Amplifier Design the maximum matching section gains as GSmax=1 1−|S11|2=2.29=3.6d B , GLmax=1 1−|S22|2=1.56=1.9d B . The gain of the mismatched transistor is G0=|S21|2=6.25=8.0d B , so the maximum unilateral transducer gain is GTUmax=3.6+1.9+8.0=13.5d B . We therefore have 2.5 dB more available gain than is required by the specifica- tions. Next, use (12.48), (12.51), and (12.52) to calculate the following data for the j , 20-2030-3040-4050-5060-6070-7080-8090-90100-100110-110120-120130-130140-140150-150160-160170-1701800.040.040.050.050.060.060.070.070.080.080.090.090.10.10.110.110.120.120.130.130.140.140.150.150.160.160.170.170.180.180.190.190.20.20.210.210.220.220.230.230.240.24 0.250.250.260.260.270.270.280.280.290.290.30.30.310.310.320.320.330.330.340.340.350.350.360.360.370.370.380.380.390.390.40.40.410.410.420.420.430.430.440.440.450.450.460.460.470.470.480.480.490.49 0.00.0AN GLE OFREFLECTION COEFFICIENTINDEGREES—>WAVELENGTHSTOWARDGENERATOR—><— W AVELENGTH STOW ARD LOAD <—II T NDUCTIVEREACTANCECOMPONEN(+X/Zo) ORCAPACTIVESUSCEPTANCE(+jB/Yo)CAPACITIVEREACTANCECOMPONENT(-jX/Zo),ORINDUCTIVESUSCEPTANCE(-jB/Yo) 0.1 0.1 0.1 0.2 0.2 0.2 0.3 0.3 0.3 0.4 0.4 0.4 0.5 0.5 0.5 0.6 0.6 0.6 0.7 0.7 0.7 0.8 0.8 0.8 0.9 0.9 0.9 1.0 1.0 1.01.21.21.21.41.41.41.61.61.61.81.81.82.02.02.03.03.03.04.04.04.05.05.05.01010102020205050500.20.2 0.2 0.20.40.4 0.4 0.40.60.6 0.6 0.60.80.8 0.8 0.81.01.0 1.0 1.0RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) ± (a)* GS = 2 dBGS = 3 dBS11GSmax *S22GLmax GL = 1 dB GL = 0 dBΓS ΓL FIGURE 12.8 Circuit design and frequency response for the transistor amplifier of Example 12.4. (a) Constant-gain circles. c12MicrowaveAmplifier Pozar September 16, 2011 14:56 12.3 Single-Stage Transistor Amplifier Design 579 3.0 3.5 4.0 4.5 5.0–10–55 0101550 Ω0.179 /H9261 0.432 /H9261 0.100 /H92610.045 /H9261 50 Ω50 Ω 50 Ω50 Ω50 Ω GT –RLGT, –RL (dB) Frequency (GHz) (c)(b) FIGURE 12.8 Continued. (b) RF circuit. (c) Transducer gain and return loss. constant-gain circles: GS=3d B gS=0.875 CS=0.706 /negationslash120◦RS=0.166 GS=2d B gS=0.691 CS=0.627 /negationslash120◦RS=0.294 GL=1d B gL=0.806 CL=0.520 /negationslash70◦RL=0.303 GL=0d B gL=0.640 CL=0.440 /negationslash70◦RL=0.440 The constant-gain circles are shown in Figure 12.8a. We choose GS=2 dB and GL=1 dB, for an overall amplifier gain of 11 dB. Then we select /Gamma1Sand/Gamma1L along these circles as shown, to minimize the distance from the center of the chart (this places /Gamma1Sand/Gamma1Lalong the radial lines at 120◦and 70◦, respectively). Thus, /Gamma1S=0.33/negationslash120◦and/Gamma1L=0.22/negationslash70◦, and the matching networks can be designed using shunt stubs as in Example 12.3. The final amplifier circuit is shown in Figure 12.8b. The response was cal- culated using CAD software, with interpolation of the given scattering parameterdata. The results are shown in Figure 12.8c, where it is seen the desired gain of 11 dB is achieved at 4.0 GHz. The bandwidth over which the gain varies by ±1 dB or less is about 25%, which is considerably better than the bandwidth ofthe maximum gain design in Example 12.3. The return loss, however, is not very good, being only about 5 dB at the design frequency. This is due to the deliberate mismatch introduced into the matching sections to achieve the specified gain. ■ c12MicrowaveAmplifier Pozar September 16, 2011 14:56 580 Chapter 12: Microwave Amplifier Design Low-NoiseAmplifie Design Besides stability and gain, another important design consideration for a microwave am- plifier is its noise figure. In receiver applications especially it is often required to have a preamplifier with as low a noise figure as possible since, as we saw in Chapter 10, the firststage of a receiver front end has the dominant effect on the noise performance of the over- all system. Generally it is not possible to obtain both minimum noise figure and maximum gain for an amplifier, so some sort of compromise must be made. This can be done byusing constant-gain circles and circles of constant noi se figure to select a usable trade-off between noise figure and gain. Here we will derive the equations for constant–noise figure circles and show how they are used in transistor amplifier design. As shown in references [1] and [2], the noise figure of a two-port amplifier can be expressed as F=F min+RN GS|YS−Yopt|2,( 12.53) where the following definitions apply: YS=GS+jBS=source admittance presented to transistor. Yopt=optimum source admittance that results in minimum noise figure. Fmin=minimum noise figure of transistor, attained when YS=Yopt. RN=equivalent noise resistance of transistor. GS=real part of source admittance. Instead of the admittance YSandYopt, we can use the reflection coefficients /Gamma1Sand/Gamma1opt, where YS=1 Z01−/Gamma1S 1+/Gamma1S, (12.54a) Yopt=1 Z01−/Gamma1opt 1+/Gamma1opt. (12.54b) /Gamma1Sis the source reflection coefficient defined in Figure 12.1. The quantities Fmin,/Gamma1opt, andRNare characteristics of the particular transistor being used, and are called the noise parameter sof the device; they may be given by the manufacturer or measured. Using (12.54), we can express the quantity |YS−Yopt|2in terms of /Gamma1Sand/Gamma1opt: |YS−Yopt|2=4 Z2 0|/Gamma1S−/Gamma1opt|2 |1+/Gamma1S|2|1+/Gamma1opt|2.( 12.55) In addition, GS=Re{YS}=1 2Z0/parenleftbigg1−/Gamma1S 1+/Gamma1S+1−/Gamma1∗ S 1+/Gamma1∗ S/parenrightbigg =1 Z01−|/Gamma1S|2 |1+/Gamma1S|2.( 12.56) Using these results in (12.53) gives the noise figure as F=Fmin+4RN Z0|/Gamma1S−/Gamma1opt|2 /parenleftbig 1−|/Gamma1S|2/parenrightbig |1+/Gamma1opt|2.( 12.57) For a fixed noise figure Fwe can show that this result defines a circle in the /Gamma1Splane. First define the noise figure parameter, N,a s N=|/Gamma1S−/Gamma1opt|2 1−|/Gamma1S|2=F−Fmin 4RN/Z0|1+/Gamma1opt|2,( 12.58) c12MicrowaveAmplifier Pozar September 16, 2011 14:56 12.3 Single-Stage Transistor Amplifier Design 581 which is a constant for a given noise figure and set of noise parameters. Then rewrite (12.58) as (/Gamma1S−/Gamma1opt)/parenleftbig /Gamma1∗ S−/Gamma1∗ opt/parenrightbig =N/parenleftbig 1−|/Gamma1S|2/parenrightbig , /Gamma1S/Gamma1∗ S−/parenleftbig /Gamma1S/Gamma1∗ opt+/Gamma1∗ S/Gamma1opt/parenrightbig +/Gamma1opt/Gamma1∗ opt=N−N|/Gamma1S|2, /Gamma1S/Gamma1∗ S−/parenleftbig /Gamma1S/Gamma1∗ opt+/Gamma1∗ S/Gamma1opt/parenrightbig N+1=N−|/Gamma1opt|2 N+1. Add|/Gamma1opt|2/(N+1)2to both sides to complete the square to obtain /vextendsingle/vextendsingle/vextendsingle/vextendsingle/Gamma1 S−/Gamma1opt N+1/vextendsingle/vextendsingle/vextendsingle/vextendsingle=/radicalBig N/parenleftbig N+1−|/Gamma1opt|2/parenrightbig (N+1).( 12.59) This result defines circles of constant noise figure with centers at CF=/Gamma1opt N+1,( 12.60a) and radii of RF=/radicalBig N/parenleftbig N+1−|/Gamma1opt|2/parenrightbig N+1.( 12.60b) EXAMPLE 12.5 LOW-NOISE AMPLIFIER DESIGN A GaAs MESFET is biased for minimum noise figure, with the following scat- tering parameters and noise parameters at 4 GHz (Z0=50/Omega1):S11=0.6/negationslash−60◦, S12=0.05/negationslash26◦,S21=1.9/negationslash81◦,S22=0.5/negationslash−60◦,Fmin=1.6d B , /Gamma1opt= 0.62/negationslash100◦, and RN=20/Omega1. For design purposes, assume the device is unilat- eral, and calculate the maximum error in GTresulting from this assumption. Then design an amplifier having a 2.0 dB noise figure with the maximum gain that iscompatible with this noise figure. Solution We first calculate that K=2.78 and /Delta1=0.37, so the device is unconditionally stable even without the approximation of a unilateral device. Next, compute the unilateral figure of merit from (12.46): U=|S 12S21S11S22|/parenleftbig 1−|S11|2/parenrightbig/parenleftbig 1−|S22|2/parenrightbig=0.059. From (12.45) the ratio GT/GTUis bounded as 1 (1+U)2<GT GTU<1 (1−U)2, or 0.891 <GT GTU<1.130. In dB, this is −0.50 <GT−GTU<0.53 dB , c12MicrowaveAmplifier Pozar September 16, 2011 14:56 582 Chapter 12: Microwave Amplifier Design where GTand GTUare now in dB. Thus, we should expect less than about ±0.5 dB error in gain. Now use (12.58) and (12.60) to compute the center and radius of the 2 dB noise figure circle: N=F−Fmin 4RN/Z0|1+/Gamma1opt|2=1.58−1.445 4(20/50)|1+0.62/negationslash100◦|2 =0.0986, CF=/Gamma1opt N+1=0.56/negationslash100◦, RF=/radicalBig N/parenleftbig N+1−|/Gamma1opt|2/parenrightbig N+1=0.24. This noise figure circle is plotted in Figure 12.9a. Minimum noise figure ( Fmin= 1.6 dB) occurs for /Gamma1S=/Gamma1opt=0.62/negationslash100◦. Next we calculate data for several input section constant-gain circles. From (12.51), we have the following results: GS(dB) gS CS RS 1.0 0.805 0.52 /negationslash60◦0.300 1.5 0.904 0.56 /negationslash60◦0.205 1.7 0.946 0.58 /negationslash60◦0.150 These circles are plotted in Figure 12.9a. We see that the GS=1.7 dB gain circle just intersects the F=2 dB noise figure circle, and that any higher gain will result in a worse noise figure. From the Smith chart the optimum solution is /Gamma1S= 0.53/negationslash75◦, yielding GS=1.7 dB and F=2.0d B . For the output section we choose /Gamma1L=S∗ 22=0.5/negationslash60◦for a maximum GLof GL=1 1−|S22|2=1.33=1.25 dB . The transistor gain is G0=|S21|2=3.61=5.58 dB, so the overall transducer gain will be GTU=GS+G0+GL=8.53 dB . A complete AC circuit for the amplifier, using open-circuited shunt stubs in the matching sections, is shown in Figure 12.9b. A computer analysis of the circuitgives a gain of 8.36 dB. ■ Low-NoiseMOSFETAmplifie MOSFETs have a relatively low AC input resistance, making them difficult to impedance match. An external series resistance can be added to the gate, but this approach increases noise power and degrades efficiency. By using a series inductor at the source of a MOSFET, c12MicrowaveAmplifier Pozar September 16, 2011 14:56 12.3 Single-Stage Transistor Amplifier Design 583 j, 20-2030-3040-4050-5060-6070-7080-8090-90100-100110-110120-120130-130140-140150-150160-160170-1701800.040.040.050.050.060.060.070.070.080.080.090.090.10.10.110.110.120.120.130.130.140.140.150.150.160.160.170.170.180.180.190.190.20.20.210.210.220.220.230.230.240.24 0.250.250.260.260.270.270.280.280.290.290.30.30.310.310.320.320.330.330.340.340.350.350.360.360.370.370.380.380.390.390.40.40.410.410.420.420.430.430.440.440.450.450.460.460.470.470.480.480.490.49 0.00.0AN GLE OFREFLECTION COEFFICIENTINDEGREES—>WAVELENGTHSTOWARDGENERATOR—><— W AVELENGTH STOW ARD LOAD <—II T NDUCTIVEREACTANCECOMPONEN(+X/Zo)ORCAPACTIVESUSCEPTANCE(+jB/Yo)CAPACITIVEREACTANCECOMPONENT(-jX/Zo),ORINDUCTIVESUSCEPTANCE(-jB/Yo) 0.1 0.1 0.1 0.2 0.2 0.2 0.3 0.3 0.3 0.4 0.4 0.4 0.5 0.5 0.5 0.6 0.6 0.6 0.7 0.7 0.7 0.8 0.8 0.8 0.9 0.9 0.9 1.0 1.0 1.01.21.21.21.41.41.41.61.61.61.81.81.82.02.02.03.03.03.04.04.04.05.05.05.01010102020205050500.20.2 0.2 0.20.40.4 0.4 0.40.60.6 0.6 0.60.80.8 0.8 0.81.01.0 1.0 1.0RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) ±Noise figure circle Gain circles * 50 Ω0.226 /H9261 0.136 /H9261 0.144 /H92610.25 /H9261 50 Ω50 Ω 50 Ω50 Ω50 Ω (b)(a)F = 2.0 dBCFΓopt ΓSS11 GS = 1.7 dB GS = 1.5 dB GS = 1.0 dBF = 1.6 dB FIGURE 12.9 Circuit design for the transistor amplifier of Example 12.5. (a) Constant-gain and constant–noise figure circles. (b) RF circuit. however, it is possible to create a resistive input impedance without adding noisy resistors. This technique is called inductive source degeneration; similar methods can be used with MESFETs and other transistors. The conceptual circuit is shown in Figure 12.10a, where the inductor Lsis placed in series with the source of the device. The equivalent circuit of the amplifier is shown in Figure 12.10b, where we have simplified the model by assuming the transistor is unilateral, and that Ri,Rds, and Cds can be ignored. For an input current Iat the gate of the transistor, the capacitor voltage is c12MicrowaveAmplifier Pozar September 16, 2011 14:56 584 Chapter 12: Microwave Amplifier Design RD LsLg RD LsLg + –G D SgmVc VcI Cgs (a) (b) FIGURE 12.10 Low-noise MOSFET amplifier. (a) Basic AC circuit. (b) Equivalent circuit using a simplified unilateral FET model. Vc=I/jωCgs. The gate voltage, relative to ground, is then V=I jωCgs+jωLs(I+gmVc) =I/parenleftbigg1 jωCgs+jωLs+gmLs Cgs/parenrightbigg . (12.61) The input impedance at the gate is Z=V I=gmLs Cgs+j/parenleftbigg ωLs−1 ωCgs/parenrightbigg ,( 12.62) showing that the circuit has produced an input resistance of gmLs/Cgs. The series inductor, Ls, can be chosen to match the input resistance of the amplifier to a source impedance, Z0. The inductor at the gate, Lg, can then be chosen to cancel the residual input reactance, which is usually capacitive. The combination of the series matching inductor, the gatecapacitance, and the effective input resistance forms a series RLC resonator. The Qof this resonator is Q=ωL gCgs gmLs.( 12.63) The bandwidth of this circuit may be relatively narrow if this Qis high. EXAMPLE 12.6 LOW-NOISE MOSFET AMPLIFIER DESIGN An Infineon BF1005 n-channel MOSFET transistor having Cgs=2.1 pF and gm=24 mS is used in a 900 MHz low-noise amplifier with inductive source degeneration, as shown in Figure 12.10. Determine the source and gate induc-tors, and estimate the bandwidth of the amplifier. Assume a source impedance of Z 0=50/Omega1. Solution From (12.62), matching the input resistance to Z0determines the source inductor as Ls=Z0Cgs gm=(50)(2.1×10−12) 0.024=4.37 nH . c12MicrowaveAmplifier Pozar September 16, 2011 14:56 12.4 Broadband Transistor Amplifier Design 585 The net reactance at the input is jX=j/parenleftbigg ωLs−1 ωCgs/parenrightbigg =− j59.5/Omega1,s ot h e required series inductance for matching is Lg=−X ω=59.5 2π/parenleftbig 900×106/parenrightbig=10.5n H . From (12.63) we can estimate the Qas Q=ωLgCgs gmLs=1.2, so the bandwidth of the amplifier could be as high as 80%. This value is probably higher than what would be obtained in practice, due to the approximations that have been made in our analysis. ■ 12.4BROADBANDTRANSISTORAMPLIFIERDESIGN The ideal amplifier would have constant gain and good input matching over the desired frequency bandwidth. As the examples of the last section have shown, conjugate match- ing will give maximum gain only over a relatively narrow bandwidth, while designing forless than maximum gain will improve the gain bandwidth, but the input and output ports of the amplifier will be poorly matched. These problems are primarily a result of the fact that microwave transistors typically are not well matched to 50 /Omega1, and large impedance mismatches are governed by the Bode–Fano gain–bandwidth criterion discussed in Chap- ter 5. Another consideration, as shown earlier in this chapter, is that |S 21|decreases with frequency at the rate of 6 dB/octave. For these reasons, special consideration must be givento the problem of designing broadband amplifiers. Some of the common approaches to this problem are listed below; note in each case that an improvement in bandwidth is achieved only at the expense of gain, complexity, or similar factors. rCompensated matching network s: Input and output matching sections can be de- signed to compensate for the gain rolloff in |S21|, but generally at the expense of the input and output matching.rResistive matching networks: Good input and output matching can be obtained by using resistive matching networks, with a corresponding loss in gain and increase in noise figure.rNegative feedback : Negative feedback can be used to flatten the gain response of the transistor, improve the input and output match, and improve the stability of the device. Amplifier bandwidths in excess of a decade are possible with this method,at the expense of gain and noise figure. rBalanced amplifier s: Two amplifiers having 90◦couplers at their input and output can provide good matching over an octave bandwidth, or more. The gain is equal tothat of a single amplifier, however, and the design requires two transistors and twice the DC power. rDistributed amplifier s: Several transistors are cascaded together along a transmis- sion line, giving good gain, matching, and noise figure over a wide bandwidth. The circuit is large, and does not give as much gain as a cascade amplifier with the samenumber of stages. rDifferential amplifier s: Driving two devices in a differential mode, with input sig- nals of opposite polarity, results in an effective series connection of device capac-itance, thus roughly doubling f T. Differential amplifiers can also provide a larger output voltage swing than a single device, and common mode noise rejection. c12MicrowaveAmplifier Pozar September 16, 2011 14:56 586 Chapter 12: Microwave Amplifier Design VB1±VA1 Z0Z0 90° hybrid90° hybridV2–GA GB± VB2±VA2± A BV1+ V1– /H9003 /H9003 FIGURE 12.11 A balanced amplifier using 90◦hybrid couplers. Below we discuss in detail the operation of balanced, differential, and distributed amplifiers. BalancedAmplifie s As we saw in Example 12.4, a fairly flat gain response can be obtained if the amplifier is designed for less than maximum gain, but the input and output matching will be poor. The balanced amplifier circuit solves this problem by using two 90◦couplers to cancel input and output reflections from two identical amplifiers. The basic circuit of a balanced amplifier is shown in Figure 12.11. The first 90◦hybrid coupler divides the input signal into two equal-amplitude components, with a 90◦phase difference, which drive the two amplifiers. The second coupler recombines the amplifier outputs. Because of the phasing properties of the hybrid coupler, reflections from the amplifier inputs cancel at the input to the hybrid, resulting in an improved impedance match; a similar effect occurs at the output of thebalanced amplifier. The gain bandwidth is not improved over that of the single amplifier sections. This type of circuit is more complex than a single-stage amplifier since it requires two hybrid couplers and two separate amplifier sections, but it has a number of interestingadvantages: rThe individual amplifier stages can be optimized for gain flatness or noise figure, without concern for input and output matching.rReflections are absorbed in the coupler terminations, improving input/output match-ing, as well as the stability of the individual amplifiers. rThe circuit provides a graceful degradation of a 6 dB loss in gain if a single amplifier section fails.rBandwidth can be an octave or more, primarily limited by the bandwidth of the couplers. In practice, balanced MMIC amplifiers often use Lange couplers, which are broadband and very compact, but quadrature hybrids and Wilkinson power dividers (with an extra 90◦line on one arm) can also be used. If we assume ideal hybrid couplers, then, with reference to Figure 12.11, the voltages incident at the amplifiers can be written as V+ A1=1√ 2V+ 1, (12.64a) V+ B1=−j√ 2V+ 1, (12.64b) c12MicrowaveAmplifier Pozar September 16, 2011 14:56 12.4 Broadband Transistor Amplifier Design 587 where V+ 1is the incident input voltage. The output voltage can be found as V− 2=−j√ 2V+ A2+1√ 2V+ B2=−j√ 2GAV+ A1+1√ 2GBV+ B1=−j 2V+ 1(GA+GB),(12.65) where GAandGBare the voltage gains of the amplifiers. Then we can write S21as S21=V− 2 V+ 1=−j 2(GA+GB),( 12.66) which shows that the overall gain of the balanced amplifier is the average of the individual amplifier voltage gains. The total reflected voltage at the input can be expressed as V− 1=1√ 2V− A1+−j√ 2V− B1=1√ 2/Gamma1AV+ A1+−j√ 2/Gamma1BV+ B1=1 2V+ 1(/Gamma1A−/Gamma1B).( 12.67) Then we can write S11as S11=V− 1 V+ 1=1 2(/Gamma1A−/Gamma1B).( 12.68) If the amplifiers are identical, then GA=GBand/Gamma1A=/Gamma1B, and (12.68) shows that S11= 0, and (12.66) shows that the gain of the balanced amplifier will be the same as the gain ofan individual amplifier. If one amplifier fails, the overall gain will drop by 6 dB, with the remaining power lost in the coupler terminations. It can also be shown that the noise figure of the balanced amplifier is F=(F A+FB)/2, where FAandFBare the noise figures of the individual amplifiers. EXAMPLE 12.7 PERFORMANCE AND OPTIMIZATION OF A BALANCED AMPLIFIER Use the amplifier of Example 12.4 in a balanced configuration operating from 3 to 5 GHz. Use quadrature hybrids, and plot the gain and return loss over this fre- quency range. Using microwave CAD software, optimize the amplifier matchingnetworks to give 10 dB gain over this band. Solution The amplifier of Example 12.4 was designed for a gain of 11 dB at 4 GHz. As seen from Figure 12.8c, the gain varies by a few dB from 3 to 5 GHz, and the return loss is no better than 5 dB. We can design a quadrature hybrid, according to thediscussion in Chapter 7, to have a center frequency of 4 GHz. Then the balanced amplifier configuration of Figure 12.11 can be modeled using a microwave CAD package, with the results shown in Figure 12.12. Note the dramatic improvementin return loss over the band as compared with the result for the original amplifier in Figure 12.8c. The input matching is best at 4 GHz since this was the design frequency of the coupler; a coupler with better bandwidth will give improved results at the band edges. Also observe that the gain at 4 GHz is still 11 dB, and that it drops by a few dB at the band edges. Most modern microwave CAD software packages have an optimization fea- ture with which a small set of design variables can be adjusted to optimize a particular performance variable. In the present example, we will reduce the gainspecification to 10 dB, and allow the CAD software to adjust the four transmission line stub and line lengths in the amplifier circuit of Figure 12.8b to give the best c12MicrowaveAmplifier Pozar September 16, 2011 14:56 588 Chapter 12: Microwave Amplifier Design 3.0 3.5–30–20–1001020 4.0 Frequency (GHz)GT, –RL (dB) 4.5 5.0–RLGTBefore optimization After optimization FIGURE 12.12 Gain and return loss, before and after optimization, for the balanced amplifier of Example 12.7. fit to this gain over the frequency range 3–5 GHz. Both amplifiers in the balanced circuit remain identical, so we should still see the improved input matching. The results of this optimization are shown in Figure 12.12, where it can be seen that the gain response is much flatter over the operating band. The inputmatch is still very good in the vicinity of the center frequency, with a slightly worse result at the low-frequency end. The optimized stub and line lengths for the amplifier matching networks are listed below: Matching Network Before After Parameter Optimization Optimization Input section stub length 0.100λ 0.109λ Input section line length 0.179λ 0.113λ Output section line length 0.045λ 0.134λ Output section stub length 0.432λ 0.461λ These represent fairly small deviations from the lengths in the original matching networks. ■ DistributedAmplifie s The concept of the distributed amplifier dates back to the 1940s, when it was used in the de- sign of broadband vacuum tube amplifiers. With recent advances in microwave integrated circuit and device processing technology, the distributed amplifier has found new applica- tions in broadband microwave amplifiers. Bandwidths in excess of a decade are possible, with good input and output matching. Distributed amplifiers are not capable of very highgains or very low noise figure, however, and generally are larger than an amplifier having comparable gain over a narrower bandwidth. The basic configuration of a microwave distributed amplifier is shown in Figure 12.13. A cascade of Nidentical FETs have their gates connected to a transmission line having a characteristic impedance Z g, with a spacing of /lscriptg, while the drains are connected to a c12MicrowaveAmplifier Pozar September 16, 2011 14:56 12.4 Broadband Transistor Amplifier Design 589 Zd GN G4 G3 G2 G1   Zd, ld Zg, lg ZgOutput Input FIGURE 12.13 Configuration of an N-stage distributed amplifier. transmission line of characteristic impedance Zd, with a spacing /lscriptd. The operation of the distributed amplifier is very similar to that of the multihole waveguide coupler discussed in Section 7.4. The input signal propagates down the gate line, with each FET tapping off some of the input power. The amplified output signals from the FETs form a travelingwave on the drain line. The propagation constants and lengths of the gate and drain lines are chosen for constructive phasing of the output signals, and the termination impedances on the lines serve to absorb waves traveling in the reverse directions. The gate and draincapacitances of the FET effectively become part of the gate and drain transmission lines, while the gate and drain resistances introduce loss on these lines. This type of circuit is also known as a traveling wave amplifier. Here we will analyze the distributed amplifier in terms of the loaded gate and drain transmission lines [8], although it is also possible to apply the concept of image parameters [9], or to simply model using CAD software. An analytical treatment has the advantage ofillustrating the underlying principles of operation of the amplifier, while the numerical CAD approach is recommended for better accuracy and optimization capabilities. The first step in the analysis of the distributed amplifier is to employ the unilateral (C gd=0)version of the FET equivalent circuit to decompose the circuit of Figure 12.13 into separate loaded transmission lines for the gate and drain terminals. These are shown inFigures 12.14 and 12.15. The gate and drain transmission lines are typically microstrip; the ground conductors are not shown in Figure 12.13, but they are in Figures 12.14 and 12.15. (a) (b)RiViInput Unit cellZg, lg Zg, lg Zg, lg Cgs VC1Ri Cgs VC2+ –+ –+ –+ –Ri Cgs VC3Ri• • • Cgs VC4ZgRi Cgs VCN+ – CgLg Rilg Cgs/lg FIGURE 12.14 (a) Transmission line circuit for the gate line of the distributed amplifier; (b) equivalent circuit of a single unit cell of the gate line. c12MicrowaveAmplifier Pozar September 16, 2011 14:56 590 Chapter 12: Microwave Amplifier Design Unit cell (a)• • • (b)Id1 Id2Rds Rds CdsCds IdNRdsCds Cd IdRdsldCds/ldIo ZdZd, ld LdZd, ld FIGURE 12.15 (a) Transmission line circuit for the drain line of the distributed amplifier; (b) equivalent circuit of a single unit cell of the drain line. The gate and drain lines are isolated except for the coupling through the dependent current sources, where Idn=gmVcn, and are matched at both ends. Figures 12.14b and 12.15b show the equivalent circuits for a single unit cell from the gate and drain lines, respectively. LgandCgare the inductance and capacitance per unit length of the gate transmission line, while Ri/lscriptgandCgs//lscriptgrepresent the equivalent per-unit-length loading due to the FET input resistance Riand gate-to-source capacitance Cgs. Similar definitions apply to the quantities Ld,Cd,Rds/lscriptd, and Cds//lscriptdfor the drain line. Thus we have taken the lumped loading of each FET and distributed its circuit parameters over the transmission lines of each unit cell. This approximation is generally valid when the electrical lengths of the unitcells are small. We can now use basic transmission line theory to find the effective characteristic impedance and propagation constants of the gate and drain lines. For the gate line, theseries impedance and shunt admittance per unit length can be written as Z=jωL g, (12.69a) Y=jωCg+jωCgs//lscriptg 1+jωRiCgs. (12.69b) If we assume that loss can be neglected for the calculation of characteristic impedance, as discussed in Section 2.7, then we have Zg=/radicalbigg Z Y=/radicalBigg Lg Cg+Cgs//lscriptg.( 12.70) For the calculation of the propagation constant we retain the resistive term since this will lead to attenuation: γg=αg+jβg=√ ZY=/radicalBigg jωLg/parenleftbigg jωCg+jωCgs//lscriptg 1+jωRiCgs/parenrightbigg . If we assume small loss, so that ωRiCgs/lessmuch1, then the above result can be simplified as c12MicrowaveAmplifier Pozar September 16, 2011 14:56 12.4 Broadband Transistor Amplifier Design 591 follows: γg=αg+jβg∼=/radicalBig −ω2Lg[Cg+Cgs(1−jωRiCgs)//lscriptg] ∼=ω2RiC2 gsZg 2/lscriptg+jω/radicalBig Lg(Cg+Cgs//lscriptg). (12.71) For the drain line, the series impedance and shunt admittance per unit length are Z=jωLd, (12.72a) Y=1 Rds/lscriptd+jω(Cd+Cds//lscriptd). (12.72b) The characteristic impedance of the drain line can be written as Zd=/radicalbigg Z Y=/radicalBigg Ld Cd+Cds//lscriptd,( 12.73) and the propagation constant can be simplified using the small-loss approximation as γd=αd+jβd=√ ZY=/radicalBigg jωLd/bracketleftbigg1 Rds/lscriptd+jω(Cd+Cds//lscriptd)/bracketrightbigg ∼=Zd 2Rds/lscriptd+jω/radicalbig Ld(Cd+Cds//lscriptd). (12.74) For an incident input voltage, Vi, the voltage on the gate-to-source capacitance of the nth FET can be written as Vcn=Vie−(n−1)γ g/lscriptg/parenleftbigg1 1+jωRiCgs/parenrightbigg ,( 12.75) for a phase reference at the first transistor. The factor in parentheses in (12.75) accounts for voltage division between RiandCgs; for typical FET parameters ωRiCgs/lessmuch1, so this factor can be approximated as unity over the bandwidth of the amplifier. The output current on the drain line can be found by recognizing that each current generator contributes wavesof the form (−1/2) I dne±γdzin each direction. Since Idn=gmVcn, the total output current at the Nth terminal of the drain line is Io=−1 2n/summationdisplay n=1Idne−(N−n)γd/lscriptd=−gmVi 2e−Nγd/lscriptdeγg/lscriptgN/summationdisplay n=1e−n(γg/lscriptg−γd/lscriptd).( 12.76) The terms in the summation will add in phase only when βg/lscriptg=βd/lscriptd, so that the phase delays on the gate and drain lines are synchronized. There is also a backward traveling wave component on the drain line, but the individual contributions to this wave will not bein phase, and therefore they at least partially cancel; the residual will be absorbed in the termination Z d. Use of the summation formula N/summationdisplay n=1xn=xN+1−x x−1 allows (12.76) to be simplified as follows: Io=−gmVi 2eγd/lscriptd/parenleftbig e−Nγg/lscriptg−e−Nγd/lscriptd/parenrightbig e−(γg/lscriptg−γd/lscriptd)−1=−gmVi 2e−Nγg/lscriptg−e−Nγd/lscriptd e−γg/lscriptg−e−γd/lscriptd.( 12.77) c12MicrowaveAmplifier Pozar September 16, 2011 14:56 592 Chapter 12: Microwave Amplifier Design For matched input and output ports, the amplifier gain can be calculated as G=Pout Pin=1 2|Io|2Zd 1 2|Vi|2/Zg=g2 mZdZg 4/vextendsingle/vextendsingle/vextendsingle/vextendsinglee−Nγg/lscriptg−e−Nγd/lscriptd e−γg/lscriptg−e−γd/lscriptd/vextendsingle/vextendsingle/vextendsingle/vextendsingle2 .( 12.78) Applying the synchronization condition that βg/lscriptg=βd/lscriptdallows this result to be further simplified to G=g2 mZdZg 4(e−Nαg/lscriptg−e−Nαd/lscriptd)2 (e−αg/lscriptg−e−αd/lscriptd)2.( 12.79) If the losses are small, the denominator in (12.79) can be approximated as (αg/lscriptg−αd/lscriptd). Several interesting aspects of the distributed amplifier can be deduced from the gain expression of (12.79). For the ideal case of a lossless amplifier (α g=αd=0),t h eg a i n reduces to G=g2 mZdZgN2 4, showing that gain increases as N2. This is in contrast to the gain of a cascade of Namplifier stages, which increases as (G0)N. When loss is present, (12.79) shows that the gain of a distributed amplifier approaches zero as N→∞ . This surprising behavior is explained by the fact that the input voltage on the gate line decays exponentially, so the FETs at the end of the amplifier receive no input signal; similarly, the amplified signals from the FETs near the beginning of the amplifier are attenuated along the drain line. The multiplicative increase in gain with Nis not enough to compensate for an exponential decay for large N. This implies that, for a given set of transistor parameters, there will be an optimum value of Nthat maximizes the gain of a distributed amplifier. This can be found by differentiating (12.79) with respect to Nand setting the result to zero to obtain Nopt=ln(α g/lscriptg/αd/lscriptd) αg/lscriptg−αd/lscriptd.( 12.80) This result depends on frequency, the device parameters, and the line lengths through the attenuation constants given in (12.71) and (12.74). EXAMPLE 12.8 DISTRIBUTED AMPLIFIER PERFORMANCE Use (12.79) to calculate the gain of a distributed amplifier from 1 to 18 GHz forN=2, 4, 8, and 16 stages. Assume Zd=Zg=Z0=50/Omega1and the follow- ing FET parameters: Ri=5/Omega1,Rds=250/Omega1,Cgs=0.30 pF, and gm=30 mS. Find the optimum value of Nthat will give maximum gain at 16 GHz. Solution We use (12.71) and (12.74) to evaluate the attenuation constants αgandαd, and then compute the gain versus frequency and Nusing (12.79). Note that the prod- uctsαg/lscriptgandαd/lscriptdare independent of /lscriptgand/lscriptd: αg/lscriptg=ω2RiC2 gsZ0 2, αd/lscriptd=Z0 2Rds. c12MicrowaveAmplifier Pozar September 16, 2011 14:56 12.4 Broadband Transistor Amplifier Design 593 2 0Gain (dB) 024681012141618 468 1 0 Frequency (GHz)12N=8 N=4 N=2 14 16 18N=16 FIGURE 12.16 Gain versus frequency for the distributed amplifier of Example 12.8. The results are shown in Figure 12.16. Observe that the gain drops off with fre- quency faster for larger N, and that at high frequencies the gain for N=16 is less than the gain for smaller N. The optimum size for maximum gain at 16 GHz can be calculated using (12.80). At 16 GHz we have αg/lscriptg=0.100 and αd/lscriptd=0.114. The optimum size is then Nopt=ln(α g/lscriptg/αd/lscriptd) αg/lscriptg−αd/lscriptd=ln(0.100/0.114 ) 0.100 −0.114=9.4, or about nine stages. Finally, note that ωRiCgs=0.17 at 18 GHz, justifying the approximation of unity for the voltage divider factor of (12.75). ■ DifferentialAmplifie s The amplifiers considered above are single-ended circuits, meaning that the input and out- put signals are referenced to a common ground. In contrast, a differential amplifier uses balanced inputs and outputs, meaning that there are two signal lines, with opposite po-larities, at each port. Figure 12.17 shows the symbols commonly used for single-ended and differential amplifiers. Differential circuits have several advantages over single-ended (a) (b)+ –+ – FIGURE 12.17 (a) Single-ended amplifier, with symbols denoting unbalanced input and output lines. (b) Differential amplifier, having balanced input and output lines. c12MicrowaveAmplifier Pozar September 16, 2011 14:56 594 Chapter 12: Microwave Amplifier Design (a) (b)+ +λ/4 λ/4– –unbalancedunbalanced balanced balanced FIGURE 12.18 Balun circuits. (a) A transformer balun. (b) The Marchand balun. circuits, including cancellation of interference that is common to both signal lines. Such common mode interference is frequently a problem with sensitive receiver circuitry on highly integrated monolithic circuits, and for this reason many of the circuits used in mod- ern RFICs use differential topologies. Another advantage of differential amplifiers is thatthey can provide output voltage swings that are approximately double that obtained with a single-ended amplifier. A disadvantage of differential circuits is that they use roughly twice the device count as the single-ended equivalents, and more associated bias power. A differential amplifier can be constructed using two single-ended amplifiers and 180 ◦ hybrids at the input and output to split and then recombine the signals (similar to the bal- anced amplifier of Section 12.4 that used 90◦hybrids). In this case the initial input and final output signals at the hybrids would be single ended (referenced to ground). Such am- plifiers are sometimes referred to as pseudo-differential [5], in contrast to fully differential amplifiers, which have balanced input and output signals. In general, a balun (balanced-to- unbalanced) circuit is used to transition from an unbalanced signal to a balanced signal (or vice versa). At low frequencies a simple transformer can be used as a balun, as shown inFigure 12.18a. At higher frequencies, a 180 ◦hybrid coupler can be used as a balun, with the unbalanced port at the difference input port of the hybrid, and the two output ports providing the balanced port. Various types of coupled line circuits can also provide a balunfunction, with one of the most popular being the Marchand balun, shown in Figure 12.18b. Figure 12.19 shows the AC circuit of a differential amplifier using two FETs. The balanced input signal is applied to the gates of the devices, and the balanced output signalis formed across the drains. In practice, an additional transistor is often used at the device RD RsVDD RD Vo+ Vi+Vo– Vi– FIGURE 12.19 Differential amplifier circuit using two FETs. The source resistance Rsmay model a current source. c12MicrowaveAmplifier Pozar September 16, 2011 14:56 12.4 Broadband Transistor Amplifier Design 595 Vi+Vo+ Vi–Vo– Ri Ri + –+ –Vc+Vc–CgsId+ Rds Rds RD RD (a) Vi+Vo+ Vi–Vo– Ri Ri + –+ –Vc+Vc–Cgs CgsId+ Rds 2Rs 2RsRds RD RD (b)CgsId – Id – FIGURE 12.20 Equivalent circuits for the differential amplifier. (a) Equivalent circuit model for the differential (odd) mode. (b) Equivalent circuit model for the common (even) mode. sources to provide a current source; this is modeled by the resistor Rs. Usually, the desired input to a differential amplifier consists of equal-amplitude signals with opposite polaritiesat the two gates, forming an odd-mode excitation. An interference signal, however, will usually appear as equal-amplitude signals with the same polarity at the inputs, forming an even-mode excitation. These modes are also referred to as the differential mode and the common mode, respectively. We can analyze the differential amplifier by decomposing an arbitrary input into the superposition of an odd mode and an even mode, similar to the analysis we used previously for the quadrature hybrid and other symmetric circuits. First consider differential (odd) mode excitation, which corresponds to the usual mode of operation for the amplifier. The equivalent circuit is shown in Figure 12.20a, where the unilateral FET model has been used. The input signals in this case are V + i=Viand V− i=−Vi; this antisymmetry establishes a zero potential at the midplane of the circuit, so there is a virtual ground at the sources, and the resistor Rscan be removed. The voltages on the capacitors are V± c=±Vi 1+jωRiCgs,( 12.81) and the output voltages on the drains are V± o=− I± dRDRds RD+Rds=∓VigmRDRds (1+jωRiCgs)(RD+Rds).( 12.82) c12MicrowaveAmplifier Pozar September 16, 2011 14:56 596 Chapter 12: Microwave Amplifier Design The voltage gain for the differential (odd) mode is then Ad=V+ o−V− o V+ i−V− i=−gmRDRds (1+jωRiCds)(RD+Rds).( 12.83) For the common (even) mode, the input signals are V+ i=V− i=Vi, and the equivalent circuit is as shown in Figure 12.20b. Due to the symmetry of the excitation, no current flows between the sources of the two devices, so the circuit can be bisected as shown, with the original resistor Rsbeing split into two resistors of 2 Rs. The voltages on the capacitors are V± c=Vi 1+jωCgs(Ri+2Rs).( 12.84) The voltage across either current source is VI=− IdRds(RD+2Rs) Rds+RD+2Rs, so the output voltages on the drains are V± o=VIRD RD+2Rs=− IdRdsRD Rds+RD+2Rs =−VigmRdsRD [1+jωCgs(Ri+2Rs)](Rds+RD+2Rs). (12.85) The voltage gain for the common (even) mode is then Ac=V+ o+V− o V+ i+V− i=−gmRdsRD [1+jωCgs(Ri+2Rs)](Rds+RD+2Rs).( 12.86) Thecommon mode rejection ratio (CMRR) of an amplifier is defined as the ratio of the differential voltage gain to the common mode voltage gain, and is a measure of how well adifferential amplifier can provide cancellation of a common mode interference signal. For the differential amplifier considered here, the common mode rejection ratio is CMRR =A d Ac=(Rds+RD+2Rs) (Rds+RD)1+jωCgs(Ri+2Rs) 1+jωCgsRi =/parenleftbigg 1+2Rs Rds+RD/parenrightbigg/parenleftbigg 1+2jωCgsRs 1+jωCgsRi/parenrightbigg . (12.87) From this result we see that if Rs=0w eh a v eC M R R= 1, which provides no common mode rejection. This is because the two circuits of Figure 12. 20 are identical when Rs=0. When Rs→∞ , however (which is the case for an ideal current source feeding the FET sources), we have CMRR →∞ , providing cancellation of the common mode signal. 12.5POWERAMPLIFIERS Power amplifiers are used in the final stages of radar and radio transmitters to increase the radiated power level. Typical output powers may be on the order of 100–500 mW for mobile voice or data communications systems, or in the range of 1–100 W for radar orfixed point radio systems. Important considerations for RF and microwave power ampli- fiers are efficiency, gain, intermodulation distortion, and thermal effects. Single transistors c12MicrowaveAmplifier Pozar September 16, 2011 14:56 12.5 Power Amplifiers 597 can provide output powers of 10–100 W at UHF frequencies, while devices at higher fre- quencies are generally limited to output powers less than 10 W. Various power-combining techniques can be used in conjunction with multiple transistors if higher output powers are required. So far we have considered only small-signal amplifier s, where the input signal power is low enough that the transistor can be assumed to operate as a linear device. The scattering parameters of linear devices are well defined and do not depend on the input power levelor output load impedance, a fact that greatly simplifies the design of fixed-gain and low- noise amplifiers. For high input powers (e.g., in the range of the 1 dB compression point or third-order intercept point), transistors do not behave linearly. In this case the impedances seen at the input and output of the transistor will depend on the input power level, and this greatly complicates the design of power amplifiers. CharacteristicsofPowerAmplifie sandAmplifie Classes The power amplifier is usually the primary consumer of DC power in most hand-held wireless devices, so amplifier efficiency is an important consideration. One measure of amplifier efficiency is the ratio of RF output power to DC input power: η=P out PDC.( 12.88) This quantity is sometimes referred to as drain efficiency (orcollector efficiency). One drawback of this definition is that it does not account for the RF power delivered at the input to the amplifier. Since most power amplifiers have relatively low gains, the efficiency of (12.88) tends to overrate the actual efficiency. A better measure that includes the effectof input power is the power added efficiency , defined as η PAE=PAE=Pout−Pin PDC=/parenleftbigg 1−1 G/parenrightbiggPout PDC=/parenleftbigg 1−1 G/parenrightbigg η, ( 12.89) where Gis the power gain of the amplifier. Silicon bipolar junction transistor amplifiers in the cellular telephone band of 800–900 MHz band have power added efficiencies on the order of 80%, but efficiency drops quickly with increasing frequency. Power amplifiers are often designed to provide the best efficiency, even if this means that the resulting gain isless than the maximum possible. Another useful parameter for power amplifiers is the compressed gain, G 1, defined as the gain of the amplifier at the 1 dB compression point. Thus, if G0is the small-signal (linear) power gain, we have G1(dB)=G0(dB)−1.( 12.90) As we have seen in Chapter 10, nonlinearities can lead to the generation of spurious fre- quencies and intermodulation distortion. This can be a serious issue in wireless transmit- ters, especially in a multicarrier system, where spurious signals may appear in adjacent channels. Linearity is also critical for nonconstant envelope modulations, such as ampli- tude shift keying and higher order quadrature amplitude modulation methods. Class A amplifiers are inherently linear circuits, where the transistor is biased to con- duct over the entire range of the input signal cycle. Because of this, class A amplifiers have a theoretical maximum efficiency of 50%. Most small-signal and low-noise amplifiers op- erate as class A circuits. In contrast, the transistor in a class B amplifier is biased to conductonly during one-half of the input signal cycle. Usually two complementary transistors are operated in a class B push-pull amplifier to provide amplification over the entire cycle. c12MicrowaveAmplifier Pozar September 16, 2011 14:56 598 Chapter 12: Microwave Amplifier Design The theoretical efficiency of a class B amplifier is 78%. Class C amplifiers are operated with the transistor near cutoff for more than half of the input signal cycle, and generally use a resonant circuit in the output stage to recover the fundamental. Class C amplifiers can achieve efficiencies near 100% but can only be used with constant envelope modulations.Higher classes, such as class D, E, F, and S, use the transistor as a switch to pump a highly resonant tank circuit, and may achieve very high efficiencies. The majority of communica- tions transmitters operating at UHF frequencies or above rely on class A, AB, or B poweramplifiers because of the need for low distortion products. Large-SignalCharacterizationofTransistors A transistor behaves linearly for signal powers well below the 1 dB compression point (IP 1d B), and so the small-signal scattering parameters should not depend on either the in- put power level or the output termination impedance. However, for power levels compara- ble to or greater than IP 1d B, where the nonlinearity of the transistor becomes apparent, the measured scattering parameters will depend on input power level and the output termina-tion impedance (as well as frequency, bias conditions, and temperature). Thus large-signal scattering parameters are not uniquely defined and do not satisfy linearity, and cannot be used in place of small-signal parameters. (For device stability calculations, however, small-signal scattering parameters can generally be used with good results.) A more useful way to characterize transistors under large-signal operating conditions is to measure the gain and output power as a function of source and load impedances. Oneway of doing this is to determine the large-signal source and load reflection coefficients, /Gamma1 SPand/Gamma1LP, (or impedances, ZSPandZLP)that maximize power gain for a particular output power (often chosen as OP 1d B), and versus frequency. Table 12.1 shows typical large-signal source and load reflection coefficients for an npnsilicon bipolar power tran- sistor, along with the small-signal scattering parameters. Another way of characterizing the large-signal behavior of a transistor is to plot con- tours of constant power output on a Smith chart as a function of the load reflection coeffi- cient,/Gamma1LP, with the transistor conjugately matched at its input. These are called load-pull contour s, and they can be obtained using an automated measurement set-up with computer- controlled electromechanical stub tuners. A typical set of load-pull contours is shown in Figure 12.21. Load-pull contours are similar in function to the constant-gain contours ofSection 12.3, but are not perfect circles due to the nonlinearities of the device. Nonlinear equivalent circuit models can also be developed and used to predict the large-signal performance of FETs and BJTs [10]. The dominant nonlinear parameters for amicrowave FET are C gs,gm,Cgd, and Rds. An important consideration in modeling large- signal transistors is the fact that most parameters are dependent on temperature, which of course increases with output power. Equivalent circuit models can be very useful whencombined with computer-aided design software. TABLE 12.1 Small-Signal Scattering Parameters and Large-Signal Reflection Coefficients (Silicon Bipolar Junction Power Transistor) f(MHz) S11 S12 S21 S22 /Gamma1SP /Gamma1LP G(dB) 800 0.76 /negationslash176◦4.10/negationslash76◦0.065 /negationslash49◦0.35/negationslash−163◦0.856 /negationslash−167◦0.455 /negationslash129◦13.5 900 0.76 /negationslash172◦3.42/negationslash72◦0.073 /negationslash52◦0.35/negationslash−167◦0.747 /negationslash−177◦0.478 /negationslash161◦12.0 1000 0.76 /negationslash169◦3.08/negationslash69◦0.079 /negationslash53◦0.36/negationslash−169◦0.797 /negationslash−187◦0.491 /negationslash185◦10.0 c12MicrowaveAmplifier Pozar September 16, 2011 14:56 12.5 Power Amplifiers 599 FIGURE 12.21 Constant–output power contours versus load impedance for a typical power FET. DesignofClassAPowerAmplifie s In this section we will discuss the use of large-signal parameters for the design of class A amplifiers. Since class A amplifiers are ideally linear, it is sometimes possible to use small- signal scattering parameters for design, but better results are usually obtained if large- signal parameters are available. As with small-signal amplifier design, the first step is tocheck the stability of the device. Since instabilities begin at low signal levels, small-signal scattering parameters can be used for this purpose. Stability is especially important for power amplifiers, as high-power oscillations can easily damage active devices and relatedcircuitry. The transistor should be chosen on the basis of frequency range and power output, ideally with about 20% more power capacity than is required by the design. Silicon bipolar transistors have higher power outputs than GaAs FETs at frequencies up to a few GHz, and are generally cheaper; GaN HBTs are becoming very popular for high-power applica-tions at RF and low microwave frequencies. Good thermal contact of the transistor package to a heat sink is essential for any amplifier with more than a few tenths of a watt power output. Input matching networks may be designed for maximum power transfer (conju-gate matching), while output matching networks are designed for maximum output power (as derived from /Gamma1 LP). The optimum values of source and load reflection coefficients c12MicrowaveAmplifier Pozar September 16, 2011 14:56 600 Chapter 12: Microwave Amplifier Design FIGURE 12.22 Photograph of a three-stage Ku-band GaN MMIC amplifier. Courtesy of Raytheon Company, Waltham, Mass. are different from those obtained from small-signal scattering parameters via (12.40). Low-loss matching elements are important for good efficiency, particularly in the output stage, where currents are highest. Internally matched chip transistors are sometimes avail- able and have the advantage of reducing the effect of parasitic package reactances, thus im-proving efficiency and bandwidth. A photograph of a GaN power amplifier chip is shown in Figure 12.22. EXAMPLE 12.9 DESIGN OF A CLASS A POWER AMPLIFIER Design a power amplifier at 2.3 GHz using a Nitronex NPT25100 GaN HEMT transistor, with an output power of 10 W. The scattering parameters of the transis- tor for VDS=28 V and ID=600 mA are as follows: S11=0.593 /negationslash178◦,S12= 0.009 /negationslash−127◦,S21=1.77/negationslash−106◦, and S22=0.958 /negationslash175◦, and the optimum large- signal source and load impedances are ZSP=10−j3/Omega1andZLP=2.5− j2.3/Omega1. For an output power of 10 W, the power gain is 16.4 dB and the drain efficiency is 26%. Design input and output impedance matching sections for thetransistor, and find the required input power, the required DC drain current, and the power added efficiency. Solution First establish the stability of the device. Using the small-signal scattering param- eters in (12.28) and (12.29) gives |/Delta1|=| S 11S22−S12S21|=0.579 <1, K=1−|S11|2−|S22|2+|/Delta1|2 2|S12S21|=2.08>1, showing that the device is unconditionally stable. Converting the large-signal source and load impedances to reflection coeffi- cients gives /Gamma1SP=0.668 /negationslash187◦, /Gamma1LP=0.905 /negationslash−175◦. c12MicrowaveAmplifier Pozar September 16, 2011 14:56 References 601 0.1690.0570.028 0.213 FIGURE 12.23 RF circuit for the amplifier of Example 12.9. For comparison, using the small-signal scattering parameters in (12.40) to find the source and load reflection coefficients for conjugate matching gives /Gamma1S=B1±/radicalBig B2 1−4|C1|2 2C1=0.508 /negationslash166◦, /Gamma1L=B2±/radicalBig B2 2−4|C2|2 2C2=0.954 /negationslash−176◦. Note that these values are approximately equal to the large-signal values /Gamma1SP and/Gamma1LP, but not exactly, due to the fact that the scattering parameters used to calculate /Gamma1Sand/Gamma1Ldo not apply for large power levels. We should use the large- signal reflection coefficients to design the input and output matching networks. The AC amplifier circuit is shown in Figure 12.23. For an output power of 10 W, the required input drive power is Pin=Pout(dBm)−G(dB)=10 log(10,000 )−16.4=23.6d B m =229 mW. The DC input power can be found from the drain efficiency as PDC=Pout/η= 38.5 W, so the DC drain current is ID=PDC/VDS=1.37 A. The power added efficiency of the amplifier can be found from (12.89) to be ηPAE=Pout−Pin PDC=10.0−0.229 38.5=25%.■ REFERENCES [1] G. D. Vendelin, A. M. Pavio, and U. L. Rohde, Microwave Circuit Design Using Linear and Nonlin- ear Techniques, John Wiley & Sons, New York, 1990. [2] G. Gonzalez, Microwave Transistor Amplifier s: Analysis and Design, 2nd edition, Prentice-Hall, Upper Saddle River, N.J., 1997. [3] R. Ludwig and P. Bretchko, RF Circuit Design: Theory and Application s, Prentice-Hall, Upper Sad- dle River, N.J., 2000. [ 4 ] T .H .L e e ,The Design of CMOS Radio-Frequency Integrated Circuits, 2nd edition, Cambridge Uni- versity Press, Cambridge, 2004. [5] M. Steer, Microwave and RF Design: A Systems Approach , SciTech, Raleigh, N.C., 2010. [6] M. Ohtomo, “Proviso on the Unconditional Stability Criteria for Linear Twoports,” IEEE Tran sac- tions on Microwave Theory and Techniques, vol. MTT-43, pp. 1197–1200, May 1995. [7] M. L. Edwards and J. H. Sinksy, “A New Criteria for Linear 2-Port Stability Using a Single Geomet- rically Derived Parameter,” IEEE Tran sactions on Microwave Theory and Techniques, vol. MTT-40, pp. 2803–2811, December 1992. [8] Y . Ayasli, R. L. Mozzi, J. L. V orhous, L. D. Reynolds, and R. A. Pucel, “A Monolithic GaAs 1– 13 GHz Traveling-Wave Amplifier,” IEEE Tran sactions on Microwave Theory and Techniques,v o l . MTT-30, pp. 976–981, July 1982. c12MicrowaveAmplifier Pozar September 16, 2011 14:56 602 Chapter 12: Microwave Amplifier Design [ 9 ] J .B .B e y e r ,S .N .P r a s a d ,R .C .B e c k e r ,J .E .N o r d m a n ,a n dG .K .H o h e n w a r t e r ,“ M E S F E TD i s - tributed Amplifier Design Guidelines,” IEEE Tran sactions on Microwave Theory and Techniques, vol. MTT-32, pp. 268–275, March 1984. [10] W. R. Curtice and M. Ettenberg, “A Nonlinear GaAs FET Model for Use in the Design of Output Circuits for Power Amplifiers,” IEEE Tran sactions on Microwave Theory and Techniques, vol. MTT- 33, pp. 1383–1394, December 1985. PROBLEMS 12.1 Consider the microwave network shown below, consisting of a 50 /Omega1source, a 50 /Omega1, 3 dB matched attenuator, and a 50 /Omega1load. (a) Compute the power gain, the available power gain, and the transducer power gain. (b) How do these gains change if the load is changed to 25 /Omega1? (c) How do these gains change if the source impedance is changed to 25 /Omega1? Zs = 50 Ω ZL = 50 Ω, 25 ΩAttenuator 3 dB 50 Ω 12.2 The Infineon BFP640F SiGe HBT has the following scattering parameters at 1.0 GHz (Z0=50/Omega1): S11=0.91/negationslash−44◦,S12=0.06/negationslash68◦,S21=3.92/negationslash149◦,a n d S22=0.93/negationslash−17◦. For the transistor in the configuration of Figure 12.1, with no matching networks: (a) Compute the power gain, availablepower gain, and transducer gain for Z S=ZL=50/Omega1. (b) Can you find ZSandZL(or/Gamma1Sand/Gamma1L) to maximize each of these gains (for this case, assume the device is unilateral, with S12=0)? 12.3 An amplifier uses a GaAs HBT device having the following scattering parameters (Z0=50/Omega1): S11=0.61/negationslash−170◦,S12=0.06/negationslash70◦,S21=2.3/negationslash80◦,a n d S22=0.72/negationslash−25◦. The input of the tran- sistor is connected to a source with Vs=2V( p e a k )a n d ZS=25/Omega1, and the output of the transistor is connected to a load of ZL=100/Omega1. (a) What are the power gain, the available power gain, the transducer power gain, and the unilateral transducer power gain? (b) Compute the available power from the source, and the power delivered to the load. 12.4 A SiGe HBT device has the following scattering parameters at 2.0 GHz: S11=0.880 /negationslash−115◦,S12= 0.029 /negationslash31◦,S21=9.40/negationslash110◦,andS22=0.328 /negationslash−67◦. Determine the stability of the device, and plot the stability circles if the device is potentially unstable. 12.5 The scattering parameters of a GaN HEMT device are given at four frequencies in Table 11.8. Use theK−/Delta1test to determine the stability of this transistor at each frequency. 12.6 Use the µ-parameter test to determine which of the following devices are unconditionally stable and, of those, which has the greatest stability: Device S11 S12 S21 S22 A 0.34 /negationslash−170◦0.06/negationslash70◦4.3/negationslash80◦0.45/negationslash−25◦ B 0.75 /negationslash−60◦0.2/negationslash70◦5.0/negationslash90◦0.51/negationslash60◦ C 0.65 /negationslash−140◦0.04/negationslash60◦2.4/negationslash50◦0.70/negationslash−65◦ 12.7 Show that for a unilateral device, where S12=0, theµ-parameter test of (12.30) implies that |S11|< 1a n d| S22|<1 for unconditional stability. 12.8 Prove that the condition for a positive discriminant in (12.40a), that is, B2 1>4|C1|2, is equivalent to the condition that K2>1. c12MicrowaveAmplifier Pozar September 16, 2011 14:56 Problems 603 12.9 Using the scattering parameter data for the GaAs MESFET given in Table 11.7, design an amplifier for maximum gain at 8.0 GHz. Design matching sections using open-circuited shunt stubs, and com- pute the gain. Check the stability of the resulting design using CAD modeling over the frequency range of 1–12 GHz, or by using stability circles at a few frequencies. 12.10 Consider the impedance matching network shown at left below, where a load ZL(/Gamma1L)at port 2 is matched to a source impedance Z0at port 1. Show that the same network will present the impedance Z=Z∗ Lat port 2 when port 1 is terminated with Z0, as shown in the figure below at right. As- sume the matching network is reciprocal and lossless. This relationship allows the impedance tuning techniques of Chapter 5 to be used to design the input and output matching networks for an amplifier. 1 2 [S] L/H9003ZL Z0 Z01 2 [S] Z, /H9003 12.11 Design an amplifier with maximum GTUusing a transistor with the following scattering parameters (Z0=50/Omega1)at 6.0 GHz: S11=0.61/negationslash−170◦,S12=0,S21=2.24/negationslash32◦,a n d S22=0.72/negationslash−83◦. Design L-section matching sections using lumped elements. 12.12 Design an amplifier to have a gain of 10 dB at 6.0 GHz, using a transistor with the following scattering parameters (Z0=50/Omega1):S11=0.61/negationslash−170◦,S12=0,S21=2.24/negationslash32◦,a n d S22=0.72/negationslash−83◦. Plot (and use) constant-gain circles for GS=1d Ba n d GL=2 dB. Use matching sections with open-circuited shunt stubs. 12.13 Compute the unilateral figure of merit for the transistor of Problem 12.4. What is the maximum error in the transducer gain if an amplifier is designed assuming the device is unilateral? 12.14 Show that the 0 dB gain circle for GS(GS=1), defined by (12.51), will pass through the center of the Smith chart. 12.15 A GaAs FET has the following scattering and noise parameters at 8 GHz (Z0=50/Omega1): S11=0.7/negationslash−110◦,S12=0.02/negationslash60◦,S21=3.5/negationslash60◦,S22=0.8/negationslash−70◦,Fmin=2.5d B , /Gamma1opt= 0.70/negationslash120◦,a n d RN=15/Omega1. Design an amplifier with minimum noise figure and maximum pos- sible gain. Use open-circuited shunt stubs in the matching sections. 12.16 A GaAs FET has the following scattering and noise parameters at 6 GHz (Z0=50/Omega1):S11= 0.6/negationslash−60◦,S12=0,S21=2.0/negationslash81◦,S22=0.7/negationslash−60◦,Fmin=2.0d B ,/Gamma1 opt=0.62/negationslash100◦,and RN=20/Omega1. Design an amplifier to have a gain of 6 dB and the minimum noise figure possible with this gain. Use open-circuited shunt stubs in the matching sections. 12.17 Repeat Problem 12.16, but design the amplifier for a noise figure of 2.5 dB and the maximum possible gain that can be achieved with this noise figure. 12.18 Repeat the analysis of the balanced amplifier of Example 12.7 using a 3 dB coupled line hybrid coupler. Use CAD software to optimize the input and output matching networks of the amplifiers to obtain a flat 10 dB gain response from 3 to 5 GHz, and compare the results with those obtained usingthe quadrature hybrid. 12.19 If the individual amplifier stages in a balanced amplifier have mismatches of /Gamma1 Aand/Gamma1Bat their output ports, show that the output mismatch of the balanced amplifier is S22=(/Gamma1A−/Gamma1B)/2. 12.20 Derive the result for the optimum size of a distributed amplifier given in (12.80). 12.21 Consider a distributed amplifier using GaAs MESFETs with the following parameters: Ri=5/Omega1, Rds=200/Omega1,Cgs=0.3p F ,a n d gm=40 mS. Calculate and plot the gain from 0 to 20 GHz for N=4, 8, and 16 sections. Find the optimum value of Nthat will give maximum gain at 16 GHz. Assume Zd=Zg=Z0=50/Omega1. 12.22 Use the transistor data given in Table 12.1 to design a power amplifier at 1 GHz with a power output of 1 W. Design the input and output matching circuits using the given large-signal reflection coefficients.Compute the required input power level. c13OscillatorsAndMixers Pozar September 16, 2011 15:44 Chapter Thirteen Oscillators and Mixers RF and microwave oscillators are found in all modern wireless communications, radar, and remote sensing systems to provide signal sources for frequency conversion and carrier generation. A solid-state oscillator uses an active nonlinear device, such as a diode or tran- sistor, in conjunction with a passive circuit to convert DC to a sinusoidal steady-state RFsignal. Basic transistor oscillator circuits can generally be used at low frequencies, often withcrystal resonators to provide improved frequency stability and low noise performance. At higher frequencies, diodes or transistors biased to a negative resistance operating point can be used with cavity, transmission line, or dielectric resonators to produce fundamental fre-quency oscillations up to 100 GHz. Alternatively, frequency multipliers, in conjunction witha lower frequency source, can be used to produce power at millimeter wave frequencies. Because of the requirement of a nonlinear active device, the rigorous analysis and design of oscillator circuits can be difficult, and is usually carried out today with sophisticated CADtools. In this chapter we begin with an overview of low-frequency transistor oscillator circuits, including the well-known Hartley and Colpitts configurations, as well as crystal-controlled os- cillators. Next we consider oscillators for use at microwave frequencies, which differ from theirlower frequency counterparts primarily due to different transistor characteristics and the abilityto make practical use of negative resistance devices and high- Qmicrowave resonators. We also discuss the important topic of oscillator phase noise. Finally, an introduction to frequency mul- tiplication techniques is given. A related topic is that of frequency conversion, or mixing,s ow e also discuss in this chapter the fundamental operations of frequency up-conversion and down-conversion. Detectors and single-ended mixers using both diodes and transistors are discussed, along with some specialized mixer circuits. Important considerations for oscillators used in RF and microwave systems include the following: rTuning range (specified in MHz/V for voltage-tuned oscillators) rFrequency stability (specified in PPM/◦C) rAM and FM noise (specified in dBc/Hz below carrier, offset from carrier) rHarmonics (specified in dBc below carrier) 604 c13OscillatorsAndMixers Pozar September 16, 2011 15:44 13.1 Rf Oscillators 605 Typical frequency stability requirements can range from 2 to 0.5 PPM/◦C, while phase noise requirements may range from −80 to −110 dBc/Hz at a 10 kHz offset from the carrier. Transistor oscillators generally have lower frequency and power capabilities than diode sources (e.g., tunnel, Gunn, or IMPATT diodes), but offer several advantages over diodes. First, oscillators using transistors are readily compatible with monolithic integrated circuitry, allow-ing easy integration with transistor amplifiers and mixers, while diode devices are often lesscompatible. In addition, a transistor oscillator circuit is much more flexible than a diode source. This is because the negative resistance oscillation mechanism of a diode is determined and lim- ited by the physical characteristics of the device itself, while the operating characteristics of atransistor can be adjusted to a greater degree by the bias point, as well as the source or loadimpedances presented to the device. Transistor oscillators usually allow more control of the frequency of oscillation, temperature stability, and output noise than do diode sources. Transis- tor oscillator circuits also lend themselves well to frequency tuning, phase or injection locking,and various modulation requirements. Transistor sources are relatively efficient but usually arenot capable of very high power outputs. Tunable sources are necessary in many types of electronic warfare systems, frequency- hopping radar and communications systems, and test systems. Transistor oscillators can bemade tunable by using an adjustable element in the resonant load, such as a varactor diode ora magnetically biased YIG sphere. Thus, a voltage-controlled oscillator (VCO) can be made by using a reverse-biased varactor diode in the tank circuit of a transistor oscillator. In a YIG- tuned oscillator (YTO), a single-crystal YIG sphere is used to control the inductance of a coil inthe tank circuit of the oscillator. Since YIG is a ferrimagnetic material, its effective permeabil-ity can be controlled with an external DC magnetic bias field, thus controlling the oscillator frequency. YIG oscillators can be made to tune over a decade or more of bandwidth, while varactor-tuned oscillators are limited to a tuning range of about an octave. YIG-tuned oscilla-tors, however, cannot be tuned as fast as varactor oscillators. 13.1RFOSCILLATORS In the most general sense, an oscillator is a nonlinear circuit that converts DC power to an AC waveform. Most RF oscillators provide sinusoidal outputs, which minimizes undesiredharmonics and noise sidebands. The basic conceptual operation of a sinusoidal oscilla- tor can be described with the linear feedback circuit shown in Figure 13.1. An amplifier with voltage gain Ahas an output voltage V o. This voltage passes through a feedback net- work with a frequency-dependent transfer function H(ω), and is added to the input Viof c13OscillatorsAndMixers Pozar September 16, 2011 15:44 606 Chapter 13: Oscillators and Mixers FIGURE 13.1 Block diagram of a sinusoidal oscillator using an amplifier with a frequency- dependent feedback path. the circuit. The output voltage can be expressed as Vo(ω)=AVi(ω)+H(ω)AVo(ω), (13.1) which can be solved to yield the output voltage in terms of the input voltage as Vo(ω)=A 1−AH(ω)Vi(ω). (13.2) If the denominator of (13.2) becomes zero at a particular frequency, it is possible to achieve a nonzero output voltage for a zero input voltage, thus forming an oscillator. This is known as the Nyquist criterion ,o rt h eBarkhausen criterion . In contrast to the design of an amplifier, where we design to achieve at least conditional stability, oscillator designdepends on an unstable circuit. The oscillator circuit of Figure 13.1 is useful conceptually but provides little help- ful information for the design of practical transistor oscillators. Thus we consider next ageneral analysis of transistor oscillator circuits. GeneralAnalysis There are a large number of possible RF oscillator circuits using bipolar or field effect transistors in either common emitter/source, base/gate, or collector/drain configurations. Various types of feedback networks lead to the well-known Hartley, Colpitts, Clapp , and Pierce oscillator circuits [1–3]. All of these variations can be represented by the general oscillator circuit shown in Figure 13.2. The equivalent circuit on the right-hand side of Figure 13.2 is used to model either a bipolar or a field effect transistor. As discussed in Chapter 10, we have assumed here aunilateral transistor, which is usually a good approximation in practice. We can simplify the analysis by assuming real input and output admittances of the transistor, defined as G iand Go, respectively, with a transistor transconductance gm. The feedback network on the left side of the circuit is formed from three admittances in a bridged-T configuration. These components are usually reactive elements (capacitors or inductors) in order to provide a frequency-selective transfer function with high Q. A common emitter/source configuration can be obtained by setting V2=0, while common base/gate or common collector/drain configurations can be modeled by setting either V1=0o rV4=0, respectively. As shown, the circuit of Figure 13.2 does not include a completed feedback path—this can be achieved by connecting node V3to node V4. Writing Kirchhoff’s equation for the four voltage nodes of the circuit of Figure 13.2 gives the following matrix equation: c13OscillatorsAndMixers Pozar September 16, 2011 15:44 13.1 Rf Oscillators 607 FIGURE 13.2 General circuit for a transistor oscillator. The transistor may be either a bipolar junction transistor or a field effect transistor. This circuit can be used for common emitter/source, base/gate, or collector/drain configurations by grounding V2,V1, orV4, respectively. Feedback is provided by connecting node V3toV4. ⎡ ⎢⎢⎢⎣(Y 1+Y3+Gi) −(Y1+Gi) −Y3 0 −(Y1+Gi+gm)(Y1+Y2+Gi+Go+gm)−Y2 −Go −Y3 −Y2 (Y2+Y3) 0 gm −(Go+gm) 0 Go⎤ ⎥⎥⎥⎦⎡ ⎢⎢⎢⎣V 1 V2 V3 V4⎤ ⎥⎥⎥⎦=0 (13.3) Recall from circuit analysis that if the ith node of the circuit is grounded, so that V i=0, the matrix of (13.3) will be modified by eliminating the ith row and column, reducing the order of the matrix by one. In addition, if two nodes are connected together, the matrix is modified by adding the corresponding rows and columns. OscillatorsUsingaCommonEmitterBJT As a specific example, consider an oscillator using a bipolar junction transistor in a com- mon emitter configuration. In this case we have V2=0, with feedback provided from the collector, so that V3=V4. In addition, the output admittance of the transistor is negligible, so we set Go=0. These conditions serve to reduce the matrix of (13.3) to the following: /bracketleftbigg (Y1+Y3+Gi)−Y3 (gm−Y3)( Y2+Y3)/bracketrightbigg/bracketleftbigg V1 V/bracketrightbigg =0,( 13.4) where V=V3=V4. If the circuit is to operate as an oscillator, then (13.4) must be satisfied for nonzero values of V1andV, so the determinant of the matrix must be zero. If the feedback net- work consists only of lossless capacitors and inductors, then Y1,Y2, and Y3must be imaginary, so we let Y1=jB1,Y2=jB2, and Y3=jB3. Also recall that the transcon- ductance gm, and transistor input conductance Gi, are real. The determinant of (13.4) then simplifies to /vextendsingle/vextendsingle/vextendsingle/vextendsingleG i+j(B1+B3) −jB3 gm−jB3 j(B2+B3)/vextendsingle/vextendsingle/vextendsingle/vextendsingle=0.( 13.5) c13OscillatorsAndMixers Pozar September 16, 2011 15:44 608 Chapter 13: Oscillators and Mixers Separately equating the real and imaginary parts of the determinant to zero gives two equations: 1 B1+1 B2+1 B3=0, (13.6a) 1 B3+/parenleftbigg 1+gm Gi/parenrightbigg1 B2=0. (13.6b) If we convert susceptances to reactances, and let X1=1/B1,X2=1/B2, and X3=1/B3, then we can write (13.6a) as X1+X2+X3=0.( 13.7a) Using (13.6a) to eliminate B3from (13.6b) reduces that equation to the following: X1=gm GiX2.( 13.7b) Since gmandGiare positive, (13.7b) implies that X1and X2have the same sign, and therefore are either both capacitors or both inductors. Equation (13.7a) then shows that X3must be opposite in sign from X1andX2, and therefore the opposite type of reactive component. This conclusion leads to two of the most commonly used oscillator circuits. IfX1andX2are capacitors and X3is an inductor, we have a Colpitts oscillator. Let X1=−1/ω0C1,X2=−1/ω0C2, and X3=ω0L3. Then (13.7a) becomes −1 ω0/parenleftbigg1 C1+1 C2/parenrightbigg +ω0L3=0, which can be solved for the frequency of oscillation, ω0,a s ω0=/radicalBigg 1 L3/parenleftbiggC1+C2 C1C2/parenrightbigg .( 13.8) Using these same substitutions in (13.7b) gives a necessary condition for oscillation of the Colpitts circuit as C2 C1=gm Gi.( 13.9) The resulting common emitter Colpitts oscillator circuit is shown in Figure 13.3a. Alternatively, if we choose X1andX2to be inductors and X3to be a capacitor, then we have a Hartley oscillator. Let X1=ω0L1,X2=ω0L2, and X3=−1/ω0C3. Then (13.7a) becomes ω0(L1+L2)−1 ω0C3=0, which can be solved for ω0to give ω0=/radicalBigg 1 C3(L1+L2).( 13.10) c13OscillatorsAndMixers Pozar September 16, 2011 15:44 13.1 Rf Oscillators 609 FIGURE 13.3 Transistor oscillator circuits using a common emitter BJT. (a) Colpitts oscillator. (b) Hartley oscillator. These same substitutions used in (13.7b) gives a necessary condition for oscillation of the Hartley circuit as L1 L2=gm Gi.( 13.11) The resulting common emitter Hartley oscillator circuit is shown in Figure 13.3b. OscillatorsUsingaCommonGateFET Next consider an oscillator using an FET in a common gate configuration. In this case V1=0, and again V3=V4provides the feedback path. For an FET the input admittance can be neglected, so we set Gi=0. Then the matrix of (13.3) reduces to /bracketleftbigg (Y1+Y2+gm+Go)−(Y2+Go) −(Go+gm+Y2)( Y2+Y3+Go)/bracketrightbigg/bracketleftbigg V2 V/bracketrightbigg =0,( 13.12) where V=V3=V4. Again we assume the feedback network is composed of lossless re- active elements, so that Y1,Y2, and Y3can be replaced with their susceptances. Setting the determinant of (13.12) to zero then gives /vextendsingle/vextendsingle/vextendsingle/vextendsingle(gm+Go)+j(B1+B2) −Go−jB2 −(Go+gm)−jB2 Go+j(B2+B3)/vextendsingle/vextendsingle/vextendsingle/vextendsingle=0.( 13.13) Equating the real and imaginary parts to zero gives two equations: 1 B1+1 B2+1 B3=0, (13.14a) Go B3+gm B1+Go B1=0. (13.14b) As before, let X1,X2, and X3be the reciprocals of the corresponding susceptances. Then (13.14a) can be rewritten as X1+X2+X3=0.( 13.15a) c13OscillatorsAndMixers Pozar September 16, 2011 15:44 610 Chapter 13: Oscillators and Mixers Using (13.14a) to eliminate B3from (13.14b) reduces that equation to X2 X1=gm Go.( 13.15b) Since gmandGoare positive, (13.15b) shows that X1andX2must have the same sign, while (13.15a) indicates that X3must have the opposite sign. If X1andX2are chosen to be negative, then these elements will be capacitive and X3will be inductive. This corresponds to a Colpitts oscillator. Since (13.15a) is identical to (13.7a), its solution gives the result for the resonant frequency for the common gate Colpitts oscillator as ω0=/radicalBigg 1 L3/parenleftbiggC1+C2 C1C2/parenrightbigg ,( 13.16) which is identical to the result obtained in (13.8) for the common emitter Colpitts oscillator. This is because the resonant frequency is determined by the feedback network, which is identical in both cases. The further condition for oscillation given by (13.15b) reduces to C1 C2=gm Go.( 13.17) If we choose X1andX2to be positive (inductive), then X3will be capacitive, and we have a Hartley oscillator. The resonant frequency of the common gate Hartley oscillator is given by ω0=/radicalBigg 1 C3(L1+L2),( 13.18) which is identical to the result of (13.10) for the common emitter Hartley oscillator. Equa- tion (13.15b) reduces to L2 L1=gm Go.( 13.19) The circuits for common gate Colpitts and Hartley oscillators are similar to the circuits shown in Figure 13.3 if the BJT is replaced with an FET device. PracticalConsiderations It must be emphasized that the above analysis is based on very idealized assumptions, and in practice successful oscillator design requires attention to factors such as the reactancesassociated with the input and output transistor ports, the variation of transistor properties with temperature, transistor bias and decoupling circuitry, and the effect of inductor losses. For these purposes computer-aided design software can be very helpful. The above analysis can be extended to account for more realistic feedback network in- ductors having series resistance, which invariably occurs in practice. For example, consider the case of a common emitter BJT Colpitts oscillator, with the impedance of the inductor given by Z 3=1/Y3=R+jωL3. Substituting into (13.4) and setting the real and imagi- nary parts of the determinant to zero gives the following result for resonant frequency: ω0=/radicalBigg 1 L3/parenleftbigg1 C1+1 C2+GiR C1/parenrightbigg =/radicalBigg 1 L3/parenleftbigg1 C/prime 1+1 C2/parenrightbigg .( 13.20) c13OscillatorsAndMixers Pozar September 16, 2011 15:44 13.1 Rf Oscillators 611 This equation is similar to the result of (13.8) for the lossless inductor, except that C/prime 1is defined as C/prime 1=C1 1+RG i.( 13.21) The corresponding condition for oscillation is R Gi=1+gm/Gi ω2 0C1C2−L3 C1.( 13.22) This result sets the maximum value of the series resistance R; the left-hand side of (13.22) should generally be chosen to be less than the right-hand side to ensure oscillation. EXAMPLE 13.1 COLPITTS OSCILLATOR DESIGN Design a 50 MHz Colpitts oscillator using a bipolar junction transistor in a com- mon emitter configuration with β=gm/Gi=30, and a transistor input resistance ofRi=1/Gi=1200/Omega1. Use an inductor with L3=0.10µH and an unloaded Q of 100. What is the minimum Qof the inductor for which oscillation will be sustained? Solution From (13.20) the series combination of C/prime 1andC2is found to be C/prime 1C2 C/prime 1+C2=1 ω2 0L3=1 (2π)2(50×106)2(0.1×10−6)=100 pF . This value can be obtained in several ways, but here we will choose C/prime 1=C2= 200 pF. From Chapter 6 we know that the unloaded Qof an inductor is related to its series resistance by Q0=ωL/R, so the series resistance of the 0.1 µH inductor is R=ω0L3 Q0=(2π)( 50×106)(0.1×10−6) 100=0.31/Omega1. Then (13.21) gives C1as C1=C/prime 1(1+RG i)=(200 pF )/parenleftbigg 1+0.31 1200/parenrightbigg =200 pF , which we see is essentially unchanged from the value found by neglecting the inductor loss. Using (13.22) with the above values gives R Gi=1+β ω2 0C1C2−L3 C1 (0.31)(1200)<1+30 (2π)2(50×106)2(200×10−12)2−0.1×10−6 200×10−12 372<7852−500=7352 which indicates that the condition for oscillation will be satisfied. This condition can be used to find the minimum unloaded inductor Qby first solving for the c13OscillatorsAndMixers Pozar September 16, 2011 15:44 612 Chapter 13: Oscillators and Mixers maximum value of series resistance R: Rmax=1 Ri/parenleftBigg 1+β ω2 0C1C2−L3 C1/parenrightBigg =7352 1200=6.13/Omega1. So the minimum unloaded Qis Qmin=ω0L3 Rmax=(2π)( 50×106)(0.1×10−6) 6.13=5.1.■ CrystalOscillators As we have seen from the above analysis, the resonant frequency of an oscillator is deter- mined from the condition that a 180◦phase shift occurs between the input and output of the transistor. If the resonant feedback circuit has a high Q, so that there is a very rapid change in the phase shift with frequency, the oscillator will have good frequency stability. Quartz crystals are useful for this purpose, especially at frequencies below a few hundred MHz, where LCresonators seldom have unloaded Qs greater than a few hundred. Quartz crys- tals may have unloaded Qs as high as 100,000 and temperature drift less than 0.001%/C◦. Crystal-controlled oscillators therefore find extensive use as stable frequency sources in RF systems; further stability can be obtained by controlling the temperature of the quartz crystal. A quartz crystal resonator consists of a small, thin sheet of quartz mounted between two metallic plates. Mechanical oscillations can be excited in the crystal through the piezo- electric effect. The equivalent circuit of a quartz crystal near its lowest resonant mode isshown in Figure 13.4a. This circuit has series and parallel resonant frequencies, ω sandωp, given by ωs=1√ LC, (13.23a) ωp=1/radicalBigg L/parenleftbiggC0C C0+C/parenrightbigg. (13.23b) The reactance of the circuit of Figure 13.4a is plotted in Figure 13.4b, where we see that the reactance is inductive in the frequency range between the series and parallel resonances. This is the usual operating point of the crystal, so that the crystal may be used in place ofthe inductor in a Colpitts or Pierce oscillator. A typical crystal oscillator circuit is shown in Figure 13.5. C0 FIGURE 13.4 (a) Equivalent circuit of a crystal resonator. (b) Input reactance of a crystal resonator. c13OscillatorsAndMixers Pozar September 16, 2011 15:44 13.2 Microwave Oscillators 613 FIGURE 13.5 Pierce crystal oscillator circuit. 13.2MICROWAVEOSCILLATORS In this section we focus on oscillator circuits that are useful at microwave frequencies, primarily employing negative resistance diodes or transistors. Figure 13.6 shows the canonical RF circuit for a one-port negative resistance oscil- lator, where Zin=Rin+jXinis the input impedance of the active device (e.g., a bi- ased diode or transistor). In general, this impedance is current (or voltage) dependent, as well as frequency dependent, which we indicate by writing Zin(I,jω)=Rin(I,jω)+ jXin(I,jω). The device is terminated with a passive load impedance, ZL=RL+jXL. Applying Kirchhoff’s voltage law gives (ZL+Zin)I=0.( 13.24) If oscillation is occurring, such that the RF current Iis nonzero, then the following two conditions must be satisfied: RL+Rin=0, (13.25a) XL+Xin=0. (13.25b) Since the load is passive, RL>0, and (13.25a) implies that Rin<0. Thus, while a positive resistance implies energy dissipation, a negative resistance implies an energy source. The condition of (13.25b) controls the frequency of oscillation. The condition in (13.24), that Negative resistance deviceI Xin RinRin Γin (Zin)ZL = RL + j XL ΓL (ZL) FIGURE 13.6 Circuit for a one-port negative resistance oscillator. c13OscillatorsAndMixers Pozar September 16, 2011 15:44 614 Chapter 13: Oscillators and Mixers ZL=− Zinfor steady-state oscillation, implies that the reflection coefficients /Gamma1Land/Gamma1in are related as /Gamma1L=ZL−Z0 ZL+Z0=−Zin−Z0 −Zin+Z0=Zin+Z0 Zin−Z0=1 /Gamma1in.( 13.26) The process of oscillation is critically dependent on the nonlinear behavior of Zin,a s follows. Initially, it is necessary for the overall circuit to be unstable at a certain frequency, that is, Rin(I,jω)+RL<0. Then any transient excitation or noise will cause an oscillation to build up at the frequency ω.A s Iincreases, Rin(I,jω)must become less negative until the current I0is reached such that Rin(I0,jω0)+RL=0, and Xin(I0,jω0)+XL(jω0)= 0. At this point the oscillator can run in a stable state. The final frequency, ω0, generally differs from the startup frequency because Xinis current dependent, so that Xin(I,jω)/negationslash= Xin(I0,jω0). Thus we see that the conditions of (13.25) are not enough to guarantee a stable state of oscillation. In particular, stability requires that any perturbation in current or frequency will be damped out, allowing the oscillator to return to its original state. This condition can be quantified by considering the effect of a small change, δI, in the current, and a small change, δs, in the complex frequency s=α+jω.I fw el e t ZT(I,s)=Zin(I,s)+ZL(s), then we can write a Taylor series for ZT(I,s)about the stable operating point I0,ω0as ZT(I,s)=ZT(I0,s0)+∂ZT ∂s/vextendsingle/vextendsingle/vextendsingle/vextendsingle s0,I0δs+∂ZT ∂I/vextendsingle/vextendsingle/vextendsingle/vextendsingle s0,I0δI=0,( 13.27) since ZT(I,s)must still equal zero if oscillation is occurring. In (13.27), s0=jω0is the complex frequency at the original operating point. Now use the fact that ZT(I0,s0)=0, and that ∂ZT/∂s=− j(∂ZT/∂ω), to solve (13.27) for δs=δα+jδω: δs=δα+jδω=−∂ZT/∂I ∂ZT/∂s/vextendsingle/vextendsingle/vextendsingle/vextendsingle s0,I0δI=−j(∂ZT/∂I)/parenleftbig ∂Z∗ T/∂ω/parenrightbig |∂ZT/∂ω|2δI.( 13.28) If the transient caused by δIandδωis to decay, we must have δα < 0 when δI>0. Equa- tion (13.28) then implies that Im/braceleftbigg∂ZT ∂I∂Z∗ T ∂ω/bracerightbigg <0, or ∂RT ∂I∂XT ∂ω−∂XT ∂I∂RT ∂ω>0.( 13.29) This relation is sometimes known as Kurokawa’s condition. For a passive load, ∂RL/∂I= ∂XL/∂I=∂RL/∂ω=0, so (13.29) reduces to ∂Rin ∂I∂ ∂ω(XL+Xin)−∂Xin ∂I∂Rin ∂ω>0.( 13.30) As discussed above, we usually have that ∂Rin/∂I>0, so (13.30) can be satisfied if ∂(XL+Xin)/∂ω/greatermuch0. This implies that a high- Qcircuit will result in maximum oscil- lator stability. Cavity and dielectric resonators are often used for this purpose. Effective oscillator design requires the consideration of several other issues, such as the selection of an operating point for stable operation and maximum power output, fre-quency pulling, large-signal effects, and noise characteristics. We leave these topics to more advanced texts [4, 5]. c13OscillatorsAndMixers Pozar September 16, 2011 15:44 13.2 Microwave Oscillators 615 Diode0.254 /H9261 0.308 /H9261 50 Ω ΓL (ZL)Γin (Zin) FIGURE 13.7 Load matching circuit for the one-port oscillator of Example 13.2. EXAMPLE 13.2 NEGATIVE RESISTANCE OSCILLATOR DESIGN A one-port oscillator uses a negative resistance diode having /Gamma1in=1.25/negationslash40◦ (Z0=50/Omega1) at its desired operating point, for f=6 GHz. Design a load match- ing network for a 50 /Omega1load impedance. Solution From either the Smith chart (see Problem 13.5) or by direct calculation, we find the input impedance of the diode as Zin=Z01+/Gamma1in 1−/Gamma1in=−44+j123/Omega1. Then, by (13.25), the load impedance must be ZL=− Zin=44−j123/Omega1. A shunt stub and series section of line can be used to convert 50 /Omega1toZL,a s shown in the circuit of Figure 13.7. ■ TransistorOscillators In a transistor oscillator, a negative resistance one-port network is effectively created by ter- minating a potentially unstable transistor with an impedance designed to drive the device in an unstable region. The circuit model of a transistor oscillator is shown in Figure 13.8. In this circuit, the RF output port is part of the load network on the output side of thetransistor, but it is also possible to use the terminating network to the left of the transistor as the output port. In the case of an amplifier, we preferred a device with a high degree Γin (Zin)ΓS (ZS)Γout (Zout)ΓL (ZL)Transistor [S]Load networkTerminating network FIGURE 13.8 Circuit for a two-port transistor oscillator. c13OscillatorsAndMixers Pozar September 16, 2011 15:44 616 Chapter 13: Oscillators and Mixers of stability—ideally, an unconditionally stable device. For an oscillator, we require a de- vice with a high degree of instability. Typically, common source or common gate FET configurations are used (common emitter or common base for bipolar junction devices), often with positive feedback to enhance the instability of the device. After the transistorconfiguration is selected, the output stability circle can be drawn in the /Gamma1 Lplane, and /Gamma1L selected to produce a large value of negative resistance at the input to the transistor. Then the terminating impedance ZS=RS+jXScan be chosen to match Zin. Because such a design often relies on the small-signal scattering parameters, and because Rinwill become less negative as the oscillator power builds up, it is often necessary to choose RSso that RS+Rin<0. Otherwise, oscillation may cease if increasing RF power increases Rinto the point where RS+Rin>0. In practice, a value of RS=−Rin 3(13.31a) is often used. The reactive part of ZSis chosen to resonate the circuit, XS=− Xin.( 13.31b) When oscillation occurs between the termination network and the transistor, oscil- lation will simultaneously occur at the output port, which we can show as follows. Forsteady-state oscillation at the input port, we must have /Gamma1 S/Gamma1in=1, analogous to the condi- tion of (13.26). Then from (12.3a) we have 1 /Gamma1S=/Gamma1in=S11+S12S21/Gamma1L 1−S22/Gamma1L=S11−/Delta1/Gamma1 L 1−S22/Gamma1L,( 13.32) where /Delta1=S11S22−S12S21. Solving for /Gamma1Lgives /Gamma1L=1−S11/Gamma1S S22−/Delta1/Gamma1 S.( 13.33) From (12.3b) we have that /Gamma1out=S22+S12S21/Gamma1S 1−S11/Gamma1S=S22−/Delta1/Gamma1 S 1−S11/Gamma1S,( 13.34) which shows that /Gamma1L/Gamma1out=1, and hence ZL=− Zout. Thus, the condition for oscillation at the load network is satisfied. Note that it is preferable to use the large-signal scatter- ing parameters of the transistor in the above development. EXAMPLE 13.3 TRANSISTOR OSCILLATOR DESIGN Design a transistor oscillator at 4 GHz using a GaAs MESFET in a common gate configuration, with a 5 nH inductor in series with the gate to increase theinstability. Choose a load network to match to a 50 /Omega1load, and an appropriate terminating network at the input to the transistor. The scattering parameters of the transistor in a common source configuration are ( Z 0=50/Omega1)S11=0.72/negationslash−116◦, S12=0.03/negationslash57◦,S21=2.60/negationslash76◦, and S22=0.73/negationslash−54◦. Solution The first step is to convert the common source scattering parameters to the scat-tering parameters that apply to the transistor in a common gate configuration with a series inductor. (See Figure 13.9a.) This is most easily done using a microwave c13OscillatorsAndMixers Pozar September 16, 2011 15:44 13.2 Microwave Oscillators 617 CAD package. The new scattering parameters are S/prime 11=2.18/negationslash−35◦, S/prime 12=1.26/negationslash18◦, S/prime 21=2.75/negationslash96◦, S/prime 22=0.52/negationslash155◦. Note that |S/prime 11|is significantly greater than |S11|, which suggests that the config- uration of Figure 13.9a is more unstable than the common source configuration. Calculating the output stability circle ( /Gamma1Lplane) parameters from (11.25) gives CL=/parenleftbig S/prime 22−/Delta1/primeS/prime 11∗/parenrightbig∗ /vextendsingle/vextendsingleS/prime 22/vextendsingle/vextendsingle2−|/Delta1/prime|2=1.08/negationslash33◦, RL=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleS/prime 12S/prime 21/vextendsingle/vextendsingleS/prime 22/vextendsingle/vextendsingle2−|/Delta1/prime|2/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=0.665. Since/vextendsingle/vextendsingleS/prime 11/vextendsingle/vextendsingle=2.18>1, the stable region is inside this circle, as shown in the Smith chart in Figure 13.9b. There is a great amount of freedom in our choice for /Gamma1L, but one objective is to make |/Gamma1in|large. We therefore try several values of /Gamma1Llocated on the opposite side of the chart from the stability circle, and select /Gamma1L=0.59/negationslash−104◦. Then we can design a single-stub matching network to convert a 50 /Omega1load to ZL= 20−j35/Omega1, as shown in Figure 13.9a. For the given value of /Gamma1L, we calculate /Gamma1inas /Gamma1in=S/prime 11+S/prime 12S/prime 21/Gamma1L 1−S/prime 22/Gamma1L=3.96/negationslash−2.4◦ orZin=−84−j1.9/Omega1. Then, from (13.31), we find ZSas ZS=−Rin 3−jXin=28+j1.9/Omega1. Using Rin/3 should ensure enough instability for the startup of oscillation. The easiest way to implement the impedance ZSis to use a 90 /Omega1load with a short length of line, as shown in the figure. It is likely that the steady-state oscillationfrequency will differ from 4 GHz because of the nonlinearity of the transistor parameters. ■ DielectricResonatorOscillators As we saw from the result of (13.30), oscillator stability is enhanced with the use of a high- Qtuning network. The unloaded Qof a resonant network using lumped elements or microstrip lines and stubs is typically limited to a few hundred (see Chapter 6), and while waveguide cavity resonators can have unloaded Qso f1 0 4or more, they are not well suited for integration in miniature microwave integrated circuitry. Another disadvantage of metal cavities is the significant frequency drift caused by dimensional expansion due to temperature variations. The dielectric cavity resonator discussed in Section 6.5 overcomesmost of these disadvantages, as it can have an unloaded Qas high as several thousand, it is compact and easily integrated with planar circuitry, and it can be made from ceramic c13OscillatorsAndMixers Pozar September 16, 2011 15:44 618 Chapter 13: Oscillators and Mixers j , CL ΓL⎪Γin⎪ = 1 Stable region for ΓL RL20-2030-3040-4050-5060-6070-7080-8090-90100-100110-110120-120130-130140-140150-150160-160170-1701800.040.040.050.050.060.060.070.070.080.080.090.090.10.10.110.110.120.120.130.130.140.140.150.150.160.160.170.170.180.180.190.190.20.20.210.210.220.220.230.230.240.24 0.250.250.260.260.270.270.280.280.290.290.30.30.310.310.320.320.330.330.340.340.350.350.360.360.370.370.380.380.390.390.40.40.410.410.420.420.430.430.440.440.450.450.460.460.470.470.480.480.490.49 0.00.0AN GLE OFREFLECTION COEFFICIENTINDEGREES—>WAVELENGTHSTOWARDGENERATOR—><— W AVELENGTH STOW ARD LOAD <—II T NDUCTIVEREACTANCECOMPONEN(+X/Zo) ORCAPACTIVESUSCEPTANCE(+jB/Yo)CAPACITIVEREACTANCECOMPONENT(-jX/Zo),ORINDUCTIVESUSCEPTANCE(-jB/Yo) 0.1 0.1 0.1 0.2 0.2 0.2 0.3 0.3 0.3 0.4 0.4 0.4 0.5 0.5 0.5 0.6 0.6 0.6 0.7 0.7 0.7 0.8 0.8 0.8 0.9 0.9 0.9 1.0 1.0 1.01.21.21.21.41.41.41.61.61.61.81.81.82.02.02.03.03.03.04.04.04.05.05.05.01010102020205050500.20.2 0.2 0.20.40.4 0.4 0.40.60.6 0.6 0.60.80.8 0.8 0.81.01.0 1.0 1.0RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) ±Γin (Zin)ΓS (ZS)ΓL (ZL)50 Ω 90 Ω0.262 /H9261 50 Ω0.319 /H9261 0.346 /H9261 5 nH (a) (b)50 Ω load50 Ω[S'] SD G FIGURE 13.9 Circuit design for the transistor oscillator of Example 13.3. (a) Oscillator circuit. (b) Smith chart for determining /Gamma1L. materials that have excellent temperature stability. For these reasons, transistor dielectric resonator oscillators (DROs) are in common use over the entire microwave, and lower millimeter wave, frequency range. A dielectric resonator is usually coupled to an oscillator circuit by positioning it in close proximity to a microstrip line, as shown in Figure 13.10a. The resonator operates in the TE 01δmode, and couples to the fringing magnetic field of the microstrip line. The c13OscillatorsAndMixers Pozar September 16, 2011 15:44 13.2 Microwave Oscillators 619 L 1 NC R Zd (a)Dielectric resonator Microstrip line (b)Z0 Z0 Γ FIGURE 13.10 (a) Geometry of a dielectric resonator coupled to a microstrip line; (b) equivalent circuit. strength of coupling is determined by the spacing, d, between the resonator and microstrip line. Because coupling is via the magnetic field, the resonator appears as a series load on the microstrip line, as shown in the equivalent circuit of Figure 13.10b. The resonator ismodeled as a parallel RLC circuit, and the coupling to the feedline is modeled by the turns ratio, N, of the transformer. Using the result of (6.19) for the impedance of a parallel RLC resonator, we can express the equivalent series impedance, Z, seen by the microstrip line as Z=N 2R 1+j2Q0/Delta1ω/ω 0,( 13.35) where Q0=R/ω0Lis the unloaded resonator Q,ω0=1/√ LCis the resonant frequency, and/Delta1ω=ω−ω0. The coupling factor, defined in (6.76), between the resonator and the feedline is the ratio of the unloaded to external Q, and can be found as g=Q0 Qe=R/ω0L RL/N2ω0L=N2R 2Z0,( 13.36) where RL=2Z0is the load resistance for a feedline with source and termination resis- tances Z0. In some cases the feedline is terminated with an open-circuit λ/4 from the res- onator to maximize the magnetic field at that point; in this case RL=Z0, and the coupling factor is twice the value given in (13.36). The reflection coefficient seen on the terminated microstrip line looking toward the resonator can be written as /Gamma1=/parenleftbig Z0+N2R/parenrightbig −Z0/parenleftbig Z0+N2R/parenrightbig +Z0=N2R 2Z0+N2R=g 1+g.( 13.37) This allows the coupling coefficient to be found from g=/Gamma1/(1−/Gamma1)after the simple pro- cedure of measuring /Gamma1at resonance; the resonant frequency and Qcan also be found by measurement. Alternatively, these quantities can be calculated using approximate analyt- ical solutions [6]. Note that this procedure leaves a degree of freedom between NandR since only the product N2Ris uniquely determined. There are many oscillator configurations using common source (emitter), common gate (base), or common drain (collector) connections of either BJTs or FETs, in addition to the optional use of series or shunt elements to increase the instability of the device [4, 5].A dielectric resonator can be incorporated into the circuit to provide frequency stability using either the parallel feedback arrangement of Figure 13.11a, or the series feedback c13OscillatorsAndMixers Pozar September 16, 2011 15:44 620 Chapter 13: Oscillators and Mixers Matching networkLoad Load Z0 Z0Dielectric resonatorDielectric resonator Matching network (b) (a) FIGURE 13.11 (a) Dielectric resonator oscillator using parallel feedback; (b) dielectric resonator oscillator using series feedback. technique shown in Figure 13.11b. The parallel configuration uses a resonator coupled to two microstrip lines, functioning as a high- Qbandpass filter that couples a portion of the transistor output back to its input. The amount of coupling is controlled by the spacingbetween the resonator and the lines, and the phase is controlled by the length of the lines. The series feedback configuration is simpler, using only a single microstrip feedline, but typically does not have a tuning range as wide as that obtained with parallel feedback.Design of an oscillator using parallel feedback is most conveniently done using microwave CAD software, but a dielectric resonator oscillator using series feedback can be designed using the same procedure that was discussed in the previous section on two-port oscillators. EXAMPLE 13.4 DIELECTRIC RESONATOR OSCILLATOR DESIGN A wireless local area network application requires a local oscillator operating at 2.4 GHz. Design a dielectric resonator oscillator using the series feedback cir- cuit of Figure 13.11b with a bipolar transistor having the following scattering parameters (Z0=50/Omega1):S11=1.8/negationslash130◦,S12=0.4/negationslash45◦,S21=3.8/negationslash36◦, and S22=0.7/negationslash−63◦. Determine the required coupling coefficient for the dielectric resonator, and the required microstrip matching network for the load. Plot the magnitude of /Gamma1outversus /Delta1f/f0for small variations in frequency about the de- sign value, assuming an unloaded resonator Qof 1000. Solution The DRO circuit is shown in Figure 13.12a. The dielectric resonator is placed λ/4 from the open end of the microstrip line; the line length /lscriptrcan be adjusted Dielectric resonator λ 4/H5129r/H5129t /H5129s Z0 Γin (Zin)( )ΓS (ZS)Γout (Zout)ΓL (ZL)ΓS′ ZS′ FIGURE 13.12 (a) Circuit for the dielectric resonator of Example 13.4. c13OscillatorsAndMixers Pozar September 16, 2011 15:44 13.2 Microwave Oscillators 621 –0.050.02.04.06.08.010.012.0 –0.04 –0.03 –0.02 –0.01 0.00 ∆f/f0 (percent)0.01 0.02 0.03 0.04 0.05out/H9003 FIGURE 13.12 Continued. (b) |/Gamma1out|vs. frequency in Example 13.4. to match the phase of the required value of /Gamma1S. In contrast to the oscillator of the previous example, the output load impedance for this circuit is part of theterminating network. The stability circles for the transistor can be plotted if desired, but are not necessary for the design since we can begin by choosing /Gamma1 Sto provide a large value of |/Gamma1out|. From (13.34) we have /Gamma1out=S22+S12S21/Gamma1S 1−S11/Gamma1S, which indicates that we can maximize /Gamma1outby making 1 −S11/Gamma1Sclose to zero. Thus we choose /Gamma1S=0.6/negationslash−130◦, which gives /Gamma1out=10.7/negationslash132◦. This corre- sponds to an impedance Zout=Z01+/Gamma1out 1−/Gamma1out=501+10.7/negationslash132◦ 1−10.7/negationslash132◦=−43.7+j6.1/Omega1. Applying the analogous startup condition of (13.31) for the output side gives the required termination impedance as ZL=−Rout 3−jXout=5.5−j6.1/Omega1. The matching network can now be designed using a Smith chart. The short- est transmission line length for matching ZLto the load impedance Z0is/lscriptt= 0.481λ, and the required open-circuit stub length is /lscripts=0.307λ. Next we match /Gamma1Sto the resonator network. From (13.35) we know that the equivalent impedance of the resonator seen by the microstrip line is real at the resonant frequency, so the phase angle of the reflection coefficient at this point, /Gamma1/prime S, must be either zero or 180◦. For an undercoupled parallel RLC resonator, R<Z0, so the proper phase will be 180◦, which can be achieved by transfor- mation through the line length /lscriptr. The magnitude of the reflection coefficient is c13OscillatorsAndMixers Pozar September 16, 2011 15:44 622 Chapter 13: Oscillators and Mixers unchanged, so we have the relation /Gamma1/prime S=/Gamma1Se2jβ/lscriptr=(0.6/negationslash−130◦)e2jβ/lscriptr=0.6/negationslash180◦, which gives /lscriptr=0.431λ. The equivalent impedance of the resonator at resonance is then Z/prime S=Z01+/Gamma1/prime S 1−/Gamma1/prime S=12.5/Omega1. The coupling coefficient can be found using (13.36), with a factor of two to ac- count for the λ/4 stub termination, as g=N2R Z0=12.5 50=0.25. The variation of |/Gamma1out|with frequency will give an indication of the frequency stability of the oscillator. We can calculate /Gamma1outfrom (13.34), after first using (13.35) to compute Z/prime S,/Gamma1/prime S, and then transforming down the line of length /lscriptr to obtain /Gamma1S. The electrical line length can be approximated as constant for the small changes in frequency associated with this calculation. A short computer program, or microwave CAD software, can be used to generate data for −0.01 < /Delta1f/f0<0.01, which is shown in the graph of Figure 13.12b. Observe that |/Gamma1out| decreases rapidly with a change in frequency as small as a few hundredths of a percent, demonstrating the sharp selectivity that can be obtained with a dielectricresonator. ■ 13.3OSCILLATORPHASENOISE The noise produced by an oscillator or other signal source is important in practice because it may severely degrade the performance of a communications or radar receiver system. Besides adding to the noise level of the receiver, a noisy local oscillator will lead to down-conversion of undesired nearby signals, thus limiting the selectivity of the receiver and how closely adjacent channels may be spaced. Phase noise refers to the short-term random fluctuation in the frequency (or phase) of an oscillator signal. Phase noise also introducesuncertainty during the detection of digitally modulated signals. An ideal oscillator would have a frequency spectrum consisting of a single delta func- tion at its operating frequency, but a realistic oscillator will have a spectrum more like that shown in Figure 13.13. Spurious signals due to oscillator harmonics or intermodulation products appear as discrete spikes in the spectrum. Phase noise, due to random fluctuations caused by thermal and other noise sources, appears as a broad, continuous distribution f0 FIGURE 13.13 Output spectrum of a typical RF oscillator. c13OscillatorsAndMixers Pozar September 16, 2011 15:44 13.3 Oscillator Phase Noise 623 localized about the output signal. Phase noise is defined as the ratio of power in one phase modulation sideband to the total signal power per unit bandwidth (1 Hz) at a particular offset, fm, from the signal frequency, and is denoted as L(fm). It is usually expressed in decibels relative to the carrier power per hertz of bandwidth (dBc/Hz). A typical oscillatorphase noise specification for a cellular radio, for example, may be −110 dBc/Hz at 25 kHz from the carrier. In the following sections we show how phase noise may be represented, and present a widely used model for characterizing the phase noise of an oscillator. RepresentationofPhaseNoise In general, the output voltage of an oscillator or synthesizer can be written as v o(t)=Vo[1+A(t)]cos[ω ot+θ(t)],( 13.38) where A(t)represents the amplitude fluctuations of the output, and θ(t)represents the phase variation of the output waveform. Of these, amplitude variations can usually be wellcontrolled, and generally have less impact on system performance. Phase variations may be discrete (due to deterministic spurious mixer products or harmonics), or random in nature (due to thermal or other random noise sources). Note from (13.38) that an instantaneousphase variation is indistinguishable from a variation in frequency. Small changes in the oscillator frequency can be represented as a frequency modula- tion of the carrier by letting θ(t)=/Delta1f fmsinωmt=θpsinωmt,( 13.39) where fm=ωm/2π is the modulating frequency. The peak phase deviation is θp=/Delta1f/fm (also called the modulation index ). Substituting (13.39) into (13.38) and expanding gives vo(t)=Vo[cosωotcos(θ psinωmt)−sinωotsin(θpsinωmt)],( 13.40) where we set A(t)=0 to ignore amplitude fluctuations. Assuming the phase deviations are small, so that θp/lessmuch1, we can use the small-argument expressions that sin x/similarequalxand cosx/similarequal1 to simplify (13.40) to vo(t)=Vo/parenleftbig cosωot−θpsinωmtsinωot/parenrightbig =Vo/braceleftbigg cosωot−θp 2[cos(ωo+ωm)t−cos(ωo−ωm)t]/bracerightbigg . (13.41) This expression shows that small phase or frequency deviations in the output of an oscilla- tor result in modulation sidebands at ωo±ωm, located on either side of the carrier signal atωo. When these deviations are due to random changes in temperature or device noise, the output spectrum of the oscillator will take the form shown in Figure 13.13. According to the definition of phase noise as the ratio of noise power in a single sideband to the carrier power, the waveform of (13.41) has a corresponding phase noise of L(f)=Pn Pc=1 2/parenleftbiggVoθp 2/parenrightbigg2 1 2V2 o=θ2 p 4=θ2 rms 2,( 13.42) where θrms=θp/√ 2 is the rms value of the phase deviation. The two-sided power spectral density associated with phase noise includes power in both sidebands: Sθ(fm)=2L(fm)=θ2 p 2=θ2 rms.( 13.43) c13OscillatorsAndMixers Pozar September 16, 2011 15:44 624 Chapter 13: Oscillators and Mixers White noise generated by passive or active devices can be interpreted in terms of phase noise by using the same definition. From Chapter 10 we know that the noise power at the output of a noisy two-port network is kT0BFG , where T0=290 K, Bis the measurement bandwidth, Fis the noise figure of the network, and Gis the gain of the network. For a 1 Hz bandwidth, the ratio of output noise power density to output signal power gives the power spectral density as Sθ(fm)=kT0F Pc,( 13.44) where Pcis the input signal (carrier) power. Note that the gain of the network cancels in this expression. Leeson’sModelforOscillatorPhaseNoise In this section we present Leeson’s model for characterizing the power spectral density of oscillator phase noise [2, 7]. As in Section 13.1, we will model the oscillator as an amplifier with a feedback path, as shown in Figure 13.14. If the voltage gain of the amplifier is included in the feedback transfer function H(ω), then the voltage transfer function for the oscillator circuit is Vo(ω)=Vi(ω) 1−H(ω).( 13.45) If we consider oscillators that use a high- Qresonant circuit in the feedback loop (e.g., Col- pitts, Hartley, Clapp, and similar oscillators), then H(ω)can be represented as the voltage transfer function of a parallel RLC resonator: H(ω)=1 1+jQ0/parenleftbiggω ω0−ω0 ω/parenrightbigg=1 1+2jQ0/Delta1ω/ω 0,( 13.46) where ω0is the resonant frequency of the oscillator, and /Delta1ω=ω−ω0is the frequency offset relative to the resonant frequency. Since the input and output power spectral densities are related by the square of the magnitude of the voltage transfer function [8], we can use (13.45)–(13.46) to write Sφ(ω)=/vextendsingle/vextendsingle/vextendsingle/vextendsingle1 1−H(ω)/vextendsingle/vextendsingle/vextendsingle/vextendsingle2 Sθ(ω)=1+4Q2 0/Delta1ω2/ω2 0 4Q2 0/Delta1ω2/ω2 0Sθ(ω) =/parenleftBigg 1+ω2 0 4Q2 0/Delta1ω2 0/parenrightBigg Sθ(ω)=/parenleftBigg 1+ω2 h /Delta1ω2 0/parenrightBigg Sθ(ω), (13.47) where Sθ(ω)is the input power spectral density, and Sφ(ω)is the output power spectral density. In (13.47) we have also defined ωh=ω0/2Q0as the half-power (3 dB) bandwidth of the resonator. FIGURE 13.14 Feedback amplifier model for characterizing oscillator phase noise. c13OscillatorsAndMixers Pozar September 16, 2011 15:44 13.3 Oscillator Phase Noise 625 FIGURE 13.15 Noise power versus frequency for an amplifier with an applied input signal. The noise spectrum of a typical transistor amplifier with an applied sinusoidal signal atf0is shown in Figure 13.15. Besides kTB thermal noise, transistors generate additional noise that varies as 1 /fat frequencies below the frequency fα.T h i s1 /f,o rflicker , noise is likely caused by random fluctuations of the carrier density in the active device. Due to the nonlinearity of the transistor, the 1 /fnoise will modulate the applied signal at f0, and appear as 1 /fnoise sidebands around f0. Since the 1 /fnoise component dominates the phase noise power at frequencies close to the carrier, it is important to include it in our model. Thus we consider an input power spectral density as shown in Figure 13.16, where K//Delta1frepresents the 1 /fnoise component around the carrier, and kT0F/P0represents the thermal noise. Thus the power spectral density applied to the input of the oscillator can be written as Sθ(ω)=kTF P0/parenleftbigg 1+Kωα /Delta1ω/parenrightbigg ,( 13.48) where Kis a constant accounting for the strength of the 1 /fnoise, and ωα=2πfαis the corner frequency of the 1 /fnoise. The corner frequency depends primarily on the type of transistor used in the oscillator. Silicon junction FETs, for example, typically have corner frequencies ranging from 50 to 100 Hz, while GaAs MESFETs have corner frequenciesranging from 2 to 10 MHz, or higher. Silicon bipolar junction transistors have corner fre- quencies that range from 5 to 50 kHz. Using (13.48) in (13.47) gives the power spectral density of the output phase noise as S φ(ω)=kT0F P0/parenleftBigg Kω2 0ωα 4Q2 0/Delta1ω3+ω2 0 4Q2 0/Delta1ω2+Kωα /Delta1ω+1/parenrightBigg =kT0F P0/parenleftBigg Kωαω2 h /Delta1ω3+ω2 h /Delta1ω2+Kωα /Delta1ω+1/parenrightBigg . (13.49) FIGURE 13.16 Idealized power spectral density of amplifier noise, including 1 /fand thermal components. c13OscillatorsAndMixers Pozar September 16, 2011 15:44 626 Chapter 13: Oscillators and Mixers f0 f0 FIGURE 13.17 Power spectral density of phase noise at the output of an oscillator. (a) Response forfh>fα(low Q). (b) Response for fh>fα(high Q). This result is sketched in Figure 13.17. There are two cases, depending on which of the middle two terms of (13.49) is more significant. In either case, for frequencies close to thecarrier at f 0, the noise power decreases as 1 /f3,o r−18 dB/octave. If the resonator has a relatively low Q, so that its 3 dB bandwidth fh>fα, then for frequencies between fαand fhthe noise power drops as 1 /f2,o r−12 dB/octave. If the resonator has a relatively high Q, so that fh<fα, then for frequencies between fhand fαthe noise power drops as 1 /f, or−6 dB/octave. At higher frequencies the noise is predominantly thermal, constant with frequency, and proportional to the noise figure of the amplifier. A noiseless amplifier with F=1( 0 dB) would produce the minimum noise floor of kT0=−174 dBm/Hz. In accordance with Figure 13.13, the noise power is greatest at frequencies closest to the carrier frequency, but(13.49) shows that the 1 /f 3component is proportional to 1 /Q2 0, so that better phase noise characteristics close to the carrier are achieved with a high- Qresonator. Finally, recall from (13.43) that the single-sideband phase noise will be one-half of the power spectral density of (13.49). These results give a reasonably good model for oscillator phase noise, and quantitatively explain the roll-off of noise power with frequency offset from the carrier. The effect of phase noise in a receiver is to degrade both the signal-to-noise ratio (or bit error rate) and the selectivity [9]. Of these, the impact on selectivity is usually the most severe. Phase noise degrades receiver selectivity by causing down conversion ofsignals located nearby the desired signal frequency. The process is shown in Figure 13.18. A local oscillator at frequency f 0is used to down convert a desired signal to an intermediate IFIF IF IFLO FIGURE 13.18 Illustrating how local oscillator phase noise can lead to the reception of undesired signals adjacent to the desired signal. c13OscillatorsAndMixers Pozar September 16, 2011 15:44 13.4 Frequency Multipliers 627 frequency (IF). Due to phase noise, however, an adjacent undesired signal can be down converted to the same IF frequency due to the phase noise spectrum of the local oscillator. The phase noise that leads to this conversion is located at an offset from the carrier equal to the IF frequency from the undesired signal. This process is called reciprocal mixing .F r o m this diagram, it is easy to see that the maximum allowable phase noise in order to achieve an adjacent channel rejection (or selectivity) of SdB (S≥0) is given by L(fm)=C(dBm) −S(dB)−I(dBm) −10 log(B),(dBc/Hz) ,( 13.50) where Cis the desired signal level (in dBm), Iis the undesired (interference) signal level (in dBm), and Bis the bandwidth of the IF filter (in Hz). EXAMPLE 13.5 GSM RECEIVER PHASE NOISE REQUIREMENTS The GSM cellular telephone standard requires a minimum of 9 dB rejection of interfering signal levels of −23 dBm at 3 MHz from the carrier, −33 dBm at 1.6 MHz from the carrier, and −43 dBm at 0.6 MHz from the carrier, for a carrier level of −99 dBm. Determine the required local oscillator phase noise at these carrier frequency offsets. The channel bandwidth is 200 kHz. Solution From (13.50) we have L(fm)=C(dBm) −S(dB)−I(dBm) −10 log(B) =−99 dBm −9d B−I(dBm) −10 log(2×105). The table below lists the required LO phase noise as computed from the above expression: Frequency Offset Interfering Signal L(fm) fm(MHz) Level (dBm) (dBc/Hz) 3.0 −23 −138 1.6 −33 −128 0.6 −43 −118 This level of phase noise requires a phase-locked synthesizer. Bit errors in GSM systems are usually dominated by the reciprocal mixing effect, while errors due to thermal antenna and receiver noise are generally negligible. ■ 13.4FREQUENCYMULTIPLIERS As frequency increases into the millimeter wave range it becomes increasingly difficult to build fundamental frequency oscillators with good power, stability, and noise character- istics. An alternative approach is to produce a harmonic of a lower frequency oscillator through the use of a frequency multiplier . As we saw in Section 10.3, a nonlinear element may generate many harmonics of an input sinusoidal signal, so frequency multiplication isa natural occurrence in circuits containing diodes and transistors. Designing a good-quality frequency multiplier, however, is a difficult task that generally requires nonlinear analysis, matching at multiple frequencies, stability analysis, and thermal considerations. We willdiscuss some of the general operational principles and properties of diode and transistor frequency multipliers, and refer the reader to the literature for more practical details [5]. c13OscillatorsAndMixers Pozar September 16, 2011 15:44 628 Chapter 13: Oscillators and Mixers Frequency multiplier circuits can be categorized as reactive diode multipliers, resis- tive diode multipliers ,o rtransistor multipliers. A reactive diode multiplier uses either a varactor or a step-recovery diode biased to present a nonlinear junction capacitance. Since losses in such diodes are small, conversion efficiencies (the fraction of RF input power thatis converted to the desired harmonic) can be relatively high. In fact, as we will show, ideal (lossless) reactive multipliers can achieve a theoretical conversion efficiency of 100%. Var- actor multipliers are most useful for low harmonic conversion (multiplier factors of 2–4),while step-recovery diodes are able to generate more power at higher harmonics. Resistive multipliers exploit the nonlinear I–Vcharacteristic of a forward-biased Schottky barrier diode. We will show that resistive multipliers have conversion efficiencies that decrease as the square of the harmonic number, and so these multipliers are only useful for low multi- plication factors. Transistor multipliers can use both bipolar junction and FET devices, andcan provide conversion gains. Transistor multipliers are limited by their cutoff frequency, however, and therefore are generally not useful at very high frequencies. A disadvantage of frequency multipliers is that noise levels are increased by the multi- plication factor. This is because frequency multiplication is effectively a phase multiplica- tion process as well, so phase noise variations get multiplied in the same way that frequency is multiplied. The increase in noise power is given by 20 log n, where nis the multiplica- tion factor. Thus a frequency doubler will increase the fundamental oscillator noise level by at least 6 dB, while a frequency tripler will lead to an increase of at least 9.5 dB. Reac- tive diode multipliers typically add little additional noise of their own since varactors andstep-recovery diodes have very low series resistances, but resistive diode multipliers can generate significant additional noise power. ReactiveDiodeMultipliers(Manley–RoweRelations) We begin our discussion with the Manley–Rowe relations, which result from a very general analysis of power conservation associated with frequency conversion in a nonlinear reac- tive element [10]. Consider the circuit of Figure 13.19, where two sources at frequencies ω 1andω2drive a nonlinear capacitor, C. The circuit also shows ideal bandpass filters to conceptually isolate powers in all harmonics of the form nω1+mω2. Since the capacitor is nonlinear, its charge Qcan be expressed as a power series in terms of the capacitor voltage, v: Q=a0+a1v+a2v2+a3v3+··· As in Section 10.3, this nonlinear relationship implies the generation of all frequency prod- ucts of the form nω1+mω2. Thus we can write the capacitor voltage as a Fourier series of /H92751/H92751 + /H92752 /H92751 – /H92752 /H927522/H92751 2/H92752 /H92751 /H92752v(t)i(t) + C –... FIGURE 13.19 Conceptual circuit for the derivation of the Manley–Rowe relations. c13OscillatorsAndMixers Pozar September 16, 2011 15:44 13.4 Frequency Multipliers 629 the form v(t)=∞/summationdisplay n=−∞∞/summationdisplay m=−∞Vnmej(nω1+mω2)t.( 13.51) Similarly, the capacitor charge and current can be written as Q(t)=∞/summationdisplay n=−∞∞/summationdisplay m=−∞Qnmej(nω1+mω2)t,( 13.52) i(t)=dQ dt=∞/summationdisplay n=−∞∞/summationdisplay m=−∞j(nω1+mω2)Qnmej(nω1+mω2)t=∞/summationdisplay n=−∞∞/summationdisplay m=−∞Inmej(nω1+mω2)t. (13.53) Since v(t)andi(t)are real functions, we must have that V−n,−m=V∗ nmandQ−n,−m= Q∗ nm. No real power can be dissipated in the lossless capacitor. If ω1andω2are not multiples of each other, there is no average power due to interacting harmonics. Then the averagepower (ignoring a factor of 4) at frequency ±|nω 1+mω2|is given as Pnm=2R e/braceleftbig VnmI∗ nm/bracerightbig =VnmI∗ nm+V∗ nmInm=VnmI∗ nm+V−n,−mI∗ −n,−m=P−n,−m. (13.54) Conservation of power can then be expressed as ∞/summationdisplay n=−∞∞/summationdisplay m=−∞Pnm=0.( 13.55) Now multiply (13.55) bynω1+mω2 nω1+mω2to obtain ω1∞/summationdisplay n=−∞∞/summationdisplay m=−∞nPnm nω1+mω2+ω2∞/summationdisplay n=−∞∞/summationdisplay m=−∞mPnm nω1+mω2=0.( 13.56) Using (13.54) and the fact that Inm=j(nω1+mω2)Qnmgives ω1∞/summationdisplay n=−∞∞/summationdisplay m=−∞n/parenleftbig −jVnmQ∗ nm−jV−n,−mQ∗ −n,−m/parenrightbig +ω2∞/summationdisplay n=−∞∞/summationdisplay m=−∞m/parenleftbig −jVnmQ∗ nm−jV−n,−mQ∗ −n,−m/parenrightbig =0 (13.57) The double summation terms in (13.57) do not depend on ω1orω2since we can always ad- just the external circuitry so that all Vnmremain constant, and the Qnmwill remain constant as well since the capacitor charge depends directly on the voltage. Thus each summation in (13.56) must be identically zero: ∞/summationdisplay n=−∞∞/summationdisplay m=−∞nPnm nω1+mω2=0, (13.58a) ∞/summationdisplay n=−∞∞/summationdisplay m=−∞mPnm nω1+mω2=0. (13.58b) c13OscillatorsAndMixers Pozar September 16, 2011 15:44 630 Chapter 13: Oscillators and Mixers Some simplification can be carried out by eliminating the negative indices of one summa- tion by using the fact that P−n,−m=Pnm. For example, from (13.58a), ∞/summationdisplay n=−∞∞/summationdisplay m=−∞nPnm nω1+mω2=∞/summationdisplay n=0∞/summationdisplay m=−∞nPnm nω1+mω2+∞/summationdisplay n=0∞/summationdisplay m=−∞−nP −n,−m −nω1−mω2 =2∞/summationdisplay n=0∞/summationdisplay m=−∞nPnm nω1+mω2=0. This results in the usual form for the Manley–Rowe relations: ∞/summationdisplay n=0∞/summationdisplay m=−∞nPnm nω1+mω2=0, (13.59a) ∞/summationdisplay n=−∞∞/summationdisplay m=0mPnm nω1+mω2=0. (13.59b) The Manley–Rowe relations express power conservation for any lossless nonlinear reac- tance, and can be useful for harmonic generation, parametric amplifiers, and frequencyconverters at RF, microwave, and optical frequencies to predict the maximum possible power gain and conversion efficiency. Reactive frequency multipliers involve a special case of the Manley–Rowe relations since only a single source is used. If we assume a source at frequency ω 1, then setting m=0 in (13.59a) gives ∞/summationdisplay n=1Pn0=0, or ∞/summationdisplay n=2Pn0=− P10,( 13.60) where Pn0represents the power associated with the nth harmonic (the DC term for n=0i s zero). In practice, P10>0 because this represents power delivered by the source, while the summation in (13.60) represents the total power contained in all the harmonics of the input signal, as generated by the nonlinear capacitor. If all harmonics but the nth are terminated with lossless reactive loads, the power balance of (13.60) reduces to /vextendsingle/vextendsingle/vextendsingle/vextendsingleP n0 P10/vextendsingle/vextendsingle/vextendsingle/vextendsingle=1,( 13.61) indicating that it is theoretically possible to achieve 100% conversion efficiency for any harmonic. Of course, in practice, losses in the diode and matching circuitry serve to reduce the achievable efficiency substantially. A block diagram of a diode frequency multiplier is shown in Figure 13.20. An input signal of frequency f 0is applied to the diode, which is terminated with reactive loads at all frequencies except nf0, the desired harmonic. If the diode junction capacitance has a square-law I–Vcharacteristic, it is often necessary to terminate unwanted harmonics with short circuits if harmonics higher than the second are to be generated. This is because volt- ages at higher harmonics may not be generated unless lower harmonic currents are allowedto flow. These currents are commonly referred to as idler currents. For example, a varactor tripler will generally require terminations to allow idler currents at 2 f 0. Typical conversion c13OscillatorsAndMixers Pozar September 16, 2011 15:44 13.4 Frequency Multipliers 631 f0f0 input n f0n f0 outpu t Low-pass filterDiode Bandpass filter FIGURE 13.20 Block diagram of a diode frequency multiplier. efficiencies for varactor multipliers range from 50 to 80% for doublers and triplers at 50 GHz. The upper frequency limit is controlled mainly by fc, the cutoff frequency of the diode, which depends on the series resistance and dynamic junction capacitance. Typical varactor cutoff frequencies can exceed 1000 GHz, but efficient frequency multiplication requires that nf0/lessmuchfc. ResistiveDiodeMultipliers Resistive multipliers generally use forward-biased Schottky-barrier diodes to provide a nonlinear I–Vcharacteristic. Resistive multipliers are less popular than reactive multipli- ers because their efficiencies are lower, especially for higher harmonic numbers. However,resistive multipliers offer better bandwidths, and more stable operation, than reactive multi- pliers. In addition, at high millimeter wave frequencies even the best varactor diodes begin to exhibit resistive properties. Since a resistive frequency multiplier is not lossless, theManley–Rowe relations do not strictly apply. However, we can derive a similar set of rela- tions for a nonlinear resistor, and demonstrate an important result for frequency conversion using nonlinear resistors. Consider the resistive multiplier circuit shown in Figure 13.21. We have simplified the analysis by specializing to the frequency multiplier case by considering only a single source frequency—the more general case of two frequency sources is treated in reference[11]. For a source frequency ω, the nonlinear resistor generates harmonics of the form nω, so the resistor voltage and current can be written as a Fourier series: v(t)= ∞/summationdisplay m=−∞Vmejmωt, (13.62a) i(t)=∞/summationdisplay m=−∞Imejmωt. (13.62b) /H9275 2/H9275 3/H9275v(t)i(t) + DC R–... /H9275 FIGURE 13.21 Conceptual circuit for the derivation of power relations in a resistive frequency multiplier. c13OscillatorsAndMixers Pozar September 16, 2011 15:44 632 Chapter 13: Oscillators and Mixers The Fourier coefficients are determined as Vm=1 TT/integraldisplay t=0v(t)e−jmωtdt, (13.63a) Im=1 TT/integraldisplay t=0i(t)e−jmωtdt. (13.63b) Sincev(t)andi(t)are real functions, we must have Vm=V∗ −mandIm=I∗ −m. The power associated with the mth harmonic is (ignoring a factor of 4) Pm=2R e/braceleftbig VmI∗ m/bracerightbig =VmI∗ m+V∗ mIm.( 13.64) Multiplying Vmof (13.63a) by −m2I∗ mand summing gives −∞/summationdisplay m=−∞m2VmI∗ m=−1 TT/integraldisplay t=0v(t)∞/summationdisplay m=−∞m2I∗ me−jmωtdt.( 13.65) Next, use the result that ∂2i(t) ∂t2=−∞/summationdisplay m=−∞m2ω2Imejmωt=−∞/summationdisplay m=−∞m2ω2I∗ me−jmωt to write (13.65) as −∞/summationdisplay m=−∞m2VmI∗ m=1 ω2TT/integraldisplay t=0v(t)∂2i(t) ∂t2dt =1 2πωv(t)∂i(t) ∂t/vextendsingle/vextendsingle/vextendsingle/vextendsingleT t=0−1 2πωT/integraldisplay t=0∂v(t) ∂t∂i(t) ∂tdt. (13.66) Since v(t)andi(t)are periodic functions (period T),w eh a v e v(0)=v(T)andi(0)= i(T). Derivatives of i(t)have the same periodicity, so the second to last term in (13.66) vanishes. In addition, we can write ∂v(t) ∂t∂i(t) ∂t=∂v(t) ∂t∂i ∂v∂v(t) ∂t=∂i ∂v/parenleftbigg∂v(t) ∂t/parenrightbigg2 . Equation (13.66) then reduces to ∞/summationdisplay m=−∞m2VmI∗ m=1 2πωT/integraldisplay t=0∂i ∂v/parenleftbigg∂v(t) ∂t/parenrightbigg2 dt=∞/summationdisplay m=0m2/parenleftbig VmI∗ m+V∗ mIm/parenrightbig =∞/summationdisplay m=0m2Pm, or ∞/summationdisplay m=0m2Pm=1 2πωT/integraldisplay t=0∂i ∂v/parenleftbigg∂v(t) ∂t/parenrightbigg2 dt.( 13.67) For positive nonlinear resistors (defined as having an I–Vcurve whose slope is always positive), the integrand of (13.67) will always be positive. Thus (13.67) can be reduced to ∞/summationdisplay m=0m2Pm≥0.( 13.68) c13OscillatorsAndMixers Pozar September 16, 2011 15:44 13.4 Frequency Multipliers 633 If all harmonics are terminated in reactive loads except for ω(the fundamental) and mω (the desired harmonic), then (13.68) reduces to P1+m2Pm>0. The power P1>0 is de- livered by the source, while Pm<0 represents harmonic power supplied by the device. The maximum theoretical conversion efficiency is then given as/vextendsingle/vextendsingle/vextendsingle/vextendsingleP m P1/vextendsingle/vextendsingle/vextendsingle/vextendsingle≤1 m2.( 13.69) This result indicates that the efficiency of a resistive frequency multiplier drops as the square of the multiplication factor. The performance of diode frequency multipliers can often be improved by using two diodes in a balanced configuration. This can lead to increased output power, improved in- put impedance characteristics, and the rejection of certain (all even or all odd) harmonics. Two diodes can be fed using a quadrature hybrid, or two diodes can be configured in anantiparallel arrangement (back-to-back with reversed polarities). The antiparallel configu- ration will reject all even harmonics of the input frequency. TransistorMultipliers Compared to diode frequency multipliers, transistor multipliers offer better bandwidth and the possibility of conversion efficiencies greater than 100% (conversion gain). FET multi- pliers also require less input and DC power than diode multipliers. In the past, before solid- state amplifiers were available at millimeter wave frequencies, high-power diode multipli-ers were one of the few ways of generating millimeter wave power. Today, however, it is possible to generate the required frequency at low power, then amplify that signal to the de- sired power level using transistor amplifiers. This approach results in better efficiency andlower DC power requirements, and it allows the separate optimization of signal generation and amplification functions. Transistor multipliers are well suited for this application. There are several nonlinearities that exist in a FET device that can be used for har- monic generation: the transconductance near pinch-off, the output conductance near pinch- off, the rectifying properties of the Schottky gate, and the varactor-like capacitances at thegate and drain. For frequency doubler operation, the most useful of these is the rectifi- cation property, where the FET is biased to conduct only during the positive half of the input signal waveform. This results in operation similar to that of a class B amplifier,and provides a multiplier circuit that is useful for low-power output (typically less than 10 dBm) at frequencies up to 60–100 GHz. Bipolar transistors can also be used for fre- quency multiplication, with the capacitance of the collector-base junction providing thenecessary nonlinearity. The basic circuit of a class B FET frequency multiplier is shown in Figure 13.22. A unilateral device is assumed here to simplify the analysis. The source is a generator offrequency ω 0, with period T=2π/ω 0, and matched to the FET with the source impedance + – SSDG IdRs Ri CgsgmVc VcVgg Vg RdsRL jXL Cds/H9275ojXs/H11001Vdd FIGURE 13.22 Circuit diagram of an FET frequency multiplier. The transistor is modeled using a unilateral equivalent circuit. c13OscillatorsAndMixers Pozar September 16, 2011 15:44 634 Chapter 13: Oscillators and Mixers Vg Vgmax VgminVt Vgg 2T Tt 2T Tt t(a) (b)Id Vd VddVdmax VdminImax /H11002/H9270–2/H11002/H9270–2 00 2T T (c)0 FIGURE 13.23 V oltage and currents in the FET multiplier (doubler) circuit of Figure 13.22. (a) Gate voltage when the transistor is biased just below pinch-off. (b) Draincurrent, which conducts when the gate voltage is above the threshold voltage. (c) Drain voltage when the load resonator is tuned to the second harmonic. Rs+jXs. The drain of the FET is terminated with a load impedance RL+jXL, which is chosen to form a parallel RLC resonator with Cdsat the desired harmonic frequency, nω0. The gate is biased at a DC voltage of Vgg<0, while the drain is biased at Vdd>0. The operation of the FET multiplier can be understood with the help of the waveforms shown in Figure 13.23. As seen in Figure 13.23a, the FET is biased below the turn-onvoltage, V t, so the transistor does not conduct until the gate voltage exceeds Vt.T h er e - sulting drain current is shown in Figure 13.23b, and is seen to be similar in form to a half-wave rectified version of the gate voltage. This waveform is rich in harmonics, so the drain resonator can be designed to present a short circuit at the fundamental and all unde- sired harmonics, and an open circuit at the desired harmonic frequency. The resulting drain voltage for n=2 is shown in Figure 13.23c. We can make an approximate analysis of the FET multiplier by representing the drain current in terms of a Fourier series. If we assume that the drain current waveform is a half-cosine function of the form id(t)=/braceleftBigg Imaxcosπt τfor|t|<τ / 2 0f orτ/2<|t|<T/2,(13.70) where τis the duration of the drain current pulse, we can find the Fourier series as id(t)=∞/summationdisplay n=0Incos2πnt T,( 13.71) c13OscillatorsAndMixers Pozar September 16, 2011 15:44 13.4 Frequency Multipliers 635 with the Fourier coefficients given by I0=Imax2τ πT, (13.72a) In=Imax4τ πTcos(nπτ/T) 1−(2nτ/T)2forn>0. (13.72b) The coefficient Inrepresents the drain current of harmonic frequency nω0, so maximizing multiplier efficiency involves maximizing In. Since (13.72b) clearly shows that the maxi- mum value of Indecreases with n, circuits of this type are generally limited to frequency doublers or triplers. For a given value of n, the maximum value of In/Imaxdepends on the ratio τ/T:f o rn =2 the optimum occurs at τ/T=0.35, while for n=3 the optimum occurs at τ/T=0.22. Because of device and biasing constraints, however, the designer usually has very little control of the pulse width τ, and practical values of τ/Tare usu- ally greater than optimum. Examination of Figure 13.23a shows that the normalized pulse duration is related to the gate voltages Vt,Vgmin, and Vgmaxas cosπτ T=2Vt−Vgmax−Vgmin Vgmax−Vgmin.( 13.73) The gate bias voltage satisfies the relation that Vgg=(Vgmax−Vgmin)/2,( 13.74) and the peak value of the AC component of the gate voltage (frequency ω0)is given by Vg=Vgmax−Vgg.( 13.75) Then the input power delivered to the FET can be expressed as Pin=1 2|Ig|2Ri=|Vg|2Ri 2|Ri−j/ω0Cgs|2.( 13.76) If the source is conjugately matched to the transistor, the input power will be equal to the available power, Pavail. On the load side, the peak value of the AC component of the drain voltage (frequency nω0)is given by VL=InRL=(Vdmax−Vdmin)/2,( 13.77) assuming resonance of XLandCds. This gives the optimal load resistance as RL=Vdmax−Vdmin 2In.( 13.78) Then the output power at the harmonic nω0is Pn=1 2|In|2RL.( 13.79) Finally, the conversion gain is given as Gc=Pn Pavail.( 13.80) c13OscillatorsAndMixers Pozar September 16, 2011 15:44 636 Chapter 13: Oscillators and Mixers EXAMPLE 13.6 FET FREQUENCY DOUBLER DESIGN A 12–24 GHz frequency doubler is designed using a GaAs MESFET with the following parameters: Vt=−2.0V , Ri=10/Omega1,Cgs=0.20 pF, Cds=0.15 pF, andRds=40/Omega1. Assume the operating point of the transistor is chosen so that Vgmax=0.2V , Vgmin=−6.0V , Vdmax=5.0V , Vdmin=1.0 V , and Imax= 80 mA. Find the conversion gain of the multiplier. Solution We first use (13.74) and (13.75) to find the peak value of the AC input voltage. The gate bias voltage is Vgg=(Vgmax−Vgmin)/2=(0.2−6.0)/ 2=−2.9V, and the peak AC input voltage is Vg=Vgmax−Vgg=0.2+2.9=3.1V. Then the input power is given by (13.76): Pin=|Vg|2Ri 2|Ri−j/ω0Cgs|2=(3.1)2(10) 2[(10)2+(1/2π( 12×109)(0.2×10−12))2] =10.7m W . The pulse width is found from (13.73) as cosπτ T=2Vt−Vgmax−Vgmin Vgmax−Vgmin=2(−2.0) −0.2+6.0 0.2+6.0=0.29, for τ T=0.406. Then the load current for the second harmonic is given by (13.72b): I2=Imax4τ πTcos(2πτ/ T) 1−(4τ/ T)2=0.262 Imax=21.0m A . The load resistance required to match the transistor is found from (13.78): RL=Vdmax−Vdmin 2I2=5−1 2(0.021)=95.2/Omega1. The output power at 24 GHz is given by (13.79): P2=1 2|I2|2RL=1 2(0.021)2(95.2) =21.0m W . Finally, the conversion gain is, assuming the input is conjugately matched, Gc=P2 Pavail=21.0 10.7=2.9d B . The load reactance required to resonate the second harmonic is XL=1/2ω 0Cds= 44.2/Omega1, which corresponds to an inductance of 0.293 nH. ■ c13OscillatorsAndMixers Pozar September 16, 2011 15:44 13.5 Mixers 637 13.5MIXERS A mixer is a three-port device that uses a nonlinear or time-varying element to achieve frequency conversion. As introduced in Section 11.1, an ideal mixer produces an output consisting of the sum and difference frequencies of its two input signals. Operation ofpractical RF and microwave mixers is usually based on the nonlinearity provided by either a diode or a transistor. As we have seen, a nonlinear component can generate a wide variety of harmonics and other products of input frequencies, so filtering must be used to selectthe desired frequency components. Modern microwave systems typically use several mix- ers and filters to perform the functions of frequency up-conversion and down-conversion between baseband signal frequencies and RF carrier frequencies. We begin by discussing some of the important characteristics of mixers, such as image frequency, conversion loss, noise effects, and intermodulation distortion. Next we discuss the operation of single-ended mixers, using either a single diode or a transistor as thenonlinear element. The balanced diode mixer circuit is then described, followed by a brief description of more specialized mixer circuits. MixerCharacteristics The symbol and functional diagram for a mixer are shown in Figure 13.24. The mixer symbol is intended to imply that the output is proportional to the product of the two input signals. We will see that this is an idealized view of mixer operation, which in actualityproduces a large variety of harmonics and other undesired products of the input signals. Figure 13.24a illustrates the operation of frequency up-conversion , as occurs in a transmit- ter. A local oscillator (LO) signal at the relatively high frequency f LOis connected to one of the input ports of the mixer. The LO signal can be represented as vLO(t)=cos 2π fLOt.( 13.81) A lower frequency baseband or intermediate frequency (IF) signal is applied to the other mixer input. This signal typically contains the information or data to be transmitted, and FIGURE 13.24 Frequency conversion using a mixer. (a) Up-conversion. (b) Down-conversion. c13OscillatorsAndMixers Pozar September 16, 2011 15:44 638 Chapter 13: Oscillators and Mixers can be expressed for our purposes as vIF(t)=cos 2π fIFt.( 13.82) The output of the idealized mixer is given by the product of the LO and IF signals: vRF(t)=KvLO(t)vIF(t)=Kcos 2π fLOtcos 2π fIFt =K 2[cos 2π( fLO−fIF)t+cos 2π( fLO+fIF)t], (13.83) where Kis a constant accounting for the voltage conversion loss of the mixer. The RF output is seen to consist of the sum and differences of the input signal frequencies: fRF=fLO±fIF.( 13.84) The spectra of the input and output signals are shown in Figure 13.24a, where we see that the mixer has the effect of modulating the LO signal with the IF signal. The sum and difference frequencies at fLO±fIFare called the sidebands of the carrier frequency fLO, with fLO+fIFbeing the upper sideband (USB), and fLO−fIFbeing the lower sideband (LSB). A double-sideband (DSB) signal contains both upper and lower sidebands, as in (13.83), while a single-sideband (SSB) signal can be produced by filtering or by using a single-sideband mixer. Conversely, Figure 13.24b shows the process of frequency down-conversion,a su s e d in a receiver. In this case an RF input signal of the form vRF(t)=cos 2π fRFt (13.85) is applied to the input of the mixer, along with the LO signal of (13.81). The output of the mixer is vIF(t)=KvRF(t)vLO(t)=Kcos 2π fRFtcos 2π fLOt =K 2[cos 2π( fRF−fLO)t+cos 2π( fRF+fLO)t]. (13.86) Thus the mixer output consists of the sum and difference of the input signal frequencies. The spectrum for these signals is shown in Figure 13.24b. In practice, the RF and LO frequencies are relatively close together, so the sum frequency is approximately twice the RF frequency, while the difference is much smaller than fRF. The desired IF output in a receiver is the difference frequency, fRF−fLO, which is easily selected by low-pass filtering: fIF=fRF−fLO.( 13.87) Note that the above discussion only considers the sum and difference outputs as generated by multiplication of the input signals, whereas in a realistic mixer many more products willbe generated due to the more complicated nonlinear behavior of the diode or transistor. These products are usually undesirable and are removed by filtering. Image frequency: In a receiver the RF input signal at frequency f RFis typically delivered from the antenna, which may receive RF signals over a relatively wide band of frequencies. For a receiver with an LO frequency fLOand IF frequency fIF, (13.87) gives the RF input frequency that will be down-converted to the IF frequency as fRF=fLO+fIF,( 13.88a) c13OscillatorsAndMixers Pozar September 16, 2011 15:44 13.5 Mixers 639 since the insertion of (13.88a) into (13.87) yields fIF(after low-pass filtering). Now con- sider the RF input frequency given by fIM=fLO−fIF.( 13.88b) Insertion of (13.88b) into (13.87) yields −fIF(after low-pass filtering). Mathematically, this frequency is identical to fIFbecause the Fourier spectrum of any real signal is sym- metric about zero frequency, and thus contains negative as well as positive frequencies. The RF frequency defined in (13.88b) is called the image response. The image response is im- portant in receiver design because a received RF signal at the image frequency of (13.88b)is indistinguishable at the IF stage from the desired RF signal of frequency (13.88a) unless steps are taken in the RF stages of the receiver to preselect signals only within the desired RF frequency band. The choice of which RF frequency in (13.88) is the desired and which is the image response is arbitrary, depending on whether the LO frequency is above or below the desired RF frequency. Another way of viewing this difference is to note that f IFin (13.88) may be negative. Observe that the desired and image frequencies of (13.88a) and (13.88b) are separated by 2 fIF. Another implication of (13.87) and the fact that fIFmay be negative is that there are two LO frequencies that can be used for a given RF and IF frequency: fLO=fRF±fIF,( 13.89) since taking the difference frequency of fRFwith these two LO frequencies gives ±fIF. These two frequencies correspond to the upper and lower sidebands when a mixer is op-erated as an up-converter. In practice, most receivers use a local oscillator set at the upper sideband, f LO=fRF+fIF, because this requires a smaller LO tuning ratio when the re- ceiver must select RF signals over a given band. Conversion loss: Mixer design requires impedance matching at three ports, complicated by the fact that several frequencies and their harmonics are involved. Ideally, each mixer port would be matched at its particular frequency (RF, LO, or IF), and undesired frequency products would be absorbed with resistive loads, or blocked with reactive terminations.Resistive loads increase mixer losses, however, and reactive loads can be very frequency sensitive. In addition, there are inherent losses in the frequency conversion process because of the generation of undesired harmonics and other frequency products. An important fig-ure of merit for a mixer is therefore the conversion loss, which is defined as the ratio of available RF input power to the available IF output power, expressed in dB: L c=10 logavailable RF input power available IF output power≥0d B.( 13.90) Conversion loss accounts for resistive losses in a mixer as well as loss in the frequency conversion process from RF to IF ports. Conversion loss applies to both up-conversionand down-conversion, even though the context of the above definition is for the latter case. Since the RF stages of receivers operate at much lower power levels than do transmit- ters, minimum conversion loss is more critical for receivers because of the importance of minimizing losses in the RF stages to maximize receiver noise figure. Practical diode mixers typically have conversion losses between 4 and 7 dB in the 1–10 GHz range. Transistor mixers have lower conversion loss, and they may even have conversion gain of a few dB. One factor that strongly affects conversion loss is the LO power level; minimum conversion loss often occurs for LO powers between 0 and 10 dBm.This power level is large enough that the accurate characterization of mixer performance often requires nonlinear analysis. c13OscillatorsAndMixers Pozar September 16, 2011 15:44 640 Chapter 13: Oscillators and Mixers Noise figure: Noise is generated in mixers by the diode or transistor elements, and by ther- mal sources due to resistive losses. Noise figures of practical mixers range from 1 to 5 dB, with diode mixers generally achieving lower noise figures than transistor mixers. The noise figure of a mixer depends on whether its input is a single-sideband signal or a double-sideband signal. This is because the mixer will down-convert noise at both sideband fre- quencies (since these have the same IF), but the power of a SSB signal is one-half that of a DSB signal (for the same amplitude). To derive the relation between the noise figure forthese two cases, first consider a DSB input signal of the form v DSB(t)=A[cos(ωLO−ωIF)t+cos(ω LO+ωIF)t].( 13.91) Upon mixing with an LO signal cos ωLOtand low-pass filtering, the down-converted IF signal will be vIF(t)=AK 2cos(ω IFt)+AK 2cos(−ω IFt)=AKcosωIFt,( 13.92) where Kis a constant accounting for the conversion loss for each sideband. The average power of the DSB input signal of (13.91) is Si=A2 2+A2 2=A2, and the average power of the output IF signal is So=A2K2 2. For noise figure, the input noise power is defined as Ni=kT0B, where T0=290 K and B is the IF bandwidth. The total output noise power is equal to the input noise plus Nadded, the noise power added by the mixer, divided by the conversion loss (assuming a reference at the mixer input): No=(KT0B+Nadded) Lc. Then using the definition of noise figure gives the DSB noise figure of the mixer as FDSB=SiNo SoNi=2 K2Lc/parenleftbigg 1+Nadded kT0B/parenrightbigg .( 13.93) The corresponding analysis for the SSB case begins with a SSB input signal of the form vSSB(t)=Acos(ωLO−ωIF)t.( 13.94) Upon mixing with the LO signal cos ωLOtand low-pass filtering, the down-converted IF signal will be vIF(t)=AK 2cos(ωIFt). (13.95) The average power of the SSB input signal of (13.94) is Si=A2 2, c13OscillatorsAndMixers Pozar September 16, 2011 15:44 13.5 Mixers 641 and the average power of the output IF signal is So=A2K2 8. The input and output noise powers are the same as for the DSB case, so the noise figure for an SSB input signal is FSSB=SiNo SoNi=4 K2Lc/parenleftbigg 1+Nadded kT0B/parenrightbigg .( 13.96) Comparison with (13.93) shows that the noise figure of the SSB case is twice that of the DSB case: FSSB=2FDSB.( 13.97) Other mixer characteristics : Since mixers involve nonlinearity, they will produce inter- modulation products. Typical values of IIP3for mixers range from 15 to 30 dBm. Another important characteristic of a mixer is the isolation between the RF and LO ports. Ideally, the LO and RF ports would be decoupled, but internal impedance mismatches and limi- tations of coupler performance often result in some LO power being coupled out of theRF port. This is a potential problem for receivers that drive the RF port directly from the antenna because LO power coupled through the mixer to the RF port will be radiated by the antenna. Because such signals can interfere with other services or users, regulatoryagencies often set stringent limits on the RF power radiated by receivers. This problem can be largely alleviated by using a bandpass filter between the antenna and mixer, or by using an RF amplifier ahead of the mixer. Isolation between the LO and RF ports is highly dependent on the type of coupler used for diplexing these two inputs, but typical values range from 20 to 40 dB. EXAMPLE 13.7 IMAGE FREQUENCY The IS-54 digital cellular telephone system uses a receive frequency band of 869– 894 MHz, with a first IF frequency of 87 MHz and a channel bandwidth of 30 kHz. What are the two possible ranges for the LO frequency? If the upper LO frequencyrange is used, determine the image frequency range. Does the image frequency fall within the receive passband? Solution By (13.89), the two possible LO frequency ranges are f LO=fRF±fIF=(869 to 894) ±87=/braceleftbigg956 to 981 MHz 782 to 807 MHz. Using the 956–981 MHz LO, we find that (13.87) gives the IF frequency as fIF=fRF−fLO=(869 to 894) −(956 to 981) =−87 MHz , so from (13.88b) the RF image frequency range is fIM=fLO−fIF=(956 to 981) +87=1043 to 1068 MHz, which is well outside the receive passband. ■ c13OscillatorsAndMixers Pozar September 16, 2011 15:44 642 Chapter 13: Oscillators and Mixers FIGURE 13.25 (a) Circuit for a single-ended diode mixer. (b) Idealized equivalent circuit. The above treatment of mixers is idealized because of the assumption that the output was proportional to the product of the input signals, thus producing only sum and difference frequencies (for sinusoidal inputs). We now discuss more realistic mixers and show that the output does indeed contain a term proportional to the product of the inputs, but it alsocontains many higher order products as well. Single-EndedDiodeMixer A basic diode mixer circuit is shown in Figure 13.25a. This type of mixer is called a single- ended mixer because it uses a single diode element. The RF and LO inputs are combined in adiplexer, which superimposes the two input voltages to drive the diode. The diplexing function can be implemented using a directional coupler or hybrid junction to providesignal combining as well as isolation between the two inputs. The diode may be biased with a DC bias voltage, which must be decoupled from the RF signal paths. This is done by using DC blocking capacitors on either side of the diode, and an RF choke between thediode and the bias voltage source. The AC output of the diode is passed through a low- pass filter to provide the desired IF output voltage. This description is for application as a down-converter, but the same mixer can be used for up-conversion since each port may beused interchangeably as an input or output port. The AC equivalent circuit of the mixer is shown in Figure 13.25b, where the RF and LO input voltages are represented as two series-connected voltage sources. Let the RF input voltage be a cosine wave of frequency ω RF: vRF(t)=VRFcosωRFt,( 13.98) and let the LO input voltage be a cosine wave of frequency ωLO: vLO(t)=VLOcosωLOt.( 13.99) c13OscillatorsAndMixers Pozar September 16, 2011 15:44 13.5 Mixers 643 Using the small-signal approximation of (11.6) gives the total diode current as i(t)=I0+Gd[vRF(t)+vLO(t)]+G/prime d 2[vRF(t)+vLO(t)]2+···.( 13.100) The first term in (13.100) is the DC bias current, which will be blocked from the IF output by the DC blocking capacitors. The second term is a replication of the RF and LO input signals, which will be filtered out by the low-pass IF filter. This leaves the third term, which can be rewritten using trigonometric identities as i(t)=G/prime d 2(VRFcosωRFt+VLOcosωLOt)2 =G/prime d 2/parenleftbig V2 RFcos2ωRFt+2VRFVLOcosωRFtcosωLOt+V2 LOcos2ωLOt/parenrightbig =G/prime d 4/bracketleftbig V2 RF(1+cos 2ω RFt)+V2 LO(1+cos 2ω LOt)+2VRFVLOcos(ω RF−ωLO)t +2VRFVLOcos(ωRF+ωLO)t/bracketrightbig . This result is seen to contain several new signal components, only one of which produces the desired IF difference product. The two DC terms again will be blocked by the blocking capacitors, and the 2ω RF,2ωLO, andωRF+ωLOterms will be blocked by the low-pass filter. This leaves the IF output current as iIF(t)=G/prime d 2VRFVLOcosωIFt,( 13.101) where ωIF=ωRF−ωLOis the IF frequency. The spectrum of the down-converting single- ended mixer is thus identical to that of the idealized mixer shown in Figure 13.24b. Single-EndedFETMixer There are several FET parameters that offer nonlinearities that can be used for mixing, but the strongest is the transconductance, gm, when the FET is operated in a common source configuration with a negative gate bias. Figure 13.26 shows the variation of transconduc-tance with gate bias for a typical FET. When used as an amplifier, the gate bias voltage is chosen near zero, or slightly positive, so the transconductance is near its maximum value, and the transistor operates as a linear device. When the gate bias is near the pinch-off re- gion, where the transconductance approaches zero, a small positive variation of gate volt- age can cause a large change in transconductance, leading to a nonlinear response. Thus the LO voltage can be applied to the gate of the FET to pump the transconductance to FIGURE 13.26 Variation of FET transconductance versus gate-to-source voltage. c13OscillatorsAndMixers Pozar September 16, 2011 15:44 644 Chapter 13: Oscillators and Mixers FIGURE 13.27 Circuit for a single-ended FET mixer. switch the FET between high- and low-transconductance states, thus providing the desired mixing function. The circuit for a single-ended FET mixer is shown in Figure 13.27. A diplexing cou- pler is again used to combine the RF and LO signals at the gate of the FET. An impedance matching network is also usually required between the inputs and the FET, which typicallypresents a very low input impedance. RF chokes are used to bias the gate at a negative voltage near pinch-off, and to provide a positive bias for the drain of the FET. A bypass ca- pacitor at the drain provides a return path for the LO signal, and a low-pass filter provides the final IF output signal. Our analysis of the mixer of Figure 13.27 follows the original work described in ref- erence [12]. The simplified equivalent circuit is shown in Figure 13.28, and is based on the unilateral equivalent circuit of a FET introduced in Section 11.3. The RF and LO input voltages are given in (13.98) and (13.99). Let Z g=Rg+jXgbe the Thevenin source impedance for the RF input port, and let ZL=RL+jXLbe the Thevenin source impedance at the IF output port. These impedances are complex to allow a conjugate match at the input and output ports for maximum power transfer. The LO port has a real genera-tor impedance of Z 0since we are not concerned with maximum power transfer for the LO signal. Since the FET transconductance is driven by the LO signal, its time variation can be expressed as a Fourier series in terms of harmonics of the LO: g(t)=g0+2∞/summationdisplay n=1gncosnω0t.( 13.102) Because we do not have an explicit formula for the transconductance, we cannot calcu- late directly the Fourier coefficients of (13.102), but must rely on measurements for these FIGURE 13.28 Equivalent circuit for the FET mixer of Figure 13.27. c13OscillatorsAndMixers Pozar September 16, 2011 15:44 13.5 Mixers 645 values. As we will see, the desired down-conversion result is due solely to the n=1t e r m of the Fourier series, so we only need the g1coefficient. Measurements typically give a value in the range of 10 mS for g1. The conversion gain of the FET mixer can be found as Gc=PIF-avail PRF-avail=/vextendsingle/vextendsingle/vextendsingleVIF D/vextendsingle/vextendsingle/vextendsingle2 RL |ZL|2 |VRF|2 4Rg=4RgRL |ZL|2/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleVIF D VRF/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle2 ,( 13.103) where VIF Dis the IF drain voltage, and the impedances ZgandZLare chosen for maximum power transfer at the RF and IF ports. The RF frequency component of the phasor volt- age across the gate-to-source capacitance is given in terms of the voltage divider between Zg,Ri, and Cgs: VRF c=VRF jωRFCgs/bracketleftbigg (Ri+Zg)−j ωRFCgs/bracketrightbigg=VRF 1+jωRFCgs(Ri+Zg).( 13.104) Multiplying the transconductance of (13.102) by vRF c(t)=VRF ccosωRFtgives terms of the form gm(t)vRF c(t)=g0VRF ccosωRFt+2g1VRF ccosωRFtcosωLOt+···.( 13.105) The down-converted IF frequency component can be extracted from the second term of (13.105) using the usual trigonometric identity gm(t)vRF c(t)/vextendsingle/vextendsingle/vextendsingle ωIF=g1VRF ccosωIFt,( 13.106) where ωIF=ωRF−ωLO. Then the IF component of the drain voltage is, in phasor form, VIF D=−g1VRF c/parenleftbiggRdZL Rd+ZL/parenrightbigg =−g1VRF 1+jωRFCgs(Ri+Zg)/parenleftbiggRdZL Rd+ZL/parenrightbigg ,( 13.107) where (13.104) has been used. Using this result in (13.103) gives the conversion gain (be- fore conjugate matching) as Gc/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsinglenot matched=/parenleftbigg2g1Rd ωRFCgs/parenrightbigg2Rg/bracketleftBigg (Ri+Rg)2+/parenleftbigg Xg−1 ωRFCgs/parenrightbigg2/bracketrightBiggRL/bracketleftbig (Rd+RL)2+X2 L/bracketrightbig. We now conjugately match the RF and IF ports to maximize the conversion gain. Thus we letRg=Ri,Xg=1/ωRFCgs,RL=Rd, and XL=0, which reduces the above result to Gc=g2 1Rd 4ω2 RFC2gsRi.( 13.108) The quantities g1,Rd,Ri, and Cgsare all parameters of the FET. Practical mixer circuits generally use matching circuits to transform the FET impedance to 50 /Omega1for the RF, LO, and IF ports. c13OscillatorsAndMixers Pozar September 16, 2011 15:44 646 Chapter 13: Oscillators and Mixers EXAMPLE 13.8 MIXER CONVERSION GAIN A single-ended FET mixer is to be designed for a wireless local area network re- ceiver operating at 2.4 GHz. The parameters of the FET are Rd=300/Omega1,Ri= 10/Omega1,Cgs=0.3 pF, and g1=10 mS. Calculate the maximum possible conver- sion gain. Solution This is a straightforward application of the formula for conversion gain given in (13.108): Gc=g2 1Rd 4ω2 RFC2gsRi=(10×10−3)2(300) 4(2π)2(2.4×109)2(10)=36.6=15.6d B . Note that this value does not include losses due to the necessary impedance match- ing networks. ■ BalancedMixer RF input matching and RF-LO isolation can be improved through the use of a balanced mixer, which consists of two single-ended mixers combined with a hybrid junction. Figure 13.29 shows the basic configuration, with either a 90◦hybrid (Figure 13.29a), or a 180◦hybrid (Figure 13.29b). As we will see, a balanced mixer using a 90◦hybrid junc- tion will ideally lead to a perfect input match at the RF port over a wide frequency range, while the use of a 180◦hybrid will ideally lead to perfect RF-LO isolation over a wide fre- quency range. In addition, both mixers will reject all even-order intermodulation products. Figure 13.30 shows a photograph of a microstrip circuit that contains several balanced mixers. We can analyze the performance of a balanced mixer using the small-signal approach that was used for the single-ended diode mixer. Here we will concentrate on the balanced FIGURE 13.29 Balanced mixer circuits. (a) Using a 90◦hybrid. (b) Using a 180◦hybrid. c13OscillatorsAndMixers Pozar September 16, 2011 15:44 13.5 Mixers 647 FIGURE 13.30 Photograph of a 35 GHz microstrip monopulse radar receiver circuit. Three bal- anced mixers using ring hybrids can be seen, along with three stepped-impedance low-pass filters, and six quadrature hybrids. Eight feedlines are aperture coupled to microstrip antennas on the reverse side. The circuit also contains a Gunn diodesource for the local oscillator. Courtesy of Millitech Inc., Northampton, Mass. mixer with a 90◦hybrid, shown in Figure 13.29a, and leave the 180◦hybrid case as a problem. As usual, let the RF and LO voltages be defined as vRF(t)=VRFcosωRFt,( 13.109) and vLO(t)=VLOcosωLOt.( 13.110) From Section 7.5, the scattering matrix for the 90◦hybrid junction is [S]=−1√ 2⎡ ⎢⎣0j10 j001 100 j 01 j0⎤ ⎥⎦,( 13.111) c13OscillatorsAndMixers Pozar September 16, 2011 15:44 648 Chapter 13: Oscillators and Mixers where the ports are numbered as shown in Figure 13.29a. The total RF and LO voltages applied to the two diodes can then be written as v1(t)=1√ 2/bracketleftbig VRFcos/parenleftbig ωRFt−90◦/parenrightbig +VLOcos/parenleftbig ωLOt−180◦/parenrightbig/bracketrightbig =1√ 2(VRFsinωRFt−VLOcosωLOt), (13.112a) v2(t)=1√ 2/bracketleftbig VRFcos/parenleftbig ωRFt−180◦/parenrightbig +VLOcos/parenleftbig ωLOt−90◦/parenrightbig/bracketrightbig =1√ 2(−VRFcosωRFt+VLOsinωLOt). (13.112b) Using only the quadratic term from the small-signal diode approximation of (11.6) gives the diode currents as i1(t)=Kv2 1=K 2/parenleftbig V2 RFsin2ωRFt−2VRFVLOsinωRFcosωLOt+V2 LOcos2ωLOt/parenrightbig , (13.113a) i2(t)=− Kv2 2=−K 2/parenleftbig V2 RFcos2ωRFt−2VRFVLOcosωRFsinωLOt+V2 LOsin2ωLOt/parenrightbig , (13.113b) where the negative sign on i2accounts for the reversed diode polarity, and Kis a constant for the quadratic term of the diode response. Adding these two currents at the input to the low-pass filter gives i1(t)+i2(t)=−K 2/parenleftbig V2 RFcos 2ω RFt+2VRFVLOsinωIFt−V2 LOcos 2ω LOt/parenrightbig , where the usual trigonometric identities have been used, and ωIF=ωRF−ωLOis the IF frequency. Note that the DC components of the diode currents cancel upon combining.After low-pass filtering, the IF output is i IF(t)=− KVRFVLOsinωIFt,( 13.114) as desired. We can also calculate the input match at the RF port and the coupling between the RF and LO ports. If we assume the diodes are matched and that each exhibits a voltage reflection coefficient /Gamma1at the RF frequency, then the phasor expression for the reflected RF voltages at the diodes will be V/Gamma11=/Gamma1V1=−j/Gamma1VRF√ 2, (13.115a) V/Gamma12=/Gamma1V2=−/Gamma1VRF√ 2. (13.115b) c13OscillatorsAndMixers Pozar September 16, 2011 15:44 13.5 Mixers 649 These reflected voltages appear at ports 2 and 3 of the hybrid, respectively, and com- bine to form the following outputs at the RF and LO ports: VRF /Gamma1=−jV/Gamma11√ 2−V/Gamma12√ 2=−1 2/Gamma1VRF+1 2/Gamma1VRF=0, (13.116a) VLO /Gamma1=−V/Gamma12√ 2−jV/Gamma11√ 2=1 2j/Gamma1VRF+1 2j/Gamma1VRF=j/Gamma1VRF. (13.116b) Thus we see that the phase characteristics of the 90◦hybrid lead to perfect cancellation of reflections at the RF port. The isolation between the RF and LO ports, however, is depen- dent on the matching of the diodes, which may be difficult to maintain over a reasonablefrequency range. ImageRejectMixer We have already discussed the fact that two distinct RF input signals at frequencies ω RF= ωLO±ωIFwill down-convert to the same IF frequency when mixed with ωLO. These two frequencies are the upper and lower sidebands of a double-sideband signal. The desired response can be arbitrarily selected as either the LSB (ω LO−ωIF)o rt h eU S B( ωLO+ωIF), assuming a positive IF frequency. The image reject mixer , shown in Figure 13.31, can be used to isolate these two responses into separate output signals. The same circuit can also be used for up-conversion, in which case it is usually called a single-sideband modulator . In this case, the IF input signal is delivered to either the LSB or the USB port of the IF hybrid, and the associated single-sideband signal is produced at the RF port of the mixer. We can analyze the image reject mixer using the small-signal approximation. Let the RF input signal be expressed as vRF(t)=VUcos(ω LO+ωIF)t+VLcos(ωLO−ωIF)t,( 13.117) where VUandVLrepresent the amplitudes of the upper and lower sidebands, respectively. Using the scattering matrix given in (13.111) for the 90◦hybrid gives the RF voltages at the diodes as vA(t)=1√ 2/bracketleftbig VUcos/parenleftbig ωLOt+ωRFt−90◦/parenrightbig +VLcos/parenleftbig ωLOt−ωIFt−90◦/parenrightbig/bracketrightbig =1√ 2[VUsin(ωLO+ωIF)t+VLsin(ωLO−ωIF)t], (13.118a) FIGURE 13.31 Circuit for an image reject mixer. c13OscillatorsAndMixers Pozar September 16, 2011 15:44 650 Chapter 13: Oscillators and Mixers vB(t)=1√ 2/bracketleftbig VUcos/parenleftbig ωLOt+ωIFt−180◦/parenrightbig +VLcos/parenleftbig ωLOt−ωIFt−180◦/parenrightbig/bracketrightbig =−1√ 2[VUcos(ω LO+ωIF)t+VLcos(ωLO−ωIF)t]. (13.118b) After mixing with the LO signal of (13.110) and low-pass filtering, the IF inputs to the IF hybrid are vA IF(t)=KVLO 2√ 2(VU−VL)sinωIFt, (13.119a) vB IF(t)=−KVLO 2√ 2(VU+VL)cosωIFt, (13.119b) where Kis the mixer constant for the squared term of the diode response. The phasor representation of the IF signals of (13.119) is VA IF=−jKV LO 2√ 2(VU−VL), (13.120a) VB IF=−KVLO 2√ 2(VU+VL). (13.120b) Combining these voltages in the IF hybrid gives the following outputs: V1=− jVA IF√ 2−VB IF√ 2=KVLOVL 2(LSB) , (13.121a) V2=−VA IF√ 2−jVB IF√ 2=−jKV LOVU 2(USB) , (13.121b) which we see are the separate sidebands of the down-converted input signal of (13.117). These outputs can be expressed in time domain form as v1(t)=KVLOVL 2cosωIFt, (13.122a) v2(t)=KVLOVU 2sinωIFt, (13.122b) which clearly shows the presence of a 90◦phase shift between the two sidebands. Also note that the image rejection mixer does not incur any additional losses beyond the usual conversion losses of the single rejection mixer. A practical difficulty with image rejection mixers is in fabricating a good hybrid at the relatively low IF frequency. Losses, and hence noise figure, are also usually greater than for a simpler mixer. DifferentialFETMixerandGilbertCellMixer The mixer shown in Figure 13.32a uses two FETs in a differential balanced configuration, similar to the differential amplifier discussed in Section 12.4. The LO input voltage and IF output voltage are balanced signals; baluns may be used at these ports to convert to single- ended signals. The RF input is single ended, and is applied to the bottom transistor. The RF and LO voltages can be written as vRF(t)=VRFcosωRFt (13.123a) v± LO(t)=± VLOcosωLOt (13.123b) c13OscillatorsAndMixers Pozar September 16, 2011 15:44 13.5 Mixers 651 RDVdd RD vIF +vIF – vLO+vLO– vRFRDVdd RD /H9275LO gmVRFvIF +vIF – –T 4T 4T 2T 3T 43T 27T 49T 42Tt 5T 25T 4vLO+(t) g (t)(a) (b) (c) FIGURE 13.32 (a) A singly balanced differential FET mixer. (b) Simplified equivalent circuit. (c) LO voltage waveform and idealized switching waveform of the top left FET. Conceptually, the circuit operates as an alternating switch, with the LO turning the top two FETs on and off with alternate half-cycles of the LO voltage. As with the differential mode of the differential amplifier, the connection between the sources of the upper FETsis a virtual ground for the LO voltage. These transistors are biased slightly above pinch- off, so each will be conducting for slightly more than half of each LO cycle. During the positive half-cycle of v + LOthe top left FET will conduct with a low resistance, and it will turn off during the negative half cycle. During the positive half cycle of v− LO(which occurs during the negative half cycle of v+ LO) the top right FET will turn on. Thus, one of the upper FETs is always conducting. The lower FET is biased into saturation and operates asa normal RF amplifier, providing RF current through the upper switches. The RF current at the drain of the bottom FET is approximately I RF=gmVRF. The RF and LO ports require impedance matching, and the IF output circuit should provide a return path to ground for the LO signal. A current source, or inductive degeneration, is often used on the source of the lower FET. A simplified equivalent circuit is shown in Figure 13.32b, with the upper FETs re- placed with ideal switches. The effect of the switches can be modeled by using a Fourier series for the idealized conductance waveform shown in Figure 13.32c. The first few termsof this Fourier series can be found as g(t)=1 2+2 πcosωLOt−2 3πcos 3ω LOt+··· (13.124) c13OscillatorsAndMixers Pozar September 16, 2011 15:44 652 Chapter 13: Oscillators and Mixers LO Switches RF Amplifiers Current Sourcev+vRFvLOvIF+ + +– – –RD RDVdd FIGURE 13.33 A Gilbert cell mixer. Then the voltages at the IF terminals can be expressed as v+ IF(t)=−g(t)IRFRD=− gmVRFRD/parenleftbigg1 2+2 πcosωLOt/parenrightbigg cosωRFt (13.125a) v− IF(t)=− [1−g(t)]IRFRD=− gmVRFRD/parenleftbigg1 2−2 πcosωLOt/parenrightbigg cosωRFt,(13.125b) where we have retained only the first two terms of (13.124). The net output IF voltage is vIF(t)=v+ IF(t)−v− IF(t)=−4 πgmVRFRDcosωLOtcosωRFt.( 13.126) Note that voltages at the RF and LO frequencies are canceled (without filtering), leaving only terms at the frequencies ωLO±ωRF. After low-pass filtering, the IF output is vIF(t)/vextendsingle/vextendsingle/vextendsingle/vextendsingle LPF=−2 πgmVRFRDcosωIFt.( 13.127) TheGilbert cell mixer, shown in Figure 13.33, uses two singly balanced FET mixers of the type shown in Figure 13.32a to form a double-balanced mixer . It has fully balanced (differential) ports for LO, RF, and IF signals. Due to the symmetry of the circuit and its ex- citations, the RF and LO signals are canceled at the IF output port. Operation is the same as the singly balanced FET mixer, with the four upper FETs operating as switches controlled by the LO voltage, and the lower two FETs operating as a balanced amplifier for the RF input signal. The circuit of Figure 13.33 includes a current source at the sources of theamplifier FETs. This mixer is frequently used in CMOS RFICs for wireless applications. OtherMixers There are a number of other mixer circuits that provide various advantages in terms of bandwidth,harmonic generation, and intermodulation products.The double-balanced mixerof Figure 13.34 uses two hybrid junctions or transformers, and provides good isolation be- tween all three ports, as well as rejection of all even harmonics of the RF and LO signals. This leads to very good conversion loss, but less than ideal input matching at the RF port.The double-balanced mixer also provides a higher third-order intercept point than either a single-ended mixer or a balanced mixer. c13OscillatorsAndMixers Pozar September 16, 2011 15:44 13.5 Mixers 653 FIGURE 13.34 Double-balanced mixer circuit. Bandpass filter for RFRF input /H9275r /H9275r /H9275i/H9275i /H92750 Lowpass filter for LO and IFLO input /H92750 =1 2(/H9275r – /H9275i)Lowpass filter for IF IF output FIGURE 13.35 Subharmonically pumped mixer using an antiparallel diode pair. Figure 13.35 shows the circuit for an antiparallel diode mixer, which is often used for subharmonically pumped millimeter wave frequency conversion. The back-to-back diodesfunction as a frequency doubler, thus requiring an LO frequency of one-half the usual value. The diode nonlinearity operates as a resistive frequency multiplier to generate the second harmonic of the LO to mix with the RF input to produce the desired output fre-quency. The antiparallel diode pair has a symmetric I–Vcharacteristic that suppresses the fundamental mixing product of the RF and LO input signals, leading to better conversion loss. A photograph of a SiGe MMIC down-converter using a subharmonic mixer is shownin Figure 13.36. Table 13.1 summarizes the characteristics of several of the mixers that we have discussed. TABLE 13.1 Mixer Characteristics Mixer Number of RF Input RF-LO Conversion Third-Order Type Diodes Match Isolation Loss Intercept Single ended 1 Poor Fair Good Fair Balanced (90◦) 2 Good Poor Good Fair Balanced (180◦) 2 Fair Excellent Good Fair Double balanced 4 Poor Excellent Excellent Excellent Image reject 2 or 4 Good Good Good Good c13OscillatorsAndMixers Pozar September 16, 2011 15:44 654 Chapter 13: Oscillators and Mixers (a) (b) LO Balun Input Balun RF InputLNA1LNA2LNA3Subharmonic Mixer LO AmpLO Input IF Filter IF Gain Adjust AmpIF Output Buffer IF Outpu tRF FilterSiGe Chip FIGURE 13.36 (a) Photograph of a monolithic integrated millimeter wave down converter using silicon germanium (SiGi). (b) Block diagram of the chip. The circuit operates from 43.5 to 45.5 GHz and includes differential LNA, LO, and IF amplifiers, a differential subharmonic mixer, and an off-chip RF filter. The noise figure is lessthan 6 dB. Courtesy of Hittite Microwave Corporation. REFERENCES [1] L. E. Larson, RF and Microwave Circuit Design for Wireless Communications , Artech House, Norwood, Mass., 1996. [2] J. R. Smith, Modern Communication Circuits , 2nd edition, McGraw-Hill, New York, 1998. [3] U. L. Rohde, Microwave and Wireless Synthesizers: Theory and Design , Wiley-Interscience, New York, 1997. c13OscillatorsAndMixers Pozar September 16, 2011 15:44 Problems 655 [4] G. D. Vendelin, A. M. Pavio, and U. L. Rohde, Microwave Circuit Design Using Linear and Nonlin- ear Techniques, John Wiley & Sons, New York, 1990. [5] S. A. Maas, Nonlinear Microwave and RF Circuits, 2nd edition, Artech House, Norwood, Mass., 2003. [6] Y . Komatsu and Y . Murakami, “Coupling Coefficient Between Microstrip Line and Dielectric Res- onator,” IEEE Transactions on Microwave Theory and Techniques , vol. MTT-31, pp. 34–40, January 1983. [7] D. B. Leeson, “A Simple Model of Feedback Oscillator Noise Spectrum,” Proceedings of the IEEE, vol. 54, pp. 329–330, 1966. [8] A. Leon-Garcia, Probability and Random Processes for Electrical Engineering, 2nd edition, Addison-Wesley, Reading, Mass., 1994. [9] M. K. Nezami, “Evaluate the Impact of Phase Noise on Receiver Performance,” Microwaves & RF Magazine, pp. 1–11, June 1998. [10] R. E. Collin, Foundations for Microwave Engineering , 2nd edition, Wiley–IEEE Press, Hoboken, N.J., 2001. [11] R. H. Pantell, “General Power Relationships for Positive and Negative Resistive Elements,” Proceed- ings of the IRE, pp. 1910–1913, December 1958. [12] R. A. Pucel, D. Masse, and R. Bera, “Performance of GaAs MESFET Mixers at X Band,” IEEE Transactions on Microwave Theory and Techniques , vol. MTT-24, pp. 351–360, June 1976. PROBLEMS 13.1 Derive the admittance matrix representation of the transistor oscillator circuit given in (13.3). 13.2 Derive the results in (13.20)–(13.22) for a Colpitts oscillator using a common emitter transistor with an inductor having a series resistance R. 13.3 Design a Colpitts oscillator operating at 200 MHz using an FET in a common gate configuration, in- cluding the effect of a lossy inductor. First derive equations for the resonant frequency and conditionrequired for sustaining oscillation for an inductor with loss, corresponding to equations (13.20)– (13.22) for the BJT case. Use these results to find the required capacitances, assuming an inductor of 15 nH with a Qof 50, and a transistor with g m=20 mS and Ro=1/Go=200/Omega1. Determine the minimum value of the inductor Qrequired to sustain oscillations. 13.4 Prove that the standard Smith chart can be used for negative resistances by plotting 1 //Gamma1∗(instead of/Gamma1). In this case, the resistance circle values are read as negative, while the reactance circles are unchanged. 13.5 Design a transistor oscillator at 1.9 GHz using a silicon BJT in a common emitter configuration driving a 50 /Omega1load on the drain side. The scattering parameters are as follows (Z0=50/Omega1):S11= 0.72/negationslash157◦,S12=0.15/negationslash56◦,S21=1.9/negationslash52◦,a n d S22=0.63/negationslash−63◦. Choose /Gamma1Lfor|/Gamma1in|/greatermuch1, and design appropriate load and terminating networks. 13.6 Repeat the oscillator design of Example 13.4 by replacing the dielectric resonator and microstrip feedline with a single-stub tuner to match /Gamma1Sto a 50 /Omega1load. Find the Qof the tuner and 50 /Omega1 load, then compute and plot |/Gamma1out|versus /Delta1f/f0. Compare with the result in Figure 13.12b for the dielectric resonator case. 13.7 Repeat the dielectric oscillator design of Example 13.4 using a GaAs MESFET having the fol- lowing scattering parameters: S11=1.2/negationslash150◦,S12=0.2/negationslash120◦,S21=3.7/negationslash−72◦,a n d S22= 1.3/negationslash−67◦. 13.8 A HEMT device in the common gate configuration has the following scattering parameters at 8 GHz (Z0=50/Omega1):S11=0.46/negationslash178◦,S12=0.045 /negationslash73◦,S21=1.41/negationslash−19◦,S22=1.02/negationslash−12◦.F o ra p - plication in an oscillator, a series inductor is added to the gate, as shown below, to increase instability. Compute and plot the µ-stability factor for Lranging from 0 to 20 nH, and determine the value that maximizes instability. (This can most easily be done with a microwave CAD package.) c13OscillatorsAndMixers Pozar September 16, 2011 15:44 656 Chapter 13: Oscillators and Mixers Port 1Port 2SD G L 13.9 An oscillator uses an amplifier with a noise figure of 6 dB and a resonator having a Qof 500, and produces a 100 MHz output at a power level of 10 dBm. If the measured fαis 50 kHz, plot the spectral density of the output noise power, and determine the phase noise (in dBc/Hz) at the following frequencies: (a) at 1 MHz from the carrier; (b) at 10 kHz from the carrier (assume K=1). 13.10 Repeat Problem 13.10 for fα=200 kHz. 13.11 Derive Equation (13.50) giving the required phase noise for a specified receiver selectivity. 13.12 Find the necessary LO phase noise specification if an 860 MHz cellular receiver with a 30 kHz channel spacing is required to have an adjacent channel rejection of 80 dB, assuming the interfering channel is at the same level as the desired channel. The final IF voice bandwidth is 12 kHz. 13.13 Apply the Manley–Rowe relations to an up-converting mixer. Assume a nonlinear reactance is ex- cited at frequencies f1(RF) and f2(LO), and terminated with open circuits at all other frequencies except f3=f1+f2. Show that the maximum possible conversion gain is given by −P11/P10= 1+ω2/ω1. 13.14 Derive the relation between pulse duration and gate voltages given in (13.73) for the FET frequency multiplier. 13.15 A double-sideband signal of the form vRF(t)=VRF[cos(ω LO−ωIF)t+cos(ωLO+ωIF)t]is ap- plied to a mixer with an LO voltage given by vLO(t)=VLOcosωLOt. Derive the output of the mixer after low-pass filtering. 13.16 A diode has an I–Vcharacteristic given by i(t)=Is(e3v(t)−1).L e tv( t)=0.1c o s ω1t+0.1c o s ω2t, and expand i(t)in a power series in v, retaining only the v,v2,a n dv3terms. For Is=1 A, find the magnitudes of the current at each frequency. 13.17 An RF input signal at 1800 MHz is down-converted in a mixer to an IF frequency of 87 MHz. What are the two possible LO frequencies and the corresponding image frequencies? 13.18 Consider a diode mixer with a conversion loss of 5 dB and a noise figure of 4 dB, and a FET mixer with conversion gain of 3 dB and a noise figure of 8 dB. If each of these mixers is followed by an IF amplifier having a gain of 30 dB and a noise figure FA, as shown below, calculate and plot the overall noise figure for both amplifier–mixer configurations for FA= 0–10 dB. 13.19 LetTSSB be the equivalent noise temperature of a mixer receiving a SSB signal, and TDSB be the temperature when it receives a DSB signal. Compute the output noise powers in each case, and show thatTSSB=2TDSB, and that therefore FSSB=2FDSB. Assume that the conversion gains for the signal and its image are identical. 13.20 If the noise power Ni=kTBis applied at the RF input port of a mixer having noise figure F(DSB) and conversion loss Lc, what is the available output noise power at the IF port? Assume the mixer is at a physical temperature T0. c13OscillatorsAndMixers Pozar September 16, 2011 15:44 Problems 657 13.21 Aphase detector produces an output signal proportional to the phase difference between two RF input signals. Let these input signals be expressed as v1=v0cosωt, v2=v0cos(ωt+θ). If these two signals are applied to a single-balanced mixer using a 90◦hybrid, show that the IF output signal, after low-pass filtering, is given by i=kv2 0sinθ, where kis a constant. If the mixer uses a 180◦hybrid, show that the corresponding output signal is given by i=kv2 0cosθ. 13.22 Analyze a balanced mixer using a 180◦hybrid junction. Find the output IF current, and the input reflections at the RF and LO ports. Show that this mixer suppresses even harmonics of the LO. Assume that the RF signal is applied to the sum port of the hybrid, and that the LO signal is appliedto the difference port. 13.23 For an image rejection mixer, let the RF hybrid have a dissipative insertion loss of L Rand the IF hybrid have a dissipative insertion loss of LI. If the component single-ended mixers each have a conversion loss Lcand noise figure F, derive expressions for the overall conversion loss and noise figure of the image rejection mixer. c14IntroToMicrowaveSystems Pozar September 26, 2011 18:40 Chapter Fourteen Introduction to Microwave Systems A microwave system consists of passive and active microwave components arranged to perform a useful function. Probably the two most important examples are microwave com- munication systems and microwave radar systems, but there are many others. In this chapter we will discuss the basic operation of several types of microwave systems to give a generaloverview of the application of microwave technology, and to show how the subjects of earlierchapters fit into the overall scheme of complete microwave systems. An important component in any radar or wireless communication system is the antenna, so we will first discuss some of the basic properties of antennas. Then we treat wireless commu-nication, radar, and radiometry systems as important applications of RF and microwave tech-nology. We also briefly discuss propagation effects, biological effects, and other miscellaneous applications. All of these topics are of sufficient depth that many books have been written for each. Our purpose here is to introduce these topics as a way of placing the earlier material in this bookin the larger context of practical system applications. The interested reader is referred to the references at the end of the chapter for more complete treatments. 14.1SYSTEMASPECTSOFANTENNAS In this section we describe some of the basic characteristics of antennas that will be needed for our study of microwave communications, radar, and remote sensing systems. We areinterested here not in the detailed electromagnetic theory of antenna operation, but rather in the systems aspect of the operation of an antenna in terms of its radiation patterns, di- rectivity, gain, efficiency, and noise characteristics. References [1] and [2] can be reviewedfor a more in-depth treatment of the fascinating subject of antenna theory and design. Figure 14.1 shows some of the different types of antennas that have been developed for commercial wireless systems. 658 c14IntroToMicrowaveSystems Pozar September 26, 2011 18:40 14.1 System Aspects of Antennas 659 FIGURE 14.1 Photograph of various millimeter wave antennas. Clockwise from top: a high-gain 38 GHz reflector antenna with radome, a prime-focus parabolic antenna, a corru- gated conical horn antenna, a 38 GHz planar microstrip array, a pyramidal hornantenna with a Gunn diode module, and a multibeam reflector antenna. A transmitting antenna can be viewed as a device that converts a guided electromag- netic wave on a transmission line into a plane wave propagating in free space. Thus, one side of an antenna appears as an electrical circuit element, while the other side provides an interface with a propagating plane wave. Antennas are inherently bidirectional, in thatthey can be used for both transmit and receive functions. Figure 14.2 illustrates the basic FIGURE 14.2 Basic operation of transmitting and receiving antennas. c14IntroToMicrowaveSystems Pozar September 26, 2011 18:40 660 Chapter 14: Introduction to Microwave Systems operation of transmitting and receiving antennas. The transmitter can be modeled as a Thevenin source consisting of a voltage generator and series impedance, delivering a power Ptto the transmitting antenna. A transmitting antenna radiates a spherical wave that, at large distances, approximates a plane wave over a localized area. A receiving antenna in-tercepts a portion of an incident plane wave, and delivers a receive power P rto the receiver load impedance. A wide variety of antennas have been developed for different applications, as summa- rized in the following categories: rWire antennas include dipoles, monopoles, loops, sleeve dipoles, Yagi–Uda arrays, and related structures. Wire antennas generally have low gains, and are most oftenused at lower frequencies (HF to UHF). They have the advantages of light weight, low cost, and simple design. rAperture antennas include open-ended waveguides, rectangular or circular horns, reflectors, lenses, and reflectarrays. Aperture antennas are most commonly used at microwave and millimeter wave frequencies, and have moderate to high gains.rPrinted antennas include printed slots, printed dipoles, and microstrip patch an- tennas. These antennas can be made with photolithographic methods, with both radiating elements and associated feed circuitry fabricated on dielectric substrates. Printed antennas are most often used at microwave and millimeter wave frequencies,and can be easily arrayed for high gain. rArray antennas consist of a regular arrangement of antenna elements with a feed network. Pattern characteristics such as beam pointing angle and sidelobe levels can be controlled by adjusting the amplitude and phase excitation of the array elements. An important type of array antenna is the phased array, in which variable-phase shifters are used to electronically scan the main beam of the antenna. FieldsandPowerRadiatedbyanAntenna While we do not require detailed solutions to Maxwell’s equations for our purposes, we do need to be familiar with the far-zone electromagnetic fields radiated by an antenna. Con-sider an antenna located at the origin of a spherical coordinate system. At large distances, where the localized near-zone fields are negligible, the radiated electric field of an arbitrary antenna can be expressed as ¯E(r,θ,φ) =/bracketleftbigˆθF θ(θ,φ) +ˆφFφ(θ,φ)/bracketrightbige−jk0r rV/m,( 14.1) where ¯Eis the electric field vector, ˆθandˆφare unit vectors in the spherical coordinate system, ris the radial distance from the origin, and k0=2π/λ is the free-space propaga- tion constant, with wavelength λ=c/f. Also defined in (14.1) are the pattern functions, Fθ(θ,φ) andFφ(θ,φ). The interpretation of (14.1) is that this electric field propagates in the radial direction with a phase variation of e−jk0rand an amplitude variation with dis- tance of 1/ r. The electric field may be polarized in either the ˆθorˆφdirection, but not in the radial direction, since this is a TEM wave. The magnetic fields associated with the electric field of (14.1) can be found from (1.76) as Hφ=Eθ η0, (14.2a) Hθ=−Eφ η0, (14.2b) c14IntroToMicrowaveSystems Pozar September 26, 2011 18:40 14.1 System Aspects of Antennas 661 where η0=377/Omega1, the wave impedance of free-space. Note that the magnetic field vector is also polarized only in the transverse directions. The Poynting vector for this wave is given by (1.90) as ¯S=¯EׯH∗W/m2,( 14.3) and the time-average Poynting vector is ¯Savg=1 2Re{¯S}=1 2Re{¯EׯH∗}W/m2.( 14.4) We mentioned earlier that at large distances the near fields of an antenna are negli- gible, and that the radiated electric field can be written as in (14.1). We can give a more precise meaning to this concept by defining the far-field distance as the distance where the spherical wave front radiated by an antenna becomes a close approximation to the ideal planar phase front of a plane wave. This approximation applies over the radiating aperture of the antenna, and so it depends on the maximum dimension of the antenna. If we call this maximum dimension D, then the far-field distance is defined as Rff=2D2 λm. (14.5) This result is derived from the condition that the actual spherical wave front radiated by the antenna departs less than π/8=22.5◦from a true plane wave front over the maximum extent of the antenna. For electrically small antennas, such as short dipoles and small loops,this result may give a far-field distance that is too small; in this case, a minimum value of R ff=2λshould be used. EXAMPLE 14.1 FAR-FIELD DISTANCE OF AN ANTENNA A parabolic reflector antenna used for reception with the direct broadcast sys- tem (DBS) is 18 inches in diameter and operates at 12.4 GHz. Find the far-field distance for this antenna. Solution The operating wavelength at 12.4 GHz is λ=c f=3×108 12.4×109=2.42 cm. The far-field distance is found from (14.5), after converting 18 inches to 0.457 m: Rff=2D2 λ=2(0.457)2 0.0242=17.3m . The actual distance from a DBS satellite to Earth is about 36,000 km, so it is safe to say that the receive antenna is in the far-field of the transmitting antenna. ■ Next, define the radiation intensity of the radiated electromagnetic field as U(θ,φ) =r2|¯Savg|=r2 2Re/braceleftbig Eθˆθ×H∗ φˆφ+Eφˆφ×H∗ θˆθ/bracerightbig =r2 2η0/bracketleftbig |Eθ|2+|Eφ|2/bracketrightbig =1 2η0/bracketleftbig |Fθ|2+|Fφ|2/bracketrightbig W, (14.6) c14IntroToMicrowaveSystems Pozar September 26, 2011 18:40 662 Chapter 14: Introduction to Microwave Systems where (14.1), (14.2), and (14.4) were used. The units of the radiation intensity are watts, or watts per unit solid angle, since the radial dependence has been removed. The radiation intensity gives the variation in radiated power versus position around the antenna. We can find the total power radiated by the antenna by integrating the Poynting vector over thesurface of a sphere of radius rthat encloses the antenna. This is equivalent to integrating the radiation intensity over a unit sphere: P rad=2π/integraldisplay φ=0π/integraldisplay θ=0¯Savg·ˆrr2sinθdθdφ=2π/integraldisplay φ=0π/integraldisplay θ=0U(θ,φ) sinθdθdφ. ( 14.7) AntennaPatternCharacteristics Theradiation pattern of an antenna is a plot of the magnitude of the far-zone field strength versus position around the antenna, at a fixed distance from the antenna. Thus the radiationpattern can be plotted from the pattern function F θ(θ,φ) orFφ(θ,φ), versus either the angleθ(for an elevation plane pattern ) or the angle φ(for an azimuthal plane pattern ). The choice of plotting either FθorFφis dependent on the polarization of the antenna. A typical antenna pattern is shown in Figure 14.3. This pattern is plotted in polar form, versus the elevation angle, θ, for a small horn antenna oriented in the vertical direction. The plot shows the relative variation of the radiated power of the antenna in dB, normalized tothe maximum value. Since the pattern functions are proportional to voltage, the radial scale of the plot is computed as 20 log |F(θ,φ)|; alternatively, the plot could be computed in terms of the radiation intensity as 10 log |U(θ,φ)|. The pattern may exhibit several distinct lobes, with different maxima in different directions. The lobe having the maximum value is called the main beam, while those lobes at lower levels are called sidelobes . The pattern of Figure 14.3 has one main beam at θ=0 and several sidelobes, the largest of which are located at about θ=±16 ◦. The level of these sidelobes is 13 dB below the level of the main beam. Radiation patterns may also be plotted in rectangular form; this is especially useful for antennas having a narrow main beam. -30 -20 -10 0 FIGURE 14.3 The E-plane radiation pattern of a small horn antenna. The pattern is normalized to 0 dB at the beam maximum, with 10 dB per radial division. c14IntroToMicrowaveSystems Pozar September 26, 2011 18:40 14.1 System Aspects of Antennas 663 A fundamental property of an antenna is its ability to focus power in a given direction, to the exclusion of other directions. Thus an antenna with a broad main beam can transmit (or receive) power over a wide angular region, while an antenna having a narrow main beam will transmit (or receive) power over a small angular region. One measure of thisfocusing effect is the 3 dB beamwidth of the antenna, defined as the angular width of the main beam at which the power level has dropped 3 dB from its maximum value (its half- power points). The 3 dB beamwidth of the pattern of Figure 14.3 is about 10 ◦. Antennas having a constant pattern in the azimuthal plane are called omnidirectional, and are useful for applications such as broadcasting or for hand-held wireless devices, where it is desired to transmit or receive equally in all directions. Patterns that have relatively narrow main beams in both planes are known as pencil beam antennas, and are useful in applications such as radar and point-to-point radio links. Another measure of the focusing ability of an antenna is the directivity , defined as the ratio of the maximum radiation intensity in the main beam to the average radiation intensity over all space: D=Umax Uavg=4πUmax Prad=4πUmax π/integraldisplay θ=02π/integraldisplay φ=0U(θ,φ) sinθdθdφ,( 14.8) where (14.7) has been used for the radiated power. Directivity is a dimensionless ratio of power, and is usually expressed in dB as D(dB)=10 log(D). An antenna that radiates equally in all directions is called an isotropic antenna. Apply- ing the integral identity that π/integraldisplay θ=02π/integraldisplay φ=0sinθdθdφ=4π to the denominator of (14.8) for U(θ,φ) =1 shows that the directivity of an isotropic ele- ment is D=1, or 0 dB. Since the minimum directivity of any antenna is unity, directivity is sometimes stated as relative to the directivity of an isotropic radiator, and written as dBi. Typical directivities for some common antennas are 2.2 dB for a wire dipole, 7.0 dB for a microstrip patch antenna, 23 dB for a waveguide horn antenna, and 35 dB for a parabolicreflector antenna. Beamwidth and directivity are both measures of the focusing ability of an antenna: an antenna pattern with a narrow main beam will have a high directivity, while a patternwith a wide beam will have a lower directivity. We might therefore expect a direct relation between beamwidth and directivity, but in fact there is not an exact relationship between these two quantities. This is because beamwidth is only dependent on the size and shape of the main beam, whereas directivity involves integration of the entire radiation pattern. Thus it is possible for many different antenna patterns to have the same beamwidth butquite different directivities due to differences in sidelobes or the presence of more than one main beam. With this qualification in mind, however, it is possible to develop approximate relations between beamwidth and directivity that apply with reasonable accuracy to a largenumber of practical antennas. One such approximation that works well for antennas with pencil beam patterns is the following: D∼=32,400 θ1θ2,( 14.9) where θ1andθ2are the beamwidths in two orthogonal planes of the main beam, in degrees. This approximation does not work well for omnidirectional patterns because there is a well-defined main beam in only one plane for such patterns. c14IntroToMicrowaveSystems Pozar September 26, 2011 18:40 664 Chapter 14: Introduction to Microwave Systems EXAMPLE 14.2 PATTERN CHARACTERISTICS OF A DIPOLE ANTENNA The far-zone electric field radiated by an electrically small wire dipole antenna oriented on the z-axis is given by Eθ(r,θ,φ) =V0sinθe−jk0r rV/m, Eφ(r,θ,φ) =0. Find the main beam position of the dipole antenna, its beamwidth, and its direc- tivity. Solution The radiation intensity for the above far-field is U(θ,φ) =Csin2θ, where the constant C=V2 0/2η 0. The radiation pattern is seen to be independent of the azimuth angle φ, and so is omnidirectional in the azimuthal plane. The pattern has a “donut” shape, with nulls at θ=0 andθ=180◦(along the z-axis), and a beam maximum at θ=90◦(the horizontal plane). The angles where the radiation intensity has dropped by 3 dB are given by the solutions to sin2θ=0.5; thus the 3 dB, or half-power, beamwidth is 135◦−45◦=90◦. The directivity is calculated using (14.8). The denominator of this expression is π/integraldisplay θ=02π/integraldisplay φ=0U(θ,φ) sinθdθdφ=2πCπ/integraldisplay θ=0sin3θdθ=2πC/parenleftbigg4 3/parenrightbigg =8πC 3, where the required integral identity is listed in Appendix D. Since Umax=C,t h e directivity reduces to D=3 2=1.76 dB . ■ AntennaGainandEfficien y Resistive losses, due to nonperfect metals and dielectric materials, exist in all practical antennas. Such losses result in a difference between the power delivered to the input of an antenna and the power radiated by that antenna. As with many other electrical components, we can define the radiation efficiency of an antenna as the ratio of the desired output power to the supplied input power: ηrad=Prad Pin=Pin−Ploss Pin=1−Ploss Pin,( 14.10) where Pradis the power radiated by the antenna, Pinis the power supplied to the input of the antenna, and Plossis the power lost in the antenna. Note that there are other factors that can contribute to the effective loss of transmit power, such as impedance mismatch at the c14IntroToMicrowaveSystems Pozar September 26, 2011 18:40 14.1 System Aspects of Antennas 665 input to the antenna, or polarization mismatch with the receive antenna. However, these losses are external to the antenna and could be eliminated by the proper use of matching networks, or the proper choice and positioning of the receive antenna. Therefore losses of this type are usually not attributed to the antenna itself, as are dissipative losses due tometal conductivity or dielectric loss within the antenna. Recall that antenna directivity is a function only of the shape of the radiation pattern (the radiated fields) of an antenna, and is not affected by losses in the antenna itself. Toaccount for the fact that an antenna having a radiation efficiency less than unity will not radiate all of its input power, we define antenna gain as the product of directivity and efficiency: G=η radD.( 14.11) Thus, gain is always less than or equal to directivity. Gain can also be computed directly, by replacing Pradin the denominator of (14.8) with Pin, since by the definition of radiation efficiency in (14.10) we have Prad=ηradPin. Gain is usually expressed in dB, as G(dB)= 10 log(G). Sometimes the effect of impedance mismatch loss is included in the gain of an antenna; this is referred to as the realized gain [1]. ApertureEfficien yandEffectiveArea Many types of antennas can be classified as aperture antennas , meaning that the antenna has a well-defined aperture area from which radiation occurs. Examples include reflector antennas, horn antennas, lens antennas, and array antennas. For such antennas, it can beshown that the maximum directivity that can be obtained from an electrically large aperture of area Ais given as D max=4πA λ2.( 14.12) For example, a rectangular horn antenna having an aperture 2 λ×3λhas a maximum direc- tivity of 24π , or about 19 dB. In practice, there are several factors that can serve to reduce the directivity of an antenna from its maximum possible value, such as nonideal amplitudeor phase characteristics of the aperture field, aperture blockage, or, in the case of reflector antennas, spillover of the feed pattern. For this reason, we define an aperture efficiency as the ratio of the actual directivity of an aperture antenna to the maximum directivity givenby (14.12). Then we can write the directivity of an aperture antenna as D=η ap4πA λ2.( 14.13) Aperture efficiency is always less than or equal to unity. The above definitions of antenna directivity, efficiency, and gain were stated in terms of a transmitting antennas, but they apply to receiving antennas as well. For a receiving antenna it is also of interest to determine the received power for a given incident plane wavefield. This is the converse problem of finding the power density radiated by a transmitting antenna, as given in (14.4). Determining received power is important for the derivation of the Friis radio system link equation, to be discussed in the following section. We expect thatreceived power will be proportional to the power density, or Poynting vector, of the incident wave. Since the Poynting vector has dimensions of W/m 2, and the received power, Pr, has dimensions of W, the proportionality constant must have units of area. Thus we write Pr=AeSavg,( 14.14) c14IntroToMicrowaveSystems Pozar September 26, 2011 18:40 666 Chapter 14: Introduction to Microwave Systems where Aeis defined as the effective aperture area of the receive antenna. The effective aperture area has dimensions of m2, and can be interpreted as the “capture area” of a receive antenna, intercepting part of the incident power density radiated toward the receive antenna. The quantity Prin (14.14) is the power available at the terminals of the receive antenna, as delivered to a conjugately matched load. The maximum effective aperture area of an antenna can be shown to be related to the directivity of the antenna as [1, 2] Ae=Dλ2 4π,( 14.15) where λis the operating wavelength of the antenna. For electrically large aperture antennas the effective aperture area is often close to the actual physical aperture area. However,for many other types of antennas, such as dipoles and loops, there is no simple relation between the physical cross-sectional area of the antenna and its effective aperture area. The maximum effective aperture area as defined above does not include the effect of lossesin the antenna, which can be accounted for by replacing Din (14.15) with G,t h eg a i n ,o f the antenna. BackgroundandBrightnessTemperature We have seen how noise power is generated by lossy components and active devices, but noise can also be delivered to the input of a receiver by the antenna. Antenna noise power may be received from the external environment, or generated internally as thermal noise due to losses in the antenna itself. While noise produced within a receiver is controllableto some extent (by judicious design and component selection), the noise received from the environment by a receiving antenna is generally not controllable, and may exceed the noise level of the receiver itself. Thus it is important to characterize the noise power deliv-ered to a receiver by its antenna. Consider the three situations shown in Figure 14.4. In Figure 14.4a we have the simple case of a resistor at temperature T, producing an available output noise power N o=kTB,( 14.16) where Bis the system bandwidth and kis Boltzmann’s constant. In Figure 14.4b we have an antenna enclosed by an anechoic chamber at temperature T. The anechoic chamber appears as a perfectly absorbing enclosure, and is in thermal equilibrium with the an-tenna. Thus the terminals of the antenna are indistinguishable from the resistor terminals of Figure 14.4a (assuming an impedance-matched antenna), and therefore it produces the FIGURE 14.4 Illustrating the concept of background temperature. (a) A resistor at temperature T. (b) An antenna in an anechoic chamber at temperature T. (c) An antenna viewing a uniform sky background at temperature T. c14IntroToMicrowaveSystems Pozar September 26, 2011 18:40 14.1 System Aspects of Antennas 667 FIGURE 14.5 Natural and man-made sources of background noise. same output noise power as the resistor of Figure 14.4a. Figure 14.4c shows the same an- tenna directed at the sky. If the main beam of the antenna is narrow enough so that it sees a uniform region at physical temperature T, then the antenna again appears as a resistor at temperature Tand produces the output noise power given in (14.16). This is true regard- less of the radiation efficiency of the antenna, as long as the physical temperature of the antenna is also T. In actuality an antenna typically sees a much more complex environment than the cases depicted in Figure 14.4. A general scenario of both naturally occurring and man- made noise sources is shown in Figure 14.5, where we see that an antenna with a relatively broad main beam may pick up noise power from a variety of origins. In addition, noisemay be received through the sidelobes of the antenna pattern or via reflections from the ground or other large objects. As in Chapter 10, where the noise power from an arbi- trary white noise source was represented as an equivalent noise temperature, we define thebackground noise temperature ,T B, as the equivalent temperature of a resistor required to produce the same noise power as the actual environment seen by the antenna. Some typ- ical background noise temperatures that are relevant at low microwave frequencies are asfollows: rSky (toward zenith) 3–5 KrSky (toward horizon) 50–100 KrGround 290–300 K The overhead sky background temperature of 3–5 K is the cosmic background radiation believed to be a remnant of the big bang at the creation of the universe. This would be the noise temperature seen by an antenna with a narrow beam and high radiation effi- ciency pointed overhead, away from “hot” sources such as the Sun or stellar radio objects.The background noise temperature increases as the antenna is pointed toward the horizon because of the greater thickness of the atmosphere, so that the antenna sees an effective background closer to that of the anechoic chamber of Figure 14.4b. Pointing the antenna toward the ground further increases the effective loss, and hence the noise temperature. Figure 14.6 gives a more complete picture of the background noise temperature, show- ing the variation of T Bversus frequency and for several elevation angles [3]. Note that the noise temperature shown in the graph follows the trends listed above, in that it is lowest for the overhead sky (θ =90◦), and greatest for angles near the horizon (θ =0◦). Also note the sharp peaks in noise temperature that occur at 22 and 60 GHz. The first is due to the resonance of molecular water, while the second is caused by resonance of molecular c14IntroToMicrowaveSystems Pozar September 26, 2011 18:40 668 Chapter 14: Introduction to Microwave Systems FIGURE 14.6 Background noise temperature of sky versus frequency. θis elevation angle mea- sured from the horizon. Data are for sea level, with surface temperature of 15◦C and surface water vapor density of 7.5 gm/m3. oxygen. Both of these resonances lead to increased atmospheric loss and hence increased noise temperature. The loss is great enough at 60 GHz that a high-gain antenna pointing through the atmosphere effectively appears as a matched load at 290 K. While loss in gen-eral is undesirable, these particular resonances can be useful for remote sensing applica- tions, or for using the inherent attenuation of the atmosphere to limit propagation distances for radio communications over small distances. When the antenna beamwidth is broad enough that different parts of the antenna pat- tern see different background temperatures, the effective brightness temperature seen by the antenna can be found by weighting the spatial distribution of background temperatureby the pattern function of the antenna. Mathematically we can write the brightness temper- ature T bseen by the antenna as Tb=2π/integraldisplay φ=0π/integraldisplay θ=0TB(θ,φ) D(θ,φ) sinθdθdφ 2π/integraldisplay φ=0π/integraldisplay θ=0D(θ,φ) sinθdθdφ,( 14.17) where TB(θ,φ) is the distribution of the background temperature, and D(θ,φ) is the di- rectivity (or the power pattern function) of the antenna. Antenna brightness temperature is referenced at the terminals of the antenna. Observe that when TBis a constant, (14.17) reduces to Tb=TB, which is essentially the case of a uniform background temperature shown in Figure 14.3b or 14.4c. Also note that this definition of antenna brightness tem- perature does not involve the gain or efficiency of the antenna, and so does not includethermal noise due to losses in the antenna. c14IntroToMicrowaveSystems Pozar September 26, 2011 18:40 14.1 System Aspects of Antennas 669 AntennaNoiseTemperatureand G/T If a receiving antenna has dissipative loss, so that its radiation efficiency ηradis less than unity, the power available at the terminals of the antenna is reduced by the factor ηrad from that intercepted by the antenna (the definition of radiation efficiency is the ratio of output to input power). This reduction applies to received noise power, as well as received signal power, so the noise temperature of the antenna will be reduced from the brightness temperature given in (14.17) by the factor ηrad. In addition, thermal noise will be generated internally by resistive losses in the antenna, and this will increase the noise temperature of the antenna. In terms of noise power, a lossy antenna can be modeled as a lossless antenna and an attenuator having a power loss factor of L=1/η rad. Then, using (10.15) for the equivalent noise temperature of an attenuator, we can find the resulting noise temperature seen at the antenna terminals as TA=Tb L+(L−1) LTp=ηradTb+(1−ηrad)Tp.( 14.18) The equivalent temperature TAis called the antenna noise temperature , and is a combi- nation of the external brightness temperature seen by the antenna and the thermal noise generated by the antenna. As with other equivalent noise temperatures, the proper interpre- tation of TAis that a matched load at this temperature will produce the same available noise power as does the antenna. Note that this temperature is referenced at the output terminals of the antenna; since an antenna is not a two-port circuit element, it does not make sense to refer the equivalent noise temperature to its “input.” Observe that (14.18) reduces to TA=Tbfor a lossless antenna with ηrad=1. If the radiation efficiency is zero, meaning that the antenna appears as a matched load and does not see any external background noise, then (14.18) reduces to TA=Tp, due to the thermal noise generated by the losses. If an antenna is pointed toward a known background tem- perature different than T0, then (14.18) can be used to determine its radiation efficiency. EXAMPLE 14.3 ANTENNA NOISE TEMPERATURE A high-gain antenna has the idealized hemispherical elevation plane pattern shown in Figure 14.7, and is rotationally symmetric in the azimuth plane. If the antennais facing a region having a background temperature T Bapproximated as given in Figure 14.7, find the antenna noise temperature. Assume the radiation efficiency of the antenna is 100%. Solution Since ηrad=1, (14.18) reduces to TA=Tb. The brightness temperature can be computed from (14.17), after normalizing the directivity to a maximum value FIGURE 14.7 Idealized antenna pattern and background noise temperature for Example 14.3. c14IntroToMicrowaveSystems Pozar September 26, 2011 18:40 670 Chapter 14: Introduction to Microwave Systems of unity: Tb=2π/integraltext φ=0π/integraltext θ=0TB(θ,φ) D(θ,φ) sinθdθdφ 2π/integraltext φ=0π/integraltext θ=0D(θ,φ) sinθdθdφ=1◦/integraltext θ=010 sin θdθ+30◦/integraltext θ=1◦0.1 sin θdθ+90◦/integraltext θ=30◦sinθdθ 1◦/integraltext θ=0sinθdθ+90◦/integraltext θ=1◦0.01 sin θdθ =−10 cos θ|1◦ 0−0.1c o s θ|30◦ 1◦−cosθ|90◦ 30◦ −cosθ |1◦ 0−0.01 cos θ|90◦ 1◦=0.00152 +0.0134 +0.866 0.0102=86.4K . In this example most of the noise power is collected through the sidelobe region of the antenna. ■ The more general problem of a receiver connected through a lossy transmission line to an antenna viewing a background noise temperature distribution TBcan be represented by the system shown in Figure 14.8. The antenna is assumed to have a radiation efficiency ηrad, and the connecting transmission line has a power loss factor of L≥1, with both at physical temperature Tp. We also include the effect of an impedance mismatch between the antenna and the transmission line, represented by the reflection coefficient /Gamma1. The equivalent noise temperature seen at the output terminals of the transmission line consists of three contri- butions: noise power from the antenna due to internal noise and the background brightnesstemperature, noise power generated from the lossy line in the forward direction, and noise power generated by the lossy line in the backward direction and reflected from the antenna mismatch toward the receiver. The noise due to the antenna is given by (14.18), but re- duced by the loss factor of the line, 1 /L, and the reflection mismatch factor, (1−|/Gamma1| 2). The forward noise power from the lossy line is given by (10.15), after reduction by the lossfactor, 1/ L. The contribution from the lossy line reflected from the mismatched antenna is given by (10.15), after reduction by the power reflection coefficient, |/Gamma1| 2, and the loss factor, 1/ L2(since the reference point for the back-directed noise power from the lossy line given by (10.15) is at the output terminals of the line). Thus the overall system noise temperature seen at the input to the receiver is given by TS=TA L(1−|/Gamma1|2)+(L−1)Tp L+(L−1)Tp L2|/Gamma1|2 =(1−|/Gamma1|2) L[ηradTb+(1−ηrad)Tp]+(L−1) L/parenleftBigg 1+|/Gamma1|2 L/parenrightBigg Tp.(14.19) Background temperature TB (/H9258, /H9278) Antenna TP, /H9257radReceiver TA TS/H9003 Lossy line TP, /H9003 FIGURE 14.8 A receiving antenna connected to a receiver through a lossy transmission line. An impedance mismatch exists between the antenna and the line. c14IntroToMicrowaveSystems Pozar September 26, 2011 18:40 14.2 Wireless Communications 671 Observe that for a lossless line ( L=1) the effect of an antenna mismatch is to reduce the system noise temperature by the factor (1−|/Gamma1|2). Of course, the received signal power will be reduced by the same amount. Also note that for the case of a matched antenna (/Gamma1=0), (14.19) reduces to TS=1 L[ηradTb+(1−ηrad)Tp]+L−1 LTp,( 14.20) as expected for a cascade of two noisy components. Finally, it is important to realize the difference between radiation efficiency and aper- ture efficiency, and their effects on antenna noise temperature. While radiation efficiencyaccounts for resistive losses, and thus involves the generation of thermal noise, aperture ef- ficiency does not. Aperture efficiency applies to the loss of directivity in aperture antennas, such as reflectors, lenses, or horns, due to feed spillover or suboptimum aperture excitation(e.g., a nonuniform amplitude or phase distribution), and by itself does not lead to any ad- ditional effect on noise temperature that would not be included through the pattern of the antenna. The antenna noise temperature defined above is a useful figure of merit for a receive antenna because it characterizes the total noise power delivered by the antenna to the input of a receiver. Another useful figure of merit for receive antennas is the G/T ratio, defined as G/T(dB)=10 logG TAdB/K, (14.21) where Gis the gain of the antenna, and TAis the antenna noise temperature. This quantity is important because, as we will see in Section 14.2, the signal-to-noise ratio (SNR) at the input to a receiver is proportional to G/TA. The ratio G/Tcan often be maximized by in- creasing the gain of the antenna, since this increases the numerator and usually minimizesreception of noise from hot sources at low elevation angles. Of course, higher gain requires a larger and more expensive antenna, and high gain may not be desirable for applications requiring omnidirectional coverage (e.g., cellular telephones or mobile data networks), so often a compromise must be made. Finally, note that the dimensions given in (14.21) for 10 log(G/T)are not actually decibels per degree kelvin, but this is the nomenclature that is commonly used for this quantity. 14.2WIRELESSCOMMUNICATIONS Wireless communications involves the transfer of information between two points without direct connection. While this may be accomplished using sound, infrared, optical, or radiofrequency energy, most modern wireless systems rely on RF or microwave signals, usually in the UHF to millimeter wave frequency range. Because of spectrum crowding and the need for higher data rates, the trend is to higher frequencies, so the majority of wirelesssystems today operate at frequencies ranging from about 800 MHz to a few gigahertz. RF and microwave signals offer wide bandwidths, and have the added advantage of being able to penetrate fog, dust, foliage, and even buildings and vehicles to some extent. Historically, wireless communication using RF energy has its foundations in the theoretical work of Maxwell, followed by the experimental verification of electromagnetic wave propagationby Hertz, and the practical development of radio techniques and systems by Tesla, Marconi, and others in the early part of the 20th century. Today, wireless systems include broadcast radio and television, cellular telephone and networking systems, direct broadcast satellite(DBS) television service, wireless local area networks (WLANs), paging systems, Global Positioning System (GPS) service, and radio frequency identification (RFID) systems [4]. c14IntroToMicrowaveSystems Pozar September 26, 2011 18:40 672 Chapter 14: Introduction to Microwave Systems These systems are beginning to provide, for the first time in history, worldwide connectivity for voice, video, and network communications. One way to categorize wireless systems is according to the nature and placement of t h eu s e r s .I na point-to-point radio system a single transmitter communicates with a single receiver. Such systems generally use high-gain antennas in fixed positions to maximize received power and minimize interference with other radios that may be operating nearby in the same frequency range. Point-to-point radios are typically used for satellite commu-nications, dedicated data communications by utility companies, and backhaul connection of cellular base stations to a central switching office. Point-to-multipoint systems connect a central station to a large number of possible receivers. The most common examples are commercial AM and FM radio and broadcast television, where a central transmitter uses an antenna with a broad azimuthal beam to reach many listeners and viewers. Multipoint- to-multipoint systems allow simultaneous communication between individual users (who may not be in fixed locations). Such systems generally do not connect two users directly, but instead rely on a grid of base stations to provide the desired interconnections betweenusers. Cellular telephone systems and some types of WLANs are examples of this type of application. Another way to characterize wireless systems is in terms of the directionality of com- munication. In a simplex system, communication occurs only in one direction—from the transmitter to the receiver. Examples of simplex systems include broadcast radio, televi- sion, and paging systems. In a half-duplex system, communication may occur in two direc- tions, but not simultaneously. Early mobile radios and citizens band radio are examples of duplex systems, and generally rely on a “push-to-talk” function so that a single channel can be used for both transmitting and receiving at different times. Full-duplex systems allow si- multaneous two-way transmission and reception. Examples include cellular telephone and point-to-point radio systems. Full-duplex transmission clearly requires a duplexing tech- nique to avoid interference between transmitted and received signals. This can be done by using separate frequency bands for transmit and receive (frequency division duplexing), or by allowing users to transmit and receive only in certain predefined time intervals (timedivision duplexing). While most wireless systems are ground based, it is also possible to use satellite sys- tems for voice, video, and data communications [5]. Satellites offer the possibility ofcommunication with a large number of users over wide areas, perhaps including the en- tire planet. Satellites in a geosynchronous earth orbit (GEO) are positioned approximately 36,000 km above Earth, and have a 24-hour orbital period. When a GEO satellite is posi-tioned above the equator, it becomes geostationary, and will remain in a fixed position rel- ative to Earth. Such satellites are useful for point-to-point radio links between widely sep- arated stations, and are commonly used for television and data communications through-out the world. At one time transcontinental telephone service relied on such satellites, but undersea fiber optics cables have largely replaced satellites for transoceanic connec- tions as being more economical and avoiding the annoying delay caused by the very longround-trip path between the satellite and Earth. Another drawback of GEO satellites is that their high altitude greatly reduces the received signal strength, making it difficult for two-way communication with handheld transceivers. Low Earth orbit (LEO) satel- lites orbit much closer to Earth, typically in the range of 500–2000 km. The shorter path length may allow line-of-sight communication between LEO satellites and hand-held ra-dios, but satellites in LEO orbits are visible from a given point on the ground for only a short time, typically between a few minutes and about 20 minutes. Effective coverage therefore requires a large number of LEO satellites in different orbital planes. The ill-fatedIridium system is probably the best-known example of a LEO satellite communications system. c14IntroToMicrowaveSystems Pozar September 26, 2011 18:40 14.2 Wireless Communications 673 Gt Gr PrPt R FIGURE 14.9 A basic radio system. TheFriisFormula A general radio system link is shown in Figure 14.9, where the transmit power is Pt,t h e transmit antenna gain is Gt, the receive antenna gain is Gr, and the received power (de- livered to a matched load) is Pr. The transmit and receive antennas are separated by the distance R. From (14.6)–(14.7), the power density radiated by an isotropic antenna (D= 1=0d B)at a distance Ris given by Savg=Pt 4πR2W/m2.( 14.22) This result reflects the fact that we must be able to recover all of the radiated power by integrating over a sphere of radius Rsurrounding the antenna; since the power is distributed isotropically, and the area of a sphere is 4π R2, (14.22) follows. If the transmit antenna has a directivity greater than 0 dB, we can find the radiated power density by multiplying by the directivity, since directivity is defined as the ratio of the actual radiation intensity to the equivalent isotropic radiation intensity. In addition, if the transmit antenna has losses, we can include the radiation efficiency factor, which has the effect of converting directivity togain. Thus, the general expression for the power density radiated by an arbitrary transmit antenna is S avg=GtPt 4πR2W/m2. (14.23) If this power density is incident on the receive antenna, we can use the concept of effective aperture area, as defined in (14.14), to find the received power: Pr=AeSavg=GtPtAe 4πR2W. Next, (14.15) can be used to relate the effective area to the directivity of the receive antenna. Again, the possibility of losses in the receive antenna can be accounted for by using thegain (rather than the directivity) of the receive antenna. Then the final result for the received power is P r=GtGrλ2 (4πR)2PtW. (14.24) This result is known as the Friis radio link formula , and it addresses the fundamental question of how much power is received by a radio antenna. In practice, the value givenby (14.24) should be interpreted as the maximum possible received power, as there are a number of factors that can serve to reduce the received power in an actual radio system. These include impedance mismatch at either antenna, polarization mismatch between theantennas, propagation effects leading to attenuation or depolarization, and multipath effects that may cause partial cancellation of the received field. c14IntroToMicrowaveSystems Pozar September 26, 2011 18:40 674 Chapter 14: Introduction to Microwave Systems Observe in (14.24) that the received power decreases as 1 /R2as the separation be- tween transmitter and receiver increases. This dependence is a result of conservation ofenergy. While it may seem to be prohibitively large for large distances, in fact the space decay of 1/ R 2is usually much better than the exponential decrease in power due to losses in a wired communications link. This is because the attenuation of power on a transmissionline varies as e −2α z(where αis the attenuation constant of the line), and at large distances the exponential function decreases faster than an algebraic dependence like 1 /R2. Thus for long-distance communications, radio links will perform better than wired links. This conclusion applies to any type of transmission line, including coaxial lines, waveguides, and even fiber optic lines. (It may not apply, however, if the communications link is landor sea based, so that repeaters can be inserted along the link to recover lost signal power.) As can be seen from the Friis formula, received power is proportional to the prod- uctP tGt. These two factors—the transmit power and transmit antenna gain—characterize the transmitter, and in the main beam of the antenna the product PtGtcan be interpreted equivalently as the power radiated by an isotropic antenna with input power PtGt. Thus, this product is defined as the effective isotropic radiated power (EIRP): EIRP=PtGtW. (14.25) For a given frequency, range, and receiver antenna gain, the received power is propor- tional to the EIRP of the transmitter and can only be increased by increasing the EIRP. This can be done by increasing the transmit power, or the transmit antenna gain, or both. LinkBudgetandLinkMargin The various terms in the Friis formula of (14.24) are often tabulated separately in a link budget, where each of the factors can be individually considered in terms of its net effect on the received power. Additional loss factors, such as line losses or impedance mismatch at the antennas, atmospheric attenuation (see Section 14.5), and polarization mismatch canalso be added to the link budget. One of the terms in a link budget is the path loss, account- ing for the free-space reduction in signal strength with distance between the transmitter and receiver. From (14.24), path loss is defined (in dB) as L 0(dB)=20 log/parenleftbigg4πR λ/parenrightbigg >0.( 14.26) Note that path loss depends on wavelength (frequency), which serves to provide a normal- ization for the units of distance. With the above definition of path loss, we can write the remaining terms of the Friis formula as shown in the following link budget: Transmit power Pt Transmit antenna line loss (−)Lt Transmit antenna gain Gt Path loss (free-space) (−)L0 Atmospheric attenuation (−)LA Receive antenna gain Gr Receive antenna line loss (−)Lr Receive power Pr We have also included loss terms for atmospheric attenuation and line attenuation. Assum- ing that all of the above quantities are expressed in dB (or dBm, in the case of Pt), we can c14IntroToMicrowaveSystems Pozar September 26, 2011 18:40 14.2 Wireless Communications 675 write the receive power as Pr(dBm)=Pt−Lt+Gt−L0−LA+Gr−Lr.( 14.27) If the transmit and/or receive antenna is not impedance matched to the transmitter/ receiver (or to their connecting lines), impedance mismatch will reduce the received power by the factor (1−|/Gamma1|2), where /Gamma1is the appropriate reflection coefficient. The resulting impedance mismatch loss, Limp(dB)=−10 log(1−|/Gamma1|2)≥0,( 14.28) can be included in the link budget to account for the reduction in received power. Another possible entry in the link budget relates to the polarization matching of the transmit and receive antennas, as maximum power transmission between transmitter and receiver requires both antennas to be polarized in the same manner. If a transmit antenna is vertically polarized, for example, maximum power will only be delivered to a vertically po-larized receiving antenna, while zero power would be delivered to a horizontally polarized receive antenna, and half the available power would be delivered to a circularly polarized antenna. Determination of the polarization loss factor is explained in references [1], [2], and [4]. In practical communications systems it is usually desired to have the received power level greater than the threshold level required for the minimum acceptable quality of service(usually expressed as the minimum carrier-to-noise ratio (CNR), or minimum SNR). This design allowance for received power is referred to as the link margin, and can be expressed as the difference between the design value of received power and the minimum thresholdvalue of receive power: Link margin (dB)=LM=P r−Pr(min)>0,( 14.29) where all quantities are in dB. Link margin should be a positive number; typical values may range from 3 to 20 dB. Having a reasonable link margin provides a level of robustness to thesystem to account for variables such as signal fading due to weather, movement of a mobile user, multipath propagation problems, and other unpredictable effects that can degrade system performance and quality of service. Link margin that is used to account for fadingeffects is sometimes referred to as fade margin. Satellite links operating at frequencies above 10 GHz, for example, often require fade margins of 20 dB or more to account for attenuation during heavy rain. As seen from (14.29) and the link budget, link margin for a given communication system can be improved by increasing the received power (by increasing transmit power or antenna gains), or by reducing the minimum threshold power (by improving the design of the receiver, changing the modulation method, or by other means). Increasing link margin therefore usually involves an increase in cost and complexity, so excessive increases in linkmargin are usually avoided. EXAMPLE 14.4 LINK ANALYSIS OF DBS TELEVISION SYSTEM The direct broadcast system in North America operates at 12.2–12.7 GHz, with atransmit carrier power of 120 W, a transmit antenna gain of 34 dB, an IF band-width of 20 MHz, and a worst-case slant angle (30 ◦) distance from the geostation- ary satellite to Earth of 39,000 km. The 18-inch receiving dish antenna has a gain of 33.5 dB and sees an average background brightness temperature of Tb=50 K, with a receiver low-noise block (LNB) having a noise figure of 0.7 dB. The re- quired minimum CNR is 15 dB. The overall system is shown in Figure 14.10. c14IntroToMicrowaveSystems Pozar September 26, 2011 18:40 676 Chapter 14: Introduction to Microwave Systems FIGURE 14.10 Diagram of the DBS system for Example 14.4. Find (a) the link budget for the received carrier power at the antenna terminals, (b)G/Tfor the receive antenna and LNB system, (c) the CNR at the output of the LNB, and (d) the link margin of the system. Solution We will take the operating frequency to be 12.45 GHz, so the wavelength is 0.0241 m. From (14.26) the path loss is L0=20 log/parenleftbigg4πR λ/parenrightbigg =20 log/parenleftBigg (4π)(39×106) 0.0241/parenrightBigg =206.2d B (a) The link budget for the received power is Pt=120 W =50.8d B m Gt=34.0 dB L0=(−)206.2 dB Gr=33.5 dB Pr=−87.9 dBm =1.63×10−12W. (b) To find G/Twe first find the noise temperature of the antenna and LNB cas- cade, referenced at the input of the LNB: Te=TA+TLNB=Tb+(F−1)T0=50+(1.175 −1)(290) =100.8K . Then G/Tfor the antenna and LNB is G/T(dB)=10 log2239 100.8=13.5d B / K . (c) The CNR at the output of the LNB is CNR=PrGLNB kTeBG LNB=1.63×10−12 (1.38×10−23)(100.8)(20 ×106)=58.6=17.7d B . Note that GLNB, the gain of the LNB module, cancels in the ratio for the output CNR. (d) If the minimum required CNR is 15 dB, the system link margin is 2.7 dB. ■ RadioReceiverArchitectures The receiver is usually the most critical component of a wireless system, having the over- all purpose of reliably recovering the desired signal from a wide spectrum of transmit- ting sources, interference, and noise. In this section we will describe some of the critical c14IntroToMicrowaveSystems Pozar September 26, 2011 18:40 14.2 Wireless Communications 677 requirements for radio receiver design and summarize some of the most common types of receiver architectures. A well-designed radio receiver must provide several different functions: rHigh gain (∼100 dB) to restore the low power of the received signal to a level near its original baseband valuerSelectivity, in order to receive the desired signal while rejecting adjacent channels, image frequencies, and interferencerDown-conversion from the received RF frequency to a lower IF frequency for processingrDetection of the received analog or digital informationrIsolation from the transmitter to avoid saturation of the receiver Because the typical signal power level from the receive antenna may be as low as −100 to−120 dBm, the receiver may be required to provide gain as high as 100 to 120 dB. This much gain should be spread over the RF, IF, and baseband stages to avoid instabilities and possible oscillation; it is generally good practice to avoid more than about 50–60 dB of gain at any one frequency band. The fact that amplifier cost generally increases with frequency is a further reason to spread gain over different frequency stages. In principle, selectivity can be obtained by using a narrow bandpass filter at the RF stage of the receiver, but the bandwidth and cutoff requirements for such a filter are usually impractical to realize at RF frequencies. It is more effective to achieve selectivity by down-converting a relatively wide RF bandwidth around the desired signal, and using a sharp- cutoff bandpass filter at the IF stage to select only the desired frequency band. In addition, many wireless systems use a number of narrow but closely spaced channels, which mustbe selected using a tuned local oscillator, while the IF passband is fixed. The alternative of using an extremely narrow band, electronically tunable RF filter is not practical. Tuned radio frequency receiver: One of the earliest types of receiving circuits to be de- veloped was the tuned radio frequency (TRF) receiver. As shown in Figure 14.11, a TRF receiver employs several stages of RF amplification along with tunable bandpass filters to provide high gain and selectivity. Alternatively, filtering and amplification may be com- bined by using amplifiers with a tunable bandpass response. At relatively low broadcast radio frequencies, such filters and amplifiers have historically been tuned using mechan- ically variable capacitors or inductors. However, such tuning is problematic because ofthe need to tune several stages in parallel, and selectivity is poor because the passband of such filters is fairly broad. In addition, all the gain of the TRF receiver is achieved at the RF frequency, limiting the amount of gain that can be obtained before oscillation oc-curs, and increasing the cost and complexity of the receiver. Because of these drawbacks TRF receivers are seldom used today, and are an especially bad choice for higher RF or microwave frequencies. FIGURE 14.11 Block diagram of a tuned radio frequency receiver. c14IntroToMicrowaveSystems Pozar September 26, 2011 18:40 678 Chapter 14: Introduction to Microwave Systems FIGURE 14.12 Block diagram of a direct-conversion receiver. Direct conversion receiver :T h e direct conversion receiver, shown in Figure 14.12, uses a mixer and local oscillator to perform frequency down-conversion with a zero IF frequency. The local oscillator is set to the same frequency as the desired RF signal, which is thenconverted directly to baseband. For this reason, the direct conversion receiver is sometimes called a homodyne receiver. For AM reception the received baseband signal would not re- quire any further detection. The direct conversion receiver offers several advantages over the TRF receiver, as selectivity can be controlled with a simple low-pass baseband filter, and gain may be spread through the RF and baseband stages (although it is difficult to ob-tain stable high gain at very low frequencies). Direct conversion receivers are simpler and less costly than superheterodyne receivers since there is no IF amplifier, IF bandpass filter, or IF local oscillator required for final down conversion. Another important advantage ofdirect conversion is that there is no image frequency, since the mixer difference frequency is effectively zero, and the sum frequency is twice the LO and easily filtered. However, a serious disadvantage is that the LO must have a very high degree of precision and stability,especially for high RF frequencies, to avoid drift of the received signal frequency. This type of receiver is often used with Doppler radars, where the exact LO can be obtained from the transmitter, but a number of newer wireless systems are being designed with directconversion receivers. Superheterodyne receiver: By far the most popular type of receiver in use today is the superheterodyne circuit, shown in Figure 14.13. The block diagram is similar to that of the direct conversion receiver, but the IF frequency is now nonzero, and is generally selected tobe between the RF frequency and baseband. A midrange IF allows the use of sharper cutoff filters for improved selectivity, and higher IF gain through the use of an IF amplifier. Tuning is conveniently accomplished by varying the frequency of the local oscillator so that the IF frequency remains constant. The superheterodyne receiver represents the culmination of over 50 years of receiver development, and is used in the majority of broadcast radios andtelevisions, radar systems, cellular telephone systems, and data communications systems. At microwave and millimeter wave frequencies it is often necessary to use two stages of down conversion to avoid problems due to LO stability. Such a dual-conversion super- heterodyne receiver employs two local oscillators, two mixers, and two IF frequencies to achieve down-conversion to baseband. FIGURE 14.13 Block diagram of a single-conversion superheterodyne receiver. c14IntroToMicrowaveSystems Pozar September 26, 2011 18:40 14.2 Wireless Communications 679 Background Antenna ReceiverTransmission line RF Amp.IF Amp.Si TbGA, ηrad, TP LT, TP Ni, SiNo, SoGRF TRFLM TMGIF TIF LO FIGURE 14.14 Noise analysis of a microwave receiver front end, including antenna and trans- mission line contributions. NoiseCharacterizationofaReceiver We can now analyze the noise characteristics of a complete antenna–transmission line– receiver front end, as shown in Figure 14.14. In this system the total noise power at the output of the receiver, No, will be due to contributions from the antenna pattern, the loss in the antenna, the loss in the transmission line, and the receiver components. This noise power will determine the minimum detectable signal level for the receiver and, for a given transmitter power, the maximum range of the communication link. The receiver components in Figure 14.14 consist of an RF amplifier with gain GRF and noise temperature TRF, a mixer with an RF-to-IF conversion loss factor LMand noise temperature TM, and an IF amplifier with gain GIFand noise temperature TIF. The noise effects of later stages can usually be ignored since the overall noise figure is dominated by the characteristics of the first few stages. The component noise temperatures can be related to noise figures as T=(F−1)T0. From (10.22) the equivalent noise temperature of the receiver can be found as TREC=TRF+TM GRF+TIFLM GRF.( 14.27) The transmission line connecting the antenna to the receiver has a loss LT, and is at a physical temperature Tp. So from (10.15) its equivalent noise temperature is TTL=(LT−1)Tp.( 14.28) Again using (10.22), we find that the noise temperature of the transmission line (TL) and receiver (REC) cascade is TTL+REC =TTL+LTTREC =(LT−1)Tp+LTTREC. (14.29) This noise temperature is defined at the antenna terminals (the input to the transmission line). As discussed in Section 14.1, the entire antenna pattern can collect noise power. If the antenna has a reasonably high gain with relatively low sidelobes, we can assume that all noise power comes via the main beam, so that the noise temperature of the antenna is given by (14.18): TA=ηradTb+(1−ηrad)Tp,( 14.30) c14IntroToMicrowaveSystems Pozar September 26, 2011 18:40 680 Chapter 14: Introduction to Microwave Systems where ηradis the efficiency of the antenna, Tpis its physical temperature, and Tbis the equivalent brightness temperature of the background seen by the main beam. (One must be careful with this approximation, as it is quite possible for the noise power collected by the sidelobes to exceed the noise power collected by the main beam, if the sidelobes are aimedat a hot background. See Example 14.3.) The noise power at the antenna terminals, which is also the noise power delivered to the transmission line, is N i=kBT A=kB[ηradTb+(1−ηrad)Tp],( 14.31) where Bis the system bandwidth. If Siis the received power at the antenna terminals, then the input SNR at the antenna terminals is Si/Ni. The output signal power is So=SiGRFGIF LTLM=SiGSYS,( 14.32) where GSYShas been defined as a system power gain. The output noise power is No=(Ni+kBT TL+REC )GSYS =kB(TA+TTL+REC )GSYS =kB[ηradTb+(1−ηrad)Tp+(LT−1)Tp+LTTREC]GSYS =kBT SYSGSYS, (14.33) where TSYShas been defined as the overall system noise temperature. The output SNR is So No=Si kBT SYS=Si kB[ηradTb+(1−ηrad)Tp+(LT−1)Tp+LTTREC].( 14.34) It may be possible to improve this SNR by various signal processing techniques. Note that it may appear to be convenient to use an overall system noise figure to calculate the degradation in SNR from input to output for the above system, but one must be very careful with such an approach because noise figure is defined only for Ni=kT0B, which is not the case here. It is often less confusing to work directly with noise temperatures and powers, as we did above. EXAMPLE 14.5 SIGNAL-TO-NOISE RATIO OF A MICROWA VE RECEIVER A microwave receiver like that of Figure 14.14 has the following parameters: f=4.0 GHz, GRF=20 dB, B=1M H z , FRF=3.0d B , GA=26 dB, LM=6.0d B , ηrad=0.90, FM=7.0d B , Tp=300 K, GIF=30 dB, Tb=200 K, FIF=1.1d B . LT=1.5d B , If the received power at the antenna terminals is Si=−80 dBm, calculate the input and output SNRs. c14IntroToMicrowaveSystems Pozar September 26, 2011 18:40 14.2 Wireless Communications 681 Solution We first convert the above dB quantities to numerical values, and noise figures to noise temperatures: GRF=1020/10=100, GIF=1030/10=1000, LT=101.5/10=1.41, LM=106/10=4.0, TM=(FM−1)T0=(107/10−1)(290) =1163 K, TRF=(FRF−1)T0=(103/10−1)(290) =289 K, TIF=(FIF−1)T0=(101.1/10−1)(290) =84 K. Then from (14.27), (14.28), and (14.30) the noise temperatures of the receiver, transmission line, and antenna are TREC=TRF+TM GRF+TIFLM GRF=289+1163 100+84(4.0) 100=304 K, TTL=(LT−1)Tp=(1.41−1)300 =123 K, TA=ηradTb+(1−ηrad)Tp=0.9(200) +(1−0.9)(300) =210 K. The input noise power, from (14.31), is Ni=kBT A=1.38×10−23(106)(210) =2.9×10−15W=−115 dBm . Then the input SNR is Si Ni=−80+115=35 dB. From (14.33) the total system noise temperature is TSYS=TA+TTL+LTTREC=210+123+(1.41)(304) =762 K. This result clearly shows the noise contributions of the various components. The output SNR is found from (14.34) as So No=Si kBT SYS, kBT SYS=1.38×10−23(106)(762) =1.05×10−14W=−110 dBm , so So No=−80+110=30 dB.■ DigitalModulationandBitErrorRate Information may be impressed upon a sinusoidal carrier using amplitude, frequency, or phase modulation. If the modulating signal is analog, as in the case of AM or FM radio, the amplitude, frequency, or phase of the carrier will undergo a continuous variation. Ifthe modulating signal represents digital data in binary form, the variation in the amplitude, frequency, or phase of the carrier will be limited to two values. These types of modulations are usually referred to as amplitude shift keying, frequency shift keying , and phase shift keying, and abbreviated as ASK, FSK, and PSK, respectively. For example, ASK may involve a carrier that is turned on for a binary “1,” and off for a binary “0.” Frequency c14IntroToMicrowaveSystems Pozar September 26, 2011 18:40 682 Chapter 14: Introduction to Microwave Systems ASK10 0 01 FSK PSKm(t) 1 t t t t FIGURE 14.15 Binary data and the resulting modulated carrier waveforms for amplitude shift keying, frequency shift keying, and phase shift keying. shift keying involves switching between two different carrier frequencies, while phase shift keying involves a 180◦phase shift of the carrier, depending on the binary data. Binary phase shift keying is also referred to as BPSK. Figure 14.15 shows the carrier waveformsthat result from binary digital modulation with ASK, FSK, and PSK methods. The majority of modern wireless systems rely on digital modulation methods due to their superior performance in the presence of noise and signal fading, lower power re-quirements, and better suitability for the transmission of data with error-correcting codes or encryption. Besides the basic binary modulation schemes described above, there are a number of other digital modulation methods. One popular method is quadrature phase shift keying (QPSK), where two data bits are used to select one of four possible phase states (0 ◦,9 0◦, 180◦, or 270◦). More generally, one can use m-ary phase shift keying, where one of 2mphase states is selected on the basis of mdata bits. It is also possible to modulate both amplitude and phase simultaneously, resulting in quadrature amplitude modulation, or QAM. Such higher order modulation methods allow higher data rates for a given channel bandwidth, but involve more system and processing complexity. In an ideal situation a receiver will detect the same binary digit that was transmitted, but the presence of noise in the communication channel introduces the possibility thaterrors will be made during the detection process. The likelihood of an error in the detection of a single bit is quantified by the bit error probability, P b, also known as the bit error rate (BER). The probability of error is dependent on the ratio of bit energy to noise powerdensity, E b/n0, where Ebis the energy received during each bit interval, and n0is the power spectral density of the noise on the channel. The probability of error decreases as bit energy increases, or as noise density decreases. If Sis the received signal (carrier) power (watts), with Tbbeing the bit period (seconds), and Rbthe bit rate (bits per second), the bit energy can be written as Eb=STb=S/Rb,(W-sec) (14.35) Then the ratio Eb/n0is Eb n0=STb n0=S n0Rb.( 14.36) c14IntroToMicrowaveSystems Pozar September 26, 2011 18:40 14.2 Wireless Communications 683 100 10–1 10–2 10–3 10–4 10–5 10–6 10–7 –10 –5 0 5 Eb /n0 (dB)BPSK QPSKASK FSK 10 15 20Probability of Bit Error, Pb FIGURE 14.16 Comparison of bit error rates for ASK, FSK, BPSK, and QPSK modulation meth- ods versus Eb/n0. (Coherent demodulation is assumed, with Gray coding for QPSK.) Since the noise power is N=n0B, where Bis the bandwidth of the receiver, the ratio of bit energy to noise power density can be expressed in terms of the SNR as Eb n0=S NBTb=S NB Rb.( 14.37) Note that this result indicates that, for a given SNR, the ratio of bit energy to noise power density will decrease (and the BER will increase) as the data rate increases. Depending on the type of modulation, the required receiver bandwidth may range from one to several times the bit rate. Figure 14.16 shows bit error probability for four types of digital modulation (ASK, FSK, BPSK, and QPSK) versus the Eb/n0ratio. The bit error rate for QPSK is the same as for BPSK, but note that QPSK involves the transmission of two bits for every one bitsent by BPSK. Each of the binary modulation methods transmits one bit during each bit period, and they are therefore said to have a bandwidth efficiency of 1 bps/Hz. Higher level modulation methods can achieve higher bandwidth efficiencies. For example, QPSK transmits two bits per period, and therefore has a bandwidth efficiency of 2 bps/Hz. Table 14.1 lists the bandwidth efficiency and the required E b/n0ratio for a bit error rate of 10−5for various digital modulation methods. EXAMPLE 14.6 LINK ANALYSIS FOR LEO SATELLITE DOWNLINK A LEO satellite at an orbital distance of 940 km uses QPSK to communicate witha handset on Earth. The satellite has a transmit power of 80 W and an antenna gainof 20 dB, while the handset has an antenna gain of 1 dB and a system temperature of 750 K. If atmospheric attenuation is 2 dB, and the required link margin is 10 dB, what is the maximum data rate for a bit error probability of 0.01? c14IntroToMicrowaveSystems Pozar September 26, 2011 18:40 684 Chapter 14: Introduction to Microwave Systems Solution The wavelength is 1.875 cm, so from (14.26) the path loss is L0=20 log/parenleftbigg4πR λ/parenrightbigg =20 log/parenleftBigg (4π)(940×103) 0.01875/parenrightBigg =176.0d B The received power is Pr=Pt+Gt−L0−LA+Gr=49+20−176−2+1=−108 dBm . For a link margin of 10 dB, this received power level should be 10 dB above the threshold level. Thus, the threshold received signal level is Smin=Pr−LM=−108−10=−118 dBm =1.58×10−15W. From Figure 14.16, the required Eb/n0for a bit error rate of 0.01 for QPSK is about 5 dB =3.16. Solving (14.36) for the maximum bit rate gives Rb=/parenleftbiggEb n0/parenrightbigg−1Smin n0=/parenleftbiggEb n0/parenrightbigg−1Smin kTsys=/parenleftbigg1 3.16/parenrightbigg1.58×10−15 /parenleftbig 1.38×10−23/parenrightbig (750)=48 kbps ■ TABLE 14.1 Summary of Performance of Various Digital Modulation Methods Modulation Eb/n0(dB) for Bandwidth Type Pb=10−5Efficiency Binary ASK 15.6 1 Binary FSK 12.6 1 Binary PSK 9.6 1 QPSK 9.6 2 8-PSK 13.0 3 16-PSK 18.7 4 16-QAM 13.4 4 64-QAM 17.8 6 WirelessCommunicationSystems We conclude this section with a summary of some of the most prevalent wireless com- munication systems in current use. Table 14.2 lists some of the commonly used frequency bands for wireless systems. Cellular telephone and data systems: Cellular voice and networking systems are in con- stant evolution, involving the use of old and new technology, existing and newly available carrier frequencies, sophisticated multiple-access techniques, international agreements,and the special interests of commercial service providers, governments, and regulatory agencies. The objective is to provide mobile users with voice and data service (including Internet access and video), with high data rates and compatibility across systems. Muchprogress has been made, but there are still technical and organizational challenges that remain. c14IntroToMicrowaveSystems Pozar September 26, 2011 18:40 14.2 Wireless Communications 685 TABLE 14.2 Wireless System Frequencies Wireless System (Country) Frequency Advanced Mobile Phone System (AMPS, United States; obsolete) U: 824–849 MHz D: 869–894 MHz GSM 850 (Americas) U: 824–849 MHz D: 869–894 MHz GSM 900 (worldwide) U: 890–915 MHz D: 935–960 MHz GSM 1800 (worldwide) U: 1710–1785 MHz D: 1805–1880 MHz GSM 1900 (Americas) U: 1850–1910 MHz D: 1930–1990 MHz Universal Mobile Telecommunications System (UMTS), U: 1920–1980 MHz band 1 (most countries) D: 2110–2170 MHz UMTS, band 2 (most countries) U: 1850–1910 MHz D: 1930–1990 MHz UMTS, band 8 (most countries) U: 880–916 MHz D: 925–960 MHz Wireless local area networks (WiFi) 902–928 MHz 2.400–2.484 GHz5.725–5.850 GHz Global Positioning System (GPS) L1: 1575.42 MHz L2: 1227.60 MHz Direct Broadcast Satellite (DBS) (Europe, Russia) 10.7–12.75 GHz (Americas) 12.2–12.7 GHz (Asia, Australia) 11.7–12.2 GHz Industrial, medical, and scientific bands (most countries) 902–928 MHz 2.400–2.484 GHz 5.725–5.850 GHz U, uplink (mobile-to-base); D, downlink (base-to-mobile). Cellular telephone systems were first proposed in the 1970s in response to the problem of providing mobile radio service to a large number of users in urban areas. Early mobile radio systems could handle only a very limited number of users due to inefficient use ofthe radio spectrum and interference between users. The cellular radio concept solved this problem by dividing a geographical area into nonoverlapping cells in which each cell has its own transmitter and receiver (the base station) to communicate with mobile users operating in that cell. Each cell site may allow as many as several hundred users to simultaneously communicate over voice and/or data channels. Frequency bands assigned to a particularcell can be reused in other, nonadjacent cells. The first cellular telephone systems were built in Japan and Europe in 1979 and 1981, and in the United States (AMPS) in 1983. These systems used analog FM modulationand divided their allocated frequency bands into several hundred channels, each of which could support an individual telephone conversation. These early services grew slowly at c14IntroToMicrowaveSystems Pozar September 26, 2011 18:40 686 Chapter 14: Introduction to Microwave Systems first, because of the initial costs of developing an infrastructure of base stations and the initial expense of handsets, but by the 1990s growth became phenomenal. Because of the rapidly growing business and consumer demand for wireless services, as well as advances in wireless technology, several second-generation standards were im-plemented in the United States, Europe, and Asia. These standards all employed digital modulation methods to provide better-quality service and more efficient use of the radio spectrum, as well as multiple-access methods that could be categorized as either time di- vision multiple access (TDMA), or code division multiple access (CDMA). Since then, most countries have made more radio spectrum available, usually as a result of freeing up frequency bands that had been used by VHF broadcast television. Today, most wireless cellular and smartphone systems have migrated to third- generation (3G) standards, or are in the process of being upgraded to 3G standards. TheInternational Mobile Telecommunications (IMT)-2000 project of the International Tele- communications Union (ITU) forms the basis for 3G standards, most of which rely on CDMA and its variations, W-CDMA and CDMA2000. At the present time, IMT-2000 sup-ports data rates of 2 Mbps for fixed users and 144 kbps for mobile users. A related effort is the 3rd Generation Partnership Project (3GPP), which is based on a collaboration of vari- ous telecommunications groups to form a migration path from existing second-generationinfrastructure to 3G, and then toward a Long-Term Evolution (LTE) goal in 2010–2011 of data rates of 100 Mbps for fixed users and 50 Mbps for mobile users. Many countries have adopted the Universal Mobile Telecommunications System standard, which is based on3GPP. Another variation is the 3GPP2 standard, which works from existing CDMA tech- nologies (including W-CDMA and CDMA2000) to provide high data rates. At present, there are many proposed standards for interim use for capitalizing on existing infrastruc-ture, as well as for new standards that will evolve into fourth-generation systems. Satellite systems for wireless voice and data: The conceptual advantage of satellite sys- tems is that a relatively small number of satellites can provide coverage to users at any location in the world, including the oceans, deserts, and mountains—areas for which itis difficult or impossible to provide cellular service. In principle, as few as three geosyn- chronous satellites can provide complete global coverage, but the very high altitude of the geosynchronous orbit makes it difficult to communicate with hand held terminals becausethe large path loss results in very low signal strength. Satellites in lower orbits can provide usable levels of signal power, but many more satellites are then needed to provide global coverage. The Iridium project, originally financed by a consortium of companies headed by Mo- torola, was the first commercial satellite system to offer worldwide hand held wireless telephone service. It consists of 66 LEO satellites in near-polar orbits, and connects mobilephone and paging subscribers to the public telephone system through a series of intersatel- lite relay links and land-based gateway terminals. Figure 14.17 shows a photo of one of the Iridium phased array antennas. The Iridium system cost was approximately $5 billion;it began service in November 1998, and filed for bankruptcy in August 1999. Iridium was acquired by the U.S. Defense Department in 2001 and is still operating at this time. One drawback of using satellites for telephone service is that weak signal levels require a line-of-sight path from the mobile user to the satellite, meaning that satellite telephones generally cannot be used in buildings, automobiles, or even in many wooded or urbanareas. This places satellite phone service at a definite performance disadvantage relative to land-based cellular services. Other commercial LEO satellite communications systems, such as Globalstar, have also ended in financial failure. Most successful satellite communications systems rely on geostationary satellites. These include the INMARSAT systems, originally used to provide communications to c14IntroToMicrowaveSystems Pozar September 26, 2011 18:40 14.2 Wireless Communications 687 FIGURE 14.17 Photograph of one of the three L-band antenna arrays for an Iridium communi- cations satellite. The Iridium system consists of 66 satellites in low Earth orbitto provide global personal satellite TDMA communications services, including voice, fax, and paging. Courtesy of Raytheon Company, Waltham, Mass. maritime shipping, but also used in remote areas. Many financial services and businesses usevery small aperture terminals (VSATs), which provide relatively low rate data com- munications to geostationary satellites with 12- to 18-inch antennas. An example of a geo-stationary satellite telephone service is the Thuraya system, which provides coverage to parts of Africa, Europe, India, and the Middle East. The subscriber link operates at L band, with a fairly compact handset. There is a noticeable conversational delay with the Thurayasystem due to the propagation time to and from the satellite. Global Positioning: The Global Positioning System (GPS) uses 24 satellites in medium Earth orbits to provide accurate position information (latitude, longitude, and elevation) to users on land, air, or sea. Originally developed as the NA VSTAR system by the U.S. Department of Defense, GPS has become one of the most pervasive applications of wirelesstechnology for consumers and businesses throughout the world. GPS receivers are used on airplanes, ships, trucks, trains, and automobiles. Advances in technology have led to substantial reductions in size and cost, so that small GPS receivers can be integrated into cellular telephones and smart phones, and hand-held GPS devices are used by hikers and sportsmen. With differential GPS, accuracies on the order of 1 cm can be achieved, acapability that has revolutionized the surveying industry. An entirely new field of study, known as geographic information systems (GIS), is based on the relation of data to location, usually obtained in conjunction with GPS. GPS positioning operates by using triangulation with a minimum of four satellites. GPS satellites are in orbits 20,200 km above Earth, with orbital periods of 12 hours. c14IntroToMicrowaveSystems Pozar September 26, 2011 18:40 688 Chapter 14: Introduction to Microwave Systems Distances from the user’s GPS receiver to these satellites are found by timing the prop- agation delay between the satellites and the receiver. The orbital positions of the satellites (ephemeris) are known to very high accuracy, and each satellite contains an extremely ac- curate clock to provide a unique set of timing pulses. A GPS receiver decodes this timinginformation and performs the necessary calculations to find the position and velocity of the receiver. The GPS receiver usually must have a line-of-sight view to at least four satel- lites in the GPS constellation, although three satellites are adequate if altitude position isknown (as in the case of ships at sea). Because of the low-gain antennas required for op- eration, the received signal level from a GPS satellite is very low—typically on the order of−130 dBm (for a receiver antenna gain of 0 dB). This signal level is usually below the noise power at the receiver, but spread-spectrum techniques are used to improve the received SNR. GPS operates at two frequency bands: L1, at 1575.42 MHz, and L2, at 1227.60 MHz, transmitting spread-spectrum signals with BPSK modulation. The L1 frequency is used to transmit ephemeris data for each satellite, as well as timing codes, which are availableto any commercial or public user. This mode of operation is referred to as the Course/ Acquisition (C/A) code. In contrast, the L2 frequency is reserved for military use, and uses an encrypted timing code referred to as the Protected (P) code (there is also a P code signal transmitted at the L1 frequency). The P code offers much higher accuracy than the C/A code. The typical accuracy that can be achieved with an L1 GPS receiver is about 100 feet. Accuracy is limited by timing errors in the clocks on the satellites and the receiver, as wellas error in the assumed position of the GPS satellites. The most significant error is generally caused by atmospheric and ionospheric effects, which introduce small but variable delays in signal propagation from the satellite to the receiver. Wireless local area networks : Wireless local area networks provide connections between computers and peripherals over short distances. Wireless networks can be found in airports,coffee shops, office buildings, college campuses, and even on commercial airliners, busses, and cruise ships. Indoor coverage is usually less than a few hundred feet. Outdoors, in the absence of obstructions and with the use of high-gain antennas, much longer ranges can beobtained. Wireless networks are especially useful when it is impossible or prohibitively ex- pensive to place network wiring in or between buildings, or when only temporary network access is needed. Mobile users, of course, can only be connected to a computer networkthrough a wireless link. Most commercial WLAN products are based on the IEEE 802.11 standards (Wi-Fi). These operate at either 2.4 or 5.7 GHz (in the industrial, scientific, and medical frequency bands), and use either frequency-hopping or direct-sequence spread-spectrum techniques. Standards 802.11a, 802.11b, and 802.11g can provide data rates up to 54 Mbps, while 802.11n (which uses multiple antennas) can achieve data rates of up to 150 Mbps. Actual data rates are often significantly lower due to nonideal propagation conditions and loading from other users. Another wireless networking standard is Bluetooth, which is intended for short-range networking of portable devices, such as cameras, printers, headsets, games, and similar applications, to resident computers or routers. Bluetooth devices operate at 2.4 GHz, withRF power in the range of 1–100 mW and corresponding operating ranges of 1–100 m. Data rates range from 1 to 24 Mbps. Millimeter wave frequencies are increasingly being considered for high speed local area networking due to the large bandwidths that are available. Figure 14.18 shows a de- velopmental model of a high-speed 60 GHz wireless networking transmitter. Direct broadcast satellites : DBS systems provide television service with continental cov- erage from geosynchronous satellites directly to home users with a relatively small 18 inch c14IntroToMicrowaveSystems Pozar September 26, 2011 18:40 14.2 Wireless Communications 689 FIGURE 14.18 Photograph of a high-speed 60 GHz wireless local area network transmitter. This WLAN operates at 59–62 GHz, with a data rate of 2.8 Gbps. It uses GaAs chips and a built-in four-element, circularly polarized microstrip antenna. Courtesy of Newlans, Inc., Acton, Mass. diameter antenna. Prior to DBS technology, satellite TV service relied on analog signals that required an unsightly dish antenna as large as 6 feet in diameter to achieve the nec- essary SNR. The smaller DBS antenna became possible through the use of digital mod- ulation techniques, which reduce the required received signal levels as compared to an analog system. DBS systems operate with carrier frequencies in the 10–12 GHz range (see Table 14.2), and typically use QPSK with digital multiplexing and error correctionto deliver digital data at a rate of 40 Mbps. Several DBS satellites are used through- out the world to provide subscriber television service, sometimes with more than one satellite per coverage area. For North America, two satellites, DBS-1 and DBS-2, arein geostationary orbit at 101.2 ◦and 100.8◦longitude, and each provides 16 channels with 120 W of radiated power per channel. These satellites use opposite circular polar- izations to minimize loss due to precipitation, and to avoid interference with each other(polarization duplexing). Point-to-point radio systems : Point-to-point radios are used to provide dedicated data con- nections between two fixed points. Electric utility companies use point-to-point radios for transmission of telemetry information for the generation, transmission, and distribution ofelectric power between generating stations and substations. Point-to-point radios are also used to connect cellular base stations to the public switched telephone network, and are at- tractive because they are generally much cheaper than running high-bandwidth fiber-opticlines below ground level. Point-to-point radios usually operate in the 18, 24, or 38 GHz bands, and use a variety of digital modulation methods to provide data rates in excess of 50 Mbps. High-gain antennas are typically used to minimize power requirements and toavoid interference with other users. Other wireless systems: Many other applications of wireless technology are being devel- oped, and we can only briefly mention some of these. One of the most pervasive may turn out to be Radio Frequency Identification (RFID) systems, which rely on small, low-cost tags that can receive an interrogatory RF signal and reply with a signal containing pre- programmed data. RFID tags can be used for retail products, inventory control, industrialmaterials, security applications, or any application that requires identification or tracking. An interesting feature of RFID tags is that they can be passively powered, whereby they store the power required for signaling by rectifying the interrogatory signal and charging asmall capacitor. This is then used to drive very low power CMOS circuitry to transmit data back to the interrogating receiver. c14IntroToMicrowaveSystems Pozar September 26, 2011 18:40 690 Chapter 14: Introduction to Microwave Systems Another area where wireless technology is beginning to experience growth is in motor vehicle and highway applications. These include toll collection, intelligent cruise control, collision avoidance radar, blind spot radar, traffic information, emergency messaging, and vehicle identification. Automatic toll collection is already in service in many parts of theUnited States and Europe. A number of automobile models are available with blind spot and collision sensors as optional equipment. 14.3RADARSYSTEMS Radar, or radio detection and ranging , is the oldest application of microwave technology, dating back to World War II. In its basic operation, a transmitter sends out a signal, whichis partly reflected by a distant target, and then detected by a sensitive receiver. If a narrow- beam antenna is used, the target’s direction can be accurately given by the angular position of the antenna. The distance to the target is determined by the time required for a pulsedsignal to travel to the target and back, and the radial velocity of the target is related to the Doppler shift of the return signal. Below are listed some of the typical applications of radar systems. Civilian applications rAirport surveillancerMarine navigationrWeather radarrAltimetryrAircraft landingrSecurity alarmsrSpeed measurement (police radar)rGeographic mapping Military applications rAir and marine navigationrDetection and tracking of aircraft, missiles, and spacecraftrMissile guidancerFire control for missiles and artilleryrWeapon fusesrReconnaissance Scientific applications rAstronomyrMapping and imagingrPrecision distance measurementrRemote sensing of the environment Early radar work in the United States and Britain began in the 1930s using very high frequency (VHF) sources. A major breakthrough occurred in the early 1940s with the British invention of the magnetron tube as a reliable source of high-power microwaves. Higher frequencies allowed the use of reasonably sized antennas with high gain, allowingmechanical tracking of targets with good angular resolution. Radar was quickly developed in Great Britain and the United States, and played an important role in World War II. Figure 14.19 shows a photograph of the phased array radar for the Patriot missile system. We will now derive the radar equation, which governs the basic operation of most radars, and then describe some of the more common types of radar systems. c14IntroToMicrowaveSystems Pozar September 26, 2011 18:40 14.3 Radar Systems 691 FIGURE 14.19 Photograph of the Patriot phased array radar. This is a C-band multifunction radar that provides tactical air defense, including target search and tracking, and missilefire control. The phased array antenna uses 5000 ferrite phase shifters to electron- ically scan the antenna beam. Courtesy of Raytheon Company, Waltham, Mass. TheRadarEquation Two basic radar systems are illustrated in Figure 14.20; in a monostatic radar the same antenna is used for both transmit and receive, while a bistatic radar uses two separate an- tennas for these functions. Most radars are of the monostatic type, but in some applications (such as missile fire control) the target may be illuminated by a separate transmit antenna.Separate antennas are also sometimes used to achieve the necessary signal isolation be- tween transmitter and receiver. Here we will consider the monostatic case, but the bistatic case is very similar. If the transmitter radiates a power P tthrough an antenna of gain G, the power density incident on the target is, from (14.23), St=PtG 4πR2,( 14.38) where Ris the distance to the target. It is assumed that the target is in the main beam direction of the antenna. The target will scatter the incident power in various directions; the ratio of the scattered power in a given direction to the incident power density is defined as the radar cross section ,σ, of the target. Mathematically, σ=Ps Stm2,( 14.39) where Psis the total power scattered by the target, and Stis the power density incident on the target. The radar cross section thus has the dimensions of area, and is a property of the target itself. It depends on the incident and reflection angles, as well as on the polarizations of the incident and reflected waves. Since the target scatters as a source of finite size, the power density of the reradiated field must decay as 1/4π R2away from the target. Thus the power density of the scattered c14IntroToMicrowaveSystems Pozar September 26, 2011 18:40 692 Chapter 14: Introduction to Microwave Systems RPt PtPrG G GReceiver/ processor Receiver/processorTarget /H9268 Target /H9268(a) (b) FIGURE 14.20 Basic monostatic and bistatic radar systems. (a) Monostatic radar system. (b) Bistatic radar system. field back at the receive antenna must be Sr=PtGσ (4πR2)2.( 14.40) Using (14.15) for the effective area of the antenna gives the received power as Pr=PtG2λ2σ (4π)3R4.( 14.41) This is the radar equation. Note that the received power varies as 1/ R4, which implies that a high-power transmitter and a sensitive low-noise receiver are needed to detect targets at long ranges. Because of noise received by the antenna and generated in the receiver, there will be some minimum detectable power that can be discriminated by the receiver. If this power is Pmin, then (14.41) can be rewritten to give the maximum range as Rmax=/bracketleftBigg PtG2σλ2 (4π)3Pmin/bracketrightBigg1/4 .( 14.42) Signal processing can effectively reduce the minimum detectable signal, and so increase the usable range. One very common processing technique used with pulse radars is pulse integration, in which a sequence of Nreceived pulses is integrated over time. The effect is to reduce the noise level, which has a zero mean, relative to the returned pulse level, resulting in an improvement factor of approximately N[6]. Of course, the above results seldom describe the performance of an actual radar sys- tem. Factors such as propagation effects, the statistical nature of the detection process, and external interference often reduce the usable range of a radar system. c14IntroToMicrowaveSystems Pozar September 26, 2011 18:40 14.3 Radar Systems 693 EXAMPLE 14.7 APPLICATION OF THE RADAR RANGE EQUATION A pulse radar operating at 10 GHz has an antenna with a gain of 28 dB and a transmitter power of 2 kW (pulse power). If it is desired to detect a target with across section of 12 m 2, and the minimum detectable signal is −90 dBm, what is the maximum range of the radar? Solution The required numerical values are G=1028/10=631, Pmin=10−90/ 10mW=10−12W, λ=0.03 m. Then the radar range equation of (14.42) gives the maximum range as Rmax=/bracketleftBigg (2×103)(631)2(12)(. 03)2 (4π)3(10−12)/bracketrightBigg1/4 =8114 m. ■ PulseRadar A pulse radar determines target range by measuring the round-trip time of a pulsed mi- crowave signal. Figure 14.21 shows a typical pulse radar system block diagram. The trans-mitter portion consists of a single-sideband mixer used to frequency offset a microwave oscillator of frequency f 0by an amount equal to the IF frequency. After power amplifi- cation, pulses of this signal are transmitted by the antenna. The transmit/receive switch is controlled by the pulse generator to give a transmit pulse width τ, with a pulse repetition frequency (PRF) of fr=1/Tr. The transmit pulse thus consists of a short burst of a mi- crowave signal at the frequency f0+fIF. Typical pulse durations range from 100 ms to 50 ns; shorter pulses give better range resolution, but longer pulses result in a better SNR after receiver processing. Typical pulse repetition frequencies range from 100 Hz to 100kHz; higher PRFs give more returned pulses per unit time, which improves performance, but lower PRFs avoid range ambiguities that can occur when R>cT r/2. In the receive mode, the returned signal is amplified and mixed with the local oscilla- tor of frequency f0to produce the desired IF signal. The local oscillator is used for both up-conversion in the transmitter and down-conversion in the receiver; this simplifies the system and avoids the problem of frequency drift, which would be a consideration if sepa-rate oscillators were used. The IF signal is amplified, detected, and fed to a video amplifier/ display. Search radars often use a continuously rotating antenna for 360 ◦azimuthal cover- age; in this case the display shows a polar plot of target range versus angle. Modern radarsuse a computer for the processing of the detected signal and display of target information. The transmit/receive (T/R) switch in the pulse radar actually performs two functions: forming the transmit pulse train, and switching the antenna between the transmitter and receiver. This latter function is also known as duplexing. In principle, the duplexing func- tion could be achieved with a circulator, but an important requirement is that a high degreeof isolation (about 80–100 dB) be provided between the transmitter and receiver to avoid transmitter leakage into the receiver, which would drown the target return (or possibly dam- age the receiver). As circulators typically achieve only 20–30 dB of isolation, some type ofswitch, with high isolation, is required. If necessary, further isolation can be obtained by using additional switches along the path of the transmitter circuit. c14IntroToMicrowaveSystems Pozar September 26, 2011 18:40 694 Chapter 14: Introduction to Microwave Systems Detector t tt tTransmit mode Pulse generator Transmit signal Detected signal/H9270 Transmitter leakageClutter and noiseTarget returnTrPulse generatorLow-noise amplifierMixer IF amplifierVideo amplifierDisplayAntennaTransmit/ receive switchPower amplifierUSB mixer f0 + fIFfIF fIFf0 Receive mode FIGURE 14.21 A pulse radar system and timing diagram. DopplerRadar If the target has a velocity component along the line of sight of the radar, the returned signal will be shifted in frequency relative to the transmitted frequency due to the Doppler effect. If the transmitted frequency is f0, and the radial target velocity is v, then the shift in frequency, or the Doppler frequency, will be fd=2vf0 c,( 14.43) where cis the velocity of light. The received frequency is then f0±fd, where the plus sign corresponds to an approaching target and the minus sign corresponds to a receding target. Figure 14.22 shows a basic Doppler radar system. Observe that it is much simpler than a pulse radar since a continuous wave signal is used, and the transmit oscillator can also be used as a local oscillator for the receive mixer because the received signal is frequencyoffset by the Doppler frequency. The filter following the mixer should have a passband corresponding to the expected minimum and maximum target velocities. It is important that the filter have high attenuation at zero frequency, to eliminate the effect of clutter return and transmitter leakage at the frequency f 0, as these signals would down-convert to zero frequency. Then a high degree of isolation is not necessary between transmitter andreceiver, and a circulator can be used. This type of filter response also helps to reduce the effect of 1/ fnoise. The above radar cannot distinguish between approaching and receding targets, as the sign of f dis lost in the detection process. Such information can be recovered, however, by using a mixer that produces separately the upper and lower sideband products. c14IntroToMicrowaveSystems Pozar September 26, 2011 18:40 14.3 Radar Systems 695 /H9268vf0 f0 f0 fd fd fd ff0 ± fd f0 ± fdAntennaCirculator Display Filter Amplifier FIGURE 14.22 Doppler radar system. Since the return of a pulse radar from a moving target will contain a Doppler shift, it is possible to determine both the range and velocity (and position, if a narrow-beam antennais used) of a target with a single radar. Such a radar is known as a pulse-Doppler radar, and it offers several advantages over pulse or Doppler radars. One problem with a pulse radar is that it is impossible to distinguish between a true target and clutter returns from the ground,trees, buildings, etc. Such clutter returns may be picked up from the antenna sidelobes. However, if the target is moving (e.g., as in an airport surveillance radar application), the Doppler shift can be used to separate its return from clutter, which is stationary relative tothe radar. RadarCrossSection A radar target is characterized by its radar cross section, as defined in (14.36), which gives the ratio of scattered power to incident power density. The cross section of a target dependson the frequency and polarizations of the incident and scattered waves, and on the incident and reflected angles relative to the target. Thus we can define a monostatic cross section (incident and reflected angles identical), and a bistatic cross section (incident and reflectedangles different). For simple shapes the radar cross section can be calculated as an electromagnetic boundary value problem; more complex targets require numerical techniques or measure-ment to find the cross section. The radar cross section of a conducting sphere can be cal- culated exactly; the monostatic result is shown in Figure 14.23, normalized to πa 2,t h e physical cross-sectional area of the sphere. Note that the cross section increases veryquickly with size for electrically small spheres ( a/lessmuchλ). This region is called the Rayleigh region, and it can be shown that σvaries as (a/λ) 4in this region. (This strong dependence on frequency explains why the sky is blue, as the blue component of sunlight scatters morestrongly from atmospheric particles than do the lower frequency red components.) For electrically large spheres, where a/greatermuchλ, the radar cross section of the sphere is equal to its physical cross section, πa 2. This is the optical region, where geometrical optics is valid. Many other shapes, such as flat plates at normal incidence, also have cross sections that approach the physical area for electrically large sizes. Between the Rayleigh region and the optical region is the resonance region, where the electrical size of the sphere is on the order of a wavelength. Here the cross section is oscillating with frequency due to phase addition and cancellation of various scattered fieldcomponents. Of particular note is the fact that the cross section may reach quite high values in this region. c14IntroToMicrowaveSystems Pozar September 26, 2011 18:40 696 Chapter 14: Introduction to Microwave Systems TABLE 14.3 Typical Radar Cross Sections Target σ(m2) Bird 0.01 Missile 0.5 Person 1 Small plane 1–2 Bicycle 2 Small boat 2 Fighter plane 3–8 Bomber 30–40 Large airliner 100 Truck 200 12 3 5 1 00.010.115 2/H9266a//H9261/H9268//H9266a2 Rayleigh region Resonance region Optical region FIGURE 14.23 Monostatic radar cross section of a conducting sphere. Complex targets such as aircraft or ships generally have cross sections that vary rapidly with frequency and aspect angle. In military applications it is often desirable to mini- mize the radar cross section of vehicles to reduce detectability. This can be accomplished by using radar-absorbing materials (lossy dielectrics) in the construction of the vehicle.Table 14.3 lists the approximate radar cross sections of a variety of different targets. 14.4RADIOMETERSYSTEMS A radar system obtains information about a target by transmitting a signal and receiving the echo from the target, and thus can be described as an active remote sensing system. Radiometry, however, is a passive technique, which develops information about a target solely from the microwave portion of the blackbody radiation (noise) that it either emitsdirectly or reflects from surrounding bodies. A radiometer is a sensitive receiver specially designed to measure this noise power. c14IntroToMicrowaveSystems Pozar September 26, 2011 18:40 14.4 Radiometer Systems 697 AtmosphereTAD TARSun TBTs EarthRadiometer antenna FIGURE 14.24 Noise power sources in a typical radiometer application. TheoryandApplicationsofRadiometry As discussed in Section 10.1, a body in thermodynamic equilibrium at a temperature T radiates energy according to Planck’s radiation law. In the microwave region this result reduces to P=kTB, where kis Boltzmann’s constant, Bis the system bandwidth, and P is the radiated power. This result strictly applies only to a blackbody, which is defined as an idealized material that absorbs all incident energy and reflects none; a blackbody also radiates energy at the same rate as it absorbs energy, thus maintaining thermal equilibrium.A nonideal body will partially reflect incident energy, and so it does not radiate as much power as would a blackbody at the same temperature. A measure of the power radiated by a body relative to that radiated by an ideal blackbody at the same temperature is theemissivity ,e, defined as e=P kTB,( 14.44) where Pis the power radiated by the nonideal body, and kTB is the power that would be emitted by a perfect blackbody. Thus, 0 ≤e≤1, and e=1 for a perfect blackbody; emissivity may be thought of as the “efficiency” of blackbody radiation. As we saw in Section 10.1, noise power can also be quantified in terms of equivalent temperature. Thus, for radiometric purposes, we can define a brightness temperature, TB,a s TB=eT,( 14.45) where Tis the physical temperature of the body. This shows that, radiometrically, a body never looks hotter than its actual temperature, since 0 ≤e≤1. Consider Figure 14.24, which shows the antenna of a microwave radiometer receiving noise powers from various sources. The antenna is pointed at a region of Earth that has an apparent brightness temperature TB. The atmosphere emits radiation in all directions; the component radiated directly toward the antenna is TAD, while the component reflected from Earth to the antenna is TAR. There may also be noise powers that enter the sidelobes of the antennas from the Sun or other sources. Thus, we can see that the total brightnesstemperature seen by the radiometer is a function of the scene under observation, as well as the observation angle, frequency, polarization, attenuation of the atmosphere, and the antenna pattern. The objective of radiometry is to infer information about the scene fromthe measured brightness temperature and an analysis of the radiometric mechanisms that relate brightness temperature to physical conditions of the scene. For example, the power c14IntroToMicrowaveSystems Pozar September 26, 2011 18:40 698 Chapter 14: Introduction to Microwave Systems FIGURE 14.25 Photograph of a stepped frequency microwave radiometer, operating at 4.7–7.2 GHz. This instrument is flown on aircraft to measure brightness temperature and infer ocean surface wind speed and rain rate estimation in hurricanes. Courtesy of ProSensing, Inc., Amherst, Mass. reflected from a uniform layer of snow over soil can be treated as plane wave reflection from a multilayer dielectric region, leading to the development of an algorithm that givesthe thickness of the snow in terms of measured brightness temperature at various frequen- cies. Figure 14.25 shows a commercial multifrequency airborne radiometer for weather applications. Microwave radiometry has developed over the last 20 years into a mature technology, one that is strongly interdisciplinary, drawing on results from fields such as electrical en- gineering, oceanography, geophysics, and atmospheric and space sciences, to name a few.Some of the more important applications of microwave radiometry are listed below. Environmental applications rMeasurement of soil moisturerFlood mappingrSnow cover/ice cover mappingrOcean surface wind speedrAtmospheric temperature profilerAtmospheric humidity profile Military applications rTarget detectionrTarget recognitionrSurveillancerMapping Astronomy applications rPlanetary mappingrSolar emission mappingrMapping of galactic objectsrMeasurement of cosmological background radiation c14IntroToMicrowaveSystems Pozar September 26, 2011 18:40 14.4 Radiometer Systems 699 fBttVRF TREFVIF V0 TB TB Observed sceneAntennaMeasureCalibrateSwitchLow-noise amplifierMixer Local oscillatorIF filterIF amplifierDetector Integrator /H9270 dt 0V0 v ~ P FIGURE 14.26 Total power radiometer block diagram. TotalPowerRadiometer The aspect of radiometry that is of most interest to the microwave engineer is the design of the radiometer itself. The basic problem is to build a receiver that can distinguish be- tween the desired external radiometric noise and the inherent noise of the receiver, even though the radiometric power is usually less than the receiver noise power. Although itis not a very practical instrument, we will first consider the total power radiometer be- cause it represents a simple and direct approach to the problem and serves to illustrate the difficulties involved in radiometer design. The block diagram of a typical total power radiometer is shown in Figure 14.26. The front end of the receiver is a standard superheterodyne circuit consisting of an RF amplifier, a mixer/local oscillator, and an IF stage. The IF filter determines the system bandwidth, B. The detector is generally a square-law device, so that its output voltage is proportional to the input power. The integrator is essentially a low-pass filter with a cutoff frequency of 1/τ , and serves to smooth out short-term variations in the noise power. For simplicity, we assume that the antenna is lossless, although in practice antenna loss will affect the apparent temperature of the antenna, as given in (14.18). If the antenna is pointed at a background scene with a brightness temperature T B, the antenna power will be PA=kTBB; this is the desired signal. The receiver contributes noise that can be characterized as a power PR=kTRBat the receiver input, where TR is the overall noise temperature of the receiver. Thus the output voltage of the radio- meter is Vo=G(TB+TR)kB,( 14.46) where Gis the overall gain constant of the radiometer. Conceptually, the system is cali- brated by replacing the antenna input with two calibrated noise sources, from which the system constants GkB andGTRkBcan be determined. (This is similar to the Y-factor method for measuring noise temperature.) Then the desired brightness temperature, TB, can be determined. Two types of errors occur with this radiometer. First is an error, /Delta1TN, in the mea- sured brightness temperature due to noise fluctuations. Since noise is a random process, the measured noise power may vary from one integration period to the next. The integrator (or low-pass filter) acts to smooth out ripples in Vowith frequency components greater than c14IntroToMicrowaveSystems Pozar September 26, 2011 18:40 700 Chapter 14: Introduction to Microwave Systems 1/τ. It can be shown that the remaining error is [7] /Delta1TN=TB+TR√ Bτ.( 14.47) This result shows that if a longer measurement time, τ, can be tolerated, the error due to noise fluctuation can be reduced to a negligible value. A more serious error is due to random variations in the system gain, G. Such varia- tions generally occur in the RF amplifier, mixer, or IF amplifier, over a period of 1 s or longer. If the system is calibrated with a certain value of G, which changes by the time a measurement is made, an error will occur, as given in reference [7] as /Delta1TG=(TB+TR)/Delta1G G,( 14.48) where /Delta1Gis the rms change in the system gain, G. It will be useful to consider some typical numbers. For example, a 10 GHz total power radiometer may have a bandwidth of 100 MHz, a receiver temperature of TR=500 K, an integrator time constant of τ=0.01 s, and a system gain variation of /Delta1G/G=0.01. If the antenna temperature is TB=300 K, (14.47) gives the error due to noise fluctuations as /Delta1TN=0.8 K, while (14.48) gives the error due to gain variations as /Delta1TG=8 K. These results, which are based on reasonably realistic data, show that gain variation is the most detrimental factor affecting the accuracy of the total power radiometer. TheDickeRadiometer We have seen that the dominant factor affecting the accuracy of the total power radiometer is the variation of gain of the overall system. Since such gain variations have a relatively long time constant ( >1 s), it is conceptually possible to eliminate this error by repeatedly calibrating the radiometer at rapid rate. This is the principle behind the operation of the Dicke null-balancing radiometer. A system diagram is shown in Figure 14.27. The superheterodyne receiver is identical to the total power radiometer, but the input is periodically switched between the antenna and a variable power noise source; this switch is called the Dicke switch . The output of the square-law detector drives a synchronous demodulator, which consists of a switch and a difference circuit. The demodulator switch operates in synchronism with the Dicke switch, so that the output of the subtractor is proportional to the difference between the noise pow-ers from the antenna, T B, and the reference noise source, TREF. The output of the subtractor is then used as an error signal to a feedback control circuit, which controls the power level of the reference noise source so that Voapproaches zero. In this balanced state, TB=TREF, andTBcan be determined from the control voltage, Vc. The square-wave sampling fre- quency, fs, is chosen to be much faster than the drift time of the system gain, so that this effect is virtually eliminated. Typical sampling frequencies range from 10 to 1000 Hz. A typical radiometer would measure brightness temperature TBover a range of about 50–300 K; this then implies that the reference noise source would have to cover this same range, which is difficult to do in practice. Thus, there are several variations on the above design, differing essentially in the way that the reference noise power is controlled or added to the system. One possible method is to use a constant TREFthat is somewhat hotter than the maximum TBto be measured. The amount of reference noise power delivered to the system is then controlled by varying the pulse width of the sampling waveform. Another approach is to use a constant reference noise power, and vary the gain of the IF stageduring the reference sample time to achieve a null output. Other possibilities, including alternatives to the Dicke radiometer, are discussed in the literature [7]. c14IntroToMicrowaveSystems Pozar September 26, 2011 18:40 14.5 Microwave Propagation 701 f TBfsTBVc V0 TREF Observed sceneDicke switchRF amplifierMixerIF filter Detector v ~ P AntennaRF noise Local oscillatorIF amplifier Square wave generatorFeedback control circuit Output voltage Control voltage Synchronous demodulator– + –Variable power noise source FIGURE 14.27 Balanced Dicke radiometer block diagram. 14.5MICROWAVEPROPAGATION In free-space, electromagnetic waves propagate in straight lines without attenuation or other adverse effects. Free-space, however, is an idealization that is only approximated when RF or microwave energy propagates through the atmosphere or in the presence ofEarth. In practice, the performance of a communication, radar, or radiometry system may be seriously affected by propagation effects such as reflection, refraction, attenuation, or diffraction. Below we discuss some specific propagation phenomenon that can influencethe operation of microwave systems. It is important to realize that propagation effects gen- erally cannot be quantified in any exact or rigorous sense, but can only be described in terms of their statistics. AtmosphericEffects The relative permittivity of the atmosphere is close to unity, but is actually a function of air pressure, temperature, and humidity. An empirical result that is useful at microwave frequencies is given by [6] /epsilon1 r=/bracketleftBigg 1+10−6/parenleftBigg 79P T−11V T+3.8×105V T2/parenrightBigg/bracketrightBigg2 ,( 14.49) where Pis the barometric pressure in millibars, Tis the temperature in kelvins, and V is the water vapor pressure in millibars. This result shows that permittivity generally de-creases (approaches unity) as altitude increases since pressure and humidity decrease with height faster than does temperature. This change in permittivity with altitude causes radio waves to bend toward Earth, as depicted in Figure 14.28. Such refraction of radio wavescan sometimes be useful since it may extend the range of radar and communication systems beyond the limit imposed by the presence of Earth’s horizon. c14IntroToMicrowaveSystems Pozar September 26, 2011 18:40 702 Chapter 14: Introduction to Microwave Systems RefractedpathLine-of-sight path EarthAtmosphere FIGURE 14.28 Refraction of radio waves by the atmosphere. If an antenna is at a height, h, above Earth, simple geometry gives the line-of-sight distance to the horizon as d=√ 2Rh,( 14.50) where Ris the radius of Earth. From Figure 14.28 we see that the effect of refraction on range can be accounted for by using an effective Earth radius kR, where k>1. A value commonly used [6] is k=4/3, but this is only an average value, which changes with weather conditions. In a radar system, refraction effects can lead to errors when determin- ing the elevation of a target close to the horizon. Weather conditions can sometimes produce a localized temperature inversion, where the temperature increases with altitude. Equation (14.49) then shows that the atmospheric permittivity will decrease much faster than normal with increasing altitude. This condition can sometimes lead to ducting (also called trapping, or anomalous propagation), where a radio wave can propagate long distances parallel to Earth’s surface via the duct created by the layer of air along the temperature inversion. The situation is very similar to propagationin a dielectric waveguide. Such ducts can range in height from 50 to 500 feet, and may be near Earth’s surface or higher in altitude. Another atmospheric effect is attenuation, caused primarily by the absorption of mi- crowave energy by water vapor or molecular oxygen. Maximum absorption occurs when the frequency coincides with one of the molecular resonances of water or oxygen, and thus atmospheric attenuation has distinct peaks at these frequencies. Figure 14.29 shows theatmospheric attenuation versus frequency. At frequencies below 10 GHz the atmosphere has very little effect on the strength of a signal. At 22.2 and 183.3 GHz, resonance peaks occur due to water vapor resonances, while resonances of molecular oxygen cause peaksat 60 and 120 GHz. Thus there are “windows” in the millimeter wave band near 35, 94, and 135 GHz where radar and communication systems can operate with minimum loss. Precipitation such as rain, snow, or fog will increase the attenuation, especially at higherfrequencies. The effect of atmospheric attenuation can be included in system design when using the Friis transmission equation or the radar equation. In some instances the system frequency may be chosen at a point of maximum at- mospheric attenuation. Remote sensing of the atmosphere (temperature, water vapor, rain rate) is often done with radiometers operating near 20 or 55 GHz to maximize the sensingof atmospheric conditions. Another interesting example is spacecraft-to-spacecraft com- munication at 60 GHz. This millimeter wave frequency band has the advantages of a large bandwidth and small antennas with high gains, and, since the atmosphere is very lossy atthis frequency, the possibilities of interference, jamming, and eavesdropping from Earth are greatly reduced. c14IntroToMicrowaveSystems Pozar September 26, 2011 18:40 14.5 Microwave Propagation 703 10 15 20 25 30 40 50 60 70 80 90 100 150 200 250 300 4000.0010.0020.0040.010.020.040.10.20.4124102040100 Sea level H2OH2O H2O 9150 m altitudeO2O2 Frequency (GHz)Attenuation (dB/km) FIGURE 14.29 Average atmospheric attenuation versus frequency (horizontal polarization). GroundEffects The most obvious effect of the presence of the ground on RF and microwave propagation is reflection from Earth’s surface (land or sea). As shown in Figure 14.30, a radar target (orreceiver antenna) may be illuminated by both a direct wave from the transmitter and a wave reflected from the ground. The reflected wave is generally smaller in amplitude than the direct wave because of the larger distance it travels, the fact that it usually radiates fromthe sidelobe region of the transmit antenna, and the fact that the ground is not a perfect reflector. Nevertheless, the received signal at the target or receiver will be the vector sum of the two wave components and, depending on the relative phases of the two waves, maybe greater or less than the direct wave alone. Because the distances involved are usually very large in terms of the electrical wavelength, even a small variation in the permittivity of the atmosphere can cause fading (long-term fluctuations) or scintillation (short-term fluctuations) in the signal strength. These effects can also be caused by reflections from inhomogeneities in the atmosphere. In communication systems fading can sometimes be reduced by making use of the fact that the fading of two communication channels having different frequencies, polarizations, EarthReflectedDirect FIGURE 14.30 Direct and reflected waves over Earth’s surface. c14IntroToMicrowaveSystems Pozar September 26, 2011 18:40 704 Chapter 14: Introduction to Microwave Systems or physical locations is essentially independent. Thus a communication link can reduce fading effects by combining the outputs of two (or more) such channels; this is called a diversity system. Another ground effect is diffraction, whereby a radio wave scatters energy in the vicin- ity of the line-of-sight boundary at the horizon, thus giving a range slightly beyond the horizon. This effect is usually very small at microwave frequencies. Of course, when ob- stacles such as hills, mountains, or buildings are in the path of propagation, diffractioneffects can be stronger. In a radar system, unwanted reflections often occur from terrain, vegetation, trees, buildings, and the surface of the sea. Such clutter echoes generally degrade or mask the return of a true target, or show up as a false target, in the context of a surveillance or tracking radar. In mapping or remote sensing applications such clutter returns may actuallyconstitute the desired signal. PlasmaEffects Aplasma is a gas consisting of ionized particles. The ionosphere consists of spherical layers of atmosphere with particles that have been ionized by solar radiation, and thus forms a plasma region. A very dense plasma is formed on a spacecraft as it reenters theatmosphere from outer space, due to the high temperatures produced by friction. Plasmas are also produced by lightning, meteor showers, and nuclear explosions. A plasma is characterized by the number of ions per unit volume; depending on this density and the frequency, a wave might be reflected, absorbed, or transmitted by the plasma medium. An effective permittivity can be defined for a uniform plasma region as /epsilon1 e=/epsilon10/parenleftBigg 1−ω2 p ω2/parenrightBigg ,( 14.51) where ωp=/radicalBigg Nq2 m/epsilon10(14.52) is the plasma frequency. In (14.52), qis the charge of the electron, mis the mass of the electron, and Nis the number of ionized particles per unit volume. By studying the solution of Maxwell’s equations for plane wave propagation in such a medium, it can be shown thatwave propagation through a plasma is only possible for ω>ω p. Lower frequency waves will be totally reflected. If a magnetic field is present, the plasma becomes anisotropic, and the analysis is more complicated. Earth’s magnetic field may be strong enough to produce such an anisotropy in some cases. The ionosphere consists of several different layers with varying ion densities; in order of increasing ion density, these layers are referred to as D,E,F1,andF2. The character- istics of these layers depend on seasonal weather and solar cycles, but the average plasma frequency is about 8 MHz. Thus, signals at frequencies less than 8 MHz (e.g., short-waveradio) can reflect off the ionosphere to travel distances well beyond the horizon. Higher frequency signals, however, will pass through the ionosphere. In the case of a spacecraft entering the atmosphere, the high velocity produces a very dense plasma around the vehicle. The electron density is high enough that, from (14.52), the plasma frequency is very high, thus inhibiting communication with the spacecraft until its velocity has decreased. Besides this blackout effect, the plasma layer may also cause a large impedance mismatch between the antenna and its feed line. c14IntroToMicrowaveSystems Pozar September 26, 2011 18:40 14.6 Other Applications and Topics 705 14.6OTHERAPPLICATIONSANDTOPICS MicrowaveHeating To the average consumer the term “microwave” connotes a microwave oven, used in many households for heating food; industrial and medical applications also exist for microwave heating. As shown in Figure 14.31, a microwave oven is a relatively simple system consist-ing of a high-power microwave source, a waveguide feed, and the oven cavity. The source is generally a magnetron tube operating at 2.45 GHz, although 915 MHz is sometimes used when greater penetration is desired. Power output is usually in the range of 500–1500 W.The oven cavity has metallic walls, and is electrically large. To reduce the effect of uneven heating caused by standing waves in the oven, a “mode stirrer,” which is just a metallic fan blade, is used to perturb the field distribution inside the oven. The food is also rotated witha motorized platter. In a conventional oven a gas or charcoal fire, or an electric heating element, generates heat external to the material to be heated. The outside portion of the material is heated by convection, and the inside of the material is warmed by conduction from the outer portion. In microwave heating, by contrast, the inside of the material is heated first. The process through which this occurs primarily involves the conduction losses in food materials hav-ing large loss tangents [8, 9]. An interesting fact is that the loss tangents of many foods decrease with increasing temperature, so that microwave heating is to some extent self- regulating. The result is that microwave cooking generally gives faster and more uniform heating of food as compared with conventional cooking. The efficiency of a microwave oven, when defined as the ratio of power converted to heat (in the food) to the power sup-plied to the oven, is generally less than 50%, but this is usually greater than the cooking efficiency of a conventional oven. The most critical issue in the design of a microwave oven is safety. Since a very high power source is used, leakage must be very small to avoid exposing the user to harm- ful radiation. Thus the magnetron, feed waveguide, and oven cavity must all be carefully shielded. The door of the oven requires particular attention; besides close mechanical tol-erances, the joint around the door usually employs RF-absorbing material and a λ/4 choke flange to reduce power leakage to an acceptable level. Power Transfer Electrical power transmission lines are a very efficient and convenient way to transfer energy from one point to another, as they have relatively low loss and initial costs, and can “Mode stirrer” Oven cavity FoodWaveguide Magnetron Power supply Rotating plate FIGURE 14.31 A microwave oven. c14IntroToMicrowaveSystems Pozar September 26, 2011 18:40 706 Chapter 14: Introduction to Microwave Systems be easily routed. There are applications, however, where it is inconvenient or impossible to use such power lines. In such cases it is conceivable that electrical power can be transmitted without wires by a well-focused microwave beam [10]. One example is a solar satellite power station, where it has been proposed that elec- tricity be generated in space by a large orbiting array of solar cells and transmitted to a receiving station on Earth by a microwave beam. We would thus be provided with a virtu- ally inexhaustible source of electricity. Placing the solar arrays in space has the advantageof power delivery uninterrupted by darkness, clouds, or precipitation, which are problems encountered with Earth-based photovoltaic arrays. To be economically competitive with other sources, a solar power satellite station would have to be very large. One proposal involves a solar array about 5 ×10 km in size feeding a 1 km diameter phased array antenna. The power output on Earth would be on theorder of 5 GW. Such a project is extremely large in terms of cost and complexity. Also of legitimate concern is the operational safety of such a scheme, in terms of both the radiation hazards associated with the system when it is operating as designed, and the risks involvedwith a malfunction of the system. These considerations, as well as the political and philo- sophical ramifications of such a large, centralized power system, have made the future of the solar power satellite station doubtful. Similar in concept, but on a much smaller scale, is the transmission of electrical power from Earth to a vehicle such as a small drone helicopter or airplane. The advantages are that such an aircraft could run indefinitely, and very quietly, at least over a limited area.Battlefield surveillance and weather prediction would be some possible applications. The concept has been demonstrated with several projects involving small pilotless aircraft. On an even smaller scale is the wireless transmission of power to RFID tags, which is feasible primarily because of the very low DC power required for appropriately designed CMOS circuitry. A related idea is the collection of ambient RF power to charge batteries of portable devices. This sounds attractive, and may be possible in principle, but it is probably not feasible in most situations, especially when other power sources are available. BiologicalEffectsandSafety The proven dangers of exposure to RF and microwave radiation are due to thermal effects. The body absorbs RF and microwave energy and converts it to heat; as in the case of a microwave oven, this heating occurs within the body and may not be felt at low levels. Such heating is most dangerous in the brain, the eye, the genitals, and the stomach organs.Excessive radiation can lead to cataracts, sterility, or cancer. It is important to determine a safe radiation level standard so that users of RF and microwave equipment will not be exposed to harmful power levels. At the time of this writing, the most recent IEEE safety standard for human exposure to electromagnetic fields is given by IEEE Standard C95.1-2005. In the RF-microwave fre- quency range of 100 MHz to 100 GHz, exposure limits are specified for the power density(W/m 2)as a function of frequency, as shown in Figure 14.32. This graph shows both the recommended limits for the general population, and for exposure in controlled environ- ments that exist for occupational workers. These limits apply to exposure averaged over either 6 minutes (for occupational workers) or 30 minutes (for the general population). The recommended safe power density limits are generally lower at lower frequencies be-cause fields penetrate the body more deeply at these frequencies. At higher frequencies most of the power absorption occurs near the skin surface, so the safe limits can be higher. At frequencies below 100 MHz electric and magnetic fields interact with the body differ-ently than higher frequency electromagnetic fields, and so separate limits are given for field components at these lower frequencies. c14IntroToMicrowaveSystems Pozar September 26, 2011 18:40 14.6 Other Applications and Topics 707 0.110 1 0.1 1.0 Frequency (GHz)Power Density (W/m2) 10.0 100.0100 Controlled environments General population1000 FIGURE 14.32 IEEE Standard C95.1-2005 recommended power density limits for human expo- sure to RF and microwave electromagnetic fields. For a controlled (occupational)environment the exposure is averaged over a 6 minute period, while for the gen- eral population the exposure is averaged over a 30 minute period. In the United States, the Federal Communications Commission (FCC) sets a separate exposure limit for hand-held wireless devices (cell phones, PDAs, smartphones, etc). These limits are given in terms of the Specific Absorption Rate (SAR), which measures how much power is dissipated as heat in a unit of tissue mass. Specific Absorption Rate is defined as SAR=σ 2ρ/vextendsingle/vextendsingle¯E/vextendsingle/vextendsingle2W/kg,( 14.53) where σis the conductivity of the tissue (S/m), ρis the density of the tissue (kg/m3), and ¯Eis the electric field in the tissue sample. For partial body exposure (typically the head or hand), the FCC limit on SAR is 1.6 W/kg, averaged over 1 g of tissue. All wireless devicessold in the United States must meet this standard. Other countries have standards that are similar in nature and scope. The European Union, for example, requires hand held wireless devices to have SAR exposure of less than 2 W/kg, averaged over 10 g of tissue. A separate standard applies to microwave ovens sold in the United States, requiring that all ovens be tested to ensure that the power level at 5 cm from any point on the ovendoes not exceed 1 mW/cm 2. Most experts feel that the above limits represent safe levels, with a reasonable margin. Some researchers, however, feel that health hazards may occur due to nonthermal effectsof long-term exposure to even low levels of microwave radiation. EXAMPLE 14.8 POWER DENSITY IN THE VICINITY OF A MICROWA VE RADIO LINK An 18 GHz common-carrier microwave communications link uses a tower- mounted antenna with a gain of 36 dB and a transmitter power of 10 W. To evalu-ate the radiation hazard of this system, calculate the power density at a distance of 20 m from the antenna. Do this for a position in the main beam of the antenna, and c14IntroToMicrowaveSystems Pozar September 26, 2011 18:40 708 Chapter 14: Introduction to Microwave Systems for a position in the sidelobe region of the antenna. Assume a worst-case sidelobe level of −10 dB. Solution The numerical gain of the antenna is Gt=1036/10=4000. From (14.23), the power density in the main beam of the antenna at a distance of R=20 m is Savg=PtGt 4πR2=(10)(4000) 4π(20)2=8W / m2. The worst-case power density in the sidelobe region is 10 dB below this value, or 0.8W / m2. Thus, the power density in the main beam at 20 m is below the U.S. standard for radiation hazard for the general population, while the power density in the sidelobe region is well below this limit. These power levels will diminish rapidly with increasing distance due to the 1/ r2dependence. ■ REFERENCES [1] C. A. Balanis, Antenna Theory: Analysis and Design , 3rd edition, John Wiley & Sons, New York, 2005. [2] W. L. Stutzman and G. A. Thiele, Antenna Theory and Design , 2nd edition, John Wiley & Sons, New York, 1998. [3] L. J. Ippolito, R. D. Kaul, and R. G. Wallace, Propagation Effects Handbook for Satellite Systems Design, 3rd edition, NASA, Washington, D.C., 1983. [ 4 ] D .M .P o z a r , Microwave and RF Design of Wireless Systems , John Wiley & Sons, New York, 2001. [5] E. Lutz, M. Werner, and A. Jahn, Satellite Systems for Personal and Broadband Communications , Springer-Verlag, Berlin, 2000. [6] M. I. Skolnik, Introduction to Radar Systems , McGraw-Hill, New York, 1962. [7] F. T. Ulaby, R. K. Moore, and A. K. Fung, Microwave Remote Sensing: Active and Passive, Volume I, Microwave Remote Sensing, Fundamentals and Radiometry. Addison-Wesley, Reading, Mass., 1981. [8] F. E. Gardiol, Introduction to Microwaves , Artech House, Dedham, Mass., 1984. [9] E. C. Okress, Microwave Power Engineering , Academic Press, New York, 1968. [10] W. C. Brown, “The History of Power Transmission by Radio Waves,” IEEE Transactions on Mi- crowave Theory and Techniques, vol. MTT-32, pp. 1230–1242, September 1984. PROBLEMS 14.1 The Iridium satellite communication system was designed with a link margin of 16 dB, and was orig- inally advertised as being capable of providing service to users with hand-held phones in vehicles, buildings, and urban areas. Today, after bankruptcy and restructuring of the company, it is rec-ommended that Iridium phones be used outdoors, with a line of sight to the satellites. Find some estimates of the link margins (due to fading) required for L-band communications into vehicles and buildings. Do you think the Iridium system would have operated reliably in these environments? Ifnot, why was the system designed with a 16 dB link margin? 14.2 An antenna has a radiation pattern function given by F θ(θ,φ) =Asin2θcosφ. Find the main beam position, the 3 dB beamwidths in the principal planes, and the directivity (in dB) for this antenna.What is the polarization of this antenna? c14IntroToMicrowaveSystems Pozar September 26, 2011 18:40 Problems 709 14.3 A monopole antenna on a large ground plane has a far-field pattern function given by Fθ(θ,φ) = Asinθfor 0≤θ≤90◦. The radiated field is zero for 90◦≤θ≤180◦. Find the directivity (in dB) of this antenna. 14.4 A DBS reflector antenna operating at 12.4 GHz has a diameter of 18 inches. If the aperture efficiency is 65%, find the directivity. 14.5 A reflector antenna used for a cellular base station backhaul radio link operates at 38 GHz, with a gain of 39 dB, a radiation efficiency of 90%, and a diameter of 12 inches. (a) Find the aperture efficiencyof this antenna. (b) Find the half-power beamwidth, assuming the beamwidths are identical in the two principal planes. 14.6 A high-gain antenna array operating at 2.4 GHz is pointed toward a region of the sky for which the background can be assumed to be at a uniform temperature of 5 K. A noise temperature of 105 K ismeasured for the antenna temperature. If the physical temperature of the antenna is 290 K, what is its radiation efficiency? 14.7 Derive equation (14.20) by treating the antenna and lossy line as a cascade of two networks whose equivalent noise temperatures are given by (14.18) and (10.15). 14.8 Consider the replacement of a DBS dish antenna with a microstrip array antenna. A microstrip array offers an aesthetically pleasing flat profile, but suffers from relatively high dissipative loss in its feednetwork, which leads to a high noise temperature. If the background noise temperature is T B=50 K, with an antenna gain of 33.5 dB and a receiver LNB noise figure of 1.1 dB, find the overall G/Tfor the microstrip array antenna and the LNB if the array has a total loss of 2.5 dB. Assume the antennais at a physical temperature of 290 K. 14.9 At a distance of 300 m from an antenna operating at 5.8 GHz, the radiated power density in the main beam is measured to be 7.5 ×10 −3W/m2. If the input power to the antenna is known to be 85 W, find the gain of the antenna. 14.10 A cellular base station is to be connected to its Mobile Telephone Switching Office located 5 km away. Two possibilities are to be evaluated: (1) a radio link operating at 28 GHz, with Gt=Gr= 25 dB, and (2) a wired link using coaxial line having an attenuation of 0.05 dB/m, with four 30 dB repeater amplifiers along the line. If the minimum required received power level for both cases is the same, which option will require the smallest transmit power? 14.11 A GSM cellular telephone system operates at a downlink frequency of 935–960 MHz, with a channel bandwidth of 200 kHz, and a base station that transmits with an EIRP of 20 W. The mobile receiver has an antenna with a gain of 0 dBi and a noise temperature of 450 K, and the receiver has a noisefigure of 8 dB. Find the maximum operating range if the required minimum SNR at the output of the receiver is 10 dB, and a link margin of 30 dB is required to account for propagation into vehicles, buildings, and urban areas. 14.12 Consider the GPS receiver system shown below. The guaranteed minimum L1 (1575 MHz) carrier power received by an antenna on Earth having a gain of 0 dBi is S i=−160 dBW. A GPS receiver is usually specified as requiring a minimum carrier-to-noise ratio, relative to a 1 Hz bandwidth, of C/N (Hz). If the receiver antenna actually has a gain GAand a noise temperature TA, derive an expression for the maximum allowable amplifier noise figure F, assuming an amplifier gain Gand a connecting line loss L. Evaluate this expression for C/N=32 dB-Hz, GA=5d B , TA=300 K, G=10 dB, andL=25 dB. LNA 14.13 A key premise in many science fiction stories is the idea that radio and TV signals from Earth can travel through space and be received by listeners in another star system. Show that this is a c14IntroToMicrowaveSystems Pozar September 26, 2011 18:40 710 Chapter 14: Introduction to Microwave Systems fallacy by calculating the maximum distance from Earth where a signal could be received with a SNR of 0 dB. Specifically, assume TV channel 4, broadcasting at 67 MHz, with a 4 MHz band- width, a transmitter power of 1000 W, transmit and receive antenna gains of 4 dB, a cosmic back- ground noise temperature of 4 K, and a perfectly noiseless receiver. How much would this dis-tance decrease if an SNR of 30 dB is required at the receiver? (30 dB is a typical value for good reception of an analog video signal.) Relate these distances to the nearest planet in our solar system. 14.14 The Mariner 10 spacecraft used to explore the planet Mercury in 1974 used BPSK with P b=0.05 (Eb/n0=1.4 dB) to transmit image data back to Earth (a distance of about 1 .6×108km). The spacecraft transmitter operated at 2.295 GHz, with an antenna gain of 27.6 dB and a carrier powerof 16.8 W. The ground station had an antenna with a gain of 61.3 dB and an overall system noise temperature of 13.5 K. Find the maximum possible data rate. 14.15 Derive the radar equation for the bistatic case where the transmit and receive antennas have gains of G tandGr, and are at distances RtandRrfrom the target, respectively. 14.16 A pulse radar has a pulse repetition frequency fr=1/Tr. Determine the maximum unambiguous range of the radar. (Range ambiguity occurs when the round-trip time of a return pulse is greater than the pulse repetition time, so it becomes unclear as to whether a given return pulse belongs to the last transmitted pulse or some earlier transmitted pulse.) 14.17 A Doppler radar operating at 12 GHz is intended to detect target velocities ranging from 1 to 20 m/sec. What is the required passband of the Doppler filter? 14.18 A pulse radar operates at 2 GHz and has a per-pulse power of 1 kW. If it is to be used to detect a target with σ=20 m2at a range of 10 km, what should be the minimum isolation between the transmitter and receiver so that the leakage signal from the transmitter is at least 10 dB below thereceived signal? Assume an antenna gain of 30 dB. 14.19 An antenna having a gain Gis shorted at its terminals. What is the minimum monostatic radar cross section in the direction of the main beam? 14.20 Consider the radiometer antenna shown below, where the antenna is at a physical temperature T p and has a radiation efficiency ηrad, and an impedance mismatch /Gamma1at its terminals. If TSis the ap- parent temperature seen by the radiometer, show that /Delta1TS//Delta1Ttrueis equal to the product of radia- tion efficiency and mismatch loss, by applying two background temperatures, TB=TpandTB= T2/negationslash=Tp. TB Tp, /H9257radRadiometer TS/H9003 14.21 The atmosphere does not have a definite thickness since it gradually thins with altitude, with a conse- quent decrease in attenuation. However, if we use a simplified “orange peel” model and assume that the atmosphere can be approximated by a uniform layer of fixed thickness, we can estimate the back-ground noise temperature seen through the atmosphere. Thus, let the thickness of the atmosphere be 4000 m, and find the maximum distance /lscriptto the edge of the atmosphere along the horizon, as shown in the figure below (the radius of Earth is 6400 km). Now assume an average atmosphericattenuation of 0.005 dB/km, with a background noise temperature beyond the atmosphere of 4 K, and find the noise temperature seen on Earth by treating the cascade of the background noise with the attenuation of the atmosphere. Do this for an ideal antenna pointing toward the zenith, and towardthe horizon. c14IntroToMicrowaveSystems Pozar September 26, 2011 18:40 Problems 711 14.22 A 28 GHz radio link uses a tower-mounted reflector antenna with a gain of 32 dB and a transmit- ter power of 5 W. (a) Find the minimum distance within the main beam of the antenna for which the U.S-recommended safe power density limit of 10 mW/cm2is not exceeded. (b) How does this distance change for a position within the sidelobe region of the antenna if we assume a worst-case sidelobe level of 10 dB below the main beam? (c) Are these distances in the far-field region of the antenna? (Assume a circular reflector, with an aperture efficiency of 60%.) 14.23 On a clear day, with the sun directly overhead, the received power density from sunlight is about 1300 W/m2. If we make the simplifying assumption that this power is transmitted via a single- frequency plane wave, find the resulting amplitude of the incident electric and magnetic fields. bapp01 Pozar September 27, 2011 17:30 Appendices Appendix A: Prefixes Appendix B: Vector Analysis Appendix C: Bessel Functions Appendix D: Other Mathematical Results Appendix E: Physical ConstantsAppendix F: Conductivities for Some Materials Appendix G: Dielectric Constants and Loss Tangents for Some Materials Appendix H: Properties of Some Microwave Ferrite MaterialsAppendix I: Standard Rectangular Waveguide Data Appendix J: Standard Coaxial Cable Data 712 bapp01 Pozar September 27, 2011 17:30 Appendix B Vector Analysis 713 APPENDIXAPREFIXES Multiplying Factor Prefix Symbol 1012tera T 109giga G 106mega M 103kilo k 102hecto h 101deka da 10−1deci d 10−2centi c 10−3milli m 10−6micro µ 10−9nano n 10−12pico p 10−15femto f APPENDIXBVECTORANALYSIS CoordinateTransformations Rectangular to cylindrical: ˆx ˆy ˆz ˆρ cosφ sinφ 0 ˆφ −sinφ cosφ 0 ˆz 001 Rectangular to spherical: ˆx ˆy ˆz ˆr sinθcosφ sinθsinφ cosθ ˆθ cosθcosφ cosθsinφ −sinθ ˆφ −sinφ cosφ 0 Cylindrical to spherical: ˆρ ˆφ ˆz ˆr sinθ 0c o s θ ˆθ cosθ 0 −sinθ ˆφ 01 0 bapp01 Pozar September 27, 2011 17:30 714 Appendices These tables can be used to transform unit vectors as well as vector components; e.g., ˆρ=ˆxcosφ+ˆysinφ Aρ=Axcosφ+Aysinφ VectorDifferentialOperators Rectangular coordinates: ∇f=ˆx∂f ∂x+ˆy∂f ∂y+ˆz∂f ∂z ∇·¯A=∂Ax ∂x+∂Ay ∂y+∂Az ∂z ∇ׯA=ˆx/parenleftbigg∂Az ∂y−∂Ay ∂z/parenrightbigg +ˆy/parenleftbigg∂Ax ∂z−∂Az ∂x/parenrightbigg +ˆz/parenleftbigg∂Ay ∂x−∂Ax ∂y/parenrightbigg ∇2f=∂2f ∂x2+∂2f ∂y2+∂2f ∂z2 ∇2¯A=ˆx∇2Ax+ˆy∇2Ay+ˆz∇2Az Cylindrical coordinates: ∇f=ˆρ∂f ∂ρ+ˆφ1 ρ∂f ∂φ+ˆz∂f ∂z ∇·¯A=1 ρ∂ ∂ρ(ρAρ)+1 ρ∂Aφ ∂φ+∂Az ∂z ∇ׯA=ˆρ/parenleftbigg1 ρ∂Az ∂φ−∂Aφ ∂z/parenrightbigg +ˆφ/parenleftbigg∂Aρ ∂z−∂Az ∂ρ/parenrightbigg +ˆz1 ρ/bracketleftbigg∂(ρAφ) ∂ρ−∂Aρ ∂φ/bracketrightbigg ∇2f=1 ρ∂ ∂ρ/parenleftbigg ρ∂f ∂ρ/parenrightbigg +1 ρ2∂2f ∂φ2+∂2f ∂z2 ∇2¯A=∇(∇·¯A)−∇×∇× ¯A Spherical coordinates: ∇f=ˆr∂f ∂r+ˆθ1 r∂f ∂θ+ˆφ rsinθ∂f ∂φ ∇·¯A=1 r2∂ ∂r(r2Ar)+1 rsinθ∂ ∂θ(sinθAθ)+1 rsinθ∂Aφ ∂φ ∇ׯA=ˆr rsinθ/bracketleftbigg∂ ∂θ(Aφsinθ)−∂Aθ ∂φ/bracketrightbigg +ˆθ r/bracketleftbigg1 sinθ∂Ar ∂φ−∂ ∂r(rAφ)/bracketrightbigg +ˆφ r/bracketleftbigg∂ ∂r(rAθ)−∂Ar ∂θ/bracketrightbigg ∇2f=1 r2∂ ∂r/parenleftbigg r2∂f ∂r/parenrightbigg +1 r2sinθ∂ ∂θ/parenleftbigg sinθ∂f ∂θ/parenrightbigg +1 r2sin2θ∂2f ∂φ2 ∇2¯A=∇ ∇· ¯A−∇×∇× ¯A bapp01 Pozar September 27, 2011 17:30 Appendix C Bessel Functions 715 Vector identities: ¯A·¯B=|A||B|cosθ, where θis the angle between ¯Aand¯B (B.1) |¯AׯB|=| A||B|sinθ, where θis the angle between ¯Aand¯B.(B.2) ¯A·¯BׯC=¯AׯB·¯C=¯CׯA·¯B (B.3) ¯AׯB=−¯BׯA (B.4) ¯A×(¯BׯC)=(¯A·¯C)¯B−(¯A·¯B)¯C (B.5) ∇(fg)=g∇f+f∇g (B.6) ∇·(f¯A)=¯A·∇f+f∇·¯A (B.7) ∇·(¯AׯB)=(∇× ¯A)·¯B−(∇× ¯B)·¯A (B.8) ∇×(f¯A)=(∇f)ׯA+f∇ׯA (B.9) ∇×(¯AׯB)=¯A∇·¯B−¯B∇·¯A+(¯B·∇)¯A−(¯A·∇)¯B (B.10) ∇·(¯A·¯B)=(¯A·∇)¯B+(¯B·∇)¯A+A×(∇× ¯B)+¯B×(∇× ¯A)(B.11) ∇·∇× ¯A=0 (B.12) ∇×(∇f)=0 (B.13) ∇×∇× ¯A=∇ ∇· ¯A−∇2¯A (B.14) Note: the term ∇2¯Ahas meaning only for rectangular components of ¯A. /integraldisplay V∇·¯Adv=/contintegraldisplay S¯A·d¯s (divergence theorem) ((B.15)) /integraldisplay S(∇× ¯A)·d¯s=/contintegraldisplay C¯A·d¯/lscript (Stokes’ theorem) ((B.16)) APPENDIXCBESSELFUNCTIONS Bessel functions are solutions to the differential equation, 1 ρd dρ/parenleftbigg ρdf dρ/parenrightbigg +/parenleftBigg k2−n2 ρ2/parenrightBigg f=0 (C.1) where k2is real and nis an integer. The two independent solutions to this equation are called ordinary Bessel functions of the first and second kind, written as Jn(kρ)andYn(kρ), and so the general solution to (C.1) is f(ρ)=AJn(kρ)+BYn(kρ) (C.2) where AandBare arbitrary constants to be determined from boundary conditions. These functions can be written in series form as Jn(x)=∞/summationdisplay m=0(−1)m(x/2)n+2m m!(n+m)!(C.3) Yn(x)=2 π/parenleftBig γ+lnx 2/parenrightBig Jn(x)−1 πn−1/summationdisplay m=0(n−m−1)! m!/parenleftbigg2 x/parenrightbiggn−2m −1 π∞/summationdisplay m=0(−1)m(x/2)n+2m m!(n+m)!/parenleftbigg 1+1 2+1 3+···+1 m+1+1 2+···+1 n+m/parenrightbigg (C.4) bapp01 Pozar September 27, 2011 17:30 716 Appendices 1.0 0.5 0 –0.5J0 J1 J2 24 6 8 1 0 x 0.5 0 –0.5 –1.0Y0Y1Y2 24 6 810x FIGURE C.1 Bessel functions of the first and second kind. where γ=0.5772 ...is Euler’s constant, and x=kρ. Note that Ynbecomes infinite at x=0, due to the ln term. From these series expressions, small argument formulas can be obtained as Jn(x)∼1 n!/parenleftBigx 2/parenrightBign (C.5) Y0(x)∼2 πlnx (C.6) Yn(x)∼−1 π(n−1)!/parenleftBigx 2/parenrightBign , n>0( C.7) Large argument formulas can be derived as Jn(x)∼/radicalbigg 2 πxcos/parenleftBig x−π 4−nπ 2/parenrightBig (C.8) Yn(x)∼/radicalbigg 2 πxsin/parenleftBig x−π 4−nπ 2/parenrightBig (C.9) Figure C.1 shows graphs of a few of the lowest order Bessel functions of each type. Recurrence formulas relate Bessel functions of different orders: Zn+1(x)=2n xZn(x)−Zn−1(x) (C.10) Z/prime n(x)=−n xZn(x)+Zn−1(x) (C.11) bapp01 Pozar September 27, 2011 17:30 Appendix C Bessel Functions 717 Z/prime n(x)=n xZn(x)−Zn+1(x) (C.12) Z/prime n(x)=1 2[Zn−1(x)−Zn+1(x)] (C.13) where Zn=JnorYn. The following integral relations involving Bessel functions are useful: /integraldisplayx 0Z2 m(kx)xd x=x2 2/bracketleftBigg Z/prime2 n(kx)+/parenleftBigg 1−n2 k2x2/parenrightBigg Z2 n(kx)/bracketrightBigg (C.14) /integraldisplayx 0Zn(kx)Zn(/lscriptx)xd x=x k2−/lscript2[kZ n(/lscriptx)Zn+1(kx)−/lscriptZn(kx)Zn+1(/lscriptx)](C.15) /integraldisplaypnm 0/bracketleftbigg J/prime2 n(x)+n2 x2J2 n(x)/bracketrightbigg xd x=p2 nm 2J/prime2 n(pnm) (C.16) /integraldisplayp/prime nm 0/bracketleftbigg J/prime2 n(x)+n2 x2J2 n(x)/bracketrightbigg xd x=(p/prime nm)2 2/parenleftbigg 1−n2 (p/primenm)2/parenrightbigg J2 n(p/prime nm)(C.17) where Jn(pnm)=0, and J/prime n(p/prime nm)=0. The zeros of Jn(x)andJ/prime n(x)are on the following two pages. Zeros of Bessel Functions of First Kind: Jn(x)=0f o r0< x<12 n 1234 0 2.4048 5.5201 8.6537 11.7915 1 3.8317 7.0156 10.17352 5.1356 8.4172 11.61983 6.3802 9.76104 7.5883 11.06475 8.77156 9.93617 11.0864 Extrema of Bessel Functions of First Kind: dJn(x)/dx=0f o r 0<x<12 n 1234 0 3.8317 7.0156 10.1735 13.3237 1 1.8412 5.3314 8.5363 11.70602 3.0542 6.7061 9.96953 4.2012 8.0152 11.34594 5.3175 9.28245 6.4156 10.51996 7.5013 11.73497 8.57788 9.64749 10.7114 10 11.7709 bapp01 Pozar September 27, 2011 17:30 718 Appendices APPENDIXDOTHERMATHEMATICALRESULTS UsefulIntegrals /integraldisplaya 0cos2nπx adx=/integraldisplaya 0sin2nπx adx=a 2, forn≥1( D . 1 ) /integraldisplaya 0cosmπx acosnπx adx=/integraldisplaya 0sinmπx asinnπx adx=0, form/negationslash=n(D.2) /integraldisplaya 0cosmπx asinnπx adx=0( D.3) /integraldisplayπ 0sin3θdθ=4 3(D.4 ) TaylorSeries f(x)=f(x0)+(x−x0)df dx/vextendsingle/vextendsingle/vextendsingle/vextendsingle x=x0+(x−x0)2 2!d2f dx2/vextendsingle/vextendsingle/vextendsingle/vextendsingle x=x0+··· (D.5) ex=1+x+x2 2!+x3 3!+··· (D.6) 1 1−x=1+x+x2+x3+···, for|x|<1( D . 7 ) √ 1+x=1+x 2−x2 8+···, for|x|<1( D . 8 ) lnx=2/parenleftbiggx−1 x+1/parenrightbigg +2 3/parenleftbiggx−1 x+1/parenrightbigg3 +···, forx>0( D.9) sinx=x−x3 3!+x5 5!+··· (D.10) cosx=1−x2 2!+x4 4!+··· (D.11) APPENDIXEPHYSICALCONSTANTS rPermittivity of free-space =/epsilon10=8.854 ×10−12F/mrPermeability of free-space =µ0=4π×10−7H/mrImpedance of free-space =η0=376.7 /Omega1rVelocity of light in free-space =c=2.998×108m/srCharge of electron =q=1.602×10−19CrMass of electron =m=9.107 ×10−31kgrBoltzmann’s constant =k=1.380×10−23J/◦KrPlanck’s constant =h=6.626 ×10−34J-secrGyromagnetic ratio =γ=1.759×1011C/Kg (for g=2) bapp01 Pozar September 27, 2011 17:30 Appendix G Dielectric Constants and Loss Tangents for Some Materials 719 APPENDIXFCONDUCTIVITIESFORSOMEMATERIALS Material Conductivity S/m (20◦C) Material Conductivity S/m (20◦C) Aluminum 3.816 ×107Nichrome 1.0×106 Brass 2.564 ×107Nickel 1.449 ×107 Bronze 1.00×107Platinum 9.52×106 Chromium 3.846 ×107Sea water 3–5 Copper 5.813 ×107Silicon 4.4×10−4 Distilled water 2×10−4Silver 6.173 ×107 Germanium 2.2×106Steel (silicon) 2×106 Gold 4.098 ×107Steel (stainless) 1.1×106 Graphite 7.0×104Solder 7.0×106 Iron 1.03×107Tungsten 1.825 ×107 Mercury 1.04×106Zinc 1.67×107 Lead 4.56×106 APPENDIXGDIELECTRICCONSTANTSANDLOSSTANGENTSFOR SOMEMATERIALS Material Frequency /epsilon1r tanδ(25◦C) Alumina (99.5%) 10 GHz 9.5–10. 0.0003 Barium tetratitanate 6 GHz 37 ±5% 0.0005 Beeswax 10 GHz 2.35 0.005 Beryllia 10 GHz 6.4 0.0003 Ceramic (A-35) 3 GHz 5.60 0.0041 Fused quartz 10 GHz 3.78 0.0001 Gallium arsenide 10 GHz 13.0 0.006 Glass (pyrex) 3 GHz 4.82 0.0054 Glazed ceramic 10 GHz 7.2 0.008 Lucite 10 GHz 2.56 0.005 Nylon (610) 3 GHz 2.84 0.012 Parafin 10 GHz 2.24 0.0002 Plexiglass 3 GHz 2.60 0.0057 Polyethylene 10 GHz 2.25 0.0004 Polystyrene 10 GHz 2.54 0.00033 Porcelain (dry process) 100 MHz 5.04 0.0078 Rexolite (1422) 3 GHz 2.54 0.00048 Silicon 10 GHz 11.9 0.004 Styrofoam (103.7) 3 GHz 1.03 0.0001 Teflon 10 GHz 2.08 0.0004 Titania (D-100) 6 GHz 96 ±5% 0.001 Vaseline 10 GHz 2.16 0.001 Water (distilled) 3 GHz 76.7 0.157 bapp01 Pozar September 27, 2011 17:30 720 Appendices APPENDIXHPROPERTIESOFSOMEMICROWAVEFERRITEMATERIALS Trans-Tech 4π Ms /Delta1HT c 4πMr Material Number G Oe /epsilon1r tanδ◦CG Magnesium ferrite TT1-105 1750 225 12.2 0.00025 225 1220 Magnesium ferrite TT1-390 2150 540 12.7 0.00025 320 1288Magnesium ferrite TT1-3000 3000 190 12.9 0.0005 240 2000Nickel ferrite TT2-101 3000 350 12.8 0.0025 585 1853Nickel ferrite TT2-113 500 150 9.0 0.0008 120 140Nickel ferrite TT2-125 2100 460 12.6 0.001 560 1426Lithium ferrite TT73-1700 1700 <400 16.1 0.0025 460 1139 Lithium ferrite TT73-2200 2200 <450 15.8 0.0025 520 1474 Yttrium garnet G-113 1780 45 15.0 0.0002 280 1277Aluminum garnet G-610 680 40 14.5 0.0002 185 515 APPENDIXISTANDARDRECTANGULARWAVEGUIDEDATA Recommended TE 10Cutoff EIA Inside Outside Frequency Frequency Designation Dimensions Dimensions Band∗Range (GHz) (GHz) WR-XX [Inches (cm)] [Inches (cm)] L 1.12–1.70 0.908 WR-650 6.500 ×3.250 6.660 ×3.410 (16.51 ×8.255) (16.916 ×8.661) R 1.70–2.60 1.372 WR-430 4.300 ×2.150 4.460 ×2.310 (10.922 ×5.461) (11.328 ×5.867) S 2.60–3.95 2.078 WR-284 2.840 ×1.340 3.000 ×1.500 (7.214 ×3.404) (7.620 ×3.810) H (G) 3.95–5.85 3.152 WR-187 1.872 ×0.872 2.000 ×1.000 (4.755 ×2.215) (5.080 ×2.540) C (J) 5.85–8.20 4.301 WR-137 1.372 ×0.622 1.500 ×0.750 (3.485 ×1.580) (3.810 ×1.905) W (H) 7.05–10.0 5.259 WR-112 1.122 ×0.497 1.250 ×0.625 (2.850 ×1.262) (3.175 ×1.587) X 8.20–12.4 6.557 WR-90 0.900 ×0.400 1.000 ×0.500 (2.286 ×1.016) (2.540 ×1.270) Ku (P) 12.4–18.0 9.486 WR-62 0.622 ×0.311 0.702 ×0.391 (1.580 ×0.790) (1.783 ×0.993) K 18.0–26.5 14.047 WR-42 0.420 ×0.170 0.500 ×0.250 (1.07×0.43) (1.27 ×0.635) Ka (R) 26.5–40.0 21.081 WR-28 0.280 ×0.140 0.360 ×0.220 (0.711 ×0.356) (0.914 ×0.559) Q 33.0–50.5 26.342 WR-22 0.224 ×0.112 0.304 ×0.192 (0.57×0.28) (0.772 ×0.488) U 40.0–60.0 31.357 WR-19 0.188 ×0.094 0.268 ×0.174 (0.48×0.24) (0.681 ×0.442) V 50.0–75.0 39.863 WR-15 0.148 ×0.074 0.228 ×0.154 (0.38×0.19) (0.579 ×0.391) E 60.0–90.0 48.350 WR-12 0.122 ×0.061 0.202 ×0.141 (0.31×0.015) (0.513 ×0.356) W 75.0–110.0 59.010 WR-10 0.100 ×0.050 0.180 ×0.130 (0.254 ×0.127) (0.458 ×0.330) F 90.0–140.0 73.840 WR-8 0.080 ×0.040 0.160 ×0.120 (0.203 ×0.102) (0.406 ×0.305) D 110.0–170.0 90.854 WR-6 0.065 ×0.0325 0.145 ×0.1125 (0.170 ×0.083) (0.368 ×0.2858) G 140.0–220.0 115.750 WR-5 0.051 ×0.0255 0.131 ×0.1055 (0.130 ×0.0648) (0.333 ×.2680) ∗Letters in parentheses denote alternative designations. bapp01 Pozar September 27, 2011 17:30 Appendix J Standard Coaxial Cable Data 721APPENDIXJSTANDARDCOAXIALCABLEDATA RG/U Impedance Inner cond. Dielectric Dielectric Cable Overall Capacitance Max. Oper. Loss at 1 GHz Type (/Omega1) Diam. (in.) Material Diam. (in.) Type Diam. (in.) (pF/ft) V oltage (dB/100 ft) RG-8A/U 52 0.0855 P 0.285 braided 0.405 29.5 5000 9.0 RG-9B/U 50 0.0855 P 0.280 braided 0.420 30.8 5000 9.0 RG-55B/U 54 0.0320 P 0.116 braided 0.200 28.5 1900 16.5 RG-58B/U 54 0.0320 P 0.116 braided 0.195 28.5 1900 17.5 RG-59B/U 75 0.0230 P 0.146 braided 0.242 20.6 2300 11.5 RG-141A/U 50 0.0390 T 0.116 braided 0.190 29.4 1900 13.0 RG-142A/U 50 0.0390 T 0.116 braided 0.195 29.4 1900 13.0 RG-174/U 50 0.0189 P 0.060 braided 0.100 30.8 1500 31.0 RG-178B/U 50 0.0120 T 0.034 braided 0.072 29.4 1000 45.0 RG-179B/U 75 0.0120 T 0.063 braided 0.100 19.5 1200 25.0 RG-180B/U 95 0.0120 T 0.102 braided 0.140 15.4 1500 16.5 RG-187/U 75 0.0120 T 0.060 braided 0.105 19.5 1200 25.0 RG-188/U 50 0.0201 T 0.060 braided 0.105 29.4 1200 30.0 RG-195/U 95 0.0120 T 0.102 braided 0.145 15.4 1500 16.5 RG-213/U 50 0.0888 P 0.285 braided 0.405 30.8 5000 9.0 RG-214/U 50 0.0888 P 0.285 braided 0.425 30.8 5000 9.0 RG-223/U 50 0.0350 P 0.116 braided 0.211 30.8 1900 16.5 RG-316/U 50 0.0201 T 0.060 braided 0.102 29.4 1200 30.0 RG-401/U 50 0.0645 T 0.215 semi-rigid 0.250 29.3 3000 — RG-402/U 50 0.0360 T 0.119 semi-rigid 0.141 29.3 2500 13.0 RG-405/U 50 0.0201 T 0.066 semi-rigid 0.0865 29.4 1500 — bapp01 Pozar September 27, 2011 17:30 Answers to Selected Problems 1.2(a)η=236/Omega1,(b)vp=1.88×108m/sec, (c)λ=0.0784 m, (d)/Delta1φ=229.5◦ 1.8(b)t/similarequal0.017 mm 1.9(a)Si=46.0W / m2,Sr=0.595 W/m2,(b)Sin=45.6W / m2 2.1(a)f=600 MHz, (b)vp=2.08×108m/sec, (c)λ=0.346 m, (d)εr=2.08, (e)I(z)=1.8e−jβz,(f)v(t,z)=0.135 cos (ωt−βz) 2.3α=0.38 dB/m 2.8Zin=203.−j5.2/Omega1 2.9Zin=19.0−j20.6/Omega1,/Gamma1L=0.62/negationslash83◦ 2.11/lscript=2.147 cm, /lscript=3.324 cm 2.12 Z0=66.7/Omega1or 150.0 /Omega1 2.16 PL=0.681 W 2.18 Pinc=0.250 W, Pref=0.010 W, Ptrans=0.240 W 2.20 (d)Zin=24.5+j20.3/Omega1,(e)/lscriptmin=0.325λ, (f)/lscriptmax=0.075λ 2.23 ZL=99−j46/Omega1 2.29 Ps=0.600 W, Ploss=0.0631 W, PL=0.1706 W 3.5 loss =0.45dB, /Delta1φ=2331◦ 3.6/lscript/similarequal10.3c m 3.9 fc=5.06 GHz 3.13 fc(TE11)=17.94 GHz, fc(TE01)=37.35 GHz 3.15 kca=3.12 3.19 W=0.217 mm, λg=4.045 cm 3.20 W=0.457 mm, λg=4.525 cm 3.21/lscript=2.0754 cm, Zin=0.27−j12.82 /Omega1 3.27vp=2.37×108m/sec, vg=1.83×108m/sec 4.4V+ 1=10/negationslash90◦,V− 1=0,Z(2) in=50/negationslash90◦ 4.14 (d)IL=10.5 dB, delay =45◦,(e)/Gamma1=0.018 /negationslash90◦ 4.18 IL =8.0 dB, delay =90◦ 4.20 PL=1.0W 4.24 VL=1/negationslash−90◦ 4.30/Delta1=0.082 cm 5.1(a)C=0.0568 pF, L=9.44 nH or L=7.10 nH, C=0.298 pF 5.3d=0.2276λ, /lscript=0.3776λ ord=0.4059λ, /lscript=0.1224λ 5.6d=0.174λ, /lscript=0.353λ ord=0.481λ, /lscript=0.147λ 722 bapp01 Pozar September 27, 2011 17:30 Answers to Selected Problems 723 5.9/lscript1=0.086λ, /lscript2=0.198λ or/lscript1=0.375λ, /lscript2=0.375λ 5.14 error =4% 5.17 Z1=1.1067 Z0,Z2=1.3554 Z0 5.21 Z1=1.095 Z0,Z2=1.363 Z0 5.24 RL <6.4 dB 6.1 f0=800 MHz, Q0=100, QL=50 6.5Q0=138 6.9 f101=9.965 GHz, Q101=6349 6.14 a=2.107 cm, d=2.479 cm, Q0=1692 6.18 f0=7.11 GHz 6.21 (c)f0=93.8 GHz, Qc=92,500 7.3 RL =20 dB, C=15 dB, D=30 dB, L =0.5 dB 7.8 change =1.2 dB 7.13 s=5.28 mm, r0=3.77 mm 7.19 s=0.20 mm, w=0.6m m 7.22 s=1.15 mm, w=1.92 mm, /lscript=6.32 mm 7.32 V− 1=V− 3=V− 4=0,V− 2=V− 5=− j0.707 8.6R=2.66, C=0.685, L=1.822 8.7N=5 8.8L1=L5=1.143 nH, C2=C4=0.928 pF, L3=0.877 nH 8.10 attenuation =11 dB 8.16β/lscript1=β/lscript5=29.3◦,β/lscript2=β/lscript4=29.4◦,β/lscript3=43.7◦ 8.18 attenuation =30 dB 8.19 bandwidth about 1.9:1 8.23 N=3 9.1 (b) µ=6.55µ 0,κ=4.95µ 0 9.4Ha=500 Oe 9.6L=1.403 cm 9.8 229 Oe <H0<950 Oe 9.12 (a)H0=2204 Oe, (b)H0=2857 Oe 9.15 L=23.5m m 9.17 L=44.5c m 9.18 L=9.2c m 10.1 F=7.0d B 10.4 Fcas=4.3d B 10.7 (a)F=6d B ,(b) F=1.76 dB, (c)F=3d B 10.14 ratio =6d B 10.15 OIP 3=20.8 dBm (coherent) 10.17 LDR =74.5d B 10.18 LDR =86.7 dB, SFDR =57.8d B 11.2 ON: IL =0.42 dB, OFF: IL =11.4 dB 11.3 ON: IL =0.044 dB, OFF: IL =18.6 dB 11.7 Ri=12.2/Omega1,Cgs=0.84 pF, Rds=213/Omega1,Cds=0.51 pF, gm=54 mS 12.1 (b)GA=0.5,GT=0.444, G=0.457 12.4 CL=4.00/negationslash96◦,RL=3.60, K=0.275 12.6 A and C are unconditionally stable12.9 G T=10.5d B bapp01 Pozar September 27, 2011 17:30 724 Answers to Selected Problems 12.13 −2.9 dB <GT−GTU<4.3 dB 12.15 GT=19.4d B 12.21 Nopt=8.4 13.3 Qmin=14 13.8 L=2.5 nH results in µ=−0.931 13.9 (a)L=−181 dBc/Hz, (b)L=−153 dBc/Hz 13.12 L=−121 dBc/Hz 13.17 fIM=1974 MHz or 1626 MHz 14.2 D=5.7d B 14.4 D=33.6d B 14.6ηrad=65% 14.8 G/T=9.7d B / K 14.11 R=15.2k m 14.13 R=1.9×109m (for SNR =0d B ) 14.17 80–1600 Hz 14.23 |E|=990 V/m bindex Pozar September 29, 2011 19:43 Index A ABCD parameters, 188–191 table for basic circuits, 190table for conversions, 192 Admittance inverter, 421–422 Admittance matrix, 174–178 table for conversions, 192 AM modulation, 528 Ampere’s law, 8 Amplifier design, 571–601 balanced, 586–588differential, 593–596 distributed, 588–593 low-noise, 580–585maximum gain, 571–575 maximum stable gain, 572 power, 596–601specified gain, 575–579 stability, 564–570 Amplitude shift keying (ASK), 681–684Anisotropic media, 11–12 Antenna aperture efficiency, 665directivity, 663 effective aperture area, 665–666 gain, 664–665G/T, 671 noise temperature, 669–671 pattern, 662–664radiation efficiency, 664 types, 659–660 Aperture efficiency, 665Aperture coupling, 215–221, 302–305 Attenuation atmospheric, 702–703transmission line, 78–85 Attenuation constant for circular waveguide, 125, 126–128coaxial line, 80 dielectric loss, 101–102 microstrip line, 149–150parallel plate waveguide, 107, 108–109 plane wave in lossy dielectric, 17–18 rectangular waveguide, 115, 116stripline, 143–144 Attenuator, 179–180 Available power gain, 559–562 B Background noise temperature, 666–668 Balanced amplifiers, 586–588 Balun, 594Bandpass filters coupled line, 426–436 coupled resonator, 437–447lumped element, 411–415 Bandstop filters coupled resonator, 437–441lumped element, 411–414 BARITT diode, 539 Bessel functions, 715–717 zeroes of, 123, 126, 717 Bethe hole coupler, 334–338 Binary phase shift keying (BPSK), 682–683Binomial coefficients, 253 Binomial filter response, 400, 402–404 Binomial matching transformer, 252–256Biological effects, 706–708 Bit error rate, 681–684 Bipolar junction transistor (BJT), 540–543Black body, 697 Bloch impedance, 384 Bode-Fano criterion, 267–269Boltzmann’s constant, 498 Bounce diagram, 87–89 Boundary conditions, 12–15 725 bindex Pozar September 29, 2011 19:43 726 Index Brewster angle, 37 Brightness temperature, 666–668 C Cavity resonators cylindrical cavity, 288–293dielectric resonator, 293–297 rectangular cavity, 284–288 Cellular telephone systems, 2–3, 684–686Characteristic impedance, 50–51, 171 coaxial line, 56 microstrip line, 148parallel plate line, 100 stripline, 142 Chebyshev filter response, 400–401, 404–405 matching transformers, 256–261 polynomials, 257–258 Chip capacitor, resistor, 233–234 Choke bias, 531, 542, 545 flange, 121 Circular cavity (see Cavity resonators)Circular polarization, 24–25, 458–460 Circular waveguide, 121–130 attenuation, 125, 126–128cutoff frequency, 124, 126–128 propagation constant, 124, 126, 128 table for, 128 Circulator ferrite junction, 487–493 general properties, 318–319, 487–488 Complementary metal oxide semiconductor (CMOS), 543 Coaxial connectors, 134Coaxial line attenuation constant, 80, 82–83 characteristic impedance, 56data for standard lines, 721 distributed line parameters, 53–54 field analysis, 54–56, 130–133higher-order modes, 131–133 power capacity, 160 propagation constant, 56 Common Mode Rejection Ratio, 596 Composite filters, 396–399 Compression point, 512–513Computer aided design (CAD), 202 Conductivity, 10 table for metals, 719 Conductor loss, 26–28 Conjugate matching, 77–78, 187, 571–575Connectors, coaxial, 134 Constant gain circles, 575–579Constant-k filters, 390–393, 397 Constant noise figure circles, 580–582 Conversion loss, mixer, 639 Coplanar waveguide, 159–160Coupled lines, 347–351 characteristic impedance, 348–351 couplers, 351–362filters, 426–436 Couplers (see Directional couplers) Coupling aperture, 215–221, 302–305 coefficient, 298–299, 619 critical, 299resonator, 297–305 Crossed-guide coupler, 371–372 Current displacement, 7 electric, magnetic, 6, 8–9 Cutoff frequency circular waveguide, 124, 126–128 parallel plate waveguide, 105, 108 rectangular waveguide, 113, 116 Cutoff wavelength, 105, 109, 117, 128 D DC block, 530–531, 642–643Decibel notation, 62–63 Demagnetization factor, 462–463 Detector, 525–529 sensitivity, 528 Dicke radiometer, 700–701 Dielectric constant, table, 719Dielectric loaded waveguide, 119–120, 153–154 Dielectric loss, 26–27 Dielectric loss tangent, table, 719Dielectric resonator oscillators, 617–622 Dielectric resonators, 293–297 Dielectric strength for air, 160–161Dielectric waveguide, 159 Differential amplifier, 593–596 Differential mixer, 650–652Digital modulation, 681–684 Diode BARITT, 539detectors, 525–529 Gunn, 538 IMPATT, 539I-V curve, 526, 538 mixer, 642–643 multipliers, 628–633 PIN, 530–531 Schottky, 525–529switches, 531–534 Varactor, 537 bindex Pozar September 29, 2011 19:43 Index727 Directional couplers, 320–323 Bethe hole, 334–338 coupled line, 351–362 Lange, 359–362Moreno crossed-guide, 372 multihole waveguide, 338–343 quadrature, 343–346Riblet short slot, 373 ring hybrid, 362–367 Schwinger reversed phase, 372–373tapered line, 367–371 Directivity antenna, 663coupler, 322–324 Discontinuities, 203–205, 209–210 microstrip, 205, 209–210waveguide, 204 Dispersion, 80, 150, 155 Distortionless line, 80–81Double sideband modulation (DSB), 638 Dynamic range, 497, 511, 519–521 E Effective aperture area, 665–666 Effective isotropic radiated power (EIRP), 674Effective permittivity, microstrip, 148 Efficiency aperture, 665power added, 597 radiation, 664 Electric energy, 25Electric field, 6 Electric flux density, 7 Electric polarizability, 217Electric potential, 98–99 Electric susceptibility, 10 Electric wall, 14–15Electromagnetic spectrum, 2 Elliptic filter, 401 Emissivity, 697Energy, electric, magnetic, 25 Energy transmission, 705–706 E-plane T-junction, 325Equal ripple filter response, 400–401, 404–405 Equivalent voltages and currents, 166–170 Even-odd mode characteristic impedance, 348–351 Exponential tapered line, 262–263 Extraordinary wave, 470–471 F Fabry-Perot resonator, 315Fade margin, 675 Far field, 661 Faraday rotation, 465–469Faraday’s law, 8Ferrite devices circulators, 487–493 gyrator, 486–487 isolators, 475–482loaded waveguide, 471–475 phase shifters, 482–486 Ferrites, 451 loss in, 460–462 permeability tensor for, 457 plane wave propagation in, 465–471table of properties, 720 Field effect transistor (FET), 543–547 Filters bandpass, 411–415, 426–447 bandstop, 411–414, 437–441 composite, 396–399constant-k, 390–393, 397 coupled line, 426–436 elliptic, 401high pass, 397, 410 high-Z, low-Z, 422–426 implementation, 415–422linear phase, 401, 406–408 low pass, 390–399, 410–412 m-derived, 393–396, 397scaling, 408–411 transformations, 410–415 Flanges, waveguide, 120–121 Flow graph, 194–198 Frequency bands, 2, 685Frequency multipliers, 627–636 Frequency shift keying (FSK), 681–683 Friis power transmission formula, 673–674 G Gain (also see Power gain) amplifier, 562–564antenna, 664–665 compression, 512–513 two-port power, 558–564 Gilbert cell mixer, 652 Global Positioning System (GPS), 687–688 Group delay, 401Group velocity, 155–157 for periodic structures, 386 for waveguide, 157 G/T, 671 Gunn diode, 538 Gyrator, 486–487Gyromagnetic ratio, 453 Gyrotropic medium (see Ferrites) H Helmholtz equations, 15–16 High electron mobility transistor (HEMT), 546–547 bindex Pozar September 29, 2011 19:43 728 Index Hertz, H., 4 Heterojunction bipolar transistor (HBT), 542–543 High pass filters constant-k, 392, 397 m-derived, 397 transformation to, 410 High-Z, low-Z filters, 422–426 History, of microwave engineering, 4–6 H-plane T-junction, 325Hybrid junctions coupled line, 351–359 quadrature, 343–346ring (rat-race), 363–367 scattering matrix, 313–314, 343, 363 tapered coupled line, 367–371waveguide magic-T, 361 I Image frequency, 638–639 Image impedance, 388–390 Image parameters, filter design using, 390–399 Image theory, 42–44 IMPATT diode, 539Impedance characteristic, 50–51, 171 concept of, 170–171 image, 388–390 intrinsic, 17wave, 17, 18, 99, 100, 101 waveguide, 100, 101 Impedance inverter, 421–422Impedance matching, 228–229 Bode-Fano criterion, 267–269 double stub, 241–246L-section, 229–233 multisection transformer, 251–261 quarter wave transformer, 72–75, 246–249single stub, 234–241 tapered line, 261–267 Impedance matrix, 174–178 table for conversions, 192 Impedance transformers (see Impedance matching) Incremental inductance rule, Wheeler, 83–85 Inductive degeneration, 583 Insertion loss, 62Insertion loss method for filter design, 399–408 Intermodulation distortion, 513–519 Inverters, admittance, impedance, 421–422Iris, waveguide, 203 Isolators field displacement, 479–482resonance, 476–479J Junction circulator, 487–493 K Kittel’s equation, 464 Klopfenstein tapered line, 264–265 Klystron, 553–554Kuroda identities, 416–419 L Lange coupler, 359–362 Linearly polarized plane waves, 15–23 Linear dynamic range, 519–521Linear phase filter, 401, 406–408 Line parameters (per unit length), 51–53 Linewidth, gyromagnetic resonance, 460Link budget, 674–676 Link margin, 675 Load pull contours, 598Loaded Q, 277 Loaded waveguide dielectric loading, 119–120, 153–154ferrite loading, 471–475 Loss (see also Attenuation constant) conductor, 26–28dielectric, 26–27 ferrite, 460–462 insertion, 62 return, 58 Loss tangent, 11 table, 719 Lossy transmission lines, 79–82 Low pass filters constant-k, 390–393, 397 high-Z, low-Z, 422–426 m-derived, 393–396, 397prototype, 401–408 L-section matching, 229–233 M Magic-T, 323, 371 Magnetic energy, 25Magnetic field, 6 Magnetic flux density, 7 Magnetic polarizability, 217Magnetic susceptibility, 11 Magnetic wall, 15 Manley-Rowe relations, 628–631Matched line, 57 Matching (see Impedance matching) Material constants table of conductivities, 719 table of dielectric constants and loss tangents, 719 table of ferrite properties, 720 bindex Pozar September 29, 2011 19:43 Index729 Maximally flat filter response, 400, 402–404 Maximum power capacity, 134, 160–161 Maximum stable gain, 572 Maxwell, J., 4–5Maxwell’s equations, 4, 6–10 m-derived filters, 393–396, 397 MEMs, 551–552Metal semiconductor FET (MESFET), 543, 544–546 Microstrip, 147–153 attenuation, 148–149 characteristic impedance, 148 coupled, 350discontinuities, 205, 209–210 effective permittivity, 148 higher order modes, 150–152propagation constant, 147–148 Microwave heating, 705 Microwave integrated circuits (MIC), 547–550 hybrid, 548 monolithic (MMIC), 548–550 Microwave oven, 705Microwave sources, 538–540, 552–556 Gunn diode, 538 IMPATT diode, 539oscillators, 605–622 tubes, 552–556 Microwave tubes, 552–556 backward wave oscillator, 554 crossed-field amplifier, 555extended interaction oscillator, 554 gyratron, 555 klystron, 553–554magnetron, 552, 554 traveling wave tube, 554 Mixers, 526, 637–654 antiparallel diode, 653 balanced, 646–649 conversion loss, 639differential FET, 650–652 diode, 642–643 double balanced, 652–653FET, 643–645 Gilbert cell, 652 image rejection, 649–650image response, 638–639 Modal analysis, 203–209 Modes cavity modes, 284–287, 288–292 circular waveguide, 121–128 parallel plate waveguide, 102–110rectangular waveguide, 110–120 Modulation, 528–529, 681–684 Metal oxide semiconductor FET (MOSFET), 543, 546Multiple reflections, on quarter wave transformer, 74–75 Multipliers (see Frequency multipliers) N Negative resistance oscillators, 613–615 Neper, 62–63Network analyzer, 188 Noise, 496–511 figure, 502–511phase, 622–627 sources, 497–498 temperature, 498–502 Noise figure, 502–511 circles, 557–561 of cascade, 504–505 of lossy line, 503–504, 508–509 of mismatched amplifier, 510–511of mixer, 640–641 of passive network, 506–508 of transistor amplifier, 580–582 O Ohm’s law for fields, 10–11Open circuit stub, impedance, 60–61 Oscillators crystal, 612–613dielectric resonator, 617–622 negative resistance, 613–615 transistor, 605–613, 615–622 P Passive intermodulation (PIM), 519 Parallel plate waveguide, 102–110 attenuation, 107, 108–109characteristic impedance (TEM), 104 table for, 109 Periodic structures analysis, 382–385 k-βdiagram, 385–386 phase and group velocities, 386 Permanent magnets, 464–465 Permeability, 7, 12 tensor, for ferrite, 457 Permittivity, 7, 11 of atmosphere, 701 Perturbation theory for attenuation, 82–83 cavity resonance, 306–312 ferrite loaded waveguide, 474 Phase constant (see Propagation constant) Phase matching, 36 Phase noise, 622–627 bindex Pozar September 29, 2011 19:43 730 Index Phase shifters Faraday rotation, 485 loaded line, 535–536 reflection, 536–537Reggia-Spencer, 486 remanent (latching), 482–485 switched line, 534–535 Phase velocity plane wave, 16 transmission line, 51waveguide, 104, 105, 109, 113, 117, 128 Phasor notation, 8–9 Physical constants, table, 718PIN diodes, 530–531 phase shifters, 534–537 switches, 531–534 Plane waves, 16–25 in conducting media, 19 in ferrites, 465–471in general lossy media, 17–18 in lossless dielectric, 16–17 reflection, 28–40 Plasma, 704 Polarizability, 217 Polarization, wave, 24Power, 25–28 Power added efficiency (PAE), 597 Power amplifiers, 596–601 Power loss, 26–27, 31–32 Power capacity of transmission line, 160–161Power divider (see also Directional coupler) resistive, 326–328 T-junction, 324–326Wilkinson, 328–333 Power gain, 558–564 Power waves, 185–188Poynting’s theorem, 24–25 Poynting vector, 26 Precession, magnetic dipole, 453–456Probe coupling, 214–215 Propagation atmospheric effects, 701–702ground effects, 703–704 plasma effects, 704 Propagation constant for circular waveguide, 124, 126, 128 coaxial line, 56 microstrip line, 147–148parallel plate guide, 98, 104, 108, 109 plane waves in a good conductor, 19, 20 plane waves in lossless dielectric, 16, 20rectangular waveguide, 112, 115, 117 stripline, 142 TEM modes, 98TM or TE modes, 100–101Q Q, 274, 277–278 for circular cavity, 291–292 for dielectric resonator, 297for rectangular cavity, 286–287 for RLC circuit, 274, 276, 278 for transmission line resonator, 280, 282, 283from resonator measurements, 305–306 Quadrature hybrid, 343–346 Quadrature phase shift keying (QPSK), 682Quarter-wave transformers multiple reflection viewpoint, 74–75 multisection, 251–261single-section, 72–75, 246–249 R Radar systems, 690–695 Radar cross section, 695–696 Radiation condition, 15 efficiency, 664hazards, 706–707 patterns, 662–663 Radiometer systems, 696–701Rat-race (ring hybrid), 363–367 Receivers, 676–680 Reciprocal networks, 175–176, 181–182Reciprocity theorem, 40–42 Rectangular cavity (see Cavity resonators) Rectangular waveguide, 110–120 attenuation, 115, 116 cutoff frequency, 113, 116 group velocity, 157 maximum power capacity, 160–161 phase velocity, 113, 117propagation constant, 112, 115, 117 table for, 117 table of standard sizes, 720 Rectification, 525–528 Reflection coefficient, 29, 57 Reflectometer, 374–375Remanent magnetization, 464, 483 Resonant circuits, 272–277 Return loss, 58Richards’ transformation, 416 Ridge waveguide, 158–159 Root-finding algorithms, 139–140 S Saturation magnetization, 455 Scattering matrix, 178–185 for circulator, 318–319, 487–488for directional coupler, 320–323 generalized, 185–188 for gyrator, 486for quadrature hybrid, 343 bindex Pozar September 29, 2011 19:43 Index731 for ring hybrid, 363 shift in reference planes, 184–185 table for conversions, 192 Schwinger reversed phase coupler, 372–373Separation of variables, 20–21, 111–112, 119–120, 122–123, 130–131, 132 Short circuit stub impedance, 59Signal flow graphs, 194–198 Single sideband modulation, 638 Skin depth, 19Slotline, 159 Slotted line, 68–72 Small reflection theory, 250–252Smith chart, 63–68 Snell’s law, 36 Sources (see Microwave sources)Scattering parameters (see Scattering matrix) Specific Absorption Ratio (SAR), 707 Spectrum analyzer, 529–530Spurious free dynamic range, 519–520 Stability amplifier, 564–570circles, 564–567 Standing wave ratio (SWR), 58 Stepped impedance filters, 422–426Stripline, 141–147 approximate analysis, 144–147 attenuation, 143 characteristic impedance, 141 coupled, 349–350propagation constant, 142 Surface current, 9–10, 13–14 Surface impedance, 33–34Surface resistance, 28, 33 Surface waves at dielectric interface, 38–40of dielectric slab, 135–139 Switches, PIN diode, 531–534 T Tapered coupled line hybrid, 367–371Tapered transmission lines exponential taper, 262–263 Klopfenstein taper, 264–265triangular taper, 263–264 Telegrapher equations, 49, 55 TEM waves and modes attenuation due to dielectric loss, 98 plane waves, 16–25propagation constant, 16, 18, 19, 98 transmission lines, 54–56 wave impedance, 17, 18, 56, 99 TE, TM modes attenuation due to dielectric loss, 101–102 propagation constant, 96–97wave impedance, 96–97Terminated transmission line, 56–62 input impedance, 59 reflection coefficient, 57 voltage maxima and minima, 58 Third-order intercept, 515–518 T-junction, 324–326 Total reflection, plane wave, 38–40Transducer power gain, 559–561 Transistor amplifier, 571–601characteristics, 540–547 mixer, 643–645, 650–652 models, 541, 544multipliers, 633–636 oscillator, 605–613, 615–622 types, 540–547 Transmission coefficient, 29, 62 Transmission line equations, 49input impedance, 59 junctions, 62 parameters, 51–54 Transmission line resonators, 278–284 Transmission lines coaxial, 53–56, 130–133microstrip, 147–153 parallel plate, 54, 102–110 stripline, 141–147 transients on, 86–89 two-wire, 54 Transverse resonance method, 153–154 Traveling wave amplifier (see Amplifier design) Traveling waves plane waves, 16 on transmission lines, 50 Through-Reflect-Line (TRL) calibration, 197–202 Two-port networks, equivalent circuits, 191–194 Two-port power gains (see Power gain) U Unilateral device, 541 Unilateral figure of merit, 576Unilateral transducer power gain, 561, 563 Unitary matrix, 181 Unit element, 417Unit matrix, 180 Unloaded Q, 274 V Varactor diode, 537Velocity (see Wave velocities) V oltage standing wave ratio (see Standing wave ratio) bindex Pozar October 5, 2011 10:58 732 Index W Wave equation, 16, 18, 20 Waveguide (see Rectangular waveguide; Circular waveguide; Loaded waveguide;Parallel plate waveguide) Waveguide components, 111 directional couplers, 334–343, 372–272discontinuities, 204 isolators, 476–482 magic-T, 323, 371phase shifters, 482–486 T-junctions, 325 Waveguide excitation by apertures, 215–221 arbitrary sources, 212–214 current sheets, 210–212 Waveguide flanges, 120–121 Waveguide impedance, 100, 101Wavelength in free-space, 16–17 on transmission line, 51 for waveguide, 105, 109, 113, 117, 128 Wave velocities group, 155–157, 386 phase, 16, 51, 104, 105, 109, 113, 117, 128 Wheeler incremental inductance rule, 83–85 Wilkinson power divider, 328–333 Wireless systems, 671–672, 684–690 Y YIG-tuned oscillator, 605Y-parameters (see Admittance matrix) Z Z-parameters (see Impedance matrix) both Pozar September 29, 2011 18:4 USEFULRESULTS Maxwell’s equations: ∇ׯE=− jωµ¯H−¯M ∇·¯D=ρ ∇ׯH=jω/epsilon1¯E+¯J ∇·¯B=0 Surface resistance and skin depth: Rs=/radicalbiggωµ 2σδs=/radicalbigg2 ωµσ Input impedance of terminated lossless transmission lines: Zin=Z0ZL+jZ0tanβ/lscript Z0+jZLtanβ/lscript(arbitrary load ) Zin=jZ0tanβ/lscript (short-circuited line) Zin=− jZ0cotβ/lscript (open-circuited line) Relations between load impedance and reflection coefficient: /Gamma1=ZL−Z0 ZL+Z0ZL=Z01+/Gamma1 1−/Gamma1 Definitions of return loss, insertion loss and SWR: RL=−20 log |/Gamma1|,IL=−20 log |T|,SWR=1+|/Gamma1| 1−|/Gamma1| Conversion between dB and nepers: 1 neper =8.686 dB Elements of the ferrite permeability tensor: µ=µ0/parenleftbigg 1+ω0ωm ω2 0−ω2/parenrightbigg ω0=µ0γH0 ωm=µ0γMs κ=µ0ωωm ω2 0−ω2(or 2.8 MHz/Oersted ) Conversion between some values of reflection coefficient, SWR, and return loss: |/Gamma1| 0.024 0.032 0.048 0.050 0.056 0.10 0.178 0.200 0.316 0.33 SWR 1.05 1.07 1.10 1.11 1.12 1.22 1.43 1.50 1.92 2.00 RL (dB) 32.3 30.0 26.4 26.0 25.0 20.0 15.0 14.0 10.0 9.6 F2 both Pozar September 29, 2011 18:4 TheABCD Parameters of Some Useful Two-Port Circuits. Circuit ABCD Parameters ZA=1 C=0B=Z D=1 YA=1 C=YB=0 D=1 Z0, /H9252A=cosβ/lscript C=jY0sinβ/lscriptB=jZ0sinβ/lscript D=cosβ/lscript N : 1 A=N C=0B=0 D=1 N Y1 Y2Y3 A=1+Y2 Y3 C=Y1+Y2+Y1Y2 Y3B=1 Y3 D=1+Y1 Y3 Z1 Z2 Z3A=1+Z1 Z3 C=1 Z3B=Z1+Z2+Z1Z2 Z3 D=1+Z2 Z3 F3 both Pozar September 29, 2011 18:4 VECTORANALYSIS CoordinateTransformations Rectangular to cylindrical: ˆx ˆy ˆz ˆρ cosφ sinφ 0 ˆφ−sinφ cosφ 0 ˆz 00 1 Rectangular to spherical: ˆx ˆy ˆz ˆr sinθcosφ sinθsinφ cosθ ˆθ cosθcosφ cosθsinφ −sinθ ˆφ−sinφ cosφ 0 Cylindrical to spherical: ˆρ ˆφ ˆz ˆr sinθ 0c o s θ ˆθ cosθ 0 −sinθ ˆφ 01 0 These tables can be used to transform unit vectors as well as vector components; e.g., ˆρ=ˆxcosφ+ˆysinφ Aρ=Axcosφ+Aysinφ B2 both Pozar September 29, 2011 18:4 VectorDifferentialOperators Rectangular coordinates: ∇f=ˆx∂f ∂x+ˆy∂f ∂y+ˆz∂f ∂z ∇·¯A=∂Ax ∂x+∂Ay ∂y+∂Az ∂z ∇ׯA=ˆx/parenleftbigg∂Az ∂y−∂Ay ∂z/parenrightbigg +ˆy/parenleftbigg∂Ax ∂z−∂Az ∂x/parenrightbigg +ˆz/parenleftbigg∂Ay ∂x−∂Ax ∂y/parenrightbigg ∇2f=∂2f ∂x2+∂2f ∂y2+∂2f ∂z2 ∇2¯A=ˆx∇2Ax+ˆy∇2Ay+ˆz∇2Az Cylindrical coordinates: ∇f=ˆρ∂f ∂ρ+ˆφ1 ρ∂f ∂φ+ˆz∂f ∂z ∇·¯A=1 ρ∂ ∂ρ(ρAρ)+1 ρ∂Aφ ∂φ+∂Az ∂z ∇ׯA=ˆρ/parenleftbigg1 ρ∂Az ∂φ−∂Aφ ∂z/parenrightbigg +ˆφ/parenleftbigg∂Aρ ∂z−∂Az ∂ρ/parenrightbigg +ˆz1 ρ/bracketleftbigg∂(ρAφ) ∂ρ−∂Aρ ∂φ/bracketrightbigg ∇2f=1 ρ∂ ∂ρ/parenleftbigg ρ∂f ∂ρ/parenrightbigg +1 ρ2∂2f ∂φ2+∂2f ∂z2 ∇2¯A=∇(∇·¯A)−∇×∇× ¯A Spherical coordinates: ∇f=ˆr∂f ∂r+ˆθ1 r∂f ∂θ+ˆφ rsinθ∂f ∂φ ∇·¯A=1 r2∂ ∂r/parenleftbig r2Ar/parenrightbig +1 rsinθ∂ ∂θ(sinθAθ)+1 rsinθ∂Aφ ∂φ ∇ׯA=ˆr rsinθ/bracketleftbigg∂ ∂θ(Aφsinθ)−∂Aθ ∂φ/bracketrightbigg +ˆθ r/bracketleftbigg1 sinθ∂Ar ∂φ−∂ ∂r(rAφ)/bracketrightbigg +ˆφ r/bracketleftbigg∂ ∂r(rAθ)−∂Ar ∂θ/bracketrightbigg ∇2f=1 r2∂ ∂r/parenleftbigg r2∂f ∂r/parenrightbigg +1 r2sinθ∂ ∂θ/parenleftbigg sinθ∂f ∂θ/parenrightbigg +1 r2sin2θ∂2f ∂φ2 ∇2¯A=∇ ∇· ¯A−∇×∇× ¯A B3