ch 10 orfanidis obs edition

10 Transmission Lines 10.1 General Properties of TEM Transmission Lines We saw in Sec. 9.3 that TEM modes are described by Eqs. (9.3.3) and (9.3.4), the latter being equivalent to a two-dimensional electrostatic problem: HT=1 ηˆz×ET ∇∇∇T×ET=0 ∇∇∇T·ET=0(TEM modes) (10.1.1) The second of (10.1.1) implies that ETcan be expressed as the (two-dimensional) gradient of a scalar electrostatic potential. Then, the third equation becomes Laplace’sequation for the potential. Thus, the electric field can be obtained from: ∇2 Tϕ=0 ET=−∇∇∇Tϕ(equivalent electrostatic problem) (10.1.2) Because in electrostatic problems the electric field lines must start at positively charged conductors and end at negatively charged ones, a TEM mode can be supportedonly in multi-conductor guides, such as the coaxial cable or the two-wire line. Hollow conducting waveguides cannot support TEM modes. Fig. 10.1.1 depicts the transverse cross-sectional area of a two-conductor transmis- sion line. The cross-section shapes are arbitrary. The conductors are equipotentials of the electrostatic solution. Let ϕa,ϕbbe the constant potentials on the two conductors. The voltage difference between the conduc-tors will be V=ϕa−ϕb. The electric field lines start perpendicularly on conductor (a) and end perpendicularly on conductor (b). The magnetic field lines, being perpendicular to the electric lines according to Eq. (10.1.1), are recognized to be the equipotential lines . As such, they close upon themselves sur- rounding the two conductors.490 10. Transmission Lines Fig. 10.1.1 Two-conductor transmission line. In particular, on the conductor surfaces the magnetic field is tangential. According to Amp `ere’s law, the line integrals of the magnetic field around each conductor will result into total currents Iand−Iflowing on the conductors in the z-direction. These currents are equal and opposite. Impedance, Inductance, and Capacitance Because the fields are propagating along the z-direction with frequency ωand wave- numberβ=ω/c, thez,tdependence of the voltage Vand currentIwill be: V(z,t)=Vejωt−jβz I(z,t)=Iejωt−jβz(10.1.3) For backward-moving voltage and current waves, we must replace βby−β. The ratio V(z,t)/I(z,t) =V/Iremains constant and independent of z. It is called the character- istic impedance of the line: Z=V I(line impedance) (10.1.4) In addition to the impedance Z, a TEM line is characterized by its inductance per unit lengthL/primeand its capacitance per unit length C/prime. For lossless lines, the three quantities Z,L/prime,C/primeare related as follows: L/prime=μZ η,C/prime=/epsilon1η Z(inductance and capacitance per unit length) (10.1.5) whereη=/radicalbig μ//epsilon1is the characteristic impedance of the dielectric medium between the conductors.†By multiplying and dividing L/primeandC/prime, we also obtain: †These expressions explain why μand/epsilon1are sometimes given in units of henry/m and farad/m. 10.1. General Properties of TEM Transmission Lines 491 Z=/radicalBigg L/prime C/prime,c=1√/epsilon1μ=1√ L/primeC/prime(10.1.6) The velocity factor of the line is the ratio c/c 0=1/n, wheren=/radicalbig /epsilon1//epsilon1 0=√/epsilon1ris the refractive index of the dielectric, which is assumed to be non-magnetic. Becauseω=βc, the guide wavelength will be λ=2π/β=c/f=c0/fn=λ0/n, whereλ0is the free-space wavelength. For a finite length lof the transmission line, the quantityl/λ=nl/λ 0is referred to as the electrical length of the line and plays the same role as the optical length in thin-film layers. Eqs. (10.1.5) and (10.1.6) are general results that are valid for any TEM line. They can be derived with the help of Fig. 10.1.2. Fig. 10.1.2 Surface charge and magnetic flux linkage. The voltageVis obtained by integrating ET·dlalong any path from (a) to (b). How- ever, if that path is chosen to be an E-field line, then ET·dl=|ET|dl, giving: V=/integraldisplayb a|ET|dl (10.1.7) Similarly, the current Ican be obtained by the integral of HT·dlalong any closed path around conductor (a). If that path is chosen to be an H-field line, such as the peripheryCaof the conductor, we will obtain: I=/contintegraldisplay Ca|HT|dl (10.1.8) The surface charge accumulated on an infinitesimal area dldz of conductor (a) is dQ=ρsdldz , whereρsis the surface charge density. Because the conductors are assumed to be perfect, the boundary conditions require that ρsbe equal to the normal component of the D-field, that is, ρs=/epsilon1|ET|. Thus,dQ=/epsilon1|ET|dldz . If we integrate over the periphery Caof conductor (a), we will obtain the total surface charge per unit z-length: Q/prime=dQ dz=/contintegraldisplay Ca/epsilon1|ET|dl But because of the relationship |ET|=η|HT|, which follows from the first of Eqs. (10.1.1), we have:492 10. Transmission Lines Q/prime=/contintegraldisplay Ca/epsilon1|ET|dl=/epsilon1η/contintegraldisplay Ca|HT|dl=/epsilon1ηI (10.1.9) where we used Eq. (10.1.8). Because Q/primeis related to the capacitance per unit length and the voltage by Q/prime=C/primeV, we obtain Q/prime=C/primeV=/epsilon1ηI⇒C/prime=/epsilon1ηI V=/epsilon1η Z Next, we consider an E-field line between points AandBon the two conductors. The magnetic flux through the infinitesimal area dldz will bedΦ=|BT|dldz=μ|HT|dldz because the vector HTis perpendicular to the area. If we integrate from (a) to (b), we will obtain the total magnetic flux linking the two conductors per unit z-length: Φ/prime=dΦ dz=/integraldisplayb aμ|HT|dl replacing|HT|=| ET|/ηand using Eq. (10.1.7), we find: Φ/prime=/integraldisplayb aμ|HT|dl=μ η/integraldisplayb a|ET|dl=μ ηV The magnetic flux is related to the inductance via Φ/prime=L/primeI. Therefore, we get: Φ/prime=L/primeI=μ ηV⇒L/prime=μ ηV I=μZ η Transmitted Power The relationships among Z,L/prime,C/primecan also be derived using energy considerations. The power transmitted along the line is obtained by integrating the z-component of the Poynting vector over the cross-section Sof the line. For TEM modes we have Pz= |ET|2/2η, therefore, PT=1 2η/integraldisplay/integraldisplay S|ET|2dxdy=1 2η/integraldisplay/integraldisplay S|∇∇∇Tϕ|2dxdy (10.1.10) It can be shown in general that Eq. (10.1.10) can be rewritten as: PT=1 2Re(V∗I)=1 2Z|I|2=1 2Z|V|2(10.1.11) We will verify this in the various examples below. It can be proved using the following Green’s identity: |∇∇∇Tϕ|2+ϕ∗∇2 Tϕ=∇∇∇T·(ϕ∗∇∇∇Tϕ) Writing ET=−∇∇∇Tϕand noting that ∇2 Tϕ=0, we obtain: |ET|2=−∇∇∇T·(ϕ∗ET) 10.1. General Properties of TEM Transmission Lines 493 Then, the two-dimensional Gauss’ theorem implies: PT=1 2η/integraldisplay/integraldisplay S|ET|2dxdy=−1 2η/integraldisplay/integraldisplay S∇∇∇T·(ϕ∗ET)dxdy =−1 2η/contintegraldisplay Caϕ∗ET·(−ˆn)dl−1 2η/contintegraldisplay Cbϕ∗ET·(−ˆn)dl =1 2η/contintegraldisplay Caϕ∗(ET·ˆn)dl+1 2η/contintegraldisplay Cbϕ∗(ET·ˆn)dl where ˆnare the outward normals to the conductors (the quantity −ˆnis the normal outward from the region S.) Because the conductors are equipotential surfaces, we have ϕ∗=ϕ∗ aon conductor (a) and ϕ∗=ϕ∗ bon conductor (b). Using Eq. (10.1.9) and noting that ET·ˆn=±| ET|on conductors (a) and (b), we obtain: PT=1 2ηϕ∗ a/contintegraldisplay Ca|ET|dl−1 2ηϕ∗ b/contintegraldisplay Cb|ET|dl=1 2ηϕ∗ aQ/prime /epsilon1−1 2ηϕ∗ bQ/prime /epsilon1 =1 2(ϕ∗ a−ϕ∗ b)Q/prime /epsilon1η=1 2V∗/epsilon1ηI /epsilon1η=1 2V∗I=1 2Z|I|2 The distribution of electromagnetic energy along the line is described by the time- averaged electric and magnetic energy densities per unit length, which are given by: W/prime e=1 4/epsilon1/integraldisplay/integraldisplay S|ET|2dxdy, W/prime m=1 4μ/integraldisplay/integraldisplay S|HT|2dxdy Using Eq. (10.1.10), we may rewrite: W/prime e=1 2/epsilon1ηPT=1 2cPT,W/prime m=1 2μ ηPT=1 2cPT Thus,W/prime e=W/prime mand the total energy density is W/prime=W/prime e+W/prime m=PT/c, which implies that the energy velocity will be ven=PT/W/prime=c. We may also express the energy densities in terms of the capacitance and inductance of the line: W/prime e=1 4C/prime|V|2,W/prime m=1 4L/prime|I|2(10.1.12) Power Losses, Resistance, and Conductance Transmission line losses can be handled in the manner discussed in Sec. 9.2. The field patterns and characteristic impedance are determined assuming the conductors are per-fectly conducting. Then, the losses due to the ohmic heating of the dielectric and theconductors can be calculated by Eqs. (9.2.5) and (9.2.9). These losses can be quantified by two more characteristic parameters of the line, the resistance and conductance per unit length, R/primeandG/prime. The attenuation coefficients due to conductor and dielectric losses are then expressible in terms R/prime,G/primeandZby: αc=R/prime 2Z,αd=1 2G/primeZ (10.1.13)494 10. Transmission Lines They can be derived in general terms as follows. The induced surface currents on the conductor walls are Js=ˆn×HT=ˆn×(ˆz×ET)/η, where ˆnis the outward normal to the wall. Using the BAC-CAB rule, we find Js=ˆz(ˆn·ET)/η. But, ˆnis parallel to ETon the surface of conductor (a), and anti parallel on (b). Therefore, ˆn·ET=±| ET|. It follows that Js=±ˆz|ET|/η=±ˆz|HT|, pointing in the +zdirection on (a) and −zdirection on (b). Inserting these expressions into Eq. (9.2.8), we find for the conductor power loss perunit z-length: P/prime loss=dPloss dz=1 2Rs/contintegraldisplay Ca|HT|2dl+1 2Rs/contintegraldisplay Cb|HT|2dl (10.1.14) Because HTis related to the total current Ivia Eq. (10.1.8), we may define the resis- tance per unit length R/primethrough the relationship: P/prime loss=1 2R/prime|I|2(conductor ohmic losses) (10.1.15) Using Eq. (10.1.11), we find for the attenuation coefficient: αc=P/prime loss 2PT=1 2R/prime|I|2 21 2Z|I|2=R/prime 2Z(10.1.16) If the dielectric between the conductors is slightly conducting with conductivity σd or loss tangent tan δ=σd//epsilon1ω, then there will be some current flow between the two conductors. The induced shunt current per unit z-length is related to the conductance by I/prime d= G/primeV. The shunt current density within the dielectric is Jd=σdET. The total shunt current flowing out of conductor (a) towards conductor (b) is obtained by integrating Jd around the periphery of conductor (a): I/prime d=/contintegraldisplay CaJd·ˆndl=σd/contintegraldisplay Ca|ET|dl Using Eq. (10.1.9), we find: I/prime d=σdQ/prime /epsilon1=G/primeV⇒G/prime=σd /epsilon1C/prime=σdη Z It follows that the dielectric loss constant (9.2.5) will be: αd=1 2σdη=1 2G/primeZ Alternatively, the power loss per unit length due to the shunt current will be P/prime d= Re(I/prime dV∗)/2=G/prime|V|2/2, and therefore, αdcan be computed from: αd=P/prime d 2PT=1 2G/prime|V|2 21 2Z|V|2=1 2G/primeZ 10.2. Parallel Plate Lines 495 It is common practice to express the dielectric losses and shunt conductance in terms of the loss tangent tan δand the wavenumber β=ω/c=ω/epsilon1η : αd=1 2σdη=1 2ω/epsilon1η tanδ=1 2βtanδandG/prime=σd /epsilon1C/prime=ωC/primetanδ(10.1.17) Next, we discuss four examples: the parallel plate line, the microstrip line, the coaxial cable, and the two-wire line. In each case, we discuss the nature of the electrostaticproblem and determine the characteristic impedance Zand the attenuation coefficients αcandαd. 10.2 Parallel Plate Lines The parallel plate line shown in Fig. 10.2.1 consists of two parallel conducting plates ofwidth wseparated by height hby a dielectric material /epsilon1. Examples of such lines are microstrip lines used in microwave integrated circuits. For arbitrary values of wandh, the fringing effects at the ends of the plates cannot be ignored. In fact, fringing requires the fields to have longitudinal components, andtherefore TEM modes are not strictly-speaking supported. Fig. 10.2.1 Parallel plate transmission line. However, assuming the width is much larger than the height, w/greatermuchh, we may ignore the fringing effects and assume that the fields have no dependence on the x-coordinate. The electrostatic problem is equivalent to that of a parallel plate capacitor. Thus, the electric field will have only a ycomponent and will be constant between the plates. Similarly, the magnetic field will have only an xcomponent. It follows from Eqs. (10.1.7) and (10.1.8) that: V=−Eyh, I=Hxw Therefore, the characteristic impedance of the line will be: Z=V I=−Eyh Hxw=ηh w(10.2.1) where we used Ey=−ηHx. The transmitted power is obtained from Eq. (10.1.10): PT=1 2η|Ey|2(wh)=1 2ηV2 h2wh=1 2ηw hV2=1 2ZV2=1 2ZI2(10.2.2)496 10. Transmission Lines The inductance and capacitance per unit length are obtained from Eq. (10.1.5): L/prime=μh w,C/prime=/epsilon1w h(10.2.3) The surface current on the top conductor is Js=ˆn×H=(−ˆy)×H=ˆzHx. On the bottom conductor, it will be Js=−ˆzHx. Therefore, the power loss per unit z-length is obtained from Eq. (9.2.8): P/prime loss=21 2Rs|Hx|2w=1 wRsI2 Comparing with Eq. (10.1.15), we identify the resistance per unit length R/prime=2Rs/w. Then, the attenuation constant due to conductor losses will be: αc=P/prime loss 2PT=R/prime 2Z=Rs wZ=Rs hη(10.2.4) 10.3 Microstrip Lines Practical microstrip lines, shown in Fig. 10.3.1, have width-to-height ratios w/h that are not necessarily much greater than unity, and can vary over the interval 0 .1<w/h< 10. Typical heights hare of the order of millimeters. Fig. 10.3.1 A microstrip transmission line. Fringing effects cannot be ignored completely and the simple assumptions about the fields of the parallel plate line are not valid. For example, assuming a propagating wavein the z-direction with z,tdependence of ejωt−jβzwith a common βin the dielectric and air, the longitudinal-transverse decomposition (9.1.5) gives: ∇∇∇TEz׈z−jβˆz×ET=−jωμ HT⇒ˆz×(∇∇∇TEz+jβET)=jωμ HT In particular, we have for the x-component: ∂yEz+jβEy=−jωμHx The boundary conditions require that the components HxandDy=/epsilon1Eybe contin- uous across the dielectric-air interface (at y=h). This gives the interface conditions: ∂yEair z+jβEair y=∂yEdiel z+jβEdiel y /epsilon10Eair y=/epsilon1Ediel y 10.3. Microstrip Lines 497 Combining the two conditions, we obtain: ∂y/parenleftbig Ediel z−Eair z/parenrightbig =jβ/epsilon1−/epsilon10 /epsilon1Eair y=jβ/epsilon1−/epsilon10 /epsilon10Ediel y (10.3.1) BecauseEyis non-zero on either side of the interface, it follows that the left-hand side of Eq. (10.3.1) cannot be zero and the wave cannot be assumed to be strictly TEM. However,Eyis small in both the air and the dielectric in the fringing regions (to the left and right of the upper conductor). This gives rise to the so-called quasi-TEM approx- imation in which the fields are assumed to be approximately TEM and the effect of the deviation from TEM is taken into account by empirical formulas for the line impedanceand velocity factor. In particular, the air-dielectric interface is replaced by an effective dielectric, filling uniformly the entire space, and in which there would be a TEM propagating mode. Ifwe denote by /epsilon1effthe relative permittivity of the effective dielectric, the wavelength and velocity factor of the line will be given in terms of their free-space values λ0,c0: λ=λ0√/epsilon1eff,c=c0√/epsilon1eff(10.3.2) There exist many empirical formulas for the characteristic impedance of the line and the effective dielectric constant. Hammerstad and Jensen’s are some of the most accurate ones [881,887]: /epsilon1eff=/epsilon1r+1 2+/epsilon1r−1 2/parenleftbigg 1+10 u/parenrightbigg−ab ,u=w h(10.3.3) where/epsilon1r=/epsilon1//epsilon1 0is the relative permittivity of the dielectric and the quantities a,bare defined by: a=1+1 49ln/bracketleftBigg u4+(u/52)2 u4+0.432/bracketrightBigg +1 18.7ln/bracketleftBigg 1+/parenleftbiggu 18.1/parenrightbigg3/bracketrightBigg b=0.564/parenleftbigg/epsilon1r−0.9 /epsilon1r+3/parenrightbigg0.053(10.3.4) The accuracy of these formulas is better than 0.01% for u<1 and 0.03% for u<1000. Similarly, the characteristic impedance is given by the empirical formula: Z=η0 2π√/epsilon1effln⎡ ⎣f(u) u+/radicalBigg 1+4 u2⎤ ⎦ (10.3.5) whereη0=/radicalbig μ0//epsilon10and the function f(u) is defined by: f(u)=6+(2π−6)exp/bracketleftBigg −/parenleftbigg30.666 u/parenrightbigg0.7528/bracketrightBigg (10.3.6) The accuracy is better than 0.2% for 0 .1≤u≤100 and/epsilon1r<128. In the limit of large ratiow/h, or,u→∞ , Eqs. (10.3.3) and (10.3.5) tend to those of the parallel plate line of the previous section:498 10. Transmission Lines /epsilon1eff→/epsilon1r,Z→η0√/epsilon1rh w=ηh w Some typical substrate dielectric materials used in microstrip lines are alumina, a ceramic form of Al 2O4wither=9.8, and RT-Duroid, a teflon composite material with /epsilon1r=2.2. Practical values of the width-to-height ratio are in the range 0 .1≤u≤10 and practical values of characteristic impedances are between 10–200 ohm. Fig. 10.3.2 shows the dependence of Zand/epsilon1effonufor the two cases of /epsilon1r=2.2 and/epsilon1r=9.8. 0 1 2 3 4 5 6 7 8 9 100255075100125150175200225Characteristic Impedance w/hZ (ohm) εr = 2.2 εr = 9.8 0 1 2 3 4 5 6 7 8 9 1012345678910Effective Permittivity w/hεeff εr = 2.2 εr = 9.8 Fig. 10.3.2 Characteristic impedance and effective permittivity of microstrip line. The synthesis of a microstrip line requires that we determine the ratio w/h that will achieve a given characteristic impedance Z. The inverse of Eq. (10.3.5)—solving for uin terms ofZ—is not practical. Direct synthesis empirical equations exist [882,887], but are not as accurate as (10.3.5). Given a desired Z, the ratiou=w/h is calculated as follows. Ifu≤2, u=8 eA−2e−A(10.3.7) and, ifu>2, u=/epsilon1r−1 π/epsilon1r/bracketleftbigg ln(B−1)+0.39−0.61 /epsilon1r/bracketrightbigg +2 π/bracketleftbig B−1−ln(2B−1)/bracketrightbig (10.3.8) whereA,B are given by: A=π/radicalbig 2(/epsilon1r+1)Z η0+/epsilon1r−1 /epsilon1r+1/parenleftbigg 0.23+0.11 /epsilon1r/parenrightbigg B=π 2√/epsilon1rη0 Z(10.3.9) The accuracy of these formulas is about 1%. The method can be improved iteratively by a process of refinement to achieve essentially the same accuracy as Eq. (10.3.5). Start-ing with ucomputed from Eqs. (10.3.7) and (10.3.8), a value of Zis computed through Eq. (10.3.5). If that Zis more than, say, 0.2% off from the desired value of the line 10.3. Microstrip Lines 499 impedance, then uis slightly changed, and so on, until the desired level of accuracy is reached [887]. Because Zis monotonically decreasing with u,i fZis less than the de- sired value, then uis decreased by a small percentage, else, uis increased by the same percentage. The three MATLAB functions mstripa ,mstrips , and mstripr implement the anal- ysis, synthesis, and refinement procedures. They have usage: [eff,Z] = mstripa(er,u); % analysis equations (10.3.3) and (10.3.5) u = mstrips(er,Z); % synthesis equations (10.3.7) and (10.3.8) [u,N] = mstripr(er,Z,per); % refinement The function mstripa accepts also a vector of several u’s, returning the correspond- ing vector of values of /epsilon1effandZ.I nmstripr , the outputNis the number of iterations required for convergence, and peris the desired percentage error, which defaults to 0.2% if this parameter is omitted. Example 10.3.1: Given/epsilon1r=2.2 andu=w/h=2,4,6, the effective permittivities and impe- dances are computed from the MATLAB call: u = [2; 4; 6]; [eff, Z] = mstripa(er,u); The resulting output vectors are: u=⎡ ⎢⎣2 4 6⎤ ⎥⎦⇒/epsilon1eff=⎡ ⎢⎣1.8347 1.9111 1.9585⎤ ⎥⎦,Z=⎡ ⎢⎣65.7273 41.7537 30.8728⎤ ⎥⎦ohm Example 10.3.2: To compare the outputs of mstrips andmstripr , we design a microstrip line with/epsilon1r=2.2 and characteristic impedance Z=50 ohm. We find: u=mstrips(2.2,50)=3.0779⇒[/epsilon1eff,Z]=mstripa(2.2,u)=[1.8811,50.0534] u=mstripr(2.2,50)=3.0829⇒[/epsilon1eff,Z]=mstripa(2.2,u)=[1.8813,49.9990] The first solution has an error of 0.107% from the desired 50 ohm impedance, and the second, a 0.002% error. As another example, if Z=100 Ω, the function mstrips results inu=0.8949,Z= 99.9495 Ω, and a 0.050% error, whereas mstripr givesu=0.8939,Z=99.9980 Ω, and a 0.002% error. /intersectionsq/unionsq In using microstrip lines several other effects must be considered, such as finite strip thickness, frequency dispersion, dielectric and conductor losses, radiation, and surfacewaves. Guidelines for such effects can be found in [881–887]. The dielectric losses are obtained from Eq. (10.1.17) by multiplying it by an effective dielectric filling factor q: αd=qω 2ctanδ=f c0πq√/epsilon1efftanδ=1 λ0πq√/epsilon1efftanδ, q=1−/epsilon1−1 eff 1−/epsilon1−1r(10.3.10)500 10. Transmission Lines Typical values of the loss tangent are of the order of 0.001 for alumina and duroid substrates. The conductor losses are approximately computed from Eq. (10.2.4): αc=Rs wZ(10.3.11) 10.4 Coaxial Lines The coaxial cable, depicted in Fig. 10.4.1, is the most widely used TEM transmission line. It consists of two concentric conductors of inner and outer radii of aandb, with the space between them filled with a dielectric /epsilon1, such as polyethylene or teflon. The equivalent electrostatic problem can be solved conveniently in cylindrical coor- dinatesρ,φ. The potential ϕ(ρ,φ) satisfies Laplace’s equation: ∇2 Tϕ=1 ρ∂ ∂ρ/parenleftBigg ρ∂ϕ ∂ρ/parenrightBigg +1 ρ2∂2ϕ ∂2φ=0 Because of the cylindrical symmetry, the potential does not depend on the azimuthal angleφ. Therefore, 1 ρ∂ ∂ρ/parenleftBigg ρ∂ϕ ∂ρ/parenrightBigg =0⇒ρ∂ϕ ∂ρ=B⇒ϕ(ρ)=A+Blnρ whereA,B are constants of integration. Assuming the outer conductor is grounded, ϕ(ρ)=0a tρ=b, and the inner conductor is held at voltage V,ϕ(a)=V, the constants A,B are determined to be B=−Vln(b/a) andA=−Blnb, resulting in the potential: ϕ(ρ)=V ln(b/a)ln(b/ρ) (10.4.1) It follows that the electric field will have only a radial component, Eρ=−∂ρϕ, and the magnetic field only an azimuthal component Hφ=Eρ/η: Eρ=V ln(b/a)1 ρ,Hφ=V ηln(b/a)1 ρ(10.4.2) IntegratingHφaround the inner conductor we obtain the current: Fig. 10.4.1 Coaxial transmission line. 10.4. Coaxial Lines 501 I=/integraldisplay2π 0Hφρdφ=/integraldisplay2π 0V ηln(b/a)1 ρρdφ=2πV ηln(b/a)(10.4.3) It follows that the characteristic impedance of the line Z=V/I, and hence the inductance and capacitance per unit length, will be: Z=η 2πln(b/a), L/prime=μ 2πln(b/a), C/prime=2π/epsilon1 ln(b/a)(10.4.4) Using Eq. (10.4.3) into (10.4.2), we may express the magnetic field in the form: Hφ=I 2πρ(10.4.5) This is also obtainable by the direct application of Amp `ere’s law around the loop of radiusρencircling the inner conductor, that is, I=(2πρ)Hφ. The transmitted power can be expressed either in terms of the voltage Vor in terms of the maximum value of the electric field inside the line, which occurs at ρ=a, that is, Ea=V//parenleftbig aln(b/a)/parenrightbig : PT=1 2Z|V|2=π|V|2 ηln(b/a)=1 η|Ea|2(πa2)ln(b/a) (10.4.6) Example 10.4.1: A commercially available polyethylene-filled RG-58/U cable†is quoted to have impedance of 50 Ω, velocity factor of 66 percent, inner conductor radius a=0.4060 mm (AWG 20-gauge wire), and maximum operating RMS voltage of 1400 volts. Determine the outer-conductor radius b, the capacitance and inductance per unit length C/prime,L/prime, the maximum power PTthat can be transmitted, and the maximum electric field inside the cable. Solution: Polyethylene has a relative dielectric constant of /epsilon1r=2.25, so thatn=√/epsilon1r=1.5. The velocity factor is c/c 0=1/n=0.667. Given that η=η0/n=376.73/1.5=251.15 Ω andc=c0/n=2.9979×108/1.5=1.9986×108m/sec, we have: Z=η 2πln(b/a)⇒b=ae2πZ/η=0.4060e2π50/251.15=1.4183 mm Therefore,b/a=3.49. The capacitance and inductance per unit length are found from: C/prime=/epsilon1η Z=1 cZ=1 1.9986×108×50=100.07 pF/m L/prime=μZ η=Z c=50 1.9986×108=0.25μH/m The peak voltage is related to its RMS value by |V|=√ 2Vrms. It follows that the maximum power transmitted is: PT=1 2Z|V|2=V2 rms Z=14002 50=39.2k W †see, for example, the 9310 Coax RG-58/U cable from www.belden.com.502 10. Transmission Lines The peak value of the electric field occurring at the inner conductor will be: |Ea|=|V| aln(b/a)=√ 2Vrms aln(b/a)=√ 2·1400 0.4060×10−3ln(1.4183/0.4060)=3.9 MV/m This is to be compared with the dielectric breakdown of polyethylene of about 20 MV/m. /intersectionsq/unionsq Example 10.4.2: Most cables have a nominal impedance of either 50 or 75 Ω. The precise value depends on the manufacturer and the cable. For example, a 50-Ω cable might actually have an impedance of 52 Ω and a 75-Ω cable might actually be a 73-Ω cable. The table below lists some commonly used cables with their AWG-gauge number of the inner conductor, the inner conductor radius ain mm, and their nominal impedance. Their dielectric filling is polyethylene with /epsilon1r=2.25 orn=√/epsilon1r=1.5. type AWGaZ RG-6/U 18 0.512 75 RG-8/U 11 1.150 50 RG-11/U 14 0.815 75 RG-58/U 20 0.406 50 RG-59/U 22 0.322 75 RG-174/U 26 0.203 50 RG-213/U 13 0.915 50 The most commonly used cables are 50-Ω ones, such as the RG-58/U. Home cable-TV uses 75-Ω cables, such as the RG-59/U or RG-6/U. The thin ethernet computer network, known as 10base-2, uses RG-58/U or RG-58A/U, which is similar to the RG-58/U but has a stranded inner copper core. Thick ethernet (10base-5) uses the thicker RG-8/U cable. Because a dipole antenna has an input impedance of about 73 Ω, the RG-11, RG-6, and RG-59 75-Ω cables can be used to feed the antenna. /intersectionsq/unionsq Next, we determine the attenuation coefficient due to conductor losses. The power loss per unit length is given by Eq. (10.1.14). The magnetic fields at the surfaces ofconductors (a) and (b) are obtained from Eq. (10.4.5) by setting ρ=aandρ=b: Ha=I 2πa,Hb=I 2πb Because these are independent of the azimuthal angle, the integrations around the peripheriesdl=adφ ordl=bdφ will contribute a factor of (2πa) or(2πb). Thus, P/prime loss=1 2Rs/bracketleftbig (2πa)|Ha|2+(2πb)|Hb|2/bracketrightbig =Rs|I|2 4π/parenleftbigg1 a+1 b/parenrightbigg (10.4.7) It follows that: αc=P/prime loss 2PT=Rs|I|2 4π/parenleftbigg1 a+1 b/parenrightbigg 21 2Z|I|2 10.4. Coaxial Lines 503 Using Eq. (10.4.4), we finally obtain: αc=Rs 2η/parenleftbigg1 a+1 b/parenrightbigg ln/parenleftbiggb a/parenrightbigg (10.4.8) The ohmic losses in the dielectric are described by Eq. (10.1.17). The total attenuation constant will be the sum of the conductor and dielectric attenuations: α=αc+αd=Rs 2η/parenleftbigg1 a+1 b/parenrightbigg ln/parenleftbiggb a/parenrightbigg+ω 2ctanδ (attenuation) (10.4.9) The attenuation in dB/m will be αdB=8.686α. This expression tends to somewhat underestimate the actual losses, but it is generally a good approximation. The αcterm grows in frequency like/radicalbig fand the term αd, likef. The smaller the dimensions a,b, the larger the attenuation. The loss tangent tan δ of a typical polyethylene or teflon dielectric is of the order of 0.0004–0.0009 up to about3 GHz. The ohmic losses and the resulting heating of the dielectric and conductors also limit the power rating of the line. For example, if the maximum supported voltage is 1400volts as in Example 10.4.2, the RMS value of the current for an RG-58/U line would be Irms=1400/50=28 amps, which would likely melt the conductors. Thus, the actual power rating is much smaller than that suggested by the maximum voltage rating. Thetypical power rating of an RG-58/U cable is typically 1 kW, 200 W, and 80 W at thefrequencies of 10 MHz, 200 MHz, and 1 GHz. Example 10.4.3: The table below lists the nominal attenuations in dB per 100 feet of the RG-8/U and RG-213/U cables. The data are from [1456]. f(MHz) 50 100 200 400 900 1000 3000 5000 α(dB/100ft) 1.31.92.74.17.5 8.016.027.0 Both are 50-ohm cables and their radii aare 1.15 mm and 0.915 mm for RG-8/U and RG- 213/U. In order to compare these ratings with Eq. (10.4.9), we took ato be the average of these two values, that is, a=1.03 mm. The required value of bto give a 50-ohm impedance isb=3.60 mm. Fig. 10.4.2 shows the attenuations calculated from Eq. (10.4.9) and the nominal ones from the table. We assumed copper conductors with σ=5.8×107S/m and polyethylene di- electric with n=1.5, so thatη=η0/n=376.73/1.5=251.15 Ω andc=c0/n=2×108 m/sec. The loss tangent was taken to be tan δ=0.0007. The conductor and dielectric attenuations αcandαdbecome equal around 2.3 GHz, and αddominates after that. It is evident that the useful operation of the cable is restricted to frequencies up to 1 GHz. Beyond that, the attenuations are too excessive and the cable may be used only for short lengths. /intersectionsq/unionsq504 10. Transmission Lines 0 1 2 3 4 5051015202530 f (GHz)dB/100 ftRG− 8/U and RG− 213/U total conductor dielectricnominal Fig. 10.4.2 Attenuation coefficient αversus frequency. Optimum Coaxial Cables Given a fixed outer-conductor radius b, one may ask three optimization questions: What is the optimum value of a, or equivalently, the ratio b/athat (a) minimizes the electric fieldEainside the guide (for fixed voltage V), (b) maximizes the power transfer PT(for fixedEafield), and (c) minimizes the conductor attenuation αc. The three quantities Ea,PT,αccan be thought of as functions of the ratio x=b/a and take the following forms: Ea=V bx lnx,PT=1 η|Ea|2πb2lnx x2,αc=Rs 2ηbx+1 lnx(10.4.10) Setting the derivatives of the three functions of xto zero, we obtain the three conditions: (a) ln x=1, (b) lnx=1/2, and (c) ln x=1+1/x, with solutions (a) b/a=e1=2.7183, (b)b/a=e1/2=1.6487 and (c) b/a=3.5911. Unfortunately, the three optimization problems have three different answers, and it is not possible to satisfy them simultaneously. The corresponding impedances Zfor the three values of b/aare 60 Ω, 30 Ω, and 76.7 Ω for an air-filled line and 40 Ω, 20 Ω, and 51 Ω for a polyethylene-filled line. The value of 50 Ω is considered to be a compromise between 30 and 76.7 Ω corre- sponding to maximum power and minimum attenuation. Actually, the minimum of αc is very broad and any neighboring value to b/a=3.5911 will result in an αcvery near its minimum. Higher Modes The TEM propagation mode is the dominant one and has no cutoff frequency. However,TE and TM modes with higher cutoff frequencies also exist in coaxial lines [865], withthe lowest being a TE 11mode with cutoff wavelength and frequency: λc=1.873π 2(a+b), fc=c λc=c0 nλc(10.4.11) 10.5. Two-Wire Lines 505 This is usually approximated by λc=π(a+b). Thus, the operation of the TEM mode is restricted to frequencies that are less than fc. Example 10.4.4: For the RG-58/U line of Example 10.4.2, we have a=0.406 mm andb=1.548 mm, resulting in λc=1.873π(a+b)/2=5.749 mm, which gives for the cutoff frequency fc=20/0.5749=34.79 GHz, where we used c=c0/n=20 GHz cm. For the RG-8/U and RG-213/U cables, we may use a=1.03 mm andb=3.60 as in Example 10.4.3, resulting in λc=13.622 mm, and cutoff frequency of fc=14.68 GHz. The above cutoff frequencies are far above the useful operating range over which the attenuation of the line is acceptable. /intersectionsq/unionsq 10.5 Two-Wire Lines The two-wire transmission line consists of two parallel cylindrical conductors of radius aseparated by distance dfrom each other, as shown in Fig. 10.5.1. Fig. 10.5.1 Two-wire transmission line. We assume that the conductors are held at potentials ±V/2 with charge per unit length±Q/prime. The electrostatic problem can be solved by the standard technique of re- placing the finite-radius conductors by two thin line-charges ±Q/prime. The locations b1andb2of the line-charges are determined by the requirement that the cylindrical surfaces of the original conductors be equipotential surfaces, the ideabeing that if these equipotential surfaces were to be replaced by the conductors, thefield patterns will not be disturbed. The electrostatic problem of the two lines is solved by invoking superposition and adding the potentials due to the two lines, so that the potential at the field point Pwill be: ϕ(ρ,φ)=−Q/prime 2π/epsilon1lnρ1−−Q/prime 2π/epsilon1lnρ2=Q/prime 2π/epsilon1ln/parenleftBigg ρ2 ρ1/parenrightBigg (10.5.1) where theρ1,ρ2are the distances from the line charges to P. From the triangles OP(+Q/prime)andOP(−Q/prime), we may express these distances in terms of the polar co- ordinatesρ,φ of the pointP:506 10. Transmission Lines ρ1=/radicalBig ρ2−2ρb1cosφ+b2 1,ρ 2=/radicalBig ρ2−2ρb2cosφ+b2 2 (10.5.2) Therefore, the potential function becomes: ϕ(ρ,φ)=Q/prime 2π/epsilon1ln/parenleftBigg ρ2 ρ1/parenrightBigg =Q/prime 2π/epsilon1ln⎛ ⎝/radicaltp/radicalvertex/radicalvertex/radicalbtρ2−2ρb2cosφ+b2 2 ρ2−2ρb1cosφ+b2 1⎞ ⎠ (10.5.3) In order that the surface of the left conductor at ρ=abe an equipotential surface, that is,ϕ(a,φ)=V/2, the ratioρ2/ρ1must be a constant independent of φ. Thus, we require that for some constant kand all angles φ: ρ2 ρ1/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle ρ=a=/radicaltp/radicalvertex/radicalvertex/radicalbta2−2ab2cosφ+b2 2 a2−2ab1cosφ+b2 1=k which can be rewritten as: a2−2ab2cosφ+b2 2=k2(a2−2ab1cosφ+b2 1) This will be satisfied for all φprovided we have: a2+b2 2=k2(a2+b2 1), b 2=k2b1 These may be solved for b1,b2in terms ofk: b2=ka, b 1=a k(10.5.4) The quantity kcan be expressed in terms of a,dby noting that because of symmetry, the charge−Q/primeis located also at distance b1from the center of the right conductor. Therefore,b1+b2=d. This gives the condition: b1+b2=d⇒a(k+k−1)=d⇒k+k−1=d a with solution for k: k=d 2a+/radicalBigg/parenleftbiggd 2a/parenrightbigg2 −1 (10.5.5) An alternative expression is obtained by setting k=eχ. Then, we have the condition: b1+b2=d⇒a(eχ+e−χ)=2acoshχ=d⇒χ=acosh/parenleftbiggd 2a/parenrightbigg (10.5.6) Becauseχ=lnk, we obtain for the potential value of the left conductor: ϕ(a,φ)=Q/prime 2π/epsilon1lnk=Q/prime 2π/epsilon1χ=1 2V This gives for the capacitance per unit length: 10.6. Distributed Circuit Model of a Transmission Line 507 C/prime=Q/prime V=π/epsilon1 χ=π/epsilon1 acosh/parenleftbiggd 2a/parenrightbigg (10.5.7) The corresponding line impedance and inductance are obtained from C/prime=/epsilon1η/Z andL/prime=μZ/η . We find: Z=η πχ=η πacosh/parenleftbiggd 2a/parenrightbigg L/prime=μ πχ=μ πacosh/parenleftbiggd 2a/parenrightbigg (10.5.8) In the common case when d/greatermucha, we have approximately k/asymptequald/a, and therefore, χ=lnk=ln(d/a) . Then,Zcan be written approximately as: Z=η πln(d/a) (10.5.9) To complete the electrostatic problem and determine the electric and magnetic fields of the TEM mode, we replace b2=akandb1=a/kin Eq. (10.5.3) and write it as: ϕ(ρ,φ)=Q/prime 2π/epsilon1ln⎛ ⎝k/radicalBigg ρ2−2akρcosφ+a2k2 ρ2k2−2akρcosφ+a2⎞ ⎠ (10.5.10) The electric and magnetic field components are obtained from: Eρ=ηHφ=−∂ϕ ∂ρ,Eφ=−ηHρ=−∂ϕ ρ∂φ(10.5.11) Performing the differentiations, we find: Eρ=−Q/prime 2π/epsilon1/bracketleftBigg ρ−akcosφ ρ2−2akρcosφ+a2k2−ρk2−akcosφ ρ2k2−2akρcosφ+a2/bracketrightBigg Eφ=−Q/prime 2π/epsilon1/bracketleftBigg aksinφ ρ2−2akcosφ+a2k2−aksinφ ρ2k2−2akρcosφ+a2/bracketrightBigg (10.5.12) The resistance per unit length and corresponding attenuation constant due to con- ductor losses are calculated in Problem 10.3: R/prime=Rs πad/radicalBig d2−4a2,αc=R/prime 2Z=Rs 2ηad acosh(d/2a)/radicalBig d2−4a2(10.5.13) 10.6 Distributed Circuit Model of a Transmission Line We saw that a transmission line has associated with it the parameters L/prime,C/primedescribing its lossless operation, and in addition, the parameters R/prime,G/primewhich describe the losses. It is possible then to define a series impedance Z/primeand a shunt admittance Y/primeper unit length by combining R/primewithL/primeandG/primewithC/prime:508 10. Transmission Lines Z/prime=R/prime+jωL/prime Y/prime=G/prime+jωC/prime(10.6.1) This leads to a so-called distributed-parameter circuit, which means that every in- finitesimal segment Δzof the line can be replaced by a series impedance Z/primeΔzand a shunt admittance Y/primeΔz, as shown in Fig. 10.6.1. The voltage and current at location z will beV(z) ,I(z) and at location z+Δz,V(z+Δz),I(z+Δz). Fig. 10.6.1 Distributed parameter model of a transmission line. The voltage across the branch a–bisVab=V(z+Δz) and the current through it, Iab=(Y/primeΔz)Vab=Y/primeΔzV(z+Δz). Applying Kirchhoff’s voltage and current laws, we obtain: V(z)=(Z/primeΔz)I(z)+Vab=Z/primeΔzI(z)+V(z+Δz) I(z)=Iab+I(z+Δz)=Y/primeΔzV(z+Δz)+I(z+Δz)(10.6.2) Using a Taylor series expansion, we may expand I(z+Δz) andV(z+Δz) to first order inΔz: I(z+Δz)=I(z)+I/prime(z)Δz V(z+Δz)=V(z)+V/prime(z)Δz andY/primeΔzV(z+Δz)=Y/primeΔzV(z) Inserting these expressions in Eq. (10.6.2) and matching the zeroth- and first-order terms in the two sides, we obtain the equivalent differential equations: V/prime(z)=−Z/primeI(z)=−(R/prime+jωL/prime)I(z) I/prime(z)=−Y/primeV(z)=−(G/prime+jωC/prime)V(z)(10.6.3) It is easily verified that the most general solution of this coupled system is express- ible as a sum of a forward and a backward moving wave: V(z)=V+e−jβcz+V−ejβcz I(z)=1 Zc/parenleftbig V+e−jβcz−V−ejβcz/parenrightbig (10.6.4) 10.7. Wave Impedance and Reflection Response 509 whereβc,Zcare the complex wavenumber and complex impedance: βc=−j/radicalBig (R/prime+jωL/prime)(G/prime+jωC/prime)=ω/radicalbig L/primeC/prime/radicalBigg/parenleftbigg 1−jR/prime ωL/prime/parenrightbigg/parenleftbigg 1−jG/prime ωC/prime/parenrightbigg Zc=/radicalBigg Z/prime Y/prime=/radicalBigg R/prime+jωL/prime G/prime+jωC/prime(10.6.5) The time-domain impulse response of such a line was given in Sec. 3.3. The real and imaginary parts of βc=β−jαdefine the propagation and attenuation constants. In the case of a lossless line, R/prime=G/prime=0, we obtain using Eq. (10.1.6): βc=ω/radicalbig L/primeC/prime=ω√μ/epsilon1=ω c=β, Zc=/radicalBigg L/prime C/prime=Z (10.6.6) In practice, we always assume a lossless line and then take into account the losses by assuming that R/primeandG/primeare small quantities, which can be evaluated by the appropriate expressions that can be derived for each type of line, as we did for the parallel-plate,coaxial, and two-wire lines. The lossless solution (10.6.4) takes the form: V(z)=V+e−jβz+V−ejβz=V+(z)+V−(z) I(z)=1 Z/parenleftbig V+e−jβz−V−ejβz/parenrightbig =1 Z/parenleftbig V+(z)−V−(z)/parenrightbig(10.6.7) This solution is identical to that of uniform plane waves of Chap. 5, provided we make the identifications: V(z)←→E(z) I(z)←→H(z) Z←→ηandV+(z)←→E+(z) V−(z)←→E−(z) 10.7 Wave Impedance and Reflection Response All the concepts of Chap. 5 translate verbatim to the transmission line case. For example, we may define the wave impedance and reflection response at locationz: Z(z)=V(z) I(z)=Z0V+(z)+V−(z) V+(z)−V−(z), Γ(z)=V−(z) V+(z)(10.7.1) To avoid ambiguity in notation, we will denote the characteristic impedance of the line byZ0. It follows from Eq. (10.7.1) that Z(z) andΓ(z) are related by: Z(z)=Z01+Γ(z) 1−Γ(z),Γ(z)=Z(z)−Z0 Z(z)+Z0(10.7.2) For a forward-moving wave, the conditions Γ(z)=0 andZ(z)=Z0are equivalent. The propagation equations of Z(z) andΓ(z) between two points z1,z2along the line separated by distance l=z2−z1are given by:510 10. Transmission Lines Z1=Z0Z2+jZ0tanβl Z0+jZ2tanβl/arrowdbllongbothΓ1=Γ2e−2jβl(10.7.3) where we have the relationships between Z1,Z2andΓ1,Γ2: Z1=Z01+Γ1 1−Γ1,Z 2=Z01+Γ2 1−Γ2(10.7.4) We may also express Z1in terms ofΓ2: Z1=Z01+Γ1 1−Γ1=Z01+Γ2e−2jβl 1−Γ2e−2jβl(10.7.5) The relationship between the voltage and current waves at points z1andz2is ob- tained by the propagation matrix: /bracketleftBigg V1 I1/bracketrightBigg =/bracketleftBigg cosβl jZ 0sinβl jZ−1 0sinβl cosβl/bracketrightBigg/bracketleftBigg V2 I2/bracketrightBigg (propagation matrix) (10.7.6) Similarly, we may relate the forward/backward voltages at the points z1andz2: /bracketleftBigg V1+ V1−/bracketrightBigg =/bracketleftBigg ejβl0 0e−jβl/bracketrightBigg/bracketleftBigg V2+ V2−/bracketrightBigg (propagation matrix) (10.7.7) It follows from Eq. (10.6.7) that V1±,V2±are related to V1,I1andV2,I2by: V1±=1 2(V1±Z0I1), V 2±=1 2(V2±Z0I2) (10.7.8) Fig. 10.7.1 depicts these various quantities. We note that the behavior of the line remains unchanged if the line is cut at the point z2and the entire right portion of the line is replaced by an impedance equal to Z2, as shown in the figure. This is so because in both cases, all the points z1to the left of z2see the same voltage-current relationship at z2, that is,V2=Z2I2. Sometimes, as in the case of designing stub tuners for matching a line to a load, it is more convenient to work with the wave admittances . DefiningY0=1/Z0,Y1= 1/Z1, andY2=1/Z2, it is easily verified that the admittances satisfy exactly the same propagation relationship as the impedances: Y1=Y0Y2+jY0tanβl Y0+jY2tanβl(10.7.9) As in the case of dielectric slabs, the half- and quarter-wavelength separations are of special interest. For a half-wave distance, we have βl=2π/2=π, which translates tol=λ/2, whereλ=2π/β is the wavelength along the line. For a quarter-wave, we haveβl=2π/4=π/2o rl=λ/4. Settingβl=πorπ/2 in Eq. (10.7.3), we obtain: 10.8. Two-Port Equivalent Circuit 511 Fig. 10.7.1 Length segment on infinite line and equivalent terminated line. l=λ 2⇒Z1=Z2,Γ 1=Γ2 l=λ 4⇒Z1=Z2 0 Z2,Γ 1=−Γ2(10.7.10) The MATLAB functions z2g.m and g2z.m computeΓfromZand conversely, by implementing Eq. (10.7.2). The functions gprop.m ,zprop.m andvprop.m implement the propagation equations (10.7.3) and (10.7.6). The usage of these functions is: G = z2g(Z,Z0); %ZtoΓ Z = g2z(G,Z0); %ΓtoZ G1 = gprop(G2,bl); % propagates Γ2toΓ1 Z1 = zprop(Z2,Z0,bl); % propagates Z2toZ1 [V1,I1] = vprop(V2,I2,Z0,bl); % propagates V2,I2toV1,I1 The parameter blisβl. The propagation equations and these MATLAB functions also work for lossy lines. In this case, βmust be replaced by the complex wavenumber βc=β−jα. The propagation phase factors become now: e±jβl−→e±jβcl=e±αle±jβl(10.7.11) 10.8 Two-Port Equivalent Circuit Any length-lsegment of a transmission line may be represented as a two-port equivalent circuit. Rearranging the terms in Eq. (10.7.6), we may write it in impedance-matrix form: /bracketleftBigg V1 V2/bracketrightBigg =/bracketleftBigg Z11Z12 Z21Z22/bracketrightBigg/bracketleftBigg I1 −I2/bracketrightBigg (impedance matrix) (10.8.1)512 10. Transmission Lines where the impedance elements are: Z11=Z22=−jZ0cotβl Z12=Z21=−jZ01 sinβl(10.8.2) The negative sign, −I2, conforms to the usual convention of having the currents coming intothe two-port from either side. This impedance matrix can also be realized in aT-section configuration as shown in Fig. 10.8.1. Fig. 10.8.1 Length-lsegment of a transmission line and its equivalent T-section. Using Eq. (10.8.1) and some trigonometry, the impedances Za,Zb,Zcof theT-section are found to be: Za=Z11−Z12=jZ0tan(βl/2) Zb=Z22−Z12=jZ0tan(βl/2) Zc=Z12=−jZ01 sinβl(10.8.3) The MATLAB function tsection.m implements Eq. (10.8.3). Its usage is: [Za,Zc] = tsection(Z0,bl); 10.9 Terminated Transmission Lines We can use the results of the previous section to analyze the behavior of a transmission line connected between a generator and a load. For example in a transmitting antennasystem, the transmitter is the generator and the antenna, the load. In a receiving system, the antenna is the generator and the receiver, the load. Fig. 10.9.1 shows a generator of voltage VGand internal impedance ZGconnected to the load impedance ZLthrough a length dof a transmission line of characteristic 10.9. Terminated Transmission Lines 513 Fig. 10.9.1 Terminated line and equivalent circuit. impedanceZ0. We wish to determine the voltage and current at the load in terms of the generator voltage. We assume that the line is lossless and hence Z0is real. The generator impedance is also assumed to be real but it does not have to be. The load impedance will have in general both a resistive and a reactive part, ZL=RL+jXL. At the load location, the voltage, current, and impedance are VL,IL,ZLand play the same role as the quantities V2,I2,Z2of the previous section. They are related by VL=ZLIL. The reflection coefficient at the load will be: ΓL=ZL−Z0 ZL+Z0/arrowdbllongbothZL=Z01+ΓL 1−ΓL(10.9.1) The quantities ZL,ΓLcan be propagated now by a distance dto the generator at the input to the line. The corresponding voltage, current, and impedance Vd,Id,Zdplay the role ofV1,I1,Z1of the previous section, and are related by Vd=ZdId. We have the propagation relationships: Zd=Z0ZL+jZ0tanβd Z0+jZLtanβd/arrowdbllongbothΓd=ΓLe−2jβd(10.9.2) where Γd=Zd−Z0 Zd+Z0/arrowdbllongbothZd=Z01+Γd 1−Γd=Z01+ΓLe−2jβd 1−ΓLe−2jβd(10.9.3) At the line input, the entire length- dline segment and load can be replaced by the impedanceZd, as shown in Fig. 10.9.1. We have now a simple voltage divider circuit. Thus, Vd=VG−IdZG=VGZd ZG+Zd,Id=VG ZG+Zd(10.9.4) Once we have Vd,Idin terms ofVG, we can invert the propagation matrix (10.7.6) to obtain the voltage and current at the load:514 10. Transmission Lines /bracketleftBigg VL IL/bracketrightBigg =/bracketleftBigg cosβd−jZ0sinβd −jZ−1 0sinβd cosβd/bracketrightBigg/bracketleftBigg Vd Id/bracketrightBigg (10.9.5) It is more convenient to express Vd,Idin terms of the reflection coefficients Γdand ΓG, the latter being defined by: ΓG=ZG−Z0 ZG+Z0/arrowdbllongbothZG=Z01+ΓG 1−ΓG(10.9.6) It is easy to verify using Eqs. (10.9.3) and (10.9.6) that: ZG+Zd=2Z01−ΓGΓd (1−ΓG)(1−Γd),ZG+Z0=2Z01 1−ΓG From these, it follows that: Vd=VGZ0 ZG+Z01+Γd 1−ΓGΓd,Id=VG ZG+Z01−Γd 1−ΓGΓd(10.9.7) whereΓdmay be replaced by Γd=ΓLe−2jβd. If the line and load are matched so that ZL=Z0, thenΓL=0 andΓd=0 andZd=Z0for any distance d. Eq. (10.9.7) then reduces to: Vd=VGZ0 ZG+Z0,Id=VG ZG+Z0(matched load) (10.9.8) In this case, there is only a forward-moving wave along the line. The voltage and current at the load will correspond to the propagation of these quantities to location l=0, which introduces a propagation phase factor e−jβd: V0=VGZ0 ZG+Z0e−jβd,I 0=VG ZG+Z0e−jβd(matched load) (10.9.9) whereV0,I0denoteVL,ILwhenZL=Z0. It is convenient also to express VLdirectly in terms ofVdand the reflection coefficients ΓdandΓL. We note that: VL=VL+(1+ΓL), VL+=Vd+e−jβd,Vd+=Vd 1+Γd It follows that the voltage VLand currentIL=VL/ZLare: VL=Vde−jβd1+ΓL 1+Γd,IL=Ide−jβd1−ΓL 1−Γd(10.9.10) ExpressingVLand alsoIL=VL/ZLdirectly in terms of VG, we have: VL=VGZ0 ZG+Z01+ΓL 1−ΓGΓde−jβd,IL=VG ZG+Z01−ΓL 1−ΓGΓde−jβd(10.9.11) 10.10. Power Transfer from Generator to Load 515 It should be emphasized that drefers to the fixed distance between the generator and the load. For any other distance, say l, from the load (or, distance z=d−lfrom the generator,) the voltage and current can be expressed in terms of the load voltageand current as follows: Vl=VLejβl1+Γl 1+ΓL,Il=ILejβl1−Γl 1−ΓL,Γl=ΓLe−2jβl(10.9.12) 10.10 Power Transfer from Generator to Load The total power delivered by the generator is dissipated partly in its internal resistance and partly in the load. The power delivered to the load is equal (for a lossless line) tothe net power traveling to the right at any point along the line. Thus, we have: Ptot=Pd+PG=PL+PG (10.10.1) This follows from VG=Vd+IdZG, which implies VGI∗ d=VdI∗ d+ZG|Id|2(10.10.2) Eq. (10.10.1) is a consequence of (10.10.2) and the definitions: Ptot=1 2Re(V∗ GId)=1 2Re/bracketleftbig (Vd+ZGId)∗Id/bracketrightbig PG=1 2Re(ZGIdI∗ d)=1 2Re(ZG)|Id|2 Pd=1 2Re(V∗ dId)=1 2Re(V∗ LIL)=PL(10.10.3) The last equality follows from Eq. (10.9.5) or from Vd±=VL±e±jβd: 1 2Re(V∗ dId)=1 2Z0/parenleftbig |Vd+|2−|Vd−|2/parenrightbig =1 2Z0/parenleftbig |VL+|2−|VL−|2/parenrightbig =1 2Re(V∗ LIL) In the special case when the generator and the load are matched to the line, so that ZG=ZL=Z0, then we find the standard result that half of the generated power is delivered to the load and half is lost in the internal impedance. Using Eq. (10.9.8) with ZG=Z0, we obtainVd=IdZG=VG/2, which gives: Ptot=|VG|2 4Z0,PG=|VG|2 8Z0=1 2Ptot,Pd=PL=|VG|2 8Z0=1 2Ptot (10.10.4) Example 10.10.1: A loadZL=50+j10 Ω is connected to a generator VG=10∠0ovolts with a 100-ft (30.48 m) cable of a 50-ohm transmission line. The generator’s internal impedance is 20 ohm, the operating frequency is 10 MHz, and the velocity factor of the line, 2/3. Determine the voltage across the load, the total power delivered by the generator, the power dissipated in the generator’s internal impedance and in the load.516 10. Transmission Lines Solution: The propagation speed is c=2c0/3=2×108m/sec. The line wavelength λ=c/f= 20 m and the propagation wavenumber β=2π/λ=0.3142 rads/m. The electrical length isd/λ=30.48/20=1.524 and the phase length βd=9.5756 radians. Next, we calculate the reflection coefficients: ΓL=ZL−Z0 ZL+Z0=0.0995∠84.29o,ΓG=ZG−Z0 ZG+Z0=−0.4286 andΓd=ΓLe−2jβd=0.0995∠67.01o. It follows that: Zd=Z01+Γd 1−Γd=53.11+j9.83,Vd=VGZd ZG+Zd=7.31+j0.36=7.32∠2.83o The voltage across the load will be: VL=Vde−jβd1+ΓL 1+Γd=−7.09+j0.65=7.12∠174.75oV The current through the generator is: Id=Vd Zd=0.13−j0.02=0.14∠−7.66oA It follows that the generated and dissipated powers will be: Ptot=1 2Re(V∗ GId)=0.6718 W PG=1 2Re(ZG)|Id|2=0.1838 W PL=Pd=1 2Re(V∗ dId)=0.4880 W We note that Ptot=PG+PL. /intersectionsq/unionsq If the line is lossy, with a complex wavenumber βc=β−jα, the powerPLat the output of the line is less than the power Pdat the input of the line. Writing Vd±= VL±e±αde±jβd, we find: Pd=1 2Z0/parenleftbig |Vd+|2−|Vd−|2/parenrightbig =1 2Z0/parenleftbig |VL+|2e2αd−|VL−|2e−2αd/parenrightbig PL=1 2Z0/parenleftbig |VL+|2−|VL−|2/parenrightbig We note that Pd>PLfor allΓL. In terms of the incident forward power at the load, Pinc=|VL+|2/2Z0, we have: Pd=Pinc/parenleftbig e2αd−|ΓL|2e−2αd/parenrightbig =Pince2αd/parenleftbig 1−|Γd|2/parenrightbig PL=Pinc/parenleftbig 1−|ΓL|2/parenrightbig (10.10.5) where|Γd|=|ΓL|e−2αd. The total attenuation or loss of the line is Pd/PL(the inverse PL/Pdis the total gain, which is less than one.) In decibels, the loss is: 10.11. Open- and Short-Circuited Transmission Lines 517 L=10 log10/parenleftbiggPd PL/parenrightbigg =10 log10/parenleftBigg e2αd−|ΓL|2e−2αd 1−|ΓL|2/parenrightBigg (total loss) (10.10.6) If the load is matched to the line, ZL=Z0, so thatΓL=0, the loss is referred to as the matched-line loss and is due only to the transmission losses along the line: LM=10 log10/parenleftbig e2αd/parenrightbig =8.686αd (matched-line loss) (10.10.7) Denoting the matched-line loss in absolute units by a=10LM/10=e2αd, we may write Eq. (10.10.6) in the equivalent form: L=10 log10/parenleftBigg a2−|ΓL|2 a(1−|ΓL|2)/parenrightBigg (total loss) (10.10.8) The additional loss due to the mismatched load is the difference: L−LM=10 log10/parenleftBigg 1−|ΓL|2e−4αd 1−|ΓL|2/parenrightBigg =10 log10/parenleftBigg 1−|Γd|2 1−|ΓL|2/parenrightBigg (10.10.9) Example 10.10.2: A 150 ft long RG-58 coax is connected to a load ZL=25+50johm. At the operating frequency of 10 MHz, the cable is rated to have 1.2 dB/100 ft of matched-line loss. Determine the total loss of the line and the excess loss due to the mismatched load. Solution: The matched-line loss of the 150 ft cable is LM=150×1.2/100=1.8 dB or in absolute units,a=101.8/10=1.51. The reflection coefficient has magnitude computed with the help of the MATLAB function z2g: |ΓL|=abs(z2g(25+50j,50)=0.62 It follows that the total loss will be: L=10 log10/parenleftBigg a2−|ΓL|2 a(1−|ΓL|2)/parenrightBigg =10 log10/parenleftBigg 1.512−0.622 1.51(1−0.622)/parenrightBigg =3.1d B The excess loss due to the mismatched load is 3 .1−1.8=1.3 dB. At the line input, we have|Γd|=|ΓL|e−2αd=|ΓL|/a=0.62/1.51=0.41. Therefore, from the point of view of the input the line appears to be more matched. /intersectionsq/unionsq 10.11 Open- and Short-Circuited Transmission Lines Open- and short-circuited transmission lines are widely used to construct resonant cir- cuits as well as matching stubs. They correspond to the special cases for the loadimpedance: ZL=∞ for an open-circuited line and ZL=0 for a short-circuited one. Fig. 10.11.1 shows these two cases. Knowing the open-circuit voltage and the short-circuit current at the end terminals a,b, allows us also to replace the entire left segment of the line, including the generator,518 10. Transmission Lines Fig. 10.11.1 Open- and short-circuited line and Th ´evenin-equivalent circuit. with a Th ´evenin-equivalent circuit. Connected to a load impedance ZL, the equivalent circuit will produce the same load voltage and current VL,ILas the original line and generator. SettingZL=∞ andZL=0 in Eq. (10.9.2), we obtain the following expressions for the wave impedance Zlat distancelfrom the open- or short-circuited termination: Zl=−jZ0cotβl Zl=jZ0tanβl(open-circuited) (short-circuited)(10.11.1) The corresponding admittances Yl=1/Zlwill be: Yl=jY0tanβl Yl=−jY0cotβl(open-circuited) (short-circuited)(10.11.2) To determine the Th ´evenin-equivalent circuit that replaces everything to the left of the terminals a,b, we must find the open-circuit voltage Vth, the short-circuit current Isc, and the Th ´evenin impedance Zth. The impedance Zthcan be determined either by Zth=Vth/Isc, or by disconnecting the generator and finding the equivalent impedance looking to the left of the terminals a,b. It is obtained by propagating the generator impedance ZGby a distance d: Zth=Z0ZG+jZ0tanβd Z0+jZGtanβd=Z01+Γth 1−Γth,Γ th=ΓGe−2jβd(10.11.3) The open-circuit voltage can be determined from Eq. (10.9.11) by setting ZL=∞, which implies that ΓL=1,Γd=e−2jβd, andΓGΓd=ΓGe−2jβd=Γth. The short- circuit current is also obtained from (10.9.11) by setting ZL=0, which gives ΓL=−1, Γd=−e−2jβd, andΓGΓd=−ΓGe−2jβd=−Γth. Then, we find: 10.11. Open- and Short-Circuited Transmission Lines 519 Vth=VGZ0 ZG+Z02e−jβd 1−Γth,I sc=VG ZG+Z02e−jβd 1+Γth(10.11.4) It follows that Vth/Isc=Zth, as given by Eq. (10.11.3). A more convenient way of writing Eq. (10.11.4) is by noting the relationships: 1−Γth=2Z0 Zth+Z0,1+Γth=2Zth Zth+Z0 Then, Eq. (10.11.4) becomes: Vth=V0Zth+Z0 Z0,I sc=I0Zth+Z0 Zth(10.11.5) whereV0,I0are the load voltage and currents in the matched case, given by Eq. (10.9.9). The intuitive meaning of these expressions can be understood by writing them as: V0=VthZ0 Zth+Z0,I 0=IscZth Zth+Z0(10.11.6) These are recognized to be the ordinary voltage and current dividers obtained by connecting the Th ´evenin and Norton equivalent circuits to the matched load impedance Z0, as shown in Fig. 10.11.2. Fig. 10.11.2 Th´evenin and Norton equivalent circuits connected to a matched load. The quantities V0,I0are the same as those obtained by connecting the actual line to the matched load, as was done in Eq. (10.9.9). An alternative way of determining the quantities VthandZthis by replacing the length-dtransmission line segment by its T-section equivalent circuit, as shown in Fig. 10.11.3. The Th ´evenin equivalent circuit to the left of the terminals a,bis easily determined by shorting the generator and finding the Th ´evenin impedance and then finding the open-circuit voltage. We have: Zth=Zb+Zc(Za+ZG) Zc+Za+ZG,V th=VGZc Zc+Za+ZG(10.11.7)520 10. Transmission Lines Fig. 10.11.3 T-section and Th ´evenin equivalent circuits. whereZa,Zb,Zcfor a length-dsegment are given by Eq. (10.8.3): Za=Zb=jZ0tan/parenleftbiggβd 2/parenrightbigg ,Zc=−jZ01 sinβd It is straightforward to verify that the expressions in Eq. (10.11.7) are equivalent to those in Eq. (10.11.3) and (10.11.4). Example 10.11.1: For the generator, line, and load of Example 10.10.1, determine the Th ´evenin equivalent circuit. Using this circuit determine the load voltage. Solution: We work with the T-section approach. The following MATLAB call gives ZaandZc, withZ0=50 andβd=9.5756: [Za,Zc]=tsection(50,9.5756)=[−661.89j,332.83j] Then, Eq. (10.11.7) gives with Zb=Za: Zth=Zb+Zc(Za+ZG) Zc+Za+ZG=20.39+j6.36 Ω Vth=VGZc Zc+Za+ZG=−10.08+j0.61=10.10∠176.52oV Alternatively, Zthcan be computed by propagating ZG=20 by a distance d: Zth=zprop(20,50,9.5756)=20.39+j6.36 Ω The load voltage is found from the Th ´evenin circuit: VL=VthZL ZL+Zth=−7.09+j0.65=7.12∠174.75oV which agrees with that found in Example 10.10.1. /intersectionsq/unionsq 10.12 Standing Wave Ratio The line voltage at a distance lfrom the load is given by Eq. (10.9.12), which can be written as follows in terms of the forward wave VL+=VL/(1+ΓL): Vl=VL+ejβl(1+Γl) (10.12.1) 10.12. Standing Wave Ratio 521 The magnitude of Vlwill be: |Vl|=|VL+||1+Γl|=|VL+||1+ΓLe−2jβl| (10.12.2) It follows that |Vl|will vary sinusoidally as a function of l. Its limits of variation are determined by noting that the quantity |1+Γl|varies between: 1−|ΓL|=1−|Γl|≤|1+Γl|≤1+|Γl|=1+|ΓL| where we used |Γl|=|ΓL|. Thus,|Vl|will vary over the limits: Vmin≤|Vl|≤Vmax (10.12.3) where Vmin=|VL+|−|VL−|=|VL+|/parenleftbig 1−|ΓL|/parenrightbig Vmax=|VL+|+|VL−|=|VL+|/parenleftbig 1+|ΓL|/parenrightbig (10.12.4) We note that the reflection coefficient at a load ZL=RL+jXLhas always magnitude less than unity, |ΓL|≤1. Indeed, this follows from the positivity of RLand the following property: ZL=Z01+ΓL 1−ΓL⇒RL=Re(ZL)=Z01−|ΓL|2 |1−ΓL|2(10.12.5) The voltage standing wave ratio (SWR) of a terminated transmission line is a measure of the degree of matching of the line to the load and is defined as the ratio of themaximum to minimum voltage along the line: S=Vmax Vmin=1+|ΓL| 1−|ΓL|/arrowdbllongboth|ΓL|=S−1 S+1(10.12.6) Because|ΓL|≤1, the SWR will always be S≥1. A matched load, ΓL=0, hasS=1. The more unmatched the load is, the larger the SWR. Indeed, S→∞ as|ΓL|→1. A matched line has Vmin=|Vl|=Vmaxat all pointsl, and is sometimes referred to as a flat line. The MATLAB function swr.m calculates the SWR from Eq. (10.12.6): S = swr(Gamma); % calculates SWR from reflection coefficient Γ The SWR can be used to quantify the amount of power delivered to the load. The percentage of reflected power from the load is |ΓL|2. Therefore, the percentage of the power delivered to the load relative to the incident power will be: PL Pinc=1−|ΓL|2=4S (S+1)2(10.12.7) The larger the SWR, the smaller the percentage of delivered power. For example, if S=9, the reflection coefficient will have magnitude |ΓL|=0.8, resulting in 1 −|ΓL|2= 0.36, that is, only 36 percent of the incident power gets transferred to the load.522 10. Transmission Lines Example 10.12.1: If the reflected wave at the load of a transmission line is 6 dB below the incident wave, what is the SWR at the load? What percentage of the incident power getstransferred to the load? Solution: The relative power levels of the reflected and incident waves will be: |ΓL|2=|V−|2 |V+|2=10−6/10=1 4⇒|ΓL|=1 2⇒S=1+0.5 1−0.5=3 The fraction of power transferred to the load is 1 −|ΓL|2=0.75, or 75 percent. /intersectionsq/unionsq If both the line and load impedances are real-valued, then the standing wave ratio is S=ZL/Z0ifZL≥Z0, andS=Z0/ZL,i fZL≤Z0. This follows from the identity: S=1+|ΓL| 1−|ΓL|=|ZL+Z0|+|ZL−Z0| |ZL+Z0|−|ZL−Z0|=max(ZL,Z0) min(ZL,Z0)(10.12.8) or, explicitly: S=1+|ΓL| 1−|ΓL|=⎧ ⎪⎪⎪⎨ ⎪⎪⎪⎩Z L Z0,ifZL≥Z0 Z0 ZL,ifZL≤Z0(10.12.9) 10.13 Determining an Unknown Load Impedance Often a transmission line is connected to an unknown impedance, and we wish to de- termine that impedance by making appropriate measurements of the voltage along theline. The SWR can be readily determined by measuring |Vl|and finding its maximum and minimum values VmaxandVmin. From the SWR, we then determine the magnitude of the reflection coefficient |ΓL|. The phase of ΓLcan be determined by finding the locations along the line at which a voltage maximum or a voltage minimum is measured. If θLis the required phase, so thatΓL=|ΓL|ejθL, then we have: |Vl|=|VL+||1+Γl|=|VL+||1+ΓLe−2jβl|=|VL+|/vextendsingle/vextendsingle1+|ΓL|ej(θL−2βl)/vextendsingle/vextendsingle At all locations lfor whichθL−2βl=±2πn, wherenis an integer, we will have Γl=|ΓL|and|Vl|will be equal to Vmax. Similarly, at all locations for which θL−2βl= ±(2n+1)π, we will have Γl=−|ΓL|and|Vl|will be equal to Vmin. We note that two successive maxima, or two successive minima, are separated by a distanceλ/2 and a maximum is separated by the next minimum by a distance λ/4, so that|lmax−lmin|=λ/4. Once such distances lmax,lminhave been determined, the full reflection coefficient can be constructed from ΓL=Γle2jβl, whereΓl=±|ΓL|depending on using a maximum- or minimum-voltage distance l. FromΓLand the knowledge of the line impedance Z0, the load impedance ZLcan be computed. Thus, we have: 10.13. Determining an Unknown Load Impedance 523 ΓL=|ΓL|ejθL=|ΓL|e2jβlmax=−|ΓL|e2jβlmin⇒ZL=Z01+ΓL 1−ΓL(10.13.1) If 0≤θL≤π, the locations for the closest maxima and minima to the load are determined from the conditions: θL−2βlmax=0,θL−2βlmin=−π resulting in the distances: lmax=θL 4πλ, l min=θL+π 4πλ,(0≤θL≤π) (10.13.2) Similarly, if−π≤θL≤0, we must solve θL−2βlmax=−2πandθL−2βlmin=−π: lmax=θL+2π 4πλ, l min=θL+π 4πλ,(−π≤θL≤0) (10.13.3) Of course, one wants to solve for θLin terms of the measured lmaxorlmin. Usinglmin is more convenient than using lmaxbecauseθLis given by the same expression in both cases. The lengths lmax,lminmay be assumed to be less than λ/2 (if not, we may subtract enough multiples of λ/2 until they are.) Expressing θLin terms of the measured lmin, we have: θL=4πlmin λ−π=2βlmin−π (10.13.4) Alternatively, we have in terms of lmax: θL=⎧ ⎪⎪⎪⎨ ⎪⎪⎪⎩4πlmax λ=2βlmax if 0≤lmax≤λ 4 4πlmax λ−2π=2βlmax−2πifλ 4≤lmax≤λ 2(10.13.5) Example 10.13.1: A 50-ohm line is connected to an unknown impedance. Voltage measure- ments along the line reveal that the maximum and minimum voltage values are 1.75 and 0.25 volts, respectively. Moreover, the closest distance to the load at which a voltage max- imum is observed is 0 .125λ. Determine the reflection coefficient ΓL, the load impedance ZL, and the closest distance to the load at which a voltage minimum is observed. For another load, the same maxima and minima are observed, but now the closest distance to the load at which a minimum is observed is 0 .125λ. DetermineΓLandZL. Solution: The SWR is determined to be S=Vmax/Vmin=1.75/0.25=7. Then, the magnitude of the reflection coefficient is found to be |ΓL|=(S−1)/(S+1)=(7−1)/(7+1)=0.75. Given that at lmax=λ/8 we observe a voltage maximum, we compute the phase from Eq. (10.13.5), θL=2βlmax=4π/8=π/2. Then, the reflection coefficient will be:524 10. Transmission Lines ΓL=|ΓL|ejθL=0.75ejπ/2=0.75j It follows that the load impedance will be: ZL=Z01+ΓL 1−ΓL=501+0.75j 1−0.75j=14+48jΩ The closest voltage minimum will occur at lmin=lmax+λ/4=0.375λ=3λ/8. Alter- natively, we could have determined the phase from Eq. (10.13.4), θL=2βlmin−π= 4π(3/8)−π=π/2. The left graph of Fig. 10.13.1 shows a plot of |Vl|versusl. 0 0.25 0.5 0.75 1 1.25 1.500.250.50.7511.251.51.752 l/ λλ|Vl|Standing Wave Pattern 0 0.25 0.5 0.75 1 1.25 1.500.250.50.7511.251.51.752 l/ λλ|Vl|Standing Wave Pattern Fig. 10.13.1 Standing wave patterns. Note the locations of the closest voltage maxima and minima to the load, that is λ/8 and 3λ/8. In the second case, we are given lmin=λ/8. It follows that θL=2βlmin−π= π/2−π=−π/2. Alternatively, we may work with lmax=lmin+λ/4=3λ/8. Because lmax>λ / 4, Eq. (10.13.5) will give θL=2βlmax−2π=4π(3/8)−2π=−π/2. The reflection coefficient and load impedance will be: ΓL=|ΓL|ejθL=0.75e−jπ/2=−0.75j⇒ZL=14−48jΩ The right graph of Fig. 10.13.1 depicts the standing wave pattern in this case. /intersectionsq/unionsq It is interesting also to determine the wave impedances at the locations along the line at which we have voltage maxima or minima, that is, at l=lmaxorlmin. The answers are expressed in terms of the SWR. Indeed, at l=lmax, we haveΓl=|ΓL|which gives: Zmax=Z01+Γl 1−Γl=Z01+|ΓL| 1−|ΓL|=SZ0 (10.13.6) Similarly, atl=lmin, we haveΓl=−|ΓL|and find: Zmin=Z01+Γl 1−Γl=Z01−|ΓL| 1+|ΓL|=1 SZ0 (10.13.7) 10.13. Determining an Unknown Load Impedance 525 We note that ZmaxZmin=Z2 0, as is expected because the points lmaxandlminare separated by a quarter-wavelength distance λ/4. Because atlmaxandlminthe wave impedances are real-valued , these points can be used as convenient locations at which to insert a quarter-wave transformer to match aline with real Z0to a complex load ZL. GivenθL, the required locations are determined from Eq. (10.13.2) or (10.13.3). We discuss this matching method later on. The MATLAB function lmin.m calculates the locations lminandlmaxfrom Eqs. (10.13.2) and (10.13.3), and the corresponding impedances ZminandZmax. It has usage: [lm,Zm] = lmin(ZL,Z0,’min’); % locations of voltage minima [lm,Zm] = lmin(ZL,Z0,’max’); % locations of voltage maxima For a lossless line the power delivered to the load can be measured at any point l along the line, and in particular, at lmaxandlmin. Then, Eq. (10.12.7) can be written in the alternative forms: PL=1 2Z0/parenleftbig |VL+|2−|VL−|2/parenrightbig =VmaxVmin 2Z0=V2 min 2Zmin=V2 max 2Zmax=V2 max 2SZ0(10.13.8) The last expression shows that for a given maximum voltage that can be supported along a line, the power transmitted to the load is Stimes smaller than it could be if the load were matched. Conversely, for a given amount PLof transmitted power, the maximum voltage will beVmax=/radicalbig 2SPLZ0. One must ensure that for a highly unmatched load, Vmaxremain less than the breakdown voltage of the line. If the line is lossy, measurements of the SWR along its length will give misleading results. Because the reflected power attenuates as it propagates backwards away from the load, the SWR will be smaller at the line input than at the load. For a lossy line with βc=β−jα, the reflection coefficient at the line input will be: Γd=ΓLe−2(α+jβ)d, which gives for the input SWR: Sd=1+|Γd| 1−|Γd|=1+|ΓL|e−2αd 1−|ΓL|e−2αd=e2αd+|ΓL| e2αd−|ΓL|=a+|ΓL| a−|ΓL|(10.13.9) where we expressed it in terms of the matched-line loss of Eq. (10.10.7). Example 10.13.2: For the RG-58 coax cable of Example 10.10.2, we find the SWRs: SL=1+|ΓL| 1−|ΓL|=1+0.62 1−0.62=4.26,Sd=1+|Γd| 1−|Γd|=1+0.41 1−0.41=2.39 If one does not know that the line is lossy, and measures the SWR at the line input, one would think that the load is more matched than it actually is. /intersectionsq/unionsq Example 10.13.3: The SWR at the load of a line is 9. If the matched-line loss is 10 dB, what is the SWR at the line input?526 10. Transmission Lines Solution: We calculate the reflection coefficient at the load: |ΓL|=S−1 S+1=9−1 9+1=0.8 The matched-line loss is a=10LM/10=1010/10=10. Thus, the reflection coefficient at the input will be |Γd|=|ΓL|/a=0.8/10=0.08. The corresponding SWR will be S=(1+0.08)/(1−0.08)=1.17. /intersectionsq/unionsq Example 10.13.4: A 50-ohm line feeds a half-wave dipole antenna with impedance of 73 +j42.5 ohms. The line has matched-line loss of 3 dB. What is the total loss of the line? What is the SWR at the load and at the line input? If the line length is doubled, what is the matched-line loss, the total loss, the input and load SWRs? Solution: The matched-line loss in absolute units is a=103/10=2. Using the MATLAB functions z2gandswr, we compute the reflection coefficient at the load and its SWR: |ΓL|=/vextendsingle/vextendsingle/vextendsingle/vextendsingleZ L−Z0 ZL+Z0/vextendsingle/vextendsingle/vextendsingle/vextendsingle=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle 73+j42.5−50 73+j42.5+50/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle= abs(z2g(73+42.5j,50))=0.3713 The SWR will be S=swr(0.3713)=2.1814. The reflection coefficient at the line input will be|Γd|=|ΓL|e−2αd=|ΓL|/a=0.1857, and its SWR, S=swr(0.1857)=1.4560. If the line length is doubled, the matched-line loss in dB will double to 6 dB, since it is given byLM=8.686αd. In absolute units, it is a=22=4. The corresponding reflection coefficient at the line input will be |Γd|=|ΓL|/a=0.0928, and its SWR, S=swr(0.0928)=1.2047. /intersectionsq/unionsq 10.14 Smith Chart The relationship between the wave impedance Zand the corresponding reflection re- sponseΓalong a transmission line Z0can be stated in terms the normalized impedance z=Z/Z 0as follows: Γ=z−1 z+1/arrowdbllongbothz=1+Γ 1−Γ(10.14.1) It represents a mapping between the complex impedance z-plane and the complex reflection coefficient Γ-plane, as shown in Fig. 10.14.1. The mapping is similar to the bilinear transformation mapping in linear system theory between the s-plane (playing the role of the impedance plane) and the z-plane of the z-transform (playing the role of theΓ-plane.) A complex impedance z=r+jxwith positive resistive part, r>0, gets mapped onto a pointΓthat lies inside the unit-circle in the Γ-plane, that is, satisfying |Γ|<1. An entire resistance line z=r(a vertical line on the z-plane) gets mapped onto a circle on the Γ-plane that lies entirely inside the unit-circle, if r> 0. Similarly, a reactance line z=jx(a horizontal line on the z-plane) gets mapped onto a circle on the Γ-plane, a portion of which lies inside the unit-circle. 10.14. Smith Chart 527 Fig. 10.14.1 Mapping between z-plane andΓ-plane. The Smith chart is a graphical representation of the Γ-plane with a curvilinear grid of constant resistance and constant reactance circles drawn inside the unit-circle. Ineffect, the Smith chart is a curvilinear graph paper. Any reflection coefficient point Γfalls at the intersection of a resistance and a reac- tance circle,r,x, from which the corresponding impedance can be read off immediately asz=r+jx. Conversely, given z=r+jxand finding the intersection of the r,x circles, the complex point Γcan be located and its value read off in polar or cartesian coordinates. To determine the centers and radii of the resistance and reactance circles, we use the result that a circle with center Cand radiusRon theΓ-plane has the following two equivalent representations: |Γ|2−C∗Γ−CΓ∗=B/arrowdbllongboth|Γ−C|=R, whereB=R2−|C|2(10.14.2) Settingz=r+jxin Eq. (10.14.1) and extracting the real and imaginary parts, we can writerandxin terms ofΓ, as follows: r=Rez=1−|Γ|2 |1−Γ|2,x=Imz=j(Γ∗−Γ) |1−Γ|2(10.14.3) In particular, the expression for the resistive part implies that the condition r>0i s equivalent to |Γ|<1. Ther,xcircles are obtained by putting Eqs. (10.14.3) in the form of Eq. (10.14.2). We have: r|Γ−1|2=1−|Γ|2⇒r/parenleftbig |Γ|2−Γ−Γ∗+1/parenrightbig =1−|Γ|2 and rearranging terms: |Γ|2−r r+1Γ−r 1+rΓ∗=1−r 1+r⇒/vextendsingle/vextendsingle/vextendsingle/vextendsingleΓ−r 1+r/vextendsingle/vextendsingle/vextendsingle/vextendsingle2 =1−r 1+r+r2 (1+r)2=/parenleftbigg1 1+r/parenrightbigg2528 10. Transmission Lines Similarly, we have x|Γ−1|2=j(Γ∗−Γ)⇒x/parenleftbig |Γ|2−Γ−Γ∗+1/parenrightbig =j(Γ∗−Γ) which can be rearranged as: |Γ|2−/parenleftbigg 1−j x/parenrightbigg Γ−/parenleftbigg 1+j x/parenrightbigg Γ∗=−1⇒/vextendsingle/vextendsingle/vextendsingle/vextendsingleΓ−/parenleftbigg 1+j x/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle2 =−1+/parenleftbigg 1+1 x2/parenrightbigg =/parenleftbigg1 x/parenrightbigg2 To summarize, the constant resistance and reactance circles are: /vextendsingle/vextendsingle/vextendsingle/vextendsingleΓ−r 1+r/vextendsingle/vextendsingle/vextendsingle/vextendsingle=1 1+r(resistance circles) /vextendsingle/vextendsingle/vextendsingle/vextendsingleΓ−/parenleftbigg 1+j x/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle=1 |x|(reactance circles)(10.14.4) The centers of the resistance circles are on the positive half of the real axis on the Γ- plane, lying between 0 ≤Γ≤1. Whenr=0, the impedance circle is the entire unit-circle with center at Γ=0. Asrincreases, the radii become smaller and the centers move towardsΓ=1. The centers of the reactance circles lie on the tangent of the unit-circle atΓ=1. Example 10.14.1: Fig. 10.14.2 depicts the resistance and reactance circles for the following values ofr,x: r=[0.2,0.5,1,2,5], x=[0.2,0.5,1,2,5] Because the point Ais at the intersection of the r=0.2 andx=0.5 circles, the corre- sponding impedance will be zA=0.2+0.5j. We list below the impedances and reflection coefficients at the points A,B,C,D,E,S,P,O : zA=0.2+0.5j, ΓA=−0.420+0.592j=0.726∠125.37o zB=0.5−j, ΓB=0.077−0.615j=0.620∠−82.88o zC=2−2j, Γ C=0.539−0.308j=0.620∠−29.74o zD=j, Γ D=j=1∠90o zE=−j, Γ E=−j=1∠−90o (short circuit) zS=0,Γ S=−1=1∠180o (open circuit) zP=∞,Γ P=1=1∠0o (matched) zO=1,Γ O=0=0∠0o The pointsSandPcorrespond to a short-circuited and an open-circuited impedance. The center of the Smith chart at point Ocorresponds to z=1, that is, an impedance matched to the line. /intersectionsq/unionsq The Smith chart helps one visualize the wave impedance as one moves away from or towards a load. Assuming a lossless line, the wave impedance and correspondingreflection response at a distance lfrom the load are given by: zl=zL+jtanβl 1+jzLtanβl/arrowdbllongbothΓl=e−2jβlΓL (10.14.5) 10.14. Smith Chart 529 Fig. 10.14.2 Smith chart example. The magnitude of Γlremains constant as lvaries, indeed, |Γl|=|ΓL|. On the Smith chart, this represents a circle centered at the origin Γ=0 of radius|ΓL|. Such circles are called constant SWR circles because the SWR is related to the circle radius by S=1+|ΓL| 1−|ΓL| The relative phase angle between ΓlandΓLis negative,−2βl, and therefore, the point Γlmoves clockwise along the constant SWR circle, as shown in Fig. 10.14.3. Conversely, iflis decreasing towards the load, the point Γlwill be moving counter-clockwise. Fig. 10.14.3 Moving towards the generator along a constant SWR circle. The rotation angle φl=2βlcan be read off in degrees from the outer periphery of the Smith chart. The corresponding length lcan also be read off in units of wavelengths530 10. Transmission Lines towards the generator (WTG) or wavelengths towards the load (WTL). Moving towards the generator by a distance l=λ/8 corresponds to a clockwise rotation by an angle of φl=2(2π/8)=π/2, that is, 90o. Moving byl=λ/4 corresponds to a 180orotation, and byl=λ/2, to a full 360orotation. Smith charts provide an intuitive geometrical representation of a load in terms of its reflection coefficient and help one design matching circuits—where matching meansmoving towards the center of the chart. However, the computational accuracy of theSmith chart is not very high, about 5–10%, because one must visually interpolate between the grid circles of the chart. Smith charts are used widely to display S-parameters of microwave amplifiers and help with the design of matching circuits. Some of the tools used in such designs are thestability circles, gain circles, and noise figure circles of an amplifier, which are intuitivelyrepresented on a Smith chart. We discuss them in Chap. 13. Various resources, including a history of the Smith chart and high-quality download- able charts in Postscript format can be found on the web site [1451]. Laursen’s Smith chart MATLAB toolbox can be used to draw Smith charts. It is avail- able from the Mathworks web site [1462]. Our MATLAB function smith.m can be used to draw simple Smith charts. 10.15 Time-Domain Response of Transmission Lines So far we discussed only the sinusoidal response of transmission lines. The response toarbitrary time-domain inputs can be obtained by writing Eq. (10.6.3) in the time domain by replacing jω→∂/∂t . We will assume a lossless line and set R/prime=G/prime=0.†We obtain then the system of coupled equations: ∂V ∂z=−L/prime∂I ∂t,∂I ∂z=−C/prime∂V ∂t(10.15.1) These are called telegrapher’s equations . By differentiating again with respect to z, it is easily verified that VandIsatisfy the uncoupled one-dimensional wave equations: ∂2V ∂z2−1 c2∂2V ∂t2=0,∂2I ∂z2−1 c2∂2I ∂t2=0 wherec=1/√ L/primeC/prime. As in Sec. 2.1, it is better to deal directly with the first-order coupled system (10.15.1). This system can be uncoupled by defining the forward and backward wave components: V±(t,z)=V(t,z)±Z0I(t,z) 2,whereZ0=/radicalBigg L/prime C/prime(10.15.2) These satisfy the uncoupled equations: ∂V± ∂z=∓1 c∂V± ∂t(10.15.3) †At RF,R/prime,G/primemay be small but cannot be assumed to be frequency-independent, for example, R/primedepends on the surface impedance Rs, which grows like f1/2. 10.15. Time-Domain Response of Transmission Lines 531 with general solutions given in terms of two arbitrary functions f(t),g(t) : V+(t,z)=f(t−z/c), V −(t,z)=g(t+z/c) (10.15.4) These solutions satisfy the basic forward and backward propagation property: V+(t,z+Δz)=V+(t−Δt,z) V−(t,z+Δz)=V−(t+Δt,z), whereΔt=Δz c(10.15.5) In particular, we have: V+(t,z)=V+(t−z/c,0) V−(t,z)=V−(t+z/c,0)(10.15.6) These allow the determination of the line voltages at any point zalong the line from the knowledge of the voltages at z=0. Next, we consider a terminated line, shown in Fig. 10.15.1, driven by a generator voltage VG(t), which is typically turned on at t=0 as indicated by the closing of the switch. Fig. 10.15.1 Transient response of terminated line. In general,ZGandZLmay have inductive or capacitive parts. To begin with, we will assume that they are purely resistive . Let the length of the line be d, so that the one- and two-way travel-time delays will be T=d/cand 2T=2d/c. When the switch closes, an initial waveform is launched forward along the line. When it reaches the load Tseconds later, it gets reflected, picking up a factor of ΓL, and begins to travel backward. It reaches the generator Tseconds later, or 2 Tseconds after the initial launch, and gets reflected there traveling forward again, and so on. The total forward- and backward-moving components V±(t,z) include all the multiple reflections. Before we sum up the multiple reflections, we can express V±(t,z) in terms of the total forward-moving component V+(t)≡V+(t,0)at the generator end, with the help of (10.15.6). In fact, we have V+(t,z)=V+(t−z/c). Applying this at the load end z=d, we haveV+ L(t)=V+(t,d)=V+(t−d/c)=V+(t−T). Because of Ohm’s law at the load,VL(t)=ZLIL(t), we have for the forward/backward components: V± L(t)=VL(t)±Z0IL(t) 2=ZL±Z0 2IL(t)⇒V− L(t)=ZL−Z0 ZL+Z0V+ L(t)=ΓLV+(t−T)532 10. Transmission Lines Therefore, we find the total voltage at the load end: VL(t)=V+ L(t)+V− L(t)=(1+ΓL)V+(t−T) (10.15.7) Using (10.15.6), the backward component at z=0 is: V−(t+T)=V−(t+d/c, 0)=V−(t,d)=V− L(t)=ΓLV+(t−T), or, V−(t)=ΓLV+(t−2T) Thus, the total line voltage at the generator end will be: Vd(t)=V+(t)+V−(t)=V+(t)+ΓLV+(t−2T) (10.15.8) More generally, the voltage at any point zalong the line will be: V(t,z)=V+(t,z)+V−(t,z)=V+(t−z/c)+ΓLV+(t+z/c−2T) (10.15.9) It remains to determine the total forward component V+(t)in terms of the multiple reflections of the initially launched wave along the line. We find below that: V+(t)=∞/summationdisplay m=0(ΓGΓL)mV(t−2mT) =V(t)+(ΓGΓL)V(t−2T)+(ΓGΓL)2V(t−4T)+···(10.15.10) whereV(t) is the initially launched waveform: V(t)=Z0 ZG+Z0VG(t) (10.15.11) Thus, initially the transmission line can be replaced by a voltage divider with Z0in series withZL. For a right-sided signal V(t) , such as that generated after closing the switch, the number of terms in (10.15.10) is finite, but growing with time. Indeed, therequirement that the argument of V(t−2mT) be non-negative, t−2mT≥0, may be solved for the limits on m: 0≤m≤M(t), whereM(t)=floor/parenleftbiggt 2T/parenrightbigg (10.15.12) To justify (10.15.10) and (10.15.11), we may start with the single-frequency case dis- cussed in Sec. 10.9 and perform an inverse Fourier transform. Defining the z-transform variableζ=ejωT=ejβd,†we may rewrite Eq. (10.9.7) in the form: Vd=V1+ΓLζ−2 1−ΓGΓLζ−2,Z 0Id=V1−ΓLζ−2 1−ΓGΓLζ−2,whereV=VGZ0 ZG+Z0 †We useζinstead ofzto avoid confusion with the position variable z. 10.15. Time-Domain Response of Transmission Lines 533 The forward and backward waves at z=0 will be: V+=Vd+Z0Id 2=V 1−ΓGΓLζ−2 V−=Vd−Z0Id 2=VΓLζ−2 1−ΓGΓLζ−2=ΓLζ−2V+ Vd=V++V−=V++ΓLζ−2V+⇒Vd(ω)=V+(ω)+ΓLe−2jωTV+(ω)(10.15.13) where in the last equation we indicated explicitly the dependence on ω. Using the delay theorem of Fourier transforms, it follows that the equation for Vd(ω) is the Fourier transform of (10.15.8). Similarly, we have at the load end: VL=VGZ0 ZG+Z01+ΓL 1−ΓGΓLζ−2ζ−1=(1+ΓL)ζ−1V+ which is recognized as the Fourier transform of Eq. (10.15.7). Next, we expand V+using the geometric series noting that |ΓGΓLζ−2|=|ΓGΓL|<1: V+=V 1−ΓGΓLζ−2=V+(ΓGΓL)ζ−2V+(ΓGΓL)2ζ−4V+··· (10.15.14) which is equivalent to the Fourier transform of Eq. (10.15.10). The same results can be obtained using a lattice timing diagram , shown in Fig. 10.15.2, like that of Fig. 5.6.1. Fig. 10.15.2 Lattice timing diagram. Each propagation segment introduces a delay factor ζ−1, forward or backward, and each reflection at the load and generator ends introduces a factor ΓLorΓG. Summing up all the forward-moving waves at the generator end gives Eq. (10.15.14). Similarly, thesummation of the backward terms at the generator, and the summation of the forward534 10. Transmission Lines and backward terms at the load, give: V−=VΓLζ−2/bracketleftbig 1+(ΓGΓL)ζ−2+(ΓGΓL)2ζ−4+···/bracketrightbig =ΓLζ−2V+ V+ L=Vζ−1/bracketleftbig 1+(ΓGΓL)ζ−2+(ΓGΓL)2ζ−4+···/bracketrightbig =ζ−1V+ V− L=ΓLVζ−1/bracketleftbig 1+(ΓGΓL)ζ−2+(ΓGΓL)2ζ−4+···/bracketrightbig =ΓLζ−1V+=ΓLV+ L ReplacingV+(t)in terms of (10.15.10), we obtain from (10.15.7) and (10.15.8): Vd(t)=V(t)+/parenleftbigg 1+1 ΓG/parenrightbigg∞/summationdisplay m=1(ΓGΓL)mV(t−2mT) VL(t)=(1+ΓL)∞/summationdisplay m=0(ΓGΓL)mV/parenleftbig t−(2m+1)T/parenrightbig(10.15.15) The line voltage at an arbitrary location zalong the line, can be determined from (10.15.9). The substitution of the series expansion of V+leads to the expression: V(t,z)=∞/summationdisplay m=0(ΓGΓL)mV(t−z/c−2mT)+ΓL∞/summationdisplay k=0(ΓGΓL)kV(t+z/c−2kT−2T) For a causal input V(t) , the allowed ranges for the summation indices m,k are: 0≤m≤floor/parenleftbiggt−z/c 2T/parenrightbigg ,0≤k≤floor/parenleftbiggt+z/c−2T 2T/parenrightbigg Example 10.15.1: A terminated line has Z0=50,ZG=450,ZL=150 Ω. The corresponding reflection coefficients are calculated to be: ΓG=0.8 andΓL=0.5. For simplicity, we takec=1,d=1,T=d/c=1. First, we consider the transient response of the line to a step generator voltage VG(t)=10u(t). The initial voltage input to the line will be: V(t)=VG(t)Z 0/(ZG+Z0)=10u(t)·50/(450+50)=u(t). It follows from (10.15.15) that: Vd(t)=u(t)+2.25∞/summationdisplay m=1(0.4)mu(t−2mT), V L(t)=1.5∞/summationdisplay m=1(0.4)mu/parenleftbig t−(2m+1)T/parenrightbig These functions are plotted in Fig. 10.15.3. The successive step levels are calculated by: Vd(t) VL(t) 1 0 1+2.25[0.41]=1.90 1.5 1+2.25[0.41+0.42]=2.26 1.5[1+0.41]=2.10 1+2.25[0.41+0.42+0.43]=2.40 1.5([1+0.41+0.42]=2.34 1+2.25[0.41+0.42+0.43+0.44]=2.46 1.5([1+0.41+0.42+0.43]=2.44 BothVdandVLconverge to the same asymptotic value: 1+2.25[0.41+0.42+0.43+0.44+···]=1.5[1+0.41+0.42+0.43+···]=1.5 1−0.4=2.5 10.15. Time-Domain Response of Transmission Lines 535 0 1 2 3 4 5 6 7 8 9 1000.511.522.53Vd(t), VL(t) t/TStep Response 1.902.262.402.46 1.502.102.342.44 generator load 0 1 2 3 4 5 6 7 8 9 1000.511.5Vd(t), VL(t) t/TPulse Response, width τ = T/ 10 0.90 0.36 0.14 0.061.50 0.60 0.24 0.10 0.04generator load Fig. 10.15.3 Transient step and pulse responses of a terminated line. More generally, the asymptotic level for a step input VG(t)=VGu(t) is found to be: V∞=V1+ΓL 1−ΓGΓL=VGZ0 ZG+Z01+ΓL 1−ΓGΓL=VGZL ZG+ZL(10.15.16) Thus, the line behaves asymptotically like a lumped circuit voltage divider with ZLin series withZG. We consider next, the response to a pulse input VG(t)=10/bracketleftbig u(t)−u(t−τ)/bracketrightbig ,s o thatV(t)=u(t)−u(t−τ), whereτis the pulse duration. Fig. 10.15.3 shows the generator and load line voltages for the case τ=T/10=1/10. The pulse levels are: [1,2.25(0.4)m]=[1.00,0.90,0.36,0.14,0.06,...] (at generator) 1.5(0.4)m=[1.50,0.60,0.24,0.10,0.04,...] (at load) The following MATLAB code illustrates the computation of Vd(t): d = 1; c=1; T = d/c; tau = T/10; VG = 10; Z0 = 50; ZG = 450; ZL = 150;V = VG * Z0 / (ZG+Z0); gG = z2g(ZG,Z0); gL = z2g(ZL,Z0); % reflection coefficients ΓG,ΓL t=0: T/1500 : 10*T; for i=1:length(t), M = floor(t(i)/2/T);Vd(i )=V* upulse(t(i), tau); i fM> =1 , m = 1:M; Vd(i) = Vd(i) + (1+1/gG)*V*sum((gG*gL).^m .* upulse(t(i)-2*m*T, tau)); end end plot(t, Vd, ’r’); where upulse(t,τ) generates the unit-pulse function u(t)−u(t−τ). The code can be adapted for any other input function V(t) .536 10. Transmission Lines The MATLAB file pulsemovie.m generates a movie of the step or pulse input as it propa- gates back and forth between generator and load. It plots the voltage V(t,z) as a function ofzat successive time instants t. /intersectionsq/unionsq Next, we discuss briefly the case of reactive terminations. These are best han- dled using Laplace transforms. Introducing the s-domain variable s=jω, we write ζ−1=e−jωT=e−sT. The terminating impedances, and hence the reflection coeffi- cients, become functions of s. For example, if the load is a resistor in series with an inductor, we have ZL(s)=R+sL. Indicating explicitly the dependence on s, we have: V+(s)=V(s) 1−ΓG(s)ΓL(s)e−2sT,whereV(s)=VG(s)Z 0 ZG(s)+Z0(10.15.17) In principle, we may perform an inverse Laplace transform on V+(s)to findV+(t). However, this is very tedious and we will illustrate the method only in the case of amatched generator, that is, when ZG=Z0, or,ΓG=0. Then,V+(s)=V(s) , where V(s)=VG(s)Z 0/2Z0=VG(s)/2. The line voltages at the generator and load ends will be from (10.15.13) and (10.15.7): Vd(s)=V(s)+ΓL(s)e−2sTV(s) VL(s)=/bracketleftbig 1+ΓL(s)/bracketrightbig e−sTV(s)(10.15.18) We consider the four typical cases of series and parallel R–Land series and parallel R–Cloads. The corresponding ZL(s)andΓL(s)are shown below, where in all cases ΓR=(R−Z0)/(R+Z0)and the parameter agives the effective time constant of the termination, τ=1/a: seriesR–L parallelR–L seriesR–C parallelR–C ZL(s)=R+sLZL(s)=RsL R+sLZL=R+1 sCZL(s)=R 1+RCs ΓL(s)=s+aΓR s+aΓL(s)=sΓR−a s+aΓL(s)=sΓR+a s+aΓL(s)=−s+aΓR s+a a=R+Z0 La=Z0R (R+Z0)La=1 (R+Z0)Ca=R+Z0 RZ0C We note that in all cases ΓL(s)has the form: ΓL(s)=(b0s+b1)/(s+a). Assuming a step-inputVG(t)=2V0u(t), we haveV(t)=V0u(t), so thatV(s)=V0/s. Then, Vd(s)=V0/bracketleftbigg1 s+ΓL(s)1 se−2sT/bracketrightbigg =V0/bracketleftbigg1 s+b0s+b1 s(s+a)e−2sT/bracketrightbigg (10.15.19) Using partial-fraction expansions and the delay theorem of Laplace transforms, we find the inverse Laplace transform: Vd(t)=V0u(t)+V0/bracketleftbiggb1 a+/parenleftbigg b0−b1 a/parenrightbigg e−a(t−2T)/bracketrightbigg u(t−2T) (10.15.20) 10.16. Problems 537 Applying this result to the four cases, we find: Vd(t)=V0u(t)+V0/bracketleftbig ΓR+(1−ΓR)e−a(t−2T)/bracketrightbig u(t−2T) (seriesR–L) Vd(t)=V0u(t)+V0/bracketleftbig −1+(1+ΓR)e−a(t−2T)/bracketrightbig u(t−2T) (parallelR–L) Vd(t)=V0u(t)+V0/bracketleftbig 1−(1−ΓR)e−a(t−2T)/bracketrightbig u(t−2T) (seriesR–C) Vd(t)=V0u(t)+V0/bracketleftbig ΓR−(1+ΓR)e−a(t−2T)/bracketrightbig u(t−2T) (parallelR–C) (10.15.21) In a similar fashion, we determine the load voltage: VL(t)=V0/bracketleftbig (1+ΓR)+(1−ΓR)e−a(t−T)/bracketrightbig u(t−T) (seriesR–L) VL(t)=V0(1+ΓR)e−a(t−T)u(t−T) (parallelR–L) VL(t)=V0/bracketleftbig 2−(1−ΓR)e−a(t−T)/bracketrightbig u(t−T) (seriesR–C) VL(t)=V0(1+ΓR)/bracketleftbig 1−e−a(t−T)/bracketrightbig u(t−T) (parallelR–C)(10.15.22) Example 10.15.2: We takeV0=1,Z0=50,R=150 Ω, and, as before, d=1,c=1,T=1. We findΓR=0.5. Fig. 10.15.4 shows the voltages Vd(t)andVL(t)in the four cases. In all cases, we adjusted LandCsuch thata=1. This gives L=200 andC=1/200, and L=37.5 andC=1/37.5, for the series and parallel cases. Asymptotically, the series R–Land the parallel R–Ccases look like a voltage divider Vd= VL=VGR/(R+Z0)=1.5, the parallel R–Lcase looks like a short-circuited load Vd= VL=0, and the series R–Clooks like and open circuit so that Vd=VL=VG=2. Using the expressions for V(t,z) of Problem 10.40, the MATLAB file RLCmovie.m makes a movie of the step input as it propagates to and gets reflected from the reactive load. /intersectionsq/unionsq 10.16 Problems 10.1 Design a two-wire line made of two AWG 20-gauge (diameter 0.812 mm) copper wires that has a 300-ohm impedance. Calculate its capacitance per unit length. 10.2 For the two-wire line shown in Fig. 10.5.1, show that the tangential component of the electric field vanishes on both cylindrical conductor surfaces. Show that the surface charge and current densities on the positively charged conductor are given in terms of the azimuthal angleφas follows: ρs(φ)=Q/prime 2πak2−1 k2−2kcosφ+1,Jsz(φ)=I 2πak2−1 k2−2kcosφ+1 Show and interpret the following: /integraldisplay2π 0ρs(φ)adφ=Q/prime,/integraldisplay2π 0Jsz(φ)adφ=I 10.3 For the two-wire line of the previous problem, show that the power loss per unit length due to ohmic conductor losses is given by:538 10. Transmission Lines 0 1 2 3 4 500.511.52 t/TVd(t), VL(t)Series R− L generator load 0 1 2 3 4 500.511.52 t/TVd(t), VL(t)Parallel R− L generator load 0 1 2 3 4 500.511.52 t/TVd(t), VL(t)Series R− C generator load 0 1 2 3 4 500.511.52 t/TVd(t), VL(t)Parallel R− C generator load Fig. 10.15.4 Transient response of reactive terminations. P/prime loss=Rs/integraldisplay2π 0|Jsz(φ)|2adφ=Rs|I|2 2πak2+1 k2−1 From this result, derive Eq. (10.5.13) for R/primeandαc. 10.4 A polyethylene-filled RG-59 coaxial cable has impedance of 75 ohm and velocity factor of 2/3. If the radius of the inner conductor is 0 .322 mm, determine the radius of the outer conductor in mm. Determine the capacitance and inductance per unit length. Assuming copper conductors and a loss tangent of 7 ×10−4for the polyethylene dielectric, calculate the attenuation of the cable in dB/100-ft at 50 MHz and at 1 GHz. Finally, calculate the cutoff frequency of higher propagating modes. 10.5 Computer Experiment : Coaxial Cable Attenuation. Consider the attenuation data of an RG- 8/U cable given in Example 10.4.3. a. Reproduce the graph of that Example. Show that with the assumed characteristics of the cable, the total attenuation may be written as a function of frequency in the form, whereαis in dB per 100 ft and fis in GHz: α(f)=4.3412f1/2+2.9131f b. Carry out a least-squares fit of the attenuation data given in the table of that Exam- ple by fitting them to a function of the form α(f)=Af1/2+Bf, and determine the fitted coefficients A,B. This requires that you find A,B by minimizing the weighted performance index: J=/summationdisplay iwi/parenleftbig αi−Af1/2 i−Bfi/parenrightbig2=min 10.16. Problems 539 where you may take the weights wi=1. Show that the minimization problem gives rise to a 2×2 linear system of equations in the unknowns A,B, and solve this system with MATLAB. Plot the resulting function of α(f) on the same graph as that of part (a). How do the fitted coefficients compare with those of part (a)? Given the fitted coefficients A,B, extract from them the estimated values of the loss tangent tanδand the refractive index nof the dielectric filling (assuming the cable radiia,band conductivity σare as given.) c. Because it appears that the 5-GHz data point is not as accurate as the others, redo part (b) by assigning only 1/2 weight to that point in the least-squares fit. Finally, redo part (b) by assigning zero weight to that point (i.e., not using it in the fit.) 10.6 Computer Experiment—Optimum Coaxial Cables. Plot the three quantities Ea,PT, andαc given in Eq. (10.4.10) versus the ratio b/a over the range 1 .5≤b/a≤4. Indicate on the graphs the positions of the optimum ratios that correspond to the minima of Eaandαc, and the maximum of PT. Moreover, write a MATLAB function that solves iteratively (for example, using Newton’s method) the equation for minimizing αc, that is, lnx=1+1/x. 10.7 LetZl=Rl+jXlbe the wave impedance on a lossless line at a distance lfrom a purely resistive load ZL. Derive explicit expressions for RlandXlin terms ofZLand the charac- teristic impedance Z0of the line for the distances l=nλ/8, wheren=1,2,3,4,5,6,7,8. Discuss the signs of Xl(inductive or capacitive) for the two cases ZL>Z 0andZL<Z 0. What happens to the above expressions when ZL=Z0? 10.8 A dipole antenna operating in the 30-meter band is connected to a transmitter by a 15-meter long lossless coaxial cable having velocity factor of 0.667 and characteristic impedance of 50 ohm. The wave impedance at the transmitter end of the cable is measured and found to be 25.5−14.9johm. Determine the input impedance of the antenna. 10.9 It is desired to measure the characteristic impedance Z0and propagation constant γ=α+jβ of a lossy line. To this end, a length lof the line is short-circuited and its input impedance Zsc is measured. Then, the segment is open-circuited and its input impedance Zocis measured. Explain how to extract the two unknown quantities Z0andγfromZscandZoc. 10.10 The wave impedances of a 100-meter long short- and open-circuited segment of a lossy transmission line were measured to be Zsc=68.45+128.13johm andZoc=4.99−16.65j ohm at 10 MHz. Using the results of the previous problem, determine the characteristicimpedance of the line Z0, the attenuation constant αin dB/100-m, and the velocity factor of the cable noting that the cable length is at least two wavelengths long. 10.11 For a lossless line, show the inequality: 1−|ΓL| 1+|ΓL|≤/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle1+ΓLe−2jβl 1−ΓLe−2jβl/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle≤1+|ΓL| 1−|ΓL| whereΓLis the load reflection coefficient. Then, show that the magnitude of the wave impedanceZlalong the line varies between the limits: Zmin≤|Zl|≤Zmax,Z min=1 SZ0,Z max=SZ0 whereZ0is the characteristic impedance of the line and S, the voltage SWR.540 10. Transmission Lines 10.12 For a lossless line, show that the current Ilat a distance lfrom a load varies between the limits: Imin≤|Il|≤Imax, whereImin=1 Z0Vmin,I max=1 Z0Vmax whereVminandVmaxare the minimum and maximum voltage along the line. Then, show that the minimum and maximum wave impedances of the previous problem can be written in the alternative forms: Zmax=Vmax Imin,Z min=Vmin Imax Recall from Sec. 10.13 that Zmax,Zmincorrespond to the distances lmaxandlmin. However, show thatIminandImaxcorrespond to lmaxandlmin, respectively. 10.13 If 500 W of power are delivered to a load by a 50-ohm lossless line and the SWR on the line is 5, determine the maximum voltage Vmaxalong the line. Determine also the quantities Vmin, Imax,Imin,Zmax, andZmin. 10.14 A transmitter is connected to an antenna by an 80-ft length of coaxial cable of characteristic impedance of 50 ohm and matched-line loss of 0.6 dB/100-ft. The antenna impedance is 30+40johm. The transmitter delivers 1 kW of power into the line. Calculate the amount of power delivered to the load and the power lost in the line. Calculate the SWR at the antenna and transmitter ends of the line. 10.15 LetSLandSdbe the SWRs at the load and at distance dfrom the load on a lossy and mismatched line. Let a=e2αdbe the matched-line loss for the length- dsegment. Show that the SWRs are related by: Sd=SL−(a−1)(S2 L−1) a(SL+1)−(SL−1)andSL=Sd+(a−1)(S2 d−1) (Sd+1)−a(Sd−1) Show that 1≤Sd≤SL. When are the equalities valid? Show also that Sd→1a sd→∞. 10.16 A 100-Ω lossless transmission line is terminated at an unknown load impedance. The line is operated at a frequency corresponding to a wavelength λ=40 cm. The standing wave ratio along this line is measured to be S=3. The distance from the load where there is a voltage minimum is measured to be 5 cm. Based on these two measurements, determine the unknown load impedance. 10.17 The wavelength on a 50 Ω transmission line is 80 cm. Determine the load impedance if the SWR on the line is 3 and the location of the first voltage minimum is 10 cm from the load. At what other distances from the load would one measure a voltage minimum? A voltage maximum? 10.18 A 75-ohm line is connected to an unknown load. Voltage measurements along the line reveal that the maximum and minimum voltage values ar e 6 V and 2 V. It is observed that a voltage maximum occurs at the distance from the load: l=0.5λ−λ 4πatan(0.75)=0.44879λ Determine the reflection coefficient ΓL(in cartesian form) and the load impedance ZL. 10.19 A load is connected to a generator by a 30-ft long 75-ohm RG-59/U coaxial cable. The SWR is measured at the load and the generator and is found to be equal to 3 and 2, respectively. Determine the attenuation of the cable in dB/ft. Assuming the load is resistive, what are all possible values of the load impedance in ohm? 10.16. Problems 541 10.20 A lossless 50-ohm line with velocity factor of 0.8 is connected to an unknown load. The operating frequency is 1 GHz. Voltage measurements along the line reveal that the maximumand minimum voltage values ar e 6 V and 2 V. It is observed that a voltage minimum occurs at a distance of 3 cm from the load. Determine the load reflection coefficient ΓLand the load impedance ZL. 10.21 The SWR on a lossy line is measured to be equal to 3 at a distance of 5 meters from the load, and equal to 4 at a distance of 1 meter from the load. a. Determine the attenuation constant of the line in dB/m. b. Assuming that the load is purely resistive, determine the two possible values of the load impedance. 10.22 A lossless 50-ohm transmission line of length d=17 m is connected to an unknown load ZLand to a generator VG=10 volts having an unknown internal impedance ZG, as shown below. The wavelength on the line is λ=8 m. The current and voltage on the line at the generator end are measured and found to be Id=40 mA andVd=6 volts. a. Determine the wave impedance Zdat the generator end, as well as the generator’s internal impedance ZG. b. Determine the load impedance ZL. c. What percentage of the total power produced by the generator is absorbed by the load? 10.23 The wavelength on a 50-ohm transmission line is 8 meters. Determine the load impedance if the SWR on the line is 3 and the location of the first voltage maximum is 1 meter from the load. At what other distances from the load would one measure a voltage minimum? A voltage maximum? 10.24 A 10-volt generator with a 25-ohm internal impedance is connected to a 100-ohm load via a 6-meter long 50-ohm transmission line. The wavelength on the line is 8 meters. Carry out the following calculations in the stated order: a. Calculate the wave impedance Zdat the generator end of the line. Then, using an equiv- alent voltage divider circuit, calculate the voltage and current Vd,Id. Then, calculate the forward and backward voltages Vd+,Vd−from the knowledge of Vd,Id. b. Propagate Vd+,Vd−to the load end of the line to determine the values of the forward and backward voltages VL+,VL−at the load end. Then, calculate the corresponding voltage and current VL,ILfrom the knowledge of VL+,VL−. c. Assuming that the real-valued form of the generator voltage is VG=10 cos(ωt) determine the real-valued forms of the quantities Vd,VLexpressed in the sinusoidal formAcos(ωt+θ).542 10. Transmission Lines 10.25 A lossless 50-ohm transmission line is connected to an unknown load impedance ZL. Voltage measurements along the line reveal that the maximum and minimum voltage values are (√ 2+1)volts and(√ 2−1)volts. Moreover, a distance at which a voltage maximum is observed has been found to be lmax=15λ/16. a. Determine the load reflection coefficient ΓLand the impedance ZL. b. Determine a distance (in units of λ) at which a voltage minimum will be observed. 10.26 A 50-ohm transmission line is terminated at a load impedance: ZL=75+j25 Ω a. What percentage of the incident power is reflected back into the line? b. In order to make the load reflectionless, a short-circuited 50-ohm stub of length dis inserted in parallel at a distance lfrom the load. What are the smallest values of the lengthsdandlin units of the wavelength λthat will make the load reflectionless? Show all work. 10.27 A load is connected to a generator by a 20-meter long 50-ohm coaxial cable. The SWR is measured at the load and the generator and is found to be equal to 3 and 2, respectively. a. Determine the attenuation of the cable in dB/m. b. Assuming that the load is resistive, what are all possible values of the load impedance in ohm? [ Hint: the load impedance can be greater or less than the cable impedance.] 10.28 A 50-ohm lossless transmission line with velocity factor of 0.8 and operating at a frequency of 15 MHz is connected to an unknown load impedance. The voltage SWR is measured to be S=3+2√ 2. A voltage maximum is found at a distance o f 1 m from the load. a. Determine the unknown load impedance ZL. b. Suppose that the line is lossy and that it is connected to the load found in part (a). Suppose that the SWR at a distance of 10 m from the load is measured to be S=3. What is the attenuation of the line in dB/m? 10.29 A lossless 50-ohm transmission line is connected to an unknown load impedance. Voltage measurements along the line reveal that the maximum and minimum voltage values are 6 V and 2 V. Moreover, the closest distance to the load at which a voltage minimum is observed has been found to be such that: e2jβlmin=0.6−0.8j.Determine the load reflection coefficient ΓLand the impedance ZL. 10.30 A resonant dipole antenna operating in the 30-meter band is connected to a transmitter by a 30-meter long lossless coaxial cable having velocity factor of 0.8 and characteristic impedance of 50 ohm. The wave impedance at the transmitter end of the cable is measured to be 40 ohm. Determine the input impedance of the antenna. 10.31 The next four problems are based on Ref. [1072]. A lossless transmission line with real characteristic impedance Z0is connected to a series RLC circuit. a. Show that the corresponding load impedance may be written as a function of frequency in the form (with f,f0in Hz): ZL=R+jRQ/parenleftBigg f f0−f0 f/parenrightBigg 10.16. Problems 543 wheref0andQare the frequency and Q-factor at resonance. Such a load impedance provides a simplified model for the input impedance of a resonant dipole antenna. Show that the corresponding SWR SLsatisfiesSL≥S0for allf, whereS0is the SWR at resonance, that is, corresponding to ZL=R. b. The SWR bandwidth is defined by Δf=f2−f1, wheref1,f2are the left and right bandedge frequencies at which the SWR SLreaches a certain level, say SL=SB, such thatSB>S 0. Often the choice SB=2 is made. Assuming that Z0≥R, show that the bandedge frequencies satisfy the conditions: f1f2=f2 0,f2 1+f2 2=2f2 0+f2 0(S0+1)2Γ2 B−(S0−1)2 Q2(1−Γ2 B),whereΓB=SB−1 SB+1 c. Show that the normalized bandwidth is given by: QΔf f0=/radicalBig (SB−S0)(S 0−S−1 B)=/radicaltp/radicalvertex/radicalvertex/radicalbt 4(Γ2 B−Γ2 0) (1−Γ0)2(1−Γ2 B),withΓ0=S0−1 S0+1 Show that the left and right bandedge frequencies are given by: f1=/radicalBigg f2 0+(Δf)2 4−Δf 2,f 2=/radicalBigg f2 0+(Δf)2 4+Δf 2 d. Show that the maximum bandwidth is realized for a mismatched load that has the following optimum SWR at resonance: S0=SB+S−1 B 2,Γ 0=Γ2 B⇒QΔfmax f0=S2 B−1 2SB=2ΓB 1−Γ2 B For example, if SB=2, we haveΓB=1/3,S0=1.25, andΔf/f 0=0.75/Q, whereas for a matched load we have S0=1 andΔf/f 0=0.50/Q. 10.32 We assume now that the transmission line of the previous problem is lossy and that the RLC load is connected to a generator by a length- dsegment of the line. Let a=e2αdbe the matched-line loss. For such lossy line, we may define the bandwidth in terms of the SWR Sd at the generator end. Show that the normalized bandwidth is given by the same expression as in the previous problem, but with the replacement ΓB→ΓLB, whereΓLB≡aΓB: QΔf f0=/radicalBig (SLB−S0)(S 0−S−1 LB)=/radicaltp/radicalvertex/radicalvertex/radicalbt 4(Γ2 LB−Γ2 0) (1−Γ0)2(1−Γ2 LB),whereSLB=1+ΓLB 1−ΓLB Show thatΓLB,SLBare the quantities ΓB,SBreferred to the load end of the line. Show that the meaningful range of the bandwidth formula is 1 ≤S0≤SLBin the lossy case, and 1≤So≤SBfor the lossless case. Show that for the same S0the bandwidth for the lossy case is always greater than the bandwidth of the lossless case. Show that this definition of bandwidth makes sense as long as the matched line loss satisfies aΓB<1. Show that the bandwidth vanishes at the S0that hasΓ0=aΓB. Show that the maximum bandwidth is realized for the optimum S0: S0=SLB+S−1 LB 2,Γ 0=Γ2 LB⇒QΔfmax f0=S2 LB−1 2SLB=2ΓLB 1−Γ2 LB=2aΓB 1−a2Γ2 B544 10. Transmission Lines Show that the optimum S0is given at the load and generator ends of the line by: S0=1+a2Γ2 B 1−a2Γ2 B,Sd0=1+aΓ2 B 1−aΓ2 B 10.33 Assume now that Z0≤Rin the previous problem. Show that the normalized bandwidth is given by: QΔf f0=/radicalBig (SLB−S−1 0)(S−1 0−S−1 LB)=/radicaltp/radicalvertex/radicalvertex/radicalbt 4(Γ2 LB−Γ2 0) (1+Γ0)2(1−Γ2 LB) Show that the maximum always occurs at S0=1. Show that the conditions aΓB<1 and 0≤S0≤SLBare still required. Show that, for the same S0, the bandwidth of the case Z0≤Ris always smaller than that of the caseZ0≥R. 10.34 Computer Experiment—Antenna Bandwidth. An 80-meter dipole antenna is resonant at f0= 3.75 MHz. Its input impedance is modeled as a series RLC circuit as in Problem 10.31. Its Q-factor isQ=13 and its resistance Rat resonance will be varied to achieve various values of the SWRS0. The antenna is connected to a transmitter with a length of 75-ohm coaxial cable with matched-line loss of a=e2αd. a. For a lossless line ( a=0 dB), plot the normalized bandwidths Q(Δf)/f 0versus the SWR at the antenna at resonance S0. Do two such plots corresponding to SWR band- width levels of SB=2 andSB=1.75. On the same graphs, add the normalized bandwidth plots for the case of a lossy line with a=2 dB. Identify on each graph the optimum bandwidth points and the maximum range of S0(for convenience, use the same vertical and horizontal scales in all graphs.) b. Assume now that S0=1.25. What are the two possible values of R? For these two cases and assuming a lossy line with a=2 dB, plot the SWR at the antenna end of the line versus frequency in the interval 3 .5≤f≤4 MHz. Then, plot the SWRs at the transmitter end of the line. Using common scales on all four graphs, add on each graph the left and right bandedge frequencies corresponding to the two SWR levels of SB=2 andSB=1.75. Note the wider bandwidth in the lossy case and for the case havingZ0≥R. 10.35 For the special case of a matched generator having ZL=Z0, or,ΓG=0, show that Eq. (10.15.15) reduces to: Vd(t)=V(t)+ΓLV(t−2T) andVL(t)=(1+ΓL)V(t−T) 10.36 A terminated transmission line may be thought of as a sampled-data linear system . Show that Eq. (10.15.15) can be written in the convolutional form: Vd(t)=/integraldisplay∞ −∞hd(t/prime)V(t−t/prime)dt/prime,VL(t)=/integraldisplay∞ −∞hL(t/prime)V(t−t/prime)dt/prime so thatV(t) may be considered to be the input and Vd(t)andVL(t), the outputs. Show that the corresponding impulse responses have the sampled-data forms: hd(t)=δ(t)+/parenleftbigg 1+1 ΓG/parenrightbigg∞/summationdisplay m=1(ΓGΓL)mδ(t−2mT) hL(t)=(1+ΓL)∞/summationdisplay m=0(ΓGΓL)mδ/parenleftbig t−(2m+1)T/parenrightbig 10.16. Problems 545 What are the corresponding frequency responses? Show that the effective time constant of the system may be defined as: τ=2Tln/epsilon1 ln|ΓGΓL| where/epsilon1is a small number, such as /epsilon1=10−2. Provide an interpretation of τ. 10.37 Computer Experiment—Rise Time and Propagation Effects . In digital systems where pulses are transmitted along various interconnects, a rule of thumb is used according to which if the rise time-constant of a pulse is tr≤2.5T, whereT=d/cis the propagation delay along the interconnect, then propagation effects must be taken into account. If tr>5T, then a lumped circuit approach may be used. Consider the transmission line of Example 10.15.1. Using the MATLAB function upulse.m , generate four trapezoidal pulses of duration td=20Tand rise times tr=0,2.5T,5T,10T. You may take the fall-times to be equal to the rise-times. For each pulse, calculate and plot the line voltages Vd(t),VL(t)at the generator and load ends for the time period 0 ≤t≤80T. Superimpose on these graphs the initial trapezoidal waveform that is launched along the line. Discuss the above rule of thumb in the light of your results. 10.38 Two coaxial transmission lines of lengths d1,d2, impedances Z01,Z02, and propagation speedsc1,c2are connected in cascade as shown below. Define the one-way travel times andz-transform variables by T1=d1/c1,T2=d2/c2,ζ1=ejωT 1, andζ2=ejωT 2. Show that the reflection response at the left of the junction is given by: Γ1=ρ+ΓLζ−2 2 1+ρΓLζ−2 2=ρ+ΓL(1−ρ2)ζ−2 2 1+ρΓLζ−2 2 whereρ=(Z02−Z01)/(Z 02+Z01)andΓLis the load reflection coefficient. Show that the forward and backward voltages at the generator end and to the right of the junction are: V+=V 1−ΓGΓ1ζ−2 1,V−=Γ1ζ−2 1V+,whereV=VGZ01 ZG+Z01 V/prime 1+=(1+ρ)ζ−1 1 1+ρΓLζ−2 2V+,V/prime 1−=(1+ρ)ΓLζ−1 1ζ−2 2 1+ρΓLζ−2 2V+ Assume a matched generator, that is, having ZG=Z01, or,ΓG=0, and a purely resistive load. Show that the time-domain forward and backward transient voltages are given by: V+(t)=V(t)=1 2VG(t) V−(t)=ρV(t−2T1)+ΓL(1−ρ2)∞/summationdisplay m=0(−ρΓL)mV(t−2mT 2−2T2−2T1) V/prime +(t)=(1+ρ)∞/summationdisplay m=0(−ρΓL)mV(t−2mT 2−T1) V/prime −(t)=ΓL(1+ρ)∞/summationdisplay m=0(−ρΓL)mV(t−2mT 2−2T2−T1)546 10. Transmission Lines Show that the line voltage V(t,z) is given in terms of the above quantities by: V(t,z)=⎧ ⎨ ⎩V+(t−z/c 1)+V−(t+z/c 1), for 0≤z≤d1 V/prime 1+/parenleftbig t−(z−d1)/c 2/parenrightbig +V/prime 1−/parenleftbig t+(z−d1)/c 2/parenrightbig ,ford1≤z≤d1+d2 10.39 Computer Experiment—Transient Response of Cascaded Lines . For the previous problem, assume the numerical values d1=8,d2=2,c1=c2=1,Z01=50,Z02=200,ZG=50, andZL=600 Ω. Plot the line voltage Vd(t)=V+(t)+V−(t)at the generator end for 0 ≤t≤5T1, in the two cases of (a) a step input VG(t)=3.25u(t), and (b) a pulse input of width τ=T1/20 defined byVG(t)=3.25/bracketleftbig u(t)−u(t−τ)/bracketrightbig . You may use the MATLAB functions ustep.m and upulse.m . For case (a), explain also the initial and final voltage levels. In both cases, explain the reasons for the time variations of Vd(t). The MATLAB file pulse2movie.m generates a movie of the pulse or step signal V(t,z) as it propagates through this structure. 10.40 Equations (10.15.21) and (10.15.22) represent the line voltages at the generator and load ends of a line terminated by a reactive load. Using inverse Laplace transforms, show that the line voltage at any point zalong such a line is given by: V(t,z)=V0u(t−z/c)+V0/bracketleftbig ΓR+(1−ΓR)e−a(t+z/c−2T)/bracketrightbig u(t+z/c−2T) (seriesR–L) V(t,z)=V0u(t−z/c)+V0/bracketleftbig −1+(1+ΓR)e−a(t+z/c−2T)/bracketrightbig u(t+z/c−2T) (parallelR–L) V(t,z)=V0u(t−z/c)+V0/bracketleftbig 1−(1−ΓR)e−a(t+z/c−2T)/bracketrightbig u(t+z/c−2T) (seriesR–C) V(t,z)=V0u(t−z/c)+V0/bracketleftbig ΓR−(1+ΓR)e−a(t+z/c−2T)/bracketrightbig u(t+z/c−2T) (parallelR–C) The MATLAB file RLCmovie.m generates a movie of these waves as they propagate to and get reflected from the reactive load. 10.41 Time-domain reflectometry (TDR) is used in a number of applications, such as determining fault locations in buried transmission lines, or probing parts of circuit that would otherwise be inaccessible. As a fault-location example, consider a transmission line of impedance Z0 matched at both the generator and load ends, having a fault at a distance d1from the source, or distanced2from the load, as shown below. The fault is shown as a shunt or series capacitor C. ButCcan equally well be replaced by an inductorL, or a resistor R. Assuming a unit-step input VG(t)=2V0u(t), show that the TDR voltageVd(t)measured at the generator end will be given by: Vd(t)=V0u(t)−V0e−a(t−2T1)u(t−2T1) (shuntC) Vd(t)=V0u(t)−V0/bracketleftbig 1−e−a(t−2T1)/bracketrightbig u(t−2T1) (shuntL) Vd(t)=V0u(t)+V0/bracketleftbig 1−e−a(t−2T1)/bracketrightbig u(t−2T1) (seriesC) Vd(t)=V0u(t)+V0e−a(t−2T1)u(t−2T1) (seriesL) Vd(t)=V0u(t)+V0Γ1u(t−2T1) (shunt or series R) 10.16. Problems 547 whereT1=d1/cis the one-way travel time to the fault. Show that the corresponding time constantτ=1/ais in the four cases: τ=Z0C 2,τ=2Z0C, τ=2L Z0,τ=L 2Z0 For a resistive fault, show that Γ1=−Z0/(2R+Z0), or,Γ1=R/(2R+Z0), for a shunt or seriesR. Moreover, show that Γ1=(Z1−Z0)/(Z 1+Z0), whereZ1is the parallel (in the shunt-Rcase) or series combination of RwithZ0and give an intuitive explanation of this fact. For a series C, show that the voltage wave along the two segments is given as follows, and also derive similar expressions for all the other cases: V(t,z)=⎧ ⎨ ⎩V0u(t−z/c)+V0/bracketleftbig 1−e−a(t+z/c−2T1)/bracketrightbig u(t+z/c−2T1),for 0≤z<d 1 V0e−a(t−z/c)u(t−z/c), ford1<z≤d1+d2 Make a plot of Vd(t)for 0≤t≤5T1, assuminga=1 for theCandLfaults, andΓ1=∓1 corresponding to a shorted shunt or an opened series fault. The MATLAB file TDRmovie.m generates a movie of the step input as it propagates and gets reflected from the fault. The lengths were d1=6,d2=4 (in units such that c=1), and the input wasV0=1.11 Coupled Lines 11.1 Coupled Transmission Lines Coupling between two transmission lines is introduced by their proximity to each other. Coupling effects may be undesirable, such as crosstalk in printed circuits, or they may be desirable, as in directional couplers where the objective is to transfer power from one line to the other. In Sections 11.1–11.3, we discuss the equations, and their solutions, describing cou- pled lines and crosstalk [1007–1024]. In Sec. 11.4, we discuss directional couplers, aswell as fiber Bragg gratings, based on coupled-mode theory [1025–1046]. Fig. 11.1.1shows an example of two coupled microstrip lines over a common ground plane, and also shows a generic circuit model for coupled lines. Fig. 11.1.1 Coupled Transmission Lines. For simplicity, we assume that the lines are lossless. Let Li,Ci,i=1,2b et h e distributed inductances and capacitances per unit length when the lines are isolated from each other. The corresponding propagation velocities and characteristic impedancesare: vi=1//radicalbig LiCi,Zi=/radicalbig Li/Ci,i=1,2. The coupling between the lines is modeled by introducing a mutual inductance and capacitance per unit length, Lm,Cm. Then, the coupled versions of telegrapher’s equations (10.15.1) become:† †C1is related to the capacitance to ground C1gviaC1=C1g+Cm, so that the total charge per unit length on line-1 is Q1=C1V1−CmV2=C1g(V1−Vg)+Cm(V1−V2), whereVg=0.