Chap 6 rewrite A REVIEWED

This is the Title PhL 3.26.05 Note that page numbering is turned on in this template. Chapter 6: Transmission Lines with Circular Conductors 6.1 A candidate transverse potential φt In the previous chapter (both Section 5.3 (b) and Section 5.4 (b)) we showed that the transverse potential of a 2-conductor balanced transmission line must have this form when viewed from far away, φt(x) ≈ ln(s22/s12) // limiting form as point x = (x,y) moves far from the conductors (5.3.11) s12 = (x-x1')2 + (y-y1')2 = |x - x1'|2 s22 = (x-x2')2 + (y-y2') = |x - x2'|2 where x1 = (x1',y1') is any point on C1 and x2 = (x2',y2') is any point on C2 . In Section 5.4.(b) we took these points to be the "center of charge" points, but any points will do when x is very far away. Suppose now we take as a candidate dimensionless transverse potential φt exactly the above limiting expression. Our candidate φt is then φt(x) = ln(s22/s12) for all values of r, close and far Certainly this meets our limiting form boundary condition (5.3.11)! We know also that this potential is a valid solution of the 2D Laplace equation, since ln(s1) and ln(s2) are each valid solutions. This fact was shown at the start of Section 5.4 (b). Since -2πlnR1 is the 2D free-space propagator, it follows that -2πlnR1 is a solution of 22D(φ) = 0 away from the point where R1 = 0, and then so is lnR1 = lns1. Then by superposition, 2lns2 - 2lns1 is also a valid solution, and this is ln(s22/s12). Thus, our φt is a valid candidate for a lossless transmission line, since for such a transmission line φt satisfies the 2D Laplace equation according to (5.4.3) t2 φt(x,y) = 0 φt(C1) = K1 φt(C2) = K2 K1-K2 = K (5.4.3) The question then becomes: what are the surfaces Ci in 2D space on which this candidate φt is a constant? Such surfaces can then serve as possible conductor cross sections for a transmission line. 6.2 Ancient Greece circa 230 BC Apollonius of Perga (262BC-190BC) [ like Joe of Chicago ] was a pretty smart guy as wiki explains. He did astronomy and therefore he did geometry. Besides giving conic sections their current names and writing eight books about them, he learned about what are now called the Apollonian Circles. These circles form the "level surfaces" for 2D bipolar orthogonal coordinates as shown in this picture http://en.wikipedia.org/wiki/Apollonian_circles When this picture is rotated around its vertical axis, the blue level circles become toroids and one then arrives at 3D toroidal coordinates, but that is another story. Our interest is in the 2D blue circles. It turns out, as the reader no doubt suspects, that the blue circles have the following simple property: s2/s1 = constant which we shall prove in a moment. Calling this constant e-B we get s2/s1 = e-B => ln(s2/s1) = - B Thus, since φt(x) = ln(s22/s12) = 2 ln(s2/s1), the blue circles are candidate equipotential surfaces for our potential φt(x) ! To show that s1/s2 = eB describes a circle, consider: |r-r2| / |r-r1| = e-B |r-r2|2 = e-2B |r-r1|2 (x-x1)2 + (y-y1)2 = e-2B [(x-x2)2 + (y-y2)2 ] This equation has the following form A(x2 + y2) + Bx + Cy + D = 0 A = (1-e-2B) or x2 + y2 + αx + βy + γ = 0 . One can then "complete the squares" to obtain the equation of a circle of radius r centered at (xc,yc) , (x - xc)2 + (y - yc)2 = r2 where -2xc = α -2yc= β xc2 + yc2 - r2 = γ . For our particular locations of r1 and r2 shown in the above figure, we have x1 = d x2 = -d y1 = y2 = 0 s12 = (x1+d)2 + y2 s22 = (x1-d)2 + y2 so s2/s1 = e-B => s1/s2 = eB => e2Bs22 = s12 => (eB/2) s22 = (e-B/2) s12 => (eB/2) [x2 - 2dx + d2 + y2] = (e-B/2) [x2 + 2dx + d2 + y2] shB (x2+d2+y2) + chB(-2dx) = 0 // shB = (eB-e-B)/2, chB = (eB+e-B)/2 (x2+d2+y2) + cothB (-2dx) = 0 x2 - 2dx cothB + y2 = -d2 x2 - 2dx cothB + d2coth2B + y2 = -d2+ d2coth2B // complete the square (x - dcothB)2 + y2 = d2csch2B . We conclude that our blue equipotential circles have this simple form (x - xc)2 + y2 = r2 xc = d cothB r = |d cschB| Using d = 5, here is a plot of these circles for 10 different B values: Since xc = d cothB, the right side curves have B > 0 while the left side have B < 0. The value B = 0 corresponds to the vertical y axis, while B = ±∞ correspond to the two focal points at d = ± 5. 6.3 Back to the Future: Reading off the Capacitance We shall now first select C2 to be a circle on the right side, so that B2 > 0. For C1 we select a second circle from either the left or the right, so B1 can have either sign. If we select C1 from the left side, we have a two-wire transmission line. If we select C2 from the right, we have an off-center coaxial transmission line. Let σ1 = sign(B1). We then have φt(x) = ln(s22/s12) = 2 ln(s2/s1) φt(C1) = 2 ln(s2/s1)|C1 = -2B1 φt(C2) = 2 ln(s2/s1)|C2 = -2B2 . Recall from (5.1.3) that φt(C1) - φt(C2) = K. Therefore, K = 2(B2-B1) = 2 (|B2| -σ1|B1|). Once we know K, we know C, G and Le for the transmission line from ****. We must now do some slightly painful algebra. First, we know from **** above that a1 = d |cschB1| => (d/a1) = sh(|B1|) => |B1| = sh-1(d/a1) a2 = d |cschB2| => (d/a2) = sh(|B2|) => |B2| = sh-1(d/a2) b = |xc2 - xc1| = |d cothB2 - dcothB1| = d |cothB2 - cothB1| From ** we write ch(K/2) = ch [|B2| -σ1|B1|] = ch|B2| ch|B1| - σ1 sh|B2| sh|B1| = - σ1 sh|B2| sh|B1| = - σ1 (d/a2) (d/a1) . Meanwhile, b = d |cothB2 - cothB1| = |d [ chB2/shB2 - chB1/shB1] | = |d [ ch|B2|/sh|B2| - σ1ch|B1|/sh|B1|] | = | d [ch|B2| sh|B1| - σ1 ch|B1| sh|B2| ] / sh|B1| sh|B2| | = | d [(d/a1) - σ1 (d/a2) ] / (d/a2) (d/a1) | = | [(1/a1) - σ1 (1/a2) ] / (1/a2) (1/a1) | = | [a2 - σ1 a1 ] | . Square this to get b2 = a22[1+(d/a2)2] + a12[1+(d/a1)2] - 2σ1a1a2 so 2 σ1a1a2 = a22[1+(d/a2)2] + a12[1+(d/a1)2] - b2 = a22 + d2 + a12 + d2 - b2 = a12 + a22 + 2d2 - b2 . The purpose of doing this is to obtain the following expression for the radical product = (a12 + a22 + 2d2 - b2) / (2 σ1a1a2 ) . We now install this into our expression above for ch(K/2) to get ch(K/2) = - σ1 (d/a2) (d/a1) = (a12 + a22 + 2d2 - b2) / (2 σ1a1a2 ) - 2d2/ (2σ1a2a1) = (a12 + a22 - b2) / (2σ1a2a1) = σ1 (1/2) (a12 + a22 - b2)/(a1a2) = σ1 (1/2) [ (a1/a2) + (a2/a1) - (b2/a1a2) ] and the annoying constant d has vanished from our expression. Therefore K = 2 ch-1 { σ1 (1/2) [ (a1/a2) + (a2/a1) - (b2/a1a2) ] } Notice that the result is symmetric under a1 ↔ a2 . We now distinguish our two cases of interest. For a twin-lead type transmission line (unequal lead diameters) we know that B1 < 0 since the C1 circle is on the left, so σ1 = sign(B1) = - 1 and then K = 2 ch-1 { (1/2) [ (b2/a1a2) - (a1/a2) - (a2/a1)] } which is an amazingly simple result. Recall from *** that Z0 = (K /) 30Ω so then Z0 = ch-1 { (1/2) [ (b2/a1a2) - (a1/a2) - (a2/a1)] } (1/) 60 Ω If the wires have diameters d1 = 2a1 and d2 = 2a2 this becomes Z0 = ch-1 { (1/2) [ (4b2/d1d2) - (d1/d2) - (d2/d1)] } (1/) 60 Ω For verification, we quote again from Reference RDE page 29-23, where our b is called D. On the other hand, if we are interested in an off-center coaxial transmission line, we select C1 from the right side of Fig *** and then σ1 = sign(B1) = + and we find K = 2 ch-1 { (1/2) [ (a1/a2) + (a2/a1) - (b2/a1a2) ] } Z0 = ch-1 { (1/2) [ (a1/a2) + (a2/a1) - (b2/a1a2) } (1/) 60 Ω = ch-1 { (1/2) [ (d1/d2) + (d2/d1) - (4b2/d1d2) } (1/) 60 Ω For verification, we quote again from Reference RDE page 29-24, where we may take d = our d1 and D = our d2 and c = our b = the center-line separation. There is one more case of interest that falls out from this analysis. If we take B1 = 0 we have which is a transmission line consisting of a round wire above an infinite flat plane. Since B1 = 0 we find from *** that K = 2B2 We know that x2c = d coth B2 = d chB2/shB2 a2 = d/sh(B2) Therefore x2c/a2 = chB2 => B2 = ch-1(x2c/a2) => K = 2 ch-1(x2c/a2) Here x2c is the distance from the wire center line to the ground plane. If we call this h and the wire radius a, we then have the following extremely simple and exact result, K = 2 ch-1(h/a) // wire radius a with center h over ground plane, exact Z0 = (K /) 30Ω = ch-1(h/a) (1/) 60 Ω where of course we must have h > a to keep the wire from touching the ground plane. Using the identity ch-1x = ln(x + ) for x ≥ 1 we can write the above as K = 2 ln [ (h/a) + ] // wire radius a with center h over ground plane, exact Z0 = ln [ (h/a) + ] (1/) 60 Ω For h >> a this becomes K = 2 ln(2h/a) // wire radius a center h over ground plane, h>> a Z0 = ln(2h/a) (1/) 60 Ω For verification, we found the following obscure web offering, http://members3.jcom.home.ne.jp/zakii/tline_e/14_microstripline_z0.htm which results are derived using an image method to handle the ground plane. For some odd reason, our usual RDE source on this subject only gives the result for h >> a . Taking d to be the wire diameter, Z0 = ln(4h/d) (1/) 60 Ω = ln(10) log (4h/d) (1/) 60 Ω ≈ log (4h/d) (1/) 138.2 Ω which then compare to RDE p 29-22