Chap 5 rewrite 3 REVIEWED

Chapter 5 rewrite #3 PhL 1.8.14 This is isntalled Chapter 5: The Transverse Problem 1 5.1 Separation of φ 1 5.2 Separation of Az 3 5.3 Development of the Transverse Problem 5 (a) kφ = kA and the transverse equations 5 (b) The scaling boundary condition on φt(x) 7 5.4 The "low-loss" approximation 9 (a) Transverse Equations for a Low-Loss transmission line 9 (b) The scaling boundary condition (5.3.11) Revisited 10 5.5 The Capacitor Problem 11 5.6 What happens if low-loss is not assumed? 17 Chapter 5: The Transverse Problem In this Chapter we define a certain "transverse" potential theory problem and a prescription for the solution of that problem to obtain K and the transmission line parameters C, G and Le. In this chapter μ, ε, ξ, β, σ are all parameters of the dielectric. 5.1 Separation of φ Let φ ≡ φ12(x) of Section 4.2. Then in the transmission line limit we found φ(x) = q(z) !Syntax Error, Idz'{ !Syntax Error, Idx1' dy1' α1(x1',y1') – !Syntax Error, Idx2' dy2' α2(x2',y2') } . (4.3.10) Rewrite the above equation as, φ(x,y,z) = q(z) φt(x,y) (5.1.1) φt(x,y) = !Syntax Error, Idz'{ !Syntax Error, Idx1' dy1' α1(x1',y1') – !Syntax Error, Idx2' dy2' α2(x2',y2') } (5.1.2) where R1 = |x - x1'|, R2 = |x - x2'|, and x is a point in the dielectric. We thus identify φt as a dimensionless "transverse potential" associated with the full potential φ. Recall that x1 and x2 are points on the surfaces of conductors C1 and C2 at the same z. Evaluate (5.1.1) at x1 then at x2 and then subtract to get the right equation below, V(z) = φ(x1) - φ(x2) = q(z) [φt(x1,y1) - φt(x2,y2)] . The left side is just V(z) according to (4.4.1). Recalling now from (4.4.7) that V(z) = q(z) K (4.4.7) we conclude that φt(x1,y1) - φt(x2,y2) = K. (5.1.3) The Helmholtz equation for φ is given by (1.5.3) for a region including dielectric and conductors, (2 + β2)φ(x,y,z) = - (1/ε) ρ(x,y,z) (1.5.3) (5.1.4) where ρ(x,y,z) exists on the boundary of the dielectric region (ie, on the conductor surfaces). Inside the dielectric there is no ρ so we then have (2 + β2)φ(x,y,z) = 0 . // dielectric region (5.1.5) Inserting (5.1.1) into (5.1.5) yields, (2 + β2) q(z) φt(x,y) = 0 or (t2 + ∂z2 + β2) q(z) φt(x,y) = 0 // t2 = 2D2 = 2 - ∂z2 or t2φt(x,y) q(z) + φt(x,y) ∂z2q(z) + β2 φt(x,y) q(z) = 0 . Divide through by φt(x,y) q(z) to get + + β2 = 0 or [ ] + = - β2 (5.1.6) which has the general form, [ h(x,y) ] + g(z) = - β2 . The only way this can be true for all x,y,z in a region is if g(z) = some constant, which call - kφ2. Then, = - kφ2 = - β2 + kφ2 . (5.1.7) We can rewrite these equations as [ t2 + (β2 - kφ2)] φt(x,y) = 0 (5.1.8) [ ∂z2 + kφ2] q(z) = 0 . (5.1.9) According to (3.8.10) and (5.1.1), for a particular z value, we expect φt(x,y) to have some constant value K1 on the entire perimeter of a cross section of conductor C1, and some other constant value K2 on the entire perimeter of a cross section of conductor C2, These facts act as boundary conditions for (5.1.7). φt(C1) = K1 φt(C2) = K2 K1 - K2 = K (5.1.10) so that (5.1.3) is realized. The second equation (5.1.9) has the following solution q(z) = q(0) e-jkz => q(z,t) = q(0) ej(ωt-kz) (5.1.11) and we find that q(z) has the form of a wave traveling down the transmission line with wavenumber kφ. The reader of Chapter 2 or of Appendix D will recognize this as the form assumed for the electric field in (2.1.1) or (D.1.1) where it was assumed as an ansatz without much a priori justification. For example, E(r,φz,t) = ej(ωt-βz) E(r,φ) . (D.1.1) where βd of Appendix D is the dielectric β in our current context. When the dust settles below, for a low-loss transmission line we shall in fact end up with kφ = β so that (D.1.1) will have the same traveling wave form as (5.1.11). 5.2 Separation of Az Let Az ≡ Az12(x) of Section 4.9. Then in the transmission line limit we found Az(x) = i(z) !Syntax Error, Idz'{ !Syntax Error, Idx1' dy1' b1(x1',y1') – !Syntax Error, Idx2' dy2' b2(x2',y2') } . (4.9.1) Rewrite the above equation as, Az(x,y,z) = i(z) Azt(x,y) (5.2.1) Azt(x,y) = !Syntax Error, Idz'{ !Syntax Error, Idx1' dy1' b1(x1',y1') – !Syntax Error, Idx2' dy2' b2(x2',y2') } (5.2.2) where R1 = |x - x1'|, R2 = |x - x2'|, and x is a point in the dielectric. We thus identify Azt as a dimensionless "transverse vector potential" associated with the full vector potential Az. Recall that x1 and x2 are points on the surfaces of conductors C1 and C2 at the same z. Evaluate (5.2.1) at x1 then at x2 and then subtract to get the right equation below, W(z) = Az(x1) - Az(x2) = i(z) [Azt(x1,y1) - Azt(x2,y2)] . The left side is just W(z) according to (4.10.1). Recalling now Le = = KL => W(z) = i(z) KL (4.10.8) we conclude that Azt(x1,y1) - Azt(x2,y2) = KL. But (4.11.26) says KL = K, so we write this last as Azt(x1t) - Azt(x2t) = K . (5.2.3) The Helmholtz equation for Az is given by (1.5.4) for a region including dielectric and conductors, (2 + β2)Az(x,y,z) = - Σi=2N μiJi . (1.5.4) (5.2.4) These Ji are currents inside the conductors. Although there is small conduction current in the dielectric, it has been absorbed into β2 as shown in (1.3.21) in the time domain with the use of the King gauge. If we take our region of interest to be the dielectric alone, we then have (2 + β2)Az(x,y,z) = 0 . // dielectric region (5.2.5) Inserting (5.2.1) into (5.2.5) yields, (2 + β2) i(z) Azt(x,y) = 0 or (t2 + ∂z2 + β2) i(z) Azt(x,y) = 0 or t2Azt(x,y) i(z) + Azt(x,y)∂z2i(z) + β2 Azt(x,y) i(z) = 0 . Now divide through by Azt(x,y) i(z) to get [] + + β2 = 0 (5.2.6) which has the general form, [ h(x,y) ] + g(z) = - β2 . The only way this can be true for all x,y,z in a region is if g(z) = some constant, which call -kA2. Then, = - kA2 = - β2 + kA2 . (5.2.7) We can rewrite these equations as [ t2 + (β2 - kA2)] Azt(x,y) = 0 (5.2.8) [ ∂z2 + kA2] i(z) = 0 . (5.2.9) According to (3.8.11) and (5.2.1), for a particular z value, we expect Azt(x,y) to have some constant value W1 on the entire perimeter of a cross section of conductor C1, and some other constant value W2 on the entire perimeter of a cross section of conductor C2, These facts act as boundary conditions for (5.1.7). Since a potential has an arbitrary zero, we shall set Azt(C1) = W1 Azt(C2) = W2 W1 - W2 = K (5.2.10) so that (5.2.3) is realized. The second equation (5.2.9) has the following solution i(z) = i(0) e-jkz => i(z,t) = i(0) ej(ωt-kz) (5.2.11) and we find that i(z) has the form of a wave traveling down the transmission line with wavenumber kA. Comparing (5.2.11) with (5.1.11), it would certainly seem odd if q(z) and i(z) had the form of traveling waves with different wavenumbers kφ ≠ kA. We will formally show in the next section that kφ = kA. 5.3 Development of the Transverse Problem (a) kφ = kA and the transverse equations The longitudinal equations from the previous two sections are these: [ ∂z2 + kφ2 ] q(z) = 0 (5.1.8) [ ∂z2 + kA2 ] i(z) = 0 . (5.2.8) But, φ(x,y,z) = q(z) φt(x,y) (5.1.1) Az(x,y,z) = i(z) Azt(x,y) . (5.2.1) Therefore, [ ∂z2 + kφ2 ] φ(x,y,z) = 0 [ ∂z2 + kA2 ] Az(x,y,z) = 0 . (5.3.1) Recall that x1 and x2 are points on the surfaces of conductors C1 and C2. If we write equations (5.3.1) first at x1 and then at x2 and then subtract, we get longitudinal equations for V(z) and W(z), [ ∂z2 + kφ2 ] V(z) = 0 // V(z) = φ(x1) - φ(x2) [ ∂z2 + kA2 ] W(z) = 0 // W(z) = Az(x1) - Az(x2) (5.3.2) where we have used the definitions V(z) and W(z) from (4.4.1) and (4.10.1). Recall now the transmission line equations (4.11.11), ∂zV(z) = - z i(z) ∂zi(z) = - yV(z) . (4.11.11) Apply ∂z to each equation and move the right side to the left side, ∂z2V(z) + z ∂z i(z) = 0 ∂z2i(z) + y ∂zV(z) = 0 . Now reuse (4.11.11) to replace ∂z i(z) = - yV(z) and ∂zV(z) = - z i(z) [∂z2 - zy ]V(z) = 0 [∂z2 - zy ] i(z) = 0 . Finally, use (4.10.8) to replace i(z) = W(z)/Le to get, [∂z2 - zy ]V(z) = 0 [∂z2 - zy ]W(z) = 0 . (5.3.3) Comparison of (5.3.3) with (5.3.2) shows that kφ2 = kA2 ≡ k2 = -zy (5.3.4) which fulfills the expectation earlier that we should have kφ = kA . Recall from (4.11.30) that z = Zs + jωLe = Zs + jω K Zs ≡ Zs1 + Zs2 y = jωC' = jω 4πξ/K (5.3.5) so k2 = -zy = -[Zs + jω K] jω 4πξ/K = -Zs jω 4πξ/K + ω2μξ = - jω Zs 4πξ/K + β2 . // see (1.5.1) for β2 (5.3.6) Therefore (β2 - k2) = jω Zs 4πξ / K = . (5.3.7) The transverse equations (5.1.7) and (5.2.7) and boundary conditions (5.1.10) and (5.2.10) may now be summarized: [ t2 + (β2-k2)] φt(x,y) = 0 φt(C1) = K1 φt(C2) = K2 K1- K2 = K (5.3.8) [ t2 + (β2-k2)] Azt(x,y) = 0 Azt(C1) = W1 Azt(C2) = W2 W1- W2 = K (5.3.9) where (β2- k2) = . (b) The scaling boundary condition on φt(x) There exists another boundary condition on φt in the case that the dielectric extends transversely to infinity. Recall (5.1.2) for φt(x,y) = φt(x), φt(x) = !Syntax Error, Idz'{ !Syntax Error, Idx1' dy1' α1(x1',y1') – !Syntax Error, Idx2' dy2' α2(x2',y2') } (5.1.2) R12 = (x-x1')2 + (y-y1')2 + (z-z')2 = s12 + (z-z')2 s12 = (x-x1')2 + (y-y1')2 (4.2.1) R22 = (x-x2')2 + (y-y2')2 + (z-z')2 = s22 + (z-z')2 s22 = (x-x2')2 + (y-y2') . If we take the point x transversely far away from the conductors, the following drawing shows the distances R1 and R2 which appear in the above integration, Fig 5.1 During the transverse integration !Syntax Error, Idx1' dy1', distance R1 does not vary much and can be replaced with a distance from x to the "center" of conductor C1 without changing the integral significantly. The same can be said for R2. We shall refer to these "center points" as x1 and x2. In this case, we obtain φt(x) ≈ !Syntax Error, Idz' { !Syntax Error, Idx1' dy1' α1(x1',y1') – !Syntax Error, Idx2' dy2' α2(x2',y2') } = !Syntax Error, Idz' { - } = !Syntax Error, Idz' ( - ) (5.3.10) where we have used the fact (4.1.3) that the transverse charge densities are normalized to unity. The dz' integral was done in (4.4.5) and equals ln(s22/s12), so then 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 s22 = (x-x2)2 + (y-y2)2 . Whatever the exact solution φt(x) might be, in the limit discussed above one must obtain φt(x) ≈ ln(s22/s12). Of course as one continues to move x away to infinity, s1 ≈ s2 and then φt(x) ≈ ln(1) = 0. Basically (5.3.11) is a boundary condition on the "scale" of the solution φt(x). If someone were to propose a possible solution φt(x) = 2.6 ln(s22/s12) for some conductor geometry, we could instantly rule out that solution since it violates the boundary condition (5.3.11). The scale of φt is restricted in this manner because the charge distributions αi appearing in (5.3.10) are normalized to unity. By the exact same argument presented above, we have Azt(x) ≈ ln(s22/s12) // limiting form as point x = (x,y) moves far from the conductors (5.3.12) We shall give an interpretation of these limiting forms in Section 5.4 (b) below. 5.4 The "low-loss" approximation (a) Transverse Equations for a Low-Loss transmission line In this approximation, we take the conductor surface impedance Zs ≈ 0. Recall from (5.3.6) that k2 = β2 - jω Zs 4πξ/K (5.3.6) Our definition of a "low-loss" transmission line is one for which k2 ≈ β2 and in this case the longitudinal wave number k as shown in (5.1.11) and (5.2.11) is k = β which is the just the β value of the dielectric. So our low-loss condition is (using (1.5.1) for β2), | jω Zs 4πξ/K| << |β2| or |Zs| << (1/4π) | β2/(ωξ)| K = (1/4π) ω | β2/(ω2ξ)| K = (1/4π) ωμ K . (5.4.1) For a symmetric-environment round wire of radius a we found in (2.4.12) that for large ω, Zs ≈ (1+j) for δ << 16a δ2 = 2/ωμσ (2.4.16) Our low-loss condition is then << ωμ (K/4π) or << ωμσ (K/4π) = (2/δ2) (K/4π) or (δ/a) << K/ . (5.4.2) We saw in the Example of Section 4.6 that K = 2 ln(a2/a1) for a coaxial cable. Even for a very large radius ratio of 100 this would be K = 2 ln(100) = 9.2. for a more typical ratio of perhaps 5, K ≈ 3.2. Then our inequality above says roughly (δ/a) << 2 which is then our ball-park estimate for applicability of the "low-loss transmission line" condition. We showed in Section 2.5 how Zs can be modified for some other geometry. If we assume this low-loss limit is in effect, then β2-k2 = jω Zs 4πξ/K ≈ 0 and our transverse equations (5.3.8) and (5.3.9) become 2D Laplace equations, t2φt(x,y) = 0 φt(C1) = K1 φt(C2) = K2 K1- K2 = K (5.4.3) t2Azt(x,y) = 0 Azt(C1) = W1 Azt(C2) = W2 W1- W2 = K (5.4.4) The parallelism between φt and Azt should not be surprising in light of Section 1.3 (b) where it was noted that A and φ are components of the same relativistic 4-vector. (b) The scaling boundary condition (5.3.11) Revisited First, a quick review. In 3D the potential (SI units) of a point charge q located at x1 is φ(x) = (q/4πε|x-x1|) = q/(4πεR1). In 2D the potential of a point charge q located at x1 is φ(x) = -(q/2πε)ln|x-x1| = -(q/2πε)lns1. The 3D φ(x) is the solution of -2(φ) = (q/ε)δ3)(x-x1) as shown in (H.1.4) and as proven in Appendix H. The quantity 1/4πR1 is the 3D free-space propagator of the 3D Laplace equation. It is the Green's function of the equation -2g(x|x1) = δ(3)(x-x1) . The 2D φ(x) is the solution of -22D(φ) = (q/ε)δ(2)(x-x1) as shown in (I.1.4) and as proven in Appendix I. The quantity -2πlns1 is the 2D free-space propagator of the 2D Laplace equation. It is the Green's function of the equation -22Dg(x|x1) = δ(2)(x-x1). With this brief review, we now examine a 2D cross section view of the transmission line of Fig 5.1 at a scale that makes the two conductors appear very small and very close together, and at the same time we imagine more elaborate cross section shapes. The dielectric is assumed non-conducting, so ξ = ε. The three points indicated by the three dots on the right all lie in the plane of paper, this is just a 2D drawing and for example x = (x,y). Fig 5.2 The dots on the left indicate the "center of charge" for each conductor and these dots appear also on the right. The claim is that when x is very far away, the variation in s1 as it moves over the perimeter of the conductor C1 cross section is so small that we can replace s1 with a distance to the center of charge of C1 and similarly for R2. Thus, on the right we end up with the 2D potential of two point charges which form a 2D electric dipole. Using the results just quoted in the above review, we find that φ(x) = φ1(x) + φ2(x) = -(q/2πε)ln|x-x1| -(-q/2πε)ln|x-x2| = -(q/2πε) lns1+(q/2πε) lns2 = (q/2πε) ln(s2/s1) = (q/4πε)ln(s22/s12) . Recalling for ξ = ε that φ(x,y,z) = q(z) φt(x,y) (5.1.1) we find that φt(x,y) = ln(s22/s12) . // for r far away Thus we have an alternate derivation and 2D dipole interpretation of our earlier "scaling boundary condition" (5.3.11). 5.5 The Capacitor Problem We have now boiled down the computation of transmission line parameters (in the transmission line limit and in the low-loss limit) to the problem of computing the capacitance of a section of transmission line. Here we assume the dielectric is non-conducting so ξ = ε and we don't have to worry about the distinction between charge densities qc and qs as discussed in (4.11.19). Solving the capacitor problem using φ A standard approach to a general 2D electrostatics capacitor problem is as follows. Start with 2D2φ(x,y) = 0 φ(C1) - φ(C2) = V = voltage between conductors (5.5.1) where we now (arbitrarily) use notation 22D in place of 2t. Since the dielectric presumably fills the region between the conductors, the dielectric is the official "region" of a Green's function problem. If we put a unit positive point charge at some location (x',y') in the dielectric region we can then formally (!) solve this 2D Green's function problem, 2D2g(x,y|x',y') = δ(x-x')δ(y-y') g(x,y|x',y') = 0 for (x,y) on both C1 and C2 g(x,y|x',y') = 0 for (x,y) = ∞ (if appropriate) (5.5.2) Here g(x,y|x',y') is specific to our geometry; it is not the 2D free-space Green's function - ln(1/R)/2π shown in (I.1.4). The free-space solution has only the lower boundary condition stated above. In very general notation, if a region contains some sources q(x) and if the potential φ is prescribed on the entire closed boundary surrounding the region by a function f, then the solution to (5.5.1) is given in Stakgold notation as (1.5.11) (which we derive in the lines following (1.5.11) for both Laplace and Helmholtz equations ), φ(x) = ∫R dx' g(x|x') q(x') – ∫σ dSξ f(ξ) ∂ξng(x|ξ) // Stakgold (6.81) . (1.5.11) where σ represents the closed boundary of the region of interest, dSξ is an integration over this boundary, and ∂ξng(x|ξ) is the derivative of the Green's function in a direction locally normal to the boundary surface. In potential theory, this type of problem is known as "the Dirichlet Problem". In our case the boundary consists of C1, C2 and the Great Circle at ∞. Stakgold deals in an arbitrary number of spatial dimensions, but of course we have only 2 dimensions here, so dSξ is a line integral around the boundary. A picture is in order, showing a cross section of the transmission line, Fig 5.3 We know that on the great circle φ = 0, so there will be no contribution from that part of the Dirichlet boundary. What we do not know are V1 and V2 which are the constant potentials on C1 and C2. If it happened that the picture had mirror symmetry in a plane separating the two conductors, we would know that V1 = V/2 and V2 = -V/2, but in the general case we don't know V1 and V2 a priori. For the moment, we leave them as to-be-determined quantities. In our application of (1.5.11) there are no charges q(x) in the dielectric region. We put one there temporarily to obtain the Green's function, but it is now gone. Thus (1.5.11) reads φ(x,y) = – C1 ds' f(C1) ∂ng(x,y|x',y') – C2 ds' f(C2) ∂ng(x,y|x',y') - GC ds' f(∞) ∂ng(x,y|x',y') = – C1 ds' V1 ∂ng(x,y|x',y') – C2 ds' V2 ∂ng(x,y|x',y') – GC ds' (0) ∂ng(x,y|x',y') = -V1 C1 ds' ∂ng(x,y|x',y') – V2 C2 ds' ∂ng(x,y|x',y') }. = V1 F1(x,y) + V2F2(x,y) (5.5.3) where the Fi(x,y) are determined by doing the line integrals for a given geometry. If y = y1(x) describes a piece of the C1 perimeter, then ds' = = dx' (5.5.4) which gives a candidate ds' for doing the line integral over that piece of the perimeter. Once φ(x,y) is known, one can compute the normal electric field En at the conductor surfaces, En(x) = - ∂n1φ(x) = -V1 [∂n1F1(x)] - V2 [∂n1F2(x)] ≡ V1 G11(x) + V2 G12(x) x on C1 En(x) = - ∂n2φ(x) = -V1 [∂n2F1(x)] - V2 [∂n2F2(x)] ≡ V1 G21(x) + V2 G22(x) . x on C2 (5.5.5) Since the conductors are different, the resulting four functions Gij will in general be different. For example, we are taking normal derivatives of the Fi at different points in space on different (1D) surfaces. We may now compute the (linear) surface charge density using (11.4.7) assuming En = 0 inside the conductor, n1(x,y) = εEn(x,y) = εV1 G11(x) + εV2 G12(x) x on C1 n2(x,y) = εEn(x,y) = εV1 G21(x) + εV2 G22(x) x on C2 (5.5.6) where ε is of course for the dielectric. One can then integrate over the boundaries of the conductors to get the total charges q1 and q2 residing on the conductors, q1 = C1 ds' n1(x',y') = εV1H11 + εV2H22 q2 = C2 ds' n2(x',y') = εV1H21 + εV2H22 (5.5.7) where the Hij are now four constants which we have computed by doing the above process. Since it turns out that H12 = H21 as shown below, we can ignore H21, G21(x), and ∂n2F1(x) in the above set of calculations. We now define some new constants cij = εHij and write the above as q1 = c11V1 + c12V2 q2 = c21V1 + c22V2 . (5.5.8) Comment: The coefficients cij are dimensionally capacitance, but they are a little strange. If we start off with the conductors holding charges q1 and q'2 and then we ground C2 to the great circle (thin wire, V2= 0), and then we measure V1 relative to the great circle, we find that V1 = q1/c11 and q2 = c21V1. So c11 is the capacitance of C1 in the presence of a grounded C2 (which is not the same as the capacitance of C1 in isolation). And c21 determines how much charge q2 is "induced" onto C2 by the presence of charged C1. Smythe (p 37) and Oughstun (p 23) refer to the cij both as "coefficients of capacitance" and "coefficients of induction". This should be distinguished from the notion of conductors C1 and C2 each having a "self-capacitance" (each in isolation) and having a "mutual capacitance" ( to be called C below). Writing the above pair of equations in matrix notation we get, = or q = c V . (5.5.9) We know all the cij because we computed them above. Then invert to get = or V = sq (5.5.10) where matrix s = c-1 is called the "mutual elastance" matrix by Smythe (p 36), and the "coefficients of potential" by Oughstun (p 21). Both authors deal with an arbitrary number of conductors. The reader will not be surprised to learn that in general cij = cji and sij = sji so the matrices c and s are in fact symmetric matrices. Smythe shows this on pages 36-37 based on what he calls "Green's Reciprocation Theorem" on page 34 (George Green once again!). This theorem can be a lifesaver in certain electrostatic problems. Now our problem as shown in Fig 5.3 is to compute the potential φ when C1 has charge q and C2 has charge -q. We then finally arrive at the appropriate values of V1 and V2 for our problem, which we said above were "to be determined". Here they are: = = q (5.5.11) so that V1 = q (s11- s12) V2 = q (s21- s22) V = V1 - V2 = q [s11+ s22 - 2s12] . // s12 = s21 as noted above (5.5.12) Finally, we have computed the (inverse) capacitance of our transmission line section, 1/C = V/q = s11 + s22 - 2s12 . But we know how to invert a simple 2x2 matrix (T = transpose, cof = cofactor, det(c) = |c| ) s = c-1 = cof(cT)/det(c) so that s = = /det(c) . (5.5.13) Then 1/C = s11 + s22 - 2s12 = ( c22 + c11 +2c12)/det(c) = and finally C = . (5.5.14) We have found verification of this result on the web from Oughstun page 27, Once we have C, we know from (4.11.30) that K = 4πε/C = 4πε . (5.5.15) Thus, we have solved "the capacitor problem" to obtain K for the transmission line. The other line parameters are then given as in (4.11.30) G = 4πσ/K Le = K . Statement of the capacitor problem in terms of φt To show that our capacitor problem is the same as (5.4.3), we first quote the capacitor problem (5.5.1), 2D2φ(x,y) = 0 φ(C1) - φ(C2) = V . (5.5.1) Then use (5.1.1) φ(x,y,z) = φt(x,y) to get 2D2φt(x,y) = 0 φt (C1) - φt(C2) = V or 2D2φt(x,y) = 0 φt (C1) - φt(C2) = V or 2D2φt(x,y) = 0 φt(C1) - φt(C2) = K which is (5.4.3). In the last step we used (4.4.7) that V(z) = K. The potentials V1 and V2 are related to constants K1 and K2 by V1 = K1 V2 = K2 (5.5.16) Solution of the capacitor problem using φt Here we just repeat the above analysis, showing how things differ. We leave out the words. The main differences are that the Vi are replaced by Ki and the factor appears on the lines where ni are computed. As before, we now start off with K1 and K2 unknown, but we find them in the end: φt(x,y) = – C1 ds' K1 ∂ng(x,y|x',y') – C2 ds' K2 ∂ng(x,y|x',y') = K1 F1(x,y) + K2F2(x,y) En(x,y) = - ∂n1φ = - ∂n1φt(x,y) = { K1 G11(x) + K2 G12(x) } x on C1 En(x,y) = - ∂n1φ = - ∂n2φt(x,y) = {K1 G21(x) + K2 G22(x) } x on C2 n1(x,y) = εEn(x,y) = εK1 G11(x) + εK2 G12(x) x on C1 n2(x,y) = εEn(x,y) = εK1 G21(x) + εK2 G22(x) x on C2 q1 = C1 ds' n1(x',y') = [εK1H11 + εK2H22] = [ c11V1 + c12V2 ] q2 = C2 ds' n2(x',y') = [εK1H21 + εK2H22] = [ c21V1 + c22V2 ] = or q = c K . = or K = sq = = 4πε K1 = 4πε (s11- s12) K2 = 4πε (s21- s22) K = K1 - K2 = 4πε [s11+ s22 - 2s12] so K = 4πε (5.5.17) Then the same capacitance shown in (5.5.14) is recovered, C = 4πε/K = . For arbitrary conductor shapes, carrying out the Green's function program just outlined is quite difficult and usually requires expanding the Green's function in some complete set of eigenfunctions and then making various approximations. Our point is that the capacitor problem is a well-posed problem and has a solution value K. Numerical evaluations are also possible. If the conductors are round, the problem can be solved exactly as we shall show in Chapter 6. 5.6 What happens if low-loss is not assumed? Let's go back to our equation before the low-loss assumption that Zs= 0, [ t2 + ] φt(x,y) = 0 φt(C1) = K/2 φt(C2) = - K/2 (5.3.8) This is now a Helmholtz equation with Helmholtz parameter , whereas with Zs = 0 we had the simpler Laplace Equation. Treating Zs as some given value ≠ 0, we could go ahead and find the Green's function for the above equation and it would be a function of K since K appears in the Helmholtz parameter. Call this Helmholtz Green's function gK(x,y|x',y'). We still have φ = q(z) φt being the full potential from which the electric field is obtained as En = -∂nφ [ recall that transverse A components are zero so this is consistent with E = - grad φ - ∂tA ]. The solution of the above PDE system then starts off φt(x,y) = – C1 ds' K1 ∂ngK(x,y|x',y') – C2 ds' K2 ∂ngK(x,y|x',y') = K1 F1(x,y,K) + K2F2(x,y,K) . (5.6.1) From this point on, every function and constant acquires and argument K: Gij(x,K), Hij(K) and then cij(K). We end up then with K = 4πε . (5.6.2) The new feature is that K appears on both sides of the last equation. This probably-complicated equation then has to be solved for K, and sometimes this is referred to as "an eigenvalue problem" for K. For example, if Zs is very small but non-zero, we would expect the solution for K to be slightly different from the value obtained with Zs = 0 and one could perhaps approach the problem using perturbation theory where the Helmholtz parameter is a "smallness parameter". Recall that k2 = β2 - jω Zs 4πξ/K (5.3.6) where now K is the "eigenvalue" of our solution above. If Zs is very small but not zero, we end up then with k = β - Δ where Δ is a small complex number. The longitudinal transmission line behavior of all z-dependent functions like φ, Azt, q, V, W, E, B is then given by (5.1.11), q(z,t) = q(z,t) = q(0) ej(ωt-kz) = q(0) ej(ωt-[β-Δ]z) = q(0) ej(ωt-[β-Re(Δ)]z) e–Im(Δ)z The real part of Δ causes a shift in the wavenumber k so the wave no longer propagates with the normal dielectric wavenumber β. Since v = ω/k, we will find that the wave is "slowed down" due to the drag effect of the non-zero surface impedance of the conductors. The imaginary part of Δ then causes an exponential decay of the wave magnitude due to ohmic losses at the conductor surface. In our Chapter 2 analysis of the round wire we found that in general Zs is itself complex, so computation of Δ is a somewhat complicated problem which we shall not attempt here. The problem of lossy transmission lines is usually approached using E and B fields, rather than potentials φ and Az, and the analysis is then similar to the way waveguides in general are treated. Due to the skin effect, the E and B fields penetrate a distance ~δ into the conductor surfaces and this results in ohmic losses and a "drag" on the propagating wave. In this approach, one ends up again with an eigenvalue problem to solve, not directly for K but for some other related parameter like k. In the 12-page Section 4.5 of his book, Matick studies a lossy-transmission line in the simplest possible case which is a stripline geometry whose gap S is small compared to the width and whose metal strips are much thicker than the skin depth δ. His parameter γ is related to our parameter k by γ = jk, and his longitudinal direction is x instead of our z. He ends up with a transcendental "eigenvalue equation" (4-66) for γ, but if loss is very small, he can approximately solve for γ with these results Im(γ) = β(1+δ/2S) Re(γ) = β (δ/2S) // Matick (4-75,76,77) which with γ = jk we translate to Im(k) = - Re(γ) = - β (δ/2S) Re(k) = Im(γ) = β(1+δ/2S) k = β(1+δ/2S) -j β (δ/2S) = β - [-(δ/2S) + j(δ/2S)] so Δ = -(δ/2S) + j(δ/2S) . Thus, for such a stripline transmission line, the longitudinal dependence of all functions has this form, q(z,t) = ej(ωt-[β-Re(Δ)]z) e–Im(Δ)z = ej(ωt-[β+δ/2S]z) e–(δ/2S)z which shows the exponential loss factor and an increased wavenumber β+δ/2S which corresponds to a decreased wavelength λ and a decreased wave velocity v = ω/k = ωλ/2π = fλ, the "drag effect". Matick has an errata in this section which is a bit confusing, so we repair it right here. His equation (4-50) should read 2E = ( + ) + ( + ) = (jωμσ - ω2με)(Ex + Ez) Matick (4-50)