Chapter 4 rewrite INSTALLED

Chapter 4 rewrite PhL 5.11.14 Chapter 4: Transmission Line Equations In this Chapter we use the potential integral expressions derived in Chapter 1 to derive the classic transmission line equations. We learn that most transmission line parameters are determined by a single geometric integral K. The approximations are clearly stated. 4.1 Computation of potential φ due to one conductor of a transmission line Our starting point is the potential φ expression given in box (1.5.23) for the potential at some arbitrary point x in the dielectric due to conductor C1 of a transmission line, φ1(x,ω) = ∫ ρ1(x',y',z',ω)dx'dy'dz' . R = |x - x'| (4.1.1) Here the point x' = (x',y',z') runs over the surface of C1 and R is the distance between the observation point x in the dielectric and the integration point x'. Parameters β and ξ are for the dielectric. Comments on ρ1: 1. ρ1 is the volume charge density associated with "surface charge" n1 according to ρ1dV' = n1dS' . 2. ρ1 is a distribution. For example, for a round wire of radius a we expect ρ1 to be proportional to δ(r'-a) where r' = . Perhaps ρ1 = f(θ')δ(r'-a) where (r',θ',z') are cylindrical coordinates with axis at the round wire center. 3. Recall from Section 1.5 (c) and (1.5.17) the fact that there are two distinct areal charge distributions called nc and ns which are related by nc = (ξ/ε)ns. Here ns is the actual surface charge distribution, whereas nc is an adjusted charge density which is directly associated with the current I in the conductor and which accounts for possible leakage in the dielectric. Our n1 and ρ1 are associated with this nc adjusted charge distribution, not with ns. That is why the external factor in (4.1.1) is 1/4πξ instead of 1/4πε. Consider now this charge density ρ1(x). Following a standard methodology, we make the assumption that its functional form may be factored in the following manner, ρ1(x,y,z) = α1(x,y) q1(z) . (4.1.2) C/m3 1/m2 C/m The dimensions of the functions in this factorization are as indicated, so the charge goes with q1. Moreover, without any loss of generality we select the relative scale of the two factors such that the integral of α1(x,y) over a slice of conductor C1 at any z is unity, !Syntax Error, Idx dy α1(x,y) = 1 . (4.1.3) Therefore, we can interpret q1(z) as the total charge per unit length on C1 at location z : !Syntax Error, Idx dy ρ1(x,y,z) = q1(z) !Syntax Error, Idx dy α1(x,y) = q1(z) • 1 = q1(z) . Assume that q2(z) is the charge on the other conductor C2 of a two-conductor transmission line. If q1(z) + q2(z) ≠ 0, then we have a net charge per unit length and the transmission line is acting as a radiating antenna as well as a transmission line. From now on, we ignore this superposed radiation problem and assume that at each value of z, the net charge on both conductors is 0 -- the line is "balanced". This means that q2(z) = - q1(z) ≡ -q(z) . (4.1.4) To simplify notation, we now dispense with the subscript and denote q1(z) = q(z). However, we maintain the subscript on α1(x,y) to emphasize that the two conductors can have completely different cross sectional shapes. The shape of the transverse distribution of charge on C1 is determined by α1(x,y), but the total charge is q(z) per unit length. How can we justify assumption (4.1.2)? This is "separation of variables". The idea is that we assume it without any justification, and then we try to find a solution to our problem which is consistent with the assumption. All we really want is to find a solution to our basic differential equations with their boundary conditions, and any assumptions we make can be justified in the end once we have found a solution. On the other hand, if an assumption like (4.1.2) does not lead to a solution, then it must have been a bad assumption. We have seen earlier how the expected EM field pattern on a transmission line has a constant transverse "shape" and this certainly motivates the assumption (4.1.2). Now insert (4.1.2) into (4.1.1) to get, φ1(x,y,z) = !Syntax Error, Idz' q(z') !Syntax Error, Idx' dy' α1(x',y') (4.1.5) R2 = (x-x')2 + (y-y')2 + (z-z')2 . 4.2 Computation of potential φ due to both conductors of a transmission line Let us now write the potential at an arbitrary point x in the dielectric due to both conductors C1 and C2 : φ12(x) = φ1(x) + φ2(x) = !Syntax Error, Idz' q(z'){ !Syntax Error, Idx1' dy1' α1(x1',y1') – !Syntax Error, Idx2' dy2' α2(x2',y2') } 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') . The minus sign between the terms is due to (4.1.4). Each conductor has its own arbitrary transverse charge distribution αi. The transverse integration variables on C1 are dx1' dy1', while those on C2 are instead dx2' dy2'. In the last two lines we introduce certain transverse distances s1 and s2 as shown. The same dz' integration variable is used for both conductors. The following drawing shows an arbitrary dielectric point x = (x,y,z) located in the z = z plane. The point x1' = (x1',y1',z') lies on C1 at some point of the C1 integration and similarly for x2' = (x2',y2',z'). The full distances R1 and R2 and the transverse distances s1 and s2 are shown. Fig 4.1 One can imagine an expression similar to (4.2.1) for a transmission line consisting of N conductors with some relationship among the qi(z) on each conductor, but we shall restrict our interest to N = 2. 4.3 The Transmission Line Limit Consider again the potential at x due to both conductors shown in (4.2.1), φ12(x) = !Syntax Error, Idz' q(z'){ !Syntax Error, Idx1' dy1' α1(x1',y1') – !Syntax Error, Idx2' dy2' α2(x2',y2') } (4.3.1) As the red-dashed z = z' plane shown in Fig 4.1 is pushed back far from the z = z plane, the vectors which are labeled by distances R1 and R2 become more aligned, and both R1 and R2 become larger. During the transverse integrations over x1' and x2', these Ri vectors then don't vary much. One could then replace the transverse charge density α1(x1',y1') with a point charge at the "center of the conductor" and not make much difference in the R1 vector and its length R1. In this situation, the {...} integrand of the above integral has this form { – } . // when |z-z'| is large (4.3.2) If we then expand the exponentials showing the first few terms, this becomes { – } = { [ - ] + [-jβ + jβ ] + (jβ)2/2 [R1-R2] + ... } = { [ - ] - (β2/2) [R1-R2] + order(β3) } (4.3.3) Since R1 ≈ R2 for large |z-z'| as just discussed, both the leading term and the β2 term are small in an absolute sense as long as β2 is not huge. When |z-z'| is large, both R1 and R2 are large and thus both 1/R1 and 1/R2 are small, and [ 1/R1 - 1/R2] is smaller still due to cancellation between the terms. So our first point is that, in the dz' integration, the main contribution to φ12(x) comes from regions of z' for which |z-z'| is small. Given then that the dz' integration in (4.3.1) is dominated by that part for which |z-z'| is small, we can see that for this controlling integration region the size of distances R1 and R2 will be on the order of the transverse dimension of the transmission line, assuming that we select the point x somewhere between the two conductors. If we vaguely define the transmission line's transverse extent as distance D, then suppose we make the following assumption concerning β : βD << 1 "small β" . (4.3.4) In this case, we can replace e-jβR = 1 and e-jβR = 1 in the integration without significantly changing the result. Then as shown in (4.3.3) there will be a correction term that is order β2 which we shall neglect, as well as higher terms of order βn with n> 2. Notice that the linear β term vanished exactly in our large |z-z'| analysis. This linear term also vanishes in the full analysis since the αi transverse charge functions are normalized to unity: { !Syntax Error, Idx1' dy1' α1(x1',y1') – !Syntax Error, Idx2' dy2' α2(x2',y2') } = (-jβ) {!Syntax Error, Idx1' dy1' α1(x1',y1') - !Syntax Error, Idx1' dy1' α1(x1',y1') ) = (-jβ) {1 - 1} - 0. (4.3.5) Thus, by setting β = 0 in (4.3.1) we are ignoring corrections on the order of β2 and higher, and if β is small, these corrections are very small. The Helmholtz parameter β for the dielectric is 2π/λ where λ is the wavelength of a wave passing down the transmission line. Thus, our "small β" assumption stated above can also be written λ >> D (4.3.6) which says the wavelength is much longer than the size of the transmission line transverse dimensions. This assumption is called the Transmission Line Limit. If we operate within this limit, then (4.3.1) may be approximated as φ12(x) = !Syntax Error, Idz' q(z'){ !Syntax Error, Idx1' dy1' α1(x1',y1') – !Syntax Error, Idx2' dy2' α2(x2',y2') } (4.3.7) We shall now use the small β assumption one more time. We assume that the linear charge density q(z') has the characteristics of a wave traveling down the transmission line ( see also Chapter 5), q(z) = q(0) e-jβz // q(z,t) = q(0,0) ej(ωt-βz) (4.3.8) so that q'(z) = -jβ q(z) q"(z) = (-jβ)2q(z) and so on. We can then write a Taylor expansion for charge density q(z') which appears in our integration, q(z') = q(z) + (z'-z) q'(z) + (1/2) (z'-z)2 q"(z) + ... = q(z) + (-jβ) q(z) (z'-z) + (1/2) (-jβ)2 q(z) (z'-z)2 + ... = q(z) [ 1 + (-jβ) (z'-z) + (1/2) (-jβ)2(z'-z)2 + ... ] . (4.3.9) Since both R1 and R2 are even functions of the quantity (z'-z), and since there is no other (z'-z) dependence in the (4.3.7) integrand, the (-jβ) term in (4.3.9) contributes nothing (this is also true more generally for (4.3.1)). Thus, if we assume small β, we can approximate q(z') ≈ q(z) where we are then ignoring a β2 size term. Once again, if β is small, β2 is very small so our error in replacing q(z') by q(z) is very small. We are only interested in the contributing region where |z-z'| is on the order of transverse dimension D, so the same βD << 1 is being assumed as earlier. We arrive then at our final result for the potential at a point x between the conductors, φ12(x) = q(z) !Syntax Error, Idz'{ !Syntax Error, Idx1' dy1' α1(x1',y1') – !Syntax Error, Idx2' dy2' α2(x2',y2') } (4.3.10) where we have thrown out terms of order β2 and higher. In this transmission line limit approximation, our Helmholtz integral (4.3.1) has been reduced to essentially an electrostatics Coulomb integral where we just sum over the contribution of each piece of charge to the total potential. As noted earlier, q(z) has the normalization of nc and not ns as discussed in Section 1.5 (c) which explains why the leading factor is and not . This allows for the dielectric to have some conductance. It should be noted that the integral of (4.3.10) converges due to the subtraction of the two terms which in turn results from the two conductors having opposite longitudinal charge densities. The individual terms in (4.3.10) do not converge and are in fact each logarithmically divergent in the sense !Syntax Error, Idz' (1/z') = ∞ . 4.4 General Calculation of V(z) We now introduce two new points x1 and x2. The point x1 lies on C1 in the z = z plane, while x2 lies on C2 in this same plane. We then evaluate φ12(x) at x = x1 and subtract from that φ12(x) at x = x2 and in this way we obtain the potential difference between the surfaces of the two conductors at z = z. Recall, Fact 2: φ ≈ constant on a conductor surface in the strong or extreme skin effect regimes within the Transmission Line Limit. (3.7.4) Thus, assuming the small δ regime and treating φ ≈ constant as an equality, the potential difference will be independent of the locations of x2 and x1 as long as they are on their respective surfaces and both have z = z. For this reason, the potential difference is a function only of z. Thus we write, using two copies of (4.3.10), V(z) ≡ φ12(x1) - φ12(x2) = q(z) !Syntax Error, Idz' { !Syntax Error, Idx1' dy1' α1(x1',y1') – !Syntax Error, Idx2' dy2' α2(x2',y2') } – q(z) !Syntax Error, Idz' { !Syntax Error, Idx1' dy1' α1(x1',y1') – !Syntax Error, Idx2' dy2' α2(x2',y2') } (4.4.1) where R112 = (x1-x1')2 + (y1-y1')2 + (z-z')2 = s112 + (z-z')2 s112 = (x1-x1')2 + (y1-y1')2 R122 = (x1-x2')2 + (y1-y2')2 + (z-z')2 = s122 + (z-z')2 s122 = (x1-x2')2 + (y1-y2')2 R222 = (x2-x2')2 + (y2-y2')2 + (z-z')2 = s222 + (z-z')2 s222 = (x2-x2')2 + (y2-y2')2 R212 = (x2-x1')2 + (y2-y1')2 + (z-z')2 = s212 + (z-z')2 s212 = (x2-x1')2 + (y2-y1')2 . (4.4.2) The vector R12 points from our new point x1 to an integration point x2' on C2. Here is a drawing of our new and more complicated situation: Fig 4.2 We next rearrange the four terms in (4.4.1) to get V(z) (4.4.3) = q(z) !Syntax Error, Idz' {!Syntax Error, Idx1' dy1' α1(x1',y1')( - ) -!Syntax Error, Idx2' dy2' α2(x2',y2') ( - ) } . It is now possible to carry out the dz' integrations. The integral of interest is the following, !Syntax Error, Idx { - } = ln(b2/a2) . (4.4.4) Since this is quite important, we confirm with Maple, The separate integrals here are logarithmically divergent, but the combination converges. Thus, !Syntax Error, Idz' ( - ) = !Syntax Error, Idz' ( - ) = ln(s212/s112) and !Syntax Error, Idz' ( - ) = ln(s222/s122) (4.4.5) so that V(z) = q(z) {!Syntax Error, Idx1' dy1' α1(x1',y1') ln(s212/s112) -!Syntax Error, Idx2' dy2' α2(x2',y2') ln(s222/s122) } s212 = (x2-x1')2 + (y2-y1')2 s222 = (x2-x2')2 + (y2-y2')2 (4.4.6) s112 = (x1-x1')2 + (y1-y1')2 s122 = (x1-x2')2 + (y1-y2')2 . The four transverse distances are shown in this figure, Fig 4.3 Equation (4.4.6) expresses the potential between the two transmission line conductors at some plane z in terms of the charge distributions on the conductors αi. In general, these charge distributions are not known, so one cannot regard (4.4.6) as a general purpose silver bullet to solve transmission line problems. On the other hand, as we shall see, equation (4.4.6) is one of a group of equations which will allow us to express several different transmission line parameters in terms the same integral, and one then obtains a relation between these parameters. For example, in analogy to what we did with a parallel plate capacitor in (1.5.19), we may define the complex capacitance C' per unit length of our transmission line using (4.4.6) as follows: = = {!Syntax Error, Idx1' dy1' α1(x1',y1') ln(s212/s112) -!Syntax Error, Idx2' dy2' α2(x2',y2') ln(s222/s122) } = K and V(z) = q(z) K (4.4.7) where K is a dimensionless real number obtained from a geometric integral of the normalized transverse charge distributions αi (recall that αi is has dimensions 1/m2 in (4.1.2)), K ≡ !Syntax Error, Idx1' dy1' α1(x1',y1') ln(s212/s112) -!Syntax Error, Idx2' dy2' α2(x2',y2') ln(s222/s122) . (4.4.8) Recall from (1.5.20) that (C', C and G are discussed further in Section 4.11 below) C' = C + 1/(jωR) = C + G/(jω) (4.4.9) where conductance (per unit length) G is associated with the imaginary part of C'. We then have 4πξ/K = C' = C + G/jω or 4π(ε+σ/jω)/K = C + G/jω // (1.5.1) for ξ so that C = 4πε/K capacitance per unit length of the transmission line G = 4πσ/K conductance per unit length of the transmission line so C/G = ε/σ . (4.4.10) Here G = 1/R' is the conductance across the dielectric between a unit length of the two conductors. This is unrelated to the longitudinal resistance R of the conductors themselves, though that parameter will arise later on in the form of surface impedance Zs. We only have G ≠ 0 if the dielectric has some conductance σ ≠ 0. Note from above and (1.2.8) that dim(C) = dim(ε) = farad/m and dim(G) = dim(σ) = mho/m . We now quote several results that will be derived later in Section 4.11. First we show below in (4.11.29) with (4.11.30) that the external inductance per unit length of our transmission line is also related to this same constant K, Le = (μ/4π)K . (4.4.11) Second, we show in (4.11.16) that the characteristic impedance of the transmission line is given by Z0 ≡ = . (4.4.12) At sufficiently large ω we can neglect the R and G terms to get this real value, Z0 = . // large ω (4.4.13) Third, we show in (4.11.36) that, for large ω, L → Le so Z0 = = = (1/4π) K = (1/4π) K Zm (4.4.14) where Zm = is the "impedance of the dielectric medium" having μ and ε. Recall that for free space we had in (1.1.29) Zfs = = 376.73032 ohms => Zfs /4π = 29.97948 ≈ 30 Ω . (4.4.15) Typically one has μ = μ0 so then (note that εrel and K are dimensionless), Z0 = (1/4π) K = (1/4π) K = (1/4π) K Zfs = K (Zfs /4π) / or Z0 ≈ (K /) 30Ω . εrel ≡ ε/ε0 . (4.4.16) We then summarize the parameters of a transmission line in terms of dimensionless real integral K : C = 4πε/K capacitance per unit length (4.4.17) G = 4πσ/K transverse conductance per unit length Le = (μ/4π) K external inductance per unit length Z0 ≈ (K /) 30Ω characteristic impedance μ = μ0, εrel ≡ ε/ε0 R = Re(Zs1+Zs2) resistance of conductors, see (4.11.34) L = Le + (1/ω) Im(Zs1+ Zs2) total inductance, see (4.11.34) Zsi = surface impedance of conductor i, see (2.4.1) and (4.11.10) The last three items are not determined by integral K and we just mention them for completeness's sake. All these equations will be more fully developed in Section 4.11 below, but we jump ahead a bit in order to display two important examples. 4.5 Example: Transmission line with widely-spaced round wires of unequal diameters Consider a transmission line made from two round wires of radii a1 and a2 and center line spacing b. In the case that b >> a1 and a2, the charge distribution on each round wire is symmetric about the wire and in this situation ( a rare one admittedly) we know the two charge distributions: α1(x,y) = α1(r,θ) = δ(r - a1)/(2πa1) α2(x,y) = α2(r,θ) = δ(r - a2)/(2πa2) . (4.5.1) The 1/(2πa1) factor is required so that the integral of α1 is unity as required by (4.1.3), !Syntax Error, Idx dy α1(x,y) = !Syntax Error, Idθ !Syntax Error, Irdr α1(r,θ) = !Syntax Error, Idθ !Syntax Error, Irdr δ(r - a1)/(2πa1) = !Syntax Error, Idθ a1/(2πa1) = 2π a1/(2πa1) = 1 . (4.5.2) Note: The reason the αi are symmetric is that the two conductors are so far apart that each one is essentially "in isolation" and so the charge assumes an axially symmetric distribution. An analogy would be that for two point charges far apart, the E field close to either point charge is spherically symmetric because close to one charge the field of the other can be neglected. Our task is then to compute the integral K shown in (4.4.8), K = !Syntax Error, Idθ1 !Syntax Error, Ir1dr1 [δ(r1 - a1)/(2πa1)] ln(s212/s112) -!Syntax Error, Idθ2 !Syntax Error, Ir2dr2 [δ(r2 - a2)/(2πa2)] ln(s222/s122) . = !Syntax Error, Idθ1 1/(2π) ln(s212/s112) -!Syntax Error, Idθ2 1/(2π) ln(s222/s122) = 1/(2π) { !Syntax Error, Idθ1 [ln(s212) - ln(s112)] - !Syntax Error, Idθ2 [ln(s222) - ln(s122)] } (4.5.3) where we then have four integrals to evaluate. We shall choose our V(z) potential-determining reference points x1 and x2 as shown in this drawing, Fig 4.4 The four sij distances can be read off from the drawing using the law of cosines, s212 = a12 + (b-a1)2 - 2 a1(b-a1) cos(θ1) s112 = a12 + a12 - 2 a1 a1 cos(θ1) = 2a12(1 - cos(θ1)) s222 = a22 + a22 - 2 a2 a2 cos(π-θ2) = 2a22(1 + cos(θ2)) s122 = a22 + (b-a2)2 + 2 a2(b-a2) cos(θ2) . (4.5.4) We then invoke the following integral from p 531 of GR7, which we rewrite as !Syntax Error, Idθ ln (A ± Bcosθ) = 2π ln[(1/2)(A + )] . (4.5.5) The four integrals are then easily evaluated: !Syntax Error, Idθ1 ln(s212) = !Syntax Error, Idθ1ln([a12 + (b-a1)2 - 2 a1(b-a1) cos(θ1)] A = a12 + (b-a1)2 B = 2 a1(b-a1) A2-B2 = [a12 + (b-a1)2]2 - 4 a12(b-a1)2 = [a12 - (b-a1)2]2 => = (b-a1)2- a12 > 0 b >> a1 => !Syntax Error, Idθ1 ln(s212) = 2π ln[(1/2)( a12 + (b-a1)2 + (b-a1)2 - a12 ) = 2π ln[(b-a1)2] . The fourth integral is the same with 1↔ 2, and the different sign of the second term in s122 makes no difference, !Syntax Error, Idθ2 ln(s122) = 2π ln[(b-a2)2] . The second integral is !Syntax Error, Idθ1 ln(s112) = !Syntax Error, Idθ1ln([2a12(1 - cos(θ1))] A = B = 2a12 , A2-B2 = 0 = 2π ln[(1/2) 2a12] = 2π ln(a12) . The third integral is similar giving !Syntax Error, Idθ2 ln(s222) = 2πln(a22) . To summarize: !Syntax Error, Idθ1 ln(s212) = 2π ln[(b-a1)2] !Syntax Error, Idθ1 ln(s112) = 2π ln(a12) !Syntax Error, Idθ2 ln(s222) = 2πln(a22) !Syntax Error, Idθ2 ln(s122) = 2π ln[(b-a2)2] . (4.5.6) Then from (4.5.3) we find K = 1/(2π) { !Syntax Error, Idθ1 [ln(s212) - ln(s112)] - !Syntax Error, Idθ2 [ln(s222) - ln(s122)] } = ln[(b-a1)2] - ln(a12) - ln(a22) + ln[(b-a2)2] = ln [] = 2 ln [] = 2 ln [] // since we assumed at the start that b >> a1, a2 = 4 ln(b/) . (4.5.7) Therefore the transmission line parameters from (4.4.17) are, K = ln(b/) C = 4πε/K = πε / ln(b/) G = 4πσ/K = πσ / ln(b/) Le = (μ/4π) K = (μ/π) ln(b/) Z0 = (K /) 30Ω = (1/) ln(b/) 120Ω . (4.5.8) Sometimes these formulas are written in terms of wire diameters di = 2ai in which case K = 4 ln[b/] = 4 ln[2b/] = 2 ln[4b2/d1d2] . (4.5.9) Since we are assuming b >> d1,d2 we know that x ≡ 2b2/d1d2 >> 1. Therefore ch-1x = ln[x + ] ≈ ln(2x) // an identity Siegel 8.56, then an approximation (4.5.10) so ch-1(2b2/d1d2) ≈ ln(4b2/d1d2) . Then we can write K as K = 2 ln[4b2/d1d2] = 2 ch-1(2b2/d1d2) (4.5.11) and so Z0 = (K /) 30Ω = (1/) ch-1(2b2/d1d2) 60Ω . (4.5.12) It is not easy to find expressions for C,G and Le for the unequal radii geometry, but Z0 does appear for example in Reference RDE page 29-23 where we find: Fig 4.5 with D = our b. For D >> d1,d1 this shows N = 2D2/(d1d2), and this then agrees with (4.5.12). This quoted result is in fact correct (with the two extra terms shown in N) even when D is not large. We shall derive this full result in Chapter 6, equation (6.3.12). In the special case that a1 = a2 ≡ a we get, K = 4 ln(b/a) C = 4πε/K = πε/ ln(b/a) G = 4πσ/K = πσ/ ln(b/a) Le = (μ/4π) K = (μ/π) ln(b/a) Z0 = (K /) 30Ω = (1/) ln(b/a) 120Ω = (1/) ln(2b/d) 120Ω d = 2a . (4.5.13) The first three results agree with King TLT p17 (30b), The expression for Z0 agrees with the RDE source quoted above, Fig 4.6 where again D = b and for "air" εrel= 1. Power Transmission Lines Ignoring proximity effects of the ground and possible ground wires, one can consider a single phase power transmission line as fitting into this example. The first interesting number is skin depth. For aluminum at f = 60 Hz we find σaluminum = 3.7 x 107 mho/m // recall σcopper ≈ 5.8 x 107 (annealed) μ0 = 4π x 10-7 henry/m δ ≡ ≈ = So δ ≈ 1 cm. Thus, skin effect could be significant for a very large diameter wire. Typically the individual strands of a 1500 amp cable are 1/6" in diameter or 0.2 cm in radius, so there is some slight non-uniformity in the current distribution. If the strands are not insulated one should think of this more in terms of the total cable diameter including all strand layers which might be 1". Usually the requirement of low power loss requires that R be relatively small compared to ωL. The 1500A cable just noted has R = .02Ω per thousand feet. Similarly, the conductance G (mostly from insulator leakage) is very small compared to ωC. Thus, (4.4.12) leads to (4.4.16) stating Z0 ≈ K 30Ω . If the full cable is 1" in diameter and the two lines are spaced 1 m apart, we can compute K from (4.5.13), K = 4 ln(b/a) = 4 ln( 1m/0.5") = 4 ln(39.37*2) = 17.5 so then from (4.4.16), Z0 ≈ K 30Ω = 17.5 * 30 Ω = 524Ω Rajput (p 554) claims power lines typically range from 400 to 600 Ω. See southwire.com for data on transmission line cables. 4.6 Example: A coaxial cable A coaxial cable is the other transmission line where we know the surface charge distribution is that given by (4.5.1). The analysis of the previous section resulting in (4.5.3) is then unchanged, and we find that K is still given by (4.5.3), K = 1/(2π) { !Syntax Error, Idθ1 [ln(s212) - ln(s112)] - !Syntax Error, Idθ2 [ln(s222) - ln(s122)] } . (4.5.3) What is different is that we have a different picture describing the various sij distances. The new picture is this, where the cross section circles have radii a2 > a1 : Fig 4.7 As we did in the previous section, we "read off" the sij expressions using the law of cosines: s212 = a12 + a22 - 2 a1a2cos(θ1) s112 = a12 + a12 - 2 a1 a1 cos(θ1) = 2a12(1 - cos(θ1)) s222 = a22 + a22 - 2 a2 a2 cos(θ2) = 2a22(1 - cos(θ2)) s122 = a22 + a12 - 2 a1a2 cos(θ2) . (4.6.1) Recalling, !Syntax Error, Idθ ln (A ± Bcosθ) = 2π ln[(1/2)(A + )] . (4.5.5) we find, !Syntax Error, Idθ1 ln(s212) = !Syntax Error, Idθ1 ln[a12 + a22 - 2a1a2cos(θ1)] A = a12 + a22 B = 2a1a2 A2-B2 = (a12 +a22)2 - 4a12a22 = (a12 -a22)2 => = (a22 -a12) > 0 since a2 > a1 so !Syntax Error, Idθ1 ln(s212) = 2π ln[(1/2)( a12 + a22 + (a22 -a12) ) = 2πln(a22) . Similarly !Syntax Error, Idθ2 ln(s122) = 2π ln[(1/2)( a12 + a22 + (a22 -a12) ) = 2πln(a22) = same as above . The other two integrals are, !Syntax Error, Idθ1 ln(s112) = !Syntax Error, Idθ1 ln[2a12(1 - cos(θ1))] A = B = 2a12 = 2π ln[(1/2)2a12] = 2πln(a12) !Syntax Error, Idθ2 ln(s222) = !Syntax Error, Idθ1 ln[2a22(1 - cos(θ2))] A = B = 2a22 = 2π ln[(1/2)2a22] = 2πln(a22) . To summarize: !Syntax Error, Idθ1 ln(s212) = 2πln(a22) !Syntax Error, Idθ1 ln(s112) = 2π ln(a12) !Syntax Error, Idθ2 ln(s222) = 2π ln(a22) !Syntax Error, Idθ2 ln(s122) = 2π ln(a22) . (4.6.2) Then from (4.5.3) we find K = 1/(2π) { !Syntax Error, Idθ1 [ln(s212) - ln(s112)] - !Syntax Error, Idθ2 [ln(s222) - ln(s122)] } = ln(a22) - ln(a12) - ln(a22) + ln(a22) = ln(a22/a12) = 2 ln(a2/a1) . (4.6.3) The coaxial transmission line parameters are then given by, C = 4πε/K = 2πε / ln(a2/a1) G = 4πσ/K = 2πσ / ln(a2/a1) Le = (μ/4π)K = (μ/2π) ln(a2/a1) Z0 = (K /) 30Ω = (1/) ln(a2/a1) 60Ω (4.6.4) We verify the C and Le parameters from http://en.wikipedia.org/wiki/Coaxial_cable , Fig 4.8 To verify the Z0 value, first recall that (the positive square root is implied here) ch-1x = ln[x + ] // Spiegel identity 8.56, valid for x ≥ +1 (4.6.5) If we set x ≡ ( + ) = and we assume a > 0 and b > 0, then x2 - 1 = - 1 = { } = => = = sign(b-a) (- ) => x + = ( + ) + sign(b-a) (- ) = => ln [x + ] = ln [ ] = sign(b-a) ln(b/a) . Thus we have shown that (note that both sides are invariant under a ↔ b ) ch-1[( + )] = sign(b-a)ln . a > 0 and b > 0 (4.6.6) With this rather elaborate fact, and since we have b > a, we can rewrite Z0 above as Z0 = (1/) ch-1[( + )] 60 Ω . (4.6.7) Again we quote from reference RDE page 29-24 Fig 4.9 In our centered case c = 0 so U = (1/2)(D/d+d/D) and we have agreement. The full off-center result is derived later in Chapter 6, equation (6.3.15). Comment: In the examples of Sections 4.5 and 4.6, the current distributions in the involved round wires are axially symmetric. Therefore all the results of Chapter 2 apply. In particular, Chapter 2 calculates the surface impedance Zs for a round wire in complete detail, including its limits for large and small ω. For example, at low frequency for a wire of radius a, Zs(ω) = + jω = Rs + jωLs // low frequency limit (2.4.12) and one sees that Rs is the expected DC resistance and Ls is the internal impedance Li as computed in Appendix C equation (C.3.5). Having presented our two Examples, we now resume development of the transmission line equations. 4.7 Computation of Az due to one conductor of a transmission line In summary box (1.5.23) we state the following Helmholtz integral for the vector potential arising from currents in a set of conductors, A(x,ω) = Σi∫μiJi(x',ω) dV' (1.5.23) where the sum Σi is over the conductors and μi is the permeability of conductor i. In our transmission line context, and as discussed in Chapter 3, the dominant current is in the z (longitudinal) direction, while transverse currents are very small. For example, in the estimate of Section 3.7 (s) we found that Jr/Jz < 1.4 x 10-4 below 10 GHz and Jr/Jz < 4.6 x 10-8 at low frequency. Looking at the above Helmholtz solution to the Helmholtz equation, if we neglect these transverse currents, we are then in effect neglecting the transverse components of A, and this is what we shall do from now on: Fact: The transverse components Ax and Ay can be neglected so that A = Az. (Appendix M) (4.7.1) Our starting point then is the following expression for the potential Az at some arbitrary point x in the dielectric due to conductor C1 of a transmission line, Az1(x,ω) = ∫ Jz1(x',y',z',ω)dx'dy'dz' . R = |x - x'| (4.7.2) Here the point x' = (x',y',z') runs over the volume of C1 and R is the distance between the observation point x in the dielectric and the integration point x'. Parameter β is for the dielectric while μ1 is for the conductor. In the analogous φ solution (4.1.1) everything has the same form as (4.7.2) but in (4.1.1) the charge density exists only on the conductor surface. Nevertheless, we represented that charge density as a volume density, and only later in examples set that volume density to a surface distribution. Thus, the parallel between the φ and the Az analysis is very close, not surprising in light of (1.3.11). Another difference is that for φ the leading factor is 1/(4πξ) where ξ was the complex dielectric constant of the dielectric. In (4.7.2) this factor is replaced by (μ1/4π) where μ1 is the magnetic permeability of the conductor. We next make the same assumption of separation of variables to write Jz1(x,y,z) = b1(x,y) i1(z) A/m2 1/m2 A (4.7.3) where i1 is scaled such that !Syntax Error, Idx dy b1(x,y) = 1 . (4.7.4) Here b1(x,y) describes the distribution of the current density across the conductor C1 cross section. At DC this density is a uniform constant, but at higher ω the density becomes non-uniform in two ways. First, it becomes concentrated away from the central region due to the skin effect. Second it is non-uniform in that it tends to concentrate on the portion of conductor C1 which is closest to conductor C2. In the corresponding equation ρ1(x,y,z) = α1(x,y) q1(z) of (4.1.2), α1(x,y) exists only on the conductor surface, and is generally non-uniform in the second sense noted above for b1(x,y). As before, we can now interpret i1(z) as the total current in C1 at z. Again assuming that there is no net superposed radiating antenna current, we have equal and opposite currents in the two conductors so the line is a balanced line, and then i2(z) = - i1(z) = -i(z) . (4.7.5) This then leads to Az1(x,y,z) = !Syntax Error, Idz' i(z') !Syntax Error, Idx' dy' b1(x',y') . (4.7.6) Comments regarding μ This is a subtle subject and is not discussed in King's transmission line theory book. If the conductor and dielectric have the same permeability so that μ1 = μ, then there exists no "magnetic boundary" between the conductor and dielectric. The solution (4.7.2) is then smooth at this boundary, and so Az1(x,y,z) "naturally" satisfies these two boundary conditions, Az1(x+) = Az1(x-) (1/μ)∂nAz1(x+) = (1/μ1) ∂nAz1(x-) (4.7.7) where x+ is just outside the conductor surface and x- is just inside. The second equation here is just (1.1.46) in the case that there is no free surface current Kfree flowing on the boundary, and indeed in our example at hand there is no such free surface current. Since we have assumed that μ = μ1, this second boundary condition just says ∂nAz1(x+) = ∂nAz1(x+). Since there is no magnetic boundary at the conductor/dielectric interface, the solution (4.7.2) is continuous and all its derivatives are also continuous at the boundary, since nothing special happens at that boundary. Thus, the Helmholtz integral solution provides the whole solution for Az1 since it meets both "boundary conditions" at this pseudo boundary. If on the other hand we have μ1 ≠ μ, then there is a magnetic boundary between conductor and dielectric which we have to worry about. In this case, (4.7.2) cannot possibly satisfy the second boundary condition of (4.7.7) since, as already noted, the Az1 of (4.7.2) satisfies ∂nAz1(x+) = ∂nAz1(x+). Thus, in this case (4.7.2) is not the full solution for Az1. One must add a homogeneous Helmholtz equation solution to (4.7.2) in order to have a proper solution for Az1 that satisfies both equations in (4.7.7). It turns out that the correct total Az1 solution can be generated by adding a certain fictitious surface current term to μ1Jz1 in (4.7.2). Since such a surface current vanishes on both sides of the boundary between μ and μ1, the Helmholtz solution due just to this surface current term is in fact a homogeneous solution to the Helmholtz equation in both the conductor and dielectric regions, away from that boundary. It turns out moreover that the correct fictitious surface current to add is in fact the magnetization surface current Jm which is created at the boundary between μ ≠ μ1. Adding this surface current is just a "trick" in order to generate the correct homogeneous adder solution so that the resulting total Az1 satisfies both boundary conditions in (4.7.7). Formally speaking, the Ji appearing in (1.5.4) and then Jz1 in (4.7.2) should not include such magnetization currents since this J is really the J in Maxwell's equation curl H = ∂tD + J, and this J does not include magnetization currents -- it includes only normal conduction currents. In our current Chapter 4, we want (4.7.2) to represent the complete solution for Az1 and for that reason we must restrict our analysis to the situation where dielectric and all conductors have the same permeability which we shall just call μ. In practice, one normally has μ = μ1 = μ0. In order to handle the more general case of μ ≠ μ1, we have to deal with the inhomogeneous adder solutions or equivalently with the abovementioned fictitious surface current, and this complicates our analysis which is already quite complicated. So, for the moment, we now make the same assumption made by King and other authors: Fact: From now on, conductors and dielectric must have the same permeability μ. (4.7.8) After fully developing this special case, we shall then extend the theory in Section 4.12 to allow for μ ≠ μ1. Appendix G shows for the round wire how the inhomogeneous adder solution is found and how it then causes the boundary conditions (4.7.7) to be met when μ1 ≠ μ. Appendix B shows how the addition of a fictitious surface current term μ0Jm provides an alternate and simpler solution to the same problem of meeting boundary conditions (4.7.7) when μ1 ≠ μ. It then shows exactly how this works in the special case of a round wire. Having now mentioned that the Helmholtz integral might not provide a total solution, the reader might fairly ask why it is that the Helmholtz integral solution φ1(x,ω) of (4.1.1) provides a complete and viable solution to the φ Helmholtz equation, given that in general the conductor (ε1) and dielectric (ε) have different ε values, so there should be an "electric boundary" where ε meets ε1. The reason is that, according to (1.1.47), the boundary condition corresponding to the second line of (4.7.7) reads [ε1En(x+) - εEn(x+)] = nfree(x) . Since we are neglecting transverse A components as stated in (4.7.1), and since our notation ∂n indicates a normal conductor derivative which is transverse (to z), we have E = - grad φ - ∂tA => En = -∂nφ (4.7.9) so we have then this set of boundary conditions for φ1, φ1(x+) = φ1(x-) [ε1∂nφ1(x+) - ε∂nφ1(x-)] = nfree(x). (4.7.10) These look a bit like (4.7.7) for Az1. The big difference is that in this case there does exist a free surface charge nfree and it simply adjusts itself to make (4.7.10) be true. Thus, the Helmholtz integral (4.1.1) does in fact meet the required electrical boundary conditions without the need for a homogeneous solution adder term. A less formal way to state this is that, in the electrical case, we can regard the surface charge as in fact lying on the dielectric side of the boundary, and then the boundary is of no interest in our problem of analyzing fields in the dielectric. 4.8 Computation of potential Az due to both conductors of a transmission line Let us now write the potential at an arbitrary point x in the dielectric due to both conductors C1 and C2. We accept the requirement of (4.7.8) and require that all conductors have the same μ as the dielectric, so then μ1 = μ and μ2 = μ. Then, Az12(x) = Az1(x) + Az2(x) = !Syntax Error, Idz' i(z') { !Syntax Error, Idx1' dy1' b1(x1',y1') – !Syntax Error, Idx2' dy2' b2(x2',y2') } R12 = (x-x1')2 + (y-y1')2 + (z-z')2 = s12 + (z-z')2 s12 = (x-x1')2 + (y-y1')2 (4.8.1) R22 = (x-x2')2 + (y-y2')2 + (z-z')2 = s22 + (z-z')2 s22 = (x-x2')2 + (y-y2') . The picture going with the above equation is identical to Fig 4.1 below (4.2.1) except the points x1' and x2' can be in the interior of the conductors, not just on the boundary of the conductor. 4.9 Transmission Line Limit Revisited In Section 4.3 we discussed the so-called transmission line limit of small β in the context of the scalar potential φ. We could (but won't) repeat the discussion verbatim here making the following substitutions: q(z) → i(z) αi → bi φ12 → Az12 → . The conclusion is that in the transmission line limit (small β, long wavelength λ = 2π/β) we may write Az12(x) = i(z) !Syntax Error, Idz'{ !Syntax Error, Idx1' dy1' b1(x1',y1') – !Syntax Error, Idx2' dy2' b2(x2',y2') } (4.9.1) which is analogous to (4.3.7). 4.10 General Calculation of W(z) We now introduce two new points x1 and x2. The point x1 lies on C1 in the z = z plane, while x2 lies on C2 in this same plane. We then evaluate Az12(x) at x = x1 and subtract from that Az12(x) at x = x2 and in this way we obtain the Az potential difference between the surfaces of the two conductors at z = z which we shall call W(z). Recall, Fact 5: On each conductor boundary, Az ≈ constant in the extreme or strong skin effect regimes. (3.7.20) Thus, assuming the small δ regime and treating Az ≈ constant as an equality, the Az potential difference will be independent of the locations of x2 and x1 as long as they are on their respective surfaces and both have z = z. For this reason, the Az potential difference is a function only of z. Thus we write, using two copies of (4.9.1), W(z) ≡ Az12(x1) - Az12(x2) = i(z) !Syntax Error, Idz' { !Syntax Error, Idx1' dy1' b1(x1',y1') – !Syntax Error, Idx2' dy2' b2(x2',y2') } – i(z) !Syntax Error, Idz' { !Syntax Error, Idx1' dy1' b1(x1',y1') – !Syntax Error, Idx2' dy2' b2(x2',y2') } (4.10.1) where R112 = (x1-x1')2 + (y1-y1')2 + (z-z')2 = s112 + (z-z')2 s112 = (x1-x1')2 + (y1-y1')2 R122 = (x1-x2')2 + (y1-y2')2 + (z-z')2 = s122 + (z-z')2 s122 = (x1-x2')2 + (y1-y2')2 R222 = (x2-x2')2 + (y2-y2')2 + (z-z')2 = s222 + (z-z')2 s222 = (x2-x2')2 + (y2-y2')2 R212 = (x2-x1')2 + (y2-y1')2 + (z-z')2 = s212 + (z-z')2 s212 = (x2-x1')2 + (y2-y1')2 . (4.10.2) The picture going with the above equation is identical to Fig 4.2 below (4.4.2) except, once again, the points x1' and x2' can be in the interior of the conductors, not just on the surface of the conductors. Also, we replace the figure's double arrow label V(z) with W(z). We then reorder the four terms to get W(z) (4.10.3) = i(z)!Syntax Error, Idz' {!Syntax Error, Idx1' dy1' b1(x1',y1')( - ) -!Syntax Error, Idx2' dy2' b2(x2',y2') (- ) } . The dz' integrals are the same as those done in Section 4.4 and we then arrive at W(z) = i(z) {!Syntax Error, Idx1' dy1' b1(x1',y1') ln(s212/s112) -!Syntax Error, Idx2' dy2' b2(x2',y2') ln(s222/s122) } s212 = (x2-x1')2 + (y2-y1')2 s222 = (x2-x2')2 + (y2-y2')2 (4.10.4) s112 = (x1-x1')2 + (y1-y1')2 s122 = (x1-x2')2 + (y1-y2')2 which is analogous to (4.4.6) for V(z). The corresponding drawing is analogous to Fig 4.3 where, once again, the integration points x1' and x2' are inside the conductor : Fig 4.10 Equation (4.10.4) expresses the Az potential between the two transmission line conductors at some plane z in terms of the current distributions bi within the conductors. Now, the Stokes theorem applied to B = curl A says curl A = B A ds = ∫S B dA . (1.1.39) Consider the red loop shown in this top view of the two transmission line conductors. The loop is intended to have a tiny width dz, and the top view obscures the fact that each conductor has an arbitrary cross section. The loop makes contact with the points x1 and x2 shown in the previous figure, Fig 4.11 Since we neglect any transverse components of A, the Stokes theorem says [Az1(top) - Az2(bottom) ] dz = [ magnetic flux through red loop] = ∫S B dA . (4.10.5) If we regard the two short dz length conductor pieces as forming a tiny "inductor", closed on the ends by the vertical red lines, we can use this definition of inductance to compute the inductance of that inductor: [magnetic flux through red loop] = (Ledz) i(z) . (4.10.6) Here (Ledz) is the inductance of our tiny loop, so Le is the transmission line inductance per unit length. We know (as in Appendix C) that there will be magnetic flux inside the conductors as well as between them, and for that reason Le as defined here only accounts for the "external" inductance of the transmission line, again see Appendix C. Since [Az1(top) - Az2(bottom) ] = W(z) according to (4.10.1), we may combine (4.10.5) and (4.10.6) to obtain W(z) = Le i(z) . (4.10.7) Therefore from (4.10.4) we have found that Le = = KL (4.10.8) where KL is the following dimensionless real number, KL ≡ !Syntax Error, Idx1' dy1' b1(x1',y1') ln(s212/s112) -!Syntax Error, Idx2' dy2' b2(x2',y2') ln(s222/s122) . (4.10.9) This number is reminiscent of the number K obtained in Section 4.4, K ≡ !Syntax Error, Idx1' dy1' α1(x1',y1') ln(s212/s112) -!Syntax Error, Idx2' dy2' α2(x2',y2') ln(s222/s122) . (4.4.8) In the next section it will be shown that these two dimensionless numbers are exactly the same. 4.11 The Classic Transmission Line Equations The results of the previous sections of this chapter may be succinctly summarized as: (4.11.1) = = K (4.4.7) Le = = KL (4.10.8) K ≡ !Syntax Error, Idx1' dy1' α1(x1',y1') ln(s212/s112) -!Syntax Error, Idx2' dy2' α2(x2',y2') ln(s222/s122) (4.4.8) KL ≡ !Syntax Error, Idx1' dy1' b1(x1',y1') ln(s212/s112) -!Syntax Error, Idx2' dy2' b2(x2',y2') ln(s222/s122) (4.10.9) Notice that we have made no assumptions whatsoever about the cross-sectional shape of the transmission line. We have only assumed that the transverse dimensions are small compared to the wavelength λ that corresponds to β -- this was the transmission line limit. There are several equations from Chapter 1 we shall now press into service: E = - grad φ - ∂tA (1.3.1) div A = - με ∂tφ - μσφ . // the King gauge (1.3.18) In the frequency domain these become, E = - grad φ - jωA div A = - j (β2/ωφ . // the King gauge, see (1.5.1) and (1.5.5) re β2 (4.11.2) According to the Fact stated in (4.7.1), potential A has only component Az, so these equations become Ez(x) = - ∂zφ(x) - jωAz(x) ∂zAz(x) = - j (β2/ωφ(x) . (4.11.3) However, as was shown at the end of Step 1 below (3.7.8), the second line of (4.11.3) can only be justified in the strong or extreme skin effect regimes, and we continue then to assume our transmission line is operating at sufficiently high ω to be in the small δ regime. The potentials in the above equations are those due to both conductors and were denoted as φ12 and Az12 in the previous sections. We then rewrite the above as Ez(x) = - ∂zφ12(x) - jωAz12(x) ∂zAz12(x) = - j (β2/ωφ12(x) (4.11.4) Recall now the conductor-surface-located points x1 and x2 as shown for example in Fig 4.10. If we evaluate each of the above equations at x = x1 and then x = x2 and then subtract, we get Ez(x1) - Ez(x2) = -∂z[φ12(x1) - φ12(x2)] - jω[Az12(x1) - Az12(x2)] ∂z[Az12(x1) - Az12(x2)] = - j (β2/ω[φ12(x1) - φ12(x2)] . (4.11.5) Then using these definitions (again, we are assuming the strong or extreme skin depth regime) V(z) ≡ φ12(x1) - φ12(x2) (4.4.1) W(z) ≡ Az12(x1) - Az12(x2) (4.10.1) we may rewrite (4.11.5) in this simple manner, Ez(x1) - Ez(x2) = - ∂zV - jωW (4.11.6a) ∂zW = - j (β2/ωV . (4.11.6b) The quantity Ez(x1) is the longitudinal electric field at point x1 on the surface of conductor C1. It is related to the conductor's at-the-surface current density by Jz(x1) = σ Ez(x1). If the conductor were "perfect", we would have σ = ∞ and Ez(x1) = 0, but real conductors are not perfect. However, since we are assuming the strong or extreme skin effect all along here in our analysis, we do know that Ez(x1) and Ez(x2) are very small. Our theory now has an inconsistency. At the start of Section 4.4 we replaced the approximate fact that φ ≈ constant on the conductor boundaries with φ = constant. Similarly at the start of Section 4.10 we replaced Az ≈ constant on the conductor boundaries with Az = constant. This allowed us to define V(z) and W(z) as if they depended only on coordinate z. We know that for a general transmission line operating at ω > 0, the current density Jz inside the conductors will not be uniformly distributed. It will be larger in the conductor region closest to the other conductor. This "proximity effect" is discussed in Appendix P from an eddy current point of view, see Fig P.13 for an example. The Jz current non-uniformity can be very dramatic as for example in a transmission line having this cross section: Fig 4.12 Since Jz is non-uniform in each conductor, so is Ez, and so we expect Ez(x1) to be a strong function of the point x1 on the perimeter of C1, certainly in the above cross section example. Recall that Ez is continuous through the boundary, so x1 being on the boundary is unambiguous. This means that the left side of (4.11.6a) is a function of (x1,y1,z) and (x2,y2,z) whereas the right side in our theory is a function only of z. To remedy this inconsistency, we now have to think of V and W as having very slight dependence on x1 and x2 which we generally ignore, but which we must face up to in (4.11.6a). This is a manifestation of the fact that in reality φ ≈ constant and Az ≈ constant on the boundaries (with ≈ and not = ). In the extreme skin effect regime (think a very good conductor), the left side of (4.11.6a) can be a violent function of x1 and x2 as in the case of the above figure, but the left side is always very small, even where it is largest, and its variation can be accommodated by the right side of (4.11.6a) which is the difference of large valued functions which vary only slightly with x1 and x2. We now proceed in our development of the transmission line equations with this understanding of the functions V and W. As shown in (2.4.1), Ez(x1) can be related to the total current in the conductor i(z) by a quantity known as the surface impedance Zs, so Ez(x1) = Zs1(x1) i1(z) Ez2(x2) = Zs2(x2) i2(z) . (4.11.7) The surface impedance of a perfect conductor is zero. Since i1(z) = -i2(z) = i(z), we rewrite (4.11.6) as, [Zs1(x1) + Zs2(x2)] i(z) = - ∂zV(z) - jωW(z) (4.11.8a) ∂zW(z) = - j (β2/ωV(z) . (4.11.8b) We just noted that Ez(x1) may vary violently around the perimeter of conductor C1, especially for a cross section like that shown above. This of course means that Zs1(x1) also varies violently on the perimeter, causing equation (4.11.8a) to have the same inconsistency as (4.11.6a). We now perform a certain slight of hand. We back up to the first equation of (4.11.4) evaluated at x = x1, Ez(x1) = - ∂zφ12(x1) - jωAz12(x1) or Zs1(x1) i1(z) = - ∂zφ12(x1) - jωAz12(x1) . (4.11.9) Multiply both sides by Jz(x1), integrate over the boundary of C1 (all at z = z) , then divide by the integral of Jz(x1) over the boundary. The left side so obtained we define as the constant Zs1 : Zs1 ≡ φ12(C1) ≡ (4.11.10a) while φ12(C1) is the weighted mean value of φ12 over the boundary of C1 and similarly for Az12(C1). We do a similar average over C2 to obtain Zs2 and φ12(C2) and Az12(C2), Zs2 ≡ φ12(C2) ≡ . (4.11.10b) Equation (4.11.9) now becomes, Zs1 i1(z) = - ∂zφ12(C1) - jωAz12(C1) . (4.11.11) We then slightly redefine V and W as follows, V(z) ≡ φ12(C1) - φ12(C2) W(z) ≡ Az12(C1) - Az12(C2) (4.11.12) Performing the same weighted average on (4.11.8b), we arrive at this adjusted version of (4.11.8), [Zs1 + Zs2] i(z) = - ∂zV(z) - jωW(z) ∂zW(z) = - j (β2/ωV(z) . (4.11.13) where now Zs1 and Zs2 are the constants just defined and V(z) and W(z) are given by (4.11.12). Comment: For transmission lines made of "thin wires", which means thin relative to their spacing, Jz is close to uniform over each conductor and the above averaging process is not necessary. The circle now closes when we insert into (4.11.13) the fact that W = Le i(z) from (4.11.1), (Zs1 + Zs2) i(z) = - ∂zV - jω Le i(z) Le ∂z i(z) = - j (β2/ωV which we then rearrange as ∂zV(z) = - [ Zs1+ Zs1+ jωLe] i(z) ∂z i(z) = - [ jβ2/(ωLe)] V(z) . (4.11.14) These are the classic transmission line equations. They are usually written in this form: = - z i(z) = - y V(z) with z = R + jωL y = G +jωC . (4.11.15) Note: We have been using bold notation only for vectors, and we now break that guideline by bolding these complex quantities z and y. Our purpose for this bolding is to distinguish them from Cartesian coordinates z and y which typically appear in the same problem. In King's books, all complex parameters are put in bold font, but we do this only for z and y. Jumping the gun a bit, if we assume now a traveling-wave z dependence ej(ωt-βz) for both V(z) and i(z), where βd is the wave's wavenumber in the dielectric, then ∂z → -jβd and the transmission line equations become -jβd V(z) = - z i(z) -jβd i(z) = - y V(z) or -jβd = - z i(z)/V(z) -jβd = - y V(z)/i(z) so equating the two expressions gives - z i(z)/V(z) = - y V(z)/i(z) => z/y = [V(z)/i(z)]2 and we then have, Z0 ≡ V(z)/i(z) = = (4.11.16) where by definition Z0 is the characteristic impedance of the transmission line. Here, z and y are called the transmission line impedance and admittance, and the four numbers R,L,G,C are defined to be the appropriate real and imaginary parts. Comparing (4.11.14) and (4.11.15), we may therefore conclude that: z = R + jωL = Zs1 + Zs2 + jωLe (4.11.17) y = G + jωC = jβ2/(ωLe) . (4.11.18) The expression for z seems quite reasonable since ωLe = XL = inductive reactance, but the expression for y seems a bit unusual. This is because we still have more work to do. There is one more equation we have not yet utilized. Recall from Chapter 1 the integral form of the equation of continuity, which in the frequency domain takes this form, div J = - jωρ -jω[∫V ρ dV] = ∫S J dS . (1.1.35) We now apply this to a Gaussian box (blue) whose faces have the same shape as the conductor cross section but are slightly larger than that cross section so as to include the conductor surface charge : Fig 4.13 Ignoring transverse current out the sides of the box (since dz is tiny), we get -jω[q(z)dz] = i(z+dz) - i(z) = total current flowing out of the box which then says ∂z i(z) = -jωq(z) . (4.11.19) From summary box (4.11.1) recall that q(z) = C'V(z) so we get ∂z i(z) = - [ jωC'] V(z) . (4.11.20) Comparison with the second equation of (4.11.14) we find the following identity, - [ jβ2/(ωLe)] = - [ jωC'] or LeC' = β2/ω2 = μξ . // see (1.5.1) regarding β2 (4.11.21) Then we can write (4.11.18) as y = G + jωC = jβ2/(ωLe) = j (β2/ω2) (ω/Le) = j (LeC') (ω/Le) = jωC' . (4.11.22) Thus, the line capacitance C is the real part of complex capacitance, C = Re(C'), and G = - ω Im(C'). Digression on the meaning of C' Back in Section 1.5 (c) we discussed the fact that nc = (ξ/ε) ns which relates actual surface charge ns to the adjusted surface charge density nc which allows for dielectric leakage. This relationship (1.5.17) was derived in two different ways. As noted in Comment 3 at the start of Section 4.1, and looking at (4.1.1) and (4.1.2), one sees that the linear charge density q(z) which appears in all our equations is in fact related to nc and not ns, so we momentarily shall refer to q(z) as qc(z). Then qc(z) = ∫nc dxdy = an integral over the conductor surface for length dz. The actual charge on the surface of this piece of conductor is qs(z) ≡ ∫ns dxdy and therefore qc/qs = nc/na = (ξ/ε). The capacitance C per unit length of our transmission line is defined by qs = C V(z) . The complex capacitance C', which includes the effect of dielectric leakage current, is defined by qc = C' V(z). Therefore C'/C = qc/qs = (ξ/ε) . (4.11.23) and so then from (4.11.22), (4.11.23) and (1.5.1), y = G + jωC = jωC' = jω(ξ/ε)C = jω (1 - jσ/εω)C = jωC + (σ/ε)C (4.11.24) so that G = (σ/ε)C . (4.11.25) We saw an example of (4.11.23) in (1.5.19) for a parallel plate capacitor, and more generally in (4.4.10). We may now rewrite the first equation in summary box (4.11.1) as = K => = K . (4.11.26) Next, combining (4.11.21) and (4.11.23) we find that LeC' = μξ (4.11.27) LeC = με = 1/v2 (4.11.28) where v is the speed of light in the dielectric. Now the second equation in (4.11.1) says that Le = KL . (4.11.29) Inserting (4.11.29) for Le and (4.11.26) for C' into (4.11.27) gives ( KL ) (4πξ/K) = μξ or KL = K . (4.11.30) This is a remarkable connection between our two seemingly unrelated constants K and KL, K ≡ !Syntax Error, Idx1' dy1' α1(x1',y1') ln(s212/s112) -!Syntax Error, Idx2' dy2' α2(x2',y2') ln(s222/s122) (4.4.8) KL ≡ !Syntax Error, Idx1' dy1' b1(x1',y1') ln(s212/s112) -!Syntax Error, Idx2' dy2' b2(x2',y2') ln(s222/s122) . (4.10.9) Since K involves a line integral of linear surface charge densities αi whereas KL involves a full cross sectional area integral of the current densities bi, it seems unlikely these integrals would be equal, but they are equal. An example of K = KL The equality even seems unlikely in a case with symmetric densities on round wires, so let's do a check using our Section 4.5 example with widely-space round wires of unequal diameters. The first thing we need is a new picture to display the "kinematics" of the KL integral ( since densities are symmetric, one should regard this picture as having b much larger than shown relative to a1 and a2), Fig 4.14 As before, we read off the four distances of interest using the law of cosines. The new distances are of course all different than they were before since x1' and x2' are now each integrated over their respective disks instead of the bounding circles. s212 = r12 + (b-a1)2 - 2 r1(b-a1) cos(θ1) s112 = r12 + a12 - 2 r1 a1 cos(θ1) s222 = r22 + a22 + 2 r2 a2 cos(θ2) s122 = r22 + (b-a2)2 + 2 r2(b-a2) cos(θ2) . The integration rule is still !Syntax Error, Idθ ln (A ± Bcosθ) = 2π ln[(1/2)(A + )] . (4.5.5) The first integral is: !Syntax Error, Idθ1 ln(s212) = !Syntax Error, Idθ1ln([r12 + (b-a1)2 - 2 r1(b-a1) cos(θ1)] A = r12 + (b-a1)2 B = 2 r1(b-a1) A2-B2 = [r12 + (b-a1)2]2 - 4 r12(b-a1)2 = [r12 - (b-a1)2]2 => = (b-a1)2- r12 > 0 b >> a1 => !Syntax Error, Idθ1 ln(s212) = 2π ln[(1/2)( r12 + (b-a1)2 + (b-a1)2 - r12 ) = 2π ln[(b-a1)2] But this integral is the same as before! The s112 integral is obtained from the above with b-a1→a1 !Syntax Error, Idθ1 ln(s112) = 2π ln(a12) which is also the same as before. The other two integrals are found from 1→ 2. Our integral summary is then exactly the same as (4.5.6), !Syntax Error, Idθ1 ln(s212) = 2π ln[(b-a1)2] !Syntax Error, Idθ1 ln(s112) = 2π ln(a12) !Syntax Error, Idθ2 ln(s222) = 2π ln(a22) !Syntax Error, Idθ2 ln(s122) = 2π ln[(b-a2)2] . (4.5.6) We now assume that the current densities bi each have radial symmetry ("widely spaced wires") , b1(r1,θ1) = b1(r1) (4.11.31) where b1(r1) is a completely arbitrary function, with the following normalization of (4.7.4), !Syntax Error, Idθ1 !Syntax Error, Ir1dr1 b1(r1) = 1 => !Syntax Error, Ir1dr1 b1(r1) = 1/2π . (4.11.32) We now proceed to calculate the constant KL KL = !Syntax Error, Idθ1 !Syntax Error, Ir1dr1 b1(r1) ln(s212/s112) -!Syntax Error, Idθ2 r2dr2 b2(r2) ln(s222/s122) = !Syntax Error, Ir1dr1 b1(r1) !Syntax Error, Idθ1 ln(s212/s112) - !Syntax Error, Ir2dr2b2(r2) !Syntax Error, Idθ2 ln(s222/s122) = 2π !Syntax Error, Ir1dr1 b1(r1) [ln[(b-a1)2]- ln(a12)] - !Syntax Error, Ir2dr2b2(r2) [ ln[(b-a2)2] - ln(a22)] = 2π [ln[(b-a1)2/a12] !Syntax Error, Ir1dr1 b1(r1) - 2π [ln[(b-a2)2/a22] !Syntax Error, Ir2dr2 b2(r2) = [ln[(b-a1)2/a12] - [ln[(b-a2)2/a22] = ln [] = K as obtained in (4.5.7) (4.11.33) and we have then shown KL = K for this particular example. The key fact is that the dθ integrals appear to be functions of ri , but the ri2 terms cancel and so the dθ integrals are independent of ri. We now summarize our results: Classical Transmission Line Equations and Parameters (4.11.34) K ≡ !Syntax Error, Idx1' dy1' α1(x1',y1') ln(s212/s112) -!Syntax Error, Idx2' dy2' α2(x2',y2') ln(s222/s122) (4.4.8) KL ≡ !Syntax Error, Idx1' dy1' b1(x1',y1') ln(s212/s112) -!Syntax Error, Idx2' dy2' b2(x2',y2') ln(s222/s122) , (4.10.9) K = KL real and dimensionless (4.11.30) = - z i(z) where z = R + jωL transmission line equations = - yV(z) y = G +jωC (4.11.15) z = Zs1 + Zs2 + jωLe (4.11.17) XL ≡ ωLe , XC ≡ 1/(ωC) y = jωC' = jωC + (σ/ε)C (4.11.24) G = (σ/ε)C (4.11.25) R = Re(Zs1+ Zs2) L = Le + (1/ω) Im(Zs1+ Zs2) Le = K (4.11.29) and (4.11.30) C' = 4πξ/K (4.11.26) C' = (ξ/ε)C (4.11.23) C = 4πε/K (4.11.26) G = 4πσ/K (4.11.26) + (4.11.25) LeC' = μξ (4.11.27) LeC = με = 1/v2 (4.11.28) Z0 = = (4.11.16) Z0 (large ω) ≈ ≈ = (1/4π) K = (K/4π) Zm // See comments below λ >> D (4.3.6) assumed transmission line limit where β = 2π/λ β2 = μεω2 - jωμσ = ω2μ ( ε - jσ/ω) = ω2μ ξ ξ ≡ ε - jσ/ω . (1.5.1) F(z) = F(0) e-az e-jbz a ≡ Re() = Re[] b = Im() see (5.3.6) attenuation phase Comments: 1. In Chapter 2 we computed the surface impedance Zs for a round wire (radius a, μc, σc) in the case of axially symmetric current and we found that, for large ω, Zs(ω) ≈ (1+j) (2.4.16) δ ≡ = skin depth (2.2.20) so that Zs(ω) ≈ (1+j) . (4.11.35) Presumably the result will be Zs(ω) ~ for any conductor cross section shape. Then L = Le + (1/ω) Im(Zs1+ Zs2) = Le + (stuff) 1/ → Le for large ω (4.11.36) For this reason, the high frequency characteristic impedance Z0 can be written as shown in (4.11.34). 2. Conductors have internal inductance Li as well as external inductance Le. In Appendix C.3 (a) we compute the low frequency internal inductance of a round wire to be Li = μc/8π = (μc/μ0) * 50 nH/m . Our Chapter 4 transmission line development makes no mention of Li. This can be traced to Figure 4.11 where only the external magnetic flux is involved. In fact, Li is accounted for in the imaginary part of the surface impedance Zs . For example, we found that for our round wire situation, Zs(ω) = + jω = Rs + jωLs // low frequency limit (2.4.12) and here one sees that Ls = = Li . 3. We have assumed that ε and μ are real. If not, the usual adjustments can be made in (4.11.34) for the interpretations of R,L,G and C. See for example (3.3.4) concerning σ being replaced by σeff if ε has an imaginary part. 4. Apart from the symmetric cases like the examples of Section 4.5 and 4.6, we do not yet have a way to compute K and the transmission line parameters since the charge and current distributions αi and bi are not known. This matter will be remedied in Chapter 5. 5. A strip transmission line of width w and separation s with s << w is the simplest example of the above summary: E = V/s n = εE = εV/s q = nw C = q/V = εw/s => K = 4πε/C = 4π(s/w) so C = 4πε/K = ε (w/s) K = 4π (s/w) G = 4πσ/K = σ (w/s) Le = (μ/4π) K = μ (s/w) Z0 ≈ (K /) 30Ω = 4π (s/w) (1/) 30Ω = (s/w) (1/) 377Ω 4.12 Modifications to account for μ ≠ μ1 ≠ μ2. These modifications only affect the Az and W(z) part of this chapter, not the first six sections which are concerned with φ and V(z). So changes start with Section 4.7. If the equality μ = μ1 = μ2 assumed in Section 4.7 is broken, the result is that surface magnetization currents appear on one or both of the conductor surfaces and these cause an alteration of the theory. Thanks to the "Jm Theorem" proven in Appendix B, this alteration can be carried through with a very minimal impact, as we now show. In Appendix B conductor magnetization surface currents are studied in some detail. The reader interested in how the magnetic modification is carried out would do well to read Appendix B at this point. A reader less interested can accept the Appendix B results and then learn below that basically nothing changes! So imagine starting with μ = μ1 = μ2 and then changing μ1 and μ2 to new values. The question is: how do the various parameters and equations of the theory change? The first modification arises in Section 4.7. As described in Appendix B.6, the modified version of (4.7.2) is this, Az1(x) = ∫ [ Jz1(x') + Jzm1(x') ] dx'dy'dz' . R = |x - x'| (4.7.2)' where Jzm1 includes only the surface component of the magnetization current on conductor C1. Appendix B.6 shows how this Jzm1 adder term in effect adds a certain homogeneous solution to the particular solution (first term above) of the Az Helmholtz equation such that the Az boundary conditions are duly satisfied at the magnetic conductor C1 boundary. According to (B.1.10), the surface current Jzm1 when expressed in surface rather than volume notation is given by Kz = - ( - ) Hθ and thus vanishes when μ1 = μ, resulting in the unmodified version of (4.7.2). We maintain the next two equations of Section 4.7 as is, involving separation of variables, Jz1(x,y,z) = b1(x,y) i1(z) A/m2 1/m2 A (4.7.3) where i1 is scaled such that !Syntax Error, Idx dy b1(x,y) = 1 . (4.7.4) This i1(z) is still the total conduction current in C1. But we now add two new equations, Jz1m(x,y,z) = b1m(x,y) i1m(z) A/m2 1/m2 A (4.12.1) where i1m is scaled such that !Syntax Error, Idx dy b1m(x,y) = 1 . (4.12.2) It is understood here that b1m(x,y) is a distribution which is restricted to the surface of C1, but we continue to write it as if it existed at all points in the cross section of C1. The integration in (4.12.2) is of course meant to include this surface distribution. We know from (B.1.11) and (B.1.12) that, for an arbitrarily shaped conductor C1, i1m(z) ≡ - ( - ) i(z) [ μ1 = conductor C1, μ = dielectric ], (4.12.3) and that the current ratio is therefore given by, f1m ≡ i1m(z)/ i(z) = - ( - ) . (4.12.4) With the above definitions, our modified (4.7.6) becomes Az1(x,y,z) = !Syntax Error, Idz' i(z') !Syntax Error, Idx' dy' [ b1(x',y') + f1m b1m(x',y') ] = !Syntax Error, Idz' i(z') !Syntax Error, Idx' dy' [ b1(x',y') + f1m b1m(x',y') ] . (4.12.5) This leads us to define a new effective transverse current density, b'1(x,y) ≡ b1(x',y') + f1m b1m(x',y') = b1(x',y') + [1-] b1m(x',y') . (4.12.6) This new transverse density b'1 is still normalized to unity, using (4.7.4) and (4.12.2) above, !Syntax Error, Idx dy b'1(x,y) = !Syntax Error, Idx dy b1(x,y) + [1-]!Syntax Error, Idx dy b'1m(x,y) = * 1 + [1-] * 1 = 1 . (4.12.7) How does b'1 differ from b1? The difference is that b1 does not include a surface current and b'1 does. We can represent equation (4.12.6) in this symbolic graphic manner: (4.12.6) Thus, from (4.12.5) and (4.12.6) we have this new version of (4.7.6), Az1(x,y,z) = !Syntax Error, Idz' i(z') !Syntax Error, Idx' dy' b'1(x,y) . (4.7.6)' The differences are that the leading factor is μ instead of μ1, and b1 is replaced by b'1. Moving into Section 4.8 we have this new version of (4.8.1), Az12(x) = Az1(x) + Az2(x) = !Syntax Error, Idz' i(z') { !Syntax Error, Idx1' dy1' b'1(x1',y1') – !Syntax Error, Idx2' dy2' b'2(x2',y2') } (4.8.1)' which is identical to (4.8.1) except bi → b'i. Then in the transmission line limit, we get this new version of (4.9.1), Az12(x) = i(z) !Syntax Error, Idz'{ !Syntax Error, Idx1' dy1' b'1(x1',y1') – !Syntax Error, Idx2' dy2' b'2(x2',y2') } (4.9.1)' From this point onward, all equations are the same apart from bi → b'i. Here are some of those equations after modification: W(z) ≡ Az12(x1) - Az12(x2) (4.10.1)' = i(z) !Syntax Error, Idz' { !Syntax Error, Idx1' dy1' b'1(x1',y1') – !Syntax Error, Idx2' dy2' b'2(x2',y2') } – i(z) !Syntax Error, Idz' { !Syntax Error, Idx1' dy1' b'1(x1',y1') – !Syntax Error, Idx2' dy2' b'2(x2',y2') } W(z) (4.10.3)' = i(z)!Syntax Error, Idz' {!Syntax Error, Idx1' dy1' b'1(x1',y1')( - ) -!Syntax Error, Idx2' dy2' b'2(x2',y2') (- ) } . W(z) = i(z) {!Syntax Error, Idx1' dy1' b'1(x1',y1') ln(s212/s112) -!Syntax Error, Idx2' dy2' b'2(x2',y2') ln(s222/s122) } (4.10.4)' KL ≡ !Syntax Error, Idx1' dy1' b'1(x1',y1') ln(s212/s112) -!Syntax Error, Idx2' dy2' b'2(x2',y2') ln(s222/s122) (4.10.9)' Le = = KL // no change (4.10.8) We then enter Section 4.11. The derivation of the transmission line equations (4.11.15) is unaffected by the above modifications; the only change is that the b'i appear in the integral KL in place of the bi. The derivation of the fact that K = KL ending in (4.11.30) is also unchanged! This at first seems strange since K has not changed, but we have apparently altered KL by the replacements bi → b'i. But KL is not an evaluation -- it is an integral equation relating KL to the b'i. In the self-consistent solution, the new functions (distributions) b'i adjust themselves so that KL does not change. KL cannot change because (4.11.30) says it must remain equal to K which is determined by the electrostatic side of the problem. It is perhaps helpful to look at (4.11.29) which says Le = KL . We know that if the dielectric μ value does not change, the external inductance Le of the transmission line cannot change so KL stays fixed. Changing μ1 and/or μ2 away from the value μ will of course change the internal inductances of the conductors, and this is duly noted below in terms of surface impedances. As μ1 is increased, the B field inside conductor C1 increases (H stays the same) so the stored B field increases, and Li increases. Finally, if we look at the example associated with Fig 4.13, we still find explicitly that KL= K because the calculation leading to (4.11.33) is unchanged when bi are replaced with b'i, since the b'i are still normalized to unity as shown in (4.12.7). The happy bottom line is that all of summary box (4.11.34) is unchanged except bi → b'i in the KL integral. The constant K can still be evaluated using the "capacitor problem" of Section 5.5 below and it is unaffected by conductors having μi ≠ μ. Having said this, let us now consider what happens to an operating transmission line which starts off with μ1 = μ2 = μ = μ0 and we then gradually turn a magic "permeability knob" so that μ1 gradually increases from μ0 to some value μ1 > μ0. That is to say, we gradually cause conductor C1 to become magnetic. The constant K (and therefore KL = K) does not change at all. This K is determined by the potential φ part of the problem in Section 4.4 and does not even know about the magnetic modification. Thus, looking at (4.11.34), C', C, G and Le do not change. In particular, Le does not change because we have not altered μ of the dielectric. The following two items shown in box (4.11.34) do change : R = Re(Zs1 + Zs2) L = Le + (1/ω) Im(Zs1 + Zs2) where Zsi is the surface impedance of conductor Ci. The non-Le term in L can be interpreted as the internal inductance of the conductors. R and L change because Zs1 changes if we change μ1. This is so because Zs1 is always a function of the skin depth δ1, and δ1 ≡ from (2.2.20). In the special case that C1 is a round wire of radius a1 with an axially symmetric current distribution (such as the center wire of a coaxial cable), we showed in (2.4.11) that the surface impedance is given by Z1s(ω) = , (2.4.11) so certainly this Zs(ω) is a function of μ1 both due to the leading constant and through the five occurrences of δ1. Both the real and imaginary parts of Z1s(ω) will change as μ1 changes, so the transmission line parameters R and L both change. In the high frequency limit , Z1s(ω) ≈ (1+j) δ << 16a . (2.4.16) so now the variation with μ1 is through the single δ1 factor shown. Again, both real and imaginary parts of Z1s(ω) vary with μ1. Since R and L change as noted above, the transmission line characteristic impedance will also change, Z0 = = (K.11) This means, for example, if we drive a semi-infinite transmission line with some fixed voltage V(z), the driving current i(z) will vary in amplitude and phase as we turn our "permeability knob" for conductor C1. This is simply because i(z) = V(z)/Z0. So the good news is that the theory of Chapter 4 is easily extended to allow for magnetic conductors and or dielectric. Once again, the summary box (4.11.34) is unchanged when μ1 = μ2 = μ is broken except for the appearance of b'i in the KL integral, and except for the fact that Zs1 and Zs2 change as noted above, causing changes in R, L and Z0. At very high frequency, one will have Z0 = and in this case Z0 is not altered, see (4.11.36).