low freq limit of E fields REVIEWED

Low Freq limit of E fields PhL 5.17.14 Early text on the low ω limits of fm etc and then of the E fields. This is where I had my slightly bogus cutoff frequency ωc that was part of a loss model. I had to undo all that stuff and it was painful. I installed a better loss model somewhere else in lines doc that has no ωc parameter. I then had to clean up Appendix M, it was all a big mess and took many hours. But basically what you see below is now in Section D.11. This loss model of course did not fix the Jz problem. (10.8.14 review). D.10 Low frequency limit of the round wire E fields 1 (a) The effect of including loss on the parameter β' 1 (b) Derivation of the simple loss model 2 (c) Low frequency evaluation of fm, gm and hm and the E fields 4 D.10 Low frequency limit of the round wire E fields Recall from (D2.2.33) that, Summary of the E field solutions : Rdc = β'2 = β2 - βd2 (D.2.33) Ez(r,m) = (1/4) ηm I Rdc (aβ') fm fm = [ - ] x = β'r Er(r,m) = (j/4) ηm I Rdc (aβd) gm gm = [ + ] xa = β'a Eθ(r,m) = (1/4) ηm I Rdc (aβd) hm hm = [ - ] Our goal here is to evaluate fm, gm and hm in the low ω limit. For low frequency we continue to use β = ej3π/4 (/δ) = ej3π/4 (2.2.20) (2.2.21) (a) The effect of including loss on the parameter β' If for the wave's wavenumber βd one uses the usual lossless βd = (ω/vd) one finds, = = constant * // no loss (D.10.1) which results in (βd/β) → 0 for the low ω limit. Since β'2 = β2 - βd2 this again gives β' = β as in the previous section. But now this β is very small at low ω, instead of very large as in the Section D.9. This is the correct limit, but we want to make sure it is still the correct limit if we include cable loss. Including in βd a model for the unavoidable loss, we find at low ω (see section (b) below) that βd = (ω-jωc)/vd ωc = (vd2/2) (RdcC) (D.10.2) E(z) = E(0) e-jβz = e-jz(ω-jω)/v = e-j(ω/v)z e-(ω/v)z where Rdc is the DC resistance per unit length of the transmission line (both conductors) and C is the capacitance per unit length. Reconsider now the low ω situation: = → constant / (D.10.3) Now (βd/β) → ∞ for the low ω limit, just the reverse of the lossless situation. Since β'2 = β2 - βd2 one finds from (D.10.2) that, for low ω, β'2 ≈ -βd2 = - (jωc)2/vd2 = (ωc/vd)2 so β' ≈ (ωc/vd) . (D.10.4) But since ωc is usually fairly small and vd is fairly large, it turns out that aβ' << 1 and we are then still interested in the small β' limit of our E field coefficients like fm, as shown in an example below. Alternatively one can write, βd = βd0 - jα βd0 = (ω/vd) α = (ωc/vd) = (vd/2) (RdcC) dim(α) = sec/m2 * (m/sec) = m-1 (D.10.5) where α is the amplitude loss factor for a propagating wave e-jβz = e-jβz e-αz Re(e-jβz) = e-αz cos(βd0z) = e-(ω/v)z cos[(ω/vd)z] . (D.10.6) (b) Derivation of the simple loss model [ here is that loss model! ] Start with this expression for power P in a wave, P = IV = U vd = [(1/2)L I 2 + (1/2)CV2 ] vd (D.10.7) Then dP/dz = [L I dI/dz + C V dV/dz ] vd = [ L I ( -y V ) + C V ( - z I ) ] vd // using (4.11.15) = - IV [ Ly + Cz] vd = - P[ L(G + jωC) + C(R + jωL)] vd = -2α P (D.10.8) where α = [ L(G + jωC) + C(R + jωL)] (vd/2) . (D.10.9) So then P(z) = e-2αz P(0) E(z) = e-αz E(0) . // electric field and all other non-energy non-power quantities (D.10.10) Ignoring G and assuming low ω, we obtain the result quoted above, α = [ L(G + jωC) + C(R + jωL)] (vd/2) ≈ (LG+RC) (vd/2) ≈ (RdcC) (vd/2) (D.10.11) The main point is that α and ωc are constant, independent of ω. As an example, consider Belden 8281 coaxial cable : For this standard 75Ω cable one finds that Rdc = (32.5 + 3.6) Ω/km = 36.1 x 10-3 ohms/m C = 69 pF/m = 69 x 10-12 farad/m vd = 0.66c = 0.66 x 3 x 108 m/sec = 2 x 108 m/sec So fc = 7770 Hz and ωc = 2πfc. So any line quantity like V(z) has this behavior, V(z) = V(0) e-(ω/v)z cos[(ω/vd)z] = V(0) e-(2πf/v)z cos[2π(f/vd)z] (D.10.12) which we can plot for V(z) over 10 km of cable for several values of f: f = 100 KHz f = 10 KHz f = 1 KHz In all cases, some signal arrives at the end (ignoring noise), but in the right two graphs one would say the cable was quite "lossy". For this example, we can compute β' for low ω: β' = (ωc/vd) = 2π * 7770 / 2*108 = .00024 = 2.4 x 10-4 m-1 a = 394μ = 394 x 10-6 m = 0.394 x 10–3 m xa = β'a = 2.4 x 10-4 * 0.394 x 10–3 = 0.95 x 10-7 (D.10.13) (c) Low frequency evaluation of fm, gm and hm and the E fields In the following we consider only m ≥ 0 since we know from (D.9.2) that f-m = fm , g-m = gm, h-m = hm . The small x limit for Jm(x) is given by NIST 10.7.3, Jn(x) = (x/2)n / n! . for n = 0,1,2,..... (D.10.14) Since Jm-1 appears in our coefficient expressions and since m = 0 is encountered, we have to deal with m = 0 as a special case, since the above limit is not valid for n = -1. To this end we use NIST 10.2.2 which is valid for integer n, J-n(x) = (-1)nJn(x) ≈ (-1)n (x/2)n / n! (D.10.15) so that J-1(x) = - J1(x) ≈ - (x/2). Our limiting forms of interest are then Jn(x) = (x/2)n / n! for n = 0,1,2,..... J-1(x) = - (x/2) for n = -1 (D.10.16) We now examine the small x limits of fm, gm, and hm . First fm for m > 0, and then for m = 0: fm = [ - ] = [ - ] = [ (m+1) (x/xa)m (2/xa) - (1/m) (x/xa)m(xa/2) ] = (x/xa)m [ (m+1) (2/xa) - (1/m) (xa/2) ] ≈ (x/xa)m (m+1) (2/xa) as xa→ 0 f0 = [ - ] = [ + ] = 2 = 2 = 4/xa First gm for m > 0, and then for m = 0: gm = [ + ] = [ - ] = (x/xa)m+1 - (x/xa)m-1 g0 = [ + ] = [ + ] = 2 = 2 (x/xa) Results for hm are then obvious since there is only a sign change between the terms in gm, hm = (x/xa)m+1 + (x/xa)m-1 h0 = 0 Small ω limit of the E field solutions : Rdc = (D.10.17) Ez(r,m) = (1/4) ηm I Rdc (aβ') fm fm = (r/a)m (m+1) (2/β'a) f0 = 4/(aβ') Er(r,m) = (j/4) ηm I Rdc (aβd) gm gm = (r/a)m+1 - (r/a)m-1 g0 = 2 (r/a) Eθ(r,m) = (1/4) ηm I Rdc (aβd) hm hm = (r/a)m+1 + (r/a)m-1 h0 = 0 Recall now these field expansions for the case that n(θ) is an even function of θ, Ez(r,θ) = (1/4) I Rdc (aβ') [ f0 + 2 Σm=1∞ fm ηm cos(mθ) ] Er(r,θ) = (j/4) I Rdc (aβd) [ g0 + 2 Σm=1∞ gm ηm cos(mθ) ] Eθ(r,θ) = (1/4) I Rdc (aβd) [ h0 + 2 Σm=1∞ hm ηm cos(mθ) ] . (D.9.4) For low ω we insert the expressions above to get Ez(r,θ) = (1/4) I Rdc [ 4 + 4 Σm=1∞ (r/a)m (m+1) ηm cos(mθ) ] Er(r,θ) = (j/4) I Rdc (aβd) [2 (r/a) + 2 Σm=1∞ [(r/a)m+1 - (r/a)m-1] ηm cos(mθ) ] Eθ(r,θ) = (1/4) I Rdc (aβd) [2 Σm=1∞ [(r/a)m+1 + (r/a)m-1] ηm cos(mθ) ] . (D.10.18) Observations on the E fields for small ω 1. The m = 0 term in Ez is just Ez = IRdc which says Jz = σ IRdc = σ I(1/σπa2) = I/(πa2). This is the current one would expect in a wire carrying DC current I. 2. Assuming a non-uniform n(θ) charge density on the wire surface, the ηm moments are non-zero and one concludes that Jz(r,θ) is asymmetric even as ω → 0. This seems to conflict with the notion that the proximity effect should go away as ω → 0 as considered in Appendix P. One item to note is that I = V/Z0 and according to (4.11.16) Z0 ≡ V(z)/i(z) = = (4.11.16) and in the limit ω→0 we get Z0 = = a very large number. As one lowers ω, one must increase the size of the termination Z0, assuming there is one, to maintain a properly terminated line. Thus, our situation here does not apply to taking the low frequency limit of a transmission line terminated with a 75Ω or 8Ω resistor. 3. The fields Er and Eθ are smaller than the field Ez by factor xa = aβd which we have assumed is small.