old Section D_11 ARCHIVE

Saving old Section D.11 PhL 6.24.14 D.11 Low frequency limit of the round wire E fields Recall from (D.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) = (j-1)/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.11.1) which results in (βd/β) → 0 for the low ω limit. Since β'2 = β2 - βd2 this again gives β' = β as in Section D.10. But now this β is very small at low ω, instead of very large as in the Section D.10. 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 where ωc = (vd2/2) (RdcC) (D.11.2) 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 / . // loss (D.11.3) Now (βd/β) → ∞ for the low ω limit, just the reverse of the lossless situation. Since β'2 = β2 - βd2 one finds from (D.11.2) that, for low ω, β'2 ≈ -βd2 = - (jωc)2/vd2 = (ωc/vd)2 so β' ≈ (ωc/vd) . (D.11.4) 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) = (m/sec)* sec/m2 = m-1 (D.11.5) 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.11.6) (b) Derivation of the simple 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 . dim(Uvd) = J/m*m/sec = watts (D.11.7) Then dP/dz = [L I dI/dz + C V dV/dz ] vd = [ L I ( -y V ) + C V ( - z I ) ] vd // using transmission line equations (4.11.14b) = - IV [ Ly + Cz] vd = - P[ L(G + jωC) + C(R + jωL)] vd = -2α P (D.11.8) where α = [ L(G + jωC) + C(R + jωL)] (vd/2) . (D.11.9) So then P(z) = e-2αz P(0) |E(z)| = e-αz |E(0)| . // electric field and all other non-power quantities (D.11.10) 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) ωc = vdα = (RdcC) (vd2/2) fc = (RdcC) (vd2/4π) (D.11.11) The main point is that in this simple model α and ωc are constants, independent of ω. Example 1: Consider Belden 8281 coaxial cable : For this standard 75Ω cable one finds from (D.11.2) and fc = ωc/2π , 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 Thus fc = 7770 Hz and ωc = 2πfc = 48,827. 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.11.12) which we can plot for V(z) over 10 km of this Belden cable for several values of f: f = 100 KHz f = 10 KHz f = 1 KHz Fig D.9 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 using (D.11.4), β' = (ωc/vd) = 48,827 / 2*108 = .00024 = 2.4 x 10-4 m-1 a = 394μ = 394 x 10-6 m = 0.394 x 10–3 m // wire radius xa = β'a = 2.4 x 10-4 * 0.394 x 10–3 = 0.95 x 10-7 (D.11.13) so at low ω it is appropriate to assume that xa = β'a and x = β'r are both << 1. Example 2: At the end of Section 4.5 we considered a power transmission line with two 1" diameter conductors separated by 1 meter. It was found that K = 17.5 and that R = .02Ω per thousand feet for each conductor. Thus Rdc ≈ 1.2 x 10-4 ohms/m for both conductors and C = 4πε0/K = 6.35 x 10-12 F/m. We then find that ωc = (c2/2) (RdcC) = 34.3 sec-1 and fc = 5.5 Hz. Such power lines are normally operated at 50 or 60Hz which is about 10fc which corresponds roughly to the left graph in Fig D.9 above. Comment: Since the transmission line equations were used in deriving the above loss model, and since the transmission line equations are only valid in the strong or extreme skin effect regimes, the loss model is not valid for "very low frequencies" which we might take to mean f < fc. In other words, the loss model is really only valid if the loss is small, and at sufficiently low ω the loss is not small and the model does not apply. In particular, the model does not apply at ω= 0 or very close to ω = 0. (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.10.2) 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.11.14) 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.11.15) so that J-1(x) = - J1(x) ≈ - (x/2). Our small-x 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.11.16) 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 The results are then: Small ω limit of the E field solutions : Rdc = (D.11.17) 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.10.4a) 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.11.18) Observations on the E fields for small ω 1. The fields Er and Eθ are smaller than the field Ez by factor (aβd) . 2. 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. 3. 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 conclusion is incorrect, and the reason it is incorrect is explained in Chapter 6.5 (e). One item to note is that I = V/Z0 and according to (4.11.16), where G = 0 since we assumed a non-conducting dielectric, Z0 ≡ V(z)/i(z) = = . (4.11.16) Thus in the limit ω→0 we get Z0 → ∞. For a finite length transmission line, as one lowers ω, one must increase the size of the termination Z0 to maintain a properly terminated line. Thus, our situation here does not apply to taking the low frequency limit of a transmission line terminated by a fixed 75Ω or 8Ω resistor. See Section 6.5 (e) for more on this subject. The major point here is that the entire model is not valid for low ω and we cannot use it to study the limit ω → 0.