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retired Appendix M 9_9_14

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Phil's retired Appendix M (dated 9.9.14) from his transmission line notes. It argues that At is below about 10^-4 of Az for frequencies from 10fc to 1000 GHz. Transverse conductor currents are shown to be under 10^-3 of the longitudinal current using the ratio of dielectric to conductor wavenumber. A qualitative angular and Bessel-phase cancellation argument in the integral for At supplies a further factor of about 10.

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Retired old Appendix M PhL retired today 9.9.14 Appendix M: Why the transverse vector potential At is small for a transmission line Claim: In the King gauge, the transverse vector potential At may be neglected for frequencies in the range 10fc to 1000 GHz, where fc is the soft loss cutoff frequency described below. (M.1) The vector potential is written below as At = ∫conductors dxdy Jt(x,y) * (stuff). We shall make the following three claims: (1) | Jt | < 10-3 |Jz| for f = 10fc Hz to 1000 GHz (which is to say: "transverse currents are small inside the conductors") (M.13) (2) in the At integral there is a cancellation effect not present in the Az integral which in effect reduces At by a factor of 10 (ballpark) relative to Az . (M.14) (3) the net ballpark result is that |At| < 10-4 |Az| for f = 10fc Hz to 1000 GHz which supports the opening claim (M.1) above. (M.22) According to (1.5.9) one can express the King gauge vector potential at all points in space in terms of the currents in the transmission line conductors in this manner, A(x,ω) = Σi∫μi Ji(x',ω) dV' R = | x - x' | (1.5.9) where Σi is a sum over all the conductors, and βd is the wavenumber in the dielectric. Therefore, the transverse part At may be written as an integral of the transverse conductor currents Jt,i : At(x) = Σi μi∫Jt,i(x',y',z') dx' dy' dz' . (M.2) Transverse refers to the x and y directions, where the infinite conductors are aligned in the z direction. The main current in a transmission line conductor is the longitudinal one Jz . When the above equation is processed in the manner of Chapter 4, and one assumes the transmission line limit, the result is At(x) = - Σi μi∫Jt,i(x',y') ln(s2) dx' dy' . s2 = (x-x')2 + (y-y'2) (M.3) In this transmission line limit, βd is small (long wavelength) and e-jβR ≈ 1. We assume the standard wave functional form such that Jt,i(x,y,z) = e-jβz Jt,i(x,y) and then again set e-jβz ≈ 1 for the contributing portion of the dz' integration and that integration produces -ln(s2) as in Ch 4 or (J.10). Meanwhile, Appendix D computes the E fields inside a round conductor for each partial wave m, and here we multiply them each by σ to get the current density components, Current Densities in a Round Wire: Rdc = β'2 = β2 - βd2 (D.2.33) Jz(r,m) = σ(1/4) ηm I Rdc (aβ') fm fm = [ - ] x = β'r Jr(r,m) = σ(j/4) ηm I Rdc (aβd) gm gm = [ + ] xa = β'a Jθ(r,m) = σ(1/4) ηm I Rdc (aβd) hm hm = [ - ] The currents are expressed in terms of a cylindrical coordinate system whose z axis runs down the center of the round conductor. Coefficient ηm is the "surface charge moment" of the mth partial wave, and the θ-space currents are given by (D.1.3a), J(r,θ) =!Syntax Error, I J(r,m) ejmθ . // partial wave expansion (M.4) The moments ηm may be obtained by solving the transmission line "capacitor problem" as outlined in Section 6.5 (a). One finds potential φ, then E, then surface charge n(θ), and finally ηm. The current components are, J = Jz + Jr + Jθ = Jz + Jt Jt = Jr + Jθ . (M.5) Rather than study these round wire internal solutions in detail, we make two observations: Observation (1): The transverse currents Jr and Jθ are very small compared to Jz. (M.6) Looking at (D.2.33) above one sees that the transverse currents Jr and Jθ are in general smaller than the longitudinal current Jz by factor |βd/β'|. In making this claim we regard the combinations of Bessel functions shown in (D.2.33) as being of the same general scale, which can be confirmed by doing plots of the various complex function magnitudes. Consider these relations : ( first line is from (1.5.1a), (3.3.2) with ε' ≈ ε, and (1.1.29) that vd = 1/ ) βd = βd0 [1 - j (1/2)tanL] ≈ βd0 = (ω/vd) = 2π/λd (M.7) β = ej3π/4 (/δ) = [(j-1)/] (/δ) = (j-1)(1/δ) δ ≡ (2.2.21) (M.8) β'2 = β2 - βd2 . (D.2.2) (M.9) Using these facts there are several ways to write the ratio | βd/β |, one of which is this: => | βd/β | = 2πδ /(λd) = π (δ/λd) (M.10) which at least suggests that | βd/β | is small since one normally thinks of skin depth δ as being much less than the wavelength of a wave on the transmission line. As shown in Appendix D.11, when losses are included one really has βd = (ω-jωc)/vd where ωc = 2πfc is a soft cutoff frequency below which line losses become intolerable. For Belden 8281 coaxial cable it is shown that fc ≈ 7 KHz, while for a typical power transmission line fc ≈ 5 Hz. Since one would never operate a transmission line with ω < ωc, that frequency region is of little interest to us. A healthy lower limit might be ω = 10ωc. A more useful expression of the ratio| βd/β | is the following, where we use fact (1.1.2) that εμ = 1/vd2 and we include the loss effect just mentioned, | |2 = ||2 δ2/2 = ||2 = | ω-jωc|2 με = | ω-jωc|2 = | 1-jωc/ω|2 => | βd/β | = | 1-jωc/ω| . (M.11) At some low frequency (perhaps ω = 10ωc) this ratio is fairly small (being ~ ), but it then increases with ω as ω1/2. If we are willing to restrict our transmission line interests to 10fc < f < 1000 GHz, and if the conductors are copper, we find that the ratio | βd/β | will always be less than the following: ≈ = = [8.85 x 10-12 * 2π * 1012 / 5.81 x 107 ]1/2 = 10-3 (M.12) Thus, all the way from f = 10fc to f = 1000 GHz, we have | βd/β| < 10-3. At 10 GHz the ratio is 10-4. Since |β| << |βd|, it follows that β'2 = β2 - βd2 ≈ β2 so then β' = β for all practical purposes and then we have shown that |βd/β'| << 10-3. Thus from (D.2.33) quoted above, | Jr | < 10-3 | Jz | | Jθ | < 10-3 | Jz | f = 10fc to 1000 GHz (M.13) The conclusion then is that the transverse currents are less than 1/1000th of the size of the longitudinal currents for the round copper conductor at all frequencies of interest below 1000 GHz, and we can reasonably assume that a similar conclusion applies to a conductor of any cross sectional shape. This then concludes our "proof" of the claim that "transverse currents are very small" inside the conductors of a transmission line." Observation (2): In the Helmholtz integration (M.3) there is a large amount of cancellation. (M.14) Let us consider the nature of this integration in the illustrative case of a two round conductors, Fig M.1 Consider the contribution to the transverse vector potential component Ax from the right conductor C2, Ax(x) = - ∫Jx(x'2,y'2) ln(s22) dx'2 dy'2 s22 = (x-x'2)2 + (y-y'22) (M.15) or Ax(x) = - ∫ [Jr(r'2,θ'2) '2 + Jθ(r'2,θ'2) '2 ] ln(s22) [a2dθ'2] dr'2 (M.16) or Ax(x) = - ∫ [Jr(r'2,θ'2) cosθ'2 - Jθ(r'2,θ'2) sinθ'2] ln(s22) [a2dθ'2] dr'2 . (M.17) The transverse currents in conductor C2 have this partial wave expansion from (M.4), Jr(r'2,θ'2) =!Syntax Error, I Jr(r'2,m) ejmθ' (M.18) and similarly for Jθ . Thus we get Ax(x,y) = - a2 Σm !Syntax Error, Idr'2 Jr(r'2,m) !Syntax Error, Idθ '2 ejmθ' cosθ'2 ln(s22) + a2 Σm !Syntax Error, Idr'2 Jθ(r'2,m) !Syntax Error, Idθ '2 ejmθ' sinθ'2 ln(s22) (M.19) where, from (D.2.33) quoted above, Jr(r,m) = σ(j/4) ηm I Rdc (aβd) [ + ] x = β'r Jθ(r,m) = σ(1/4) ηm I Rdc (aβd) [ - ] xa = β'a . It is in theory possible to first do the dθ'2 integration in (M.19) and then do the dr'2 integration and get an analytic result for Ax(x,y). We have dealt with similar angle integrations elsewhere in this document. Rather then attempt this task, we instead consider the portion of the 2D integration represented by the red ring in Fig M.1. On this ring, r'2 is constant, and our interest is the θ'2 integration. For any value of m (except for ±1) the trigonometric functions like ejmθ' cosθ'2 integrate to 0, for example, Fig M.2 For these values of m, were it not for the fact that s22 varies around the red circle, Ax(x,y) would be identically 0. Although s22 does vary on the red circle, ln(s2) varies very little, and we expect to still have this strong cancellation in the θ'2 integral so Ax(x,y) is then small. It is true that if x and x'2 were to approach the conductor boundary from opposite sides, then ln(s2) would vary a lot more and the cancellation would be less, but we ignore this detail in our qualitative argument. For m = ± 1 this smallness argument fails since for example cos2(θ'2) does not average to 0 around the red ring. Ignoring the θ'2 variation in s22 we get in this case (setting e±jθ' ~ cosθ'2) Ax(x,y) ≈ - a2 Σm !Syntax Error, Idr'2 Jr(r'2,m) ln(s22) !Syntax Error, Idθ '2 cos2θ'2 ≈ - a2 Σm !Syntax Error, Idr'2 Jr(r'2,±1) ln(s22) π (M.20) Now we make a different argument which concerns the behavior of the complex Bessel functions as a function of r'2. As studied in Chapter 2, these functions have a dramatically oscillating phase even in the soft skin depth limit, and we expect then to get cancellation due to this phase as we integrate on the radial segment shown blue in Fig M.1, and again ln(s22) varies slowly on this ray due to the nature of ln. The arguments made above for Ax(x,y) also apply to Ay(x,y), and it seems reasonable to assume that the arguments are generally valid for an arbitrary conductor cross section. Admittedly our analysis here is imprecise and qualitative, but we think it is convincing that there is in fact much cancellation when the transverse currents are integrated over the conductors. This stands in stark contrast to the longitudinal situation where Az, being the Helmholtz integral of Jz, involves a generally non-cancelling integration (per conductor) over a generally large current component. We now wish to compare the following two integrals, where we pick component Ar to represent a transverse component of At, Ar(x) = - Σi μi∫Jr,i(x',y') ln(s2) dx' dy' . s2 = (x-x')2 + (y-y'2) (M.3) Az(x) = - Σi μi∫Jz,i(x',y') ln(s2) dx' dy' . s2 = (x-x')2 + (y-y'2) (M.21) We have shown in (M.13) that | Jr | < 10-3 | Jz |. Without any mathematical rigor, and allowing a factor of 10 "gain" from the cancellation effect of Observation (2), we make the following ballpark estimate, |At| < 10-4 |Az| f = 10fc to 1000 GHz. (M.22) It is assumed that as ω increases, the transmission line geometry is appropriately shrunk so the transmission line limit remains operative.