averaging repair

This was my 2nd effort at averaging repair, I got rid of the Jz weighting, and this was installed some time perhaps around June 16. (give or take a week). Averaging Repair. Our theory now has an inconsistency which needs to be fixed. 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, where Jz will be large near the gap and small far from the gap: 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 for the above cross section example. This means that the left side of (4.11.6a) is a function of x1 = (x1,y1,z) and x2 = (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). In reality we have V(x1,x2) and W(x1,x2). 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. So first rewrite (4.11.6a) as Ez(x1) - Ez(x2) = - ∂z V(x1,x2) - jω W(x1,x2) (4.11.6a)' Backing up another step, we write out of (4.11.4a) for the two perimeter points x1 and x2, (1/σ)Jz(x1) = Ez(x1) = - ∂zφ12(x1) - jωAz12(x1) x1 on perimeter of C1 (1/σ)Jz(x2) = Ez(x2) = - ∂zφ12(x2) - jωAz12(x2) x2 on perimeter of C2 (4.11.4a)' Calling the perimeter distances of the conductors P1 and P2, we then average each of these equations around its appropriate perimeter. Apply (1/P1) ∫C1 ds1 to the first equation and (1/P2) ∫C1 ds2 to the second to get [ ds1 is a distance element along the perimeter of C1 ] , (1/σ)<Jz(x1)>C1 = <Ez(x1) >C1 = - ∂z<φ12(x1) >C1 - jω<Az12(x1) >C1 (1/σ)<Jz(x2)>C2 = <Ez(x2) >C2 = - ∂z<φ12(x2) >C2 - jω<Az12(x2) >C2 . Subtract the second line from the first to get, [<Ez(x1) >C1 - <Ez(x2) >C2] = - ∂z[<φ12(x1) >C1 - <φ12(x2) >C2] - jω [<Az12(x1) >C1 - <Az12(x2) >C2 ] . We now redefine V and W to be the averages appearing in these equations, along with Ez1 and Ez2 : Ez1(z) ≡ <Ez(x1) >C1 = (1/P1) ∫C1 ds1 Ez(x1) Ez2(z) ≡ <Ez(x2) >C2 = (1/P2) ∫C2 ds2 Ez(x2) V(z) ≡ <φ12(x1) >C1 - <φ12(x2) >C2 = <V(x1,x2)>C1,C2 W(z) ≡ <Az12(x1) >C1 - <Az12(x2) >C2 = <W(x1,x2)>C1,C2 (4.11.7) with this result [Ez1(z) - Ez2(z)] = - ∂z V(z) - jω W(z) (4.11.8) Meanwhile, the surface impedances on C1 and C2 are defined by (see C.2.1) , Ez1(x1) = Zs1(x1) i1(z) Ez2(x2) = Zs2(x2) i2(z) (4.11.9) which we average in the same way to obtain Ez1(z) = Zs1 i1(z) Zs1 ≡ (1/P1) ∫C1 ds1 Zs1(x1) Ez2(z) = Zs2 i2(z) Zs2 ≡ (1/P2) ∫C2 ds2 Zs2(x2) (4.11.10) Then using i(z) = i1(z) = -i2(z) we may rewrite (4.11.8) and (4.11.6b) as [Zs1 + Zs2] i(z) = - ∂z V(z) - jω W(z) ∂zW(z) = - j (β2/ωV(z) (4.11.11) where the second equation above is the < >C1,C2 average of (4.11.6b). Continuing this repair effort, we back up to box (4.11.1) and write V(x1,x2) = q(z) / C'(x1,x2) = q(z) [ K(x1,x2) ] W(x1,x2) = i(z) Le(x1,x2) = i(z) [ KL(x1,x2)] (4.11.12) which we average in the same way to get V(z) = q(z) W(z) = i(z) Le ≡ (1/P1) ∫C1 ds1 (1/P2)∫C2 ds2 = < >C1,C2 Le ≡ (1/P1) ∫C1 ds1 (1/P2)∫C2 ds2 Le(x1,x2) = < Le(x1,x2)>C1,C2 . (4.11.13) The "constants" K and KL in (4.11.1) are similarly replaced with their <>C1,C2 averages. Inserting the equations on the first line of (4.11.13) into (4.11.11) then gives (Zs1 + Zs2) i(z) = - ∂zV(z) - jω Le i(z) Le ∂z i(z) = - j (β2/ωV(z) 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 classical 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) ********************************************8 Now let's try to compute Zs1 for the two cylinder geometry! From Bipolar I know that E = (V/a) (chξ–cosu) but this is the transverse field in the dielectric, not what I want! Where do I talk about Ez and n(θ) and tracking? How about (6.5.19) which says Ez(a,θ) = (-jω/σ) (β/βd) n(θ) n1(ξ1,θ) = (q/2π)[ 1 + 2 !Syntax Error, I (-1)m e-m|ξ| cos(mθ) ] . Bipolar (A.13) (6.5.5) n(θ) = n(θ)/a Then we seem to have Ez(a,θ) = (-jω/σ) (β/βd)(1/a) (q/2π)[ 1 + 2 !Syntax Error, I (-1)m e-m|ξ| cos(mθ) ] Ez(a,θ)max = (-jω/σ) (β/βd)(1/a) (q/2π)[ 1 + 2 !Syntax Error, I e-m|ξ| ] <Ez(a,θ)> = (1/2π) ∫dθ Ez(a,θ) = (-jω/σ) (β/βd)(1/a) (q/2π)[1] P = 2πa Then we get (my formula) p = (2πa) * <Ez(a,θ)> / Ez(a,θ)max = (2πa) 1/ [ 1 + 2 !Syntax Error, I e-m|ξ| ] ξ1 = - sh-1 (a/a) = - sh-1 [ ] sh ξ1 = - (a/a) = - Now back up 1 + 2 !Syntax Error, I e-m|ξ| 1 + Σm=1∞ xm = 1 + x + x2 + = 1/(1-x) Σm=1∞ xm = 1/(1-x) - 1 = x / (1-x) Maple confirms 1 + 2 Σm=1∞ xm = 1 + 2x/(1-x) = (1+x)/(1-x) Maple confirms, this is > 1 Let x = e-ξ so then have 1 + 2 !Syntax Error, I e-mξ = = = = coth(ξ/2) = = We then seem to have p = (2πa) 1/ [ 1 + 2 !Syntax Error, I e-mξ] = (2πa) Check limit first. As ξ→∞, get p = 2πa which is correct, good! Now assume that ξ1 = - ξ2 and ξ2 > 0 so we then have b = a( cothξ2 - cothξ1) = 2a cothξ2 = 2a chξ2/shξ2 shξ2 = (a/a) so b = 2a chξ2/( (a/a) ) = 2a chξ2 chξ2 = b/(2a) This seems to simple, but I see in (11.2) of bipolar b = xc2 - xc1 = a/thξ2 - a/thξ1 = 2a/thξ2 = 2a coth ξ2 agrees with above So OK we then have = = a = (1/2) from below (6.5.7) so we get p = (2πa) = (2πa) = (2πa) = (2πa) = (2πa) = (2πa) = (2πa) [ ]1/2 If b >> a, this becomes p = (2πa) [ 1- 4a/b]1/2 ≈ (2πa) (1 - 2a/b) p = 2πa ≈ 2πa which is certainly very similar but not the same. What happens if I square my ratio to get something more power oriented, so then p1 ≡ (Zs1/Zs1,max)2P1 Then the above would say p = (2πa) = (2πa) [ ]2 = (2πa) and this also disagrees with the King quoted result. = Reader Exercise. First, recall these high-frequency results for a transmission line consisting of two round conductors, Ez(a,θ) = (-jω/σ) (β/βd) n(θ) // just below the surface (6.5.19) n(θ) = (q/2π)(1/a)[ 1 + 2 !Syntax Error, I (-1)m e-mξ cos(mθ) ] (6.5.5) where we use ξ2 > 0 for the right conductor in Fig. 6.7 and the fact that n = n/a. Thus, the longitudinal E field on the right-side conductor's boundary is given by, Ez(a,θ) = (-jω/σ) (β/βd)(1/a) (q/2π)[ 1 + 2 !Syntax Error, I (-1)m e-mξ cos(mθ) ] . (a) Using our model definition of the effective active perimeter p given in (4.11.10), p ≡ P = P show that p = (2πa) = (2πa) [ Hint: Show that 1 + 2 Σm=1∞ xm = (1+x)/(1-x) = coth(ξ2/2) where x = e-ξ .] (b) For cylindrical conductors both of radius a, using (6.5.6) that shξ2 = (d/a) b = 2d chξ2/shξ2 show that chξ2 = b/(2a) and therefore that p = (2πa) = (2πa) . This is similar to (but different from) the result we quoted from King, p = 2πa . a = wire radius, b = center line separation (2.5.2) King in turn quotes the result from this reference which we have not been able to examine, J.R. Carson, "Wave Propagation over Parallel Wires: The Proximity effect", Philosophical Magazine, volume IXLI, June 1921, pages 607–633. King does not specify how his active perimeter region is defined. What he says (TLT p 30) is that the surface impedance is given with radius a replaced by an effective radius aeff = a . For wide-spaced conductors, both formulas give p = 2πa as we would expect. When the conductors almost touch so b → 2a, both formulas give p→0 again as we would expect. For b >> a, our result gives p = (2πa) (1-2a/b) while King's gives p = (2πa) (1-2a2/b2) . Fat twinlead Fig 2.16