loss1 REVIEWED

Repair of old ωc loss model PhL 6.22.14 Here I review the old ωc loss model. My focus here is on the usual complex averaging business, and I completely overlook the point that vd is not the same at all ω which is the real reason this loss model is no good. The phase velocity is not always vd. Somewhere else I cleared this up, but I don't offhand know where that is done. (10.8.14). In Appendix D.11 I came up with a loss model that I now think is wrong. Here is that model _______________________________________________________ (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. __________________________________________________________ According to the above model, since it says |E(z)| = e-αz |E(0)|, we can identify α = jβd of Appendix D, and then we have βd = -jα = -j[ L(G + jωC) + C(R + jωL)] (vd/2) = -j[ L(G + jωC) + C(R + jωL)] (vd/2) I am not exactly sure what is wrong with this model, but I do know that it disagrees with the model developed in Chapter 4 which is this from (5.3.4) βd = k = -j = -j and which is verified in both HM and Matick. One thing I notice is that the two models agree at large ω: βd1 = -j[ L(G + jωC) + C(R + jωL)] (vd/2) ≈ -j[ L(jωC) + C(jωL)] (vd/2) = -j[ 2L(jωC)] (vd/2) = ωLC vd = ω/vd if we assume βd is just the β of the dielectric, then we have => LC = 1/vd2 // another equation I need to work into things! Meanwhile βd2 = -j = -j = -j = -j j ω = ω and then again we set this to ω/vd to get vd = 1/. So vd is the correct phase velocity at high ω regardless of the values of the 4 parameters. The models are completely different however at ω = 0 We have βd1 = -j[ L(G + jωC) + C(R + jωL)] (vd/2) = -j[ L(G) + C(R)] (vd/2) = -j[ L(G) + C(R)] (1/2) 1/ = (-j/2) [G + R] βd2 = -j = -j Question: What is wrong with my model bracketed above? Let's look at the opening statement P = IV = U vd = [(1/2)L I 2 + (1/2)CV2 ] vd Here (1/2)L I 2 is the total mag energy stored in the inductance of the line per unit length. Here (1/2)CV 2 is the total elec energy stored in the capacitance of the line per unit length. So I take a little dz of the line, and U is how much energy is stored in that dz. Energy is being burned in that dz as well, and maybe that is what I have omitted? Well, think of this as p(z) = i(z)V(z) = U(z) vd = [(1/2)L i(z) 2 + (1/2)CV(z)2 ] vd and at some z, this IS the total stored energy regardless of loss rate in that dz. Ah, but is this correct for complex i(z) and V(z)? OK, I will go through that song and dance. We have i(z) = ej(ωt-βz) i(0) iphy(z) = Re[ej(ωt-βz) i(0)] V(z) = ej(ωt-βz)V(0) Vphy(z) = Re[ej(ωt-βz) V(0)] The actual power burned is pph(z) = iph(z)Vph(z) = Re[i(z)] Re[V(z)] = Re[ej(ωt-βz) i(0)] Re[ej(ωt-βz) V(0))] = { cos(ωt-βd)Re[i(0)] - sin(ωt-βd)Im[i(0)] } { cos(ωt-βd)Re[V(0)] - sin(ωt-βd)Im[V(0)] } = cos2(ωt-βd) Re[i(0)] Re[V(0)] + sin2(ωt-βd) Im[i(0)] Im[V(0)] - sin(ωt-βd) cos(ωt-βd) { Re[i(0)] Im[V(0)] + Im[i(0)] Re[V(0)] The trick I think is now to average this result over time, so the squared trig terms So we then get < pph(z) > = 1/2 Re[i(0)] Re[V(0)] + 1/2 Im[i(0)] Im[V(0)] = 1/2 { Re[i(0)] Re[V(0)] + Im[i(0)] Im[V(0)] } This has the form f = Re(a)Re(b) + Im(a)Im(b) But notice that Re(ab*) = Re [ Re(a)+j Im(a)][Re(b)-jIm(b)] = Re(a)Re(b) + Im(a)Im(b) = Re(a*b) Therefore we have shown that < pph(z) > = (1/2) Re{ i(0)V*(0) } But putting in the phases, they then cancel and this becomes < pph(z) > = (1/2) Re{ i(z)V*(z) } watts What