Including Loss in the Appendix D Model REVIEWED

Including Loss in the Appendix D Model PhL 3.17.14 Reviewed 5.14.14. The main item here is a derivation of the decay parameter α for small ω: βd → (ω/vd) - j [(Rdc1+ Rdc2)C ] (vd/2) = (ω/vd) - jα α = (Rdc1+ Rdc2)C ] (vd/2) This is then the replacement you should make in App D wherever βd appears "solo". I then use this in other docs. The second item here was done wrong, trying to get β'2 for low ω. I repaired things and it now says β'2 → α2 which is a constant and agrees with elsewhere, but only for super super low ω is this relevant as I show in "conclusions on low freq". Here we look at the power in a wave, and at its rate of loss per distance down the line. We need to know how much power is in the wave in the first place. Maybe get that from L and C based on field energy storage. UL = (1/2)L I2 L = per unit length UL = energy/unit length UC = (1/2)CV2 If the wave travels at vD roughly speaking, then Pwave = U(joules/m) * vd (m/sec) = joules/sec = watts Then we have Pwave = [(1/2)L I2 + (1/2)CV2 ] vd Since the current is uniform, I think we can argue that KL = K without using any King theory. Then let's just use our standard forms (ignoring Li for the moment) C = 4πε/K Le = K then we have so far Pwave = [(1/2) K I2 + (1/2) 4πε/K V2 ] vd Now write this for the rate of power loss going down the line, dPwave/dz = [ K I dI/dz + 4πε/K V dV/dz ] vd Now I can see that dV/dz = - IRdc (low ω) in each conductor, so perhaps dV/dz = - I(Rdc1+ Rdc2). But I don't see how I can decrease! Current in a resistor does not decrease with length. But this seems to conflict with I = exp(jωt - jβd) if βd has a complex part. Is this the AC vs DC issue? Well think of Current is shunted through C, so even without G, I can reduce to do this loss through C. How much current goes through C? ZC = 1/(jωC) and So dI(z) = -V(z)/Zc = -V(z) jωCdz so dI/dz = -jωC V(z) where C is per unit length OK, then I get dPwave/dz = -[ K I jωC V(z) + 4πε/K V I(Rdc1+ Rdc2) ] vd = -I V [ K jωC + 4πε/K (Rdc1+ Rdc2) ] vd = -I V [ K jωC + C (Rdc1+ Rdc2) ] vd = -I V C [ K jω + (Rdc1+ Rdc2) ] vd = -Pwave C [ K jω + (Rdc1+ Rdc2) ] vd So as ω → 0 this says (in this limit, there is in fact no dI/dz so first term is 0) dPwave/dz = - Pwave [(Rdc1+ Rdc2)C ] vd For units, we have to show that [...] vd = 1/m, but RC = sec/m2 * m/sec = 1/m so OK. Now go back to I(z) = I0 exp(-jkz) V(z) = V0 exp(-jkz) P(z) = P0 exp(-2jkz) = P0 exp(-2j[kR+jkI]z) = P0 exp(-2jkRz) exp(+2kIz) |P(z)| = P0 exp(+2kIz) d|P(z)|/dz = 2kI P0 Compare this to dPwave/dz = - Pwave [(Rdc1+ Rdc2)C ] vd and we conclude that 2kI = - [(Rdc1+ Rdc2)C ] vd // units = (sec/m2)*(m/sec) = 1/m, correct This is the first time (I think) that I have ever attempted this estimate of loss for low ω. The implication is that the wave is really doing this: exp(-jkz) where k ≈ βd +jkI = βd - j [(Rdc1+ Rdc2)C ] (vd/2) This means that in Appendix D right from the start we should replace βd → βd - j [(Rdc1+ Rdc2)C ] (vd/2) [ this is the main point here ] or βd → (ω/vd) - j [(Rdc1+ Rdc2)C ] (vd/2) [βd → (ω/vd) - j α , note def of α ] at least for very low ω, since this is what the line is really doing !!! Now we always think of the imaginary loss part as being small compared to the real βd part. BUT, this is not true as ω → 0 as this last equation makes clear!!! As ω → 0, we have "all loss and no wave". I think this might change the low ω limit of my Appendix D fields and perhaps resolve that conundrum. Notice that βd2 → [(ω/vd) - j [(Rdc1+ Rdc2)C ] (vd/2)]2 = (ω/vd)2 - [(Rdc1+ Rdc2)C ]2 (vd/2)2 - 2j [(Rdc1+ Rdc2)C ] (vd/2) (ω/vd) [ = (ω/vd)2 -α2 -2jα (ω/vd) ] β'2 = β2 - βd2 → -jωμσ - (ω/vd)2 + [(Rdc1+ Rdc2)C ]2 (vd/2)2 + 2j [(Rdc1+ Rdc2)C ] (vd/2) (ω/vd) [ = -jωμσ - (ω/vd)2 + α2 + 2jα(ω/vd) ] Now we take ω → 0 we get β'2 → [(Rdc1+ Rdc2)C ]2 (vd/2)2 [ = α2 ] whereas it used to go to 0 ! If resistances are low, perhaps we have [(Rdc1+ Rdc2)C ] (vd/2) << 1 so we can then approximate [ I had errors above, fixed them on 5.15.14, new conclusion is just ] β'2 → [(Rdc1+ Rdc2)C ]2 (vd/2)2 [ = α2 ] C = 4πε/K [ Hold the phone here: this disagrees with what I say in my 3.20.14 "Conclusions on low freq" doc. OK, I had errors above and they are now fixed and things agree. The above limit is for ω < 10-9 Hz ! ] ] OK, then what happens to this Ez(r,m) = (1/4) ηm I Rdc (aβ') [ - ] fm = [ - ] Notice that small Rdc implies small β' but also small δ . But I still don't see a nice way to claim that the fields Ez(r,m) → 0 as ω → 0. I know that x→ 0 does not do it, and x→∞ involves this This is a very poorly defined limit and I cannot believe it would apply. [ happily I stopped here. This stuff is done more correctly in "conclusions on low freq limit ". ]