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On Mar 21, 9:32*pm, Roy Lewallen wrote:
Keith Dysart wrote: . . . The equation * Pg(t) = Pf.g(t) + Pr.g(t) is more interesting. The basis for this is superposition. The forward and reverse voltage and current are superposed to derive the actual voltage and current. It would seem invalid to also sum the powers. To me it was a complete surprise that summing the voltages produces the correct total voltages and, at the same time, summing the powers (which are a squared function of the voltage) also produce the correct result. But by starting with the equations used to derive forward and reverse voltage and current, it can be easily shown with appropriate substitution that Ptot is always equal to Pforward + Preverse (Or Pf - Pr if you use the other convention for the direction of the energy flow)s. It simply falls out from the way that Vf and Vr are derived from Vactual and Iactual. It only holds true when Z0 is purely real. Of course, when it isn't, time domain analysis becomes very much more cumbersome. But it's not hard to show the problem using steady state sinusoidal analysis, and that's where the cos term appears and is appropriate. * * So * * Pg(t) = Pf.g(t) + Pr.g(t) * is always true. For any arbitrary waveforms. Inclusion * of cos(theta) terms would be incorrect. Thanks for providing the limitation. But I am having difficulty articulating where the math in the following derivation fails. Starting by measuring the actual voltage and current at a single point on the line, and wishing to derive Vf and Vr we have the following four equations: V = Vf + Vr I = If - Ir Zo = Vf / If Zo = Vr / Ir rearranging and substituting Vf = V - Vr = V - Zo * Ir = V - Zo * (If - I) = V - Zo * (Vf/Zo - I) = V - Vf + Zo * I = (V + Zo * I)/2 similarly Vr = (V - Zo * I)/2 Pf = Vf * If = Vf**2 / Zo = ((V + Zo * I)(V + Zo * I)/4)/Zo = (V**2 + 2(V * Zo * I) + Zo**2 * I**2) / (4 * Zo) Pr = Vr * Ir = (V**2 - 2(V * Zo * I) + Zo**2 * I**2) / (4 * Zo) So, comtemplating that P = Pf - Pr and substituting P = (V**2 + 2(V * Zo * I) + Zo**2 * I**2) / (4 * Zo) - (V**2 - 2(V * Zo * I) + Zo**2 * I**2) / (4 * Zo) = 4(V * Zo * I) / (4 * Zo) = V * I as required. So when Zo is real, i.e. can be represented by R, it is clear that P always equals Pf - Pr. And it does not even matter which value of R is used for R. It does not have to be the characteristic impedance of the transmission line, the subtraction of powers still produces the correct answer. But when Zo has a reactive component, it still cancels out of the equations. So why is this not a proof that also holds for complex Zo. I suspect it has to do with complex Zo being a concept that only works for single frequency sinusoids, but am having difficulty discovering exactly where it fails. And if it is related to Zo and single frequency sinusoids, does that mean that P = Pf - Pr also always works for single frequency sinusoids? ...Keith |
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