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Owen Duffy wrote in
: As an exercise, think of a generator that has a Thevenin equivalent of some voltage V and a series impedance of R+j0, connected to a half wave of lossless transmission line where Zo=R. To give a numerical example, lets make V=100 and R=50, so Vr=50 and "reflected power"=50. How much of the "reflected power" is dissipated in the generator. In this case, the generator dissipates less heat than were it terminated in 50 ohms. Ok, the solution: Lets examine the matched load scenario for a start, the 100V generator with 50 ohms internal resistance and a matched 50 ohm load. The current is 100/(50+50) or 1A, the power in the load is 1^2*50 or 50W, the power dissipated in the source is 1^2*50 or 50W. No lets look at the scenario with the o/c half wave lossless line attached to the generator. In the steady state, current from the generator is zero, dissipation in the generator is 0^2*50 or 0, voltage at the generator terminals and at the o/c (load end) of the line is 100V. At the o/c load end, the complex reflection coefficient is 1, so Vf=50V, Vr=50V and "reflected power"=50^2/50 or 50W. But, wait a minute, there is 50W of "reflected power" on the line, the line is matched to the source, and there is zero dissipation in the source, less than when it has a matched load. Don't take anything above to mean that I represent that a simple linear model is a good representation of a transmitter PA. This simple example that shows that existence of "reflected power" on a transmission line does not necessarily result in some or all of the "reflected power" being dissipated in the generator. I will leave it to Cecil to take to confuse this simple example with some photon based complication. Owen |
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