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![]() "Roy Lewallen" wrote in message ... Tarmo Tammaru wrote: Roy, You are cheating. In the steady state there is no load on your source. Regardless of what the Bird meter reads. Do one of the following: Why is this cheating? There is reverse power on the line. The source is not absorbing the reverse power. You and others have said, without qualification, that it does. I've shown a case where it doesn't. What you have done is to put a resonant parallel tuned circuit on the output end of the 50 Ohm resistor. After 1/2 cycle of RF, the voltage on the two ends of the resistor is equal. Hence no power is delivered from the source, provided it is lossless coax. Tell me if I am wrong, but you could put an HP oscillator and a transformer that delivers 70.7V into an open circuit in place of the 100W amp, and in the steady state you would reach the same conditions. It would just take longer. If the Bird read 100W before, it would still read 100W. Assume it is a perfect WM that does not absorb any power. If you buy the tuned circuit analogy, the WM is in effect measuring circulating current in the tuned circuit. Here is what I think the problem with trying to understand reflected power in transmitters is. On the one hand, you have forward and reflected power in and out of a black box. On the other hand, open up the black box and measure the IMPEDANCE back towards the load. Without knowledge of reflections, we know from the Smith chart how a misterminated line is going to change the load on the collector of the transistor, and hence cause it to change its power output. So, in the impedance analysis there is no concept of absorbing reflected power. You don't like talking about pulses, but that is the way I was taught reflections. We used "Electromagnetic Energy Transmission and Radiation" by Adler, Chu, and Fano. 1960 (gulp!) Tam/WB2TT |
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Conservation of Energy | Antenna |