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On Dec 30, 10:30*am, Cecil Moore wrote:
Keith Dysart wrote: No need for translation, though this is not quite what was said above. Note the words "automatically" in the first quote and "may be quite different" in the second. The original authors allow for the possibility that it might be the same, while your "translation" removes that possibility. They engaged in typical author-speak. I think not. They wrote with precision in an attempt to prevent themselves being quoted out of context. It didn't work, or course. My university professors had no such limitations. They were quite harsh on anyone who tried to figure out where the power goes inside a Thevenin or Norton equivalent source. It is not I who wants it both ways. For me it is clear that there is no reflection when the output (source) impedance is the same as Z0. And when it is not equal to Z0, there is a reflection. Apparently that is NOT clear to you. In the earlier example, there is no impedance discontinuity at the '+' points, yet you require reflections at those points. That's what you cannot have both ways. If there's no traveling wave energy flowing through the '+' points, there must exist reflections. Well, there is no energy flowing through the '+' points. And I have no issue if you wish to claim that there are reflections at these points, though I might use 'bouncing' to differentiate from reflections occuring at points with non-zero reflection coefficients. If reflections exist, there must exist an impedance discontinuity. There is no impedance discontinuity. You do seem to be trying to have it both ways. No energy is flowing (q.v. IEEE definition of instantaneous power), and yet you want energy to be flowing. Not when the output (source) impedance is known. It is then easy to compute the magnitude of the reflection using the standard rules for reflection coeficient. Although many have tried to prove that the output (source) impedance is the impedance encountered by the reflected waves, all of those numerous experiments have failed. You, Cecil, are the only one who believes this. Any good book on transmission lines will tell you otherwise. Therefore, there is a high probability that the impedance encountered by the reflected waves is *NOT* the output (source) impedance. The argument has raged loud and long since at least the 1980's. You are not going to resolve it by hand-waving. Keith, if you can prove that the reflected waves encounter the output (source) impedance, you are a better man than all of the many others who have tried and failed. Web references and Spice models which agree that "the output (source) impedance is the impedance encountered by the reflected waves" have been previously provided, but you refused to explore them. If I recall correctly, this was because they did not model the complexity of an average ham transmitter so they were not of interest to you. Since you have refused to explore the question, you should refrain from making pronouncements. ...Keith |
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