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Keith Dysart wrote:
On Dec 30, 9:51 am, Cecil Moore wrote: Roger wrote: Roy Lewallen wrote: Yes, that is correct. The impedance of the source (a perfect voltage source) is zero, so the reflection coefficient seen by the reverse traveling wave is -1. The logic of this assumption eludes me. In fact, it seems completely illogical and counter to the concept of how voltage and current waves are observed to move on a transmission line. The reflection coefficient for a short is -1, is it not? One way of viewing a short is that it is a perfect voltage source (i.e. 0 output impedance) set to 0 volts. Good point. The power disappears here, but energy transfer is evident (we have current). We both have power and do not have power. At the source, when the reflect wave returns and re-reflects, we have 2v from the reflected wave matched with 1v from the source. We have both 1v and 2v. OK. Setting it to some other voltage (or function describing the voltage) does not alter its output impedance. It there fore creates the same reflection of the travelling wave, regardless of the voltage function it is generating. A real world voltage source has an output impedance. Use this impedance to compute the reflection coefficient. The reflection will be the same regardless of the voltage function being generated by the source. ...Keith Thanks for pointing out the short circuit voltage view. I had not thought of that. What would you think of using a perfect current source for these thought experiments? My suggested perfect current source would supply only as much current as could be absorbed by the load, so no power would be used by the source, and current would be limited by the load. 73, Roger, W7WKB |
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