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wrote:
That's what you get when you use the surge impedance and compute surge rho. Try steady state impedance for steady state rho. rho = (0-50)/(0+50) = -1 as expected Heh, heh, the steady state impedances are V/I ratios which are results incapable of causing anything. Your logic is a closed loop. The V/I ratios cause the rho to be -1. Therefore, rho=-1 causes the appropriate V/I ratios. Can you name any other result in reality that causes itself? Hint: Only physical impedance discontinuities can cause reflections. Image impedances are incapable of causing anything since they are an end result. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
#2
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Cecil Moore wrote:
wrote: That's what you get when you use the surge impedance and compute surge rho. Try steady state impedance for steady state rho. rho = (0-50)/(0+50) = -1 as expected Heh, heh, the steady state impedances are V/I ratios which are results incapable of causing anything. Your logic is a closed loop. The V/I ratios cause the rho to be -1. Therefore, rho=-1 causes the appropriate V/I ratios. Can you name any other result in reality that causes itself? Hint: Only physical impedance discontinuities can cause reflections. Image impedances are incapable of causing anything since they are an end result. Was it not only a few days ago that there was agreement that If it is acceptable to claim that sometimes there is no reflection at an impedance discontinuity, then it must also be acceptable to claim that sometimes there is a reflection where there is no discontinuity. ? ....Keith |
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