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#1
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W5DXP wrote:
http://www.gmi.edu/~drussell/Demos/s....html#standing Another web page which correctly uses superposition only for amplitude; not power. In an ideal line terminated by Zo, ... That configuration is not covered by my statement above which applies only to standing waves on lossless unterminated lines. True. I had moved on to a different configuration, the one originally being discussed. ....Keith |
#3
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Cecil Moore wrote:
wrote: W5DXP wrote: http://www.gmi.edu/~drussell/Demos/s....html#standing Another web page which correctly uses superposition only for amplitude; not power. Nobody is using superposition for power. It may have been my misinterpretation, but I understood you were claiming that the net power distribution on the line should be computed by summing Pf and Pr. If you weren't saying this, then the other way to compute power is to use p(t) = v(t) * i(t) (NET) which leads to no energy crossing the voltage and current zeroes. Do you deny the fact that two 100W light bulbs put out more irradiance (power) than one? It is my policy never to deny facts. I leave that for others. You argue that there must be reflections at a voltage node. No. Merely that if there were, the picture would be no different. The above web page indicates such doesn't exist. Visualize the picture if there were reflections at the minima and maxima. The plots would look exactly the same. Curiousity -- why are you so sure there aren't? The picture would be the same! Those forward and reflected waves flow smoothly right through each other. Another way of viewing the picture and as long as you stick to voltages and currents; no problem. It is only with the claim that energy flows past a point with a constant voltage of 0 that I have a problem. Take another look. In the simulation they show voltage waves. No problem. ....Keith |
#4
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wrote:
Visualize the picture if there were reflections at the minima and maxima. The plots would look exactly the same. Curiousity -- why are you so sure there aren't? The picture would be the same! Because the energy in bright ring interference patterns is NOT trapped between the dark rings. That you choose to remain ignorant of that centuries-old fact of physics is not my problem. -- 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! =----- |
#5
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Cecil Moore wrote:
wrote: Visualize the picture if there were reflections at the minima and maxima. The plots would look exactly the same. Curiousity -- why are you so sure there aren't? The picture would be the same! Because the energy in bright ring interference patterns is NOT trapped between the dark rings. That you choose to remain ignorant of that centuries-old fact of physics is not my problem. Seems to me there must be power in the light rings and none in the dark rings. I am not sure where you get this notion of 'trapped', but if there is no power in the dark rings, then the power from the light rings is certainly not in the dark rings. ....Keith |
#6
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wrote:
Seems to me there must be power in the light rings and none in the dark rings. Yes I am not sure where you get this notion of 'trapped', :-) Good one :-) You say the power in a transmission line is trapped between the zero power points and cannot cross the boundary. The transmission line is virtually identical to the bright and dark rings in a light interference pattern. but if there is no power in the dark rings, then the power from the light rings is certainly not in the dark rings. Point is, the energy in the bright rings is not trapped there. It continues to flow in a straight line. The bright rings and dark rings are the result of interference between two beams of light traveling in straight lines at the speed of light. The voltage-zero and voltage-maximum points on a transmission line are the result of interference between two waves traveling in straight lines at the speed of light. In the absence of a physical impedance discontinuity, there is nothing to change their momentum. Their momentum allows them to "coast" across a voltage-zero point. -- 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! =----- |
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