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#11
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Owen Duffy wrote:
Dan Bloomquist wrote in news:4khKh.5006$ya1.3770 @news02.roc.ny: Owen Duffy wrote: Dan Bloomquist wrote in : Are the principles that apply to this example inconsistent with the general case (eg dc, ac, transient, steady state etc)? Yes. I would have thought the answer was NO, the principles are *not* inconsistent with the general case. Hi Owen, See the example I posted where a pulse is sent down the line. In that case all the energy is dissipated in the source resistor. Dan, I don't deny there may be cases where that may happen, but it is not true that reflected power is necessarily dissipated (or partly dissipated) in the equivalent source resistance. And I answered accordingly when you asked the very question, I agreed with you. And I showed just that in the step example of a previous post. Owen Best, Dan. |
#12
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Cecil Moore wrote in news:KGhKh.7646$yW.5893
@newssvr11.news.prodigy.net: Owen Duffy wrote: You could devise an example with an AC source that had exactly the same outcome, more complicated mathematics to solve it, but the exactly the same outcome. But the point is that I can devise an example with an AC source that has the opposite outcome simply by using a Norton equivalent source. When the reflection arrives, the dissipation doubles. Your Norton equivalent doesn't pretend to replicate my (real) source in every way, it only pretends to supply the same voltage and current to its load... and it does that. Owen |
#13
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Dan Bloomquist wrote in news:xYhKh.5014$ya1.2122
@news02.roc.ny: But your underlying principle is that the source steps and doesn't change. So any general case where the source is at some different level is now not covered. It semantics Dan, the excitation is part of the scenario or example, it is not a principle of circuits or transmisssion lines. Owen |
#14
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Owen Duffy wrote:
I would have thought the answer was NO, the principles are *not* inconsistent with the general case. Even considering the dissipation within a Thevenin equivalent circuit is inconsistent with any valid significance according to Ramo and Whinnery. -- 73, Cecil, w5dxp.com |
#15
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Owen Duffy wrote:
Dan Bloomquist wrote in news:xYhKh.5014$ya1.2122 @news02.roc.ny: But your underlying principle is that the source steps and doesn't change. So any general case where the source is at some different level is now not covered. It semantics Dan, the excitation is part of the scenario or example, it is not a principle of circuits or transmisssion lines. Like saying the impedance of a capacitor is the same no matter what the frequency...... From here, I'll let you have the last word. I'm not into hand waving.... Owen Best, Dan. |
#16
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Dan Bloomquist wrote in
: Owen Duffy wrote: Dan Bloomquist wrote in news:4khKh.5006$ya1.3770 @news02.roc.ny: Owen Duffy wrote: Dan Bloomquist wrote in : Are the principles that apply to this example inconsistent with the general case (eg dc, ac, transient, steady state etc)? Yes. I would have thought the answer was NO, the principles are *not* inconsistent with the general case. Hi Owen, See the example I posted where a pulse is sent down the line. In that case all the energy is dissipated in the source resistor. Dan, I don't deny there may be cases where that may happen, but it is not true that reflected power is necessarily dissipated (or partly dissipated) in the equivalent source resistance. And I answered accordingly when you asked the very question, I agreed with you. And I showed just that in the step example of a previous post. Dan we are agreed! My conclusion is that the view put by some that reflected power is (necessarily) fully or partly dissipated in the PA equivalent source resistance, (possibly overheating the PA,) is a simplistic view, the explanation doesn't apply in general and although apparently appealing, it is wrong. Owen |
#17
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Owen Duffy wrote:
Dan, I don't deny there may be cases where that may happen, but it is not true that reflected power is necessarily dissipated (or partly dissipated) in the equivalent source resistance. The question is whether reflected power is ever dissipated in the source. I can come up with a black box source that dissipates 100% of the reflected power. All it takes is a circulator and a load resistor. How does your DC principles handle a circulator? -- 73, Cecil, w5dxp.com |
#18
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Owen Duffy wrote:
Your Norton equivalent doesn't pretend to replicate my (real) source in every way, ... I thought we were discussing general principles, not replicating a special case example in every way. -- 73, Cecil, w5dxp.com |
#19
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Owen Duffy wrote:
My conclusion is that the view put by some that reflected power is (necessarily) fully or partly dissipated in the PA equivalent source resistance, (possibly overheating the PA,) is a simplistic view, the explanation doesn't apply in general and although apparently appealing, it is wrong. The question is: Does any reflected joules/second ever get dissipated in the PA? To ascertain the answer, one must calculate the interference patterns in both directions. If the energy is not dissipated anywhere else, it is dissipated in the PA. -- 73, Cecil, w5dxp.com |
#20
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Owen Duffy wrote in
: .... When the switch is closed, current flows into the line. Until t=2T seconds, the current If that flows into the line equals 100V/(100+100 ohms) = 0.5A, during which time the resistor dissipates heat at the rate of I^2*R = 100W. The voltage of the wave Vf travelling from the source is I*Zo = 50V. This situation is constant until t=2T seconds. .... David Ryeburn is awake, even if I wasn't at 5.38 when I wrote this. should be ...during which time the resistor dissipates heat at the rate of I^2*R = 25W... It doesn't materially change anything, but I messed up! Owen |
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