I see that you are convinced that you've explained where the bouncing
waves of average power go and what they do. Did anyone understand it? If so, would someone else please try to explain it to me? Where does that 18 watts of "reverse power" go, and why? Roy Lewallen, W7EL Cecil Moore wrote: [More generalizations with no numbers] |
Roy Lewallen wrote:
I see that you are convinced that you've explained where the bouncing waves of average power go and what they do. Did anyone understand it? If so, would someone else please try to explain it to me? Where does that 18 watts of "reverse power" go, and why? Wouldn't it go to the circulator load which must always be placed at the source in order to clearly illustrate what happens when a circulator isn't in place at the source? ac6xg |
Jim Kelley wrote:
Roy Lewallen wrote: I see that you are convinced that you've explained where the bouncing waves of average power go and what they do. Did anyone understand it? If so, would someone else please try to explain it to me? Where does that 18 watts of "reverse power" go, and why? Wouldn't it go to the circulator load which must always be placed at the source in order to clearly illustrate what happens when a circulator isn't in place at the source? ac6xg Sorry, I was asking about the simple example I posted (which doesn't have a circulator), not the problem posted by Cecil. How does placing a circulator at the source illustrate what happens when it isn't in place? Isn't it possible to explain what happens to the "reverse power" without a circulator? If not, why not? Roy Lewallen, W7EL |
Roy Lewallen wrote:
Please tell me which of the numbers I posted disagree with yours, and which numbers you got. Once again, mine a It's not your numbers, Roy, it's your premises that violate the conservation of momentum principle among other principles of physics. RF waves possesss momentum and that momentum MUST be preserved. Your premises simply violate the conservation of momentum principle. When you assert that the reflected waves possess no energy and it is stored in some magic place at sub-light speeds, you are in violation of the principles of physics. You can resolve all of this by telling us where the energy is stored, besides in reflected waves, when we are dealing with light in free space and no transmission line because exactly the same thing happens with EM light waves as happens with EM RF waves. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- |
Roy Lewallen wrote: Jim Kelley wrote: Roy Lewallen wrote: I see that you are convinced that you've explained where the bouncing waves of average power go and what they do. Did anyone understand it? If so, would someone else please try to explain it to me? Where does that 18 watts of "reverse power" go, and why? Wouldn't it go to the circulator load which must always be placed at the source in order to clearly illustrate what happens when a circulator isn't in place at the source? ac6xg Sorry, I was asking about the simple example I posted (which doesn't have a circulator), not the problem posted by Cecil. It was a tongue-in-cheek reply, Roy. These problems always seem to end up with a circulator in them at some point - clearly illustrating what happens under entirely different circumstances. :-) 73, Jim AC6XG |
Roy Lewallen wrote:
Did anyone understand it? If so, would someone else please try to explain it to me? Where does that 18 watts of "reverse power" go, and why? Bob Lay, w9dmk, could explain it but he's off on a camping trip. I think the QEX editors can now explain it also. 18 watts is rejected by the mismatched load. It was part of a forward wave that contained energy and momentum. That energy and momentum MUST be conserved according to the laws of physics. The reflected wave heads back toward the source at the speed of light and part is re-reflected as Pref2*rho^2. The remaining Pref(1-rho^2) part engages with Pfor1*rho^2 in an equal magnitude/opposite phase wave cancellation as explained by Walter Maxwell on page 23-9 of "Reflections II". Since energy and momentum cannot be cancelled, the two cancelled waves conserve energy and momentum by heading back toward the load at the speed of light as a re-reflection of energy which joins the forward wave energy. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- |
Jim Kelley wrote:
Roy Lewallen wrote: Did anyone understand it? If so, would someone else please try to explain it to me? Where does that 18 watts of "reverse power" go, and why? Wouldn't it go to the circulator load which must always be placed at the source in order to clearly illustrate what happens when a circulator isn't in place at the source? When a Z0-match is in place at the source, everything is also clear since zero reflected energy reaches the source. The Z0-match case, using an antenna tuner, is the most likely case to be encountered in ham radio. 100% of the reflected energy and momentum is re-reflected back toward the load, just as Walter Maxwell has been saying as long as I can remember. This stuff is not new. It is explained in "Fields and Waves ..." by Ramo and Whinnery, copyright 1950's. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- |
Roy Lewallen wrote:
Isn't it possible to explain what happens to the "reverse power" without a circulator? It is certainly possible in a Z0-matched system where zero reflected energy reaches the source and the source is feeding its designed-for load. If no reflected energy reaches the source, the source impedance doesn't matter, except for efficiency, which doesn't affect the rest of the system. A one ohm source providing 100v to a 50 ohm load and a one megohm source providing 100v to a 50 ohm load will result in identical external conditions when driving a 50 ohm load. The picture is not as clear when reflected current and voltage are allowed to flow into the source. We usually don't know what the source impedance is and that certainly handicaps any analysis. Modern designers simply resort to protection circuitry and don't worry about the energy analysis. We do know that reflections reaching the source are not benign. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- |
Jim Kelley wrote:
It was a tongue-in-cheek reply, Roy. These problems always seem to end up with a circulator in them at some point - clearly illustrating what happens under entirely different circumstances. :-) The circulator, lossless feedlines of unreasonable length, Time Domain Reflectometers, TV ghosting, etc. are all tools for illustration purposes. However, a Z0-matched system is an ordinary configuration in ham radio and is easy to analyze since no reflected energy is allowed to reach the source. The conservation of energy and momentum rules dictate where the energy must go in such a case. We can debate why the reflected energy is 100% re-reflected but there is no question that it *is* 100% re-reflected because none reaches the source and there are only two directions in a transmission line. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- |
I've explained how I calculated how much energy is stored in the
transmission line. The movement of energy within the line is complex; in the abbreviated analysis I've had time to do so far, it sloshes back and forth in regions within the line. It does not travel in waves of average power, bouncing back and forth, and believing so isn't necessary in order comply with energy conservation. Your view of power and energy is oversimplified, and it fails when you're pressed to explain what happens at the interface between the line and the outside world. Momentum is conserved in mechanical elastic collisions, but not in inelastic ones, e.g., when energy is being extracted. I wouldn't begin to try to apply this to a transmission line, but I see it doesn't bother you. I understand and believe the fundamental principles of physics and thermodynamics -- I'm just careful not to misapply them. Roy Lewallen, W7EL Cecil Moore wrote: Roy Lewallen wrote: Please tell me which of the numbers I posted disagree with yours, and which numbers you got. Once again, mine a It's not your numbers, Roy, it's your premises that violate the conservation of momentum principle among other principles of physics. RF waves possesss momentum and that momentum MUST be preserved. Your premises simply violate the conservation of momentum principle. When you assert that the reflected waves possess no energy and it is stored in some magic place at sub-light speeds, you are in violation of the principles of physics. You can resolve all of this by telling us where the energy is stored, besides in reflected waves, when we are dealing with light in free space and no transmission line because exactly the same thing happens with EM light waves as happens with EM RF waves. |
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