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#1
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![]() W5DXP wrote: Jim Kelley wrote: Waves that propagate to a dissipative load transfer energy to that dissipative load. How does the wave about to be dissipated differ from the wave that is about to be canceled? Hint: they are identical. The waves themselves are indeed identical, i.e. you can't measure a difference between them on a transmission line. But one might indeed be transferring energy while the other is not. For example, take a length of 50 ohm transmission line with a short at one end. Think about this circuit with and without a circulator at the source. With the circulator in circit, energy is transferred from the source to the circulator load. Without the circulator, the source transfers no energy to a load. If there are no re-reflections from the source, is there a measureable difference? How does the wave know ahead of time whether it is going to encounter a dissipative load or not? It obviously cannot know ahead of time so all waves with the same V and I in phase carry the same amount of energy. This is a misconception on your part. I don't claim anything has to know anything. That is your claim, and I don't agree. I think it's silly. A wave about to encounter a dissipative load or the same wave about to be canceled carry the same amount of energy. They have the same potential to transfer energy. The do not necessarily transfer that energy. That depends on boundary conditions. The wave gives up that energy in both cases. In the first case it gives up its energy as heat. In the second case, it gives its energy to the constructive interference. I understand that is your theory. I even like the sound of your theory. But you need to understand that it is only your theory. It is an unproven, and unsupported theory. I've tried to come up with support for it, but find none. Yes, the energy which would have been reflected does appear in the transmitted direction. We know this from conservation of energy - energy incident equals energy transmitted plus energy reflected. What does not appear to happen is all the bouncing around you describe. It can be shown that at the first boundary, two reflected waves destructively interfere, producing zero reflected energy. It can also be seen at that boundary that the two waves traveling in the forward direction yield all of the forward energy due to their constructive interference. I understand that if you take the interference term and change its sign you get the same number, and its not just a coincidence. Yes, you can say the amount cancelled in the reflected direction equals the amount of enhancement in the forward direction. But that doesn't mean energy had to turn around in order to accomplish that. That would only be true if energy were indeed traveling in the reverse direction to begin with. When we realize it's not, there's nothing left to account for Cecil. There's no conservation of energy problem to solve. It's all there, just as it says in the physics books, and enegineering books as well. If you want to believe that energy bounces all over the place like, so be it. Just don't expect to be able to sell paraphysical phenomena as science. But as I've tried to tell you many times before, when you can show that Poynting vector solution changing direction solely as a result of destructive interference, you'll have a case - and my vote. Until then, it looks like a non-starter. 73, ac6xg |
#2
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Jim Kelley wrote:
The waves themselves are indeed identical, i.e. you can't measure a difference between them on a transmission line. But one might indeed be transferring energy while the other is not. How can two waves that are identical possibly be doing different things. It doesn't make any sense at all. Waves cannot exist without intrinsic energy!!! If any wave is destroyed, it gives up its energy. There is simply no other place for the energy to go. How does the wave know ahead of time whether it is going to encounter a dissipative load or not? It obviously cannot know ahead of time so all waves with the same V and I in phase carry the same amount of energy. This is a misconception on your part. I don't claim anything has to know anything. That is your claim, and I don't agree. I think it's silly. That the wave has to be smart falls out from your assertions. Your waves appear to exist without intrinsic energy so they have to know ahead of time whether to exist without energy or not. Only a wave existing without energy could be destroyed and not give up any energy in the process. But you need to understand that it is only your theory. It is an unproven, and unsupported theory. I've tried to come up with support for it, but find none. Please reference the cross-posting from sci.physics.electromag. Somebody predicted that those guys would tear me apart. On the contrary, they tend to agree with me. It can be shown that at the first boundary, two reflected waves destructively interfere, producing zero reflected energy. It can also be seen at that boundary that the two waves traveling in the forward direction yield all of the forward energy due to their constructive interference. Those reflected waves not only destructively interfere, they are destroyed so they MUST give up any intrinsic energy in the waves. Nothing else is possible. Yes, you can say the amount cancelled in the reflected direction equals the amount of enhancement in the forward direction. But that doesn't mean energy had to turn around in order to accomplish that. Huh??? "Reflected" to "forward" isn't a turn around? Can I have a hit off whatever you are smokin'? :-) That would only be true if energy were indeed traveling in the reverse direction to begin with. When we realize it's not, there's nothing left to account for Cecil. If reflected energy from a mismatched load is not traveling in the reverse direction, pray tell what direction is it traveling in? I can see you haven't referenced the reflected power flow vector. There's no conservation of energy problem to solve. There certainly is in your scenario. The reflected energy doesn't ever change directions but it somehow joins the forward wave anyway. Can you spell B-A-F-F-L-E-G-A-B? We know reflected energy is traveling rearward from a mismatched load. Your mantras cannot change that fact of physics. -- 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 80,000 Newsgroups - 16 Different Servers! =----- |
#3
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![]() W5DXP wrote: Jim Kelley wrote: The waves themselves are indeed identical, i.e. you can't measure a difference between them on a transmission line. But one might indeed be transferring energy while the other is not. How can two waves that are identical possibly be doing different things. It doesn't make any sense at all. Waves cannot exist without intrinsic energy!!! If any wave is destroyed, it gives up its energy. There is simply no other place for the energy to go. Here's a quote from Born and Wolfe, pg. 48, talking about waves encountering the interface between dense and less dense media: "Although there is a field in the second medium, it is easy to see that no energy flows across the boundary. More precisely it will be shown that, although the components of the Poynting vector in the direction normal to the boundary is in general finite, its time average vanishes; this implies that the energy flows to and fro, but that there is no lasting flow into the second medium." Please reference the cross-posting from sci.physics.electromag. Somebody predicted that those guys would tear me apart. On the contrary, they tend to agree with me. I tended to agree with what you said there too. You can be very agreeable when you want to be. :-) I notice you didn't mention anything about photons dematerializing and rematerializing over there. Give that one a try. :-) Those reflected waves not only destructively interfere, they are destroyed so they MUST give up any intrinsic energy in the waves. Nothing else is possible. If you mean they give up their energy to the two waves traveling in the other direction, then I agree. Can I have a hit off whatever you are smokin'? :-) I've got a pack of Camel lights out in the car. You can have 'em. That would only be true if energy were indeed traveling in the reverse direction to begin with. When we realize it's not, there's nothing left to account for Cecil. If reflected energy from a mismatched load is not traveling in the reverse direction, pray tell what direction is it traveling in? P2 is energy that had been traveling in the reflected direction and was re-reflected. That's it. Good thing too, because any more than P2 and P1 and you'd have a perpetual motion machine! :-) I can see you haven't referenced the reflected power flow vector. I've seen you reference it, but I have seen you define one - or write it out mathematically. There's no conservation of energy problem to solve. There certainly is in your scenario. The reflected energy doesn't ever change directions but it somehow joins the forward wave anyway. Can you spell B-A-F-F-L-E-G-A-B? You wrote the bafflegab - whatever that is. I certainly never said reflected energy joins the forward energy. That's your invention. 73, ac6xg |
#4
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Cecil, W5DXP wrote:
"They", above, are reflected waves. The two waves traveling in the other direction are forward waves." Forward waves? Being, I suppose, a voltage wave and a current wave, as it is impossible to have two voltage waves or two current waves of identical frequency traveling in the same direction on the same line. Two identical frequency waves, voltage or current, form a resultant and merge. They can`t maintain a separate existence and become inseparable in a single waveform. Best regards, Richard Harrison, KB5WZI |
#5
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Richard Harrison wrote:
Cecil, W5DXP wrote: "They", above, are reflected waves. The two waves traveling in the other direction are forward waves." Forward waves? Being, I suppose, a voltage wave and a current wave, as it is impossible to have two voltage waves or two current waves of identical frequency traveling in the same direction on the same line. Two identical frequency waves, voltage or current, form a resultant and merge. They can`t maintain a separate existence and become inseparable in a single waveform. Of course they superpose. In the s-parameter equation, b1=s11*a1+s12*a2, those two reflected waves superpose to b1. In the s-parameter equation, b2=s21*a1+s22*a2, those two forward waves superpose to b2. They superpose at the point where they both appear together for the first time. But the point is that s11*a1 and s12*a2 are flowing in the direction of the source when they cancel (interfere destructively). s21*a1 and s22*a2 are flowing in the direction of the load when they interfere constructively. -- 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 80,000 Newsgroups - 16 Different Servers! =----- |
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