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#1
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wrote:
Can you sum up the problem with conservation of energy that modern RF textbooks get wrong? They don't get it wrong - they just don't discuss it at all. But here is an example of the problem: http://eznec.com/misc/food_for_thought/ First article - last paragraph. W7EL considers steady-state conditions while ignoring the previous transient state conditions. He implies that the energy in the reflected wave cannot be recovered but it is indeed dissipated as power in the system after power is removed from the source. The source supplies exactly the amount of energy during the transient power up conditions needed to support the forward and reflected waves during steady-state. This is easy to prove. But W7EL's Ivory Tower protects Him from peons like me. -- 73, Cecil http://www.qsl.net/w5dxp |
#2
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The net power flux in the line gets smaller as the reflected wave gets
stronger while maintaining a constant electric field (constant voltage as in Roy's example). If you can match to the new impedance at the line input; that is, make the electric fields both stronger, you can get a larger net power flux even in the presence of some elevated SWR. See LaTeX formatted math at http://en.wikipedia.org/wiki/User ![]() The flux of stored power in the line, interestingly enough, is a sinusoidal function of position. I'm still thinking what to make of it, but I thought I'd post the math for people to look at (and check, please!!!!) ... I'll be back later. 73, Dan |
#3
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#4
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Cecil Moore wrote:
If I understand correctly, Roy's argument is that since the source is not supplying any steady-state energy to the lossless stub, there is no energy in the reflected wave within the stub. That sounds right... if the reflection coefficient is 1 then there's no net power flux into/through the line in steady state, and this can be described if you like by counterpropagating waves each carrying the same amount of energy. The problem is, in your other example where you say 200 joules in the forward wave + 100 joules in the reflected wave = 300 joules in the line total, you're neglecting the vector character of the power flux. Yes, the waves carry energy, but they carry it in different directions. The net power flux in the line with 200W forward power and 100W reflected power is 100W net power flowing to the load from the source. The real part of the Poynting vector of the reflected wave opposes that of the forward wave, as long as I got all the signs right. I don't think we can neglect the imaginary part of the Poynting vector, though. It's not zero and I think it represents the flow of the power in the stored fields in the line, and if we want to get the total energy in the line, we have to include the stored fields. Dan |
#6
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Cecil,
You've set up a false dichotomy here. When I, and others, write "The electric field is the superposition of a forward and reverse traveling wave" maybe it would be better to say "The electric field has two terms, one that appears to be a forward traveling wave and one that appears to be a reverse traveling wave." or something like that. There's one electric field vector and one Poynting vector. Or there are two. The structure of the electric field and the structure of the real part of the Poynting vector both admit BOTH explanations of what's happening. You're not gonna get 300J in your one second line.... the stored energy flux in the line depends on the wavelength of the incident RF, and in retrospect, you might expect this from the fact that a misterminated line goes through cyclical impedance variations as you change its length (something that I know you're quite familiar with :-) ) I think the energy density per unit length in the line is proportional to the Poynting vector (or it's integral over the cable cross section, and the proportionality constant is the group velocity of the waves, I think) I left Jackson at work, so I'm not certain right now. What I am certain of is that you can't take the energy in the forward wave and add it to the energy of the reflected wave and get that there are 300J in a 1 second line carrying a 200W forward wave and a 100W reverse wave. Rather, there's a 100W net forward power flux and THAT will give you the energy contained in the part of the field that's actually moving from source to load. The energy contained in the reactive part has an integral that's going to cyclically vary with the length of the line, and sometimes goes through zero (kL or kL - phi equal to an integer multiple of Pi... or any integer multiple of a half wavelength, which happens to be the length of an impedance repeating line, eh?) Dan |
#7
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wrote:
When I, and others, write "The electric field is the superposition of a forward and reverse traveling wave" maybe it would be better to say "The electric field has two terms, one that appears to be a forward traveling wave and one that appears to be a reverse traveling wave." or something like that. There's one electric field vector and one Poynting vector. Or there are two. The structure of the electric field and the structure of the real part of the Poynting vector both admit BOTH explanations of what's happening. I know and accept both explanations. The problem is the other side refuses to acknowledge the validity of the wave reflection model. If you have gotten the idea that I reject the superposed wave model, you are mistaken. I fully accept both models. The problem is that others have rejected the non-superposed component wave model. I contend that one gets the same results using the components of superposition, i.e. the forward wave and the reflected wave, that one obtains after the superposition of those two waves. Others say that is an invalid treatment because superposition causes the reflected wave to cease to exist and the energy just "sloshes" around inside the transmission line. (Never mind that RF energy must necessarily travel at the speed of light and only reverses direction at an impedance discontinuity.) Rather, there's a 100W net forward power flux and THAT will give you the energy contained in the part of the field that's actually moving from source to load. The energy contained in the reactive part has an integral that's going to cyclically vary with the length of the line, and sometimes goes through zero (kL or kL - phi equal to an integer multiple of Pi... or any integer multiple of a half wavelength, which happens to be the length of an impedance repeating line, eh?) That's one model. The other model is, assuming a purely resistive Z0, the forward voltage is in phase with the forward current and therefore there are no reactive vars in the forward wave. The reflected voltage is in phase with the reflected current and therefore there are no reactive vars in the reflected wave. This model works just as well as the one above, sometimes better because of simplicity. It has the advantage of being easily able to track the real energy because there is no "unreal" energy in the model. :-) If the forward wave component is analyzed separately, there are no vars because the forward voltage is in zero phase with the forward current (assuming a perfectly resistive Z0). The same is true for the reflected wave. So we are easily able to calculate how much energy is contained in those two waves devoid of any calculation of vars. Assume that we have a one megahertz signal into a transmission line that is electrically 360 degrees long, near lossless, the forward power is 200W, and the reflected power is 100W. I am willing to bet the energy contained in the feedline during steady-state is very close to 300 microjoules no matter how complicated the math used to get the answer that I just came up with off the top of my head. Note that the transmission line is one millionth of a second long and therefore contains one millionth of the energy of a one second long line. -- 73, Cecil http://www.qsl.net/w5dxp |
#8
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Cecil Moore wrote:
[snip] Some people on this newsgroup say that the wave reflection model is invalid, that forward and reflected waves don't have a separate existence. From QEX: "Contrary to popular views, the forward and reverse waves on a transmission line are not separate fields." It would follow that a laser beam normally incident upon an ideal mirror results in a beam of light not superposed from separate forward and reverse fields. I have challenged people holding those concepts to create a standing wave without superposing separate forward and reverse waves and have gotten zero responses. Cecil, I believe Dan has addressed this issue, and I am sure that I have on many occasions. When superposition applies, as it does in this linear, non-pathological case, there is no difference between the reality of the components vs. the reality of the sum. In other words, there is no more information from your separation of a standing wave into forward and reverse components than there is in the standing wave itself. The standing wave is a perfectly good and complete solution to the wave equations applicable to this steady-state problem. It is possible to sub-divide in many ways, but there is no new information in doing so. If you want to specifically address transients then another set of equations will be needed. ad hominem You have railed against seduction by math models on many occasions. However, that is exactly what you are doing here. Trying to create some new physical reality by manipulating the numbers. /ad hominem 73, Gene W4SZ |
#9
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Gene Fuller wrote:
When superposition applies, as it does in this linear, non-pathological case, there is no difference between the reality of the components vs. the reality of the sum. I accept both methods of analysis. What I object to is people rejecting the wave reflection model. e.g. "Reflected waves don't really exist and don't really contain any energy." As in the earlier discussion about standing wave phase, you and I wind up on the same side of the argument. What I object to is the irrational conclusions drawn by proponents of the standing wave analysis. Here are some of the past assertions: Forward waves and reflected waves cease to exist after they have been superposed. The standing wave is all that is left. Question: If the components of superposition cease to exist, doesn't that mean the products of superposition also cease to exist? (No answer) Standing waves don't require a forward wave and a reflected wave. Question: How does one obtain a standing wave without a forward traveling wave and a reverse traveling wave? (No answer) You have railed against seduction by math models on many occasions. I have railed against the assertions that superposition destroys forward and reflected waves such that they cease to exist in reality. From a recent QEX article: "I wish to emphasize the fact that the forward and reverse waves really do not exist separately, ..." So I ask you, Gene, if forward waves and reverse waves do not exist separately, how can they possibly be superposed? Seems to me, the above statement from QEX is a violation of cause and effect. If forward waves and reverse waves do not exist separately, how can they possibly be separated by a circulator? -- 73, Cecil http://www.qsl.net/w5dxp |
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