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#101
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K7ITM wrote:
So, for example, if I send 50 watts of a sinusoid down a 50 ohm line, and there's a transition to a 291.4 ohm line that's half a wave long at the sinusoid's frequency, terminated in 50 ohms, there's no reflected power on the 50 ohm line. Cool. I knew that. Make that no *NET* reflected power. Two reflected waves had to engage in wave cancellation for there to be no net reflected power. RF waves respond to real-world physical impedance discontinuities. In re- reading what Roy wrote, I see NO disagreement with that. But in the 291.4 ohm line, there's 100 watts forward and 50 watts reverse. At the interface between the two lines, there's a total of 100 watts coming in: 50 from the 50 ohm line and 50 from the 291 ohm line. And wonder of wonder, there's 100 watts going out; it happens to all be in the 291 ohm line. Let's analyze that example: 50W--50 ohm line--+--1/2WL 291.4 ohm line---50 ohm load Pfor1=50w-- Pfor2=100w-- --Pref1=0 --Pref2=50w The physical power reflection coefficient is 0.5. So Pfor1 splits into two parts, P1=25w being transmitted and P3=25w being reflected. Pref2 likewise sees a power reflection coefficient of 0.5 and splits into two parts, P2=25w re- reflected and P4=25w not re-reflected. Since Pref1=0, total destructive interference exists toward the source. Pref1 = P3 + P4 - 2*SQRT(P3*P4) Pref1 = 25w + 25w - 2*SQRT(25w*25w) = 0 The associated S-Parameter equation is: b1 = s11(a2) + s12(a2) = 0 P3 and P4 are the reflected wave components associated with wave cancellation. On the constructive interference side toward the load: Pfor2 = P1 + P2 + 2*SQRT(P1*P2) Pfor2 = 25w + 25w + 2*SQRT(25w*25w) = 100w The associated S-Parameter equation is: b2 = s21(a1) + s22(a2) where |b2|^2 = 100w Everything is perfectly consistent. Where is an error? If you go back to Roy's posting in this thread and look at the WHOLE paragraph where he issued the challenge (if you want to call it a challenge), you'll see that you have to come up with an example where there's a node with different power coming out than going in, to be disagreein' with him. Well, if Roy is asking for proof of violation of the conservation of energy principle, his concept of "sloshing" energy comes pretty close. -- 73, Cecil http://www.w5dxp.com |
#102
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Owen Duffy wrote:
. . . Your workup is correct enough for the case you describe (though for Pf/Pref=2, VSWR=5.8). . . . Thanks for the correction. I apologize for the error and stand corrected. Roy Lewallen, W7EL |
#103
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Cecil Moore wrote:
Jim Kelley wrote: Cecil Moore wrote: Of course, I regularly obtain 200 watts of forward power from my 100 watt IC-706. It's all due to constructive interference. Nice try, but you're kidding yourself if you think you're getting 200 watts out of an IC706. As anyone can readily see, I did NOT say I was "getting 200 watts out of my IC706" so your attempt at obfuscation is obvious. I said "I regularly obtain 200 watts of forward power from my 100 watt IC-706". Here is my exact configuration for my 33' rotatable dipole based on actual measurements on 20m. Then your observation was unrelated to the topic of discussion, which was your claim that 2 Joules per second could be obtained from a 1 watt laser. It must have been offered as a diversion. Heck, on 17m, I regularly obtain 350 watts of forward power using a 100W IC-706 as the source. It's certainly an impressive meter reading. It is your contention then that interference caused a 250 Joule per second increase in the amount of energy being produced by your radio? If not, then this must also have been offered as a diversion. Back to the laser example, the answer you can't seem to get right is that, recombing the split beam back into one beam will at best recover 1 watt of laser power. That's the limit allowed by conservation of energy as it happens. That's true for average power, Jim, and I have never said otherwise. You can't average power 'going' one direction with power 'going' in some other direction. That doesn't even make sense. If you integrate all of the energy, from all the bright fringes it still doesn't exceed the energy coming from the source. There is no amount of hand waving that can cause a one watt laser to deliver 2 Joules of energy per second. But if we observe interference rings, the bright rings can contain all the power while the dark rings contain none. Thus, the bright rings represent *double the average power* just as Born and Wolf report. Born and Wolf doesn't discuss power and interference, and they certainly don't interchange intensity and power in their discussions like you do. There can be no more 'power gain' along a transmission line than there can be along a beam of light. With respect to antennas one can obviously observe an increase in power in a particular direction given the fact that antennas do in fact redistribute energy by generating an interference pattern. But even with all the interference and power averaging taken into consideration, the radiated power will still not exceed the input power. You can't get 400 watts to a load using two 100 watt PA's no matter how much interfering and averaging you do, Cecil. How can you even make that claim with a straight face? Shirley, that is not beyond your comprehension. Quite honestly it does seem to be beyond yours. And please stop calling me Shirley. :-) 73, Jim AC6XG |
#104
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Jim Kelley wrote:
Then your observation was unrelated to the topic of discussion, which was your claim that 2 Joules per second could be obtained from a 1 watt laser. You know that I never claimed that, Jim. As bright a guy as you are, why do you have to stoop to falsifying what I have said? If the average intensity is one watt per square inch and we cause interference, the intensity of one particular square inch can go to zero in the case of total destructive interference. Since that energy has to go somewhere, it goes into constructive interference in some other square inch. If it is total constructive interference, we will indeed double the average intensity to two watts per square inch. The total power is still the same but the power density has doubled in half the area while going to zero in the other half of the area. Hint: (2 watts/in^2 + 0 watts/in^2)/2 in^2 = 1 watt It's certainly an impressive meter reading. It is your contention then that interference caused a 250 Joule per second increase in the amount of energy being produced by your radio? Of course not. That's another false implication of yours. The interference causes the forward power to be 250 watts higher than the output of the source. Such is the nature of constructive interference. The reflected power is 250 watts. |Forward power| = |source power| + |reflected power| You can't average power 'going' one direction with power 'going' in some other direction. I didn't know we had two directions in the laser example. I thought both beams were going the same direction. If the beams are going in opposite directions, we get standing waves and that changes things. If your split beams were going in opposite directions, I misunderstood the example and you can forget everything I have said about it. If you integrate all of the energy, from all the bright fringes it still doesn't exceed the energy coming from the source. That's what I have said all along, Jim. You are obviously trying to set up some sort of straw man. The energy in the bright fringes exceeds the average energy. That is all I have ever said. There is no amount of hand waving that can cause a one watt laser to deliver 2 Joules of energy per second. Of course not and I never said it would. Please stop trying to imply that I said something that I never said. It is downright unethical to debate in such an unfair manner. Born and Wolf doesn't discuss power and interference, and they certainly don't interchange intensity and power in their discussions like you do. Intensity is power density. I have a habit of saying power when I should have said power density. For that I apologize and will try to break myself of that habit. But you can always tell by the context that I was talking about power density. There can be no more 'power gain' along a transmission line than there can be along a beam of light. On the contrary, reflection gain from a tuner is an accepted concept within the wave reflection model. But even with all the interference and power averaging taken into consideration, the radiated power will still not exceed the input power. Nobody believes the conservation of energy principle can be violated so that is just another one of your straw men. Go ahead, be my guest, and knock it down if it makes you feel better. Nobody cares. You can't get 400 watts to a load using two 100 watt PA's no matter how much interfering and averaging you do, Cecil. How can you even make that claim with a straight face? I didn't make that claim. I said that Keith would have to prove it was possible before his assertions made any sense. I challenged Keith to prove that was possible. I certainly have never made that claim. You usual lack of ethics is showing. If you are forced to lie to make your point, your point is not worth making. -- 73, Cecil http://www.w5dxp.com |
#105
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Jim Kelley wrote:
You can't get 400 watts to a load using two 100 watt PA's no matter how much interfering and averaging you do, Cecil. How can you even make that claim with a straight face? I didn't make that claim. I challenged Keith to prove it was possible. Only then could he claim that sources obey the rules of superposition. But let's talk about superposing two 100 watt waves in a 50 ohm transmission line. Never mind exactly how it is accomplished for now. Wave1 = 70.7V at 0 deg and is 70.7^2/50 = 100 joules/sec Wave2 = 70.7V at 0 deg and is 70.7^2/50 = 100 joules/sec Please superpose those two coherent waves within a 50 ohm transmission line and see what answer is obtained for total voltage and total power. Hint: Power = V^2/Z0 You can visualize the superposition by going to the following web page and setting both phases to zero. http://micro.magnet.fsu.edu/primer/j...ons/index.html -- 73, Cecil http://www.w5dxp.com |
#106
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On Apr 9, 6:34 pm, Owen Duffy wrote:
Roy Lewallen wrote : Owen Duffy wrote: If in fact the power delivered by the "100 watt IC706" radio was indeed 100W, and some directional wattmeter correctly indicated 200W forward, it must indicate 200W-100W reflected which is indicative of a VSWR of 5.8, which should have reduced power output from the IC706 markedly. . . . Nah, no problem. Connect your rig through a half wavelength of 250 ohm ladder line to a 50 ohm load. Presto, 200 watts "forward power" and 5:1 SWR on the line, and the poor ignorant Icom doesn't have any hint that all those waves of power or energy or whatever are bouncing around on the line, trying desperately but unsuccessfully to overheat the final or whatever they're supposed to do. Of course, it would take a 250 ohm directional wattmeter to read that "forward power" or SWR. But we don' need no steenkin' meter -- we know it's there, don't we? Roy Lewallen, W7EL Roy, I was assuming that the instrument was a nominal 50 ohm instrument measuring conditions adjacent to the transmitter. Your workup is correct enough for the case you describe (though for Pf/Pref=2, VSWR=5.8). Of course, if you had a coaxial reflectometer calibrated (nulled) for 8.6 ohms or 290 ohms then you would get the same indications on a 50 ohm load, you don't actually need the 8.6 ohm or 290 ohm transmission line. These are just examples that question the reality of these "component powers" when you can change their magnitude by choosing the reference impedance for measurement or calculation. They reinforce the view that whilst Pf-Pr has meaning (irrespective of Z), Pf and Pr each alone have no meaning. Owen Hi Owen, (Noted your other response, to my posting, about the assumption of 50 ohm line. I obviously didn't make that assumption...) I've been harping in this forum for YEARS about the need to calibrate an SWR meter or a return loss bridge to the impedance you're wanting to use it for. I'm always somewhat amazed that some folk just "don't get it." Thanks for helping reinforce the need for the proper calibration. Those bridges are not magic; they can only measure voltages or currents produced by (hopefully) linear combinations of transduced line voltages and currents. I think it was about ten years ago in this group that I posted the derivation of the equations to determine forward and reflected from measurement of voltage and current at a point on a TEM line, IF you know the impedance of the line. Cheers, Tom |
#107
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On Apr 9, 6:08 pm, Cecil Moore wrote:
K7ITM wrote: So what? Roy didn't say that was difficult at all. Now answer the REST of Roy's parapgraph which you so conveniently failed to include in your quoted material. I will be glad to if you will tell me specifically what "REST of Roy's paragraph" you are talking about. -- 73, Cecil http://www.w5dxp.com Go look it up yourself, lad. At this point, how am I supposed to know what you're talking about, since you've dropped all the original reference. If you can't bother to quote the whole posting you're replying to, don't be askin' anybody else to go fetch it for you. |
#108
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"K7ITM" wrote in
oups.com: .... transduced line voltages and currents. I think it was about ten years ago in this group that I posted the derivation of the equations to determine forward and reflected from measurement of voltage and current at a point on a TEM line, IF you know the impedance of the line. Hi Tom, I guess it is the misconceptions about these things that drive us to write such stuff. I drafted an article recently showing what the typical SWR meter samples, and the use of that information, including treatment of Pf and Pr as stand alone values. The article is at http://www.vk1od.net/VSWR/VSWRMeter.htm . Owen |
#109
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On Apr 9, 11:05 pm, Owen Duffy wrote:
"K7ITM" wrote groups.com: ... transduced line voltages and currents. I think it was about ten years ago in this group that I posted the derivation of the equations to determine forward and reflected from measurement of voltage and current at a point on a TEM line, IF you know the impedance of the line. Hi Tom, I guess it is the misconceptions about these things that drive us to write such stuff. I drafted an article recently showing what the typical SWR meter samples, and the use of that information, including treatment of Pf and Pr as stand alone values. The article is athttp://www.vk1od.net/VSWR/VSWRMeter.htm. Owen Hi Owen, I had a quick look at your article. Though I didn't try to proof-read it for accuracy, I was reminded that the equations I posted those long years ago said that if you know the _instantaneous_ voltage and current at a point on a line, and know its impedance (as a frequency- independent quantity), the equations apply, and you can resolve that instantaneous pair of values into forward and reverse. That's something that's not immediately obvious when people think only about sine waves. Cheers, Tom |
#110
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K7ITM wrote:
Go look it up yourself, lad. At this point, how am I supposed to know what you're talking about, since you've dropped all the original reference. I did go back and read it more than once and I am still have no idea what you are talking about. Maybe this? w7el wrote: I think that no one will be able to draw a diagram of such a summing system which doesn't also produce, due solely to the reflection and transmission of the mirrors, a beam or beams containing exactly the amount of energy "missing" from the summed beam. If it is this part, this is just another one of Roy's straw men. Nobody in the world is saying that the wave reflection model violates the conservation of energy principle. In fact, the wave reflection model does more to preserve the conservation of energy principle than Roy's notion of "sloshing" energy going nowhere. -- 73, Cecil http://www.w5dxp.com |
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