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Constructive interference in radiowave propagation
Jim Kelley wrote:
So do you get the point, or not? I've always gotten the point. The question is since you seem to have blown the dust off of Born and Wolf, do you get the point? Born and Wolf agrees in every way with Hecht. Total constructive interference yields an intensity four times the intensity of one individual wave. -- 73, Cecil http://www.w5dxp.com |
Constructive interference in radiowave propagation
Mike Lucas wrote:
Keith, I really like when junk science gurus exibit constructive interference, the bafflegab sums to four times the normal power! Exactly in accordance with Born and Wolf's equation (17) in their chapter on "Interference and Interferometers". Quoting: "In the special case where I1 = I2, (15) reduces to I = 4*I1" [for in phase waves]. I'm sorry if you disagree with the laws of physics. You are not alone in being ignorant of what occurs when two coherent waves are superposed in phase. The kicker is that we cannot have two coherent waves superposed in phase at one location without having two coherent waves superposed out of phase somewhere else (assuming no local source). -- 73, Cecil http://www.w5dxp.com |
Constructive interference in radiowave propagation
Cecil Moore wrote:
Jim Kelley wrote: So according to your theory I can take a 1 watt laser, split the beam into two coherent beams, recombine the beams in-phase together along the same path thus creating constructive interference, and obtain 2 watts of laser power. Or would it be 4 watts? If it were total constructive interference, two 1/2W beams would yield an intensity of 2 watts. Of course, at another location, total destructive interference would have to occur where the intensity was zero. And so that's your explanation for how 2 Joules per second can be obtained from a source which is in fact producing only 1 Joule per second. As I said before, this ain't rocket science. That is for damn sure. :-) Jim AC6XG |
Constructive interference in radiowave propagation
"Cecil Moore"
The kicker is that we cannot have two coherent waves superposed in phase at one location without having two coherent waves superposed out of phase somewhere else (assuming no local source). _____________ As shown by the real-world result for the combined r-f system configuration I posted earlier (and which result I, and many others have measured in real hardware systems) -- one output port of the 3 dB hybrid combiner when powers there are equal and in phase (ie, fully coherent) contains twice the average output power of either tx alone, while the other output port has zero power. These are exactly the conditions in your quote above. The conflict here occurs because of your belief that the combined, average power in this scenario is not 2X, but 4X that of the individual transmitters. Or have you changed your mind from what you first posted? RF |
Constructive interference in radiowave propagation
Jim Kelley wrote:
And so that's your explanation for how 2 Joules per second can be obtained from a source which is in fact producing only 1 Joule per second. Of course, I regularly obtain 200 watts of forward power from my 100 watt IC-706. It's all due to constructive interference. If you have a bright ring exhibiting 2 joules/sec/unit-area and a dark ring exhibiting zero joules/sec/unit-area, don't you see how that averages out to the average source power of one joule/sec? As I said before, this ain't rocket science. That is for damn sure. :-) Glad you (finally after all these years) agree. -- 73, Cecil http://www.w5dxp.com |
Constructive interference in radiowave propagation
Richard Fry wrote:
The conflict here occurs because of your belief that the combined, average power in this scenario is not 2X, but 4X that of the individual transmitters. No, no, no! The combined average power is 2X. The point of total constructive interference is 4X. The point of total destructive interference is 0X. The average of the total constructive interference and the total destructive interference is 2X. Seems to me that you have not been reading and understanding my postings. -- 73, Cecil http://www.w5dxp.com |
Constructive interference in radiowave propagation
Cecil Moore wrote in
t: Jim Kelley wrote: And so that's your explanation for how 2 Joules per second can be obtained from a source which is in fact producing only 1 Joule per second. Of course, I regularly obtain 200 watts of forward power from my 100 watt IC-706. It's all due to constructive interference. 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. Taking another view, if the IC706 will tolerate VSWR of 2 before reducing power (ie and still deliver exactly 100W), then the forward power would be just 113W. More likely, the IC706 levels its power output on the forward detector, and runs 100W "forward" until the reflected power reaches about 12W whereupon it reduces drive so maintain maximum reflected power =12W. Did you make this example up on the fly, or is it the result of actual observation on one or many occasions? Owen |
Constructive interference in radiowave propagation
Cecil Moore wrote: Jim Kelley wrote: And so that's your explanation for how 2 Joules per second can be obtained from a source which is in fact producing only 1 Joule per second. 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. 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. 73, ac6xg |
Constructive interference in radiowave propagation
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 |
Constructive interference in radiowave propagation
On Apr 9, 4:59 pm, Owen Duffy wrote:
Cecil Moore wrote . net: Jim Kelley wrote: And so that's your explanation for how 2 Joules per second can be obtained from a source which is in fact producing only 1 Joule per second. Of course, I regularly obtain 200 watts of forward power from my 100 watt IC-706. It's all due to constructive interference. 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. Taking another view, if the IC706 will tolerate VSWR of 2 before reducing power (ie and still deliver exactly 100W), then the forward power would be just 113W. More likely, the IC706 levels its power output on the forward detector, and runs 100W "forward" until the reflected power reaches about 12W whereupon it reduces drive so maintain maximum reflected power =12W. Did you make this example up on the fly, or is it the result of actual observation on one or many occasions? Owen Let's see... 600 ohm line feeding a nominally 60 ohm dipole. (It's not very high above the ground.) 600 ohm line is 1/2 wave long, and essentially lossless. I feed 60 watts in, the 60 ohm dipole absorbs 60 watts. I suppose the forward power on the line is a bit more than 60 watts. My transmitter doesn't seem to have too much trouble with the 60 ohm load, though a balun between the unbalanced transmitter output and the balanced line is nice. Cheers, Tom |
Constructive interference in radiowave propagation
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. Did you make this example up on the fly, or is it the result of actual observation on one or many occasions? I made it up but let's take the actual example of my 33' rotatable dipole on 20m. IC-706---1WL 300 ohm twinlead---51.5 ohm antenna The IC-706 is delivering 100 watts. The forward power is 200 watts. The reflected power is 100 watts. Why in the world would a VSWR of 5.8:1 on the 300 ohm twinlead "reduce the output from the IC706 markedly" when the IC-706 is perfectly matched to a 51.5 ohm virtual impedance? -- 73, Cecil http://www.w5dxp.com |
Constructive interference in radiowave propagation
On Apr 9, 5:48 am, Cecil Moore wrote:
Roy Lewallen wrote: Next, do the same for a transmission line. Show how two coherent traveling waves can be produced which will propagate together in the same direction but out of phase with each other, resulting in a net zero field at all points beyond some summing point. Roy, it is done all the time represented by the S-Parameter equation for the reflected wave toward the source. Assuming an impedance discontinuity in a transmission line: b1 = s11(a1) + s12(a2) = 0 If b1=0, then s11(a1) and s12(a2) are indeed two coherent traveling waves propagating together in the same direction but out of phase with each other. The net reflected wave field is zero. -- 73, Cecil http://www.w5dxp.com 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. |
Constructive interference in radiowave propagation
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. 100W IC-706--choke--1 WL 300 ohm twinlead---51.5 ohm antenna Pfor=200W-- --Pref=100W The forward power is indeed 200 watts. I have measured it with my 300 ohm SWR meter. Heck, on 17m, I regularly obtain 350 watts of forward power using a 100W IC-706 as the source. 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. 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. Shirley, that is not beyond your comprehension. If there is zero intensity in the dark rings, there must be double the average intensity in the bright rings to keep from violating the conservation of energy principle. -- 73, Cecil http://www.w5dxp.com |
Constructive interference in radiowave propagation
K7ITM wrote:
Let's see... 600 ohm line feeding a nominally 60 ohm dipole. (It's not very high above the ground.) 600 ohm line is 1/2 wave long, and essentially lossless. I feed 60 watts in, the 60 ohm dipole absorbs 60 watts. I suppose the forward power on the line is a bit more than 60 watts. My transmitter doesn't seem to have too much trouble with the 60 ohm load, though a balun between the unbalanced transmitter output and the balanced line is nice. The SWR is 10:1 making rho = 9/11 = 0.818 and rho^2 = 0.67 So the SWR is close to 10:1 and the forward power with a 100 watt IC-706 source is about 303 watts. Such is the nature of constructive interference. -- 73, Cecil http://www.w5dxp.com |
Constructive interference in radiowave propagation
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. Thanks, but I'm not interested in Cecil's response, since I know for certain that he'll never directly answer the question. Evasion, subject changing, diversion, misquoting, and anything else but solid reasoning or direct confrontation of a problem is all you'll ever get, as each new batch of people joining this group sadly discover sooner or later. I won't see his imaginative but evasive responses anyway unless quoted by someone, since he's one of the four people in my kill file. ("Jesus" is another of the four. I haven't seen any postings by him for quite some time, but there's no harm in leaving him listed in case he's resurrected someday.) But I'd really like to hear from anyone else who's subscribed to Cecil's unique views of superposition, wave interaction, and bouncing waves of average power (although admittedly these views seem to rapidly morph as necessary to avoid any direct confrontation with rational thought). Or even anyone who believes (as I think you've said you do, Walt) that traveling waves can and do affect each other under any circumstances in a linear medium. Show me how waves can modify each other simply by passing, and any circumstance under which they do. And I'll show you how exactly the same result can be explained without this supposed interaction. Roy Lewallen, W7EL |
Constructive interference in radiowave propagation
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 |
Constructive interference in radiowave propagation
Roy Lewallen wrote:
Show me how waves can modify each other simply by passing, and any circumstance under which they do. STRAW MAN ALERT! Nobody has said that passing waves can modify each other. The requirements for wave modification are very well defined. The two waves must be coherent and traveling in the same direction in the same path. Under those circumstances wave modification is impossible to avoid. When the reflected waves are canceled in the direction of the source by an antenna tuner or other matching means, those waves permanently disappear in the direction of the source. Seems permanent disappearance meets the definition of being modified. -- 73, Cecil http://www.w5dxp.com |
Constructive interference in radiowave propagation
On Apr 9, 8:31 am, Cecil Moore wrote:
Keith Dysart wrote: The junk science is often presented with very rational sounding arguments and it can be difficult to detect the flaws. This example was a case for me and you expose the flaw nicely. Hint to omniscient gurus: One cannot use ignorance for exposing flaws. Roy says in his Food for Thought article: I personally don't have a compulsion to understand where this power "goes". Seemingly, that feeling of his is supposed to be enough incentive to discourage the rest of us to give up on our quest for tracking the energy through the system. Roy has ploinked me for disagreeing with him. What does that say about his inability to technically defend his concepts? The S-Parameter equations completely debunk what Roy posted. b1 = s11(a1) + s12(a2) = 0 |b1|^2, the reflected power, equals zero because of wave cancellation involving those components of a1 (forward normalized voltage) and a2 (reflected normalized voltage). If s11, a1, s12, and a2 are all non-zero, then wave cancellation has occurred between s11(a1) and s12(a2) proving Roy's statements to be false. The above wave cancellation happens every time a ham adjusts his antenna tuner for zero reflected power. -- 73, Cecil http://www.w5dxp.com 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. 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. 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. |
Constructive interference in radiowave propagation
Roy Lewallen wrote in
: 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 |
Constructive interference in radiowave propagation
"K7ITM" wrote in
oups.com: On Apr 9, 4:59 pm, Owen Duffy wrote: Cecil Moore wrote . net: Jim Kelley wrote: And so that's your explanation for how 2 Joules per second can be obtained from a source which is in fact producing only 1 Joule per second. Of course, I regularly obtain 200 watts of forward power from my 100 watt IC-706. It's all due to constructive interference. 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. Taking another view, if the IC706 will tolerate VSWR of 2 before reducing power (ie and still deliver exactly 100W), then the forward power would be just 113W. More likely, the IC706 levels its power output on the forward detector, and runs 100W "forward" until the reflected power reaches about 12W whereupon it reduces drive so maintain maximum reflected power =12W. Did you make this example up on the fly, or is it the result of actual observation on one or many occasions? Owen Let's see... 600 ohm line feeding a nominally 60 ohm dipole. (It's not very high above the ground.) 600 ohm line is 1/2 wave long, and essentially lossless. I feed 60 watts in, the 60 ohm dipole absorbs 60 watts. I suppose the forward power on the line is a bit more than 60 watts. My transmitter doesn't seem to have too much trouble with the 60 ohm load, though a balun between the unbalanced transmitter output and the balanced line is nice. Tom, as I noted in another response, I assumed that the observation was based on measurement with a nominal 50 ohm instrument adjacent to the transmitter. You can of course re-evaluate the forward and reflected power components in a 50 ohm load by changing the reference impedance, and so make them anything you want. Cecil didn't suggest a specific reference impedance, so knowing that the IC706 is designed for a nominal 50 ohm load, it seemed reasonable to make the assumptions that I did. Owen |
Constructive interference in radiowave propagation
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 |
Constructive interference in radiowave propagation
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 |
Constructive interference in radiowave propagation
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 |
Constructive interference in radiowave propagation
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 |
Constructive interference in radiowave propagation
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 |
Constructive interference in radiowave propagation
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 |
Constructive interference in radiowave propagation
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. |
Constructive interference in radiowave propagation
"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 |
Constructive interference in radiowave propagation
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 |
Constructive interference in radiowave propagation
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 |
Constructive interference in radiowave propagation
On Apr 9, 11:49 pm, Cecil Moore wrote:
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 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. It would be highly valuable if you would point to any step in my solutions that require that 100 + 100 equal 400. I certainly have never stated this. I know I have previously offered solutions that you have declared incorrect but you have been unwilling to point to the error. I have also offered to check your solution but you have never offered one, apparently because your analysis techniques are inadequate to solve the problem. ....Keith PS Should you wish to try, recall that the problem was a signal generator with a 450 Ohm output impedance driving a 450 Ohm line connected to a 75 Ohm load. What is the magnitude of the re-reflected signal at the generator? That is, what is the magnitude of the ghost created by the reflected signal? (And you can't make changes to the problem to solve it.) |
Constructive interference in radiowave propagation
On Apr 10, 12:26 am, Cecil Moore wrote:
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 Oh yes, let us do this. 70.7 + 70.7 - 141.4 141.4**2/50 - 400 100 + 100 - 200 But 200 is not equal to 400. So which was the invalid step? Adding the voltages together? (Might mean superposition is invalid?) Or adding the powers together? (Might mean no conservation of energy?) Or perhaps V**2/Z0 is invalid? (Maybe this is the best choice?) Or 200 is equal to 400? (This would be messy!) For full marks, explain your answer. ....Keith |
Constructive interference in radiowave propagation
On Apr 9, 9:50 pm, Cecil Moore wrote:
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 You are very comfortable with this style of example, you analyze it well, and the numbers confirm your expectation. But a good theory works for more than just one example. Why don't you analyze my example: Generator with 291.4 Ohm ------------ 291.4 Ohm line ---- 50 Ohm load Source Impedance Pfor2=100w-- --Pref2=50w What is the power emitted by the generator into the line? Where does Pref2=50W go? Are there ghosts? What is the magnitude? What power would the generator emit if the line was terminated with 291.4 Ohms? Please do not modify the example for analysis since this may change the results. There is much to be learned by trying examples that may challenge your expectations. ....Keith |
Constructive interference in radiowave propagation
This thread has been a hoot - I just about snorted my coffee out my
nose a couple of times, reading this in the early AM... Just to muddy the waters nicely, here is an excellent demonstration of something for nothing, i.e. faster than light propagation by constructive wave interference called "supraluminal propagation".. http://gregegan.customer.netspace.ne...ETS/20/20.html Lets see, for starters assume that the constructively interfered wave is propagationg at 1.2C it means that E = ( MC^2 )^1.2 right???? denny gawd, I can't wait to see how you guys handle this - worm holes, anyone? |
Constructive interference in radiowave propagation
Keith Dysart wrote:
It would be highly valuable if you would point to any step in my solutions that require that 100 + 100 equal 400. I certainly have never stated this. I went over this before. In order for a source to obey the rules for superposition, it must be linear enough to engage in constructive interference. That means it must be able to double its voltage and double its current at the same time in order to supply the total constructive interference energy requirements required by superposition of coherent waves. No commercially available amateur radio transmitter that I know of will do that. Therefore, your simple proposed method will not work on the average amateur radio transmitter because it is NOT LINEAR. All you can do is invent sources in your head. I'll bet you can even leap tall buildings in a single bound, in your head. I have also offered to check your solution but you have never offered one, apparently because your analysis techniques are inadequate to solve the problem. Yes, they are as are everyone else's, including yours. That's why there has *never been a solution* and the argument continues to rage after decades. If anyone had ever figured it out, we wouldn't be having this discussion. You and I are looking at the moon. You assert that it is 1000 miles away. I say that's wrong. You ask me how far away is it? I answer that I don't know but I know it is not 1000 miles away. You say that since I don't know how far away the moon is then it has to be 1000 miles away. That's where our present argument stands. -- 73, Cecil http://www.w5dxp.com |
Constructive interference in radiowave propagation
Keith Dysart wrote:
On Apr 10, 12:26 am, Cecil Moore wrote: 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 Oh yes, let us do this. 70.7 + 70.7 - 141.4 141.4**2/50 - 400 100 + 100 - 200 But 200 is not equal to 400. So which was the invalid step? There was no invalid step. The two coherent waves engaged in total constructive interference just as they do in a Z0-matched transmission line in the direction of the source. The extra 200 watts is supplied by total destructive interference at the Z0-match in the direction of the source. 200 watts of reflections have been eliminated toward the source and join the forward power wave toward the load. In ham terms, all the reflected power has been re-reflected. The S-Parameter equation yields the same results where b2 is the normalized voltage flowing toward the load. s21(a1) = 70.7V/SQRT(50) = 10 [s21(a1)]^2 = 100W s22(a2) = 70.7V/SQRT(50) = 10 [s22(a2)]^2 = 100W b2 = s21(a1) + s22(a2) = 141.4V/SQRT(50) = 20 |b2|^2 = 20^2 = 400W = the forward power constructive & destructive interference is built right into the S-Parameter analysis. Forward and reflected waves are built right into the S-Parameter analysis. There was no invalid step. -- 73, Cecil http://www.w5dxp.com |
Constructive interference in radiowave propagation
Keith Dysart wrote:
Why don't you analyze my example: Generator with 291.4 Ohm ------------ 291.4 Ohm line ---- 50 Ohm load Source Impedance Pfor2=100w-- --Pref2=50w Any example where reflected energy is allowed to reach the source cannot be analyzed in any valid real-world way. That is what the decades-old argument between experts is all about. There has never been a valid way to analyze such a real-world system. Of course, you can analyze this system in your mind but real-world results with real-world ham transmitters will prove your analysis invalid. If there existed a valid way to analyze real-world ham transmitters with incident reflections, everyone would be using that method but, so far, bench measurements prove that no valid analysis exists. And so the argument continues to wage on. You are not going to resolve it with a simple-minded model that exists only in your mind and not in reality. Replace the above purely imaginary generator with an IC-706 and get back to us. -- 73, Cecil http://www.w5dxp.com |
Constructive interference in radiowave propagation
Cecil Moore wrote:
There was no invalid step. The two coherent waves engaged in total constructive interference just as they do in a Z0-matched transmission line in the direction of the source. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Darn, that should have been "in the direction of the load". -- 73, Cecil http://www.w5dxp.com |
Constructive interference in radiowave propagation
On Apr 10, 10:00 am, Cecil Moore wrote:
Keith Dysart wrote: On Apr 10, 12:26 am, Cecil Moore wrote: 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 Oh yes, let us do this. 70.7 + 70.7 - 141.4 141.4**2/50 - 400 100 + 100 - 200 But 200 is not equal to 400. So which was the invalid step? There was no invalid step. The two coherent waves engaged in total constructive interference just as they do in a Z0-matched transmission line in the direction of the source. The extra 200 watts is supplied by total destructive interference at the Z0-match in the direction of the source. 200 watts of reflections have been eliminated toward the source and join the forward power wave toward the load. In ham terms, all the reflected power has been re-reflected. The S-Parameter equation yields the same results where b2 is the normalized voltage flowing toward the load. s21(a1) = 70.7V/SQRT(50) = 10 [s21(a1)]^2 = 100W s22(a2) = 70.7V/SQRT(50) = 10 [s22(a2)]^2 = 100W b2 = s21(a1) + s22(a2) = 141.4V/SQRT(50) = 20 |b2|^2 = 20^2 = 400W = the forward power constructive & destructive interference is built right into the S-Parameter analysis. Forward and reflected waves are built right into the S-Parameter analysis. There was no invalid step. Now just above, you started with 100W + 100W and ended with 400W. And you wonder why readers think you advocate this position. Does this not cause you some discomfort? It clearly violates conservation of energy. For me, it is pretty clear. If you start with 200W, you can not end up with 400W. ....Keith |
Constructive interference in radiowave propagation
On Apr 10, 10:16 am, Cecil Moore wrote:
Keith Dysart wrote: Why don't you analyze my example: Generator with 291.4 Ohm ------------ 291.4 Ohm line ---- 50 Ohm load Source Impedance Pfor2=100w-- --Pref2=50w Any example where reflected energy is allowed to reach the source cannot be analyzed in any valid real-world way. A strange assertion. Consider two wire phone lines; transmitter and receiver at each end. Consider cable modems; ditto. Consider computer busses; ditto. I am sure you will find a "good" reason why either these are special cases or amateur RF is a special case and therefore something does not apply. Oh well. ....Keith |
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