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#1
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#2
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k4wge wrote:
Look like Steve's been busy. He told me he needed to rewrite his QEX article based on what he had learned on this newsgroup. Maybe that is it. -- 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 100,000 Newsgroups - 19 Different Servers! =----- |
#3
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Hi Cecil,
I went to the url k4wge indicated for Steve's work and found that he has a book on antennas published by Noble Publications going for $207 after a 25% discount. I would need to know the quality of editorial review it survived before spending that much on a new book. I'm wondering if the book had ANY technical review before printing. I'm also wondering how much of the stuff from his three-part QEX article will appear in his book. That article is chuck full of erroneous material. I know you and others on this thread tried to correct some of his incorrect notions to no avail. And you know that he and I went at it pretty strong, but he wouldn't listen. He told me early on that my concepts of matching appearing in Reflections are wrong, and said that using the statements I made there he could prove me technically incompetent. In two Communications Quarterly issues he said as much, referring to my writings in Reflections and in QEX.. During our last email communication he reiterated his comment concerning incompetence, and said he was going to write a definitive article concerning matching that would prove it, but that this time he wasn't going to mention me by name. That 'definitive' article is the three-parter in QEX.. Well, he didn't mention me by name, but he presented one of my examples, number for number, and then erroneously tore it to pieces. Anyone having read my work would recognize that he was targeting me even though not by name. Enough of that. I've been reading a little mail on the rraa, and I see you're still hangin' in there. Hope you're well, but I guess you must be, cuz you still have that characteristic sarcastic (no that's not quite the word for it) response at the right moment. Take care, Walt |
#4
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Walter Maxwell wrote:
In two Communications Quarterly issues he said as much, referring to my writings in Reflections and in QEX.. During our last email communication he reiterated his comment concerning incompetence, and said he was going to write a definitive article concerning matching that would prove it, but that this time he wasn't going to mention me by name. Walt, as you know, I tangled with Steve over the subject of interference which he claimed didn't exist at a match point. He later changed his mind and told me in an email that he needed to rewrite part 3 of his article. Maybe he got it right on his CDs. _Optics_, by Hecht proves Steve's equations in part 3 to be the actual interference equations from optics with RF Power substituted for light Irradiance. IMO, you and Steve were much closer in principles than either one of you realized. He said in Part 3: "... the two rearward traveling waves at the match point (rearward waves 1 and 2) are 180 degrees out of phase with respect to each other and a complete cancellation of both waves occurs." You said in Reflections II: "With equal magnitudes and opposite phase at the same point (Point A, the matching point), the sum of the two (rearward- traveling) waves is zero." This agrees with J. C. Slater, from _Microwave_Transmissions_, "The fundamental principle behind the elimination of reflections is then to have each reflected wave canceled by another wave of equal amplitude and opposite phase." All three above appear to me to be in agreement so the disagreements are really about the down-in-the-noise details. Some gurus on this newsgroup disagree with you, Dr. Best, and J. C. Slater. Others on this newsgroup have been asking about your opinion of conjugately- matched transmitters. I have no interest in that particular discussion but you might point out some references. -- 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 100,000 Newsgroups - 19 Different Servers! =----- |
#5
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On Sun, 30 May 2004 16:45:03 -0500, Cecil Moore wrote:
Walter Maxwell wrote: In two Communications Quarterly issues he said as much, referring to my writings in Reflections and in QEX.. During our last email communication he reiterated his comment concerning incompetence, and said he was going to write a definitive article concerning matching that would prove it, but that this time he wasn't going to mention me by name. Walt, as you know, I tangled with Steve over the subject of interference which he claimed didn't exist at a match point. He later changed his mind and told me in an email that he needed to rewrite part 3 of his article. Maybe he got it right on his CDs. _Optics_, by Hecht proves Steve's equations in part 3 to be the actual interference equations from optics with RF Power substituted for light Irradiance. IMO, you and Steve were much closer in principles than either one of you realized. Sorry Cecil, I don't think so. Steve has missed the most vital aspect of the phenomenon--what happens to the energy, or power in the reflected waves on return to the match point. He said in Part 3: "... the two rearward traveling waves at the match point (rearward waves 1 and 2) are 180 degrees out of phase with respect to each other and a complete cancellation of both waves occurs." Yes, but Cecil, the cancellation of the waves is only in the rearward direction, because at the match point the waves and the energy they carry (volts x amps) are totally reversed. Now to continue what Steve said is: "The result of this wave cancellation is that the total steady-state rearward-traveling wave has a net voltage of 0 V nd 0 A, respectively, and an impedance match occurs." No No No. As we've discussed earlier, voltage and current cannot both go to zero simultaneously, except in the rearward direction. When voltage goes to zero at the match point because the two returning voltages are equal magnitude and of opposite phase, the current is doubled and the V x I energy in the rearward traveling waves is totally re-reflected in the forward direction. Steve totally ignores the energy in the reflected waves, except to say, "A total re-reflection of the reflected voltage, current and power does not occur at the match point and it (re-reflection) is not necessary for the impedance match to occur." This statement is totally untrue, because as I said above, all of the power in the reflected waves of voltage AND current is totally re-reflected in the forward direction, the same as if the E and H fields had encountered a physical short. We all know what happens in this case. The only difference is that the virtual short established by the wave interference is one way only--to the rearward traveling waves. I know you don't agree with me that a one-way virtual short is what causes the re-reflection, but in a short time I'll be able to prove it to you in a manner you'll not be able to rebut. Stay tuned. You said in Reflections II: "With equal magnitudes and opposite phase at the same point (Point A, the matching point), the sum of the two (rearward- traveling) waves is zero." Which means zero impedance, the boundary condition causing the total re-reflection. This exactly what Slater is implying. This agrees with J. C. Slater, from _Microwave_Transmissions_, "The fundamental principle behind the elimination of reflections is then to have each reflected wave canceled by another wave of equal amplitude and opposite phase." Cecil, the Slater reference is where I originally obtained this concept for my QST article that appeared in Oct 1973, nearly 30 years ago. Check the ref number in Reflections--No. 35. All three above appear to me to be in agreement so the disagreements are really about the down-in-the-noise details. Some gurus on this newsgroup disagree with you, Dr. Best, and J. C. Slater. Those who disagree with Slater need to refresh their memories with a review of transmission lines 101. They ain't gonna win. Others on this newsgroup have been asking about your opinion of conjugately- matched transmitters. I have no interest in that particular discussion but you might point out some references. Cecil, I don't have a particular reference handy, but I can quote some of my own measurements the others you mention might find of interest.. IMO they'll have a hard time disgreeing with the data if they don't already believe a transmitter is conjugately matched to its load when it's delivering all of its available power at an arbitrarily selected drive level within the normal operating range. Those who don't believe will get quite a surprise when I reveal what the output source resistance of the xmtr really is under this condition. Waddya think? Walt |
#6
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Walter Maxwell wrote:
Cecil Moore wrote: IMO, you and Steve were much closer in principles than either one of you realized. Sorry Cecil, I don't think so. Steve has missed the most vital aspect of the phenomenon--what happens to the energy, or power in the reflected waves on return to the match point. Well, once Steve admitted that the two reflected waves completely cancel each other in a matched system, what happens to the pre-existing energy in those two waves before they cancel is obvious. Energy cannot be destroyed and if it doesn't flow toward the source, it must flow toward the load as explained at the bottom of the Melles Griot Web Page: http://www.mellesgriot.com/products/optics/oc_2_1.htm "Clearly, if the wavelength of the incident light and the thickness of the film are such that a phase difference exists between reflections of p, then REFLECTED WAVEFRONTS INTERFERE DESTRUCTIVELY, and overall reflected intensity is a minimum. If the two reflections are of equal amplitude, then this amplitude (and hence intensity) minimum will be ZERO." (emphasis mine) "In the absence of absorption or scatter, THE PRINCIPLE OF CONSERVATION OF ENERGY indicates ALL (rearward-traveling) "LOST" REFLECTED INTENSITY will appear as ENHANCED INTENSITY IN THE (forward-traveling) TRANSMITTED BEAM. The sum of the reflected and transmitted beam intensities is always equal to the incident intensity. This important fact has been confirmed experimentally." (emphasis mine) Steve at first said the energy in the canceled waves continues to flow toward the source without a voltage and current and that interference was not involved. He later changed his mind. All that should be archived on r.r.a.a on Google for the summer of 2001. Here's an excerpt. Steve said: "The total forward power increases as a direct result of the vector superposition of forward voltage and current. This DOES NOT require a corresponding destructive interference process ..." thus contradicting Hecht in _Optics_ who says any constructive interference process must be accompanied by an equal magnitude of destructive interference. Now to continue what Steve said is: "The result of this wave cancellation is that the total steady-state rearward-traveling wave has a net voltage of 0 V nd 0 A, respectively, and an impedance match occurs." No No No. As we've discussed earlier, voltage and current cannot both go to zero simultaneously, except in the rearward direction. But that's what he said above. The rearward-traveling wave indeed does have a net voltage of 0 V and 0 A and the reflections toward the source disappear at the match point. I think you and Steve really agree on about 98% of this match point stuff but you two obviously disagree on the definition of "re-reflection". I know you don't agree with me that a one-way virtual short is what causes the re-reflection, but in a short time I'll be able to prove it to you in a manner you'll not be able to rebut. Stay tuned. You might want to check your work against an s-parameter analysis. The s-parameter equations a b1 = s11(a1) + s12(a2) and b2 = s21(a1) + s22(a2) Given that a match point in a transmission line can be considered to be a two- port network, |s22|^2 is the Power (re)reflected from the network output divided by the (reflected) Power incident on the network output. s22 is the reflection coefficient looking into port 2 and is *not equal to 1.0 or zero*. In a matched system with nothing but resistances, it is often the negative of the reflection coefficient at the load. This is covered in HP's AN 95-1 available on the web. Those who don't believe will get quite a surprise when I reveal what the output source resistance of the xmtr really is under this condition. Waddya think? I think I am too ignorant of the subject to venture an opinion. Which of the following systems do you think I would prefer, the conjugately-matched one at 50% efficiency or the non-conjucately one at 98% efficiency? 1 ohm XMTR----100V out---tuner---1/2WL 450 ohm line--50 ohm load 50 ohm XMTR---100V out---tuner---1/2WL 450 ohm line--50 ohm load Arguing that Tesla/Westinghouse would have to conjugately match their 60 Hz AC generators is what shot down Edison's dream of an all DC power distribution system for the USA. I would love to have a transmitter with a zero ohm internal impedance, completely conjugately unmatched. :-) -- 73, Cecil http://www.qsl.net -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
#7
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Cecil:
[snip] I would love to have a transmitter with a zero ohm internal impedance, completely conjugately unmatched. :-) -- 73, Cecil http://www.qsl.net [snip] I would love it as well! :-) I fear that Walter is far too hung up on that conjugate match stuff... Walt, you need to move on to more important things. Don't continue to fuss with others over such a trivial and ultimately unimportant point. Have you left for the North yet, or still in Deland? Sorry I've been far too busy with consulting to get over to see you this past winter. BTW, Cecil... Faced with the possibility of owning such an ideal transmitter and given the choice... I wonder, considering the condition of most amateur antennas... which you would love better in practice... A completely mismatched zero Ohm impedance voltage transmitter -OR- A completely mismatched infinite Ohm impedance current transmitter. -- Peter K1PO Indialantic By-the-Sea, FL. |
#8
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On Mon, 31 May 2004 08:06:43 -0500, Cecil Moore wrote:
Walter Maxwell wrote: Cecil Moore wrote: IMO, you and Steve were much closer in principles than either one of you realized. Sorry Cecil, I don't think so. Steve has missed the most vital aspect of the phenomenon--what happens to the energy, or power in the reflected waves on return to the match point. Well, once Steve admitted that the two reflected waves completely cancel each other in a matched system, what happens to the pre-existing energy in those two waves before they cancel is obvious. Energy cannot be destroyed and if it doesn't flow toward the source, it must flow toward the load as explained at the bottom of the Melles Griot Web Page: http://www.mellesgriot.com/products/optics/oc_2_1.htm "Clearly, if the wavelength of the incident light and the thickness of the film are such that a phase difference exists between reflections of p, then REFLECTED WAVEFRONTS INTERFERE DESTRUCTIVELY, and overall reflected intensity is a minimum. If the two reflections are of equal amplitude, then this amplitude (and hence intensity) minimum will be ZERO." (emphasis mine) In that case, of course all of the incident energy is transmitted. "In the absence of absorption or scatter, THE PRINCIPLE OF CONSERVATION OF ENERGY indicates ALL (rearward-traveling) "LOST" REFLECTED INTENSITY will appear as ENHANCED INTENSITY IN THE (forward-traveling) TRANSMITTED BEAM. The sum of the reflected and transmitted beam intensities is always equal to the incident intensity. This important fact has been confirmed experimentally." (emphasis mine) Cecil, you're preaching to the choir here--see my above statement. Steve at first said the energy in the canceled waves continues to flow toward the source without a voltage and current and that interference was not involved. He later changed his mind. All that should be archived on r.r.a.a on Google for the summer of 2001. Here's an excerpt. Steve said: "The total forward power increases as a direct result of the vector superposition of forward voltage and current. This DOES NOT require a corresponding destructive interference process ..." thus contradicting Hecht in _Optics_ who says any constructive interference process must be accompanied by an equal magnitude of destructive interference. Superposition of forward voltage and current? I didn't realize that voltage and current superpose. But if Steve says so. I agree on the interference, but doesn't the destructive have to occur before there can be constructive interference? Now to continue what Steve said is: "The result of this wave cancellation is that the total steady-state rearward-traveling wave has a net voltage of 0 V nd 0 A, respectively, and an impedance match occurs." No No No. As we've discussed earlier, voltage and current cannot both go to zero simultaneously, except in the rearward direction. But that's what he said above. The rearward-traveling wave indeed does have a net voltage of 0 V and 0 A and the reflections toward the source disappear at the match point. I think you and Steve really agree on about 98% of this match point stuff but you two obviously disagree on the definition of "re-reflection". I don't recall Steve ever mentioning current. He simply says the voltages cancel, resulting in 0 V. What Steve apparently doesn't understand is how the energy direction is reversed when the rearward voltages and currents go to zero. Since the energy cannot go to zero what happens at the match point is that the E fields go to zero and the disappearing E field energy merges into the H field energy, raising both the H field energy and its associated current to double their values relative to those prior to re-reflection. The energy now propagates forward and the E and H fields resume their normal relationship, half the total energy in each. This is precisely what happens to the EM field when it encounters a physical short circuit. But in our case the reflected EM fields encounter a virtual short circuit, with the same result as with a physical short except that the virtual short to the reflected waves is transparent to the source wave. I know you don't agree with me that a one-way virtual short is what causes the re-reflection, but in a short time I'll be able to prove it to you in a manner you'll not be able to rebut. Stay tuned. As I said above I'll prove to you conclusively that the virtual short circuit is established by the wave interference, contrary to what you and many on this rraa thread believe. You might want to check your work against an s-parameter analysis. The s-parameter equations a b1 = s11(a1) + s12(a2) and b2 = s21(a1) + s22(a2) Given that a match point in a transmission line can be considered to be a two- port network, |s22|^2 is the Power (re)reflected from the network output divided by the (reflected) Power incident on the network output. s22 is the reflection coefficient looking into port 2 and is *not equal to 1.0 or zero*. In a matched system with nothing but resistances, it is often the negative of the reflection coefficient at the load. This is covered in HP's AN 95-1 available on the web. I have no disagreement with the S parameter analysis, Cecil. Those who don't believe will get quite a surprise when I reveal what the output source resistance of the xmtr really is under this condition. Waddya think? I think I am too ignorant of the subject to venture an opinion. Which of the following systems do you think I would prefer, the conjugately-matched one at 50% efficiency or the non-conjucately one at 98% efficiency? 1 ohm XMTR----100V out---tuner---1/2WL 450 ohm line--50 ohm load 50 ohm XMTR---100V out---tuner---1/2WL 450 ohm line--50 ohm load Cecil, your comparison above lacks logic. Your are assuming that the XMTR is a classical generator, where the maximum efficiency cannot exceed 50% because the internal resistance is dissipative. OTH, the souce resistance at the output of the XMTR is non-dissipative, in which case it wouldn't matter which of the two XMTRs you choose. The only dissipative resistance in the amp is that which heats the plate. That dissipation is the only dissipation in the source--the other dissipation is only in the load. If you don't agree with this concept please review Chapter 19 in Reflections 2. I also have other measurements that prove the concept is true. Arguing that Tesla/Westinghouse would have to conjugately match their 60 Hz AC generators is what shot down Edison's dream of an all DC power distribution system for the USA. Doncha just love Tesla? Walt |
#9
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On Mon, 31 May 2004 13:26:58 GMT, "Peter O. Brackett"
wrote: Cecil: [snip] I fear that Walter is far too hung up on that conjugate match stuff... Walt, you need to move on to more important things. Don't continue to fuss with others over such a trivial and ultimately unimportant point. Hi Peter, I agree, but if you had read the last portion of Cecil's post you'd have seen that apparently others have again asked my opinion of the subject. Have you left for the North yet, or still in Deland? Yep, been back in Michigan since April 12--coooold up here--wish we were still in DeLand. Lot's a thunderstorms too. Sorry I've been far too busy with consulting to get over to see you this past winter. We'll be back in DeLand in Nov. so please continue planning to come over. Walt |
#10
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Walter Maxwell wrote:
wrote: Steve at first said the energy in the canceled waves continues to flow toward the source without a voltage and current and that interference was not involved. He later changed his mind. All that should be archived on r.r.a.a on Google for the summer of 2001. Here's an excerpt. Steve said: "The total forward power increases as a direct result of the vector superposition of forward voltage and current. This DOES NOT require a corresponding destructive interference process ..." thus contradicting Hecht in _Optics_ who says any constructive interference process must be accompanied by an equal magnitude of destructive interference. Superposition of forward voltage and current? I'm sure he meant "superposition of forward voltages and superposition of forward currents." I don't recall Steve ever mentioning current. I think you are right re his article. The above quote is from an r.r.a.a. posting circa Summer 2001. What Steve apparently doesn't understand is how the energy direction is reversed when the rearward voltages and currents go to zero. "How" is not explained in any of the physics references. The closest physics reference that explains it is _Optics_, by Hecht where he says something like, at a point some distance from a source, constructive interference must be balanced by an equal magnitude of destructive interference. In a matched system, there is "complete destructive interference" toward the source side of the match point and "complete constructive interference" toward the load side of the match point. Energy is always displaced from the "complete destructive interference" event to the "complete constructive interference" event. (That's what you call a "virtual short" or "virtual open" capable of re-reflecting the reflected energy.) In s-parameter terms, b1 is the reflected voltage from port 1 toward the source. Port 1 is the input to a matched tuner (transmatch). The equation is: rearward-traveling voltage reflected toward the source b1 = s11(a1) + s12(a2) For b1 to be zero, i.e. zero reflections toward the source, s11(a1) must be equal in magnitude and opposite in phase to s12(a2). That is "complete destructive interference". Since there are only two directions, "complete constructive interference" must occur in the direction of b2 = s21(a1) + s22(a2) toward the load which is the opposite direction from b1. s11 is the port 1 reflection coefficient. a1 is the port 1 incident voltage. s21 is the port 2 to port 1 transmission coefficient. a2 is the voltage reflected from the load that is incident upon port 2. Match-Point Port1 Port2 Source------Z01--------x------------Z02------------load a1-- --a2 --b1 b2-- The only dissipative resistance in the amp is that which heats the plate. That dissipation is the only dissipation in the source--the other dissipation is only in the load. Why isn't the source impedance a negative resistance, i.e. a source of power Vs a positive resistance, a sink of power? -- 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 100,000 Newsgroups - 19 Different Servers! =----- |
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