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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! =----- |
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 |
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! =----- |
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 |
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! =----- |
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. |
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 |
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 |
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! =----- |
On Wed, 02 Jun 2004 08:37:30 -0500, Cecil Moore wrote:
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.) Cecil, I explained the 'how', both in Reflections and in QEX. My explantion of 'how' is what Steve is continually stating is incorrect, especially in his last 3-part QEX article. Statements in that article prove he doesn't understand the wave mechanism that reverses the direction of the reflected energy. Evidence of this is that by simply saying the voltages cancel is insufficient description of how the energies reverse direction. In fact, in his Oct 99 ComQuart article he specifically states that both voltages and power cancel. This tell me that he doesn't understand the wave action he's attempting to teach. MIT's Slater and Harvard's Alford both explain it brilliantly, but Steve rejects those references as 'irrelevant', and says I mistakenly used them as references in Reflections. What is really perplexing to me is that several posters on this subject said that Steve's 3-parter is the best and most illuminating article they ever read on the subject. How can they have missed some of the most egregious errors appearing in that paper is unbelievable! 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. Cecil, if s11(a1) is equal in magnitude but in opposite phase with s12(a2) this constitutes a short circuit. Assume two generators delivering harmonically related output voltages equal to the two 's' voltages. When the generators are connected with their output terminals reversed, causing their voltages to be 180 degrees out of phase--this configuration is a SHORT CIRCUIT. What I've been trying to say is that this is the same condition as when the reflected waves of voltage and current from a mismatched termination are of equal magnitude and opposite phase with the voltage and current waves reflected by a matching device such as a stub, the opposing voltages in those two sets of waves constitute a short circuit the same as the voltages delivered by the two opposing generators. 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? Cecil, the source impedance is often correctly referred to as a negative resistance. But it must be remembered that the source resistance of Class B and C amps is non-dissipative, and thus totally re-reflect incident reflected power. By this I mean that the dissipative resistance that heats the plate is entirely separate from the output resistance represented by the load line. Remember, the DC power goes to only two places: that which is dissipated as heat, and that which is delivered to the load. The reflected power incident on the output terminals of the tank has no effect on the power dissipated as heat. Here's an example. First, adjust an amp to deliver 100 watts into a 50-ohm resistive load. Second, change the load to a reactive 50 + j50 load and readjust the pi-network to again deliver 100 watts into the new load. The plate current will be exactly as in the first case, and the heat dissipated will be the same. The difference is that in the first case the output impedance of the amp was 50 + j0, while in the second case the output impedance is 50 - j50, due to readjusting the reactive components in the pi-network to match the 50 + j50-ohm load. Whether one likes it or not, this constitutes a conjugate match. As for the plate temperature remaining the same in both cases, first, the readjustment of the pi-network returned the input resistance of the network to the same value as in the first case. Thus the plates saw no different condition between the two cases. And second, Eric Nichols, KL7AJ, has measured calorimetrically the temperature of the water cooling the tubes of megawatt transmitters with greatly differing values of reflected power incident on the xmtr. He has shown that the water temperature remains constant whatever the value of the reflected power. Since you mentioned earlier that some posters would like my opinion on the nature of the source resistance in rf amps I'll put a paragraph or two together with measurement data to support my opinion. Walt |
On Wed, 02 Jun 2004 16:49:09 GMT, Walter Maxwell wrote:
On Wed, 02 Jun 2004 08:37:30 -0500, Cecil Moore wrote: 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.) Cecil, I explained the 'how', both in Reflections and in QEX. My explantion of 'how' is what Steve is continually stating is incorrect, especially in his last 3-part QEX article. Statements in that article prove he doesn't understand the wave mechanism that reverses the direction of the reflected energy. Evidence of this is that by simply saying the voltages cancel is insufficient description of how the energies reverse direction. In fact, in his Oct 99 ComQuart article he specifically states that both voltages and power cancel. This tell me that he doesn't understand the wave action he's attempting to teach. MIT's Slater and Harvard's Alford both explain it brilliantly, but Steve rejects those references as 'irrelevant', and says I mistakenly used them as references in Reflections. What is really perplexing to me is that several posters on this subject said that Steve's 3-parter is the best and most illuminating article they ever read on the subject. How can they have missed some of the most egregious errors appearing in that paper is unbelievable! snip Cecil, the following is a direct quote from Steve's Comm Quart Article, Oct 1999: "For the impedance matching network to 'work', this analysis must demonstrate that the steady-state traveling backward power developed at the matching network input is equal in magnitude but 180 degrees out of phase with the initial power reflected at the matching network input. For this to occur Vback must be the negative of VR. In this case ALL POWER TRAVELING BACKWARD TOWARDS THE TRANSMISTTER WILL BE CANCELED, resulting in the steady-state matched condition." Emphasis mine. Walt |
Walter Maxwell wrote:
Cecil Moore wrote: "How" is not explained in any of the physics references. Cecil, I explained the 'how', both in Reflections and in QEX. Yes, I know you did, Walt. By "physics references" above, I meant books like college physics textbooks, e.g. _Optics_, by Hecht. What is really perplexing to me is that several posters on this subject said that Steve's 3-parter is the best and most illuminating article they ever read on the subject. How can they have missed some of the most egregious errors appearing in that paper is unbelievable! Not recognizing his power equations as classical EM physics interference terms was a pretty huge mistake in Part 3. But alleged gurus on this newsgroup have done the same thing. Apparently, power is simply ignored in present-day transmission line theory. Cecil, if s11(a1) is equal in magnitude but in opposite phase with s12(a2) this constitutes a short circuit. I agree it constitutes a "short circuit" for superposed rearward- traveling voltages. But exactly the same thing happens to the current as happens to the voltage. And an "open circuit" is what causes the rearward-traveling currents to superpose to zero. The two rearward-traveling superposing voltages might be: (100v at zero degrees) superposed with (100v at 180 degrees) The superposed sum of the two rearward-traveling voltages is zero. This indeed acts like a short where voltages go to zero. The two corresponding rearward-traveling superposing currents might be: (2a at 180 degrees) superposed with (2a at zero degrees) The superposed sum of the two rearward-traveling currents is zero. This acts like an open where currents go to zero. Or if you prefer, both the E-fields and the H-fields cancel to zero when complete destructive interference occurs. In a transmission line, it causes a surge of constructive interference energy in the opposite direction, something you have called "re-reflection from a virtual short". -- 73, Cecil http://www.qsl.net/w5dxp |
Walter Maxwell wrote:
Cecil, the following is a direct quote from Steve's Comm Quart Article, Oct 1999: "For the impedance matching network to 'work', this analysis must demonstrate that the steady-state traveling backward power developed at the matching network input is equal in magnitude but 180 degrees out of phase with the initial power reflected at the matching network input. For this to occur Vback must be the negative of VR. In this case ALL POWER TRAVELING BACKWARD TOWARDS THE TRANSMISTTER WILL BE CANCELED, resulting in the steady-state matched condition." Joules/sec possesses phase? Joules/sec can be canceled? -- 73, Cecil http://www.qsl.net/w5dxp |
Cecil Moore wrote: Walter Maxwell wrote: Cecil, the following is a direct quote from Steve's Comm Quart Article, Oct 1999: "For the impedance matching network to 'work', this analysis must demonstrate that the steady-state traveling backward power developed at the matching network input is equal in magnitude but 180 degrees out of phase with the initial power reflected at the matching network input. For this to occur Vback must be the negative of VR. In this case ALL POWER TRAVELING BACKWARD TOWARDS THE TRANSMISTTER WILL BE CANCELED, resulting in the steady-state matched condition." Joules/sec possesses phase? Joules/sec can be canceled? Therein lies part of the problem with thinking that the unit (Joules/sec) moves along a transmission line. Energy in Joules moves. (Joules/sec) of power does not. 73, Jim AC6XG |
On Thu, 03 Jun 2004 14:03:26 -0500, Cecil Moore
wrote: Walter Maxwell wrote: Cecil Moore wrote: "How" is not explained in any of the physics references. Cecil, I explained the 'how', both in Reflections and in QEX. Yes, I know you did, Walt. By "physics references" above, I meant books like college physics textbooks, e.g. _Optics_, by Hecht. What is really perplexing to me is that several posters on this subject said that Steve's 3-parter is the best and most illuminating article they ever read on the subject. How can they have missed some of the most egregious errors appearing in that paper is unbelievable! Not recognizing his power equations as classical EM physics interference terms was a pretty huge mistake in Part 3. But alleged gurus on this newsgroup have done the same thing. Apparently, power is simply ignored in present-day transmission line theory. Cecil, if s11(a1) is equal in magnitude but in opposite phase with s12(a2) this constitutes a short circuit. I agree it constitutes a "short circuit" for superposed rearward- traveling voltages. But exactly the same thing happens to the current as happens to the voltage. And an "open circuit" is what causes the rearward-traveling currents to superpose to zero. The two rearward-traveling superposing voltages might be: (100v at zero degrees) superposed with (100v at 180 degrees) The superposed sum of the two rearward-traveling voltages is zero. This indeed acts like a short where voltages go to zero. Cecil, this is exactly what I've been trying to persuade you of, but always said no, there is no short developed. But you must also agree that under this condition the current doubles. The two corresponding rearward-traveling superposing currents might be: (2a at 180 degrees) superposed with (2a at zero degrees) The superposed sum of the two rearward-traveling currents is zero. This acts like an open where currents go to zero. Of course, but the voltage doubles. Or if you prefer, both the E-fields and the H-fields cancel to zero when complete destructive interference occurs. In a transmission line, it causes a surge of constructive interference energy in the opposite direction, something you have called "re-reflection from a virtual short". Well, Cecil, here's where we part company to a degree. Unlike voltage and current that can go to zero simultaneously only in the rearward direction, E and H fields can never go to zero simultaneously. At a short circuit the E field collaples to zero, but its energy temporarily merges with the H field, making the H field double it normal value. But the changing H field immediately reestablishes the E field, both now traveling in the forward direction. And yes, this is called re-reflection. If Steve understands the action of the fields in the EM wave it's hard to understand why he finds it so erroneous to associate voltage and current with the their respective fields in impedance matching. Apparently he can't conceive that the voltages and currents in reflected waves can be considered to have been delivered by separate generators connected with opposing polarities. Walt |
Cecil, H. Adam Stevens sent you an email earlier today, with a cc to me. I
replied to all, but the copy to you came back host unknown. But your address appeared as . What's the ONEDOT? I now see it in your return address on the rraa. What's going on? Walt |
On Thu, 03 Jun 2004 12:44:10 -0700, Jim Kelley wrote:
Cecil Moore wrote: Walter Maxwell wrote: Cecil, the following is a direct quote from Steve's Comm Quart Article, Oct 1999: "For the impedance matching network to 'work', this analysis must demonstrate that the steady-state traveling backward power developed at the matching network input is equal in magnitude but 180 degrees out of phase with the initial power reflected at the matching network input. For this to occur Vback must be the negative of VR. In this case ALL POWER TRAVELING BACKWARD TOWARDS THE TRANSMISTTER WILL BE CANCELED, resulting in the steady-state matched condition." Joules/sec possesses phase? Joules/sec can be canceled? Hi Jim, Of course you're right, but that's not the point. The point is that reflected energy is not canceled, nor does it disappear at the matching point, instead it is re-reflected into the forward direction. This is the point that Steve apparently doesn't understand. And this is the reason his power budget is incorrect in his 3-part article, he ignored the energy appearing at the match point, assuming that it disappeared, though his word is 'canceled'. I thought my emphasis with capitalization would contain the necessary info, but I can see now that I should have made the emphasis show that the 'canceled energy' was erroneous, because energy cannot be canceled. In this case it is re-reflected, a concept Steve ignores. Therein lies part of the problem with thinking that the unit (Joules/sec) moves along a transmission line. Energy in Joules moves. (Joules/sec) of power does not. 73, Jim AC6XG Sorry, Jim, I put my response in the wrong place. Walt |
Jim Kelley wrote:
Therein lies part of the problem with thinking that the unit (Joules/sec) moves along a transmission line. Energy in Joules moves. (Joules/sec) of power does not. What about the Poynting Vector and Power Flow Vectors? What about the 60 Hz "power generation" and "power distribution" system? Are you saying that the trailing edge of an ExH wave is not moving? Are you saying that the ExB power in the light from Alpha Centauri didn't come from Alpha Centauri? -- 73, Cecil http://www.qsl.net/w5dxp |
Walter Maxwell wrote:
Cecil, this is exactly what I've been trying to persuade you of, but always said no, there is no short developed. But you must also agree that under this condition the current doubles. Nope, for complete destructive interference, both the E-field and H-field associated with the two interfering waves collapse to zero. For the resulting complete constructive interference, the ratio of the E-field to the H-field equals the characteristic impedance of the medium. The two corresponding rearward-traveling superposing currents might be: (2a at 180 degrees) superposed with (2a at zero degrees) The superposed sum of the two rearward-traveling currents is zero. This acts like an open where currents go to zero. Of course, but the voltage doubles. Nope, again here are the two sets of reflected waves. #1 100v at zero degrees and 2a at 180 degrees = 200W #2 100v at 180 degrees and 2a at zero degrees = 200W Superposing those two reflected waves yields zero volts and zero amps. Well, Cecil, here's where we part company to a degree. Unlike voltage and current that can go to zero simultaneously only in the rearward direction, E and H fields can never go to zero simultaneously. For "complete destructive interference" as explained in _Optics_, by Hecht, the E-field and B-field (H-field) indeed do go to zero simultaneously. That is what causes a completely dark ring in a set of interference rings. Of course, a resulting corresponding complete constructive interference causes the brightest of rings. If Steve understands the action of the fields in the EM wave it's hard to understand why he finds it so erroneous to associate voltage and current with the their respective fields in impedance matching. Apparently he can't conceive that the voltages and currents in reflected waves can be considered to have been delivered by separate generators connected with opposing polarities. He pretty much ignored current. His power equations are exact copies of the light irradiance interference equations from optics, but he apparently didn't realize it until I pointed it out to him. -- 73, Cecil http://www.qsl.net/w5dxp |
Cecil Moore wrote:
Jim Kelley wrote: Therein lies part of the problem with thinking that the unit (Joules/sec) moves along a transmission line. Energy in Joules moves. (Joules/sec) of power does not. What about the Poynting Vector and Power Flow Vectors? What about them? What about the 60 Hz "power generation" and "power distribution" system? What are you trying to imply about power generation? "Power distribution system" is really a misnomer. In the present day venacular it would be called exactly what it is - an "energy distribution system". Are you saying that the trailing edge of an ExH wave is not moving? I don't recall ever expressing the opinion that traveling waves don't travel. However mathematical formulas do not propagate along transmission lines. Fields do, but there is no such thing as an ExB "field". 73, Jim AC6XG |
On Thu, 03 Jun 2004 16:18:36 -0500, Cecil Moore
wrote: Walter Maxwell wrote: Cecil, this is exactly what I've been trying to persuade you of, but always said no, there is no short developed. But you must also agree that under this condition the current doubles. Nope, for complete destructive interference, both the E-field and H-field associated with the two interfering waves collapse to zero. For the resulting complete constructive interference, the ratio of the E-field to the H-field equals the characteristic impedance of the medium. The two corresponding rearward-traveling superposing currents might be: (2a at 180 degrees) superposed with (2a at zero degrees) The superposed sum of the two rearward-traveling currents is zero. This acts like an open where currents go to zero. Of course, but the voltage doubles. Nope, again here are the two sets of reflected waves. #1 100v at zero degrees and 2a at 180 degrees = 200W #2 100v at 180 degrees and 2a at zero degrees = 200W Superposing those two reflected waves yields zero volts and zero amps. Well, Cecil, here's where we part company to a degree. Unlike voltage and current that can go to zero simultaneously only in the rearward direction, E and H fields can never go to zero simultaneously. For "complete destructive interference" as explained in _Optics_, by Hecht, the E-field and B-field (H-field) indeed do go to zero simultaneously. That is what causes a completely dark ring in a set of interference rings. Of course, a resulting corresponding complete constructive interference causes the brightest of rings. If Steve understands the action of the fields in the EM wave it's hard to understand why he finds it so erroneous to associate voltage and current with the their respective fields in impedance matching. Apparently he can't conceive that the voltages and currents in reflected waves can be considered to have been delivered by separate generators connected with opposing polarities. He pretty much ignored current. His power equations are exact copies of the light irradiance interference equations from optics, but he apparently didn't realize it until I pointed it out to him. Well, Cecil, perhaps I don't understand Melles-Griot. However, when an EM wave encounters a short circuit the E field goes to zero, but in its change from normal level to zero, that changing field develops a corresponding H field that adds to its original value, causing it to double--the H field does NOT go to zero. But as it returns to its normal value that change in turn develops a new E field propagating in the opposite direction. From there on the EM field reconstitutes itself with normal E and H fields of equal value, each supporting one half of the propagating energy. This sequence must also prevail in optics--are you sure you are interpreting your double zero at the correct point in the circles? Walt |
Jim Kelley wrote:
... there is no such thing as an ExB "field". Good grief, Jim, ExB is proportional to the irradiance of a light beam. I'm sorry if my ASCII-limited character set hairlips you. -- 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! =----- |
Walter Maxwell wrote:
Well, Cecil, perhaps I don't understand Melles-Griot. However, when an EM wave encounters a short circuit the E field goes to zero, but in its change from normal level to zero, that changing field develops a corresponding H field that adds to its original value, causing it to double--the H field does NOT go to zero. But as it returns to its normal value that change in turn develops a new E field propagating in the opposite direction. From there on the EM field reconstitutes itself with normal E and H fields of equal value, each supporting one half of the propagating energy. I absolutely agree, Walt, "when an EM wave encounters a short circuit ...". But when "complete destructive interference" occurs, something else happens. The E-field goes to zero AND the H-field (B-field) goes to zero at the same time. If you concentrate on the voltage, it looks like a short. If you concentrate on the current, it looks like an open. It is both or neither. Complete destructive interference requires that both fields go to zero simultaneously and emerge as constructive interference in the opposite direction obeying the rule that E/H=V/I=Z0. It is an energy reflection that is also an impedance transformation at an impedance discontinuity. Let's assume that 100v at zero degrees with a current of 2a at 180 degrees encounters another wave traveling in the same rearward direction of 100v at 180 degrees with a current of 2a at zero degrees. These two waves cancel. The voltage goes to zero AND the current goes to zero. Each wave was associated with 200 watts. So a total of 400 watts reverses directions. Assuming the destructive interference occurred in a 50 ohm environment and the resulting constructive interference occurred in a 300 ohm environment, the reflected wave would be 346 volts at 1.16 amps. It's pretty simple math. 346*1.16 = 400 watts 346/1.16 = 300 ohms The above quantities represent the destructive/constructive interference. These quantities must be added to the other voltages and currents that are present to obtain the net voltage and net current. -- 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! =----- |
It didn't bounce.
hmmmmm I use spam filters. Oh well, ........... takes all kinds. 73 BTW Walt, I've had one of your baluns for almost 20 years. Damn thing just won't die. H. "Walter Maxwell" wrote in message ... Cecil, H. Adam Stevens sent you an email earlier today, with a cc to me. I replied to all, but the copy to you came back host unknown. But your address appeared as . What's the ONEDOT? I now see it in your return address on the rraa. What's going on? Walt |
Cecil Moore wrote: Jim Kelley wrote: ... there is no such thing as an ExB "field". Good grief, Jim, ExB is proportional to the irradiance of a light beam. Good grief, Cecil, irradiance isn't a field and doesn't propagates either! 73, Jim AC6XG |
On Fri, 04 Jun 2004 09:55:56 -0700, Jim Kelley
wrote: Cecil Moore wrote: Good grief, Jim, ExB is proportional to the irradiance of a light beam. Good grief, Cecil, irradiance isn't a field and doesn't propagates either! Hi Jim, Nearly every posting that Cecil pens with his rustic understanding of Optics suffers from the obvious lack of experience. There is a certain amount of pretense in this irradiance (an archaic radiometric term used incorrectly for photometrics in an argument that calls for luminous flux), much like quoting I²R and not knowing what Ohms or Amperes are. He's lucky there are so very few that appreciate the gaffs of his unintended humor. 73's Richard Clark, KB7QHC |
Jim Kelley wrote:
Good grief, Cecil, irradiance isn't a field and doesn't propagates either! As I said earlier, irradiance is proportional to the cross product of the E-field and B-field. Why do you think we call them fields if they are not fields? Exactly how does the irradiance from Alpha Centauri get to us without propagating? Methinks you are playing word games. -- 73, Cecil http://www.qsl.net/w5dxp |
Richard Clark wrote:
Nearly every posting that Cecil pens with his rustic understanding of Optics suffers from the obvious lack of experience. Everything I write about light/optics comes straight from _Optics_, by Hecht. I assume he is more of an expert than you are and he uses the term "irradiance" for the power contained in a light beam. Hecht's interference equations involving irradiance are identical in concept to Dr. Best's power equations in his QEX article. -- 73, Cecil http://www.qsl.net/w5dxp |
On Fri, 04 Jun 2004 13:11:55 -0500, Cecil Moore
wrote: Everything I write about light/optics comes straight from A Xerox shows as much precision and no comprehension. Still enjoy the gaffs tho' Great show! ;-) |
On Fri, 04 Jun 2004 12:59:02 -0500, Cecil Moore
wrote: Exactly how does the irradiance from Alpha Centauri get to us without propagating? A smile a minute ;-) How does a money wire transfer get into your account without moving from the mint? [Hint to the literate: irradiance does not move, neither does the money in a wire transfer.] |
Richard Clark wrote: On Fri, 04 Jun 2004 12:59:02 -0500, Cecil Moore wrote: Exactly how does the irradiance from Alpha Centauri get to us without propagating? A smile a minute ;-) How does a money wire transfer get into your account without moving from the mint? [Hint to the literate: irradiance does not move, neither does the money in a wire transfer.] Cecil says that a credit card is money, and presumably so is a wire transfer. All things are indeed, equal - apparently. Amazing analogy though - a power debit from one account, a power credit at the other - without power (or irradiance) having actually traveled the distance. All owing to the facility that transfers energy. 73, Jim AC6XG |
On Fri, 4 Jun 2004 06:24:24 -0500, "H. Adam Stevens, NQ5H"
wrote: It didn't bounce. hmmmmm I use spam filters. Oh well, ........... takes all kinds. 73 BTW Walt, I've had one of your baluns for almost 20 years. Damn thing just won't die. H. Thanks, H, I appreciate knowing that. Walt |
On Fri, 04 Jun 2004 12:31:59 -0700, Jim Kelley
wrote: Richard Clark wrote: On Fri, 04 Jun 2004 12:59:02 -0500, Cecil Moore wrote: Exactly how does the irradiance from Alpha Centauri get to us without propagating? A smile a minute ;-) How does a money wire transfer get into your account without moving from the mint? [Hint to the literate: irradiance does not move, neither does the money in a wire transfer.] Cecil says that a credit card is money, and presumably so is a wire transfer. It doesn't look like reciprocity would work. Is money a credit card (promise to pay instead of actual payment)? All things are indeed, equal - apparently. Ah yes, appearance, the bane of bandwidth limited perception. Amazing analogy though - a power debit from one account, a power credit at the other - without power (or irradiance) having actually traveled the distance. All owing to the facility that transfers energy. The power of analogy can be stretched further. Actually money is the energy (potential for trade) and is exhausted (power dissipated) when transferred to a debit account. If you tender more money than required (mismatch) some is returned to you to dissipate later. If you tender less than required (mismatch) some is expected from you to make the economic engine go. If you start trying to substitute, like say explaining in terms of optics why your bills are chartreuse where most are green - they push a small hidden button and you are Abu Ghraib bound. 73's Richard Clark, KB7QHC |
Richard Clark wrote:
[Hint to the literate: irradiance does not move, I've never said that power flows but a lot of pretty smart people think it does. The Poynting Vector works just as well for light as it works for RF. The Power Flow Vector works just as well for light as it does for RF. -- 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! =----- |
Jim Kelley wrote:
Cecil says that a credit card is money, ... My dictionary says that a credit card is money. Please channel your objections to Barnes and Noble. -- 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! =----- |
Richard Clark wrote:
Jim Kelley wrote: Cecil says that a credit card is money, and presumably so is a wire transfer. It doesn't look like reciprocity would work. OTOH, Jim says that since his beginning balance and his ending balance are the same, that proves that absolutely no transactions (no debits and no credits) have occurred during the entire month. What are the chances of that happening in an active account? Zero? -- 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! =----- |
On Fri, 04 Jun 2004 16:51:49 -0500, Cecil Moore
wrote: OTOH, Jim says that since his beginning balance and his ending balance are the same, that proves that absolutely no transactions (no debits and no credits) have occurred during the entire month. What are the chances of that happening in an active account? Zero? Gosh, financially illiterate too! I suppose you could quote Keynesian Economics texts to prove this quaint notion of yours however. ;-) That's entertainment as they used to say. |
On Fri, 04 Jun 2004 16:30:22 -0500, Cecil Moore
wrote: Richard Clark wrote: [Hint to the literate: irradiance does not move, I've never said that power flows but a lot of pretty smart people think it does. A conundrum, does that put you outside the crowd looking in? There are a lot of things you've never said. Can you prove it? Definition by negative example is next to worthless. |
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