|
On Fri, 04 Jun 2004 16:37:33 -0500, Cecil Moore
wrote: My dictionary says that a credit card is money. Try stop payment on a $20 bill you spent yesterday. |
On Thu, 03 Jun 2004 21:40:02 -0500, Cecil Moore wrote:
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. Cecil, at this point I'm not clear what's happening in your above example. Where and what is the phase reference for these two waves? It appears to me that the reference phase must be that of the source wave, because the voltage and current in both rearward traveling waves are 180 degrees out of phase. Educate me on how the cancellation takes place and how the energy reverses direction. It seems to me the out of phase voltage yields a short circuit, while the out of phase current yields an open circuit. How can both exist simultaneously? And further, what circuit can produce these two waves simultaneously? In addition, I believe your example has changed the subject. My discourse concerns what occurs to an EM wave when it encounters a short circuit. In this case you're going to have to prove to me that both E and H fields go to zero. IMO it can't happen. The E field collapses to zero while the H field doubles, because the energy in the changing E field merges into the H field. Before the EM wave encountered the short both fields contained the same energy, thus the E-field energy adding to the original H field energy, doubles it, and while that H field was changing it was developing a new E field that launches a new wave in the opposite direction, the reflected wave. So my argument with you, Cecil, is that I maintain the H field doubles on encountering a short circuit, while you maintain that both E and H fields go to zero. What's your answer to this dilemma? Walt Walt |
Walter Maxwell wrote:
Cecil, at this point I'm not clear what's happening in your above example. Where and what is the phase reference for these two waves? It appears to me that the reference phase must be that of the source wave, because the voltage and current in both rearward traveling waves are 180 degrees out of phase. Educate me on how the cancellation takes place and how the energy reverses direction. J. C. Slater explains how the cancellation takes place. "The method of eliminating reflections is based on the interference between waves. Two waves half a wavelength apart are in opposite phases, and the sum of them, if their amplitudes are numerically equal, is zero. 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." Note that the above applies to both voltage waves and current waves. Both voltage and current go to zero during complete destructive interference, i.e. both E-field and H-field go to zero during complete destructive interference. It seems to me the out of phase voltage yields a short circuit, while the out of phase current yields an open circuit. How can both exist simultaneously? They can't, and that's my argument. It's easy to understand. If the two rearward- traveling voltages are 180 degrees out of phase then the two rearward-traveling currents MUST also be 180 degrees out of phase, since the reflected current is ALWAYS 180 degrees out of phase with the reflected voltage. If one looks only at the voltages, one will say it's a short-circuit. If one looks only at the currents, one will say it's an open-circuit. And further, what circuit can produce these two waves simultaneously? According to J. C. Slater (and Reflections II, page 23-9) a match point produces these two waves simultaneously, two reflected voltages 180 degrees out of phase and two reflected currents 180 degrees out of phase. In addition, I believe your example has changed the subject. My discourse concerns what occurs to an EM wave when it encounters a short circuit. There is no argument about what happens at a short circuit. What I am saying is that a match point is NOT a short circuit. In this case you're going to have to prove to me that both E and H fields go to zero. IMO it can't happen. J. C. Slater says that's what happens in the above quote. Voltages 1/2WL apart in time cancel to zero. Currents 1/2WL apart in time cancel to zero. So my argument with you, Cecil, is that I maintain the H field doubles on encountering a short circuit, while you maintain that both E and H fields go to zero. What's your answer to this dilemma? My argument is that it is NOT a short circuit. It is "complete destructive interference" as explained in _Optics_, by Hecht where both the E-field and B-field go to zero. J. C. Slater says that the two rearward-traveling voltages are 180 degrees out of phase and the two rearward-traveling currents are 180 degrees out of phase. So whatever happens to the voltage also happens to the current, i.e. destructive interference takes both voltage and current to 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 23:06:16 -0500, Cecil Moore
wrote: And further, what circuit can produce these two waves simultaneously? According to J. C. Slater (and Reflections II, page 23-9) a match point produces these two waves simultaneously, two reflected voltages 180 degrees out of phase and two reflected currents 180 degrees out of phase. You know, Walt, just like a wave breaking against the reef. The coral reflects a wave of water, and a wave of salt, and perhaps a wave of brine shrimp... If you correspond along these lines long enough, you may rummage up a chowder from Cecil, but I wouldn't expect Chateâubriand. 73's Richard Clark, KB7QHC |
On Sat, 05 Jun 2004 05:14:49 GMT, Richard Clark wrote:
On Fri, 04 Jun 2004 23:06:16 -0500, Cecil Moore wrote: And further, what circuit can produce these two waves simultaneously? According to J. C. Slater (and Reflections II, page 23-9) a match point produces these two waves simultaneously, two reflected voltages 180 degrees out of phase and two reflected currents 180 degrees out of phase. You know, Walt, just like a wave breaking against the reef. The coral reflects a wave of water, and a wave of salt, and perhaps a wave of brine shrimp... If you correspond along these lines long enough, you may rummage up a chowder from Cecil, but I wouldn't expect Chateâubriand. 73's Richard Clark, KB7QHC Well, Richard, if I ain't gonna git Chateaubriand I ain't goin'. Walt |
On Fri, 04 Jun 2004 23:06:16 -0500, Cecil Moore wrote:
Walter Maxwell wrote: Cecil, at this point I'm not clear what's happening in your above example. Where and what is the phase reference for these two waves? It appears to me that the reference phase must be that of the source wave, because the voltage and current in both rearward traveling waves are 180 degrees out of phase. Educate me on how the cancellation takes place and how the energy reverses direction. Cecil, your reference below to Chapter 23 in Reflections has re-oriented me and I'm now on your page. This is the case of a 1/4wl matching transformer. I didn't recognize my own writing. Sorry. J. C. Slater explains how the cancellation takes place. "The method of eliminating reflections is based on the interference between waves. Two waves half a wavelength apart are in opposite phases, and the sum of them, if their amplitudes are numerically equal, is zero. 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." Yes, Cecil, this quote is my Ref 35, which I used to support this concept in my QST article in Oct 1974, so I am well familiar with it. And yes, you are correct in that the waves reflected from points A and B in Fig 6 are in the phase you state. Note that the above applies to both voltage waves and current waves. Both voltage and current go to zero during complete destructive interference, i.e. both E-field and H-field go to zero during complete destructive interference. But Cecil, take another look at Fig 6 on page 23-5 to note that those two waves arrive 180 out of phase at point A, which means only that the E and H fields cancel in the rearward direction only, resulting in a Zo match to the source. But what's important (and I inadvertantly omitted this fact in the book, which will be corrected in III) is that when the waves reflected from B reach A they find a discontinuity at A of an open-circuit type condition, because the line Zo now goes from low to high (in the rearward direction). Thus when the waves reflected at B arrive at A the voltage which is already at 360 (0) degrees does not change phase, but the current which is at 180 degrees on arrival changes by 180 due to the open circuit at A to rearward traveling waves. Thus the current is now also at 0 degrees. With both voltage and current at 0 degrees relative to the source, all the power in the waves reflected at B are re-reflected at A and add to the source power. Consequently, Cecil, both the E and H fields go to zero only in rearward direction at the match point, which is necessary for no reflections to travel rearward of the match point, but the H field goes to zero only temporarily while the E field is doubled temporarily, as they should when encountering an open circuit as they do on arriving at A. It seems to me the out of phase voltage yields a short circuit, while the out of phase current yields an open circuit. How can both exist simultaneously? I've answered my own dumb question here. They can't, and that's my argument. It's easy to understand. If the two rearward- traveling voltages are 180 degrees out of phase then the two rearward-traveling currents MUST also be 180 degrees out of phase, since the reflected current is ALWAYS 180 degrees out of phase with the reflected voltage. If one looks only at the voltages, one will say it's a short-circuit. If one looks only at the currents, one will say it's an open-circuit. I think I've addressed this paragraph above. And further, what circuit can produce these two waves simultaneously? As I embarrassingy discovered it's the 1/4wl transformer--dumb me. According to J. C. Slater (and Reflections II, page 23-9) a match point produces these two waves simultaneously, two reflected voltages 180 degrees out of phase and two reflected currents 180 degrees out of phase. In addition, I believe your example has changed the subject. My discourse concerns what occurs to an EM wave when it encounters a short circuit. I was confused here, Cecil, because you did change the subject from discussing what happens to the E and H fields when encountering a short to the 1/4wl transformer example without my catching on to the change. As I said, dumb me. There is no argument about what happens at a short circuit. What I am saying is that a match point is NOT a short circuit. In this case the match point at A is an open circuit. In this case you're going to have to prove to me that both E and H fields go to zero. IMO it can't happen. J. C. Slater says that's what happens in the above quote. Voltages 1/2WL apart in time cancel to zero. Currents 1/2WL apart in time cancel to zero. Yep, but only in the rearward direction. So my argument with you, Cecil, is that I maintain the H field doubles on encountering a short circuit, while you maintain that both E and H fields go to zero. What's your answer to this dilemma? We've answered this. My argument is that it is NOT a short circuit. It is "complete destructive interference" as explained in _Optics_, by Hecht where both the E-field and B-field go to zero. J. C. Slater says that the two rearward-traveling voltages are 180 degrees out of phase and the two rearward-traveling currents are 180 degrees out of phase. So whatever happens to the voltage also happens to the current, i.e. destructive interference takes both voltage and current to zero. I'll not argue with the way Hecht describes the phenomena. G'nite, Cecil, it's 3 :20 AM and I've got to go back to bed. |
Walter Maxwell wrote:
But Cecil, take another look at Fig 6 on page 23-5 to note that those two waves arrive 180 out of phase at point A, which means only that the E and H fields cancel in the rearward direction only, resulting in a Zo match to the source. Yes, and that is exactly my point. EXACTLY the same thing happens to the E-fields and H-fields. That means exactly the same thing that happens to the rearward- traveling voltages also happens to the rearward-traveling currents. Two equal- magnitude/opposite-phase voltages cancel. Two equal-magnitude/opposite-phase currents cancel. That doesn't happen at either an open or a short. If one looks at just the voltages, it looks like a short. If one looks at just the currents, it looks like an open. With both voltage and current at 0 degrees relative to the source, all the power in the waves reflected at B are re-reflected at A and add to the source power. Reflected waves ALWAYS have the voltage and current 180 degrees out of phase. "With both voltage and current at 0 degrees" implies a forward wave which is the *EFFECT* of re-reflection, i.e. after re-reflection. The rearward-traveling voltage and current CANNOT be in phase before the re-reflection. After the re- reflection, they are no longer rearward-traveling so they are in-phase and forward- traveling. It's a cause and effect thing. 1. Proper adjustment of the tuner/stub/quarter-wave-transformer CAUSES all voltages and currents at the perfect match point to be in-phase or 180 degrees out of phase. 2. Destructive interference between two rearward-traveling waves of equal magnitudes and opposite phases causes the E-field voltage and H-field current to go to zero simultaneously, i.e. the E-fields superpose to zero in the rearward direction and the H-fields superpose to zero in the rearward direction. EXACTLY the same thing happens to the H-fields as happens to the E-fields in the rearward direction. Neither a short nor an open acts like that. "... 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." From the Melles-Groit web page. 3. "... the principle of conservation of energy indicates all "lost" reflected intensity will appear as enhanced intensity in the transmitted beam." Quoted from the Melles Groit web page. They don't say how but they do say why. In a perfectly matched system, all rearward-traveling energy changes direction at the match point and joins the forward wave as constructive interference. That constructive interference energy is equal in magnitude to the energy involved in the destructive interference event. It's a simple conservation of energy process. There is no argument about what happens at a short circuit. What I am saying is that a match point is NOT a short circuit. In this case the match point at A is an open circuit. Destructive interference causes the rearward-traveling voltage to go to zero. Voltage doesn't go to zero at an open circuit. Exactly the same thing happens to the rearward-traveling voltages and rearward-traveling currents. Voltage goes to zero at a short-circuit. Current goes to zero at an open circuit. A match point causes both to go to zero in the rearward direction. That means the voltage sees a short-circuit looking rearward and the current sees an open-circuit looking rearward. "Consequently, all corresponding voltage and current phasors are 180 degrees out of phase at the matching point. ... With equal magnitudes and opposite phase at the (matching) point, the sum of the two waves is zero." _Reflections_II_, page 23-9. This is true for both step-up or step-down impedance discontinuities. At a perfectly matched point, the two rearward-traveling voltages are ALWAYS of equal magnitude and opposite phase. The two rearward-traveling currents are ALWAYS of equal magnitude and opposite phase. That's the only way complete destructive interference of rearward-traveling waves can occur. J. C. Slater says that's what happens in the above quote. Voltages 1/2WL apart in time cancel to zero. Currents 1/2WL apart in time cancel to zero. Yep, but only in the rearward direction. The rearward direction is what we are talking about. The point is that EXACTLY the same thing happens to the two rearward-traveling current waves as happens to the two rearward-traveling voltage waves. A short-circuit doesn't affect voltages and currents in the same way. An open-circuit doesn't affect voltages and currents in the same way. A match point affects the rearward- traveling voltages and rearward-traveling currents in EXACTLY the same way. The re-reflection at a match point is a conservation of energy reflection where the rearward destructive interference energy supplies energy to constructive interference in the opposite direction. For light, the equation a Destructive Interference Irradiance = I1 + I2 - 2{SQRT[(I1)(I2)]} (9.16) Constructive Interference Irradiance = I1 + I2 + 2{SQRT[(I1)(I2)]} (9.15) _Optics_, by Hecht, fourth edition, page 388 Note the similarities to equations 13 and 15 in Dr. Best's QEX article, Part 3. PFtotal = P1 + P2 - 2{SQRT[(P1)(P2)]} (Eq 15) PFtotal = P1 + P2 + 2{SQRT[(P1)(P2)]} (Eq 13) Too bad he didn't label them as Hecht did, as "total destructive interference" and "total constructive interference" equations. -- 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! =----- |
Cecil Moore wrote:
Walter Maxwell wrote: But Cecil, take another look at Fig 6 on page 23-5 to note that those two waves arrive 180 out of phase at point A, which means only that the E and H fields cancel in the rearward direction only, resulting in a Zo match to the source. Yes, and that is exactly my point. EXACTLY the same thing happens to the E-fields and H-fields. That means exactly the same thing that happens to the rearward-traveling voltages also happens to the rearward-traveling currents. In my class in secondary school counseling, I learned a technique that might be helpful here. It's called, "Be the thing." Whatever it is that you are trying to understand, mentally become that thing. In other words, assume that you are the reflected current to find out what you would experience. Obviously, it is just a mental exercise, but one that I have found quite useful throughout the years. First, assume that you are the reflected voltage from a mismatched load. What do you encounter back at the match point? You encounter another reflected voltage with equal magnitude and opposite phase traveling in the same rearward direction. What happens to you? Your momentum in the rearward direction is reversed and your energy starts flowing toward the load. As a reflected voltage, based on your necessarily limited knowledge, you assume that you must have encountered a virtual short circuit. Second, assume that you are the reflected current from a mismatched load. What do you encounter back at the match point? You encounter another reflected current with equal magnitude and opposite phase traveling in the same rearward direction. What happens to you? Your momentum in the rearward direction is reversed and your energy starts flowing toward the load. As a reflected current, based on your necessarily limited knowledge, you assume that you must have encountered a virtual open circuit. There exists an apparent contradiction. A match point cannot simultaneously be a virtual short and a virtual open. How is the apparent contradiction resolved? Is there anything else in physics that can cause a total reflection of energy besides a short, open, or pure reactance? The answer is, "yes", and it happens all the time in the field of optics. In a system with only two directions of energy travel available, total destructive interference in one direction has to result in total constructive interference in the other direction. That's the way perfect non-glare thin-film coated glass works in the presence of a coherent single-frequency laser beam. -- 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 Sat, 05 Jun 2004 13:05:18 -0500, Cecil Moore wrote:
Cecil Moore wrote: Walter Maxwell wrote: But Cecil, take another look at Fig 6 on page 23-5 to note that those two waves arrive 180 out of phase at point A, which means only that the E and H fields cancel in the rearward direction only, resulting in a Zo match to the source. Yes, and that is exactly my point. EXACTLY the same thing happens to the E-fields and H-fields. That means exactly the same thing that happens to the rearward-traveling voltages also happens to the rearward-traveling currents. In my class in secondary school counseling, I learned a technique that might be helpful here. It's called, "Be the thing." Whatever it is that you are trying to understand, mentally become that thing. In other words, assume that you are the reflected current to find out what you would experience. Obviously, it is just a mental exercise, but one that I have found quite useful throughout the years. First, assume that you are the reflected voltage from a mismatched load. What do you encounter back at the match point? You encounter another reflected voltage with equal magnitude and opposite phase traveling in the same rearward direction. What happens to you? Your momentum in the rearward direction is reversed and your energy starts flowing toward the load. As a reflected voltage, based on your necessarily limited knowledge, you assume that you must have encountered a virtual short circuit. Second, assume that you are the reflected current from a mismatched load. What do you encounter back at the match point? You encounter another reflected current with equal magnitude and opposite phase traveling in the same rearward direction. What happens to you? Your momentum in the rearward direction is reversed and your energy starts flowing toward the load. As a reflected current, based on your necessarily limited knowledge, you assume that you must have encountered a virtual open circuit. There exists an apparent contradiction. A match point cannot simultaneously be a virtual short and a virtual open. How is the apparent contradiction resolved? Is there anything else in physics that can cause a total reflection of energy besides a short, open, or pure reactance? The answer is, "yes", and it happens all the time in the field of optics. In a system with only two directions of energy travel available, total destructive interference in one direction has to result in total constructive interference in the other direction. That's the way perfect non-glare thin-film coated glass works in the presence of a coherent single-frequency laser beam. Yes, Cecil, I understand. However I don't particularly like the notion of saying both fields go to zero, or both fields go to zero in the rearward direction. Confusing. Remember, weeks ago I swore that both fields could never go to zero simultaneously? The reason I disagreed with you is that you didn't mention the 'direction'. The reason I dislike hearing that both fields go to zero is that it's really not true. Like I've said many times, on encountering a short,circuit voltage and the E field go to zero and the current and H field doubles AND REVERSES DIRECTION. To me, Reversing direction is more meaningful and less confusing than both going to zero, and it still says there is no energy propagating rearward of the match point. Going now to the cancellation process when the voltages and currents of both waves are mutually out of phase. You say that voltages 180 out yields a short (agreed) and that currents 180 out yields an open. Sounds good, and I mistakenly agreed a coupla days ago. But I don' think so. I believe voltage 180 out defines a short--period. Look at it this way. Take a zip cord and put male plugs on both ends. Plug one end into an outlet, say the top one, and then plug the other end into the bottom outlet with the polarities reversed. With respect to voltage we have a 'circuit breaker' short circuit, because the voltages entering the zip cord at each end were 180 out. But so were the currents initially. Then why the short circuit current flow? Certainly not because the circuit is open to current. Another scenario with the same initial conditions and results: Take two identical generators delivering the same level of harmonically related output voltages. Connect their terminals in phase.Voltages in phase--currents in phase. Result? No current flow. Why? Zero voltage differential. Open circuit to voltage--open circuit to current. Now reconnect their terminals in the opposite manner. Voltages 180 out--currents 180 out. Do we have current flow? You bet--dead short! Because current results from voltage, if voltages are 180 out of phase we have a short to both voltage and curent. No open circuit to current. Cecil, I hope we're both still on the same page on this one; Walt |
Walter Maxwell wrote:
Yes, Cecil, I understand. However I don't particularly like the notion of saying both fields go to zero, or both fields go to zero in the rearward direction. But Walt, that's exactly what happens when total destructive interference occurs as explained by J. C. Slater in _Microwave_Transmission_. I believe voltage 180 out defines a short--period. That same belief is what got Dr. Best into trouble. He never considered what happens to the reflected current waves. In a sense, your and his disagreements are because you both made the same conceptual mistake and arrived at different conclusions because of that common mistake. If you and he had not made that shared mistake, you both would have arrived at the same conclusions. Another scenario with the same initial conditions and results: Take two identical generators delivering the same level of harmonically related output voltages. Connect their terminals in phase.Voltages in phase--currents in phase. Result? No current flow. Why? Zero voltage differential. Open circuit to voltage--open circuit to current. Now reconnect their terminals in the opposite manner. Voltages 180 out--currents 180 out. Do we have current flow? You bet--dead short! Because current results from voltage, if voltages are 180 out of phase we have a short to both voltage and curent. No open circuit to current. This is the problem with trying to use circuit analysis to replace network analysis. Put the two sources at the two ends of a transmission line and please reconsider the outcome. Equip the two sources with circulators and dummy loads so the outcome cannot be in doubt. -- 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 Sat, 05 Jun 2004 10:14:09 -0500, Cecil Moore wrote:
Walter Maxwell wrote: But Cecil, take another look at Fig 6 on page 23-5 to note that those two waves arrive 180 out of phase at point A, which means only that the E and H fields cancel in the rearward direction only, resulting in a Zo match to the source. Yes, and that is exactly my point. EXACTLY the same thing happens to the E-fields and H-fields. That means exactly the same thing that happens to the rearward- traveling voltages also happens to the rearward-traveling currents. Two equal- magnitude/opposite-phase voltages cancel. Two equal-magnitude/opposite-phase currents cancel. That doesn't happen at either an open or a short. If one looks at just the voltages, it looks like a short. If one looks at just the currents, it looks like an open. Snip J. C. Slater says that's what happens in the above quote. Voltages 1/2WL apart in time cancel to zero. Currents 1/2WL apart in time cancel to zero. Yep, but only in the rearward direction. The rearward direction is what we are talking about. The point is that EXACTLY the same thing happens to the two rearward-traveling current waves as happens to the two rearward-traveling voltage waves. A short-circuit doesn't affect voltages and currents in the same way. An open-circuit doesn't affect voltages and currents in the same way. A match point affects the rearward- traveling voltages and rearward-traveling currents in EXACTLY the same way. The re-reflection at a match point is a conservation of energy reflection where the rearward destructive interference energy supplies energy to constructive interference in the opposite direction. For light, the equation a Destructive Interference Irradiance = I1 + I2 - 2{SQRT[(I1)(I2)]} (9.16) Constructive Interference Irradiance = I1 + I2 + 2{SQRT[(I1)(I2)]} (9.15) _Optics_, by Hecht, fourth edition, page 388 Note the similarities to equations 13 and 15 in Dr. Best's QEX article, Part 3. PFtotal = P1 + P2 - 2{SQRT[(P1)(P2)]} (Eq 15) PFtotal = P1 + P2 + 2{SQRT[(P1)(P2)]} (Eq 13) Too bad he didn't label them as Hecht did, as "total destructive interference" and "total constructive interference" equations. Sorry, Cecil, in spite of their similarity with Hecht's, these equations are totally invalid. Steve derived them from his Eq 9, which is also totally invalid for use with reflected power. This equation is correct and valid when there are two separate and individual sources. But here there is only one source, the transceiver. When connecting two batteries in series Eq 9 works, because there is enough energy there to support the additional current demanded with the increased voltage. But not when the transceiver is the sole source of power. With the transmission line system Steve's voltage V2 comes from the same source as V1. The problem is that when the total forward power resulting from the addition of reflected power and source power the total forward power is never absorbed in the load, the power resulting from the reflection is subtracted from the total power. This limitation does not occur when there are two separate sources to maintain the increased current. Because Steve used Eq 9 in an invalid way to derive Eqs 10 through 15, all of these derived equations are also invalid. Try Eq 13 for example. It says 75 w plus 8.33 w = 133.33 w, as you well know. This is absurd! In addition, because the powers don't add up correctly using V1 and V2 at zero phase relationship, he concocted the ruse that they must add vectorially, and he goes through several values of phase relationships to show what the forward power would be with the various phases. This is poppycock, because the phase relationship between the source (V1) and re-reflected voltage (V2) is ALWAYS ZERO on lossless lines. His initial problem is that he misinterpreted Eq 6 in Part 1 to yield the forward voltage Vfwd, where it actually yields the voltage E of the standing wave at any point on the line, where the point on the line is determined by the 'L' term in the exponents on the right-hand side of the equation. In other words, the summation of terms on the right-hand side of his Eq 6 does not equal forward voltage Vfwd, as it indicates incorrectly, but instead equals the voltage of the standing wave. In addition to other errors, the entire right-hand column of page 46 is invalid. Walt |
On Sat, 05 Jun 2004 15:23:37 -0500, Cecil Moore wrote:
Walter Maxwell wrote: Yes, Cecil, I understand. However I don't particularly like the notion of saying both fields go to zero, or both fields go to zero in the rearward direction. But Walt, that's exactly what happens when total destructive interference occurs as explained by J. C. Slater in _Microwave_Transmission_. I believe voltage 180 out defines a short--period. That same belief is what got Dr. Best into trouble. He never considered what happens to the reflected current waves. In a sense, your and his disagreements are because you both made the same conceptual mistake and arrived at different conclusions because of that common mistake. If you and he had not made that shared mistake, you both would have arrived at the same conclusions. Cecil, how do you figure I made a mistake in this issue? I have always considered voltage 180 out as a short. And my writings show voltage at 180 as a short, as stated on page 23-9. I agree that the opposite phases of both voltage and current in that discussion resulted in the cancelation of reflected power traveling in the 225-ohm section of line. And during the last day or two I leaned toward thinking the out of phase current implied an open circuit. But you can see from my words above that voltage rules--when the voltages are 180 out of phase it defines a short circuit. My zip cord example is evidence to that. Consequently, I don't agree that Steve and I made the same mistake. My writings delivered the correct mathematical answers--Steve's does not. The mistake I made on page 23-9 is in overlooking that it is the effective open circuit condition seen looking in the rearward direction by the reflected waves at point A is what gave both the voltage and current waves the reversal and phase change to zero relative to the source waves. Another scenario with the same initial conditions and results: Take two identical generators delivering the same level of harmonically related output voltages. Connect their terminals in phase.Voltages in phase--currents in phase. Result? No current flow. Why? Zero voltage differential. Open circuit to voltage--open circuit to current. Now reconnect their terminals in the opposite manner. Voltages 180 out--currents 180 out. Do we have current flow? You bet--dead short! Because current results from voltage, if voltages are 180 out of phase we have a short to both voltage and curent. No open circuit to current. This is the problem with trying to use circuit analysis to replace network analysis. Put the two sources at the two ends of a transmission line and please reconsider the outcome. Equip the two sources with circulators and dummy loads so the outcome cannot be in doubt. Cecil, I don't believe the outcome is in doubt. Walt |
Walter Maxwell wrote:
Sorry, Cecil, in spite of their similarity with Hecht's, these equations are totally invalid. Before we go any farther, Walt, please reference a copy of _Optics_, by Hecht. Dr. Best's equations are valid. He just didn't understand what he was dealing with and presented them improperly. Hecht presents them properly. Dr. Best's equations are the classical physics equations for the destructive interference and constructive interference. With the transmission line system Steve's voltage V2 comes from the same source as V1. Yes, V1 comes from the generator and V2 comes from reflections from the load which come originally from the generator. Dr. Best didn't understand the S-parameter analysis and presented his material in an invalid way. But even though he didn't understand what he was saying, his equations are valid. He was completely off base in his explanations. It was like Einstein coming up with E = MC^2 and then completely blowing the explanation. In an S-parameter analysis, b2 = s21(a1) + s22(a2) In Dr. Best's analysis, VFtotal = V1 + V2 whe V1 = V1F*(transmission coefficient) + V2R*(reflection coefficient) V1F is the voltage incident upon the impedance discontinuity from the left. (port1) V2R is the voltage incident upon the impedance discontinuity from the right. (port2) There's a one-to-one correspondence above. If the S-parameter analysis is valid, then Dr. Best's equations are valid. He just didn't present them in a valid manner. Because Steve used Eq 9 in an invalid way to derive Eqs 10 through 15, all of these derived equations are also invalid. Try Eq 13 for example. It says 75 w plus 8.33 w = 133.33 w, as you well know. This is absurd! His equations are valid. His knowledge is what was invalid. It is true that 75w + 8.33W + 2*SQRT(75W*8.33W) = 133.33W P1 + P2 + interference power = PFtotal I presented this to Dr. Best 9 months before his Part 3 was published. He simply didn't pay any attention. From a voltage standpoint where ci means constructive interference: V1^2/Z02 + V2^2/Z02 + Vci^2/Z02 = VFtotal^2/Z02 For a Z02 equal to 150 ohms (if I remember correctly) 106.07v^2 + 35.35v^2 + 86.6v^2 = 141.42v^2 So V1^2 + V2^2 + Vci^2 = VFtotal^2 This is exactly what you have been saying all along, something that Dr. Best simply didn't understand. He completely ignored the interference term without which the voltage equation cannot balance. 2*SQRT(75W*8.33W) is the constructive interference term supplied by the destructive interference event on the other side of the match point which Dr. Best completely ignored in his article. In addition, because the powers don't add up correctly using V1 and V2 at zero phase relationship, he concocted the ruse that they must add vectorially, and he goes through several values of phase relationships to show what the forward power would be with the various phases. This is poppycock, because the phase relationship between the source (V1) and re-reflected voltage (V2) is ALWAYS ZERO on lossless lines. Only if the system is perfectly matched. If the system is not matched, V1 and V2 can have any phase relationship. Dr. Best's equations are valid but he just didn't comprehend their meaning. When I called him on it, he seemed never to have heard of destructive/constructive interference. That's what set me to researching EM waves in the arena of optics. I looked at the situation assuming that you two guys are both knowledgeable and intelligent and I arrived at the conclusion that the two of you are only two inches apart. But (IMO) neither one of you is willing to move that one inch to bridge the gap. (I have said this before in private email to Walt.) -- 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:
But you can see from my words above that voltage rules--when the voltages are 180 out of phase it defines a short circuit. This is exactly the same mistake that Dr. Best made. *VOLTAGE DOESN'T RULE!* Current is *equally* important to voltage. If you had assumed that "current rules", you would be saying - "when the currents are 180 out of phase it defines an open circuit". My argument is actually a minor point but bridges part of the gap between you and Dr. Best. (And absolutely nothing being discussed here concerns the source impedance of a transmitter. All we are discussing is what happens at a match point in a transmission line or at a tuner.) -- 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 Sat, 05 Jun 2004 19:11:32 -0500, Cecil Moore wrote:
Walter Maxwell wrote: Sorry, Cecil, in spite of their similarity with Hecht's, these equations are totally invalid. Before we go any farther, Walt, please reference a copy of _Optics_, by Hecht. Dr. Best's equations are valid. He just didn't understand what he was dealing with and presented them improperly. Hecht presents them properly. Dr. Best's equations are the classical physics equations for the destructive interference and constructive interference. I'm trying to locate my Hecht paper, but can't at the moment. Let's make sure we're talking about the same set of of equations. The ones I'm saying are invalid as stated in his article appear on Page 46, Col 2. These equations are all being used invalidly as a result of his Eq 9, Part 3 being invalid for use with reflected power with only one source. This a concept you're just going to have to learn to accept. This equation is valid ONLY when there are two separate sources--the rearward traveling reflected wave is NOT a second source, and it will NOT work on power contained in the reflected wave. You must understand why the invalidity of his Eq 6 in Part 1is responsible for the entire problem. Once you understand this point you'll understand why the equations on Page 46 are invalid. Cecil, Steve's Eq 9 is valid only if there is more than one source. In this case there is only one, the transceiver. This a concept that you apparentely aren't getting, and Steve didn't either. Because Steve derived his Eqs 10 thru 15 from an invalid premise concerning Eq 6 those equations are invalid. Just plug his values for P1 and P2 into any of those equations and you'll get invalid answers. With the transmission line system Steve's voltage V2 comes from the same source as V1. Yes, V1 comes from the generator and V2 comes from reflections from the load which come originally from the generator. But Cecil, you CAN'T add V1 and V2 in any manner to obtain forward power, because adding V1 and V2 does not yield forward voltage. This is the first place Steve erred in Part 1. Go back and review my previous msg where I explain why his Eq 6 in Part 1 is invalid. Dr. Best didn't understand the S-parameter analysis and presented his material in an invalid way. But even though he didn't understand what he was saying, his equations are valid. He was completely off base in his explanations. It was like Einstein coming up with E = MC^2 and then completely blowing the explanation. In an S-parameter analysis, b2 = s21(a1) + s22(a2) In Dr. Best's analysis, VFtotal = V1 + V2 whe V1 = V1F*(transmission coefficient) + V2R*(reflection coefficient) How many times do I have to explain that V1 + V2 does NOT equal VF total? This concept cannot be fudged into the S- parameter analysis, because V1 + V2 = VF total is an invalid premise. V1F is the voltage incident upon the impedance discontinuity from the left. (port1) V2R is the voltage incident upon the impedance discontinuity from the right. (port2) I repeat, Cecil, V1 + V2 yields only the swr, not VF total forward voltage. There's a one-to-one correspondence above. If the S-parameter analysis is valid, then Dr. Best's equations are valid. He just didn't present them in a valid manner. The S-parameter analysis would be valid if the premise that V1 + V2 = VF were valid, which it is not. Please review my explanation why his Eq 6 in Part 1 is invalid. This is the crux of the entire case, which has made the Eqs that I say are invalid, invalid. Because Steve used Eq 9 in an invalid way to derive Eqs 10 through 15, all of these derived equations are also invalid. Try Eq 13 for example. It says 75 w plus 8.33 w = 133.33 w, as you well know. This is absurd! His equations are valid. His knowledge is what was invalid. It is true that 75w + 8.33W + 2*SQRT(75W*8.33W) = 133.33W This is Steve's Eq 12, which appears to be correct when theta = zero, but he qualifies this equation, saying tghat when V1 and V2 are in phase the total power can be determined by Eq 13. NOT SO. The ironic part of this is that when the system is matched V1 and V2 are always in phase on lossless lines. P1 + P2 + interference power = PFtotal This fact is precisely what Steve emphatically denies, I presented this to Dr. Best 9 months before his Part 3 was published. He simply didn't pay any attention. From a voltage standpoint where ci means constructive interference: I don't understand the ci term--please explain. V1^2/Z02 + V2^2/Z02 + Vci^2/Z02 = VFtotal^2/Z02 V1^2/Zo + V2^2/ Zo = total forward power. The left -hand terms of the above equation are power terms that do not equal VFtotal^2/Zo, because V1 + V2 does not equal VFtotal. For a Z02 equal to 150 ohms (if I remember correctly) 106.07v^2 + 35.35v^2 + 86.6v^2 = 141.42v^2 So V1^2 + V2^2 + Vci^2 = VFtotal^2 Again, if V1 + V2 does not equal VFtotal, then the sum of the squares of V1 and V2 cannot equal the square of VFtotal, unless Vci^2 is a large negative number. This is exactly what you have been saying all along, something that Dr. Best simply didn't understand. He completely ignored the interference term without which the voltage equation cannot balance. 2*SQRT(75W*8.33W) is the constructive interference term supplied by the destructive interference event on the other side of the match point which Dr. Best completely ignored in his article. In addition, because the powers don't add up correctly using V1 and V2 at zero phase relationship, he concocted the ruse that they must add vectorially, and he goes through several values of phase relationships to show what the forward power would be with the various phases. This is poppycock, because the phase relationship between the source (V1) and re-reflected voltage (V2) is ALWAYS ZERO on lossless lines. Only if the system is perfectly matched. If the system is not matched, V1 and V2 can have any phase relationship. Yeah, but if V1 and V2 have any relationship other than zero the system is NOT perfectly matched. Dr. Best's equations are valid but he just didn't comprehend their meaning. When I called him on it, he seemed never to have heard of destructive/constructive interference. That's what set me to researching But Cecil, why would we be considering any condition other than perfectly matched? All of his equations from 9 thru 15 specifiy the route to obtain either PF or VF. This means conditions are for a matched system. EM waves in the arena of optics. I looked at the situation assuming that you two guys are both knowledgeable and intelligent and I arrived at the conclusion that the two of you are only two inches apart. But (IMO) neither one of you is willing to move that one inch to bridge the gap. (I have said this before in private email to Walt.) Sorry, Cecil, we're miles apart, and will be until the erroneous Eq 6 in Part 1 is corrected. I know I've rambled all over the place with redundancies, but I tried to respond to each of your paragraph statements as they occurred. Walt |
On Sat, 05 Jun 2004 19:20:18 -0500, Cecil Moore wrote:
Walter Maxwell wrote: But you can see from my words above that voltage rules--when the voltages are 180 out of phase it defines a short circuit. This is exactly the same mistake that Dr. Best made. *VOLTAGE DOESN'T RULE!* Current is *equally* important to voltage. If you had assumed that "current rules", you would be saying - "when the currents are 180 out of phase it defines an open circuit". Then how can you explain what happens when you reverse the zip cord, plugging one end in one way and the other with the prongs reversed ? Are you saying the currents in this condition are seeing an open circuit? My argument is actually a minor point but bridges part of the gap between you and Dr. Best. (And absolutely nothing being discussed here concerns the source impedance of a transmitter. All we are discussing is what happens at a match point in a transmission line or at a tuner.) |
Walter Maxwell wrote:
I'm trying to locate my Hecht paper, but can't at the moment. Let's make sure we're talking about the same set of of equations. The ones I'm saying are invalid as stated in his article appear on Page 46, Col 2. Walt, I'm sorry, all I have is the QEX CD and the pages are not the same as they were in QEX magazine. Dr. Best's equations 13 and 15 are the classical physics interference equations virtually identical to the irradiance equations in _Optics_, by Hecht. But Cecil, you CAN'T add V1 and V2 in any manner to obtain forward power, because adding V1 and V2 does not yield forward voltage. Yes, it does, Walt. Consider the following matched system similar to the example in Dr. Best's article. 100w XMTR-----50 ohm line---x---1/2WL 150 ohm line---50 ohm load P1 = 100W(1-rho^2) = 100(1-.25) = 75W P2 = 33.33W(rho^2) = 33.33W*.25 = 8.33W P1 = 75W, P2 = 8.33W, PFtotal = 133.33W V1 = 106.07V, V2 = 35.35V, VFtotal = 141.4V V1 + V2 = VFtotal, 106.07V + 35.35V = 141.4V How many times do I have to explain that V1 + V2 does NOT equal VF total? But it does, Walt. See above. 75w + 8.33W + 2*SQRT(75W*8.33W) = 133.33W (Equation 13) This is Steve's Eq 12, which appears to be correct when theta = zero, but he qualifies this equation, saying that when V1 and V2 are in phase the total power can be determined by Eq 13. NOT SO. Yes, it is, Walt. Equation 13 is valid for any matched system. That's equation 13 just above. Perhaps you don't understand what Dr. Best means by P1 and P2. PFtotal = P1 + P2 + 2*SQRT(P1*P2) total forward power 133.33W = 75W + 8.33W + 50W There's 50W of constructive interference and the equation balances perfectly. -- 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:
Then how can you explain what happens when you reverse the zip cord, plugging one end in one way and the other with the prongs reversed ? Are you saying the currents in this condition are seeing an open circuit? I'm sorry, Walt, you lost me. What zip cord? -- 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 Sat, 05 Jun 2004 23:39:35 -0500, Cecil Moore
wrote: What zip cord? :-) A point offered that is not even half a day old... Symptoms of two conversations running under the impression they are on the same planet. |
On Sat, 05 Jun 2004 20:06:48 GMT, Walter Maxwell wrote:
Cecil, I hope we're both still on the same page on this one; Same page, different books.... Hi Walt, Let's see, the score is one argument, two correspondents, and three explanations. Do you know where Cecil is? Used to be "Find Waldo." 73's Richard Clark, KB7QHC |
On Sat, 05 Jun 2004 23:39:35 -0500, Cecil Moore wrote:
Walter Maxwell wrote: Then how can you explain what happens when you reverse the zip cord, plugging one end in one way and the other with the prongs reversed ? Are you saying the currents in this condition are seeing an open circuit? I'm sorry, Walt, you lost me. What zip cord? Cecil, you must not be reading my posts. Go back one or two to get the drift. Walt |
On Sat, 05 Jun 2004 23:36:07 -0500, Cecil Moore wrote:
Walter Maxwell wrote: I'm trying to locate my Hecht paper, but can't at the moment. Let's make sure we're talking about the same set of of equations. The ones I'm saying are invalid as stated in his article appear on Page 46, Col 2. Walt, I'm sorry, all I have is the QEX CD and the pages are not the same as they were in QEX magazine. Dr. Best's equations 13 and 15 are the classical physics interference equations virtually identical to the irradiance equations in _Optics_, by Hecht. But Cecil, you CAN'T add V1 and V2 in any manner to obtain forward power, because adding V1 and V2 does not yield forward voltage. Yes, it does, Walt. Consider the following matched system similar to the example in Dr. Best's article. 100w XMTR-----50 ohm line---x---1/2WL 150 ohm line---50 ohm load P1 = 100W(1-rho^2) = 100(1-.25) = 75W P2 = 33.33W(rho^2) = 33.33W*.25 = 8.33W P1 = 75W, P2 = 8.33W, PFtotal = 133.33W V1 = 106.07V, V2 = 35.35V, VFtotal = 141.4V V1 + V2 = VFtotal, 106.07V + 35.35V = 141.4V How many times do I have to explain that V1 + V2 does NOT equal VF total? But it does, Walt. See above. Well, Cecil, so far I haven't been able to follow the logic in the lines above. Perhaps they are correct for that particular example, but I'm not so sure. For example, if 100 w is available only 75 w will enter the 150-ohm line initially, so initially only 56.25 w is absorbed in the 50-ohm load and 18.75 w is reflected, which then sees a 3:1 mismatch on return to the 50-ohm line, making the re-reflected power 4 .69 w . I haven't had time to work through the remaining steps to the steady state. So before I do I notice that you state that Eqs 10 thru 15 are valid for any matched system. So let's see if this is so by using the example Steve used earlier in his Part 3, the one he took from Reflections and then called it a 'fallacy'. This is that example: 50-ohm lossles line terminated with a 150 + j0 load. 100 w available from the source at 70.71 v. Matched at the line input, in the steady state the total forward power Pfwd = 133.33 w and power reflected is Pref = 33.333 w. With these steady state powers the forward voltage is 81.65 v and reflected voltage is 40.82 v. Due to the integration of the reflected waves the source voltage 70.71 increased 10.94 v to 81.65 v. Therefore, in the steady state V1 = 81.65 v and V2 = 40.82 v. Now using V1 and V2 in Eqs 7 and 8 we get P1 = 133.33 w and P2 = 33.333 w. So far so good, these Eqs are correct and valid. But now let's plug V1 and V2 into Eq 9: (V1 + V2) = 122.47 v. Then 122.47^2/50 = 300 w. Something's amiss here. So let's go a step further and plug P1 and P2 into Eq 13: P1 = 133.33 w and P2 = 33.333 w. Then using Eq 13, PF total = (sqrt 133.33 + sqrt 33.333)^2 = 300 w. We get the same incorrect value as with Eq 9. Now why can this be? For starters, P1 derived from Eq 7 is already the total forward power. So why is the forward power value P1 plugged into Eq 13 do derive the total foward power PFtotal? Do you see where this is going, Cecil? So now let's go one step backward to use Eq 11: Plugging V1 and V2 into Eq 11 we get VFtotal^2 = 14 ,999.6983 (call it 15,000). Therefore, VFtotal = sqrt VFtotal^2 = 122.4733 v. This is the same incorrect value obtained using Eq 9. So, Cecil, do you still believe these equations are valid for every matched situation? Walt |
On Sun, 06 Jun 2004 15:33:31 GMT, Walter Maxwell wrote:
On Sat, 05 Jun 2004 23:36:07 -0500, Cecil Moore wrote: Walter Maxwell wrote: Well, Cecil, so far I haven't been able to follow the logic in the lines above. Perhaps they are correct for that particular example, but I'm not so sure. For example, if 100 w is available only 75 w will enter the 150-ohm line initially, so initially only 56.25 w is absorbed in the 50-ohm load and 18.75 w is reflected, which then sees a 3:1 mismatch on return to the 50-ohm line, making the re-reflected power 4 .69 w . I haven't had time to work through the remaining steps to the steady state. So before I do I notice that you state that Eqs 10 thru 15 are valid for any matched system. So let's see if this is so by using the example Steve used earlier in his Part 3, the one he took from Reflections and then called it a 'fallacy'. This is that example: 50-ohm lossles line terminated with a 150 + j0 load. 100 w available from the source at 70.71 v. Matched at the line input, in the steady state the total forward power Pfwd = 133.33 w and power reflected is Pref = 33.333 w. With these steady state powers the forward voltage is 81.65 v and reflected voltage is 40.82 v. Due to the integration of the reflected waves the source voltage 70.71 increased 10.94 v to 81.65 v. Therefore, in the steady state V1 = 81.65 v and V2 = 40.82 v. Now using V1 and V2 in Eqs 7 and 8 we get P1 = 133.33 w and P2 = 33.333 w. So far so good, these Eqs are correct and valid. But now let's plug V1 and V2 into Eq 9: (V1 + V2) = 122.47 v. Then 122.47^2/50 = 300 w. Something's amiss here. So let's go a step further and plug P1 and P2 into Eq 13: P1 = 133.33 w and P2 = 33.333 w. Then using Eq 13, PF total = (sqrt 133.33 + sqrt 33.333)^2 = 300 w. We get the same incorrect value as with Eq 9. Now why can this be? For starters, P1 derived from Eq 7 is already the total forward power. So why is the forward power value P1 plugged into Eq 13 do derive the total foward power PFtotal? Do you see where this is going, Cecil? So now let's go one step backward to use Eq 11: Plugging V1 and V2 into Eq 11 we get VFtotal^2 = 14 ,999.6983 (call it 15,000). Therefore, VFtotal = sqrt VFtotal^2 = 122.4733 v. This is the same incorrect value obtained using Eq 9. So, Cecil, do you still believe these equations are valid for every matched situation? Walt Cecil, one point I forgot in the msg above--the standing wave. Remember I said earlier that V1 + V2 yields the standing wave, not forward voltage Vfwd? We know that V1 + V2 = 122.47 v. This is the max value of the standing wave. Now V1 - V2 = 40.82 v. Notice that (V1+ V2)/(V1 - V2) = 122.47/40.82 = 3. 0. Doesn't this look something like the SWR? Walt |
Walter Maxwell wrote:
Cecil Moore wrote: Walter Maxwell wrote: Then how can you explain what happens when you reverse the zip cord, plugging one end in one way and the other with the prongs reversed ? Are you saying the currents in this condition are seeing an open circuit? I'm sorry, Walt, you lost me. What zip cord? Cecil, you must not be reading my posts. Go back one or two to get the drift. Are you talking about the two generators hooked up back to back. If so, where are the two EM waves traveling in the same direction which is a prerequisite for complete destructive/constructive interference? The generated EM waves, in your example, are moving in opposite directions which is simply not relevant to this discussion. Please conjure up an experiment where two coherent waves are moving in the same direction without affecting the generators. -- 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 in message ... snip Now V1 - V2 = 40.82 v. Notice that (V1+ V2)/(V1 - V2) = 122.47/40.82 = 3. 0. Doesn't this look something like the SWR? Walt Walt You really should write a book. (rim shot) =]8*O 73 H. |
Walter Maxwell wrote:
Well, Cecil, so far I haven't been able to follow the logic in the lines above. Perhaps they are correct for that particular example, but I'm not so sure. Please pick any example of your choice of matched systems. Dr. Best's equation 13 will be valid for any matched system. VFtotal = V1 + V2 will be valid for any system, matched or not. For example, if 100 w is available only 75 w will enter the 150-ohm line initially, so initially only 56.25 w is absorbed in the 50-ohm load and 18.75 w is reflected, which then sees a 3:1 mismatch on return to the 50-ohm line, making the re-reflected power 4 .69 w . I haven't had time to work through the remaining steps to the steady state. When you do, you will get the same values as I. So before I do I notice that you state that Eqs 10 thru 15 are valid for any matched system. So let's see if this is so by using the example Steve used earlier in his Part 3, the one he took from Reflections and then called it a 'fallacy'. As I have said before, Steve's equations are correct but he simply drew the wrong conclusions from them. His "fallacy" is a fallacy but has nothing to do with his equations. He simply drew the wrong conclusions from valid equations. This is that example: 50-ohm lossles line terminated with a 150 + j0 load. 100 w available from the source at 70.71 v. Matched at the line input, in the steady state the total forward power Pfwd = 133.33 w and power reflected is Pref = 33.333 w. With these steady state powers the forward voltage is 81.65 v and reflected voltage is 40.82 v. Due to the integration of the reflected waves the source voltage 70.71 increased 10.94 v to 81.65 v. Therefore, in the steady state V1 = 81.65 v and V2 = 40.82 v. Nope, you simply misunderstood what Steve said. V1 and V2 are *NOT* on the 50 ohm line. In fact, the 1 WL 50 ohm line is irrelevant and just serves to confuse. V1 and V2 are voltages existing *at the match point* at the INPUT of the tuner. Steve chose an example that is virtually impossible to explain or understand. Let's take it step by step starting with Steve's example: 100W XMTR---50 ohm line---tuner---1WL 50 ohm line---150 ohm load The 1WL 50 ohm lossless line is irrelevant except for power measurements so eliminate it. 100W XMTR---50 ohm line---tuner---150 ohm load The tuner can now be replaced by 1/4WL of 86.6 ohm feedline. 100W XMTR---50 ohm line---x---1/4WL 86.6 ohm line---150 ohm load 100W-- 107.76W-- --0W --7.76W V1-- V2-- Now we have an example that is understandable and we haven't changed any of the conditions. The magnitude of the reflection coefficient is 0.268. That makes the magnitude of the transmission coefficient equal to 1.268 (Rule of thumb for matched systems with single step-function impedance discontinuities) So V1 = 70.7 * 1.268 = 89.6V V2 = VF2 * 0.268 = 6.95V VFtotal = V1 + V2 = 89.6V + 6.95V = 96.6V PFtotal = 96.6V^2/86.6 = 107.76W So, Cecil, do you still believe these equations are valid for every matched situation? Yes, they are, Walt, once one understands them. I don't blame you for being confused about Steve's example. He chose the worst example possible and didn't explain it very well at all. I still maintain that you two are two inches apart and neither one of you will budge an inch. -- 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:
Remember I said earlier that V1 + V2 yields the standing wave, not forward voltage Vfwd? Dr. Best's V1 + V2 are moving in the same direction toward the load and do NOT yield the standing wave. You have misunderstood what he was trying to say (which is super easy to do). Dr. Best's V1 is proportional to the S-parameter term, s21(a1) The only difference is that all S-parameter voltages are normalized to the square root of Z0. Dr. Best's V2 is proportional to the S-parameter term, s22(a2) Consider this easy-to-understand generalized example. XMTR---Z01---x---1/2WL Z02---load=Z01 VF1-- VF2-- --VR1 --VR2 VF1 is the forward voltage in the Z01 section. VF2 is the forward voltage in the Z02 section. VR1 is the reflected voltage in the Z01 section. VR2 is the reflected voltage in the Z02 section. RHO is the physical reflection coefficient and TAU is the physical transmission coefficient which is equal to (1+RHO) at an impedance discontinuity. Dr. Best's V1 = VF1(TAU) His V2 = VR2(RHO) Dr. Best's V1 is the part of VF1 that makes it through the impedance discontinuity (coming from the source side). It is traveling toward the load. Dr. Best's V2 is the reflected voltage re-reflected from the match point. It is also traveling toward the load. V1 and V2 superpose into VF2 which Dr. Best calls VFtotal. And that is indeed what happens. That Dr. Best was somewhat incapable of explaining the equation is his fault, not yours. His explanations were so obtuse, I'm not sure he understood them himself. He certainly didn't recognize the classical interference equations as such. -- 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:
Now V1 - V2 = 40.82 v. Notice that (V1+ V2)/(V1 - V2) = 122.47/40.82 = 3. 0. Doesn't this look something like the SWR? Well, let's see. V1 = VF1(TAU) V2 = VR2(RHO) [VF1(TAU) + VR2(RHO)]/[VF1(TAU) - VR2(RHO)] Does that look something like SWR? -- 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, 06 Jun 2004 12:44:38 -0500, Cecil Moore wrote:
Walter Maxwell wrote: Well, Cecil, so far I haven't been able to follow the logic in the lines above. Perhaps they are correct for that particular example, but I'm not so sure. Please pick any example of your choice of matched systems. Dr. Best's equation 13 will be valid for any matched system. VFtotal = V1 + V2 will be valid for any system, matched or not. For example, if 100 w is available only 75 w will enter the 150-ohm line initially, so initially only 56.25 w is absorbed in the 50-ohm load and 18.75 w is reflected, which then sees a 3:1 mismatch on return to the 50-ohm line, making the re-reflected power 4 .69 w . I haven't had time to work through the remaining steps to the steady state. When you do, you will get the same values as I. So before I do I notice that you state that Eqs 10 thru 15 are valid for any matched system. So let's see if this is so by using the example Steve used earlier in his Part 3, the one he took from Reflections and then called it a 'fallacy'. As I have said before, Steve's equations are correct but he simply drew the wrong conclusions from them. His "fallacy" is a fallacy but has nothing to do with his equations. He simply drew the wrong conclusions from valid equations. This is that example: 50-ohm lossles line terminated with a 150 + j0 load. 100 w available from the source at 70.71 v. Matched at the line input, in the steady state the total forward power Pfwd = 133.33 w and power reflected is Pref = 33.333 w. With these steady state powers the forward voltage is 81.65 v and reflected voltage is 40.82 v. Due to the integration of the reflected waves the source voltage 70.71 increased 10.94 v to 81.65 v. Therefore, in the steady state V1 = 81.65 v and V2 = 40.82 v. Nope, you simply misunderstood what Steve said. V1 and V2 are *NOT* on the 50 ohm line. In fact, the 1 WL 50 ohm line is irrelevant and just serves to confuse. V1 and V2 are voltages existing *at the match point* at the INPUT of the tuner. Steve chose an example that is virtually impossible to explain or understand. Cecil, Steve's equations 4 thru 8 are correct, valid, and completely GENERAL. In the text preceding the Eqs he even specifies Zo as 50 ohm, not that it matters. He correctly defines V1 in Eq 7 and he correctly defines V2 in Eq 8. However, he makes a vital error in the paragraph preceding these two equations. He says incorrectly, "When two forward traveling waves add, general superposition theory and Kirchoff's voltage law require that the total forward-traveling voltage be the vector sum of the individual forward-traveling voltages such that VFtotal = V1 + V2." This statement is TOTALLY FALSE, and this error is the basis for the remainder of his equations to be invalid. Kirchoff's voltage law does not apply in this case of transmission line practice. There are times when circuit theory doesn't hold and transmission line theory is required. You are still ignoring his Eq 6 in Part 1, which is false and invalid because he mistook the expression for determining the standing wave for the total forward wave. Look back at my previous msg where I've shown that the addition of the forward and reflected waves yields the standing wave, not the forward wave. The standing wave is NOT the forward wave. Refer to the text on your CD at the paragraph that begins: "A transmission line system analysis must be performed with voltage and current, from which the power is then derived." Read on from this point to where it reads, "If total re-reflection of power occurs at the T-network, the re-reflected voltage must have the same magnitude as the reflected voltage." Please note the values of voltages and currents that appear in that entire section, because this section is a direct copy of my example he took from Reflections in an attempt to disprove it. Which he does in the next sentence: "Therefore, based on the assumption that total power re-reflection and in-phase forward-wave addition, the total forward-traveling wave of 81.65 v must be the result of a voltage having a magnitude of 70.711 v adding in phase with a voltage having a magnitude of 40.825 v. Two in-phase complex voltages having magnitudes of 70.711 v and 40.825 v cannot add together such that the resulting voltage has a magnitude 81.65 v." OF COURSE IT CAN'T, because these two voltages CANNOT BE ADDED TOGETHER TO DERIVE THE FORWARD VOLTAGE.. This is further evidence that he didn't realize that Eq 6 in Part 1 was wrong, because it derives the standing wave voltage, NOT THE FORWARD VOLTAGE. He then goes on in an unsuccessful attempt to prove my method wrong, in which he gets himself into trouble with those equations that don't work in general. If you still don't see the problem, Cecil, please go back and review my analysis in the earlier post and explain to me why you disagree with my results that show conclusively that the equations don't work. To conclude, I have shown you why I have not used his values of V1 and V2 incorrectly, as you say. If you can show that I'm wrong I'll take the time to study the step-by-step in your example below. Walt Let's take it step by step starting with Steve's example: 100W XMTR---50 ohm line---tuner---1WL 50 ohm line---150 ohm load The 1WL 50 ohm lossless line is irrelevant except for power measurements so eliminate it. 100W XMTR---50 ohm line---tuner---150 ohm load The tuner can now be replaced by 1/4WL of 86.6 ohm feedline. 100W XMTR---50 ohm line---x---1/4WL 86.6 ohm line---150 ohm load 100W-- 107.76W-- --0W --7.76W V1-- V2-- Now we have an example that is understandable and we haven't changed any of the conditions. The magnitude of the reflection coefficient is 0.268. That makes the magnitude of the transmission coefficient equal to 1.268 (Rule of thumb for matched systems with single step-function impedance discontinuities) So V1 = 70.7 * 1.268 = 89.6V V2 = VF2 * 0.268 = 6.95V VFtotal = V1 + V2 = 89.6V + 6.95V = 96.6V PFtotal = 96.6V^2/86.6 = 107.76W So, Cecil, do you still believe these equations are valid for every matched situation? Yes, they are, Walt, once one understands them. I don't blame you for being confused about Steve's example. He chose the worst example possible and didn't explain it very well at all. I still maintain that you two are two inches apart and neither one of you will budge an inch. Cecil, I've shown where we're apart. It's a lot more than 2 inches. |
Walter Maxwell wrote:
To conclude, I have shown you why I have not used his values of V1 and V2 incorrectly, as you say. If you can show that I'm wrong I'll take the time to study the step-by-step in your example below. Steve is essentially doing an S-parameter analysis without the (square root of Z0) normalization. Since we know that an S-parameter analysis of a match point is indeed valid, a lot of Steve's equations are valid by association. Assuming the S-parameter equation is valid, we can use it to prove that VFtotal = V1 + V2 V1/SQRT(Z0) = s21(a1) in the S-parameter analysis. V2/SQRT(Z0) = s22(a2) in the S-parameter analysis. VFtotal/SQRT(Z0) = b2 in the S-parameter analysis. Given: b2 = s21(a1) + s22(a2) is a valid S-parameter equation. Therefore, VFtotal/SQRT(Z0) = V1/SQRT(Z0) + V2/SQRT(Z0) is a valid equation because there is an EXACT one-to-one correspondence to the S-parameter equation. Therefore, if we multiply both sides by SQRT(Z0) we get VFtotal = V1 + V2 The only way for the above equation to be wrong is if the S-parameter equation is wrong. The only difference in the S-parameter equation and Dr. Best's equation is the normalization by [SQRT(Z0)]. Given a generalized matched system: XMTR---Z01---x---1/4WL Z02---load VF1-- VF2-- --VR1 --VR2 VF2 = VF1(TAU) + VR2(RHO) = V1 + V2 Dr. Best's equation b2 = s21(a1) + s22(a2) S-parameter equation Dr. Best is essentially quoting an S-parameter analysis -- 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, 06 Jun 2004 17:32:24 -0500, Cecil Moore wrote:
Walter Maxwell wrote: To conclude, I have shown you why I have not used his values of V1 and V2 incorrectly, as you say. If you can show that I'm wrong I'll take the time to study the step-by-step in your example below. Steve is essentially doing an S-parameter analysis without the (square root of Z0) normalization. Since we know that an S-parameter analysis of a match point is indeed valid, a lot of Steve's equations are valid by association. snip Dr. Best is essentially quoting an S-parameter analysis Cecil, if the S-parameteri analysis is applied correctly the results of the S-parameter analysis should agree with the results of mine that appears in my earlier posts. You have not responded to the results of my analysis that proves Steve's use of the equations 9 thru 15 is incorrect. I've proved that these equations do not work in general. Referring to my analysis, please show me where I went wrong, if that's your position. Walt |
Walter Maxwell wrote:
Cecil, if the S-parameteri analysis is applied correctly the results of the S-parameter analysis should agree with the results of mine that appears in my earlier posts. You have not responded to the results of my analysis that proves Steve's use of the equations 9 thru 15 is incorrect. I've proved that these equations do not work in general. Referring to my analysis, please show me where I went wrong, if that's your position. I thought I did that, Walt. Your V1 and Dr. Best's V1 are NOT the same quantity. Your V2 and Dr. Best's V2 are NOT the same quantity. It is no wonder that you didn't get the same results. The 1WL 50 ohm line in Dr. Best's example is absolutely irrelevant. Calculating anything on that line is a waste of effort. Please center your calculations around the match point. Plug any values into the following generalized matched system: Z0-match XMTR-----Z01-----x-----1/4WL Z02-----load VF1-- VF2-- --0V --VR2 VF2 = VFtotal in Dr. Best's article traveling toward the load VF1(TAU) = V1 in Dr. Best's article traveling toward the load VR2(RHO) = V2 in Dr. Best's article traveling toward the load VR2 will always equal VF1(TAU) + VR2(RHO) = V1 + V2 just like b2 will always equal s21(a1) + s22(a2) -- 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, 06 Jun 2004 18:03:31 -0500, Cecil Moore wrote:
Walter Maxwell wrote: Cecil, if the S-parameteri analysis is applied correctly the results of the S-parameter analysis should agree with the results of mine that appears in my earlier posts. You have not responded to the results of my analysis that proves Steve's use of the equations 9 thru 15 is incorrect. I've proved that these equations do not work in general. Referring to my analysis, please show me where I went wrong, if that's your position. I thought I did that, Walt. Your V1 and Dr. Best's V1 are NOT the same quantity. Your V2 and Dr. Best's V2 are NOT the same quantity. It is no wonder that you didn't get the same results. The 1WL 50 ohm line in Dr. Best's example is absolutely irrelevant. Calculating anything on that line is a waste of effort. Cecil, it seems like we're going around in cirles. If Steve's equations are valid they should work in general. It doesn't matter whether we use the values from his T network section that comes later, or the values in my example that he attempts to prove incorrect. What does matter is that the equations must deliver the correct answers regardless of the values used in the equations. I have proved that valid values plugged into his equations don't yield the correct answers. Cec;il, why are you avoiding trying to understand the basis for his erroneous concept of adding forward and reflected voltages to obtain total forward voltage? You don't even respond to my discussion on this point. Walt |
Walter Maxwell wrote:
Cecil, why are you avoiding trying to understand the basis for his erroneous concept of adding forward and reflected voltages to obtain total forward voltage? You don't even respond to my discussion on this point. I'm not trying to avoid it, Walt. Dr. Best simply doesn't do that. V1 is a *forward-traveling voltage*. V2 is a *forward-traveling voltage*. Their sum is VFtotal, the *total forward-traveling voltage*. He does NOT add a forward voltage to a reflected voltage. V2 is the *forward-traveling* re- reflected voltage equal to VR2(RHO). When the reflected voltage is acted upon by the reflection coefficient, it becomes a forward-traveling voltage. That you think Dr. Best is adding forward and reflected voltages, is the source the present misunderstanding. The individual Poynting Vector for V1 points toward the *load*. The individual Poynting Vector for V2 points toward the *load*. V1 and V2 are coherent component waves, both flowing toward the load so, of course, they superpose. Again, consider the following *matched* configuration where RHO is the reflection coefficient and TAU is the transmission coefficient. XMTR---Z01---x---1/4WL Z02---load VF1-- VF2-- --VR1 --VR2 There are four superposition components that occur. Two of them are traveling toward the load and two of them are traveling toward the source. V1 = VF1(TAU) traveling toward the load V2 = VR2(RHO) traveling toward the load Adding these two forward-traveling voltages yields VF2 = V1 + V2 V3 = VF1(RHO) traveling toward the source V4 = VR2(TAU) traveling toward the source Adding these two rearward-traveling voltages yields VR1 = V3 + V4 which, in a matched case is zero because V3 = -V4. VF1 breaks up into two components, V1 toward the load and V3 toward the source. VR2 breaks up into two components, V2 toward the load and V4 toward the source. Collect and superpose the two forward-traveling terms and you get the total forward-traveling voltage. Collect and superpose the two rearward-traveling terms and you get the total rearward-traveling voltage. -- 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, 06 Jun 2004 20:44:46 -0500, Cecil Moore wrote:
Walter Maxwell wrote: Cecil, why are you avoiding trying to understand the basis for his erroneous concept of adding forward and reflected voltages to obtain total forward voltage? You don't even respond to my discussion on this point. I'm not trying to avoid it, Walt. Dr. Best simply doesn't do that. V1 is a *forward-traveling voltage*. V2 is a *forward-traveling voltage*. Their sum is VFtotal, the *total forward-traveling voltage*. He does NOT add a forward voltage to a reflected voltage. V2 is the *forward-traveling* re- reflected voltage equal to VR2(RHO). When the reflected voltage is acted upon by the reflection coefficient, it becomes a forward-traveling voltage. That you think Dr. Best is adding forward and reflected voltages, is the source the present misunderstanding. The individual Poynting Vector for V1 points toward the *load*. The individual Poynting Vector for V2 points toward the *load*. V1 and V2 are coherent component waves, both flowing toward the load so, of course, they superpose. Cecil, I know V2 is the re-reflected voltage, but what I'm trying to persuade you is that they do NOT superpose to form the forward voltage--they superpose only to form the standing wave. You've go to accept that the standing wave voltage is NOT the forward voltage. If you can't come to realize this is the key to the problem I'm going to have to give up. Incidentally, you say tau is 1+ rho as the transmission coefficient, which when muliplied by input voltage yields forward voltage. I thought (1 - rho^2) is the transmission coefficient. These two terms are not equal. Can you explain the difference? Walt |
Walter Maxwell wrote:
Cecil, I know V2 is the re-reflected voltage, but what I'm trying to persuade you is that they do NOT superpose to form the forward voltage--they superpose only to form the standing wave. You've go to accept that the standing wave voltage is NOT the forward voltage. If you can't come to realize this is the key to the problem I'm going to have to give up. I'm sorry, Walt, Your belief that V2 is a reflected wave is the root of the misunderstanding. V2 is a re-reflected wave and is therefore forward-traveling toward the load. V2 is equal to the reflected wave voltage multiplied by the reflection coefficient. V1 and V2 are traveling in the same direction, toward the load. Incidentally, you say tau is 1+ rho as the transmission coefficient, which when muliplied by input voltage yields forward voltage. I thought (1 - rho^2) is the transmission coefficient. These two terms are not equal. Can you explain the difference? (1-rho^2) is the POWER transmission coefficient. For a single impedance discontinuity situation, 1+rho is the VOLTAGE transmission coefficient. From the IEEE Dictionary: "transmission coefficient, ... Note 2, An interface is a special case of a network where the reference planes associated with the incident and transmitted waves become coincident; in this case the voltage transmission coefficient is equal to one plus the voltage reflection coefficient." -- 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, 06 Jun 2004 23:18:37 -0500, Cecil Moore
wrote: Walter Maxwell wrote: voltage is NOT the forward voltage. If you can't come to realize this is the key to the problem I'm going to have to give up. I'm sorry, Walt, Your belief that V2 is a reflected wave is the root of the misunderstanding. V2 is a re-reflected wave Another way of saying "You are right, Walt, you are wrong." Let's see, you two have passed this SAME thing back and forth 39 times, cannot agree about what each thinks about ONE particular, and each of you insist you know what a third party meant. Well, I'm off to Foggy Bottom (D.C.) again to where they do this kind of thing for a living and call it law. ;-) Let's see if a week improves this chowder. 73's Richard Clark, KB7QHC |
Richard Clark wrote,
Let's see, you two have passed this SAME thing back and forth 39 times, cannot agree about what each thinks about ONE particular, and each of you insist you know what a third party meant. Well, I'm off to Foggy Bottom (D.C.) again to where they do this kind of thing for a living and call it law. ;-) Let's see if a week improves this chowder. 73's Richard Clark, KB7QHC The triumph of hope over experience. 73, Tom Donaly, KA6RUH |
On Sun, 06 Jun 2004 23:18:37 -0500, Cecil Moore wrote:
Walter Maxwell wrote: Cecil, I know V2 is the re-reflected voltage, but what I'm trying to persuade you is that they do NOT superpose to form the forward voltage--they superpose only to form the standing wave. You've go to accept that the standing wave voltage is NOT the forward voltage. If you can't come to realize this is the key to the problem I'm going to have to give up. I'm sorry, Walt, Your belief that V2 is a reflected wave is the root of the misunderstanding. V2 is a re-reflected wave and is therefore forward-traveling toward the load. V2 is equal to the reflected wave voltage multiplied by the reflection coefficient. V1 and V2 are traveling in the same direction, toward the load. Cecil, please read me in the first paragraph. By Steve's own words he says the re-reflected wave must equal the reflected wave. This means the system is matched in his account. Therefore, V2, is not the root of any misunderstanding in my part. You are still not getting the picture concerning that V1 and V2 cannot be added to establish the forward wave, as Steve incorrectly believes, they add only to form the standing wave. I'm sorry that I didn't think of this earlier that Steve copied his Eq 6 in Part 1 from Johnson, except he placed Vfwd, the forward voltage, instead of E for the standing wave voltage. Look it up in your Johnson on Pages 99 and 100. He derives Eq 4.23 (the Eq Steve misunderstands) on Page 99, and expresses it on Page 100. However, note on Page 98, the beginning of Sec 4.2: "The equations for E and I along the line can be expressed...." So Cecil, please understand that this equation does NOT yield the forward voltage, as Steve believes, which is the root of his misunderstanding throughout his entire paper. Concerning tau, I've seen it described in an HP App Note, which I didn't bring to Michigan, but I've never used it. However, if the power transmission coefficient is (1 - rho^2) the coefficient is 0.75 for rho = 0.5. Therefore, for 100 w forward only 75 w are delivered. This condition is shown valid experimentally. Now let's use tau = 1 + rho as the voltage transmission coefficient. I interpret this to mean tau x input voltage = forward voltage arriving at a mismatched load. For a 100 w source at 50 ohms with the same rho as above, we have 70.71 x 1.5 = 106.07 v. But we know that the forward voltage on a matched 50-ohm line with rho = 0.5 is 81.65 v. Why the difference? I should have been more aware of the explanation in the HP App Note--there must be a reason shown there to explain the difference. Walt |
Richard Clark wrote:
On Sun, 06 Jun 2004 23:18:37 -0500, Cecil Moore wrote: I'm sorry, Walt, Your belief that V2 is a reflected wave is the root of the misunderstanding. V2 is a re-reflected wave Another way of saying "You are right, Walt, you are wrong." It's a very minor mistake, Richard, and one easily made. If you have a copy of Dr. Best's QEX article, please feel free to express your take on this discussion. Let's see if I can present superposition in ASCII graphics. rho is the voltage reflection coefficient and tau is the voltage transmission coefficient. Assume VF1 has a phase angle of zero degrees. Phase angles are important in the following but since the system is matched, all phase angles are either at zero degrees or at 180 degrees at the match point. So a sign change is equivalent to a 180 degree phase shift. 100W XMTR---50 ohm line---x---1/2WL 150 ohm line---50 ohm load VF1=70.7V-- VF2=141.4V-- --VR1=0V --VR2=70.7V According to the rules of superposition, the two voltages incident upon 'x', VF1 and VR2, can be considered separately and then added. ---------------------------------------------------------------- Breaking VF1 down into its two superposition components yields: x | VF1=70.7V (100W)--| |-- V1=106.06V (75W) V3=35.35V (25W)--| | VF1 = 70.7V at zero degrees (100W) V1 = VF1(forward-tau) = 70.7(1.5) = 106.06V at zero degrees (75W) V3 = VF1(forward-rho) = 70.7(0.5) = 35.35V at zero degrees (25W) Note that PF1 = 100W = P1 + P3 = 75W + 25W ----------------------------------------------------------------- Breaking VR2 down into its two superposition components yields. x | |-- VR2=70.7V (33.33W) V4=35.35V (25W)--| |-- V2=35.35V (8.33W) | VR2 = 70.7V at 180 degrees (33.33W) V2 = VR2(reverse-rho) = 70.7(-0.5) = 35.35V at zero degrees (8.33W) V4 = VR2(reverse-tau) = 70.7(0.5) = 35.35V at 180 degrees (25W) Note that PR2 = 33.33W = P2 + P4 = 25W + 8.33W ----------------------------------------------------------------- Now, following the rules of superposition: To get the total forward voltage, add V1 + V2 VF2 = V1 + V2 = 106.06V + 35.35V = 141.4V (133.33W) To get the total reflected voltage, add V3 and V4 VR1 = V3 + V4 = 35.35V - 35.35V = zero volts (0W) Note: Dr. Best neglected to mention P3 and P4 in his QEX article. P3+P4 is the interference joules/sec. They are scalar values. All voltages are consistent and all powers are consistent. So, in this matched system, all reflected power is re-reflected: PF2 = 133.33W = P1 + P2 + P3 + P4 = 75W + 8.33W + 25W + 25W PF2 = P1 + P2 + (complete constructive interference) = 133.33W PR1 = P3 + P4 - (complete destructive interference) = zero watts Note that the voltage forward-rho = (150-50)/(150+50) = +0.5 reverse-rho = (50-150)/(50+150) = -0.5 (180 deg phase shift) -- 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! =----- |
| All times are GMT +1. The time now is 09:26 PM. |
Powered by vBulletin® Copyright ©2000 - 2025, Jelsoft Enterprises Ltd.
RadioBanter.com