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Walter Maxwell wrote:
Cecil, please read me in the first paragraph. By Steve's own words he says the re-reflected wave must equal the reflected wave. No, he says that is a fallacy. He says the re-reflected wave equals the reflected wave multiplied by the reflection coefficient. His example is: reflected power = 33.33W, re-reflected power = 8.33W re-reflected power = 33.33W(rho^2) = 33.33W(0.25) = 8.33W Here's what Dr. Best said in his QEX article, Part 3: "When two forward-traveling waves add, general superposition theory ... require(s) that the total forward traveling voltage be the vector sum of the individual forward-traveling voltages such that VFtotal = V1 + V2." He clearly implies that V2 is a forward-traveling wave and it is. Numerically, it is equal to the voltage reflected from the load multiplied by the reverse reflection coefficient. In S-parameter terms, V2 is the s22(a2) term. 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. Yes, 100W(0.75) is perfectly consistent with 70.7V(1.5)^2/150 where Z0=150 ohms 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. No, no, no. Mistaken interpretation. tau x input voltage = V1 (not the total forward voltage) rho x reflected voltage = V2 (re-reflected) V1 + V2 = forward voltage arriving at a mismatched load V1 is proportional to the S-parameter term, s21(a1) where s21 is the forward-transmission coefficient. V2 is proportional to the S-parameter term, s22(a2) where s22 is the reverse-reflection coefficient. b2 is the total forward voltage arriving at a mismatched load. VFtotal = V1 + V2 is virtually identical to b2 = 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 Mon, 07 Jun 2004 09:56:04 -0500, Cecil Moore wrote:
Walter Maxwell wrote: Cecil, please read me in the first paragraph. By Steve's own words he says the re-reflected wave must equal the reflected wave. No, he says that is a fallacy. He says the re-reflected wave equals the reflected wave multiplied by the reflection coefficient. His example is: reflected power = 33.33W, re-reflected power = 8.33W But when the system is matched the reflection coefficient is 1.0. re-reflected power = 33.33W(rho^2) = 33.33W(0.25) = 8.33W Here's what Dr. Best said in his QEX article, Part 3: "When two forward-traveling waves add, general superposition theory ... require(s) that the total forward traveling voltage be the vector sum of the individual forward-traveling voltages such that VFtotal = V1 + V2." He clearly implies that V2 is a forward-traveling wave and it is. Numerically, it is equal to the voltage reflected from the load multiplied by the reverse reflection coefficient. In S-parameter terms, V2 is the s22(a2) term. Cecil, it is clear that you are not reading my posts!!! You quoted Steve above, but I quoted the SAME quote earlier, explaining that he is WRONG. Please reread my quote. General superposition theory does NOT require that the forward voltage be the vector sum of the individual forward-traveling voltages. When are you going to understand that that superposition yields the standing wave, NOT the forward wave? I've told you this over and over again, but you apparently aren't listening. Cecil, please go to Johnson as I pointed out earlier and become educated as to where Steve screwed up. Walt |
Walter Maxwell wrote: On Mon, 07 Jun 2004 09:56:04 -0500, Cecil Moore wrote: Walter Maxwell wrote: Cecil, please read me in the first paragraph. By Steve's own words he says the re-reflected wave must equal the reflected wave. No, he says that is a fallacy. He says the re-reflected wave equals the reflected wave multiplied by the reflection coefficient. His example is: reflected power = 33.33W, re-reflected power = 8.33W But when the system is matched the reflection coefficient is 1.0. But if rho = 1.0, then tau = 2.0. Wasn't it 1.5 just a minute ago? 73, Jim AC6XG |
Walter Maxwell wrote:
But when the system is matched the reflection coefficient is 1.0. Not in an S-parameter analysis which is essentially what Dr. Best is doing. He is dealing with the PHYSICAL reflection coefficient, (Z2-Z1)/Z2+Z1), not the virtual reflection coefficient based on reflected power being zero. This may be the source of the misunderstanding between you two. His reflection coefficients, like the S-parameter reflection coefficients, don't change from startup to steady-state. They remain constant even when no signal is present. General superposition theory does NOT require that the forward voltage be the vector sum of the individual forward-traveling voltages. Superposition of two individual forward-traveling voltages requires that the sum be the vector sum as long as the interference energy requirements are met. When are you going to understand that that superposition yields the standing wave, NOT the forward wave? I've told you this over and over again, but you apparently aren't listening. I am listening, Walt, and still telling you that you are wrong about that concept. Superposition of two waves traveling in opposite directions yields the standing wave. Superposition of two coherent waves traveling in the same direction yields DESTRUCTIVE/CONSTRUCTIVE INTERFERENCE, not standing waves. Cecil, please go to Johnson as I pointed out earlier and become educated as to where Steve screwed up. I know where Steve screwed up, Walt, and he did screw up. But it wasn't in the area of an S-parameter analysis in the forward direction. It was in the rearward direction that he screwed up royally. In the afore-mentioned example, the four powers associated with the four superposition voltages are P1, P2, P3, and P4. P1=75W, P2=8.33W, P3=25W, and P4=25W. P1+P3 = 100W = generated power P2+P4 = 33.33W = power reflected from the load Dr. Best tell us that 75W+8.33W+2*SQRT(75*8.33)=133.33W and that is mathematically true. What Dr. Best doesn't tell us is that 2*SQRT(75*8.33) = P3+P4 and that they are the interference component powers from classical optics theory. You are grouping the powers together in this manner, (P1+P3)+(P2+P4)=133.33W and that is perfectly true mathematically. Both of you are doing essentially the same thing but neither one of you realizes it. You guys are two inches apart and don't know it. It feels like I am listening to two flatlanders argue about the third dimension. -- 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:
Walter Maxwell wrote: But when the system is matched the reflection coefficient is 1.0. But if rho = 1.0, then tau = 2.0. Wasn't it 1.5 just a minute ago? Dr. Best is using physical reflection coefficients, aka an S-parameter analysis, which are never 1.0 in a system with a mismatched load. -- 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 Mon, 07 Jun 2004 15:28:01 -0500, Cecil Moore wrote:
Walter Maxwell wrote: But when the system is matched the reflection coefficient is 1.0. Not in an S-parameter analysis which is essentially what Dr. Best is doing. He is dealing with the PHYSICAL reflection coefficient, (Z2-Z1)/Z2+Z1), not the virtual reflection coefficient based on reflected power being zero. This may be the source of the misunderstanding between you two. His reflection coefficients, like the S-parameter reflection coefficients, don't change from startup to steady-state. They remain constant even when no signal is present. General superposition theory does NOT require that the forward voltage be the vector sum of the individual forward-traveling voltages. Superposition of two individual forward-traveling voltages requires that the sum be the vector sum as long as the interference energy requirements are met. When are you going to understand that that superposition yields the standing wave, NOT the forward wave? I've told you this over and over again, but you apparently aren't listening. I am listening, Walt, and still telling you that you are wrong about that concept. Superposition of two waves traveling in opposite directions yields the standing wave. Superposition of two coherent waves traveling in the same direction yields DESTRUCTIVE/CONSTRUCTIVE INTERFERENCE, not standing waves. Cecil, please go to Johnson as I pointed out earlier and become educated as to where Steve screwed up. I know where Steve screwed up, Walt, and he did screw up. But it wasn't in the area of an S-parameter analysis in the forward direction. It was in the rearward direction that he screwed up royally. Cecil, you're not even close to knowing where Steve screwed up. And you won't know until you follow my directions and go to Johnson and read what I said to read. I can't understand why you refuse to go in that direction. Because you refuse we're going around in circles. This has to stop. What Johnson and I are trying to tell you is that the superposition of the source voltage and the re-reflected voltages DO NOT establish the forward voltage. Steve used Johnson's Eq for determining the standing wave voltage at any point on the line incorrectly as the forward re-reflected voltage wave V2. THIS IS WHERE HE SCREWED UP, CECIL., SCREWING UP THE ENTIRE ARTICLE !! Walt |
Walter Maxwell wrote:
What Johnson and I are trying to tell you is that the superposition of the source voltage and the re-reflected voltages DO NOT establish the forward voltage. If Johnson really said that, he screwed up because he is in direct contradiction to the S-parameter analysis described in Ramo, Whinnery, and Van Duzer in _Fields_and_Waves_in_Communications_Electronics_, (c) 1965, page 603, Section 11.09 "Scattering and Transmission Coefficients". You and Dr. Best obtain the same component powers. You just group them differently. You guys are two inches away from agreeing. Please provide the location of the Johnson quote details again. I sort by date and cannot find your Johnson reference even when sorting by threads. -- 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 Mon, 07 Jun 2004 16:52:15 -0500, Cecil Moore wrote:
Walter Maxwell wrote: What Johnson and I are trying to tell you is that the superposition of the source voltage and the re-reflected voltages DO NOT establish the forward voltage. If Johnson really said that, he screwed up because he is in direct contradiction to the S-parameter analysis described in Ramo, Whinnery, and Van Duzer in _Fields_and_Waves_in_Communications_Electronics_, (c) 1965, page 603, Section 11.09 "Scattering and Transmission Coefficients". No Cecil, Johnson didn't screw up. Incidentally, the equation in question is his Eq. 4.23, derived on Pages 98 and 99, and displayed on Page 100. Steve screwed up because, as I've been repeating, he misinterpreted the equation to determine the forward wave instead of the standing wave. From all of your statements so far, it appears you may have misunderstood what the forward wave really is. Here's a clue. When a tuner is properly adjusted in matching a 50-ohm line to a 150 + j0 resistance and the output voltage of the source is 70.71 v. the forward voltage is 70.71 x 1.1547 = 81.65 v. Do you recognize it or know where the 1.1547 came from ? Steve doesn't have a clue. Walt |
Cecil,
I am happy to see that we are in agreement on this long-standing argument. I explained this to Walt in some private e-mails about 18 months ago, and I wrote similar explanations in this group. I was vilified by both Walt and you. I will summarize once again. Both the Walter Maxwell model and the Steve Best model for this transmission line problem work correctly. They are internally self consistent, and they give the same physical answers. They obey the standard laws of physics and mathematics. However it is not possible to mix-and-match the models, using equations and definitions from one model in the other. Walt continues to mis-read Steve Best's QEX articles. He needs to throw away all of his pre-conceived definitions about the meaning of specific Vx components and read exactly what Steve wrote. 73, Gene W4SZ Cecil Moore wrote: You and Dr. Best obtain the same component powers. You just group them differently. You guys are two inches away from agreeing. |
Walter Maxwell wrote:
No Cecil, Johnson didn't screw up. Incidentally, the equation in question is his Eq. 4.23, derived on Pages 98 and 99, and displayed on Page 100. Steve screwed up because, as I've been repeating, he misinterpreted the equation to determine the forward wave instead of the standing wave. On page 99, in between equations 4.22 and 4.23 is an equation presented by Johnson that is close to Steve's equation to which you are objecting. From all of your statements so far, it appears you may have misunderstood what the forward wave really is. Here's a clue. When a tuner is properly adjusted in matching a 50-ohm line to a 150 + j0 resistance and the output voltage of the source is 70.71 v. the forward voltage is 70.71 x 1.1547 = 81.65 v. Do you recognize it or know where the 1.1547 came from ? Steve doesn't have a clue. The forward voltage wave equals s21(a1)+s22(a2) which is virtually the same as V1+V2 because V1/[SQRT(Z0)] + V2/[SQRT(Z0)] = s21(a1) + s22(a2). It's the old one-to-one correspondence rule. As I said before, Walt, the 1WL section of 50 ohm line in Steve's example is *completely irrelevant* to the discussion. Steve only included that line so the forward/reflected power could be measured and the impedance reproduced. As long as you allow that 1WL of 50 ohm line to enter into your thinking and calculations, you will never understand what he was trying to say. Please remove the 1WL of 50 ohm line and please re-think your position. All the conditions discussed by Steve occur at the *MATCH POINT* which, I think you will agree, is NOT on the 50 ohm line. -- 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! =----- |
Gene Fuller wrote:
I explained this to Walt in some private e-mails about 18 months ago, and I wrote similar explanations in this group. I was vilified by both Walt and you. I still believe that neither model is perfectly consistent. I do believe that an S-parameter analysis is perfectly consistent. Both the Walter Maxwell model and the Steve Best model for this transmission line problem work correctly. They are internally self consistent, and they give the same physical answers. But how can Steve deny the 100% re-reflected energy premise when all the energy winds up at the load? How did it get to the load if it was not re-reflected at the match point? TV ghosting proves that it is indeed re-reflected at the match point. -- 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 Mon, 07 Jun 2004 23:40:03 GMT, Gene Fuller wrote:
Cecil, I am happy to see that we are in agreement on this long-standing argument. I explained this to Walt in some private e-mails about 18 months ago, and I wrote similar explanations in this group. I was vilified by both Walt and you. I will summarize once again. Both the Walter Maxwell model and the Steve Best model for this transmission line problem work correctly. They are internally self consistent, and they give the same physical answers. They obey the standard laws of physics and mathematics. However it is not possible to mix-and-match the models, using equations and definitions from one model in the other. Walt continues to mis-read Steve Best's QEX articles. He needs to throw away all of his pre-conceived definitions about the meaning of specific Vx components and read exactly what Steve wrote. 73, Gene W4SZ Hello again Gene, What Steve wrote exactly, Gene, and I have read it exactly, over and over again, is based entirely on an invalid premise that you and all the rest who, like you, have expressed the notion that Steve's paper is the 'most brilliant and definitive paper ever written on the subject. Unfortunately, it's both ironic and tragic that a person so brilliant a mathematician, and knowledgeable in transmission line technique, should make such a grievous error in formulating the basis for his entire paper. It's also tragic that all of you who think his article is so great also made the same error in not recognizing the problem in the First Part of the article, his Eqs 6, 7, and 8, because they are invalid. Steve used an equation in Johnson's "Transmission Lines and Networks" as the foundation for the entire paper. But unfortunately he misunderstood the meaning of the equation. He misunderstood it as early as 1998 when I pointed out why he misunderstands it, but he denied that he misunderstands it , stating that it is I who misunderstands it. Not a chance. The equation he misunderstands is Eq 4.23 on Page 100 in Johnson, which is explained and derived on Pages 98 and 99, to derive the voltage E of the standing wave for any position along a mismatched transmission line. Steve's misunderstanding is that he believes the equation expresses the value of the forward voltage on the line. Consequently, his Eq 6 in Part 1 says that 'Vfwd = the terms on the right-hand side of the equation copied from Johnson', while the correct version is 'E = the terms on the righ-hand side'. The difference between 'E' and Vfwd is so significantly different as to be unbelievable that he, of all people, used it for this purpose, and it is also unbelievable and shameful that you experienced engineers failed to recognize, not only the error in Steve's equation, but the ramifications it had on the domino effect on the remainder of his paper, especially making his equations 9 thru 15 in Part 3 invalid for general use. I'd like for you and Cecil, and any other reader who believes Steve's paper is correct, to rethink the definitions of the standing wave and the forward voltage with the view toward another review of the paper. Then make that review and see if you still believe the material appearing there is correct. If you then still believe the paper is correct I'd like for you to let me know the technical reason why you believe it's correct so that I can discuss it with you. It's really worth that effort--don't just blow it off as another rant and rave from Walt Maxwell. Thanks for listening to what I think is an important issue. Walt |
Good luck. There are a few "engineers" here I would let touch my toaster.
tom K0TAR Walter Maxwell wrote: On Mon, 07 Jun 2004 23:40:03 GMT, Gene Fuller wrote: Cecil, I am happy to see that we are in agreement on this long-standing argument. I explained this to Walt in some private e-mails about 18 months ago, and I wrote similar explanations in this group. I was vilified by both Walt and you. I will summarize once again. Both the Walter Maxwell model and the Steve Best model for this transmission line problem work correctly. They are internally self consistent, and they give the same physical answers. They obey the standard laws of physics and mathematics. However it is not possible to mix-and-match the models, using equations and definitions from one model in the other. Walt continues to mis-read Steve Best's QEX articles. He needs to throw away all of his pre-conceived definitions about the meaning of specific Vx components and read exactly what Steve wrote. 73, Gene W4SZ Hello again Gene, What Steve wrote exactly, Gene, and I have read it exactly, over and over again, is based entirely on an invalid premise that you and all the rest who, like you, have expressed the notion that Steve's paper is the 'most brilliant and definitive paper ever written on the subject. Unfortunately, it's both ironic and tragic that a person so brilliant a mathematician, and knowledgeable in transmission line technique, should make such a grievous error in formulating the basis for his entire paper. It's also tragic that all of you who think his article is so great also made the same error in not recognizing the problem in the First Part of the article, his Eqs 6, 7, and 8, because they are invalid. Steve used an equation in Johnson's "Transmission Lines and Networks" as the foundation for the entire paper. But unfortunately he misunderstood the meaning of the equation. He misunderstood it as early as 1998 when I pointed out why he misunderstands it, but he denied that he misunderstands it , stating that it is I who misunderstands it. Not a chance. The equation he misunderstands is Eq 4.23 on Page 100 in Johnson, which is explained and derived on Pages 98 and 99, to derive the voltage E of the standing wave for any position along a mismatched transmission line. Steve's misunderstanding is that he believes the equation expresses the value of the forward voltage on the line. Consequently, his Eq 6 in Part 1 says that 'Vfwd = the terms on the right-hand side of the equation copied from Johnson', while the correct version is 'E = the terms on the righ-hand side'. The difference between 'E' and Vfwd is so significantly different as to be unbelievable that he, of all people, used it for this purpose, and it is also unbelievable and shameful that you experienced engineers failed to recognize, not only the error in Steve's equation, but the ramifications it had on the domino effect on the remainder of his paper, especially making his equations 9 thru 15 in Part 3 invalid for general use. I'd like for you and Cecil, and any other reader who believes Steve's paper is correct, to rethink the definitions of the standing wave and the forward voltage with the view toward another review of the paper. Then make that review and see if you still believe the material appearing there is correct. If you then still believe the paper is correct I'd like for you to let me know the technical reason why you believe it's correct so that I can discuss it with you. It's really worth that effort--don't just blow it off as another rant and rave from Walt Maxwell. Thanks for listening to what I think is an important issue. Walt |
Walter Maxwell wrote:
The equation he misunderstands is Eq 4.23 on Page 100 in Johnson, which is explained and derived on Pages 98 and 99, to derive the voltage E of the standing wave for any position along a mismatched transmission line. Steve's misunderstanding is that he believes the equation expresses the value of the forward voltage on the line. Consequently, his Eq 6 in Part 1 says that 'Vfwd = the terms on the right-hand side of the equation copied from Johnson', while the correct version is 'E = the terms on the righ-hand side'. Johnson's equation contains all possible components - both forward and reflected, so Johnson's equation predicts the total net standing-wave voltage. (Note that Johnson's equation contains both positive and negative exponents.) Dr. Best's equation contains only the forward components of Johnson's equation. Therefore, Dr. Best's equation predicts *only* the forward- traveling voltage. (Note that Dr. Best's equation contains only negative exponents and represents the E+ half of Johnson's equation while omitting the E- half of Johnson's equation.) In other words, the two equations do *not* predict the same quantity and they are *not* supposed to be the same equations. As I said before, Dr. Best made a lot of conceptual errors but his equations seem to be valid. If you break Johnson's equation into two parts representing E+ and E-, the E+ part will be Dr. Best's Eq 6. Please consider the following matched system with a 1/2 second long lossless transmission line. The physical rho is 0.5. Assume VF1 is a constant 70.7V. 100W XMTR---50 ohm line---x---1/2 second long lossless 150 ohm line---50 ohm load VF1=70.7V-- VF2-- --VR1 --VR2 V1 is equal to 70.7(1.5) = 106.06V and is a constant value V2 starts out at zero and builds up to 35.35V VF2=V1+V2 assuming they are in phase Here is (conceptually simplified) how VF2 builds up to its steady-state value. VR2(t+1) is 1/2 of VF2(t) and V2(t+1) is 1/4 of VF2(t) Time in Seconds V1 VR2 V2 VF2 0 106.06 0 0 106.06 1 106.06 53.03 26.5 132.56 2 106.06 66.26 33.13 139.19 3 106.06 69.58 34.79 140.85 4 106.06 70.4 35.2 141.26 5 106.06 70.63 35.32 141.38 6 106.06 70.69 35.34 141.39 7 106.06 70.7 35.35 141.4 very close to steady-state My rounding is a little off but above is a close approximation to what happens. Note that VF2 is equal to V1 + V2 and converges on the proper value of 141.4V. Steady-state VF2 = 106.06 + 26.5 + 6.63 + 1.66 + 0.41 + ... This is indeed of the form 106.06(1 + 1/4 + 1/16 + 1/64 + 1/256 + ...) or VF2 = V1(1 + A + A^2 + A^3 + A^4 + ...) -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
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Hash: SHA1 "Tom" == Tom Ring writes: Tom Good luck. There are a few "engineers" here I would let touch my Tom toaster. tom K0TAR I had a similar thought, but it involved a bathtub. Jack. - -- Jack Twilley jmt at twilley dot org http colon slash slash www dot twilley dot org slash tilde jmt slash -----BEGIN PGP SIGNATURE----- Version: GnuPG v1.2.4 (FreeBSD) iD8DBQFAxTnMGPFSfAB/ezgRAsQBAKDC+1XNvN/fBNwTD9Naxm7wimVOVwCg8KJu Zjl+lNxr+FMpZCv1C2GftGg= =nAYU -----END PGP SIGNATURE----- |
On Mon, 07 Jun 2004 22:22:49 -0500, Cecil Moore wrote:
Walter Maxwell wrote: The equation he misunderstands is Eq 4.23 on Page 100 in Johnson, which is explained and derived on Pages 98 and 99, to derive the voltage E of the standing wave for any position along a mismatched transmission line. Steve's misunderstanding is that he believes the equation expresses the value of the forward voltage on the line. Consequently, his Eq 6 in Part 1 says that 'Vfwd = the terms on the right-hand side of the equation copied from Johnson', while the correct version is 'E = the terms on the righ-hand side'. Johnson's equation contains all possible components - both forward and reflected, so Johnson's equation predicts the total net standing-wave voltage. (Note that Johnson's equation contains both positive and negative exponents.) Dr. Best's equation contains only the forward components of Johnson's equation. Therefore, Dr. Best's equation predicts *only* the forward- traveling voltage. (Note that Dr. Best's equation contains only negative exponents and represents the E+ half of Johnson's equation while omitting the E- half of Johnson's equation.) In other words, the two equations do *not* predict the same quantity and they are *not* supposed to be the same equations. As I said before, Dr. Best made a lot of conceptual errors but his equations seem to be valid. If you break Johnson's equation into two parts representing E+ and E-, the E+ part will be Dr. Best's Eq 6. Please consider the following matched system with a 1/2 second long lossless transmission line. The physical rho is 0.5. Assume VF1 is a constant 70.7V. 100W XMTR---50 ohm line---x---1/2 second long lossless 150 ohm line---50 ohm load VF1=70.7V-- VF2-- --VR1 --VR2 V1 is equal to 70.7(1.5) = 106.06V and is a constant value V2 starts out at zero and builds up to 35.35V VF2=V1+V2 assuming they are in phase Here is (conceptually simplified) how VF2 builds up to its steady-state value. VR2(t+1) is 1/2 of VF2(t) and V2(t+1) is 1/4 of VF2(t) Time in Seconds V1 VR2 V2 VF2 0 106.06 0 0 106.06 1 106.06 53.03 26.5 132.56 2 106.06 66.26 33.13 139.19 3 106.06 69.58 34.79 140.85 4 106.06 70.4 35.2 141.26 5 106.06 70.63 35.32 141.38 6 106.06 70.69 35.34 141.39 7 106.06 70.7 35.35 141.4 very close to steady-state My rounding is a little off but above is a close approximation to what happens. Note that VF2 is equal to V1 + V2 and converges on the proper value of 141.4V. Sorry to spoil your fun, Cecil, but 141.4 v is not the forward voltage, it's the max voltage of the standing wave. This is the same mistake that Steve made. I asked you to rethink what the forward voltage and you have come up wrong. You are using circuit theory for superposition, as Steve did, when circuit theory fails to apply in certain transmission line cases. This is one of those cases. In circuit theory you can superpose the voltages from two sources and V1 + V2 equals Vtotal. But the re-reflected voltage CANNOT be added to the source voltage in the transmission line case to obtain Vfwd, because there is only ONE source. I'll say it again, Cecil, V1 + V2 = maxV of the standing wave--V1 + V2 does NOT equal Vfwd because V2 is not a second source, it came fr om ONE source, the transceiver, and therefore superposition of V1 and V2 does not apply to establish Vfwd. Therefore, it is true that maxV may be incident on the load if the relative phase between the reflected and forward waves permits, The only time the forward wave is incident on the load is when the load = Zo. Please, Cecil, go back to the drawing board and come up with the correct Vfwd--it is not equal in any way to the right-hand side of Johnson's Eq 4.23;. Remember, I said earlier that when rho = 0.5 and the circuit is matched, Vfwd = Vsource x 1.1547. When you discover where the 1.1547 came from when rho = 0.5 you will have discovered the source of Vfwd. Steady-state VF2 = 106.06 + 26.5 + 6.63 + 1.66 + 0.41 + ... This is indeed of the form 106.06(1 + 1/4 + 1/16 + 1/64 + 1/256 + ...) or VF2 = V1(1 + A + A^2 + A^3 + A^4 + ...) Wrong, Cecil. Change VF2 to 'E', the max value of the standing wave, and the values obtained from the sum of the terms in the geometric series will be correct. Walt |
On Tue, 08 Jun 2004 15:54:50 GMT, Walter Maxwell wrote:
On Mon, 07 Jun 2004 22:22:49 -0500, Cecil Moore wrote: Walter Maxwell wrote: The equation he misunderstands is Eq 4.23 on Page 100 in Johnson, which is explained and derived on Pages 98 and 99, to derive the voltage E of the standing wave for any position along a mismatched transmission line. Steve's misunderstanding is that he believes the equation expresses the value of the forward voltage on the line. Consequently, his Eq 6 in Part 1 says that 'Vfwd = the terms on the right-hand side of the equation copied from Johnson', while the correct version is 'E = the terms on the righ-hand side'. snip Sorry to spoil your fun, Cecil, but 141.4 v is not the forward voltage, it's the max voltage of the standing wave. This is the same mistake that Steve made. I asked you to rethink what the forward voltage and you have come up wrong. You are using circuit theory for superposition, as Steve did, when circuit theory fails to apply in certain transmission line cases. This is one of those cases. In circuit theory you can superpose the voltages from two sources and V1 + V2 equals Vtotal. But the re-reflected voltage CANNOT be added to the source voltage in the transmission line case to obtain Vfwd, because there is only ONE source. I'll say it again, Cecil, V1 + V2 = maxV of the standing wave--V1 + V2 does NOT equal Vfwd because V2 is not a second source, it came fr om ONE source, the transceiver, and therefore superposition of V1 and V2 does not apply to establish Vfwd. I'll make it a little easier for you, Cecil. I' ll give you two ways to determine Vfwd. Way 1. Vfwd = sqrt(Power fwd x Zo) Way 2. Vfwd = Vsource x sqrt[1/(1- rho^2)] Now plug rho = 0.5 into the expression for Way 2 and see what comes up. Walt Therefore, it is true that maxV may be incident on the load if the relative phase between the reflected and forward waves permits, The only time the forward wave is incident on the load is when the load = Zo. Please, Cecil, go back to the drawing board and come up with the correct Vfwd--it is not equal in any way to the right-hand side of Johnson's Eq 4.23;. Remember, I said earlier that when rho = 0.5 and the circuit is matched, Vfwd = Vsource x 1.1547. When you discover where the 1.1547 came from when rho = 0.5 you will have discovered the source of Vfwd. Steady-state VF2 = 106.06 + 26.5 + 6.63 + 1.66 + 0.41 + ... This is indeed of the form 106.06(1 + 1/4 + 1/16 + 1/64 + 1/256 + ...) or VF2 = V1(1 + A + A^2 + A^3 + A^4 + ...) Wrong, Cecil. Change VF2 to 'E', the max value of the standing wave, and the values obtained from the sum of the terms in the geometric series will be correct. Walt |
I'll make it a little easier for you, Cecil. I' ll give you two ways to determine Vfwd. Way 1. Vfwd = sqrt(Power fwd x Zo) Way 2. Vfwd = Vsource x sqrt[1/(1- rho^2)] Now plug rho = 0.5 into the expression for Way 2 and see what comes up. Walt Cecil, let's take it one step further and include current, forward current, that is, Ifwd. Ifwd = Isouce x sqrt[1/(1 - rho^2)] Now let's see what happens with a souce voltage of 70.71 v at 100 w on a 50-ohm line. The source current is 1.414 a. Now find Ifwd x Vfwd and see what comes up. Walt |
Walter Maxwell wrote:
On Mon, 07 Jun 2004 22:22:49 -0500, Cecil Moore wrote: Walter Maxwell wrote: The equation he misunderstands is Eq 4.23 on Page 100 in Johnson, which is explained and derived on Pages 98 and 99, to derive the voltage E of the standing wave for any position along a mismatched transmission line. Steve's misunderstanding is that he believes the equation expresses the value of the forward voltage on the line. Consequently, his Eq 6 in Part 1 says that 'Vfwd = the terms on the right-hand side of the equation copied from Johnson', while the correct version is 'E = the terms on the righ-hand side'. Johnson's equation contains all possible components - both forward and reflected, so Johnson's equation predicts the total net standing-wave voltage. (Note that Johnson's equation contains both positive and negative exponents.) Dr. Best's equation contains only the forward components of Johnson's equation. Therefore, Dr. Best's equation predicts *only* the forward- traveling voltage. (Note that Dr. Best's equation contains only negative exponents and represents the E+ half of Johnson's equation while omitting the E- half of Johnson's equation.) In other words, the two equations do *not* predict the same quantity and they are *not* supposed to be the same equations. As I said before, Dr. Best made a lot of conceptual errors but his equations seem to be valid. If you break Johnson's equation into two parts representing E+ and E-, the E+ part will be Dr. Best's Eq 6. Please consider the following matched system with a 1/2 second long lossless transmission line. The physical rho is 0.5. Assume VF1 is a constant 70.7V. 100W XMTR---50 ohm line---x---1/2 second long lossless 150 ohm line---50 ohm load VF1=70.7V-- VF2-- --VR1 --VR2 V1 is equal to 70.7(1.5) = 106.06V and is a constant value V2 starts out at zero and builds up to 35.35V VF2=V1+V2 assuming they are in phase Here is (conceptually simplified) how VF2 builds up to its steady-state value. VR2(t+1) is 1/2 of VF2(t) and V2(t+1) is 1/4 of VF2(t) Time in Seconds V1 VR2 V2 VF2 0 106.06 0 0 106.06 1 106.06 53.03 26.5 132.56 2 106.06 66.26 33.13 139.19 3 106.06 69.58 34.79 140.85 4 106.06 70.4 35.2 141.26 5 106.06 70.63 35.32 141.38 6 106.06 70.69 35.34 141.39 7 106.06 70.7 35.35 141.4 very close to steady-state My rounding is a little off but above is a close approximation to what happens. Note that VF2 is equal to V1 + V2 and converges on the proper value of 141.4V. Sorry to spoil your fun, Cecil, but 141.4 v is not the forward voltage, it's the max voltage of the standing wave. This is the same mistake that Steve made. I asked you to rethink what the forward voltage and you have come up wrong. Nope, I haven't, Walt. The maximum voltage of the standing wave equals the maximum forward voltage plus the maximum reflected voltage. The forward voltage is 141.4V. The reflected voltage is 70.7V. The maximum standing-wave voltage is 141.4V + 70.7V = 212.1V. The minimum standing-wave voltage is 141.4V - 70.7V = 70.7V. The VSWR = 212.1V/70.7V = 3:1. Rho = 0.5 SWR = (1+Rho)/(1-Rho) = 1.5/0.5 = 3:1 Everything is perfectly consistent. So the maximum standing-wave voltage is 212.1V, not 141.4V. Please reconsider - here's the steady-state powers for the above matched example: 100W XMTR-----50 ohm line---x---150 ohm line---50 ohm load PF1=100W-- PF2=133.33W-- --PR1=0 --PR2=33.33W The forward power is 133.33W and Z0=150 ohms. Last time I looked, the square root of [133.33W(150 ohms)] was 141.4V. 141.4V^2/150 ohms equals 133.33W. The forward voltage is indeed 141.4V. Until you discover your mental block, whatever it is, this discussion is not going to progress. The above example is exceptionally simple so your mistake must also be simple. You, yourself, have used the above example and always agreed that 100W + 33.33W = 133.33W, i.e. forward power equals generated power plus reflected 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! =----- |
Walter Maxwell wrote:
I'll make it a little easier for you, Cecil. I' ll give you two ways to determine Vfwd. OK, here's the example: 100W XMTR---50 ohm line---x----1/2WL 150 ohm line---50 ohm load PF1=100W-- PF2=133.33W-- --PR1=0W --PR2=33.33W Way 1. Vfwd = sqrt(Power fwd x Zo) So Vfwd = sqrt(133.33W x 150ohms) = 141.4 volts (that's what I said) Way 2. Vfwd = Vsource x sqrt[1/(1- rho^2)] Now plug rho = 0.5 into the expression for Way 2 and see what comes up. OK, Vfwd = 70.7V x sqrt[1/0.75] = 81.6 volts (something wrong there) We know Vfwd has to be 141.4V as calculated above using Ohm's law. Walt, your two ways don't yield the same value. Perhaps that is the source of your confusion about what Dr. Best said. Your Way 2 appears to be a calculation of the voltage delivered to a mismatched load. -- 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:
Cecil, let's take it one step further and include current, forward current, that is, Ifwd. Ifwd = Isouce x sqrt[1/(1 - rho^2)] This equation for voltage was wrong. Perhaps, we should resolve that problem first. Forward current can be calculated from the square root of (forward power divided by Z0). So IF2 = SQRT(133.33W/150ohms) = 0.943 amps Funny thing is that 141.4v*0.943a = 133.33 watts, the known forward power on the 150 ohm line. -- 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, W2DU wrote:
"I`ll give you two ways to determine Vfwd." Walter makes it too easy by using the power relationships, but that`s the way it is. The forward (incident) wave is opposed by Ro in a practical line. The reward (reflected) wave is also opposed by Ro (the surge impedance) of the line. Power from the transmitter is nearly the same as that delivered to the load as loss is small and no significant room exists in the line to store RF. Power to the load is the difference between forward power and reflected power. The voltage at any point on a mismatched line is the sum of the forward and reflected waves at that point but is merely a manifestatoion of SWR and has little practical value. At a current or a voltage loop (maximum), the forward and reflected amps or volts are in-phase. So, if a line is opened at a loop point, the impedance looking toward the load is a pure resistance regardless of the nature of the load (see page 37 of "Transmission Lines, Antennas, and Wave Guides" by King, Mimno, and Wing). Best regards, Richard Harrison, KB5WZI |
Richard Harrison wrote:
The voltage at any point on a mismatched line is the sum of the forward and reflected waves at that point but is merely a manifestatoion of SWR and has little practical value. All true and completely irrelevant to Dr. Best's article since he was *never* talking about the net voltage. He is talking about the forward voltage. Any notion that Dr. Best was talking about net voltage is a mistaken notion. (I can't believe I'm defending Dr. Best but he said what he said.) -- 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 Tue, 08 Jun 2004 13:29:02 -0500, Cecil Moore wrote:
Walter Maxwell wrote: I'll make it a little easier for you, Cecil. I' ll give you two ways to determine Vfwd. OK, here's the example: 100W XMTR---50 ohm line---x----1/2WL 150 ohm line---50 ohm load PF1=100W-- PF2=133.33W-- --PR1=0W --PR2=33.33W Way 1. Vfwd = sqrt(Power fwd x Zo) So Vfwd = sqrt(133.33W x 150ohms) = 141.4 volts (that's what I said) Way 2. Vfwd = Vsource x sqrt[1/(1- rho^2)] Now plug rho = 0.5 into the expression for Way 2 and see what comes up. OK, Vfwd = 70.7V x sqrt[1/0.75] = 81.6 volts (something wrong there) We know Vfwd has to be 141.4V as calculated above using Ohm's law. Walt, your two ways don't yield the same value. Perhaps that is the source of your confusion about what Dr. Best said. Your Way 2 appears to be a calculation of the voltage delivered to a mismatched load. Cecil, you've hit the problem on the nose. Until a few posts past I didn't realize you were using a 150-ohm line. I've been using 50 ohms for the line Zo terminated with 150 ohms, and you'll find my numbers are correct for that Zo. Vfwd = 81.6 v is correct for 100 w on a matched 50 ohm line with 3:1 mismatched load. You'll also get 81.6 using my Way 1. However, on a 50-ohm line with a 3:1 mismatch and Vfwd = 81.6v (actually it's 81.65 v), the reflected wave is 40.83 v. When the system is matched the phase of the re-reflected wave is 0° and is also 40.83 v. Therefore, V1 = 81.65 v and V2 = 40.83,v, which means V1 + V2 cannot equal Vfwd. Yet, in the paragraph preceding Steve's Eq 13 he says that when the phase is 0° Eq. 13 derives PFtotal. It does not. Using his Eq 7 V1 yields P1 = 133.33 w, and using his Eq 8 V2 P2 = = 33.33 w. Using these values of P1 and P2 in his Eq. 12 yields PFtotal = 238 watts, which is also what we get when plugging V1 and V2 into his Eq 9. We know 238 watts is incorrect, because it need to be 133.33 w. It makes no sense to plug P1 from Eq 7 into Eq 12 to find PFtotal, because P1 is already PF total. This clearly proves that Eq 12 is invalid, because Eq 7 and 8 are valid. I am studying your case with the 150-ohm line, but I haven't yet corralled it. Walt re-reflected voltage is also 40.83 v. |
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On Tue, 08 Jun 2004 12:58:17 -0500, Cecil Moore wrote:
Walter Maxwell wrote: On Mon, 07 Jun 2004 22:22:49 -0500, Cecil Moore wrote: snip. Please consider the following matched system with a 1/2 second long lossless transmission line. The physical rho is 0.5. Assume VF1 is a constant 70.7V. 100W XMTR---50 ohm line---x---1/2 second long lossless 150 ohm line---50 ohm load VF1=70.7V-- VF2-- --VR1 --VR2 V1 is equal to 70.7(1.5) = 106.06V and is a constant value V2 starts out at zero and builds up to 35.35V VF2=V1+V2 assuming they are in phase Here is (conceptually simplified) how VF2 builds up to its steady-state value. VR2(t+1) is 1/2 of VF2(t) and V2(t+1) is 1/4 of VF2(t) Time in Seconds V1 VR2 V2 VF2 0 106.06 0 0 106.06 1 106.06 53.03 26.5 132.56 2 106.06 66.26 33.13 139.19 3 106.06 69.58 34.79 140.85 4 106.06 70.4 35.2 141.26 5 106.06 70.63 35.32 141.38 6 106.06 70.69 35.34 141.39 7 106.06 70.7 35.35 141.4 very close to steady-state My rounding is a little off but above is a close approximation to what happens. Note that VF2 is equal to V1 + V2 and converges on the proper value of 141.4V. Sorry to spoil your fun, Cecil, but 141.4 v is not the forward voltage, it's the max voltage of the standing wave. This is the same mistake that Steve made. I asked you to rethink what the forward voltage and you have come up wrong. OK Cecil, I now understand why 141.4 v is the forward voltage. But there is more below. Nope, I haven't, Walt. The maximum voltage of the standing wave equals the maximum forward voltage plus the maximum reflected voltage. The forward voltage is 141.4V. The reflected voltage is 70.7V. The maximum standing-wave voltage is 141.4V + 70.7V = 212.1V. The minimum standing-wave voltage is 141.4V - 70.7V = 70.7V. The VSWR = 212.1V/70.7V = 3:1. Rho = 0.5 SWR = (1+Rho)/(1-Rho) = 1.5/0.5 = 3:1 Everything is perfectly consistent. So the maximum standing-wave voltage is 212.1V, not 141.4V. And yes, Cecil I now concur that the standing wave voltage is 212.1 v. There is still more. Please reconsider - here's the steady-state powers for the above matched example: 100W XMTR-----50 ohm line---x---150 ohm line---50 ohm load PF1=100W-- PF2=133.33W-- --PR1=0 --PR2=33.33W The forward power is 133.33W and Z0=150 ohms. Last time I looked, the square root of [133.33W(150 ohms)] was 141.4V. 141.4V^2/150 ohms equals 133.33W. The forward voltage is indeed 141.4V. Yep, I agree, but as I said above, there is still more. Until you discover your mental block, whatever it is, this discussion is not going to progress. The above example is exceptionally simple so your mistake must also be simple. You, yourself, have used the above example and always agreed that 100W + 33.33W = 133.33W, i.e. forward power equals generated power plus reflected power. I assure you, Cecil, there is no mental block. Returning to Steve's paper, and as I said in the preceding post, Steve precedes his Eq 13 stating this equation works under the conditions where V1 and V2 are in phase. This means that the system is matched and thus the re-reflected voltage equals the reflected voltage. With forward voltage V1 = 141.4 and reflected voltage V2 = 70.71, and using these values in his Eqs 7 and 8 and then plugging the resulting values of P1 and P2 in Eq 13, we get PFtotal = 300 w. This again proves that Eq 13 is invalid, because P1 from Eq 7 already derived the total forward power, yet he uses P2 in Eq 13 to obtain the total forward power. Do you not see what's wrong here? Walt |
Walter Maxwell wrote:
Cecil, you've hit the problem on the nose. Until a few posts past I didn't realize you were using a 150-ohm line. I've only explained and presented it about ten times now. The impedances are irrelevant, Walt. The equations have to work no matter what the impedances. I have presented ASCII schematics of what I was discussing. Would you be so kind as to do likewise? I am getting sick and tired of this run-around. -- 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:
This means that the system is matched and thus the re-reflected voltage equals the reflected voltage. No! NO! NO! The voltage re-reflected from any physical impedance discontinuity equals the reflected voltage times the physical reflection coefficient. That is the S-parameter term s22(a2). s22 is the physical reflection coefficient. a2 is the reflected voltage incident upon port 2. If the reflected voltage is 70.7V and the physical voltage reflection coefficient is 0.5, then the re-reflected voltage is 35.35V. This is all explained in HP's AN 95-1 on 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 Tue, 08 Jun 2004 16:43:04 -0500, Cecil Moore
wrote: Walter Maxwell wrote: This means that the system is matched and thus the re-reflected voltage equals the reflected voltage. No! NO! NO! The voltage re-reflected from any physical impedance discontinuity equals the reflected voltage times the physical reflection coefficient. That is the S-parameter term s22(a2). s22 is the physical reflection coefficient. a2 is the reflected voltage incident upon port 2. If the reflected voltage is 70.7V and the physical voltage reflection coefficient is 0.5, then the re-reflected voltage is 35.35V. This is all explained in HP's AN 95-1 on S-parameter analysis. Cecil, when a mismatched line is matched at the input with a tuner the reflection coefficient at the matching point in the tuner is 1.0 when the tuner is matched, thus the re-reflected voltage wave must be equal to that of the reflected wave. My statements are all based in these conditions. These conditions are in effect when using an antenna tuner, so any equations that work in general must work with the tuner setup. When the tuner is matched the re-reflected voltage is in phase with the source wave. Then, according to the conditions Steve set forth to use Eq 13 this equation should be viable for use with conditions set up with the tuner, but it doesn't. You say this equation is valid, so can you come up with the correct answers using it with a tuner matching the mismatched line? Unfortunately, my S-parameter texts are behind in Florida, and I've forgotten the significance of the 'a' and 'b' terms. |
Walter Maxwell wrote:
Cecil, when a mismatched line is matched at the input with a tuner the reflection coefficient at the matching point in the tuner is 1.0 when the tuner is matched, thus the re-reflected voltage wave must be equal to that of the reflected wave. My statements are all based in these conditions. That is NOT true of an S-parameter analysis, Walt, which is essentially what Dr. Best is using in his article. In a system with reflections the S-parameter reflection coefficients are NEVER equal to 1.0. How can I make you understand that you and Dr. Best are using entirely different systems of analysis and yours has no bearing on his. Here's your arguments: Steve: "It's a plant." Walt: "No, it's a tree." Steve: "No, no, it's a plant." Walt: "No, no, it's a tree." In the following matched system, the reflection coefficient in Dr. Best's system of analysis is ALWAYS |0.5| and is NEVER 1.0. This is also true for an S-parameter analysis. The only difference between Dr. Best's analysis and an S-parameter analysis is that he didn't normalize his voltages to SQRT(Z0). I presented that fact a couple of postings ago. 100W XMTR---50 ohm line---x---1/2WL 150 ohm line---50 ohm load. Nowhere at no time is Dr. Best's reflection coefficient equal to anything except |0.5|. It is an absolute constant whether the XMTR is off, in the transient state, or in the steady-state. Please read this paragraph over and over until it soaks in. In Dr. Best's system of analysis, the only time that the reflection coefficient is 1.0 is in a system with a short, open, or pure reactance. In his system of analysis, a reflection coefficient of 1.0 NEVER occurs at a matched or mismatched impedance discontinuity if the impedances are not zero or infinite (or purely reactive). Unfortunately, my S-parameter texts are behind in Florida, and I've forgotten the significance of the 'a' and 'b' terms. The HP AN 95-1 ap note is available at: http://www.sss-mag.com/pdf/hpan95-1.pdf The S-parameter reflection coefficients are constant and do NOT change as the reflected power changes. They are the same with no signal applied, or during the transient state, or during the steady-state. The S-parameter reflection coefficient is *NEVER* 1.0 when there exists an impedance discontinuity. You and Dr. Best are NOT USING THE SAME REFLECTION COEFFICIENTS!!!! You and Dr. Best are NOT USING THE SAME SYSTEM OF ANALYSIS!!!! Nothing you say about your system of analysis is relevant to the system of analysis that Dr. Best used. -- 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, Steve's equations are out there for all to see as general equations.
Nowhere has he made any caviats to the contrary, or that they only work for reflection coefficients of 0.5. If they only work for one specific coefficient it should be so noted that they are not for general use. And if equations are not valid for general use they are then invalid. Can you explain this? Walt |
Walter Maxwell wrote:
Cecil, Steve's equations are out there for all to see as general equations. Nowhere has he made any caviats to the contrary, or that they only work for reflection coefficients of 0.5. If they only work for one specific coefficient it should be so noted that they are not for general use. And if equations are not valid for general use they are then invalid. Can you explain this? They do work for any reflection coefficient, Walt. He chose 0.5 for his examples because a 0.5 voltage reflection coefficient reflects half of the voltage and that's easy to do in your head. The S-parameter analysis will work for any reflection coefficient. If you will take the time to perform an S-parameter analysis, you will understand Steve's equations. His analysis is completely different from yours. Here's one with a voltage reflection coefficient of 0.707. You can pick any Z0 for the 1/2WL line below and therefore choose any reflection coefficient you want. 100W XMTR---50 ohm line---x---1/2WL 291.5 ohm line---50 ohm load VF1=70.7V-- VF2=241.45V-- --VR1=0V --VR2=170.7V Dr. Best's V1 = VF1(1+rho) = 70.7(1.707) = 120.7V Dr. Best's V2 = VR2(rho) = 170.7(0.707) = 120.7V VF2 = V1 + V2 = 120.7V + 120.7V = 241.4V (200W forward) This is just an S-parameter analysis without normalizing to the square root of Z0. In fact, we can *derive* Dr. Best's equations from the S-parameter equations simply by multiplying by the square root of Z0. b2 = s21(a1) + s22(a2) b2*Sqrt(Z0) = s21(a1)Sqrt(Z0) + s22(a2)Sqrt(Z0) VF2 = b2*Sqrt(Z0), V1 = s21(a1)Sqrt(Z0), V2 = s22(a2)Sqrt(Z0) VF2 = V1 + V2 -- 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, how about doing me a favor and humor me a little.
How about solving my example, the one that Steve took from Reflections and then trashed. To repeat, 100 w, 70.71 v source, 50-ohm lossline terminated in a 3:1 mismatch and perfectly matched at the input with a rho = 1.0. Vfwd = 81.65 v and Vref = 40.83 V. Please solve this using Steve's method of solution, Eqs 5 thru 13. If this problem can't be solved with Steve's method, his method must be invalid, not general. Walt |
Walter Maxwell wrote:
Cecil, how about doing me a favor and humor me a little. How about solving my example, the one that Steve took from Reflections and then trashed. To repeat, 100 w, 70.71 v source, 50-ohm lossline terminated in a 3:1 mismatch and perfectly matched at the input with a rho = 1.0. Here's the configuration. 100W XMTR--50 ohm line--x-tuner-y--1wl 50 ohm line---150 ohm load The tuner function is replaced by a 1/4WL series section which performs *exactly* the same function so we can see what is really happening inside the tuner. The match point is at 'x', not at 'y'. The system is NOT matched at 'y'. Therefore, Dr. Best's system is not matched at the output of the tuner. His choice to use a tuner for that illustration, instead of a 1/4WL transformer, was a very bad one and has obviously led to a lot of confusion and misunderstanding. rho=0.268 rho=-0.268 rho=0.5 100W XMTR---50 ohm line---x---1/4WL 86.6 ohm line--y--1wl 50 ohm line---150 ohm load VF1=70.7V-- VF2=96.58V-- VF3=81.65V-- --VR1=0V --VR2=25.9V --VR3=40.82V PF1=100W-- PF2=107.76W-- PF3=133.33W-- --PR1=0W --PR2=7.76W --PR3=33.33W The mismatched impedance discontinuity at 'y' is interesting because we have forward power and reflected power flowing to/from both directions. That's what happens at a tuner output. The actual match point is at the tuner input at 'x'. At point 'y' on the 50 ohm line we have a V1y of 70.7V corresponding to the 100W forward power that made it through the impedance discontinuity. This can also be calculated by using the transmission coefficient. Since the impedance is a step-down impedance, tau = 1-rho V1y = VF2(1-rho) = 96.58(0.732) = 70.7V V2y will be VR3(rho) = 40.82(0.268) = 10.94V VF3 = 70.7V + 10.94V = 81.65V. (133.33W) The modified S-parameter analysis works once again and this time it is at an unmatched impedance discontinuity at point 'y'. We could also do the modified S-parameter analysis at point 'x'. V1x = VF1(1+rho) = 70.7V(1.268) = 89.6V V2x = VR2(rho) = 25.9V(0.268) = 6.94V VF2 = V1 + V2 = 89.64V + 6.94V = 96.58V (107.7W) -- 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:
Continuing and calculating interference energy at point 'y': V1y = VF2(1-rho) = 96.58(0.732) = 70.7V (100W) V2y will be VR3(rho) = 40.82(0.268) = 10.94V (2.39W) VF3 = 70.7V + 10.94V = 81.65V. (133.33W) The power equation at point 'y' looking toward the load is: PF3 - P1y - P2y = constructive interference Note that P2y + constructive interference = re-reflected power. 133.33W - 100W - 2.39W = +30.94W A '+' sign on the interference term indicates constructive interference. There is 30.49 watts of constructive interference at 'y' looking toward the load. That energy must come from somewhere. Therefore, there is 30.49 watts of destructive interference at 'y' looking toward the source. The flow of energy from the destructive interference event to the constructive interference event is an interference-caused reflection. The total re-reflected power at 'y' is 2.39W + 30.94W = 33.33W In this case, everything except P1y is re-reflected. This is in agreement with Walter Maxwell and in disagreement with Dr. Best. V1x = VF1(1+rho) = 70.7V(1.268) = 89.6V (88.09W) V2x = VR2(rho) = 25.9V(0.268) = 6.94V (0.556W) VF2 = V1 + V2 = 89.64V + 6.94V = 96.58V (107.7W) The power equation at point 'x' looking toward the load is: PF2 - P1 - P2 = constructive interference 107.7W - 88.09W - 0.556W = +18.42W The '+' sign tells us that there is 18.42 watts of constructive interference at 'x' looking toward the load. That energy must come from somewhere. Therefore, there is 18.42 watts of destructive interference at 'x' looking toward the source. This is also an interference reflection. The total re-reflected power at 'x' is 0.556W + 18.42W = 18.98W Everything except P1 is re-reflected at a match point. Using the S-parameter equations: b1 = s11(a1) + s12(a2) b2 = s21(a1) + s22(a2) s21(a1) is the only voltage term that doesn't get reflected at least once. s11(a1), s12(a2), and s22(a2) are all reflected terms. If the energy in these terms winds up at the load, they have all been re-reflected. -- 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 Tue, 08 Jun 2004 23:09:32 -0500, Cecil Moore
wrote: Walter Maxwell wrote: Cecil, how about doing me a favor and humor me a little. How about solving my example, the one that Steve took from Reflections and then trashed. To repeat, 100 w, 70.71 v source, 50-ohm lossline terminated in a 3:1 mismatch and perfectly matched at the input with a rho = 1.0. Thanks, Cecil, I'll be studying it carefully. Using a 1/4 wl transformer seems like a great idea. I hadn't thought of that. Walt |
Hi Cecil,
Man, when I ask for you to humor me you certainly do it up brown!! You've performed a magnificent job of detailing all the reflections and re-reflections in the 1/4 wl transformer replacing the tuner. That performance really took a lot of work, and I greatly appreciate it. I haven't yet finished reviewing it to the point of verifying in my mind all the various voltages, but I appreciate that the numbers came out correct. However, I have another request for you. Can you find any forward voltages that are appropriate to plug into Steve's Eq 9 and obtain 133.33 watts PFtotal ? Walt |
On Wed, 09 Jun 2004 19:16:39 GMT, Walter Maxwell wrote:
Hi Cecil, Man, when I ask for you to humor me you certainly do it up brown!! You've performed a magnificent job of detailing all the reflections and re-reflections in the 1/4 wl transformer replacing the tuner. That performance really took a lot of work, and I greatly appreciate it. I haven't yet finished reviewing it to the point of verifying in my mind all the various voltages, but I appreciate that the numbers came out correct. However, I have another request for you. Can you find any forward voltages that are appropriate to plug into Steve's Eq 9 and obtain 133.33 watts PFtotal ? Walt Cecil, I found the two Vfwd's we're looking for. They are derived from V1y and V2y. 70.7 v and 10.94 v. The interesting part is that 10.94 v is the exact increase in voltage resulting from adding 33.333 w to 100 w. This I knew many moons ago. But one would never find this Vfwd without performing the complete reflection analysis you performed. With this, plus the fact that a reflection analysis on a stub matching configuration would not reveal this forward voltage still makes Steve's Eq 9 invalid for general use. This is because the re-reflected forward voltage there, 40.82 v added to the source voltage 70.71 v does not equal the real total forward voltage of 81.65 v. Therefore, my final comment on Eq 9 is that it works in specific cases but it certainly is not valid in general. Thanks again, Cecil, for the elegant reflection analysis, bravo! Walt |
Walter Maxwell wrote:
On Tue, 08 Jun 2004 23:09:32 -0500, Cecil Moore wrote: Walter Maxwell wrote: Cecil, how about doing me a favor and humor me a little. How about solving my example, the one that Steve took from Reflections and then trashed. To repeat, 100 w, 70.71 v source, 50-ohm lossline terminated in a 3:1 mismatch and perfectly matched at the input with a rho = 1.0. Thanks, Cecil, I'll be studying it carefully. Using a 1/4 wl transformer seems like a great idea. I hadn't thought of that. It's easier to do it in steps. Step 1. 100W XMTR---50 ohm line---x---1/4WL 86.6 ohm line---150 ohm load Calculate all the forward and reflected components. Step 2. 100W XMTR---50 ohm line---x---1/4WL 86.6 ohm line---y---1WL 50 ohm line---150 ohm load All other forward and reflected components remain the same. Calculate the forward and reflected components on the 50 ohm line going to the load. -- 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:
However, I have another request for you. Can you find any forward voltages that are appropriate to plug into Steve's Eq 9 and obtain 133.33 watts PFtotal ? Here is a quote from an earlier posting: ---------------------- Continuing and calculating interference energy at point 'y': V1y = VF2(1-rho) = 96.58(0.732) = 70.7V (100W) V2y will be VR3(rho) = 40.82(0.268) = 10.94V (2.39W) VF3 = 70.7V + 10.94V = 81.65V. (133.33W) --------------- So --- PFtotal = |VFtotal|^2/Z0 = 81.65V^2/50ohms = 133.33W PFtotal = |V1+V2|^2/Z0 = (70.7V+10.94V)^2/Z0 = 133.33W How's that? -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
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