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Reg Edwards wrote:
But where have you hidden this remarkable transmission line which is long enough to mug and hoodwink so-called SWR meters? It does not exist! Most resonant 50-ohm-coax-fed dipoles would meet those requirements. All that is needed is for it to be long enough to force the forward V/I ratio to Z0. Given the small diameter of a piece of coax compared to a wavelength at HF, it doesn't seem that it take a very long length. Waveguides may be a different story. -- 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, 04 Oct 2003 10:46:41 -0500, Cecil Moore
wrote: Reg Edwards wrote: SWR meters are designed to operate and provide indications of SWR, Rho, Fwd Power, Refl.Power, on the ASSUMPTION that the internal impedance of the transmitter is 50 ohms. I believe that to be an incorrect statement, Reg. The assumption is that a Z0 of 50 ohms exists and the transmission line is long enough to force the ratio of V/I to be 50 ohms for the forward wave and the reflected wave. The phase between the forward voltage and current is assumed to be zero. The phase between the reflected voltage and current is assumed to be zero. Given all those assumptions, the internal impedance of the transmitter is irrelevant. I'm not saying all those assumptions are always met. Sorry, Cecil, the phase between reflected voltage and current is always 180 degrees, not zero. If it were not for this phenomenon the standing wave would not be established as forward and reflected waves of both voltage and current pass through each other otherwise undisturbed. Walt, w2du Put a 50 ohm dummy load on an SWR meter and feed it with a transmitter of unknown source impedance. The SWR meter will always read 1:1 because the dummy load forces the V/I ratio to be 50 no matter what the source impedance. That's what the 50 ohm characteristic impedance of the transmission is supposed to do to make the source impedance irrelevant. PS: In the whole of his excellent 236-page exceedingly comprehensive volume, Chipman, in 1969, makes not the slightest mention of SWR meters. In 1969, virtually all ham transmitters had an adjustable pi-net output so an SWR meter was not needed. When I started out as a ham in the 1950's, just as many hams used 75 ohm coax as used 50 ohm coax, maybe more. The pi-net output of a typical ham transmitter back then didn't care what the Z0 was. I didn't own an SWR meter until the 1980's when I bought an IC-745. In 1969, the "antenna tuner" was built into the transmitter. If wide- range antenna tuners were built into transmitters today, there would be little need for the SWR meter. I don't know of anyone who puts an SWR meter between an SGC-230 and the antenna. |
Walter Maxwell wrote:
Sorry, Cecil, the phase between reflected voltage and current is always 180 degrees, not zero. Yep, I know better, I just mis-spoke. Did you know that there is no such convention for light? It's Kirchhoff's current convention that dictates a 180 degree phase between reflected voltage and reflected current. EM light doesn't follow Kirchhoff's convention. For EM light, there is no phase shift in the reflection if the index of refraction is higher. If the index of refraction is lower, there is a 180 degree phase shift in both E and H fields. -- 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! =----- |
Reg Edwards wrote:
But where have you hidden this remarkable transmission line which is long enough to mug and hoodwink so-called SWR meters? It does not exist! Your argument falls flat at the start. I don't know the answer to this question but perhaps some lurker does. On each side of my SWR meter, I have three feet of RG-400 coax. The spacing between conductors must be about 0.1 inches. With a spacing of 0.1 inch between conductors in the coax, what length of coax is required to force a V/I ratio of 50 ohms? The ratio of 3 feet to 0.1 inches is 360. I suspect that three feet is plenty long enough. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
"Richard Clark" wrote in message ... .................................... Richard, how can you possibly believe that the output impedance of the source has any effect on the SWR on a transmission line? Stephen Adam of HP using Beatty describes it quite well. The data you have by email and has been posted here demonstrates it equally well. It takes no more than two resistors and a length of line to confirm or deny. My data confirms it, absolutely no one has offered negative evidence, simply denials................................... 73's Richard Clark, KB7QHC Richard, Please humor me. If the source impedance has an effect on SWR, surely an equation exists that spells that out. Perhaps you can divulge it for everybody. Tam/WB2TT |
"Reg Edwards" wrote in message ... Reg, that can't possibly be you. Someone has hijacked your e-mail. =========================== Ian, it IS me! Please calm yourself. Let me put what I said into somewhat different words. SWR meters are designed to operate and provide indications of SWR, Rho, Fwd Power, Refl.Power, on the ASSUMPTION that the internal impedance of the transmitter is 50 ohms. It makes the same INCORRECT assumption as a lot of people do. This should not be surprising because it was people who designed it. So SWR meters nearly always give FALSE indications about what actually exists. Perfectionists may be upset at the repercussions of this alarming statement. PS: In the whole of his excellent 236-page exceedingly comprehensive volume, Chipman, in 1969, makes not the slightest mention of SWR meters. ---- Reg, G4FGQ Reg, The SWR meter knows only by how much the load deviates from 50 Ohms. Somebody has very cleverly interpreted this, and calculated how the meter scale should be labeled. Actually, the SWR/Power meter has information that is not displayed. For instance, for a 2:1 SWR it has the needed information to differentiate between a 25 and 100 Ohm koad, but to display this information would require adding extra parts BTW, I did force the SWR meter to see a different source impedance. There was no difference in SWR readings for either the 1:1 or 2:1 case. |
On Sat, 04 Oct 2003 10:08:40 -0500, Cecil Moore
wrote: Here's a quote from Chipman. .... Would you like to pose a complex Z0? Hi Cecil, What for? Random selections of quotation are hardly the basis for discussion. -Eh- let me take that back, it seems it's the ONLY basis for discussion. I note, as posed before, that Ian who "might" hold a copy of Chipman has yet to respond to my points about its contents. My having offered chapter and verse, it would seem less than a monumental task to simply turn to the page and verify my selection, or refute Chipman's statements. Those who hold him dear seem the least likely to respond to his teachings. 73's Richard Clark, KB7QHC |
On Sat, 4 Oct 2003 14:26:37 -0400, "Tarmo Tammaru"
wrote: Richard, Please humor me. If the source impedance has an effect on SWR, surely an equation exists that spells that out. Perhaps you can divulge it for everybody. Tam/WB2TT Hi Tam, I could, but I won't. If you wish to be humored, email me as has Walt. That venue is far more productive than me providing yet another citation, its quote, its elaboration, its demonstration at the bench with data, to only observe it won't "change my mind" sclerosis mentality exhibited in this forum's sneer review. I have offered several such matters direct to the point to have those who could respond from actual verification by observing it in their copy simply vanish from the debate. As I recall, you did exactly the same thing with a Motorola application note about source Z that you never confirmed or denied after introducing it as evidence against my position. As for humoring, I find that in the obvious logical disconnects and exhibiting it here. 73's Richard Clark, KB7QHC |
SWR meters are designed to operate and provide indications of SWR, Rho,
Fwd Power, Refl.Power, on the ASSUMPTION that the internal impedance of the transmitter is 50 ohms. It makes the same INCORRECT assumption as a lot of people do. This should not be surprising because it was people who designed it. So SWR meters nearly always give FALSE indications about what actually exists. ------------------------------------------------------ Reg, BTW, I did force the SWR meter to see a different source impedance. There was no difference in SWR readings for either the 1:1 or 2:1 case. ------------------------------------------------------ Tarmo, And of course, as you and I know, on whatever line there is between the meter and transmitter, the swr is neither the indicated 1:1 nor 2:1 because the input impedance looking back towards the tranmitter is not the assumed 50 ohms. Both readings are false, even meaningless. There may in fact be no standing waves to measure. To avoid confusing novices and budding engineers, retarding education, rename the meter the TLI (Transmitter Loading Indicator) which is what it really is. ---- Reg, G4FGQ |
Richard Clark wrote:
I note, as posed before, that Ian who "might" hold a copy of Chipman has yet to respond to my points about its contents. I have ordered a copy. From what I have read of Chipman so far, everything can be explained by achieving a conjugate match at one point on the transmission line when the reactance looking in either direction is at a maximum. This is simply a resonance effect. Here's a bench experiment that might shed some light on this problem. source--50 ohm coax--(-j500)--SWR meter--(+j500)--50 ohm coax--50 ohm load There is a localized high reactive energy exchange through the SWR meter between the capacitance and the coil but nowhere else on the transmission line. That has got to have an effect on the SWR reading which is probably not good. What, exactly, is the big deal? It is just another distributed circuit problem. -- 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! =----- |
Whenever you're dealing with current, you have to pay attention to the
definition of positive direction. If you define the positive direction of forward current as being toward the load and of reflected current toward the source, then Vf is in phase with If and Vr is in phase with Ir. I suspect that a similar caution needs to be heeded when dealing with optics. Roy Lewallen, W7EL Cecil Moore wrote: Walter Maxwell wrote: Sorry, Cecil, the phase between reflected voltage and current is always 180 degrees, not zero. Yep, I know better, I just mis-spoke. Did you know that there is no such convention for light? It's Kirchhoff's current convention that dictates a 180 degree phase between reflected voltage and reflected current. EM light doesn't follow Kirchhoff's convention. For EM light, there is no phase shift in the reflection if the index of refraction is higher. If the index of refraction is lower, there is a 180 degree phase shift in both E and H fields. |
On Sat, 04 Oct 2003 18:00:39 -0700, Roy Lewallen wrote:
Whenever you're dealing with current, you have to pay attention to the definition of positive direction. If you define the positive direction of forward current as being toward the load and of reflected current toward the source, then Vf is in phase with If and Vr is in phase with Ir. I suspect that a similar caution needs to be heeded when dealing with optics. Roy Lewallen, W7EL Well, Roy, if what you say above is true then why does the phase of reflected voltage change 180 degrees and reflected current does not change when the forward waves encounter a perfect short-circuit termination? And on the other hand, why does the phase of reflected current change 180 degrees and reflected voltage does not change when the forward waves encounter a perfect open-circuit termination? How then can the reflected voltage and current be other than 180 degrees regardless of the load? If what you say is true then my explanation in Reflections concerning the establishment of the standing wave must be all wrong. Is this what you're saying? Walt, W2DU Cecil Moore wrote: Walter Maxwell wrote: Sorry, Cecil, the phase between reflected voltage and current is always 180 degrees, not zero. Yep, I know better, I just mis-spoke. Did you know that there is no such convention for light? It's Kirchhoff's current convention that dictates a 180 degree phase between reflected voltage and reflected current. EM light doesn't follow Kirchhoff's convention. For EM light, there is no phase shift in the reflection if the index of refraction is higher. If the index of refraction is lower, there is a 180 degree phase shift in both E and H fields. |
Roy Lewallen wrote:
Whenever you're dealing with current, you have to pay attention to the definition of positive direction. If you define the positive direction of forward current as being toward the load and of reflected current toward the source, then Vf is in phase with If and Vr is in phase with Ir. I suspect that a similar caution needs to be heeded when dealing with optics. Optics doesn't have the luxury of only two directions so Kirchhoff's conventions are meaningless for light. With light scattering in any number of directions in 3D space, 3D optical engineers must be a little more careful than 1D RF engineers. :-) Obviously, light and RF waves obey the same physics but an RF transmission line is essentially a one-dimensional environment with a plus and minus direction. I assume fiber-optics is subject to that same simplification. -- 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:
How then can the reflected voltage and current be other than 180 degrees regardless of the load? Consider a transmission line driven by two identical sources which are signal generators with circulators+loads designated by SGCL. SGCL1-----------50 ohm coax----------SGCL2 All the power sourced by SGCL1 is dissipated in SGCL2 and all the power sourced by SGCL2 is dissipated in SGCL1. The system is perfectly symmetrical. Will there be ordinary standing waves? Of course. The voltage and current from SGCL1 are in phase. The voltage and current from SGCL2 are in phase. The only difference between the two currents is Kirchhoff's convention. When the voltages are maximum at the same point, they superpose to 2*V. When the two currents are maximum at the same point, they superpose to zero because they are traveling in opposite directions. We would say that SGCL2's voltage and current are 180 degrees out of phase. Someone looking at the experiment from the other side of the screen would say that SGCL1's voltage and current are 180 degrees out of phase. It is only a convention, one that doesn't exist for 3D light. -- 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! =----- |
I'm not at all saying your explanation is wrong. I'm just pointing out
the effect of defining the direction of current. You say that the phase of the reflected current changes 180 degrees. Another way to say the exactly the same thing is that the current doesn't change phase but reverses direction. The phase of the current depends on its direction, so is affected by how we define "positive" direction. The phase of voltage, on the other hand, isn't. Consider a short circuit. At that point, Vr = -Vf. That, we know from the requirement that the total V, the sum of Vr and Vf, has to be zero. But how about the current? The magnitude of the reflected current equals the magnitude of the forward current. At the short, the current isn't zero -- it's twice If. Assuming that If is always defined as being positive toward the load, let's first define the positive direction of Ir as also being toward the load. Then the total current at any point is If + Ir. At the short, it's If + Ir = 2 * If, which says that Ir = If. You can say that the phase of the current hasn't changed as a result of the reflection. But if we define the positive direction of Ir as being toward the source, then the total current at any point on the line is If - Ir. At the short it's If - Ir = 2 * If, so Ir = -If. So the phase of Ir is 180 degrees relative to the phase of If. Of course, it's also traveling in the opposite direction, by definition. So you have your choice. You can say that the reflected current is flowing in the same direction as the forward current, and with the same phase. Or you can equally correctly say that the current has reversed both direction and phase due to the reflection. They're exactly equivalent, both give correct mathematical results, and are equally valid. The same reasoning applied at an open circuit, where the total current is zero, shows that when Ir is defined as positive toward the load, the total current = If + Ir = 0 means Ir = -If. In other words, the reflected current, defined as being in the same direction as the forward current, has undergone a 180 degree phase shift. But if Ir is deemed positive toward the source, then the total current is If - Ir = 0, so we say that the reflected current has undergone a reversal of direction but no change in phase. As long as we always calculate the total current by using Kirchoff's principle as If + Ir if Ir is positive toward the load, or If - Ir if it's positive toward the source, all results are valid. Among the consequences of the two possible definitions of positive direction for Ir is that the current reflection coefficient Ir/If can be either equal to the voltage reflection coefficient, or its negative. And, as in earlier postings, one can conclude that Vr/Ir can equal either Z0 or -Z0. Both depend on the definition of the positive direction of Ir (assuming that If is consistently defined as positive toward the load, which is a good assumption). The need to be careful with the definition, and always making it clear, is illustrated by the fact that of the first four fields/transmission line texts I pulled off my shelf, two (Holt and Johnson) defined the positive direction of If toward the load, and two (Johnk and Kraus) toward the source. So you can't make an assumption that the definition is even usually one way or the other. Roy Lewallen, W7EL Walter Maxwell wrote: On Sat, 04 Oct 2003 18:00:39 -0700, Roy Lewallen wrote: Whenever you're dealing with current, you have to pay attention to the definition of positive direction. If you define the positive direction of forward current as being toward the load and of reflected current toward the source, then Vf is in phase with If and Vr is in phase with Ir. I suspect that a similar caution needs to be heeded when dealing with optics. Roy Lewallen, W7EL Well, Roy, if what you say above is true then why does the phase of reflected voltage change 180 degrees and reflected current does not change when the forward waves encounter a perfect short-circuit termination? And on the other hand, why does the phase of reflected current change 180 degrees and reflected voltage does not change when the forward waves encounter a perfect open-circuit termination? How then can the reflected voltage and current be other than 180 degrees regardless of the load? If what you say is true then my explanation in Reflections concerning the establishment of the standing wave must be all wrong. Is this what you're saying? Walt, W2DU Cecil Moore wrote: Walter Maxwell wrote: Sorry, Cecil, the phase between reflected voltage and current is always 180 degrees, not zero. Yep, I know better, I just mis-spoke. Did you know that there is no such convention for light? It's Kirchhoff's current convention that dictates a 180 degree phase between reflected voltage and reflected current. EM light doesn't follow Kirchhoff's convention. For EM light, there is no phase shift in the reflection if the index of refraction is higher. If the index of refraction is lower, there is a 180 degree phase shift in both E and H fields. |
Roy Lewallen wrote:
I'm not at all saying your explanation is wrong. I'm just pointing out the effect of defining the direction of current. You say that the phase of the reflected current changes 180 degrees. Another way to say the exactly the same thing is that the current doesn't change phase but reverses direction. The phase of the current depends on its direction, so is affected by how we define "positive" direction. The phase of voltage, on the other hand, isn't. It might be easier to visualize using E & H fields and the right hand rule. Point your thumb in the direction of wave travel (North). The orthogonal index finger (up) represents the E-field and the orthogonal middle finger (East) represents the H-field. Then turn the thumb in the opposite direction (South) while keeping the index finger (E-field) pointed in the same direction (up). The middle finger (H-field) will reverse direction by 180 degrees (West) but the orthogonal relationships between the direction of travel and the fields are still identical. Just another way of visualizing what you said above. -- 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, 05 Oct 2003 00:27:28 -0700, Roy Lewallen wrote:
I'm not at all saying your explanation is wrong. I'm just pointing out the effect of defining the direction of current. You say that the phase of the reflected current changes 180 degrees. Another way to say the exactly the same thing is that the current doesn't change phase but reverses direction. The phase of the current depends on its direction, so is affected by how we define "positive" direction. The phase of voltage, on the other hand, isn't. Consider a short circuit. At that point, Vr = -Vf. That, we know from the requirement that the total V, the sum of Vr and Vf, has to be zero. But how about the current? The magnitude of the reflected current equals the magnitude of the forward current. At the short, the current isn't zero -- it's twice If. Assuming that If is always defined as being positive toward the load, let's first define the positive direction of Ir as also being toward the load. Then the total current at any point is If + Ir. At the short, it's If + Ir = 2 * If, which says that Ir = If. You can say that the phase of the current hasn't changed as a result of the reflection. But if we define the positive direction of Ir as being toward the source, then the total current at any point on the line is If - Ir. At the short it's If - Ir = 2 * If, so Ir = -If. So the phase of Ir is 180 degrees relative to the phase of If. Of course, it's also traveling in the opposite direction, by definition. So you have your choice. You can say that the reflected current is flowing in the same direction as the forward current, and with the same phase. Or you can equally correctly say that the current has reversed both direction and phase due to the reflection. They're exactly equivalent, both give correct mathematical results, and are equally valid. The same reasoning applied at an open circuit, where the total current is zero, shows that when Ir is defined as positive toward the load, the total current = If + Ir = 0 means Ir = -If. In other words, the reflected current, defined as being in the same direction as the forward current, has undergone a 180 degree phase shift. But if Ir is deemed positive toward the source, then the total current is If - Ir = 0, so we say that the reflected current has undergone a reversal of direction but no change in phase. As long as we always calculate the total current by using Kirchoff's principle as If + Ir if Ir is positive toward the load, or If - Ir if it's positive toward the source, all results are valid. Among the consequences of the two possible definitions of positive direction for Ir is that the current reflection coefficient Ir/If can be either equal to the voltage reflection coefficient, or its negative. And, as in earlier postings, one can conclude that Vr/Ir can equal either Z0 or -Z0. Both depend on the definition of the positive direction of Ir (assuming that If is consistently defined as positive toward the load, which is a good assumption). The need to be careful with the definition, and always making it clear, is illustrated by the fact that of the first four fields/transmission line texts I pulled off my shelf, two (Holt and Johnson) defined the positive direction of If toward the load, and two (Johnk and Kraus) toward the source. So you can't make an assumption that the definition is even usually one way or the other. Thanks, Roy, for the lucid explanation. I had not previously thought of the interplay between direction and polarity, as you have so clearly pointed out. I have always considered a minus rho for current when rho is positive for voltage, bu I didn'[t carry the thought through far enough. Walt, W2DU Roy Lewallen, W7EL Walter Maxwell wrote: On Sat, 04 Oct 2003 18:00:39 -0700, Roy Lewallen wrote: Whenever you're dealing with current, you have to pay attention to the definition of positive direction. If you define the positive direction of forward current as being toward the load and of reflected current toward the source, then Vf is in phase with If and Vr is in phase with Ir. I suspect that a similar caution needs to be heeded when dealing with optics. Roy Lewallen, W7EL Well, Roy, if what you say above is true then why does the phase of reflected voltage change 180 degrees and reflected current does not change when the forward waves encounter a perfect short-circuit termination? And on the other hand, why does the phase of reflected current change 180 degrees and reflected voltage does not change when the forward waves encounter a perfect open-circuit termination? How then can the reflected voltage and current be other than 180 degrees regardless of the load? If what you say is true then my explanation in Reflections concerning the establishment of the standing wave must be all wrong. Is this what you're saying? Walt, W2DU Cecil Moore wrote: Walter Maxwell wrote: Sorry, Cecil, the phase between reflected voltage and current is always 180 degrees, not zero. Yep, I know better, I just mis-spoke. Did you know that there is no such convention for light? It's Kirchhoff's current convention that dictates a 180 degree phase between reflected voltage and reflected current. EM light doesn't follow Kirchhoff's convention. For EM light, there is no phase shift in the reflection if the index of refraction is higher. If the index of refraction is lower, there is a 180 degree phase shift in both E and H fields. |
On Sat, 04 Oct 2003 17:31:31 -0500, Cecil Moore
wrote: everything can be explained by achieving a conjugate match at one point on the transmission line when the reactance looking in either direction is at a maximum. Hi Cecil, I don't know which is funnier: that you have a one-solution-answers-every-question; or that you have so many of them. Reach into your bag and present us the conjugate for: source50---50 ohm feedline---+---150 ohm feedline---load150 or the rather more terse (and simpler - bound to confound): source=200Ohm(resistive)---50 ohm feedline---load=200Ohm(resistive) 73's Richard Clark, KB7QHC |
On Sat, 4 Oct 2003 19:49:19 +0000 (UTC), "Reg Edwards"
wrote: SWR meters are designed to operate and provide indications of SWR, Rho, Fwd Power, Refl.Power, on the ASSUMPTION that the internal impedance of the transmitter is 50 ohms. It makes the same INCORRECT assumption as a lot of people do. This should not be surprising because it was people who designed it. So SWR meters nearly always give FALSE indications about what actually exists. ------------------------------------------------------ Reg, BTW, I did force the SWR meter to see a different source impedance. There was no difference in SWR readings for either the 1:1 or 2:1 case. ------------------------------------------------------ Tarmo, And of course, as you and I know, on whatever line there is between the meter and transmitter, the swr is neither the indicated 1:1 nor 2:1 because the input impedance looking back towards the tranmitter is not the assumed 50 ohms. Both readings are false, even meaningless. There may in fact be no standing waves to measure. To avoid confusing novices and budding engineers, retarding education, rename the meter the TLI (Transmitter Loading Indicator) which is what it really is. ---- Reg, G4FGQ Reg, can you furnish a a mathematical expression that includes source resistance as a required parameter for determining SWR? Walt, W2DU |
Correction:
On Sun, 05 Oct 2003 00:27:28 -0700, Roy Lewallen wrote: . . . (Last paragraph): The need to be careful with the definition, and always making it clear, is illustrated by the fact that of the first four fields/transmission line texts I pulled off my shelf, two (Holt and Johnson) defined the positive direction of If toward the load, and two (Johnk and Kraus) ^^ toward the source. . . That should be Ir, not If. I've never seen the positive direction of If defined as anything other than toward the load. Roy Lewallen, W7EL |
Reg, can you furnish a a mathematical expression that includes source
resistance as a required parameter for determining SWR? -------------------------------------------- No Walt. Can you ? Why do you ask ? Reg. |
Richard Clark wrote:
wrote: everything can be explained by achieving a conjugate match at one point on the transmission line when the reactance looking in either direction is at a maximum. I don't know which is funnier: that you have a one-solution-answers-every-question; or that you have so many of them. It's not my solution, Richard, it's Chipman's solution. "These large reflection coefficients are an example of the phenomenon of 'resonant rise of voltage' in series resonant circuits." -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Richard Clark wrote:
source=200 Ohm(resistive)---50 ohm feedline---load=200 Ohm(resistive) Are you saying that the SWR will vary up and down the line when the feedline is lossless? Is the above example in Chipman? -- 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, 05 Oct 2003 22:50:17 -0500, Cecil Moore
wrote: Richard Clark wrote: source=200 Ohm(resistive)---50 ohm feedline---load=200 Ohm(resistive) Are you saying that the SWR will vary up and down the line when the feedline is lossless? Cecil, you can be so thick. Do you inhabit the center of the universe? We've waded through this long ago. Is the above example in Chipman? No. He does have a snippet of math that will provide the same answer found for similar (differing only by magnitude of R's) examples by other authors. These issues are new only to folks here. Hi Cecil, It seems that whenever I challenge you to one of your comments such as: everything can be explained by achieving a conjugate match you fly from it to prove or question some remote issue. Instead of going over old material that you abandoned (and will only abandon again), why not simply offer the group the conjugate of: source=200 Ohm(resistive)---50 ohm feedline---load=200 Ohm(resistive) which was my query this time (or are you abandoning that too?). 73's Richard Clark, KB7QHC |
Reg Edwards wrote:
Reg, that can't possibly be you. Someone has hijacked your e-mail. =========================== Ian, it IS me! Oh yes? Whoever he is, he *would* say that, wouldn't he? (Catching-up after a hectic long weekend...) -- 73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB) Editor, 'The VHF/UHF DX Book' http://www.ifwtech.co.uk/g3sek |
Reg, can you furnish a a mathematical expression that
includes source resistance as a required parameter for determining SWR? -------------------------------------------- No Walt. Can you ? Why do you ask ? Reg. |
Richard Clark wrote:
wrote: Are you saying that the SWR will vary up and down the line when the feedline is lossless? Cecil, you can be so thick. Do you inhabit the center of the universe? We've waded through this long ago. You cloud the issue because you refuse to answer simple questions. I don't remember what your answer was and I can't find your previous answer on Google. Is a simple "yes" or "no" too much to ask? Instead of going over old material that you abandoned (and will only abandon again), why not simply offer the group the conjugate of: source=200 Ohm(resistive)---50 ohm feedline---load=200 Ohm(resistive) which was my query this time (or are you abandoning that too?). If the lossless 50 ohm feedline is a multiple of 1/2WL long, the system is conjugately matched. Chipman says the extra power term only exists when the reactance of the feedline is opposite in sign to the reactance of the load but your load is purely resistive. So I don't know what you are trying to say. Therefore, I don't know whether to agree with you or not. -- 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, 06 Oct 2003 09:50:17 -0500, Cecil Moore
wrote: Instead of going over old material that you abandoned (and will only abandon again), why not simply offer the group the conjugate of: source=200 Ohm(resistive)---50 ohm feedline---load=200 Ohm(resistive) which was my query this time (or are you abandoning that too?). So I don't know what you are trying to say. Therefore, I don't know whether to agree with you or not. Hi Cecil, You don't have to know as it is not a matter of agreeing, it is a matter of your statement offering: everything can be explained by achieving a conjugate match and I see nothing about that in a halfwave line that instead achieves a Zo match, not a conjugate. A conjugate has very specific properties and you cannot provide an expression that offers the conjugate for the situation: source=200 Ohm(resistive)---50 ohm feedline---load=200 Ohm(resistive) Hence, the generality you impart to Chipman, due to your limitations, reveals it is neither a generality nor is it necessarily even a derivation of Chipman. Your two pages of copy are 230-odd pages shy of understanding. Let's just juggle the notion of Zo matching out with a slight boundary change: source=200 Ohm(resistive)---50 ohm feedline---load=600 Ohm(resistive) What is the expression you offer to support your statement that yields the conjugate? Barring an answer, it follows your statement that everything can be explained by achieving a conjugate match is yet another in a long list of absurdities. Perhaps you should await Chipman's arrival (Waiting for Godot?) before continuing on. However, given the consequences of that arrival for others in this group, that could mean total abstinence in discussion as so many seem to read him in the closet and find themselves locked in a small, dark room. 73's Richard Clark, KB7QHC |
On Sat, 4 Oct 2003 19:49:19 +0000 (UTC), "Reg Edwards"
wrote: SWR meters are designed to operate and provide indications of SWR, Rho, Fwd Power, Refl.Power, on the ASSUMPTION that the internal impedance of the transmitter is 50 ohms. Reg, G4FGQ Well, Reg, the reason I asked for an expression that includes the source resistance in measuring SWR is that you said above that the internal impedance of the transmitter is ASSUMED to be 50 ohms. This implies that the SWR is dependent on the internal impedance of the source, does it not ? AsI have understood Richard C., he also asserts that SWR is dependent the internal impedance of the source. This concept is foreign to me, so if I'm wrong I'd like to have some proof that the source impedance can have any influence on SWR. Walt |
On Mon, 06 Oct 2003 16:40:27 GMT, Walter Maxwell wrote:
This concept is foreign to me, so if I'm wrong I'd like to have some proof that the source impedance can have any influence on SWR. Walt Hi Walt, I have not seen any correspondence from you. However, as to proof, that has been tendered to you, you simply lack the facilities to test it at this time. That much has already been established and further elaboration is unnecessary. Debate can continue without resolution, a simple hour's work at the bench can put the cap to it. My data stands un-refuted (barring the usual cackle of nay-saying sneer review), and even more, without test at ANY other bench. I can only conclude that: 1.) My data is bullet-proof; 2.) others lack the ability to perform the task; 3.) 1&2 above, but narcissistic debate is the real focus of critics. The triumph of the nay-sayers is in my admission that I know that I am in error. They undoubtedly grasp that statement as the chalice of their noble musings leaving them undisturbed to step up to the bench. It also is revealed in their piteous cries of the calamity of Amateur Radio's future that awaits us. This last comes as no surprise to the rising tide in the kulture of institutionalized ignorance where the supreme technical achievement is enacted by pushing a credit card across the display case. I am bound to be in error through my own admission, but my admission comes with a bounds of accuracy. To others here, my error is absolute and demonstration to attest that is unnecessary. This unsullied nobility is then undercut by the jejune debate they indulge in over issues of a philosophical nature - actually a mystical assignation with metaphysics. I suppose I frustrate many because I am not afraid to be wrong. The frustration is often railed in terms of my style (their being outgunned on two fronts) and compounded by their inertia for doing simple things well (the loss of yet another, third front). You guys need more threads devoted to the definition of weight so you can devastate the farmer's mud-logic. ;-) 73's Richard Clark, KB7QHC |
Walt, W2DU wrote:
"This ratio (reflection coefficient) is determined only by the load and the line, not by the generator.." Terman is squarely in Walt`s corner. He says on page 87 of his 1955 edition: "Reflection coefficient = rho = E2 / E1 = (ZL/Zo) -1 / (ZL/Zo) +1. No Zsource appears in the equation, only ZL and Zo. " Were it not so, Terman would have told you! There is a nice photo of Walt, W2DU in the April 1973 edition of QST. That`s the edition with (2) Bird wattmeters on the cover. This edition initiated a series of articles on "Reflections" by Walt. Best regards, Richard Harrison, KB5WZI |
"Walter Maxwell" wrote in message ... This concept is foreign to me, so if I'm wrong I'd like to have some proof that the source impedance can have any influence on SWR. Walt Walt, I don't think the source impedance has any effect on SWR. In fact I have changed the source impedance and saw no change in SWR. But Since tha SWR meter is a really dumb bunny, I wonder of the meter can be mislead by a reactive source impedance that forces the current to be out of phase with the voltage. Perhaps a case where the source and load both are reactive? Tam/WB2TT |
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Tam, WB2TT wrote:
"--I wonder if the meter can be mislead by a reactive source impedance that forces the current to be out of phase with the voltage." In the usual h-f transmission line, Zo appears as an Ro. This means that you put volts across it and the resulting current in the line travels according to Ohm`s law controlled by the surge impedance of the line. Ro means the current is in-phase with the volts across the line in both directions of travel. The funny stuff seen on a line with reflections comes from looking at both directions of travel at the same time. That`s not the best way to look at the line and that`s why the Bird wattmeter uses a directional coupler to extract information on the traveling wave in one direction at a time. Surely maximum power transfer is enabled by a conjugate match. "Dumb bunny " SWR meters may indicate anything. Terman says on page 76 of his 1955 edition regarding maximum power transfer: "This is accomplished by making the load the conjugate of its generator impedance as defined by thevenin`s theorem." It is Walt who is "bullet-proof" on this. Best regards, Richard Harrison, KB5WZI |
Richard Clark wrote:
everything can be explained by achieving a conjugate match ... and I see nothing about that in a halfwave line that instead achieves a Zo match, not a conjugate. A conjugate has very specific properties and you cannot provide an expression that offers the conjugate for the situation: You conveniently trimmed off the rest of my statement. When the line is lossy, it is possible to achieve a conjugate match at a point but nowhere else. The requirement of a conjugate match for a lossy line is that the impedance looking in either direction is the conjugate of the other direction. That can be achieved at a single point in a lossy system, e.g. at the load. The rule that if a conjugate match exists at one point, then a conjugate match exists at all points, is *ONLY* true for lossless systems. Let's just juggle the notion of Zo matching out with a slight boundary change: source=200 Ohm(resistive)---50 ohm feedline---load=600 Ohm(resistive) What is the expression you offer to support your statement that yields the conjugate? Barring an answer, it follows your statement that everything can be explained by achieving a conjugate match ... Again, please note that you deliberately snipped the context of that statement, not a very ethical thing to do. is yet another in a long list of absurdities. Well, since you changed the contextual conditions away from a possible conjugate match, nothing in the new example cannot be explained by achieving a conjugate match, since a conjugate match is impossible in the new example. What do you think changing the context proves? Nothing that you have said is true at the center of the sun. How's that for a context change? -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Richard Clark wrote:
My data stands un-refuted (barring the usual cackle of nay-saying sneer review), and even more, without test at ANY other bench. I can only conclude that: 1.) My data is bullet-proof; 2.) others lack the ability to perform the task; 3.) 1&2 above, but narcissistic debate is the real focus of critics. I offered an experiment that might prove you right, Richard, but so far you, nor anyone else, has offered any response. Source---coax---(+j500)---SWR meter---(-j500)---50 ohm load Seems to me the resonant reactances, in series or parallel, on each side of the SWR meter, might add an equal magnitude of energy to the forward energy and reflected energy seen by the SWR meter thus changing the SWR reading. I suspect the above condition is representative of what you are seeing in your measurements. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Richard Harrison wrote:
In the usual h-f transmission line, Zo appears as an Ro. RG-174 has about 6dB loss per 100ft on 12m. Its Z0 is equal to Sqrt[(R+jXL)/(G+jXC)]. Does (R+jXL)/(G+jXC) really equal 2500 for RG-174 on 12m? The specs say the Z0 of RG-174 is a nominal 50 ohms. -- 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, 06 Oct 2003 14:25:33 -0500, Cecil Moore
wrote: Richard Clark wrote: everything can be explained by achieving a conjugate match ... and I see nothing about that in a halfwave line that instead achieves a Zo match, not a conjugate. A conjugate has very specific properties and you cannot provide an expression that offers the conjugate for the situation: You conveniently trimmed off the rest of my statement. You are welcome. Let's just juggle the notion of Zo matching out with a slight boundary change: source=200 Ohm(resistive)---50 ohm feedline---load=600 Ohm(resistive) What is the expression you offer to support your statement that yields the conjugate? Barring an answer, it follows your statement that everything can be explained by achieving a conjugate match ... is yet another in a long list of absurdities. Again, please note that you deliberately snipped the context of that statement, not a very ethical thing to do. This time, everyone welcomes it. Nothing that you have said is true at the center of the sun. How's that for a context change? Hi Cecil, About average from you except this time you offered no solution for either context change. As such, it appears your statement everything can be explained by achieving a conjugate match ... has no meaning outside of the center of the sun. I will leave that you cannot demonstrate your statement anywhere in the known universe. 73's Richard Clark, KB7QHC |
Richard Clark wrote:
As such, it appears your statement everything can be explained by achieving a conjugate match ... has no meaning outside of the center of the sun. None of my statements has any meaning outside of the context in which they are offered (which you conventiently attempt to obscure). It appears that you do not want to resolve anything. If so, I hope you won't mind if others resolve your technical problems at the very time that you are 100% resistant to any resolution. -- 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, 06 Oct 2003 15:16:04 -0500, Cecil Moore
wrote: None of my statements has any meaning outside of the context in which they are offered Hi Cecil, Well, as this all thread started from one context: everything can be explained by achieving a conjugate match being so encompassing as to enlarge beyond your capacity to explain, we find ourselves with shortfalls of example to any context. Your response of a Zo match is an embarrassing example of poor application for conjugation, so it would appear that even your single context is meaningless. 73's Richard Clark, KB7QHC |
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