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Jim Kelley wrote:
To my way of thinking, rho is entirely dependent upon the impedances, and the voltages (reflected voltages in particular) are dependent upon rho. Not the other way around. Even when the impedances are only V/I ratios? Seems like circular logic to me. The V/I ratio causes rho which causes the voltage??? -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Tom Bruhns wrote:
Cecil Moore wrote: Does it really hurt anything to remind everyone that +1 at 180 degrees equals -1 at zero degrees? No, and I already agreed with that in another posting in this thread. When a piece of coax is shorted, we can calculate: rho = (Z1-Z0)/(Z1+Z0) = (0-50)/(0+50) = -1 After that, we can say it really means +1 at 180 degrees but it is mathematically consistent in either case. -- 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, 29 Sep 2003 17:14:38 -0500, Cecil Moore
wrote: I give up trying to communicate with you Hi Cecil, You did that long ago and were in denial only till now. 73's Richard Clark, KB7QHC |
Cecil Moore wrote: Jim Kelley wrote: To my way of thinking, rho is entirely dependent upon the impedances, and the voltages (reflected voltages in particular) are dependent upon rho. Not the other way around. The V/I ratio causes rho which causes the voltage??? Nope. Rho is not dependent upon V/I ratios other than those at real physical impedance discontinuities. I think you know that. V/I ratios can vary with position along the line and are not constrained to equalling Z0 or Zl. I'm curious why you would ask. 73, Jim AC6XG |
Cecil Moore wrote:
Tom Bruhns wrote: Cecil Moore wrote: Does it really hurt anything to remind everyone that +1 at 180 degrees equals -1 at zero degrees? No, and I already agreed with that in another posting in this thread. When a piece of coax is shorted, we can calculate: rho = (Z1-Z0)/(Z1+Z0) = (0-50)/(0+50) = -1 After that, we can say it really means +1 at 180 degrees but it is mathematically consistent in either case. This has been repeated so frequently of late, without qualification, that readers may begin to believe that it is true in general. It is only true for the special case of single frequency sinusoidal waveforms. For more complex waveforms (consider square, sawtooth, step, for example), negation and 180 degree phase shift are not the same operation. Since reflection coefficients (at least for lines with approximately real Z0) work perfectly fine for these more complex waveforms, it seems unwise to be thinking that negation and 180 degree phase shift are the same. For the example above, stick with rho = -1. It will help you solve more problems that way. ....Keith |
Jim Kelley wrote:
Cecil Moore wrote: The V/I ratio causes rho which causes the voltage??? Nope. Rho is not dependent upon V/I ratios other than those at real physical impedance discontinuities. I think you know that. V/I ratios can vary with position along the line and are not constrained to equalling Z0 or Zl. Consider the following: Source---50 ohm feedline---+---1/2WL 150 ohm---50 ohm load Isn't the 50 ohms that causes rho=0 on the 50 ohm feedline simply the V/I ratio at 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! =----- |
wrote:
It is only true for the special case of single frequency sinusoidal waveforms. Which is the general case for a key-down ham transmitter. -- 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:
wrote: I give up trying to communicate with you You did that long ago and were in denial only till now. You created your own reality based on your feelings. Nobody else is capable of experiencing it. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
On Mon, 29 Sep 2003 23:19:47 -0500, Cecil Moore
wrote: You created your own reality based on your feelings. Nobody else is capable of experiencing it. Hi Cecil, See? You haven't given up afterall! Feelings hmmm? Yeah, I suppose I get pretty emotional over your wild nonsense of sqrt of 1 could be -1. It's called laughing until your sides ache (not really, more a chuckle). You need to ride your bike more and skip the library. Practice smiling once a day, read an eh posting from Stefano to break the ice. 73's Richard Clark, KB7QHC |
Cecil Moore wrote:
wrote: It is only true for the special case of single frequency sinusoidal waveforms. Which is the general case for a key-down ham transmitter. It is indeed the usual case, but limiting your thinking to the usual case reduces your opportunity for understanding. Believing too strongly in the usual case will inhibit your ability to understand when you begin to explore more general cases. First you will have to unlearn your beliefs. If you have repeated them to yourself for too long without understanding their limitations, you can find it very difficult to let go of them even when they no longer serve. It is therefore useful to occasionally remind yourself of the limitations applicable to your assertions. ....Keith |
Richard Clark wrote:
Feelings hmmm? Yeah, I suppose I get pretty emotional over your wild nonsense of sqrt of 1 could be -1. It is well known that (-1)^2 = +1. You sometimes quote Johnson as a reference. From Johnson, section 1.6, page 16: "The ratio 'k' is called the reflection coefficient." "The terminating impedance is zero at this end, provided the internal resistance of the generator is negligible; hence for the generator end" the reflection coefficient, "k(g) = -Z0/Z0 = -1" -- 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! =----- |
wrote:
Cecil Moore wrote: wrote: It is only true for the special case of single frequency sinusoidal waveforms. Which is the general case for a key-down ham transmitter. It is indeed the usual case, but limiting your thinking to the usual case reduces your opportunity for understanding. Can you give me an example of a key-down CW transmitter that is not single frequency sinusoidal waveforms? -- 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, 30 Sep 2003 08:41:06 -0500, Cecil Moore
wrote: Richard Clark wrote: Feelings hmmm? Yeah, I suppose I get pretty emotional over your wild nonsense of sqrt of 1 could be -1. It is well known that (-1)^2 = +1. You cannot show that any two powers used to compute Rho are negative to fulfill this shift of your logic. I probably should have said rho^2 = Pref/Pfwd. When Pref = Pfwd, rho can be plus or minus one. Rho can never be plus or minus one on the basis of sqrt (Pref/Pfwd). To insist otherwise is the joke that gets me emotional with the chuckles and gets you into a huff claiming you can't talk about it. You sometimes quote Johnson Hi Cecil, This again demonstrates how you are an unreliable correspondent. You ascribe an action to me that is simply not true. From Johnson, section 1.6, page 16: Is NOT a citation. Who is Johnson? Certainly you feel free to associate my name with him, but Dr. Samuel Johnson never said any such thing in his life. Note that I always give full names and complete titles to my citations - unreliable correspondents are lazy correspondents who throw statement after statement against the wall until one sticks and they call that their authority. Read any of Gene Nygaard's postings for boundless examples. Cecil, this laziness of yours is part and parcel to your poor recitations and flawed logic. You squirm to pull away from your absurd example of imparting direction of power flow based on an erroneous concept of finding negativity extracted from a dependant variable based on negative power ratios. You are attempting to recast that argument into other equivalent terms of Rho, while maintaining this charade of that same sign inversion supporting your un-referenced direction issue. You cannot demonstrate the direction flow sign being constructed from a negative Rho on the basis of sqrt(Pref/Pfwd). If you feel you cannot communicate with me, you certainly have that right; but for this issue I am not the only one and it is not due to my lack of communication ability (as I am probably the only one here credentialed to that matter). 73's Richard Clark, KB7QHC |
Richard Clark wrote:
You cannot show that any two powers used to compute Rho are negative to fulfill this shift of your logic. It doesn't require either power to be negative. All it requires is a short circuit. +1 simply has two square roots. rho = -1 for a short circuit and rho = +1 for an open circuit. All (rho = -1) requires is a short at the end of a transmission line as explained in _Transmission_ Lines_and_Networks_, by Walter C. Johnson when he was chairman of the Princeton EE Dept. Here's how he calculated rho for a short: rho = (Z1-Z0)/(Z1+Z0) = (0-Z0)/(0+Z0) = -Z0/Z0 = -1 So your argument is with Dr. Johnson whom I am merely quoting. The (rho = -1) simply indicates a 180 degree phase shift in the reflected voltage at the short. From Johnson, section 1.6, page 16: Is NOT a citation. Who is Johnson? It doesn't surprise me a bit that you are ignorant of Johnson. In 1950, his book was one of the series of McGraw-Hill Electrical and Electronic Engineering Series with Terman as the consulting editor and containing textbooks by Kraus, Skilling, Terman, and others. ... it is not due to my lack of communication ability (as I am probably the only one here credentialed to that matter). Must be really difficult to communicate when you are so special as to be the "only one here" who is "credentialed to that matter". Many of the posters to this newsgroup have written books and articles which I find to be communicated rather well. -- 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, 30 Sep 2003 12:48:32 -0500, Cecil Moore
wrote: Richard Clark wrote: You cannot show that any two powers used to compute Rho are negative to fulfill this shift of your logic. Walter C. Johnson when he was chairman of the Princeton EE Dept. Here's how he calculated rho for a short: rho = (Z1-Z0)/(Z1+Z0) = (0-Z0)/(0+Z0) = -Z0/Z0 = -1 So your argument is with Dr. Johnson whom I am merely quoting. No my argument is with your perversion of yet another source in your vain attempt to draw a faulty conclusion in applying direction as the basis of the -1 drawn from your observation I probably should have said rho^2 = Pref/Pfwd. When Pref = Pfwd, rho can be plus or minus one. I notice you continually flee from your assertion to prove a different statement "Dr. Johnson" made. It doesn't surprise me a bit that you are ignorant of Johnson. The unreliable correspondent once again, in laziness, again fails to offer which Johnson. Presumably Walter, but you don't say, and several Dr. Johnsons have been employed as sources in this group. Your characteristic failure to attend boundary conditions is consistent with your inability to preserve your assertion that somehow a negative association is made with the sqrt(Pref/Pfwd) and that it proves a change of direction (wholly unsubstantiated by any but your own Johnson). Must be really difficult to communicate when you are so special as to be the "only one here" who is "credentialed to that matter". Many of the posters to this newsgroup have written books and articles which I find to be communicated rather well. And with whom you have such difficulty communicating with. 73's Richard Clark, KB7QHC |
Cecil, W5DXP wrote:
'I probably should have said rho^2 = Pref / Pfwd, rho can be plus or minus 1." Terman mentions a power ratio at the bottom of page 97 of his 1955 edition: "This definition of standing-wave ratio is sometimes called voltage standing-wave ratio (VSWR) to distinguish it from the standing-wave ratio expressed as a power ratio, which is (Emax / Emin) squared. Best regards, Richard Harrison, KB5WZI |
Cecil Moore wrote: Consider the following: Source---50 ohm feedline---+---1/2WL 150 ohm---50 ohm load Isn't the 50 ohms that causes rho=0 on the 50 ohm feedline simply the V/I ratio at point '+'? The nature of things a point '+' are undefined, so I can't address that. But according to the way you defined the problem, the characteristic impedance of the 50 ohm feedline is 50 ohms. That sets the V/I ratio. The impedance is determined by the distributed capacitances and inductances of the transmission line - not by the voltage you put across it. Is there some other way I'm supposed to look at it? 73, Jim AC6XG |
Cecil Moore wrote:
Isn't the 50 ohms that causes rho=0 on the 50 ohm feedline simply the V/I ratio at point '+'? I see now. Your interested in something else here, I think. The rho for the whole network which includes both impedance discontinuities is indeed zero. We've talked about that before. But the rho for the single discontinuity at '+' is not equal to zero. The reflected impedance (the load impedance, repeated a half wavelength away) is not considered in the evaluation of rho at '+'. It is the characteristic impedance of the line that is considered. You would agree, no? 73, Jim AC6XG |
Cecil Moore wrote:
All (rho = -1) requires is a short at the end of a transmission line as explained in _Transmission_ Lines_and_Networks_, by Walter C. Johnson when he was chairman of the Princeton EE Dept. Here's how he calculated rho for a short: rho = (Z1-Z0)/(Z1+Z0) = (0-Z0)/(0+Z0) = -Z0/Z0 = -1 So your argument is with Dr. Johnson whom I am merely quoting. The (rho = -1) simply indicates a 180 degree phase shift in the reflected voltage at the short. Quite false. Negation is not simply a 180 degree phase shift. And if Walter C. Johnson is worthy of the respect he receives here, he has certainly never said it is. ....Keith |
Jim Kelley wrote:
Cecil Moore wrote: Consider the following: Source---50 ohm feedline---+---1/2WL 150 ohm---50 ohm load Isn't the 50 ohms that causes rho=0 on the 50 ohm feedline simply the V/I ratio at point '+'? The nature of things a point '+' are undefined, Nope, they are not. The V/I ratio at '+' equals 50 ohms. so I can't address that. But according to the way you defined the problem, the characteristic impedance of the 50 ohm feedline is 50 ohms. That sets the V/I ratio. The impedance is determined by the distributed capacitances and inductances of the transmission line - not by the voltage you put across it. Is there some other way I'm supposed to look at it? There are no reflections on the 50 ohm feedline because it "sees" 50 ohms at point '+'. The 50 ohms seen at point '+' is a V/I ratio equal to 50 ohms. So V affects rho. And rho causes that same V? See the circular logic? -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Jim Kelley wrote:
Cecil Moore wrote: Isn't the 50 ohms that causes rho=0 on the 50 ohm feedline simply the V/I ratio at point '+'? I see now. Your interested in something else here, I think. The rho for the whole network which includes both impedance discontinuities is indeed zero. We've talked about that before. But the rho for the single discontinuity at '+' is not equal to zero. The reflected impedance (the load impedance, repeated a half wavelength away) is not considered in the evaluation of rho at '+'. It is the characteristic impedance of the line that is considered. You would agree, no? I would agree if you were talking about s11. But rho on the coax is zero. The impedance at '+' is 50 ohms. rho = (50-50)/(50+50) = sqrt(0/Pfwd) = 0 at point '+'. And that 50 ohms is a V/I ratio which, I assume, you would agree cannot cause a voltage. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Cecil Moore wrote:
But rho on the coax is zero. Coax doesn't have a 'rho' - unless it's broken or damaged coax. The impedance at '+' is 50 ohms. By virtue of reflection from the 50 ohm load. rho = (50-50)/(50+50) = sqrt(0/Pfwd) = 0 at point '+'. The same is true at points to the left of point '+' as well. So what? And that 50 ohms is a V/I ratio which, I assume, you would agree cannot cause a voltage. As far as I know, V/I ratios don't "cause" anything. 73, Jim AC6XG |
Cecil Moore wrote: Jim Kelley wrote: The nature of things a point '+' are undefined, Nope, they are not. There can exist no real point where the characteristic line impedance is both 50 ohms and 150 ohms. There are no reflections on the 50 ohm feedline because it "sees" 50 ohms at point '+'. Oh my gawd, somebody has brainwashed our Cecil! Maybe the pod people have invaded Texas! :-) What about cancelled reflected waves, destructive interference and all that? 73, Jim AC6XG |
Jim, AC6XG wrote:
"As far as I know, V/I ratios don`t cause anything." As symbols for load impedance, V/I ratios cause reflections when they differ from the r-f transmission line surge impedance which feeds them. Best rergards, Richard Harrison, KB5WZI |
Richard Harrison wrote: Jim, AC6XG wrote: "As far as I know, V/I ratios don`t cause anything." As symbols for load impedance, V/I ratios cause reflections when they differ from the r-f transmission line surge impedance which feeds them. Best rergards, Richard Harrison, KB5WZI Hi Richard, Would it be fair to say then that you don't agree that only physical boundaries cause reflections? 73, Jim AC6XG |
Jim Kelley wrote:
As far as I know, V/I ratios don't "cause" anything. They sometimes cause 'rho' which then becomes an end result and not the cause of anything. At a two-port network with reflections, rho usually cannot be calculated from the physical impedances involved. The moral is be careful about saying that rho causes anything. Rho may be only the end result of everything. The s11 reflection coefficient doesn't suffer from that characteristic. -- 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:
There can exist no real point where the characteristic line impedance is both 50 ohms and 150 ohms. I agree, and using the same logic, there also can be no such thing as real steady-state conditions. That doesn't keep us from using it as a real concept. +---------------- | ---------+ 50 ohms 150 ohms ---------+ | +---------------- You can draw a vertical line through the transition point. That vertical line has zero width. Ergo, the 50 ohm to 150 ohm transition requires zero length, conceptually, of course. But you knew that already. While you were at it, why didn't you point out that there cannot exist a real characteristic impedance exactly equal to 50 ohms or 150 ohms? What about cancelled reflected waves, destructive interference and all that? I should have said there are no net reflections on the 50 ohm feedline. There are two component reflections at point '+' that cancel each other as illustrated by the s-parameter equation, b1 = s11*a1 + s12*a2 = 0 s11 is a reflection coefficient for a1 and s12 is a transmission coefficient for a2, the voltage reflected from the load. Those two components have to cancel for the b1 net reflections to equal zero. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Richard Harrison wrote:
Jim, AC6XG wrote: "As far as I know, V/I ratios don`t cause anything." As symbols for load impedance, V/I ratios cause reflections when they differ from the r-f transmission line surge impedance which feeds them. So can a V/I ratio cause reflections in the absence of a physical impedance? For instance: Source---50 ohm feedline---+---150 ohm feedline---450 ohms The V/I ratio at point '+' is equal to 1350 ohms and the reflections on the coax are the same as if a 1350 ohm resistor existed at point '+'. So does the V/I=1350 ohms ratio cause the reflections at point 'x'? Or is the reflected voltage, b1, equal to s11*a1 + s12*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! =----- |
Jim, AC6XG wrote:
"Would it be fair to say that you don`t agree that only physical boundaries cause reflections?" A transmission line may be terminated in its surge impedance, yet somewhere along the line suffer a discontinuity producing an impedance irregularity and subsequent rteflection. The line may be matched to its final load, yet have reflections on the transmitter side of the irregularity. Terman illustrates this on page 118 of his 1955 edition. Terman suggests several reflection possibilities:: "sharp bends, insulating supports, resistive joints, coupled circuits and extraneous objects" as typical irregularities. I am not persuaded that reflection is caused by anything other than a physical discontinuity, but Terman includes "coupled circuits" in his list, and I think this indicates a physical discontinuity may be referred to a reflection point from elsewhere. Best regards, Richard Harrison, KB5WZI |
Cecil Moore wrote:
Jim Kelley wrote: As far as I know, V/I ratios don't "cause" anything. They sometimes cause 'rho' which then becomes an end result and not the cause of anything. I disagree. Reflections are caused by real impedances not reflected ones. Have you changed your mind about this? At a two-port network with reflections, rho usually cannot be calculated from the physical impedances involved. It can certainly be done using the optical formulas for a pair of boundaries. For the two boundaries as a network, and we call rho at the first boundary r12 and rho at the second boundary r23 then rho(network) = (r12 + r23)/(1 + r12*r23) = 0. Note that if we use your value for r12, the network generates a reflection. I note the utility of negative rho in this example. But, with a transmission line at odd multiples of lambda/4, rho for the network would be at a maximum and the network equation would be (r12 - r23)/(1 - r12*r23). In such a case you'd want to use a load impedance that would provide a r23 of +.5. (x - 150)/(x + 150) = .5, so x = 450. The moral is be careful about saying that rho causes anything. Rho may be only the end result of everything. No question that rho is the end result of a ratio of impedances. It's been my view that, like the V/I ratios we were speaking about, rho is not a cause but a result. 73, Jim AC6XG |
Cecil Moore wrote: Jim Kelley wrote: There can exist no real point where the characteristic line impedance is both 50 ohms and 150 ohms. I agree, and using the same logic, there also can be no such thing as real steady-state conditions. That doesn't keep us from using it as a real concept. No, but it made it difficult for me to anticipate every possible observation you might want to make about what goes on at such a point. 73, Jim |
Richard Harrison wrote: I am not persuaded that reflection is caused by anything other than a physical discontinuity, but Terman includes "coupled circuits" in his list, and I think this indicates a physical discontinuity may be referred to a reflection point from elsewhere. I see, yes. Interesting that he would distinguish and include "coupled circuits". Why do you suppose he felt "coupled circuits" were somehow different than other circuits? 73, Jim AC6XG |
Jim Kelley wrote:
Cecil Moore wrote: Jim Kelley wrote: As far as I know, V/I ratios don't "cause" anything. They sometimes cause 'rho' which then becomes an end result and not the cause of anything. I disagree. Reflections are caused by real impedances not reflected ones. Have you changed your mind about this? Where did I say rho causes reflections? I didn't! Your statement does NOT disagree with my statement. At a two-port network with reflections, rho usually cannot be calculated from the physical impedances involved. It can certainly be done using the optical formulas for a pair of boundaries. Unfortunately, only the index of refraction of one of the boundaries is known in my statement above. The index of refraction of the second boundary is unknown, i.e. only the impedances at the two-port network are known - the load is unknown. No question that rho is the end result of a ratio of impedances. It's been my view that, like the V/I ratios we were speaking about, rho is not a cause but a result. But earlier, I thought you said rho caused a result. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Jim Kelley wrote:
Cecil Moore wrote: Jim Kelley wrote: There can exist no real point where the characteristic line impedance is both 50 ohms and 150 ohms. I agree, and using the same logic, there also can be no such thing as real steady-state conditions. That doesn't keep us from using it as a real concept. No, but it made it difficult for me to anticipate every possible observation you might want to make about what goes on at such a point. All that proves is that you are not omniscient. :-) -- 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:
I see, yes. Interesting that he would distinguish and include "coupled circuits". Why do you suppose he felt "coupled circuits" were somehow different than other circuits? Additionally, exactly what does he mean by "coupled circuits"? Is a perfect transformer a coupled circuit? I don't see why there would be any reflections at a perfect transformer. -- 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:
All that proves is that you are not omniscient. :-) Do you know anybody who is? :-) 73 de ac6xg |
Cecil Moore wrote: Unfortunately, only the index of refraction of one of the boundaries is known in my statement above. The index of refraction of the second boundary is unknown, i.e. only the impedances at the two-port network are known - the load is unknown. Only if you've forgotten what you said it was. :-) If it's unknown, how could you have known what it was a half wavelength away? We are speaking about the problem you posed yesterday, right? No question that rho is the end result of a ratio of impedances. It's been my view that, like the V/I ratios we were speaking about, rho is not a cause but a result. But earlier, I thought you said rho caused a result. I'll repost what I said. If you find something in error, please advise. Thanks. "To my way of thinking, rho is entirely dependent upon the impedances, and the voltages (reflected voltages in particular) are dependent upon rho. Not the other way around." 73, Jim AC6XG |
Jim Kelley wrote:
Cecil Moore wrote: All that proves is that you are not omniscient. :-) Do you know anybody who is? :-) I know people who think they are. :-) -- 73, Cecil, W5DXP |
Jim Kelley wrote:
If it's unknown, how could you have known what it was a half wavelength away? We are speaking about the problem you posed yesterday, right? No, we are speaking about a statement I made unrelated to the problem I posted. For that statement, the length of the feedline is unknown and the load is unknown. What is known is the forward power and reflected power on each side of the impedance discontinuity. No question that rho is the end result of a ratio of impedances. It's been my view that, like the V/I ratios we were speaking about, rho is not a cause but a result. But earlier, I thought you said rho caused a result. "To my way of thinking, rho is entirely dependent upon the impedances, and the voltages (reflected voltages in particular) are dependent upon rho. You said "rho is not a cause but a result" but then implied that voltages are caused by (dependent upon) rho. Seems to me, rho cannot both cause a voltage and be caused by a (voltage divided by a current) which is an impedance upon which rho is dependent. -- 73, Cecil, W5DXP |
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