Home |
Search |
Today's Posts |
#51
![]() |
|||
|
|||
![]()
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! =----- |
#52
![]() |
|||
|
|||
![]()
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! =----- |
#53
![]() |
|||
|
|||
![]()
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 |
#54
![]() |
|||
|
|||
![]()
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! =----- |
#55
![]() |
|||
|
|||
![]()
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 |
#56
![]() |
|||
|
|||
![]()
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 |
#57
![]() |
|||
|
|||
![]() 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 |
#58
![]() |
|||
|
|||
![]()
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 |
#59
![]() |
|||
|
|||
![]()
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 |
#60
![]() |
|||
|
|||
![]()
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! =----- |
Reply |
Thread Tools | Search this Thread |
Display Modes | |
|
|