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Roy wrote -
If you have some other "rho" you want to argue about, please call it something else. =========================== - - - and while you are about it change the name of the SWR meter. |
On Wed, 03 Sep 2003 18:50:08 -0500, Cecil Moore
wrote: Since you snipped my posting, I have no idea what it was all about. Hi Cecil, I didn't this time, and I doubt you are any further ahead. Perhaps you suffer from the Motorola syndrome of confusion. ;-) So, is your response to the offer of the bridge description Yes? or No? (Consult google to fill the short attention span problems.) 73's Richard Clark, KB7QHC |
So, do you want the bridge description or not?
Richard Clark, KB7QHC ============================ Rich, pleased to receive a message from you succinct enough to read. ;o) But it looks like there's at least two contestants who have now vanished from the thread. Keep stirring it up. |
On Thu, 4 Sep 2003 00:38:54 +0000 (UTC), "Reg Edwards"
wrote: So, do you want the bridge description or not? Richard Clark, KB7QHC ============================ Rich, pleased to receive a message from you succinct enough to read. ;o) But it looks like there's at least two contestants who have now vanished from the thread. Keep stirring it up. How do you mix an ingredient of one? That query alone will bring in at least a dozen recipes. ;-) |
Cecil Moore wrote:
wrote: And yes, |rho| can be greater than unity for a passive load. But the power reflection coefficient cannot be greater than 1.0 which is what the argument is all about. Which is entirely consistent with my previous statement: It follows that when rho is greater than unity, it is not 'physically meaningful to separate the total power as the sum of the incident and reflected power' so the equation |rho| = Sqrt(Pref/Pfwd) has no meaning. I suppose one might phrase it as 'there is no such thing as a power reflection coefficient' when it is not 'physically meaningful to separate the total power as the sum of the incident and reflected power'. ....Keith |
Cecil Moore wrote:
Reg Edwards wrote: Roy wrote - If you have some other "rho" you want to argue about, please call it something else. - - - and while you are about it change the name of the SWR meter. Trouble is, (Z2-Z1)/(Z2+Z1) is not always equal to Sqrt(Pref/Pfwd) What then? The equality was always iffy when you don't take the absolute value. But once you do, the equality may hold depending on the equations you use to derive Pref and Pfwd. Whether Pref or Pfwd represent something physically meaningful is another question, also dependent on how you derive them. ....Keith |
Keith wrote:
"I suppose one might phrase it as "There is no such thing as a power reflection coefficient" when it is not physically meaningful to separate the total power as the sum of the incident and reflected power so the equatiomn: [rho] = sq. rt. (Pref / Pfwd) has no meaning." We don`t have a choice of options on a menu to select or reject from. Reality is whatever it is and we accept it and describe it as best we can. Terman says on page 97 of his 1955 edition: "{rho} = (SWR-1) / SWR + 1." Power varies as the equare of the voltage, because when you increase the volts you also automatically increase the amps (Ohm`s law). Thus, Terman has a subscript at the bottom of page 97 which is relevant: "The 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." In my long rxperience, I`ve found it`s never profitable to argue with Terman. He is as close to infallible as any wrirter I`ve ever read. Best regards, Richard Harrison, KB5WZI |
Cecil Moore wrote in message ...
Dr. Slick wrote: And how do you explain the rho 1 for a passive network? Shouldn't be possible. And neither should a negative SWR. This seems to me to be somewhat akin to the fact that s11 and rho can have different values at an impedance discontinuity where a 'third power' is commonplace. Roy's 'third power' at the load appears to be analogous to a re-reflection of some sort as the inductive load tries and fails to dump energy back into the Z0=68-j39 transmission line. A re-reflection is another component of forward power. The ratio of reflected Poynting vector to forward Poynting vector is |rho|^2. In Roy's example, the total average Poynting vector points toward the load indicating that (Pz+ - Pz-) 0. That means |rho|^2 cannot be greater than 1.0. Cecil, The ratio Pref/Pfwd is directly related to the ratio [rho]. Consider that after the absolute value brackets, the phase information is gone. But since we are going to a ratio of average (RMS) values OR peak values of power, it doesn't matter. In other words, if you use V**2/R, the "V" can be either peak or RMS, it doesn't matter, because it is a ratio. And of course, the "R" doesn't matter either. And of course, the phase information is gone with the absolute value brackets. If you agree that the Pref/Pfwd ratio cannot be greater than 1 for a passive network, then neither can the [Vref/Vfwd]= rho be greater than 1 either. Some people wanna rewrite some books here. Slick |
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"Reg Edwards" wrote in message ...
When the line is not lossless, ie., it has appreciable attenuation in dB per 1/4-wavelength, then the ratio is 'distorted' and has a phase angle. So negative values of indicated SWR can be expected at some values of | Vmax | / | Vmin | What are you talking about? If it have losses, and they are dissipative losses, the amplitude of the voltage will decrease due to voltage drops. That would be moving AWAY from having a greater reflected voltage than an incident one. But, that's impossible anyways with a passive network. The concept of Negative SWRs is rubbish. SWR is calculated from the square of | rho |. As I've said before, immediately | rho | is squared, half the information it contains is junked. Any discussion/argument about power waves following rho-squared on a lossy (a real ) line is meaningless piffle. Anybody who writes books about power waves, selling them to make a living, is obtaining money under false pretences. On the other hand we should be kind to otherwise unemployed Ph.D's. They too have wive's and kid's to clothe, feed and provide a roof over their heads. That's life! --- Reg. Remind me not to be YOUR book when it comes out! The ratio Pref/Pfwd is directly related to the ratio [rho]. Pref/Pfwd = [rho]**2 Absolute value brackets are a must! Consider that after the absolute value brackets, the phase information is gone. But since we are going to a ratio of average (RMS) values OR peak values of power, it doesn't matter. In other words, if you use V**2/R, the "V" can be either peak or RMS, it doesn't matter, because it is a ratio. And of course, the "R" doesn't matter either. And of course, the phase information is gone with the absolute value brackets. If you agree that the Pref/Pfwd ratio cannot be greater than 1 for a passive network, then neither can the [Vref/Vfwd]= rho be greater than 1 either. Some people wanna rewrite some books here. Slick |
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"David Robbins" wrote in message ...
sorry, no scanner here. how do you get rho1? please give me the Zo and Zl to try out, i have been playing for a while with the basic equations and haven't found a case where either formulat gives rho1. and of course if |rho|=1 then swr can never be negative. I think Reg put it best: "Dear Dr Slick, it's very easy. Take a real, long telephone line with Zo = 300 - j250 ohms at 1000 Hz. Load it with a real resistor of 10 ohms in series with a real inductance of 40 millihenrys. The inductance has a reactance of 250 ohms at 1000 Hz. If you agree with the following formula, Magnitude of Reflection Coefficient of the load, ZL, relative to line impedance = ( ZL - Zo ) / ( ZL + Zo ) = 1.865 which exceeds unity, and has an angle of -59.9 degrees. The resulting standing waves may also be calculated. Are you happy now ?" --- Reg, G4FGQ If it were not for Reg pointing out this example, i wouldn't have researched and corrected my original, "purely real" Zo post with the more general conjugate Zo formula. And i researched it because i knew that you cannot have a R.C. greater than one for a passive network (you can only have a R.C. greater than one for an active network, which would be a "return gain" instead of a "return loss"), so i knew that when Zo is complex, my original post must have been wrong. Intelligent people can be close-minded, that is for certainly, in which case, their intelligence is blunted. Slick |
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Richard Harrison wrote:
Keith wrote: "I suppose one might phrase it as "There is no such thing as a power reflection coefficient" when it is not physically meaningful to separate the total power as the sum of the incident and reflected power so the equatiomn: [rho] = sq. rt. (Pref / Pfwd) has no meaning." We don`t have a choice of options on a menu to select or reject from. Reality is whatever it is and we accept it and describe it as best we can. Terman says on page 97 of his 1955 edition: "{rho} = (SWR-1) / SWR + 1." Power varies as the equare of the voltage, because when you increase the volts you also automatically increase the amps (Ohm`s law). Thus, Terman has a subscript at the bottom of page 97 which is relevant: "The 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." In my long rxperience, I`ve found it`s never profitable to argue with Terman. He is as close to infallible as any wrirter I`ve ever read. Terman may be infallible, but I often find it unwise to trust his interpreters. The mention of SWR strongly implies lossless lines since VSWR varies along a lossy line. Perhaps in prose previous to the equation above he has limited his discussion to the lossless case. Quotes out of context must be interpreted with great care. ....Keith |
"Dr. Slick" wrote:
wrote in message ... Cecil Moore wrote: wrote: And yes, |rho| can be greater than unity for a passive load. But the power reflection coefficient cannot be greater than 1.0 which is what the argument is all about. Which is entirely consistent with my previous statement: It follows that when rho is greater than unity, it is not 'physically meaningful to separate the total power as the sum of the incident and reflected power' so the equation |rho| = Sqrt(Pref/Pfwd) has no meaning. It certainly does, because the ratio Pref/Pfwd is directly related to the ratio [rho]. Consider that after the absolute value brackets, the phase information is gone. But since we are going to a ratio of average (RMS) values OR peak values of power, it doesn't matter. Are you gonna re-write some books? Don't think I need to. But many need to read their books with more care. The root question is: Do Prev and Pfwd have physical meaning? or Are Prev and Pfwd just poorly named quantities which are useful in certain (common) circumstances? Once you settle on the latter definition, your difficulties will disappear and you will be free to experience rho through its full range of values. The proof that Prev and Pfwd are not, in general, physical things has been offerred on many recent threads. Although useful, it is unfortunate that Pnet does equal Pfwd-Prev for many common situations since people are tempted to generalize. The belief then becomes so ingrained, that when examples are presented which demonstrate the lack of generality, the example is rejected, or the equations modified, rather than examining the incorrect beliefs which lead to the difficulties. ....Keith |
"Dr. Slick" wrote:
wrote in message ... And yes, |rho| can be greater than unity for a passive load. ...Keith Absolute Rubbish.. Could you produce a passive circuit that will reflect a greater voltage than what you feed it? I'd LOVE to see that... Several examples have been presented, but rather than accepting them, you changed the definition of rho. Perhaps you could build one of these circuits to determine if modifying the definition of rho was appropriate. The ratio Pref/Pfwd is directly related to the ratio [rho]. Pref/Pfwd = [rho]**2 Absolute value brackets are a must! Consider that after the absolute value brackets, the phase information is gone. But since we are going to a ratio of average (RMS) values OR peak values of power, it doesn't matter. In other words, if you use V**2/R, the "V" can be either peak or RMS, it doesn't matter, because it is a ratio. And of course, the "R" doesn't matter either. And of course, the phase information is gone with the absolute value brackets. If you agree that the Pref/Pfwd ratio cannot be greater than 1 Which I haven't since Pref and Pfwd are just computed numbers and the result for some circuits is that Pref/Pfwd is greater than 1. Of course, Pnet is not equal to Pfwd-Pref in these circumstances so there is no violation of basic physics. It is just that the computation of Pfwd and Pref does not really produce real powers (though, again unfortunately, the dimension of the quantity produced is power). for a passive network, then neither can the [Vref/Vfwd]= rho be greater than 1 either. ....Keith |
Richard Harrison wrote:
In my long rxperience, I`ve found it`s never profitable to argue with Terman. He is as close to infallible as any wrirter I`ve ever read. All of the handy-dandy transmission line formulas that we have been using for many years apply specifically to lossless lines. A line with loss has a complex value of Z0. If the imaginary part of Z0 is more than a few percent of the real part we should use different methods. One famous example: Pload = Pforward - Preflected is one that has to be treated with suspicion if the line has appreciable loss (complex Z0). Another is : SWR = [1+|rho|]/[1-|rho|] At high values of rho close to 1.0, SWR becomes a totally useless concept. This is true regardless of which formula for rho that we use. We use the Smith chart outer circle to plot lengths of transmission line, for example stubs and matching transformers. We assume these lines taken by themselves are lossless and have infinite SWR (the outer circle of the Smith chart is the "locus" of infinite SWR). If we know the matched loss of a particular coax (dB per 100 ft) it is far better to use a math program and calculate everything, if the matched loss is not negligible. The computer is much more revealing than the Smith chart when line loss is significant. Bill W0IYH |
Dr. Slick wrote:
If you agree that the Pref/Pfwd ratio cannot be greater than 1 for a passive network, then neither can the [Vref/Vfwd]= rho be greater than 1 either. But apparently rho=(Z2-Z1)/(Z2+Z1) can be greater than unity. So those are not the same reflection coefficients. A physical rho and an image rho are quite often different values. s11 is a physical rho. Vref/Vfwd is an image rho. -- 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! =----- |
Dr. Slick wrote:
If you agree that the Pref/Pfwd ratio cannot be greater than 1 for a passive network, then neither can the [Vref/Vfwd]= rho be greater than 1 either. Sqrt(Pref/Pfwd) cannot be greater than one. (Z2-Z1)/(Z2+Z1) can be greater than one. Both are defined as 'rho' but they are not always equal. (Z2-Z1)/(Z2+Z1) is a physical reflection coefficient. Sqrt(Pref/Pfwd) is an image reflection coefficient. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
wrote:
It is just that the computation of Pfwd and Pref does not really produce real powers (though, again unfortunately, the dimension of the quantity produced is power). Funny how they can heat resistors both at the load end and at the source end when the source is equipped with a circulator+load. Incidentally, I've come up with a proof that there are no reflections at a voltage null in a homogeneous transmission line. Consider a 50 ohm lossless feedline. At a voltage null - rho = (50-50)/(50+50) = 0 i.e. no reflections in either direction -- 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 in message ... What all you experts have forgotten is that SWR on a lossless line is the ratio of two voltages, max and min, SPACED APART BY 1/4-WAVELENGTH. That is if the line is long enough to contain both a max and a min. When the line is not lossless, ie., it has appreciable attenuation in dB per 1/4-wavelength, then the ratio is 'distorted' and has a phase angle. So negative values of indicated SWR can be expected at some values of | Vmax | / | Vmin | SWR is calculated from the square of | rho |. As VSWR is defined as |Vmax|/|Vmin| and so can never be negative. in lossless lines this expression can be reduced to a function of rho, but that method is not valid in lossy lines. VSWR is not a constant in lossy lines and probably doesn't really mean much of anything as each voltage maximum and minimum is a different value, so which ones do you use??? |
Cecil Moore wrote:
wrote: It is just that the computation of Pfwd and Pref does not really produce real powers (though, again unfortunately, the dimension of the quantity produced is power). Funny how they can heat resistors both at the load end and at the source end when the source is equipped with a circulator+load. Not surprising. The energy comes from the source. Incidentally, I've come up with a proof that there are no reflections at a voltage null in a homogeneous transmission line. Consider a 50 ohm lossless feedline. At a voltage null - rho = (50-50)/(50+50) = 0 i.e. no reflections in either direction That's what you get when you use the surge impedance and compute surge rho. Try steady state impedance for steady state rho. rho = (0-50)/(0+50) = -1 as expected ....Keith |
David Robbins wrote:
btw, for whom ever has it... i am still waiting to see the derivation of the conjugate rho formula. i published one on here for the 'classical' version, where is the other one??? It exists in the Kurokawa paper, "Power Waves and the Scattering Matrix". He defines a new kind of wave, different from traveling waves, and calls them "Power Waves". That conjugate term is apparently the result of this new definition of waves. He says, "... when the main interest is in the relation between various circuits in which the sources are uncorrelated, the traveling waves are not considered as the best independent variables to use for the analysis." Seems he is not talking about a system where all the waves are coherent and has defined a new concept of a "Power Wave" which includes an alternate definition of a reflection coefficient which includes a conjugate term. -- 73, Cecil, W5DXP |
"W5DXP" wrote in message ... David Robbins wrote: btw, for whom ever has it... i am still waiting to see the derivation of the conjugate rho formula. i published one on here for the 'classical' version, where is the other one??? It exists in the Kurokawa paper, "Power Waves and the Scattering Matrix". He defines a new kind of wave, different from traveling waves, and calls them "Power Waves". That conjugate term is apparently the result of this new definition of waves. He says, "... when the main interest is in the relation between various circuits in which the sources are uncorrelated, the traveling waves are not considered as the best independent variables to use for the analysis." Seems he is not talking about a system where all the waves are coherent and has defined a new concept of a "Power Wave" which includes an alternate definition of a reflection coefficient which includes a conjugate term. -- 73, Cecil, W5DXP is that paper on the web somewhere?? i figured it had to be something with computing powers that was getting mixed in here some how, i think that is the only place you can end up with conjugates in transmission lines. so i assume its not a simple 1 page derivation from basic root principles, it must take a whole new language to express it. |
VSWR is not a constant in lossy lines and
probably doesn't really mean much of anything as each voltage maximum and minimum is a different value, so which ones do you use??? ------------------------------------------------------- --------- Dear David, You have expressed my sentiments exactly. I have never used either or any of them. What does anybody do with value of SWR when they imagine they know it? I'm pleased to make your acquaintance! For some years I have mildly advertised the idea of changing the name the name of the common-or-garden, so called SWR meter / combined forward-and-reflected power meter, to the TLI (Transmitter Loading Indicator) which is all it does. Although I must admit, at the present state of the art, it is a very useful instrument when changing antennas. Is the transmitter loaded with a resistance of 50 ohms or is it not? { Actually, the meter on my top-band transmitter indicates relative to 75 ohms } And there HAS to be SOMETHING more than the weather to talk about in QSO's and, of course, on this newsgroup. ;o) ---- Reg, G4FGQ |
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ok, Keith, i look forward with great interest on your
imaginary passive circuit which can reflect more power than what you feed it (incident power). I can't wait to hook it up to see more reflected power than incident on my DAIWA meter, that would be very interesting. Slick |
"Reg Edwards" wrote in message ...
Dear Cec, Your arithmetic is abominable. ;o) Dr Slick's vanishing-act was a better tactic. Your only avenue of escape is to prove the | rho | meter gives incorrect meter readings. ok, Reg, i look forward with great interest on your imaginary passive circuit which can reflect more power than what you feed it (incident power). I can't wait to hook it up to see more reflected power than incident on my DAIWA meter, that would be very interesting. Slick |
If the loss per unit wavelength is large enough, and you produced a plot of
voltage vs. distance x. The voltage maximum would be at the source, and the voltage minimum at the load. Try a thousand miles or so of RG58 at 60 Hz. I suspect that to see anything that looks like a standing wave you would have to look at dV/dx. Remember, I can always define a lossier line. Tam/WB2TT |
wrote:
That's what you get when you use the surge impedance and compute surge rho. Try steady state impedance for steady state rho. rho = (0-50)/(0+50) = -1 as expected Heh, heh, the steady state impedances are V/I ratios which are results incapable of causing anything. Your logic is a closed loop. The V/I ratios cause the rho to be -1. Therefore, rho=-1 causes the appropriate V/I ratios. Can you name any other result in reality that causes itself? Hint: Only physical impedance discontinuities can cause reflections. Image impedances are incapable of causing anything since they are an end 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! =----- |
wrote:
On the other hand... if Vrev = rho * Vfwd then rho^2 = (Vrev^2/Z0) / (Vfwd^2/Z0) So (Vrev^2/Z0) / (Vfwd^2/Z0) can be greater than one. If Sqrt(Pref/Pfwd) can not be greater than 1 then either Pfwd is not equal to (Vfwd^2/Z0) or Prev is not equal to (Vrev^2/Z0) Intriguing result, is it not? Only to the uninitiated. Contradictions don't exist in reality. They only exist in human minds. That should be a clue. |rho1| is not always equal to |rho2|. -- 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! =----- |
David Robbins wrote:
is that paper on the web somewhere?? Perhaps someone will offer it to you as a .pdf file. -- 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! =----- |
Dr. Slick wrote:
The definition of Rho has been set for "God-knows-how-long!" Actually, 'rho' has contradictory definitions. (Z2-Z1)/(Z2+Z1) is not always the same value as Sqrt(Pref/Pfwd) because of interference energy. -- 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 . ..
On 3 Sep 2003 23:16:39 -0700, (Dr. Slick) wrote: I think Reg put it best: Hi OM, If you are going to quote him as an authority supporting you, you should at least accept his offer of a bridge to settle this hash shouldn't you? You can't run far on one legged stilts. 73's Richard Clark, KB7QHC Who ever said he was an authority? All he did was open up a can-o-worms by showing how the "normal" Gamma equation is not always less than 1 for passive networks, which is good because i wouldn't have found out about the correct conjugate equation (when Zo is complex). But it is wise to pick and choose your fights, eh? And i'd rather argue with someone who is making sense at any particular point in time. Show us a passive circuit that reflects more power than you feed it (incident), on my Daiwa meter, and i will be VERY impressed. I'm be waiting a long time for that schematic.... Slick |
Dr. Slick wrote:
I can't wait to hook it up to see more reflected power than incident on my DAIWA meter, that would be very interesting. If that's really what you want to observe, connect the meter backwards. :-) -- 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:
wrote: That's what you get when you use the surge impedance and compute surge rho. Try steady state impedance for steady state rho. rho = (0-50)/(0+50) = -1 as expected Heh, heh, the steady state impedances are V/I ratios which are results incapable of causing anything. Your logic is a closed loop. The V/I ratios cause the rho to be -1. Therefore, rho=-1 causes the appropriate V/I ratios. Can you name any other result in reality that causes itself? Hint: Only physical impedance discontinuities can cause reflections. Image impedances are incapable of causing anything since they are an end result. Was it not only a few days ago that there was agreement that If it is acceptable to claim that sometimes there is no reflection at an impedance discontinuity, then it must also be acceptable to claim that sometimes there is a reflection where there is no discontinuity. ? ....Keith |
"Dr. Slick" wrote in message m... Cecil Moore wrote in message ... Dr. Slick wrote: If you agree that the Pref/Pfwd ratio cannot be greater than 1 for a passive network, then neither can the [Vref/Vfwd]= rho be greater than 1 either. Sqrt(Pref/Pfwd) cannot be greater than one. (Z2-Z1)/(Z2+Z1) can be greater than one. Both are defined as 'rho' but they are not always equal. (Z2-Z1)/(Z2+Z1) is a physical reflection coefficient. Sqrt(Pref/Pfwd) is an image reflection coefficient. I agree that Sqrt(Pref/Pfwd) cannot be greater than one for a passive network. (Z2-Z1)/(Z2+Z1) can be greater than one, for passive networks and certain combinations of complex Z1 and Z2. I feel this is incorrect usage of this formula, which should be limited to purely real Zo. A [rho] that is greater than one gives meaningless negative SWR data, and is limited to active devices. it only gives negative swr values if you incorrectly use the lossless line approximation to calculate vswr from rho. that is the incorrectly applied formula in this case. that formula is not valid for a lossy line, you must go back to the original definition of VSWR=|Vmax|/|Vmin|. which as we have also noted is not meaningful on a lossy line as Vmax and Vmin are different at each max and min point because of the losses in the line affecting both the forward and reverse waves. |
is that paper on the web somewhere?? i figured it had to be something
with computing powers that was getting mixed in here some how, i think that is the only place you can end up with conjugates in transmission lines. so i assume its not a simple 1 page derivation from basic root principles, it must take a whole new language to express it. ===================================== Yeah ! Trouble is nobody has yet dug up the ancient stone on which the language is carved and translated. There are too many unjustified * 's to make any sense out of thse recently discovered hieroglyphics. I am reminded of my old dear maths master, Mr Stevens. God had blessed him. He was a rare survivor of the machine gun bullets, shrapnel, flame-throwers, and chlorine-gas breathed in without a gas mask while hanging on the barbed wire in no-man's land between the trenches, Shell-fire-Corner, Ypres, Belgium, 1917. He spoke in a hoarse whisper and I always sat in the front row of desks in his classes so I could better hear him. He kept people awake at the back of the class by throwing missiles - sticks of chalk of which he was amply stocked. He referred to Factorial(x) = x! = x Exclamation mark, as "x By Jove" and so endowed on me a lifelong love of the beauty of mathematics. He also taught History in similar vein. --- Reg. |
On 4 Sep 2003 15:36:28 -0700, (Dr. Slick) wrote:
Richard Clark wrote in message . .. On 3 Sep 2003 23:16:39 -0700, (Dr. Slick) wrote: I think Reg put it best: snip Who ever said he was an authority? snip I'm be waiting a long time for that schematic.... Slick You're been thinkin' he done put it best and is no authority - uh huh. Are all such sources of your support held in similar esteem? I suppose the translation to the greater bulk of what you had to say is no. 73's Richard Clark, KB7QHC |
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