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Complex Z0 [Corrected]
Complex Z0 [Corrected]
- Maybe it has already been noticed, but anyway, it seems [this time] that, given a uniform transmission line with complex characteristic impedance, the magnitude of the reflection coefficient for any passive terminal load is lower or equal than - Sqrt([1 + Sin(Abs[t0])]/[1 - Sin(Abs[t0])]), - where t0 is the argument of Z0. - Sincerely, - pez,SV7BAX & yin,SV7DMC |
Complex Z0 [Corrected]
- Maybe it has already been noticed, but anyway, it seems [this time] that, given a uniform transmission line with complex characteristic impedance, the magnitude of the reflection coefficient for any passive terminal load is lower or equal than - Sqrt([1 + Sin(Abs[t0])]/[1 - Sin(Abs[t0])]), - where t0 is the argument of Z0. - =============================== It is true the formula gives the greatest possible magnitude of the reflection corfficient, Rho, for any given value of the angle of Zo. As t0 approaches -45 degrees, Rho approaches 1+Sqrt(2) = 2.414 As you must know, the formula is obtained by differentiating Rho with respect to the angle of Zo and then equating to zero. It provides proof that values of Rho greater than unity do exist. But a worship of mathematical logic is not part of any religion on this newsgroup. --- Reg, G4FGQ |
Reg Edwards wrote:
I challenge anyone to find a reflectometer calculator that shows rho 1. First of all, please define precisely what is a 'reflectometer calculator' ? Is it hardware or is it software? Heh, heh, if it's software, Reg will have one by tomorrow. :-) -- 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 ...
I challenge anyone to find a reflectometer calculator that shows rho 1. First of all, please define precisely what is a 'reflectometer calculator' ? Is it hardware or is it software? Hehe, well, this really is an amateur group, isn't it! Sorry, i keep forgetting. If you work in the RF field long enough, you eventually come across these cardboard slide-rules that will give you SWR versus rho versus mismatch loss. Every one that i have seen (HP, Roos, etc.) have a scale for rho that goes from zero to one. Also, i have never see a negative SWR in my life. Besser does mention that when you have an active device, that you can have a rho 1, and actually a Return GAIN instead of a Return Loss. Some people here seem to incorrectly think you can have a return gain with a passive network... Slick |
Dr. Slick wrote:
Some people here seem to incorrectly think you can have a return gain with a passive network... Does anyone remember what is the absolute value of a complex number? -- 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:
Some people here seem to incorrectly think you can have a return gain with a passive network... Does anyone remember what is the absolute value of a complex number? Found the answer in, "Higher Mathematics for Engineers and Physicists". I suspect the square of the absolute value of the voltage reflection coefficient is the volt-amp reflection coefficient, not the power reflection coefficient. With a complex characteristic impedance, what is being reflected is volt-amps. I suspect the reflected volt-amps can be higher than the incident volt-amps. I seriously doubt that the reflected watts can be higher than the incident watts. The correct *power* reflection coefficient therefore may be something like |Re(rho)|^2 where 'Re' means "the real part of". The simpler |rho|^2 may be the volt-amp reflection coefficient when Z0 is complex. Using deductive reasoning, since the real part of the voltage reflection coefficient cannot be greater than 1.0, it seems to me that |1.0|^2 may be the maximum power reflection coefficient. The complex voltage reflection coefficient squared may be the volt-amp reflection coefficient which can be greater than 1.0. In a transmission line with a complex characteristic impedance, the reflected voltage and reflected current would not be in phase. Therefore, their product would be volt-amps, not watts. Reflected watts could be obtained from Vref*Iref*cos(theta) which would always be less than (or equal to) Vref*Iref. -- 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! =----- |
In the numerical example I posted, I calculated the average real power
incident at the load (that is, the power calculated from the forward voltage and current waves), and the average real reflected power at the load (that is, the power calculated from the reverse voltage and current waves). The "reflected power" is greater than the "incident power". However, the net power exiting the line and entering the load is a positive value. That's because the net power isn't equal to the "forward power" minus the "reverse power" at that point. I gave the equation for total power in that analysis, and if you plug in the numbers, you'll see that the total power is correct. If you are interested in calculating the "reactive power" for some reason, you can easily do so from the complex voltages and currents which have been calculated for you. And for those who are wondering about your question, the absolute value of a complex number is the magnitude of that number. In the example I gave, all the complex values were given in polar form, with the first part being the magnitude. Roy Lewallen, W7EL Cecil Moore wrote: Cecil Moore wrote: Some people here seem to incorrectly think you can have a return gain with a passive network... Does anyone remember what is the absolute value of a complex number? Found the answer in, "Higher Mathematics for Engineers and Physicists". I suspect the square of the absolute value of the voltage reflection coefficient is the volt-amp reflection coefficient, not the power reflection coefficient. With a complex characteristic impedance, what is being reflected is volt-amps. I suspect the reflected volt-amps can be higher than the incident volt-amps. I seriously doubt that the reflected watts can be higher than the incident watts. The correct *power* reflection coefficient therefore may be something like |Re(rho)|^2 where 'Re' means "the real part of". The simpler |rho|^2 may be the volt-amp reflection coefficient when Z0 is complex. Using deductive reasoning, since the real part of the voltage reflection coefficient cannot be greater than 1.0, it seems to me that |1.0|^2 may be the maximum power reflection coefficient. The complex voltage reflection coefficient squared may be the volt-amp reflection coefficient which can be greater than 1.0. In a transmission line with a complex characteristic impedance, the reflected voltage and reflected current would not be in phase. Therefore, their product would be volt-amps, not watts. Reflected watts could be obtained from Vref*Iref*cos(theta) which would always be less than (or equal to) Vref*Iref. |
Nope.
I'm glad you're finding the time to look over the example. I see you've stumbled into the first problem with assigning a power to each individual wave. I'm afraid you'll encounter additional dilemmas as you dig deeper into it. Roy Lewallen, W7EL Cecil Moore wrote: Roy Lewallen wrote: The "reflected power" is greater than the "incident power". So if the load is put into a black box, there is more power coming out of the box than is going in? |
Roy Lewallen wrote:
I see you've stumbled into the first problem with assigning a power to each individual wave. I've stumbled upon the first problem in your solution. :-) What are Z0 and ZLoad again? Is Z0 physically possible? Is ZLoad physically possible? -- 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! =----- |
Roy Lewallen wrote:
No, the average Poynting vector points toward the load. That automatically says Pz- is not larger than Pz+. There are only two component Poynting vectors, 'Pz+' forward and 'Pz-' reflected. If so, surely you came up with the same result, including the third power term. If you haven't done the derivation, or if you'd like to compare your derivation of total average power with mine, I'll be glad to post it. Assuming coherent waves, all wave components flowing toward the load superpose into the forward wave and all wave components flowing away from the load superpose into the reflected wave. Since there are only two directions, there cannot exist a third wave. If your average Poynting vector points toward the load, Pz- cannot possibly be larger than Pz+. But feel free to post the derivation. -- 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! =----- |
"Roy Lewallen" wrote in message ... Z0 = 68 - j39 ohms. Zl = 10 + j50 ohms. Zl is certainly physically possible. I believe Z0 is also. According to A/C/F the angle of Zo is constrained to +/- 45 degrees. |
Cecil Moore wrote:
Using deductive reasoning, since the real part of the voltage reflection coefficient cannot be greater than 1.0, ... So much for my deductive reasoning - A kind soul has furnished proof by email that the real part of the voltage reflection can be greater than 1.0. Z_0 = 50 - 25j (which is well within the - 45 to 45 degree angle bounds) Z_L = 50 + 250j (chosen to make the arithmetic easy; there are lots more) Then Z_L - Z_0 = 275j and Z_L + Z_0 = 100 + 225j so that gamma = 11j/(4 + 9j) = 11j(4 - 9j)/(16 + 81) = (99 + 44j)/97. -- 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! =----- |
Well, Cecil, I think we're zeroing in on the flaw in your perception of
how the powers add. I hope you won't just keep saying it's impossible, and will instead sharpen your pencil to show, as I have, the forward, reverse, and total voltages, currents, and powers at both ends of the line. And how everything can work together consistently to fit into your view of power addition and subtraction. A number of people have been trying for a long time to convince you there's a flaw in your logic, but so far you haven't been able to see it. Hopefully, in the process of deriving the values for this circuit, you'll see where your logic has gone astray. Or, perhaps, you'll come up with a completely consistent set of voltages, currents, and powers that do fit within your view. And we'll all learn from it as we see where the difference arises between your analysis and mine. Until you come up with your analysis, though, I won't pay much attention to your complaints that it's wrong unless you're able to show where in the analysis the error lies. I've posted the derivation of the total power formula on this thread. In going through it, I found an error in the formula posted with my numerical example. I've posted a correction for that on the same thread as the example. In the correction posting, I also show how the formula produces the same result as I got by directly calculating the total power from the load voltage and current. A closing quotation, from Johnson's _Transmission Lines and Networks_: "[For a low loss line] P = |E+|^2 / Z0 - |E-|^2 / Z0. We can regard the first term in this expression as the power associated with the forward-traveling wave, and the second term as the reflected power. This simple separation of power into two components, each associated with one of the traveling waves, can be done only when the characteristic impedance is a pure resistance. Otherwise, the interaction of the two waves gives rise to a third component of power. Thus, the concept applies to low-loss lines and to distortionless lines, but not to lossy lines in general." Something for you to think about. Or maybe you subscribe to Reg's view that these texts are written by marketeers and salesmen. After all, as Chairman of Princeton's EE department, I suppose Johnson's job was primarily PR. I'm quite sure that if you look carefully at any text where the author subtracts "reverse power" from "forward power" to get total power, that somewhere prior to that the assumption is made that loss is zero and/or the line's characteristic impedance is purely real. Roy Lewallen, W7EL Cecil Moore wrote: Roy Lewallen wrote: No, the average Poynting vector points toward the load. That automatically says Pz- is not larger than Pz+. There are only two component Poynting vectors, 'Pz+' forward and 'Pz-' reflected. If so, surely you came up with the same result, including the third power term. If you haven't done the derivation, or if you'd like to compare your derivation of total average power with mine, I'll be glad to post it. Assuming coherent waves, all wave components flowing toward the load superpose into the forward wave and all wave components flowing away from the load superpose into the reflected wave. Since there are only two directions, there cannot exist a third wave. If your average Poynting vector points toward the load, Pz- cannot possibly be larger than Pz+. But feel free to post the derivation. |
I believe that, and the value I used is within that range.
Roy Lewallen, W7EL Tarmo Tammaru wrote: "Roy Lewallen" wrote in message ... Z0 = 68 - j39 ohms. Zl = 10 + j50 ohms. Zl is certainly physically possible. I believe Z0 is also. According to A/C/F the angle of Zo is constrained to +/- 45 degrees. |
Roy Lewallen wrote:
Well, Cecil, I think we're zeroing in on the flaw in your perception of how the powers add. I don't see how this thread is relevant to the treatment of powers in lossless lines. Perhaps you have misunderstood what I said. A number of people have been trying for a long time to convince you there's a flaw in your logic, but so far you haven't been able to see it. If there's a flaw for lossless lines with purely resistive characteristic impedances, please present it. So far, nobody has. Here's what I said in my magazine article on my web page" "For the purpose of an energy analysis involving *LOSSLESS* transmission lines, we do not need to know anything about the source or the load or the length of the transmission lines." I'm quite sure that if you look carefully at any text where the author subtracts "reverse power" from "forward power" to get total power, that somewhere prior to that the assumption is made that loss is zero and/or the line's characteristic impedance is purely real. Of course, that's why my previous assertions have been only about lossless lines. Do you happen to have a lossless example that proves my concepts about lossless lines are wrong? I have no concepts about lossy lines except that they obey the conservation of energy principle. -- 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! =----- |
Roy Lewallen wrote:
P1 = fP - rP + (|fE1|^2 / |Z0|) * rho * exp(-2 * ax) * 2 * sin(delta) * sin(2 * bx - 2 * psi). Seems to me, all the terms with a '+' sign would be forward power, by definition, and all the terms with a '-' sign would be reflected power, by definition. I don't see any violation of the conservation of energy principle. The power equation balances. -- 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 apologize. I thought your view of power waves was alleged to hold true
even with loss. If it's restricted to lossless lines (which have purely real Z0), then the total average power does equal "forward power" minus "reverse power". So please don't bother yourself with trying to explain the component of power that's neither the "forward power" nor "reverse power". Roy Lewallen, W7EL Cecil Moore wrote: Roy Lewallen wrote: Well, Cecil, I think we're zeroing in on the flaw in your perception of how the powers add. I don't see how this thread is relevant to the treatment of powers in lossless lines. Perhaps you have misunderstood what I said. A number of people have been trying for a long time to convince you there's a flaw in your logic, but so far you haven't been able to see it. If there's a flaw for lossless lines with purely resistive characteristic impedances, please present it. So far, nobody has. Here's what I said in my magazine article on my web page" "For the purpose of an energy analysis involving *LOSSLESS* transmission lines, we do not need to know anything about the source or the load or the length of the transmission lines." I'm quite sure that if you look carefully at any text where the author subtracts "reverse power" from "forward power" to get total power, that somewhere prior to that the assumption is made that loss is zero and/or the line's characteristic impedance is purely real. Of course, that's why my previous assertions have been only about lossless lines. Do you happen to have a lossless example that proves my concepts about lossless lines are wrong? I have no concepts about lossy lines except that they obey the conservation of energy principle. |
Roy:
[snip] There are a lot of opportunities for typos in a derivation like this, especially when restricted to plain ASCII characters. I'd appreciate very much if anyone finding an error, either in concept, fact, assumption, or just typo, to call it to my attention so it can be corrected. Roy Lewallen, W7EL [snip] Wow! You said it Roy. BTW... thanks for all of your nice work. But for my taste it's far too detailed and seems filled with gratuitously long symbols for ASCII text consumers. My eyes glazed over and I nearly fell asleep and had to stop following after a couple of screens of what seemed to turn into gibberish before my eyes. Not your fault mind you, it's mine. On the other hand, ASCII NewsGroup postings are hardly the media for sharing such detailed algebraic/numeric developments! I just don't get the point of all of your wonderful efforts! Thoughts, comments, -- Peter |
"Peter O. Brackett" wrote:
On another whole level it simply DOES NOT MATER which defiinition of the reflection coefficient one uses to make design calculations though, as long as the definition is used consistently throughout any calculations. One can convert any results based on the non-conjugate version of rho to results based on the conjugate version of rho and vice versa. In other words, neither version is "RIGHT" or "WRONG" as long as the results from using that particular definition are interperted correctly in terms of the original definition. While true, this is not what is occuring in the 'revised rho' debate. Their claim is simply that 'classical rho' has been mis-calculated all these years and we should start using the 'proper' calculation. There is no acknowledgement that 'revised rho' will have different properties than 'classical rho' and that, therefore, they are introdcing a new entity. Their claim of incorrectness derives from the fact that 'classical rho' can have a magnitude greater than 1 and a belief that this means reflected power is greater than incident. This belief is inconsistent with generally accepted knowledge, so rather than modifying the belief, the derivation of 'classical rho' is rejected. Their second difficulty derives from not being able to separate the behaviour at a particular interface from the system behaviour. They do not recognize that a reflection at a particular interface (which would reduce energy transfer at that interface), can improve overall system energy transfer by improving the energy transfer at another interface. This being what a transmission line transformer does, for example. Once they overcome these two hurdles, they will have no problems with the classical definition of rho. So... who gives a damm about the defintion of rho as long as you are consistent in it's use. It simply doesn't matter! [Unless you choose M to be singular. ;-) ] There is no problem if this is what the 'revised rho' crowd really is attempting to do, but they should clearly state this and have the courtesy to pick a new name (despite Humpty-Dumpty's assertions) to facilitate clear communication. Really though, you are thinking several levels above them when you hypothesize the existence of other, self-consistent, definitions of rho. They are still at the 'classical rho computation is just plain wrong' level. ....Keith |
Roy Lewallen wrote:
Well, shucks, that makes it easy. Just being logical. There are only two directions in a transmission line, forward and reverse. If all the waves are coherent, all forward waves superpose to one wave and all reverse waves superpose to one other wave. Your net forward power is greater than your net reflected power by the net amount of power accepted by the load. This happens locally at the load no matter what is happening elsewhere in the transmission line. Cecil Moore wrote: Seems to me, all the terms with a '+' sign would be forward power, by definition, and all the terms with a '-' sign would be reflected power, by definition. I don't see any violation of the conservation of energy principle. The power equation balances. -- 73, Cecil http://www.qsl.net/w5dxp "One thing I have learned in a long life: that all our science, measured against reality, is primitive and childlike ..." Albert Einstein -----= 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! =----- |
Peter O. Brackett wrote:
My eyes glazed over and I nearly fell asleep and had to stop following after a couple of screens of what seemed to turn into gibberish before my eyes. Not your fault mind you, it's mine. What helps for me is to print it out. I evolved looking at printed pages, not computer screens. I can scribble the correct math operators down so I don't have to remember what e** means. I just don't get the point of all of your wonderful efforts! Roy may have explained Richard's data. In any case, it's good to know that we cannot use the simplified wave reflection model on very lossy lines. It appears that a lossy line doesn't yield a smooth spiral on a Smith Chart. -- 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! =----- |
It obviously doesn't bother you that your new "forward power" isn't the
product of forward voltage and current, or that the new "reverse power" isn't the product of reverse voltage and current. But then I guess we shouldn't be surprised. Roy Lewallen, W7EL Cecil Moore wrote: Roy Lewallen wrote: Well, shucks, that makes it easy. Just being logical. There are only two directions in a transmission line, forward and reverse. If all the waves are coherent, all forward waves superpose to one wave and all reverse waves superpose to one other wave. Your net forward power is greater than your net reflected power by the net amount of power accepted by the load. This happens locally at the load no matter what is happening elsewhere in the transmission line. Cecil Moore wrote: Seems to me, all the terms with a '+' sign would be forward power, by definition, and all the terms with a '-' sign would be reflected power, by definition. I don't see any violation of the conservation of energy principle. The power equation balances. |
I've given the equation. With that and a spreadsheet or plotting program
(or graph paper) of your choice, you can have the plot in minutes. Note that x is the distance from the load. Roy Lewallen, W7EL Cecil Moore wrote: . . . What does a plot of that extra power look like up and down the line? |
Note also that if the equation for total average power is used to find
average power at any point along the line, fE1 becomes the forward voltage at the observation point, not necessarily the forward voltage at the input end of the line. So to calculate the total power as a function of position along the line, it's probably best to use a voltage at a fixed point, such as fE2, the forward voltage at the load, in its place. Make the substitution |fE1|^2 = |fE2|^2 * exp(2 * ax) for |fE1|^2 to make the power equation more usable for this purpose. Roy Lewallen, W7EL Roy Lewallen wrote: I've given the equation. With that and a spreadsheet or plotting program (or graph paper) of your choice, you can have the plot in minutes. Note that x is the distance from the load. Roy Lewallen, W7EL Cecil Moore wrote: . . . What does a plot of that extra power look like up and down the line? |
As derived, the equation for total average power at any point along the
line contains the term |fE1|^2, which is the square of the magnitude of the forward voltage at that point. It might be more useful to replace fE1 with a value which doesn't vary with position along the line, such as fE2, the forward voltage at the load. The substitution is: |fE1|^2 = |fE2|^2 * exp(2 * ax) which gives the alternate formula P1 = (|fE2|^2 / |Z0|) * (exp(2 * ax) * (1 - rho^2 * exp(-4 * ax)) * cos(delta) + rho * (2 * sin(delta) * sin(2 * bx - 2 * psi))) And, subtituting values for "forward power" and "reverse power": P1 = fP - rP + (|fE2|^2 / |Z0|) * rho * 2 * sin(delta) * sin(2 * bx - 2 * psi). Roy Lewallen, W7EL Roy Lewallen wrote: Here's the calculation of total average power P1 at any point on a transmission line. The point on the line is called point 1, and the location of the load is called point 2. The distance between them is x. . . . = (|fE1|^2 / |Z0|) * ((1 - rho^2 * exp(-4 * ax)) * cos(delta) + rho * exp(-2 * ax) * (2 * sin(delta) * sin(2 * bx - 2 * psi))) Subtituting values for "forward power" and "reverse power", we have: P1 = fP - rP + (|fE1|^2 / |Z0|) * rho * exp(-2 * ax) * 2 * sin(delta) * sin(2 * bx - 2 * psi). . . . |
Cecil:
[snip] In any case, it's good to know that we cannot use the simplified wave reflection model on very lossy lines. It appears that a lossy line doesn't yield a smooth spiral on a Smith Chart. -- 73, Cecil http://www.qsl.net/w5dxp [snip] Amen brother and... heh, heh... especially for broad band signals. Smith Charts are for mono-chomatic signals. Most tough transmission problems are broad band and the Smith Chart yeilds no useful insight in those problems. A widely applied practical example is the transmission of bi-directional broad band digital subscriber loop (DSL) signals over telephone twisted pair. Telephone twisted pair is very lossy... at the "standard" 18,000 foot length you can barely tell what is connected on the other end, short or open. In fact it might just as well be "semi-infinite"! The longest spans we have built chips for were up to 47,000 feet of #24 AWG full duplex data transmission at the basic rate with digital echo cancellation on both ends using trellis coded pulse amplitude modulation. I can assure you that 47,000 feetof #24 AWG definitely has a complex and lossy Zo! I had quite a few big "spools" of such cable in my lab for the beta tests! The big problem with such designs is not maximum power transfer, rather it is hearing the remote end in the presence of the local transmitter blasting away on the same pair as the receiver [talker echo] and so one needs to "image match" the transmitter to eliminate as much talker echo as possible and just take whatever power reaches the receiver at the other end. Of course you have some control over the spectrum of the power that reaches the other end by "pre-coding" at the transmitter, still the optimum strategy at the transmitter is to get an "image match". i.e. make the generator internal impedance as close to the complex Zo as you can make it! And... in those problems you need to differentiate two forward waves and two reflected waves. Heh, heh... hard to do that using just the two symbols Vfwd and Vref or V_+ and V_-, you need symbols for at least two each... Say Vfwd_1 and Vfwd_2 and Vref_1 and Vref_2, etc... messy to say the least! For this reason I much prefer the Scattering Formalism symbols "a" for incident and "b" for reflected, a1 for indicdent on port 1 and a2 for incident on port 2, then b1, b2, etc... Sometime, when I get some free time from my current consulting gig, I'll prepare a short example for the group of the problems inherent in full duplex signalling over complex Zo lines in situations where the "best" Engineering solution is "image match" not "conjugate match". ;-) -- Peter K1PO Indialantic By-the-Sea, FL. |
Reg:
[snip] Peter, the stage is now set to introduce Eigenvectors, Eigenvalues and Sylvesters theorem for square matrices. ;o) --- Reg [snip] Ahhhh... Sylvester, I knew him well! Roy even fusses at me and insists that the transformation matrix M between v, i and a, b include a factor of 1/2 as: a = v/2 + Ri/2 = 1/2 [v + Ri] b = v/2 - Ri/2 = 1/2 [v - Ri] i.e. M is: |1/2 R/2| |1/2 -R/2| so that the M is chosen for the voltage and current values to match the values of v and i found in solutions of the Telegraphists Equation. That's fine, but with that simple multiplier common to all elements of the linear combination of electricals to make the waves, it just don't matter since as you know, when you form rho = b/a the factor of 1/2 just drops out. I suppose introducing the "outer product" of the two vectors B = [b1, b2] and [A^-1]'[ = [1/a1, 1/a2]' as Bx[A^-1]' = S to form the two by two Scattering Matrix and showing that the factor of 1/2 disappears there as well would be far far too much for this group to assimilate! I can hear Oliver rolling over in his grave! Long live Sylvester! ;-) -- Peter K1PO Indialantic By-the-Sea, FL. |
Roy Lewallen wrote:
It obviously doesn't bother you that your new "forward power" isn't the product of forward voltage and current, or that the new "reverse power" isn't the product of reverse voltage and current. But then I guess we shouldn't be surprised. I guess I will quote Aristotle on that one, Roy. A thing is what it is - It's not something else. Two directions are all that exist, forward and backwards. Do I need to publish a binary truth table? -- 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! =----- |
"Peter O. Brackett" wrote in message link.net... Sometime, when I get some free time from my current consulting gig, I'll prepare a short example for the group of the problems inherent in full duplex signalling over complex Zo lines in situations where the "best" Engineering solution is "image match" not "conjugate match". ;-) -- Peter K1PO Indialantic By-the-Sea, FL. Peter, You have just answered the question that people have been arguing about the last few weeks. I hope you have time to describe this in more detail. Tam/WB2TT |
Peter O. Brackett wrote: A widely applied practical example is the transmission of bi-directional broad band digital subscriber loop (DSL) signals over telephone twisted pair. Peter K1PO Indialantic By-the-Sea, FL. As Peter notes, telephone transmission line impedance is always complex. The parameters R, L, G & C per unit length (series resistance, series inductance, shunt conductance, shunt capacitance) are NOT CONSTANT with frequency, or temperature. So the cable impedance is not constant either. Signal spectra extend from nearly DC (a few kHz) up to 12MHz or more - many octaves. Even over voiceband, 400Hz to 2800 Hz, the cable impedance changes a _lot_. Lengths vary from several feet to 10s of kft. There are often/usually open-ended shunt cable sections, a.k.a. bridged taps, along the cable. Other things, like series lumped loading coils (inductors), may appear if not removed from longer cables. Signals at the DSL receiver ends are umpteen dB below the transmitter signal levels on the same pair of wires, and can be in the same band if separate to-the-customer and to-the-network bands are not used. Smith Charts, as much as I like them for ham purposes, are of no help. This subject is addressed in T1.417-2001, Issue 1 "Spectrum Management For Loop Transmission Systems" January, 2001 Developed by Sub-Committee T1E1.4 which develops the xDSL standards (DSL, HDSL, ADSL, VDSL,...) for North America. Annex B of T11.417 deals with the modeling of cables for such cases: formulas, RLCG vs. freq. and other data for common AWG and metric cables... Software packages are available offline and online. http://net3.argreenhouse.com:8080/dsl-test/index.htm (A free registration is needed.) (The other 200+ pages are left for the reader.) The latest working draft of Issue 2 is available free* as document T1E1.4/2003-002R3 from http://www.t1.org/filemgr/filesearch.taf Do a "Simple Search" for filename 3e140023 When the "Results of Simple Search" page appears, click on the blue full name T1E1.4/2003-002R3 under the Contributions column. When the next page appears, click on the blue 3e140023 after "File Prefix" to finally download the document. (2.1MB) I just tried this procedure to be sure it works. * The official Issue 1 is over US$300. In 1995 I was the first editor and wrote the first draft of what became T1.417. Much of what I wrote is intact word-for-word as the first half of Annex B (to my amazment) - the general descriptive part before the nitty-gritty models and numbers. Have fun. There will be an exam. :) Cheers, 73, Ron McConnell Retired Secretary T1E1.4 N 40º 46' 57.9" W 74º 41' 21.9" FN20ps77GU46 [FN20ps77GV75] http://home.earthlink.net/~rcmcc |
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"The Third Term" - Roy Lewallen wrote: - | No, the average Poynting vector points toward the load. | Power leaves the line and enters the load, as it should. | ... | I imagine your problem | with the solution is your notion that | the total average power | is the difference between the | "forward power" and "reverse power". | | But it's not. | | I gave the equation showing what the total power | is, and as you can see, | there's a third term involved. | When this is | taken into consideration, you see that there's a net power flow out of | the line into the load, as there should be. | ... - As usually, Mr. Roy Lewallen, points the right direction. And this time, it is of: - "The Third Term". - In the whole of the book by R.A.Chipman, a phrase, less than a printed line, is proved enough to cause a major upset: - "The third term on the right represents interaction between the two waves." - But when there is such a steadfast loyalty to the existence of some kind of "interference" between two, rather clearly distinct waves, the incident and the reflected one, it is difficult for anybody to compromise himself and accept that the same two waves, so clearly distinct until now, when are coming along a line with complex Z0, have to bear in addition some kind of "interaction". - Very difficult, indeed. - Sincerely, - pez SV7BAX - |
Johnson uses the same term of "interaction" to describe the origin of
the extra power term. However, you can hopefully see from the analysis I posted that only ordinary superposition of the forward and reverse voltage and current waves is necessary for the term to appear. So I don't feel that "interaction" is entirely appropriate. The extra term actually is a result of the calculation of average power. I've said many times that it's risky to abandon the time information in the power waveform and deal only with averages. If the voltage and current in each wave aren't in phase with each other, there are components of the total VI product that add together but don't show up in the averages of the individual forward and reverse average powers. There's no mystery or true "interaction" involved. The "problem" lies simply in calculating average "forward power" and "reverse power" separately, throwing away all time related information, then expecting them to add or subtract to get the total. Roy Lewallen, W7EL pez wrote: - "The Third Term" - Roy Lewallen wrote: - | No, the average Poynting vector points toward the load. | Power leaves the line and enters the load, as it should. | ... | I imagine your problem | with the solution is your notion that | the total average power | is the difference between the | "forward power" and "reverse power". | | But it's not. | | I gave the equation showing what the total power | is, and as you can see, | there's a third term involved. | When this is | taken into consideration, you see that there's a net power flow out of | the line into the load, as there should be. | ... - As usually, Mr. Roy Lewallen, points the right direction. And this time, it is of: - "The Third Term". - In the whole of the book by R.A.Chipman, a phrase, less than a printed line, is proved enough to cause a major upset: - "The third term on the right represents interaction between the two waves." - But when there is such a steadfast loyalty to the existence of some kind of "interference" between two, rather clearly distinct waves, the incident and the reflected one, it is difficult for anybody to compromise himself and accept that the same two waves, so clearly distinct until now, when are coming along a line with complex Z0, have to bear in addition some kind of "interaction". - Very difficult, indeed. - Sincerely, - pez SV7BAX - |
Roy Lewallen wrote:
The "problem" lies simply in calculating average "forward power" and "reverse power" separately, throwing away all time related information, then expecting them to add or subtract to get the total. The s-parameter analysis doesn't have a problem doing that, Roy. Your analysis won't either when you include all the appropriate terms. -- 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! =----- |
No, no, Cecil. *You* need to include all the separate voltages and
currents, to show us. I'm glad that you find the s-parameter analysis to be more trouble free. So do it, and when you're done, substitute back in for transmission line characteristic impedance, length, and loss; and load impedance, then show us the resulting voltages, currents, and powers. Should be easy, no? Roy Lewallen, W7EL Cecil Moore wrote: Roy Lewallen wrote: The "problem" lies simply in calculating average "forward power" and "reverse power" separately, throwing away all time related information, then expecting them to add or subtract to get the total. The s-parameter analysis doesn't have a problem doing that, Roy. Your analysis won't either when you include all the appropriate terms. |
Roy Lewallen wrote:
No, no, Cecil. *You* need to include all the separate voltages and currents, to show us. Do you think I am capable of inventing something so complex just to pull your leg, Roy? What I am saying is grounded in physics. You simply made a simple mistake in assuming one term was forward voltage when actually, there are two terms when added together, yield the forward voltage. You didn't realize that you were doing a 2-port analysis. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Cecil:
[snip] forward voltage. You didn't realize that you were doing a 2-port analysis. -- 73, Cecil http://www.qsl.net/w5dxp [snip] I was just castigated by Roy and Dave Robbins on another thread for making postings about "ports", apparently we are not allowed to discuss "ports" when we are discussing transmission lines since ports have only to do with networks and transmission lines have no ports!!! ;-) -- Peter K1PO [A guy who believes that transmission lines have two ports.] Indialantic By-the-Sea, Fl |
Peter O. Brackett wrote:
I was just castigated by Roy and Dave Robbins on another thread for making postings about "ports", apparently we are not allowed to discuss "ports" when we are discussing transmission lines since ports have only to do with networks and transmission lines have no ports!!! Darn Peter, I was hoping you could help me explain to Roy what is wrong with his analysis - that he is using a 2-port analysis and getting four power terms as a result, two of which have to be added to get forward power and the other two of which have to be added to get reflected power. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Dave Shrader wrote:
Or, how about: one with an in'port' and the second with an out'port'? Speaking of which, I just thought of a way to alleviate the confusion about the earlier example which was: ------lossy feedline---x----10+j60 ohm load I hope my memory is correct about the load value. If we add one wavelength of lossless feedline to the experiment, we don't change anything but things become a lot clearer. ---lossy feedline--+--1WL lossless 50 ohm feedline--10+j60 ohm load Pfwd1-- Pfwd2-- --Pref1 --Pref2 It is readily apparent that Pref2 cannot be greater than Pfwd2. If my math is correct, at the load, |rho|=0.85 and SWR=12.3:1 |rho|^2=0.7225 so Pref2 = 0.7225(Pfwd2) The main thing to gather from the above example is that: (Pfwd2 - Pref2) = (Pfwd1 - Pref1) Therefore if Pfwd2 Pref2 then Pfwd1 Pref1 Conclusion: This is NOT an example of Pfwd Pref. -- 73, Cecil, W5DXP |
"Cecil Moore" wrote in message ... Peter O. Brackett wrote: I was just castigated by Roy and Dave Robbins on another thread for making postings about "ports", apparently we are not allowed to discuss "ports" when we are discussing transmission lines since ports have only to do with networks and transmission lines have no ports!!! Darn Peter, I was hoping you could help me explain to Roy what is wrong with his analysis - that he is using a 2-port analysis and getting four power terms as a result, two of which have to be added to get forward power and the other two of which have to be added to get reflected power. i like ports, i use them all the time in network analysis, they are an important part of the tcp/ip protocol!. for transmission lines all you really need is voltage OR current waves, everything else falls out from those. computing power and trying to reflect power can only lead to confusion, because unless you use the complete formula for complex powers you are losing important information.... and don't even start on computing rms or other average powers, then you have totally lost the physical significance of the waves. |
Dave:
[snip] Is that one on the East Coast and one on the West Coast?? Or, how about: one with an in'port' and the second with an out'port'? :-) Deacon Dave, W1MCE [snip] I always have a port before dinner. And often I have one sitting on the operating desk while I am transmitting. I like the fortified ports best. I always love the hotels in Portugal, cuz they always seem to leave a nice decanter of port on the night stand every day. Helps to put you to sleep. -- Peter K1PO Indialantic By-the-Sea, FL |
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