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#121




Peter O. Brackett wrote:
I am afraid that Cecil has been seduced by the "age" of some of the optical publications into believing that the opticians are ahead of the elecricians, but that is simply not so... we electricians know lots more about wave dynamics than opticians. So, Peter, what mechanism of physics causes the reversal in the momentum of the reflected wave when a Z0match exists?  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! = 
#122




On Wed, 27 Aug 2003 00:27:27 GMT, "Peter O. Brackett"
wrote: Richard: [snip] I know you eschew academic references in favor of "first principles," but others may want more material than the simple puzzle aspect. They can consult "Transmission Lines & Networks," Walter C. Johnson, Chapter 13, "Insertion Loss and Reflection Factors." But lest those who go there for the answer, I will state it is from another reference, Johnson simply is offered as yet another reference to balance the commonly unstated inference of SWR mechanics being conducted with a Z matched source. "Transmission Lines," Robert Chipman is another (and not the source of the puzzle either). 73's Richard Clark, KB7QHC [snip] I hear you... I don't eschew academic references, but when it comes to systems Engineering, I do try to follow what our great President Regan once said, "Trust, but Verify!" Hi Peter, I presume by this you have neither a solution nor an analysis by first principles. Just as well, this stumper has baffled many. Two resistors and a hank of line.... 73's Richard Clark, KB7QHC 
#123




And why "Blackfaced"? It's a classical allusion  a test. You failed. Now I know more about you. And many of the other "adults" in this NG act infantile IN THEIR OWN WAY. Witness how Roy will compare someone to his adolescent son, when they don't agreed with him. And then look at Cecil. Same sh**. Classic case of projection. And you know, you could be a highschool drop out for all i care. The great thing about the NGs is that people don't judge you by your age or looks, only by what you write. In that case, you're in trouble. If you wrote logically, like Richard, then i would respect you...but you don't. I don't care where Richard went to school, he knows his sh**. Because you think he agreed with you. Anyway, you've satisfied my curiosity. Thank you. If the other members of the newsgroup want to read and respond to your posts, fine. Most of us, though, went through middle school once and don't want to repeat the experience. Good luck with your job search. 73, Tom Donaly, KA6RUH 
#124




I read the analysis earlier, and didn't and don't see how it constitutes
the proof I proposed. Permit me to repeat the entirety of what I said, and not just the last sentence. I said: I'll restate something I mentioned before (first incorrectly, then corrected). Connecting a load to a transmission line which is the complex conjugate of the transmission line Z0 does *not* guarantee maximum power delivery from the source, or to the load. The load impedance which provides maximum load power is the complex conjugate of the impedance looking back from the load toward the source. That impedance is the source impedance transformed through the transmission line between source and load, and it's not generally the same as the line's Z0, or its complex conjugate. When this condition of maximum load power is met, there will almost certainly be voltage and current wave reflections on the line  there would be none only if the optimum load impedance coincidentally happened to be equal to the line Z0. So the argument that there can be no reflection of the voltage wave under the condition of maximum power transfer is wrong. You didn't show differently in your analysis, and no one has stepped forward with a contrary proof, derivation from known principles, or numerical example that shows otherwise. Now let's see if this constitutes such a proof. Peter O. Brackett wrote: Roy: [snip] You didn't show differently in your analysis, and no one has stepped forward with a contrary proof, derivation from known principles, or numerical example that shows otherwise. Roy Lewallen, W7EL [snip] Yes I did. I guess that you missed that post. I'll paste a little bit of that posting here below so that you can see it again. [begin paste] We are discussing *very* fine points here, but... [snip] ratio of the reflected to incident voltage as rho = b/a would yeild the usual formula: rho = b/a = (Z  R)i/(Z + R)i = (Z  R)/(Z+ R). Here I'll assume that you've defined a and b as equalling the forward and reverse voltages in a transmission line, which is a different definition than in your other posting. With that definition of a and b, and if R is the transmission line characteristic impedance, then the equation is valid for a transmission line, and ok so far. Otherwise, you're talking about something other than a transmission line, so the "proof" doesn't apply at all. In which no conjugates appear! Now if we take the internal/reference impedance R to be complex as R = r + jx then for a "conjugate match" the unknown Z would be the conjugate of the internal/reference impedance and so that would be: Z = r  jx So the load impedance is the complex conjugate of the transmission line characteristic impedance. Ok. Thus the total driving point impedance faced by the incident voltage a would be 2r: R + Z = r + jx + r  jx = 2r Here you've lost me. a is the forward voltage in the transmission line. What can be the meaning of its facing a driving point impedance? The forward wave sees only the characteristic impedance of the line; at all points the ratio of forward voltage to forward current is simply the Z0 of the line. So I don't believe that it sees 2r anywhere. This is where I'm stuck. If you can show where along the line the forward voltage wave "faces" 2r, that is, Vf/If = 2r, I can continue. . . . Roy Lewallen, W7EL 
#126




Roy:
[snip] Here you've lost me. a is the forward voltage in the transmission line. What can be the meaning of its facing a driving point impedance? The forward wave sees only the characteristic impedance of the line; at all points the ratio of forward voltage to forward current is simply the Z0 of the line. So I don't believe that it sees 2r anywhere. This is where I'm stuck. If you can show where along the line the forward voltage wave "faces" 2r, that is, Vf/If = 2r, I can continue. Roy Lewallen, W7EL [snip] I appologize if I am "going to fast" for the limitations inherent in NewsGroup postings. Let me take it a little more slowly here... To see that there is essentially no difference between my (apparently two) different definitions of the incident and reflected waves a and b, consider the following scenario of an actual transmission line having surge impedance [characteristic impedance] Zo. Let Zo be in general complex. i.e. Zo = sqrt[(R + jwL)/(G + jwC)] where R, L, G, and C are the primary parameters of R Ohms, L Henries, G Siemens, and C Farads per unit length. Now at any particular frequency w = 2*p*f you will find that this general complex surge impedance Zo evaluates to a complex number, say Zo(jw) = r + jx. Later let's let r = 50 Ohms and x = 5 Ohms so that we can work out a numerical example. Consider either a semiinfinite length of this Zo line, or even a finite length of the Zo line terminated in an impedance equal to Z0. I am sure that you will agree that both the semiinfinite Zo line or the finite length Zo line terminated in Zo have the same driving point impedance namely Zo. Now excite this semiinfinte Zo line by an ideal generator of open circuit voltage Vi = 2*a behind an impedance equal to the surge impedance Zo. In other words this is a Thevenin generator of ideal constant voltage 2a behind a complex impedance of Zo. And so... since an ideal voltage source has zero impedance, the termination at the source end of this semiinfinite Zo line is Zo, and so the line which has a driving point impedance of Zo is terminated in Zo. Thus the port or reference plane between the generator and the semiinfinite line is an "image match" point. For reference, a rough ASCII style schematic of this situation would be as follows. Generator = Vi = 2aVolts  generator impedance Z0  impedance Zo of semiinfinite line. Clearly the the generator impedance of Zo and the line's driving point impedance of Zo constitute a voltage divider hanging off the generator with open circuit volts Vi = 2a Volts, and since Zo = Z0 they divide the generator voltage, which equals 2a volts exactly by 2, and so the voltage across the input of the semiinfinte line is exactly a volts. This a is the incident voltage wave of the Scattering Formalism and it is exactly the same as the classical forward voltage for transmission lines. Note that it is exactly half of the generator's open circuit volts. i.e. the voltage across the perfectly terminated [image matched] line is exactly a Volts or 1/2 of Vi Volts. [Aside: That is why I said in an earlier post that your forward wave and my incident wave only differ by a factor of two! They are really no different it's only a scale factor. You can compensate as I have by calling the Zo generator voltage 2a, but for simplicity I often just call it a to eliminate the factors of 2 that occur all over the place. Sorry if that is confusing, I'll try not to do that any more. :). Now everybody knows and everybody will agree that at the junction of the Zo line terminated in Zo [NOT a conjugate match] where the Thevenin generator is connected to the semiinfinite Zo line there is an image match AND there will be no reflected voltage waves. What is more since the Zo line is semiinfinite, or the alternative of a finite length and terminated in Zo at the far end, there will be no reflections at the "far end" either. i.e. in this situation there are no reflections and the incident wave "a" just propagates into the line and there is no reflected wave b to interfere with it. Note that this does not mean that maximum power is transmitted into the line. You and I both agree that when a generator is image matched [Cecil likes to call "image match" a Zo match] that in the general case of complex Zo maximum power is not transferred but there is NO reflected voltage when measured with a reflectometer which uses Zo as it's reference impedance. This scenario at the Zo generator driving the Zo line is the same situation found when you cut into an infinite length of Zo line and insert a reflectometer with reference impedance Zo, the reflectometer will read out the reflected voltage b as: b = rho*a = (ZoZo)/(Zo+ Zo) = 0 Zero, nada, nil... no reflections. Now lets do a simple numerical example. Consider that at some frequency the surge impedance evaluated to say Zo = r + jx = 50 + j5 Ohms and that we set the generator open circuit voltage to be Vi = 2a = 2 volts. Thus the "incident voltage wave is a = 1 Volt. The maximum power available from this Zo generator will occur when the load, call it Z, on the generator is a conjugate match to the generator impedance Zo. Thus for maximum power transfer Z must be the complex conjugate of Zo, i.e. we must let Z = 50  j5, to extract maximum power not Z = 50 +j5. Under these conjugate matched conditions, the total impedance faced by the generator is the sum of its' internal impedance Zo and the external Z, i.e. the total impedance is Zo + conj(Zo) = 50 + j5 + 50  j5 = 100 Ohms And so the generator impresses it's 2 Volts of open circuit volts across the resulting totalof 100 Ohms supplying a current of I = 2/100 = 20mA to the real part of the load of 50 Ohms for a maximum power of: Pmax = I*I*R = 0.02*0.02*50 = 0.0004*50 = 0.02 = 20 mWatts. Under this conjugate matched condition, where the maximum power of 20mW is transferred, if we use the classical definition of rho = (Z  Zo)/(Z + Zo) with Z = 50  j5 and Zo = 50 + j5 we get: rho = b/a = ((50  j5)  (50 + j5))/((50  j5) + (50 + j5)) = (j5  j5)/(100) = j10/100 = j/10 Since the incident wave is a = 1 Volt, then: b = rho* a = j/10 Volts. With a conjugate match, and using the classical definition of rho, as measured by a reflectometer using Zo as it's reference impedance, there will be a reflected voltage of magnitude 0.1 Volts, or one tenth of a volt at a phase lag of ninety degrees. It is interesting to see what value of reflected voltage would be indicated by a Bird Model 43 in this circumstance of a perfect conjugate match. A Bird Model 43 uses an internal reference impedance of R = 50 Ohms and it implements the "classical" definition of rho. i.e. The Bird Model 43 calculates the reflected voltage b as b(Bird): b(BIRD) = rho * a = (Z  50)/(Z + 50) * 1 = (50  j5  50)/(50  j5 + 50) = j5/(100  j5) Actually the Bird cannot indicate phase angles, rather it just computes the approximate magnitude of the reflected voltage which in this case would be b(Bird) = 5/sqrt(100*100 + 5*5) = 5/sqrt(1025) = 0.156 Volts. Compare this to the magnitude of the "true" reflected voltage which is 0.1 Volts as computed by a true reflectometer which used Zo as its' internal reference impedance. And so... contrary to popular opinion, the Bird does not indicate zero reflected power when it is inserted into a perfect conjugate matched [maximum power transfer] situation when the Zo of the system is not a pure 50 Ohms! Now if we choose to use a different definition of rho, say the one proposed by Slick why then we will get different results for the reflected voltage, in fact with his somewhat erroneous formula for rho he will read rho as zero in the conjugate matched situation. However. even though one can use any definition as long as one consistently uses it in all theoretical developments and measurement, as in one of my other posts to this thread, I do not believe Slick's or anybody elses redefinition of rho to be approved by "Mother Nature". Mother Nature uses rho = (Z  Zo)/(Z + Z0) simply because that is the reflection coefficient that exists at any point along any infinitely long transmission line of constant surge impedance Zo where looking to the right and to the left at any point in the line one sees the same driving point impedance Zo in both directions and there are no discontinuities in Zo to cause reflectons, and so the reflection coefficient must be zero and (Zo  Zo)/(Zo + Zo) is the only formulation of a reflection factor that supports that condition. I hope this "treatise" helps you to understand my thoughts on this.  Peter K1PO Indialantic BytheSea, FL. .. 
#127




Peter O. Brackett wrote:
There is no reflection at a Zomatch! The wave reflection model works just fine at a Z0match. There are no "waves", only particles, photons and electrons to be precise! Please don't forget the necessary virtual photons and your inability to even prove that they exist. Check out Feynman's Lectures for an "undergraduate" introduction to QED, and find out why "waves" are just plain WRONG! Does he prove that virtual photons exist? Quoting Hecht: "We've learned from Quantum Mechanics that at base interference is one of the most fundamental mysteries in physics." There is an interesting book out in the bookstores recently, written for the layman, which details the strange life of quantum mechanics, the book is a great read: Thanks, I'll look for it. Right now, I'm about half way through _Quarks,_ Leptons,_and_the_Big_Bang_.  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! = 
#128




On Thu, 28 Aug 2003 00:33:08 0000, "David Robbins"
wrote: you have also hit on the basic problem in this discussion. there are two totally different topics in play here and they are getting mixed up back and forth all over this thread. case 1: transmission line with a load.. case 2: generator feeding a load... keep your conditions straight or these discussions will go on for ever with both sides knowing they are right and that the other one is wrong.... but only because they are discussing two different problems. Hi David, If you don't respond with an answer, I will mark it up to your having lost track of the thread through your computer swapping. I quoted only as much of your post to offer that I have no interest in the conjugation issue whatever logic it is cloaked in. This is a simple, practical example expressed in heat lost in the transmission line. Very simple conditions, two resistors and a hank of transmission line. The scenario begins: "A 50Ohm line is terminated with a load of 200+j0 ohms. The normal attenuation of the line is 2.00 decibels. What is the loss of the line?" Having stated no more, the implication is that the source is matched to the line (source Z = 50+j0 Ohms). This is a half step towards the full blown implementation such that those who are comfortable to this point (and is in fact common experience) will observe their answer and this answer a "A = 1.27 + 2.00 = 3.27dB" "This is the dissipation or heat loss...." we then proceed: "...the generator impedance is 100+0j ohms, and the line is 5.35 wavelengths long." 73's Richard Clark, KB7QHC 
#129




Peter O. Brackett wrote:
There are no "waves", only particles, photons and electrons to be precise! Given a coherent light source, like a laser, and exactly the right indices of refraction, can you describe in a nutshell, how nonglare thinfilm coating works using only particles? Do out of phase photons cancel each other when they are traveling in the same 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! = 
#130




Thanks for hanging in there. Let's see where this one goes. . .
Peter O. Brackett wrote: Roy: [snip] Here you've lost me. a is the forward voltage in the transmission line. What can be the meaning of its facing a driving point impedance? The forward wave sees only the characteristic impedance of the line; at all points the ratio of forward voltage to forward current is simply the Z0 of the line. So I don't believe that it sees 2r anywhere. This is where I'm stuck. If you can show where along the line the forward voltage wave "faces" 2r, that is, Vf/If = 2r, I can continue. Roy Lewallen, W7EL [snip] I appologize if I am "going to fast" for the limitations inherent in NewsGroup postings. Let me take it a little more slowly here... To see that there is essentially no difference between my (apparently two) different definitions of the incident and reflected waves a and b, consider the following scenario of an actual transmission line having surge impedance [characteristic impedance] Zo. Let Zo be in general complex. i.e. Zo = sqrt[(R + jwL)/(G + jwC)] where R, L, G, and C are the primary parameters of R Ohms, L Henries, G Siemens, and C Farads per unit length. Now at any particular frequency w = 2*p*f you will find that this general complex surge impedance Zo evaluates to a complex number, say Zo(jw) = r + jx. Later let's let r = 50 Ohms and x = 5 Ohms so that we can work out a numerical example. The numbers aren't necessary, but that's fine. Consider either a semiinfinite length of this Zo line, or even a finite length of the Zo line terminated in an impedance equal to Z0. I am sure that you will agree that both the semiinfinite Zo line or the finite length Zo line terminated in Zo have the same driving point impedance namely Zo. The terminated line I'm comfortable with. "Semiinfinite" isn't in my lexicon, but I'll see where it goes. Now excite this semiinfinte Zo line by an ideal generator of open circuit voltage Vi = 2*a behind an impedance equal to the surge impedance Zo. In other words this is a Thevenin generator of ideal constant voltage 2a behind a complex impedance of Zo. And so... since an ideal voltage source has zero impedance, the termination at the source end of this semiinfinite Zo line is Zo, . . . Maybe the trouble is with the "semiinfinite" aspect, but here we part company again. The termination at the source end of the line is, by definition, the source impedance, which is zero, not Z0. Can't you do this analysis with a plain, ordinary, transmission line of finite length? Lossless is ok, lossy is ok. Or is your proof true only if the line is "semiinfinite" in length (whatever that is)? Roy Lewallen, W7EL 
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