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#21




I have made a living as a professional Engineer designing transmission
equipment over the past four decades, currently more than $4BB gross shipped to world wide =============================== There's an old adage "With radio any bloody thing will work". 
#22




Let's run a quick calculation as Tom has suggested several times.
Consider a transmission line with Z0 = 50  j10 ohms connected to a 50 + j0 ohm load. As I hope will be evident, it doesn't matter what's connected to the source end of the line or how long it is. We'll look at the voltage V and current I at a point within the cable, but very, very close to the load end. Conditions are steady state. When numerical V or I is required, assume it's RMS. I hope we can agree of the following. If not, it's a waste of time to read the rest of the analysis. 1. Vf / If = Z0, where Vf and If are the forward voltage and current respectively. 2. Vr / Ir = Z0, where Vr and Ir are the reverse voltage and current respectively. The minus sign is due to my using the common definition of positive Ir being toward the load. (If this is too troublesome to anyone, let me know, and I'll rewrite the equations with positive Ir toward the source.) 3. Gv = Vr / Vf, where Gv is the voltage reflection coefficient. This is the usual definition. Likewise, 4. Gi = Ir / If 5. V = Vf + Vr 6. I = If + Ir Ok so far? Combine 1 and 2 to get 7. Vf / If = (Vr / Ir) = Vr / Vf = (Ir / If) From 3, 4, and 7, 8. Gv = Gi From 3 and 5, 9. V = Vf(1 + Gv) and similarly from 4 and 6, 10. I = If(1 + Gi) From 8 and 10, 11. I = If(1  Gv) Dividing 9 by 10, 12. V / I = (Vf / If) * [(1 + Gv) / (1  Gv)] Combining 12 with 1, 13. V / I = Z0 * [(1 + Gv) / (1  Gv)] And finally we have to observe that, by inspection, the voltage V just inside the line has to equal the voltage Vl just outside the line and, by Kirchoff's current law, the current I just inside the line has to equal the current Il just outside the line: 14. Vl = V and Il = I, where Vl and Il are the voltage and current at the load. From 13 and 14, and noting that Zl = Vl / Il, then, 15. Zl = Z0 * [(1 + Gv) / (1  Gv)] Now let's test the formulas for Gv that have been presented. We have three to choose from: A. The one just posted by Peter, (Zl  Z0conj) / (Zl + Z0conj) B. Slick's, (Zl  Z0conj) / (Zl + Z0) C. The one in all my texts and used by practicing engineers, (Zl  Z0) / (Zl + Z0) We'll plug the numbers into Peter's (A) first, giving us: Gv = (50  50  j10) / (50 + 50 + j10) = j10 / (100 + j10) = 0.009901  j0.09901 Plugging this into the right side of 15, we get: Zl = 46.15 j19.23 This obviously isn't correct  Zl is known to be 50 + j0. Let's try Slick's (B): Gv = (50  50  j10) / (50 + 50  j10) = j10 / (100  j10) = 0.009901  j0.09901 from which we calculate, from 15, Zl = 48.00  j20.00 Again, obviously not right. Finally, using the universally accepted formula (C): Gv = (50  50 + j10) / (50 + 50  j10) = j10 / (100  j10) = 0.009901 + j0.09901 from which, from 15, we get Zl = 50.00 + j0 This is the value we know to be Zl. Getting the correct result for one numerical example does *not* prove that an equation is correct. However, getting an incorrect result for even one numerical example *does* prove that an equation isn't correct. So we can conclude that variations A and B aren't correct.  Actually, the accepted formula (C) can be derived directly from equation 15, so if all the steps to that point are valid, so is the accepted formula. Why aren't Peter's or Slick's formulas correct? The real reason is that they aren't derived from known principles by an orderly progression of steps like the ones above. There's simply no way to get from known voltage and current relationships to the conjugate equations. They're plucked from thin air. That's simply not adequate or acceptable for scientific or engineering use. (I challenge anyone convinced that either of those equations is correct to present a similar development showing why.) What about the seemingly sound logic that the accepted formula doesn't work for complex Z0 because it implies that a conjugate match results in a reflection? The formula certainly does imply that. And it's a fact  a conjugate match guarantees a maximum transfer of power for a given source impedance. But it doesn't guarantee that there will be no reflection. We're used to seeing the two conditions coincide, but that's just because we're used to dealing with a resistive Z0, or at least one that's close enough to resistive that it's a good approximation. The fact that the conditions for zero reflection and for maximum power transfer are different is well known to people accustomed to dealing with transmission lines with complex Z0. But doesn't having a reflection mean that some power is reflected and doesn't reach the load, reducing the load power from its maximum possible value? As you might know from my postings, I'm very hesitant to deal with power "waves". But what's commonly called forward power doesn't stay constant as the load impedance is changed, nor does the forward voltage. So it turns out that if you adjust the load for a conjugate match, there is indeed reflected voltage, and "reflected power". But the forward voltage and power are greater when the load is Z0conj than when Zl = Z0 and no reflection takes place  enough greater that maximum power transfer occurs for the conjugate match, with a reflection present. I'd welcome any corrections to any statements I've made above, any of the equations, or the calculations. The calculations are particularly subject to possible error, so should undergo particular scrutiny. I'll be glad to correct any errors. Anyone who disagrees with the conclusion is invited and encouraged to present a similar development, showing the derivation of the alternate formula and giving numerical results from an example. That's how science, and good engineering, are done. And what it takes to convince me. Roy Lewallen, W7EL Peter O. Brackett wrote: Slick: [snip] [snip] You are absolutely correct Roy, that formula given by "Slick" is just plain WRONG! rho = (Z  R)/(Z + R) Always has been, always will be. [snip] After consideration, I must agree with Slick. Slick is RIGHT and I was WRONG! Slick please accept my apologies!!! I was wrong, and I admit it! Indeed, the correct formula for the voltage reflection coefficient "rho" when computed using a "reference impedance" R, which is say the, perhaps complex, internal impedance R = r + jx of a generator/source which is loaded by a perhaps complex load impedance Z = ro + j xo must indeed be: rho = (Z  conj(R))/(Z + conj(R)) = (Z  r + jx)/(Z + r  jx) For indeed as Slick pointed out elsewhere in this thread, how else will the reflected voltage equal zero when the load is a conjugate match to the generator. Slick thanks for directing the attention of this "subtlety" to the newsgroup, and again... Slick, please accept my apologies, I was too quick to criticize! Good work, and lots of patience... :) Regards,  Peter K1PO Indialantic BytheSea, FL. 
#23




Methinks someone is confused between the conditions for maximum power
transfer and no reflections. ....Keith "Peter O. Brackett" wrote: Slick: [snip] [snip] You are absolutely correct Roy, that formula given by "Slick" is just plain WRONG! rho = (Z  R)/(Z + R) Always has been, always will be. [snip] After consideration, I must agree with Slick. Slick is RIGHT and I was WRONG! Slick please accept my apologies!!! I was wrong, and I admit it! Indeed, the correct formula for the voltage reflection coefficient "rho" when computed using a "reference impedance" R, which is say the, perhaps complex, internal impedance R = r + jx of a generator/source which is loaded by a perhaps complex load impedance Z = ro + j xo must indeed be: rho = (Z  conj(R))/(Z + conj(R)) = (Z  r + jx)/(Z + r  jx) For indeed as Slick pointed out elsewhere in this thread, how else will the reflected voltage equal zero when the load is a conjugate match to the generator. Slick thanks for directing the attention of this "subtlety" to the newsgroup, and again... Slick, please accept my apologies, I was too quick to criticize! Good work, and lots of patience... :) Regards,  Peter K1PO Indialantic BytheSea, FL. 
#24




Roy Lewallen wrote:
Let's run a quick calculation as Tom has suggested several times. Consider a transmission line with Z0 = 50  j10 ohms connected to a 50 + j0 ohm load. As I hope will be evident, it doesn't matter what's connected to the source end of the line or how long it is. We'll look at the voltage V and current I at a point within the cable, but very, very close to the load end. Conditions are steady state. When numerical V or I is required, assume it's RMS. I hope we can agree of the following. If not, it's a waste of time to read the rest of the analysis. 1. Vf / If = Z0, where Vf and If are the forward voltage and current respectively. 2. Vr / Ir = Z0, where Vr and Ir are the reverse voltage and current respectively. The minus sign is due to my using the common definition of positive Ir being toward the load. (If this is too troublesome to anyone, let me know, and I'll rewrite the equations with positive Ir toward the source.) 3. Gv = Vr / Vf, where Gv is the voltage reflection coefficient. This is the usual definition. Likewise, 4. Gi = Ir / If 5. V = Vf + Vr 6. I = If + Ir Ok so far? I am looking at W.C. Johnson, pages 15 and 16, Equations 1.22 through 1.25, where he shows, very concisely and elegantly, that the reflection coefficient is zero only when the complex terminating impedance is identically equal to the the complex value of Z0. And not the complex conjugate of Z0. In other words (read carefully), the complex terminating impedance whose value is Z0 is equivalent to an *infinite extension* of the coax whose complex value is Z0. This is the "clincher". Observe also the the reflection coefficient for current is the negative of the reflection coefficient for voltage. Bill W0IYH 
#25




Roy Lewallen wrote:
But doesn't having a reflection mean that some power is reflected and doesn't reach the load, reducing the load power from its maximum possible value? As you might know from my postings, I'm very hesitant to deal with power "waves". But what's commonly called forward power doesn't stay constant as the load impedance is changed, nor does the forward voltage. So it turns out that if you adjust the load for a conjugate match, there is indeed reflected voltage, and "reflected power". But the forward voltage and power are greater when the load is Z0conj than when Zl = Z0 and no reflection takes place  enough greater that maximum power transfer occurs for the conjugate match, with a reflection present. The "primary" waves are voltage and current, and the power value involved is "derived" from the primary waves. Pick a point on the line and calculate the instantaneous power at that point. Over time, that value of power "travels" along the line in concert with the voltage and current waves. There is nothing wrong with this. All of the hulabaloo over this subject isn't worth a hill of beans. Everyone understands what power "flow" and "power outage" mean. Bill W0IYH 
#26




William E. Sabin wrote:
I am looking at W.C. Johnson, pages 15 and 16, Equations 1.22 through 1.25, where he shows, very concisely and elegantly, that the reflection coefficient is zero only when the complex terminating impedance is identically equal to the the complex value of Z0. And not the complex conjugate of Z0. In other words (read carefully), the complex terminating impedance whose value is Z0 is equivalent to an *infinite extension* of the coax whose complex value is Z0. This is the "clincher". That's fine. I agree entirely, and it follows from my analysis and my conclusion. A similar analysis can be found in many texts. My offering to provide a large number of references has brought forth no interest from the most vocal participants, and they've also showed a lack of willingness to work through the simple math themselves. So I felt that it might be a good idea to post the derivation before more converts are made to this religion of proofbygutfeelandflawedlogic. Observe also the the reflection coefficient for current is the negative of the reflection coefficient for voltage. Likewise. Bill W0IYH One other thing. If the reflection coefficient is zero, then all of the real power that is dumped into the coax is being delivered to the load, minus the losses in the coax. The system is as good as it can get. Bill W0IYH I agree with this only in the sense that the "system is as good as it can get" means only that the loss in the coax is minimized for a given delivered power. It doesn't guarantee that the maximum possible power will be delivered to the load. After some thought, I see I was in error in stating that terminating the line in its complex conjugate necessarily results in the maximum power transfer to the load. If the 50  j0 had been a source impedance instead of a transmission line impedance, that would be true. However, what results in the maximum power from the source in a system like this is that the *source* be conjugately matched to the impedance seen at the input of the line. Although terminating the line in its characteristic impedance minimizes the line loss, it doesn't guarantee maximum load power. Of course, the conjugate matching theorem says that if the line is lossless, a load which is the conjugate of the impedance looking back toward the source from the load will result in a conjugate match at the source and everywhere else along the line, so that will effect maximum power transfer. But the impedance looking back toward the source from the load isn't by any means necessarily equal to the Z0 of the line, so the conjugate of Z0 isn't necessarily the optimum impedance for power transfer, as I erroneously stated. And I don't believe that the conjugate match theorem applies to a lossy line. I certainly don't want to start up an argument about this topic, though, and will simply state for certain that the maximum net power will be delivered to the line when the impedance seen looking into the line is equal to the complex conjugate of the source impedance. It should be easy to set up a couple of simple numerical examples to illustrate this. Unfortunately, I'm pressed for time at the moment and have to run. I apologize for the error regarding terminating impedance and maximum power transfer. Thanks for spurring me to rethink the conditions for maximum power transfer. I apologize for the error. Roy Lewallen, W7EL 
#27




"William E. Sabin" [email protected] wrote in message ...
.... In other words (read carefully), the complex terminating impedance whose value is Z0 is equivalent to an *infinite extension* of the coax whose complex value is Z0. This is the "clincher". Exactly, Bill! And with that, you don't need any equations. You only need to realize that if you cut a line which had no reflection on it, you can weld it back together without introducing reflections (assuming you do a perfect job restoring it to its original state). That should be obvious and easy to grasp even for a mathophobe. And thanks to Roy for taking the trouble to actually go through the algebra and post it. I honestly hoped that S. would do that, and still cling to the belief that there may have been a lurker or two who also did it. Cheers, Tom 
#28




"William E. Sabin" [email protected] wrote in message ...
.... The usage of complex conjugate Z0* becomes significant when calculating very large values of VSWR, according to some authors. But for these very large values of standing waves, the concept of VSWR is a useless numbers game anyway. For values of VSWR less that 10:1 the complex Z0 is plenty good enough for good quality coax. My working definition for SWR is (1+rho)/(1rho). (Note the overall absolute value in the denominator, so it never goes negative.) Rho, of course, is Vr/Vf = (ZloadZo)/(Zload+Zo), no conjugates. In that way, when rho=1, that is, when Vr=Vf, SWR becomes infinite. When rho1, SWR comes back down, and corresponds _exactly_ to the Vmax/Vmin you would observe _IF_ you could propagate that Vr and Vf without loss to actually establish standing waves you could measure. I agree with Bill that all this is academic, but my working definition seems to me to be consistent with the _concept_ of SWR, and does not require me to be changing horses in midstream and remembering when to use a conjugate and when not to. I _NEVER_ use Zo* in finding rho. To me it's not so much a matter of nitpicking an area of no real practical importance as coming up with a firm definition I don't have to secondguess. Except on r.r.a.a., I don't seem to ever get into discussions about such things, but my definitions are easy to state up front so anyone I'm talking with can understand where I'm coming from on them. W.C. Johnson points out on page 150 that the concept: Pload = Pforward  Preflected is strictly correct only when Z0 is pure resistance. But the calculations of real power into the coax and real power into the load are valid and the difference between the two is the real power loss in the coax. For these calculations the complex value Z0 for moderately lossy coax is useful and adequate. The preoccupation with VSWR values is unfortunate and excruciatingly exact answers involve more nitpicking than is sensible. Agreed! And thanks for the reference re powers. Cheers, Tom 
#29




Roy Sez 
That's fine. I agree entirely, and it follows from my analysis and my conclusion. A similar analysis can be found in many texts. My offering to provide a large number of references has brought forth no interest from the most vocal participants, and they've also showed a lack of willingness to work through the simple math themselves. So I felt that it might be a good idea to post the derivation before more converts are made to this religion of proofbygutfeelandflawedlogic. =============================== YOUR analysis ! Oliver Heaviside worked it all out 120 years back. For those antagonists who are not willing, or are unable, or are too frightened, or are just too plain lazy to work through the simple ARITHMETIC themselves, download in a few seconds program COAXPAIR from website below and run immediately. It is designed to do exactly the calculations everybody is at war about. (And many more). Use any ordinary coax with an overal length of 30 dB or more at 1000 Hz (to ensure a nice negative angle of Zo) and terminate the line with its own calculated input impedance = Zo = Ro+jXo. Then terminate the line with its conjugate. Observe what happens to the reflection coefficient, SWR and actual overall loss. Observe what happens as line length is reduced. Program COAXPAIR (with other transmission line progs) has been sitting there for 2 years or more.  ======================= Regards from Reg, G4FGQ For Free Radio Design Software go to http://www.g4fgq.com ======================= 
#30




"William E. Sabin" [email protected] wrote in message ...
I am looking at W.C. Johnson, pages 15 and 16, Equations 1.22 through 1.25, where he shows, very concisely and elegantly, that the reflection coefficient is zero only when the complex terminating impedance is identically equal to the the complex value of Z0. And not the complex conjugate of Z0. This is ABSOLUTELY WRONG! The reflection coefficient is zero only when the Zload is the conjugate of the Zo. Go look it up in any BASIC RF book! Slick 
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