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Analyzing Stub Matching with Reflection Coefficients
On Fri, 13 Apr 2007 19:10:18 -0700, Roy Lewallen wrote:
Walter Maxwell wrote: Thank you, Roy, I appreciate your comments, as always. However, I knew that you have always considered that virtual opens and shorts cannot cause reflections, and I was hoping my discussion would have persuaded you otherwise. snip I'd think that this diode-like property of virtual shorts would be a major clue that they're not real, but a mathematical convenience. The virtual short is a point where the sum of the voltages of all waves, forward and reflected, add to zero. If this condition causes waves to reflect when struck from one direction, what possible physical explanation could there be for it to do absolutely nothing to waves traveling the other way? So I repeat the question: If a virtual short circuit cannot cause reflections, then what causes the reflection at the stub point? My answer is this: There is no total re-reflection at the stub point. It only looks that way. As you've observed, the waves (traveling in one direction, anyway) behave just as though there was such a re-reflection. But the waves actually are reflecting partially or totally from the end of the stub and other more distant points of impedance discontinuity, not from a "virtual short". The sum of the forward wave and those reflections add up to zero at the stub point to create the "virtual short", and to create waves which look just like they're totally reflecting from the stub point. This has some parallels to a "virtual ground" at an op amp input. From the outside world, the point looks just like ground. But it isn't really. The current you put into that junction isn't going to ground, but back around to the op amp output. Turn off the op amp and the "virtual ground" disappears. Likewise, waves arriving at the virtual short look just like they're reflecting from it. But they aren't. They're going right on by -- from either direction --, not having any idea that there's a "virtual short" there -- that is, not having any idea what the values or sum of other waves are at that point. They go right on by, reflect from more distant discontinuities, and the sum of those reflections arrives at the virtual short with the same phase and amplitude the wave would have if it had actually reflected from the virtual short. Like with the op amp, you can "turn off" the virtual short by altering those distant reflection points such as the stub end. Please let me emphasize again that not I or anyone else who has posted is disputing the validity of your matching methods or the utility of the "virtual short" concept. The only disagreement is in the contention that the "virtual short" actually *effects* reflections rather than being solely a consequence of them. snip I maintain that such an example can't be found, because in fact reflection takes place only at physical discontinuities and not at "virtual shorts". Waves in a linear medium simply don't reflect from or otherwise affect each other. I'm not saying that you can't apply the analytical concept of "virtual shorts" to arrive at the same, valid, result. Or that the "virtual short" approach won't be easier. But I am saying that it's not necessary in order to fully analyze any transmission line problem, simply because it's not real. Can you come up with such an example? Roy Hi R oy, Consider my two explanations, or definitions of what I consider a virtual short--perhaps it should have a different name, because of course 'virtual' implies non-existence. The short circuit evident at the input of the two line examples I presented---do you agree that short circuits appear at the input of the two lines? If so, what would you call them? Roy, I'd like for you to take another, but perhaps closer look at the summarizing of the reflection coefficients below. I originally typed in the wrong value for the magnitude of the resultant coefficients. With the corrected magnitudes in place, the two paragraphs following the summarization now make more sense, because the short circuit established at the stub point leads correctly to the wave action that occurs there. Summarizing reflection coefficient values at stub point with stub in place: Line coefficients: voltage 0.5 at +120°, current -60° (y = 1 + j1.1547) Stub coefficients: voltage 0.5 at -120°, current +60° (y = 1 - j1.1547) Resultant coefficients: voltage 0.5 at 180°, current 0.5 at 0° WRONG Resultant coefficients: voltage 1.0 at 180°, current 1.0 at 0° CORRECT Repeating from my original post for emphasis: These two resultant reflection coefficients resulting from the interference between the load-reflected wave at the stub point and the reflected wave produced by the stub define a virtual short circuit established at the stub point. The following paragraph shows how the phases of the reflected waves become in phase with the source waves so that the reflected waves add directly to the source waves, establishing the forward power, which we know exceed the source power when the reflected power is re-reflected. The same concept applies to antena tuners. Again repeating for emphasis: Let's now consider what occurs when a wave encounters a short circuit. We know that the voltage wave encounters a phase change of 180°, while the current wave encounters zero change in phase. Note that the resultant voltage is at 180°, so the voltage phase changes to 0° on reflection at the short circuit, and is now in phase with the source voltage wave. In addition, the resultant current is already at 0°, and because the current phase does not change on reflection at the short circuit, it remains at 0° and in phase with source current wave. Consequently, the reflected waves add in phase with the source waves, thus increasing the forward power in the line section between the stub and the load. Keep in mind that the short at the stub point is a one-way short, diode like, as you say, because in the forward direction the voltage reflection coefficient rho is 0.0 at 0°, while in the reverse direction, rho at the stub point is 1.0 at 180°, which is why it's a one-way short. You say that no total re-reflection occurs at the stub point. However, with a perfect match the power rearward of the stub is zero, and all the source power goes to the load in the forward direction. Is that not total reflection? Using the numbers of my bench experiment, assuming a source power of 1 watt, and with the magnitude rho of 0.04, power going rearward of the stub is 0.0016 w, while the power absorbed by the load is 0.9984 w, the sum of which is 1 w. The SWR seen by the source is 1.083:1, and the return loss in this experiment is 27.96 dB, while the power lost to the load is 0.0070 dB. From a ham's practical viewpoint the reflected power is totally re-reflected. In my example using the 49° stub the capacitive reactance it established at its input is Xc = -57.52 ohms. Thus its inductive susceptance B = 0.0174 mhos, which cancels the capacitive line susceptance B = -0.0174 mhos appearing at the stub point. My point is that the 49° stub can be replaced with a lumped capacitance Xc = -57.52 ohms directly on the line with the same results as with the stub--with the same reflection coefficients. In this case one cannot say that the re-reflection results from the physical open circuit terminating the stub line. Various posters have termed my approach as a 'short cut'. I disagree. I prefer to consider it as the wave analysis to the stub-matching procedure, in contrast to the traditional method of simply saying that the stub reactance cancels the line reactance at the point on the line where the line resistance R = Zo. In my mind the wave analysis presents a more detailed view of what's actually happening to the pertinent waves while the impedance match is being established. Walt |
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