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Analyzing Stub Matching with Reflection Coefficients
Walt, before digging into your recent posting, I'd really like to get
one issue settled. I think it would be helpful in our discussion. The issue is: Can you find even one example of any transmission line problem which cannot be solved, or a complete analysis done, without making the assumption that waves reflect from a "virtual short" or "virtual open"? That is, any example where such an assumption is necessary in order to find the currents, voltages, and impedances, and the magnitude and phase of forward and reverse voltage and current waves? Roy Lewallen, W7EL |
#2
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Analyzing Stub Matching with Reflection Coefficients
On Sat, 14 Apr 2007 16:04:55 -0700, Roy Lewallen wrote:
Walt, before digging into your recent posting, I'd really like to get one issue settled. I think it would be helpful in our discussion. The issue is: Can you find even one example of any transmission line problem which cannot be solved, or a complete analysis done, without making the assumption that waves reflect from a "virtual short" or "virtual open"? That is, any example where such an assumption is necessary in order to find the currents, voltages, and impedances, and the magnitude and phase of forward and reverse voltage and current waves? Roy Lewallen, W7EL No Roy, of course not. I am not attempting to assert that reflection coefficients should be used in such an analysis. I'm only asserting that it's another way of performing an analysis, one that I believe paints a more visible picture of the how the pertinent waves behave in the circuit. If I still haven't persuaded you that it's a viable way of analyzing the impedance matching function then I'll back off and not pursue the issue any further. Incidentally, you didn't answer my questions. Walt |
#3
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Analyzing Stub Matching with Reflection Coefficients
Walter Maxwell wrote:
On Sat, 14 Apr 2007 16:04:55 -0700, Roy Lewallen wrote: Walt, before digging into your recent posting, I'd really like to get one issue settled. I think it would be helpful in our discussion. The issue is: Can you find even one example of any transmission line problem which cannot be solved, or a complete analysis done, without making the assumption that waves reflect from a "virtual short" or "virtual open"? That is, any example where such an assumption is necessary in order to find the currents, voltages, and impedances, and the magnitude and phase of forward and reverse voltage and current waves? Roy Lewallen, W7EL No Roy, of course not. I am not attempting to assert that reflection coefficients should be used in such an analysis. I'm only asserting that it's another way of performing an analysis, one that I believe paints a more visible picture of the how the pertinent waves behave in the circuit. We're certainly not communicating well! I have never questioned that the use of "virtual shorts" is another way of performing an analysis, nor that it helps visualize some of the things going on. If I still haven't persuaded you that it's a viable way of analyzing the impedance matching function then I'll back off and not pursue the issue any further. Nor have I questioned that it's a viable way of analyzing the impedance matching function. If you'll read what I've written, you'll hopefully see that my only point of contention is with your claim that waves reflect from a "virtual short". They do not. And the lack of a single example of a system whose analysis requires this to happen is evidence that they do not. If you back off and not pursue the issue any further, you'll continue with your belief that "virtual shorts" cause reflections. And I'm afraid that will detract from the wealth of accurate and useful things you do say. So please continue. But don't waste time arguing that the concept of "virtual shorts" is a useful analytical tool. I've always agreed with that, and haven't seen any postings indicating anyone else doesn't. Incidentally, you didn't answer my questions. I wanted to get an answer to mine, first. Now that I have, I'll answer yours. Roy Lewallen, W7EL |
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Analyzing Stub Matching with Reflection Coefficients
Roy Lewallen, W7EL wrote:
"If you`ll read what I`ve written, you`ll hopefully see that my only point of contention is with your claim that waves reflect from a "virtual short". They do not." Seems to me they do. If you are lucky enough to have a copy of Terman`s 1955 opus, we can reason together. On page 91 is found Fig. 4-3 Vector (phasor) diagrams showing manner in which incident and reflected waves combined to produce a voltage distribution on the transmission line. At an open circuit, the voltage phasors are in-phase. E2, the reflected phasor, rotates clockwise as it travels back toward the source. E1, the incident phasor, rotates counter-clockwise as we look back toward the source. Looking 1/4-wavelength back from the open-circuit, E2 and E1, each having rotated 90-degrees, but in opposite directions, are now 180-degrees out-of-phase. On page 92, Fig. 4-4 shows the current, which summed to zero at the open circuit, has risen to its maximum value at 1/4-wavelength back from the open-circuit while the voltage dropped to its minimum, nearly zero, maybe close enough to declare a "virtual short-circuit", 1/4-wavelength back from the open-circuit. What`s a short-circuit? Little voltage and much current. What`s the difference between a physical short and the virtual short? Nothing except the shunting conductor. Is there current flowing at the open-circuit end of the 1/4-wave line segment? No, the open-circuit won`t support current. If a high-impedance generator of the same frequency were connected to the virtual short point on the line, would it also be shorted? Yes. Where? At the virtual short, not the open-circuit at the end of the line. Best regards, Richard Harrison, KB5WZI |
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Analyzing Stub Matching with Reflection Coefficients
I wrote:
"Where? At the virtual short, not the open-circuit at the end of the line." If the virtual short were replaced with a real short, would anything change? Not a thing except the line voltage distribution diagram would lose its final 1/4-wavelength. Best regards, Richard Harrison, KB5WZI |
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Analyzing Stub Matching with Reflection Coefficients
Richard Harrison wrote:
I wrote: "Where? At the virtual short, not the open-circuit at the end of the line." If the virtual short were replaced with a real short, would anything change? Not a thing except the line voltage distribution diagram would lose its final 1/4-wavelength. Don't you consider it a significant difference that no voltage, current, or power would reach the load? Roy Lewallen, W7EL |
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Analyzing Stub Matching with Reflection Coefficients
Roy Lewallen wrote:
Richard Harrison wrote: If the virtual short were replaced with a real short, would anything change? Not a thing except the line voltage distribution diagram would lose its final 1/4-wavelength. Don't you consider it a significant difference that no voltage, current, or power would reach the load? :-) :-) That's what I asked when you said the virtual load on a transmitter could be replaced with a lumped circuit "without changing anything". -- 73, Cecil http://www.w5dxp.com |
#8
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Analyzing Stub Matching with Reflection Coefficients
Roy Lewallen wrote:
"Don`t you consider it a significant difference that no voltage, current, or power would reach the load?" No, because the example`s load is a perfect open-circuit despite Richard Clark`s disdain for good insulators. From the shorting point all the way back to the generator, the voltage distribution is unchanged, real short or virtual short. (I did not say unchanging.) Best regards, Richard Harrison, KB5WZI |
#9
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Analyzing Stub Matching with Reflection Coefficients
I'm not sure how many times it's worthwhile to keep repeating this, but
I guess I'll give it another couple of tries before giving up. Richard Harrison wrote: Roy Lewallen, W7EL wrote: "If you`ll read what I`ve written, you`ll hopefully see that my only point of contention is with your claim that waves reflect from a "virtual short". They do not." Seems to me they do. If you are lucky enough to have a copy of Terman`s 1955 opus, we can reason together. Sorry, I'm not. All I have is a 1947 Third Edition of _Radio Engineering_. I'm unfortunately stuck with having to think for myself. But I trust you to quote him accurately, and Terman is to be trusted. On page 91 is found Fig. 4-3 Vector (phasor) diagrams showing manner in which incident and reflected waves combined to produce a voltage distribution on the transmission line. I'm sure they're correct, and similar diagrams can be found in many of my other texts. At an open circuit, the voltage phasors are in-phase. Yes. And the current phasors are out of phase. E2, the reflected phasor, rotates clockwise as it travels back toward the source. E1, the incident phasor, rotates counter-clockwise as we look back toward the source. Yes. These of course follows from the mathematical analysis of transmission lines, found in many texts, and with which I'm very familiar. Looking 1/4-wavelength back from the open-circuit, E2 and E1, each having rotated 90-degrees, but in opposite directions, are now 180-degrees out-of-phase. On page 92, Fig. 4-4 shows the current, which summed to zero at the open circuit, has risen to its maximum value at 1/4-wavelength back from the open-circuit while the voltage dropped to its minimum, nearly zero, maybe close enough to declare a "virtual short-circuit", 1/4-wavelength back from the open-circuit. Yes, this is universally known. What`s a short-circuit? Little voltage and much current. Well, at a short circuit you'll find zero volts and any current. You'll also find this at other places which aren't short circuits, such as where multiple voltage waves add to zero and at the summing junction of a perfect op amp. These aren't short circuits, but they are points of zero voltage. Saying they are all the same is like saying that because you find water in a creek, any place you find water must be a creek. What sort of logic is that? What`s the difference between a physical short and the virtual short? Nothing except the shunting conductor. Well, yes. For one thing, waves won't reflect from a virtual short. They will, from a real short. Another difference is that a real short will prevent any waves from proceeding beyond it; they pass right through a virtual short. Good thing, too, or you wouldn't get any power to your load. Another is behavior at other frequencies and with other waveshapes. Walt has mentioned another, that a virtual short acts like a real short only in one direction, even when all the other conditions for similarity are met. Is there current flowing at the open-circuit end of the 1/4-wave line segment? No, the open-circuit won`t support current. Correct, of course. If a high-impedance generator of the same frequency were connected to the virtual short point on the line, would it also be shorted? Yes. Where? At the virtual short, not the open-circuit at the end of the line. Well, yes and no. When you first hook it to the virtual short, it won't be shorted -- it'll see just the Z0 of the cable. Only when its output reaches the end of the stub, reflects back, and adds to the forward wave will it be short circuited. So it's the open end of the line which is essential to creating the apparent short at the generator. Now let us, as you say, reason together. You're pointing out some similarities between a virtual short and a real one, and giving that as evidence that waves reflect from a virtual short. So consider a point on a 50 ohm line at which the forward and reverse waves add to a V/I of, say, 10 ohms, purely resistive to keep it simple. If you connect a generator (of the correct frequency) at that point, it will see 10 ohms after things settle down to steady state, just like your generator saw a short circuit at the "virtual short" in steady state. So can we conclude that a traveling wave will partially reflect when it encounters the 10 ohm point? The effective or "virtual" reflection coefficient can be calculated as -2/3, from which the reflected wave can be calculated. And, in fact, if we assume that such a reflection takes place, we can calculate the magnitude and phase of the resulting wave and, sure enough, there it will really be. But if the wave does really reflect from this "virtual discontinuity", we might have a problem. That point is a ways away from the "virtual short" point (check your Terman diagram if you don't follow), so we have a partial reflection occurring at this point as well as the full reflection from the "virtual short". In fact, unless the line is matched, we'll have reflections from every point along the line, or at best everywhere except an infinitesimally short spot every half wavelength! What a mess! Does Terman describe this problem in his book? A diagram, perhaps, showing the infinite number of partial reflections taking place all along the line? No? Well, then, maybe it takes a perfect "virtual short" to get a reflection, and even a tiny, tiny imperfection will prevent it. So that would mean that you'd get no reflection at all from a "virtual almost-short" on even a very slightly lossy line, right? The whole idea goes to pot when you add even a tiny amount of loss? Or is a little loss ok? Then we get a full reflection from a good or pretty good "virtual short", but nothing if it gets too far from perfection. Do me a favor and check your Terman for an equation or graph which shows just where this abrupt transition point is (that is, at what "virtual resistance" the reflection ceases), and why it exists. Help me out here with my reasoning. Roy Lewallen, W7EL |
#10
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Analyzing Stub Matching with Reflection Coefficients
Richard Harrison wrote:
Roy Lewallen, W7EL wrote: "If you`ll read what I`ve written, you`ll hopefully see that my only point of contention is with your claim that waves reflect from a "virtual short". They do not." Seems to me they do. I wonder if we can agree that if there is no physical impedance discontinuity, there can be no reflections? For instance, given a piece of 50 ohm open-ended coax with a driving source: source-----50 ohm coax-----+---1/4WL 50 ohm coax--open There is a virtual short at point '+' and that virtual short exists at a point where there is no physical impedance discontinuity. Can we agree that the forward wave is unaffected by that virtual short? Can we agree that the reflected wave is unaffected by that virtual short? After all, there is absolutely nothing there that can physically disrupt any waves. Or given one wavelength of coax being driven by a signal generator equipped with a circulator load. SGCL---1/4 WL---x---1/4WL---y---1/4WL---x---1/4WL---open There are obviously reflections at the open. Are there any reflections at the virtual open at 'y'? Are there any reflections at the virtual shorts at 'x'? I would submit that in the above example, the *only* reflections in the entire system are happening at the open end of of the coax and that the virtual shorts and opens are themselves effects and not the cause of anything (except maybe arguments) :-) -- 73, Cecil http://www.w5dxp.com |
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