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#121
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Analyzing Stub Matching with Reflection Coefficients
Cecil Moore wrote:
x a1=10----| |----s21(a1)=5 toward the load s11(a1)=5----| For those of you who are unfamiliar with the way s-parameter voltages are normalized, here are the *actual measured* voltages at the impedance discontinuity at t=0 + delta-t. All three voltages can easily be seen on an o'scope a short time after t=0. During the buildup to steady- state Vref1 goes from 50V to 0V. How that is possible without interaction is the question for ac6xg. x Vfor1=70.7V----| |----120.7V=Vfor2 toward the load Vref1=50V----| rho1-- | --tau2(Vref2)=0 The corresponding equation used by RF engineers is: Vref1 = rho1(Vfor1) + tau2(Vref2) In this t=0 case, Vref2=0, so Vref1 = rho1(Vfor1) at t=0 -- 73, Cecil http://www.w5dxp.com |
#122
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Analyzing Stub Matching with Reflection Coefficients
Cecil Moore wrote:
Gene Fuller wrote: You don't believe in superposition, do you? It is discussed in lots of books if you want to understand. Do you believe Jim's argument that two coherent EM waves of equal magnitudes and opposite phases traveling collinearly in the same direction in a transmission line can never be canceled? If Jim is right, we can toss the s-parameter analysis in the garbage can and join Roy in calling it gobbledigook (sic). Cecil, Never is a long time. And I am sure you would slip away from free space or a linear medium to provide some counter example as soon as I agreed. I agree with Jim and Roy, and most of the rest of the world. Electromagnetic waves, or photons if you prefer, simply do not interact without the assistance of interfaces, discontinuities, or a non-linear medium. Interference is a result from linear superposition. No waves are harmed in the process. At interfaces and discontinuities lots of things can happen. There are well-established techniques for analyzing those things. There is no law of "conservation of waves", however. There is also no law that says all of the individual component waves you may choose to create need to have some sort of detailed energy balance. I have explained several times how the conservation of energy law works, but you seem to disbelieve me. (Hint: I did not make this stuff up. I gave you direct quotes from very reliable sources.) Since you keep bringing up s-parameters, with the implication that they provide some new truth, perhaps you might go back and re-read AN-95-1. From page 7 of the slide version: "If other independent and dependent variables had been chosen, the network would have been described, as before, by two linear equations similar to equations 1 and 2, except that the variables and the parameters describing their relationships would be different. However, all parameter sets contain the same information about a network, and it is always possible to calculate any set in terms of any other set." The other variables described earlier in the note include voltage and current. Again, we come to my old standby, mathematical convenience. S-parameters are very useful, but they bring nothing new to the physical reality. 73, Gene W4SZ |
#123
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Analyzing Stub Matching with Reflection Coefficients
Cecil Moore wrote:
Cecil Moore wrote: x a1=10----| |----s21(a1)=5 toward the load s11(a1)=5----| For those of you who are unfamiliar with the way s-parameter voltages are normalized, here are the *actual measured* voltages at the impedance discontinuity at t=0 + delta-t. All three voltages can easily be seen on an o'scope a short time after t=0. During the buildup to steady- state Vref1 goes from 50V to 0V. How that is possible without interaction is the question for ac6xg. x Vfor1=70.7V----| |----120.7V=Vfor2 toward the load Vref1=50V----| rho1-- | --tau2(Vref2)=0 The corresponding equation used by RF engineers is: Vref1 = rho1(Vfor1) + tau2(Vref2) In this t=0 case, Vref2=0, so Vref1 = rho1(Vfor1) at t=0 Cecil, I am impressed! That's a pretty fancy o'scope you got. Measurement of voltages to four significant digits at t = 0 + delta-t is definitely world-class. That 291.4 ohm line is pretty special as well. Is there some non-standard definition of *actual measured* that we should consider to help interpret your results? 8-) 73, Gene W4SZ |
#124
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Analyzing Stub Matching with Reflection Coefficients
Gene Fuller wrote:
Never is a long time. Waves never interact *is* a long time. I agree with Jim and Roy, and most of the rest of the world. Electromagnetic waves, or photons if you prefer, simply do not interact without the assistance of interfaces, discontinuities, or a non-linear medium. Interference is a result from linear superposition. No waves are harmed in the process. "No waves are harmed in the process" implies that waves can never be canceled. Yet, all the textbooks and all the web pages say that when two coherent collinear waves of equal magnitudes and opposite phases meet, they only appear to be destroyed but their energy components are actually "redistributed" in different directions. That's exactly what happens to reflected waves toward the source when a Z0-match is achieved. The energy components in the reflected waves, s11(a1) and s12(a2), are redistributed back toward the load. How does b1 ever go permanently to zero without s11(a1) and s12(a2) canceling each other? There is also no law that says all of the individual component waves you may choose to create need to have some sort of detailed energy balance. Yes, there is, Gene. It is called the conservation of energy principle. You cannot create energy in one place, have it destroyed in another place, and then argue that everything is all right because the net energy balance remains the same. "If other independent and dependent variables had been chosen, the network would have been described, as before, by two linear equations similar to equations 1 and 2, except that the variables and the parameters describing their relationships would be different. However, all parameter sets contain the same information about a network, and it is always possible to calculate any set in terms of any other set." The other variables described earlier in the note include voltage and current. Again, we come to my old standby, mathematical convenience. S-parameters are very useful, but they bring nothing new to the physical reality. Then why are you so afraid to discuss an s-parameter analysis? Please respond to my example posting. How does s11(a1) go from 5 to 0 without interacting with something? -- 73, Cecil http://www.w5dxp.com |
#125
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Analyzing Stub Matching with Reflection Coefficients
On Apr 16, 8:35 pm, Cecil Moore wrote:
No they don't. If the waves themselves changed, then their resultant superposition would also change. It's a completely unfounded notion, If what you say is true, then if we measure field strengths far enough away from an antenna to get outside the range of interference, then all antennas are isotropic. "Outside the range of interference"? Yes, please call NASA and tell them about that. :-) ac6xg |
#126
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Analyzing Stub Matching with Reflection Coefficients
Gene Fuller wrote:
I am impressed! That's a pretty fancy o'scope you got. Measurement of voltages to four significant digits at t = 0 + delta-t is definitely world-class. That 291.4 ohm line is pretty special as well. Your non-technical diversions are noted. How about discussing the technical question that was posed? The voltages can be viewed on an o'scope. Their magnitudes agree with the above calculated magnitudes. Do you know of any reason that those voltages would not obey the rules of the reflection model? When does a reflection coefficient of 0.707 not reflect 0.707 of the incident voltage? The 291.4 ohm line is chosen to make rho=s11=0.707. My "450" ohm ladder-line measures to be actually 380 ohms. Nothing special about 291.4 ohms except that: (291.4-50)/(291.4+50) = 0.707 We could actually design a feedline with Z0 = 291.4 ohms. Want me to show you how? :-) If you don't like that value, use 300 ohms and 51.5 ohms for the coax. I'm sure between the choices of 50 ohm coax and 52 ohm coax, there must be a 51.5 ohm coax in there somewhere. -- 73, Cecil http://www.w5dxp.com |
#127
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Analyzing Stub Matching with Reflection Coefficients
Jim Kelley wrote:
"Outside the range of interference"? Yes, please call NASA and tell them about that. :-) You said that EM waves cannot be perfectly collinear. Therefore, there has to exist a distance where they diverge and stop interfering. Your words, not mine. -- 73, Cecil http://www.w5dxp.com |
#128
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Analyzing Stub Matching with Reflection Coefficients
On Apr 16, 9:20 pm, Cecil Moore wrote:
Do you believe Jim's argument that two coherent EM waves of equal magnitudes and opposite phases traveling collinearly in the same direction in a transmission line can never be canceled? I asked you to show me the two waves of equal magnitude and opposite phase travelling in the same direction in a transmission line. Show me the waves, Cecil. AC6XG |
#129
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Analyzing Stub Matching with Reflection Coefficients
On Apr 17, 8:30 am, Cecil Moore wrote:
You said that EM waves cannot be perfectly collinear. Therefore, there has to exist a distance where they diverge and stop interfering. Your words, not mine. You are mistaken. In order to prevent errors of this sort in the future, please quote the words you intend to refer to. ac6xg |
#130
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Analyzing Stub Matching with Reflection Coefficients
Cecil Moore wrote:
Gene Fuller wrote: I am impressed! That's a pretty fancy o'scope you got. Measurement of voltages to four significant digits at t = 0 + delta-t is definitely world-class. That 291.4 ohm line is pretty special as well. Your non-technical diversions are noted. How about discussing the technical question that was posed? The voltages can be viewed on an o'scope. Their magnitudes agree with the above calculated magnitudes. Do you know of any reason that those voltages would not obey the rules of the reflection model? When does a reflection coefficient of 0.707 not reflect 0.707 of the incident voltage? The 291.4 ohm line is chosen to make rho=s11=0.707. My "450" ohm ladder-line measures to be actually 380 ohms. Nothing special about 291.4 ohms except that: (291.4-50)/(291.4+50) = 0.707 We could actually design a feedline with Z0 = 291.4 ohms. Want me to show you how? :-) If you don't like that value, use 300 ohms and 51.5 ohms for the coax. I'm sure between the choices of 50 ohm coax and 52 ohm coax, there must be a 51.5 ohm coax in there somewhere. Cecil, You rarely reply directly to anything, and this is no exception. I was commenting on the *actual measurement* that you claimed. You added the emphasis, not me. Perhaps it was a *virtual measurement* instead? Do you find that you can achieve the desired results more often with such measurements? 8-) 73, Gene W4SZ |
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