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#1
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Owen Duffy wrote:
Cecil Moore wrote: Reg, I'm curious how you would solve this stub problem without a Smith Chart. | 45 deg | 45 deg | Source====Z01=========Z02====open Stub sections are lossless. Z01 = 600 ohms and is 45 degrees long. Z02 = 50 ohms and is 45 degrees long. What is the impedance looking into the stub from the source? I missed the significance of this problem Cecil. Is it principally a theoretical (being lossless) problem that a Smith chart can solve, or does it have some other significance? It's just a mental exercise with a hidden significance. This is the type of problem that I would solve with a Smith Chart. How about a solution? What impedance does the source see? The physical length of the stub is 90 degrees. What is the electrical length of the stub in degrees? -- 73, Cecil http://www.qsl.net/w5dxp |
#2
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Cecil Moore wrote:
It's just a mental exercise with a hidden significance. This is the type of problem that I would solve with a Smith Chart. How about a solution? What impedance does the source see? The physical length of the stub is 90 degrees. What is the electrical length of the stub in degrees? KISS But if you did that you couldn't get this thread to last forever. tom K0TAR |
#3
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Tom Ring wrote:
KISS Keep It Simple Stupid? But if you did that you couldn't get this thread to last forever. I have kept it as simple as possible. Wonder why nobody has ventured an answer? -- 73, Cecil http://www.qsl.net/w5dxp |
#4
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![]() "Cecil Moore" wrote in message . .. Owen Duffy wrote: Cecil Moore wrote: Reg, I'm curious how you would solve this stub problem without a Smith Chart. | 45 deg | 45 deg | Source====Z01=========Z02====open Stub sections are lossless. Z01 = 600 ohms and is 45 degrees long. Z02 = 50 ohms and is 45 degrees long. What is the impedance looking into the stub from the source? I missed the significance of this problem Cecil. Is it principally a theoretical (being lossless) problem that a Smith chart can solve, or does it have some other significance? It's just a mental exercise with a hidden significance. This is the type of problem that I would solve with a Smith Chart. How about a solution? What impedance does the source see? The physical length of the stub is 90 degrees. What is the electrical length of the stub in degrees? -- 73, Cecil http://www.qsl.net/w5dxp Hi Cecil Is it posible that the length of the "stubs" change? I'd have thought the length of the stub is always the same. 45 degrees should always be 45 degrees, shouldnt it?? An open circuit, 45 degrees back along a 50 ohm line looks like 50 ohms capacitive. That 50 ohms looks like something like 500 ohms inductive as viewed 45 degrees back along a 600 ohm line. I'd guess your point is that 500 ohms of pure inductive reactance is never seen 90 degrees back from an open, no matter what the Zo of the line Jerry |
#5
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Jerry Martes wrote:
Is it posible that the length of the "stubs" change? I'd have thought the length of the stub is always the same. 45 degrees should always be 45 degrees, shouldnt it?? An open circuit, 45 degrees back along a 50 ohm line looks like 50 ohms capacitive. That 50 ohms looks like something like 500 ohms inductive as viewed 45 degrees back along a 600 ohm line. I'd guess your point is that 500 ohms of pure inductive reactance is never seen 90 degrees back from an open, no matter what the Zo of the line The point I'm eventually going to make is about loading coils in mobile antennas but let's stick with the above stub example. | 45 deg | 45 deg | Source====Z01=========Z02====open Z01 = 600 ohms, Z02 = 50 ohms If the Z0 were constant and the stub was 90 degrees long, the source would see zero ohms. Yet in our above example the stub is physically 90 degrees long and the source sees +j500 ohms. The above stub is electrically 130 degrees long. There is a 45 degree delay through the Z01 section of stub. There is a 45 degree delay through the Z02 section of stub. There is a 40 degree phase shift at the Z01 to Z02 junction. If we want to turn the above stub into a functional 1/4WL open stub such that the source sees zero ohms, we can remove 40 degrees from the Z02 section. If we make the Z02 section 5 degrees long, the entire stub will be electrically 90 degrees long, and 1/4WL resonant. There will be a 45 degree delay through the Z01 section There will be a 5 degree delay through the Z02 section There will be a 40 degree phase shift at the Z01 to Z02 junction. -- 73, Cecil http://www.qsl.net/w5dxp |
#6
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![]() "Cecil Moore" wrote in message m... Jerry Martes wrote: Is it posible that the length of the "stubs" change? I'd have thought the length of the stub is always the same. 45 degrees should always be 45 degrees, shouldnt it?? An open circuit, 45 degrees back along a 50 ohm line looks like 50 ohms capacitive. That 50 ohms looks like something like 500 ohms inductive as viewed 45 degrees back along a 600 ohm line. I'd guess your point is that 500 ohms of pure inductive reactance is never seen 90 degrees back from an open, no matter what the Zo of the line The point I'm eventually going to make is about loading coils in mobile antennas but let's stick with the above stub example. Hi Cecil I sure dont want to get involved with any mobil antenna loading coil discussions, I admit that I'm not qualified. | 45 deg | 45 deg | Source====Z01=========Z02====open Z01 = 600 ohms, Z02 = 50 ohms If the Z0 were constant and the stub was 90 degrees long, the source would see zero ohms. Yeah, but is isnt a line with a constant Zo Yet in our above example the stub is physically 90 degrees long and the source sees +j500 ohms. The above stub is electrically 130 degrees long. I'd disagree with a conclusion that, just because the impedance seen by the source is 500 ohms, the line connecting it to a load is 90 degrees long. There is a 45 degree delay through the Z01 section of stub. There is a 45 degree delay through the Z02 section of stub. There is a 40 degree phase shift at the Z01 to Z02 junction If I disagree, do I have to get involved with some lengthy mathmatical discussion? I'm not skilled enough to argue with you Cecil. I'm not even smart. But, I sure dont see how anyone can conclude there is a phase shift at the junction of two transmission lines. There is a shunt capacitive reactance that results from that abrupt change in dimensions, but I wouldnt have thought it would be enough to result in 40 degrees of phase shift. .. If we want to turn the above stub into a functional 1/4WL open stub such that the source sees zero ohms, we can remove 40 degrees from the Z02 section. If we make the Z02 section 5 degrees long, the entire stub will be electrically 90 degrees long, and 1/4WL resonant. There will be a 45 degree delay through the Z01 section There will be a 5 degree delay through the Z02 section There will be a 40 degree phase shift at the Z01 to Z02 junction. I think I agree with the concept you are using. You are giving the condition where a purely capacitive reactor of 50 ohm impedance is required to be resonated by introducing a series inductor. A short length of higher impedance transmission line will sure do that. The higher the line Zo, the shorter it needs to be to resonate that 50 ohm capacitor. -- 73, Cecil http://www.qsl.net/w5dxp Jerry |
#7
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Jerry Martes wrote:
"Cecil Moore" wrote in message Yet in our above example the stub is physically 90 degrees long and the source sees +j500 ohms. The above stub is electrically 130 degrees long. I'd disagree with a conclusion that, just because the impedance seen by the source is 500 ohms, the line connecting it to a load is 90 degrees long. Well, it is physically 90 degrees long because the two physical pieces are physically 45 degrees each. That's a given. However, the +j500 result tells us that it is electrically 130 degrees removed from the open circuit at the far end. There is a 45 degree delay through the Z01 section of stub. There is a 45 degree delay through the Z02 section of stub. There is a 40 degree phase shift at the Z01 to Z02 junction If I disagree, do I have to get involved with some lengthy mathmatical discussion? I'm not skilled enough to argue with you Cecil. I'm not even smart. But, I sure dont see how anyone can conclude there is a phase shift at the junction of two transmission lines. There is an abrupt change in the Gamma angle of the reflection coefficient at the impedance discontinuity. I can show you why on a phasor graphic. Simplified, it goes something like this. Itotal = 21.5*sin(25) = 10*sin(65) where 21.5 is the phasor amplitude of the current in the 50 ohm section at the junction and 10 is the phasor amplitude of the current in the 600 ohm section at the junction. The values must be the same even though the magnitude of Z0, which controls the amplitude of the current, has changed. If those values must be equal and the amplitude changes because the Z0 changed, the only other thing that can change is the phase angle. -- 73, Cecil http://www.qsl.net/w5dxp |
#8
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![]() "Cecil Moore" wrote in message . .. Jerry Martes wrote: "Cecil Moore" wrote in message Yet in our above example the stub is physically 90 degrees long and the source sees +j500 ohms. The above stub is electrically 130 degrees long. I'd disagree with a conclusion that, just because the impedance seen by the source is 500 ohms, the line connecting it to a load is 90 degrees long. Well, it is physically 90 degrees long because the two physical pieces are physically 45 degrees each. That's a given. However, the +j500 result tells us that it is electrically 130 degrees removed from the open circuit at the far end. There is a 45 degree delay through the Z01 section of stub. There is a 45 degree delay through the Z02 section of stub. There is a 40 degree phase shift at the Z01 to Z02 junction If I disagree, do I have to get involved with some lengthy mathmatical discussion? I'm not skilled enough to argue with you Cecil. I'm not even smart. But, I sure dont see how anyone can conclude there is a phase shift at the junction of two transmission lines. There is an abrupt change in the Gamma angle of the reflection coefficient at the impedance discontinuity. I can show you why on a phasor graphic. Simplified, it goes something like this. Itotal = 21.5*sin(25) = 10*sin(65) where 21.5 is the phasor amplitude of the current in the 50 ohm section at the junction and 10 is the phasor amplitude of the current in the 600 ohm section at the junction. The values must be the same even though the magnitude of Z0, which controls the amplitude of the current, has changed. If those values must be equal and the amplitude changes because the Z0 changed, the only other thing that can change is the phase angle. -- 73, Cecil http://www.qsl.net/w5dxp Hi Cecil Thanks for pointing me toward learning about reflection coefficient. I am really surprised that there is such a large amount of phase shift at that junction. Jerry |
#9
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Jerry Martes wrote:
Thanks for pointing me toward learning about reflection coefficient. I am really surprised that there is such a large amount of phase shift at that junction. W8JI has theorized an even larger phase shift there at the junction of a loading coil and the stinger of a mobile antenna. Unfortunately, he attributes 100% of the phase shift below the stinger to that single junction point, while ignoring the phase shift provided by the loading coil. This thread is my attempt at setting the technical record straight and correcting W8JI's new old wives' tale. This discussion started years ago on QRZ.com where both sides were wrong and nobody was 100% technically correct, including myself. -- 73, Cecil http://www.qsl.net/w5dxp |
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