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#111
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#112
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On Dec 12, 8:44 pm, "AI4QJ" wrote:
"Cecil Moore" wrote in message . net... wrote: Although we know that the 200 ohm line is longer, there is no indication that the length of the 600 ohm line must or must not change. The phase shift at the impedance discontinuity depends upon the *ratio of Z0High/Z0Low*. The following two examples have the same phase shift at the impedance discontinuity. Z0High Z0Low ---43.4 deg 600 ohm line---+---10 deg 100 ohm line---open ---43.4 deg 300 ohm line---+---10 deg 50 ohm line---open So how long does the 600 ohm line have to be in the following example for the stub to exhibit 1/4WL of electrical length? ---??? deg 600 ohm line---+---10 deg 50 ohm line---open Based on what you say above, Zo high/Zo low = 6 resul;ts in 37 degrees. Then for phase shift, (Zo high/Zo low)*37 deg = 6*D where D = phase shift. If the above ratio is literally true, then in the above example, 12*37deg/6 = 74 degrees and length of the 600 ohm line = 90-74-10 = 6 degrees. However, I would like to find a reference for the math showing the characteristic impedance ratio relationship with phase shift. I am reluctant to accepting formulas without seeing them derived at least once. In another thread, "Calculating a (fictitious) phase shift; was : Loading Colis", David Ryeburn has provided the arithmetic for the general case which includes the following expression: -j*(Z_2)*cot(alpha + beta) = -j*(Z_1)*cot(alpha) "alpha" is the length of the open line "beta" is the "phase shift" at the joint "Z_2" is the impedance of the open line "Z_1" is the impedance of the driven line It can be seen, as noted by Cecil, that beta will be the same if the impedances of the two lines are scaled proportionally. But beta is dependant not only on the two impedances, but also on the length of the open line. There are no simple relationships here. It does not seem to be a concept that is particularly useful for the solving of problems. ....Keith |
#113
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It does not seem to be a concept that is
particularly useful for the solving of problems. Looks like you haven't thought it through. If one wants to create a shortened dual-Z0 stub with equal lengths of each section of Z0High and Z0Low, here is the corresponding chart for different ratios of Z0High/Z0Low. http://www.w5dxp.com/DualZ0.gif As an example, one can create an electrical 1/4WL stub that is 1/3 the normal physical length by using 600 ohm line and 50 ohm line. -- 73, Cecil http://www.w5dxp.com |
#114
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On Dec 13, 5:57 am, Cecil Moore wrote:
It does not seem to be a concept that is particularly useful for the solving of problems. Looks like you haven't thought it through. If one wants to create a shortened dual-Z0 stub with equal lengths of each section of Z0High and Z0Low, here is the corresponding chart for different ratios of Z0High/Z0Low. http://www.w5dxp.com/DualZ0.gif As an example, one can create an electrical 1/4WL stub that is 1/3 the normal physical length by using 600 ohm line and 50 ohm line. There are many ways to create the impedances for matching, each with different advantages. As you point out, one of the benefits of using two different impedance lines is a reduction in material, though, you could go all the way to just using a lumped capacitor and save even more. This reality, however, does not demonstrate any value for the *concept* of phase shift at a discontinuity. For the concept to be useful it should facilitate understanding or problem solving. As far as I can tell, you always solve the problem in the conventional way (change the angle on the Smith chart, un-normalize the impedance, re-normalize the impedance to the new Z0, measure the angle to get to the desired impedance) and then work out the "phase shift" at the discontinuity. It sure looks like additional work that adds no value since the important and useful information has already been derived before computing the "phase shift". As such, I declare it "not particularly useful". ....Keith |
#115
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Cecil Moore wrote:
It does not seem to be a concept that is particularly useful for the solving of problems. Looks like you haven't thought it through. If one wants to create a shortened dual-Z0 stub with equal lengths of each section of Z0High and Z0Low, here is the corresponding chart for different ratios of Z0High/Z0Low. http://www.w5dxp.com/DualZ0.gif As an example, one can create an electrical 1/4WL stub that is 1/3 the normal physical length by using 600 ohm line and 50 ohm line. Cecil, So how do you make that 12:1 connection, say 50 ohm coax to 600 ohm open line? Do you s'pose the connection bits add any phase shift all by themselves? Do you have a model for that extra phase shift? 8-) 73, Gene W4SZ |
#116
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Keith Dysart wrote:
This reality, however, does not demonstrate any value for the *concept* of phase shift at a discontinuity. It may indeed have little value for stubs. But for loaded mobile antennas the value is obvious. The value is that it explains the phase shift through a loading coil in a loaded mobile antenna and the phase shift at the coil to stinger junction. Using dual-Z0 transmission line stubs we are ready to understand loaded mobile antennas, the phase shift through the loading coil, and the "missing degrees" at the coil to stinger junction. According to Dr. Corum, my 75m Texas Bugcatcher coil has a Z0 of ~4000 and a VF of ~0.02. The stinger has a Z0 in the ballpark of 400 ohms and a VF close to 1.0 Knowing what we know about a dual-Z0 1/4WL stub, we can now use that knowledge to analyze a base- loaded mobile antenna with coil and stinger. ---Z0=4000 ohm coil---+---10 deg 400 ohm stinger Now it's a piece of cake. How many degrees of loading coil do we need to make the configuration 90 electrical degrees long? Arctan((400/4000)*cot(10)) = ~30 degrees What is the impedance at the coil to stinger junction? 400*cot(10) = ~ -j2300 ohms What is the phase shift at the coil to stinger junction? 90 - 30 - 10 = ~50 degrees I stumbled upon the dual-Z0 stub idea in trying to understand the phase shifts and delays in a loaded mobile antenna. The same general principles apply. Using traveling-wave current to measure the delay through my Texas Bugcatcher coil agreed within 15% with these calculated values. One side said the coil had to make up the missing 80 degrees of antenna that necessarily had to be there with a 10 degree stinger. This side did not understand the phase shift at the coil to stinger junction. The other side said the coil, like a lumped inductor, has ~zero phase shift through it. This side did not understand the limitations of the lumped circuit model. The delay through a coil is what it measures and calculates to be within a certain accuracy. It is not 80 degrees and it is not ~zero degrees. -- 73, Cecil http://www.w5dxp.com |
#117
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Cecil Moore wrote:
... 400*cot(10) = ~ -j2300 ohms ... I am in the middle of a brain fog/block. I am attempting to get an equation to obtain the cotangent of x (or 10 in the above.) I HAVE DONE THIS BEFORE-got something wrong here ... (1/tan(10)) = 5.67128 and (400*(1/tan(10)) = 2268.51273 There is no "built in" cotan function on my ti-86, ti-83, etc. Help me out Cecil, anyone? Regards, JS |
#118
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John Smith wrote:
There is no "built in" cotan function on my ti-86, ti-83, etc. Poor guy - why can't you do cotangent functions in your head? :-) Help me out Cecil, anyone? How about: cot(x) = tan(90-x) cot(10) = tan(80) = 5.67 -- 73, Cecil http://www.w5dxp.com |
#119
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In article
, Keith Dysart wrote: In another thread, "Calculating a (fictitious) phase shift; was : Loading Colis", David Ryeburn has provided the arithmetic for the general case which includes the following expression: -j*(Z_2)*cot(alpha + beta) = -j*(Z_1)*cot(alpha) "alpha" is the length of the open line "beta" is the "phase shift" at the joint Yes. "Z_2" is the impedance of the open line "Z_1" is the impedance of the driven line Backwards. Z_1 is the characteristic impedance of the open-circuited line (100 ohms, in some of the examples previously discussed). Z_2 is the characteristic impedance of line from the junction point back to the source (600 ohms, in those examples). But beta is dependant not only on the two impedances, but also on the length of the open line. There are no simple relationships here. It does not seem to be a concept that is particularly useful for the solving of problems. I couldn't agree more. The angle alpha + beta is useful; the angle beta is not. But I thought I would provide the formulas in an attempt to set things straight. In article . Keith Dysart wrote: There are many ways to create the impedances for matching, each with different advantages. As you point out, one of the benefits of using two different impedance lines is a reduction in material, though, you could go all the way to just using a lumped capacitor and save even more. Agreed. I've always liked capacitors better than transmission line segments. It takes a pretty crummy capacitor to have as low a Q as a transmission line section is likely to have. Even inductors are often better than transmission line segments. But W5DXP was trying to explain how a loaded mobile antenna worked (using transmission line concepts). However, loaded mobile antennas presumably radiate, at least a little, and my analysis (and W5DXP's discussion of angle lengths of transmission lines and "phase shift" at their junction) is for *LOSSLESS* transmission lines. This makes me wonder. David, ex-W8EZE -- David Ryeburn To send e-mail, use "ca" instead of "caz". |
#120
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Cecil Moore wrote:
... Poor guy - why can't you do cotangent functions in your head? :-) ... Cecil: The extremly difficult will take me a couple of minutes ... The impossible takes just a bit more time. :-) I AM NOT A GURU! yanno? LOL Regards, JS |
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