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#2
September 7th 03, 12:32 AM
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(David or Jo Anne Ryeburn) wrote in message .. .
In article , (Dr. Slick) wrote: Hello, Consider a source impedance of Zo=50+j200 and Zl=0-j200. ****** (1) A *source* impedance of Z_0 = 50 + 200j is easily arranged. A *transmission line surge impedance* of Z_0 = 50 + 200j is impossible; surge impedances of transmission lines must have angles between - Pi/4 radians and + Pi/4 radians. Ok, a source impedance then. I don't fully understand why your last statement needs to be so. Since these are both series equivalent impedances, Zo is like a 50 ohm resistor with a series inductor, and Zl is like a series capacitor. At ONE test frequency, the inductive and capacitive reactances will cancel out (series resonance). When this happens, is will be equivalent to Zo=50 and Zl=0, which is a short. ********** (2) Not equivalent in any reasonable sense. 50 and 50 + 200j aren't equal, nor are - 200j and 0 equal. I understand your point, but the reactances WILL cancel. And if you are feeding from a lossless 50 ohm transmission line, the circuit won't know the difference. If you incorrectly use the "normal" equation for rho (when Zo is complex), you will get: Rho = (Zload-Zo)/(Zload+Zo) = (-50-j400)/50 = 403.1 /_ -97 degrees (3) You forgot the factor of 50 in the denominator. The quantity you are calculating above is approximately a magnitude of 8.062257748 at an angle of about - 97.12501636 degrees. Of course this is silly for a value of rho (but not as silly as 403.1 at an angle of - 97 degrees). However see my comment (1) above. My mistake. Wrote too quickly. A gain of about 8 is STILL insane for a passive network! (4) I hope most readers believe the way to calculate rho when Z_L = 0 is rho = (Z_L - Z_0)/(Z_L + Z_0) = (0 - Z_0)/(0 + Z_0) = - 1. rho = (Z_L - Z_0*)/(Z_L + Z_0) I agree with you. But the incident voltage in this case will be coming out of a series inductor of +j200 reactance at the test frequency. It will be charging up a capacitor, but the reflected voltage will not be 8 times the incident. Again, the reactances will cancel at the series resonance, so in effect, if you are feeding a lossless 50 ohm tranmission line, you will not be able to tell the difference. It will appear exactly like a 50 ohm line shorted at the end. Where do you stand David? Slick |
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#3
September 7th 03, 02:14 AM
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In article ,
(Dr. Slick) wrote: (David or Jo Anne Ryeburn) wrote in message .. . In article , (Dr. Slick) wrote: Hello, Consider a source impedance of Zo=50+j200 and Zl=0-j200. ****** (1) A *source* impedance of Z_0 = 50 + 200j is easily arranged. A *transmission line surge impedance* of Z_0 = 50 + 200j is impossible; surge impedances of transmission lines must have angles between - Pi/4 radians and + Pi/4 radians. Ok, a source impedance then. In that case you shouldn't be using a formula intended to apply to the surge impedance of a transmission line. I don't fully understand why your last statement needs to be so. I assume that by "last statement" you mean "A *transmission line surge impedance* of Z_0 = 50 + 200j is impossible; surge impedances of transmission lines must have angles between - Pi/4 radians and + Pi/4 radians." This follows immediately from the formula Z_0 = sqrt((R + jwL)/(G + jwC)), the facts that none of w, R, L, G, or C are negative, the way angles work when one divides complex numbers and takes square roots, and the fact that the real part of Z_0 can't be negative (which decides which of the two square roots should be used). Where do you stand David? I believe that algebra speaks for itself. I believe that whether a model accurately depicts reality has to be tested by experiment. And I believe that when many such experiments have been previously carried out, all confirming the accuracy of the depiction, any claim that the model is inaccurate and that another one is accurate has to be supported with extraordinarily strong empirical evidence. David, ex-W8EZE -- David or Jo Anne Ryeburn To send e-mail, remove the letter "z" from this address. |
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#4
September 7th 03, 06:12 AM
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(David or Jo Anne Ryeburn) wrote in message .. .
In article , (Dr. Slick) wrote: (David or Jo Anne Ryeburn) wrote in message .. . In article , (Dr. Slick) wrote: Hello, Consider a source impedance of Zo=50+j200 and Zl=0-j200. ****** (1) A *source* impedance of Z_0 = 50 + 200j is easily arranged. A *transmission line surge impedance* of Z_0 = 50 + 200j is impossible; surge impedances of transmission lines must have angles between - Pi/4 radians and + Pi/4 radians. Ok, a source impedance then. In that case you shouldn't be using a formula intended to apply to the surge impedance of a transmission line. You mean i can't use Zo=50 + j200 with Rho = (Zload-Zo*)/(Zload+Zo), for complex Zo? Only up to Zo=50 + j50? Ok, well, the conjugate formula still makes more sense to me. Where do you stand David? I believe that algebra speaks for itself. I believe that whether a model accurately depicts reality has to be tested by experiment. And I believe that when many such experiments have been previously carried out, all confirming the accuracy of the depiction, any claim that the model is inaccurate and that another one is accurate has to be supported with extraordinarily strong empirical evidence. David, ex-W8EZE If the algebra speaks for itself, what does it say to you? Is Besser and Kurokawa and the ARRL incorrect? If you're not too sure and you don't wanna say, i wouldn't blame you. Slick |
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