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Rho = (Zload-Zo*)/(Zload+Zo), for complex Zo
Hello,
Consider a source impedance of Zo=50+j200 and Zl=0-j200. 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. 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 So some silly people on this NG think that a short will reflect a voltage 403.1 times the incident voltage, which is absolutely insane. Now try the correct "conjugate" equation (for complex Zo): Rho = (Zload-Zo*)/(Zload+Zo) = -50/50= -1 Which is exactly what you should get for a short, a full reflection with a phase shift of 180 degrees, but the ratio can never be more than one for a passive network. And consequently, the ratio of the reflected to incident powers can also never be more than 1 for a passive network: If you use ratios, it doesn't matter whether you use peak or RMS voltages. ([Vpeak.incident/Vpeak.reflected])= ([Vrms.incident/Vrms.reflected])=sqrt(Pi/Pr)=[rho] This is because the sqrt(2) is in the numerator and denominator. And the 2 (after squaring) is also factored out in the Power RC! And...the Zo is also factored out for Power RC! The Zo is not needed for the Power RC, because the impedance of the source is identical to the load for the reflected power! Sure, you use the Zo in relation to Zl to get rho, but once you get rho, you have the power RC. Hah! the plot thickens a bit.... Slick |
#2
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#4
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"Dr. Slick" wrote:
Hello, Consider a source impedance of Zo=50+j200 and Zl=0-j200. An excellent example. 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. 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 Corrected arithmetic error - -1-j8 = 8.062/_ -97.125 So some silly people on this NG think that a short will reflect a voltage 403.1 times the incident voltage, which is absolutely insane. Now try the correct "conjugate" equation (for complex Zo): Rho = (Zload-Zo*)/(Zload+Zo) = -50/50= -1 Which is exactly what you should get for a short, a full reflection with a phase shift of 180 degrees, but the ratio can never be more than one for a passive network. So for this example using the 'revised' rho Vr = -Vi so the voltage across the capacitor would be Vi + Vr = 0 . Let us do some circuit analysis. As you say above, the equivalent circuit is 3 elements in series: a 50 ohm resister, a +j200 ohm inductor and -j200 ohm capacitor. Let us apply 1 volt to this circuit... Total impedance 50+j200+0-j200 = 50 ohms Total current (volts/impedance) 1/50 = .02 A Voltage across resistor .02 * 50 = 1 V Voltage across inductor .02 * (0+j200) = 4/_ 90 Volts Voltage across capacitor (the load) .02 * (0-j200) = 4/_ -90 Now for the check... Vi = 0.5 V With classic rho Vr = 0.5 * 8.062/_ -97.125 = 4.031/_ -97.125 Vload = Vi+Vr = 0.5 + 4.031/_ -97.125 = 4/_ -90 The same as computed using circuit theory. With revised rho Vr = -.5 Vload = 0.5 -0.5 = 0 Which is more useful? classic rho which properly predicts the voltage across the load or revised rho which just provides some number I know which one I'd pick. The thing to remember is that in circuits with inductors it is very easy to achieve voltages greater than any of the supplies. ....Keith |
#5
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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 |
#6
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"Dr. Slick" wrote in message om... (Dr. Slick) wrote in message . com... Rho = (Zload-Zo)/(Zload+Zo) = (-50-j400)/50 = 403.1 /_ -97 degrees Oops! that should be Rho = 8.06 /_ -97 degrees. Forgot to divide by 50. A multiple of 8 is still insane for a short. Slick you really don't understand resonances do you. and you obviously haven't read and understood the paper that derived the 'power wave reflection coefficient' that you are misapplying to voltage and current waves. |
#7
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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. |
#8
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#10
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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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