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#101
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W5DXP wrote in message ...
Dr. Slick wrote: The concept of phase is meaningless for power. Actually, in a transmission line, it's not. Reference Dr. Best's QEX article. He introduces a term familiar to the field of optics: P1 + P2 + Sqrt(P1*P2)cos(delta) where delta is the angle between V1 and V2. Sqrt(P1*P2)cos(delta) is the interference term whose magnitude depends on the phase between two voltages. If the phase between V1 and V2 is less than 90 deg, the sign of that term is positive and the interference is constructive. If the phase angle between V1 and V2 is between 90 deg and 180 deg, the sign of that term is negative and the interference is destructive. Ok, well perhaps Dr. Best is talking about constructive and destructive interference, but i'm just talking about incident and reflected power. Page 32 of "Electronic Applications of the Smith Chart": "The power RC is defined simply as the ratio of the reflected to the incident power in a waveguide. Numerically it is equivalent to the square of the voltage reflection coefficient (atually, this should be Magnitude-Slick). However, unlike the voltage RC, the power RC has magnitude only, since 'phase' as applied to power is meaningless." Slick |
#102
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Dr. Slick wrote:
But his original formulas for ai and bi (incident and reflected voltage waves) are quite confusing in terms of subscripts, and he doesn't show us the derivation for the conjugate RC formula, unfortunately. He has simply invented a new model. There are a lot of them. From now on, go through a binary decision tree to see if someone is repeating the past or is offering something new. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
#103
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Dr. Slick wrote:
Ok, well perhaps Dr. Best is talking about constructive and destructive interference, ... Yes, he was, but unfortunately he didn't realize it until a few weeks after I told him. :-) -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
#104
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"Dr. Slick" wrote:
ok, Keith, i look forward with great interest on your imaginary passive circuit which can reflect more power than what you feed it (incident power). I can't wait to hook it up to see more reflected power than incident on my DAIWA meter, that would be very interesting. I would never claim there is a passive circuit which violated the rules of conservation of energy but if you define A = Vf**2/Z0 and B = Vr**2/Z0 then there are circuits for which B is greater than A and if you accept the often held view that Pfwd = A and Prev = B then YOU are also claiming the Pref is greater than Pfwd. I claim that A and B don't, in reality, have much to do with power at all. On the other hand Vr Vf has been demonstrated by several examples of which Roy's August 20 post http://groups.google.ca/groups?dq=&hl=en&lr=&ie=UTF-8 &threadm=vl88o381tks0cc%40corp.supernews.com &prev=/groups%3Fdq%3D%26num%3D25%26hl%3Den%26lr%3D%26ie%3 DUTF-8 %26group%3Drec.radio.amateur.antenna%26start%3D25 is just one example. This example had a line with Z0 of 68-j39 ohms connected to a load with impedance 10+j50 ohms. While this example demonstrated lossy lines, for this analysis we can simplify the line to its Thevenin equivalent: - an ideal voltage source - producing a 10 kHz sinusoid - at 52.68 V - with a source impedance of 68-j39 ohms connected to a load of 10+j50 ohms. The incident voltage is 26.34 V. Using polar notation... - source voltage: 52.68/_ 0.0 [52.68+j0.0] V - source impedance: 78.4/_ -29.8 [68-j39] ohms - load impedance: 51.0/_ 8.7 [10+j50] ohms Total circuit impedance is source + load impedance: 78.4/_ -29.8 + 51.0/_ 8.7 = 78.8/_ 8.03 [78+j11] Circuit current (voltage/impedance) 52.68/_ 0.0 / 78.8/_ 8.03 = 0.669/_ -8.03 [0.662-j0.0934] Voltage at load (current * impedance) 0.669/_ -8.03 * 51.0/_ 8.7 = 34.1/_ 70.7 [11.3+j32.2] which agrees with Roy's. Reflected voltage (load voltage - incident) 34.1/_ 70.7 - 26.34/_ 0.0 = 35.5/_ 115.1 [-15.0+j32.2] which also agrees. So reflected voltage is greater than incident voltage which leads to rho being greater than unity. Now about that Daiwa.... Directional wattemeters compute the Vf and Vr using Vf = (V + I*Z0) /2 Vr = (V - I*Z0) /2 Your Daiwa is probably calibrated for Z0 = 50 ohms, but let's assume we can recalibrate for Z0 = 78.4/_ -29.8 [68-j39] ohms. Then it will obtain Vf = (34.1/_ 70.7 + 0.669/_ -8.03 * 78.4/_ -29.8) /2 = 26.34/_ 0.0 [26.34+j0.0] Vr = (34.1/_ 70.7 - 0.669/_ -8.03 * 78.4/_ -29.8) /2 = 35.5/_ 115.1 [-15.0+j32.2] as expected. Please note that your Daiwa DOES think that Vr is greater than Vf. Assuming your Daiwa works like most directional wattmeters it will feed these voltages (appropriately scaled) to a meter which will move linearly in response to the voltage. So if your Daiwa had a linear scale it would have no difficulty showing Vf and Vr (except that it would need to be adjusted for the different Z0) and it would show a greater Vr than Vf (i.e. rho 1). But displaying power (even though meaningless in this case) is somewhat more difficult. Your Daiwa likely computes power by having non-linear markings on the meter representing V**2/Z0. This works fine for real Z0, but will not do for complex Z0. For this, you need more sophosticated computation than is possible with just a non-linear scale, so the power indicated by your Daiwa will be quite incorrect. But it does get Vf and Vr correct (assuming it is adjusted for the different Z0). ....Keith |
#105
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David:
[snip] "David Robbins" wrote in message ... "Peter O. Brackett" wrote in message .net... Slick: What about a negative inductance for a load? A negative capacitance? And don't tell me those are not passive! a negative inductance is a capacitance, and a negative capacitance is an inductance! and yes, they are both passive! [snip] You are only half correct David. They are both passive, but a negative inductance certainly is NOT a capacitor. And a negative capacitor certainly is NOT an inductor. Compare the expressions below to convince yourself of the truth of this assertion. The inductive reactance for a positive inductor is: X = j*w*L of a negative inductor X = -j*w*L of a positive capacitor X = 1/j*w*C = -j/w*C of a negative capacitor X = 1/-j*w*C = j/w*C The magnitude of the reactance of a negative inductor goes up with frequency while the magnitude of a capacitor goes down with frequency, etc... I have synthesized filters with both negative inductors and negative capacitors, they are theoretically passive and they do work. -- Peter K1PO Indialantic By-the-Sea, FL |
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