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Where does it go? (mismatched power)
On Fri, 11 Jun 2010 17:35:34 -0700 (PDT), K1TTT
wrote: you can qualify an impedance as non-dissipative It's called reactance. not always. there is a non-dissipative resistance. a lossless transmission line has a pure real impedance, but no dissipation. Hi David, Ah! The appeal of the infinite, lossless transmission line. You forgot radiation. I think we can both agree that it can be measured, even by the inference of a chain of measurements and those measurements would consistently show the absence of heat. I think you would enjoy the absurdity of its appeal to being the plate "resistance" when the 1400 Ohms there would be embodied in 28 infinitely long lossless cables being tucked inside the tube. For others, If it were so (where those 28 infinitely long lines are tucked into another convenient dimension that sprang into being to satisfy argument), then what is the source of the heat of the plate that is in excess of that of the DC operating point? If some feel compelled to jump on the losslessness of radiation loss (now, if that isn't a paradox) - all fine and well for Art's short wound self-resonant radiators that would eventually devolve in size to the plate tank. If plate resistance was due to radiation, what need for antennas? Of course, we could add another rhetorical dimension that absorbs radiation.... 73's Richard Clark, KB7QHC |
Where does it go? (mismatched power)
On Jun 12, 5:51*pm, Richard Clark wrote:
On Fri, 11 Jun 2010 17:35:34 -0700 (PDT), K1TTT wrote: you can qualify an impedance as non-dissipative It's called reactance. not always. *there is a non-dissipative resistance. *a lossless transmission line has a pure real impedance, but no dissipation. Hi David, Ah! *The appeal of the infinite, lossless transmission line. *You forgot radiation. a lossless line would not be lossless if it lost energy due to radiation! |
Where does it go? (mismatched power)
Richard Clark wrote in
: .... I observed how you violated the adjusted-for part of "designed/adjusted for, and expecting a 50 + j 0 ohm load." The IC7000 has no relevant user adjustments, it is designed for and adjusted for in its design, manufacture, and alignment. As no claim has been made by anyone about a source being constant in Z nor in Power across all loads and all frequencies, your response does not conform to your own reprise of the "proposition." To be usable, any pretence of linearity at least over a limited range of loads, power, frequency, Zeq must be sufficiently constant within that range. What you have performed is a load pull which constructs a curve of complex source impedances around the point at which the transmitter was adjusted for a 50 Ohm load. All well and good. However, No, I have performed a go/nogo test on whether the transmitter delivers sufficiently constant forward power into a quite limited range of loads. The meaning of "sufficiently" is proposed in the reference article describing the test and providing the mathematical basis for the test. Thevenin's theorem says nothing of this. The correct test, to the letter of the theorem is a test no one performs: the measured open circuit voltage divided by the measured short circuit current. No, Thevenin's theorem does not speak at all of how to test a source. For a linear source, your proposed test of o/c voltage and s/c current would provide sufficient data, as would ANY two points on the (complex) v/i relationship. I expect that a typical HF ham transmitter output is not linear over the entire v/i characteristic, but my test just focuses on whether it is approximately linear over a limited range of loads. My recollection of Walt's tests were that they tested at points other than Zl=0 and Zl=infinity. ******* If I were to return to another statement from your link offered in my quote above: the transmitter is 50+j0Ohm is in all likelihood incorrect. I am speaking strictly to what is reported and to the implied accuracy of 50 ±0.5 Ohms. I seriously doubt that you have the means to achieve the absolute accuracy of 1%. You dwell at some further length on the implicit accuracy of the stated quantity, but ignore the explicit discussion about a reasonable tolerance for the test. Owen |
Where does it go? (mismatched power)
....
As a courtesy to me, a foreigner tourist ham, would you mind stop for a brief moment your more general differences and tell me if you agree on the behavior of a Thevenin generator with a series resistance of 50 ohms in relation to changes in impedance of a lossless TL predicted by the Telegrapher's equations solutions in terms of the power dissipated on the load resistance and series resistence of Thevenin source? I am pretty serious about this: until today I could not know if you agree in that!! :) sure, if you properly apply the telegrapher's equations and the thevenin equivalent methods. *The real problem is that if you try to do that for most amateur radio transmitters the source impedance is not linear, and even worse may be time varying, which renders the thevenin equivalent source substitution invalid. OK. Thank you very much. This clarify so much the issue to me. Please, another question: On the same system-example, who does not agree with the notion that the reflected power is never dissipated in Thevenin Rs? (I am referring to habitual posters in these threads, of course) |
Where does it go? (mismatched power)
On Sat, 12 Jun 2010 10:57:08 -0700 (PDT), K1TTT
wrote: a lossless line would not be lossless if it lost energy due to radiation! Hmmm, You want to reconsider that? 1. How could you tell if a lossless line lost energy by radiation? B. Radiation was the second in your incomplete list, not a line characteristic. 73's Richard Clark, KB7QHC |
Where does it go? (mismatched power)
On Sat, 12 Jun 2010 10:43:37 -0700 (PDT), K1TTT
wrote: that assumes the source is linear... with a non-linear source it is much more complicated to describe the full range of it's response. you would have to do a series of output voltage vs current curves for a range of impedances. Ah! But the difference is that no one is going to go to the bench to see if this holds true and roll with the punch if they are surprised that it works out perfectly. 73's Richard Clark, KB7QHC |
Where does it go? (mismatched power)
On Jun 12, 12:39*pm, K1TTT wrote:
i don't do s stuff so i have no idea what you just proved... give me the impedances and voltages/currents. Too bad about the "s stuff". Here are the RF equations for a Z01 to Z02 impedance discontinuity in a transmission line. The forward voltage on the Z01 side is Vfor1 and the reflected voltage from the impedance discontinuity (back toward the source) is Vref1. The forward voltage on the Z02 side is Vfor2 and the reflected voltage (from the load) is Vref2. Hopefully, the reflection and transmission coefficients are self-explanatory. rho1 is the reflection coefficient encountered by Vfor1, etc. Vref1 = Vfor1(rho1) + Vref2(tau2) = 0 That is wavefront cancellation in action. The external reflection phasor, Vfor1(rho1), is equal in magnitude and 180 degrees out of phase with the internal reflection phasor, Vref2(tau2), arriving from the mismatched load. Vfor2 = Vfor1(tau1) + Vref2(rho2) If these RF equations are normalized to SQRT(Z0), they are the same as the s-parameter equations. -- 73, Cecil, w5dxp.com |
Where does it go? (mismatched power)
On Jun 12, 8:59*pm, Richard Clark wrote:
On Sat, 12 Jun 2010 10:57:08 -0700 (PDT), K1TTT wrote: a lossless line would not be lossless if it lost energy due to radiation! Hmmm, You want to reconsider that? 1. *How could you tell if a lossless line lost energy by radiation? there would be missing energy? |
Where does it go? (mismatched power)
On Jun 12, 9:17*pm, Cecil Moore wrote:
On Jun 12, 12:39*pm, K1TTT wrote: i don't do s stuff so i have no idea what you just proved... give me the impedances and voltages/currents. Too bad about the "s stuff". Here are the RF equations for a Z01 to Z02 impedance discontinuity in a transmission line. The forward voltage on the Z01 side is Vfor1 and the reflected voltage from the impedance discontinuity (back toward the source) is Vref1. The forward voltage on the Z02 side is Vfor2 and the reflected voltage (from the load) is Vref2. Hopefully, the reflection and transmission coefficients are self-explanatory. rho1 is the reflection coefficient encountered by Vfor1, etc. Vref1 = Vfor1(rho1) + Vref2(tau2) = 0 That is wavefront cancellation in action. The external reflection phasor, Vfor1(rho1), is equal in magnitude and 180 degrees out of phase with the internal reflection phasor, Vref2(tau2), arriving from the mismatched load. Vfor2 = Vfor1(tau1) + Vref2(rho2) If these RF equations are normalized to SQRT(Z0), they are the same as the s-parameter equations. -- 73, Cecil, w5dxp.com ok, so you defined a case where the superposition of the reflected and refracted waves at a discontinuity results in a zero sum. is that supposed to prove something? did i ever say that you could not define such a case?? |
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