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On Nov 18, 7:22 am, Cecil Moore wrote:
K7ITM wrote: On Nov 17, 4:03 pm, Cecil Moore wrote: The waves are launched by the external reflection from a Z0-match and the internal reflection from the load. So the waves are going opposite directions along the line?? No, all reflections travel toward the source and therefore, are traveling in the same direction. Their Poynting vectors are all toward the source. Given the following Z0-match impedance discontinuity in a transmission line with the source to the left and the load to the right: Z0-match ------Z01---+---Z02------ Pfor1-- Pfor2-- --Pref1=0 --Pref2 The power reflection coefficient is rho^2 = [(Z02-Z01)/(Z02+Z01)]^2 Pref1 is a combination of two reflected waves 1. P1 = Pfor1(rho^2) "the external reflection from the Z0-match" 2. P2 = Pref2(1-rho^2) "the internal reflection from the load" Pref1 = P1 + P2 + 2*SQRT(P1*P2)cos(A) Pref1 equals zero at a Z0-match so P1+P2 and A=180 deg. -- 73, Cecil http://www.w5dxp.com Ah, finally you get around to telling us the setup. So to get to the conditions in the original posting, we must have a total power coming into that Z01:Z02 junction exactly equal to the power leaving it. For example, if Z01 = 50 ohms as implied by your numbers, and Z02 = 100 ohms, barring stupid math errors, I make out that the left-to-right power on the Z01 line is 450 watts, and the right-to-left power on the Z02 line is 56.25 watts, for a total of 506.25 watts. Since you've only accounted for 171 watts, the remainder must be going off to the right from that junction. Change the phases, and the power will split differently. This seems to all agree with standard superpostion. So what the heck was the point of the original posting in this thread? Or, why do I even bother reading these things in the first place, since they all turn out to be pretty boring? Once again, we see that everything interesting going on in the system is happening right at the discontinuity where waves arrive and are reflected. Same in a Wilkinson combiner, same in a "magic T" (which I suppose the Wilkinson is, if you look at it the right way), same as in a resistive combining network (if you account for power dissipated in the resistors), ... |
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