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Derivation of Reflection Coefficient vs SWR
Cecil Moore wrote:
Keith Dysart wrote: Cecil Moore wrote: Rs Vg Pfor--25w Vl +----/\/\/-----+----------------------+ | 50 ohm ~23w--Pref | | 0.04w / 1.92w Vs 90 degrees \ Rl 100 cos(wt) 1.96w 50 ohm line / 1 ohm | \ load | | +--------------+----------------------+ gnd When you find (as you will), that Pfor and Vfor do not change when the reflected wave returns, it should be difficult to assert that the reflected wave is re-reflected at Vg. Not difficult at all. I've added the powers to the diagram above. The reflected power is ~23 watts. The power dissipated in the source resistor is ~0.04 watts. The reflected power is obviously *NOT* being dissipated in the source resistor so where is it going instead? The answer is destructive interference at the source resistor is redistributing the reflected energy back toward the load as an equal magnitude of constructive interference. I call that a reflection. The redistribution of energy due to interference is a well known and well understood phenomenon in optical physics but has been virtually ignored in the field of RF engineering. It is not a conventional reflection. It is wave cancellation in action. It is obeying the following equation: Ptot = P1 + P2 - 2*SQRT(P1*P2) where -2*SQRT(P1*P2) is the destructive interference term. Is this the total applied power or the remaining power (Prem)(which is really the reflected power), where Prem = Z0(If - Ir)^2 = Z0((If^2) +(Ir^2) - 2(If*Ir)) I think these (yours and mine) are both the same equation except that I changed the terms so that we could use it on a transmission line. Z0 is transmission line impedance, If is forward current, Ir is reflected current Just to clarify, 2*SQRT(P1*P2) = 2*SQRT((Z*I1^2)*(Z*I2^2) = 2*Z(I1*I2) The equation should apply at all locations on the transmission line. If the reflected wave was re-reflected at Vg, there would necessarily be a change to Vfor and Pfor. Nope, not true. The phase angle between the voltages determines which direction the energy flows. If the above example is changed to 1/2WL instead of 1/4WL, the forward and reflected voltages and powers remain the same but the interference at the source resistor changes from destructive to constructive and the source resistor heats up. 73, Roger, W7WKB |
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