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Derivation of Reflection Coefficient vs SWR
"Cecil Moore" wrote in message
. net... Roger Sparks wrote: The equation should apply at all locations on the transmission line. What you say is true but there is a caveat. Remember that I have previously said that interference can exist without wave cancellation. As long as Z0 remains constant up and down a transmission line, there is no forward and reflected wave interaction. They pass each other like "ships in the night". The resistor is a physical object. When the forward and reflected waves superpose at the *resistor*, there is an obvious interaction, i.e., a permanent interference. One only has to observe the power dissipated in the resistor to ascertain that it is not the same as the forward power plus the reflected power unless the E-fields of the two waves are 90 degrees apart, a condition for which zero interference exists. What is interesting is a procedure for determining the current through Rs using only powers. We have to be pretty accurate with the powers to do that. We know the forward power is 25w. The power reflection coefficient at the load is [(50-1)/(50+1)]^2 = 0.92311. That makes the reflected power from the load equal to 23.077 watts. When the forward wave and reflected wave superpose at the 50 ohm source resistor, they are 180 degrees out of phase which makes the interference term minus and therefore destructive. 25w + 23.077w - 2*SQRT(25w*23.077w) = 0.03845w That's the dissipation in the 50 ohm source resistor. So the RMS current is SQRT(0.03845w/50ohm)= 0.02773a Ig(t) = 1.414(0.02773)cos(wt) = 0.039216 cos(wt) And that is, indeed, the current through the 50 ohm source resistor, Rs, using only an energy analysis. All energy, except 0.03845w, is "redistributed", i.e. reflected back toward the load. This is all explained in my energy analysis article at: http://www.w5dxp.com/energy.htm -- 73, Cecil http://www.w5dxp.com I disagree. If the current through the 50 ohm source resistor is 0.039216, then the power in the 50 ohm source resistor cannot be 0.03845w. Namely because 0.039216 * 0.039216 * 50 = 0.076895w (eye-squared-are, matey). Either your power is wrong or your current is wrong. My own calculations show that your current is correct. Therefore, your energy analysis appears to be in error. John |
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