Roy Lewallen wrote:

rho at the source is zero; the source matches the transmission line Z0.
So the "part" which is re-reflected is zero, by your calculations.

Not my calculations. The measured SWR at the source terminals
is 4:1. rho = (SWR-1)/(SWR+1) = 3/5 = 0.6

That part of Walt's text is talking about voltage and current waves. I'm
familiar, thanks, with how they interact. You're the one talking about
waves of average power

Walt's text talks about RMS (average) voltage and current. When
you multiply RMS voltage by RMS current, you get average power.
V+*I+ = forward power, V+/I+ = Z0, (PV+)=(E+)x(H+)
V-*I- = reflected power, V-/I- = Z0, (PV-)=(E-)x(H-)

-- since none of the reverse power wave is
re-reflected, what happens to it?

On the contrary, it is 100% re-reflected by wave cancellation,
the fourth thing that can cause 100% re-reflection. Optical
engineers understand that. Most RF engineers don't. Adding
the one wavelength of lossless 200 ohm feedline reveals the
wave cancellation mechanism.

What two waves head back toward the load? None of the alleged reverse
power wave is reflected, by your own calculation.

Not my calculation. All of the reverse power wave is reflected
by wave cancellation at the source terminals. No reflected power
is dissipated in the source resistor. Therefore, it is all
reflected back toward the load. The one wavelength of 200 ohm
lossless feedline reveals exactly what happens.
--
73, Cecil http://www.qsl.net/w5dxp


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