Cecil Moore wrote:
Roy Lewallen wrote:

Did anyone understand it? If so, would someone else please try to
explain it to me? Where does that 18 watts of "reverse power" go, and
why?


Bob Lay, w9dmk, could explain it but he's off on a camping trip.
I think the QEX editors can now explain it also. 18 watts is
rejected by the mismatched load. It was part of a forward wave
that contained energy and momentum. That energy and momentum
MUST be conserved according to the laws of physics. The reflected
wave heads back toward the source at the speed of light and part
is re-reflected as Pref2*rho^2.

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.

The remaining
Pref(1-rho^2) part

(which is all of it, since none was re-reflected)

engages with Pfor1*rho^2 in an equal
magnitude/opposite phase wave cancellation as explained by Walter
Maxwell on page 23-9 of "Reflections II".

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 -- since none of the reverse power wave is
re-reflected, what happens to it? Where does the power go? Out the
source resistor? I believe you said that 11.52 watts of it did. What
happened to the rest?

Since energy and momentum
cannot be cancelled, the two cancelled waves conserve energy and
momentum by heading back toward the load at the speed of light as
a re-reflection of energy which joins the forward wave energy.

What two waves head back toward the load? None of the alleged reverse
power wave is reflected, by your own calculation. Where did it go?
What's the other "cancelled wave" and where did it come from? What's its
magnitude?

Roy Lewallen, W7EL