Jim Kelley wrote:
Cecil Moore wrote:
The energy in the canceled waves is why the forward
power is often greater than the source power.

But that's in the wrong direction, Cecil. You claim there's energy in
the canceled waves. You need to be able to measure that for the proof
of your belief. Obviously there's energy in the forward waves. They're
not canceled.

I certainly do *NOT* claim there's energy in the
canceled waves after they are canceled. If you could
win the argument without bearing false witness, you
would already have done so.

You seem to be acting obtuse just for the hell of it.
There is energy in the canceled waves before they are
canceled. It's equal to 2*ExB. After they are canceled,
that 2*ExB energy heads in the opposite direction toward
the load. Consider that the following s-parameter
equation is a steady-state process.

b1 = s11*a1 + s12*a2 = 0

This is a continuous steady-state process. There is
energy in s11*a1. The power is equal to |s11*a1|^2
because s11 and a1 are not zero. There is energy in
s12*a2. The power is equal to |s12*a2|^2 because s12
and a2 are not zero. The power in b1 is equal to |b1|^2.

|s11*a1|^2 + |s12*a2|^2 - destructive interference = 0

For instance,

|s11*a1|^2 = 100 watts, |s12*a2| = 100 watts

Is that calculation beyond your math capabilities?

100w + 100w + 2*SQRT(100w*100w) = 0

There is a total of 200 watts in the two waves just
before they cancel during steady-state. Those 200 watts
head back toward the load after they cancel. That is
*NOT* a one time occurrence. It is a continuous steady-
state occurrence. The two waves are being continuously
canceled during a steady-state wave cancellation process.

Please read the HP ApNote 95-1 until you understand it.
Please study the irradiance equation until you understand
that it is a steady-state process.
--
73, Cecil http://www.w5dxp.com