Keith Dysart wrote:
Pr.g(t) = -50 + 50 cos(2wt)
= -50 -50
= -100
which would appear to be the 100 watts needed to
heat the source resistor.

Which, contrary to your previous assertions, agrees
with what I have said previously. This is the
destructive interference energy stored in the
feedline 90 degrees earlier and being returned
as constructive interference to the source
resistor. It's impossible to sweep it under the
rug when the source power is zero, huh?

All power comes from the source. Since power is
being delivered to the source resistor at a time
when the source is delivering zero power, it must
have been previously been stored in the reactance
of the transmission line.
--
73, Cecil http://www.w5dxp.com