Keith Dysart wrote:
. . .
But do not expect the power dissipated in the resistor
to increase by the same amount as the "reflected power".
In general, it will not. This is what calls into question
whether the reflected wave actually contains energy.

Do some simple examples with step functions. The math
is simpler than with sinusoids and the results do not
depend on the phase of the returning wave, but simply
on when the reflected step arrives bach at the source.

Examine the system with the following terminations on
the line: open, shorted, impedance greater than Z0,
and impedance less than Z0.

Because excitation with a step function settles to
the DC values, the final steady state condition is
easy to compute. Just ignore the transmission line
and assume the termination is connected directly
to the Thevenin generator. When the line is present,
it takes longer to settle, but the final state will
be the same with the line having a constant voltage
equal to the voltage output of the generator which
will be the same as the voltage applied to the load.

Then do the same again, but use a Norton source. You
will find that conditions which increase the dissipation
in the resistor of the Thevenin equivalent circuit
reduce the dissipation in the resistor of the Norton
equivalent circuit and vice versa.

This again calls into question the concept of power
in a reflected wave, since there is no accounting
for where that "power" goes.

I heartily second Keith's recommendations.

For some simple illustrations of one problem with chasing "power waves"
around, see http://eznec.com/misc/Food_for_thought.pdf, particularly the
"Forward and Reverse Power" section beginning on p. 6 and the table on
p. 8. This was originally written and posted more than five years ago
and, to my knowledge, the problems it raises with the concept of "power
waves" still haven't been addressed in the thousands of postings on the
topic in the intervening time.

Roy Lewallen, W7EL