Roger Sparks wrote:
Vrs(t) = Vs(t) - Vg(t) + Vref(t)

Good job, Roger. Let's simplify the equation through
substitution.

Let V1(t) = [Vs(t) - Vg(t)]

Let V2(t) = Vref(t)

Vrs(t) = V1(t) + V2(t)

The power to Rs would be (Vrs^2)/50 = (70.7^2)/50 = 100w, all coming
from the reflection.

Yes, because when Vs(t)=0, there is zero instantaneous
interference.

In summary, power to the resistor Rs comes via two paths, one longer
than the other by 90 degrees (in this example). The short path is the
series path of two resistors composed of the source resistor Rs and the
input to the 50 ohm transmission line measured by Vg. The long path is
the series path of one resistor Rs and one capacitor composed of the
shorted transmission line. Both paths are available at all times. Power
flows through both paths to Rs at all times, but because of the time
differential in arrival timing, at some point Rs will receive power only
from Vs, and at another point, receive power only from Vref.

What Keith is missing is that:

Vrs(t)^2 = [V1(t) + V2(t)]^2 NOT= V1(t)^2 + V2(t)^2

In his math, Keith is asserting that since

Prs(t) NOT= P1(t) + P2(t), then the reflected power is not
being dissipated in the source resistor. But every sophomore
EE knows NOT to try to superpose powers like that. Roger,
I'll bet you know not to try to superpose powers?

Since Keith doesn't listen to me, would you pass that
technical fact on to him?

When Keith uses the correct equation:

Prs(t) = P1(t) + P2(t) +/- 2*SQRT[P1(t)*P2(t)]

he will see that the equation balances and therefore
100% of the reflected energy is dissipated in the
source resistor since the interference term averages
out to zero over each cycle.
--
73, Cecil http://www.w5dxp.com