On Mar 5, 11:11*am, Cecil Moore wrote:
Keith Dysart wrote:
On Mar 4, 10:27 pm, Cecil Moore wrote:
Keith Dysart wrote:
Thus these examples do not demonstrate that the reflected power
is dissipated in the source resistor.
There are no reflections at the source so the reflected
energy flows through the source resistor. There is no
interference to redistribute any energy. There is no
other place for the reflected energy to go.

That is the conundrum, isn't it?

And yet the analysis of instantaneous energy flows definitely
shows that the reflected energy is not the energy being dissipated
in the source resistor.

Your analysis seems to be flawed. You are adding average
power terms to instantaneous power terms which is mixing
apples and oranges.

I do not think that is the case.

The expression for instantaneous power in Rs before the reflection
(or, if you prefer, when a 50 ohm load is used), is

Prs(t) = 50 + 50cos(2wt)

It is trivial to compute the average of this since the average
of a sine wave is 0, but that does not make the expression the
sum of an average and an instantaneous power.

As an exercise, compute the power in a 50 ohm resistor that has
a 100 volt sine wave across at, that is
V(t) = 100 cos(wt)

You will find the result is of the form shown above.

So when you add the instantaneous power in Rs before the reflection
arrives with the instantaneous power from the reflection it will
not sum to the instantaneous power dissipated in Rs after the
reflection returns.

Thus conveniently showing that for this example, the reflected
power is not dissipated in Rs.

...Keith