Cecil Moore wrote:
Roy Lewallen wrote:

In my third example, where does the other 10 watts of reflected power
go? If it goes to the load and back, why does it reflect off the
source resistor?


I have read the third example, which is NOT steady-state,
and I don't understand the question. Give me some steady-
state values and I will discuss it.

It most certainly is steady state, as are the other two examples.
Perhaps I wasn't clear in saying I was referring to my recent posting.
Here it is again:

[The setup is a 100 volt zero impedance voltage source in series with a
50 ohm resistor driving a half wavelength of 50 ohm transmission line.]

---------

Well, shoot, maybe the source resistor dissipates all the reverse power
*plus* some more power that comes from somewhere else. So let's try a
200 ohm load. Now the current is 0.4 amp, the power in the 200 ohm load
resistor is 32 watts, and the power in the 50 ohm source resistor is 8
watts. The SWR is 4:1, the forward power is 50 watts, and the reverse
power is 18 watts. Oops, the source resistor is only dissipating 8 watts
but the reverse power is 18 watts. Not only isn't it dissipating all the
reverse power, but it isn't even dissipating that extra power that came
from somewhere else when we connected the 16.67 ohm resistor. Wonder
where the other 10 watts of reverse power went?

---------

Must have bounced off the forward wave, then? If so, did it bounce only
when it reached the source, and not all along the line? Why? How much of
it bounced back at the load and why? Can you show me how you calculate
the various source resistor dissipations for this and the other two
examples using your bouncing average power model?

I'd also welcome H.'s comments on this.

Roy Lewallen, W7EL