On Thu, 01 Mar 2007 20:45:44 GMT, Owen Duffy wrote:

Any notion that reflected power *must* flow back to the source and is
dissipated as heat will not lead to the correct solution of this problem.

Hi Owen,

Why must it flow back, when the generator itself presents a huge
mismatch to the line? The laws of reflection work at both ends of the
line and to force the presumption that initial reflected power would
flow through this discontinuity without reflection is a strained
expectation. However, I see by the alteration that follows, that we
now have a fully matching source:

As an exercise, think of a generator that has a Thevenin equivalent of
some voltage V and a series impedance of R+j0, connected to a half wave
of lossless transmission line where Zo=R. To give a numerical example,
lets make V=100 and R=50, so Vr=50 and "reflected power"=50. How much of
the "reflected power" is dissipated in the generator. In this case, the
generator dissipates less heat than were it terminated in 50 ohms.

"reflected power"= 50 what? Watts? So be it. Your intent may not to
be incomplete nor difficult, but what is wrong with specifying the
load? Do we get a line length?

Let's see, I start with your termination of 50 Ohms to find 1 Amp
flowing through both source and load resistances. 50 Watts each. Now
I move to a load that will reflect 50 Watts - the former load's
complete contribution. I presume by this you mean either an open or a
short at the load end. As you left it unspecified, and it being my
choice, I can show that the source resistor will get hotter than a $5
pistol. I will, however, be complete and discuss how the source
resistance dissipates the reverse power for both short and open.

To cut to the chase, and not knowing the length of the line, we have a
spectrum of choices, but we may as well force the situation with
another resonant line any number of lossless halfwaves.

It is obvious that the two powers combine constructively or
destructively. One is with the energy reflecting a zero voltage
component and all the current (or now twice the current through the
source resistance), the other is with the energy reflecting an
identical voltage component, and none of the current flows. As I've
offered, the spectrum of possible answers yields to the same analysis,
this was simpler.

Both times the reflected power had to be absorbed by the source
resistance. It does not mean that contribution has to always resolve
to more heat (haven't I already demonstrated this in this thread?).

I will at this point re quote Chipman to roughly this scenario (being
more general, he didn't specify the reflection).

"At the signal source end of the line ... none of the power
reflected by the terminal load impedance is re-reflected on
returning to the input end of the line."
The ellipsis reveals that the source Z matches the line Z.

73's
Richard Clark, KB7QHC