Interesting question and actually BEFORE I read R. Clark's post I decided to
say that I will begg off at this point. I may be overcome with curiosity on
this question in the near future (like I did for the QST WATTMETER article
calculations), but for now I prefer not to take the challenge. Joules are
much more than I want to consider. Job, Wife, Mother, Mother-in-law,
sister-in-law all needing care and a job that is full of crap. I use this
for enjoyable discussions that I can contribute to and this a way more brain
power that I want to devote. I suspect you, Cecil, can explain it. I know
that a 1/2 wave repeats the load Z, smith Chart and all that and can work
with that.

you win. 73

--
Steve N, K,9;d, c. i My email has no u's.
"Cecil Moore" wrote in message
...
Steve Nosko wrote:
100W XMTR---50 ohm coax---x---1/2 WL 450 ohm ladder-line---50 ohm load

Assuming yes... I think you are saying that we have a 450 ohm line and a
50
ohm load and therefore there is a reflection. Is that where you are?

Yes, there are reflections on the 450 ohm ladder line and none on the 50
ohm coax.

There
is no reflected energy on the 50 ohm coax and therefore no momentum in
the reflected waves on the 50 ohm coax.

I think you contradicted yourself here. Are you saying that there
IS or
IS NOT reflected energy/waves on the 50 ohm section (to the left of the
"--x--"??

No contradiction. There's no reflected energy on the 50 ohm coax because
those two reflections are cancelled at the Z0-match point 'x'. They are
equal in magnitude and opposite in phase.

So what changes the direction
and momentum of the energy wave reflected from the load?

You lost me here. Perhaps you are asking; "If there is a wave
reflected at the end with the 50 ohm, back toward the left, how does
this
power then get absorbed IN that load?" Is this the question?

No, there is 178 joules/sec rejected by the load on the 450
ohm line. That energy possesses direction and momentum toward the source.
What reverses the direction and momentum of that reflected energy?

The wave, in the 450 section, reflected from the load is NOT the power
being
delivered TO the load, so it does not have to be "changed' to be
delivered
there.

There is 100 joules/sec traveling to the right and none to the left on the
50 ohm
coax. There is 278 joules/sec traveling to the right and 178 joules/sec to
the left
on the 450 ohm ladder-line. How does the rearward traveling 178
joules/sec, rejected
by the load, get turned around and join the forward traveling 100 source
joules/sec
in order to add up to 278 joules/sec of forward power on the ladder line
that is
incident upon the load?

It's a simple question: How does energy rejected by the load become energy
incident
upon the load?
--
73, Cecil http://www.qsl.net/w5dxp



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