"Steve Nosko" wrote in message
...

"Tam/WB2TT" wrote in message
...

"Henry Kolesnik" wrote in message
.. .
Richard
...I got
out the book REFLECTIONS II by Walt Maxwell W2DU. I'm typing
verbatim
from
page 2-2 and 23-1

I paraphrase / quote the quote:
Power reflected from a mismatch back into the PA is not absorbed
there.
He writes:
"when the pi-network tank is tuned to resonance, a virtual
short circuit to rearward traveling waves is created at the input
of the network. Consequently, instead of the reflected power
reaching the tubes of the amplifier, it is totally re-reflected
toward the load by the virtual short "

Hank, I'm sorry, but I believe this is not correct as stated. The
word
"totally", I believe is misleading and contrary to what I think all will
agree. Namely that there MUST be SOME real part (although the true amount
is disputed here) to the Zout of the PA and therefore SOME power MUST be
absorbed there. I believe the point of contention truely is just how much
is absorbed and how much is reflected...

...Also what happens in a transistor final with no pi?

AGAIN. Be careful that you DO NOT keep assuming that the full power is
the
incident power. Incident meaning forward power or that which the
transmitter is sending toward the load. I have a real corker below. Be
careful. Do not try this at home, I am a professional. (yea, I know
this'll
spawn all kinds of grief)


Tam sez:
Henry,
Here is an example of what you just said. Take a sine wave source, and
connect it to a 1/4 wave section of shorted transmission line through a
series resistor R. The reflected wave will reach this resistor 1/2 cycle
later, and will be in phase with the source. For a lossless transmission
line, there will be *0 Volts across the resistor*. There will be 0
current
through the resistor, and the reflected wave will be re reflected for
all
values of R, including R=Z0, because the reflected wave will not "know"
what
R is. You can get the same answer from knowing that the impedance
looking
into a 1/4 wave section of shorted transmission line is infinite.
Tam/WB2TT

Tam,
Although I believe you have digressed somewhat, I will follow this
path
since it contains a closely related concept. You obviously have a pretty
good handle on much of this (as others do, up to a point, and struggle to
make their mental models fit the reality - or to make other's mental
models
fit theirs)...However, there is a paradox here which appears to be the
root
of this disagreement and your example hit it right on the head. You have
some implicit assumptions here.

1) Zero volts across the resistor = (not said, but implied) the t-line
acts
like an open at the input end, therefore there is no current into the line
which results in the zero V across R. V=I^2 * R. I think everybody will
agree with this. (shorted 1/4 wave acts like an open and an open 1/4 wave
looks like a short - mantra of all, no?).

2) Resistor current = zero therefore reverse traveling wave gets reflected
toward the load end. The implication clearly is that with zero resistor
current, no energy can be flowing out that way, so it must be reflected.

HOWEVER...
Been a while since I went to this depth and interesting to do so,
though
unnecessary, I will anyway...

If the input to this stub acts like an open, there can be NO current,
thus no power entering, therefore there can be no forward wave, no
reflected
wave and no summation of waves to make the open in the first place and no
need to re-reflect the reflected wave from the resistor who (or is it
whom)
has no current. This appears to be the root cause of this problem.
Now, we can say that (and I think it has been said) that it makes no
difference how large these two waves (which cancel each other to form the
open circuit at the stub input) can be any absolute magnitude. 1 amp 10
amps 100 amps doesn't matter - they cancel. So what are they really?
There are two things to consider as you work out how you will resolve
this paradox in your mental model.
1) I believe the most important -- If the Z looking in to the stub is
high,
how can you send a large amount of power down the line. If you say it is
an
infinite Z then you have a really big problem explaining how it got there
in
the first place.
2) There is a tendency to assume that the forward power is the same power
as
when the load is not a short, but Zo.

Interesting puzzle, but I don;t need to go further. At some point you
must
'believe' something and I can comfortably stop there...for now.

Oh well...
--
Steve N, K,9;d, c. i My email has no u's.


Steve,
I think there is an analogy here to a high Q resonant parallel tuned
circuit. It takes very little to exite this, but there can be large
circulating current in the L and C.

BTW, on my shorted transmission line: I had looked at the current on the
transmission line - at the wrong end of the TL. There is current at the
shorted end, none on the source end in steady state. Didn't even know the
program was smart enough to know the difference. Seems to me that once you
have put all the energy into the TL that it can store, you don't have to
supply any more energy (?)

Tam/WB2TT