On Fri, 23 Mar 2007 15:10:40 -0800, Richard Clark wrote:

On Fri, 23 Mar 2007 19:30:56 GMT, Walter Maxwell
wrote:

I don't understand what you mean by 'taking only one
of two degrees of the 360.'

Hi Walt,

I offered:
they take only one or two degrees of the 360.

Arguments that are confined only at 0 or 180 (the one OR two degrees)
and are submitted as proofs as though they boxed the compass. All too
often I've seen one condition (at one phase angle) offered as a
negation of internal heating to prove the source lacks its own
internal resistance. When I've taken exactly the same circuit and
explored the 180th degree alternative, I've demonstrated melt-down
clear and simple. The dissipation of energies does not always lead to
this consequence, but if we were to average the analysis over a
complete 360 degrees, we can only arrive at the obvious evidence of
resistance and calories expended.

But even for coherent reflections, if the PA tank circuit has very low loss
for incident power (which it does), why does it not have ~ equally low loss
for load reflections of that power? Such would mean that load reflections
would pass through the tank to appear at the output element of the PA, where
they can add to its normal power dissipation.

The paragraph above seems to me to imply that RF doesn't understand the destructive and constructive
interference phenomena involved with re-reflection.

Then asking a question to clarify would be in order. To me, it reads
quite ordinarily as a statement of symmetry. In my own words, it
would say that if a tank circuit can pass energy from source to load
in one direction, it can certainly perform the same transformation in
the opposite direction. After all, that is the function of
transformation and a passive circuit composed of L and C is strictly
linear. Circuit analysis allows us to treat a load as a source in the
complete circuit description.

Richard, assume a mismatched load has produced both voltage and current refleftions on the line that result in
a particular reactive impedance at the line input. The line input is connected to the output of the
transceiver that has a pi-network output coupling circuit. When the network has been adjusted to deliver all
the available power into the line the output source impedance is the complex conjugate of the line input
impedance. In this condition the reflected voltage and current waves are totally re-reflected back into the
line, while adding in phase to the voltage and current waves from the source, respectively. Consequently, the
reflected waves do not pass rearward through the network to be incident on the plate. Only if the network is
mistuned, such as being connected to the reactive input of the line without being retuned to resonance, in
which case the excessive plate current due to being mistuned will result in an inordinate amount of heating of
the plate.

This is the symmetry of the illustration of external signals. You
used external signals yourself as part of your case study; hence the
relevance has been made by you.

Whoa, Richard! You'll have to point out where I've discussed external signals in any case study involving
phase relationships between forward and reflected waves. I've never done so knowingly.

It seems to me that in your initial post in the original thread (that
was largely ignored for comment) you made mention of injecting a
signal from an external source into the mouth of the dragon for the
purposes of measuring the source Z. Am I wrong?

Yes, you are wrong here, because I made no mention of the 'mouth of a dragon'. That comment must have come
from another poster, twarn't I.

And we return to the sine non quo for the discussion: phase.

That's true, but although RF apparently realizes that the phase relationship is relevant, he doesn't seem to
understand the details of the phase requirements that achieve the necessary interferences that accomplish the
impedance matching.

That is not what I read.

It seems he is on the face of it, doesn't it? Afterall, he is quite
explicit to this in the statement you are challenging.

No Richard, I don't believe he is. I don't see the 'explicitness' you seem to find. It's the complete lack of
the explicitness that makes me believe he doesn't quite get it.

That has not been my impression of the complete post.

Nothing here contradicts anything either of you have to say.

True, but RF just hasn't said it all, because, as I said above, I don't believe he understands the details of
the phase requirements to achieve the match.

That has not been my impression of the complete post.

Richard, try this on for size and then determine whether you believe RF understands the function of the
phasing in impedance matching:

Assume a 150-ohm pure resistance terminates a 50-ohm line, producing a voltage reflection coefficient rhoV
=0.5 at 0°, a current reflection coefficient rhoi, yielding a 3:1 mismatch. We want to place a series stub at
the appropriate position on the line to yield a match at that point. The appropriate position on the line is
where the real portion of the line impedance is 50 ohms, with a residual reactance, which in this case is
-j57.7 ohms, determined by the 3:1 mismatch. A series stub having a terminal impedance of +j57.7 ohms cancels
the residual reactance, achieving a match at the stub point.

Now to the phase relationships that occur here that I believe RF has not considered.

First, the line rhoV at the stub is 0.5 at -60°, and the stub rhoV is 0.5 at +60°, with the resultant rho =
0°.
Second, the line rhoi at the stub is 0.5 at +120°, and the stub rhoi is 0.5 at -120°, with the resultant rho =
180°.
Third, with resultant voltage and current at 0° and 180°, respectively, we have achieved a virtual open
circuit to waves reflected from the 150-ohm mismatch, causing total re-reflection of the reflected waves at
the stub point. Proof that total re-reflection has occurred is by observing that there is no evidence of any
reflected waves rearward from the stub point to the source.

Now, when adjusting the output network of a tube-type transceiver to deliver all the available power into a
line having reflections, the adjustment of the network accomplishes the same function as the stub on the line
in the above discussion. Consequently, this is the reason why the reflected power is totally re-reflected at
the output terminals of the network, and is never seen at the plate of the amp to cause heating.

This is the concept I believe Richard Fry is not appreciating. If I'm wrong on this I hope he'll straighten me
out.

Walt