Revisiting the Power Explanation
 

On Mar 24, 12:40 pm, Cecil Moore wrote:
Keith Dysart wrote:
Standard analysis can tell us quite a bit about the situation
without the need for further information. The information
we have:
- Generator with 450 Ohm output impedance

There are a number of techniques to ensure that the
reflected waves encounter an impedance of 450 ohms
but none of those techniques are implemented in
amateur radio transmitters. Your generator is
obviously equipped with a circulator load of 450
ohms, 450 ohm attenuation pads, or some active
feedback.

While it might be, none of the above are necessary nor
included in the example I have analyzed.
A simple voltage source in series with 450 Ohms resistor
or a current source in parallel with one will produce
exactly the same results.

Look carefully at the analysis and you will find no
evidence of any of the items you have mentioned above,
unless you consider the 450 Ohm resistor to be an
attenuation pad. (Though it is this resistor which matches
the line and ensures there is no reflection).

Given that, your analysis is correct but
moot. Your point is already known and accepted by
every initiated person including me.

Not you, I suspect, since you seem to find it necessary
to add some components which were not in the description
of the setup.

I freely admit that if the proper circulator load
exists or if 450 ohm attenuator pads exist, they
will dissipate the reflected power by heating
up the circulator load resistor or the pads.

Yes, indeed they would. The more challenging case for
you is the example I presented without these components.
How do you explain the absence of ghosts? And the
unaccountability of the 'reflected power'.

While an example with a circulator makes it easy to
locate and account for the 'reflected power', there is
more to be learned by studying the example without.

If you can not demonstrate a flaw in the analysis then you
may wish to explore why the example without a circulator
(or other additional components) causes you difficulty.

But it's tough to argue that there is no energy in
the reflected waves while they are being used to boil
water.

While you can construct examples where the 'reflected
power' will boil water, there are alternate examples where
the total system dissipation drops in the presence of
'reflected power'. So sometimes 'reflected power' boils
water and sometimes it does not. Not a very reliable source
of energy it seems.

But I do suggest going back to the example. If there is no
flaw in the analysis, please explain where the 'reflected
power' goes.

....Keith

Revisiting the Power Explanation
 

Cecil Moore wrote in news:9wcNh.95$Q23.73
@newssvr17.news.prodigy.net:


I freely admit that if the proper circulator load
exists or if 450 ohm attenuator pads exist, they
will dissipate the reflected power by heating
up the circulator load resistor or the pads.

Considering only the attenuator case, you suggest in such a general
statement that reflected power always increases the dissipation in a real
source that includes an attenuator for the purpose of constraining the
equivalent source impedance.

Such a suggestion is incorrect in the steady state, some loads may increase
the generator's internal dissipation, some may decrease it depending on the
attenuator circuit.

Owen

Revisiting the Power Explanation
 

In message , Cecil Moore
writes
Richard Clark wrote:
Cecil Moore wrote:
your analysis is correct but moot
Is he stealing your style?

"Moot" is an interesting word, Richard. From
Webster's - "moot - 1. a: debatable, b: disputed"

Have a look here.
http://en.wikipedia.org/wiki/Moot_hall
I remember 'Moot Hall' from my days at primary school (some 60 years
ago), learning about the Anglo-Saxons. I guess the word may possibly be
associated with 'meet', ie a meeting hall where things were debated.
However, my Anglo-Saxon is a bit rusty (not much call for it these
days).
Ian.
--

Revisiting the Power Explanation
 

On Sat, 24 Mar 2007 17:08:48 GMT, Cecil Moore
wrote:

Richard Clark wrote:
Cecil Moore wrote:
your analysis is correct but moot

Is he stealing your style?

"Moot" is an interesting word, Richard. From
Webster's - "moot - 1. a: debatable, b: disputed"

Correct but "disputed." Hmmm.
Correct but "debatable." Hmmm.

Sounds like the definition of masturbation without orgasm. Yes, I
recognize your style there.

Revisiting the Power Explanation
 

Ian Jackson wrote:
In message , Cecil
Moore writes
Richard Clark wrote:
Cecil Moore wrote:
your analysis is correct but moot
Is he stealing your style?

"Moot" is an interesting word, Richard. From
Webster's - "moot - 1. a: debatable, b: disputed"

Have a look here.
http://en.wikipedia.org/wiki/Moot_hall
I remember 'Moot Hall' from my days at primary school (some 60 years
ago), learning about the Anglo-Saxons. I guess the word may possibly be
associated with 'meet', ie a meeting hall where things were debated.
However, my Anglo-Saxon is a bit rusty (not much call for it these days).

Cecil was using "moot" in its legal sense: that a point had become
irrelevant, or no longer needed to be decided because of a change in
circumstances.

Or at least, Cecil tried to claim that a point made by Keith had become
moot. But Keith disputed that... and so it rumbles on.


--

73 from Ian GM3SEK

Revisiting the Power Explanation
 

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

Revisiting the Power Explanation
 

Keith Dysart wrote:
While an example with a circulator makes it easy to
locate and account for the 'reflected power', there is
more to be learned by studying the example without.

You are not allowed to deny the existence of the 450 ohm
real world load while asserting that it still exists.
If it exists, it dissipates the reflected energy. If it
doesn't exist, it reflects the reflected energy. Please
choose one or the other - obviously, you cannot have both
at the same time.

While you can construct examples where the 'reflected
power' will boil water, there are alternate examples where
the total system dissipation drops in the presence of
'reflected power'. So sometimes 'reflected power' boils
water and sometimes it does not. Not a very reliable source
of energy it seems.

If reflected energy is not dissipated, it undergoes destructive
interference and is redirected back toward the load as constructive
interference instead of being incident upon the source. Why
are you having difficulty with that concept from page 388
of "Optics", by Hecht, 4th edition?

But I do suggest going back to the example. If there is no
flaw in the analysis, please explain where the 'reflected
power' goes.

Again, reflected power doesn't flow so it doesn't go anywhere.
Until you understand that simple fact of physics, further
discussion is unlikely to yield valid results. "Reflected power
that goes" is only one of your many conceptual flaws. So your
first logical step would be proving that reflected power actually
flows. After you do that, we can continue to your other conceptual
flaws.
--
73, Cecil http://www.w5dxp.com

Revisiting the Power Explanation
 

Owen Duffy wrote:
Considering only the attenuator case, you suggest in such a general
statement that reflected power always increases the dissipation in a real
source that includes an attenuator for the purpose of constraining the
equivalent source impedance.

Owen, why do you feel compelled to lie about what I said?

If the reflected energy is dissipated within the source, it
increases the dissipation. If the reflected energy is not
dissipated within the source, it does not increase the
dissipation. Why is that so hard for you to understand?
--
73, Cecil http://www.w5dxp.com

Revisiting the Power Explanation
 

Richard Clark wrote:
Correct but "disputed." Hmmm.

i.e. 92% correct.

Correct but "debatable." Hmmm.

i.e. 85% correct.
--
73, Cecil http://www.w5dxp.com

Revisiting the Power Explanation
 

Cecil Moore wrote:

*** 1 ***

If reflected energy is not dissipated, it undergoes destructive
interference and is redirected back toward the load as constructive
interference instead of being incident upon the source. Why
are you having difficulty with that concept from page 388
of "Optics", by Hecht, 4th edition?


*** 2 ***


Again, reflected power doesn't flow so it doesn't go anywhere.
Until you understand that simple fact of physics, further
discussion is unlikely to yield valid results. "Reflected power
that goes" is only one of your many conceptual flaws. So your
first logical step would be proving that reflected power actually
flows. After you do that, we can continue to your other conceptual
flaws.


Cecil,

Paragraphs 1 and 2 appear to declare exactly the opposite behavior for
energy (power).

Is there some subtle re-definition going on to allow "redirected back
toward the load" and "it doesn't go anywhere" in the same message?

What is it that you are trying to say?

73,
Gene
W4SZ