On Sun, 25 Mar 2007 22:20:25 GMT, Walter Maxwell
wrote:

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.

I call this condition 1. It exhibits a mismatch and it exhibits the
probability of the reflected energy being absorbed by the source to
the degree of the phase relationships.

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.

I call this condition 2. It exhibits just what you describe:

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.

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.

From March 14:
"2. The amplifier is now powered down and the load resistance RL is
measured across the input terminals of the resonant pi-network tank
circuit (from plate to ground) with an HP-4815 Vector Impedance
Meter."

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

....

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.

If you follow my separation of arguments above, you will find it
demands that the source MUST have the intervention of an outside agent
to resolve its probability of facing destructive energy. In the arts,
this is called Deus ex Machina. In the world of science, we would
have to say that the source is extremely non-linear when its internal
state of Z (a distinctly non-Thevenin characteristic) changes to
follow the load.

In short, of course no energy finds its way in, we twisted knobs to
make that a self-fulfilling prophecy. Without this intervention
reflected energies present the real probability of destruction by
heat.

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

I will let him speak for himself.

73's
Richard Clark, KB7QHC

On Mon, 26 Mar 2007 00:35:46 GMT, Cecil Moore
wrote:

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

i.e. 92% correct.

Correct but "debatable." Hmmm.

i.e. 85% correct.

This is from someone who's experience an orgasm in the 80 percentile?
Now we venture into almost pregnancies.

Gene Fuller wrote:
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.

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

Paragraph 1 is about energy and doesn't mention power at all.
Paragraph 2 is about power and doesn't mention energy at all.

Energy is as different from power as length is different from
velocity. Do you also get length and velocity confused?

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?

Here's a quote from my web page:

"The term "power flow" has been avoided in favor of "energy flow".
Power is a measure of that energy flow per unit time through a plane."

The dimensions of power flow would be watts/sec. I'm not sure
what quantity watts/sec would represent or where it would go.
--
73, Cecil http://www.w5dxp.com

Richard Clark wrote:
This is from someone who's experience an orgasm in the 80 percentile?

At my age, I wish it was up to 80%. :-)
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote in
:

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?

You are not being paranoid here are you Cecil?


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?

Cecil, I can understand the above paragraph ok, it is self defining
though so I don't consider it to be any great advance of the art.

Whilst I can understand the paragraph, the content is so trivial I can't
understand why you felt it necessary to state it.

Owen

Owen Duffy wrote:
You are not being paranoid here are you Cecil?

The word would be inebriated to the point of poor
judgment. It was a rough day loading and unloading
a full 6x12 U-Haul trailer all by myself. I apologize.

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?

Cecil, I can understand the above paragraph ok, it is self defining
though so I don't consider it to be any great advance of the art.

Didn't say it was an advance. It is my position that
you questioned and it really seems obvious to me.

Whilst I can understand the paragraph, the content is so trivial I can't
understand why you felt it necessary to state it.

It was apparently necessary to get you to understand my
position. Wherever the circulator load resistor is
located is where the dissipation of the energy in the
reflected wave will occur. It can be either inside the
source box or outside the source box.
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:
Gene Fuller wrote:
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.

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

Paragraph 1 is about energy and doesn't mention power at all.
Paragraph 2 is about power and doesn't mention energy at all.

Energy is as different from power as length is different from
velocity. Do you also get length and velocity confused?

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?

Here's a quote from my web page:

"The term "power flow" has been avoided in favor of "energy flow".
Power is a measure of that energy flow per unit time through a plane."

The dimensions of power flow would be watts/sec. I'm not sure
what quantity watts/sec would represent or where it would go.


Cecil,

I understand the physics quite well, thank you.

What I neglected to fully pick up was the silly semantic game you were
playing. We all know that "power flow" is a bit of an oxymoron, but it
still is widely used as an equivalent to "energy flow". I was just
trying to understand the phase of the moon in Texas for this particular
thread.

73,
Gene
W4SZ

On Mar 25, 3:48 am, Ian White GM3SEK wrote:
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


Rumbles, yes indeed. But does it retain any entertainment value, or
has it simply gotten painful to watch?

Cheers,
Tom

Gene Fuller wrote:
We all know that "power flow" is a bit of an oxymoron, but it
still is widely used as an equivalent to "energy flow".

From your last posting it was not readily apparent
that you knew the difference between energy and power.

There is no reflected power wave but there is a
reflected RF energy traveling wave, the power of which
can be measured at a fixed point. A Bird directional
wattmeter indirectly senses the reflected power at the
fixed location of the Bird.
--
73, Cecil http://www.w5dxp.com

On Sun, 25 Mar 2007 19:14:07 -0800, Richard Clark wrote:

On Sun, 25 Mar 2007 22:20:25 GMT, Walter Maxwell
wrote:

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.

I call this condition 1. It exhibits a mismatch and it exhibits the
probability of the reflected energy being absorbed by the source to
the degree of the phase relationships.

Richard, although it exhibits a mismatch, and thus detunes the source, the probability of the reflected energy
being absorbed by the source is zero. The additional power dissipated in the source is due to lowered
impedance of the network resulting from off-resonance operation, thus increasing the plate current. The
reflected energy does not enter the network, but only results in a decrease in the power delivered relative to
that when the reactance in the load is cancelled by correct retuning of the source network.

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.

I call this condition 2. It exhibits just what you describe:

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.

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.

From March 14:
"2. The amplifier is now powered down and the load resistance RL is
measured across the input terminals of the resonant pi-network tank
circuit (from plate to ground) with an HP-4815 Vector Impedance
Meter."

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

...

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.

If you follow my separation of arguments above, you will find it
demands that the source MUST have the intervention of an outside agent
to resolve its probability of facing destructive energy. In the arts,
this is called Deus ex Machina. In the world of science, we would
have to say that the source is extremely non-linear when its internal
state of Z (a distinctly non-Thevenin characteristic) changes to
follow the load.

In short, of course no energy finds its way in, we twisted knobs to
make that a self-fulfilling prophecy. Without this intervention
reflected energies present the real probability of destruction by
heat.

Twisting the knobs is what adjusts the output impedance of the network to the complex conjugate of the load.
But without this intervention the real probability of destruction by heat is not resulting from the reflected
energy reaching the plate--it is only that the reflected energy detuned the network, causing the plate current
to rise because the network is off resonance.

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

I will let him speak for himself.

OK, Richard F, speak up.

73's
Richard Clark, KB7QHC