james wrote:

Agreed that a rig cannot detect the difference between forward and
reflected power. If the reflection coeffiecient of the source is zero
then final stage of a transmiter will look purely resistive to any
power reflected by the load. Thereby that refelcted power is
dissapated as heat. Other reflection coefficients at the source will
yield lesser amounts of reflected power from the load as heat.

james

I've posted many, many times on this topic and have shown a number of
cases where the load is perfectly matched but the power dissipated in
the source resistor is less than or greater than the "reverse power",
clearly demonstrating that this concept is incorrect. There are several
examples at Food for thought.txt available at
http://eznec.com/misc/food_for_thought/.

Because I've posted so much on the topic I won't do it all again. But I
know at least one person on this newsgroup would be glad to have an
opportunity to express his views once again. I'll leave this discussion
to those who want to revisit it; I don't. But I do want to caution
readers that this view of "reflected power" is demonstrably incorrect.

Roy Lewallen, W7EL

Roy Lewallen wrote:
But I do want to caution
readers that this view of "reflected power" is demonstrably incorrect.

So is your concept of "sloshing" energy. Reflected energy
waves are demonstrably real. One can find out exactly where
the reflected power goes by taking the interference power
terms into account. Optics engineers figured it out a long
time ago.
--
73, Cecil http://www.qsl.net/w5dxp

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On Wed, 29 Jun 2005 23:00:17 -0500, Cecil Moore
wrote:
Optics engineers figured it out a long time ago.
And you have consistently failed in its demonstration - so what?

Richard Clark wrote:

Cecil Moore wrote:
Optics engineers figured it out a long time ago.

And you have consistently failed in its demonstration - so what?

I can lead you to water but I can't make you drink.
--
73, Cecil http://www.qsl.net/w5dxp

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Roy Lewallen wrote:
There are several
examples at Food for thought.txt available at
http://eznec.com/misc/food_for_thought/.

Regarding errors in the first food_for_thought:

A 100w source equipped with a circulator and load while
looking into an open line, will generate 100w and dissipate
100w in the circulator load. That 100w is definitely not free
power. It can be demonstrated to have made a round trip to
the open end of the feedline and then back to the circulator
load.

The error in your thinking is that the source would see an open
circuit when it is equipped with a circulator and load. It won't.
It will *always* see the Z0 of the feedline as its load (assuming
the circulator load equals Z0). That's the purpose of using
the circulator and load - to allow the source to see a fixed
load equal to Z0.
--
73, Cecil http://www.qsl.net/w5dxp

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Much noise has been radiated. I speculate that a reminder about what
linearity means might get things back on track. In a linear network (lumped
or distributed) superposition (of linear signals) produces correct results.

The last statement works in both directions. (The degree to which a
network is linear is the same as the degree to which superposition is
valid.) (If one supplies a large enough signal to any network, it will
become non-linear - as in letting-out-the-smoke-put-in-at-the-factory.)

The catch in all of the above is that superposition only applies to
linear signals and power (however indicated) is not a linear signal.
Power, which could be complex power S = V*I* (the phasor voltage time the
conjugate of the phasor current) or the magnitude of S (apparent power) or
the real part of S ("real" power), simply does not obey superposition even
in a network that is linear.

Bottom line: assuming the use of networks (lumped or distributed) that
are essentially linear, one is only allowed to combine phasor voltages or
phasor currents (but not their product nor the square of such linear
signals). Once combined, the resultant voltage and the resultant current
may be used to find a measure of power. (The "combined" mentioned must be
a linear, additive process.)

It seems to me that Roy, and others, have plowed this ground many times.

73 Mac N8TT

--
J. Mc Laughlin; Michigan U.S.A.
Home:

"Roy Lewallen" wrote in message
snip

I've posted many, many times on this topic and have shown a number of
cases where the load is perfectly matched but the power dissipated in
the source resistor is less than or greater than the "reverse power",
clearly demonstrating that this concept is incorrect. There are several
examples at Food for thought.txt available at
http://eznec.com/misc/food_for_thought/.

Because I've posted so much on the topic I won't do it all again. But I
know at least one person on this newsgroup would be glad to have an
opportunity to express his views once again. I'll leave this discussion
to those who want to revisit it; I don't. But I do want to caution
readers that this view of "reflected power" is demonstrably incorrect.

Roy Lewallen, W7EL

J. Mc Laughlin wrote:
Bottom line: assuming the use of networks (lumped or distributed) that
are essentially linear, one is only allowed to combine phasor voltages or
phasor currents (but not their product nor the square of such linear
signals).

Power can certainly be combined, just not superposed. Here is
the irradiance equation from _Optics_, by Hecht.

Itot = I1 + I2 + Sqrt(I1*I2)cos(theta)

Irradiance is power/unit-area. The last term is known as the
"interference" term. Hecht provides separate chapters for
interference and superposition, the best treatment of those
two subjects of which I am aware.

Here is the transmission line forward power equation from Dr.
Best's QST transmission line article.

Ptot = P1 + P2 + Sqrt(P1*P2)cos(theta)

It is virtually identical to the irradiance equation above.
The last term is known to be the "interference" term and
for a Z0-matched system, THETA EQUALS ZERO, so for a Z0-
matched system, a complete analysis can be done using only
the forward and reflected power magnitudes. This is something
I and others have been saying for years and it has been
called "gobbledegook" (sic) by Roy (and worse by others) even
though Roy admits that he doesn't care to understand where the
power goes.

It seems to me that Roy, and others, have plowed this ground many times.

Yes, and they are still not 100% correct. Roy has said:
'I personally don't have a compulsion to understand where
this power "goes".' Too bad he doesn't have that compulsion
because, with a small amount of mental effort, he surely would
have figured it out before me - if by no other means, simply by
referencing the chapter on EM wave interference in _Optics_.
--
73, Cecil, http://www.qsl.net/w5dxp

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On Thu, 30 Jun 2005 09:41:05 -0500, Cecil Moore
wrote:
Power can certainly be combined, just not superposed. Here is
the irradiance equation from _Optics_, by Hecht.
Itot = I1 + I2 + Sqrt(I1*I2)cos(theta)
According to this formula, for two laser beams of 100W each, without
any phase difference, we can illuminate a target with
Itot = 100 + 100+ Sqrt(100*100)cos(0)
Itot = 300W

Every CBer's dream....

Hecht should sue you for copyright Unfair Use.

Richard Clark wrote:

On Thu, 30 Jun 2005 09:41:05 -0500, Cecil Moore
wrote:

Power can certainly be combined, just not superposed. Here is
the irradiance equation from _Optics_, by Hecht.
Itot = I1 + I2 + Sqrt(I1*I2)cos(theta)

According to this formula, for two laser beams of 100W each, without
any phase difference, we can illuminate a target with
Itot = 100 + 100+ Sqrt(100*100)cos(0)
Itot = 300W

Sorry, I made a mistake in the equation. Please forgive
my omission. Here's the correct equation:

Itot = I1 + I2 + 2*Sqrt(I1*I2)cos(theta)

So if theta equals zero, Itot would be 400w, not 300w.

This is the power of superposition of coherent waves. When
you superpose two 100w coherent laser beams, the resultant
power is indeed 400w and must be supplied by the sources
or supplied by destructive interference from somewhere else.
This is all explained in _Optics_, by Hecht. Hows about
reading it so I won't have to explain superposition to you?
--
73, Cecil, http://www.qsl.net/w5dxp

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On Thu, 30 Jun 2005 11:18:37 -0500, Cecil Moore
wrote:
Sorry, I made a mistake in the equation.
....
Hows about reading it so I won't have to explain superposition to you?
Certainly no one stands any chance of understanding your explanations
given the continuing goof-ups.