The rig has no way of detecting any alleged "reflected power". It can't
tell the difference between a feedline with a lot of "reflected power",
a feedline with no "reflected power", and a plain resistor. It behaves
exactly the same in all cases, provided only that the impedance that
each provides to it is the same.

Anyone not convinced of this should put a couple or more dummy loads in
series or parallel, make up a few lengths of transmission line of
various impedances, and see for himself.

Roy Lewallen, W7EL

james wrote:
On Wed, 29 Jun 2005 16:42:49 -0500, Cecil Moore
wrote:


Sometimes yes, sometimes no. If the reflected current arrives out
of phase with the forward current, then the final dissipation can
actually be *reduced* by the mismatch.

*****

Power is power. Phase is not a problem. Take the mafnitude of the
transmitted power and teh magnitude of the reflected power. The
results are phaseless. The magnitudes add linearly.

QED

james

On Wed, 29 Jun 2005 16:29:26 -0700, Roy Lewallen
wrote:

The rig has no way of detecting any alleged "reflected power". It can't
tell the difference between a feedline with a lot of "reflected power",
a feedline with no "reflected power", and a plain resistor. It behaves
exactly the same in all cases, provided only that the impedance that
each provides to it is the same.

*****

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

Anyone not convinced of this should put a couple or more dummy loads in
series or parallel, make up a few lengths of transmission line of
various impedances, and see for himself.

Roy Lewallen, W7EL

james wrote:
On Wed, 29 Jun 2005 16:42:49 -0500, Cecil Moore
wrote:


Sometimes yes, sometimes no. If the reflected current arrives out
of phase with the forward current, then the final dissipation can
actually be *reduced* by the mismatch.

*****

Power is power. Phase is not a problem. Take the mafnitude of the
transmitted power and teh magnitude of the reflected power. The
results are phaseless. The magnitudes add linearly.

QED

james

james wrote:

On Wed, 29 Jun 2005 16:29:26 -0700, Roy Lewallen
wrote:


The rig has no way of detecting any alleged "reflected power". It can't
tell the difference between a feedline with a lot of "reflected power",
a feedline with no "reflected power", and a plain resistor. It behaves
exactly the same in all cases, provided only that the impedance that
each provides to it is the same.


*****

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


Anyone not convinced of this should put a couple or more dummy loads in
series or parallel, make up a few lengths of transmission line of
various impedances, and see for himself.

Roy Lewallen, W7EL

james wrote:

On Wed, 29 Jun 2005 16:42:49 -0500, Cecil Moore
wrote:



Sometimes yes, sometimes no. If the reflected current arrives out
of phase with the forward current, then the final dissipation can
actually be *reduced* by the mismatch.

*****

Power is power. Phase is not a problem. Take the mafnitude of the
transmitted power and teh magnitude of the reflected power. The
results are phaseless. The magnitudes add linearly.

QED

james




You need to read _Reflections II, Transmission Lines and Antennas_
by M. Walter Maxwell, W2DU. Even better, get a book on electromagnetics.
You might be able to puzzle some of it out although much of
the math might be too esoteric for you.
73,
Tom Donaly, KA6RUH

On Thu, 30 Jun 2005 00:19:22 GMT, "Tom Donaly"
wrote:

You need to read _Reflections II, Transmission Lines and Antennas_
by M. Walter Maxwell, W2DU. Even better, get a book on electromagnetics.
You might be able to puzzle some of it out although much of
the math might be too esoteric for you.
73,
Tom Donaly, KA6RUH
*****

Whats wrong with stating that power is reflected by the load? Isn't
power delivered to the load from the source? Elementary electronics
states that power is voltage time current.

Currents in a transmission line are induced currents. They are induced
from the E and H fields of the TEM wave. I hope that you don't think
that current races up and down the coax millions a times per second?

james

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