Keith Dysart wrote:
What drove me to look at alternate explanations for these kinds of
examples
was that the 'reverse power' explanation fails miserably when the
power
gets back to the generator.

The reverse energy wave follows the principles of
conservation of energy and superposition. That we
have no clue what the generator looks like to the
reflected energy wave is not a good reason to
abandon those principles. In fact, allowing
destructive interference to be accompanied by
an energy reflection solves all the problems.
--
73, Cecil http://www.w5dxp.com

Keith Dysart wrote:
I have yet to question the reflection of EM radiation, just the
existence of "reverse power" in transmission lines.

Then you are certainly engaging in the proverbial Red Herring.

A simple example that I can never make add up is a 50 Watt generator
with a 50 ohm output impedance, driving a 50 ohm line which is open at
the end. Using the "reverse power" explanation, 50 W of "forward
power"
from the generator is reflected at the open end, providing 50 W of
"reverse
power". Since the generator is matched to the line there is no
reflection
when this "reverse power" reaches the generator so it disappears into
the generator. If this is truly power, it must go somewhere else, be
dissipated, transformed into some other form or stored (based on the
conservation of energy principle). Where did it go?

This is a lot like the 1/2WL W7EL example in his food for
thought articles. The generator is *NOT* matched to the line
as it sees an open circuit and cannot continue to stuff 50
watts into the open circuit. The generator is as mismatched
as it can possibly be. The reflected wave also sees that open
circuit and is 100% reflected. Since the generator is not
delivering any power and there is a forward power and a
reflected power, the reflected power is supplying the
forward power. Anything else violates the conservation
of energy principle.

Most correspondents agree that what happens depends on the design
of generator; dissipation either increases, decreases or stays the
same (compared to when the line was terminated in 50 Ohms and the
power going down the line is dissipated in the termination). This
does
not make an easy explanation for where that supposedly real power
goes. Of course, if it is not real power, then there is no issue,
which
leads one back to looking for explanations other than "reverse power".

Any level of interference is possible depending upon the
phase angle between the forward E-field and the reflected
E-field. All this is explained in "Optics" by Hecht which
some people have apparently avoided reading/understanding.
Optical physicists solved this problem a century ago. They
don't have the luxury of dealing with voltages and currents
and are forced to deal with power densities. You should
try trodding their paths and enlightening yourself.
--
73, Cecil http://www.w5dxp.com

Owen Duffy wrote:
If one takes measurements with the instrument, it is true that the power
at a point is "forward power" less "reflected power", and the
manufacturer has scaled the instrument in Watts to facilitate that
calculation, but that does not imply that the value of "forward power" or
"reflected power" has any stand alone value, the ratio of the two is
meaningful, the difference of the two is meaningful, but one alone is
meaningless.

How do you explain the fact that a transmission line
contains exactly the amount of energy needed to support
the actual forward power and reflected power? If it
is not moving at the speed of light (modified by VF)
it is not EM energy.
--
73, Cecil http://www.w5dxp.com

Owen Duffy wrote:
The voltage at the point may be higher than under matched conditions for
the same load power, and that may cause insulation breakdown.

Caused by the in-phase superposition of forward and
reflected voltages.

The current at the point may be higher than under matched conditions for
the same load power, and that would cause higher loss in conductors and
may result in damage.

Caused by the in-phase superposition of forward and
reflected currents.

None of these explanations require designating "reflected power" at a
point, or implying that it is the energy in "reflected power" that is
totally and solely responsible for the physical damage.

EM wave energy necessarily travels at the speed of light.
There is exactly the amount of EM wave energy contained
in a transmission line to support the forward power and
reflected power.
--
73, Cecil http://www.w5dxp.com

Keith Dysart wrote:
And you have done an excellent job of doing so by providing
an explanation that refers only to forward and reflected
voltage and current. Not a mention of "reverse power" in
the explanation..

The forward voltage is in phase with the forward current.
The reflected voltage is in phase with the reflected
current. Vref*Iref*cos(0) = reflected power
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote in
t:

....
EM wave energy necessarily travels at the speed of light.
There is exactly the amount of EM wave energy contained
in a transmission line to support the forward power and
reflected power.

You are not suggesting that the energy contained in a transmission line in
the steady state in the general case is constant, are you?

Owen

Cecil Moore wrote:
Gene Fuller wrote:
It is interesting that you can be so precise at times and so sloppy
at other times. I very carefully limited my discussion to steady state
conditions, which is what everyone is already talking about in this
case. You then conveniently inject modulation into the mix, completely
ignoring what I said.

*Every* real world system has noise modulation
that can be tracked through the system riding on
the forward and reflected traveling waves. Thus
steady-state is never reached in reality and your
argument is therefore just a mind game.

Cecil, every time someone *almost* succeeds in making you stick to the
point, you accuse them of messing with your mind.

Nobody is trying to restrict your freedom of thought or speech. Nobody
wants to, and nobody can. But lots of people are hoping, begging,
pleading that you develop the SELF-discipline to follow an argument all
the way through to its conclusion, without jumping outside of the
boundaries you laid down at the start.

This may indeed trouble your mind; but the process of scientific inquiry
wasn't ever *intended* to feel comfortable. Its only goal is to find
things out and get them right... and that involves staying on track,
even when it takes you outside of the comfort zone. In fact, then most
of all, because it's a sure sign that the track is leading somewhere.



--

73 from Ian GM3SEK 'In Practice' columnist for RadCom (RSGB)
http://www.ifwtech.co.uk/g3sek

Owen Duffy wrote:
Cecil Moore wrote in
EM wave energy necessarily travels at the speed of light.
There is exactly the amount of EM wave energy contained
in a transmission line to support the forward power and
reflected power.

You are not suggesting that the energy contained in a transmission line in
the steady state in the general case is constant, are you?

Obviously, a leading question. :-) I'm not talking
about instantaneous values here. All my statements
apply only to values averaged over an integer
number of RF cycles in one second.

What I am saying
is that a transmission line obeys the conservation
of energy principle. Whatever energy has not gone
somewhere else is still in the transmission line
and is exactly the sum of the energy required by the
forward wave plus the energy required by the reflected
wave. The steady-state energy stored in a transmission
line with reflections is greater than the steady-
state energy delivered to the load. The extra energy
was sourced during the transient build-up state and
has not yet been delivered to the load. The argument
that 100 watts in and 100 watts out (during steady-
state) doesn't leave any energy left over for the
reflected waves is invalid. Below I give an example
of 100 watts in and 100 watts out with 300 joules
stored in the forward and reflected waves.

I think it would be safe to say that for any one-
wavelength section of line, the average energy
content is constant during steady-state and is
exactly the amount of energy required to support
the forward traveling wave and reflected traveling
wave.

Here's a graphic that I earlier provided that
illustrates the energy buildup to steady-state.

http://www.w5dxp.com/1secsgat.gif
--
73, Cecil http://www.w5dxp.com

Ian White GM3SEK wrote:
Nobody is trying to restrict your freedom of thought or speech. Nobody
wants to, and nobody can. But lots of people are hoping, begging,
pleading that you develop the SELF-discipline to follow an argument all
the way through to its conclusion, without jumping outside of the
boundaries you laid down at the start.

Sorry Ian, I don't trust you guys enough to roll
dice in the dark with you and then let you tell
me what value was rolled. :-) You are perfectly
free to play mashed potatoes with RF joules but
please don't ask me to join in. Sometimes I think
you have to be just pulling my leg.

It was Gene who first pointed out the difference
between a traveling wave and a standing wave. Now
he says there is no difference.

Gene Fuller, W4SZ wrote:
In a standing wave antenna problem, such as the one you describe,
there is no remaining phase information. Any specific phase
characteristics of the traveling waves died out when the startup
transients died out.

Phase is gone. Kaput. Vanished. Cannot be recovered. Never to be
seen again.

One can send two coherent light beams in opposite
directions almost collinear to each other and observe
the standing waves. Hecht has a graphic of such in
"Optics". The two light beams, forward and reverse,
emerge past the standing wave space undisturbed and
unaffected by the superposition.

How does your theory hold up when measurement of voltage
and current is impossible and everything occurs in free
space for anyone to observe with his/her own eyes? If it
doesn't work for EM light waves then it also doesn't
work for EM RF waves except as a shortcut.

So please take the above example from optics and show
me how the reverse traveling wave is different on
either side of the standing wave space.
--
73, Cecil http://www.w5dxp.com

On Wed, 21 Mar 2007 08:18:14 -0500, "Richard Fry" wrote:

"Walter Maxwell" wrote
(RF): And if so, would that also mean that such a tx would not be prone
to producing r-f intermodulation components when external signals
are fed back into the tx from co-sited r-f systems?

This issue is irrelevant, because the signals arriving from a co-sited
system would not be coherent with the local source signals, while load-
reflected signals are coherent. The destructive and constructive
interference that occurs at the output of a correctly loaded and tuned
PA requires coherence of the source and reflected waves to achieve
the total re-reflection of the reflected waves back into the direction
toward the load.

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.

Also, does not the result of combining the incident and reflected waves in
the tx depend in large part on the r-f phase of the reflection there
relative to the r-f phase of the incident wave? And the r-f phase of the
reflection is governed mostly by the number of electrical wavelengths of
transmission line between the load reflection and the plane of
interest/concern -- which is independent of how the tx has been
tuned/loaded.

If the ham transmitter designs that your paper applies to produce a total
re-reflection of reverse power seen at their output tank circuits, then
there would be no particular need for "VSWR foldback" circuits to protect
them. Yet I believe these circuits are fairly common in ham transmitters,
aren't they? They certainly are universal in modern AM/FM/TV broadcast
transmitters, and are the result of early field experience where PA tubes,
tx output networks, and the transmission line between the tx and the antenna
could arc over and/or melt when reflected power was sufficiently high.

RF
Richard, your statement above begs the question, "Are you aware of the phase relationships between forward and
reflected voltages and between forward and reflected currrents that accomplish the impedance-matching effect
at matching points such as with stub matching and also with antenna tuners?

When the matching is accomplished the phase relationship between the foward and reflected voltages can become
either 0° or 180°, resulting in a total re-reflection of the voltage. If the resultant voltage is 0°, then the
resultant current is 180°, thus voltage sees a virtual open circuit and the current sees a virtual short
circuit. The result is that the reflected voltage and current are totally re-reflected IN PHASE with the
source voltage and current. This is the reason the forward power in the line is greater than the source power
when the line is mismatched at the load, but where the matching device has re-reflected the reflected waves.

This phenomenon occurs in all tube transmitters in the ham world when the tank circuit is adjusted for
delivering all available power at a given drive level. When this condition occurs the adjustment of the
pi-network has caused the relationship between the forward and reflected voltages to be either 0° or 180° and
vice versa for currents, as explained above. When this condition occurs, destructive interference between the
forward and reflected voltages, as well as between the forward and reflected currents, causes the reflected
voltage and current to cancel. However, due to the conservation of energy, the reflected voltage and current
cannot just disappear, so the resulting constructive interference following immediately, causes the reflected
voltage and current to be reversed in direction, now going in the foward direction along with and in phase
with the forward voltage and current.

In transmitters with tubes and a pi-network output coupling circuit there is no 'fold back' circuitry to
protect the amp, because none is needed, due to the total re-reflection of the reflected power. It is only in
solid-state transmitters that have no circuitry to achieve destructive and constructive interference that
requires fold back to protect the output transistors.

This has been a quick and dirty explanation of the phase relations that accomplish impedance matching.
However, I have explained it in much more detail in my book "Reflections--Transmission Lines and Antennas."
Yes, I know the book has been sold out and now unavailable, but I have put several chapters on my web page
avaliable for downloading. The pertinent chapters covering this issue are Chapters 3, 4, and 23, available at
www.w2du.com. I hope that reviewing these chapters will be helpful in clearing up some of the
misunderstandings that are clearly evident in some of the postings on this thread.

Walt, W2DU