Cecil Moore wrote:
Roy Lewallen wrote:

Did anyone understand it? If so, would someone else please try to
explain it to me? Where does that 18 watts of "reverse power" go, and
why?


Bob Lay, w9dmk, could explain it but he's off on a camping trip.
I think the QEX editors can now explain it also. 18 watts is
rejected by the mismatched load. It was part of a forward wave
that contained energy and momentum. That energy and momentum
MUST be conserved according to the laws of physics. The reflected
wave heads back toward the source at the speed of light and part
is re-reflected as Pref2*rho^2.

rho at the source is zero; the source matches the transmission line Z0.
So the "part" which is re-reflected is zero, by your calculations.

The remaining
Pref(1-rho^2) part

(which is all of it, since none was re-reflected)

engages with Pfor1*rho^2 in an equal
magnitude/opposite phase wave cancellation as explained by Walter
Maxwell on page 23-9 of "Reflections II".

That part of Walt's text is talking about voltage and current waves. I'm
familiar, thanks, with how they interact. You're the one talking about
waves of average power -- since none of the reverse power wave is
re-reflected, what happens to it? Where does the power go? Out the
source resistor? I believe you said that 11.52 watts of it did. What
happened to the rest?

Since energy and momentum
cannot be cancelled, the two cancelled waves conserve energy and
momentum by heading back toward the load at the speed of light as
a re-reflection of energy which joins the forward wave energy.

What two waves head back toward the load? None of the alleged reverse
power wave is reflected, by your own calculation. Where did it go?
What's the other "cancelled wave" and where did it come from? What's its
magnitude?

Roy Lewallen, W7EL

Jim Kelley wrote:

It was a tongue-in-cheek reply, Roy. These problems always seem to end
up with a circulator in them at some point - clearly illustrating what
happens under entirely different circumstances. :-)

Thanks for the explanation. When Cecil has difficulty explaining a
circuit, he comes up with one which he can explain and tries to deflect
the discussion to it. This happens over and over and over -- it's really
a pain to try to keep steering the discussion back to the original circuit.

That leaves us with no one who claims to understand Cecil's
"explanation" of my simple circuit. Except, I guess, one guy who's on
vacation and some folks who don't read this newsgroup.

Roy Lewallen, W7EL

Cecil Moore wrote:
Jim Kelley wrote:

Roy Lewallen wrote:

Did anyone understand it? If so, would someone else please try to
explain it to me? Where does that 18 watts of "reverse power" go, and
why?


Wouldn't it go to the circulator load which must always be placed at
the source in order to clearly illustrate what happens when a
circulator isn't in place at the source?


When a Z0-match is in place at the source, everything is also clear
since zero reflected energy reaches the source.

But there's a Z0 match at the source in my example. Simple wave
mechanics show that the initial traveling voltage and current waves
reflect from the load, return to the source, and don't reflect any
further -- steady state is reached after a single round trip. That's a
Z0 source match. No circulator is required.

The Z0-match case,
using an antenna tuner, is the most likely case to be encountered
in ham radio. 100% of the reflected energy and momentum

Walt's book mentions momentum? Where?

is re-reflected
back toward the load, just as Walter Maxwell has been saying as long
as I can remember.

Hm. You said earlier that 11.52 watts of the reverse power wave reached
the source. In a posting a short while ago you said that none of it is
re-reflected (since rho at the source = 0), and now you say it all is.

Guess that covers just about any possibility -- no matter how things
come out, you can say you gave the right answer. You might consider
tossing out a couple more just in case.

This stuff is not new. It is explained in "Fields and Waves ..."
by Ramo and Whinnery, copyright 1950's.

Did they give three different answers to a simple question?

Roy Lewallen, W7EL

If you put the perfect voltage source and the source resistor into a box
and label it "Source", you have a Source whose impedance perfectly
matches the transmission line. It's a Z0-matched system.

The source impedance of my circuit is as simple as it can get. If you
can't explain how it works, it reveals a deficiency of your theory; I
can easily calculate the voltage, current, and power in any component at
any moment of time. Without requiring bouncing waves of average power.
Or a circulator.

Roy Lewallen, W7EL

Cecil Moore wrote:
Roy Lewallen wrote:

Isn't it possible to explain what happens to the "reverse power"
without a circulator?


It is certainly possible in a Z0-matched system where
zero reflected energy reaches the source and the source
is feeding its designed-for load. If no reflected energy
reaches the source, the source impedance doesn't matter,
except for efficiency, which doesn't affect the rest of
the system. A one ohm source providing 100v to a 50 ohm
load and a one megohm source providing 100v to a 50 ohm
load will result in identical external conditions when
driving a 50 ohm load.

The picture is not as clear when reflected current and
voltage are allowed to flow into the source. We usually
don't know what the source impedance is and that
certainly handicaps any analysis. Modern designers simply
resort to protection circuitry and don't worry about
the energy analysis. We do know that reflections reaching
the source are not benign.

Roy Lewallen wrote:
I've explained how I calculated how much energy is stored in the
transmission line. The movement of energy within the line is complex; in
the abbreviated analysis I've had time to do so far, it sloshes back and
forth in regions within the line.

From where to where at what speed? Sloshing sounds like a
violation of the conservation of momentum. What force changes
the direction of the slosh? How much efficiency is lost in
the energy required to change sloshing directions?

It does not travel in waves of average
power, bouncing back and forth, and believing so isn't necessary in
order comply with energy conservation.

Nothing RF travels in waves of average power. Average power is
just our simplified shorthand way of dealing with it. The AC
voltage is equal to the RMS value at only two points in
the cycle. Sticking with instantaneous values for everything
AC would complicate things beyond belief.

The values of voltage in your chart were RMS (average) values.
You don't seem to have a problem dealing with RMS values of
voltages and currents. Why do you have a problem with average
power associated with those average RMS values of voltage and
current.

Your view of power and energy is
oversimplified, and it fails when you're pressed to explain what happens
at the interface between the line and the outside world.

What is located at that interface? Anybody is pressed to explain
things when the source impedance is unknown and cannot be measured.

Momentum is conserved in mechanical elastic collisions, but not in
inelastic ones, e.g., when energy is being extracted. I wouldn't begin
to try to apply this to a transmission line, but I see it doesn't bother
you.

Hecht, in "Optics" certainly applies it to EM waves. RF EM waves
differ from light waves only in frequency. Conservation of
momentum is a cornerstone of EM light physics and is included
in the rules of relativity.
--
73, Cecil http://www.qsl.net/w5dxp


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Roy Lewallen wrote:

rho at the source is zero; the source matches the transmission line Z0.
So the "part" which is re-reflected is zero, by your calculations.

Not my calculations. The measured SWR at the source terminals
is 4:1. rho = (SWR-1)/(SWR+1) = 3/5 = 0.6

That part of Walt's text is talking about voltage and current waves. I'm
familiar, thanks, with how they interact. You're the one talking about
waves of average power

Walt's text talks about RMS (average) voltage and current. When
you multiply RMS voltage by RMS current, you get average power.
V+*I+ = forward power, V+/I+ = Z0, (PV+)=(E+)x(H+)
V-*I- = reflected power, V-/I- = Z0, (PV-)=(E-)x(H-)

-- since none of the reverse power wave is
re-reflected, what happens to it?

On the contrary, it is 100% re-reflected by wave cancellation,
the fourth thing that can cause 100% re-reflection. Optical
engineers understand that. Most RF engineers don't. Adding
the one wavelength of lossless 200 ohm feedline reveals the
wave cancellation mechanism.

What two waves head back toward the load? None of the alleged reverse
power wave is reflected, by your own calculation.

Not my calculation. All of the reverse power wave is reflected
by wave cancellation at the source terminals. No reflected power
is dissipated in the source resistor. Therefore, it is all
reflected back toward the load. The one wavelength of 200 ohm
lossless feedline reveals exactly what happens.
--
73, Cecil http://www.qsl.net/w5dxp


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Roy Lewallen wrote:
But there's a Z0 match at the source in my example. Simple wave
mechanics show that the initial traveling voltage and current waves
reflect from the load, return to the source, and don't reflect any
further -- steady state is reached after a single round trip. That's a
Z0 source match. No circulator is required.

You obviously don't understand a Z0-match, Roy. All the reflected
power is re-reflected back toward the load at a Z0-match. What you
have in your example is a Z0-match to 200 ohms. You can prove it
by observing that the reflections disappear on a piece of 200 ohm
feedline.

------200 ohm feedline---+---1/2WL 50 ohm feedline---200 ohm load

The SWR on the 50 ohm feedline is 4:1. The SWR on the 200 ohm
feedline is 1:1. That is a 200 ohm Z0-match and rho is 0.6

Hm. You said earlier that 11.52 watts of the reverse power wave reached
the source.

Reflected by wave cancellation *at* the source , certainly not
flowing back through the source. What I said is that the two 11.52
watt reflected wave components undergo wave cancellation that results
in 23.04 watts being re-reflected.

In a posting a short while ago you said that none of it is
re-reflected ...

Sorry, Roy, you know and I know that is a lie. Please
apologize and retract that false statement.
--
73, Cecil http://www.qsl.net/w5dxp


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Roy Lewallen wrote:
If you put the perfect voltage source and the source resistor into a box
and label it "Source", you have a Source whose impedance perfectly
matches the transmission line. It's a Z0-matched system.

Sorry, Roy, a 50 ohm source driving a 200 ohm load is NOT a
Z0-matched system by any stretch of the imagination. From the
ARRL Antenna Book: "The Z0 mismatch (at the load) creates a
reflection having a magnitude of rho = (ZL-Z0)/(ZL+Z0)
causing a reflection loss rho^2 that is referred back along
the line to the generator. This in tern causes the generator to
see the same magnitude of Z0 mismatch at the line input."

This exactly describes your example which has a 50 ohm Z0
mismatch (and a 200 ohm Z0 match). The mismatch is easy to
see in the following:

50 ohm source--1WL 200 ohm line--+--1/2WL 50 ohm line--200 ohm load

Point '+' is a Z0-match to 200 ohms. A Z0-match to 50 ohms
doesn't exist anywhere.

The source impedance of my circuit is as simple as it can get. If you
can't explain how it works, it reveals a deficiency of your theory;

You assume I don't know how it works because you don't even know
what a Z0-match is???????? How very typical of you.
--
73, Cecil http://www.qsl.net/w5dxp


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Jim Kelley wrote:
Wouldn't it go to the circulator load which must always be placed at
the source in order to clearly illustrate what happens when a
circulator isn't in place at the source?

ac6xg
Sorry, I was asking about the simple example I posted (which
doesn't have a circulator), not the problem posted by Cecil.

It was a tongue-in-cheek reply, Roy. These problems always seem to end
up with a circulator in them at some point - clearly illustrating what
happens under entirely different circumstances. :-)


Cecil imagines a circulator is a device that separates forward and
reflected waves of average power, because that is what Cecil's theory
says they must do.

In other words, Cecil's circulator is a device that takes a perfectly
straight-forward argument... and makes it circular.


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

Ian White GM3SEK wrote:
In other words, Cecil's circulator is a device that takes a perfectly
straight-forward argument... and makes it circular.

Cecil's circulator is a device that gets hot when reflected
waves are present, proving that reflected waves possess
energy and power.

Cecil's TDR allows everything about a reflected pulse to
be known and measured.

Cecil's TV ghosting allows one to see the reflections with
one's own eyes.

Cecil's one foot long piece of 200 ohm feedline allows one
to observe the elimination of reflections thus proving a
200 ohm Z0-match where the one who offered the example
thought there was a 50 ohm Z0-match.

I'm a professional teacher who uses lots of visual aids.
They work well on open-minded people. But they do often
pi$$ off closed-minded people because they leave nothing
worth arguing about.
--
73, Cecil http://www.qsl.net/w5dxp

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