W5DXP wrote:

|rho| is the reflection coefficient when a2=a2

Prophetic!

Yes, Magid explains it thoroughly and well. His analysis includes
traveling waves from the beginning to the end. It's basically the same
analysis I've done myself.

What he *didn't* do was to throw the time-dependent information out the
window at the very beginning of the analysis, as you've done in all your
analyses involving only average power. That allows him to actually show
where the energy is at every instant of time. It looks to me like his
analysis is valid, so if your method leads to a different conclusion, I
feel there's good reason to doubt its validity.

Y'know, if every author seems to have it wrong, considering how many
really, really intelligent and learned people there are out there, and
how many times their work has been carefully reviewed, have you
considered for just a moment the possibility that maybe, just maybe,
they have it right and you have it wrong? But if you're absolutely
convinced you've discovered something that all those other folks are
mistaken about, why not write your own textbook? Or at least a paper or
two for the professional publications. The world will thank you for
straightening them out.

Roy Lewallen, W7EL

W5DXP wrote:
Roy Lewallen wrote:

You've now heard of Magid, ...


Unfortunately, every author and guru that I have ever encountered,
at some point, confuses cause and effect. I'm sorry I can't get over
to the Texas A&M library right now but Magid seems to believe that
standing waves can be sustained without a forward wave and a reflected
wave. Does he explain how that is possible? In all honesty, it is an easy
mistake to make. Even Ramo, Whinnery, and Van Duzer make the same
mistake. Not exactly a quote but: The reflection coefficient is caused
by the ratio of the reflected power to the forward power. Therefore,
the ratio of the reflected power to the forward power is caused by
the reflection coefficient. "Logic" like this seems to abound in
the field of transmission lines.

(Dr. Slick) wrote in message . com...
(Tom Bruhns) wrote in message om...

As others have noted, the magnitude of the reflection coefficient can
be greater than unity with a passive line and load. Don't try to read
too much physical significance into that, however.


That's impossible, unless you have a free source of energy, in
which case you should send me the schematic!

Do you believe everything people tell you? How did they set this
up? I don't think i will get an answer to this.

Excuse me? As a matter of fact, the way I discovered that you can
have a reflection coefficient greater than unity was to set up the
boundary conditions for a real transmission line feeding a moderately
high Q inductor and discover that |rho|1. Up to that point, nobody
had told me it was possible. At the time, it was a somewhat
surprising result to me. Try it yourself. On a line, Vf/If = -Vr/Ir
= Zo. At the end of the line connected to an load whose impedance is
Zl, the current is V/Zl, but V = Vf+Vr, and the current there must
also be If+Ir. You should be able to reduce those to the well-known
equation for reflection coefficient, and if you plug in numbers such
as Zo=50-j5 and Zl=1+j100, you'll see that |Vr/Vf|1. It only takes
three or four lines of simple linear algebra and perhaps plugging the
complex numbers into a calculator to find the result for that example.
Again, TRY it for yourself. DO NOT just believe what I tell you. If
you have trouble with the algebra, I'll be happy to help you out.

Cheers,
Tom

Thanks for the analogy.

One can mathematically and conceptually conceive two opposite-traveling
waves that add up to the observed standing wave, and that's fine. The
problem comes with assigning power or energy to the waves. Then you run
into the problem of how one wave got the energy over the barrier into
the pocket and the other wave took the same amount back out, without
transfering any air molecules across the barrier in the process.

The average power analysis looks to me something like this. Suppose you
have two batteries each with exactly 2 volts potential and zero internal
resistance, with a 2 ohm resistor connected between their positive
terminals. The negative terminals are connected together. You replace
the battery on the left with a short (turning it off), and observe that
the current through the resistor is one amp to the left. Then you hook
the left hand battery back up and turn the right hand battery "off" by
replacing it with a short. You observe that there's one amp now flowing
through the resistor to the right. Finally, turn both batteries "on" by
putting them both in place. You can use superposition to conclude,
correctly, that there's zero current flowing in the resistor.

But it's silly to insist that there's a forward two watt "power wave"
flowing to the right, and another two watt wave flowing to the left. You
subtract one from the other and, sure enough, get zero. But are the
"power waves" real? Studying and analyzing these imaginary waves is
surely a lot more interesting than simply looking at the circuit and
noting that the "boring" (as Cecil calls it) net power is zero. But
aren't you studying ghosts?

Even more risky is adding the things. This time hook two one volt
batteries in series with the 2 ohm resistor and energize one at a time.
With the upper one on and the lower one "off" (replaced with a short)
you get 1/2 amp. You've got a "power wave" of I^2 * R = 1/2 watt. Turn
the lower one on and the upper one "off", and you get another "power
wave" of 1/2 watt, in the same direction. Turn them both on, and you
have a power flow of, um, 2 watts. Welcome to the new math.

Roy Lewallen, W7EL

Jack Smith wrote:

Roy:

Interesting point and I don't recall reading or hearing it elsewhere.

The following is dashed off without fully thinking it through, so no
warranty on its accuracy.

If you think of a sound wave (longitudinal transmission, of course) in
a lossless acoustic transmission line terminated with a short, the
individual air molecules within each 1/4 wave section are likewise
trapped since at the 1/4 wave points there is zero sound pressure.
This may be a useful analogy for the electromagnetic transverse
propagating T-line.

Jack K8ZOA

Roy Lewallen wrote:
Y'know, if every author seems to have it wrong, considering how many
really, really intelligent and learned people there are out there, and
how many times their work has been carefully reviewed, have you
considered for just a moment the possibility that maybe, just maybe,
they have it right and you have it wrong?

"It" is not always the same thing. What I am saying is that nobody's
perfect, nobody knows everything, and 1000 years from now, most of what
you and I believe today will be obsolete.

For instance, Ramo & Whinnery use the square root of the ratio of reflected
power to forward power to define the reflection coefficient. Shortly
thereafter, they use the reflection coefficient as the cause of that
power ratio. So in essence, they are saying that the reflection
coefficient causes itself.

Another example from this newsgroup. After standing waves have been caused
by the superposition of forward waves and reflected waves, they somehow
become self-sustaining without the forward waves and reflected waves.
--
73, Cecil http://www.qsl.net/w5dxp



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Roy Lewallen wrote:
One can mathematically and conceptually conceive two opposite-traveling
waves that add up to the observed standing wave, and that's fine. The
problem comes with assigning power or energy to the waves.

Proposing waves that exist without energy seems more problematical
to me. And nobody has come up with an explanation of how standing
waves can exist without forward waves and reflected waves in a
single source, single feedline, single load system. Seems to me,
that is an absolutely necessary first step in proposing self-
sustaining standing waves.

The average power analysis looks to me something like this. Suppose you
have two batteries ...

In a typical ham installation, there is only one source and all system
energy is supplied by that single source. Please limit your examples
to one source to adhere to reality. Having to change the example
away from reality to multiple sources to make a point is a weakness,
not a strength, in the argument. In a typical ham installation, instant
energy is NOT available since the source is nanoseconds away. Any
energy needed instantaneously must already be there.
--
73, Cecil http://www.qsl.net/w5dxp



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Richard Harrison wrote:

Keith wrote:
"At the quarter wave points where voltage and current are always zero,
there is no energy flowing. Period."

Yes there is energy flowing, and it is flowing in both directions if it
is flowing in one direction. Otherwise there would be no standing wave.

This view that there is energy flowing in both directions at the same
time
leads to some strange conclusions.

The directional wattmeter uses the following expressions to compute its
displayed value:
Vf = (V + IR)/2
Vr = (V - IR)/2
where V and I are the instantaneous voltage and current at the same
point
on the line and then computes power from:
Pf = average(Vf**2/R)
Pr = average(Vr**2/R)
Appropriate scaling (which we'll ignore for simplicity) is needed
depending on the characteristics of the voltage and current sensors.

Let's apply these expressions to some simple examples.

Connect a length of 50 Ohm transmission line to a 9 Volt battery and
wait
for the transient to die:
Vf = (9 + 0)/2 - 4.5 V
Vr = (9 - 0)/2 - 4.5 V
Pf = 0.405 Watt
Pr = 0.405 Watt
Disconnect the battery and the capacitance of the line remains charged
to
9 Volts. Are you really quite comfortable with the notion that this
line,
charged to 9 Volts, with 0 current everywhere has 0.405 Watts flowing
forward canceling at all points the 0.405 Watts flowing in reverse?

Others have stated that lamp cord has an impedance of about 100 Ohms.
Assume this to be correct.
With a table lamp plugged in but the lamp turned off:
Vf = (120 + 0)/2 - 60 V
Vr = (120 - 0)/2 - 60 V
Pf = 36 Watt
Pr = 36 Watt
Again, are you comfortable with this conclusion?

How about phone lines (phone on hook) at 600 Ohms and 48 Volts:
Vf = 24 V
Vr = 24 V
Pf = 0.96 W
Pr = 0.96 W
In a 400 pair cable there is 384 Watts flowing in the forward and
reverse direction. From a hundred thousand line central office we
get 960 Kilowatts forward and reverse, when the current is zero,
everywhere. Comfort level?

With the forward and reverse power view, to completely understand
the behaviour of the circuit, we need to know Z0 or we can not
compute these forward and reverse powers which seem to be
fundamental.

Consider a flashlight with the lamp off: the wire twists and turns
and has a very non-constant Z0. While computationally tedious, it
is possible to determine Pf and Pr at every point along the wire.
Isn't it necessary to do this to obtain a complete understanding
of what is happening in the circuit, if Pf and Pr are flowing?
If not, why not? The necessity for doing this is the logical
conclusion of the Pf and Pr approach.

Presented with a wire charged to high static voltage, what is Pf and
Pr? Presented with a sphere charge to a high static voltage, what
is the Pf and Pr?
Presented with a simple length of wire, how much power is flowing?

Similarly for a capacitor. The plates of a capacitor must have power
reflecting in all directions even when it is just charged to a
constant voltage (according to the Pf and Pr theory).

Given these bizarre results it would seem wise to limit the application
of the watts indicated by a directional 'watt'meter to those things
which are proper and not assume that it necessarily means that there
is real power flowing.

....Keith

How can you get a reflection coefficient greater than one
into a passive network? I'd really like to know.

=====================================

Dear Dr Slick, it's very easy.

Take a real, long telephone line with Zo = 300 - j250 ohms at 1000 Hz.

Load it with a real resistor of 10 ohms in series with a real inductance of
40 millihenrys.

The inductance has a reactance of 250 ohms at 1000 Hz.

If you agree with the following formula,

Magnitude of Reflection Coefficient of the load, ZL, relative to line
impedance

= ( ZL - Zo ) / ( ZL + Zo ) = 1.865 which exceeds unity,

and has an angle of -59.9 degrees.

The resulting standing waves may also be calculated.

Are you happy now ?
---
Reg, G4FGQ

Keith wrote:
"Connect a length of 50 Ohm transmission line to a 9 Volt battery and
wait for the transient to die."

The transient is the only part of your proposal for which the 50-ohm
surge impedance has meaning. That`s why it`s called the "surge"
impedance.

The 50 ohms is Zo and equals the sq.rt. of Z/Y, or in a perfect line
this is sq. rt. of L/C.

L and C require an alternatng current (changing, not static) to produce
reactance, so the d-c calculations are meaningless to Zo.

Best regards, Richard Hsarrison, KB5WZI

W5DXP wrote in message ...

[s11]=rho, rho being just the magnitude of the s11.

I just told you that is not always true.

s11 is the reflection coefficient when a2=0

|rho| is the reflection coefficient when a2=a2

In a Z0-matched system with reflections, they are NOT equal.


You're right, but we are talking about a one-port network, the
antenna and transmission line.


Slick