W5DXP wrote:

http://www.gmi.edu/~drussell/Demos/s....html#standing

Another web page which correctly uses superposition only for amplitude;
not power.

In an ideal line terminated by Zo, ...

That configuration is not covered by my statement above which
applies only to standing waves on lossless unterminated lines.

True. I had moved on to a different configuration, the one originally
being discussed.

....Keith

W5DXP wrote:
The forward wave is carrying Vfwd^2*Z0 watts associated with the
forward Poynting vector and the reflected wave is carrying Vref^2*Z0
watts associated with the reflected Poynting vector.

These, of course, should be Vfwd^2/Z0 and Vref^2/Z0.
--
73, Cecil http://www.qsl.net/w5dxp



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wrote:

W5DXP wrote:

http://www.gmi.edu/~drussell/Demos/s....html#standing

Another web page which correctly uses superposition only for amplitude;
not power.

Nobody is using superposition for power. Do you deny the fact that
two 100W light bulbs put out more irradiance (power) than one?

You argue that there must be reflections at a voltage node. The
above web page indicates such doesn't exist. Those forward and
reflected waves flow smoothly right through each other. Take
another look.
--
73, Cecil http://www.qsl.net/w5dxp



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W5DXP wrote:

wrote:
We have a choice of two rho for this situation:

Correction: We have a choice of two reflection coefficients
each with its own unique definition.

black box - 0, computed from the surge impedance of the line and
the steady state impedance of the load

Actually, Sqrt(Pref/Pfwd), the definition of rho.

This definition seems incomplete. There is a choice of sign. How
do you pick?

open box - 0.5, computed from the surge impedance of the line and
the surge impedance of the load

Actually, (150-50)/(150+50), the definition of s11.

In any case, what we have in this experiment is a case where there
IS an impedance discontinuity and yet there is no reflection (if
you use the "black box" rho, as is often done).

This is technically not true.

None-the-less, no one seems to have difficulty treating it as if it is.

The NET reflections are zero. There
are two non-zero component reflections as seen from the s-parameter
equation:

b1 = s11*a1 + s12*a2

These three terms are all reflections. b1 is the NET reflections
toward the source. Since b1 = zero, s11*a1 = -s12*a2, i.e. the two
component reflections are of equal magnitude and opposite phase and
therefore cancel. This is explained in the last three paragraphs on
the Melles-Griot web page.

Yes, but only if the box is open. How do you analyze the black box
when you are only permitted to know the impedance at the inteface
looking
towards the load? rho is just computed from the only information
available. No one complains that the problem can't be solved.

So, if we are allowed to say in the first experiment that rho is 0
despite an impedance discontinuity, we are equally allowed to say
for the second that rho is -1 despite the absence of a discontinuity.

There is NOT an absence of a discontinuity.

There is no discontinuity at the interface in question.

There is NO physical discontinuity at the black box.

Exactly, and people argue that it is inappropriate to claim that a
reflection occurs at this interface.

And yet, in the symmetrical case where there IS a discontinuity,
many are quite comfortable talking about the lack of reflections
at the inteface.

So, if you don't wish to permit reflections at interfaces without
an impedance discontinuity, please never speak of an absence
of reflections at an interface with a discontinuity.

If it's good for the goose, it should be good for the gander.

....Keith

Cecil Moore wrote:

wrote:

W5DXP wrote:

http://www.gmi.edu/~drussell/Demos/s....html#standing

Another web page which correctly uses superposition only for amplitude;
not power.

Nobody is using superposition for power.

It may have been my misinterpretation, but I understood you were
claiming that the net power distribution on the line should be
computed by summing Pf and Pr. If you weren't saying this,
then the other way to compute power is to use p(t) = v(t) * i(t) (NET)
which leads to no energy crossing the voltage and current zeroes.

Do you deny the fact that two 100W light bulbs put out more
irradiance (power) than one?

It is my policy never to deny facts. I leave that for others.

You argue that there must be reflections at a voltage node.

No. Merely that if there were, the picture would be no different.

The above web page indicates such doesn't exist.

Visualize the picture if there were reflections at the minima
and maxima. The plots would look exactly the same.
Curiousity -- why are you so sure there aren't? The picture
would be the same!

Those forward and reflected waves flow smoothly right through
each other.

Another way of viewing the picture and as long as you stick to
voltages and currents; no problem.

It is only with the claim that energy flows past a point with
a constant voltage of 0 that I have a problem.

Take another look.

In the simulation they show voltage waves. No problem.

....Keith

wrote:
Visualize the picture if there were reflections at the minima
and maxima. The plots would look exactly the same.
Curiousity -- why are you so sure there aren't? The picture
would be the same!

Because the energy in bright ring interference patterns is NOT
trapped between the dark rings. That you choose to remain ignorant
of that centuries-old fact of physics is not my problem.
--
73, Cecil http://www.qsl.net/w5dxp



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Cecil Moore wrote:

wrote:
Visualize the picture if there were reflections at the minima
and maxima. The plots would look exactly the same.
Curiousity -- why are you so sure there aren't? The picture
would be the same!

Because the energy in bright ring interference patterns is NOT
trapped between the dark rings. That you choose to remain ignorant
of that centuries-old fact of physics is not my problem.

Seems to me there must be power in the light rings and none in
the dark rings. I am not sure where you get this notion of 'trapped',
but if there is no power in the dark rings, then the power from
the light rings is certainly not in the dark rings.

....Keith

Ian White, G3SEK wrote:
If a problem can be solved without knowing what's in the box, you have
found a very powerful solution that will work for *anything* inside there.

Such general solutions don't always exist; but if you make a habit of
opening "black boxes" when it isn't necessary, you never will find a
general solution - and that's guar-an-teed.

The other side of the coin is that a black box can be used to obscure the
truth. It's easy to use a TDR to find out if the apparent steady-state
short at the black box terminals is physical or not. But such an experimental
act is prohibited by the steady-state religion. A large percentage of the
black box examples on this newsgroup are designed to obscure, not enlighten.
If obscuring is the goal, let's perform our black box experiments in total
darkness to make it even more challenging for truth-seekers.
--
73, Cecil http://www.qsl.net/w5dxp



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wrote:

Cecil Moore wrote:
Should have said |rho|. |rho| is often used without its Gamma angle.

It still remains incomplete. The sign is needed to know Vr.

The point was that Pref=0 and therefore, there is no sign to deal
with for rho=Sqrt(Pref/Pfwd). +0 = -0

And, of course, the sign information was lost as soon as Vf and Vr
were converted to power.

In the example case, Pref=0 and therefore rho=0. What difference does
the sign make when rho=0?

Thinking of black boxes is an excellent tool to assist in the search
for knowledge. As a tool, it can help identify which information is
critical to a solution and which is not.

That's true, but refusing to open the black box when it is possible to
do so to gain knowledge is sheer nonsense for truth-seekers. OTOH, it
is a great tool for someone who is seeking to obscure the truth and
promote the steady-state religion at all costs.

As a person searching for knowledge through thought experiments you
decide when to open the black box.

Yes, and what is the person, who then objects to the opening,
searching for?

Do consider adding this thinking tool to your toolbox. It is quite
powerful.

It is in my toolbox but it does not dictate my attitudes toward
reality. If it is finally possible to open the black box in
order to ascertain the truth, I will certainly open it. Before
that, I will subject it to TDR experiments, something you would
no doubt object to since it is not "steady-state". With a TDR,
it is duck soup to prove that an open-circuit 1/4WL stub is
not shorted at the mouth of the stub.

And yet you feel quite free to state that sometimes there is NO
reflection when there IS an impedance discontinuity. All I want
is some consistency.

I said there is no NET reflection at a Z0-match. I provided the
following s-parameter equation to illustrate that the two
reflections at a Z0-match will superpose and cancel each other.
b1 = s11*a1 + s12*a2 Do you understand the s-parameter equations?
If not, HP's AN 95-1 is available for download from the web.

If you wish to claim that there are no NET reflections at some
impedance discontinuities, then, for consistency, we should also
be able to claim that there are NET reflections when there is
no impedance discontinuity.

Goose and gander.

I suggest you study and understand the difference between a logically
inclusive statement and a logically exclusive statement. The difference
lies in the use of the word "some" Vs "all". Above you performed a
logical switcharoo when you went from "some" to an implication of "all".
That is known in logic as an "argumentum ad absurdum".

Sometimes, there are NET reflections at a physical impedance
discontinuity. If a single RF source signal exists, there are
always component reflections at a physical impedance discontinuity.

Sometimes, there are no NET reflections at a physical impedance
discontinuity because the two component reflections cancel. This
is true of Z0-matched systems with reflections.

Sometimes, there are net reflections existing where there is no
impedance discontinuity. They are the result of reflections at
a physical impedance discontinuity somewhere else. Reflected waves
are traveling waves. They travel from their point of origin to other
places in the system.

Sometimes, there are no net reflections where there is no impedance
discontinuity, e.g. a flat system.
--
73, Cecil http://www.qsl.net/w5dxp



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wrote:
Seems to me there must be power in the light rings and none in
the dark rings.

Yes

I am not sure where you get this notion of 'trapped',

:-) Good one :-) You say the power in a transmission line is trapped
between the zero power points and cannot cross the boundary. The
transmission line is virtually identical to the bright and dark
rings in a light interference pattern.

but if there is no power in the dark rings, then the power from
the light rings is certainly not in the dark rings.

Point is, the energy in the bright rings is not trapped there. It
continues to flow in a straight line. The bright rings and dark
rings are the result of interference between two beams of light
traveling in straight lines at the speed of light. The voltage-zero
and voltage-maximum points on a transmission line are the result of
interference between two waves traveling in straight lines at the
speed of light. In the absence of a physical impedance discontinuity,
there is nothing to change their momentum. Their momentum allows them
to "coast" across a voltage-zero point.
--
73, Cecil http://www.qsl.net/w5dxp



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