Analyzing Stub Matching with Reflection Coefficients
 

Jim Kelley wrote:
Cecil Moore wrote:

How do two waves cancel without interacting?

No interaction is required in order for fields to cancel. Only that
they occupy the same space at the same time and be of the correct
amplitude and phase.

Interaction is certainly required for them to cancel forever.
Otherwise, we have all these energyless phantom ghost waves
existing forever.
--
73, Cecil http://www.w5dxp.com

Analyzing Stub Matching with Reflection Coefficients
 

On Apr 19, 2:59 pm, Richard Clark wrote:
On Thu, 19 Apr 2007 14:11:21 -0700, Jim Kelley
wrote:

Cecil Moore wrote:

How do two waves cancel without interacting?

No interaction is required in order for fields to cancel. Only that
they occupy the same space at the same time and be of the correct
amplitude and phase.

And are put into a load.

As waves are completely independant, then they never interact. A load
is required to reveal the cancellation. No load, and any issue of
cancelling fields is strictly limited to what goes on between the
ears.

73's
Richard Clark, KB7QHC


If nobody is there to hear it, does a tree that falls in the middle of
the forest (or anywhere else) make a noise? If there's a field and
nobody is there to measure it, is it there? If there's a summation to
zero and there's nobody there to verify that the summation really
occured, did it occur? Wheeee! Where will this take us next? Beaker
is soooo excited by the possibilities.

From the labs,
Bunsen

Analyzing Stub Matching with Reflection Coefficients
 

On Apr 19, 4:47 pm, Cecil Moore wrote:
Keith Dysart wrote:
Are those travelling waves really crossing the
voltage maxima? Or are they being reflected?

There is no impedance discontinuity - therefore no
reflections.

But there is no discontinuity at the source, ...

You are still making the same mistake as you have from
the beginning. There is a discontinuity at the source.
In your example, it is called the load-line. The load-
line in the previous example is source voltage divided
by zero, i.e. infinity. That's what the reflected wave
sees.

But the load line is the same for the generator equipped
with a circulator so it too must reflect the wave. If this is
the case, what heats the circulator load resistor?

I notice you skipped the tougher questions related to a
transmission line that is 3/8 wavelength long? Is there a
zero to divide by in this case? Apply analysis equally
to Experiment A and B.

....Keith

Analyzing Stub Matching with Reflection Coefficients
 

Dr. Honeydew wrote:
you were not disallowing the fact that the source, which is matched to
the line, sucks up the entire reverse wave from the line. I'm happy
to see we agree on that little point after all.

Old diversionary trick - doesn't work. We do NOT agree on "that
little point". Agreeing on that little point requires negative
energy which doesn't exist. The reflected energy is 100% re-
reflected. Nothing is coming out of the source and nothing
is going into the source.

But that requires complete reflection of the return
wave at the interface between the source and the line.

Yes, reality requires that. However, wet dreams do not
so be my guest.

I repeat: [In the case of a line connected to a source, with the
impedance of the two matched--not conjugate] It's obvious that the
source is sourcing the forward voltage wave, and it's sucking up the
entire reverse voltage wave from the line.

All with the voltage waves being completely devoid of energy.
Will wet dreams never cease?
--
73, Cecil http://www.w5dxp.com

Analyzing Stub Matching with Reflection Coefficients
 

Richard Clark wrote:
A load is required to reveal the cancellation.

What's the load for b1 = s11(a1) + s12(a2) = 0 ?
--
73, Cecil http://www.w5dxp.com

Analyzing Stub Matching with Reflection Coefficients
 

Keith Dysart wrote:
But the load line is the same for the generator equipped
with a circulator so it too must reflect the wave.

ABSOLUTELY NOT!!! The load line for a circulator equipped
generator is *ALWAYS* 50 ohms (for a 50 ohm generator).
The generator *always* sees 50 ohms as a load. How could
the load-line possibly be anything else besides 50 ohms?
--
73, Cecil http://www.w5dxp.com

Analyzing Stub Matching with Reflection Coefficients
 

On 19 Apr 2007 15:25:41 -0700, "Dr. Honeydew"
wrote:

If there's a summation to
zero

Hi Tom,

And how are you going to sum it without a load?

and there's nobody there to verify that the summation really
occured, did it occur?

Nobody, I presume to mean no body of a load. No load hence no
combination, hence no interference.

Fairly simple stuff, but mystics here still demand that energy
disappears in these situations to be "shuffled" around like balance
ledgers done by Arthur Andersen for billing electricity to California.
Some things never change.

If you (a body) happen to be there, then yes you "might" detect a
null, a peak, or any combination in between. You (the body) are the
load. Even that is a stretch of the imagination, but your
Doppelganger can no doubt spread it out. I suppose if the fields were
viciously strong you might feel a prickle of heat (or cool, some RF
waves lend that perception). Otherwise the waves simply go on their
way without any notice of the other, and certainly to no literal
notice at all. Why would it be otherwise?

Wheeee! Where will this take us next?
Beaker
is soooo excited by the possibilities.

Really? Don't hold back now on our account. :-)

73's
Richard Clark, KB7QHC

Analyzing Stub Matching with Reflection Coefficients
 

On Apr 19, 12:02 am, Cecil Moore wrote:
Keith Dysart wrote:
I have stated that there is no re-reflection of the reflected wave at
the
source. Since the source is matched to the line, the reflection
coefficient is 0 and the wave just .... Well it must go into the
source
since tau is one. But at least it is not reflected when rho is zero.

But you are missing the point. You say the source is matched
to the line but the source is obviously re-reflecting 100% of the
reflected energy.

But the same can be said for experiment A, can't it? Measured
conditions on the line are identical.

Your special magic source is doing exactly
the opposite of what you claim it is doing. The calculated
physical reflection coefficient may be 0 but the virtual
reflection coefficient, SQRT(Pref/Pfor), is 1.0.

Please apply this calculation to Experiment A and check
your result. No difference. Oooopppps.

This is
the point I have been making ever since you started posting.

As you observe for Experiment B, the current is zero so as you
say "The source is not only not sourcing any forward power, it is
also not sinking any reflected power."

Of course the current is also zero at the same point for
Experiment A, so there as well, the source is not only not sourcing
any forward power, it is also not sinking any reflected power.

That's again where you are wrong. In Experiment A, the circulator
load resistor is sinking 100W, i.e. 100% of the reflected power.
A bit of modulation will show that the power being sunk by the
circulator load resistor has made a round trip to the end of
the transmission line short and back.

And if you did the same test with Experiment B you would get the
same result.

A bit more analysis for Experiment A yields some more questions.
Terminate the line with a 50 Ohm resistor. The source is now
providing power to the line, there is no reflection on the line and
the circulator dissipates nothing.
Remove the resistor. The reflection returns. The circulator once
again dissipates 100 W. But as you said, in this condition,
"The source is not only not sourcing any forward power, it is also
not sinking any reflected power." So where did that 100 W being
dissipated in the circulator come from?

In Experiment A, the source is sourcing 100 watts and the
circulator load resistor is sinking 100 watts after the
round trip delay to the end of the line and back. If the
source signal is modulated, the delay between the source
signal and the dissipated signal is obvious and can be
measured.

No different for B.

I suggest a further extension to both Experiment A and
Experiment B. Replace the 1/4 WL stub with a 1 and 1/4 WL
stub. Now, at each 1/4 WL along the line coming back from
the load, no energy is flowing because either the current is
0 or the voltage is 0. So this absence of energy flow happens
not just at the source but repeatedly along the line. This
makes it difficult to accomodate the thought that the
forward or reflected travelling waves are transporting energy
along the line (at least at the quarter wave points).

The "absence of energy flow" is an illusion. There is 100
joules/sec in the forward wave and 100 joules/sec in the
reflected wave. Since the waves are flowing in opposite
directions, you can argue that there is no *net* energy
flow, but the component wave energy flow is alive and well.

Well this is the point you lock up on and it is one of the root
causes of all subsequent errors. I suspect your refusal to
accept the common knowledge that the output impedance
of the generator can be well known, that this controls the
amount of reflection at the generator and that superposition
works is the realization that this will conflict with "energy in
the waves". I encountered the same dilemma when first
dealing with these questions, but came to the answer that
works.

Now back to the quibble. You said: "The source sources 100 watts
and the circulator resistor dissipates 100 watts which is all of the
reflected power."

Yes, in Experiment A but obviously not in Experiment B.
Your source has failed to perform the way you said it would.
As I said in the beginning, there will be re-reflections from
your source. In this case, there is 100% re-reflection.
Real world conditions are not as simple-minded as you say.

But, of course, your proof of re-reflections was your apriori
knowledge of the interior of the generator. I present you with
two 50 Ohm generators; one constructed with a circulator and
one constructed with a 50 Ohm resistor. How do you tell which
is which? The line conditions will be the same, so you can't.
Do you really want a theory of reflections that is dependent on
knowing the internals of the generator?

It would be more precise to say "The source sources 100 watts
and the circulator resistor dissipates 100 watts which is numerically
equal to the reflected power." I contend that it is this "numerical
equality" that has led many astray into believing that the
circulator is dissipating the "reflected power".

No, modulation on the reflected wave proves that it has made
a round trip to the end of the line and back. There is no
getting around that fact. There is also no getting around
the fact that the energy content of the stub is identical
in both experiments. The number of joules in the stub, in
both cases, is exactly the magnitude needed to support the
100W forward wave and the 100W reflected wave. The energy
in the stub in Experiment A is obviously real. The energy
in the stub in Experiment B is identical to Experiment A.

Yes indeed. All line conditions are exactly the same. The same
energy is stored. The output impedance is the same. Modulation
has exactly the same effects. The reflected modulated signal is not
re-reflected at the source for either case. They are
indistinguishable unless you dissassemble the generator.

But as we have seen,
no energy crosses the 0 current node into the generator so the
"reflected power" can not make it to the circulator (or the source
resistance, if the generator happens to have one).

Your "no energy crosses the 0 current node" is just an ignorant
illusion.

You can not possible be arguing that P is not equal to V times I,
can you?

And are you disputing that if V or I is at all times 0, there must be
no energy flowing.
If so, back to grade 11 science please.

The forward current and reflected current are alive
and well and simply superpose to a net current of zero at that
point. We are discussing EM wave energy and a boundary condition
for EM waves to exist is that they must travel at c(VF). If they
don't, they are no longer EM waves.

At a current node, forward current equals 1.414 amps at 0 deg.
Reflected current equals 1.414 amps at 180 deg. Of course, the
*net* current is zero but there is no physical impedance discontinuity
to cause any change in the forward and reflected waves at that
point.

Well it is pretty clear to me that if the net current is always 0,
then
no current is ever flowing and the power must be zero.

It seems to be a bit of a common fallacy to assign meaning to the
intermediate currents computed for the superposition solution to
a problem. Connect two equivalent batteries in parallel. The only
sensible answer is that no current is flowing since the voltages
are the same. (In another post you seem to accept this), though
if you use superposition to solve the problem, you get a very large
current flowing in one direction and an equally large current flowing
in the other. Only the resultant current is in any sense real. Same
for those waves on a transmission line. Just superposition. Just a
convenience to help reach the final solution. Don't over extend
and assign them meaning (or energy). In the end it will cause
deep conceptual difficulties.

....Keith

Analyzing Stub Matching with Reflection Coefficients
 

Keith Dysart wrote:
Cecil Moore wrote:
But you are missing the point. You say the source is matched
to the line but the source is obviously re-reflecting 100% of the
reflected energy.

But the same can be said for experiment A, can't it? Measured
conditions on the line are identical.

But conditions in the sources are exactly opposite. One
is sourcing and sinking power. The other is completely
powerless.

Please apply this calculation to Experiment A and check
your result. No difference. Oooopppps.

Oooopppps means you obviously made a mistake. The conditions
are opposite not alike.

And if you did the same test with Experiment B you would get the
same result.

Obviously not since source B is at room temperature.

No different for B.

Source A is dissipating 200 watts. Source B is dissipating
zero watts. How is that not different?

Well this is the point you lock up on and it is one of the root
causes of all subsequent errors. I suspect your refusal to
accept the common knowledge that the output impedance
of the generator can be well known, that this controls the
amount of reflection at the generator ...

The example proves this to be a wrong-headed concept.

I present you with
two 50 Ohm generators; one constructed with a circulator and
one constructed with a 50 Ohm resistor. How do you tell which
is which?

The one with the rapidly rising temperature is the one with
the circulator. The one that remains at room temperature
is your ten cent resistor version.

Yes indeed. All line conditions are exactly the same.

But the source conditions are radically different. One
source is sourcing 100 watts and sinking 100 watts. The
other source is sourcing 0 watts and sinking 0 watts.

You can not possible be arguing that P is not equal to V times I,
can you?

Pfor - Pref = 0 where those Ps are Poynting vectors.
Pfor is not zero. Pref is not zero. Pnet is zero
only because of a directional convention.

And are you disputing that if V or I is at all times 0, there must be
no energy flowing.

Make that at all times AND AT ALL PLACES and I will agree.
If V is zero it only means that all the energy has migrated
into the magnetic field and indeed, when V is zero, I is
at a maximum.

Well it is pretty clear to me that if the net current is always 0,
then no current is ever flowing and the power must be zero.

True if the net current everywhere is zero. Not true if the
current is only zero at a point. Current on each side of
a zero current point is prima facie evidence of energy flow
in both directions.
--
73, Cecil http://www.w5dxp.com

Analyzing Stub Matching with Reflection Coefficients
 

Richard Clark wrote:

As waves are completely independant, then they never interact.

Perhaps we can agree that they could at least share a common media
through which to propagate.

A load
is required to reveal the cancellation.

A load - like a transducer of some sort. That would indeed be
required in order to convert the effect to a form that we can more
readily observe.

No load, and any issue of
cancelling fields is strictly limited to what goes on between the
ears.

I hope you're not suggesting that if we cannot observe the
interference effect, it does not occur. There is good reason to
believe that just the opposite is true, Richard. It's a real paradox.
;-)

73, Jim AC6XG