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Old April 20th 07, 01:57 AM posted to rec.radio.amateur.antenna
Keith Dysart Keith Dysart is offline
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First recorded activity by RadioBanter: Mar 2007
Posts: 124
Default Analyzing Stub Matching with Reflection Coefficients

On Apr 19, 7:20 pm, Cecil Moore wrote:
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.


Of course we can turn that around by building the generator for
Experiment B in the Norton style instead of the Thevenin style.
Then it will dissipate 4 times the power it would when terminated
with 50 Ohms.

Or how about the variant that will dissipate exactly the same
as the circulator without having one. Its dissipation always
numerically rises by the same amount as the "reflected power".
You should work on this generator since it would really help
convince people that the reflected power is dissipated in the
generator.
(Hint: There are two ways to build it and you need a current
source, a voltage source and two resistors.)

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.


In your characteristic style, you have removed the description of
what is being compared so the reader can not refer back and notice
that you are just contradicting. Oh well.

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.


Minor changes to the generator for B can give you a very hot one
or one that is exactly the same temperature as the one with
the circulator. Your simple contradiction is a an exceedingly weak
argument.

No different for B.


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


The conditions on the lines are indistinguishable and yet you
claim one is reflecting and one is not. How did the line know
whether it should reflect or not? Or is the wave that knows?

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.


You really should crack a basic text book, although you are
in kind of deep now to back track so that path does carry
high risk. You could also google or try the simulation. All
will support the statement I made above.

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.


But if we use a Norton style, which one gets hotter?
Or the third style (have you worked out how to build it yet?),
which will dissipate exactly the same as the circulator.

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.


I won't repeat myself.

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.


Actually Pnet is zero because of basic circuit theory and the
universally accepted understanding that P = VI.

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.


In circuits, we measure the power at a particular place.
I certainly want my power company to do this. If the voltage
or current is 0 at a particular place in the circuit then no
energy is flowing at that place in the circuit. If you are disputing
this, I contend that you do not accept that P = VI.

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.


Inventive. But it doesn't fly. P = VI or it doesn't.

In fact what happens is the energy flows towards the voltage
and current nulls from both sides at the same time, then it flows
away. But it never crosses. Of course, the above statement
applies to sinusoidal excitation. The story is a bit different for
pulse and step. If the line is excited by a step function, for
example, after the reflection makes it back the generator, the
current is zero everywhere, though the 'Bird Watts' are still
'flowing' (but it does take a DC coupled directional wattmeter
to observe this).

....Keith