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Roy Lewallen wrote:
And I can show exactly
where every erg of energy is at every instant in any component and at
every point along the transmission line. I submit that any analysis
technique which can't do this without knowing what happens to energy
entering the source is inferior.

I nominate this for "Quote of the year", by acclamation.
Anyone who responds to the question: Where does the energy
go? - with "We do not know and we do not care." does not
deserve to be in the discussion.
--
73, Cecil http://www.w5dxp.com

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On Fri, 18 Apr 2008 10:57:24 -0500
Cecil Moore wrote:

Roger Sparks wrote:
This is much too arbitrary for me.

It's not arbitrary at all - it's the result of very
carefully chosen boundary conditions including specifying
a 100% 50 ohm system with ideal components. It agrees with
Roy's posting about ideal source impedances and short-circuits.

It is arbitrary because when using the short circuit model, you limit the discussion to only one example of transmission line termination--the example of a low impedance on one end of the line and a higher impedance on the other. As you well know, two additional examples of transmission line termination are possible--the transmission line impedance is higher (or lower) than the termination at either end.

It is also arbitrary because the reflected wave information is useless. Any reflection from the source is folded into the sine wave to form one wave, which becomes the forward wave. The only real accomplishment from the exercise to to shift the frequency for a one cycle "bump", and that accomplishment does more to introduce confusion than contribute to better understanding.

--
73, Roger, W7WKB

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On Apr 18, 10:48*am, Cecil Moore wrote:
Keith Dysart wrote:
So you are saying that ideal voltage sources work differently
in distributed networks than they do in lumped circuits.

"Work differently" is a loaded expression. Since both
lumped circuit and distributed network models exist,
it is safe to say that the lumped circuit model and
the distributed network model indeed "work differently".
If they didn't "work differently", there would be no
need for both of them to exist.

Your previous error is obvious. You were using the
lumped circuit model on the left side of Rs and
using the distributed network model on the right
side of Rs. If it is necessary to use the distributed
network model for part of the network, then it is
absolutely necessary to be consistent for all of
the network.

When you switched to the lumped circuit model on
the right side of Rs, the energies balanced. When
you switch to the distributed network model on
the left of Rs, the energies will also balance.

The lumped circuit model is a subset of the distributed
network model. See:http://www.ttr.com/corum/andhttp://w...1-MASTER-1.pdf

Here's some quotes: "Lumped circuit theory fails because
it's a *theory* whose presuppositions are inadequate.
Every EE in the world was warned of this in their first
sophomore circuits course. ... Lumped circuit theory
isn't absolute truth, it's only an analytical *theory*.
... Distributed theory encompasses lumped circuits and
always applies."

It is a good thing I checked the original references, otherwise
I would have had to assign Corum and Corum immediately to the
flake bucket where they could join some of the other contendors
on this group. But no, it turns out they have the appropriate
qualifications on all their statements about when it is
appropriate to use a lumped analysis and when it is not. And
when is lumped okay, when the physical dimensions of the
elements can be measured in small fractions of a wavelength.
Nothing new there, most of us know that.

Let me remind you of the circuit at hand:

50 ohms
+----------\/\/\/\/-----------+
+| +|
Vsl=100 VDC Vsr=50 VDC
| |
+-----------------------------+

Firstly, there are no inductors to cause any sort of difficulty.
Secondly, it is constructed of ideal components, which have the
luxury of being infinitely small.
And thirdly, it is a DC circuit so the two points above do not
matter any way.

So we are back to the question you keep dodging...

Where does the energy being absorbed by these ideal voltage
sources go?

0+j0 ohms cannot absorb energy.

Now that is a non-sequitor. The element absorbing energy is an
ideal voltage source, not a resistor.

As Eugene Hecht said:
"If the quantity to be measured is the net energy per unit
area received, it depends on 'T' and is therefore of limited
utility. If however, the 'T' is now divided out, a highly
practical quantity results, one that corresponds to the
average energy per unit area per unit time, namely 'I'."
'I' is the irradiance (*AVERAGE* power density).

It is a DC circuit. This means the instantaneous value is
the average value.

(But poor Hecht, here he is saying power is more useful than
cumulative energy, and you misinterpret him to be comparing
instantaneous to average. And is it the distribution over
the area that is being averaged, or the distribution over
time?)

You seem to have discovered that "limited utility".
You have an instantaneous power calculated over an
infinitesimally small amount of time being dissipated
or stored in an impedance of 0+j0. Compared to that
assertion, the Virgin Birth seems pretty tame. :-)

Again, it is a DC circuit since we have had to go back
to learning the fundamentals of ideal voltage sources.

But back to the simple question...

Where does the energy that is flowing into the ideal
DC voltage source on the right go?

If no energy is flowing into the ideal voltage source
on the right, where is the energy that is being
continuously provided by the ideal voltage source on
left going? Of the 100 W being provided by the ideal
source on the left, only 50 W is being dissipated in
the resistor. Where goes the remaining 50 W, if not
into the ideal voltage source on the right?

...Keith

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On Apr 18, 4:23*pm, Cecil Moore wrote:
Roy Lewallen wrote:
Anyone interested in learning more about this and its application can
look up "dynamic resistance" on the web or in an appropriate text.

Another name for "dynamic resistance" is "virtual resistance".
It is an *EFFECT* of superposition, not a cause of anything.

Are you suggesting that the negative "dynamic resistance" of a
tunnel diode is an *EFFECT* of superposition?

Please feel free to expand on this claim.

...Keith

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Roger Sparks wrote:
It is arbitrary because when using the short circuit
model, you limit the discussion to only one example of
transmission line termination ...

It is not arbitrary - it is what it is. All of the
source impedance, Rs, is separated from Vs. That
leaves only a 0+j0 dead short impedance possible
for the series source. If it was some other value,
the distributed network model would still work.

Any reflection from the source is folded into the sine
wave to form one wave, which becomes the forward wave.

Yes, in exact accordance with the distributed network
model. What it means is that the source is not delivering
the folded-in reflected energy at the time the reflected
energy joins the source signal. Since I don't see that
energy in Keith's equations, chances are that is why
they didn't balance.
--
73, Cecil http://www.w5dxp.com

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Keith Dysart wrote:
On Apr 18, 4:23 pm, Cecil Moore wrote:
Another name for "dynamic resistance" is "virtual resistance".
It is an *EFFECT* of superposition, not a cause of anything.

Are you suggesting that the negative "dynamic resistance" of a
tunnel diode is an *EFFECT* of superposition?

Sorry, I was mistaken about the definition of "dynamic"
in the context of physical components.
--
73, Cecil http://www.w5dxp.com

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On Sat, 19 Apr 2008 13:02:08 GMT
Cecil Moore wrote:

Roger Sparks wrote:

Any reflection from the source is folded into the sine
wave to form one wave, which becomes the forward wave.

Yes, in exact accordance with the distributed network
model. What it means is that the source is not delivering
the folded-in reflected energy at the time the reflected
energy joins the source signal. Since I don't see that
energy in Keith's equations, chances are that is why
they didn't balance.
--

You are missing an important observation here. When the energy from the reflected wave folds into the forward wave, any further analysis is a replication of the first analysis. Nothing new is learned.

I think what you want to see is a source matched to the load, so that all the energy flows one way, to the resistor Rs. Your proposal is to allow/force reflectons from the source Vs so that effectively, the resistor Rs becomes the only load. This is then a demonstration/proof for what?

--
73, Roger, W7WKB

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Roger Sparks wrote:
You are missing an important observation here. When the
energy from the reflected wave folds into the forward wave,
any further analysis is a replication of the first analysis.
Nothing new is learned.

I'm posting steady-state values. If it is already
steady-state, nothing new is needed. The point is
that some of the steady-state forward energy is
not being delivered by the source.

Your proposal is to allow/force reflectons from the source
Vs so that effectively, the resistor Rs becomes the only load.

No, the resistor Rs is not the only load. The resistor
plus the rest of the network is the load. With a 45 deg
shorted stub the load is 50+j50 as you previously
reported.
--
73, Cecil http://www.w5dxp.com

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Keith Dysart wrote:
So we are back to the question you keep dodging...

I'm not dodging anything. I am ignoring irrelevant
examples. The context under discussion is configurations
of single source systems with reflections where the
average interference is zero. So please explain how
your example applies to what we are discussing. Where
are the reflections? Where is the average interference
equal to zero?
--
73, Cecil http://www.w5dxp.com

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On Apr 19, 11:23*pm, Cecil Moore wrote:
Keith Dysart wrote:
So we are back to the question you keep dodging...

I'm not dodging anything. I am ignoring irrelevant
examples. The context under discussion is configurations
of single source systems with reflections where the
average interference is zero. So please explain how
your example applies to what we are discussing. Where
are the reflections? Where is the average interference
equal to zero?

Your explanations are predicated on a misunderstanding of
the behaviour of ideal voltage sources. Despite your
protests to the contrary, ideal voltage sources can, and
do, absorb energy.

The discussion needs to digress to address this fundamental
misunderstanding since this misunderstanding renders all
else moot. And so...

Let me remind you of the circuit at hand:

50 ohms
+----------\/\/\/\/-----------+
+| +|
Vsl=100 VDC Vsr=50 VDC
| |
+-----------------------------+

And we are back to the question you keep dodging...

Where does the energy that is flowing into the ideal
DC voltage source on the right go?

If no energy is flowing into the ideal voltage source
on the right, where is the energy that is being
continuously provided by the ideal voltage source on
left going? Of the 100 W being provided by the ideal
source on the left, only 50 W is being dissipated in
the resistor. Where goes the remaining 50 W, if not
into the ideal voltage source on the right?

Once it is agreed that ideal DC voltage sources can indeed
absorb energy, we can discuss the same for ideal AC voltage
sources. Once that is agreed, we can return to the
discussion of your 'interference free' circuit.

...Keith