On Apr 16, 11:56*pm, Cecil Moore wrote:
Keith Dysart wrote:
Q1. Where does this energy come from?
Q2. Where does this energy go?

Answer to both. We do not know and we do not care.

Well, that certainly puts an end to this
discussion.

An intriguing response.

It was such a good example that it just stopped you in
your tracks.

But was this because you have learned that you were in error
and now better understand the behaviour of sources when
they are absorbing energy?

Or was it because you have realized the contradictions
inherent in your previous explanations but can not
work out how to resolve them?

Or was it because you have detected just a hint of the
contradictions to come and want to give up before having
to realize their full impact?

...Keith

Keith Dysart wrote:
It was such a good example that it just stopped you in
your tracks.

Not at all. Everything I have said applies to a
distributed network. Since you insist on using a
lumped circuit source in a distributed network
example, further discussion is futile because
you are mixing apples and oranges.

What you perceive as the source absorbing energy is
the reverse traveling wave energy flowing back through
the source unimpeded. After all, how much energy can
be absorbed by 0+j0 ohms? The illusion of energy
absorption is the direct result of you refusing to
deal with the component wave energies.
--
73, Cecil http://www.w5dxp.com

On Apr 17, 7:35*am, Cecil Moore wrote:
Keith Dysart wrote:
It was such a good example that it just stopped you in
your tracks.

Not at all. Everything I have said applies to a
distributed network. Since you insist on using a
lumped circuit source in a distributed network
example, further discussion is futile because
you are mixing apples and oranges.

So you are saying that ideal voltage sources work differently
in distributed networks than they do in lumped circuits.

Perhaps you could expand on the differences using the
following two circuits. This one is lumped.

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

And this one is similar but includes a transmission line.

50 ohms
+----------\/\/\/\/----+----------------+-------+
+| arbitrary +|
Vsl=100 VDC length, any Vsr=50 VDC
| impedance line |
+----------------------+----------------+-------+

After the circuit has settled how will the ideal voltage
sources on the right be behaving differently in the two
examples? If you think it matters, try 50 ohm line.

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

What you perceive as the source absorbing energy is
the reverse traveling wave energy flowing back through
the source unimpeded. After all, how much energy can
be absorbed by 0+j0 ohms?

None. But pushing a current against a voltage (regardless of
the cause of the voltage) will definitely use energy.

The illusion of energy
absorption is the direct result of you refusing to
deal with the component wave energies.

These circuits are DC, but Vf and Vr still work. The line settles
to a constant (over length) 50 V regardless of the line impedance,
but the value of Vf and Vr will depend on the impedance of the line.
Why does it settle to 50 V? What else can it do? It is connected
to a 50 VDC ideal voltage source.

Using a line impedance of 50 ohms, compute Vr. Ooopppps. Vr is
equal to 0: no reflected power.

The ideal voltage source on the right, after the circuits settle,
will be absorbing 50 joules/s in both cases.

...Keith

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/ and
http://www.ttr.com/TELSIKS2001-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."

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

0+j0 ohms cannot absorb energy. 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).

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. :-)
--
73, Cecil http://www.w5dxp.com

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

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

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