On Apr 15, 12:47*pm, Cecil Moore wrote:
Keith Dysart wrote:
Is there some reason why you think that it is only
necessary to account for the energy removed from the
circuit by heating?

Heating is the only way that the energy can be removed
from the ideal closed system.

And that you can ignore energy being removed in other ways?

No energy is removed in other ways. There is no other
way for energy to leave the ideal closed system.

And how do you know the ideal source does not dispose
of the energy it receives by getting warm?

An ideal source has zero source impedance, by definition.
All of the source impedance is contained in Rs, the
ideal source resistor.

I see we need to back up a bit further.

Consider a 10 VDC ideal voltage source.

When 2 amps are flowing out of the positive terminal, the
ideal voltage source is delivering 20 joules per second
to the circuit.

Q1. Where does this energy come from?

When 1.5 amps is flowing into the positive terminal, the
ideal voltage source is absorbing 15 joules per second
from the circuit.

Q2. Where does this energy go?

Answer to both. We do not know and we do not care. An
ideal voltage source can deliver energy to a circuit
and it can remove energy from a circuit; that is part
of the definition of an ideal voltage source. It does
not matter how it does it.

But just as easily as it can supply energy, it can
remove it.

Without understanding these basics of the ideal voltage
source, it will be impossible to correctly analyze
circuits that include them.

Perhaps this has been the root cause of the misunderstandings.

...Keith

On Tue, 15 Apr 2008 14:40:28 -0700 (PDT)
Keith Dysart wrote:

On Apr 15, 12:47*pm, Cecil Moore wrote:
Keith Dysart wrote:
Is there some reason why you think that it is only
necessary to account for the energy removed from the
circuit by heating?

Heating is the only way that the energy can be removed
from the ideal closed system.

And that you can ignore energy being removed in other ways?

No energy is removed in other ways. There is no other
way for energy to leave the ideal closed system.

And how do you know the ideal source does not dispose
of the energy it receives by getting warm?

An ideal source has zero source impedance, by definition.
All of the source impedance is contained in Rs, the
ideal source resistor.

I see we need to back up a bit further.

Consider a 10 VDC ideal voltage source.

When 2 amps are flowing out of the positive terminal, the
ideal voltage source is delivering 20 joules per second
to the circuit.

Q1. Where does this energy come from?

When 1.5 amps is flowing into the positive terminal, the
ideal voltage source is absorbing 15 joules per second
from the circuit.

Q2. Where does this energy go?

Answer to both. We do not know and we do not care. An
ideal voltage source can deliver energy to a circuit
and it can remove energy from a circuit; that is part
of the definition of an ideal voltage source. It does
not matter how it does it.

But just as easily as it can supply energy, it can
remove it.

Without understanding these basics of the ideal voltage
source, it will be impossible to correctly analyze
circuits that include them.

Perhaps this has been the root cause of the misunderstandings.

...Keith

This WIKI article mentions the ability of an ideal voltage source to absorb power.

http://en.wikipedia.org/wiki/Voltage_source
--
73, Roger, W7WKB

Roger Sparks wrote:
This WIKI article mentions the ability of an ideal voltage source to absorb power.

It says: "A primary voltage source can supply (or absorb)
energy ..."

That's easy to comprehend for a battery source. Not so
easy for an ideal RF source with a zero series impedance.
If we define an RF source as a coherent RF battery, anything
is possible (at least in our minds). Which of the following
makes more sense?

1. Destructive interference energy is stored somewhere
in the system and delivered back to the system 90 degrees
later in the cycle just as it is by a physical inductor
or capacitor.

2. An RF battery inside the ideal source stores the extra energy
in coherent RF form and delivers it back to the system as needed.

http://en.wikipedia.org/wiki/Voltage_source

It also says: "The internal resistance of an ideal voltage
source is zero;" So exactly how does something with an
internal resistance of zero absorb any power?
--
73, Cecil http://www.w5dxp.com

On Wed, 16 Apr 2008 09:31:32 -0500
Cecil Moore wrote:

Roger Sparks wrote:
This WIKI article mentions the ability of an ideal voltage source to absorb power.

It says: "A primary voltage source can supply (or absorb)
energy ..."

That's easy to comprehend for a battery source. Not so
easy for an ideal RF source with a zero series impedance.

Unless we allow it absorb energy with the same ease that it emits energy.

If we define an RF source as a coherent RF battery, anything
is possible (at least in our minds). Which of the following
makes more sense?

1. Destructive interference energy is stored somewhere
in the system and delivered back to the system 90 degrees
later in the cycle just as it is by a physical inductor
or capacitor.

No, the problem here is that the individual switched batteries that would make up this system would not be equally discharged. Some would actually gain charge, and never deliver power to the system. So I don't like this choice.

2. An RF battery inside the ideal source stores the extra energy
in coherent RF form and delivers it back to the system as needed.

This sounds identical to the 1st choice. You can make such a circuit with many individually switched batteries so it becomes a good test of the theory. Again, some of the batteries would actually gain energy.

http://en.wikipedia.org/wiki/Voltage_source

It also says: "The internal resistance of an ideal voltage
source is zero;" So exactly how does something with an
internal resistance of zero absorb any power?

How do we deal with the I x E power when the reflected voltage is greater than the forward voltage from the source? One way is to ignore it and only deal with averages. Another is to allow the ideal voltage source to absorb power. A third way is to allow the energy to be stored in constructive and destructive interference some place in the system. A fourth way is to allow reflections from the ideal voltage source with the result that power will build internally and the phase shift of the resultant system wave will shift farther away from the driving voltage, until some stable power input to disipation rate is reached.

I like your RF battery idea. It could actually be built to a close approximation of an actual sine wave.
--
73, Roger, W7WKB

Roger Sparks wrote:
Cecil Moore wrote:
That's easy to comprehend for a battery source. Not so
easy for an ideal RF source with a zero series impedance.

Unless we allow it absorb energy with the same ease that it emits energy.

How can a device with a zero impedance, i.e. zero resistance,
zero capacitance, and zero inductance, absorb energy? We can
certainly allow it to magically absorb energy but of what
use is that?

A third way is to allow the energy to be stored in constructive
and destructive interference some place in the system.

As you know, this is the one I prefer. Another thing that
neither you nor Keith has done is to account for the
reverse-flowing energy through the source. I suspect if
that was done, every thimbleful of energy would be
accounted for. So far, net energy calculations have been
used on one side of Rs and component energy calculations
on the other. That would work only if Rs was not dissipating
power.
--
73, Cecil http://www.w5dxp.com

On Wed, 16 Apr 2008 13:36:05 -0500
Cecil Moore wrote:

Roger Sparks wrote:
Cecil Moore wrote:
That's easy to comprehend for a battery source. Not so
easy for an ideal RF source with a zero series impedance.

Unless we allow it absorb energy with the same ease that it emits energy.

How can a device with a zero impedance, i.e. zero resistance,
zero capacitance, and zero inductance, absorb energy? We can
certainly allow it to magically absorb energy but of what
use is that?

It gives a tool to check our work inside the system. We think we know what happens once the energy arrives into the known components, but we don't know what exactly happens in the voltage source itself. We solve that by equating "energy in equals energy out". Then we refine that equation to "energy in equals energy returned plus energy disipated plus energy stored during some time period". Ein = Er + Ed + Es

This is the "conservation of energy" principle at work.

A third way is to allow the energy to be stored in constructive
and destructive interference some place in the system.

"A third way is to allow the energy to be stored in constructive
and destructive interference SOME PLACE IN THE SYSTEM." In this quote, I capitalized "SOME PLACE IN THE SYSTEM" because we need to understand where interference is located. If interference is located within the voltage source, it is really located *OUTSIDE* the system because we do not completely understand the voltage source itself.

As you know, this is the one I prefer. Another thing that
neither you nor Keith has done is to account for the
reverse-flowing energy through the source. I suspect if
that was done, every thimbleful of energy would be
accounted for. So far, net energy calculations have been
used on one side of Rs and component energy calculations
on the other. That would work only if Rs was not dissipating
power.

Are you expecting additional reflections? Any additonal reflections from the source would only be mechanical copies of the reflection at the short, and would add or subtract to the forward wave following the rules of sine wave addition.

But how can we have a source with zero resistance, zero capacitance, and zero inductance because in the real world, any source has impedance? The short has "zero resistance, zero capacitance, and zero inductance but it does not emit energy nor have a reverese voltage, both properties of the voltage source. It is not reasonable to assign the properties of the short to the voltage source, ignoring the reverse voltage situation, and expect reflectons from the source to be identical to reflections from a short.

--
73, Roger, W7WKB

Roger Sparks wrote:
"A third way is to allow the energy to be stored in constructive
and destructive interference SOME PLACE IN THE SYSTEM." In this
quote, I capitalized "SOME PLACE IN THE SYSTEM" because we need
to understand where interference is located.

Actually, all we need to realize is that, in a
distributed network, the current through the source
is NOT equal to the current being provided by the
source. It is the source current superposed with
reflected current and re-reflected current. The fact
that the *NET* power is negative is NOT because the
source is sinking power. It is caused by the other
"sources" of energy in the system, i.e. the energy
in the reflected waves.

Are you expecting additional reflections?

Once steady-state is reached, the magnitude of the
reflections are also steady-state. The total energy
flowing toward the load is the forward energy. The
total energy flowing in the other direction is the
reflected energy. Both of those magnitudes are fixed
during steady-state.

I have asked multiple times that this simple mental
experiment be performed and my request has been ignored.

Please install an ideal 50 ohm directional wattmeter
at point 'x' and tell us what are those readings
for forward power and reflected power.

GND--Vs--x--Rs--------45deg 50 ohm----short
100v 50ohm

Once the forward and reverse energy flows at point
'x' are understood, it will become clear that all
of the net current through the source is NOT being
provided by the source. That the source is sinking
energy is an illusion caused by reflections.
--
73, Cecil http://www.w5dxp.com

Roger Sparks wrote:
. . . But how can we have a source with zero resistance, zero
capacitance, and zero inductance because in the real world, any
source has impedance? The short has "zero resistance, zero
capacitance, and zero inductance but it does not emit energy nor have
a reverese voltage, both properties of the voltage source. It is not
reasonable to assign the properties of the short to the voltage
source, ignoring the reverse voltage situation, and expect reflectons
from the source to be identical to reflections from a short.

None of the ideal components we use for linear circuit analysis exist in
the real world. We use ideal resistances which have no inductance or
capacitance, capacitances which have no resistance or inductance, ideal
inductances which have no resistance or capacitance, ideal controlled
sources which operate over an infinite range of control and output
values. And try making anything even vaguely resembling an ideal
transformer. So what's the problem in accepting an ideal voltage source
as another model element? If you want a better approximation of
something you can build in the real world, add an ideal resistance to
the ideal voltage source, and you'll have a much better representation
of most real sources.

As for the way the source reacts to an impinging wave, note that the
voltage across a short circuit doesn't change when a wave hits it.
Neither does the voltage across an ideal voltage source. Consequently,
they do have exactly the same effect on waves.

Roy Lewallen, W7EL

This discussion about ideal sources and power absorption would be rather
amazing if I hadn't seen so much of the long sad history of this whole
thread. It's just another diversion to deflect the discussion away from
some of the sticky problems with alternative theories.

It looks like Cecil could benefit from momentarily abandoning his power
waves, virtual reflections, photons, s parameters, and constructive
interference, and go back to basic first or second semester electric
circuit theory. Connect an ideal voltage source to a series RL or RC --
all elements lumped, not distributed. Find the power P(t) at each of the
three nodes. (It's easy. Just calculate v(t) and i(t) and multiply the
two.) Remember that the sign of P(t) shows the direction of energy flow.
If I had the patience and self-control to participate in this endless
thread (and I admittedly have neither), I'd refuse to say another word
about it until Cecil shows that he knows what the power waveforms are at
those three nodes. Why argue with someone about distributed networks who
hasn't mastered lumped circuit analysis?

Roy Lewallen, W7EL

Correction: Finding the voltage requires a reference node. So the power
needs to be found at only two of the nodes, with the third being the
reference (ground) node.

Roy Lewallen, W7EL

Roy Lewallen wrote:
This discussion about ideal sources and power absorption would be rather
amazing if I hadn't seen so much of the long sad history of this whole
thread. It's just another diversion to deflect the discussion away from
some of the sticky problems with alternative theories.

It looks like Cecil could benefit from momentarily abandoning his power
waves, virtual reflections, photons, s parameters, and constructive
interference, and go back to basic first or second semester electric
circuit theory. Connect an ideal voltage source to a series RL or RC --
all elements lumped, not distributed. Find the power P(t) at each of the
three nodes. (It's easy. Just calculate v(t) and i(t) and multiply the
two.) Remember that the sign of P(t) shows the direction of energy flow.
If I had the patience and self-control to participate in this endless
thread (and I admittedly have neither), I'd refuse to say another word
about it until Cecil shows that he knows what the power waveforms are at
those three nodes. Why argue with someone about distributed networks who
hasn't mastered lumped circuit analysis?

Roy Lewallen, W7EL