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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

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In article vfqdnbEBq5VjiZrVnZ2dnUVZ_gudnZ2d@easystreetonline , Roy
Lewallen wrote:

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.

Hello, all, and then there's that circuit topology problem found in some
circuit theory textbooks:

There is an infinite planar square lattice of connected 1-ohm resistors
that extends to infinity in all directions. A DC ohmmeter placed across
any resistor would register what value? The answer is 0.5 ohm but how do
you go about solving for it? (Hint: There is an easy way and a much more
difficult (general) way to attack this problem. This problem is a discrete
example that in the limit would indicate the distribution of a DC electric
field (and currents) in a conducting, isotropic metal plate as one moves
away from the source of excitation. Sincerely, and 73s from N4GGO,

John Wood (Code 5550) e-mail:
Naval Research Laboratory
4555 Overlook Avenue, SW
Washington, DC 20375-5337

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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

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Roy Lewallen wrote:
Why argue with someone about distributed networks who
hasn't mastered lumped circuit analysis?

Roy, after your use of standing-wave current with
its unchanging reference phase (as reported by EZNEC)
to measure the delay through a 75m bugcatcher coil
as virtually zero delay, it is obvious that your
"lumped circuit analysis" cannot be trusted. Want
to read about "... the Failure of Lumped-Element
Circuit Theory" and diagnose your errors regarding
those loading coil measurements?

http://www.ttr.com/corum/
--
73, Cecil http://www.w5dxp.com

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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

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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

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On Thu, 17 Apr 2008 09:46:55 -0500
Cecil Moore wrote:

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.

Yes, in nature the returning power will affect the frequency of the source. The ideal voltage source absorbs the reflected power rather than allowing a change in frequency.

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.

This circuit has an impedance of 73.5 + J44.1 ohms at the point x. Energy sinking into the source occurs during the cycle. It is not an illusion.

Think of things this way. Any signal must contain energy. A signal without energy is not a signal. Nature must have some method of signaling between source and load when the load does not match the impedance of power flow. This is simple logic based on the known fact that the apparent impedance of a transmission line depends upon the load at the end of the transmission line.

Power returning to the source is the signal nature requires to throttle the output of the source. Power returning to the source will change the frequency of the source in nature, but in our ideal voltage source, nature is suspended and the frequency is held constant. Our problem as engineers is to isolate each effect and to make valid predictions so that we can build more intelligently.
--
73, Roger, W7WKB

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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.

Cecil Moore wrote:
Well, that certainly puts an end to this discussion.

Then, Roy Lewallen wrote:
... 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.....

Just when you think you're out of it, they pull you back in.

K7JEB

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Roger Sparks wrote:
Yes, in nature the returning power will affect the
frequency of the source.

If the reflected wave is a sine wave coherent with
the forward wave, why would the frequency change
when they are superposed?

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

This circuit has an impedance of 73.5 + J44.1 ohms
at the point x. Energy sinking into the source occurs
during the cycle. It is not an illusion.

I got a different impedance value but for purposes
of discussion, all that matters is that the impedance
is NOT 50 ohms.

Let's assume your impedance is correct. That means
the ratio of reflected power to forward power at
point 'x' is 0.1452 in the 50 ohm environment.
Why would reflected energy flowing back through
the source be considered "energy sinking". Why
cannot the reflected energy flow back through the
source unimpeded by the 0+j0 impedance and be
reflected by the short to ground on the other side?
That same thing happens all the time to standing
wave current on the outside of the coax braid.
Why do the laws of physics change inside a source?

In this example, we are delivering 100 watts to Rs.
With a power reflection coefficient of 0.1452, the
forward power has to be 117 watts making the reflected
power equal to 17 watts at point 'x'. That reflected
power of 17 watts is not absorbed by the source. It
flows back through the source, unimpeded by the 0+j0
impedance, and is reflected by the short to ground.
It joins the 100 watts being generated by the source
to become the forward wave of 117 watts. No absorbing
of energy required - just good old distributed network
physics.

Once this 17 watt wave is tracked back through the
source, reflected, and superposed with the 100 watts
being generated by the source, all will become clear
without any power absorption by the source being
necessary. Here's a diagram of what's happening.

100w 100w
GND--------Vs----------Rs--------45deg----------short
17w-- 117w-- 50w--
--17w --17w --50w

All energy flows balance perfectly and there is no
absorption by the source.
--
73, Cecil http://www.w5dxp.com

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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