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On Apr 9, 3:03*pm, Roy Lewallen wrote:
Keith Dysart wrote:
On Apr 8, 8:51 am, Cecil Moore wrote:
. . .
The
forward wave flows unimpeded through the node as does
the equal magnitude reflected wave. The net energy flow
is zero. The average energy flow is zero.

Anyone who believes there is zero energy at a standing-
wave current node should touch that point on a transmission
line (which just happens to be the same point as the
maximum voltage anti-node).

No one has said there is zero energy. Only that there is
zero energy flow. For energy flow, one needs simultaneous
voltage and current.
. . .

In the interesting case of a current node on an infinite-SWR line, it
appears we do have energy flow without any current, and therefore
without power. Energy flows into the node from both directions in equal
amounts at the same time, and out to both directions in equal amounts at
the same time.

I think I would be tempted to cut the node down the middle creating
two nodes with no current flow or energy flow between them.

What we don't have is *net* energy flow at the node.
Likewise, there's charge flowing into the node from both directions, and
out in both directions, which results in the zero net current. I don't
believe that's the same as saying there's no energy or charge flow at
all, even though the power and current are zero. And it's not necessary
to separately consider forward and reverse waves of current, energy, or
power in order to observe this -- it can be seen from looking only at
the total charge or energy.

One thought experiment I rather like is the infinite SWR ideal line.
Cut the line at all the current zeroes. The voltage, current, and,
I'd suggest, the energy distribution do not change. The line can
be re-assembled, again with no change.

Those who argue that energy is crossing the node, have to explain that
the cut has completely changed the mechanism that keeps the net energy
in its place, but the voltage and current distributions remain the
same. I prefer the basic circuit theory tenet that one can cut a
line carrying no current without impacting the circuit. This leads
to a model with no energy crossing the node.

...Keith

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On Apr 9, 12:29*pm, Cecil Moore wrote:
Keith Dysart wrote:
Thus I strongly suggest that Vg, Ig, Pg, represent reality. The
others are a convenient alternative view for the purposes of
solving problems.

Of course they represent *net* reality but we are trying
to determine what is happening at a component wave level.
Defining the component waves out of existence is an un-
acceptable substitute for ascertaining what is happening
in reality.

Typically we see Vg split into Vf and Vr, but why stop at two.
Why not 3, or 4?

Because two is what a directional wattmeter reads. The
two superposed waves, forward and reverse, can be
easily distinguished from one another. Two superposed
coherent forward waves cannot be distinguished from
each other. That's why we stop at two - because it is
foolish to go any farther.

You sometimes use three. Discussions of ghosts have at
least three. Not so foolish, methinks.

There is power coming from the transmission line. Looking at Pg(t),
some of the time energy flows into the line, later in the cycle
it flows out. The energy transfer would be exactly the same if the
transmission line was replaced by a lumped circuit element. And
we don't need Pf and Pr for an inductor.

OTOH, the distributed network model is a superset of
the lumped circuit model so the inadequate lumped
circuit model might confuse people. Hint: changing
models to make waves disappear from existence doesn't
make the waves disappear.

The model is not inaccurate when the question is framed
with the model, as you do for Fig 1-1.

The lumped circuit model is adequate for lumped circuits.
It is inadequate for a lot of distributed network problems.
If the lumped circuit model worked for everything, we
wouldn't ever need the distributed network model.

True, but the question at hand, based on Fig 1-1 is lumpy.

I suggest that you take your circuit and apply distributed
network modeling techniques to it including reflection
coefficients and forward and reflected voltages, currents,
and powers at all points in the circuit. Note that the
reflections are *same-cycle* reflections. If the lumped
circuit model analysis differs from the distributed network
model analysis, the lumped circuit analysis is wrong.

Ummm. It was your circuit.

It goes up because the impedance presented by the transmission
changes when the reflection returns. This change in impedance
alters the circuit conditions and the power in the various
elements change. Depending on the details of the circuit,
these powers may go up, or they may go down when the reflection
arrives.

That is true, but the impedance is *VIRTUAL*, i.e. not an
impedor, and is therefore only an *EFFECT* of superposition.
We are once again left wondering about the *CAUSE* of the
virtual impedance, i.e. the details of the superposition
process. Ignoring those details will not solve the problem.

Actually, the transmission line input impedance is quite real,
formed from distributed capacitance and inductance. Like most two
terminal circuits, it can be reduced to simpler form.

...Keith

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On Apr 9, 12:59*pm, Cecil Moore wrote:
Keith Dysart wrote:
Cecil Moore wrote:
There is no capacitance or inductance in the source to
store energy.

"In" is an oxymoron for the lumped circuit model.
The lumped reactance exists *at* the same point as
the source because everything is conceptually lumped
into a single point.

In the real world, circuits are never single points
and there exists a frequency at which distributed
network effects cannot be ignored. In reality,
distributed network effects occur for all real
circuits but they can often be ignored as negligible.

The two inches of wire connecting the source to the
source resistor has a characteristic impedance and
is a certain fraction of a wavelength long. If it is
not perfectly matched, reflections will occur, i.e.
there will exist forward power and reflected power on
that two inches of wire.

It was your Fig 1-1, made of ideal elements with none
of these issues.

For the 1/8WL
shorted line, there appears to be 125 watts of forward
power and 25 watts of reflected power at points on each
side of the source.

Not if there is no transmission line.

Aha, there's your error. What would a Bird directional
wattmeter read for forward power and reflected power?
Consider that short pieces of 50 ohm coax are used
to connect the real-world components together.

It would read something completely different if it was
calibrated for 75 ohms, though the difference between
Pf and Pr would be the same.

But that is not the circuit of your Fig 1-1.

Or chose any characteristic impedance and do the math.
You will discover something about the real world, i.e.
that you have been seduced by the lumped circuit model.

It was your circuit; Fig 1-1.

Perhaps. *But I don't need more examples where the
powers balance. I already have the one example where
they don't.

And that one example is outside the scope of the
preconditions of my Part 1 article. Let me help you
out on that one. There are an infinite number of
examples where the reflected power is NOT dissipated
in the source resistor but none of those examples,
including yours, satisfies the preconditions specified
in my Part 1 article. Therefore, they are irrelevant
to this discussion.

As long as you agree that the imputed energy in the
reflected wave is not dissipated in the source
resistor; and only claim that the imputed average
power in the reflected wave is numerically equal
to the increase in the dissipation.


But there are no component powers in the source. It
is a simple circuit element.

No wonder your calculations are in error.
Perform your calculations based on the readings of
an ideal 50 ohm directional wattmeter and get back to us.

Well there's a plan. Measure everything in a
circuit with a directional wattmeter. You first.
Start with Fig 1-1.

But you'll have to choose the calibration impedance.

I'd suggest 100 ohms for the section between the
source and the source resistor because the source
resistor and the line initially present a 100 ohm
impedance and you would not want any reflections
messing up the measurements.

Hint: Mismatches cause reflections, even in real-world
circuits. The reflections happen to be *same-cycle*
reflections. The simplified lumped circuit model, that
exists in your head and not in reality, ignores those
reflections and thus causes confusion among the
uninitiated who do not understand its real-world
limitations.

We should explore this new world. Please discard your
voltmeter, ammeter, oscilloscope, .... All you need
is a Bird wattmeter.

...Keith

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Keith Dysart wrote:
On Apr 9, 12:51 pm, Roger Sparks wrote:
. . .
So far as breaking Vg into many sequential/different Vf and Vr, we usually need to do that. Cecil chose our simple example to prevent re-reflection (reflection of the reflection) but even then it is apparent that the voltage source will have a reactive component.

I still think of a voltage source as just being a voltage source, not
something
with resitance, reactance or impedance.
. . .

An ideal voltage source has, by definition, zero impedance, which means
zero resistance and zero reactance. No amount of current you put into it
or take out of it will alter its voltage. The ratio of voltage to
current at its terminals is the impedance of the load which the source
sees, not the impedance of the source. If a source used in an analysis
has finite resistance, it's not an ideal voltage source.

Roy Lewallen, W7EL

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Keith Dysart wrote:
One thought experiment I rather like is the infinite SWR ideal line.
Cut the line at all the current zeroes. The voltage, current, and,
I'd suggest, the energy distribution do not change. The line can
be re-assembled, again with no change.

When you cut the line, you introduce an open circuit
where none existed before. Everything obeys the
distributed network model, at least conceptually.

Before you cut the line, there is no physical
impedance discontinuity and therefore no
reflections. You have to invent complete new
laws of physics for that one. I'm not against
you inventing new laws of physics but you need
to invent them before, not after, you use them. :-)
--
73, Cecil http://www.w5dxp.com

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On Apr 9, 1:22*pm, Cecil Moore wrote:
Keith Dysart wrote:
On Apr 8, 8:51 am, Cecil Moore wrote:
Roy Lewallen wrote:
Now, I don't know of any way to assign "ownership" to bundles of energy.
One way is to add a unique bit of modulation to each
bundle of wave energy. I am fond of using a TV signal
and observing ghosting on the screen. This, of course,
assumes that the modulation stays with the same component
wave to which it was originally associated.

But as soon as you modulate, you no longer have sinusoidal
steady state.

You know and I know that is a copout diversion to avoid
your having to face the technical facts.

You seem to be the one who knows this. I don't.

Consider the 1 second interval from 4.5 to 5.5 seconds.
In this second 0.016393 joules flow for an average
power of 0.016393 W. But the sum of the imputed power
in the two spectral components is 1 W. Where did the
missing energy go?

Hint: Missing energy is impossible except in your mind.
Just because you are ignorant of where the energy goes
doesn't mean it is missing. It just means that you fail
to understand interference. Have you not read Hecht's
Chapter 9 on "Interference"?

Obviously, interference is present and there is *NO*
missing energy. I have previously listed the possibilities
at least four times so will not bother listing them again.

Between 9.5 and 10.5 seconds, 1.983607 J (average 1.983607 W)
of energy flows. By 'interference', I think you are suggesting
the missing power from 4.5 to 5.5 appears as excess power
between 9.5 and 10.5, thus satisfying your conservation of
energy requirement. But where was the energy stored for
5 seconds until it could be delivered.

Or, more intriguing, between 0 and 1 second, 1.935466 J
(average 1.935466 W) of energy flowed, but the sum of
the powers of the two constituents was only 1 J (average
1 W) in this interval. Where did the extra energy come from?
Was it borrowed from the future? It did not come from
the past since the generator was not yet on.

Just another example of why assigning too much reality
to the imputed powers of the components of superposition
is misleading.

Just another example of ignorance in action. Waves
possess energy that cannot be destroyed. Just because
you cannot track it doesn't mean it cannot be tracked.

That is why I pose the question, hoping for someone to
describe the mechanism that the energy for the flow that
is happening now can be borrowed from the future.

But what happens if the generator is turned off before
the future arrives? Where did the extra energy come from
then?

In other examples, you have suggested inserting a zero length
transmission line to aid analysis. Why not insert a zero length
transmission line with an impedance to produce the desired
reflection?

What would be the characteristic impedance of a length of
transmission that caused a reflection coefficient of 1.0?

Exactly. With the source impedance being zero, you can use
any impedance line you like.

No one has said there is zero energy. Only that there is
zero energy flow. For energy flow, one needs simultaneous
voltage and current.

Vfor/Ifor = Z0, Vfor*Ifor = Pfor = EforxHfor
If an EM wave exists, it is moving at the speed of
light and transferring energy. For Z0 purely resistive,
Vfor cannot exist without Vfor/Z0 = Ifor. Vfor is
always in phase with Ifor.

Assigning too much reality to component signals is
seriously misleading.

Assigning reality to the components of superposition
is seriously misleading???? Can we therefore throw
out the entire principle of superposition?

Well, you can solve these problems in other ways.
Superposition is not required, but it is certainly
convenient. I would not throw it out just because
the constituent powers do not necessarily have
any real meaning. I would just make sure I used
it in ways that did not mislead.

Until one can grasp the simplicity of a transmission line,
moving to the complexity of free space offers nothing but
obfuscation.

It is obvious that you have many things you desire to
hide inside that black transmission line to which we
are not even allowed to attach a directional wattmeter.

Since you are incapable of explaining what happens in
free space for all to see, why should we believe that
you have figured out what is happening inside a
transmission line where everything is hidden from view?

Actually, I have a pretty good grasp of what is happening
in free space, and it is all available to you by extension
from the behaviours of the one dimensional transmission
line. But there is little point in going there until the
transmission line is understood. Three dimensional free
space has much too much wiggle room. And you are an
expert at wiggling.

...Keith

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Keith Dysart wrote:
Actually, the transmission line input impedance is quite real,
formed from distributed capacitance and inductance. Like most two
terminal circuits, it can be reduced to simpler form.

I didn't say virtual impedances are not real. I
said they are not causes of anything and are,
instead, effects of superposition incapable of
causing anything in the complete absence of a
physical impedance.

You have for a long while now, confused cause
and effect. Maybe you should review the three
separate definitions of "impedance" given in
the IEEE Dictionary.
--
73, Cecil http://www.w5dxp.com

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On Apr 9, 9:33*pm, Cecil Moore wrote:
Keith Dysart wrote:
One thought experiment I rather like is the infinite SWR ideal line.
Cut the line at all the current zeroes. The voltage, current, and,
I'd suggest, the energy distribution do not change. The line can
be re-assembled, again with no change.

When you cut the line, you introduce an open circuit
where none existed before. Everything obeys the
distributed network model, at least conceptually.

Before you cut the line, there is no physical
impedance discontinuity and therefore no
reflections. You have to invent complete new
laws of physics for that one. I'm not against
you inventing new laws of physics but you need
to invent them before, not after, you use them. :-)

And yet....

The voltage distribution on the line, the current
distribution on the line and the energy distribution
on the line has not changed one iota.

This new physical impedance discontinuity has not
had any observable effect. All it seems to change
is the reflection of the unobservable forward and
reflected waves.

But the voltage distribution on the line, the current
distribution on the line and the energy distribution
on the line has not changed one iota.

These reflections can not be so important if they
can not be observed.

...Keith

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Keith Dysart wrote:
As long as you agree that the imputed energy in the
reflected wave is not dissipated in the source
resistor;

My ethical standards will not allow me to lie about
technical facts in evidence. You cannot bully me
into doing so.

When the average interference is zero, all of the
average reflected energy is dissipated in the source
resistor. It is true for all examples of Fig. 1-1.
You have not presented even one example where
that is not a true statement.

We should explore this new world.

There's no new world. All I am presenting is the
distributed network model which is lots older than
I. What you should present are your new laws of
physics that contradict the distributed network model.
--
73, Cecil http://www.w5dxp.com

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Roger Sparks wrote:
Of course one way would be if Vf actually did reflect from Vr.

That would require a brand new set of laws of physics.
--
73, Cecil http://www.w5dxp.com

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Keith Dysart wrote:
That is why I pose the question, hoping for someone to
describe the mechanism that the energy for the flow that
is happening now can be borrowed from the future.

Destructive interference would have to happen first.

But what happens if the generator is turned off before
the future arrives? Where did the extra energy come from
then?

There is no extra energy. Constructive interference is
impossible without that supply of energy.

Actually, I have a pretty good grasp of what is happening
in free space, and it is all available to you by extension
from the behaviours of the one dimensional transmission
line. But there is little point in going there until the
transmission line is understood.

It is the exact opposite. There is no point in inventing
new laws of physics for transmission lines if those new
laws don't work in free space.

So please present an example of EM waves reflecting off
of other EM waves in free space. Do you really think the
energy in the standing wave beam of a laser is reversing
direction and momentum every cycle?
--
73, Cecil http://www.w5dxp.com

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Keith Dysart wrote:
The voltage distribution on the line, the current
distribution on the line and the energy distribution
on the line has not changed one iota.

In one case the wave energy changes direction and
momentum at the physical discontinuity. In the
other case, there exists nothing to change the
wave direction and momentum.

This new physical impedance discontinuity has not
had any observable effect. All it seems to change
is the reflection of the unobservable forward and
reflected waves.

Yes, exactly in agreement with the laws of physics.

But the voltage distribution on the line, the current
distribution on the line and the energy distribution
on the line has not changed one iota.

Please present your new laws of physics that allow
EM waves to reflect off of EM waves in the complete
absence of a physical discontinuity. And please
demonstrate such in free space so we can see the
results.
--
73, Cecil http://www.w5dxp.com

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On Apr 9, 9:40*pm, Cecil Moore wrote:
Keith Dysart wrote:
Actually, the transmission line input impedance is quite real,
formed from distributed capacitance and inductance. Like most two
terminal circuits, it can be reduced to simpler form.

I didn't say virtual impedances are not real. I
said they are not causes of anything and are,
instead, effects of superposition incapable of
causing anything in the complete absence of a
physical impedance.

But the distributed capacitance and inductance
are physical impedances.

You have for a long while now, confused cause
and effect. Maybe you should review the three
separate definitions of "impedance" given in
the IEEE Dictionary.

Neither 'virtual impedance' nor 'impedance, virtual'
are in the dictionary (at least the 7th Edition).

...Keith

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On Apr 9, 9:48*pm, Cecil Moore wrote:
Keith Dysart wrote:
As long as you agree that the imputed energy in the
reflected wave is not dissipated in the source
resistor;

My ethical standards will not allow me to lie about
technical facts in evidence. You cannot bully me
into doing so.

When the average interference is zero, all of the
average reflected energy is dissipated in the source
resistor. It is true for all examples of Fig. 1-1.
You have not presented even one example where
that is not a true statement.

But all you have demonstrated is that the imputed
average power in the reflected wave is *numerically
equal* to the average increase in the dissipation
of the source resistor. Which is good, as long as
that is all you claim. Which it some times seems
to be, especially when you qualify with "interference
is zero".

Finer grained analysis shows that the imputed
energy (not average) in the reflected wave is not
dissipated in the source resistor. The trouble
is, sometimes you agree with this (when you
invoke that interference is present), but other
times you don't (see your response to the opening
paragraph). It is this flip-flop that makes your
actual position difficult to discern.

...Keith

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On Apr 9, 9:57*pm, Cecil Moore wrote:
Keith Dysart wrote:
That is why I pose the question, hoping for someone to
describe the mechanism that the energy for the flow that
is happening now can be borrowed from the future.

Destructive interference would have to happen first.

For the example under discussion, the signals start with
'constructive interference'. There has not yet been an
opportunity for desctructive interference, which happens
later. So where does the extra energy at the start come
from?

But what happens if the generator is turned off before
the future arrives? Where did the extra energy come from
then?

There is no extra energy. Constructive interference is
impossible without that supply of energy.

Exactly. This is the problem with your model. The extra
energy (i.e. the energy greater than that in the sum of
the spectral components) is present, but your model does
not have somewhere for this energy to come from.

Actually, I have a pretty good grasp of what is happening
in free space, and it is all available to you by extension
from the behaviours of the one dimensional transmission
line. But there is little point in going there until the
transmission line is understood.

It is the exact opposite. There is no point in inventing
new laws of physics for transmission lines if those new
laws don't work in free space.

There are no new laws of physics. There is just the
opportunity for a better understanding of what is
happening. This better understanding applies in free
space as well. It is just much easier to obtain this
better understanding on the transmission line and then
move to free space.

So please present an example of EM waves reflecting off
of other EM waves in free space.

That was someone elses suggestion, not mine.

Do you really think the
energy in the standing wave beam of a laser is reversing
direction and momentum every cycle?

Why does this thought make you uncomfortable? Is it because
you are trying to commingle the wave explanation with the
partical explanation? These are a duality. You use one
or the other, but not bits from each at the same time.

...Keith

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On Apr 9, 10:02*pm, Cecil Moore wrote:
Keith Dysart wrote:
The voltage distribution on the line, the current
distribution on the line and the energy distribution
on the line has not changed one iota.

In one case the wave energy changes direction and
momentum at the physical discontinuity. In the
other case, there exists nothing to change the
wave direction and momentum.

As I expected, you claim that the situations are
*completely* different. And yet the voltage, current
and energy distributions are identical. There are
no observable differences. And yet you claim they
are *completely* different. And yet there are no
observable differences. And yet....

Tis a puzzle, isn't it.

And we know from circuit theory that we can cut
a conductor carrying no current without affecting
the circuit. Why should it be different here?

This new physical impedance discontinuity has not
had any observable effect. All it seems to change
is the reflection of the unobservable forward and
reflected waves.

Yes, exactly in agreement with the laws of physics.

At least with your interpretation of the laws.

But the voltage distribution on the line, the current
distribution on the line and the energy distribution
on the line has not changed one iota.

Please present your new laws of physics that allow
EM waves to reflect off of EM waves in the complete
absence of a physical discontinuity.

Again, not my claim. But using your previous approach
for analysis, perhaps we should insert a zero length
line of the appropriate impedance to provide the cause
for the reflection, if you insist on a reflection.

And please
demonstrate such in free space so we can see the
results.

...Keith

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Keith Dysart wrote:
But the distributed capacitance and inductance
are physical impedances.

But they are constant, i.e. there is no physical
impedance *discontinuity*. The reflection coefficient
inside a homogeneous piece of transmission line is
(Z0-Z0)/(Z0+Z0)=0, i.e. there can be no reflections.
The reflection coefficient in free space is
(1.0-1.0)/(1.0+1.0)=0, i.e. there can be no reflections
in free space.

Neither 'virtual impedance' nor 'impedance, virtual'
are in the dictionary (at least the 7th Edition).

"Virtual" essentially means that no physical impedor
exists. The virtual impedance definition is covered
by definition (B), the ratio of voltage to current
which *causes* the impedance. A virtual impedance
is an *effect*, not a cause.
--
73, Cecil http://www.w5dxp.com

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Keith Dysart wrote:
Finer grained analysis shows that the imputed
energy (not average) in the reflected wave is not
dissipated in the source resistor.

It is the joules in instantaneous power that must
be conserved, not the instantaneous power. There
is no such thing as a conservation of power
principle yet all you have presented are power
calculations. "Where's the beef?"

How many joules are there in 100 watts of
instantaneous power?
--
73, Cecil http://www.w5dxp.com

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Keith Dysart wrote:
For the example under discussion, the signals start with
'constructive interference'.

If the source is local and capable of supplying energy,
all is well and good as I have said many times before.
But constructive interference in the absence of any
source of energy is impossible.

That is exactly why you need to perform your calculations
with the source removed from the source resistor by one
wavelength of ideal 50 ohm transmission line. If you come
up with a violation of the conservation of energy principle,
something is wrong with your math.

If two coherent signals need constructive interference and
energy is not available, the two signals react as if they
were not coherent, i.e. Ptot = P1 + P2. Physics 201.

Exactly. This is the problem with your model. The extra
energy (i.e. the energy greater than that in the sum of
the spectral components) is present, but your model does
not have somewhere for this energy to come from.

Yes it does - as I have explained about 5 times now. If a
*local source* is present, constructive interference energy
can and often does come from the source. Why do you find
that fact so hard to comprehend? Sources supply energy -
that's what sources do. An ideal local source can react
instantaneously to any energy requirement.

There are no new laws of physics.

On the contrary - there are no existing laws of physics
that allow EM waves to bounce off each other yet that's
what you are proposing. You are inventing new laws of
physics to support your (magical) thinking. So please
produce the theory and proof that EM waves can bounce
off of each other.

That was someone elses suggestion, not mine.

Copout alert! How can *your* reflections at a passive
node occur without EM waves reflecting off of other EM
waves?

These are a duality. You use one
or the other, but not bits from each at the same time.

Keith, you have been willy-nilly mixing bits of the
distributed network model with bits of the lumped circuit
model ensuring that your energy equations will not balance.
You are the absolute worst offender of your own advice.
--
73, Cecil http://www.w5dxp.com

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Keith Dysart wrote:
As I expected, you claim that the situations are
*completely* different. And yet the voltage, current
and energy distributions are identical. There are
no observable differences. And yet you claim they
are *completely* different. And yet there are no
observable differences. And yet....

I did NOT claim that the situations are *completely*
different. I said that some conditions are different
and some conditions are the same. Voltages and currents
are the same yet there is certainly a difference between
an open circuit and a short circuit. Besides, in the
real world, cutting the line would certainly cause
observable differences.

Tis a puzzle, isn't it.

Nope, if you were born without your five senses,
you would feel that way about everything in existence.
Why do you deliberately choose to remain handicapped
by ignorance?

A bit of modulation would cure up the mystery for
you. If any modulation crosses the node, it is a
good bet that wave energy is carrying the
modulation. If phase locked TV signal generators
equipped with circulator load resistors are
installed at each end of a transmission line,
the TV signals can be observed on normal TV
sets crossing the standing wave nodes as if
they didn't exist. Removing the modulation
is unlikely to reverse the laws of physics.

And we know from circuit theory that we can cut
a conductor carrying no current without affecting
the circuit. Why should it be different here?

Please prove that a short circuit and an open circuit
are identical.

Please present your new laws of physics that allow
EM waves to reflect off of EM waves in the complete
absence of a physical discontinuity.

Again, not my claim.

Seems your theory requires such. Please explain how
reflections can occur at a passive standing wave node
without EM waves bouncing off of each other.

Energy and momentum both must be conserved. A causeless
reversal of energy and momentum is impossible whether
it is a bullet or an EM wave.

But using your previous approach
for analysis, perhaps we should insert a zero length
line of the appropriate impedance to provide the cause
for the reflection, if you insist on a reflection.

Please produce an example of
a real world transmission line that would support your
100% reflection. Hint: what would be the Z02
characteristic impedance in the reflection
coefficient equation, (50-Z02)/(50+Z02) = 1.0 ???
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Apr 10, 7:52*am, Cecil Moore wrote:
Keith Dysart wrote:
But the distributed capacitance and inductance
are physical impedances.

But they are constant, i.e. there is no physical
impedance *discontinuity*. The reflection coefficient
inside a homogeneous piece of transmission line is
(Z0-Z0)/(Z0+Z0)=0, i.e. there can be no reflections.
The reflection coefficient in free space is
(1.0-1.0)/(1.0+1.0)=0, i.e. there can be no reflections
in free space.

Neither 'virtual impedance' nor 'impedance, virtual'
are in the dictionary (at least the 7th Edition).

"Virtual" essentially means that no physical impedor
exists. The virtual impedance definition is covered
by definition (B), the ratio of voltage to current
which *causes* the impedance. A virtual impedance
is an *effect*, not a cause.

The transmission line definitely falls into
definition (C), "A physical device or combination of
devices whose impedance as defined in definition (A) or
(B) can be determined." The TL is a combination of
devices, a lot of very small ones, and its impedance
can be determined.

Using 26 pf/ft as a representative value for
RG-58, dividing the 45 degree section into 45
pieces, applying the normal rules for parallel and
series circuit elements, the impedance at the
entry to the line is trivially (using Excel) calculated
to be 50.443 /_ 90. Subdividing into smaller
elements would increase accuracy. If I could remember
my calculus, the exact answer could be derived.

There is no need for forward or reflected waves
at all; just basic AC circuit theory.

...Keith

The Rest of the Story
 

On Apr 10, 8:01*am, Cecil Moore wrote:
Keith Dysart wrote:
Finer grained analysis shows that the imputed
energy (not average) in the reflected wave is not
dissipated in the source resistor.

It is the joules in instantaneous power that must
be conserved, not the instantaneous power. There
is no such thing as a conservation of power
principle yet all you have presented are power
calculations. "Where's the beef?"

The computation using energy instead of power has
also been done (and published here) and found also
to demonstrate that the reflected is not dissipated
in the source resistor.

How many joules are there in 100 watts of
instantaneous power?

Obviously. It depends on how long you let the
100 W of instantaneous power flow. Integrate and
the answer shall be yours.

...Keith

The Rest of the Story
 

On Apr 10, 9:01*am, Cecil Moore wrote:
Keith Dysart wrote:
As I expected, you claim that the situations are
*completely* different. And yet the voltage, current
and energy distributions are identical. There are
no observable differences. And yet you claim they
are *completely* different. And yet there are no
observable differences. And yet....

I did NOT claim that the situations are *completely*
different. I said that some conditions are different
and some conditions are the same. Voltages and currents
are the same yet there is certainly a difference between
an open circuit and a short circuit. Besides, in the
real world, cutting the line would certainly cause
observable differences.

Tis a puzzle, isn't it.

Nope, if you were born without your five senses,
you would feel that way about everything in existence.
Why do you deliberately choose to remain handicapped
by ignorance?

A bit of modulation would cure up the mystery for
you. If any modulation crosses the node, it is a
good bet that wave energy is carrying the
modulation.

When you modulate the carrier, the resulting signal
has many frequencies. Unless great care is taken
with the choice of modulation frequencies and
carrier frequencies, they will each have nodes
at different places on the transmission line. Without
a common node, there is no place that energy
does not cross, hence modulation makes the
question moot. In effect, the standing waves
of the various frequency components do not line
up. (Remember superposition works, for voltages).

If care is taken with the selection of modulation
frequencies with regards to the carrier, then nodes
can be created on the transmission line and neither
the carrier nor the modulation will cross such a
node.

If phase locked TV signal generators
equipped with circulator load resistors are
installed at each end of a transmission line,
the TV signals can be observed on normal TV
sets crossing the standing wave nodes as if
they didn't exist.

Which they don't, as explained above.

Removing the modulation
is unlikely to reverse the laws of physics.

True, but it can change whether there are nodes.

And we know from circuit theory that we can cut
a conductor carrying no current without affecting
the circuit. Why should it be different here?

Please prove that a short circuit and an open circuit
are identical.

An open circuit is just as useful as a short circuit
when no current is flowing.

Please present your new laws of physics that allow
EM waves to reflect off of EM waves in the complete
absence of a physical discontinuity.

Again, not my claim.

Seems your theory requires such. Please explain how
reflections can occur at a passive standing wave node
without EM waves bouncing off of each other.

Energy and momentum both must be conserved. A causeless
reversal of energy and momentum is impossible whether
it is a bullet or an EM wave.

What causes energy to flow into and then out of a capacitor?
Look for your answer there.

But using your previous approach
for analysis, perhaps we should insert a zero length
line of the appropriate impedance to provide the cause
for the reflection, if you insist on a reflection.

Please produce an example of
a real world transmission line that would support your
100% reflection. Hint: what would be the Z02
characteristic impedance in the reflection
coefficient equation, (50-Z02)/(50+Z02) = 1.0 ???

That is just too easy...
(50-Z02)/(50+Z02) = 1.0
50 - Z02 = 50 + Z02
-2 * Z02 = 0
Z02 = 0
That wasn't so hard, was it?

...Keith

The Rest of the Story
 

Keith Dysart wrote:
There is no need for forward or reflected waves
at all; just basic AC circuit theory.

Now do it in free space. EM waves are EM waves
no matter what the medium.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Keith Dysart wrote:
The computation using energy instead of power has
also been done (and published here) and found also
to demonstrate that the reflected is not dissipated
in the source resistor.

Well, that certainly violates the conservation of
energy principle. We know the reflected energy is
not dissipated in the load resistor, by definition.

The only other device in the entire system capable
of dissipation is the source resistor. Since the reflected
energy is not dissipated in the load resistor and you say
it is not dissipated in the source resistor, it would
necessarily have to magically escape the system or build
up to infinity (but it doesn't). You keep digging your
hole deeper and deeper.

How many joules are there in 100 watts of
instantaneous power?

Obviously. It depends on how long you let the
100 W of instantaneous power flow. Integrate and
the answer shall be yours.

I'm not the one making the assertions. How many joules
of energy exist in *YOUR* instantaneous power calculations?
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Keith Dysart wrote:
If care is taken with the selection of modulation
frequencies with regards to the carrier, then nodes
can be created on the transmission line and neither
the carrier nor the modulation will cross such a
node.

Please prove your assertion on the bench. Until you
do, there is little left to discuss.

Please produce an example of
a real world transmission line that would support your
100% reflection. Hint: what would be the Z02
characteristic impedance in the reflection
coefficient equation, (50-Z02)/(50+Z02) = 1.0 ???

That is just too easy...
(50-Z02)/(50+Z02) = 1.0
50 - Z02 = 50 + Z02
-2 * Z02 = 0
Z02 = 0
That wasn't so hard, was it?

Now build one. Be sure to verify that you can transfer
energy from end to end. Until you do, there is little
left to discuss.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Cecil Moore wrote:
Keith Dysart wrote:
If care is taken with the selection of modulation
frequencies with regards to the carrier, then nodes
can be created on the transmission line and neither
the carrier nor the modulation will cross such a
node.

Please prove your assertion on the bench. Until you
do, there is little left to discuss.

And if you do, the distributed network model will
have to be overhauled.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Fri, 11 Apr 2008 13:25:40 GMT
Cecil Moore wrote:

Keith Dysart wrote:
The computation using energy instead of power has
also been done (and published here) and found also
to demonstrate that the reflected is not dissipated
in the source resistor.

Well, that certainly violates the conservation of
energy principle. We know the reflected energy is
not dissipated in the load resistor, by definition.

The only other device in the entire system capable
of dissipation is the source resistor. Since the reflected
energy is not dissipated in the load resistor and you say
it is not dissipated in the source resistor, it would
necessarily have to magically escape the system or build
up to infinity (but it doesn't). You keep digging your
hole deeper and deeper.


You write "The only other device in the entire system capable
of dissipation is the source resistor." which is a correct statement. Unfortunately, the circuit is intended to illustrate the absence of interference under special circumstances but an instant analysis shows that all the power can not be accounted for. We can only conclude that interference is present. Not good because the circuit was intended to illustrate a case of NO interference.

Our choice of a voltage source is incomplete because we did not assign it a mechanism to provide a reactive voltage, allowing the source to only apply a sinsoidal voltage without specifying the current or current timing. As a result, reflected power will return to the source resulting in an apparent loss of power to the system and resistor Rs. It is not a magical loss of power, only the result of interference acting within the cycle.

The circuit is very useful to investigate interference more carefully because on the AVERAGE, the interference IS zero. Using spreadsheets, we can see how the interference both adds and subtracts from the instantaneous applied voltage, resulting in cycling variations in the power applied to the resistor and other circuit elements. A very instructive exercise.

--
73, Roger, W7WKB

The Rest of the Story
 

Roger Sparks wrote:
You write "The only other device in the entire system capable
of dissipation is the source resistor." which is a correct statement.

Therefore, all power dissipated in the circuit must be dissipated
in the load resistor and the source resistor because there is
nowhere else for it to go. Since the reflected power is not
dissipated in the load, by definition, it has to be dissipated
in the source resistor but not at the exact time of its arrival.
There is nothing wrong with delaying power dissipation for 90
degrees of the cycle. In Parts 2 and 3 of my articles, I will show
how the source decreases it power output to compensate for destructive
interference and increases it power output to compensate for
constructive interference.

Unfortunately, the circuit is intended to illustrate the absence of

[AVERAGE] interference under special circumstances but an instant analysis shows

that all the power can not be accounted for.

Not surprising since there is no conservation of power principle.

We can only conclude that

[instantaneous] interference is present. Not good because the circuit was intended to

illustrate a case of NO [AVERAGE] interference.

I took the liberty of adding adjectives in brackets[*] to your
above statements. It doesn't matter about the instantaneous values
of power since not only do they not have to be conserved, but they
are also "of limited usefulness", according to Eugene Hecht, since
the actual energy content of instantaneous power is undefined even
when the instantaneous power is defined.

The circuit is very useful to investigate interference more carefully because on the AVERAGE,

the interference IS zero. Using spreadsheets, we can see how the
interference both adds and

subtracts from the instantaneous applied voltage, resulting in cycling
variations in the power

applied to the resistor and other circuit elements. A very instructive
exercise.

Instructive as long as we remember that a conservation of power
principle doesn't exist and therefore, equations based on instantaneous
powers do not have to balance. The joules, not the watts, are what must
balance.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Cecil Moore wrote:
Roger Sparks wrote:
You write "The only other device in the entire system capable
of dissipation is the source resistor." which is a correct statement.

Therefore, all power dissipated in the circuit must be dissipated
in the load resistor and the source resistor because there is
nowhere else for it to go. Since the reflected power is not
dissipated in the load, by definition, it has to be dissipated
in the source resistor but not at the exact time of its arrival.
There is nothing wrong with delaying power dissipation for 90
degrees of the cycle. In Parts 2 and 3 of my articles, I will show
how the source decreases it power output to compensate for destructive
interference and increases it power output to compensate for
constructive interference.

Unfortunately, the circuit is intended to illustrate the absence of

[AVERAGE] interference under special circumstances but an instant
analysis shows

that all the power can not be accounted for.

Not surprising since there is no conservation of power principle.

The concept of a wave is energy located at a predicted place after some
time period. That is a concept of conservation of power.

We can only conclude that

[instantaneous] interference is present. Not good because the circuit
was intended to

illustrate a case of NO [AVERAGE] interference.

I took the liberty of adding adjectives in brackets[*] to your
above statements. It doesn't matter about the instantaneous values
of power since not only do they not have to be conserved, but they
are also "of limited usefulness", according to Eugene Hecht, since
the actual energy content of instantaneous power is undefined even
when the instantaneous power is defined.

The circuit is very useful to investigate interference more carefully
because on the AVERAGE,

the interference IS zero. Using spreadsheets, we can see how the
interference both adds and

subtracts from the instantaneous applied voltage, resulting in cycling
variations in the power

applied to the resistor and other circuit elements. A very instructive
exercise.

Instructive as long as we remember that a conservation of power
principle doesn't exist and therefore, equations based on instantaneous
powers do not have to balance. The joules, not the watts, are what must
balance.

Forget the conservation of power at your own peril, because we need to
depend upon the predictability of waves of energy acting over time to
solve these problems. When the instantaneous powers do not balance, we
know that we do not yet have the complete solution or complete circuit.


73, Roger, W7WKB

The Rest of the Story
 

On Apr 11, 9:25*am, Cecil Moore wrote:
Keith Dysart wrote:
The computation using energy instead of power has
also been done (and published here) and found also
to demonstrate that the reflected is not dissipated
in the source resistor.

Well, that certainly violates the conservation of
energy principle. We know the reflected energy is
not dissipated in the load resistor, by definition.

The only other device in the entire system capable
of dissipation is the source resistor. Since the reflected
energy is not dissipated in the load resistor and you say
it is not dissipated in the source resistor, it would
necessarily have to magically escape the system or build
up to infinity (but it doesn't).

You seem to have forgotten that a voltage source can
absorb energy. This happens when the current flows
into it rather than out.

Recall the equation
Ps(t) = Prs(t) + Pg(t)

When the voltage source voltage is greatr than the
voltage at the terminals of the line (Vg(t)), energy
flows from the source into the resistor and the line.
When the voltage at the line terminals is greater
than the voltage source voltage, energy flows from
the line into the resistor and the voltage source.

At all times
Ps(t) = Prs(t) + Pg(t)
holds true.

Conservation of energy at work. No lost energy.

gartuitous comment snipped

How many joules are there in 100 watts of
instantaneous power?

Obviously. It depends on how long you let the
100 W of instantaneous power flow. Integrate and
the answer shall be yours.

I'm not the one making the assertions. How many joules
of energy exist in *YOUR* instantaneous power calculations?

We have been down that path; the spreadsheet has been
published. The flows of energy described by
Ps(t) = Prs(t) + Pg(t)
always balance.

The integration of these energy flows over any interval
also balance.

Energy is conserved. The world is as it should be.

...Keith

The Rest of the Story
 

On Apr 11, 9:32*am, Cecil Moore wrote:
Keith Dysart wrote:
If care is taken with the selection of modulation
frequencies with regards to the carrier, then nodes
can be created on the transmission line and neither
the carrier nor the modulation will cross such a
node.

Please prove your assertion on the bench. Until you
do, there is little left to discuss.

Speculating that this is your way of asking for an
explantion...
With a shorted line, nodes occur every 90 degrees. These
90 degree nodal points are at different places on
the line for different frequencies because the
wavelengths are different. With the nodes at
different physical locations for the different
frequencies, when you sum the responses for all of
the frequencies there are no longer well defined
nodes.

If the frequencies are rational numbers, then
nodes for the total response will exist at the lowest
common multiples of the wavelengths, but it will
typically take a very long line to find one.

...Keith

The Rest of the Story
 

On Apr 11, 3:30*pm, Cecil Moore wrote:
Roger Sparks wrote:
You write "The only other device in the entire system capable
of dissipation is the source resistor." which is a correct statement.

Therefore, all power dissipated in the circuit must be dissipated
in the load resistor and the source resistor because there is
nowhere else for it to go.

Please do not forget the source. It can absorb energy.

Since the reflected power is not
dissipated in the load, by definition, it has to be dissipated
in the source resistor but not at the exact time of its arrival.
There is nothing wrong with delaying power dissipation for 90
degrees of the cycle.

If you can't identify where the energy is stored for those 90
degrees you do not have a complete story. Or you are violating
conservation of energy and therefore have no story what-so-ever.

In Parts 2 and 3 of my articles, I will show
how the source decreases it power output to compensate for destructive
interference and increases it power output to compensate for
constructive interference.

Unfortunately, the circuit is intended to illustrate the absence of
[AVERAGE] interference under special circumstances but an instant analysis shows
that all the power can not be accounted for. *

Not surprising since there is no conservation of power principle.

Conservation of energy means that energy flows must be conserved.
Therefore, conservation of power.

We can only conclude that
[instantaneous] interference is present. Not good because the circuit was intended to
illustrate a case of NO [AVERAGE] interference.

I took the liberty of adding adjectives in brackets[*] to your
above statements. It doesn't matter about the instantaneous values
of power since not only do they not have to be conserved, but they
are also "of limited usefulness", according to Eugene Hecht, since
the actual energy content of instantaneous power is undefined even
when the instantaneous power is defined.

Are you sure that is why Hecht wrote what he did? He would, in all
likelihood, have an apoplexy if he knew how his words were being used.

The circuit is very useful to investigate interference more carefully because on the AVERAGE,

the interference IS zero. *Using spreadsheets, we can see how the
interference both adds and

subtracts from the instantaneous applied voltage, resulting in cycling
variations in the power

applied to the resistor and other circuit elements. *A very instructive
exercise.

Instructive as long as we remember that a conservation of power
principle doesn't exist and therefore, equations based on instantaneous
powers do not have to balance. The joules, not the watts, are what must
balance.

Since the total energies in your equations do not balance either,
there is still a problem with your hypothesis.

It would be helpful, however, if you could actually demonstrate a
system where the energies balance, but the flows do not. This would
settle the matter once and for all. (You won't find one, since
balanced flows are a consequence of conservation of energy).

...Keith

The Rest of the Story
 

Keith Dysart wrote:
It would be helpful, however, if you could actually demonstrate a
system where the energies balance, but the flows do not.

That's obviously easy to demonstrate in a distributed
network system. We can have energy flowing into both
ends of a loading coil at the same time and 180 degrees
later, energy flowing out of both ends at the same time.
The energies balance but the flows are completely
unbalanced and indeed defy the lumped circuit model.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Apr 12, 3:39*pm, Cecil Moore wrote:
Keith Dysart wrote:
It would be helpful, however, if you could actually demonstrate a
system where the energies balance, but the flows do not.

That's obviously easy to demonstrate in a distributed
network system. We can have energy flowing into both
ends of a loading coil at the same time and 180 degrees
later, energy flowing out of both ends at the same time.
The energies balance but the flows are completely
unbalanced and indeed defy the lumped circuit model.

You are not quite looking at the system correctly.

It is a system with two ports (bottom and top) where
energy can enter or leave, and one element (coil)
which can store energy.

The energy that flows in the bottom either flows out
the top or increases the energy stored in the coil.
The energy flowing into the bottom is equal to
the sum of the energy flowing out the top plus the
increase in the energy stored in the coil.
Expressed arithmetically
Pbottom(t) = Pcoil(t) + Ptop(t)

For the specific situation you describe above:
"energy flowing out of both ends at the same time"
means that the energy stored in the ooil is being
reduced to supply the energy leaving the top and
the bottom. The sum of the energy flows out of
the top and the bottom is exactly equal to the rate
at which the stored energy is being reduced.

Lumped or not lumped is moot.
The same analysis can be applied to a transmission
line. The energy flow into the left is exactly
equal to the energy flow out on the right plus
the rate of increase in the energy stored in the
line.

Energy flows (aka power) do indeed balance, though
you certainly have to correctly pick the flows that
should balance.

...Keith

The Rest of the Story
 

Keith Dysart wrote:
For the specific situation you describe above:
"energy flowing out of both ends at the same time"
means that the energy stored in the ooil is being
reduced to supply the energy leaving the top and
the bottom. The sum of the energy flows out of
the top and the bottom is exactly equal to the rate
at which the stored energy is being reduced.

Yes, the energy obviously balances but the instantaneous
powers are in opposite directions and therefore cannot
balance.

Lumped or not lumped is moot.

Energy cannot flow out of both ends of a lumped circuit
inductor. The current is, by definition, exactly the
same at both ends as it is for the lumped inductors in
EZNEC. You might find these class notes informative.

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

The Rest of the Story
 

On Apr 13, 9:45*am, Cecil Moore wrote:
Keith Dysart wrote:
For the specific situation you describe above:
"energy flowing out of both ends at the same time"
means that the energy stored in the ooil is being
reduced to supply the energy leaving the top and
the bottom. The sum of the energy flows out of
the top and the bottom is exactly equal to the rate
at which the stored energy is being reduced.

Yes, the energy obviously balances but the instantaneous
powers are in opposite directions and therefore cannot
balance.

Lumped or not lumped is moot.

Energy cannot flow out of both ends of a lumped circuit
inductor. The current is, by definition, exactly the
same at both ends as it is for the lumped inductors in
EZNEC. You might find these class notes informative.

It is well known that if one builds the wrong model one
will get the wrong answer. You build the wrong model,
then claim that flows do not balance. Unbalanced flows
are the expected result from incomplete models.

Your imcompleteness is that you forgot to include the
energy flow into the electric and magnetic fields around
the coil. When one does not forget this flow, all of
the flows will balance at every instant.

...Keith

The Rest of the Story
 

Keith Dysart wrote:
Your imcompleteness is that you forgot to include the
energy flow into the electric and magnetic fields around
the coil. When one does not forget this flow, all of
the flows will balance at every instant.

Sorry, it may or may not be a coil. It is in a black box
whose contents are unknown. Including the energy flows
inside the black box is impossible. The instantaneous
power into the black box does not balance the instantaneous
power out of the black box.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Apr 13, 8:41*pm, Cecil Moore wrote:
Keith Dysart wrote:
Your imcompleteness is that you forgot to include the
energy flow into the electric and magnetic fields around
the coil. When one does not forget this flow, all of
the flows will balance at every instant.

Sorry, it may or may not be a coil. It is in a black box
whose contents are unknown. Including the energy flows
inside the black box is impossible. The instantaneous
power into the black box does not balance the instantaneous
power out of the black box.

Black boxes are an excellent way to set problems which help us
learn about the meaning of theories.

Conservation of energy and its corollary, conservation of power,
is used in a different way for analyzing black boxes than it
is when we analyzed the fully specified circuit in your Fig 1-1.

With the black box, knowing the power function on the two ports,
we can compute the energy flow into the storage elements within
the box. If the flow out of one port is not always exactly
balanced by the flow into the other, then we know that the black
box is storing some energy and therefore that it has some elements
which store energy. In a more typical situation, we do not have
a completely black box, but we know some of its elements. We can
use the balance of energy flows to help us decide if we have all
the elements. If some of the energy flow is unaccounted for, then
we have not yet found all the elements.

If the box is truly opague, then all we can say is that it has
some energy storage elements and that collectively, the flow
into these elements is described by
Pport1(t) - Pport2(t)

The situation is somewhat different in Fig 1-1. All the elements
of the system are completely specified in Fig 1-1 and we used
circuit theory to compute the energy flows. Not surprisingly, they
completely balanced:
Ps(t) = Prs(t) + Pg(t)
Associated with Fig 1-1, there is a secondary hypothesis that it
should be possible to account for another energy flow, the imputed
flow in the reflected wave on the line. The inability to account
for this flow, given the conservation of power corollary to the
conservation of energy law, is a very strong indicator that the
energy flow imputed to the reflected wave is not an actual energy
flow.

...Keith

The Rest of the Story
 

Keith Dysart wrote:
All the elements
of the system are completely specified in Fig 1-1 and we used
circuit theory to compute the energy flows. Not surprisingly, they
completely balanced:
Ps(t) = Prs(t) + Pg(t)

Yes, but that is only *NET* energy flow and says nothing
about component energy flow. Everything is already known
about net energy flow and there are no arguments about it
so you are wasting your time. Your equation above completely
ignores reflections which is the subject of the thread.

You object to me being satisfied with average energy flow
while you satisfy yourself with net energy flow. I don't see
one iota of conceptual difference between our two positions.

After hundreds of postings, all you have proved is that
Eugene Hecht was right when he said instantaneous powers
are "of limited utility", such that you cannot even tell
me how many joules there are in 100 watts of instantaneous
power when it is the quantity of those very joules that
are required to be conserved and not the 100 watts.

The limit in your quest for tracking instantaneous energy
is knowing the position and momentum of each individual
electron. Good luck on that one.

I am going to summarize the results of my Part 1 article
and be done with it.

In the special case presented in Part 1, there are only
two sources of power dissipation in the entire system,
the load resistor and the source resistor. None of the
reflected energy is dissipated in the load resistor
because the chosen special conditions prohibit reflections
from the source resistor. Therefore, all of the energy not
dissipated in the load resistor is dissipated in the source
resistor because there is no other source of dissipation
in the entire system. Only RL and Rs exist. Pr is not
dissipated in RL. Where is Pr dissipated? Even my ten year
old grandson can solve that problem and he's no future
rocket scientist.
--
73, Cecil http://www.w5dxp.com