Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
Could you better describe how you determine that the source has a Z0
equal to the line Z0? I can guess that you use a Thévenin equivalent
circuit and set the series resistor to Z0.

This will do it.

We have a black box. We use a 12vdc battery and a current
meter to measure the impedance of the black box. The
current meter reads zero when we connect the 12vdc battery
to the black box terminals. What is the impedance inside
the black box since test V/I = infinity?

What is actually in the black box is a very low source
impedance battery in series with a 50 ohm resistor.

This is what you are up against when reflections arrive
at your source. The ideal voltage source does *NOT* exhibit
zero ohms. Gary Coffman once likened it to spitting down
a fire hose.

If the reflections are in phase with the source voltage,
the source exhibits infinite ohms to the reflections and
of course, 100% re-reflection results. The reflection
coefficient encountered by the reflected waves is variable
and depends upon the relative phase between the source and
the reflections. The series resistor has very little effect
on the reflection coefficient.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Roy Lewallen wrote:
A perfect voltage source has a zero impedance, so if it's connected to a
transmission line with no series resistance, it presents a reflection
coefficient of -1.

But that is only when the source voltage is zero. What gives
you the right to assert such a thing when the source is turned
on? Please quote references and be specific.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Roy Lewallen wrote:
Roger wrote:

Could you better describe how you determine that the source has a Z0
equal to the line Z0? I can guess that you use a Thévenin equivalent
circuit and set the series resistor to Z0.

Probably the simplest way is to put the entire source circuitry into a
black box. Measure the terminal voltage with the box terminals open
circuited, and the current with the terminals short circuited. The ratio
of these is the source impedance. If you replace the box with a Thevenin
or Norton equivalent, this will be the value of the equivalent circuit's
impedance component (a resistor for most of our examples).

If the driving circuitry consists of a perfect voltage source in series
with a resistance, the source Z will be the resistance; if it consists
of a perfect current source in parallel with a resistance, the source Z
will be the resistance. You can readily see that the open circuit V
divided by the short circuit I of these two simple circuits equals the
value of the resistance.

The power output of the Thévenin equivalent circuit follows the load.

Sorry, I don't understand this. Can you express it as an equation?


There seems to be some confusion as to the terms "Thévenin equivalent
circuit", "ideal voltage source", and how impedance follows these
sources.

Two sources we all have access to are these links:

Voltage source:
http://en.wikipedia.org/wiki/Voltage_source
Thévenin equivalent circuit:
http://en.wikipedia.org/wiki/Th%C3%A9venin%27s_theorem

Three important observations about the ideal voltage source:

1. The voltage is maintained, no matter what current flows through the
source. Presumably, this would also mean that the voltage would be
maintained if a negative current flowed through the source. Thus, if we
set the voltage to 1 volt, the voltage would remain 1v even if we
supplied an infinite amount of power to the source.

2. The idea voltage source can ADSORB an infinite amount of power while
maintaining voltage. In this capacity, it is like a variable resistor,
capable of changing resistance to maintain a design voltage, no matter
what current is supplied to it.

3. The idea voltage source has an infinite impedance at the design
voltage, not a zero impedance as many have suggested. Zero internal
resistance is assumed to reasonably allow the ideal voltage source to
supply or adsorb current without changing voltage internally. It is not
zero impedance with the result that voltage drops to zero if external
power FLOWS INTO the ideal source.

Current flows into the ideal voltage source when the applied voltage
exceeds the design voltage. At the point the current reverses, we have
voltage/zero current, which is infinite impedance.

Two things to notice about the Thévenin equivalent circuit:

1. It contains an ideal voltage source "IN SERIES" with a resistor.
This has important implications when externally supplied voltage exceeds
the design voltage. Any returning power would not only reverse the
current flow in the ideal voltage source, it would develop voltage
across the internal series resistor. The output Thevenin voltage would
be the design voltage plus the voltage developed in the resistor.

2. The impedance of the Thevenin equivalent source would be infinite at
the design voltage because a voltage will existed from the ideal source
but current does not flow. This is true no matter what the design
impedance is for the Thevenin source.

Please notice in the link about the Thevenin circuit, a reference to a
"Thévenin-equivalent resistance". This resistance uses the ideal
voltage source set to zero. This appears to be the circuit that entered
the discussion at some point, justifying a negative 1 reflection factor.


Therefore, when the load delivers power, the Thévenin equivalent
circuit adsorbs power. Right?

Certainly, any energy leaving the transmission line must enter the
circuitry to which it's connected. Is that what you mean?

Roy Lewallen, W7EL

If the rules found in the links are acceptable, I could agree that
energy be allowed to enter the connected circuitry. I think we will
find that the returning power from a reflected wave is either completely
reflected with no change in energy, or adds to the voltage, thus
increasing the current flow and power contained in the system.

73, Roger, W7WKB

Standing-Wave Current vs Traveling-Wave Current
 

Roger wrote:
If the rules found in the links are acceptable, I could agree that
energy be allowed to enter the connected circuitry. I think we will
find that the returning power from a reflected wave is either completely
reflected with no change in energy, or adds to the voltage, thus
increasing the current flow and power contained in the system.

I think you are on the right track, Roger. Another way of
saying it is: If the principles of superposition are used,
the superposition must necessarily be implemented. It appears
to me that the principles of superposition are being used but
the results of the ensuing necessary superposition are, for
some ulterior motive, being completely ignored.

When the superposition interference patterns between the source
waves and the reflected waves are taken into account, everything
becomes perfectly clear (including the deliberate obfuscations).
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
On Jan 1, 1:20 pm, Roger wrote:
Discussing forward and reflecting waves, when is stability reached.





Roy Lewallen wrote:
If "stability" means steady state, a transmission line with any
resistance at either end or both ends is less complicated to analyze
than the particularly difficult lossless case I used for my analysis
which never reaches a true steady state. The presence of resistance
allows the system to settle to steady state, and that process can easily
and quantifiably be shown. And in two special cases, the process from
turn-on to steady state is trivially simple -- If the line is terminated
with Z0 (technically, its conjugate, but the two are the same for a
lossless line since Z0 is purely resistive), steady state is reached
just as soon as the initial forward wave arrives at the far end of the
line. No reflections at all are present or needed for the analysis. The
second simple case is when the source impedance equals Z0, resulting in
a source reflection coefficient of zero. In that case, there is a single
reflection from the far end (assuming it's not also terminated with Z0),
but no re-reflection from the source, and steady state is reached as
soon as the first reflected wave arrives at the source.
Roy Lewallen, W7EL
Could you better describe how you determine that the source has a Z0
equal to the line Z0? I can guess that you use a Thévenin equivalent
circuit and set the series resistor to Z0.

This will do it. As will a Norton with the parallel
resistor set to Z0.

The power output of the Thévenin equivalent circuit follows the load.
Therefore, when the load delivers power, the Thévenin equivalent circuit
adsorbs power. Right?

This apparently simple question has a very complicated
answer that depends on what precisely is meant by
"load delivers power" and "circuit absorbs power".

If by "load delivers power", you mean the reflected
wave, then this may or may not (depending on the
phase), mean that energy is transfered into the
generator.

If you mean that the time averaged product of
the actual voltage and current at the generator
terminals show a transfer of energy into the
generator, then energy is indeed flowing into
the generator.

If by "circuit absorbs power", you mean that
there is an increase in the energy dissipated
in the generator, this can not be ascertained
without detailed knowledge of the internal
arrangement of the generator and also depends
the meaning of "load delivers power", discussed
above.

...Keith

I think we will find it simpler than that.

I posted some links to descriptions of ideal voltage sources. After
reading them, I am taking another look at how we define impedance from
an ideal voltage source in a separate posting. You may wish to read the
links, and hopefully, my posting.

Here are the links:

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

Thévenin equivalent circuit:
http://en.wikipedia.org/wiki/Th%C3%A9venin%27s_theorem

73, Roger, W7WKB

Standing-Wave Current vs Traveling-Wave Current
 

Roger wrote:
Roy Lewallen wrote:
Roger wrote:

The power output of the Thévenin equivalent circuit follows the load.

Sorry, I don't understand this. Can you express it as an equation?


There seems to be some confusion as to the terms "Thévenin equivalent
circuit", "ideal voltage source", and how impedance follows these
sources.

Two sources we all have access to are these links:

Voltage source:
http://en.wikipedia.org/wiki/Voltage_source
Thévenin equivalent circuit:
http://en.wikipedia.org/wiki/Th%C3%A9venin%27s_theorem

As far as I can see, I've used the terms entirely consistently with the
definitions and descriptions in the linked articles. If you can find any
instance in which I haven't, please point it out so I can correct it.

Three important observations about the ideal voltage source:

1. The voltage is maintained, no matter what current flows through the
source. Presumably, this would also mean that the voltage would be
maintained if a negative current flowed through the source.

Actually, the wikipedia article correctly states this explicitly.

Thus, if we
set the voltage to 1 volt, the voltage would remain 1v even if we
supplied an infinite amount of power to the source.

You have to be very careful with using infinite amounts of anything
because the underlying mathematics runs into problems. For example, look
up "impulses" in any circuits text, and you'll find situations where a
pulse has zero width and infinite height yet finite area. So let's just
say the voltage stays at 1 volt if you apply any finite amount of current.

2. The idea voltage source can ADSORB an infinite amount of power while
maintaining voltage.

Yes, that's correct.

In this capacity, it is like a variable resistor,
capable of changing resistance to maintain a design voltage, no matter
what current is supplied to it.

No, that's not correct, unless you're willing to work with negative
resistances. V/I is a negative number when you supply current to a
voltage source. So it doesn't in any way act like a resistance, changing
or not.

3. The idea voltage source has an infinite impedance at the design
voltage, not a zero impedance as many have suggested.

No, that's not correct.

Zero internal
resistance is assumed to reasonably allow the ideal voltage source to
supply or adsorb current without changing voltage internally.

Yes, that's correct.

It is not
zero impedance with the result that voltage drops to zero if external
power FLOWS INTO the ideal source.

Sorry, I don't understand that statement. The voltage never changes,
regardless of the current flowing in or out. Review the wikipedia
article or any circuits text.

Current flows into the ideal voltage source when the applied voltage
exceeds the design voltage.

You can't "apply" a voltage to an ideal voltage source, except through
an impedance. If you did, for example by connecting it to another
voltage source of different value, the conditions defined for the
voltage sources are contradictory and can't exist at the same time. If
you do apply a voltage through an impedance, the current is simply the
voltage across the impedance divided by the impedance.

At the point the current reverses, we have
voltage/zero current, which is infinite impedance.

It sounds like you're trying to calculate some sort of "instantaneous
impedance". This isn't a generally accepted concept, and you'll have to
develop a fairly complete set of rules and mathematics to cover it.

Two things to notice about the Thévenin equivalent circuit:

1. It contains an ideal voltage source "IN SERIES" with a resistor.
This has important implications when externally supplied voltage exceeds
the design voltage.

Not really. The current simply equals the voltage across the resistor
divided by the resistance, in accordance with Ohm's law.

Any returning power would not only reverse the
current flow in the ideal voltage source, it would develop voltage
across the internal series resistor. The output Thevenin voltage would
be the design voltage plus the voltage developed in the resistor.

Your problem here, and I suspect elsewhere, is that you've embraced the
idea that there are waves of flowing and reflecting power. Your
otherwise reasoned questions and arguments are a good illustration into
the traps this concept leads a person into. You're seeing that in order
to support this mistaken concept, you have to reject some very
fundamental principles of electrical circuit operation. So you end up
with only two choices:

1. Re-write a great deal of fundamental circuit theory and reject the
foundation which has consistently given demonstrably correct results for
over a century, or
2. Realize that the flowing-power concept is flawed and abandon it.

The choice is yours.

2. The impedance of the Thevenin equivalent source would be infinite at
the design voltage because a voltage will existed from the ideal source
but current does not flow. This is true no matter what the design
impedance is for the Thevenin source.

You're trying to define the impedance of a source as the ratio of its
voltage to the current being drawn from it. That's not the impedance of
the source, it's the impedance of the load. The two are not the same.
The impedance of the source is its open circuit voltage divided by its
short circuit current. Another way to determine the impedance of the
source is to apply various loads and see how much the voltage drops. Any
test you run will confirm that the ideal voltage source has a zero
resistance, as my postings, the wikipedia article, or any circuits text
tell you.

Please notice in the link about the Thevenin circuit, a reference to a
"Thévenin-equivalent resistance".

This is the resistance component of the Thevenin equivalent circuit.
It's equal to the resistance of the circuitry for which the Thevenin
equivalent is being substituted.

This resistance uses the ideal
voltage source set to zero.

Can you explain this? There is no requirement that the source be set to
zero in determining or using the Thevenin equivalent resistance.

This appears to be the circuit that entered
the discussion at some point, justifying a negative 1 reflection factor.

My analysis did not contain a Thevenin equivalent to any circuit. It
contained only an ideal voltage source.

Therefore, when the load delivers power, the Thévenin equivalent
circuit adsorbs power. Right?

Certainly, any energy leaving the transmission line must enter the
circuitry to which it's connected. Is that what you mean?

Roy Lewallen, W7EL

If the rules found in the links are acceptable, I could agree that
energy be allowed to enter the connected circuitry. I think we will
find that the returning power from a reflected wave is either completely
reflected with no change in energy, or adds to the voltage, thus
increasing the current flow and power contained in the system.

You're building a logical argument from a flawed premise. I'm afraid
you've only just begun to see the problems you'll be having as you
pursue this path.

Roy Lewallen, W7EL

Standing-Wave Current vs Traveling-Wave Current
 

"Cecil Moore" wrote in message
.net...
Roger wrote:
If the rules found in the links are acceptable, I could agree that
energy be allowed to enter the connected circuitry. I think we will
find that the returning power from a reflected wave is either
completely
reflected with no change in energy, or adds to the voltage, thus
increasing the current flow and power contained in the system.

I think you are on the right track, Roger. Another way of
saying it is: If the principles of superposition are used,
the superposition must necessarily be implemented. It appears
to me that the principles of superposition are being used but
the results of the ensuing necessary superposition are, for
some ulterior motive, being completely ignored.

When the superposition interference patterns between the source
waves and the reflected waves are taken into account, everything
becomes perfectly clear (including the deliberate obfuscations).
--
73, Cecil http://www.w5dxp.com

I have been down wrong tracks before! I sure hope this is the right
track. Thanks.

The principles of superposition are mathematically usable, not too hard,
and I think very revealing. Yes, if we use part of the model, we must
use it all the way. To do otherwise would be error, or worse.

You have been a supporter of this theory for a long time.

73, Roger, W7WKB

Standing-Wave Current vs Traveling-Wave Current
 

Roy Lewallen wrote:
Your problem here, and I suspect elsewhere, is that you've embraced the
idea that there are waves of flowing and reflecting power.

Roy, your problem is that you have embraced the idea that
reflected waves contain zero energy such that zero reflected
energy can be measured as zero reflected power passing a
measurement point. However, zero energy waves are impossible!!!

2. Realize that the flowing-power concept is flawed and abandon it.

Why don't you realize that reflected waves of zero energy
are impossible?

Can you explain this? There is no requirement that the source be set to
zero in determining or using the Thevenin equivalent resistance.

It is a requirement of the superposition principle that the
source voltage be set to zero for determining the voltage and
current due only to the reflected wave. I'm surprised you
are trying to sweep that fact under the rug.

My analysis did not contain a Thevenin equivalent to any circuit. It
contained only an ideal voltage source.

An ideal voltage source plus a series resistor *IS* a Thevenin
equivalent circuit. Nice try at obfuscation.

You're building a logical argument from a flawed premise.

Nobody's premise is more flawed than yours.

You are only performing 1/2 of the pre-superposition phase.
Everything can be explained if you actually go through with
the superposition of which you are so afraid.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Roger wrote:
The principles of superposition are mathematically usable, not too hard,
and I think very revealing. Yes, if we use part of the model, we must
use it all the way. To do otherwise would be error, or worse.

Roy and Keith don't seem to realize that the zero source
impedance for the ideal voltage source is only when the
source is turned off for purposes of superposition. They
conveniently avoid turning the source voltage on to complete
the other half of the superposition process. When the
source signal and the reflected wave are superposed at
the series source resistor, where the energy goes becomes
obvious. Total destructive interference in the source
results in total constructive interference toward the load.
See below.

You have been a supporter of this theory for a long time.

Yes, I have. I am a supporter of the principles and laws of
physics. Others believe they can violate the principle of
conservation of energy anytime they choose because the
principle of conservation of energy cannot be violated -
go figure.

Roger, I have explained all of this before. If you are
capable of understanding it, I take my hat off to you.
Here's what happens in that ideal voltage source using
the rules of superposition.

1. With the reflected voltage set to zero, the source
current is calculated which results in power in the
source resistor. Psource = Isource^2*Rsource

2. With the source voltage set to zero, the reflected
current is calculated which results in power in the
source resistor. Pref = Iref^2*Rsource

Now superpose the two events. The resultant power
equation is: Ptot = Ps + Pr + 2*SQRT(Ps*Pr)cos(A)
where 'A' is the angle between the source current
and the reflected current. Note this is NOT power
superposition. It is the common irradiance (power
density) equation from the field of optics which
has been in use in optical physics for a couple of
centuries so it has withstood the test of time.

For the voltage source looking into a 1/2WL open
circuit stub, angle 'A' is 180 degrees. Therefore
the total power in the source resistor is:

Ptot = Ps + Pr + 2*SQRT(Ps*Pr)cos(180) or

Ptot = Ps + Pr - 2*SQRT(Ps*Pr) = 0

Ptot in the source resistor is zero watts but we
already knew that. What is important is that the
last term in that equation above is known as the
"interference term". When it is negative, it indicates
destructive interference. When Ptot = 0, we have
"total destructive interference" as defined by
Hecht in "Optics", 4th edition, page 388.

Whatever the magnitude of the destructive interference
term, an equal magnitude of constructive interference
must occur in a different direction in order to
satisfy the conservation of energy principle. Thus
we can say with certainty that the energy in the
reflected wave has been 100% re-reflected by the
source. The actual reflection coefficient of the
source in the example is |1.0| even when the source
resistor equals the Z0 of the transmission line.
I have been explaining all of this for about 5 years
now. Instead of attempting to understand these relatively
simple laws of physics from the field of optics, Roy
ploinked me. Hopefully, you will understand.

All of this is explained in my three year old Worldradio
energy article on my web page.

http://www.w5dxp.com/energy.htm
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

On Jan 1, 3:59*pm, Richard Clark wrote:
On Tue, 1 Jan 2008 06:12:48 -0800 (PST), Keith Dysart

wrote:
To illustrate some of these weaknesses, consider an example
where a step function from a Z0 matched generator is applied
to a transmission line open at the far end.

Hi Keith,

It would seem we have either a Thevenin or a Norton source (again, the
ignored elephant in the living room of specifications). *This would
have us step back to a Z0 in series with 2V or a Z0 in parallel with V
- it seems this would be a significant detail in the migration of what
follows:

I do not think so. Regardless of the output impedance of the
generator, the step impresses some voltage V which begins
to propagate down the line. Since the experiment ends before
the reflection returns to the generator, the impedance
of the generator is irrelevant.

The step function eventually reaches the open end where
the current can no longer flow. The inductance insists
that the current continue until the capacitance at the
end of the line is charged to the voltage which will stop
the flow. This voltage is double the voltage of the step
function applied to the line (i.e 2*V).

Fine (with omissions of the fine grain set-up)

However, what follows is so over edited as to be insensible:Once the
infinitesimal capacitance at the end of the line is
charged,

energy has reached the "end of the line" so to speak; and yet:the current now has to stop just a bit earlier

TIME is backing up? *

Ah. The challenge of written precisely. Consider the
generator to be on the left end of the line and the open
to be at the right end. When the capacitance at the right
end charges to 2 * V, the current now has to stop a little
bit more to the left.

Are we at the edge of an event horizon?and this charges the inifinitesimal capacitance a bit
further from the end.

BEYOND the end of the line? *Just how long can this keep up?

Again, the poor writing meant a bit further to left
from the end.

Very strange stuff whose exclusion wouldn't impact the remainder:So a step in the voltage propagates
back along the line towards the source. In front of this
step, current is still flowing. Behind the step, the

behind the reflected step, rather?current is zero and voltage is 2*V.

Want to explain how you double the stored voltage in the distributed
capacitance of the line without current? *

The energy formerly present in the inductance of the line
has been transferred to the capacitance.

The definition of capacitance is explicitly found in the number of
electrons (charge or energy) on a surface; which in this case has not
changed.The charge that
is continuing to flow from the source is being used
to charge the distributed capacitance of the line.

It would appear now that charge is flowing again, but that there is a
confusion as to where the flow comes from. *

Using the new reference scheme, charge to the left
of the leftward propagating step continues to flow,
while charge to the right has stopped flowing.

Why would the source at
less voltage provide current to flow into a cap that is rising in
potential above it? *

The beauty of inductance. You get extra voltage when
you try to stop the flow.

Rolling electrons uphill would seem to be
remarkable.

Well yes, but they don't roll for long. And you don't
get more out than you put in. Pity. Or I could be quite rich.

Returning to uncontroversial stuff:The voltage that is propagating backwards along the
line has the value 2*V, but this can also be viewed as
a step of voltage V added to the already present voltage
V. The latter view is the one that aligns with the "no
interaction" model; the total voltage on the line is
the sum of the forward voltage V and the reverse
voltage V or 2*V.

If this is the "latter view" then the former one (heavily edited
above?) is troubling to say the least.

Challenges with the referent.
The former view is that a voltage of 2*V is propagating
back along the line. The latter view is that it is a
step of V above the already present V.

In this model, the step function has propagated to the
end, been reflected and is now propagating backwards.
Implicit in this description is that the step continues
to flow to the end of the line and be reflected as
the leading edge travels back to the source.

This is a difficult read. *You have two sentences. *Is the second
merely restating what was in the first, or describing a new condition
(the reflection)?

Agreed. What is a good word to describe the constant
voltage that follows the actual step change in
voltage? The "tread" perhaps?

"The tread continues to flow to the end of the line
and be reflected as the leading edge travels back to
the source."

I am not sure that that is any clearer.

And this is the major weakness in the model.

Which model? *

The "no interaction" model.

The latter? or the former? It claims
the step function is still flowing in the portion of
the line that has a voltage of 2*V and *zero* current.

Does a step function flow? *

Perhaps the "tread"? But then should the step change
be called the "riser"?

As for "zero" current, that never made
sense in context here.

Somewhat clarified, I hope. But for clarity, the
current to the right of the leftward propagating
step is zero.

Now without a doubt, when the voltages and currents
of the forward and reverse step function are summed,
the resulting totals are correct. *

In this thread, that would be unique.

Or a miracle?

But it seems to
me that this is just applying the techniques of
superposition. And when we do superposition on a
basic circuit, we get the correct totals for the
voltages and currents of the elements but we do
not assign any particular meaning to the partial
results.

Amen.

Unfortunately, more confusion:A trivial example is connecting to 10 volt batteries
in parallel through *a .001 ohm resistor.

Parallel has two outcomes, which one? *"Through" a resistor to WHERE?
In series? *In parallel? *

Much to ambiguous.

I know. Trying to conserve words leads to confusion.
Try: negative to negative, positives connected using
the resistor.


The partial
results show 10000 amps flowing in each direction
in the resistor with a total of 0.

This would suggest in parallel to the parallel batteries, but does not
resolve the bucking parallel or aiding parallel battery connection
possibilities. *The 0 assignment does not follow from the description,
mere as one of two possible solutions.

But I do not
think that anyone assigns significance to the 10000
amp intermediate result. Everyone does agree that
the actual current in the resistor is zero.

Actually, no. *Bucking would have 0 Amperes. *Aiding would have 20,000
Amperes.

However, by this forced march through the math, it appears there are
two batteries in parallel; (series) bucking; with a parallel resistor.

So in the end, successful communication of the schematic.

The "no interaction" model,

Is this the "latter" or former model?while just being
superposition, seems to lend itself to having
great significance applied to the intermediate
results.

Partially this may be due to poor definitions.

Certainly as I read it.If the
wave is defined as just being a voltage wave, then
all is well.

Still ambiguous.

And then deeper:But, for example, when looking at a solitary pulse,
it is easy (and accurate) to view the wave as having
more than just voltage. One can compute the charge,
the current, the power, and the energy.

It would seem if you knew the charge, you already know the energy; but
the power?

Just energy per unit time. We know the energy distribution
on the line, so we know the power at any point and time.

But when
two waves are simultaneously present, it is only
legal to superpose the voltage and the current.

And illegal if only one is present? *

No. Legal to also compute the power.

Odd distinction. *Is there some
other method like superposition that demands to be used for this
instance?But it is obvious that a solitary wave has voltage,
current, power, etc. But when two waves are present
it is not legal to.... etc., etc.
The "no interaction" model does not seem to resolve
this conflict well, and some are lead astray.

I was lost on a turn several miles back.

Perhaps try again, with the clarifications.

And it was this conflict that lead me to look for
other ways of thinking about the system.

I can only hope for clarity from this point on.

But given the history, disappointment will be no surprise.
Eh?

Earlier you asked for an experiment. How about this
one....

Take two step function generators, one at each end
of a transmission line. Start a step from each end
at the same time. When the steps collide in the
middle, the steps can be viewed as passing each
other without interaction, or reversing and
propagating back to their respective sources.

Why just that particular view?

Those seem to be the common alternatives.
If there are more, please share.

We
can measure the current at the middle of the line
and observe that it is always 0.

Is it? *When?

Always.

If, for some infinitesimal line section, there is no current through
it, then there is no potential difference across it.

Or did you mean along it?

Hence, the when is some infinitesimal time before the waves of equal
potential meet - and no current flow forever after.

Therefore the
charge that is filling the capacitance and causing
the voltage step which is propagating back towards
each generator

How did that happen? *No potential difference across an infinitesimal
line section, both sides at full potential (capacitors fully charged,
or charging at identical rates). *Potentials on either side of the
infinitesimal line section are equal to each other and to the sources,
hence no potential differences anywhere, *No potential differences, no
current flow, no charge change, no reflection, no more wave.

The last bit of induction went to filling the last capacitance element
with the last charge of current. *Last gasp. *No more gas. *Nothing
left. Finis.

must be coming from the generator
to which the step is propagatig because no charge
is crossing the middle of the line.

Do you like it?

Not particularly. *What does it demonstrate?

That they bounce rather than pass silently.

...Keith

73's
Richard Clark, KB7QHC

...Keith