On Tue, 1 Jan 2008 18:44:17 -0800 (PST), Keith Dysart
wrote:

Since the experiment ends before
the reflection returns to the generator,

Is it a step, or is it a pulse?

Makes a huge difference in the analysis.

the impedance
of the generator is irrelevant.

The impedance perhaps, but not the voltage. I specifically offered a
difference of sources (Norton vs. Thevenin). The impedance is the
same either way, the voltage is not. As you introduce power later in
this response, power at the source is even more contentious. As such,
we can drop this altogether.

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.

Still doesn't make sense, and I wouldn't attribute it to precision
seeing that you've written the same thing twice.

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

Yes, if you are in a left-right progression, further is more to the
right. There is the ambiguity of "end effect."

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.

That is always so. You haven't offered anything differentiable from
always, where previously (and later in your response, here) you
suggested current stopped flowing even when voltage was doubling.

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,

What new reference scheme?

charge to the left
of the leftward propagating step continues to flow,
while charge to the right has stopped flowing.
If precision was ever called for, now is the time.

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.

You don't explain where the extra voltage is. There is a double
voltage to be sure, but you have not exactly drawn a cause-and-effect
relationship.

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.

Sounds more like a problem for the challenged (i.e. there is no
difference between 2*V and V+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?

Does it matter? It is another step. That should be obvious for the
sake of superposition.

"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.

Not particularly, since we now have three things moving:
1. Step;
2. Tread;
3. Leading Edge.
It would seem to me that both 1 and 2 have a 3; so what are you trying
to say with the novel introduction of two more terms?

And this is the major weakness in the model.

Which model? *

The "no interaction" model.

And which model is the "no interaction" model? The former, or the
latter?

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"?

So now we have four things moving:
1. Step;
2. Tread;
3. Leading Edge.
4. Riser
It would seem to me that both 1,2 and 4 have a 3; so what are you
trying to say with the novel introduction of three more terms?

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.

So now the leftward propagating step is
1. Step;
2. Tread;
3. Leading Edge;
4. Riser?

If I simply discard the last three invented terms; went out on the
thin limb of interpretation; then, in my mind's eye I would see that,
yes, no current is flowing to the physical right of the transient (the
only thing that is moving - as evidenced through voltage).

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.

Makes quite a difference. Perhaps not in the math, but certainly in
the concept.

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.

Actually no. You describe a resistor in a series loop; I describe a
resistor in parallel. Consider the repetition of your last statement
above:
positives connected using the resistor.
Positives connected through the resistor (the sense of "using" the
resistor as connector)?
Or positives connected, then using the resistor (where it is also
connected, but unstated as such) to the negatives?

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.

Time has not been quantified, whereas voltage, hence charge, hence
energy has. Why introduce a new topic without enumerating it, when
its inclusion adds nothing anyway? You don't develop anything that
explains the step in terms of power. You don't use power anywhere. In
fact we then step into the philosophy of does a line actually move
power, or energy? That debate would cloud any issue of a step's
migration, reflection, or any of a spectrum of characteristics that
power has no sway over.

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.

Please note you start your sentence with a "But." Sentences (much
less paragraphs) are not started with coordinating conjunctions. When
you use "but," the logical implication is that you are coordinating:
two-waves with an equal ranking and unexpressed: one-wave.

One may also use "but" at the beginning of a sentence as a logical
connector between equal ideas (a transitional adverb); however, you
don't have anything in the preceding, original paragraph other than a
single wave.

I do not see anything distinctive about two waves over one wave until
your recent injection of power (not at all in the original), and to no
effect except to raise the objection (rather circular and unnecessary
inclusions to abandon the discussion of reflection in rhetorical
limbo). Apparently your transition spans many postings to a festering
point by Cecil (this is, of course, another one of my
interpretations).

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.

Neither view works.

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

Is it? *When?

Always.

You seem to be in self-contradiction when you describe voltage doubles
without current flow.

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?

A point well made in the scheme of precise language. Yet and all,
taken singly (one/either conductor) or doubly (a transmission line
section); then the identical statement remains true with both
interpretations for the step condition.

Of course, a lot may be riding on whether you are speaking of a step
function, or a pulse. Seeing the original was explicit to step, and
never introduces pulse; then there is no current flow as I responded:
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.

How do you introduce recoil or maintain momentum without energy? Odd.

Please remove my need to perform interpretation and give me a more
succinct accounting. Further, limit this to one scenario (this last
failed example is certainly dead in the water as far as collisions
go).

73's
Richard Clark, KB7QHC