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