Standing-Wave Current vs Traveling-Wave Current
 

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

Standing-Wave Current vs Traveling-Wave Current
 

On Tue, 01 Jan 2008 15:00:54 -0800, Roger wrote:

There seems to be some confusion as to the terms
....
Two things to notice about the Thévenin equivalent circuit:
1. It contains an ideal voltage source "IN SERIES" with a resistor.

Hi Roger,

One confusion would seem to originate in your reliance in resistors.

A resistor (R) is NOT the series (or for Norton the parallel) passive
device - it is an Impedance (Z).

This misapplication (R) was perpetuated by Edison when he battled
Westinghouse for funding of power generation projects. He would
invariably craft the resistor (R) into the equation for the bankers to
prove DC was more efficient that AC systems, when in fact the
engineering (Z) proves quite the contrary.

I suppose the bankers bought it (R) at the outset, but market
economics (Z) hammered Edison into the ground when the bookkeeping of
AC made their investors rich at the expense of DC.

This also raises issues of the confusion over conjugate matching and
Z0 matching where many correspondents here freely intermix the two's
characteristics as though they belong to one or the other (or both, or
neither).

73's
Richard Clark, KB7QHC

Standing-Wave Current vs Traveling-Wave Current
 

Correction:

Roy Lewallen wrote:
. . .
Reflection coefficients are complex numbers, so they can't properly be
described as "positive" or "negative" except in the special cases of +1
and -1. In all other cases, the can only be described by their magnitude
and angle, or real and imaginary component. . .

It's also reasonable to talk about positive or negative reflection
coefficients when you're restricting the possibilities to ones having an
angle of zero or 180 degrees. This would be the case, for example, when
dealing with lossless lines (purely real Z0) connected to other lossless
lines or to a purely resistive source or load.

Roy Lewallen, W7EL

Standing-Wave Current vs Traveling-Wave Current
 

On Jan 2, 1:41*am, Roger wrote:
Keith Dysart wrote:
On Jan 1, 6:00 pm, Roger wrote:
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

I don't disagree with anything I read there.

But you may not quite have the concept of impedance
correct.

The impedance of the Thevenin/Norton equivalent source
is not V/I but rather the slope of the line representing
the relationship of the voltage to the current.
When there is a source present, this
line does not pass through the origin. Only for
passive components does this line pass through
the origin in which case it becomes V/I.

Because the voltage to current plot for an ideal
voltage source is horizontal, the slope is 0 and hence
so is the impedance. For an ideal current source
the slope is vertical and the impedance is infinite.

Hoping this helps clarify....

...Keith

Yes, I can understand that there is no change in voltage no matter what
the current load is, so there can be no resistance or reactive component
in the source. *The ideal voltage link also said that the ideal source
could maintain voltage no matter what current was applied. *Presumably a
negative current through the source would result in the same voltage as
a positive current, which is logical if the source has zero impedance.

This is true.

During the transition from negative current passing through the source
to positive current passing through the source, the current at some time
must be zero. *

This would occur, for example, when the output terminals
are open circuited. Or connected to something that had
the same open-circuit voltage as your source.

How is the impedance of the perfect source defined at
this zero current point? *

This is not a different question than: How is the
resistance of a resistor defined when it has no
current flowing in it?

It continues to satisfy the relation V = I * R,
though one can not compute R from the measured
V and I.

How is the impedance of the attached system
defined?

It is unknown, but has a voltage equal to the
voltage of the source.

The resistor in the Thevenin/Norton equivalent source is selected with
some criteria in mind. *What I would like to do is to design a Thevenin
* source to provide 1v across a 50 ohm transmission line INFINATELY
long, ignoring ohmic resistance. I would like the source resistor to
absorb as much power as the line, so that if power ever returns under
reflected wave conditions, it can all be absorbed by the resistor. *I
think the size of such a resistor will be 50 ohms.

That is the correct value to not have any reflections
at the source.

But do not expect the power dissipated in the resistor
to increase by the same amount as the "reflected power".
In general, it will not. This is what calls into question
whether the reflected wave actually contains energy.

Do some simple examples with step functions. The math
is simpler than with sinusoids and the results do not
depend on the phase of the returning wave, but simply
on when the reflected step arrives bach at the source.

Examine the system with the following terminations on
the line: open, shorted, impedance greater than Z0,
and impedance less than Z0.

Because excitation with a step function settles to
the DC values, the final steady state condition is
easy to compute. Just ignore the transmission line
and assume the termination is connected directly
to the Thevenin generator. When the line is present,
it takes longer to settle, but the final state will
be the same with the line having a constant voltage
equal to the voltage output of the generator which
will be the same as the voltage applied to the load.

Then do the same again, but use a Norton source. You
will find that conditions which increase the dissipation
in the resistor of the Thevenin equivalent circuit
reduce the dissipation in the resistor of the Norton
equivalent circuit and vice versa.

This again calls into question the concept of power
in a reflected wave, since there is no accounting
for where that "power" goes.

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

On Jan 2, 2:59*am, Richard Clark wrote:
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?

As stated, a step.

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.

The generality of the specified voltage (V) would seem
to adequately cover both the Thevenin and Norton generators.

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.

Given your previous writings, I suspect that you have
a solid understanding of the behaviour of an open-circuited
transmission line excited with a step function.

Perhaps you could make an attempt at writing a clear
description of the behaviour of such a system in terms
of charge flow and storage. Since "wave" is a word
overloaded with meanings, it would be good not to use
it in the description.

Once a clear description exists, I can extend it
using the same clear terminology to illustrate
the points of interest.

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
It happens without effort because the output impedance
of the Norton/Thevenin equivalent circuit is the same
as the circuit it replaces, otherwise it is not an
equivalent.

False. There can be a large difference in the output
impedance of an amplifier designed to drive a 50 ohm
load and a 50 ohm Thevenin equivalent circuit.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

On Jan 1, 9:03*pm, Cecil Moore wrote:
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.

I am not sure you have the methodology quite correct.
The source is not turned off; its output is set to 0.

It does what every ideal voltage source will do when
set to a voltage; maintain that voltage. Through all
of this, the impedance of the ideal source remains 0.

Now it turns out that an ideal voltage source set
to zero volts can be replaced by a short which also
has an impedance of 0 and produces no volts. But this
does not alter that the ideal source always has an
impedance of 0.

Analogously, an ideal current source always has an
infinite impedance. When set to 0 amps, it behaves
exactly like an open circuit.

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.

You should really stop repeating this to yourself. No
one is attempting to violate the principle of conservation
of energy.

By continually repeating this mantra, you convince
yourself that you do not need to examine the claims
of those who disagree with you. So you do not
examine and understand their claims. This seriously
limits your capability to learn.

If you truly wish to demolish the claims, you should
study them in great detail, then write an even better
and more persuasive description of the claim than did
the original author. Then identify and point out the
flaws.

As it stands, you do not examine the claims, but
immediately coat them with the tar of "violates
conservation of energy" or some other mantra and
walk away.

It does not lead to learning.

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

Gene Fuller wrote:
For future reference, however, just remember: Fields first, then power
or energy. That's the way superposition really works.

Way back before optical physicists could measure light
wave fields, they were dealing with reflectance,
transmittance, and irradiance - all involving power
or energy. They are still using those concepts today
proven valid over the past centuries. Optical physicists
calculate the fields *AFTER* measuring the power density
and they get correct consistent answers.

Use whatever method works for you but don't try to change
or replace the body of the laws of physics that was in place
before your grandfather was born. Your rejection of those
laws of physics from the past centuries is why you are so
confused today by your steady-state short cuts. It's why
Keith doesn't recognize a 1.0 reflection coefficient when
it is staring him in the face. It's why Roy rejects energy
in reflected waves. Optical physicists have known for
centuries where the energy goes. That RF engineers are
incapable of performing an energy analysis is sad.

Irradiance (intensity) is a power density. Many problems
in physics can be solved without even knowing or caring
about the strength of the fields. Here is one such lossless
line problem for you.

100w--50 ohm line--+--1/2WL 300 ohm line--50 ohm load
Pfor1--|--Pfor2
Pref1--|--Pref2

Without using fields, voltages, or currents: Calculate
the magnitudes of the four P terms above. Using the RF
power reflection-transmission coefficients, please explain
the magnitude of Pref1. If you cannot do that, you really
need to broaden your horizons and alleviate your ignorance.

Quoting HP AN 95-1: "The previous four equations show that
s-parameters are simply related to power gain and mismatch
loss, quantities which are often of more interest than the
corresponding voltage functions."
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

On Jan 1, 4:16*pm, Cecil Moore wrote:
Keith Dysart wrote:
The Norton or Thévenin equivalent circuits seem *capable of positive
reflection coefficients. *

Either can be positive, negative, or zero depending
on the value of the output impedance compared to Z0.

Would you please quote a reference that addresses
the subject of reflection coefficients from Thevenin
or Norton equivalent sources?

Seshardi, "Fundamentals of transmission lines and
electoromagnetic fields", page 19-21, explains
reflection diagrams and the generator is styled
after Thevenin. Page 21 says

RHOg = (Rg-R0)/(Rg+R0)

RHOg is the reflection coefficient at the generator
and Rg is the value of the output resistor. The
exposition has not yet generalized to Z so is still
in terms of R.

But do not feel limited to Seshardi. Any decent
book on transmission lines will cover the subject.

Or even easier, google ''"lattice diagram" reflection',
"reflection diagram" or "bounce diagram". You will
find many examples using Thevenin sources.

But this same information has been repeatedly provided
and ignored. Will this time be different?

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

On Jan 2, 8:39*am, Cecil Moore wrote:
Keith Dysart wrote:
It happens without effort because the output impedance
of the Norton/Thevenin equivalent circuit is the same
as the circuit it replaces, otherwise it is not an
equivalent.

False. There can be a large difference in the output
impedance of an amplifier designed to drive a 50 ohm
load and a 50 ohm Thevenin equivalent circuit.

Then your Thevenin circuit is not an equivalent
for the amplifier, is it?

Please study "equivalent circuit" and report back.

...Keith