Standing-Wave Current vs Traveling-Wave Current
 

On Mon, 31 Dec 2007 05:42:16 -0800, Roger wrote:

Several responders discussed the Norton or Thévenin equivalent circuits.
These circuits assume steady state conditions, and are intended to be
replacements for groups of components. A very useful concept, but not
appropriate for startup excitation of a transmission line unless the
intent was to investigate the source AND transmission line as a system.

Hi Roger,

Quite a bizarre stance. Do you regularly come to these points of view
without further thought, elaboration, or analysis? It seems you have
assigned some remarkable characteristics to the Norton and Thevenin
source and you don't offer us anything substantive beyond what would
be called "Arthur's proposition."

To cut to the chase, Norton and Thevenin sources are appropriate to
network analysis irrespective of your perception. You should look
closer at your issues of difficulty because it does not reside in
their classic application to classic situations.

More later. I as still thinking about the "perfect power source".

That's fine, but it is now turning into a mantra rather than a
productive examination. You would do well to spill your guts as to
why you commit so much of the calendar to this thought. Alone, it is
as meaningful as announcing that one is thinking about "cold fusion."

73's
Richard Clark, KB7QHC

Standing-Wave Current vs Traveling-Wave Current
 

Gene Fuller wrote:
Cecil Moore wrote:
Lots of energy is flowing in both directions.
Only the *NET* energy flow is zero.

I guess you still have not gone back to the books to try to understand
what electromagnetic energy is all about. A good review of the Poynting
theorem would help to minimize the sort of nonsense you spouted above.

You mean like Ramo & Whinnery in "Fields and Waves ..."
2nd edition, page 291? (begin quote)

In such problems we are often most interested in the ratio
of power in the reflected wave to that in the incident wave,
and this ratio is given by the square of the magnitude of
[rho], as can be shown by considering the Poynting vectors:

Pz-/Pz+ = |rho|^2

end quote.

Pz+ is the forward power Poynting vector. Pz- is the reflected
power Poynting vector. The *NET* Poynting vector is the
difference between those two Poynting vectors.

If Pz+ = Pz-, then the net Poynting vector is zero but
the component Poynting vectors still exist. If you disagree,
please take it up with Ramo & Whinnery.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Gene Fuller wrote:
It is interesting that you ridicule zero impedance ideal sources, and
then you provide an example with a circulator. Do you really think a
circulator is more ideal than a good voltage source?

If models alleviate ignorance, they are good.
If models promote ignorance, they are bad.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Richard Clark wrote:
To cut to the chase, Norton and Thevenin sources are appropriate to
network analysis irrespective of your perception.

All my references indicate that those sources are only
appropriate for *steady-state* use. Roger is searching
for a transient state source.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Cecil Moore wrote:
Roger wrote:
The problem is that "We want to investigate a 1/2 wave length of
transmission line, excited at one end. How soon is stability reached?"

I guess the answer depends upon your definition of
"stability" above.

You might start with a loaded version:

http://www.w5dxp.com/1secsgat.gif

Yes, this is the idea, exactly.

The loaded version is much more complicated than the unloaded version.

Stability is always reached provided power input contains a maximum.

73, Roger, W7WKB

Standing-Wave Current vs Traveling-Wave Current
 

Cecil Moore wrote:
Richard Clark wrote:
To cut to the chase, Norton and Thevenin sources are appropriate to
network analysis irrespective of your perception.

All my references indicate that those sources are only
appropriate for *steady-state* use. Roger is searching
for a transient state source.

Right!

73, Roger, W7WKB

Standing-Wave Current vs Traveling-Wave Current
 

Gene Fuller wrote:
Cecil Moore wrote:
Lots of energy is flowing in both directions.
Only the *NET* energy flow is zero.

I guess you still have not gone back to the books to try to understand
what electromagnetic energy is all about. A good review of the Poynting
theorem would help to minimize the sort of nonsense you spouted above.

Another example from HP's AN 95-1

b1 = s11*a1 + s12*a2 b2 = s21*a1 + s22*a2

Z01 Z02
-------------+-------------
Pfor1--|--Pfor2
Pref1--|--Pref2

|a1|^2 = Pfor1 |b2|^2 = Pfor2
|b1|^2 = Pref1 |a2|^2 = Pref2
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Roger wrote:
The loaded version is much more complicated than the unloaded version.

The problem is that with the unloaded version, an ideal
autotuner might have trouble achieving a 50 ohm match
to an infinite impedance. :-)
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Cecil Moore wrote:
Gene Fuller wrote:
Cecil Moore wrote:
Lots of energy is flowing in both directions.
Only the *NET* energy flow is zero.

I guess you still have not gone back to the books to try to understand
what electromagnetic energy is all about. A good review of the
Poynting theorem would help to minimize the sort of nonsense you
spouted above.

You mean like Ramo & Whinnery in "Fields and Waves ..."
2nd edition, page 291? (begin quote)

In such problems we are often most interested in the ratio
of power in the reflected wave to that in the incident wave,
and this ratio is given by the square of the magnitude of
[rho], as can be shown by considering the Poynting vectors:

Pz-/Pz+ = |rho|^2

end quote.

Pz+ is the forward power Poynting vector. Pz- is the reflected
power Poynting vector. The *NET* Poynting vector is the
difference between those two Poynting vectors.

If Pz+ = Pz-, then the net Poynting vector is zero but
the component Poynting vectors still exist. If you disagree,
please take it up with Ramo & Whinnery.

Cecil,

As usual, you have taken something generally accepted as true, and then
you have added your own special spin. You might have noticed that Ramo &
Whinnery did not go into all of the "net" baloney, and neither did HP in
AN-95. That stuff is only in your imagination.

The last I heard, energy is a scalar quantity. In the cases we are
considering, energy is only positive or zero, not negative. How do two
non-negative scalar quantities "flow" past each other, adding up to a
"net" of zero, unless the initial quantities themselves are zero?

I agree that the concepts included in R&W and in AN-95 are widely used.
However, they were never intended to be distorted into your dream world.

The bouncing traveling wave model is an extremely useful mathematical
device. It does not trump physical reality.

73,
Gene
W4SZ

Standing-Wave Current vs Traveling-Wave Current
 

Gene Fuller wrote:
As usual, you have taken something generally accepted as true, and then
you have added your own special spin. You might have noticed that Ramo &
Whinnery did not go into all of the "net" baloney, and neither did HP in
AN-95. That stuff is only in your imagination.

Translation: I don't have the balls to argue with
Ramo & Whinnery and HP over issues that are more than
obvious to any casual initiated observer.

The last I heard, energy is a scalar quantity. In the cases we are
considering, energy is only positive or zero, not negative.

As you well know, the convention is to apply a negative
sign to positive energy flowing in the opposite direction
from the "forward" energy which is arbitrarily assigned
a plus sign. When Ramo & Whinnery say, using Poynting
vectors, that Pz(net) = Pz+ - Pz-, they are not implying
negative power. They are just using the Poynting vector
convention of direction of energy flow in a transmission
line. We are saying the same thing when we say:

P(load) = P(forward) - P(reflected) = P(net)

Transmission lines have the advantage of having only two
directions so '+' can be assigned to one direction and
'-' assigned to the other. It is completely arbitrary -
the signs can be swapped and the results remain the same.

I agree that the concepts included in R&W and in AN-95 are widely used.
However, they were never intended to be distorted into your dream world.

See "Translation" above.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Cecil Moore wrote:


As you well know, the convention is to apply a negative
sign to positive energy flowing in the opposite direction
from the "forward" energy which is arbitrarily assigned
a plus sign.

Still using that vivid imagination, I see.

Let's see even one reference that mentions explicitly the concept of
applying a negative sign to positive energy. Not power, not voltage, not
current, not waves, but energy.



73,
Gene
W4SZ

Standing-Wave Current vs Traveling-Wave Current
 

Gene Fuller wrote:
Cecil Moore wrote:
As you well know, the convention is to apply a negative
sign to positive energy flowing in the opposite direction
from the "forward" energy which is arbitrarily assigned
a plus sign.

Let's see even one reference that mentions explicitly the concept of
applying a negative sign to positive energy. Not power, not voltage, not
current, not waves, but energy.

If you keep feigning ignorance like that Gene, you are
going to lose all respect. If the Poynting vector has
a negative sign, as used by Ramo & Whinnery, that sign
is an indication of the *direction of energy flow*,
see quote below.

From Ramo & Whinnery:

The Poynting vector is "the vector giving *direction* and
magnitude of *energy flow*". When Ramo & Whinnery hang a
sign on a Poynting vector in a transmission line, it is
an indication of the direction of energy flow.

For pure standing waves,
"The average [NET] value of Poynting vector is zero
at every cross-sectional plane; this emphasizes the
fact that on the average as much energy is carried
away by the reflected wave as is brought by the
incident wave."

What? Reflected waves "carrying" energy? Shame on
Ramo & Whinnery for contradicting the rraa gurus.

It is impossible to satisfy you, Gene. When I quote
reference after reference about reflected power, you
say power doesn't reflect. When I change it to reflected
energy, you ask for a reference.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

On Mon, 31 Dec 2007 08:11:49 -0800, Roger wrote:

Cecil Moore wrote:
Richard Clark wrote:
To cut to the chase, Norton and Thevenin sources are appropriate to
network analysis irrespective of your perception.

All my references indicate that those sources are only
appropriate for *steady-state* use. Roger is searching
for a transient state source.

Right!

Really, the wail and torment of your two's grief is no more
sophisticated than: the dog ate my homework. This is a noble quest in
search of a "free lunch."

In the end, it serves absolutely no purpose in trying to resolve the
"confusion" of why a traveling wave antenna has standing wave energy
where the purpose of this thread was strangled in the crib by Cecil
when I dropped that one. ;-)

73's
Richard Clark, KB7QHC

Standing-Wave Current vs Traveling-Wave Current
 

Richard Clark wrote:
Really, the wail and torment of your two's grief is no more
sophisticated than: the dog ate my homework.

Nope, it's more like the gurus ate reality.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Roger wrote:
Cecil Moore wrote:
Roger wrote:
The problem is that "We want to investigate a 1/2 wave length of
transmission line, excited at one end. How soon is stability reached?"

I guess the answer depends upon your definition of
"stability" above.

You might start with a loaded version:

http://www.w5dxp.com/1secsgat.gif

Yes, this is the idea, exactly.

The loaded version is much more complicated than the unloaded version.

Stability is always reached provided power input contains a maximum.

73, Roger, W7WKB

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

Standing-Wave Current vs Traveling-Wave Current
 

Roy Lewallen wrote:
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.

Assuming that reflection cannot exist without energy,
where does the ExH*t energy in the reflected wave go?

(OK now, everyone hold their breath for an answer.)
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

On Mon, 31 Dec 2007 14:13:43 -0600, Cecil Moore
wrote:

You've got a battery in one pocket, and a mouse in the other. Where
did the energy go?

(OK now, everyone hold their breath for an answer.)

Standing-Wave Current vs Traveling-Wave Current
 

Richard Clark wrote:
You've got a battery in one pocket, and a mouse in the other. Where
did the energy go?

Unfortunately for your argument, neither a mouse nor
a battery exists inside a transmission line. Outside
of usually minor heat losses, nothing exists besides
EM waves traveling at the speed of light.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Cecil, W5DXP wrote:
"In such problems we are often most interested in the ratio of power in
the reflected wave to that in the incident wave, and the ratio is given
by the square of the magnitude of [rho] as can be shown by considering
the Poynting vectors:"

Terman agrees. On page 98 of his 1955 opus, Terman shows the value of
[rho] vs. SWR.

On page 97 Terman writes:
"This definition of standing-wave ratio is sometimes called the voltage
standing-wave ratio (VSWR) to distinguish it from the standing-wave
ratio expressed as a power ratio which is (Emax/Emin)squared."

Best regards, Richard Harrison, KB5WZI

Standing-Wave Current vs Traveling-Wave Current
 

On Mon, 31 Dec 2007 14:30:55 -0600, Cecil Moore
wrote:

Richard Clark wrote:
You've got a battery in one pocket, and a mouse in the other. Where
did the energy go?

Unfortunately for your argument, neither a mouse nor
a battery exists inside a transmission line. Outside
of usually minor heat losses, nothing exists besides
EM waves traveling at the speed of light.

So, you refuse, or cannot account for the energy? Energy is conserved
in every system, even pockets and when the units of volts in one case
are conveniently offered, your theory fails to balance the pockets.

(OK now, everyone hold their breath for an answer.)

Standing-Wave Current vs Traveling-Wave Current
 

On Dec 31, 11:11*am, Roger wrote:
Cecil Moore wrote:
Richard Clark wrote:
To cut to the chase, Norton and Thevenin sources are appropriate to
network analysis irrespective of your perception.

All my references indicate that those sources are only
appropriate for *steady-state* use. Roger is searching
for a transient state source.

Right!

73, Roger, W7WKB

Odd. Cecil has not named his "references" which is
quite unusual for he truly likes to name-drop: Ramo,
Whinnery, Hecht, IEEE, ...

You would be well served to google "reflection diagram"
or "bounce diagram" where you will find fine examples
of computing re-reflection using the source impedance
of generators modelled using the Thevenin equivalent
circuit. It is not particularly complicated, though
it can be tedious.

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
On Dec 31, 11:11 am, Roger wrote:
Cecil Moore wrote:
Richard Clark wrote:
To cut to the chase, Norton and Thevenin sources are appropriate to
network analysis irrespective of your perception.
All my references indicate that those sources are only
appropriate for *steady-state* use. Roger is searching
for a transient state source.
Right!

73, Roger, W7WKB

Odd. Cecil has not named his "references" which is
quite unusual for he truly likes to name-drop: Ramo,
Whinnery, Hecht, IEEE, ...

You would be well served to google "reflection diagram"
or "bounce diagram" where you will find fine examples
of computing re-reflection using the source impedance
of generators modelled using the Thevenin equivalent
circuit. It is not particularly complicated, though
it can be tedious.

...Keith

One of Cecil's common techniques is to declare any combination of
perfect voltage source and resistance to be a "Thevenin equivalent",
which he then claims relieves him of any obligation to consider the
power supplied by the source or dissipated in the resistor.

Of course, that combination of components has no special properties or
restrictions, and must conform to the same rules as any linear
components. It is, in fact, an excellent choice for many examples and
illustrations because of its bare simplicity. Only when a substitution
is made for some other combination of linear components does the perfect
voltage source and resistor become an "equivalent", and in that case you
can correctly state that the power supplied by the perfect source and
dissipated by the resistor aren't necessarily the same as for the
circuit being replaced. But any analysis which isn't valid when driven
by a perfect voltage source in series with a resistance (or current
source in parallel with a resistance) is fundamentally flawed. Waving
your hands and declaring it a "Thevenin equivalent" and therefore not
subject to the rules all linear circuits must abide by is simply a way
of saying your theory can't handle simple cases.

Moving on, my electrical circuits texts abound with examples in which
some initial condition is assumed, then a source is "turned on" or
connected with an imaginary switch at t = 0, and the transition from the
initial state to steady or final state is studied -- exactly as I did in
my analysis. The source is most often a perfect voltage or current
source and, in the sections dealing with sine wave AC circuits, produces
a sine wave. The assumption of an initial condition (usually, but not
always, that all voltages and currents in the circuit are zero) is
absolutely required when solving the fundamental integro-differential
equations which result from circuits containing inductances and
capacitances.

I'll be glad to give page references from Pearson & Maler and Van
Valkenburg, but it's really not necessary since anyone having any
electric circuits text can find abundant examples. It appears that some
of the posters either never took a basic course in electrical circuits,
or forgot some very fundamental principles which were taught. But used
circuits texts can be purchased for a very modest price, so there's
little excuse for remaining ignorant if a person is truly interested in
learning about the topic. (There's also the Internet, but you have to be
more careful in sorting out the good information from the bad, and this
can be difficult if you're not already pretty well acquainted with the
topic.) I'd really like it, too, since I'd be able to present an
analysis now and then without having the most basic principles of linear
circuit analysis questioned and debated.

I'm afraid that the fuss about the source is primarily a way to avoid
confronting the facts, which are apparently disturbing to some of the
imaginative alternative theories being promoted.

The SPICE simulation of the circuit I analyzed was, of course, a
transient analysis. The source was a perfect voltage source which
produced a sine wave beginning at t = 0 and continuously after that,
just as in my analysis. For anyone having SPICE, here's the netlist:

* 360 degree transmission line with open end and voltage source,
* transient response showing runup

..TRAN .05 30

v0 1 0 sin(0 1 1hz)
tl1 1 0 2 0 td=1 z0=50
tl4 2 0 3 0 td=4 z0=50
rl 3 0 1meg

..PROBE
..END

TL_1_sec.gif is a plot of v(2), which is the junction of tl1 and tl4.
TL_5_sec.gif is a plot of v(3), which is the open far end of the line. A
plot of v(1) would of course show the source voltage, a constant 1 volt
peak sine wave beginning at t = 0. As I mentioned, the 1 megohm
terminating resistor rl is necessary to keep the version of SPICE I have
from blowing up; it can be any value that's large enough to not have an
appreciable effect on the result.

Roy Lewallen, W7EL

Standing-Wave Current vs Traveling-Wave Current
 

On Dec 31 2007, 11:37*pm, Keith Dysart wrote:
On Dec 31, 11:11*am, Roger wrote:

Cecil Moore wrote:
Richard Clark wrote:
To cut to the chase, Norton and Thevenin sources are appropriate to
network analysis irrespective of your perception.

All my references indicate that those sources are only
appropriate for *steady-state* use. Roger is searching
for a transient state source.

Right!

73, Roger, W7WKB

Odd. Cecil has not named his "references" which is
quite unusual for he truly likes to name-drop: Ramo,
Whinnery, Hecht, IEEE, ...

You would be well served to google "reflection diagram"
or "bounce diagram" where you will find fine examples
of computing re-reflection using the source impedance
of generators modelled using the Thevenin equivalent
circuit. It is not particularly complicated, though
it can be tedious.

...Keith

Also, googling '"lattice diagram" reflection' will yield
a different set of interesting examples.

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

On Dec 30 2007, 6:18 pm, Roy Lewallen wrote:
Keith Dysart wrote:

I predict that the pulse arriving at the left end will
have the same voltage, current and energy profile as
the pulse launched at the right end and the pulse
arriving at the right end will be similar to the
one launched at the left.

They will appear exactly AS IF they had passed
through each other.

The difficulty with saying THE pulses passed
through each other arises with the energy. The
energy profile of the pulse arriving at the left
will look exactly like that of the one launched
from the right so it will seem that the energy
travelled all the way down the line for delivery
at the far end. And yet, from the experiment above,
when the pulses arriving from each end have the
same shape, no energy crosses the middle of the
line.

So it would seem that the energy that actually
crosses the middle during the collision is
exacly the amount of energy that is needed to
reconstruct the pulses on each side after the
collision.

If all the energy that is launched at one end
does not travel to the other end, then I am
not comfortable saying that THE pulse travelled
from one end to the other.

But I have no problem saying that the system
behaves AS IF the pulses travelled from one
end to the other.

You said that:

What will happen? Recall one of the basics about
charge: like charge repel. So it is no surprise
that these two pulses of charge bounce off each
and head back from where they came.

Yet it sounds like you are saying that despite this charge repulsion and
bouncing of waves off each other, each wave appears to be completely
unaltered by the other? It seems to me that surely we would detect some
trace of this profound effect.

. . .

Is there any test you can conceive of which would produce different
measurable results if the pulses were repelling and bouncing off each
other or just passing by without noticing the other?

There are equations describing system behavior on the assumption of no
wave interaction, and these equations accurately predict all measurable
aspects of line behavior without exception. Have you developed equations
based on this charge interaction which predict line behavior with
equal accuracy and universal applicability?

No equations. I expect that such equations would be more complex
than those describing the behaviour using superposition. Since
the existing equations and techniques for analysis are tractable
and produce accurate results, I am not motivated to develop an
alternate set with lower utility.

And yet the "no interaction" model, while accurately predicting
the behaviour has some weaknesses with explaining what is
happening. It is, I suggest, these weaknesses that help lead
some so far astray.

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. Charge begins to
flow into the line. The ratio of the current to voltage on
the line is defined by the distributed inductance and
capacitance. The inductance is resisting the change in current
which causes a voltage to charge the capacitance. A voltage
step (call this V for later use) propagates down the line
at the speed of light. In front of this step, the voltage,
current and charge in the line is zero. After the step, the
capacitance is charged to the voltage and charge is flowing
in the inductance.

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). Once the
infinitesimal capacitance at the end of the line is
charged, the current now has to stop just a bit earlier
and this charges the inifinitesimal capacitance a bit
further from the end. 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
current is zero and voltage is 2*V. The charge that
is continuing to flow from the source is being used
to charge the distributed capacitance of the line.

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.

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.

And this is the major weakness in the model. It claims
the step function is still flowing in the portion of
the line that has a voltage of 2*V and *zero* current.

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

A trivial example is connecting to 10 volt batteries
in parallel through a .001 ohm resistor. The partial
results show 10000 amps flowing in each direction
in the resistor with a total of 0. 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.

The "no interaction" 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. If the
wave is defined as just being a voltage wave, then
all is well.

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

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

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. We
can measure the current at the middle of the line
and observe that it is always 0. Therefore the
charge that is filling the capacitance and causing
the voltage step which is propagating back towards
each generator 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?

....Keith

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
Odd. Cecil has not named his "references" which is
quite unusual for he truly likes to name-drop: Ramo,
Whinnery, Hecht, IEEE, ...

I get ragged on for giving too many references and
not giving enough references. You guys please make
up your minds.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

On Jan 1, 10:42*am, Cecil Moore wrote:
Keith Dysart wrote:
Odd. Cecil has not named his "references" which is
quite unusual for he truly likes to name-drop: Ramo,
Whinnery, Hecht, IEEE, ...

I get ragged on for giving too many references and
not giving enough references. You guys please make
up your minds.

A reference backing up your claim that a well-defined
source impedance can not be used to compute transient
reflections at the source would be entirely appropriate.

And you might find googling '"lattice diagram" reflection'
to assist in understanding the counter-claim.

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
And yet the "no interaction" model, while accurately predicting
the behaviour has some weaknesses with explaining what is
happening. It is, I suggest, these weaknesses that help lead
some so far astray.

Everything can be understood within the context of
superposition and conservation of energy. No need
to invent any new laws of physics like EM waves
bouncing off each other in free space.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Roy Lewallen wrote:
But any analysis which isn't valid when driven
by a perfect voltage source in series with a resistance (or current
source in parallel with a resistance) is fundamentally flawed.

Any model that violates the laws of physics is fundamentally
flawed. Your model has EM energy sloshing around like water.
Your model has EM energy neither flowing into the source
nor being reflected. That is a violation of the conservation
of energy principle.

The SPICE simulation of the circuit I analyzed was, of course, a
transient analysis. The source was a perfect voltage source which
produced a sine wave beginning at t = 0 and continuously after that,
just as in my analysis. For anyone having SPICE, here's the netlist:

What is it that you think you have proved? That there
is no energy in reflected waves? That EM waves don't
move at the speed of light? That the conservation of
energy principle is invalid? What?
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
A reference backing up your claim that a well-defined
source impedance can not be used to compute transient
reflections at the source would be entirely appropriate.

I made no such claim. My claim is that *your* source
is *NOT* well-defined and is just a result of your
hand-waving fantasies. When you stop refusing to
provide a schematic, we can discuss whether it is
well-defined or not.

Again, I freely admit that you can leap tall buildings
at a single bound in your mind.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
On Dec 30, 5:30 pm, Roger wrote:


I don't recall any examples using perfect CURRENT sources. I think a
perfect current source would supply a signal that could respond to
changing impedances correctly. It should solve the dilemma caused by
the rise in voltage which occurs when when a traveling wave doubles
voltage upon encountering an open circuit, or reversing at the source.

What do you think?

A perfect current source has an output impedance of
infinity, just like an open circuit. The reflection
coefficient is 1.

Similar to the reflected voltage for the perfect
voltage source, the reflected current cancels leaving
just the current from the perfect current source.

...Keith

This disagrees with Roy, who assigns a -1 reflection coefficient when
reflecting from a perfect voltage source.

The Norton or Thévenin equivalent circuits seem capable of positive
reflection coefficients. That is all that I am looking for.

Your search suggestion from a different posting '"lattice diagram"
reflection'yields some examples that demonstrate positive reflection
coefficients.

I must have missed something, because I can't understand why there is an
insistence that a negative reflection coefficient must exist at the
source for the 1/2 or 1 wavelength long transmission line fed at one end.

73, Roger, W7WKB

Standing-Wave Current vs Traveling-Wave Current
 

On Jan 1, 12:20*pm, Roger wrote:
Keith Dysart wrote:
On Dec 30, 5:30 pm, Roger wrote:

I don't recall any examples using perfect CURRENT sources. *I think a
perfect current source would supply a signal that could respond to
changing impedances correctly. *It should solve the dilemma caused by
the rise in voltage which occurs when when a traveling wave doubles
voltage upon encountering an open circuit, or reversing at the source.

What do you think?

A perfect current source has an output impedance of
infinity, just like an open circuit. The reflection
coefficient is 1.

Similar to the reflected voltage for the perfect
voltage source, the reflected current cancels leaving
just the current from the perfect current source.

...Keith

This disagrees with Roy, who assigns a -1 reflection coefficient when
reflecting from a perfect voltage source.

I don't think there is disagreement...
- perfect current source, infinite output impedance,
equivalent to open circuit, RC = 1
- perfect voltage source, zero output impedance,
equivalent to short circuit, RC = -1
- output impedance equal to Z0, RC = 0
- output impedance greater than Z0, RC 0
- output impedance less than Z0, RC 0

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.

That is all that I am looking for.

Your search suggestion from a different posting '"lattice diagram"
reflection'yields some examples that demonstrate positive reflection
coefficients.

This would only be because the examples happened to use
output impedances greater than Z0.

I must have missed something, because I can't understand why there is an
insistence that a negative reflection coefficient must exist at the
source for the 1/2 or 1 wavelength long transmission line fed at one end.

The reflection coefficient depends on the values of line
characteristic impedance (Z0) and the output impedance of the
source.

Recall that RC = (Z2-Z1)/(Z2+Z1)

For the reflected wave arriving back at the source,
Z1 is the characteristic impedance of the line (Z0)
and Z2 is the output impedance of the source.

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

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.

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

73, Roger, W7WKB

Standing-Wave Current vs Traveling-Wave Current
 

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:

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? 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?

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 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. Why would the source at
less voltage provide current to flow into a cap that is rising in
potential above it? Rolling electrons uphill would seem to be
remarkable.

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.

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)?

And this is the major weakness in the model.
Which 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? As for "zero" current, that never made
sense in context here.

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.

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.

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.

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?

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? 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.

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.

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?

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

Is it? When?

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

...Keith

73's
Richard Clark, KB7QHC

Standing-Wave Current vs Traveling-Wave Current
 

On Tue, 01 Jan 2008 10:20:19 -0800, Roger wrote:

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

Hi Roger,

That doesn't describe an equivalency, merely a proportionality.

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

What about phase?

73's
Richard Clark, KB7QHC

Standing-Wave Current vs Traveling-Wave Current
 

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?

To the best of my knowledge, there is absolutely no
requirement that a Thevenin or Norton equivalent
circuit exhibit the same reflection coefficient
at the circuit it replaces.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

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

The mere concept of a Thevenin equivalent circuit
absorbing power is contrary to the rules for the
use of a Thevenin equivalent circuit. No significance
can be automatically assigned to the power conditions
inside a Thevenin equivalent source.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

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?

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

Standing-Wave Current vs Traveling-Wave Current
 

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

Standing-Wave Current vs Traveling-Wave Current
 

Roy Lewallen wrote:
If the driving circuitry consists of a perfect voltage source in series
with a resistance, the source Z will be the resistance;

This is obviously not true and easily illustrated using
a battery.

| |
Gnd--||---\/\/\/\/---+---(I)---||--Gnd
| 50 ohms |
12v 12v

Using a 12v source and current meter (I) to test the
impedance to ground from point '+', what does that
V/I impedance measure? Can you spell infinity?

The same principle holds true for an AC source when
phase is considered. Replace the 12v battery with
a 12vac source and test the impedance at '+' with
a phase-locked 12vac source. The measured V/I will
be infinite just as it is in the DC case.

It would certainly appear that the reflection coefficient
seen by the reflections changes depending upon the phase
of the reflections. In any case, the source Z is obviously
not what the reflections encounter.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Roger wrote:
Keith Dysart wrote:
On Dec 30, 5:30 pm, Roger wrote:


I don't recall any examples using perfect CURRENT sources. I think a
perfect current source would supply a signal that could respond to
changing impedances correctly. It should solve the dilemma caused by
the rise in voltage which occurs when when a traveling wave doubles
voltage upon encountering an open circuit, or reversing at the source.

What do you think?

A perfect current source has an output impedance of
infinity, just like an open circuit. The reflection
coefficient is 1.

Similar to the reflected voltage for the perfect
voltage source, the reflected current cancels leaving
just the current from the perfect current source.

...Keith

This disagrees with Roy, who assigns a -1 reflection coefficient when
reflecting from a perfect voltage source.

It appears you're confusing perfect voltage and current sources.

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. A perfect current source has an infinite impedance,
so if it's connected to a transmission line with no parallel resistance,
the reflection coefficient is +1, as Keith says.

The Norton or Thévenin equivalent circuits seem capable of positive
reflection coefficients. That is all that I am looking for.

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. Under normal circumstances,
reflection coefficients can have any magnitude from zero to one, and any
angle. A Thevein equivalent, like the circuit it's replacing, can
present any possible reflection coefficient. For example, a Thevenin
equivalent circuit having an impedance of 19 - j172 (that is, the
equivalent consists of a perfect voltage source in series with a 19 ohm
resistance and a capacitance of 172 ohms reactance) will present a
reflection coefficient of about 0.8 - j0.5 to a 50 ohm transmission
line. This is also true of a Norton equivalent consisting of a perfect
current source shunted by an impedance of 19 - j172 ohms.

Of course, a voltage source and series impedance not acting as an
equivalent for any other circuit can also be used to drive the line, and
the source impedance will equal the impedance that's in series with the
perfect source. Likewise a current source with a parallel impedance.

. . .

I must have missed something, because I can't understand why there is an
insistence that a negative reflection coefficient must exist at the
source for the 1/2 or 1 wavelength long transmission line fed at one end.

No one has insisted on that at all. As I've said, any reflection
coefficient can exist at the source. It depends solely on the impedances
of the transmission line and the source. Both Keith and I have given the
equation describing the simple relationship, and you can find it also in
many references. It was -1 for my example only because I used a perfect
voltage source with no series impedance.

Roy Lewallen, W7EL