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

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

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

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

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

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

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

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

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

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