Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
On Dec 30, 9:42 am, Cecil Moore wrote:
Bottom line: At points '+' in the example before any
cutting, either reflections exist or they don't.

If reflections exist, there has to exist an impedance
discontinuity to cause the reflections. There is no
impedance discontinuity.

If reflections don't exist, there is nothing to change
the direction of the flowing energy. Therefore, energy
is flowing both ways through the '+' points.

Now there's a bit of a pickle. You previously agreed
that
P(x,t) = V(x,t) * I(x,t)
but now you claim that there is energy flowing.

No pickle at all. Any rational person would agree
that if we have 10 joules flowing in one direction
and 10 joules flowing in the other direction that
the *NET* power is zero and the *NET* energy flow
is zero.

When you look at yourself in the mirror, there is
just as much energy flowing toward the mirror as
is flowing away from the mirror yet you see your
image. Do you really think that the EM waves
flowing toward the mirror are interacting with
the EM waves flowing away from the mirror?

But it is easily proven that energy is flowing
from one SGCL to the other.

It would be wonderful if you could provide this
easy PROOF since such a proof would settle the
question.

I already presented the proof. Establish a data
transfer between the signal generator and the
other resistor and then cut the connection. If
you assert that the data connection doesn't
go away, I feel sorry for you.

If the purpose of your argument is to confuse,
confound, and obfuscate, I fully understand why
you are presenting it. If your goal is technical
knowledge, then your approach is pathological.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

On Dec 30, 10:30*am, Cecil Moore wrote:
Keith Dysart wrote:
No need for translation, though this is not quite what
was said above. Note the words "automatically" in the
first quote and "may be quite different" in the second.
The original authors allow for the possibility that it
might be the same, while your "translation" removes
that possibility.

They engaged in typical author-speak.

I think not. They wrote with precision in an attempt
to prevent themselves being quoted out of context.
It didn't work, or course.

My university
professors had no such limitations. They were quite
harsh on anyone who tried to figure out where the
power goes inside a Thevenin or Norton equivalent
source.

It is not I who wants it both ways. For me it is clear
that there is no reflection when the output (source)
impedance is the same as Z0. And when it is not equal
to Z0, there is a reflection.

Apparently that is NOT clear to you. In the earlier
example, there is no impedance discontinuity at the
'+' points, yet you require reflections at those
points. That's what you cannot have both ways.

If there's no traveling wave energy flowing through
the '+' points, there must exist reflections.

Well, there is no energy flowing through the '+' points.
And I have no issue if you wish to claim that there
are reflections at these points, though I might use
'bouncing' to differentiate from reflections occuring
at points with non-zero reflection coefficients.

If
reflections exist, there must exist an impedance
discontinuity. There is no impedance discontinuity.

You do seem to be trying to have it both ways.
No energy is flowing (q.v. IEEE definition of
instantaneous power), and yet you want energy
to be flowing.

Not when the output (source) impedance is known. It is
then easy to compute the magnitude of the reflection
using the standard rules for reflection coeficient.

Although many have tried to prove that the output (source)
impedance is the impedance encountered by the reflected waves,
all of those numerous experiments have failed.

You, Cecil, are the only one who believes this. Any good
book on transmission lines will tell you otherwise.

Therefore,
there is a high probability that the impedance encountered
by the reflected waves is *NOT* the output (source) impedance.
The argument has raged loud and long since at least the 1980's.
You are not going to resolve it by hand-waving.

Keith, if you can prove that the reflected waves encounter
the output (source) impedance, you are a better man than all
of the many others who have tried and failed.

Web references and Spice models which agree that "the
output (source) impedance is the impedance encountered
by the reflected waves" have been previously provided,
but you refused to explore them. If I recall correctly,
this was because they did not model the complexity of
an average ham transmitter so they were not of
interest to you. Since you have refused to explore
the question, you should refrain from making
pronouncements.

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
My model requires nothing of EM waves, in-so-far-as
it is completely described in terms of charge. Just
the basics for current flow.

Yes, I agree. Your model doesn't even require the
EM waves to obey the laws of physics. It's a lot
like my mother's model which requires only God.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

On Dec 30, 2:33*pm, Cecil Moore wrote:
Roger wrote:
The "perfect voltage source" controls or better, overwhelms any effect
that might be caused by the reflected wave. *It completely defeats any
argument or description about reflected waves.

The impedance of the perfect voltage source is supposed
to be zero.

The impedance of a perfect voltage source IS zero.

How close humans can come to achieving such a device
depends largely on the budget.

Bench power supplies come quite close within their
current limits and single quadrant operation.

Spending more money will get you closer.

The fact that 0 can not be achieved does not detract
from their utility to assist understanding.

I, like you, suspect that is a complete
impossibility.

What I would like to see is a TV signal
fed through a source only to emerge unscathed on the
other side and viewable on a TV screen without distortion
just as the theory predicts.

A voltage source has two sides? Explain!

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

Cecil Moore wrote:
Roger wrote:
If the impedance of the perfect source is zero, what limits the power
output from the source?

Exactly, just make the output impedance zero and the
perfect source will deliver infinite power. Does that
sound like it is related to reality?

I think I would say it differently. I would describe the "perfect
voltage source" as a variable impedance, constant voltage source, with
all the impedance supplied by external loads. To my thinking, to define
the perfect voltage source as having no impedance, and then using "no
impedance" as an excuse to assign a negative reflection factor so that
voltage from another source is inverted, is beyond belief. Not reality
at all, as you say.

I can accept the concept of infinite power, and recognize the
impossibility at the same time. It would do very bad things to a real
circuit!

However, I can see the dilemma faced by a purist who sees 2v from a
reflected wave (because the reflected wave has returned to the source
and reflected as if it were an open end) and the 1v from the source at
exactly the same location. Something must be wrong.

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?

73, Roger, W7WKB

Standing-Wave Current vs Traveling-Wave Current
 

On Sun, 30 Dec 2007 12:28:29 -0800, Roger wrote:

When we define both the source voltage and the source impedance, we also
define the source power. Two of the three variables in the power
equation are defined, so power is defined.

Hi Roger,

I see a free mixing of "perfect" voltage and current sources, source
impedances, black boxes, and what appears (above) to be a forced
presumption of source power.

As is typical within these debates, something must be broken. For
one, these "perfect" sources paired with an impedance specification
necessarily describes a Thenvenin source (for some reason, no one sees
the elephant in their living room here). For every Thevenin source,
there is an equivalent Norton source; that, for either hidden within a
black box, is indistinguishable from the other.

Given this equivalency, the forced power presumption collapses. Power
in the black box (if in fact that is the intent of this coy
"perfection") cannot be known.

73's
Richard Clark, KB7QHC

Standing-Wave Current vs Traveling-Wave Current
 

On Sun, 30 Dec 2007 12:40:31 -0800, Roger wrote:

If the impedance of the perfect source is zero, what limits the power
output from the source?

Hi Roger,

What does it matter?

If the limit is supplied by the load, then the
perfect source has evolved into a real source as soon as the circuit has
been defined.

That is a stretch of the imagination as real sources don't have any of
the characteristics of a fictitious "perfect" source.

73's
Richard Clark, KB7QHC

Standing-Wave Current vs Traveling-Wave Current
 

On Sun, 30 Dec 2007 14:21:44 -0800 (PST), Keith Dysart
wrote:

A voltage source has two sides? Explain!

The In side, and the Out side.

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
On Dec 30, 9:51 am, Cecil Moore wrote:
Roger wrote:
Roy Lewallen wrote:
Yes, that is correct. The impedance of the source (a perfect voltage
source) is zero, so the reflection coefficient seen by the reverse
traveling wave is -1.
The logic of this assumption eludes me. In fact, it seems completely
illogical and counter to the concept of how voltage and current waves
are observed to move on a transmission line.

The reflection coefficient for a short is -1, is it not?

One way of viewing a short is that it is a perfect
voltage source (i.e. 0 output impedance) set to 0
volts.

Good point. The power disappears here, but energy transfer is evident
(we have current). We both have power and do not have power.

At the source, when the reflect wave returns and re-reflects, we have 2v
from the reflected wave matched with 1v from the source. We have both
1v and 2v.

OK.

Setting it to some other voltage (or function describing
the voltage) does not alter its output impedance. It
there fore creates the same reflection of the travelling
wave, regardless of the voltage function it is generating.

A real world voltage source has an output impedance. Use
this impedance to compute the reflection coefficient.
The reflection will be the same regardless of the
voltage function being generated by the source.

...Keith

Thanks for pointing out the short circuit voltage view. I had not
thought of that.

What would you think of using a perfect current source for these thought
experiments? My suggested perfect current source would supply only as
much current as could be absorbed by the load, so no power would be used
by the source, and current would be limited by the load.

73, Roger, W7WKB

Standing-Wave Current vs Traveling-Wave Current
 

On 30 Dec, 14:13, Cecil Moore wrote:
Keith Dysart wrote:
My model requires nothing of EM waves, in-so-far-as
it is completely described in terms of charge. Just
the basics for current flow.

Yes, I agree. Your model doesn't even require the
EM waves to obey the laws of physics. It's a lot
like my mother's model which requires only God.
--
73, Cecil *http://www.w5dxp.com

Cecil
For $5.00 I can start another thread
and take Richard away from bothering you !
Art