Standing-Wave Current vs Traveling-Wave Current
 

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. 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.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Roy Lewallen wrote:
Roger wrote:
Roy Lewallen wrote:

No interpretation necessary. Plug in a time t and distance x from the
input end of the line in degrees, and the result is the value of the
forward voltage wave at that time and place. For example, if the
frequency is 1 MHz, w = 2.828 X 10^6/sec. So 100 ns from the time you
connected the source (or 100 ns from the beginning of any source
cycle, since the signal is periodic), the value of the forward
voltage wave 20 degrees from the input of the cable is sin(2.828 X
10^6 * 100 X 10^-9 radians + 20 degrees) = sin(36.2 degrees) ~ 0.59
volts.

Sorry, I can not follow the numbers. For frequency of 1 MHz, I would
expect w = 2*pi* 10^6 = approximately 6.28 * 10^6 radians per second.
100 ns would be about 0.628 radians = 36 degrees.

The wave would have moved 36 degrees.

I apologize. I made an error and you're correct.

A point in 20 degrees from the input end would be found 16 degrees
behind the leading edge of the wave. The voltage should be sin(16) =
0.276v.

I wonder why we can not get the same results?

If you mean for the calculation of the voltage at 100 ns and 20 degrees
down the line, it's because of my error. It should be sin(36 + 20
degrees) ~ 0.83.

From the equation vf(t,x) = sin(wt-x), I am getting
vf(36 degrees, 20 degrees) = sin(36-20) = sin(16) = 0.276v.

Could we look at five points on the example? (The example has frequency
of 1 MHz, entered the transmission line 100 ns prior to the time of
interest, and traveled 36 degrees into the line) Using zero voltage on
the leading edge as a reference point on the sine wave, and the input
point on the transmission line as the second reference point, find the
voltage at (36, -20), (36, 0), (36, 20), (36, 36)and (36, 40). All
points are defined in degrees.

Not on the line yet, at -20 degrees, sin(36+20) = 0.83.
At line input, at 0 degrees, sin(36+0) = 0.59v.
On the line, at +20 degrees, sin(36-20) = 0.276v.
On the line, at +36 degrees, sin(36-36) = 0v.
On the line, at +40 degrees, sin(undefined, -40) = (wave has not arrived)

Each of us must be using a different reference point because we are
getting different results.


As I mentioned, time domain analysis of transmission lines is relatively
rare. But there's a very good treatment in Johnk, _Engineering
Electromagnetic Fields and Waves_. Another good reference is Johnson,
_Electric Transmission Lines_. Near the end of Chapter 14, he actually
has an example with a zero-impedance source: "Let us assume that the
generator is without impedance, so that any wave arriving at the
transmitting end of the line is totally reflected with reversal of
voltage; the reflection factor at the sending end is thus -1." And he
goes through a brief version of essentially the same thing I did to
arrive at the steady state.

I wonder if it is possible that Johnson, _Electric Transmission Lines_,
presented an incorrect example? That occasionally happens, but rarely.

Or maybe some subtle condition assumption is different making the
example unrelated to our experiment?

clip.......
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.

If you'd like, I can pretty easily modify my analysis to include a
finite source resistance of your choice. It will do is modify the source
reflection coefficient and allow the system to converge to steady state.
Would you like me to? There's no reason the choice of a perfect voltage
source should interfere with the understanding of what's happening --
none of the phenomena require it.

Just for conversation, we could place a 50 ohm resistor in parallel with
the full wave 50 ohm transmission line, which is open ended. At
startup, the "perfect voltage source" would see a load of 50/2 = 25
ohms. At steady state, the "perfect voltage source" would see a load of
50 ohms in parallel with "something" from the transmission line. The
power output from the "perfect voltage source" would be reduced below
the startup output. We would arrive at that conclusion using traveling
waves, tracing the waves as they move toward stability.

Would it be acceptable to use a perfect CURRENT source, along with a
parallel resistor. Then CURRENT would remain constant, but voltage
would vary. Again, power into the test circuit would vary.

When you use a "perfect voltage source" with a -1 reflection factor,
you are saying that a perfect polarity reversing plane (or
discontinuity) exists which reflects and reverses the reflected wave.

Yes, it behaves exactly like a short circuit to arriving waves.

However, the reflecting plane (or discontinuity) is one way because
it does allow passage of the forward wave. This is equivalent to
passing the forward wave through a rectifier. Is it fair to our
discussion to insist on using a voltage source that passes through a
rectifier?

The voltage source isn't acting like a rectifier, but a perfect source.
It resists as strongly as it can any change to what it's putting out.
I'm using superposition to separate what it's putting out from the
effects of other waves.

We could use a "black box" wave source. The only thing we would know
for sure about this source is that when it fed a Zo resistor through a
Zo feed line, there would be no swr on the feed line. Would that be
an acceptable voltage source for our discussions?

I have no problem with a "black box" source for which we know only the
voltage (as a function of time, e.g. Vs * sin(wt) where Vs is a
constant) and source impedance. The voltage is the open terminal voltage
of the black box, and the source impedance is that voltage divided by
the short circuit terminal current. One circuit (of an infinite number
of possibilities) having these characteristics is the Thevenin
equivalent, which is a perfect voltage source like I've been using in
the example, in series with the specified impedance. But for the
analysis I won't need to know what's in the box, just its voltage and
impedance. Analysis with any finite source (or load) resistance is
easier than the no-loss case because it allows convergence to steady state.

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.

Now if the we use such a source, a reflection would bring additional
power back to the input. We would need to begin the analysis all over
again as if we were restarting the experiment, this time with two
voltages applied (the source and reflected voltages). Then we would
have the question: Should the two voltages should be added in series, or
in parallel?

Your answer has been to use a reflection factor of -1, which would be to
reverse the polarity. This presents a dilemma because when the
reflected voltage is equal to the forward voltage, the sum of either the
parallel or series addition is zero. You can see what that does to our
analysis. Power just disappears so long as the reflective wave is
returning, as if we are turning off the experiment during the time the
reflective wave returns.

Would a "perfect CURRENT source" without any restrictions about
impedance work as an initial point for you? That would be a source that
supplied one amp but rate of power delivery could vary.

73, Roger, W7WKB

Standing-Wave Current vs Traveling-Wave Current
 

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. 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.

My thinking keeps evolving, so new twist emerge.

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

73, Roger, W7WKB

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
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.

Good grief, Keith, just blame God for everything
and get it over with.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Roy Lewallen wrote:
No, for two reasons. One: Contrary to Cecil's theories, one wave doesn't
cause any change in another.

One wave being able to cancel another wave is not a change?
Shirley, you jest. In the following application, dial up
an identical frequency and amplitude and an opposite phase
and see what happens.

http://micro.magnet.fsu.edu/primer/j...ons/index.html

Does anybody else detect the supernatural content of
Roy's postings?
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Roger wrote:
Now if the we use such a source, a reflection would bring additional
power back to the input.

An important point is that a reflection would bring
additional *energy* back to the input, not necessarily
power to the source. If the energy is re-reflected, it
will *not* appear as power in the source.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

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?
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

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.

If you do not accept that P = V * I, please clearly
state so. The discussion could then continue with
the more basic issue.

The *NET* energy flow is zero.

NET energy flow is not defined in your favourite
reference: the IEEE dictionary.

How does it differ from "instantaneous power" for
which the IEEE does have a definition (P=V*I, if
you are interested)?

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.

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

On Dec 30, 9:47*am, Cecil Moore wrote:
Keith Dysart wrote:
So do the travelling waves "reflect" off each
other? Save the term "reflect" for those cases
where there is an impedance discontinuity and
use "bounce" for those cases where no energy
is crossing a point and even Cecil may be
happy. But bounce it does.

If your model requires EM waves to bounce off of
each other, it doesn't represent reality. What
you may be seeing is the case where destructive
interference in one direction has to equal constructive
interference in the other direction. This is simple
wave cancellation, not bouncing.

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.

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

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.

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