Standing-Wave Current vs Traveling-Wave Current
 

On Dec 29, 7:10*pm, Cecil Moore wrote:
Keith Dysart wrote:
When the source impedance is the same as Z0 there
is no impedance discontinuity to produce a reflection.

Here's a quote from "Fields and Waves ..." by
Ramo and Whinnery: "... significance cannot be
automatically attached to a calculation of power
loss in the internal impedance of the equivalent
circuit."

True indeed.

And from "T-Lines and Networks" by Walter C. Johnson:
"Although the Thevenin equivalent produces the correct
exterior results, its internal power relations may be
quite different from those of the network it replaces.
The power dissipated in the equivalent impedance of the
Thevenin circuit is not the same as the power dissipated
in the resistance of the actual network."

Different words, but still true.

Translation: Do not use a Thevenin source impedance
to try to track power dissipation within the Thevenin
equivalent box - it won't work.

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.

Your argument that
there is zero power dissipation in the source resistor
inside the Thevenin equivalent box is bogus.

Nowhere has that argument been made.

Is the reflected wave reflected even without a
discontinuity?

The incident reflected wave is either reflected by the
source or it isn't. You cannot have it both ways.

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.

Since
the internal conditions inside a Thevenin equivalent
circuit are a complete unknown, either choice is
a possibility.

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.

What you will find is that, for a real-world source,
the destructive interference on one side of the source
equals the constructive interference on the other side
of the source. The conservation of energy principle
prohibits having it any other way.

Except when the output (source) impedance is equal to
Z0, in which case there is no reflection, and, if I
understand your claim, no interference, destructive
or constructive.

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

Roger wrote:
Roy Lewallen wrote:

No. x is referenced to the input end of the line. This is very
important. I'm sorry my statement that it is "any point x, in degrees,
along the line" didn't make this clear.
Yes, it is critical. I am sorry that I misunderstood this.

In our example then, "x" will always be positive. How am I to interpret
the meaning of vf(t, x) = sin(wt-x)?

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.

I understand that the convention for displaying a sin wave is that one
rotation is 2pi radians with positive rotation being counter clockwise
beginning with sin(x) = 0. Sin(90) = 1. Counter clockwise rotation
would indicate the the sin immediately becomes negative,
so sin(-90) = -1.

I usually display sine waves on a linear graph of value vs. time, as in
the .gif files I referenced. There is no rotation involved. Apparently
you're referring to the display of phase angle on a polar graph, but I
don't see where you're getting the values you're describing. A phase
angle of 90 degrees is at 0 + j1 on the graph, or 90 degrees CCW from
the real axis; -90 degrees is 90 degrees CW from the real axis, at 0 - j1.

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.

Consider that it can be derived various ways, agrees with all published
information and, as I've demonstrated, can be applied to get the correct
answer to a transmission line problem.

Maybe my concept of voltage being a concentration of positive (or
negative) charges is leading me astray. If two waves move in opposite
directions, but both of a positive character, at the time of crossing
paths, the voltages add.

The from two waves always add, vectorially, no matter what the
direction, value, or polarity, as long as they're in a linear medium.

It happens at the open ends when the direction
reverses.

Any time the Z0 of the medium or transmission line changes, a reflection
takes place. The magnitude and angle of the reflected wave compared to
the original wave is known as the reflection coefficient. An open end is
only an extreme case, where the reflection coefficient is +1.

It MUST happen identically when the reflected positive wave
returns to the source (at time 720 degrees in our one wavelength
example) and encounters the next positive wave just leaving the source.

No, for two reasons. One: Contrary to Cecil's theories, one wave doesn't
cause any change in another. Although they vectorially add, each can be
treated completely independently as if the other doesn't exist. If
you'll look carefully at my analysis, I did just that. Two: The input in
the example isn't an open circuit, but exactly the opposite case: it's a
perfect voltage source, which has a zero impedance.

In my model, the source voltage must change when the returning wave
hits the input end.

Then we've been using a different model. The one I've been using is
the one proposed by "Dave" -- a half wavelength open circuited line
driven by a voltage source -- except with your change in line length
to one wavelength. You cannot cause the voltage of a perfect voltage
source to change.

Are you assuming that vr is always propagating from the source as if the
source always supplied vf and vr simultaneously? As if vf was supplied
for time = 4pi, and then vr was applied?

I am assuming that the source provides vf. All other waves result from that.

Your idea of a 5 wavelength long example was a good one Roy. It may
provide a way out of what seems to be a logical impasse (reversal at the
voltage source may be uncompromisable).

The analysis is identical with a one wavelength line, and nearly so with
a half wavelength one. It's just that the various forward and reverse
waves exist independently long enough on the 5 wavelength line that you
can see the effect on the total which each one has.

We could allow our future discussions (if any) to consider an extremely
long line, but consider only the 1/2 or 1 wavelength at the end for our
discussions. Thus, the source (and source for major disagreement) is
far removed from our discussion section. We could then consider the
input source as just another node for as long as we wanted.

I'm happy to entertain an alternative analysis. The result should be the
same as mine, however, which the SPICE model shows to be correct.

Perhaps some readers don't realize that the SPICE model isn't just a
graph of the equations I derived. It's a circuit simulator which uses
fundamental laws to show the behavior of circuits. The SPICE model
consisted only of two transmission lines both having the same impedance
and connected in tandem (so I could show the voltage one wavelength from
the input), a perfect voltage source, and a 1 megohm terminating load
which is necessary because SPICE has problems with a completely open
circuited transmission line. It knew nothing of my analysis or
equations, yet it produced an identical result.

Traveling waves easily explain standing waves on a 1/2 wavelength
section, as you demonstrated. Maybe they can explain or clarify more
things if we can get past "hang ups" such as the " -1 reflection at a
perfect voltage source".

Who's "we"?

The analysis procedure I illustrated can be used to derive all the
steady state transmission line formulas, including ones describing
standing waves; voltage, current, and impedance transformation; delay;
and so forth. It can even be used when loss is present, although the
math gets a lot stickier. Only one additional step is necessary to find
the steady state solution, and that's to find the sum of all forward
waves to get a single combined forward wave and likewise combine all the
reverse waves into a single reverse wave. This can be done with a simple
formula for summing an infinite series, because each reflected wave
bears the same relationship to its original. There are usually much
easier ways to get a steady state solution, but this approach allows
seeing just what happens as the line is charged and the waves are created.

Roy Lewallen, W7EL

Standing-Wave Current vs Traveling-Wave Current
 

On Dec 29, 7:47*pm, Roger wrote:
Keith Dysart wrote:
[snipped]
I completely concur with your analysis.

No doubt you have fine tuned the analysis to notice that the current
stops (meaning becomes unobservable) at the identical instant that the
voltage spike (to double) is observed. *You would have noticed that the
zone of unmeasurable current spreads equally both ways from the
collision point at the velocity of the wave(s). The voltage spike
spreads in lock step with the loss of current detection. *The maximum
width of the loss of current and voltage spike is the width of either of
the pulses.

Now did the two pulses reflect, or pass through one another? *I have
considered the question and can not discern a difference in my analysis
either way. *IT SEEMS TO MAKE NO DIFFERENCE!

Agreed. Either view produces the same results.

It is a minor extension to have this model deal
with sinusoidal excitation.

What happens when these pulses arrive back at the
generator? This depends on generator output
impedance. If it is 50 ohms (i.e. equal to Z0),
then there is no reflection and 1 joule is
dissipated in each generator. Other values
of impedance result in more complicated
behaviour.

Roy and I are talking about this on other postings. *I guess the purest
might point out that a 50 ohm generator only has a voltage to current
ratio of 50, but we don't know if it also has a resistor to absorb
energy. *

All true.

It is like a black box where the only thing we know about it is
that when we connect a 50 ohm resistor to it through a 50 ohm
transmission line, there are no standing waves.

In this case, a reflected wave could be used like a radar pulse to learn
what might be inside the box.

Or slightly more precisely... An equivalent circuit that
will provide the same behaviour.

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.

...Keith

We certainly think similarly *Keith. *Thanks for the posting.

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

On Dec 29, 8:05*pm, Roger wrote:
Cecil Moore wrote:
Roger wrote:
Are there reflections at point "+"? *Traveling waves going in opposite
directions must pass here, therefore they must either pass through one
another, or reflect off one another.

In the absence of a real physical impedance discontinuity,
they cannot "reflect off one another". In a constant Z0 transmission
line, reflections can only occur at the ends of the line and only
then at an impedance discontinuity.

Cecil, this sounds more like a pronouncement from God than like an
conclusion from observations.



Does this match your own concept of the traveling waves acting at the
{+} point Cecil? *If not, where do we differ?

Where we differ is that you allow traveling waves to "reflect
off one another". There are no laws of physics which allow
that in the absence of a physical impedance discontinuity.
EM waves simply do not bounce off each other.

I am not aware of any laws of physics that prevent it either. *I don't
see any evidence that it happens in open space, like light bouncing off
light. *It might happen on transmission lines however. *I just cannot
find any convincing evidence either way. *What I have deduced so far
indicates that it makes no difference which happens.

Maybe both things happen (both reflect and pass). *This because the EM
field travels very close to the speed of light. *It is a little hard to
see how one wave could "see" the other coming. *On the other hand, the
charges move slowly, far below the speed of light. *It is easy to see
how they might "see or feel" each other coming.

Consider the electric field, for example. Like charge,
two fields of the same polarity will exert a force on
each other. Arguably, this is the same force that
charge exerts.

Also consider that field lines never cross. It would
therefore seem impossible for the two electric fields
associated with two EM waves to pass each other
without interaction. Just like the places on the
transmission line where the voltage or current (and
therefore power) is always zero, there are places in
space where the E or H field (and therefore power) are
always zero.

And whether the wave in space is viewed as passing
or bouncing, the results should be the same, just
as it is on a transmission line.

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

Roy Lewallen wrote:

...
No interpretation necessary. Plug in a time t and distance x from the
...

Oh gawd, this gripes me most of all "t", as in rotations of the earth,
now what the hell has that got to do with rf? Distance is OK ...

JS

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
. . .
Notice a key point about this description. It is
completely in terms of charge. There is not a single
mention of EM waves, travelling or otherwise.

Now we expand the experiment by placing a pulse
generator at each end of the line and triggering
them to each generate a 50V one second pulse at
the same time. So after one second a pulse has
completely entered each end of the line and these
pulse are racing towards each other at the speed
of light (in the line). In another second these
pulses will collide at the middle of the line.

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. At the center
of the line, for one second the voltage is 100 V
(50 V from each pulse), while the current is
always zero. No charge crossed the mid-point. No
energy crossed the mid-point (how could it if
the current is always zero (i.e. no charge
moves) at the mid-point.

. . .

That's an interesting and compelling argument. With the conditions you
describe, I don't see how it would be possible to tell whether the waves
reflected from each other or simply passed by without interacting.

But suppose we launch waves of different shapes from the two directions,
say a triangular wave and a rectangular one. Or perhaps make them
asymmetrical in some fashion. It seems to me that then we should be able
to tell which of the two possibilities happened. Being different, you
could argue that the reflection wouldn't be complete. But shouldn't we
expect some distortion of any part of a pulse that was acted upon by the
other?

I'll put my money on each of the waves arriving at the opposite end
unchanged. What do you predict will happen?

Roy Lewallen, W7EL

Standing-Wave Current vs Traveling-Wave Current
 

Roy Lewallen wrote:
Roger wrote:
Roy Lewallen wrote:

No. x is referenced to the input end of the line. This is very
important. I'm sorry my statement that it is "any point x, in
degrees, along the line" didn't make this clear.
Yes, it is critical. I am sorry that I misunderstood this.

In our example then, "x" will always be positive. How am I to
interpret the meaning of vf(t, x) = sin(wt-x)?

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.

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?

I understand that the convention for displaying a sin wave is that one
rotation is 2pi radians with positive rotation being counter clockwise
beginning with sin(x) = 0. Sin(90) = 1. Counter clockwise rotation
would indicate the the sin immediately becomes negative,
so sin(-90) = -1.

I usually display sine waves on a linear graph of value vs. time, as in
the .gif files I referenced. There is no rotation involved. Apparently
you're referring to the display of phase angle on a polar graph, but I
don't see where you're getting the values you're describing. A phase
angle of 90 degrees is at 0 + j1 on the graph, or 90 degrees CCW from
the real axis; -90 degrees is 90 degrees CW from the real axis, at 0 - j1.

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.

Consider that it can be derived various ways, agrees with all published
information and, as I've demonstrated, can be applied to get the correct
answer to a transmission line problem.

I have never seen the derivation that supports a negative 1, and still
can't get the same numbers that you get. The derivations that I have
seen say otherwise.

Maybe my concept of voltage being a concentration of positive (or
negative) charges is leading me astray. If two waves move in opposite
directions, but both of a positive character, at the time of crossing
paths, the voltages add.

The from two waves always add, vectorially, no matter what the
direction, value, or polarity, as long as they're in a linear medium.

I don't think you mean vector addition here. As you said previously,
the traveling waves add voltage when they pass. That would be scaler
addition.

It happens at the open ends when the direction reverses.

Any time the Z0 of the medium or transmission line changes, a reflection
takes place. The magnitude and angle of the reflected wave compared to
the original wave is known as the reflection coefficient. An open end is
only an extreme case, where the reflection coefficient is +1.

It MUST happen identically when the reflected positive wave returns
to the source (at time 720 degrees in our one wavelength example) and
encounters the next positive wave just leaving the source.

No, for two reasons. One: Contrary to Cecil's theories, one wave doesn't
cause any change in another. Although they vectorially add, each can be
treated completely independently as if the other doesn't exist. If
you'll look carefully at my analysis, I did just that. Two: The input in
the example isn't an open circuit, but exactly the opposite case: it's a
perfect voltage source, which has a zero impedance.

I agree that one wave does not change the other. It looked to me like
you were using SCALER addition in your analysis, which I agreed with.

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.

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. 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?

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?

In my model, the source voltage must change when the returning wave
hits the input end.

Then we've been using a different model. The one I've been using is
the one proposed by "Dave" -- a half wavelength open circuited line
driven by a voltage source -- except with your change in line length
to one wavelength. You cannot cause the voltage of a perfect voltage
source to change.

Are you assuming that vr is always propagating from the source as if
the source always supplied vf and vr simultaneously? As if vf was
supplied for time = 4pi, and then vr was applied?

I am assuming that the source provides vf. All other waves result from
that.

Your idea of a 5 wavelength long example was a good one Roy. It may
provide a way out of what seems to be a logical impasse (reversal at
the voltage source may be uncompromisable).

The analysis is identical with a one wavelength line, and nearly so with
a half wavelength one. It's just that the various forward and reverse
waves exist independently long enough on the 5 wavelength line that you
can see the effect on the total which each one has.

We could allow our future discussions (if any) to consider an
extremely long line, but consider only the 1/2 or 1 wavelength at the
end for our discussions. Thus, the source (and source for major
disagreement) is far removed from our discussion section. We could
then consider the input source as just another node for as long as we
wanted.

I'm happy to entertain an alternative analysis. The result should be the
same as mine, however, which the SPICE model shows to be correct.

Perhaps some readers don't realize that the SPICE model isn't just a
graph of the equations I derived. It's a circuit simulator which uses
fundamental laws to show the behavior of circuits. The SPICE model
consisted only of two transmission lines both having the same impedance
and connected in tandem (so I could show the voltage one wavelength from
the input), a perfect voltage source, and a 1 megohm terminating load
which is necessary because SPICE has problems with a completely open
circuited transmission line. It knew nothing of my analysis or
equations, yet it produced an identical result.

Traveling waves easily explain standing waves on a 1/2 wavelength
section, as you demonstrated. Maybe they can explain or clarify more
things if we can get past "hang ups" such as the " -1 reflection at a
perfect voltage source".

Who's "we"?

The analysis procedure I illustrated can be used to derive all the
steady state transmission line formulas, including ones describing
standing waves; voltage, current, and impedance transformation; delay;
and so forth. It can even be used when loss is present, although the
math gets a lot stickier. Only one additional step is necessary to find
the steady state solution, and that's to find the sum of all forward
waves to get a single combined forward wave and likewise combine all the
reverse waves into a single reverse wave. This can be done with a simple
formula for summing an infinite series, because each reflected wave
bears the same relationship to its original. There are usually much
easier ways to get a steady state solution, but this approach allows
seeing just what happens as the line is charged and the waves are created.

Roy Lewallen, W7EL

I followed your analysis and thought it very well done. My only concern
was the "perfect voltage source". I think that using a source voltage
that has effectively passed through a diode destroys the results of a
good analysis.

73, Roger, W7WKB

Standing-Wave Current vs Traveling-Wave Current
 

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.

Consider that it can be derived various ways, agrees with all
published information and, as I've demonstrated, can be applied to get
the correct answer to a transmission line problem.

I have never seen the derivation that supports a negative 1, and still
can't get the same numbers that you get. The derivations that I have
seen say otherwise.

It's not clear to me what the problem is. Do you mean that you can't get
the same end result for the voltages and currents on the line at various
times? The results I got were confirmed with the SPICE model, so I have
high confidence they're correct. If you're using some reflection
coefficient other than -1 at the source, it's not a surprise that your
final results would be different.

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.

Maybe my concept of voltage being a concentration of positive (or
negative) charges is leading me astray. If two waves move in
opposite directions, but both of a positive character, at the time of
crossing paths, the voltages add.

The from two waves always add, vectorially, no matter what the
direction, value, or polarity, as long as they're in a linear medium.

I don't think you mean vector addition here. As you said previously,
the traveling waves add voltage when they pass. That would be scaler
addition.

I should have said phasor rather than vector addition, a mistake I've
made before. You can add the waves point by point at each instant of
time, in which you're doing simple scalar addition. Or you can add the
phasors, which have magnitude and phase like vectors, of the whole
time-varying waveforms at once.

It happens at the open ends when the direction reverses.

Any time the Z0 of the medium or transmission line changes, a
reflection takes place. The magnitude and angle of the reflected wave
compared to the original wave is known as the reflection coefficient.
An open end is only an extreme case, where the reflection coefficient
is +1.

It MUST happen identically when the reflected positive wave returns
to the source (at time 720 degrees in our one wavelength example) and
encounters the next positive wave just leaving the source.

No, for two reasons. One: Contrary to Cecil's theories, one wave
doesn't cause any change in another. Although they vectorially add,
each can be treated completely independently as if the other doesn't
exist. If you'll look carefully at my analysis, I did just that. Two:
The input in the example isn't an open circuit, but exactly the
opposite case: it's a perfect voltage source, which has a zero impedance.

I agree that one wave does not change the other. It looked to me like
you were using SCALER addition in your analysis, which I agreed with.

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.

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.

But the source resistor has no impact whatsoever on the transmission
line SWR -- it's dictated solely by the line and load impedances. So
you'll have to think of some other criterion to base your choice on.

If you want "no SWR" (by which I assume you mean SWR = 1) on the line,
the only way to get it is to change the load to the complex conjugate of
Z0. But then the analysis becomes trivial: the initial forward wave gets
to the end, and that's it -- steady state has been reached. Finished.

I followed your analysis and thought it very well done. My only concern
was the "perfect voltage source". I think that using a source voltage
that has effectively passed through a diode destroys the results of a
good analysis.

Well, the results are correct. But as I said, this doesn't guarantee
that the analysis is. I welcome alternative analyses which also produce
the correct result.

I'm sorry you can't get around the concept of separating the source
output from its effect on returning waves. Superposition is a powerful
technique without which an analysis like this would be nearly hopeless
to do manually. The source isn't "rectifier" at all and, in fact,
introducing any nonlinear device such as a rectifier would invalidate
most if not all the analysis, and eliminate any hope that it would
produce the correct answer. The fact that it does produce the right
answer is strong evidence that no nonlinear devices are included. But if
adding a source resistance would help, I'll do the same analysis with a
source resistance of your choice(*). Shoot, you can do it too. Just
change the source reflection coefficient from -1 to (Zs - Z0) / (Zs +
Z0) using whatever Zs you choose. The steps are the same, but you'll see
that the reflections get smaller each time, allowing the system to
converge to steady state.

(*) You could, of course, put a pure reactance at the source. But then
we'd end up with a source reflection coefficient having a magnitude of 1
but a phase angle of + or -90 degrees, and still get a full amplitude
reflection and no convergence. A complex source impedance (having both R
and X) would give us a reflection coefficient of less than one, and
convergence, but make the math a little messier. So I'd prefer a plain
resistance if it's all the same to you.

Roy Lewallen, W7EL

Standing-Wave Current vs Traveling-Wave Current
 

On Dec 30, 2:12*am, Roy Lewallen wrote:
Keith Dysart wrote:
. . .
Notice a key point about this description. It is
completely in terms of charge. There is not a single
mention of EM waves, travelling or otherwise.

Now we expand the experiment by placing a pulse
generator at each end of the line and triggering
them to each generate a 50V one second pulse at
the same time. So after one second a pulse has
completely entered each end of the line and these
pulse are racing towards each other at the speed
of light (in the line). In another second these
pulses will collide at the middle of the line.

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. At the center
of the line, for one second the voltage is 100 V
(50 V from each pulse), while the current is
always zero. No charge crossed the mid-point. No
energy crossed the mid-point (how could it if
the current is always zero (i.e. no charge
moves) at the mid-point.

* . . .

That's an interesting and compelling argument. With the conditions you
describe, I don't see how it would be possible to tell whether the waves
reflected from each other or simply passed by without interacting.

But suppose we launch waves of different shapes from the two directions,
say a triangular wave and a rectangular one. Or perhaps make them
asymmetrical in some fashion. It seems to me that then we should be able
to tell which of the two possibilities happened. Being different, you
could argue that the reflection wouldn't be complete. But shouldn't we
expect some distortion of any part of a pulse that was acted upon by the
* other?

I'll put my money on each of the waves arriving at the opposite end
unchanged. What do you predict will happen?

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.

On the other, it is completely intriguing that
a directional voltmeter could be placed anywhere
on the line and the voltage profile of the two
pulses can be recovered. And this is true even
at the middle of the line where, in the experiment
with identical pulse shapes, no current flows and
no energy crosses. But the shape of the two pulses
can still be recovered.

So in the end I say it is AS IF the voltage and
current pulses pass through each other, but the
energy does not necessarily do so. That way I
am not left with having to account for where the
reflected energy goes when it arrives back at
the source.

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

On Dec 30, 7:33*am, Keith Dysart wrote:
On Dec 30, 2:12*am, Roy Lewallen wrote:
What do you predict will happen?

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.

On the other, it is completely intriguing that
a directional voltmeter could be placed anywhere
on the line and the voltage profile of the two
pulses can be recovered. And this is true even
at the middle of the line where, in the experiment
with identical pulse shapes, no current flows and
no energy crosses. But the shape of the two pulses
can still be recovered.

So in the end I say it is AS IF the voltage and
current pulses pass through each other, but the
energy does not necessarily do so. That way I
am not left with having to account for where the
reflected energy goes when it arrives back at
the source.

...Keith

While I wrote the above about the energy in the
pulse not flowing all the way across the line,
the same can be said for the charge. The charge
which describes the pulse does not all go across
the middle of the line; just enough charge
flows to reconstruct the pulse on the other
side.

So the similar question is: Is it the same pulse
if it is not the same charge? Or is it another
pulse that just happens to have the same shape?

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
Well something causes you to latch up and bail with
"its not real world", ...

Don't feel unique - I do the same thing when someone
says God created the heavens and earth in 6 days.

You are either confused about what I said (or deliberately
bearing false witness). Please correct your confusion
(or lack of ethics) or I will stop responding.

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. The *NET*
energy flow is zero. But it is easily proven that
energy is flowing from one SGCL to the other.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

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

Standing-Wave Current vs Traveling-Wave Current
 

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 reason that the logic eludes you, Roger, is that it
can exist only in the human mind where anything is possible.
Roy's above statement requires a belief in the supernatural.

The impedance of a real world source is never zero and
Roy cannot leap tall buildings at a single bound except
in his mind. :-)
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Roger wrote:
Roy and I are talking about this on other postings. I guess the purest
might point out that a 50 ohm generator only has a voltage to current
ratio of 50, but we don't know if it also has a resistor to absorb
energy. It is like a black box where the only thing we know about it is
that when we connect a 50 ohm resistor to it through a 50 ohm
transmission line, there are no standing waves.

Good for you, Roger, every reference I have on Thevenin
equivalent boxes say that what is happening inside the
box is irrelevant because it bears no resemblance to
reality.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Roger wrote:
Cecil Moore wrote:
In the absence of a real physical impedance discontinuity,
they cannot "reflect off one another". In a constant Z0 transmission
line, reflections can only occur at the ends of the line and only
then at an impedance discontinuity.

Cecil, this sounds more like a pronouncement from God than like an
conclusion from observations.

Sorry, it wasn't meant that way. It is just a fact of
physics that is obvious. EM waves possess energy and
momentum. Both of those values must be preserved. In
a passive constant-Z0 transmission line, if there are
no impedance discontinuities to cause reflections, then
reflections cannot exist. If reflections don't exist,
the momentum of the EM waves doesn't change.

Where we differ is that you allow traveling waves to "reflect
off one another". There are no laws of physics which allow
that in the absence of a physical impedance discontinuity.
EM waves simply do not bounce off each other.

I am not aware of any laws of physics that prevent it either.

Why doesn't a reflection coefficient equal to zero
imply no bouncing?

I don't
see any evidence that it happens in open space, like light bouncing off
light. It might happen on transmission lines however. I just cannot
find any convincing evidence either way. What I have deduced so far
indicates that it makes no difference which happens.

If it doesn't happen with light waves, it probably doesn't
happen with RF waves - they are identical except for
wavelength.

I am stating a negative premise, that EM waves do not bounce
off of each other. A negative premise cannot be proved. It
can only be disproved and it only takes one case to do so.

If anyone can offer just one proven case of one EM wave bouncing
off of another EM wave in the complete absence of a change in
the media, then I will be proven wrong.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

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

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

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
Agreed. Either view produces the same results.

Either view, Creationism or the Big Bang, produces
the same results. :-)
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
Also consider that field lines never cross. It would
therefore seem impossible for the two electric fields
associated with two EM waves to pass each other
without interaction.

They certainly superpose so the field lines don't
cross but superposition doesn't cause interaction
in free space or in a constant-Z0 transmission line
where neither medium changes. Free space superposition
results in zero reflections.

And whether the wave in space is viewed as passing
or bouncing, the results should be the same, just
as it is on a transmission line.

Any "view" is possible in your mind but not every
view of yours that is possible is associated with
reality. Your view could just as easily be the
same as my mother's (RIP), i.e. that God is
responsible for everything.

If your view is disassociated from reality, it is
no better than a religion.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

John Smith wrote:
Oh gawd, this gripes me most of all "t", as in rotations of the earth,
now what the hell has that got to do with rf? Distance is OK ...

What do minutes and seconds have to do with spherical
angular rotation? Shirley, you jest.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Roy Lewallen wrote:
I'll put my money on each of the waves arriving at the opposite end
unchanged. What do you predict will happen?

I predict that Keith will say that square waves and
triangular waves are not steady-state. :-)
--
73, Cecil http://www.w5dxp.com

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

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