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