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

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

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

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

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

Richard Clark wrote:
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.

Every reference I have on Thevenin and Norton equivalent
circuits say essentially that same thing.
--
73, Cecil http://www.w5dxp.com

Roger wrote:
Roy Lewallen wrote:
Roger wrote:
Roy Lewallen wrote:

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.

I should quit before I get farther behind! In my haste I made a second
error by adding rather than subtracting the 20 degrees. Again, you're
correct and I apologize for the error.

I'll take extra care to try and avoid goofing up again.

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)

Ok.

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.

I'm a little confused by your dual reference points, both of which seem
to refer to physical positions. In my analysis, I give an equation for a
voltage as a function of both time and physical position. The time (t)
reference point is the time at which the source is connected. At that
instant, the sine wave source voltage is zero. The reference point for
position (x) is the input end of the line. The voltage at any time and
any position can be determined from the equation, provided that the wave
being referred to has reached the position on the line at or before the
time being evaluated (and provided that I don't make a stupid arithmetic
error -- or two). You'll note that in my analysis, I usually give a time
at which the equation is valid -- this is to insure that the wave has
reached any point of interest on the line.

Not on the line yet, at -20 degrees, sin(36+20) = 0.83.

The equation vf = sin(wt - x) isn't valid until the wave has reached
point x on the line. It's also not valid at any point not on the line.
So it can't be used under those conditions.

At line input, at 0 degrees, sin(36+0) = 0.59v.

100 ns after turn-on, the wave has propagated 36 degrees down the line,
so it's present at the input end (x = 0) and the equation is valid. The
voltage at the line input is as you calculated, 0.59 v. at that time.

On the line, at +20 degrees, sin(36-20) = 0.276v.

Correct.

On the line, at +36 degrees, sin(36-36) = 0v.

Correct.

On the line, at +40 degrees, sin(undefined, -40) = (wave has not arrived)

Correct.

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

Without my careless errors, we get the same result except for the first
case.

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.

Certainly it happens, but both he and I get the same, correct result
using the same method. And it seems unlikely that both he and Johnk made
the same error.

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

I'm not sure why you see this as necessary. Do you have an analysis
which correctly predicts the voltage at all times on the line after
startup but which uses some different interaction of waves returning to
the source? If so, please present it. I've presented mine and shown that
I arrived at the correct result.

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.

No, at startup, the source always sees Z0, regardless of the load
termination. The load termination has no effect at the source end until
the reflected wave returns. And if the line is terminated in Z0, then
there is no reflected wave and the source sees Z0 forever. To the load,
a terminated line looks exactly like a plain resistor of value Z0. (I'm
assuming a lossless line as we have been all along.)

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.

With the line terminated in 50 ohms, steady state is reached as soon as
the initial forward wave reaches the load.

If the termination is some value other than Z0, the steady state
impedance seen by the transmission line will be the load impedance (for
our example one wavelength line or any line an integral number of half
wavelengths long). When starting, the initial impedance is always Z0 for
one round trip, for any line length. Then it will jump or down. With
each successive round trip, the impedance will either monotonically step
in the direction of the final value, or oscillate around it but with the
excursion decreasing each time. Which behavior you'll see and the rate
at which it converges are dictated by the relationship among source,
load, and characteristic impedances.

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.

The transmission line will react to a perfect current source in parallel
with a resistance exactly the same as for a perfect voltage source with
a series resistance. Put them in boxes and there's no test you can
devise which can distinguish them by tests of the terminal
characteristics. So yes, that's fine, or just a black box about which
the contents are completely unknown except for any two of the open
circuit voltage, short circuit current, or impedance. (It's assumed that
the box contains some linear circuit that doesn't change during the
analysis.) The results will be identical for any of the three choices.

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.

We define the maximum amount of power which can be extracted from the
box, but not the power that it's delivering. If we open circuit or short
circuit the box or terminate it with a pure reactance, it's delivering
no power at all. If we terminate it with the complex conjugate of its
impedance, it's delivering V^2 / (2 * Rp) where Rp is the parallel
equivalent load resistance and V is the open circuit voltage. With any
other termination it's delivering less than that but more than zero.

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

This presumes that there are power waves bouncing around on the line.

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?

That's not a question I can answer, since it's a consequence of a
premise I don't believe. If your premise has merit, you should be able
to express it mathematically and arrive at the answer to the question.
Feel free to do an analysis using multiple sources and traveling waves
of power. If you get the correct answer for how a line actually behaves,
we'll try it out with a number of different conditions and see if it
holds up.

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.

That's correct but no dilemma.

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.

By "our" analysis do you mean my analysis or the one you propose?
There's no rule saying power can't disappear -- power is not conserved.
In fact, look at any point along the open circuited transmission line
and you'll find that the power "disappears" -- goes to zero -- twice
each V or I cycle. When you first turn on the source, the source
produces average power. In steady state, the average power is zero. It
"disappeared", without any dissipative elements in the circuit. Energy
*is* conserved, however, and there is no energy disappearing in my
analysis. Feel free to try and conserve power in your analysis if you
want. But energy had better be conserved in the process.

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.

Sure. The only difference to the analysis procedure would be to change
the source reflection coefficient from -1 to +1, since a perfect current
source has an infinite impedance. I don't know how it would affect the
final outcome without going through the steps. Of course, the reflection
coefficient for the current wave would go from +1 to -1, so I imagine
you'd run into the same conceptual problem if you tried doing an
analysis of the current waves.

Roy Lewallen, W7EL

Roy Lewallen wrote:
Roger wrote:
Roy Lewallen wrote:
Roger wrote:
Roy Lewallen wrote:

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.

I should quit before I get farther behind! In my haste I made a second
error by adding rather than subtracting the 20 degrees. Again, you're
correct and I apologize for the error.

I'll take extra care to try and avoid goofing up again.

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)

Ok.

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.

I'm a little confused by your dual reference points, both of which seem
to refer to physical positions. In my analysis, I give an equation for a
voltage as a function of both time and physical position. The time (t)
reference point is the time at which the source is connected. At that
instant, the sine wave source voltage is zero. The reference point for
position (x) is the input end of the line. The voltage at any time and
any position can be determined from the equation, provided that the wave
being referred to has reached the position on the line at or before the
time being evaluated (and provided that I don't make a stupid arithmetic
error -- or two). You'll note that in my analysis, I usually give a time
at which the equation is valid -- this is to insure that the wave has
reached any point of interest on the line.

Not on the line yet, at -20 degrees, sin(36+20) = 0.83.

The equation vf = sin(wt - x) isn't valid until the wave has reached
point x on the line. It's also not valid at any point not on the line.
So it can't be used under those conditions.

At line input, at 0 degrees, sin(36+0) = 0.59v.

100 ns after turn-on, the wave has propagated 36 degrees down the line,
so it's present at the input end (x = 0) and the equation is valid. The
voltage at the line input is as you calculated, 0.59 v. at that time.

On the line, at +20 degrees, sin(36-20) = 0.276v.

Correct.

On the line, at +36 degrees, sin(36-36) = 0v.

Correct.

On the line, at +40 degrees, sin(undefined, -40) = (wave has not arrived)

Correct.

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

Without my careless errors, we get the same result except for the first
case.

I make errors as well, so no criticism from here. I am just glad that
we are on the same page on how we use the equation.

As for the (36, -20) point, I was indeed specifying a point that had not
been supplied to the transmission line. It was a prediction, as if the
wave existed physically and was moving toward our experiment. We could
not actually observe it until time = 56 degrees.

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.

Certainly it happens, but both he and I get the same, correct result
using the same method. And it seems unlikely that both he and Johnk made
the same error.

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

I'm not sure why you see this as necessary. Do you have an analysis
which correctly predicts the voltage at all times on the line after
startup but which uses some different interaction of waves returning to
the source? If so, please present it. I've presented mine and shown that
I arrived at the correct result.

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.

No, at startup, the source always sees Z0, regardless of the load
termination. The load termination has no effect at the source end until
the reflected wave returns. And if the line is terminated in Z0, then
there is no reflected wave and the source sees Z0 forever. To the load,
a terminated line looks exactly like a plain resistor of value Z0. (I'm
assuming a lossless line as we have been all along.)

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.

With the line terminated in 50 ohms, steady state is reached as soon as
the initial forward wave reaches the load.

If the termination is some value other than Z0, the steady state
impedance seen by the transmission line will be the load impedance (for
our example one wavelength line or any line an integral number of half
wavelengths long). When starting, the initial impedance is always Z0 for
one round trip, for any line length. Then it will jump or down. With
each successive round trip, the impedance will either monotonically step
in the direction of the final value, or oscillate around it but with the
excursion decreasing each time. Which behavior you'll see and the rate
at which it converges are dictated by the relationship among source,
load, and characteristic impedances.

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.

The transmission line will react to a perfect current source in parallel
with a resistance exactly the same as for a perfect voltage source with
a series resistance. Put them in boxes and there's no test you can
devise which can distinguish them by tests of the terminal
characteristics. So yes, that's fine, or just a black box about which
the contents are completely unknown except for any two of the open
circuit voltage, short circuit current, or impedance. (It's assumed that
the box contains some linear circuit that doesn't change during the
analysis.) The results will be identical for any of the three choices.

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.

We define the maximum amount of power which can be extracted from the
box, but not the power that it's delivering. If we open circuit or short
circuit the box or terminate it with a pure reactance, it's delivering
no power at all. If we terminate it with the complex conjugate of its
impedance, it's delivering V^2 / (2 * Rp) where Rp is the parallel
equivalent load resistance and V is the open circuit voltage. With any
other termination it's delivering less than that but more than zero.

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

This presumes that there are power waves bouncing around on the line.

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?

That's not a question I can answer, since it's a consequence of a
premise I don't believe. If your premise has merit, you should be able
to express it mathematically and arrive at the answer to the question.
Feel free to do an analysis using multiple sources and traveling waves
of power. If you get the correct answer for how a line actually behaves,
we'll try it out with a number of different conditions and see if it
holds up.

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.

That's correct but no dilemma.

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.

By "our" analysis do you mean my analysis or the one you propose?
There's no rule saying power can't disappear -- power is not conserved.
In fact, look at any point along the open circuited transmission line
and you'll find that the power "disappears" -- goes to zero -- twice
each V or I cycle. When you first turn on the source, the source
produces average power. In steady state, the average power is zero. It
"disappeared", without any dissipative elements in the circuit. Energy
*is* conserved, however, and there is no energy disappearing in my
analysis. Feel free to try and conserve power in your analysis if you
want. But energy had better be conserved in the process.

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.

Sure. The only difference to the analysis procedure would be to change
the source reflection coefficient from -1 to +1, since a perfect current
source has an infinite impedance. I don't know how it would affect the
final outcome without going through the steps. Of course, the reflection
coefficient for the current wave would go from +1 to -1, so I imagine
you'd run into the same conceptual problem if you tried doing an
analysis of the current waves.

Roy Lewallen, W7EL

I agree that we would have the same problem with a perfect current
source which had an infinite impedance.

How about using a perfect POWER source, that could not absorb power.
Output power would be limited by the external impedance.
Mathematically, the perfect POWER source would be described by

Ps = (Vp^2)/Zvp = Zip * Ip^2 where Ps = maximum power output, Vp =
voltage from perfect voltage source, Zvp = 0, Zip = infinity, and Ip =
current from a perfect current source.

The power output could be infinite, but power could never be absorbed by
the source.

Just as for both perfect voltage and perfect current sources, the actual
power output would be limited by external loads.

The impedance of the perfect POWER source would be Vp/Ip, both
controlled by external loads. To me that means that the output power
from the perfect POWER source would follow the impedance presented by
the load, but power going into the source would be defined as being
zero.

Would the "perfect power source" be acceptable to you?

73, Roger, W7WKB

Roger wrote:

I agree that we would have the same problem with a perfect current
source which had an infinite impedance.

How about using a perfect POWER source, that could not absorb power.
Output power would be limited by the external impedance. Mathematically,
the perfect POWER source would be described by

Ps = (Vp^2)/Zvp = Zip * Ip^2 where Ps = maximum power output, Vp =
voltage from perfect voltage source, Zvp = 0, Zip = infinity, and Ip =
current from a perfect current source.

The power output could be infinite, but power could never be absorbed by
the source.

Just as for both perfect voltage and perfect current sources, the actual
power output would be limited by external loads.

The impedance of the perfect POWER source would be Vp/Ip, both
controlled by external loads. To me that means that the output power
from the perfect POWER source would follow the impedance presented by
the load, but power going into the source would be defined as being zero.

Would the "perfect power source" be acceptable to you?

No. As I mentioned on another thread, this is a nonlinear device, which
would not permit using the linear circuit analysis I used. It would
require reverting to fundamental differential equations, which I'm not
willing to do. I'll be willing to use absolutely any linear source
(which doesn't change during the analysis period) you can devise. Any
linear source can be reduced to a Thevenin or Norton equivalent to
produce identical results.

Can't your concept of transmission lines deal with linear sources? If
not, why not?

Roy Lewallen, W7EL

Roy Lewallen wrote:
Roger wrote:

I agree that we would have the same problem with a perfect current
source which had an infinite impedance.

How about using a perfect POWER source, that could not absorb power.
Output power would be limited by the external impedance.
Mathematically, the perfect POWER source would be described by

Ps = (Vp^2)/Zvp = Zip * Ip^2 where Ps = maximum power output, Vp =
voltage from perfect voltage source, Zvp = 0, Zip = infinity, and Ip =
current from a perfect current source.

The power output could be infinite, but power could never be absorbed
by the source.

Just as for both perfect voltage and perfect current sources, the
actual power output would be limited by external loads.

The impedance of the perfect POWER source would be Vp/Ip, both
controlled by external loads. To me that means that the output power
from the perfect POWER source would follow the impedance presented by
the load, but power going into the source would be defined as being zero.

Would the "perfect power source" be acceptable to you?

No. As I mentioned on another thread, this is a nonlinear device, which
would not permit using the linear circuit analysis I used. It would
require reverting to fundamental differential equations, which I'm not
willing to do. I'll be willing to use absolutely any linear source
(which doesn't change during the analysis period) you can devise. Any
linear source can be reduced to a Thevenin or Norton equivalent to
produce identical results.

Can't your concept of transmission lines deal with linear sources? If
not, why not?

Roy Lewallen, W7EL

No, my concept of transmission lines deals fine with linear sources, it
is the non-linear constant voltage source and constant current source
that can not handle a second source of power arriving at a time later
than the original pulse.

Thank you for much thoughtful discussion. I have learned a lot, and
sharpened my skills. I won't drop the topic, because it can bear great
fruit, but let's you and I drop it for a while.

You may be right about differential equations needed for an perfect
POWER source, but so far, I don't think so. I am being honest that I
think your analysis applies up to the point of using the -1 reflection
factor at the source. In fact, I will probably use it in the future,
with an attribute to you.

Following Keith's post discussing zero voltage as a current source, I
can see why you might insist on using the -1 reflection factor. As time
passes, I will try to improve the concept of the perfect POWER source to
see if that can bridge the conceptual differences, but retain the
relative simplicity of sine waves adding.

73, Roger, W7WKB