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