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