Roy Lewallen wrote:
Roger wrote:

I see your example as identical to Keith's example of two wave pulses
traveling in opposite directions.

Not quite. The voltage following a pulse is zero; with my sine wave
example the source continues producing a sine wave at all times while
wave interaction is occurring.

At the point of interaction, Keith's example has a reflection factor
of 1 or zero, depending upon whether the waves bounce or pass.
Keith's example is not a short circuit because two pulses of identical
polarity are interacting so a reflection factor of -1 could never exist.

Because there are no mathematics and, as far as I can see, no mechanism
for wave interaction, I can't discuss Keith's example except from the
standpoint that the waves don't interact. (Excluding, of course,
superposition from the meaning of "interact".)

In your example, the presence of voltage from the ideal source creates
conditions identical to Keith's example for the returning reflected
wave. Accepting this premise, then the reflection factor must be
either 1 or zero, depending upon whether the waves bounce or pass.

Sorry, I don't accept the premise.

By assuming that the waves reflect at the ideal source, you proved
that the reflection factor is -1, which is the factor for a short
circuit.

Yes, that is correct.

This can not be the case, so waves must not reflect at the ideal
voltage source, they must pass.

Well, yes it can be the case, and is. But let's see where rejection of
this fact leads.

vtot(t, 0) = vf(t, 0) + vr(t, 0)

= sin(wt)tot = sin(wt) + sin(wt)

= 2*sin(wt)

Someone will certainly say the the vtot(t,0) at the source location is
the source voltage, because it is defined that way.

I certainly say that.

The conditions at point (t,0) itself is actually unknown (because vf
mysteriously appears, and vr disappears by going off the transmission
line), but point (t, 0) is defined by assumptions. Therefore, at

vtot(t, 0) = vf(t, 0) + vr(t, 0)

= sin(wt)tot = sin(wt) + sin(wt)

= 2*sin(wt)

So by rejecting the fact that the waves reflect at the ideal source,
you're forced to conclude that the voltage at the output of a perfect
source is unknown, with waves "mysteriously" appearing and disappearing.
It also appears that you've violated Kirchoff's voltage law at the
input. My analysis requires no such contradictions, unknown voltages, or
mysteries. As an engineer, I like my analysis a whole lot better,
because it can give me answers which are testably correct.

Can you use your theory to show the voltage at all times at the input
or, if you choose, just inside the input of the line, as I've done using
conventional theory?
Yes. Here is the difference. I am using a short cut by looking at the
way waves travel and superimpose. My assumption is that the input
impedance matches the impedance of the line. Thus 1v generates a 1v
sine wave. For a simple example, such as any multiple of 1/2 wave, this
causes current proportional to applied voltage and impedance of the
line. For example, 50v applied to a 50 ohm line gives a 1 amp current.

We know a line 1/2 wavelength long will absorb energy for the length of
time it takes for a wave to travel the length of line and return, so
power must be applied during the entire time. Obviously, 2 half waves
will be applied over that time period so the amount of energy on the
line is equal to the amount of energy contained in the power applied
over time by two half waves, so total power applied is 2 * initial power.

Equally obvious, by leaving the source connected after the reflected
wave has returned, the question of how the source reacts to the external
application of power must be answered. This is a second question,
unrelated to traveling waves on a transmission line. My answer to this
question is to disconnect the source and replace the source with a
connection identical to the condition at the far end of the transmission
line. You can see that this leaves the transmission line isolated with
power circulating on the line (is is POWER, time is involved).

Your solution (as presented in Analysis 2) and the Power Analysis was to
keep the source supplied at all times in an attempt to create continuous
real world conditions, as contrasted to my method of a quick and
correctly timed disconnect. Both experiments are real world simulations
and can be performed.

Your analysis is identical to what happens at a real world transmitter.
A very good, automatically tuned transmitter! Your choice of an
ideal voltage source amounts to setting conditions that automatically
adjust for impedance changes as fast as the change happens. The
transmission line sees an outgoing impedance change going from 150 ohms
to infinity in your example. The source sees an outgoing impedance
change from 200 ohms to infinity during the same analysis period.

It is interesting to compare the final voltage ratios ( directly
proportional to energy levels and directly proportional to power levels)
on the transmission line to the SWR ratio. You used a SWR ratio of 3:1
and I used 1:1. You found that the total energy stored on the line was
4 times the initial energy applied, and I predicted that the total
energy stored was 2 times the initial. The link here is that the power
storage factor(SF) (I know you dislike the term) is found from

SF = SWR + 1.

Your storage factor was SF = 3 + 1 = 4. My storage factor was
SF = 1 + 1 = 2.

If we can accept that the storage factor is always SF = SWR + 1 for all
situations, then we have a very convient way to predict the voltage
increase that results form impedance mismatch at a discontinuity. I am
not yet ready to accept this generalization, but we should look at it
more carefully.

Ending comment. Both your and my analysis are correct. Congratulations
to both us!

73, Roger, W7WKB