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?

I've used SPICE to verify my analysis. If I modify the SPICE model to
show the voltage just inside the line, do you think it will show the
voltage to be 2*sin(wt)? SPICE makes no assumptions about waves,
bouncing or not, so if it shows sin(wt) (as it certainly will), where do
you think its error is?

Roy Lewallen, W7EL