Keith Dysart wrote:
The key point is that since 62.5 nsec represents 1/4WL at the
frequency of interest, if the line was 1/4WL long, you would
find things that happen every 62.5 nsec.

No, that is not a key point. If we figure out some way
to get a 90 degree phase shift to happen in 10 nsec, so
be it. There is nothing that forces a phase shift to
take the same amount of time as it does in a transmission
line.

There is no doubt that all the variants discussed do share one
property with a 1/4WL open stub; the input impedance after a
long settling time is the same. But other properties, particularly
settling time, are quite different.

I don't have a problem with that statement. Everything I
have talked about is in regards to steady-state conditions.

If you are saying that along the transmission line there is a point
every 125 nsec where the impedance is 0, I would agree for
a constant impedance line, but if you change the impedance then
the points on either side of the discontinuity are not 125 nsec
apart.

If I never said that, your statement would still be true.
I agree and never said otherwise.

But 46.4 + 10 does not equal 90.

Yet we know for certain that the reflected wave has undergone
a 90 degree phase shift or else it wouldn't be in phase with
the forward wave at the feedpoint. That additional phase shift
has no time constraint. Deduction will tell one where it is
located. It is not in the 600 ohm section. It is not in the
100 ohm section. It is not at the open end of the stub. Since
the only location left is the impedance discontinuity, it must
occur at that point and indeed the Smith Chart shows us that
is exactly where it occurs in an exactly predictable value.
"Elementary, my dear Watson".
--
73, Cecil http://www.w5dxp.com