I'm not sure how many times it's worthwhile to keep repeating this, but
I guess I'll give it another couple of tries before giving up.

Richard Harrison wrote:
Roy Lewallen, W7EL wrote:
"If you`ll read what I`ve written, you`ll hopefully see that my only
point of contention is with your claim that waves reflect from a
"virtual short". They do not."

Seems to me they do.

If you are lucky enough to have a copy of Terman`s 1955 opus, we can
reason together.

Sorry, I'm not. All I have is a 1947 Third Edition of _Radio
Engineering_. I'm unfortunately stuck with having to think for myself.
But I trust you to quote him accurately, and Terman is to be trusted.

On page 91 is found Fig. 4-3 Vector (phasor) diagrams showing manner in
which incident and reflected waves combined to produce a voltage
distribution on the transmission line.

I'm sure they're correct, and similar diagrams can be found in many of
my other texts.

At an open circuit, the voltage phasors are in-phase.

Yes. And the current phasors are out of phase.

E2, the reflected phasor, rotates clockwise as it travels back toward
the source.

E1, the incident phasor, rotates counter-clockwise as we look back
toward the source.

Yes. These of course follows from the mathematical analysis of
transmission lines, found in many texts, and with which I'm very familiar.

Looking 1/4-wavelength back from the open-circuit, E2 and E1, each
having rotated 90-degrees, but in opposite directions, are now
180-degrees out-of-phase.

On page 92, Fig. 4-4 shows the current, which summed to zero at the open
circuit, has risen to its maximum value at 1/4-wavelength back from the
open-circuit while the voltage dropped to its minimum, nearly zero,
maybe close enough to declare a "virtual short-circuit", 1/4-wavelength
back from the open-circuit.

Yes, this is universally known.

What`s a short-circuit? Little voltage and much current.

Well, at a short circuit you'll find zero volts and any current. You'll
also find this at other places which aren't short circuits, such as
where multiple voltage waves add to zero and at the summing junction of
a perfect op amp. These aren't short circuits, but they are points of
zero voltage. Saying they are all the same is like saying that because
you find water in a creek, any place you find water must be a creek.
What sort of logic is that?

What`s the difference between a physical short and the virtual short?
Nothing except the shunting conductor.

Well, yes. For one thing, waves won't reflect from a virtual short. They
will, from a real short. Another difference is that a real short will
prevent any waves from proceeding beyond it; they pass right through a
virtual short. Good thing, too, or you wouldn't get any power to your
load. Another is behavior at other frequencies and with other
waveshapes. Walt has mentioned another, that a virtual short acts like a
real short only in one direction, even when all the other conditions for
similarity are met.

Is there current flowing at the open-circuit end of the 1/4-wave line
segment? No, the open-circuit won`t support current.

Correct, of course.

If a high-impedance generator of the same frequency were connected to
the virtual short point on the line, would it also be shorted? Yes.
Where? At the virtual short, not the open-circuit at the end of the
line.

Well, yes and no. When you first hook it to the virtual short, it won't
be shorted -- it'll see just the Z0 of the cable. Only when its output
reaches the end of the stub, reflects back, and adds to the forward wave
will it be short circuited. So it's the open end of the line which is
essential to creating the apparent short at the generator.

Now let us, as you say, reason together.

You're pointing out some similarities between a virtual short and a real
one, and giving that as evidence that waves reflect from a virtual
short. So consider a point on a 50 ohm line at which the forward and
reverse waves add to a V/I of, say, 10 ohms, purely resistive to keep it
simple. If you connect a generator (of the correct frequency) at that
point, it will see 10 ohms after things settle down to steady state,
just like your generator saw a short circuit at the "virtual short" in
steady state. So can we conclude that a traveling wave will partially
reflect when it encounters the 10 ohm point? The effective or "virtual"
reflection coefficient can be calculated as -2/3, from which the
reflected wave can be calculated. And, in fact, if we assume that such a
reflection takes place, we can calculate the magnitude and phase of the
resulting wave and, sure enough, there it will really be.

But if the wave does really reflect from this "virtual discontinuity",
we might have a problem. That point is a ways away from the "virtual
short" point (check your Terman diagram if you don't follow), so we have
a partial reflection occurring at this point as well as the full
reflection from the "virtual short". In fact, unless the line is
matched, we'll have reflections from every point along the line, or at
best everywhere except an infinitesimally short spot every half
wavelength! What a mess! Does Terman describe this problem in his book?
A diagram, perhaps, showing the infinite number of partial reflections
taking place all along the line?

No? Well, then, maybe it takes a perfect "virtual short" to get a
reflection, and even a tiny, tiny imperfection will prevent it. So that
would mean that you'd get no reflection at all from a "virtual
almost-short" on even a very slightly lossy line, right? The whole idea
goes to pot when you add even a tiny amount of loss? Or is a little loss
ok? Then we get a full reflection from a good or pretty good "virtual
short", but nothing if it gets too far from perfection. Do me a favor
and check your Terman for an equation or graph which shows just where
this abrupt transition point is (that is, at what "virtual resistance"
the reflection ceases), and why it exists.

Help me out here with my reasoning.

Roy Lewallen, W7EL