On Fri, 04 Jun 2004 23:06:16 -0500, Cecil Moore wrote:

Walter Maxwell wrote:
Cecil, at this point I'm not clear what's happening in your above example. Where
and what is the phase reference for these two waves? It appears to me that the
reference phase must be that of the source wave, because the voltage and current
in both rearward traveling waves are 180 degrees out of phase. Educate me on how
the cancellation takes place and how the energy reverses direction.

Cecil, your reference below to Chapter 23 in Reflections has re-oriented me and
I'm now on your page. This is the case of a 1/4wl matching transformer. I didn't
recognize my own writing. Sorry.

J. C. Slater explains how the cancellation takes place.
"The method of eliminating reflections is based on the interference
between waves. Two waves half a wavelength apart are in opposite
phases, and the sum of them, if their amplitudes are numerically
equal, is zero. The fundamental principle behind the elimination of
reflections is then to have each reflected wave canceled by another
wave of equal amplitude and opposite phase."

Yes, Cecil, this quote is my Ref 35, which I used to support this concept in my
QST article in Oct 1974, so I am well familiar with it. And yes, you are
correct in that the waves reflected from points A and B in Fig 6 are in the
phase you state.

Note that the above applies to both voltage waves and current waves.
Both voltage and current go to zero during complete destructive interference,
i.e. both E-field and H-field go to zero during complete destructive
interference.

But Cecil, take another look at Fig 6 on page 23-5 to note that those two waves
arrive 180 out of phase at point A, which means only that the E and H fields
cancel in the rearward direction only, resulting in a Zo match to the source.
But what's important (and I inadvertantly omitted this fact in the book, which
will be corrected in III) is that when the waves reflected from B reach A they
find a discontinuity at A of an open-circuit type condition, because the line Zo
now goes from low to high (in the rearward direction). Thus when the waves
reflected at B arrive at A the voltage which is already at 360 (0) degrees does
not change phase, but the current which is at 180 degrees on arrival changes by
180 due to the open circuit at A to rearward traveling waves. Thus the current
is now also at 0 degrees. With both voltage and current at 0 degrees relative to
the source, all the power in the waves reflected at B are re-reflected at A and
add to the source power.

Consequently, Cecil, both the E and H fields go to zero only in rearward
direction at the match point, which is necessary for no reflections to travel
rearward of the match point, but the H field goes to zero only temporarily while
the E field is doubled temporarily, as they should when encountering an open
circuit as they do on arriving at A.

It seems to me the out of phase voltage yields a short circuit, while the out of
phase current yields an open circuit. How can both exist simultaneously?

I've answered my own dumb question here.

They can't, and that's my argument. It's easy to understand. If the two rearward-
traveling voltages are 180 degrees out of phase then the two rearward-traveling
currents MUST also be 180 degrees out of phase, since the reflected current is
ALWAYS 180 degrees out of phase with the reflected voltage. If one looks only
at the voltages, one will say it's a short-circuit. If one looks only at the
currents, one will say it's an open-circuit.

I think I've addressed this paragraph above.

And further, what circuit can produce these two waves simultaneously?

As I embarrassingy discovered it's the 1/4wl transformer--dumb me.

According to J. C. Slater (and Reflections II, page 23-9) a match point
produces these two waves simultaneously, two reflected voltages 180 degrees
out of phase and two reflected currents 180 degrees out of phase.

In addition, I believe your example has changed the subject. My discourse
concerns what occurs to an EM wave when it encounters a short circuit.

I was confused here, Cecil, because you did change the subject from discussing
what happens to the E and H fields when encountering a short to the 1/4wl
transformer example without my catching on to the change. As I said, dumb me.

There is no argument about what happens at a short circuit. What I am saying
is that a match point is NOT a short circuit.

In this case the match point at A is an open circuit.

In this
case you're going to have to prove to me that both E and H fields go to zero.
IMO it can't happen.

J. C. Slater says that's what happens in the above quote. Voltages 1/2WL apart
in time cancel to zero. Currents 1/2WL apart in time cancel to zero.

Yep, but only in the rearward direction.

So my argument with you, Cecil, is that I maintain the H field doubles on
encountering a short circuit, while you maintain that both E and H fields go to
zero. What's your answer to this dilemma?

We've answered this.

My argument is that it is NOT a short circuit. It is "complete destructive
interference" as explained in _Optics_, by Hecht where both the E-field and
B-field go to zero. J. C. Slater says that the two rearward-traveling voltages
are 180 degrees out of phase and the two rearward-traveling currents are 180
degrees out of phase. So whatever happens to the voltage also happens to the
current, i.e. destructive interference takes both voltage and current to zero.

I'll not argue with the way Hecht describes the phenomena.

G'nite, Cecil, it's 3 :20 AM and I've got to go back to bed.