Roger Sparks wrote:
Well, I would like to think I could understand it, but maybe something is
wrong like Roy suggests. I think it is all falling into place, but our
readers are not all in agreement.

Anyone who will take the time to understand and is capable
of understanding, will understand. What I am reporting are
centuries-old proven scientific concepts. Readers need only
reference an optics textbook for irradiance, reflectance,
and transmittance. "Optics", by Hecht is easy reading.

Is this a Thevenin source? If so, what is the internal resistor set to in
terms of Z0?

The example was a Thevenin source driving a 1/2WL open
line through a source resistor equal to Z0, e.g. 50 ohms.

For condition number 2 below, is this a Thevenin equivalent resistance?

Yes, all the rules of superposition and Thevenin equivalency
are being followed.

1. To add the total power of each wave and then also add the power
reduction of the phase relationship, would be creating power unless the
cos(A) is negative. Something seems wrong here, probably my understanding.

Yep, I haven't given the other corresponding equation. The
pair of equations that satisfy the conservation of energy
requirements are (in Ramo&Whinnery Poynting vector notation):

|Pz+| = P1 + P2 + 2*SQRT(P1*P2)cos(A) eq.1

|Pz-| = P3 + P4 + 2*SQRT(P3*P4)cos(-A) eq.2

In a transmission line, the Poynting vector for Pz+ and the
Poynting vector for Pz- are pointed in opposite directions.

The destructive interference, as you observed, carries a
negative sign. This negative sign does NOT create energy.
It means that some excess amount of destructive interference
energy must go somewhere. It naturally goes into constructive
interference somewhere else. Please see:

http://www.mellesgriot.com/products/optics/oc_2_1.htm

http://micro.magnet.fsu.edu/primer/j...ons/index.html

2. The two voltages should be equal, therfore the power delivered by each
to the source resister would be equal.

Under each of the two superposition steps, yes. However, when
the signals are combined, interference results, and the power
in the source resistor is reduced to zero. That "excess" energy
must go somewhere and it appears as constructive interference
redistributed toward the load. See the web pages above.

3. The power delivered to the source resistor will arrive a different times
due to the phase difference between the two waves.

We are talking average power. Please don't get bogged down in
instantaneous power which would be the same when integrated.

4. If the reflected power returns at the same time as the delivered power
(to the source resistor), no power will flow. This because the resistor in
a Thevenin is a series resistor. Equal voltages will be applied to each side
of the resitor. The voltage difference will be zero.

Yes, the result is destructive interference.

5. If the reflected power returns 180 degrees out of phase with the applied
voltage (to the source resistor), the voltage across the resistor will
double with each cycle, resulting in an ever increasing current.(and power)
into the reflected wave.

Good time to switch over to a Norton equivalent.

This is correct for this condition. The problem with the equation comes
when cos(90) which can easily happen.

That's no problem at all. It is just constructive interference
which implies destructive interference somewhere else. Destructive
interference must always equal constructive interference to avoid
violation of the conservation of energy principle.

Light, especially solar energy, is a collection of waves of all magnitudes
and frequencies. We should see only about 1/2 of the power that actually
arrives due to superposition. The reflection from a surface should create
standing waves in LOCAL space that will contain double the power of open
space. I think your equation Ptot = Ps + Pr + 2*SQRT(Ps*Pr)cos(A) would be
correct for a collection of waves, but incorrect for the example.

A collection of waves is not likely to be coherent so the equation
would not work. That equation works for any coherent EM wave. CW RF
waves are coherent.

I tried to read your article a couple of weeks ago, but I found myself not
understanding. I have learned a lot since then thanks to you, Roy, and
others. I will try to read it again soon.

You might want to pick up a copy of "Optics", by Hecht
available from http://www.abebooks.com Please pay close
attention to the chapters on interference and superposition.

So what do you think Cecil?

Also try HP's AN 95-1, an s-parameter app-note available on the
web. Pay close attention to what they say about power and what
happens when you square the s-parameter equations. Surprise!
You get the power interference equation above.
--
73, Cecil http://www.w5dxp.com