Correction. Please note the change below. I apologize for the error.

Roger Sparks wrote:
"Cecil Moore" wrote in message
...
Roger wrote:
The principles of superposition are mathematically usable, not too hard,
and I think very revealing. Yes, if we use part of the model, we must
use it all the way. To do otherwise would be error, or worse.

clip some

Roger, I have explained all of this before. If you are
capable of understanding it, I take my hat off to you.
Here's what happens in that ideal voltage source using
the rules of superposition.


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.

Could I ask a couple of questions to make sure I am understanding your
preconditions?

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

1. With the reflected voltage set to zero, the source
current is calculated which results in power in the
source resistor. Psource = Isource^2*Rsource


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

2. With the source voltage set to zero, the reflected
current is calculated which results in power in the
source resistor. Pref = Iref^2*Rsource

Now superpose the two events. The resultant power
equation is: Ptot = Ps + Pr + 2*SQRT(Ps*Pr)cos(A)
where 'A' is the angle between the source current
and the reflected current. Note this is NOT power
superposition. It is the common irradiance (power
density) equation from the field of optics which
has been in use in optical physics for a couple of
centuries so it has withstood the test of time.

OK. A few reactions.

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.

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

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

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.

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.

For the voltage source looking into a 1/2WL open
circuit stub, angle 'A' is 180 degrees. Therefore
the total power in the source resistor is:

Ptot = Ps + Pr + 2*SQRT(Ps*Pr)cos(180) or

Ptot = Ps + Pr - 2*SQRT(Ps*Pr) = 0

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

I would suggest the equation Ptot = Ps+ Pr*(sine(A) where angle 'A' is the
angle between positive peaks of the two waves. This angle will rotate twice
as fast as the signal frequency due to the relative velocity between the
waves..
No, this angle is fixed by system design. If the system changes, this
angle will rotate twice as fast as the angle change of a single wave.
The equation should be OK.
Ptot in the source resistor is zero watts but we
already knew that. What is important is that the
last term in that equation above is known as the
"interference term". When it is negative, it indicates
destructive interference. When Ptot = 0, we have
"total destructive interference" as defined by
Hecht in "Optics", 4th edition, page 388.

Whatever the magnitude of the destructive interference
term, an equal magnitude of constructive interference
must occur in a different direction in order to
satisfy the conservation of energy principle. Thus
we can say with certainty that the energy in the
reflected wave has been 100% re-reflected by the
source. The actual reflection coefficient of the
source in the example is |1.0| even when the source
resistor equals the Z0 of the transmission line.
I have been explaining all of this for about 5 years
now. Instead of attempting to understand these relatively
simple laws of physics from the field of optics, Roy
ploinked me. Hopefully, you will understand.

All of this is explained in my three year old Worldradio
energy article on my web page.

http://www.w5dxp.com/energy.htm
--
73, Cecil http://www.w5dxp.com

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.

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.

So what do you think Cecil?

73, Roger, W7WkB