Follow up and more corrections.
Roger wrote:
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.

The equation can NOT be OK for anything except a single interaction
point. It can not be used to plot the interaction between waves because
the wave location information is not present.

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


In the last 24 hours, Roy posted a revised analysis that contains
results useful here. He presented a voltage example that resulted in a
steady state with steady state voltages 4 time initial value. Under
superposition, this would equate to 4 times the initial power residing
on the transmission line under the conditions presented. This concurs
with other authors who predict power on the transmission line may exceed
the delivered power due to reflected waves.

Your equation "Ptot = Ps + Pr + 2*SQRT(Ps*Pr)cos(A)" is taken from
illumination and radiation theory to describe power existing at a point
in space near a reflecting surface. If we consider space to be a
transmission media, and the reflecting surface to be a discontinuity in
the transmission media, then we have a situation very similar to an
electrical transmission line near a line discontinuity.

It is entirely reasonable to consider that the reflection ratio between
the space transmission media and the reflective surface would result in
an storage factor equaling 4 times the peak power of the initial forward
wave. By storage factor, I simply mean the ratio of forward power to
total power on the transmission media under standing wave conditions.

Under open circuit conditions, a half wavelength transmission line will
have a storage factor of 2. Roy presented an example where the storage
factor was 4. Perhaps the space transmission media also has a storage
factor of 4 under some conditions, as described by "Ptot = Ps + Pr +
2*SQRT(Ps*Pr)cos(A)".

73, Roger, W7WKB