On May 16, 6:03*am, Wimpie wrote:
I am not ignoring a problem (as you suggested), I am just using the
right tool to solve a problem.

I'm sorry, Wim, that is just not true. When you convert a V/I ratio to
a lumped circuit impedance, you are switching models in mid-example
and it changes everything in one direction (while keeping conditions
the same in the other direction). Switching to a model that doesn't
recognize reflections at all when a question about reflections arises
is an obvious logical diversion. The lumped-circuit model does not
recognize reflected energy and therefore does not allow the tracking
of reflected energy. The distributed network/wave reflection model
does allow for the tracking of reflected energy and was developed
because of the limitations of the lumped circuit model.

I will buy your assertion that one can use voltage and current to
achieve the same thing if one is careful not to violate the known laws
of EM wave physics. In my example, the reflected power on the 100 ohm
line is 12.5 watts. The reflected voltage is 35.35 volts. The
reflected current is 0.3535. The reflected voltage and reflected
current are 180 degrees out of phase so the power is real, i.e.
cos(180)=1.0, and the reflected wave has a Poynting vector magnitude
of 12.5 watts (per coax cross-sectional area).

In my very simple example, there is no reflected power on the 50 ohm
feedline yet there is (35.35)(0.3535)=12.5 watts of reflected power
from the load incident upon the 100/50 ohm impedance discontinuity. I
ask you again: Exactly what phenomenon of EM wave physics causes the
reflected wave to *reverse* its momentum and direction of energy flow
when the magnitude of the reflection coefficient is 0.3333? As long as
you refuse to answer this simple question about such a simple example,
this discussion will go nowhere.

Cecil, I think you have sufficient knowledge to form an opinion
without hiding behind others. You also have the equipment to figure
out some things yourself, and I gave some hints to help you. *The only
question is, are you willing to do this?

If you cannot answer the simple question about what happens to the
reflected energy at a simple passive impedance discontinuity, I am not
about to trust your assertions about what happens inside an active
source. Trying to introduce a more complicated example while refusing
to deal with the very simple example that I provided is an obvious and
typical logical diversion. Again, I am not going to cooperate in your
attempts at diversions. If you don't know why reflected energy
reverses momentum and direction at a passive Z0-match, just say so.

Here's an easier example: Two EM waves superpose in a Z0=100 ohm
environment. Each wave is 100 volts at 1 amp = 100 watts. The phase
angle on wave1 is +60 degrees and the phase angle on wave2 is -60
degrees. The superposition results in a new wave of 100 volts at 1 amp
= 100 watts with a phase angle of zero degrees.

We superposed two 100 watt waves and the result was one 100 watt wave.
What happened to the other 100 watts?
--
73, Cecil, w5dxp.com
"Halitosis is better than no breath at all.", Don, KE6AJH/SK