Roy Lewallen wrote:
Well, let's see. Begin with two identical, phase locked generators with
fixed 50 ohm output resistances. Connect the output of generator A to a
one wavelength 50 ohm transmission line, and the output of generator B
to a half wavelength 50 ohm line. Connect the far ends of the lines
together, and to a third transmission line of any length. Let's properly
terminate the third line for simplicity. Superposition should work with
this system, so begin by turning off generator A. The one wavelength
line is now perfectly terminated and looks just like a 50 ohm resistor
across the third line. Generator B puts half its power into generator
A's output resistance and half into the third line's load.

If generator A has 100 watts available to a 50 ohm load, how
much power is being dissipated in the resistor at the end of
the third transmission line?

Did you account for the fact that the generator sees 25 ohms,
not 50 ohms? Are you ignoring the reflections on generator
A's feedline?

There's a
wave traveling down that line. Now turn off B and turn on A, and note
that half of A's power is going to B's source resistance and half into
the third line's load.

With either source turned off, the voltage reflection coefficient at
the junction of the three lines is rho = (25-50)/(25+50) = -0.33.
Did you account for the resulting reflections?

If you believe as I do that waves don't interact in a linear medium and
believe in the validity of superposition in such a medium, then you
believe that when both generators are on there are two waves going down
that third line. They're exactly equal but out of phase, so they add to
zero everywhere along the line. With the system on and in steady state,
there's absolutely no way you can tell the difference between this sum
of two waves and no waves at all. *They are the same.* If you look at
the input to the third line, you'll find a point with zero voltage
across the line, and zero current entering or leaving it. Where you will
get into serious trouble is if you assign a power to each of the
original waves.

Would you agree that the waves are EM waves? Would you agree
that the waves each have an E-field and a B-field? Would you
agree that the joules/sec in each wave is proportional to
ExB and that the waves could not exist without those joules/sec?
There is absolutely no problem assigning joules/sec to each
EM wave. In fact, the laws of physics demands it.

Then you'll have a real job explaining where the power
in one of the waves went when you turned on the second generator --
among other problems.

It's no problem at all. Optical physicists have been doing it
for over a century. The energy analysis at the feedline junction
point is very straight forward. It simply obeys the wave reflection
model, the superposition principle, and the conservation of
energy principle.

A solution to the problem based on the assumption that there are no
waves on the third line and one which claims there are two canceling
waves are equally valid, and both should give identical answers.

EM waves cannot exist without ExB joules/sec, i.e. EM waves
cannot exist devoid of energy. Those two waves engaged in
destructive interference which redirected the sum of their
energy components back toward the sources as constructive
interference. They interacted at the physical impedance
discontinuity and ceased to exist in the third feedline.

In your example, with both sources on, the SWR on the two
generator feedlines is infinite. There is exactly enough joules
stored in each line to support the forward and reflected powers
measured by a Bird directional wattmeter.
--
73, Cecil, w5dxp.com