wrote:
Do you mean the mashed-potatoes energy equations where
the wave energy components are completely ignored?

Calling the steady state solution "mashed-potatoes" is ridiculous.

I'm not calling the solutions ridiculous. I am calling some
of the conceptual conclusions ridiculous. For instance, since
the *net* Poynting vector equals delivered source watts, there
are zero watts in the reflected wave even though there are
joules in the reflected wave moving at the speed of light.

If you want to know what happens during the transient, you MUST start
from the original differential equation that describes how the fields
were set up in the line in the first place, but it doesn't matter in
the steady state. The answer is the same.

Not according to some of the gurus posting here. The joules
supplied during the initial transient state are being swept
under the steady-state rug. Witness "Food For Thought #1"
on www.eznec.com

I don't want to have to do it but one of these days I think it might be
necessary.

What you will find is that there is exactly the energy in the
transmission line required to support the real speed-of-light
forward traveling wave and the real speed-of-light reflected
traveling wave - no more and no less. The argument that there
is no more energy in a transmission line with reflections than
is being supplied by the source is simply false.

The Poynting vector in a mismatched line has been worked out by me,
here, and I'd appreciate comments on the MATH.

http://en.wikipedia.org/wiki/Useran_Zimmerman/Sandbox

What about the forward Poynting vector, Pz+, and the reflected
Poynting vector, Pz-. Reference page 291, "Fields and Waves ...",
Ramo and Whinnery, 2nd edition?

My claim is that that's it. That's all. No handwaving, no setting up
steady state startup and claiming that that energy remains in the line.
The real part of the Poynting vector represents real power flux and
the imaginary part represents reactive power flux. To get the energy
that exists in the line you just add it all up...

The forward wave's average Poynting vector is real. The reflected
wave's average Poynting vector is real. Adding two real power flow
vectors is a lot easier than integrating a power flow vector with
real and imaginary parts. And, bottom line, one obtains exactly
the same results with 10% of the work. Subtract the reverse
Poynting vector from the forward Poynting vector to determine
the net power. Add the two Poynting vectors and multiply by the
length in seconds of the feedline to determine the total energy.
What could be simpler?

Am I wrong, Cecil et. al?

No, you are correct but you are going around the world to get
from California to New York. My method, supported by Ramo and
Whinnery, can be done in much less time and can be more easily
understood by people not familiar with your method. By the time
you get out your calculator, I can have your answer waiting for
you and it will be the same answer you get after wasting a lot
of time.

When the SWR is infinite, there is no energy in the line associated
with flowing power. None at all. The reflection coefficient is 1, and
all energy in the line is associated with circulating power (reactance
only, as you would expect from a stub)

You are, of course, talking about *net* "flowing" power. But your
answer is exactly the same as component forward power in real
joules/sec plus component reflected power in real joules/sec.
And thinking in terms of those real component powers is a lot
easier than your imaginary stuff.

Pz+ - Pz- = Pz-tot sourced and delivered

(Pz+ + Pz-)* feedline length in seconds = total feedline energy

There is also a physics problem with the imaginary concept. EM
wave energy must necessarily move at the speed of light. A small
amount of TV modulation will prove that those forward and reflected
waves are still moving end to end at the speed of light and still
transferring information.
--
73, Cecil http://www.w5dxp.com