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. All
other terms besides the steady-state ones have left the picture in the
steady state. This is true of the waves, their amplitudes, the energy
in the line, everything. That's what makes steady-state analysis
useful... you can do it and be right.

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. If you start with the full,
time dependent equations for turning the source on into some line,
whatever it is that happens with the power and where it goes will
become clear from the transient solution, which I know you haven't
worked out. Neither have I.

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

This is not a problem that will be solved in with Logic and English,
Cecil. There is no argument if you start with a full , time dependent
mathematical description of the waves. That is what will answer the
question "Where Does the Power Go" You'll end up with a time dependent
Poynting vector that will tell you. I hypothesize, for no particularly
good reason, that such a description will lead to the excess power
having been delivered to the load through the interaction of the
transient solutions on the line.

- - - - - -

Copied and pasted from the rest of my post at eHam:

In a matched line, none of us would disagree that the power flux out of
one end of the line is the power flux into the other end.

From what I understand, the electromagnetic energy contained in that
line is related to the Poynting vector. The integral of the Poynting
vector over the cross sectional area of the line gives you the time
averaged power flowing in the line. The power flows at the group
velocity of the waves in the line, for normal transmission lines, this
is somewhere between 0.6 - 1.0 times the speed of light.

The power flow divided by the group velocity gives you the energy
density per unit length in the line.

The power flow is the Poynting vector integrated over the cross
sectional area of the line.

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

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...

Take the real part, divide it by the propagation velocity on the line,
integrate it over the area of the line (I think for TEM, the fields
are uniform and so multiplying by the area of the line is sufficient)

This gives you the energy density per unit length for the power that's
flowing from source to load. Integrate that over the length of the
line. Set aside.

Take the imaginary part, do the same thing. This gives you the energy
contained in the line of the reactive, circulating power. Add in the
previously calculated energy of the flowing power, and you're done.

Am I wrong, Cecil et. al? If so, Cecil, could you please write down
the description mathematically or get someone to help you out, and
could you please look at and check my math?

I started with the assumption that there are two counterpropagating
waves with some arbitrary electric field amplitudes and an arbitrary
phase relationship. I didn't use anything about Thevenin. I didn't
use anything about virtual open circuits.

The forward and reverse waves have been included, from the beginning.
They're both there.

When the SWR is 1:1, there is no energy in the line associated with
reactive stored power. None at all. The reflection coefficient is
zero, and all energy in the line is associated with flowing power.

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)


Dan

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

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. All
other terms besides the steady-state ones have left the picture in the
steady state. This is true of the waves, their amplitudes, the energy
in the line, everything. That's what makes steady-state analysis
useful... you can do it and be right.

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. If you start with the full,
time dependent equations for turning the source on into some line,
whatever it is that happens with the power and where it goes will
become clear from the transient solution, which I know you haven't
worked out. Neither have I.

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

This is not a problem that will be solved in with Logic and English,
Cecil. There is no argument if you start with a full , time dependent
mathematical description of the waves. That is what will answer the
question "Where Does the Power Go" You'll end up with a time dependent
Poynting vector that will tell you. I hypothesize, for no particularly
good reason, that such a description will lead to the excess power
having been delivered to the load through the interaction of the
transient solutions on the line.

- - - - - -

Copied and pasted from the rest of my post at eHam:

In a matched line, none of us would disagree that the power flux out of
one end of the line is the power flux into the other end.

From what I understand, the electromagnetic energy contained in that
line is related to the Poynting vector. The integral of the Poynting
vector over the cross sectional area of the line gives you the time
averaged power flowing in the line. The power flows at the group
velocity of the waves in the line, for normal transmission lines, this
is somewhere between 0.6 - 1.0 times the speed of light.

The power flow divided by the group velocity gives you the energy
density per unit length in the line.

The power flow is the Poynting vector integrated over the cross
sectional area of the line.

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

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...

Take the real part, divide it by the propagation velocity on the line,
integrate it over the area of the line (I think for TEM, the fields
are uniform and so multiplying by the area of the line is sufficient)

This gives you the energy density per unit length for the power that's
flowing from source to load. Integrate that over the length of the
line. Set aside.

Take the imaginary part, do the same thing. This gives you the energy
contained in the line of the reactive, circulating power. Add in the
previously calculated energy of the flowing power, and you're done.

Am I wrong, Cecil et. al? If so, Cecil, could you please write down
the description mathematically or get someone to help you out, and
could you please look at and check my math?

I started with the assumption that there are two counterpropagating
waves with some arbitrary electric field amplitudes and an arbitrary
phase relationship. I didn't use anything about Thevenin. I didn't
use anything about virtual open circuits.

The forward and reverse waves have been included, from the beginning.
They're both there.

When the SWR is 1:1, there is no energy in the line associated with
reactive stored power. None at all. The reflection coefficient is
zero, and all energy in the line is associated with flowing power.

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)


Dan


Don't expect much math from Cecil. He always neglects the phase
difference of the reflected wave, and won't do the algebra, no matter
what.
73,
Tom Donaly, KA6RUH

Tom Donaly wrote:
Don't expect much math from Cecil. He always neglects the phase
difference of the reflected wave, and won't do the algebra, no matter
what.

Simply not true, Tom. For a Z0-match, the phase is always
zero or 180 degrees. Only addition or subtraction is
ever required at a Z0-match. I've got the answer while
you guys are still trying to figure out the cosine of
zero degrees. :-)
--
73, Cecil http://www.w5dxp.com