Cecil Moore wrote:
Gene Fuller wrote:
This says that steady state depends on something else, namely the
beginning and the end of the steady state condition. That is simply
incorrect. In steady state conditions there is no concept of beginning
or end.

A 12VDC battery is sitting there with a 200 amp*hour charge. Are
you asserting that there is no concept of where the 200 amp*hours
came from? Please tell me you are not that stupid.

Consider the one second long transmission line with 200W of forward
power and 100W of reflected power. That requires 300 joules of
energy during steady-state. If the 300 joules was not supplied
during the transient state, then it must have magically appeared
out of thin air in violation of the conservation of energy principle?
Is that what you are trying to tell us?

Cecil,

You can wave your hands all you want, but it won't have much impact on
the correct math and physics.

Try writing the appropriate equations for your puzzler, in steady state
conditions, and then figure out where to insert the transient behavior.
Good luck.

This is basic stuff taught in numerous math and technical courses. If
don't accept the basic math, then I guess we will not agree.

73,
Gene
W4SZ

Gene Fuller wrote:
Try writing the appropriate equations for your puzzler, in steady state
conditions, and then figure out where to insert the transient behavior.

I have already provided the equations, Gene. In a one second long
lossless transmission line, 200 watts of forward power equals 200
joules of energy in the forward wave. 100 watts of reflected power
equals 100 joules in the reflected wave. Total joules in the
transmission line equals 200 + 100 = 300 joules. The equations
are trivial.
--
73, Cecil, W5DXP

wrote:
I'm still thinking what to make of it, but I thought I'd post the math
for people to look at (and check, please!!!!)

Thanks, Dan.

If I understand correctly, Roy's argument is that since
the source is not supplying any steady-state energy to
the lossless stub, there is no energy in the reflected
wave within the stub.

If we make Roy's lossless stub one second long, the
picture becomes clearer. For the first two seconds,
the 100 watt source pours 200 joules of RF energy
into the stub. After that, during steady-state, the
source supplies zero energy. But those 200 joules
are still in the stub, 100 joules in the forward
wave and 100 joules in the reflected wave, and they must
obey the conservation of energy principle.

There is *always* the exact amount of energy in the
forward and reflected waves as required to achieve
the forward and reflected power readings in a Z0
calibrated environment.
--
73, Cecil, W5DXP

On Sat, 26 Aug 2006 02:33:03 GMT, Gene Fuller
wrote:

Cecil Moore wrote:
Gene Fuller wrote:
In steady state conditions there is no concept of beginning
or end.

If the 300 joules was not supplied
during the transient state, then it must have magically appeared
out of thin air in violation of the conservation of energy principle?

Hi Gene,

The bare contradiction is enough to condemn this thread.

However, it does have its amusing character of "Who's on first?"

Continuing that metaphor, Cecil would believe having been born on
third base, that he had hit a triple to be there. ;-)

73's
Richard Clark, KB7QHC

On Fri, 25 Aug 2006 20:10:14 -0700, Richard Clark
wrote:

Continuing that metaphor, Cecil would believe having been born on
third base, that he had hit a triple to be there. ;-)

For the concept challenged,

Being on third is the steady state.

The transient state is one of:
Being born;
hitting a triple;
hitting a double and then a batter advancing the runner(s);
hitting a single (then like wise with the batter's assist);
stealing a base;
or two.

The steady state also has to satisify other conditions that were taken
up by transients like outs and innings. e.g. being on third with
three outs does not mean you can stay on third.

Thus the next transient is
The side is retired (state change)
or, as in this case of the bottom of the ninth and an untied score
The game is over (solution).

73's
Richard Clark, KB7QHC

A correction - insert dt instead of dz.

The fundamental partial differential equations of transmission lines
are -

- dv/dz = R + L*di/dt

- di/dz = G + C*dv/dt

where volts v and current i are incremental functions of distance and
time, and z is incremental distance along line.

Everything else follows.

Similar equations can be written in terms of frequency.

It is often easier to think in terms of Time and Distance rather than
Frequency and Impedance.
----
Reg.

Cecil Moore wrote:

If I understand correctly, Roy's argument is that since
the source is not supplying any steady-state energy to
the lossless stub, there is no energy in the reflected
wave within the stub.

That sounds right... if the reflection coefficient is 1 then there's no
net power flux into/through the line in steady state, and this can be
described if you like by counterpropagating waves each carrying the
same amount of energy.

The problem is, in your other example where you say 200 joules in the
forward wave + 100 joules in the reflected wave = 300 joules in the
line total, you're neglecting the vector character of the power flux.

Yes, the waves carry energy, but they carry it in different directions.
The net power flux in the line with 200W forward power and 100W
reflected power is 100W net power flowing to the load from the source.
The real part of the Poynting vector of the reflected wave opposes that
of the forward wave, as long as I got all the signs right.

I don't think we can neglect the imaginary part of the Poynting vector,
though. It's not zero and I think it represents the flow of the power
in the stored fields in the line, and if we want to get the total
energy in the line, we have to include the stored fields.


Dan

Richard Clark wrote:
Continuing that metaphor, Cecil would believe having been born on
third base, that he had hit a triple to be there. ;-)

That is actually the other side of the argument. When
an observer arrived after the game started, Cecil was
on third base. Using steady-state logic, the newcomer
assumes that Cecil is there without ever having to swing
a bat.

Someone looks at a transmission line during steady-state.
The source is supplying 100 watts. The load is dissipating
100 watts. The forward power is 200 watts. The reflected
power is 100 watts. The incorrect assumption is that the
source is incapable of delivering the 200 watts of forward
power and the 100 watts of reflected power. But the exact
amount of energy required to support those values
was provided to the transmission line before steady-
state was reached. It was rejected by the load and is
still there in the transmission line during steady-state.
--
73, Cecil http://www.qsl.net/w5dxp

wrote:
Cecil Moore wrote:

If I understand correctly, Roy's argument is that since
the source is not supplying any steady-state energy to
the lossless stub, there is no energy in the reflected
wave within the stub.

That sounds right... if the reflection coefficient is 1 then there's no
net power flux into/through the line in steady state, and this can be
described if you like by counterpropagating waves each carrying the
same amount of energy.

If the counterpropagating waves each carry the same amount
of energy then there cannot be zero energy in the reflected
wave.

Some people on this newsgroup say that the wave reflection
model is invalid, that forward and reflected waves don't
have a separate existence. From QEX: "Contrary to popular
views, the forward and reverse waves on a transmission line
are not separate fields." It would follow that a laser beam
normally incident upon an ideal mirror results in a beam of
light not superposed from separate forward and reverse fields.

I have challenged people holding those concepts to create a
standing wave without superposing separate forward and
reverse waves and have gotten zero responses.

The problem is, in your other example where you say 200 joules in the
forward wave + 100 joules in the reflected wave = 300 joules in the
line total, you're neglecting the vector character of the power flux.

I apologize if I accidentally gave that wrong impression.
I subscribe to Ramo/Whinnery's notion that the power reflection
coefficient is equal to the reflected Poynting vector divided by
the forward Poynting vector. Pz-/Pz+ = |rho|^2

Yes, the waves carry energy, but they carry it in different directions.
The net power flux in the line with 200W forward power and 100W
reflected power is 100W net power flowing to the load from the source.
The real part of the Poynting vector of the reflected wave opposes that
of the forward wave, as long as I got all the signs right.

Yes, that's what I am saying. The other side would assert that
there is no reflected Poynting vector and no forward Poynting
vector - that there only exists the net Poynting vector. This
is a change away from mainstream RF engineering taught in the
1950's. The "modern" concept seems to be that forward and
reflected waves don't exist. All that exists are the standing
wave and the net forward traveling wave with nothing moving
backwards.
--
73, Cecil http://www.qsl.net/w5dxp

Cecil,

You've set up a false dichotomy here. When I, and others, write "The
electric field is the superposition of a forward and reverse traveling
wave" maybe it would be better to say "The electric field has two
terms, one that appears to be a forward traveling wave and one that
appears to be a reverse traveling wave." or something like that.
There's one electric field vector and one Poynting vector. Or there
are two. The structure of the electric field and the structure of the
real part of the Poynting vector both admit BOTH explanations of what's
happening.

You're not gonna get 300J in your one second line.... the stored energy
flux in the line depends on the wavelength of the incident RF, and in
retrospect, you might expect this from the fact that a misterminated
line goes through cyclical impedance variations as you change its
length (something that I know you're quite familiar with :-) )

I think the energy density per unit length in the line is proportional
to the Poynting vector (or it's integral over the cable cross section,
and the proportionality constant is the group velocity of the waves, I
think) I left Jackson at work, so I'm not certain right now. What I
am certain of is that you can't take the energy in the forward wave and
add it to the energy of the reflected wave and get that there are 300J
in a 1 second line carrying a 200W forward wave and a 100W reverse
wave. Rather, there's a 100W net forward power flux and THAT will give
you the energy contained in the part of the field that's actually
moving from source to load. The energy contained in the reactive part
has an integral that's going to cyclically vary with the length of the
line, and sometimes goes through zero (kL or kL - phi equal to an
integer multiple of Pi... or any integer multiple of a half wavelength,
which happens to be the length of an impedance repeating line, eh?)



Dan