Hey Cecil,

Can you sum up the problem with conservation of energy that modern RF
textbooks get wrong?

Dan

wrote:
Can you sum up the problem with conservation of energy that modern RF
textbooks get wrong?

They don't get it wrong - they just don't discuss it at all.
But here is an example of the problem:

http://eznec.com/misc/food_for_thought/

First article - last paragraph. W7EL considers
steady-state conditions while ignoring the previous
transient state conditions. He implies that the
energy in the reflected wave cannot be recovered but
it is indeed dissipated as power in the system after
power is removed from the source. The source supplies
exactly the amount of energy during the transient power
up conditions needed to support the forward and reflected
waves during steady-state. This is easy to prove. But
W7EL's Ivory Tower protects Him from peons like me.
--
73, Cecil http://www.qsl.net/w5dxp

The net power flux in the line gets smaller as the reflected wave gets
stronger while maintaining a constant electric field (constant voltage
as in Roy's example). If you can match to the new impedance at the
line input; that is, make the electric fields both stronger, you can
get a larger net power flux even in the presence of some elevated SWR.

See LaTeX formatted math at
http://en.wikipedia.org/wiki/Useran_Zimmerman/Sandbox

The flux of stored power in the line, interestingly enough, is a
sinusoidal function of position.

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!!!!)

... I'll be back later.

73,
Dan

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

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

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

Cecil Moore wrote:

[snip]

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.

Cecil,

I believe Dan has addressed this issue, and I am sure that I have on
many occasions.

When superposition applies, as it does in this linear, non-pathological
case, there is no difference between the reality of the components vs.
the reality of the sum. In other words, there is no more information
from your separation of a standing wave into forward and reverse
components than there is in the standing wave itself.

The standing wave is a perfectly good and complete solution to the wave
equations applicable to this steady-state problem. It is possible to
sub-divide in many ways, but there is no new information in doing so.

If you want to specifically address transients then another set of
equations will be needed.


ad hominem

You have railed against seduction by math models on many occasions.
However, that is exactly what you are doing here. Trying to create some
new physical reality by manipulating the numbers.

/ad hominem

73,
Gene
W4SZ