I've been told that this discussion continues in Letters
to the Editor in QEX magazine, to which I don't subscribe
any more. Without knowing the context of those present
discussions, I composed a letter to the QEX editors.
Here it is:

“Where Does the Power Go?” was answered in my magazine article,
“An Energy Analysis at an Impedance Discontinuity in a Transmission
Line”, published by Worldradio magazine and available on my web
page at:

http://www.w5dxp.com/energy.htm

Here is a thought experiment that should lay the subject to rest.
Assume a one-second long lossless 50 ohm open-circuit transmission
line being driven by a 141.4 volt, 50 ohm Thevenin source. At time,
t = 0, source power is supplied to the transmission line. After one
second, 100 joules will have been delivered to the line’s forward
wave and the forward wave will have just reached the open-circuit at
the other end of the transmission line. What happens next simply
follows the laws of physics.

The conservation of energy principle tells us that the 100 joules
in the forward wave must be conserved. The conservation of momentum
principle tells us that the momentum in the forward wave must be
conserved. The open-circuit prohibits the forward wave from continuing
in the forward direction and the forward wave ceases to exist beyond
the open-circuit point. The forward wave ceases to exist so what
happens to the energy and momentum in the forward wave? Physically,
there is only one possibility.

The energy and momentum in the forward electromagnetic wave is moving
at the speed of light and is indeed conserved by transferring its
energy and momentum to the reflected wave at the open-circuit point
in accordance with the rules of the distributed-network/wave-reflection
model. This starts to happen at t = 1 second at which time 100 joules
of electromagnetic energy exists in the forward wave in the transmission
line. (One can actually calculate the force being exerted on the open-
circuit by the forward wave. It is akin to the pressure of sunlight.)

The transfer of energy and momentum from the forward wave to the
reflected wave continues throughout the t =1 to t = 2 second time
frame. At the end of two seconds, the forward wave contains 100
joules and the reflected wave contains 100 joules. Thus 200 joules
are "stored" in the transmission line during the first two seconds.
After two seconds, steady-state has been reached and the Thevenin
source ceases to supply any energy. No additional energy is needed
since the system is lossless. But please note that 200 joules exist
in the transmission line all during steady-state, exactly enough
energy to support the 100 joules/second forward watts and the 100
joules/second reflected watts.

Any theory of where the energy goes must account for the 200 joules
of energy in the transmission line. Those 200 joules cannot be
destroyed by sweeping them under the steady-state rug. An ideal
(lossless) directional wattmeter will tell us that forward
watts = 100 watts and that reflected watts = 100 watts. 200 joules
is required to support that number of forward and reflected watts.
Is it just a coincidence that the 200 joules existing in the
transmission line during steady-state is *exactly* the magnitude
of energy required to support 100 watts forward and 100 watts
reflected?

The number of joules in a transmission line is *always* exactly the
number required to support the forward watts and reflected watts.
Since EM waves cannot stand still, it seems logical to leave the
energy right where it is at the beginning of steady-state - 200 joules
per second being exchanged between the forward wave and the reflected
wave. The energy exchanges at the two ends of the transmission line
are balanced and equal so one might argue that there is no net energy
exchange between the forward and reflected waves during steady-state.
But since the two energy transfers are occurring thousands of miles
apart, that would seem to be a moot point.

The transmission line is charged with this energy during the power
on transient phase before steady-state is reached. The transmission
line is discharged (energy dissipated) during the power down transient
phase after steady-state. It is simply conservation of energy at work.
It is interesting to note that if the transmission is discharged
through a 50 ohm resistor at the far open-end, it will be discharged
at a rate of 100 watts for two seconds.
--
73, Cecil http://www.w5dxp.com

Gosh Cecil... In a lossless line if you put in 100 joules, at the rate
of 100 joules per second, and get 200 joules back, I predict that in
less than 30 seconds (30 doublings) there will be a titanic explosion
exceeding all the Hydrogen bombs on the planet... Do the math!

OK, tongue out of cheek...
Assume that we put a 1 second squirt at the rate of 100 Joules per
second, into the input end of the line, it is an open line at the far
end, and yank the input end of the line out of the transmitter at
exactly 1 second... Now we have a lossless line with a wave going
forward and a reflected wave coming back... Do you still claim 200
joules?

We are going to do this in quarters (25% of time, 25% of 100 joules and
25% of line length)
Start pushing energy into the line at T=0 and jump ahead to T = .25
second... The is 25 joules total in the .00 to .25 line section and 0
joules in the other 3 sections... At T=.50 there are 25 joules in each
of the first two .25 sections and nothing in the last two .25
sections... Total 50 joules... The next two time periods is more of
the same so let's jump to T=1.00
The line is now just 'full', with 4 'quarters' of energy having been
inserted into the line, at that instant the leading edge of the wave
reaches 1.00 of line (.25 x 4) and reflects off the open terminal,
folding back and going towards the input... The total power in the
line is 100 joules at T= 1.00.... No surprise there...
At 1.25 seconds we have a quarter of the wave reflected back and
re-reaching the .75 line point (25 joules in that quarter) laid over
top of the quarter of the wave front still going outwards in the .75 to
1.00 section (25 joules in that quarter).
So at T = 1.25 in that 0.75 to 1.00 portion of line we have 50 joules
of energy (first quarter coming back plus second quarter still going
out, passing each other... )
In the .50 to .75 line we have 25 joules of energy...
In the .25 to .50 line we have 25 joules of energy...
In the .00 to .25 line we have 0 joules of energy...

At T = 1.50 we have a wave going out in the .50 to .75 section and also
in the .75 to 1.00 section... And reflected wave coming back in both
those sections for 50 watts in each quarter section... The .00 to .25,
and .25 to .50 sections are empty... For a total of 100 joules...

T=1.75 and 2.00 is just more of the same... 4 quarters of 25 joules
each lined up a row with 4 quarters of 0 joules each... Chasing each
other up and down the line, folding back and passing each other at each
..25 time period... A given quarter line section can have 0 joules, 25
joules or 50 joules, present at the instantaneous T of any given time
block..

In a lossless line this power will slosh back and forth forever...
Never having more than 50 joules in any quarter section of line and
certainly not doubling to 200 joules... If you have ever held a slinky
toy between your hands and watched the wave bouncing from hand to hand,
you have seen what happens in the 1 second of line section...

BTW, this can be visualized by cutting 4 paper strips 3" long and
marking them as 25 joules each and pushing them onto your 1 foot ruler
(or ugly 250 mm stick) one at a time and just like box cars on a train
see what they do...

Awww righty, now lets leave the transmitter on the line instead of
yanking it... Sort of like non coitus interruptus...
At T=1.00 nothing is happening as far as the transmitter knows... The
line is still taking power like a hungry calf taking milk from a cow...

At T = 1.25 we have put 125 joules into the line... The line section
..75 to 1.00 is doubled up with 25 joules going and 25 joules coming...
At T = 1.50 we have put in 150 joules... Line sections 3 and 4 are
doubled up to 50 joules each and line sections 1 and 2 have 25 joules
each on board...
At t = 1.75 sections 2,3,4 are doubled up and only section one has 25
joules currently on board...
At T = 2.00 we have 50 joules in each quarter section of line for a
total of 200 joules, and Cecil is fat and happy... Thinks he has been
vindicated...

But wait! There's more... These Ginsu knifes will slice and dice and
uuunhhh, - oops, wrong topic...
Well, as a thought experiment there is nothing to inhibit me from
pumping another 100 joules into the line in the T = 2.00 to 3.00 time
block, and again in the 3.00 to 4.00 time block, ad infinitum until we
come to the Big Bang... So Cecil, why doesn't this happen? You can't
have it both ways, a theoretical thought experiment of putting energy
into a lossless line and at T = 2.00 suddenly revert to some mumble
jumbo about the energy going away at the same rate it is entering, or
the transmitter absorbing the reflected wave (those bottles gonna
vaporize on the first long dah)... There is nothing in our theoretical
world to differentiate T = 2 from any T = n + 1... Inquiring minds,
and all that...

But let me suddenly play spoiler here... After 2 seconds, the reflected
returning slug of energy present in each quarter line section is 180
degrees out of phase with the out going energy slug in each respective
section... The two slugs vectorally neutralize each other... No energy
present in that quarter section at any given T... So for each .25 T
after T = 2.00 you are merely replacing the energy that has been
neutralized by the returning wave.... Remember, energy can neither be
created nor destroyed... So, where did the energy go?

denny / k8do

Denny wrote:
Assume that we put a 1 second squirt at the rate of 100 Joules per
second, into the input end of the line, it is an open line at the far
end, and yank the input end of the line out of the transmitter at
exactly 1 second... Now we have a lossless line with a wave going
forward and a reflected wave coming back... Do you still claim 200
joules?

No, 100 joules/second * one second would be 100 joules in your
above example. In my previous example the 100 joule/sec "squirt"
was two seconds long. Thus, in my example, 200 joules made it
into the one-second long line.

Never having more than 50 joules in any quarter section of line and
certainly not doubling to 200 joules... If you have ever held a slinky
toy between your hands and watched the wave bouncing from hand to hand,
you have seen what happens in the 1 second of line section...

True for your example, but not true for my example. In my example
the 100 watt source was delivering 100 watts for two seconds. Last
time I checked 100 joules/sec * 2 seconds = 200 joules.

At T = 2.00 we have 50 joules in each quarter section of line for a
total of 200 joules, and Cecil is fat and happy... Thinks he has been
vindicated...

But wait! There's more... These Ginsu knifes will slice and dice and
uuunhhh, - oops, wrong topic...
Well, as a thought experiment there is nothing to inhibit me from
pumping another 100 joules into the line in the T = 2.00 to 3.00 time
block, ...

On the contrary - if the frequency is an integer number of
cycles and the feedline is exactly one second long, the source
will see an infinite impedance after two seconds and further
sourcing of energy will be impossible. So I am ignoring the
rest of your posting until you understand that fact. Under
the stated the boundary conditions, it is impossible to shove
any more than 200 joules into the line.

At the end of two seconds, the forward current and the newly
arrived reflected current are 180 degrees out of phase and
cancel each other, i.e. after 2 seconds, Ifor + Iref = ZERO
The reflected voltage arrives back in phase with the forward
voltage and so the voltage at that point equals the voltage
in the 141.4 volt source and zero current flows between those
two equal potentials. There is 200 joules in the transmission
line with no possibility of adding any more. The boundary
conditions limit the energy content of the transmission line
to 200 joules, exactly the number of joules needed to support
the steady-state 100 watts forward and 100 watts reflected.
The laws of physics strike again.

Please rethink your position.
--
73, Cecil http://www.w5dxp.com

On Mon, 02 Oct 2006 19:09:56 GMT, Cecil Moore
wrote:
So I am ignoring
:-)

Richard Clark wrote:
Cecil Moore wrote:
So I am ignoring

:-)

Yep, when someone begins their argument with:

Assume 1 = 2, I am inclined to ignore it.
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:
I've been told that this discussion continues in Letters
to the Editor in QEX magazine, to which I don't subscribe
any more. Without knowing the context of those present
discussions, I composed a letter to the QEX editors.
Here it is:

“Where Does the Power Go?” was answered in my magazine article,
“An Energy Analysis at an Impedance Discontinuity in a Transmission
Line”, published by Worldradio magazine and available on my web
page at:

http://www.w5dxp.com/energy.htm

The problem with your paper, Cecil, is the part where you try to
invent the "4th Mechanism of Reflection".

73, Jim, AC6XG

On Mon, 02 Oct 2006 20:18:57 GMT, Cecil Moore
wrote:
Assume 1 = 2
:-)

the big confusion factor is using power and energy at all. they are both
derived from the much simpler to handle and understand voltage or current
waves. the biggest problem is that once you change from voltage or current
to power you lose the information necessary to calculate superposition
because you no longer have the phase information from the basic wave
components. this is partly a result of the common use of the swr meter that
measures forward and reflected 'power', everyone thinks they understand how
it works, but very few really do.

"Jim Kelley" wrote in message
...
Cecil Moore wrote:
I've been told that this discussion continues in Letters
to the Editor in QEX magazine, to which I don't subscribe
any more. Without knowing the context of those present
discussions, I composed a letter to the QEX editors.
Here it is:

“Where Does the Power Go?” was answered in my magazine article,
“An Energy Analysis at an Impedance Discontinuity in a Transmission
Line”, published by Worldradio magazine and available on my web
page at:

http://www.w5dxp.com/energy.htm

The problem with your paper, Cecil, is the part where you try to invent
the "4th Mechanism of Reflection".

73, Jim, AC6XG

Jim Kelley wrote:
The problem with your paper, Cecil, is the part where you try to invent
the "4th Mechanism of Reflection".

I wish I had invented it, Jim, but the mechanism of wave
reflection due to interference was well known and under-
stood by optical engineers long before I was born. It's
how non-reflective glass works. Ideally, interference
at the thin-film coating reflects all of the light back
toward the picture behind the glass.
--
73, Cecil http://www.w5dxp.com

Dave wrote:
the big confusion factor is using power and energy at all. they are both
derived from the much simpler to handle and understand voltage or current
waves.

But it is hard to answer the "Where does the power go?"
question without using power and energy. BTW, I didn't
start that question. Jon Bloom asked that question in a
Dec. 1994 QEX article as a rebuttal to the information
in "Reflections", by Walter Maxwell.

the biggest problem is that once you change from voltage or current
to power you lose the information necessary to calculate superposition
because you no longer have the phase information from the basic wave
components.

If one knows the length of the transmission line and the
velocity factor, the phases can be deduced. If one is
dealing with a Z0-match, which is most common in amateur
radio, the phase information is trivial because all the
voltages and all the currents are either in-phase or
180 degrees out of phase at the Z0-match point.
--
73, Cecil http://www.w5dxp.com