Jim Kelley wrote:

Cecil Moore wrote:
Of course, my statement is related to steady-state. I don't
see anything worth responding to, Jim. Where's the beef?

The problem is that there should only be a 1 second lapse of time
between the beginning of gozinta at 100 Joules/sec and the beginning of
comezouta at 100 Joules/sec. At what point is the additional 2 seconds
worth of energy fed into the system?

During the power-on transient phase. The load rejects half the incident
power. To keep things simple, assume a very smart fast tuner. After
one second, the feedline will contain 100 joules. The load will have
accepted zero joules. After two seconds, the feedline will contain the
100 joules generated plus 50 joules rejected by the load and the load
will have accepted 50 joules. Already the feedline contains 150 joules
while the source is putting out 100 joules per second. After 'n'
seconds, the line contains 300 joules, 100 from the source and 200
rejected by the load during the power-on transient stage.

seconds forward energy reflected energy load power
1 100 0 0
2 100 50 50
3 150 50 50
4 150 75 75
5 175 75 75
6 175 87.5 87.5
7 187.5 87.5 87.5
8 187.5 93.75 93.75
9 193.75 93.75 93.75
10 193.75 96.875 96.875
n 200 100 100

After 10 seconds the source has output 1000 joules. The load
has accepted 709.375 joules. 290.625 joules are already stored
in the feedline on the way to 300 joules during steady-state.
This is simple classical reflection model stuff.

If a load in rejecting half its incident power, the steady-
state reflected power will equal the steady-state load
power. The steady-state forward power will be double
either one of those.
--
73, Cecil http://www.qsl.net/w5dxp

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Cecil Moore wrote:
Ian White GM3SEK wrote:

Roy has posed a test problem that is very easy to understand, and can
be solved unambiguously by simple arithmetic. Solving it using
S-parameters will take time and some depth of understanding, but we
can be confident that they WILL give exactly the same result in the end.


It's not Roy's results that are flawed. It's his premises. If
one has a 100v source with a 50 ohm series impedance feeding
a 200 ohm resistor, Roy's results are perfect. But when we add
that 1/2WL of 200 ohm line, it changes things from a circuit
analysis to a distributed network analysis. Much more energy
is stored in the system, using the transmission line, than has
reached the load during steady-state. Roy tries to completely
ignore the stored energy and alleges that there is no energy in
the reflected waves. But there is *exactly* the same amount of
energy stored in the feedline as is required for the forward
waves and reflected waves to posssess the energy predicted by
the classical wave reflection model or an S-parameter analysis
or an analysis by Walter Maxwell of "Reflections" fame.
. . .

Ah, the drift and misattribution has begun. I'll butt in just long
enough to steer it back.

I made no premises, and have not made any statement about energy in
reflected waves. I only reported currents and powers which I believe are
correct. Nothing you or anyone has said has indicated otherwise. I do
question the notion of bouncing waves of average power, and have
specifically shown that H's statement about the source resistor
absorbing all the reflected power, when its value is equal to the line
impedance, is clearly false. (The "reflected power" is 18 watts; the
resistor dissipates 8.) I haven't seen any coherent explanation of the
observable currents and power dissipations that's consistent with the
notion of bouncing current waves. Perhaps your dodging and hand-waving
has convinced someone (the QEX editor?), but certainly not me.

It's not a 200 ohm line, it's a 50 ohm line. (I see that I neglected to
state this when giving my example, and I apologize. But it can be
inferred from the load resistance and SWR I stated.)

It baffles me how you think you can calculate the line's stored energy
without knowing its time delay. The calculation of stored energy is
simple enough, but it requires knowledge of the line's time delay. A
half wavelength line at 3.5 MHz will store twice as much energy as a
half wavelength line at 7 MHz, all else being equal. Even if you knew
the frequency (which I didn't specify), you'd also need to know the
velocity factor to determine the time delay and therefore the stored
energy. I'm afraid your methods of calculating stored energy are in error.

But if you think the stored energy is important and you find (by
whatever calculation method you're using) that it's precisely the right
value to support your interesting theory, modify the example by doubling
the line length to one wavelength. The forward and reverse powers stay
the same, power dissipation in source and load resistors stay the same,
impedances stay the same -- there's no change at all to my analysis or
any of the values I gave. But the energy stored in the line doubles.
(Egad, I hope your stored energy calculation method isn't so bizarre
that it allows doubling the line length without doubling the stored
energy. But I guess I wouldn't be surprised.) So if the stored energy
was precisely the right amount before, now it's too much by a factor of
two. And if you find you like that amount of stored energy, double the
line length again.

I can see why you avoid the professional publications.

Roy Lewallen, W7EL

Roy Lewallen wrote:
Ah, the drift and misattribution has begun. I'll butt in just long
enough to steer it back.

I made no premises, and have not made any statement about energy in
reflected waves. I only reported currents and powers which I believe are
correct. Nothing you or anyone has said has indicated otherwise. I do
question the notion of bouncing waves of average power, ...

That's exactly the false premise I am talking about, Roy. If you assume
waves of reflected power don't exist, you will find a way to rationalize
proof of that premise. You are looking under the streetlight where
the light is better instead of in the dark spot where you lost your
keys (keys being analogous to reflected power).

It's not a 200 ohm line, it's a 50 ohm line. (I see that I neglected to
state this when giving my example, and I apologize. But it can be
inferred from the load resistance and SWR I stated.)

Yes, you are right about that. But one can visualize the interference
by inserting 1WL of lossless 200 ohm feedline between the source and
the 50 ohm line which changes virtually nothing outside of the 200
ohm line.

100v/50ohm source with 80v at the output terminals.

80v--1WL lossless 200 ohm line--+--1/2WL 50 ohm line--200 ohm load

The source sees the same impedance as before. The same impedance as
before is seen looking back toward the source. Voltages, currents,
and powers remain the same. But now the interference patterns are
external to the source and can be easily analyzed.

It baffles me how you think you can calculate the line's stored energy
without knowing its time delay.

The time delay was given at one second, Roy. I really wish you would
read my postings. Here is the quote from the earlier posting:

************************************************** *****************
* If we make Roy's lossless 50 ohm feedline one second long (an *
* integer number of wavelengths), during steady-state, the source *
* will have supplied 68 joules of energy that has not reached the *
* load. That will continue throughout steady state. The 68 joules *
* of energy will be dissipated by the system during the power-off *
* transient state. *
************************************************** *****************

I DIDN'T EVER TRY TO CALCULATE THE STORED ENERGY IN YOUR LINE. But a
VF could be assumed for your lossless line and a delay calculated
from the length. Or you can just scale my one second line down to a
one microsecond line. The results will conceptually be the same. Of
course, the one microsecond line would have to be defined as an integer
number of half wavelengths but the frequency could be chosen for that
result.

So 68 microjoules would be be stored in that one microsecond feedline,
50 microjoules in the forward wave and 18 microjoules in the reflected
wave. The source is still supplying 32 microjoules per microsecond and
there is exactly enough energy stored in the feedline to support the
energy in the forward wave and reflected wave as predicted by the wave
reflection model or S-parameter analysis.

The calculation of stored energy is
simple enough, but it requires knowledge of the line's time delay.

The time delay was given, Roy, at one second. See the above quote.
That technique changes watts to joules. Buckets of joules are not
as easy to hide as watts.

A
half wavelength line at 3.5 MHz will store twice as much energy as a
half wavelength line at 7 MHz, all else being equal. Even if you knew
the frequency (which I didn't specify), you'd also need to know the
velocity factor to determine the time delay and therefore the stored
energy. I'm afraid your methods of calculating stored energy are in error.

I'm afraid you don't read my postings. THE FREQUENCY AND VELOCITY FACTOR
DO NOT MATTER WHEN THE FEEDLINE IS SPECIFIED TO BE ONE SECOND LONG.
Wavelength and VF are automatically taken into account by the assumption
of a one second long feedline. I have used that example before for
that very reason. A one second long feedline is filled with joules.
Joules are harder to sweep under the rug than watts are.

But if you think the stored energy is important and you find (by
whatever calculation method you're using) that it's precisely the right
value to support your interesting theory, modify the example by doubling
the line length to one wavelength. The forward and reverse powers stay
the same, power dissipation in source and load resistors stay the same,
impedances stay the same -- there's no change at all to my analysis or
any of the values I gave. But the energy stored in the line doubles.
(Egad, I hope your stored energy calculation method isn't so bizarre
that it allows doubling the line length without doubling the stored
energy. But I guess I wouldn't be surprised.)

If the forward power and reflected power remain the same, doubling the
length of the feedline must necessarily double the amount of energy
stored in the forward and reflected waves. That fact supports my side
of the argument, not yours.

So if the stored energy
was precisely the right amount before, now it's too much by a factor of
two. And if you find you like that amount of stored energy, double the
line length again.

Feeble attempt at obfuscation. The amount of energy stored in a feedline
is proportional to its length assuming the same forward and reflected
power levels and assuming integer multiples of a wavelength. Again,
that supports my side of the argument 100% - and doesn't support yours.
It appears to me that you have just admitted your mistake but don't
realize it yet.

I can see why you avoid the professional publications.

Actually, I have had a lot more articles published in professional
publications than I will ever have published in amateur publications.
I was an applications engineer for Intel for 13 years and professional
publication was required in the job description.
--
73, Cecil http://www.qsl.net/w5dxp

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Cecil Moore wrote:

Jim Kelley wrote:

Cecil Moore wrote:

Of course, my statement is related to steady-state. I don't
see anything worth responding to, Jim. Where's the beef?


The problem is that there should only be a 1 second lapse of time
between the beginning of gozinta at 100 Joules/sec and the beginning
of comezouta at 100 Joules/sec. At what point is the additional 2
seconds worth of energy fed into the system?


During the power-on transient phase. The load rejects half the incident
power. To keep things simple, assume a very smart fast tuner. After
one second, the feedline will contain 100 joules. The load will have
accepted zero joules. After two seconds, the feedline will contain the
100 joules generated plus 50 joules rejected by the load and the load
will have accepted 50 joules. Already the feedline contains 150 joules
while the source is putting out 100 joules per second. After 'n'
seconds, the line contains 300 joules, 100 from the source and 200
rejected by the load during the power-on transient stage.

seconds forward energy reflected energy load power
1 100 0 0
2 100 50 50
3 150 50 50
4 150 75 75
5 175 75 75
6 175 87.5 87.5
7 187.5 87.5 87.5
8 187.5 93.75 93.75
9 193.75 93.75 93.75
10 193.75 96.875 96.875
n 200 100 100

After 10 seconds the source has output 1000 joules. The load
has accepted 709.375 joules. 290.625 joules are already stored
in the feedline on the way to 300 joules during steady-state.
This is simple classical reflection model stuff.

If a load in rejecting half its incident power, the steady-
state reflected power will equal the steady-state load
power. The steady-state forward power will be double
either one of those.

It really is an interesting theory. And I'm willing to concede on a
certain point here. If we were to fit a curve to the data in your far
right side column, what we have is a dispersion curve. That is a
predictable phenomenon, most easily observable on long transmission
lines. However as this is not actual data, an important column is
missing. A column marked 'energy from source' is crucial to proving
your point. Without running the experiment and taking the data we can't
really know how much energy would be in any of the columns at any given
time. When we assume what that energy might be, we run the risk of
making an ass out of u and me. Well, mostly u. :-)

73, AC6XG

Jim Kelley wrote:
A column marked 'energy from source' is crucial to proving
your point.

Jim, I was hoping you were capable of multiplying 100 joules/sec
by the number of seconds to get the total number of joules
delivered to the system over time by the source. My 1000 joules
after ten seconds is 100 joules/sec multiplied by ten seconds.
Is that math too difficult for you? :-)

Maybe you need a simpler example. Here it is:

100w SGCL source----one second long feedline----load

The SGCL source is a signal generator equipped with a circulator
and circulator load. The circulator load dissipates all the
reflected power incident upon the signal generator. The signal
generator outputs a constant 100 watts.

The load is chosen such that the power reflection coefficient
is equal to 0.5, i.e. half the power incident upon the load
is reflected and half accepted by the load.

This configuration reaches steady-state in 2+ seconds. After 2+
seconds, the forward wave contains 100 joules and the reflected
wave contains 50 joules. 50 watts is being dissipated by the
load and 50 watts is being dissipated by the circulator load.
The source has output 150 joules of energy that has not been
dissipated by the load or the circulator load. 150 joules is
exactly the amount of energy to support the energy levels of
the forward wave and the reflected wave.

What could be simpler than that if you really believe in the
conservation of energy principle?
--
73, Cecil, http://www.qsl.net/w5dxp


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Cecil Moore wrote:

Jim Kelley wrote:

A column marked 'energy from source' is crucial to proving your point.


Jim, I was hoping you were capable of multiplying 100 joules/sec
by the number of seconds to get the total number of joules
delivered to the system over time by the source. My 1000 joules
after ten seconds is 100 joules/sec multiplied by ten seconds.
Is that math too difficult for you? :-)

:-) My contention is that it's too remedial. What you require is faith,
not math. Is the source supposed to be a virtual fire hydrant of
constant energy, or is it more like a real system? You seem to be
assuming a constant 100 Joules per second input, regardless of the fact
that the impedance the source sees is changing over the interval.
That's not particularly realistic, hence a need for the empirical. But
we could assume that the source is constant, and continue.

Maybe you need a simpler example. Here it is:

100w SGCL source----one second long feedline----load

The SGCL source is a signal generator equipped with a circulator
and circulator load. The circulator load dissipates all the
reflected power incident upon the signal generator. The signal
generator outputs a constant 100 watts.

The load is chosen such that the power reflection coefficient
is equal to 0.5, i.e. half the power incident upon the load
is reflected and half accepted by the load.

This configuration reaches steady-state in 2+ seconds. After 2+
seconds, the forward wave contains 100 joules and the reflected
wave contains 50 joules. 50 watts is being dissipated by the
load and 50 watts is being dissipated by the circulator load.
The source has output 150 joules of energy that has not been
dissipated by the load or the circulator load.

You have provided a lot of detail about where it all resides and in what
proportions, but you still haven't shown how much energy a source would
actually produce under such circumstances. Further, you're assuming
that energy would move forward in a transmission line at a rate higher
than the rate at which it is provided by the source. This is highly
speculative and suspect. What we know for sure is, once steady state is
achieved, energy is absorbed by the load(s) at the same rate at which it
is generated, all the energy from the source goes to the load(s). Given
that, there's very little impetus to believe that there need be any more
than one second's worth of energy held within a one second long
transmission line. It is therefore reasonable to contend that in the
first scenario, 100 Joules of energy is held within the transmission
line as it propagates toward the load. And in this latest scenario, 50
Joules is heading toward the load, and 50 is in the path to the
circulator for a total of 100 Joules stored within the one second long
transmission line.

The way to prove that there's any greater surplus of energy held within
the transmission line would be to make the energy vs. time measurements
at each end of such a transmission line. Absent that, it's purposeful
speculation.

73, AC6XG

Cecil Moore wrote:
. . .
Yes, you are right about that. But one can visualize the interference
by inserting 1WL of lossless 200 ohm feedline between the source and
the 50 ohm line which changes virtually nothing outside of the 200
ohm line.

100v/50ohm source with 80v at the output terminals.

80v--1WL lossless 200 ohm line--+--1/2WL 50 ohm line--200 ohm load

The source sees the same impedance as before. The same impedance as
before is seen looking back toward the source. Voltages, currents,
and powers remain the same. But now the interference patterns are
external to the source and can be easily analyzed.

Hm, I wasn't having any trouble analyzing the system without the 200 ohm
line. Why do you have to make the system more complex in order to apply
your theory? Looking back at previous postings, it appears that any time
anyone presents a model that gives you difficulty, you simply modify it
to suit yourself, and deflect the discussion. I'm not interested in
whether you can explain your theories in models of your choice. What
remains to be shown is whether you can do so for the extremely simple
model I proposed. You might recall that the first example in my posting
in response to H's claim did indeed have source resistor dissipation
equal to the "reverse power" -- it's much easier to apply a defective
theory if you're free to choose special cases that support it. A valid
theory should be able to work on all models, unless you clearly give the
boundaries of its validity and why it has those limitations.

It baffles me how you think you can calculate the line's stored energy
without knowing its time delay.


The time delay was given at one second, Roy. I really wish you would
read my postings. Here is the quote from the earlier posting:

************************************************** *****************
* If we make Roy's lossless 50 ohm feedline one second long (an *
* integer number of wavelengths), during steady-state, the source *
* will have supplied 68 joules of energy that has not reached the *
* load. That will continue throughout steady state. The 68 joules *
* of energy will be dissipated by the system during the power-off *
* transient state. *
************************************************** *****************

I DIDN'T EVER TRY TO CALCULATE THE STORED ENERGY IN YOUR LINE. But a
VF could be assumed for your lossless line and a delay calculated
from the length. Or you can just scale my one second line down to a
one microsecond line. The results will conceptually be the same. Of
course, the one microsecond line would have to be defined as an integer
number of half wavelengths but the frequency could be chosen for that
result.

So 68 microjoules would be be stored in that one microsecond feedline,
50 microjoules in the forward wave and 18 microjoules in the reflected
wave. The source is still supplying 32 microjoules per microsecond and
there is exactly enough energy stored in the feedline to support the
energy in the forward wave and reflected wave as predicted by the wave
reflection model or S-parameter analysis.

The calculation of stored energy is simple enough, but it requires
knowledge of the line's time delay.


The time delay was given, Roy, at one second. See the above quote.
That technique changes watts to joules. Buckets of joules are not
as easy to hide as watts.

I apologize for not having read your posting more carefully. When I get
snowed or when the discussion deviates from the point in question, I do
tend to not read the rest. Because the stored energy has nothing to do
with what happens to the waves of bouncing average power in steady
state, I did ignore your details about it. Double my line length and the
waves of bouncing average power have the same values as before, although
the stored energy has doubled.


If the forward power and reflected power remain the same, doubling the
length of the feedline must necessarily double the amount of energy
stored in the forward and reflected waves. That fact supports my side
of the argument, not yours.

I'm not sure what you think my side of the argument is. That a
transmission line doesn't contain stored energy? Of course it does. (See
below, where I calculate it using conventional wave mechanics.) I'm
simply asking where your imaginary waves of bouncing average power go in
a painfully simple steady state system.

Your argument is that there are waves of bouncing average power. I asked
where they went in the simple circuit I described. Calculation of the
energy stored in the line does nothing to explain it. All it does is
create the necessary diversion to deflect the discussion from the fact
that you don't know or at best have only a vague and general idea.

Here's a derivation of the energy stored in the line using conventional
wave mechanics:

T is the time it takes a wave to traverse the line in one direction. The
analysis begins at t = 0, with the line completely discharged, at which
time the source is first turned on.

1. From time t = 0 to T, the impedance seen looking into the line is 50
ohms, so the 100 volt source sees 100 ohms. Source current = 1 amp,
source is delivering 100 watts, source resistor is dissipating 50 watts,
and load resistor is dissipating zero. Therefore the energy being put
into the line is 100 - 50 = 50 watts * t.
2. From time t = T to 2T, all the conditions external to the line are
the same as above, except that the load resistor is now dissipating 32
watts, so the net energy being put into the line from the source is 100
- 50 - 32 = 18 watts * (t - T).
3. After time = 2T, steady state occurs, with the conditions I gave
originally. From that time onward, the power into the line equals the
power out, so no additional energy is being stored.

It's obvious from the above that the energy stored in the line from 0 to
T = 50 * T joules, and from T to 2T = 18 * T joules, for a total energy
storage of 68 * T joules.

No bouncing waves of average power are required; this can be completely
solved, as can all other transmission line problems, by looking at
voltage and current waves, along with simple arithmetic. I didn't bother
doing this earlier simply because it's irrelevant; it has nothing to do
with steady state conditions any more than the DC charge on a capacitor
has to do with an AC circuit analysis. The only reason it's important is
that it provided you with yet another way to divert the discussion from
the question of what happens to those imaginary waves of bouncing
average power in the steady state.

In the steady state we've got energy stored in the line (of course), and
18 watts of "reverse power". 8 watts is being dissipated in the source
resistor. We can store just as much or little energy in the line as we
choose by changing its length; as long as the line remains an integral
number of half wavelengths long, there's no change to the line's
"forward power", "reverse power" or any external voltage, current, or
power. In fact, if we choose a line length that's not an integral number
of half wavelengths long, we change the dissipation in the source and
load resistors without any change in the "forward power", "reverse
power", or reflection coefficients at either end of the line.

Your postings indicate that in your mind you've completely explained
where the bouncing waves of average power are going, what they do, and
why. If I'm correct and this is indeed all you have to offer, I'll once
again bow out. I look forward to a careful reading of the QEX article.

Roy Lewallen, W7EL

Jim Kelley wrote:
You seem to be
assuming a constant 100 Joules per second input, regardless of the fact
that the impedance the source sees is changing over the interval.

It's a mental exercise, Jim. I told you it was equipped with
a very fast very smart autotuner. If you are cool with a one
second long lossless feedline, you should be cool with a
very fast very smart autotuner.

You have provided a lot of detail about where it all resides and in what
proportions, but you still haven't shown how much energy a source would
actually produce under such circumstances.

It's too simple to mention. The signal generator is putting out a
constant 100 watts. Hint: multiply the watts (joules/sec) by the
number of seconds to get the total joules. Dimensional analysis
indicates the product will be joules.

Further, you're assuming
that energy would move forward in a transmission line at a rate higher
than the rate at which it is provided by the source.

Nope, I'm not. All wave energy moves at the speed of light. You
are confused. 100 joules per second is headed toward the load.
50 joules per second is headed away from the load.

This is highly speculative and suspect.

Easy to understand given your level of confusion. To get the forward
power, divide the load power by one minus the power reflection
coefficient. That's 50w/0.5 = 100 watts. That's how you calculate
forward power.

... there's very little impetus to believe that there need be any more
than one second's worth of energy held within a one second long
transmission line.

Jim, if you have 1.5 gallons in a tank with one gallon/sec flowing
in and one gallon/sec flowing out, how many gallons are in the tank?
You have to have enough energy in the feedline to support the
forward power and the reflected power. More or less than that
amount would violate the conservation of energy principle.

It is therefore reasonable to contend that in the
first scenario, 100 Joules of energy is held within the transmission
line as it propagates toward the load.

Yes, half is headed into the load and half will be rejected by
the load.

And in this latest scenario, 50 Joules is heading toward the load,

50 joules are destined for the load but 100 joules are heading
toward the load. Remember to get 50 watts into the load, you
must hit the load with 100 watts. 100 watts for one second is
100 joules, not 50.

and 50 is in the path to the
circulator for a total of 100 Joules stored within the one second long
transmission line.

Your math or model or both are faulty. The forward power must
be 100 watts to get 50 into the load. Therefore, the forward
wave energy in a one second feedline is 100 joules. The reflected
wave energy is half of that. Therefore, there's 150 joules in
the feedline.
--
73, Cecil http://www.qsl.net/w5dxp

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Roy Lewallen wrote:
Hm, I wasn't having any trouble analyzing the system without the 200 ohm
line.

Yes, you were, Roy. That's why your results didn't agree with
mine AND the wave reflection model analysis AND the S-parameter
analysis AND the conservation of energy principle AND the
conservation of momentum principle.

Those RF waves are virtually identical to a laser beam in free
space. Where do you store the reflected energy when the reflected
wave is a light beam? An RF reflected wave *IS* a light beam, just
at a low frequency.

Why do you have to make the system more complex in order to apply
your theory?

Why are you afraid of using the correct model in order to
achieve correct results? DISTRIBUTED NETWORK THEORY IS SIMPLY
MORE COMPLEX THAN LUMPED CIRCUIT THEORY AND FOR GOOD REASON. Your
lumped circuit theory will NOT solve distributed network problems.
You have already proved that more than once in the past.

When I get snowed or when the discussion deviates from the point
in question, I do tend to not read the rest.

How do you give yourself permission to assert that your adversary
is speaking gobbledygook when you have deliberately not read his
postings?

Because the stored energy has nothing to do
with what happens to the waves of bouncing average power in steady
state, I did ignore your details about it.

This is your false premise in action. The "waves of bouncing
energy" exist precisely because of the stored energy. The
SWR circle impedance transformation depends upon those
"bounding waves". Take away the "bouncing waves" and the
feedline is incapable of transforming impedances.

No bouncing waves of average power are required;

It logically directly follows that transmission lines are
incapable of transforming impedances and all transmission
lines are therefore matched. Reminds me of your point-sized
mobil loading coils. :-)

The wave energy emerges from the source traveling at the speed of
light. A principle of physics says the momentum in that generated
wave must be conserved. Exactly where does the momentum go when
you put the brakes on the wave energy and store it in your magic
transmission line? Where do you hide the wave momentum?
--
73, Cecil http://www.qsl.net/w5dxp

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Cecil Moore wrote:
Roy Lewallen wrote:

Hm, I wasn't having any trouble analyzing the system without the 200
ohm line.


Yes, you were, Roy. That's why your results didn't agree with
mine AND the wave reflection model analysis AND the S-parameter
analysis AND the conservation of energy principle AND the
conservation of momentum principle.

Please tell me which of the numbers I posted disagree with yours, and
which numbers you got. Once again, mine a

Power from the source = 40 watts
Power dissipated in the source resistor = 8 watts
Power dissipated in the load resistor = 32 watts
Line SWR = 4:1
Line "Forward power" = 50 watts
Line "Reverse power" = 18 watts

Those are the only results I posted. What results did you get which are
different?

Cecil's results:

Power from the source =
Power dissipated in the source resistor =
Power dissipated in the load resistor =
Line SWR =
Line "Forward power" =
Line "Reverse power" =

Roy Lewallen, W7EL