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.

The challenge for Cecil is to make his own theory do the same.

"My theory" gives the exact same results as an S-parameter analysis
or a classical wave reflection model analysis. That's why I know it's
correct. Roy's (and Dr. Best's) models give the correct results for
voltage and current but not for power/energy. Remember Dr. Best's
assertion that 75w + 8.33w = 133.33w? It's been four years since
I told him here on this newsgroup that was ridiculous and that
75w + 8.33w + 50w of constructive interference = 133.33w
He responded that no interference existed or was necessary. That
can be verified by accessing Google, summer 2001.

Interference is built into the S-parameter model and the classical
wave reflection model but a lot of RF people don't recognize it. Dr.
Best's term, 2*SQRT(P1)*SQRT(P2), is known in the field of optics as
the "interference term" but he didn't know that at the time of
publication of his QEX article.
--
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:
. . .
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

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

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

It's not your numbers, Roy, it's your premises that violate
the conservation of momentum principle among other principles
of physics. RF waves possesss momentum and that momentum MUST
be preserved. Your premises simply violate the conservation
of momentum principle. When you assert that the reflected
waves possess no energy and it is stored in some magic place
at sub-light speeds, you are in violation of the principles
of physics.

You can resolve all of this by telling us where the energy
is stored, besides in reflected waves, when we are dealing with
light in free space and no transmission line because exactly
the same thing happens with EM light waves as happens with
EM RF waves.
--
73, Cecil http://www.qsl.net/w5dxp


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I see that you are convinced that you've explained where the bouncing
waves of average power go and what they do.

Did anyone understand it? If so, would someone else please try to
explain it to me? Where does that 18 watts of "reverse power" go, and why?

Roy Lewallen, W7EL

Cecil Moore wrote:

[More generalizations with no numbers]

Roy Lewallen wrote:

I see that you are convinced that you've explained where the bouncing
waves of average power go and what they do.

Did anyone understand it? If so, would someone else please try to
explain it to me? Where does that 18 watts of "reverse power" go, and why?

Wouldn't it go to the circulator load which must always be placed at the
source in order to clearly illustrate what happens when a circulator
isn't in place at the source?

ac6xg

Roy Lewallen wrote:
Did anyone understand it? If so, would someone else please try to
explain it to me? Where does that 18 watts of "reverse power" go, and why?

Bob Lay, w9dmk, could explain it but he's off on a camping trip.
I think the QEX editors can now explain it also. 18 watts is
rejected by the mismatched load. It was part of a forward wave
that contained energy and momentum. That energy and momentum
MUST be conserved according to the laws of physics. The reflected
wave heads back toward the source at the speed of light and part
is re-reflected as Pref2*rho^2. The remaining
Pref(1-rho^2) part engages with Pfor1*rho^2 in an equal
magnitude/opposite phase wave cancellation as explained by Walter
Maxwell on page 23-9 of "Reflections II". Since energy and momentum
cannot be cancelled, the two cancelled waves conserve energy and
momentum by heading back toward the load at the speed of light as
a re-reflection of energy which joins the forward wave energy.
--
73, Cecil http://www.qsl.net/w5dxp


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