Richard,

This has been an entertaining monologue on electron tubes. (Must be
International Fractured Physics Week on RRAA.)

I won't attempt to address the numerous howlers, since Henry seemed
satisfied with your tale, but you might want to rethink the item you
"correct" in this message.

Secondary electron emission also occurs at much lower incident electron
energies. As a metrologist you are most likely familiar with electron
microscopy. Secondary electron emission is the typical mode of detector
operation. The incident electron energy in modern SEM's is now in the
range of 500 to 700 volts.

One of the reasons for adding more grids in vacuum tubes is to manage
secondary emission from the plate. This occurs at much less than 17 kV.

73,
Gene
W4SZ



Richard Clark wrote:
On Tue, 07 Jun 2005 11:44:29 -0700, Richard Clark
wrote:


When that electron stream strikes the plate and raises the
temperature, it is just short enough power to present this secondary
emission. But if we were to run at 17KV or so, then the electron
stream would be so aggressive as to produce high energy effects such
as X-Ray emission.


Something felt wrong here. I should have said:
...just short enough energy to present this secondary emission.

On Tue, 07 Jun 2005 20:17:28 GMT, Gene Fuller
wrote:

One of the reasons for adding more grids in vacuum tubes is to manage
secondary emission from the plate. This occurs at much less than 17 kV.

Hi Gene,

Sure, but not X-Rays which attend Cold Cathode tubes (skipping the gas
filled tubes I've already touched upon) and do require elevated
potentials to be so useful (that they become dangerously useful).

Glad you won't attempt the other howlers.

73's
Richard Clark, KB7QHC

Cecil Moore wrote:

Roy Lewallen wrote:

Finally, an actual answer. So of the 18 watts of "reverse power",
11.52 watts is making it to the source to "engage in constructive
interference". Does any of it get dissipated in the source resistor,
or does it just slide through unscathed?


It never encounters the source resistor as it is re-reflected by
wave cancellation, not by an impedance discontinuity. I have a QEX
article coming soon that will explain the details.

The Journal of Irreproducible Results could also be persuaded to publish
that claim, Cecil. Any implication that almost half of Maxwell's
equations are superfluous should easily qualify as an irreproducible
result. :-)

ac6xg

Give me a break, Reg. Download it for free and enjoy from:

http://www.sss-mag.com/pdf/hpan95-1.pdf

In all likelihood they won't be able to make any sense out of it
anyway.

=========================

I gave you a break. The whole lot of it is nothing but Scattering
parameters. Might be fine for SHF and above. Useless for mobile
matching at HF. As predicted I couldn't understand one line of it. I
doubt if anybody except the several authors have ever worked right
through it. Have you? (2 smileys).

It's mid-day here. There's not a drop of wine in the house.
----
Your old pal, Reg.

Reg Edwards wrote:

Give me a break, Reg. Download it for free and enjoy from:

http://www.sss-mag.com/pdf/hpan95-1.pdf

In all likelihood they won't be able to make any sense out of it
anyway.

=========================

I gave you a break. The whole lot of it is nothing but Scattering
parameters. Might be fine for SHF and above. Useless for mobile
matching at HF.

"Useless" only in the sense of being unnecessarily complicated and
inconvenient to use. But not incorrect!

ALL valid methods will give the same correct result, if they are applied
correctly. That's how we confirm that a method is valid - by checking
its results for a series of test problems that can be solved by other
methods too.

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.

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


--
73 from Ian GM3SEK 'In Practice' columnist for RadCom (RSGB)
http://www.ifwtech.co.uk/g3sek

Cecil Moore wrote:

Jim Kelley wrote:

Cecil Moore wrote:

It can easily be shown that 300 joules of energy have been
generated that have not been delivered to the load, i.e.
those 300 joules of energy are stored in the feedline.


Not easy if t 2 sec. :-)


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?

73, AC6XG

Jim Kelley wrote:

Cecil Moore wrote:
It never encounters the source resistor as it is re-reflected by
wave cancellation, not by an impedance discontinuity. I have a QEX
article coming soon that will explain the details.

The Journal of Irreproducible Results could also be persuaded to publish
that claim, Cecil. Any implication that almost half of Maxwell's
equations are superfluous should easily qualify as an irreproducible
result. :-)

Don't know exactly what you are inferring but the editors of QEX
have seen the light, :-) even if at RF frequencies. Quite a few
sources from the field of optics indicate that the phenomenon is well
known in that field even if not well understood in the field of RF.

It is very simple physics, Jim. When two coherent EM waves of equal
amplitudes and opposite phases attempt to travel in the same direction
in the same path, they cancel each other in their original direction
of travel. This is explained under "total destructive interference"
in "Optics" by Hecht. Since the energy in the two waves cannot be
canceled, that energy goes somewhere else. In a transmission line,
there are only two directions. If two waves cancel in one direction,
their combined energy components head back in the only other direction.
Everyone has seen that light interference pattern with his/her own eyes
and some just never realized what was happening.

Here's an example. If you wade through it, you will be forced to admit
that the destructive interference/wave cancellation at the non-glare
surface 'A' causes a reversal in the direction of the reflected
irradiance. That, my friend, is a 100% re-reflection, just as Walter
Maxwell has been saying for decades. 'n' is the index of refraction:

n=1.0 | n=1.2222 | n=1.4938
Laser-------air-------|---1/4WL thin film---|---infinite glass----...
| |
A B

The reflection from surface 'A' is canceled by an equal magnitude
and opposite phase reflection from surface 'B'. The energy components
in those two canceled waves join the forward wave because energy
cannot be destroyed (even though that concept seems to have serious
consequences to your mental health :-). Here's a quote from the
following web page.

http://micro.magnet.fsu.edu/primer/j...ons/index.html

“When two waves of equal amplitude and wavelength that are 180-degrees
out of phase with each other meet, they are not actually annihilated.
All of the photon energy present in these waves must somehow be
recovered or redistributed in a new direction, according to the law of
energy conservation ... Instead, upon meeting, the photons are
redistributed to regions that permit constructive interference, so the
effect should be considered as a redistribution of light waves and
photon energy rather than the spontaneous construction or destruction
of light.”

In an RF transmission line, since there are only two possible
directions, the only “regions that permit constructive interference”
and "redistribution in a new direction" at an impedance discontinuity
is the opposite direction from the direction of destructive
interference/wave cancellation.

The above laser example is virtually identical to the following:

RF XMTR--50 ohm coax--+--1/4WL 61 ohm coax--+--infinite 75 ohm coax

If we use a coherent laser beam, no coax is required, so the
behavior of the actual EM waves is relatively easy to analyze.
--
73, Cecil http://www.qsl.net/w5dxp

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Reg Edwards wrote:
I gave you a break. The whole lot of it is nothing but Scattering
parameters.

Exactly. There's nothing better for analyzing a transmission
line discontinuity. And there's nothing better for understanding
reflections.

It's mid-day here. There's not a drop of wine in the house.

Have some wine and you'll be able to understand the HP Ap Note.
That's how I first understood it. :-)
--
73, Cecil http://www.qsl.net/w5dxp

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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:

Jim Kelley wrote:
The Journal of Irreproducible Results could also be persuaded to
publish that claim, Cecil. Any implication that almost half of
Maxwell's equations are superfluous should easily qualify as an
irreproducible result. :-)


Don't know exactly what you are inferring but the editors of QEX
have seen the light, :-) even if at RF frequencies. Quite a few
sources from the field of optics indicate that the phenomenon is well
known in that field even if not well understood in the field of RF.

Cecil,

The Journal of Irreproducable Results is a hilarious journal that has
had a number of interesting articles in it. One I remember had to do
with Peanut Butter and the Three Stooges and the Precession of the
Earth's Axis (someone correct me if I misremembered). I believe it has
been referred to as Mad Magazine for Stephen Hawking.

Another had to do with a nice compression algorithm that eventually
reduced the input to 1 bit, no matter the input, and since that bit was
predictable as just a 1, we could eliminate that also.

tom
K0TAR

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

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

Jim Kelley wrote:
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

Sorry, I was asking about the simple example I posted (which doesn't
have a circulator), not the problem posted by Cecil.

How does placing a circulator at the source illustrate what happens when
it isn't in place? Isn't it possible to explain what happens to the
"reverse power" without a circulator? If not, why not?

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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Roy Lewallen wrote:
Jim Kelley wrote:

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


Sorry, I was asking about the simple example I posted (which doesn't
have a circulator), not the problem posted by Cecil.

It was a tongue-in-cheek reply, Roy. These problems always seem to end
up with a circulator in them at some point - clearly illustrating what
happens under entirely different circumstances. :-)

73, Jim 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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Jim Kelley wrote:

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?

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?

When a Z0-match is in place at the source, everything is also clear
since zero reflected energy reaches the source. The Z0-match case,
using an antenna tuner, is the most likely case to be encountered
in ham radio. 100% of the reflected energy and momentum is re-reflected
back toward the load, just as Walter Maxwell has been saying as long
as I can remember.

This stuff is not new. It is explained in "Fields and Waves ..."
by Ramo and Whinnery, copyright 1950's.
--
73, Cecil http://www.qsl.net/w5dxp


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Roy Lewallen wrote:
Isn't it possible to explain what happens to the
"reverse power" without a circulator?

It is certainly possible in a Z0-matched system where
zero reflected energy reaches the source and the source
is feeding its designed-for load. If no reflected energy
reaches the source, the source impedance doesn't matter,
except for efficiency, which doesn't affect the rest of
the system. A one ohm source providing 100v to a 50 ohm
load and a one megohm source providing 100v to a 50 ohm
load will result in identical external conditions when
driving a 50 ohm load.

The picture is not as clear when reflected current and
voltage are allowed to flow into the source. We usually
don't know what the source impedance is and that
certainly handicaps any analysis. Modern designers simply
resort to protection circuitry and don't worry about
the energy analysis. We do know that reflections reaching
the source are not benign.
--
73, Cecil http://www.qsl.net/w5dxp


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Jim Kelley wrote:
It was a tongue-in-cheek reply, Roy. These problems always seem to end
up with a circulator in them at some point - clearly illustrating what
happens under entirely different circumstances. :-)

The circulator, lossless feedlines of unreasonable length,
Time Domain Reflectometers, TV ghosting, etc. are all tools
for illustration purposes. However, a Z0-matched system is
an ordinary configuration in ham radio and is easy to analyze
since no reflected energy is allowed to reach the source. The
conservation of energy and momentum rules dictate where the
energy must go in such a case. We can debate why the reflected
energy is 100% re-reflected but there is no question that it
*is* 100% re-reflected because none reaches the source and
there are only two directions in a transmission line.
--
73, Cecil http://www.qsl.net/w5dxp


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I've explained how I calculated how much energy is stored in the
transmission line. The movement of energy within the line is complex; in
the abbreviated analysis I've had time to do so far, it sloshes back and
forth in regions within the line. It does not travel in waves of average
power, bouncing back and forth, and believing so isn't necessary in
order comply with energy conservation. Your view of power and energy is
oversimplified, and it fails when you're pressed to explain what happens
at the interface between the line and the outside world.

Momentum is conserved in mechanical elastic collisions, but not in
inelastic ones, e.g., when energy is being extracted. I wouldn't begin
to try to apply this to a transmission line, but I see it doesn't bother
you. I understand and believe the fundamental principles of physics and
thermodynamics -- I'm just careful not to misapply them.

Roy Lewallen, W7EL

Cecil Moore wrote:
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.

Cecil Moore wrote:
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.

rho at the source is zero; the source matches the transmission line Z0.
So the "part" which is re-reflected is zero, by your calculations.

The remaining
Pref(1-rho^2) part

(which is all of it, since none was re-reflected)

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".

That part of Walt's text is talking about voltage and current waves. I'm
familiar, thanks, with how they interact. You're the one talking about
waves of average power -- since none of the reverse power wave is
re-reflected, what happens to it? Where does the power go? Out the
source resistor? I believe you said that 11.52 watts of it did. What
happened to the rest?

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.

What two waves head back toward the load? None of the alleged reverse
power wave is reflected, by your own calculation. Where did it go?
What's the other "cancelled wave" and where did it come from? What's its
magnitude?

Roy Lewallen, W7EL

Jim Kelley wrote:

It was a tongue-in-cheek reply, Roy. These problems always seem to end
up with a circulator in them at some point - clearly illustrating what
happens under entirely different circumstances. :-)

Thanks for the explanation. When Cecil has difficulty explaining a
circuit, he comes up with one which he can explain and tries to deflect
the discussion to it. This happens over and over and over -- it's really
a pain to try to keep steering the discussion back to the original circuit.

That leaves us with no one who claims to understand Cecil's
"explanation" of my simple circuit. Except, I guess, one guy who's on
vacation and some folks who don't read this newsgroup.

Roy Lewallen, W7EL

Cecil Moore wrote:
Jim Kelley wrote:

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?


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?


When a Z0-match is in place at the source, everything is also clear
since zero reflected energy reaches the source.

But there's a Z0 match at the source in my example. Simple wave
mechanics show that the initial traveling voltage and current waves
reflect from the load, return to the source, and don't reflect any
further -- steady state is reached after a single round trip. That's a
Z0 source match. No circulator is required.

The Z0-match case,
using an antenna tuner, is the most likely case to be encountered
in ham radio. 100% of the reflected energy and momentum

Walt's book mentions momentum? Where?

is re-reflected
back toward the load, just as Walter Maxwell has been saying as long
as I can remember.

Hm. You said earlier that 11.52 watts of the reverse power wave reached
the source. In a posting a short while ago you said that none of it is
re-reflected (since rho at the source = 0), and now you say it all is.

Guess that covers just about any possibility -- no matter how things
come out, you can say you gave the right answer. You might consider
tossing out a couple more just in case.

This stuff is not new. It is explained in "Fields and Waves ..."
by Ramo and Whinnery, copyright 1950's.

Did they give three different answers to a simple question?

Roy Lewallen, W7EL

If you put the perfect voltage source and the source resistor into a box
and label it "Source", you have a Source whose impedance perfectly
matches the transmission line. It's a Z0-matched system.

The source impedance of my circuit is as simple as it can get. If you
can't explain how it works, it reveals a deficiency of your theory; I
can easily calculate the voltage, current, and power in any component at
any moment of time. Without requiring bouncing waves of average power.
Or a circulator.

Roy Lewallen, W7EL

Cecil Moore wrote:
Roy Lewallen wrote:

Isn't it possible to explain what happens to the "reverse power"
without a circulator?


It is certainly possible in a Z0-matched system where
zero reflected energy reaches the source and the source
is feeding its designed-for load. If no reflected energy
reaches the source, the source impedance doesn't matter,
except for efficiency, which doesn't affect the rest of
the system. A one ohm source providing 100v to a 50 ohm
load and a one megohm source providing 100v to a 50 ohm
load will result in identical external conditions when
driving a 50 ohm load.

The picture is not as clear when reflected current and
voltage are allowed to flow into the source. We usually
don't know what the source impedance is and that
certainly handicaps any analysis. Modern designers simply
resort to protection circuitry and don't worry about
the energy analysis. We do know that reflections reaching
the source are not benign.

Roy Lewallen wrote:
I've explained how I calculated how much energy is stored in the
transmission line. The movement of energy within the line is complex; in
the abbreviated analysis I've had time to do so far, it sloshes back and
forth in regions within the line.

From where to where at what speed? Sloshing sounds like a
violation of the conservation of momentum. What force changes
the direction of the slosh? How much efficiency is lost in
the energy required to change sloshing directions?

It does not travel in waves of average
power, bouncing back and forth, and believing so isn't necessary in
order comply with energy conservation.

Nothing RF travels in waves of average power. Average power is
just our simplified shorthand way of dealing with it. The AC
voltage is equal to the RMS value at only two points in
the cycle. Sticking with instantaneous values for everything
AC would complicate things beyond belief.

The values of voltage in your chart were RMS (average) values.
You don't seem to have a problem dealing with RMS values of
voltages and currents. Why do you have a problem with average
power associated with those average RMS values of voltage and
current.

Your view of power and energy is
oversimplified, and it fails when you're pressed to explain what happens
at the interface between the line and the outside world.

What is located at that interface? Anybody is pressed to explain
things when the source impedance is unknown and cannot be measured.

Momentum is conserved in mechanical elastic collisions, but not in
inelastic ones, e.g., when energy is being extracted. I wouldn't begin
to try to apply this to a transmission line, but I see it doesn't bother
you.

Hecht, in "Optics" certainly applies it to EM waves. RF EM waves
differ from light waves only in frequency. Conservation of
momentum is a cornerstone of EM light physics and is included
in the rules of relativity.
--
73, Cecil http://www.qsl.net/w5dxp


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Roy Lewallen wrote:

rho at the source is zero; the source matches the transmission line Z0.
So the "part" which is re-reflected is zero, by your calculations.

Not my calculations. The measured SWR at the source terminals
is 4:1. rho = (SWR-1)/(SWR+1) = 3/5 = 0.6

That part of Walt's text is talking about voltage and current waves. I'm
familiar, thanks, with how they interact. You're the one talking about
waves of average power

Walt's text talks about RMS (average) voltage and current. When
you multiply RMS voltage by RMS current, you get average power.
V+*I+ = forward power, V+/I+ = Z0, (PV+)=(E+)x(H+)
V-*I- = reflected power, V-/I- = Z0, (PV-)=(E-)x(H-)

-- since none of the reverse power wave is
re-reflected, what happens to it?

On the contrary, it is 100% re-reflected by wave cancellation,
the fourth thing that can cause 100% re-reflection. Optical
engineers understand that. Most RF engineers don't. Adding
the one wavelength of lossless 200 ohm feedline reveals the
wave cancellation mechanism.

What two waves head back toward the load? None of the alleged reverse
power wave is reflected, by your own calculation.

Not my calculation. All of the reverse power wave is reflected
by wave cancellation at the source terminals. No reflected power
is dissipated in the source resistor. Therefore, it is all
reflected back toward the load. The one wavelength of 200 ohm
lossless feedline reveals exactly what happens.
--
73, Cecil http://www.qsl.net/w5dxp


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Roy Lewallen wrote:
But there's a Z0 match at the source in my example. Simple wave
mechanics show that the initial traveling voltage and current waves
reflect from the load, return to the source, and don't reflect any
further -- steady state is reached after a single round trip. That's a
Z0 source match. No circulator is required.

You obviously don't understand a Z0-match, Roy. All the reflected
power is re-reflected back toward the load at a Z0-match. What you
have in your example is a Z0-match to 200 ohms. You can prove it
by observing that the reflections disappear on a piece of 200 ohm
feedline.

------200 ohm feedline---+---1/2WL 50 ohm feedline---200 ohm load

The SWR on the 50 ohm feedline is 4:1. The SWR on the 200 ohm
feedline is 1:1. That is a 200 ohm Z0-match and rho is 0.6

Hm. You said earlier that 11.52 watts of the reverse power wave reached
the source.

Reflected by wave cancellation *at* the source , certainly not
flowing back through the source. What I said is that the two 11.52
watt reflected wave components undergo wave cancellation that results
in 23.04 watts being re-reflected.

In a posting a short while ago you said that none of it is
re-reflected ...

Sorry, Roy, you know and I know that is a lie. Please
apologize and retract that false statement.
--
73, Cecil http://www.qsl.net/w5dxp


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Roy Lewallen wrote:
If you put the perfect voltage source and the source resistor into a box
and label it "Source", you have a Source whose impedance perfectly
matches the transmission line. It's a Z0-matched system.

Sorry, Roy, a 50 ohm source driving a 200 ohm load is NOT a
Z0-matched system by any stretch of the imagination. From the
ARRL Antenna Book: "The Z0 mismatch (at the load) creates a
reflection having a magnitude of rho = (ZL-Z0)/(ZL+Z0)
causing a reflection loss rho^2 that is referred back along
the line to the generator. This in tern causes the generator to
see the same magnitude of Z0 mismatch at the line input."

This exactly describes your example which has a 50 ohm Z0
mismatch (and a 200 ohm Z0 match). The mismatch is easy to
see in the following:

50 ohm source--1WL 200 ohm line--+--1/2WL 50 ohm line--200 ohm load

Point '+' is a Z0-match to 200 ohms. A Z0-match to 50 ohms
doesn't exist anywhere.

The source impedance of my circuit is as simple as it can get. If you
can't explain how it works, it reveals a deficiency of your theory;

You assume I don't know how it works because you don't even know
what a Z0-match is???????? How very typical of you.
--
73, Cecil http://www.qsl.net/w5dxp


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Jim Kelley wrote:
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
Sorry, I was asking about the simple example I posted (which
doesn't have a circulator), not the problem posted by Cecil.

It was a tongue-in-cheek reply, Roy. These problems always seem to end
up with a circulator in them at some point - clearly illustrating what
happens under entirely different circumstances. :-)


Cecil imagines a circulator is a device that separates forward and
reflected waves of average power, because that is what Cecil's theory
says they must do.

In other words, Cecil's circulator is a device that takes a perfectly
straight-forward argument... and makes it circular.


--
73 from Ian GM3SEK 'In Practice' columnist for RadCom (RSGB)
http://www.ifwtech.co.uk/g3sek

Ian White GM3SEK wrote:
In other words, Cecil's circulator is a device that takes a perfectly
straight-forward argument... and makes it circular.

Cecil's circulator is a device that gets hot when reflected
waves are present, proving that reflected waves possess
energy and power.

Cecil's TDR allows everything about a reflected pulse to
be known and measured.

Cecil's TV ghosting allows one to see the reflections with
one's own eyes.

Cecil's one foot long piece of 200 ohm feedline allows one
to observe the elimination of reflections thus proving a
200 ohm Z0-match where the one who offered the example
thought there was a 50 ohm Z0-match.

I'm a professional teacher who uses lots of visual aids.
They work well on open-minded people. But they do often
pi$$ off closed-minded people because they leave nothing
worth arguing about.
--
73, Cecil http://www.qsl.net/w5dxp

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