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wrote:
So how are you taking into account the stored energy in the ideal
autotuner?

The ideal autotuner is lossless. The delay through the
tuner compared to one second is negligible. The energy
in the tuner components is negligible compared to 300
joules.

The principles are just as easy to comprehend if the
tuner is replaced by a circulator and load.
--
73, Cecil http://www.qsl.net/w5dxp

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Cecil Moore wrote:
wrote:
So how are you taking into account the stored energy in the ideal
autotuner?

The ideal autotuner is lossless. The delay through the
tuner compared to one second is negligible. The energy
in the tuner components is negligible compared to 300
joules.

Since the Z0-match is achieved at the ideal autotuner
input, just consider the lossless autotuner to be part
of the one second lossless transmission line system.
--
73, Cecil http://www.qsl.net/w5dxp

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What's the complex Poynting vector in the tuner? Why should I believe
that the delay through the tuner is negligible? Why should I believe
that the stored fields are negligible? It's not obvious to me.

For that matter, why does the ideal autotuner not reciprocal? The
impedance in the presence of the standing waves at the tuner-line
junction is R+jX. That is transformed by the autotuner so the source
sees 50 ohms, but that also means that the output impedance of the
generator is transformed as viewed by a wave coming down the line.

It's not transformed to infinity. If it is a circulator and load, this
is a distinctly nonreciprocal system. If it's a tuner affecting a
match, it is reciprocal. If your definition of an ideal autotuner
includes the nonreciprocity, please send me a schematic of such a
device.

I think I calculated the Poynting vector for this situation, I think
the Poynting vector divided by the group velocity of the waves and
integrated over the area of the coax gives the energy density per unit
length in the steady state, and if you've tuned the line length so that
the imaginary part of the Poynting vector goes to zero then the
explanation you give falls apart.

The tuner matters here. This is the problem with English language
discussions, especially disagreements, about electrical systems.
"There exists an ideal autotuner at the input" is insufficient
specification of the mathematical problem. If you place a device that
does everything you say it does at the source-line junction, maybe you
do get 300J in the one-second line.

Why isn't it reciprocal? Why is its phase shift negligible? Is it
just because it's physically small? I seem to remember a certain hotly
contested paper that I would think would have certainly put that sort
of idea out of your head. I think the crux of the misunderstanding is
in the tuner. I think it has stored energy and nonnegligible phase
delay, else it is non-reciprocal and dissipative (circulator+load).

What *IS* an ideal autotuner? I think you've stuffed too many
conditions into its operation for it to be a physically realizable
device. I'm pretty sure that it breaks the time reversal symmetry of
the system. Send out a pulse into your one second misterminated line
via your ideal autotuner. Wait half a second. FREEZE! Now run the
film backward while obeying the rules of your autotuner. The pulse
doesn't go back into the source. It bounces. What's up?

73,
Dan

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wrote:
What's the complex Poynting vector in the tuner? Why should I believe
that the delay through the tuner is negligible? Why should I believe
that the stored fields are negligible? It's not obvious to me.

The delay calculates out to be in nanoseconds compared to
the one second delay in the lossless line. The stored
fields calculate out to be in microjoules compared to
the 300 joules stored in the one second long feedline.

The real question is, allowing for all the things to which
you have objected, why is there very close to the number
of joules in the feedline necessary to support the forward
and reflected waves? What makes you think there are not
enough joules in the line to support the forward and
reflected waves?

For that matter, why does the ideal autotuner not reciprocal? The
impedance in the presence of the standing waves at the tuner-line
junction is R+jX. That is transformed by the autotuner so the source
sees 50 ohms, but that also means that the output impedance of the
generator is transformed as viewed by a wave coming down the line.

It can be proved that a 50 ohm Z0-match, because of destructive
interference on the 50 ohm side, reflects 100% of the reflected
energy back toward the load.

It's not transformed to infinity. If it is a circulator and load, this
is a distinctly nonreciprocal system. If it's a tuner affecting a
match, it is reciprocal. If your definition of an ideal autotuner
includes the nonreciprocity, please send me a schematic of such a
device.

Please read the Worldradio magazine article from my web page.
What you are calling nonreciprocity is merely destructive
interference on the source side of the Z0-match and constructive
interference on the load side of the Z0-match.

I think I calculated the Poynting vector for this situation, I think
the Poynting vector divided by the group velocity of the waves and
integrated over the area of the coax gives the energy density per unit
length in the steady state, and if you've tuned the line length so that
the imaginary part of the Poynting vector goes to zero then the
explanation you give falls apart.

Not if one deals with the forward Poynting vector and reflected
Poynting vector *separately* as does Ramo/Whinnery.

Pz+ = 200 watts of forward power in a matched line.

Pz- = 100 watts of reverse power in a matched line.

Superpose the two coherent waves and count the joules in the
superposed waves.

The tuner matters here. This is the problem with English language
discussions, especially disagreements, about electrical systems.
"There exists an ideal autotuner at the input" is insufficient
specification of the mathematical problem. If you place a device that
does everything you say it does at the source-line junction, maybe you
do get 300J in the one-second line.

Why isn't it reciprocal?

Because a Z0-match allows energy flow in
only the forward direction. That's how antenna tuners work.
Destructive interference keeps reflected energy from flowing
past the Z0-match toward the source. That's also how 1/4WL
thin-film anti-reflective glass works.

Why is its phase shift negligible? Is it
just because it's physically small?

All it has to do is shift the impedance from 300 ohms to 50 ohms.
A few microhenries and a few picofarads will do that. Its phase
shift would be in the nanoseconds.

What *IS* an ideal autotuner?

It's a concept to be thought about and discussed. If one accepts
a one second lossless feedline, why not an ideal autotuner for the
sake of conceptual discussion? Then a real autotuner can be
substituted and the losses, delays, and energy components estimated.

But like I said, the concept that I am presenting doesn't require
the ideal autotuner. It is actually easier to understand using
a circulator and load. There is exactly the amount of joules in
the transmission line needed to support the forward and reflected
energy waves.
--
73, Cecil http://www.qsl.net/w5dxp

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wrote:
Send out a pulse into your one second misterminated line
via your ideal autotuner. Wait half a second. FREEZE! Now run the
film backward while obeying the rules of your autotuner. The pulse
doesn't go back into the source. It bounces. What's up?

I certainly didn't say that. The steady-state voltage and
current component values have to exist for 100% destructive
interference to occur at a Z0-match. On the source side of
the Z0-match, in order to eliminate the reflected energy
wave, two reflected wave components are required to be
traveling in the same direction, be equal in magnitudes
and opposite in phase. Short pulses make those conditions
impossible.

In your above pulse example, there is no interference at
the impedance discontinuity and therefore a Z0-match doesn't
even exist. A Z0-match depends upon a steady-state supply
of RF waves.

A Z0-match at the input of a tuner works exactly like the
1/4WL thin-film anti-reflected glass coating. If one sends
a 1/2WL pulse of light at that piece of glass, there will
be reflections. Anti-reflective glass depends upon a
steady-state supply of light waves. One of the cancellation
components is 180 degrees older than the other one.

Let's return to the Signal Generators equipped with Circulators
and Loads (SGCL). Would you agree that the feedline contains
300 joules in the following example?

200W SGCL----one second long lossless feedline----100W SGCL
200W-- --100W
--
73, Cecil http://www.qsl.net/w5dxp

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Cecil Moore wrote:

In your above pulse example, there is no interference at
the impedance discontinuity and therefore a Z0-match doesn't
even exist. A Z0-match depends upon a steady-state supply
of RF waves.

A Z0-match at the input of a tuner works exactly like the
1/4WL thin-film anti-reflected glass coating. If one sends
a 1/2WL pulse of light at that piece of glass, there will
be reflections. Anti-reflective glass depends upon a
steady-state supply of light waves. One of the cancellation
components is 180 degrees older than the other one.

You really do need to choose your words a bit more carefully, Cecil.

Let's return to the Signal Generators equipped with Circulators
and Loads (SGCL). Would you agree that the feedline contains
300 joules in the following example?

200W SGCL----one second long lossless feedline----100W SGCL
200W-- --100W

Yes. But how much is there with the circulator load removed? (Not
replaced by an autotuner.) Numbers please.

Why you suppose I keep asking this question, Cecil?

73, ac6xg

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Jim Kelley wrote:
You really do need to choose your words a bit more carefully, Cecil.

Your statement is a personal opinion without technical content.

200W SGCL----one second long lossless feedline----100W SGCL
200W-- --100W

Yes. But how much is there with the circulator load removed? (Not
replaced by an autotuner.) Numbers please.

OK, let's remove the circulator loads and replace them
with unknown devices or objects inside black boxes.

Here's the same one second long lossless feedline with
unknown black boxes connected at each end (where BB stands
for black box).

BB#1------one second long lossless feedline------BB#2
200W-- --100W

Given: all conditions on the one second long lossless feedline
are identical in both of the above examples.

The laws of physics tell us that the joules in the two
lines with identical conditions are equal, i.e. 300
joules in these two cases. It doesn't matter what exists
or doesn't exist at the two ends of the line. 200 watts
forward and 100 watts reverse requires 300 joules/sec.

In other words, given a purely resistive Z0 and a properly
calibrated wattmeter, it is impossible to get 200 watts
of forward power and 100 watts of reverse power without
there being 300 joules in the one second long lossless line.
If there wasn't 300 joules in that line, we wouldn't
measure 200 watts forward and 100 watts reverse.
--
73, Cecil, http://www.qsl.net/w5dxp

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Jim Kelley wrote:
Why you suppose I keep asking this question, Cecil?

Because you believe in intelligent waves that somehow
sense their ultimate fate and therefore know how much
energy to carry or not?

Wave#1: "I am going to be dissipated in a circulator
load and therefore, I must contain N joules of energy."

Wave#2: "I am not going to be dissipated in a circulator
load and therefore, I must contain zero energy."

Given the following two configurations where BB stands
for Black Box:

BB#1----one second long lossless feedline----BB#2
200W-- --100W

BB#3----one second long lossless feedline----BB#4
200W-- --100W

Given: All voltages, currents, fields, and powers
are identical in both systems.

Please present us with a set of Black Boxes that will
satisfy your assertions and result in different energy
magnitudes in the two lines.
--
73, Cecil http://www.qsl.net/w5dxp

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Cecil Moore wrote:
Jim Kelley wrote:

Why do you suppose I keep asking this question, Cecil?


Because you believe in intelligent waves that somehow
sense their ultimate fate and therefore know how much
energy to carry or not?

No. That's something you keep insisting on talking about. Try to
keep your mind focussed on the fact that we're discussing the steady
state. (Or as you might proffer, after the waves have "decided" what
they're going to do.)

How about we just get back to the question. Energy in your 1 sec.
transmission line, only with no load on the circulator (and no
autotuner). How much? Explain. Try it!

73 de ac6xg

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Jim Kelley wrote:
How about we just get back to the question. Energy in your 1 sec.
transmission line, only with no load on the circulator (and no
autotuner). How much? Explain. Try it!

I honestly don't know what happens when the third port
of the circulator is unterminated but it is irrelevant
because the steady-state forward and reflected powers
would not be the same. I concede to you the fact that
one can always make an example too complicated to
analyze presumably for the purpose of diverting the
issue. But why would you want to divert the issue?

The real question is: Do two identical feedlines with
identical signal characteristics contain the same
amount of energy as in the following two examples.
Again SGCL is a Signal Generator with a Circulator and
Load resistor. Z0 is 50 ohms and the load is 291.42 ohms.

200W SGCL-----one second long lossless feedline---load
200W-- --100W
lossless
100W---tuner--one second long lossless feedline---load
200W-- --100W

It is easy to prove that the first example contains 300
joules in the feedline after two seconds.

It is easy to specify that the conditions in the feedline
in the two examples are identical during steady-state.

Are you asserting that two identical feedlines with identical
conditions contain a different number of joules?

I have another example in another posting.
--
73, Cecil http://www.qsl.net/w5dxp

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Jim Kelley wrote:
Try to keep
your mind focussed on the fact that we're discussing the steady state.

If one wants to know where the energy in the transmission
line came from, one must consider the initial transient
state. Let's take a simple example like the one W7EL uses
in his food for thought #1, a simple stub, except let's
make it one second long and lossless.

100W------one second long lossless feedline------open

During the first two seconds, the source pours 200
joules into the stub. After two seconds, steady-state
is reached and there is no net transfer of energy in
either direction at the source. Where are the 200 joules
that were sourced and not dissipated?
--
73, Cecil http://www.qsl.net/w5dxp

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Cecil Moore wrote:
Jim Kelley wrote:

How about we just get back to the question. Energy in your 1 sec.
transmission line, only with no load on the circulator (and no
autotuner). How much? Explain. Try it!


I honestly don't know what happens when the third port
of the circulator is unterminated

You should think about it because amateur radios don't usually have
circulators on their output, and the result illuminates a corner of
the discussion that you've been pretty closed minded about.

Without a load on the circulator, you must know that no energy can be
flowing to it or through it. No power is being dissipated by it.
What effect should that have on the total energy in the transmission
line, and why?

I hope it's not true about Reg. That would be a huge loss. Perhaps
somebody will archive his website.

73, ac6xg

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On Wed, 30 Aug 2006 10:17:44 -0700, Jim Kelley
wrote:

I hope it's not true about Reg. That would be a huge loss. Perhaps
somebody will archive his website.

Hi Jim,

The announcement came from his account, and his computer.

I've archived his site, but for others who wish to do the same, the
best tool for that purpose can be found at:
http://www.httrack.com

This is a website harvesting robot that will replicate an entire
website into the directory of your choice (changing links so that you
can browse it on your system).

73's
Richard Clark, KB7QHC

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Jim Kelley wrote:

Cecil Moore wrote:
I honestly don't know what happens when the third port
of the circulator is unterminated

You should think about it because amateur radios don't usually have
circulators on their output, and the result illuminates a corner of the
discussion that you've been pretty closed minded about.

I know what happens if the circulator is not present. I
just don't don't know what happens with an unterminated
circulator.

Maybe you can help me out here. If the load resistors
are removed from the circulators in the previous experiment,
what is the forward power and reflected power readings on
the one second long lossless transmission line during
steady-state?

Without a load on the circulator, you must know that no energy can be
flowing to it or through it. No power is being dissipated by it. What
effect should that have on the total energy in the transmission line,
and why?

Let's remove the circulators and try to figure out what
happens. Let's assume the signal generator is a Thevenin
equivalent with an internal resistance of 50 ohms and not
equipped with a circulator.

200W SG---one second long 50 ohm lossless feedline---100W SG

Would you agree with me that during the first second, the
feedline is loaded with 300 joules because there have been
no reflections? What happens after that depends upon the
reflection coefficients at the signal generators. But the
$64k question is what happens to the original 300 joules?

I hope it's not true about Reg. That would be a huge loss. Perhaps
somebody will archive his website.

One of the posters to the news
of Reg's death was whom I suspect is
Reg's son (or nephew). Humans are mortal and can die at any time.
My son died when he was 8 months old. I wish he had lived as long
as Reg who certainly had a full, useful, and colorful life.
--
73, Cecil http://www.qsl.net/w5dxp

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On Wed, 30 Aug 2006 12:12:20 -0700, Richard Clark wrote:

On Wed, 30 Aug 2006 10:17:44 -0700, Jim Kelley
wrote:

I hope it's not true about Reg. That would be a huge loss. Perhaps
somebody will archive his website.

Hi Jim,

The announcement came from his account, and his computer.

I've archived his site, but for others who wish to do the same, the
best tool for that purpose can be found at:
http://www.httrack.com

This is a website harvesting robot that will replicate an entire
website into the directory of your choice (changing links so that you
can browse it on your system).

73's
Richard Clark, KB7QHC

Richard, I've just reviewed the url above, and found that I don't know how to
use it to download Reg's web page. I see that you've already downloaded it, so
could I download it from your copy?

Walt, W2DU

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On Wed, 30 Aug 2006 15:52:52 -0400, Walter Maxwell
wrote:

On Wed, 30 Aug 2006 12:12:20 -0700, Richard Clark wrote:

On Wed, 30 Aug 2006 10:17:44 -0700, Jim Kelley
wrote:

I hope it's not true about Reg. That would be a huge loss. Perhaps
somebody will archive his website.

Hi Jim,

The announcement came from his account, and his computer.

I've archived his site, but for others who wish to do the same, the
best tool for that purpose can be found at:
http://www.httrack.com

This is a website harvesting robot that will replicate an entire
website into the directory of your choice (changing links so that you
can browse it on your system).

73's
Richard Clark, KB7QHC

Richard, I've just reviewed the url above, and found that I don't know how to
use it to download Reg's web page. I see that you've already downloaded it, so
could I download it from your copy?

Walt, W2DU

Hi Walt,

The paths are tied intimately to my file system hierarchy (which is
pretty deep). Rather, I will give you a walk-thru. In spite of the
apparent complexity (it is a technician's tool), it is quite simple to
use with only two or three particulars to satisfy:
1. As directed on the front page, press the NEXT button;
2. On the next page for Project Name, enter Reg Edwards G4FGQ;
3. Below that (skip the category), click the ellipses button to open
a storage path and select an existing folder the website will be
stored here in a folder named Reg Edwards G4FGQ;
4. Press the NEXT button at the bottom;
5. leave the ACTION selection at "Download web site(s)";
6. past Reggie's top level page,
http://www.btinternet.com/~g4fgq.regp, into the Web Addresses text
box;
7. Press the NEXT button at the bottom;
8. At the next page, Press the FINISH button at the bottom.

This will start the robots harvesting with a view of them on a new
page that shows each robot in its own thread - about half a dozen of
them running simultaneously. Depending on the load at the server, the
entire process should take 5 minutes or so at T1 speeds. The total
download is 4.33MB.

The robot activity screen will disappear at the end of the harvesting.

I can zip up a copy (sorry Reg) and mail it for those who want a copy
that is located at the drive root (I did this down load again to
confirm the steps described above).

73's
Richard Clark, KB7QHC

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Jim Kelley wrote:
You should think about it because amateur radios don't usually have
circulators on their output, and the result illuminates a corner of the
discussion that you've been pretty closed minded about.

Here's a brainteaser for you, Jim. Assume the following example
under steady-state conditions:

XMTR--X--tuner--Y--one second long lossless feedline---291.42 ohms
Ps=100W Pfor=200W-- --100W=Pref PL=100W

The lossless tuner is tuned for a Z0-match so there is zero
reflected power at point 'X'. The Z0 of the feedline is 50
ohms. The voltage reflection coefficient at the load is
0.707 making the power reflection coefficient 0.5

We suddenly disconnect the source at point 'Y'. After the
source is disconnected, how many joules of energy are delivered
to the load? _____

How many seconds does it take to dissipate all the energy in
the feedline? _____

On a second by second basis, how much power is delivered to the
load after the disconnect? _____ _____ _____ _____ _____
--
73, Cecil, http://www.qsl.net/w5dxp

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Cecil Moore wrote:
Jim Kelley wrote:

You should think about it because amateur radios don't usually have
circulators on their output, and the result illuminates a corner of
the discussion that you've been pretty closed minded about.


Here's a brainteaser for you, Jim. Assume the following example
under steady-state conditions:



XMTR--X--tuner--Y--one second long lossless feedline---291.42 ohms
Ps=100W Pfor=200W-- --100W=Pref PL=100W

If you aren't going to solve the problem I posed, then I don't see why
I should feel obliged to bother with any more of yours. Only seems
fair. There is a significant point to my question that you still
have not addressed. Would you have us believe that some new
revelation is being presented by this latest problem of yours? Or
perhaps it's simply intended to divert attention away, once again,
from my question.

Or maybe it was just too hard for you. How about a simpler problem then:

XMTR--X--one second long lossless feedline---infinite ohms
PS=100W

How many Joules are stored in the transmission line? Why?

73, ac6xg

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On Fri, 01 Sep 2006 11:03:39 -0700, Jim Kelley
wrote:

How many Joules are stored in the transmission line?

100

Why?

Spillage. The cup is two halves full (or no halves empty for the
pessimists in the crowd)

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Richard Clark wrote:

On Fri, 01 Sep 2006 11:03:39 -0700, Jim Kelley
wrote:


How many Joules are stored in the transmission line?


100

That may be the correct solution to a somewhat different problem.

Why?


Spillage. The cup is two halves full (or no halves empty for the
pessimists in the crowd)

Pessimists would report that "there are a disasterous and
insurmountable number of full halves", when in fact there are just two.

73, ac6xg

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Jim Kelley wrote:
Or maybe it was just too hard for you.

It was impossible for anyone to solve without having
a math model for the sources.

How about a simpler problem then:

This is not simpler - it is contradictory. It provides a
constant power source with nowhere for the source power
to go during steady-state. If you change the source to
a Thevenin equivalent then it can be solved.

XMTR--X--one second long lossless feedline---infinite ohms
PS=100W

Please explain how the XMTR can supply 100W during steady-
state? Where does the 100 joules/sec go? As stated, this
problem, like your other one, is impossible to resolve.

If you change the source to a Thevenin equivalent 141.42V
and 50 ohm source impedance and specify Z=50 ohms feedline
impedance then the problem becomes solvable. The answer is
that 200 joules exist in the feedline and, contrary to what
W7EL states, will dissipate in the source resistance after
the source voltage goes to zero. Note the source resistance
is the only resistance in the entire circuit.
--
73, Cecil http://www.qsl.net/w5dxp

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On Fri, 01 Sep 2006 12:49:43 -0700, Jim Kelley
wrote:

Richard Clark wrote:

On Fri, 01 Sep 2006 11:03:39 -0700, Jim Kelley
wrote:


How many Joules are stored in the transmission line?


100

That may be the correct solution to a somewhat different problem.


Hi Jim,

So, with a wrong answer (and a direct answer at that, imagine!). Do I
warrant the "correct" answer?

73's
Richard Clark, KB7QHC

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Richard Clark wrote:
So, with a wrong answer (and a direct answer at that, imagine!). Do I
warrant the "correct" answer?

With Jim's lackadaisical approach to conservation of
energy, there is no "correct" answer. He didn't even
tell you what the forward and reflected power readings
were. He also doesn't seem to realize that a 100W XMTR
cannot force any power into an infinite load. I am trying
to apply some boundary conditions that will remedy that
problem.

So Richard, what do you think happens when a constant
power output XMTR is facing an infinite load? What
happens when an irresistible force meets an immovable
object?
--
73, Cecil http://www.qsl.net/w5dxp

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On Fri, 01 Sep 2006 22:13:32 GMT, Cecil Moore
wrote:
With Jim's lackadaisical approach to conservation of
energy, there is no "correct" answer.

A long winded answer for you don't know. Talk about lackadaisical
with all the fluff cut off the end of this too.

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Richard Clark wrote:
Cecil Moore wrote:
With Jim's lackadaisical approach to conservation of
energy, there is no "correct" answer.

A long winded answer for you don't know.

Only a fool would venture an answer under the boundary
conditions defined by Jim. 100 watts into an infinite
load is impossible. The answer is the same as the
number of angels that can dance on the head of a pin.
--
73, Cecil http://www.qsl.net/w5dxp

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On Fri, 01 Sep 2006 22:49:12 GMT, Cecil Moore
wrote:

Richard Clark wrote:
Cecil Moore wrote:
With Jim's lackadaisical approach to conservation of
energy, there is no "correct" answer.

A long winded answer for you don't know.

Only a fool would venture an answer under the boundary
conditions defined by Jim.

a lackadaisical 14 word apology replacing a 3 word admission of not
knowing - and across two postings. No doubt this will be followed by:
1. more lackadaisical postings;
2. more lackadaisical words;
3. no answer.

It's curious that you haven't denied that 100 Joules is NOT the right
answer! At least Jim went that far, if no further. In a two man race
you have managed to put yourself in third place.

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Cecil Moore wrote:

Richard Clark wrote:

Cecil Moore wrote:

With Jim's lackadaisical approach to conservation of
energy, there is no "correct" answer.


A long winded answer for you don't know.


Only a fool would venture an answer under the boundary
conditions defined by Jim. 100 watts into an infinite
load is impossible. The answer is the same as the
number of angels that can dance on the head of a pin.

I may be a relative simpleton here, but the transmitter won't know that
for 2 seconds, Cecil. I bet some power may manage to get into the
transmission line.

tom
K0TAR

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Richard Clark wrote:
It's curious that you haven't denied that 100 Joules is NOT the right
answer!

I also haven't denied that 100 angels cannot dance on
the head of a pin for exactly the same reasons.
--
73, Cecil http://www.qsl.net/w5dxp

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Tom Ring wrote:
I may be a relative simpleton here, but the transmitter won't know that
for 2 seconds, Cecil. I bet some power may manage to get into the
transmission line.

Of course, I said that it was possible from a Thevenin
equivalent circuit. But Jim said the XMTR was putting
out 100 watts into an infinite load during steady-
state, obviously an impossible mental boundary condition.

A Thevenin equivalent circuit will supply 100 joules/sec
into the feedline and then stop supplying power. That
is not an impossible mental boundary condition. I'm
ready to discuss that configuration.
--
73, Cecil http://www.qsl.net/w5dxp

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On Sat, 02 Sep 2006 02:12:19 GMT, Cecil Moore
wrote:

Richard Clark wrote:
It's curious that you haven't denied that 100 Joules is NOT the right
answer!

I also haven't denied that 100 angels cannot dance on
the head of a pin for exactly the same reasons.

So, you have a reason to agree that 100 Joules is the right answer?

Boy, talk about getting respect for a straight answer to a straight
question. I'll let you two get back to your burlesque act. (I would
have called it vaudeville, but its like waiting for the stripper's
pasties to fall off.)

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Richard Clark wrote:
Cecil Moore wrote:
I also haven't denied that 100 angels cannot dance on
the head of a pin for exactly the same reasons.

So, you have a reason to agree that 100 Joules is the right answer?

As the problem was stated, I have NO reason to include 100
joules in the set of correct answers. I also have NO reason
to exclude 100 joules from the set of correct answers.

If Jim had specified that the forward power and reflected
power in the one-second long lossless stub was 50 watts
each during steady-state, then we could conclude that
there is 100 joules in the feedline.
--
73, Cecil http://www.qsl.net/w5dxp

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Cecil Moore wrote:
If you change the source to a Thevenin equivalent 141.42V
and 50 ohm source impedance and specify Z=50 ohms feedline
impedance then the problem becomes solvable.

Going with this configuration which is similar to the 1/2
WL stub suggested by W7EL in his Food For Thought #1:

SW
Source nc c Pfor=100W-- --Pref=100W
141.4V---o---o---one second long lossless 50 ohm stub--open
50 ohm
o---/\/\/\/\/\--Gnd
no 50 ohms

For the first two seconds after power up, the Thevenin
source delivers 200 joules into the stub while dissipating
200 joules as heat in the source resistance. Those 200
joules of energy in the stub must be conserved.

During steady-state, assuming the stub is an exact integral
number of wavelengths, the source will see an open circuit
and be delivering zero power. A wattmeter calibrated for
50 ohms will read 100 watts forward power and 100 watts
reflected power during steady-state.

After steady-state is reached, we throw the switch and connect
the stub to a 50 ohm dummy load. 100 watts will be supplied
to the dummy load for two seconds for a total of 200 joules.

Who was it who said that reflected power cannot be
recovered? (Food For Thought #1 on www.eznec.com).
--
73, Cecil http://www.qsl.net/w5dxp

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On Sat, 02 Sep 2006 13:15:02 GMT, Cecil Moore
wrote:

As the problem was stated, I have NO reason to
.... blah, blah, blah, blah

What a tedius excuse.

My answer stands, contradicted but without ANY correction. There's no
challenge in saying "'t'ain't so," or offering mea culpas, regrets,
plea bargains, or reams of Xerox copies of plate tectonics as applied
to transmission line theory.

What a boring strip-tease....

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Richard Clark wrote:
My answer stands, contradicted but without ANY correction.

I have corrected it in another posting. With a 50
ohm, 141.4V Thevenin source, the answer is 200 joules,
not your 100 joules. It takes two seconds for a one
second long lossless feedline to "fill up" with energy.
--
73, Cecil http://www.qsl.net/w5dxp

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On Sat, 02 Sep 2006 18:08:52 GMT, Cecil Moore
wrote:

I have corrected it in another posting...

Blah, blah, blah, tailored of course to a different problem. SOS

Hi All,

I saw Albert Brooks movie "Looking for Comedy in the Muslim World"
last night that had a routine by him that reminds me of this
boot-shuffling, toe-scrubbing side-step:

Brooks (to an audience of mixed Muslims, Hindus, Sikhs....):
"We are going to investigate the humor of improvisation and I need you
to give me a nationality"

"German!"

"OK, German. Now how about a occupation?"

"Farmer!"

"OK, a German Farmer. What does he raise?"

"Turnips!"

"Is he married?"

"Yes, with 5 children!"

"OK, is he rich or poor?"

"POOR!"

"OK, a poor German Farmer, married with 5 kids who raises Turnips.
That's great, we are going to have a lot of fun with this. (starting
to rummage up a routine, and then changing his mind) No, I don't like
him being German. We are going to make him Chinese, but Chinese don't
raise beets, they raise rice and they are doing pretty good, so he's
not poor.

"So, we have a middle class Chinese rice farmer with 5 kids - NO! I
read somewhere they have a population limit, and I don't want him
married. Let's make him divorced with 1 kid.

"No - raising rice is such a stereotype, so let's make him a bus
driver.

"OK, a divorced middle class Chinese bus driver with - No, I hear
Japan is really getting into this divorce thing. OK, a divorced
middle class Japanese auto worker with no children! Imagine him
asking for a raise:
"I want more money! Chinese make more money raising rice than
I do! Even Germans eating turnips do. Give me money!"

Yeah - right.....

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Richard Clark wrote:
Cecil Moore wrote:
I have corrected it in another posting...

Blah, blah, blah, tailored of course to a different problem. SOS

Yes, tailored to a configuration that is possible
(Thevenin source) rather than the previous impossible
one (100 watts into an infinite impedance).

So far, Jim has refused to discuss all possible
configurations and seems to prefer only the impossible
ones. One wonders why.
--
73, Cecil http://www.qsl.net/w5dxp