The Rest of the Story
 

On Mar 7, 10:24*am, Cecil Moore wrote:
Dave wrote:
"Keith Dysart" wrote:
What is the mechanism that creates the effect we call interference?

superposition.

Not disagreeing - just expanding:
Superposition is certainly necessary but superposition
alone is not sufficient. Superposition can occur with
or without interference.
--
73, Cecil *http://www.w5dxp.com

So superposition is the underlying mechanism that can be used to
explain all.
Interference is a phenomon that is observed for certain particular
conditions of superposition.

Works for me.

...Keith

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Cecil Moore wrote:
Gene Fuller wrote:
You may or may not already know this, but a lot of detailed optical
analysis these days is done with full 3-D electromagnetic simulation,
starting from Maxwell equations and boundary conditions. Interference,
coherence, energy flow, and all of the other stuff you like to discuss
can be *output* from that analysis, but those items are not part of
the input. The "centuries old" optics simply does not get the job
done. The "centuries old" stuff may work in the (impossible) cases
where everything is completely lossless and ideal, but it doesn't give
the right answers in the real world.

Ideal examples are time-honored ways of discussing concepts
and getting away from the vagaries of the real world. If one
understands the ideal examples, one is in a position to then
proceed to understanding the real world. If one fails to
understand the conceptual principles underlying the ideal
examples, one cannot possibly understand the real world.

Your posting seems to reflect your usual sour grapes attitude.
I will expect you to object to every example that uses lossless
transmission lines from now on including ones by Ramo & Whinnery,
Walter Johnson, Walter Maxwell, J. C. Slater and Robert Chipman.

I don't know why you would choose to accuse me of "sour grapes". That is
a characteristic of someone who has lost an argument. 8-)

You keep referring to the optical masters of old as being a huge
resource that is largely unknown to the RF crowd. I am merely
introducing the 21st century into the discussion.

73,
Gene
W4SZ

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Cecil Moore wrote:
Gene Fuller wrote:
The definition of irradiance, according to NIST, is power per unit
area. The standard units are W/m2 or lumen/m2.

Exactly how much power can exist in a zero unit of
time?

You previously objected to things that don't match
the real world. Instantaneous irradiance would rely
on an infinitesimally small amount of time, something
that doesn't match reality very well. One would think
you would therefore object to the concept of instantaneous
irradiance since it cannot be measured in reality and
exists only in the math model in the human mind.

Still spinning the words, huh? Do you really think that "time-averaged"
and "zero" time are the only possible choices?

Do you disagree with NIST?

73,
Gene
W4SZ

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Keith Dysart wrote:
My understanding of your claim was that for the special case of
a 45 degree line supplied from a matched source, the energy
in the reflected wave is dissipated in the source resistor.

I have told you time and again that your understanding is
wrong. My claim is that for the special case of a 45 degree
phase difference between the forward wave and reflected wave,
the *average* power in the reflected wave is dissipated in the
source resistor. Irradiance is an *average* power density as
defined by Hecht, in "Optics". I have told you previously (many
times) that when I use the word "power", I am talking about
*average* power. I agree with Hecht that instantaneous power
is "of limited utility" and is therefore mostly irrelevant.

For the record - for the umteenth time: When I say "power", I
am talking about "*average* power". If I ever talk about
instantaneous power, I will say "instantaneous power". If
you still don't understand, you need professional help.

This sentence fragment from your document suggests this:
"reflected energy from the load is flowing through the source
resistor, RS, and is being dissipated there".

I left no doubt as to what I meant in my document. Here is
a quote from the second paragraph in my document:
"Please note that any power referred to in this paper
is an *average power*. Nothing is being asserted or implied about
instantaneous powers. In fact, instantaneous powers are completely
irrelevant to the following discussion."

I simply don't know how to say it any plainer than that.

I really resent your lack of ethics in this matter.
If you are forced to create a Big Lie about what I have
said in order to try to win, is it really worth it?
--
73, Cecil http://www.w5dxp.com

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Cecil Moore wrote:
K7ITM wrote:
Certainly, those who design and build FTIR spectrometers know
perfectly well that interference does not depend on a narrow-band
coherent source.

How narrow-band? How coherent? In the irradiance (power
density) equation, Ptot = P1 + P2 + 2*sqrt(P1*P2)cos(A),
if the angle 'A' is varying rapidly, what value do you
use for cos(A)?

A constant average sustained level of destructive
interference cannot be maintained between two waves
unless they are coherent. If they are not coherent
the interference will average out to zero.

Gee, I wonder if the experts may have moved beyond the elementary optics
textbook descriptions?

Are you suggesting that FTIR cannot work unless one has your nice 1-D
configurations with perfectly monochromatic waves? Does everything need
to be collinear and coherent?

73,
Gene
W4SZ

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Keith Dysart wrote:
So superposition is the underlying mechanism that can be used to
explain all.
Interference is a phenomon that is observed for certain particular
conditions of superposition.

And permanent redistribution of energy is a phenomena that is
observed for certain particular conditions of interference.

Seems like a no-brainer to me.
--
73, Cecil http://www.w5dxp.com

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Gene Fuller wrote:
Do you disagree with NIST?

Could be that NIST didn't consider the fact that someone
could stupid enough to believe in a non-zero irradiance
averaged over an instantaneous time period of zero. :-)

Official Notice: For the purposes of my postings, I am using
Hecht's definition of "irradiance", i.e. "the average energy
per unit area per unit time".
--
73, Cecil http://www.w5dxp.com

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Gene Fuller wrote:
Are you suggesting that FTIR cannot work unless one has your nice 1-D
configurations with perfectly monochromatic waves?

Have you stopped beating your wife? Please cease and
desist with your diversions in the form of innuendo.

It is not my fault that a transmission line is essentially
one-dimensional but I am willing to take technical advantage
of that fact of physics. It is not my fault that CW transmitters
emit essentially monochromatic waves but I am willing to take
technical advantage of that fact of physics.
--
73, Cecil http://www.w5dxp.com

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Gene Fuller wrote:
Cecil Moore wrote:
Gene Fuller wrote:
The definition of irradiance, according to NIST, is power per unit
area. The standard units are W/m2 or lumen/m2.

Exactly how much power can exist in a zero unit of
time?

You previously objected to things that don't match
the real world. Instantaneous irradiance would rely
on an infinitesimally small amount of time, something
that doesn't match reality very well. One would think
you would therefore object to the concept of instantaneous
irradiance since it cannot be measured in reality and
exists only in the math model in the human mind.

Still spinning the words, huh? Do you really think that "time-averaged"
and "zero" time are the only possible choices?

Do you disagree with NIST?

73,
Gene
W4SZ

Cecil must have flunked calculus. Also, differential calculus is
the life-blood of classical electromagnetic theory. The ancients used
it liberally. Cecil probably doesn't believe in speed, because speed is
also an instantaneous quantity. Maybe he can use his argument on the
next ham-fisted Texas trooper who stops him for speeding on his motorcycle.
73,
Tom Donaly, KA6RUH

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On Mar 7, 5:31*pm, Cecil Moore wrote:
Keith Dysart wrote:
My understanding of your claim was that for the special case of
a 45 degree line supplied from a matched source, the energy
in the reflected wave is dissipated in the source resistor.

I have told you time and again that your understanding is
wrong. My claim is that for the special case of a 45 degree
phase difference between the forward wave and reflected wave,
the *average* power in the reflected wave is dissipated in the
source resistor. Irradiance is an *average* power density as
defined by Hecht, in "Optics". I have told you previously (many
times) that when I use the word "power", I am talking about
*average* power. I agree with Hecht that instantaneous power
is "of limited utility" and is therefore mostly irrelevant.

For the record - for the umteenth time: When I say "power", I
am talking about "*average* power". If I ever talk about
instantaneous power, I will say "instantaneous power". If
you still don't understand, you need professional help.

This sentence fragment from your document suggests this:
"reflected energy from the load is flowing through the source
resistor, RS, and is being dissipated there".

I left no doubt as to what I meant in my document. Here is
a quote from the second paragraph in my document:
"Please note that any power referred to in this paper
is an *average power*. Nothing is being asserted or implied about
instantaneous powers. In fact, instantaneous powers are completely
irrelevant to the following discussion."

I simply don't know how to say it any plainer than that.

I really resent your lack of ethics in this matter.
If you are forced to create a Big Lie about what I have
said in order to try to win, is it really worth it?

OK. I think I've got it now.

You are *not* claiming that the *energy* from the reflected wave
is dissipated in the source resistor, because for the *energy*
in the reflected wave to be dissipated in the source resistor,
the *energy* would have to dissipate at the same time that the
reflected wave delivered the *energy*, and the analysis of
instantaneous *energy* flows shows that this is not the case.

Rather, you are saying that the average reflected power is
numerically equal to the increase in the average dissipation
in the source resistor.

I can accept that as correct.

You might consider rewriting the sentence "reflected energy
from the load is flowing through the source resistor, RS, and
is being dissipated there" since it refers to the energy in
the reflected wave and may mislead others in the same way it
mislead me.

...Keith

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OK Cecil. I understand that none of your discussion applies to instantaneous values, only average values.

I am examining Keith's spread sheet carefully. I found that Open Calc will open the spreadsheet so I can read Keith's cell directions. (Thanks for making the spread sheet available Keith.) I want to make sure that I fully understand what is happening here, to the best of my ability and time available.

On Fri, 07 Mar 2008 12:13:45 -0600
Cecil Moore wrote:

Roger Sparks wrote:
On Fri, 07 Mar 2008 15:15:08 GMT
Cecil Moore wrote:
My formula applies *only to average power*.

I find myself surprised at your insistance that the instantaneous
case must be seperated from the average case in our example of waves
on a transmission line.

You have misunderstood what I am objecting to - which is:
Please don't say or imply that I have said or implied
anything about instantaneous values. I have NOT done so!
The formulas and concepts being used in the discussion
of instantaneous values are NOT mine! My objection is
to the false statements being made about what I have
posted.

Please go ahead and have your instantaneous discussion with
Keith but please don't say or imply that anything associated
with that discussion is about anything I have posted. Saying
or implying such a thing is simply false.

It is informative to look at each problem from many angles.

I agree - just stop saying that it is based on my assertions.
You or Keith may be right or wrong but either way, it is not
associated with anything I have posted. Please leave my name
out of any discussion concerning instantaneous values.

Did I misunderstand your premise, and you were really trying to say that
the inclusion of a 50 ohm source resistor would prevent the source from
ever 'seeing' anything but a 100 ohm load? I don't think that was your
intent.

Please add a dimension to your thinking. No matter what the
value of the load resistor, the source delivers 100 watts.
There are an infinite number of loads that the source could
"see" besides 100+j0 ohms that will make that condition true.

When the load is 0 ohms, the source "sees" 50+j50 ohms.
When the load is 12.5 ohms, the source "sees" 73.5+j44.1 ohms.
When the load is 25 ohms, the source "sees" 90+j30 ohms.
Only when the load is 50 ohms, does the source "see" 100+j0 ohms.
--
73, Cecil http://www.w5dxp.com


--
73, Roger, W7WKB

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On Mar 7, 12:14 pm, Cecil Moore wrote:
....
OK Tom, here's a challenge for you.

The only thing more foolish than the "challenge" that followed that
would be the person who accepted it. Since I don't suppose you think
me a fool, I'll just ignore it.

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On Mar 7, 2:34 pm, Gene Fuller wrote:
Cecil Moore wrote:
K7ITM wrote:
Certainly, those who design and build FTIR spectrometers know
perfectly well that interference does not depend on a narrow-band
coherent source.

How narrow-band? How coherent? In the irradiance (power
density) equation, Ptot = P1 + P2 + 2*sqrt(P1*P2)cos(A),
if the angle 'A' is varying rapidly, what value do you
use for cos(A)?

A constant average sustained level of destructive
interference cannot be maintained between two waves
unless they are coherent. If they are not coherent
the interference will average out to zero.

Gee, I wonder if the experts may have moved beyond the elementary optics
textbook descriptions?

Are you suggesting that FTIR cannot work unless one has your nice 1-D
configurations with perfectly monochromatic waves? Does everything need
to be collinear and coherent?

73,
Gene
W4SZ

So--I have a classic Michelson interferometer, and I see the classic
ring pattern on the screen at the "output" port. I also have a
sensitive microchannel plate detector system that I propose to put in
place of the screen, so that I can reduce the light amplitude to where
it makes sense to be observing it with the very sensitive detector.
In fact, I propose to reduce the light level to the point that the
short wavelength light I'm using is only putting a few photons per
second into the interferometer. I'll count a significant fraction of
those photons and identify where they landed on the microchannel
plate. Do you suppose, Gene, that I'll still see the same
interference pattern that I saw with the much higher intensity light?
Is there any limit to how low a light level I can use and still see
the pattern?

If I do still see the pattern, there must be yet another "dimension" I
need to add to my understanding of the situation -- not rooted in
classical Maxwell e&m.

And of course a dimension that is removed if you think only of average
quantities is time; one who thinks only in terms of averages removes
the possibility of the deeper understanding that resolution as a
function of time allows.

Cheers,
Tom

The Rest of the Story
 

Keith Dysart wrote:
You are *not* claiming that the *energy* from the reflected wave
is dissipated in the source resistor, because for the *energy*
in the reflected wave to be dissipated in the source resistor,
the *energy* would have to dissipate at the same time that the
reflected wave delivered the *energy*, and the analysis of
instantaneous *energy* flows shows that this is not the case.

I clearly stated that my claim is based on a special case
zero interference condition - it's even in the title of the
article. The instantaneous energy that you (not I) introduced,
does not meet the zero interference precondition. Therefore,
anything that does not meet the zero interference precondition
that I enumerated is an irrelevant diversion. You introduced
that irrelevant straw man and tried to make hay out it. :-)

you are saying that the average reflected power is
numerically equal to the increase in the average dissipation
in the source resistor.

I can accept that as correct.

Finally, after a million words. :-)

You might consider rewriting the sentence "reflected energy
from the load is flowing through the source resistor, RS, and
is being dissipated there" since it refers to the energy in
the reflected wave and may mislead others in the same way it
mislead me.

Since "average reflected power" is dependent upon the
"reflected energy", I don't see any problem. Would
"average reflected energy" work for you? It should be
obvious that, associated with interference during each
cycle, destructive interference energy is stored during
part of the cycle and delivered back as constructive
interference energy during another part of the cycle.
The intra-cycle interference averages out to zero.

Many have objected to the term "reflected power" saying
it is not power that is reflected but instead is "reflected
energy". So I stopped talking about "reflected power" and
started talking about "reflected energy". Now you object
to the use of the term "reflected energy". Would you and
the rest of the guru attack gang please get together on
what term you would like for me to use?
--
73, Cecil http://www.w5dxp.com

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Roger Sparks wrote:
OK Cecil. I understand that none of your discussion applies
to instantaneous values, only average values.

Indeed, instantaneous values are a diversion away from
my discussion/article/example using average values.

I am examining Keith's spread sheet carefully. I found that
Open Calc will open the spreadsheet so I can read Keith's cell
directions. (Thanks for making the spread sheet available
Keith.) I want to make sure that I fully understand what is
happening here, to the best of my ability and time available.

Given the superposition of two voltages, when the two
voltages have the same sign, constructive interference
is present. When the two voltages have opposite signs,
destructive interference is present. The destructive
interference during each cycle exactly equals (and
supplies) the constructive interference during the
same cycle so the average interference is zero.
--
73, Cecil http://www.w5dxp.com

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K7ITM wrote:
Cecil Moore wrote:
OK Tom, here's a challenge for you.

The only thing more foolish than the "challenge" that followed that
would be the person who accepted it. Since I don't suppose you think
me a fool, I'll just ignore it.

Tom, you are the one who implied that coherency or
incoherency doesn't matter. The challenge would have
proved it doesn't matter. Since you refused the
challenge, can we assume that it does matter?
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

K7ITM wrote:
On Mar 7, 2:34 pm, Gene Fuller wrote:
Cecil Moore wrote:
K7ITM wrote:
Certainly, those who design and build FTIR spectrometers know
perfectly well that interference does not depend on a narrow-band
coherent source.
How narrow-band? How coherent? In the irradiance (power
density) equation, Ptot = P1 + P2 + 2*sqrt(P1*P2)cos(A),
if the angle 'A' is varying rapidly, what value do you
use for cos(A)?
A constant average sustained level of destructive
interference cannot be maintained between two waves
unless they are coherent. If they are not coherent
the interference will average out to zero.
Gee, I wonder if the experts may have moved beyond the elementary optics
textbook descriptions?

Are you suggesting that FTIR cannot work unless one has your nice 1-D
configurations with perfectly monochromatic waves? Does everything need
to be collinear and coherent?

73,
Gene
W4SZ

So--I have a classic Michelson interferometer, and I see the classic
ring pattern on the screen at the "output" port. I also have a
sensitive microchannel plate detector system that I propose to put in
place of the screen, so that I can reduce the light amplitude to where
it makes sense to be observing it with the very sensitive detector.
In fact, I propose to reduce the light level to the point that the
short wavelength light I'm using is only putting a few photons per
second into the interferometer. I'll count a significant fraction of
those photons and identify where they landed on the microchannel
plate. Do you suppose, Gene, that I'll still see the same
interference pattern that I saw with the much higher intensity light?
Is there any limit to how low a light level I can use and still see
the pattern?

If I do still see the pattern, there must be yet another "dimension" I
need to add to my understanding of the situation -- not rooted in
classical Maxwell e&m.

And of course a dimension that is removed if you think only of average
quantities is time; one who thinks only in terms of averages removes
the possibility of the deeper understanding that resolution as a
function of time allows.

Cheers,
Tom

Tom,

One step at a time. Cecil has not yet accepted the real world in the
classical state. The quantum state will need to wait.

8-)

73,
Gene
W4SZ

The Rest of the Story
 

Gene Fuller wrote:
Cecil has not yet accepted the real world in the
classical state.

I know there is no such thing as a lossless transmission
line, Gene. That doesn't prohibit me from using lossless
transmission lines in an example, does it? Every textbook
on transmission lines that I have ever seen does the same
thing.
--
73, Cecil http://www.w5dxp.com

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On Sat, 8 Mar 2008 10:12:04 -0800 (PST), K7ITM wrote:

Is there any limit to how low a light level I can use and still see
the pattern?

Hi Tom,

The Quantum Efficiency of the eye is between 40% and 50%. The time to
convert one photon in a visual receptor is roughly 14 femtoseconds.

I have seen no component that can match the bandwidth, dynamic range,
AND sensitivity of the eye.

73's
Richard Clark, KB7QHC

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On Sat, 08 Mar 2008 14:21:59 -0800, Richard Clark wrote:

On Sat, 8 Mar 2008 10:12:04 -0800 (PST), K7ITM wrote:

Is there any limit to how low a light level I can use and still see the
pattern?

Hi Tom,

The Quantum Efficiency of the eye is between 40% and 50%. The time to
convert one photon in a visual receptor is roughly 14 femtoseconds.

I have seen no component that can match the bandwidth, dynamic range,
AND sensitivity of the eye.

Side note:

Back in the day, when I did a lot of color processing and printing, I
would spend log hours in the darkroom in total darkness.

The only light shedding items were the old Gra-Lab timers, and
luminescent tape on the dangerous corners. After several hours, the light
output from them was just about nil.

My eyes were totally adapted, full visual purple. At this time, the
luminescent tape and dials seemed to scintillate discretely, not the
typical overall glow.

I always wondered if I was possibly seeing individual photon effects.

- 73 de Mike N3LI -

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On Sat, 08 Mar 2008 19:12:01 -0600, Mike Coslo
wrote:

I always wondered if I was possibly seeing individual photon effects.

At the risk of over-extending the discussion: one in two events, with
the photoreceptors doing what we would call pulse-stretching as the
mind is wholly incapable of discerning a visual event of less than,
roughly, 10mS duration.

73's
Richard Clark, KB7QHC

The Rest of the Story
 

On Mar 8, 2:30 pm, Cecil Moore wrote:
Keith Dysart wrote:
You are *not* claiming that the *energy* from the reflected wave
is dissipated in the source resistor, because for the *energy*
in the reflected wave to be dissipated in the source resistor,
the *energy* would have to dissipate at the same time that the
reflected wave delivered the *energy*, and the analysis of
instantaneous *energy* flows shows that this is not the case.

I clearly stated that my claim is based on a special case
zero interference condition - it's even in the title of the
article. The instantaneous energy that you (not I) introduced,
does not meet the zero interference precondition. Therefore,
anything that does not meet the zero interference precondition
that I enumerated is an irrelevant diversion. You introduced
that irrelevant straw man and tried to make hay out it. :-)

you are saying that the average reflected power is
numerically equal to the increase in the average dissipation
in the source resistor.

I can accept that as correct.

Finally, after a million words. :-)

You might consider rewriting the sentence "reflected energy
from the load is flowing through the source resistor, RS, and
is being dissipated there" since it refers to the energy in
the reflected wave and may mislead others in the same way it
mislead me.

Since "average reflected power" is dependent upon the
"reflected energy", I don't see any problem. Would
"average reflected energy" work for you? It should be
obvious that, associated with interference during each
cycle, destructive interference energy is stored during
part of the cycle and delivered back as constructive
interference energy during another part of the cycle.
The intra-cycle interference averages out to zero.

Now you have me quite confused. One moment you agree that
your claim is mere numerical eqivalency and the next you
seem to again be claiming that the energy in the reflected
wave is dissipated in the source resistor.

Can you clarify which is really your claim?

Many have objected to the term "reflected power" saying
it is not power that is reflected but instead is "reflected
energy". So I stopped talking about "reflected power" and
started talking about "reflected energy". Now you object
to the use of the term "reflected energy". Would you and
the rest of the guru attack gang please get together on
what term you would like for me to use?

I have never been particularly fussy about the terminology.
"Energy flow", "power flow": The latter is often used when,
strictly, the former is meant, but there is seldom confusion,
except for those excessive pedantics who choose to be
confused.

My issue was that you seemed, in that sentence, to be saying
that the reflected energy was dissipated in the source
resistor. But earlier you had stated that was not your claim.
See my request for clarification above.

....Keith

The Rest of the Story
 

Keith Dysart wrote:
Now you have me quite confused. One moment you agree that
your claim is mere numerical eqivalency and the next you
seem to again be claiming that the energy in the reflected
wave is dissipated in the source resistor.

I have never used the words "mere numerical equivalency"
so I have never claimed any such thing. The energy in
the reflected wave is dissipated in the source resistor
but you are confused about the timing of that dissipation.
It doesn't happen when you are saying it happens. Some
magnitude of energy is stored and dissipated later in
the same cycle.

Your analysis so far has completely ignored
*one additional source of instantaneous energy*, namely
the reactive component in the network, i.e. the reactance
of the transmission line. When you account for the temporary
storing of the intra-cycle destructive interference energy
in the transmission line followed by its later release as
constructive interference, you will find that all of the
energy in the reflected wave is dissipated in the source
resistor.

During part of a cycle, energy is lost from the source
resistor into the transmission line. During the following
part of the same cycle, that same energy is recovered
from the transmission line back to the source resistor.

instantaneous power dissipated in the source resistor =
1. instantaneous forward power plus
2. instantaneous reflected power plus
3. instantaneous interference (can be plus or minus)

Your calculations so far have completely ignored that
third term which is a source (or sink) of energy depending
upon what part of the cycle exists.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

"Keith Dysart" wrote in message
...
On Mar 8, 2:30 pm, Cecil Moore wrote:
Keith Dysart wrote:
You are *not* claiming that the *energy* from the reflected wave
is dissipated in the source resistor, because for the *energy*
in the reflected wave to be dissipated in the source resistor,
the *energy* would have to dissipate at the same time that the
reflected wave delivered the *energy*, and the analysis of
instantaneous *energy* flows shows that this is not the case.

I clearly stated that my claim is based on a special case
zero interference condition - it's even in the title of the
article. The instantaneous energy that you (not I) introduced,
does not meet the zero interference precondition. Therefore,
anything that does not meet the zero interference precondition
that I enumerated is an irrelevant diversion. You introduced
that irrelevant straw man and tried to make hay out it. :-)

you are saying that the average reflected power is
numerically equal to the increase in the average dissipation
in the source resistor.

I can accept that as correct.

Finally, after a million words. :-)

You might consider rewriting the sentence "reflected energy
from the load is flowing through the source resistor, RS, and
is being dissipated there" since it refers to the energy in
the reflected wave and may mislead others in the same way it
mislead me.

Since "average reflected power" is dependent upon the
"reflected energy", I don't see any problem. Would
"average reflected energy" work for you? It should be
obvious that, associated with interference during each
cycle, destructive interference energy is stored during
part of the cycle and delivered back as constructive
interference energy during another part of the cycle.
The intra-cycle interference averages out to zero.

Now you have me quite confused. One moment you agree that
your claim is mere numerical eqivalency and the next you
seem to again be claiming that the energy in the reflected
wave is dissipated in the source resistor.

Can you clarify which is really your claim?

Many have objected to the term "reflected power" saying
it is not power that is reflected but instead is "reflected
energy". So I stopped talking about "reflected power" and
started talking about "reflected energy". Now you object
to the use of the term "reflected energy". Would you and
the rest of the guru attack gang please get together on
what term you would like for me to use?

NEITHER! they are both confusing. use the most fundamental things that you
can measure, either voltage or current. either one is completely defined in
the basic maxwell equations, and either one is completely sufficient to
describe ALL effect on a cable or in any circuit.

The Rest of the Story
 

Dave wrote:
"Keith Dysart" wrote in message
Cecil Moore wrote:
So I stopped talking about "reflected power" and
started talking about "reflected energy". Now you object
to the use of the term "reflected energy". Would you and
the rest of the guru attack gang please get together on
what term you would like for me to use?

NEITHER! they are both confusing. use the most fundamental things that you
can measure, either voltage or current. either one is completely defined in
the basic maxwell equations, and either one is completely sufficient to
describe ALL effect on a cable or in any circuit.

If Maxwell's equations could be used to answer the
questions that we are asking, why haven't they been
answered a long time ago?

How can Maxwell's equations be used to track the path
and fate of the energy in a reflected wave?
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

K7ITM wrote:
On Mar 5, 1:27 pm, Cecil Moore wrote:
The rules change between non-coherent,
non-interfering sources and coherent, interfering sources.

And exactly which part of "linear system" do you fail to understand?

Tom, I am ready to eat crow and apologize to you on this
one. My false statement above was a rash assumption based
on my aversion to instantaneous values.

Last night, while puppy siting my girlfriend's dog in
a Walmart parking lot, I realized that the power density
(irradiance) equation as presented by Hecht and others
does work at the instantaneous level. Mea Culpa.

The difference between coherent and incoherent signals
is not in the rules but in the timing. Keith and I
were making essentially the same mistake. Thanks for
making me think this one through.
--
73, Cecil http://www.w5dxp.com

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Keith Dysart wrote:
My issue was that you seemed, in that sentence, to be saying
that the reflected energy was dissipated in the source
resistor. But earlier you had stated that was not your claim.

My earlier claim was that the average power in a reflected
wave is dissipated in the source resistor when the forward
wave is 90 degrees out of phase with the reflected wave at
the source resistor. In that earlier claim, I didn't care
to discuss instantaneous power and thus excluded instantaneous
power from that claim.

For instantaneous values, it will be helpful to change the
example while leaving the conditions at the source resistor
unchanged. Here's the earlier example:

Rs Vg Vl
+----/\/\/-----+----------------------+
| 50 ohm |
| 1/8 WL |
Vs 45 degrees 12.5 ohm
100v RMS 50 ohm line Load
| |
| |
+--------------+----------------------+
gnd

Here's the present example:

Rs Vg Vl
+----/\/\/-----+----------------------+
| 50 ohm |
| 1 WL |
Vs 360 degrees 23.5+j44.1
100v RMS 50 ohm line ohm Load
| |
| |
+--------------+----------------------+
gnd

If I haven't made some stupid mistake, the conditions at
the source resistor are identical in both examples. But
in the second example, it is obvious that energy can
be stored in the transmission line during part of a cycle
(thus avoiding dissipation at that instant in time) and be
delivered back to the source resistor during another part
of the cycle (to be dissipated at a later instant in time).
That is the nature of interference energy and is exactly
equal to the difference between the two powers that you
calculated. You neglected to take into account the ability
of the network reactance to temporarily store energy and
dissipate it later in time.

All of the reflected energy is dissipated in the source
resistor, just not at the time you thought it should be.

While performing this analysis, you will discover the ability
of superposing waves to redistribute energy (even in the
complete absence of ordinary reflections). Not possible in
the above special case examples, but under other conditions,
the redistribution of energy can attain perpetual steady-state
status (even in the complete absence of ordinary reflections).

There are two mechanisms involved in the re-routing of
steady-state reflected energy back toward the load.

1. Ordinary reflection of single EM waves governed by the
rules of the wave reflection model.

2. A redistribution of reflected energy back toward the
load as a result of superposition of waves accompanied
by an average level of steady-state interference.
--
73, Cecil http://www.w5dxp.com

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On Sun, 09 Mar 2008 14:35:25 GMT
Cecil Moore wrote:

Keith Dysart wrote:
My issue was that you seemed, in that sentence, to be saying
that the reflected energy was dissipated in the source
resistor. But earlier you had stated that was not your claim.

My earlier claim was that the average power in a reflected
wave is dissipated in the source resistor when the forward
wave is 90 degrees out of phase with the reflected wave at
the source resistor. In that earlier claim, I didn't care
to discuss instantaneous power and thus excluded instantaneous
power from that claim.

For instantaneous values, it will be helpful to change the
example while leaving the conditions at the source resistor
unchanged. Here's the earlier example:

Rs Vg Vl
+----/\/\/-----+----------------------+
| 50 ohm |
| 1/8 WL |
Vs 45 degrees 12.5 ohm
100v RMS 50 ohm line Load
| |
| |
+--------------+----------------------+
gnd

Here's the present example:

Rs Vg Vl
+----/\/\/-----+----------------------+
| 50 ohm |
| 1 WL |
Vs 360 degrees 23.5+j44.1
100v RMS 50 ohm line ohm Load
| |
| |
+--------------+----------------------+
gnd

If I haven't made some stupid mistake, the conditions at
the source resistor are identical in both examples. But
in the second example, it is obvious that energy can
be stored in the transmission line during part of a cycle
(thus avoiding dissipation at that instant in time) and be
delivered back to the source resistor during another part
of the cycle (to be dissipated at a later instant in time).
That is the nature of interference energy and is exactly
equal to the difference between the two powers that you
calculated. You neglected to take into account the ability
of the network reactance to temporarily store energy and
dissipate it later in time.

All of the reflected energy is dissipated in the source
resistor, just not at the time you thought it should be.

I think we can all agree to what would happen in the turn off situation following a long period of stable power flow. When the source voltage steps from 100v to 0v, the power on the transmission line reflects to the 50 ohm source resistor and is all absorbed.

On the other hand, in the case of steady power flow, the source is presented with a reactive load of 73.5 + J44.1 ohms. The reactive part of this load results from returning power from the transmission line.

For the resistor Rs, it will have power applied from two sources, the source and the reflected power from the transmission line, i.e., the source power and reflected power are in series when considered in relationship with the resistor Rs. The problem is that 'what looks like two sources to resistor Rs, is really only one source, Vs'.

I think you are looking for solutions that show how power to Rs peaks at a different time from when power into the transmission line peaks, which is yet a different time from when power from Vs peaks. You are looking for the instantaneous timing of 3 peaks.


While performing this analysis, you will discover the ability
of superposing waves to redistribute energy (even in the
complete absence of ordinary reflections). Not possible in
the above special case examples, but under other conditions,
the redistribution of energy can attain perpetual steady-state
status (even in the complete absence of ordinary reflections).

There are two mechanisms involved in the re-routing of
steady-state reflected energy back toward the load.

1. Ordinary reflection of single EM waves governed by the
rules of the wave reflection model.

2. A redistribution of reflected energy back toward the
load as a result of superposition of waves accompanied
by an average level of steady-state interference.
--
73, Cecil http://www.w5dxp.com


--
73, Roger, W7WKB

The Rest of the Story
 

On Mar 8, 12:49 pm, Cecil Moore wrote:
....
Tom, you are the one who implied that coherency or
incoherency doesn't matter.

No, you're the one who inferred that from what I actually did say,
which is (in assorted different wordings) that we're dealing with a
linear system, and as such we can fully analyze it and indeed get MORE
insight into what's going on by using instantaneous voltages and
currents than we can in looking only at averaged powers. Indeed, in
setting up the analysis, it matters not a whit what relationship the
reverse wave has to the forward wave. There is neither a need nor an
advantage to special-case something like coherency; it all falls out
of the much more general analysis. There may be interesting things to
observe in the outcome of the analysis; that's a different issue.

Note that, as far as I've been able to determine, Michelson did not
have a coherent light source to shine into his interferometer, but
still he saw interference patterns. Perhaps he had invented lasers
but failed to tell the world about them, but I rather doubt it.
Carried to something more of an extreme, FTIRs tell you about the
spectral content of relatively broadband light, a few octaves, by
analyzing the interference pattern produced by a Michelson
interferometer. But the instruments and models I can reasonably apply
to our slow-as-molasses radio-frequency waves let me get even more
insights into what's going on.

As far as I can see, this is all in agreement with what Gene, Roger
and Keith, and perhaps some others I haven't been following, have been
saying in this thread, and what many others have said for years in
this group. If you get the idea that people aren't listening to you,
maybe it's because they've heard your story and see no advantage in it
over the more general analysis that's readily available.

The Rest of the Story
 

On Sun, 9 Mar 2008 15:07:26 -0700 (PDT), K7ITM wrote:

snip


Note that, as far as I've been able to determine, Michelson did not
have a coherent light source to shine into his interferometer, but
still he saw interference patterns. Perhaps he had invented lasers

snip


It is said he used sodium vapor gas light (~589 nm). Coherent enough.

Chuck

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Chuck wrote in
:

On Sun, 9 Mar 2008 15:07:26 -0700 (PDT), K7ITM wrote:

snip


Note that, as far as I've been able to determine, Michelson did not
have a coherent light source to shine into his interferometer, but
still he saw interference patterns. Perhaps he had invented lasers

snip


It is said he used sodium vapor gas light (~589 nm). Coherent enough.

I have a wonderful old book, named "A Treatise on Optics", written sometime
in the late 1800's. I have to look it up here, Perhaps Cecil would be
interested in a pdf copy if I locate it again.

Those old boys could do an awful lot that we do today, and I was amazed at
the precise measurements that they could do with almost entirely mechanical
devices. Coherent light is not the half of it!

We are both more and less advanced than we might think.

- 73 de Mike -

The Rest of the Story
 

K7ITM wrote:
Cecil Moore wrote:
Tom, you are the one who implied that coherency or
incoherency doesn't matter.

No, you're the one who inferred that from what I actually did say,
which is (in assorted different wordings) that we're dealing with a
linear system, and as such we can fully analyze it and indeed get MORE
insight into what's going on by using instantaneous voltages and
currents than we can in looking only at averaged powers.

Tom, I have apologized, eaten crow, and said Mea Culpa.
What more do you want from me?
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Roger Sparks wrote:
For the resistor Rs, it will have power applied from two sources, the source and the reflected power from the transmission line, i.e., the source power and reflected power are in series when considered in relationship with the resistor Rs. The problem is that 'what looks like two sources to resistor Rs, is really only one source, Vs'.

Actually, it will have energy applied from three sources, the
source, the reflected energy, and the reactive energy stored
in the transmission line. The energy that Keith is missing
comes from the reactance in the transmission line.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Mon, 10 Mar 2008 02:38:05 GMT
Cecil Moore wrote:

Roger Sparks wrote:
For the resistor Rs, it will have power applied from two sources, the source and the reflected power from the transmission line, i.e., the source power and reflected power are in series when considered in relationship with the resistor Rs. The problem is that 'what looks like two sources to resistor Rs, is really only one source, Vs'.

Actually, it will have energy applied from three sources, the
source, the reflected energy, and the reactive energy stored
in the transmission line. The energy that Keith is missing
comes from the reactance in the transmission line.
--
73, Cecil http://www.w5dxp.com

Here is a link to a discussion of AC power. http://en.wikipedia.org/wiki/AC_power

How do you propose to seperate reactive power from reflected power?

--
73, Roger, W7WKB

The Rest of the Story
 

Cecil Moore wrote:
K7ITM wrote:
Cecil Moore wrote:
Tom, you are the one who implied that coherency or
incoherency doesn't matter.

No, you're the one who inferred that from what I actually did say,
which is (in assorted different wordings) that we're dealing with a
linear system, and as such we can fully analyze it and indeed get MORE
insight into what's going on by using instantaneous voltages and
currents than we can in looking only at averaged powers.

Tom, I have apologized, eaten crow, and said Mea Culpa.
What more do you want from me?

Rending of hair and wearing of sackcloth? 8^)

- 73 de Mike N3LI -

The Rest of the Story
 

Roger Sparks wrote:
How do you propose to seperate reactive power from reflected power?

For average power, it's easy using phasors.
real power = V*I*cos(theta)
reactive power = V*I*sin(theta)

I know very little about instantaneous power having
avoided thinking about it for most of my life.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Cecil Moore wrote:
Dave wrote:
"Keith Dysart" wrote in message
Cecil Moore wrote:
So I stopped talking about "reflected power" and
started talking about "reflected energy". Now you object
to the use of the term "reflected energy". Would you and
the rest of the guru attack gang please get together on
what term you would like for me to use?

NEITHER! they are both confusing. use the most fundamental things
that you can measure, either voltage or current. either one is
completely defined in the basic maxwell equations, and either one is
completely sufficient to describe ALL effect on a cable or in any
circuit.

If Maxwell's equations could be used to answer the
questions that we are asking, why haven't they been
answered a long time ago?

How can Maxwell's equations be used to track the path
and fate of the energy in a reflected wave?

Cecil,

There may be some terminology confusion here.

Maxwell's equations are really the only relevant physical equations
there are to work with, at least in the classical regime. The discussion
about constructive, destructive, superposition, linearity, etc.
represents merely mathematical manipulation of the basic physical
entities embodied in Maxwell's equations.

All of this *math* is of course very important and very useful. However,
it is not a replacement for the *physical* laws known as the Maxwell
equations.

(And we all know that these endless threads are closely parallel to the
blind men describing an elephant puzzle.)

73,
Gene
W4SZ

The Rest of the Story
 

On Mar 9, 10:35 am, Cecil Moore wrote:
Keith Dysart wrote:
My issue was that you seemed, in that sentence, to be saying
that the reflected energy was dissipated in the source
resistor. But earlier you had stated that was not your claim.

My earlier claim was that the average power in a reflected
wave is dissipated in the source resistor when the forward
wave is 90 degrees out of phase with the reflected wave at
the source resistor. In that earlier claim, I didn't care
to discuss instantaneous power and thus excluded instantaneous
power from that claim.

For instantaneous values, it will be helpful to change the
example while leaving the conditions at the source resistor
unchanged. Here's the earlier example:

Rs Vg Vl
+----/\/\/-----+----------------------+
| 50 ohm |
| 1/8 WL |
Vs 45 degrees 12.5 ohm
100v RMS 50 ohm line Load
| |
| |
+--------------+----------------------+
gnd

Here's the present example:

Rs Vg Vl
+----/\/\/-----+----------------------+
| 50 ohm |
| 1 WL |
Vs 360 degrees 23.5+j44.1
100v RMS 50 ohm line ohm Load
| |
| |
+--------------+----------------------+
gnd

If I haven't made some stupid mistake, the conditions at
the source resistor are identical in both examples.

No silly mistakes. This number is computed by the spreadsheet at
http://keith.dysart.googlepages.com/...oad,reflection
as 23.529411764706+44.1176470588235j, so you are pretty close.

But
in the second example, it is obvious that energy can
be stored in the transmission line during part of a cycle
(thus avoiding dissipation at that instant in time) and be
delivered back to the source resistor during another part
of the cycle (to be dissipated at a later instant in time).
That is the nature of interference energy and is exactly
equal to the difference between the two powers that you
calculated. You neglected to take into account the ability
of the network reactance to temporarily store energy and
dissipate it later in time.

I don't think I did.

The power delivered by the generator to the line is
Pg(t) = 32 + 68 cos(2wt)
The average power delivered to the line is 32W while
the peak power is 100W towards the load and 36W from
the line to the generator.
The exact same function describes the power delivered
to the line for 12.5 ohm load 45 degree line example,
and the 23.5+44.1j ohm load 360 degree line example.

And if the load was connected directly to the generator,
the same power would be delived directly to the load.

In all cases the power dissipated in the source resistor
is
Prs(t) = 68 + 68 cos(2wt-61.92751306degrees)
This power varies between 0.0W and 136W.

This is true even when the load is connected directly to
the generator without a line to created reflected power.

The power provided by the source is
Ps(t) = 100 + 116.6190379 cos(2wt-30.96375653degrees)
= Prs(t) + Pg(t)
so all of the energy delivered by source is nicely
accounted for. All of the energy dissipated in the
source resistor and delivered to the line originates
in the voltage source.

Looking at the line where it connects to the generator,
we find that the forward power is
Pf.g(t) = 50 + 50cos(2wt)
and the reflected power is
Pr.g(t) = -18 + 18cos(2wt)

Pf.g(t) + Pr.g(t) = 32 + 68cos(2wt)
= Pg(t)
As expected, the sum of the forward and reflected power
at the generator terminals is exactly the power delivered
by the generator to the line.

So you are correct, energy is stored in and returned from
the line on each cycle, but this is the net energy, not
the forward or reflected power since these are both
computed into non-reactive impedances, their respective
voltage and current are always in phase, so their powers
always flow in only one direction.

All of the reflected energy is dissipated in the source
resistor, just not at the time you thought it should be.

So as yet, there is no mechanism to explain storage of
reflected power so it can be dissipated at a different
time in the source resistor.

I stand by my contention that the reflected power is not
being dissipated in the source resistor because the
dissipation does not occur at the correct time.

You could convince me to change my contention by locating
the entity that is storing the energy, and showing some
analysis that explains why the amount of the energy
time shift is a function of the magnitude of the
reflected power.

....Keith

The Rest of the Story
 

Keith Dysart wrote:
So as yet, there is no mechanism to explain storage of
reflected power so it can be dissipated at a different
time in the source resistor.

Of course there is, Keith. That's what reactances do.
A reactance stores energy during part of the cycle
and gives it back during a different part of the cycle.
The energy stored during part of the cycle is the
destructive interference energy that you are missing
from you equation. It is delivered back during the
next part of the cycle and dissipated 90 degrees
later.

When the signs of the two superposed voltages are opposite,
there is "excess" energy available which is stored in the
transmission line. 90 degrees later, when the voltages
have the same sign thus requiring constructive interference
energy, that "excess" energy is delivered back to the source
resistor to be dissipated.

Since we already know that the interference energy averages
out to zero, the energy imbalance that you discovered is
obviously energy being displaced in time by the reactance
whether it is from a coil or from the transmission line.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Mar 11, 7:18*pm, Cecil Moore wrote:
Keith Dysart wrote:
So as yet, there is no mechanism to explain storage of
reflected power so it can be dissipated at a different
time in the source resistor.

Of course there is, Keith. That's what reactances do.
A reactance stores energy during part of the cycle
and gives it back during a different part of the cycle.
The energy stored during part of the cycle is the
destructive interference energy that you are missing
from you equation. It is delivered back during the
next part of the cycle and dissipated 90 degrees
later.

When the signs of the two superposed voltages are opposite,
there is "excess" energy available which is stored in the
transmission line. 90 degrees later, when the voltages
have the same sign thus requiring constructive interference
energy, that "excess" energy is delivered back to the source
resistor to be dissipated.

Since we already know that the interference energy averages
out to zero, the energy imbalance that you discovered is
obviously energy being displaced in time by the reactance
whether it is from a coil or from the transmission line.

I can see that words describe a somewhat plausable conjecture,
but to be convincing, a mathematical exposition is needed.

When is the energy stored where? Energy stored in a capacitor,
for example, should correlate with the voltage on the
capacitor. Similarly for current in an inductor.

To be convincing, the various functions of time need to
align appropriately.

So far, the only convincing expressions I have seen
a
Ps(t) = Prs(t) + Pg(t)
and
Pg(t) = Pf.g(t) + Pr.g(t)
and, of course, all the corresponding voltage and
current functions line up as expected.

To show that the reflected power is dissipated in
the resistor will require the derivation of some
X(t) such that
Prs(t) = 50cos(2wt) + Pr.g(t) + X(t)
and X(t) needs to be in terms of some known
physical quantities of the circuit.

...Keith