Why are the 'Fixed Stars' so FIXED?

"Henri Wilson" HW@.... wrote in message
...
On 6 May 2007 10:28:13 -0700, George Dishman
wrote:


"Henri Wilson" HW@.... wrote in message

Yes it does, that is basic probability theory.

Probability is not a cause of anything. It's a result.

Nobody said anything about probability being causal.

George, like many others, you are completely misinterpreting the role of
statistics, which is a science dealing with the outcome of multiple
events.

Henry, I'm not talking about statistics, I'm talking
about probability. There is a subtle distinction.

there is.

Mathematics, on the other hand, is designed to analyse or predict single
events.

Maybe you should study probability a bit before
trying to discuss it.

I have already studied it George.

Then get out your notes and do some revision, I
don't intend to waste my time explaining it to you.

You can say that BEFORE the bullet is fired...because the conditions
that
cause
the bullet to land where it does are random.
However, that does not alter the fact that each bullet hits where it
does
for
specific physical reasons that are theoretically capable of being
mathematically analysed and explained.

Whether or not true randomicity exists is a big question.

No, it's not a question at all, it is proven beyond
any doubt.

You are jumping in too early again George.

No, this was proven decades ago.

It IS a big question that includes things like 'free will' if human action
is
involved..

Indeed, without true randomness as an intrinsic property
'free will' could not exist, and with randomness it is
seen to be nothing more than noise in the system. Leave
the philosophers and religiously minded to deal with that.

If I fire a bullet that misses the target, PROBABILITY says, 'that's OK,
it is
a statistical fact that no matter how good the shooter, occasionally one
WILL
miss'.
However, I say, it missed simply because I didn't aim in the right
direction.

dx.dp = h_bar/2

....
Nobody has demonstrated that true randomicty exists, at any level.

Sorry Henry, your decades out of date again.

No. even at the atomic level, this has never been completely resolved.

For instance, consider radioactive decay.
We know all about its exponential rate. ..but we don't know why each event
occurs exactly when it does.

What we know is that if it is exactly exponential, then
every decay event is independent of all others and occurs
with no cause.

Is there a unique physical explanation for each
one. Likewise, we don't know why emitted particles move in the directions
they
do even though the angular distibution is statistical predictable.

We know dx.dp = h_bar/2

Yes. A grating deflects an individual photon depending on
the colour of that beam, not the rate at which photons
arrive. I'm thinking of say a dim red laser with a flux
of a few photons per minute. Like the coin tosses, each
one is deflected purely on its intrinsic properties.

If all the photons are identical, should they all be deflected by the
same
amount?

To within the intrinsic uncertainty of the energy property.
That means there is a fundamental lower limit to line width.
You can think of that either as the (gaussian) spectrum of
the line showing the power in each frequency that you get
from a Fourier transform of the received sine wave or as a
histogram of the photon energies (which will produce a small
spread of deflection angles) or by transforming to the time
domain as the phase jitter on the RF sine wave. They are all
just different coneptual models of the same feature.

If E=h.nu there is no distribution at all.

dt.dE = h_bar/2

If you measure with some certainty when a photon arrives,
you increase the spread of the energy. That is one reason
why a monochromatic laser line cannot have zero width.

I would like to think that the diffraction angle depends on the actual
phase of
the photon's INTRINSIC oscillation when it strikes the grating..

Frequency (or equivalently wavelength), not phase.

In the case of monochromatic light, the theory says energy is relfected
equally
at all angles but is reinforced only at one angle. Destructive
interference
occurs at all other angles thus nullifying energy transfer at those
angles.

Yes but it is the frequency that determines the angle, not
the phase.

Try to explain THAT with the particle model George. How actually do photon
'particles' cancel each other out?

Read QED Henry, that is exactly what it does.

Yes. When it hits a grating each photon deflects depending
only on its own properties and not the properties of other
photons that arrive some seconds earlier or later.

yes. That would have to be right.

Excellent. That is a major agreement Henry.

not really...

What? Didn't you just agree? "yes. That would have
to be right." sounds like an agreement to me.

Consider microwaves hitting a wire grid.
Each photon in the wave is deflected by an angle that depends
only on its own properties independent of any others.

But there is also a second diffraction based on the microwave
'wavelength'.

Same thing.

No it isn't. If you modulate a laser beam with a 100000hz signal, you get
two
entirely different diffraction patterns.

Suppose the unmodulated light has a frequency of fc.
If you fire the modulated light at a grating There
are two obvious possibilities, either you get a line
with time varying intensity at an angle corresponding
to fc, or you get a signal which has a carrier fc and
two sidebands at +/- fm

fc
|
fc-fm | fc+fm
______|____|____|______

and each frequency produces a line of constant intensity.
Either way, you don't get any power at the angle
corresponding to fm itself.

My understanding is that the stream contains a mixture
of three frequencies of photons and if you have the
resolving power in the grating, you get three lines
but a lower resolution will cause the lines to overlap
and the interference then causes the time varying
intensity.

Sorry George, I cannot imagine a single photon that is maybe 1
lightsecond in
length and expands as a radio signal diverges. Do you think it expands
forever?

Photons are particles Henry. Look at the example I gave
of the sodium doublet. The line width has to be less than
6A while the mean wavelength is 5893A. The Zeeman effect
produces individual lines with far smaller spacing. A line
of 5893A wavelength and width of 0.003A must contain more
than 1.7 million cycles so would be more than 1 light
second long in a classical wave model, yet it is absorbed
instantly by a single electron in the photo-electric effect.

You have never seen zeeman lines from ONE transition.
there are always millions involved.

For the sodium doublet, I believe there are just four
for one line and two for the other.

....
I believe the sagnac effect is due to an entirely different
factor...such
as a
local EM frame that behaves like an aether.

I don't care what you belive, it is a fact that the measued
speed is independent of the speed of the source.

Nobody has ever measured OWLS at all George, let alone from a moving
source.

Still in denial Henry? Nothing you say will be treated
with anything but contempt as long as you are unable to
face reality. Sagnac does precisely that.

Come on George, you don't have any kind of model for a photon. You think
it's just a couple of sinewaves drawn at right angles on paper.

No, I think it is a fundamental particle like an electron
which has the property of carrying energy (and others).

'the property of carrying energy'
That doesn't really tell us much does it George...hardly a model...

Of course not, the model is the equations of QED. I'm
only giving you a hand-waving overview.

I think when the charge is taken to some destination, the car
also arrives at the same place. You can't send the car to
Boston and have the charge arrive in Cairo which is what you
are suggesting. Beyond that discussions of their length are
irrelevant, the length has no analog in the photon.

How do you know.

Because your suggestion is equivalent to saying the heat
produced by friction in an ocean wave can be deposited
inland.

George, you know how water waves can be diffracted, for instance by a
row
of vertical bars.

Yes, and the energy of the waves is then carried in
another direction to be deposited where the waves go.

If photon are particles that are reflected over 360 degrees from each
line, how
do you explain all that destructive interference over the 359.9 degrees.

A grating has to have a spacing that gives a sensible
deflection Henry or it becomes a mirror. That's why
chicken wire can be used as an RF reflector.

Do you really believe that the water molecules that go up and down near
the
bars are the ones that end up making the diffraction pattern maybe 100
metres
away?

No Henry, exactly my point. That is what you are telling
me, that the grating angle for the wave is not the same
as that for the photons composing the wave.

Your theory has to rely 100% on the wave model of light to expain
gratings.

No, "my theory" is QED which is purely a particle theory.

..and then it fails.

.... and it works perfectly, it is one of the most accurate
theories in the whole of science.

My model of photons as independent vibrating quanta
explains it all.

Rubbish, it can't even explain the photo-electric effect.

Henry, I think we have maybe got a handle on this, in
your grating equation if you have red laser light
arriving at a level of one photon per second, would you
use the frequency of the red light or the 1Hz rate of
one photon per second to work out the deflection angle.
I say it is that of the light regardless of the arrival
rate, you are telling me the wave energy goes to one
place at an angle determined by the 1Hz figure while
the photons themselves go to the location given by the
red light frequency.

the should be another very weak energy build up where the 1 hz is
diffracted.
How about modifying your experiment to make the 1 Hz sinusoidal.

How about you calculate how much energy BaTh says is in
this extra mode you have invented. For a fairly bright
source with random arrival times (e.g. a sodium lamp
where the photons are emitted thermally) there should
be a background continuum under the lines. Make your
prediction of that level and then research the literature.

I'm too busy...how about YOU do it.

Easy, zero, each photon is independent.

The concept matches the data very well.

It makes no sense though, how can the energy go anywhere
other than where the photons go?

Strange things happen.

Perhaps, but for your bizarre idea to 'match the data
very well' requires _all_ the energy to go where the
1Hz deflection predicts and none to go with the photons.
As I said, it makes no sense.

How do you explain destructive interference with the particle model
George?

Look up "sum over histories".

George

On Mon, 7 May 2007 09:13:54 +0100, "George Dishman"
wrote:


"Henri Wilson" HW@.... wrote in message
.. .


You are jumping in too early again George.

No, this was proven decades ago.

It IS a big question that includes things like 'free will' if human action
is
involved..

Indeed, without true randomness as an intrinsic property
'free will' could not exist, and with randomness it is
seen to be nothing more than noise in the system. Leave
the philosophers and religiously minded to deal with that.

No, it's a physics matter.

If I fire a bullet that misses the target, PROBABILITY says, 'that's OK,
it is
a statistical fact that no matter how good the shooter, occasionally one
WILL
miss'.
However, I say, it missed simply because I didn't aim in the right
direction.

dx.dp = h_bar/2

Yes, Heisenberg got in on the action too...but it's not really relevant to
bullets hitting a target.

No. even at the atomic level, this has never been completely resolved.

For instance, consider radioactive decay.
We know all about its exponential rate. ..but we don't know why each event
occurs exactly when it does.

What we know is that if it is exactly exponential, then
every decay event is independent of all others and occurs
with no cause.

Yes the rate depends on how many are left.....fair enough.. but that doesn't
tell us much either.
I say there is an exact physical reason why each atom decays when it does...put
it all together and you end up with an exponential decay pattern....but I could
be wrong if TRUE RANDOMNESS EXISTS.

Is there a unique physical explanation for each
one. Likewise, we don't know why emitted particles move in the directions
they
do even though the angular distibution is statistical predictable.

We know dx.dp = h_bar/2

George that doesn't tell us why photons move in the directions they do.
There has to be a physical reason that leads to the random distribution of
directions.


To within the intrinsic uncertainty of the energy property.
That means there is a fundamental lower limit to line width.
You can think of that either as the (gaussian) spectrum of
the line showing the power in each frequency that you get
from a Fourier transform of the received sine wave or as a
histogram of the photon energies (which will produce a small
spread of deflection angles) or by transforming to the time
domain as the phase jitter on the RF sine wave. They are all
just different coneptual models of the same feature.

If E=h.nu there is no distribution at all.

dt.dE = h_bar/2

If you measure with some certainty when a photon arrives,
you increase the spread of the energy. That is one reason
why a monochromatic laser line cannot have zero width.

Yes George, we all know what Heisenberg said...and I don't disagree with
it...but what's really behind the stats?


Frequency (or equivalently wavelength), not phase.

In the case of monochromatic light, the theory says energy is relfected
equally
at all angles but is reinforced only at one angle. Destructive
interference
occurs at all other angles thus nullifying energy transfer at those
angles.

Yes but it is the frequency that determines the angle, not
the phase.

That's a big statement.

Try to explain THAT with the particle model George. How actually do photon
'particles' cancel each other out?

Read QED Henry, that is exactly what it does.

I think it is clutching at straws. It doesn't offer any physical solutions.

Yes. When it hits a grating each photon deflects depending
only on its own properties and not the properties of other
photons that arrive some seconds earlier or later.

yes. That would have to be right.

Excellent. That is a major agreement Henry.

not really...

What? Didn't you just agree? "yes. That would have
to be right." sounds like an agreement to me.

I'm not 100% convinced. Your classical wave theory say it deflects in all
directions. How come?

Consider microwaves hitting a wire grid.
Each photon in the wave is deflected by an angle that depends
only on its own properties independent of any others.

But there is also a second diffraction based on the microwave
'wavelength'.

Same thing.

No it isn't. If you modulate a laser beam with a 100000hz signal, you get
two
entirely different diffraction patterns.

Suppose the unmodulated light has a frequency of fc.
If you fire the modulated light at a grating There
are two obvious possibilities, either you get a line
with time varying intensity at an angle corresponding
to fc, or you get a signal which has a carrier fc and
two sidebands at +/- fm

fc
|
fc-fm | fc+fm
______|____|____|______

and each frequency produces a line of constant intensity.
Either way, you don't get any power at the angle
corresponding to fm itself.

My understanding is that the stream contains a mixture
of three frequencies of photons and if you have the
resolving power in the grating, you get three lines
but a lower resolution will cause the lines to overlap
and the interference then causes the time varying
intensity.

See you don't really know the answer. You're just speculating.
I think there is a worthwhile experiment awaiting to be done here.


than 1.7 million cycles so would be more than 1 light
second long in a classical wave model, yet it is absorbed
instantly by a single electron in the photo-electric effect.

You have never seen zeeman lines from ONE transition.
there are always millions involved.

For the sodium doublet, I believe there are just four
for one line and two for the other.

OK ,....but I meant the lines you see are made of of many photons from the same
transition.

local EM frame that behaves like an aether.

I don't care what you belive, it is a fact that the measued
speed is independent of the speed of the source.

Nobody has ever measured OWLS at all George, let alone from a moving
source.

Still in denial Henry? Nothing you say will be treated
with anything but contempt as long as you are unable to
face reality. Sagnac does precisely that.

Sagnac is as big a mystery to you as it ever was. I still reckon there is an EM
reference frame involved. YTour explanation is certainly straight out of the
LET book.
What's the betting Sagnac wont work in deep space...below the WDT.

Come on George, you don't have any kind of model for a photon. You think
it's just a couple of sinewaves drawn at right angles on paper.

No, I think it is a fundamental particle like an electron
which has the property of carrying energy (and others).

'the property of carrying energy'
That doesn't really tell us much does it George...hardly a model...

Of course not, the model is the equations of QED. I'm
only giving you a hand-waving overview.

well QED is not worth worrrying about either...


George, you know how water waves can be diffracted, for instance by a
row
of vertical bars.

Yes, and the energy of the waves is then carried in
another direction to be deposited where the waves go.

If photons are particles that are reflected over 360 degrees from each
line, how
do you explain all that destructive interference over the 359.9 degrees.

A grating has to have a spacing that gives a sensible
deflection Henry or it becomes a mirror. That's why
chicken wire can be used as an RF reflector.

That doesn't answer the question.
How can you get destructive interference for particles?

Do you really believe that the water molecules that go up and down near
the
bars are the ones that end up making the diffraction pattern maybe 100
metres
away?

No Henry, exactly my point. That is what you are telling
me, that the grating angle for the wave is not the same
as that for the photons composing the wave.

Your theory has to rely 100% on the wave model of light to expain
gratings.

No, "my theory" is QED which is purely a particle theory.

and a very vague one...

..and then it fails.

... and it works perfectly, it is one of the most accurate
theories in the whole of science.

Which aspect?

My model of photons as independent vibrating quanta
explains it all.

Rubbish, it can't even explain the photo-electric effect.

Of course it can. Each photon has to possess enough intrinsic energy to knock
out an electron.
I tell you something though. I reckon a really intense beam of low energy
photons would knock out electrons that they should not, purely because there is
a chance two would hit at the same instant....another worthwhile experiment....


How about you calculate how much energy BaTh says is in
this extra mode you have invented. For a fairly bright
source with random arrival times (e.g. a sodium lamp
where the photons are emitted thermally) there should
be a background continuum under the lines. Make your
prediction of that level and then research the literature.

I'm too busy...how about YOU do it.

Easy, zero, each photon is independent.


That's settled then...


Perhaps, but for your bizarre idea to 'match the data
very well' requires _all_ the energy to go where the
1Hz deflection predicts and none to go with the photons.
As I said, it makes no sense.

How do you explain destructive interference with the particle model
George?

Look up "sum over histories".

What the hell is that?


George





www.users.bigpond.com/hewn/index.htm

Einstein's Relativity - the greatest HOAX since jesus christ's virgin mother.

"George Dishman" wrote in news:f1mmtj$tr3$1
@news.freedom2surf.net:

Suppose the unmodulated light has a frequency of fc.
If you fire the modulated light at a grating There
are two obvious possibilities, either you get a line
with time varying intensity at an angle corresponding
to fc, or you get a signal which has a carrier fc and
two sidebands at +/- fm

fc
|
fc-fm | fc+fm
______|____|____|______

and each frequency produces a line of constant intensity.
Either way, you don't get any power at the angle
corresponding to fm itself.

IF the amplifiers following the mixer were flat from DC through light, you
WOULD also have output at fm.

Normally, however fm would be lost because it is far from the frequencies
of interest.

My understanding is that the stream contains a mixture
of three frequencies of photons and if you have the
resolving power in the grating, you get three lines

correct.

but a lower resolution will cause the lines to overlap
and the interference then causes the time varying
intensity.

No.

A detector follows the 'envelope' of the modulated signal and
'demodulates' it, producing fm.

[all the above assumes A3A modulation commonly called AM or amplitude
modulation].





--
bz

please pardon my infinite ignorance, the set-of-things-I-do-not-know is an
infinite set.

remove ch100-5 to avoid spam trap

"bz" wrote in message
98.139...
"George Dishman" wrote in news:f1mmtj$tr3$1
@news.freedom2surf.net:

Suppose the unmodulated light has a frequency of fc.
If you fire the modulated light at a grating There
are two obvious possibilities, either you get a line
with time varying intensity at an angle corresponding
to fc, or you get a signal which has a carrier fc and
two sidebands at +/- fm

fc
|
fc-fm | fc+fm
______|____|____|______

and each frequency produces a line of constant intensity.
Either way, you don't get any power at the angle
corresponding to fm itself.

I am assuming fully linear mixing with modulation
index 1, no sideband or carrier suppression.

IF the amplifiers following the mixer were flat from DC through light, you
WOULD also have output at fm.

sin(a)*sin(b) = (cos(a-b) - cos(a+b))/2

So

sin(fc.t)*(1+M*sin(fm.t)) =
sin(fc.t) + M/2*(cos((fc-fm).t) - cos(((fc+fm).t))

There is no component at fm, only the three I listed.

Normally, however fm would be lost because it is far from the frequencies
of interest.

My understanding is that the stream contains a mixture
of three frequencies of photons and if you have the
resolving power in the grating, you get three lines

correct.

but a lower resolution will cause the lines to overlap
and the interference then causes the time varying
intensity.

No.

A detector follows the 'envelope' of the modulated signal and
'demodulates' it, producing fm.

Following the envelope is essentially peak detection
with low pass filtering. Consider the effect of that
on the diagram for 50% modulation:

http://en.wikipedia.org/wiki/Amplitu...dulation_index

Would you not describe that as a time varying amplitude
or light intensity in the context? The amplitude of the
carrier varies at rate fm.

[all the above assumes A3A modulation commonly called AM or amplitude
modulation].

A3A is single sideband, suppressed carrier. I was
describing A3 mode, both sidebands, M1 and no carrier
suppression.

George

"George Dishman" wrote in
:


"bz" wrote in message
98.139...
"George Dishman" wrote in news:f1mmtj$tr3$1
@news.freedom2surf.net:

Suppose the unmodulated light has a frequency of fc.
If you fire the modulated light at a grating There
are two obvious possibilities, either you get a line
with time varying intensity at an angle corresponding
to fc, or you get a signal which has a carrier fc and
two sidebands at +/- fm

fc
|
fc-fm | fc+fm
______|____|____|______

and each frequency produces a line of constant intensity.
Either way, you don't get any power at the angle
corresponding to fm itself.

I am assuming fully linear mixing with modulation
index 1, no sideband or carrier suppression.

IF the amplifiers following the mixer were flat from DC through light,
you WOULD also have output at fm.

sin(a)*sin(b) = (cos(a-b) - cos(a+b))/2

So

sin(fc.t)*(1+M*sin(fm.t)) =
sin(fc.t) + M/2*(cos((fc-fm).t) - cos(((fc+fm).t))

There is no component at fm, only the three I listed.

You may be correct. That math would seem to support it.

I would have sworn that the modulating signal also WOULD appear if it were
not filtered out. And it normally is because it is normally audio and the
carrier is RF.

For instance, in the case of mixing two signals that are close to the same
frequency, as in a hetrodyne receiver, you get the sum, the difference and
both f1 and f2.

Normally, however fm would be lost because it is far from the
frequencies of interest.

My understanding is that the stream contains a mixture
of three frequencies of photons and if you have the
resolving power in the grating, you get three lines

correct.

but a lower resolution will cause the lines to overlap
and the interference then causes the time varying
intensity.

No.

A detector follows the 'envelope' of the modulated signal and
'demodulates' it, producing fm.

Following the envelope is essentially peak detection
with low pass filtering. Consider the effect of that
on the diagram for 50% modulation:

http://en.wikipedia.org/wiki/Amplitu...dulation_index

Would you not describe that as a time varying amplitude
or light intensity in the context? The amplitude of the
carrier varies at rate fm.

2 points for you.

[all the above assumes A3A modulation commonly called AM or amplitude
modulation].

A3A is single sideband, suppressed carrier. I was
describing A3 mode, both sidebands, M1 and no carrier
suppression.

Looks like either the designations have changed since I took my exams
(first class radio telephone, 2nd class radio telegraph, amateur extra
class) or my memory has played a nasty trick on me.

A check of my handbooks shows that memory is the offending agent.
A3 is double side band, A3A single, reduced carrier. A3J single,
suppressed carrier. I notice that A3E is the current designation for
double sideband with full carrier.

Well, hopefully, it won't be the last time I am wrong about something.




--
bz

please pardon my infinite ignorance, the set-of-things-I-do-not-know is an
infinite set.

remove ch100-5 to avoid spam trap

"bz" wrote in message
98.139...
"George Dishman" wrote in
:
"bz" wrote in message
98.139...
"George Dishman" wrote in news:f1mmtj$tr3$1
@news.freedom2surf.net:

Suppose the unmodulated light has a frequency of fc.
If you fire the modulated light at a grating There
are two obvious possibilities, either you get a line
with time varying intensity at an angle corresponding
to fc, or you get a signal which has a carrier fc and
two sidebands at +/- fm

fc
|
fc-fm | fc+fm
______|____|____|______

and each frequency produces a line of constant intensity.
Either way, you don't get any power at the angle
corresponding to fm itself.

I am assuming fully linear mixing with modulation
index 1, no sideband or carrier suppression.

IF the amplifiers following the mixer were flat from DC through light,
you WOULD also have output at fm.

sin(a)*sin(b) = (cos(a-b) - cos(a+b))/2

So

sin(fc.t)*(1+M*sin(fm.t)) =
sin(fc.t) + M/2*(cos((fc-fm).t) - cos(((fc+fm).t))

There is no component at fm, only the three I listed.

You may be correct. That math would seem to support it.

I would have sworn that the modulating signal also WOULD appear if it were
not filtered out. And it normally is because it is normally audio and the
carrier is RF.

The carrier apears because of the "(1+" term and likewise
the modulating signal will slip through if there is a DC
bias on the RF. In practical terms there usually will be
in a physical circuit but not in the case of light incident
on an optical modulator. As you say, it always gets filtered
out.

For instance, in the case of mixing two signals that are close to the same
frequency, as in a hetrodyne receiver, you get the sum, the difference and
both f1 and f2.

For a perfect four-quadrant multiplier with no DC offsets,
you only get sum and difference. f1 gets through if there
is DC on the f2 signal and vice versa.

[all the above assumes A3A modulation commonly called AM or amplitude
modulation].

A3A is single sideband, suppressed carrier. I was
describing A3 mode, both sidebands, M1 and no carrier
suppression.

Looks like either the designations have changed since I took my exams
(first class radio telephone, 2nd class radio telegraph, amateur extra
class) or my memory has played a nasty trick on me.

A check of my handbooks shows that memory is the offending agent.
A3 is double side band, A3A single, reduced carrier. A3J single,
suppressed carrier. I notice that A3E is the current designation for
double sideband with full carrier.

I was never a ham though I had some friends who were,
my interest was always in the digital side. I Googled
and randomly got this page:

http://jproc.ca/rrp/coverdale_ddr5k.html

Well, hopefully, it won't be the last time I am wrong about something.

No problem, I've learned more than you from this :-)

George

"George Dishman" wrote in news:f1o0ir$30m$1
@news.freedom2surf.net:

I was never a ham though I had some friends who were,
my interest was always in the digital side. I Googled
and randomly got this page:

http://jproc.ca/rrp/coverdale_ddr5k.html

For something a bit more 'state of the art'
http://www.elecraft.com/K3/K3_Data%20Sheet_rev06.pdf


Well, hopefully, it won't be the last time I am wrong about something.

No problem, I've learned more than you from this :-)

Wanna bet?




--
bz

please pardon my infinite ignorance, the set-of-things-I-do-not-know is an
infinite set.

remove ch100-5 to avoid spam trap

bz wrote:

For something a bit more 'state of the art'
http://www.elecraft.com/K3/K3_Data%20Sheet_rev06.pdf

What kind of funky microphone connector is on the front panel
of this baby? Looks non-standard.

Steve

"bz" wrote in message
98.139...
"George Dishman" wrote in news:f1mmtj$tr3$1
@news.freedom2surf.net:

Suppose the unmodulated light has a frequency of fc.
If you fire the modulated light at a grating There
are two obvious possibilities, either you get a line
with time varying intensity at an angle corresponding
to fc, or you get a signal which has a carrier fc and
two sidebands at +/- fm

fc
|
fc-fm | fc+fm
______|____|____|______

and each frequency produces a line of constant intensity.
Either way, you don't get any power at the angle
corresponding to fm itself.

I am assuming fully linear mixing with modulation
index 1, no sideband or carrier suppression.

IF the amplifiers following the mixer were flat from DC through light, you
WOULD also have output at fm.

sin(a)*sin(b) = (cos(a-b) - cos(a+b))/2

So

sin(fc.t)*(1+M*sin(fm.t)) =
sin(fc.t) + M/2*(cos((fc-fm).t) - cos(((fc+fm).t))

There is no component at fm, only the three I listed.

Normally, however fm would be lost because it is far from the frequencies
of interest.

My understanding is that the stream contains a mixture
of three frequencies of photons and if you have the
resolving power in the grating, you get three lines

correct.

but a lower resolution will cause the lines to overlap
and the interference then causes the time varying
intensity.

No.

A detector follows the 'envelope' of the modulated signal and
'demodulates' it, producing fm.

Following the envelope is essentially peak detection
with low pass filtering. Consider the effect of that
on the diagram for 50% modulation:

http://en.wikipedia.org/wiki/Amplitu...dulation_index

Would you not describe that as a time varying amplitude
or light intensity in the context? The amplitude of the
carrier varies at rate fm.

[all the above assumes A3A modulation commonly called AM or amplitude
modulation].

A3A is single sideband, suppressed carrier. I was
describing A3 mode, both sidebands, M1 and no carrier
suppression.

George