"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