Why are the 'Fixed Stars' so FIXED?

"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

(Steve Pope) wrote in news:f1o818$rcd$1
@blue.rahul.net:

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.

Elecraft uses that connector on the K2 and the K3. It allows use of a
headphone with boom mike, ptt and some control lines, if desired. Seems to
work well enough.



--
bz

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

"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