>>cos(cos(kt)) looks strange to me. Is there another formulation?

No this is correct, trust me. Its a signal detected with a photodiode.
"Fred Marshall" <fmarshallx@remove_the_x.acm.org> wrote in message news:<M9Cdnam16MKgbmbdRVn-sw@centurytel.net>...

> cos(cos(kt)) looks strange to me. Is there another formulation?
>
> "?ine Canby" <aine_canby@yahoo.com> wrote in message
> news:57ed59a.0407170951.2b593774@posting.google.com...
> > Hi all,
> >
> > I hope you can help me with this problem because I'm really confused
> > by it at this stage.
> >
> > I have implemented a demodulation scheme on a dsp processor and I'm
> > currently trying to establish the sensitivity in terms of radians per
> > root Hz.
> >
> > The signal before demodulation is -
> >
> > cos(cos(2*PI*8kHz*t) + s(t))
> >
> > The demodulated signal is of the form -
> >
> > s(t)
> >
> > The signal prior to demodulation from the ADC is sampled at 40kHz. The
> > pre-demodulation signal therefore has a bandwidth of 20kHz with
> > carriers at 8kHz and 16kHz, and the demodulated signal s(t) has a
> > bandwidth of 4kHz.
> >
> > To find the sensitivity of the system I made s(t) =
> > PI/10^n*cos(2*PI*1kHz); n having values of between 1 and 6. I then
> > took the power spectral density of the demodulated signal. Then, I
> > took the ratio of the power in the 1kHz bin (signal)and divided its
> > value by the sum of the powers in all the other bins(noise). This
> > division gave me a value of 1 at s(t) = PI/10^4*cos(2*PI*1kHz). I then
> > described the sensitivity as -
> >
> > (PI/10^4)/(20k)^0.5 = 2.22*10-6, where 20kHz is the pre-demodulation
> > bandwidth
> >
> > Does this sound correct?
> >
> > The primary source of the noise is in fact the carrier harmonics which
> > are not fully filtered during the demodulation process. Increasing the
> > filter length will improve this obviously. Also, after I demodulate
> > the signal, I'm downsampling it by a factor of 5, giving the signal a
> > sampling frequency of 8kHz. This signal is transferred from the DSP to
> > my PC, which only has a transmission rate of 10kHz (hence the
> > downsampling). This means that what remains of the carrier harmonics
> > (noise) will now alias down into the signal band.
> >
> > So perhaps I should define my bandwidth as 4kHz instead of 20kHz? I
> > guess I should.
> >
> > And use -
> >
> > (PI/10^4)/(4k)^0.5 rads/Hz^0.5
> >
> > Thanks for your help,
> >
> > Aine.

Reply by Fred Marshall●July 19, 20042004-07-19

cos(cos(kt)) looks strange to me. Is there another formulation?
"?ine Canby" <aine_canby@yahoo.com> wrote in message
news:57ed59a.0407170951.2b593774@posting.google.com...

> Hi all,
>
> I hope you can help me with this problem because I'm really confused
> by it at this stage.
>
> I have implemented a demodulation scheme on a dsp processor and I'm
> currently trying to establish the sensitivity in terms of radians per
> root Hz.
>
> The signal before demodulation is -
>
> cos(cos(2*PI*8kHz*t) + s(t))
>
> The demodulated signal is of the form -
>
> s(t)
>
> The signal prior to demodulation from the ADC is sampled at 40kHz. The
> pre-demodulation signal therefore has a bandwidth of 20kHz with
> carriers at 8kHz and 16kHz, and the demodulated signal s(t) has a
> bandwidth of 4kHz.
>
> To find the sensitivity of the system I made s(t) =
> PI/10^n*cos(2*PI*1kHz); n having values of between 1 and 6. I then
> took the power spectral density of the demodulated signal. Then, I
> took the ratio of the power in the 1kHz bin (signal)and divided its
> value by the sum of the powers in all the other bins(noise). This
> division gave me a value of 1 at s(t) = PI/10^4*cos(2*PI*1kHz). I then
> described the sensitivity as -
>
> (PI/10^4)/(20k)^0.5 = 2.22*10-6, where 20kHz is the pre-demodulation
> bandwidth
>
> Does this sound correct?
>
> The primary source of the noise is in fact the carrier harmonics which
> are not fully filtered during the demodulation process. Increasing the
> filter length will improve this obviously. Also, after I demodulate
> the signal, I'm downsampling it by a factor of 5, giving the signal a
> sampling frequency of 8kHz. This signal is transferred from the DSP to
> my PC, which only has a transmission rate of 10kHz (hence the
> downsampling). This means that what remains of the carrier harmonics
> (noise) will now alias down into the signal band.
>
> So perhaps I should define my bandwidth as 4kHz instead of 20kHz? I
> guess I should.
>
> And use -
>
> (PI/10^4)/(4k)^0.5 rads/Hz^0.5
>
> Thanks for your help,
>
> Aine.

Reply by Rune Allnor●July 18, 20042004-07-18

aine_canby@yahoo.com (?ine Canby) wrote in message news:<57ed59a.0407170951.2b593774@posting.google.com>...

> Hi all,
>
> I hope you can help me with this problem because I'm really confused
> by it at this stage.
>
> I have implemented a demodulation scheme on a dsp processor and I'm
> currently trying to establish the sensitivity in terms of radians per
> root Hz.
>
> The signal before demodulation is -
>
> cos(cos(2*PI*8kHz*t) + s(t))
>
> The demodulated signal is of the form -
>
> s(t)

So this is a phase modulated signal, right? I have forgotten most,
if not all, of what little I ever knew about phase and frequency
modulation, but I think you should be very careful when analyzing
these types of systems. The simple formulas for bandwidth do no
longer apply, the modulated signal has a much greater banwidth than
the unmodulated signal.
Rune

Reply by ?ine Canby●July 17, 20042004-07-17

Hi all,
I hope you can help me with this problem because I'm really confused
by it at this stage.
I have implemented a demodulation scheme on a dsp processor and I'm
currently trying to establish the sensitivity in terms of radians per
root Hz.
The signal before demodulation is -
cos(cos(2*PI*8kHz*t) + s(t))
The demodulated signal is of the form -
s(t)
The signal prior to demodulation from the ADC is sampled at 40kHz. The
pre-demodulation signal therefore has a bandwidth of 20kHz with
carriers at 8kHz and 16kHz, and the demodulated signal s(t) has a
bandwidth of 4kHz.
To find the sensitivity of the system I made s(t) =
PI/10^n*cos(2*PI*1kHz); n having values of between 1 and 6. I then
took the power spectral density of the demodulated signal. Then, I
took the ratio of the power in the 1kHz bin (signal)and divided its
value by the sum of the powers in all the other bins(noise). This
division gave me a value of 1 at s(t) = PI/10^4*cos(2*PI*1kHz). I then
described the sensitivity as -
(PI/10^4)/(20k)^0.5 = 2.22*10-6, where 20kHz is the pre-demodulation
bandwidth
Does this sound correct?
The primary source of the noise is in fact the carrier harmonics which
are not fully filtered during the demodulation process. Increasing the
filter length will improve this obviously. Also, after I demodulate
the signal, I'm downsampling it by a factor of 5, giving the signal a
sampling frequency of 8kHz. This signal is transferred from the DSP to
my PC, which only has a transmission rate of 10kHz (hence the
downsampling). This means that what remains of the carrier harmonics
(noise) will now alias down into the signal band.
So perhaps I should define my bandwidth as 4kHz instead of 20kHz? I
guess I should.
And use -
(PI/10^4)/(4k)^0.5 rads/Hz^0.5
Thanks for your help,
Aine.