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.

# System Sensitivity based on Power Spectral Density

Started by ●July 17, 2004

Reply by ●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 ●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 ●July 26, 20042004-07-26

>>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.