comparing fourierspectra of signal and noise
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From: Candide Voltaire on 17 Dec 2009 03:35 In the literature concerning noise it is written that as the fourier transform is not defined for a noise signal, one should first take the autocorrelation of the noise signal and then apply the fourier transform. When I look at the units of these operations this will give me a result in Volt^2/Hz. If however I have an ordinary deterministic signal I don't have to take the autocorrelation first but just the fourier transform so this will give me a result in Volt/Hz, now then how should one compare two results in different units? candide p.s. I'm especially thinking of low frequency circuits in which noise- units are not expressed as power-units
From: MooseFET on 17 Dec 2009 10:02 On Dec 17, 12:35 am, Candide Voltaire <candideguev...(a)gmail.com> wrote: > In the literature concerning noise it is written that as the fourier > transform is not defined for a noise signal, one should first take the > autocorrelation of the noise signal and then apply the fourier > transform. When I look at the units of these operations this will give > me a result in Volt^2/Hz. If however I have an ordinary deterministic > signal I don't have to take the autocorrelation first but just the > fourier transform so this will give me a result in Volt/Hz, now then > how should one compare two results in different units? If you have the anti-alias filter before the ADC, the noise you are measuring is all within the band. Although the Fourier transform is not defined for a random function, once you have a finite bunch of numbers from the ADC it really is no longer random. It is now exactly those numbers. The transform is now defined. It is the step of taking a finite number of values from the infinitely long that has made the change to the data. In the infinite random string, there were an infinite number of frequencies. Now there are a finite number. If the noise's amplitude didn't vary too quickly with frequency you can assume that each bin out of the FFT is a flat section of the spectrum exactly as wide as the difference between bins. The results will be "noisy" because you have taken a random chunk of the infinite data. You can rate the noise as Volts/sqrt(Hz) based on the FFT of the data. Once you have this, you can adjust it for any bandwidth that you wish to assume and then compare volts to volts to figure a SNR.
From: Vladimir Vassilevsky on 17 Dec 2009 10:27 Candide Voltaire wrote: > In the literature concerning noise it is written that as the fourier > transform is not defined for a noise signal, one should first take the > autocorrelation of the noise signal and then apply the fourier > transform. Fourier transform is defined for any square integrable function. Since gaussian noise is not square integrable, you can't take Fourier directly. However you can still find PSD as Fourier of the autocorrelation function (Wiener Khintchine theorem). > When I look at the units of these operations this will give > me a result in Volt^2/Hz. Yes. Power per unity bandwidth. > If however I have an ordinary deterministic > signal I don't have to take the autocorrelation first but just the > fourier transform so this will give me a result in Volt/Hz, now then > how should one compare two results in different units? No. Volt/sqrt(Hz) > candide > p.s. I'm especially thinking of low frequency circuits in which noise- > units are not expressed as power-units Never mind. The stuff above is mathematical nit-picking. When dealing with the real signals, you don't have to worry about inifinities. Vladimir Vassilevsky DSP and Mixed Signal Design Consultant http://www.abvolt.com
From: Candide Voltaire on 19 Dec 2009 08:16
On Dec 17, 4:27 pm, Vladimir Vassilevsky <nos...(a)nowhere.com> wrote: > CandideVoltaire wrote: > > In the literature concerning noise it is written that as the fourier > > transform is not defined for a noise signal, one should first take the > > autocorrelation of the noise signal and then apply the fourier > > transform. > > Fourier transform is defined for any square integrable function. Since > gaussian noise is not square integrable, you can't take Fourier > directly. However you can still find PSD as Fourier of the > autocorrelation function (Wiener Khintchine theorem). > > > When I look at the units of these operations this will give > > me a result in Volt^2/Hz. > > Yes. Power per unity bandwidth. > > > If however I have an ordinary deterministic > > signal I don't have to take the autocorrelation first but just the > > fourier transform so this will give me a result in Volt/Hz, now then > > how should one compare two results in different units? > > No. Volt/sqrt(Hz) How then?: F(v(t))=int(v(t)*e^(jwt))dt (integration from -inf to +inf) this gives Volts*secs or Volts/Hz not Volt/sqrt(Hz) candide > >candide > > p.s. I'm especially thinking of low frequency circuits in which noise- > > units are not expressed as power-units > > Never mind. The stuff above is mathematical nit-picking. When dealing > with the real signals, you don't have to worry about inifinities. > > Vladimir Vassilevsky > DSP and Mixed Signal Design Consultanthttp://www.abvolt.com |