If I connect my sensor to a FFT analyser and make a noise analyses between 0-12,5 Hz, the noise at 1 Hz is 500 nV/sqrt(Hz). On the other hand, by disconnecting the sensor from FFT analyser and make noise analys (i.e. nothing is connected to the FFT analyser) the noise level at 1 Hz is 200 nV/sqrt(Hz). So, to determine the exact noise of my sensor what must I do? is it correct to substract these two values?
Yes, you are in the right track to estimate noise level... But I suggest you short the input when you disconnecy the sensor. If not the cable might pick up more noise than you actually have, especially if the input has high impedance.
The FFT of the noise is not equivalent to the time course of the noise; is is just a measurement of noise properties over a short interval - in other words the FFT just gives a different view of the signal, it doesn't make the noise and signal inherently more separable. In the time domain, you see the statistics of the noise component, design an order statistic filter. In the frequency domain, you see the band limits of the signal - design a band-pass filter.
Adaptive approaches, or model based estimators of the signal or noise processes, may give you a more satisfactory noise filter, but simple observation of the FFT is only a start. The key is providing the model of the system which makes the components separable. That may be a K-L or principle components model, may be a linearity or continuity model, or a physical model of your problem, but you need some further assumptions to pull the problem apart.
Dear James,
I don't think that Ugur would subtract the noise of the FFT at each istante in time but would kown if he can estimate the noise level of the sensor from the two measurements.
Remember if you have two uncorrelated noise sources, as we can assume the noise of the sensor and the noise floor of FFT analyser, their variances add up, not standard deviations.
So your result should be 458 nV/Hz.
Dear Simone
Sorry but I couldnt understand how you found 458 nV/Hz. In fact, I think that 500-200=300 will be my sensor noise because 500 already contains noise floor coming from the FFT analyser. Is it correct?
sqrt(500^2 - 200^2) e-9.
You can add the variances not standard deviations.
Since the number you presented had dimensions of V/sqrt(Hz) they clearly are the square root of the power spectral densities. The power spectral density (assumed as constant) multiplied for BW gives the variance.
Now you can put BW = 1 and obtain the above calculation.
Dear Ugur,
The uncorrelated noise sources are added by adding their power, which is the mean square value of the signal. Therefore if the root mean square value of one is sigma1 and the other is sigma2 then the the total noise sigma = square root of ( sigma1^2 +sigma2^2).
Accordingly sigma 2= square root of (500^2-200^2)=458.2575 nV/square root Hz.
The answer of Simone is okay. This means your sensor is much more noisy than the FFT analyser.
Sorry. Much thanks also to other contributions made by Varun, James and Fernando.
First you have to short circuit the input of the FFT, to avoid external
Pick up noise but also internal noise generated by current noise at the input of the FFT.
Second you have to subtract the Spectral density (voltage squared / Hz) that you measured from the overall Spectral density measured with your sensor connected.
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