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I have a signal $x(t)$. I want to find the Fourier Transform of it, $X(f)$, and then extract a narrow frequency range from $X(f)$ by use of a Band Pass Filter (BPF) in frequency domain.

Can I instead filter $x(t)$ by using a BPF in time domain and then find the Fourier Transform of the filtered signal?

I believe these two are equivalent.

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  • $\begingroup$ To be more specific, I do a Power Spectral Density (PSD), and not just a FT, but the PSD is nothing more than a normalized and squared FT. $\endgroup$ – student1 Mar 15 '15 at 0:19
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    $\begingroup$ by "extract", do you mean multiplying the spectrum of X by some BPF function? multiplying in the frequency domain is the same as filtering (me might say instead "convolution") in the time domain. $\endgroup$ – robert bristow-johnson Mar 15 '15 at 0:58
  • $\begingroup$ Yes, that is what I mean. The convolution theorem is what make be believe they are equivalent, I just wanted to be sure that this is still the case even if the operation is a PSD and not simply an FFT (although as mentioned above this should not be a problem). $\endgroup$ – student1 Mar 15 '15 at 1:02
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    $\begingroup$ your BPF will have an impulse response $h(t)$ and the Fourier Transform of that impulse response is $H(f)$. convolving the input $x(t)$ with the impulse response $h(t)$ $$ y(t) = \int\limits_{-\infty}^{\infty} h(u) \ x(t-u) du $$ will result in the output $y(t)$ (which still needs to be Fourier Transformed to see the spectrum of the output). the convolution property of the Fourier Transform says that the spectra or transforms of these three signals is related as $$ Y(f) = H(f) \ X(f) $$ . $\endgroup$ – robert bristow-johnson Mar 15 '15 at 1:07
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There may be some slight differences due to the band-pass filtering of FFT results in the frequency domain being a circular convolution (with some wrap-around artifacts) rather than a pure linear convolution.

If you filter first, starting in time before your FFT window, any windowing artifacts from any out-of-band spectrum will be reduced before leaking into the FFT filter's pass-band. Or alternatively, you could zero-pad the FFT window by the length of your filter's impulse response.

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