4 ms·

I'm not quite sure I follow: evaluating the fourier transform (whether continuous or discrete) at a given frequency omega gives you a complex value that encodes

I'm not quite sure I follow: evaluating the fourier transform (whether continuous or discrete) at a given frequency omega gives you a complex value that encodes both the magnitude and the phase. Usually for the sake of plotting we just take the magnitude of this. Given these values for all frequencies, the FT et al. are perfectly invertible and you'll get back to your original sampled signal. But it seems like you want to go from the spectrum of the windowed signal to the spectrum of the "true" signal. I.e. undo the effects of the sinc ringing, which would involve reconstructing a posteriori the most probable signal. I don't know much here, but for DTFT maybe deconvolution techniques are something to look at? Alternatively for the DFT since we only have information about discrete frequencies, some sort of technique that tries to interpolate samples at the edges to ensure the periodic summed waveform doesn't have discontinuities?

krackers 4y ago
Oh also this is only semi-related, but the following is also really unintuitive to me: given a windowed function, it could have been constructed from any number of possible functions. But in the fourier domain, we're guaranteed that the spectrum of the windowed function is simply the spectrum of the original function convolved with sinc (which I just think of as "smoothing"). So the paradox is that the portions of the original function outside of the window can be "anything", even something completely wonky, and yet somehow when convolving its spectrum with sinc we always end up with the same thing. Trying to puzzle this out: Let the "original" function be a superposition of the windowed portion like f1 and a non-windowed portion like f2: orig = rect * f1 + (1 - rect) * f2 Then we have F{orig} = F{rect * f1} + F{(1 - rect) * f2} Now F{1 - rect} is (dirac - sinc) so we have F{orig} = F{rect * f1} + F{(1 - rect) * f2} = F{f1} ⊛ sinc + (dirac - sinc) ⊛ F{f2} Since convolution is distributive F{orig} = F{f1} ⊛ sinc + (dirac - sinc) ⊛ F{f2} = (F{f1} ⊛ sinc) - (F{f2} ⊛ sinc) + F{f2} The above I've plotted to verify, and it's intuitive so far. rect * orig = f1 F{orig} ⊛ sinc = sinc ⊛ ((F{f1} ⊛ sinc) - (F{f2} ⊛ sinc) + F{f2}) = sinc ⊛ (F{f1} ⊛ sinc) - sinc ⊛ (F{f2} ⊛ sinc) + F{f2}) = sinc ⊛ sinc ⊛ F{f1} - sinc ⊛ F{f2} + sinc ⊛ sinc ⊛ F{f2} But sinc ⊛ sinc = sinc (easiest to see in the fourier domain), so the second terms cancel out and we get = sinc ⊛ F{f1} So I think the real magic here is that for any function f(x), f(x) - sinc ⊛ f(x) is a function that gives 0 when again convolved with sinc. The math works out so that sinc is precisely the function which captures that "uncertainty" as to what the portions outside the window are. Really mind-blowing. --- But back on topic to your question, note that even truncating your data doesn't get rid of the sinc ringing, it's always there as an artifact of the windowing, it's just not always revealed by the discrete sampling of the spectrum. The sinc ringing itself tells you the measure of uncertainty. In terms of how to practically use this information of a DFT though, probably just taking the spectrum values of magnitude above sinc main-lobe/side-lobe ratio would be good enough? Or interpolate the DFT values to reconstruct the DTFT spectrum and then resample it yourself? I can't be of much help here, I'm more comfortable with the theory than practical usage.

krackers 4y ago

Oh also this is only semi-related, but the following is also really unintuitive to me: given a windowed function, it could have been constructed from any number of possible functions. But in the fourier domain, we're guaranteed that the spectrum of the windowed function is simply the spectrum of the original function convolved with sinc (which I just think of as "smoothing"). So the paradox is that the portions of the original function outside of the window can be "anything", even something completely wonky, and yet somehow when convolving its spectrum with sinc we always end up with the same thing. Trying to puzzle this out: Let the "original" function be a superposition of the windowed portion like f1 and a non-windowed portion like f2: orig = rect * f1 + (1 - rect) * f2 Then we have F{orig} = F{rect * f1} + F{(1 - rect) * f2} Now F{1 - rect} is (dirac - sinc) so we have F{orig} = F{rect * f1} + F{(1 - rect) * f2} = F{f1} ⊛ sinc + (dirac - sinc) ⊛ F{f2} Since convolution is distributive F{orig} = F{f1} ⊛ sinc + (dirac - sinc) ⊛ F{f2} = (F{f1} ⊛ sinc) - (F{f2} ⊛ sinc) + F{f2} The above I've plotted to verify, and it's intuitive so far. rect * orig = f1 F{orig} ⊛ sinc = sinc ⊛ ((F{f1} ⊛ sinc) - (F{f2} ⊛ sinc) + F{f2}) = sinc ⊛ (F{f1} ⊛ sinc) - sinc ⊛ (F{f2} ⊛ sinc) + F{f2}) = sinc ⊛ sinc ⊛ F{f1} - sinc ⊛ F{f2} + sinc ⊛ sinc ⊛ F{f2} But sinc ⊛ sinc = sinc (easiest to see in the fourier domain), so the second terms cancel out and we get = sinc ⊛ F{f1} So I think the real magic here is that for any function f(x), f(x) - sinc ⊛ f(x) is a function that gives 0 when again convolved with sinc. The math works out so that sinc is precisely the function which captures that "uncertainty" as to what the portions outside the window are. Really mind-blowing. --- But back on topic to your question, note that even truncating your data doesn't get rid of the sinc ringing, it's always there as an artifact of the windowing, it's just not always revealed by the discrete sampling of the spectrum. The sinc ringing itself tells you the measure of uncertainty. In terms of how to practically use this information of a DFT though, probably just taking the spectrum values of magnitude above sinc main-lobe/side-lobe ratio would be good enough? Or interpolate the DFT values to reconstruct the DTFT spectrum and then resample it yourself? I can't be of much help here, I'm more comfortable with the theory than practical usage.