16 ms·

I thought this was really well done without any misleading oversimplifications. I do think the author missed the central trick of the Fourier Transform though.

• HelloNurse 4y ago
  Citing from that page > In other words, $ J_k(\beta)$ is the amplitude of the $ k$ th harmonic in the Fourier-series expansion of the periodic signal $ x_m(t)$ . Or from the parent page about FM > It is well known that sinusoidal frequency-modulation of a sinusoid creates sinusoidal components that are uniformly spaced in frequency by multiples of the modulation frequency, with amplitudes given by the Bessel functions of the first kind [15]. As a special case, frequency-modulation of a sinusoid by itself generates a harmonic spectrum in which the $ k$ th harmonic amplitude is proportional to $ J_k(\beta)$ , where $ k$ is the order of the Bessel function and $ \beta $ is the FM index. In other words, Bessel functions come up in Fourier transforms of frequency modulation signals, because integrals. The calculations have been figured out centuries ago. Regarding phase and arbitrary frequencies, it's easier to understand in the discrete time case: a vector of N complex numbers (time domain samples) goes in, it's multiplied by a matrix, and a vector of N complex numbers (Fourier coefficients) comes out. Regardless of the input signal, such a transform can be inverted (up to rounding errors and the like) by multiplication with the inverse of the forward matrix: the difference between signals of different frequencies is that if periods are a whole fraction of N samples the projections fall on one Fourier coefficient instead of spreading around all of them. Aligning the signal's period with the window length M allows reproducing the whole infinite duration signal from a N coefficient slice because all slices are identical.
  • geokon 4y ago
    Oh yikes yikes yikes. Yes, I wrote that question way too fast and mixed things up. You're totally right, Bessel functions are related to FM and not this ringing. I got things confused. Sorry about that "Aligning the signal's period with the window length M allows reproducing the whole infinite duration signal from a N coefficient slice because all slices are identical." Yeah but this isn't really a solution. I mean, yes.. you can cut your data to be a multiple of the frequency you're looking for, do an FFT, look at the phase. If you didn't know the exact frequency you could also do a "search" truncating your data at different values and looking for a peak. But that's.. ugly. The problem is you end up throwing away data. In my usecase I have maybe 2-3 periods of the frequency I'm interested in. Truncating my data throws away a lot of information. I'd like to do an FFT of all the data I have (it's actually not evenly spaces, so I'd probably have to interpolate it) then I'd like to look at the ringing directly and estimate the original frequency and phase. Ideally after that I'd somehow calculate some uncertainties on those values
• krackers 4y ago
  >anyone has a more rigorous explanation of the in-between frequencies part I think it's because the DFT is one period of the sampled version of the DTFT of your discrete input signal, or equivalently one period of the DTFT of the periodic continuation of your discrete input. (Equivalence here is that convolving by a dirac comb in time domain is multiplying by a dirac comb = sampling in the frequency domain) When you use a windowing function (e.g. multiplication by a rect) to convert an infinite continuous signal into a time-limited one, the fourier spectrum of the original becomes convolved by the spectrum of the window (e.g. sinc). For a given sample length (window size), I guess if you capture the signal at a period boundary then your discretization of the DTFT is done exactly at peaks and zeroes. Otherwise if there's a mismatch, then you're not sampling the DTFT exactly at the peaks/zeroes so you get the spectral leakage. See [1] for a nice interactive example. Figure 3 shows as input a pure sin. If this was infinite, the fourier transform would be a dirac delta at the correpsonding frequency. But since it's windowed by a fixed window of 12 samples, the DTFT is the dirac convolved with a sinc. Another perspective might be that if you don't capture it at a period boundary, when you create a periodic summation of your input you're introducing higher frequencies at the discontinuity. [1] https://jackschaedler.github.io/circles-sines-signals/dft_leakage.html https://jackschaedler.github.io/circles-sines-signals/dft_le... [2] https://geo-ant.github.io/blog/2021/dft-spectral-leakage-and-windowing https://geo-ant.github.io/blog/2021/dft-spectral-leakage-and... [3] https://dspillustrations.com/pages/posts/misc/spectral-leakage-zero-padding-and-frequency-resolution.html https://dspillustrations.com/pages/posts/misc/spectral-leaka... [4] https://dsp.stackexchange.com/questions/34211/is-spectral-leakage-due-to-windowing-different-for-the-dtft-and-dft https://dsp.stackexchange.com/questions/34211/is-spectral-le... Or you can just grind through the computation and see that the power spectrum is indeed sinc-like https://ccs.fau.edu/~bressler/EDU/STSA/Modules/VIII.pdf https://ccs.fau.edu/~bressler/EDU/STSA/Modules/VIII.pdf
  • krackers 4y ago
    Can't edit, but I just realized that when you go from continuous to discrete via the sampling step you also introduce replication in the frequency domain. So you have Continuous sin wave: Spectrum is a single dirac Sampling: Infinite samples of sin wave - spectrum is periodic train of impulses Windowing: Fixed number of samples - spectrum is sinc convolved with that impulse train. The DTFT would end here. DFT: Windowed samples are then periodically extended - spectrum is spectrum of DTFT multiplied by a comb. Even assuming nyquist is satisfied, the tails of the sinc would still interact with each other, so the shape of the DTFT isn't _just_ a sinc, right? It'd be almost a sinc but with some additional wonkiness from the tails?