4 ms·

Did I understand this correctly, what you are doing is essentially: X[n] = F[x[k]][n/2] if (n even) else F[x'[k]][(n+1)/2] With F[x[k]] the DFT of the time-do

Did I understand this correctly, what you are doing is essentially: X[n] = F[x[k]][n/2] if (n even) else F[x'[k]][(n+1)/2] With F[x[k]] the DFT of the time-domain signal x[k], x'[k] = x[k]·exp(2·pi·i·k·alpha) and this alpha some constant which yields a frequency-domain shift by 25Hz. If so: How does this method compare to zero-padding the time-domain signal (i.e. sinc-interpolating the frequency domain)? It is an interesting concept, but alas it's not immediately clear to me how to analyze this...

gbh444g 5y ago
This sounds about right. I assume your (n+1)/2 is really n+1/2. The idea, like you've said, is to get Y[k+1/2] values where Y = FFT[X]. Whether this is mathematically sound is another question. I presume that it is, for two reasons. First, FFT essentially convolves X with a bunch of sinusoids with frequencies from a fixed set: 0 Hz, 50 Hz, 100 Hz and so on. There's nothing wrong with manually convolving X with a 57.3 Hz sinusoid, it's just FFT isn't designed for this (it's designed for rapid computation). The other reason is that combining such shifted FFTs we get what looks almost exactly like a CWT (i.e. wavelet transform). As for sinc-interpolation, I think it's mathematically equivalent. Say we shift the input X with Z[k] = exp(ik/N...) and get XZ. Then we transform it to FFT[XZ] = FFT[X] conv FFT[Z], so it's convolving FFT[X] with FFT[Z] where FFT[Z] is probably that sinc kernel. I certainly know from experiments that FFT of exp(2·pi·i·k·alpha) where alpha doesn't precisely align with the 1024 grid produces a fuzzy function with a max around alpha and a bell-shaped curved around it, the width of the curve depends on how precisely alpha fits into one of the 1024 grid points.
- ggerganov 5y ago
  Instead of combining 2 FFTs of 1024 bins (one shifted + one non-shifted), could you not just calculate 1 FFT of 2048 bins? Isn't it the same result?
  - gbh444g 5y ago
    Larger FFT window has undesired side effects because the estimated frequencies are averaged over the entire window. Moreover, the FFT output always spans from 0 Hz to 24 kHz (with 48 kHz sample rate), so to zoom into the 0..6 kHz region we'll need a window with 8192 bins or about 150 ms. With such window it would be impossible to capture rapid volume oscillations.

gbh444g 5y ago

This sounds about right. I assume your (n+1)/2 is really n+1/2. The idea, like you've said, is to get Y[k+1/2] values where Y = FFT[X]. Whether this is mathematically sound is another question. I presume that it is, for two reasons. First, FFT essentially convolves X with a bunch of sinusoids with frequencies from a fixed set: 0 Hz, 50 Hz, 100 Hz and so on. There's nothing wrong with manually convolving X with a 57.3 Hz sinusoid, it's just FFT isn't designed for this (it's designed for rapid computation). The other reason is that combining such shifted FFTs we get what looks almost exactly like a CWT (i.e. wavelet transform). As for sinc-interpolation, I think it's mathematically equivalent. Say we shift the input X with Z[k] = exp(ik/N...) and get XZ. Then we transform it to FFT[XZ] = FFT[X] conv FFT[Z], so it's convolving FFT[X] with FFT[Z] where FFT[Z] is probably that sinc kernel. I certainly know from experiments that FFT of exp(2·pi·i·k·alpha) where alpha doesn't precisely align with the 1024 grid produces a fuzzy function with a max around alpha and a bell-shaped curved around it, the width of the curve depends on how precisely alpha fits into one of the 1024 grid points.

ggerganov 5y ago
Instead of combining 2 FFTs of 1024 bins (one shifted + one non-shifted), could you not just calculate 1 FFT of 2048 bins? Isn't it the same result?
- gbh444g 5y ago
  Larger FFT window has undesired side effects because the estimated frequencies are averaged over the entire window. Moreover, the FFT output always spans from 0 Hz to 24 kHz (with 48 kHz sample rate), so to zoom into the 0..6 kHz region we'll need a window with 8192 bins or about 150 ms. With such window it would be impossible to capture rapid volume oscillations.