I'm looking for a way to get a wavelet-like resolution in the frequency domain. So I thought of applying the Fourier transform on the results of a wavelet transform.

Say we have a time-domain signal with a length of 8.

From the viewpoint of tiling of the time and frequency axis, this signal would have vertical lines creating 8 bins, meaning that there is no frequency resolution in any of the coefficients. If we take the Fourier transform, then the vertical lines would be transposed into horizontal lines, having no time resolution but a frequency resolution of 8 bins.

We now perform a wavelet transform with a 3-level decomposition on the time-domain signal, so that the coefficients are [4,2,1,1], where the level with 4 coefficients would have a resolution of 2 temporal bins and 4 spectral bins each, level with 2 coefficients would have a resolution of 4 temporal bins and 2 spectral bins each, and so on. (Similar to half of the image on the right)

If we take the Fourier transform of these individual levels, is the result mathematically meaningful?

Edit: Modulated complex lapped transform (which I just found out) has the tiling that I expected my question to have.

  • $\begingroup$ I don't understand your question. What do you mean with "similar to the Fourier transform" ? You should look at the undecimated discrete Wavelet transform, and think to the discrete Wavelet transform and to the short-time Fourier transform as filter-bank transforms (similar to an audio equalizer) en.wikipedia.org/wiki/Filter_bank $\endgroup$ – reuns May 19 '17 at 21:19
  • $\begingroup$ @user1952009 I think my question is a bit too ambiguous. A better question would be, if I wanted a wavelet-like resolution scheme in the frequency domain, does the Fourier transform of the wavelet transform achieve such desired qualities? I've plotted some plots comparing the Fourier transform vs. the Fourier transform of the wavelet transform, and indeed the latter looks like the Fourier transform in different scales, and I was wondering if this is mathematically justified. $\endgroup$ – stock username May 20 '17 at 1:56
  • $\begingroup$ I told you the answer : both the DWT and the STFT are filter-banks. The main difference is that (with respect to the frequency) the bandwith of the filters are linearly spaced in the case of the STFT, and they are log-linearly spaced in the case of the DWT. For making this rigorous, you need to write them as filter-banks, and plot the FT of each filter's impulse response. $\endgroup$ – reuns May 20 '17 at 2:06
  • $\begingroup$ @user1952009 I do understand that STFT provides a rectangular time-frequency resolution, since a signal is split into frames. But I'm referring to the case for a single frame. DWT maintains a bit of both temporal and spectral resolution, whereas the DFT of a single frame only contains spectral data. $\endgroup$ – stock username May 20 '17 at 2:23
  • $\begingroup$ That's the point, there are no such "single frame" in filter-banks transforms ($= $ time/frequency representation). $\endgroup$ – reuns May 20 '17 at 2:25

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