Why does each node in the wavelet scattering transform split into multiple paths as in this figure from Deep Scattering Spectrum?

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I understand roughly what’s happening along a single path, but I don’t understand why each node splits into three.


1 Answer 1


We take a wavelet transform over other wavelet transforms, which involves multiple wavelets. E.g.:

  1. First-order uses 100 wavelets to tile the frequency domain and produce 100 coefficients
  2. Second-order uses 16 wavelets to tile the frequency domain and produces 8 coefficients per first-order coefficient, yielding a total of 800
  3. Third-order uses 9 wavelets ... etc

(Note second-order wouldn't actually produce 800 in implementation since xi2 > xi1 wavelets are discarded, where xi == center frequency).

The idea is, first-order coefficients capture temporal modulations (AM-FM). Second-order capture the modulations of those modulations, extracting higher order structure, and the process repeats. Progressively we require less wavelets as there's less fine-grained structure to capture (also energy and non-redundancy reasons), but it's still more than one wavelet (i.e. path), as just one wavelet cannot cover entire frequency domain without losing information (unless it's very wide, but that's a bad wavelet).

  • 1
    $\begingroup$ Thank you. Actually I already watched that lecture series. Mallat gives great talks! When you say multiple wavelets, are you talking about dilations/translations of a single mother wavelet, or multiple mother wavelets? Sorry, my background isn't in DSP. It might be helpful if you could contrast the scattering wavelet transform and the discrete wavelet transform. The latter splits the input signal into the low and high frequencies, and then the low frequency content is passed into the next iteration. The scattering transform seems to pass multiple signals on instead of just one. $\endgroup$
    – Churchjm
    Commented Apr 25, 2021 at 22:56
  • $\begingroup$ "dilations/translations of a single mother wavelet" is correct, I recommend this tutorial which also contrasts with STFT and DWT. You can also find helpful visuals here - also interactively. $\endgroup$ Commented Apr 25, 2021 at 23:03
  • $\begingroup$ "pass multiple signals on instead of just one" - sort of. So e.g. first order's first wavelet will capture the highest frequencies; that's passed as input to second order (namely, its absolute value, the wavelet is complex in time domain). Same for first order's second wavelet, and so on. It's abs(CWT) on top of abs(CWT) on top of abs(CWT)... $\endgroup$ Commented Apr 25, 2021 at 23:08
  • $\begingroup$ I’m sorry, I don’t think I’m asking my question clearly. The tutorial discusses the DWT, which I understand fairly well (at least from a high level), but it doesn’t mention the scattering transform. My hang up is that each iteration of the DWT produces two new signals (one from the high pass filter, and one from the low pass filter), but each iteration of the scattering transform produces 4+, including the low frequency outputs and a bunch of high frequency signals that are passed to the next level. Does the scattering transform depth have the same meaning as the DWT (e.g. scale)? $\endgroup$
    – Churchjm
    Commented Apr 25, 2021 at 23:30
  • $\begingroup$ Didn’t see your second reply. How about we try this way; what is the difference between the first order first wavelet, and first order second wavelet? $\endgroup$
    – Churchjm
    Commented Apr 25, 2021 at 23:33

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