In 1, analyzing the equivalence between MFCC and wavelet transform, the scaled wavelets are truncated for $\lambda < 2\pi Q/T$ to fit the time spread of the wavelets within the length $T$ of the window. The lower part of the spectrum is divided into $Q-1$ filters of equal band $2 \pi/T$ plus one more low pass filter $\phi$.

In 2 (pag 106), the scaling function gives a solution for a complete representation of a function $f$ when $wf(u,s)$ is not known for $s>s_0$. The scaling function is defined as:

$|\hat\phi(\omega)|^2 = \int_1^{+\infty} \frac{|\hat\psi(s\omega)|^2}{s} ds$

My question is: What's the reason to divide the lower spectrum in $Q-1$ $+1$ filters instead of using a single low pass filter built with a single scaling function? I guess that, by using the solution described in 1, it is possible to control the length of the window $T$ and to keep a good resolution at very low frequencies. However, I'm not completely sure and I'd appreciate some feedback.

Thanks in advance for the answer.


1 Answer 1


It keeps the scale of the largest scale wavelet $\leq T$ while still tiling the entire frequency axis.

using a single low pass filter built with a single scaling function

would not achieve this: $\phi$ of scale $T$, and $\psi$ whose temporal widths don't taper, will result either in $\psi$ whose width is $>T$, or in incomplete tiling.

Note that a jump from CQT to STFT isn't the only possible scheme: scattering.m transitions wavelet widths smoothly. Advantage of an abrupt transition is that more of the overall representation is CQT, while that of a smooth transition is the smoothness itself - best option varies by application. Also see "Subsampling" and "Filterbank implementation" here.

  • $\begingroup$ Thanks for your answer. It's the way I understood it, not sure if it was clear in the original question. My doubt came from thinking that the lowest part of the spectrum rarely carries any useful information. However, as you remarked, it's important to take the length $T$ into account. Furthermore, the procedure to build the filter bank (except Q) is the same for higher-order scatter where I think having a good low-frequency resolution is important. $\endgroup$
    – dac
    Commented Dec 24, 2021 at 3:33
  • 1
    $\begingroup$ @dac "lowest part ... rarely carries any useful information" - varies strongly by application; it's often true for audio. "higher-order scatter ... good low-frequency resolution is important" definitely, always. The idea's to have a complete transform - if we suspect some parts are useless, can always tailor to application. -- The other two should stand as a separate question, yes. $\endgroup$ Commented Dec 24, 2021 at 5:34
  • $\begingroup$ Thanks! Good to point out also that issue about the smooth transition. $\endgroup$
    – dac
    Commented Dec 24, 2021 at 16:04

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