Building the low pass filter in the wavelet scattering transform

Question

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.

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.