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MATLAB has a library of wavelet functions, showing their "continuous forms" as well as the the decomposition and reconstruction filters.

In decimated wavelet transform the filter size remains the same and the data points are downsampled by $2^j$ at every level $j$. A question that I cannot find an answer is how are the wavelet filters and the actual continuous forms are related? I am chemist, exploring wavelets for some applications.

For example, MATLAB has db9. Its continuous estimation, wavelet function, psi or phi do not look similar to the decomposition filter points. One text says that some wavelets start out as filter and later their continuous forms are estimated, but why in the case of db9 and many other, the filter points do not match the shape of the wavelet function?

Thanks.

MATLAB wavelet library

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  • $\begingroup$ Thank you for the interaction. Your question dug deep on the root of wavelets and their discretization, far more than SE posts can drill $\endgroup$ – Laurent Duval Dec 25 '19 at 22:46
  • $\begingroup$ Self learning process by an outsider of the field is difficult but it goes deeper and deeper. Thanks for the useful hints. $\endgroup$ – M. Farooq Dec 25 '19 at 23:03
  • $\begingroup$ Sure, because there is a gap between theory and practice, mathematical background and understanding. Harmonic analysis and wavelets are however rewarding. Go deeper $\endgroup$ – Laurent Duval Dec 25 '19 at 23:06
  • $\begingroup$ Thanks, btw, do you happen to know in DWT denoising what Matlab means by scaled white noise vs. unscaled white noise. I have searched their entire reference guide but could not find the difference. $\endgroup$ – M. Farooq Dec 25 '19 at 23:18
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    $\begingroup$ I'm currently answering the related question $\endgroup$ – Laurent Duval Dec 25 '19 at 23:19
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I will start with an averaging filter analogy. The simplest (non trivial) and cheapest way to average a discrete signal $x[n]$ is to compute the mean of two adjacent samples, and remplace the two samples by this average. This amounts to compute the discrete system:

$$\frac{x[2n]+x[2n+1]}{2}\,.$$

This can be recast as the convolution $$y[n] = x[n]\ast \frac{ [1\; 1]}{2}$$

You might remember that $\frac{ [1\; 1]}{2}\ast \frac{ [1\; 1]}{2} = \frac{ [1\;2\; 1]}{4}$. And that $\frac{ [1\;2\; 1]}{4}\ast \frac{ [1\; 1]}{2} = \frac{ [1\;3\;3\; 1]}{8}$

This should remind you of the Pascal or binomial triangle, that evolve to a Gaussian shape:

Pascal binomial Gaussian

Here, starting with a crude, and a simple finite-impulse response two-point averaging filters, via iterations, you can end up with an IIR, not recursive, infinite-support Gaussian filter. This example is more a metaphor of how discrete filters generate a continuous function.

To be able to add downsampling within the filters, without losing information, contraints the data smoothing picture. In the discrete context, this requires to have a downsampled smoothed version, and a difference version that records when the data is not smooth enough to look like its smooth version. This is the root to most practical discrete wavelet transforms. The Multi Resolution Analysis relates an (yet unknown) scaling function $\phi$ at two-scales, via a linear combination, with weights corresponding to the filter $H$ coefficients. This is explained for instance in Signal Processing: Fourier and Wavelet Representations, 6.1. Introduction, Scaling Function and its Properties, page 191. Finally, the spectrum of the scaling function is given by: $$\Phi^{(\infty)}(\omega)= \prod_{k=1}^\infty \frac {1} {\sqrt 2} H\left( \frac {\omega} {2^k}\right) \Phi^{(\infty)}(0)$$ where iterated half-cut-off frequencies are apparent. The take-away message is: the spectrum of the scaling function is related to the that of the low-pass filter. And in the same manner, the spectrum of the wavelet function is related to the that of the high-pass filter. Some more pointers are given in:

The limit is that, if you start from continuous wavelets, very few of them can be faithfully implemented as discrete schemes as filters, and those filters, when they exist, don't look like wavelets. Because they don't have too. Their resemblance with wavelets only make sense when they are repeated over and over at different scales, across many dyadic sub-sampling.

And most discrete wavelets (like Daubechies') don't have an analytic formula. But if you look at the bottom line, from afar, the filters and the wavelets look similar in terms of envelope (red line below) and zero-crossings. So the global shape is not so different, as shown below with appropriate scaling of the wavelets and filters:

wavelet scaling function, mother wavelet and corresponding filters, with envelope

This similarity seems common for smooth-enough wavelets. Honestly, I have not investigated this into more details, because the actual ressemblance of filters and wavelets is not so important, their properties are way more of interest.

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    $\begingroup$ Nice illustration, and I believe that is a result of central limit theorem. So you are saying that repeated [1 1] convolution is essentially a Gaussian filter. I think the very critical part of your answer is "discrete schemes as filters, and those filters, when they exist, don't look like wavelets." Q_1 Probably that is too mathematical an question in DWT, what comes first if someone is designing a $new$ discrete wavelet decomposition ? The discrete filter points or the discrete wavelet function? $\endgroup$ – M. Farooq Dec 21 '19 at 3:35
  • $\begingroup$ Q_2: Do db9 wavelet and scaling functions have any analytic form or are they just closely connected points which make it look like a continuous function? $\endgroup$ – M. Farooq Dec 21 '19 at 3:37
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    $\begingroup$ I have not been clear enough, probably. Most discrete wavelets, especially the Daubechies ones, don't have known analytical expression. I'll expand the answer later. $\endgroup$ – Laurent Duval Dec 21 '19 at 11:37
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    $\begingroup$ Thanks for expanding the answer, in the end you wrote "And most discrete wavelets (like Daubechies') don't have any analytic formula. But if you look at the bottom line, from afar, the filters and the wavelets look similar in terms of envelop and zero-crossings. So the global shape is not so different." Have a look at the filter in the photograph I attached in the OP. The decomposition filters and the wavelets do not have the same zero crossings, nor there is any resemblance in terms of overall shape. Am I missing something? $\endgroup$ – M. Farooq Dec 24 '19 at 1:11
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    $\begingroup$ I've add a graph, which I hope can give you an hint. The envelope of the wavelet (resp. scaling function) and the corresponding filters are somehow similar. This is not a general rule, but can be observed for smooth enough filters $\endgroup$ – Laurent Duval Dec 25 '19 at 19:33

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