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By reading some books about wavelets, it seems that Discretized CWT is not the same thing as DWT.

Can we give a classification of Discretized CWT vs DWT vs FWT etc. ?

A few remarks :

  • All I read about DWT or MRA (multi-resolution analysis) involves a dyadic factor (2) for $a$ in $\frac{1}{\sqrt{a}} \Psi(\frac{t-b}{a})$, ie $a=2^m$. Is it always true ? Remark : This is good in order to have a non-redondant transform (see B Z's answer here Scalogram (and related nomenclatures) for DWT?), but the graphical representation of such a transform isn't very satisfying for visual analysis of a signal. The fact that there is $ a=2^m$ involved implies that the frequency bands studied [in the case where the mother-wavelet has a narrow-band fourier transform, e.g. Morlet wavelet http://ieeexplore.ieee.org/ieee_pilot/articles/06/ttg2009061375/assets/img/article_1/fig_3/large.gif] are things like [20hz,40hz], [40hz,80hz], [80hz, 160hz], etc. Thus we cannot hope a good graphical representation with precise frequency resolution. It seems that this kind of transform is more suited for "approximation + details" than for visual frequency analysis (scalogram, etc.)

  • On the other hand, a discretization of Continuous wavelet transform allows to take $a=1.0004^m$ for example, and thus have a more precise frequency resolution, for one who wants to do a graphical representation (scalogram).

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Basj
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  • hi, where did you find that discretization of Continuous wavelet transform allows to take $a=1.0004^{m}$ what the reference for that,plz? @basj – SAH Nov 20 '13 at 16:50
  • The continuous wavelet transform is the collection of $Tf(a,b) = \frac{1}{\sqrt{a}} \int_{-\infty}^{+\infty} f(t) \psi((t-b)/a) d t$ for $a, b \in R$. (found in any reference about wavelet). So I think there nothing wrong in discretizing $a$ with such a sequence. I don't claim it will be possible to invert the process by keeping these $a$ only, but for analysis purpose only, I think there's nothing wrong at choising $a$ like this. What do you think Electricman ? – Basj Nov 20 '13 at 16:56
  • I agree, but where did you find 1.0004? for example we know in DWT they use 2. but what does matlab use for cwt? exactly 1.0004? @basj – SAH Nov 20 '13 at 17:15
  • I just said $1.0004$ randomly ;) just to have something smaller than 2 ... @Electricman – Basj Nov 21 '13 at 19:08
  • yeah! but i am very ciouros to know what matlab uses. @bsaj – SAH Nov 21 '13 at 19:10

2 Answers2

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For the CWT the values in your definition can be any $a\in \mathbb R^+$ and $b\in \mathbb R$ if applied to functions. The CWT can also be applied to discrete signals (discretized CWT) with $b$ restricted to appropriate discrete values (and any $a$, but typically $2^{j/v}$ where $v$ is the "voices/octave" of the transformation). Now when we restrict $a$ to some discrete values $2^j$ we get the DWT, which is an orthogonal transformation.

(Small edit to correct the scale : $2^{j/v}$ instead of $a^{j/v}$)

Basj
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Gummi F
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The wavelet function in a continuous WT needs to satisfy certain, relatively weak, admissibility conditions.

For the discretized CWT to work, one needs that the shifted and scaled functions corresponding to the choice of $a$ and $b$ form a frame. This gives slightly stronger conditions for the mother wavelet, but essentially this can always be satisfied if $a-1$ and $b$ are chosen small enough.

Even when $a=2$ and $b=1$ still satisfies the frame conditions, there is no guarantee that there exists a father wavelet or scaling function that complements the wavelet frame to a multiresolution analysis.

You need a MRA with finite filters to implement a fast wavelet transform with a theoretically sound sampling mechanism valid for all scales and depths of the transform. Essentially, you need a continuous father wavelet with a compact peak as its main feature to justify the initialization of the FWT by just sampling the signal.

In the opposite direction, not all filter bank systems that have satisfying properties for a fast discrete WT (for a small depth of the FWT recursion) lead back to a MRA, the scaling equation may have no continuous solution.

Lutz Lehmann
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