Power Efficiency Guide

16 Theorem of Calderon, Grossmann, Morlet.

Given the Fourier transform of y(x). Morlet and Grossmann proved that if the wavelet y(x) has a Fourier transform equal to zero at o = 0, then the wavelet transform satisfies an energy conservation equation and f (x) can be reconstructed from its wavelet transform. The wavelet y(x) can be understood as the impulse response of a band-pass filter and the wavelet transform as a convolution with a band-pass filter which is dilated. When the scale s is large, Wsf (x) detects the lower frequency components of the signal f (x). When the scale s decreases, the support of ys(x) decreases so the wavelet transform Wsf (x) is sensitive to finer details. The scale s determines the size and the regularity of signal features extracted by the wavelet transform.

5.5.4.2 Dyadic Wavelet Transform and Reconstruction Formula

The wavelet transform depends on two parameters s and x which vary continuously over the set of real numbers. For practical applications, s and x must be discretized. For a specific class of wavelets, the scale parameter can be sampled along the dyadic sequence [2j]jeZ, without modifying the properties of the transform. The wavelet transform at the scale 2j is given by:

At each scale 2j, the function W2jf(x) is continuous since it is equal to the convolution of two functions in L2(R). The Fourier transform of W2 jf (x) is

When the following constraint is imposed:

then the whole frequency axis is covered by a dilation of y(rn) by the scales factors [2j] jeZ. Any wavelet satisfying the constraint is called a dyadic wavelet. The dyadic wavelet transform corresponds also to the sequence of functions:

Let "W" be the dyadic wavelet operator defined by Wf = [W2jf (x)] j6Z. From the equation of the Fourier transform of W2j f(x) with its constraint (supra), and by using the Parseval theorem, we can write an energy conservation equation:11

17 Remark: The norm (energy) of f(x) e L2 (R) is given by || f ||2 = f+Z | f(x)\2 dx.

We pose y2j(x) = V2j{~x)- The function f(x) is reconstructed from its dyadic wavelet transform by the reconstruction formula:

Given V, the space of the dyadic wavelet transforms [W2jf (x)] jeZ, for all the functions f (x) e L2(R).18 Then, l2(L2) the Hilbert space of all sequences of functions [hj(x)]jeZ, such that hj(x) e L2(R) and I+—\\hj(x)\\2 < The energy conservation equation (see supra) proves that V is a subspace of l2 (L2). Then, W—1 denotes the operator from l2(L2) to L2 (R) defined by: W—1 [hj (x)] jeZ = I+— hj * \j/2j(x). The reconstruction fonnula (see supra) shows that the restriction of //' 1 to the wavelet space V is the inverse of the dyadic wavelet transform operator W. Every sequence of functions [hj(x)] jeZ e l2(L2) is not-a priori-the dyadic wavelet transform of some function f (x) e L2 (R). In fact, if there is a function f (x) e L2 (R) such that [hj(x)] jeZ = Wf, then we should have W W—1 [hj(x)]jeZ = [hj(x)]jeZ.

When we replace the operators W and W—1 by their expression given in the equations W2jf (x) = f * y2j (x) (supra: wavelet transform) and W—1 [hj(x)] jeZ = hj * V2j {x) (see supra), then it comes:

ZtZJn*Klj{x) = hj{x) with je Zand K,j(x) = y2j *y2j(x). (5.34)

The sequence [hj (x)] jeZ is a dyadic wavelet transform if and only if the aforementioned equations L/tTl«,hi * Kij(x) = hj{x) and Kjj(x) = \j/2j * V2j{x) hold. These equations are known as reproducing kernel equations, and show the correlation between the functions W2jf (x) of a dyadic wavelet transform. The correlation between the functions W2jf (x) and W2lf (x) at two different scales 2j and 2l can be understood by observing their Fourier transform: W2jf (®) = f(a)v(2j03) and W2lf (o) = f(o)y(2la). The redundancy of W2jf(o) and W2lf (o) depend on the overlap of the functions ^(2jo) and ^(2lo). The energy of this overlap is equal to the energy of the kernel Kl} j (x). (It is maximum for l = j — 1, l = j + 1.) Let PV be an operator defined by

this operator is a projector from l2 (L2) on the Vspace. It is possible to prove that any sequence of functions [hj(x)] jeZ e l2 (L2) satisfies PV [hj(x)] jeZ e V, and any element of V is invariant under the action of this operator.

18 Recall: The following spaces correspond to the respective functions or associated signals: (1) L2(R): Finite energy functions: /1 f (t)\2dt < and consequently is the space of integrable square functions. (2) LP(R): Functions such that / \f (t)\pdt < (3) l2(Z): Discrete finite energy signals: \ f (t )\2 < (4) lP(Z): Discrete signals such that: \ f (t )\p <

A signal is usually measured with a finite resolution which imposes a finer scale when computing the wavelet transform. For practical purposes, the scale parameter must also vary on a finite range. We are going to explain how to interpret mathematically a dyadic wavelet transform on a finite range. In both previous sections (for the sake of simplification) the model was based on functions of a continuous parameter x, but we have to discretize the abscissa x and explain efficient algorithms for computing a discrete wavelet transform and its inverse.

For practical purposes, it is not possible to compute the wavelet transform at all scales 2j for j varying between —^ and +«>. In fact, we are limited by a finite larger scale and a non-zero finer scale. In order to normalize, we suppose that the finer scale is equal to 1 and 2j is the largest scale. (With f (x) e L2). Between the scales 1 and 2j, the wavelet transform [W2jf (x)]1<< can be interpreted as the details available when smoothing f (x) at the scale 1 but which have disappeared when f (x) at the larger scale 2j. At this stage, we introduce a function 0 (x) whose Fourier transform is:

Because we know that the wavelet y(x) verifies \v(2ja)\2 = 1, so we have lim \0(a)\2 = 1. In addition, the energy distribution of the Fourier transform 0(a)

is localized in the low frequencies, thus 0 (x) is a smoothing function. Given S2j the smoothing operator defined as follows:

The larger the scale 2j, the more details of f (x) are removed by the smoothing operator S2j. The dyadic wavelet transform [W2jf (x)]1<j<J between the scale 1 and 2j give the details available in S1 f (x) but not for S2jf (x). The Fourier transforms

S1 f (a), S2jf (a), W2jf (a) of S1 f (x), S2jf (x), W2jf (x) are respectively given by:

W f (a) = 0(a) f(a), Wjf (a) = 0(2Ja) f(a), W^f (a) = W(2ja) fW(a).

The first equation \W(a)\2 = 1+=! \y(2ja)\2 (i.e. Fourier transform of 0(x)) gives:

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