Multiresolution and Wavelets

Abstract

Time domain and frequency domain analysis suffer the drawback of a scarce resolution, in frequency and in time, respectively. In several applications a segmented resolution (multiresolution) is required, where relevant parts of the signal are detailed, and other parts are only roughly represented. The modern technique that achieves this goal is provided by wavelets that are the target of this chapter. Wavelets are strongly related to the topics developed in the previous chapter, since they can be effectively formulated in the framework of generalized transforms (wavelet transform) and their practical implementation is by means of filter banks. In the final part of the chapter, multiresolution analysis is illustrated with examples of applications.

Appendices

Appendix A: Proof of Proposition 15.7 on Coefficients at Steps m and m+1

We first establish the following statement:

Proposition 15.14

A ℤ _m →ℝ interpolator followed by an ℝ→ℤ _m+1 decimator is equivalent to a ℤ _m →ℤ _m+1 decimator. To calculate the global impulse response $\tilde{g}_{12}(t)$, first calculate the convolution of the component impulse response g ₁₂(t)=g ₂(t)∗g ₁(t), t∈ℝ, and then apply the ℝ→ℤ _m down-sampling, that is,

$$\tilde{g}_{12}(v)=\int_{\mathbb{R}}\mathrm{d}\tau\,g_1(v-\tau)g_2(\tau) ,\quad v\in\mathbb{Z}.$$

The statement, illustrated in Fig. 15.38, is a consequence of the general result on the cascade of QIL tf, developed in Chap. 7, as soon as we note that ℤ _m ⊃ℤ _m+1 (see Problem 15.6). The conclusion is that the combination of the two blocks, each one operating on continuous times, gives a block operating on discrete times, which is the crucial point for the full discretization.

Fig. 15.38

Interpretation of Proposition 15.14: the impulse response $\tilde{g}_{12}(u)$ of the ℤ _m →ℤ _m+1 decimator is the sampled version of the convolution g ₁∗g ₂(t), t∈ℝ

We now prove Proposition 15.7. The link S _m (u)→S _m+1(v) is given by ℤ _m →ℝ interpolator followed by a decimator with impulse response respectively

$$c_{m+1}(t)=\varphi^{(m+1)\ast}(-t) ,\qquad e_m(t)=2^{-m}\varphi ^{(m)}(t) . $$

(15.93)

Then, we apply Proposition 15.14 with g ₁=c _m+1 and g ₁=e _m and evaluate the continuous convolution

$$g_{12}(t)=2^{-m}\int_{\mathbb{R}}\mathrm{d}\tau\,\varphi^{(m+1)\ast}(\tau -t) \varphi^{(m)}(\tau) , $$

(15.94)

where, by (15.22),

$$\begin{aligned}[c]&\varphi^{(m+1)}(t)=2^{-(m+1)} \varphi\bigl(2^{-(m+1)}t\bigr)=2^{-m}2^{-1/2}\varphi\bigl(2^{-m}(t/2)\bigr) ,\\&\varphi^{(m)}(t)=2^{-m} \varphi\bigl(2^{-m}t\bigr) .\end{aligned}$$

Hence,

$$ \begin{array}{*{20}l} \displaystyle {g_{12} (t) = } \hfill & { = 2^{ - m} 2^{ - m} 2^{ - 1/2} \int_{\rm \mathbb{R}} {\rm d} \tau \,\varphi *\left( {\frac{{2^{ - m} \tau- 2^{ - m} t}}{2}} \right)\varphi (2^{ - m} \tau )} \hfill\\ {} \hfill & { = 2^{ - m} 2^{ - 1/2} \int_{\rm \mathbb{R}} {{\rm d}\tau } \,\varphi *\left( {\frac{{\tau- 2^{ - m} t}}{2}} \right)\varphi (\tau ),} \hfill\\ \end{array} $$

where we have make the variable change τ→2^−m τ. Next, we use the two-scale relation (15.46a) with t→(τ−2^−m t)/2 and obtain

$$g_{12}(t)=2^{-m}\sum_n g_0^\ast(n)\int_{\mathbb{R}}\mathrm{d}\tau\,\varphi^\ast\bigl(\tau -n-2^{-m}t\bigr) \varphi(\tau) ,$$

where in general we cannot use the orthogonality condition. But, restricting t to ℤ _m , that is, t=2^m k, the integral gives δ _n−k . Hence,

$$\tilde{g}_{12}\bigl(2^m k\bigr)=g_{12}\bigl(2^m k\bigr )=2^{-m}g_0^\ast(-k) .$$

To get the impulse response of the link S _m (u)→D _m+1(v), it is sufficient to replace c _m+1 with d _m+1 in (15.93) and the two-scale equation (15.46a) with (15.47). The result is $\tilde{g}_{12}(2^{m}k)=2^{-m}g_{1}^{\ast}(-k)$.

Appendix B: Interpretation of the Expansion of ϕ(t) and ψ(t)

We consider the expansion of the functions ϕ(t) and ψ(t) obtained with the bases $\boldsymbol{\varPhi}_{-1}=\{\sqrt{2} \varphi (2t-n)\mid n\in\mathbb{Z}\}$

$$\begin{aligned}[c]g_0(n)&=\sqrt{2}\int_{\mathbb{R}}\mathrm{d}t \,\varphi(t) \varphi ^\ast(2t-n),\qquad \varphi(t)=\sqrt{2} \sum_{n=-\infty}^{+\infty}g_0(n) \varphi(2t-n), \\g_1(n)&=\sqrt{2}\int_{\mathbb{R}}\mathrm{d}t\, \varphi(t) \varphi ^\ast(2t-n),\qquad \psi(t)=\sqrt{2} \sum_{n=-\infty}^{+\infty}g_1(n) \varphi(2t-n) ,\end{aligned} $$

(15.95)

where the right-hand sides represent the two-scale equations.

The system interpretation of these relations shown in Fig. 15.39. With the introduction of the auxiliary signals

$$y(t)=\varphi(t/2),\qquad g(t)=\bigl(1/\sqrt{2}\,\bigr)\varphi ^\ast(-t)\,{\buildrel\varDelta\over=} \,\bigl(1/\sqrt{2}\,\bigr)\varphi _-^\ast(t) ,$$

the first of (15.95) can be written as

$$g_0(n)=\int_{-\infty}^{+\infty}\bigl(1/\sqrt{2}\,\bigr) \varphi (u/2) \varphi(u-n)\, \mathrm{d}u=\int_{-\infty}^{+\infty}q(u-n) y(u) \,\mathrm{d}u .$$

Then, starting from ϕ(t), g ₀(n) is obtained with scale change of $a=\frac{1}{2}$ followed by an ℝ→ℤ decimator with impulse response q(t), t∈ℝ. In the inverse relation (15.46a), (15.46b) the coefficients g ₀(n), n∈ℤ, are ℤ→ℝ interpolated with impulse response $\sqrt{2} \varphi(t)$ and give the intermediate signal y(t); then, y(t) is dilated by $a=\frac{1}{2}$ to give the scaling function ϕ(t), t∈ℝ.

Fig. 15.39

Above: generation of expansion coefficients g ₀(n) from the scaling function ϕ(t) and recovery of ϕ(t) from g ₀(n). Below: generation of g ₁(n) from the mother wavelet ψ(t) and recovery of ψ(t) from g ₁(n)

A similar interpretation, shown in Fig. 15.39, holds for the generation of the coefficients g ₁(n) from the mother wavelet ψ(t) and the reconstruction of ψ(t) from g ₁(n).

Using the rules of the Fourier analysis of interpolators and decimators, we find

$$ \begin{array}{l} Y(f) = \sqrt {2\Phi } (f)G_0 \left( f \right),\quad \Phi (f) = \frac{1}{2}Y\left( {\frac{1}{2}f} \right) = \frac{{\sqrt 2 }}{2}\Phi \left( {\frac{1}{2}f} \right)G_0 \left( {\frac{1}{2}f} \right), \\ {\rm references} \\ \begin{array}{*{20}l} {Y(f)} \hfill & { = 2\Phi (2f),} \hfill\\ {G_0 (f)} \hfill & { = {\rm rep}_1 \left[ {\left( {1/\sqrt 2 } \right)\Phi *(f)Y(f)} \right] = \sqrt 2 {\rm rep}_1 \left[ {\Phi *(f)\Phi (2f)} \right],} \hfill\\ \end{array} \\ \end{array} $$

which give (15.21).

We now prove Proposition 15.11. Combination of (15.63) and (15.66) gives

$$ \begin{array}{*{20}l} 2 \hfill & { = \sum\limits_{k =- \infty }^{ + \infty } {\left| {\Phi \left( {\frac{1}{2}f - \frac{1}{2}k} \right)} \right|^2 \left| {G_0 \left( {\frac{1}{2}f - \frac{1}{2}k} \right)} \right|} ^2 } \hfill\\ {} \hfill & { = \sum\limits_{k =- \infty }^{ + \infty } {\left| {\Phi \left( {\frac{1}{2}f - h} \right)} \right|^2 \left| {G_0 \left( {\frac{1}{2}f - h} \right)} \right|} ^2 } \hfill\\ {} \hfill & { + \sum\limits_{k =- \infty }^{ + \infty } {\left| {\Phi \left( {\frac{1}{2}f - h - \frac{1}{2}} \right)} \right|^2 \left| {G_0 \left( {\frac{1}{2}f - h - \frac{1}{2}} \right)} \right|^2 .} } \hfill\\ \end{array} $$

Next, considering that G ₀(f) has period 1 and using again (15.66), we have

$$ \begin{array}{*{20}l} 2 \hfill & { = \left| {G_0 \left( {\frac{1}{2}f} \right)} \right|^2 \sum\limits_{h =- \infty }^{ + \infty } {\left| {\Phi \left( {\frac{1}{2}f - h} \right)} \right|^2 \left| {G_0 \left( {\frac{1}{2}f - \frac{1}{2}} \right)} \right|^2 } \sum\limits_{h =- \infty }^{ + \infty } {\left| {\Phi \left( {\frac{1}{2}f - h - \frac{1}{2}} \right)} \right|^2 } } \hfill\\ {} \hfill & { = \left| {G_0 \left( {\frac{1}{2}f} \right)} \right|^2+ \left| {G_0 \left( {\frac{1}{2}f - \frac{1}{2}} \right)} \right|^2 ,} \hfill\\ \end{array} $$

which is equivalent to (15.67). In fact, from (15.21) we obtain $2 |\varPhi(f)|^{2}=|\varPhi(\frac{1}{2}f)|^{2} |G_{0}(\frac{1}{2}f)|^{2}$. Taking the periodic repetition and using (15.66), we get (15.67).