The Scaling Function as a MSA Identifier

Once a scaling function has been chosen and constructed, the corresponding MSA is uniquely determined. However, not all square-integrable functions qualify as scaling functions. In fact, to qualify, the definition of a MSA on page  implies that a scaling function must satisfy two key properties. They are (i) Eq.(2.107),

$$\int\limits_{-\infty}^\infty \overline\phi (u-\ell')\phi(u-\ell'')du=\delta_{\ell'\ell''}$$ (2110)

and (ii) Eq.(2.108) on page , or equivalently with $k=-1$

$$\boxed{ \phi(t)=\sqrt{2}\sum_{-\infty}^\infty h_\ell\phi(2t-\ell)~. }$$ (2111)

where

$$h_\ell=\langle \phi(2t-\ell),\phi(u-\ell)\rangle~.$$

The boxed is known as the scaling equation for the scaling function $\phi(t)$ . Both Eqs.(2.110) and (2.111) put strong restrictions on the collection of square-integrable functions. The first constitutes a discrete infinitude of constraints and is equivalent to

$$\boxed{ \int\limits_{-\infty}^\infty \overline\phi (u)\phi(u-\ell)du=\delta_{0\ell}~. }$$

The second is a statement about the dilation operator $D$ ,

$$L^2 \stackrel{D}{\longrightarrow}L^2$$

$$f(t) \sim\!\leadsto Df(t)=\sqrt{2^{-1}}f(2^{-1}t) ~.$$

That second constraint, Eq.(2.111), demands that, even though $D$ changes the shape of the graph of $f$ , the resulting function still lies in the subspace

$$\mathbf V_{0}=span\{\phi(t-\ell):~\ell=0,\pm 1,\cdots\}$$

spanned by the discretely tranlated functions $\phi(t-\ell)$ . In other words, the function $\phi(t)$ is such that the subspace $\mathbf{V}_0$ generated from this function is invariant under $D$ .

Both boxed equations put severe restrictions on the collection of square-integrable functions, but these restrictions are not strong enough to single out a unique function. Instead, they narrow the field of candidates to those $L^2$ -functions which qualify as scaling functions for MSAs. The nature of these restrictions becomes more transparent if one expresses them in the Fourier domain, instead of the given domain. Thus by introducing the Fourier transform of $\phi(t)$ ,

$$\frac{1}{\sqrt{2\pi}} \int\limits_{-\infty}^\infty e^{-i\omega t}\phi(t)\,dt=\hat{\phi}(\omega)~,$$

one finds for the first equation that

$$\delta_{0\ell} =\int\limits_{-\infty}^\infty \overline\phi (t)\phi(t-\ell)dt$$

$$=\int\limits_{-\infty}^\infty \overline{\hat{\phi} (\omega)} \hat\phi(\omega)e^{i\omega\ell}d\omega$$

$$=\int\limits_0^{2\pi} \sum_{n=-\infty}^\infty \left\vert \hat\phi(\omega+2\pi n\right\vert^2 e^{i\omega\ell}d\omega$$

This equation says that the sum of the squared magnitude is a function all of whose Fourier coefficients vanish - except for the one corresponding to $\ell =0$ . Such a function is a constant, namely

$$\boxed{ \sum_{n=-\infty}^\infty\left\vert \hat\phi(\omega+2\pi n)\right\vert^2=\frac{1}{2\pi}~. }$$ (2112)

This condition on the Fourier transform of $\phi$ is Mallat's necessary and suffient condition for the discrete translates of $\phi(t)$ to form an o.n. basis for $\mathbf{V}_0$ .

In the Fourier domain the second equation, the scaling equation (2.111), has a simple form. Taking the Fourier transform of the equivalent equation

$$\phi\left(\frac{t}{2}\right)= \sqrt{2}\sum_{-\infty}^\infty h_\ell\phi(t-\ell)~,$$

one finds

$$\boxed{ \hat{\phi} (2\omega)=H(\omega)\hat{\phi}(\omega)~,<tex2html_comment_mark>62 }$$ (2113)

where

$$H(\omega)=\frac{\sqrt{2}}{2}\sum_{\ell=-\infty}^\infty h_\ell e^{i\omega\ell}$$

is a $2\pi$ -periodic function of $\omega$ :

$$H(\omega+2\pi)=H(\omega)~.$$

Equation (2.113) is a linear equation. It expresses in the Fourier domain the relation between the input $\hat{\phi}(\omega)$ and the output $\hat{\phi}(2\omega)$ of a time-invariant linear system. In the theory of such systems the function $H(\omega)$ is therefore known as a filtering function or filter in brief. Its periodicity is an important but fairly mild restriction on $H(\omega)$ . That condition can be strengthened considerably by incorporating the Fourier normalization condition, Eq.(2.112) into Eq.(2.113). One does this by inserting Eq.(2.112) into Eq.(2.113). The result is that $H(\omega)$ satisfy the additional normalization condition

$$\left\vert H(\omega)\right\vert^2 +\left\vert H(\omega+\pi)\right\vert^2=1~.$$ (2114)

Thus the scaling equation is a linear equation, and to qualify as a scaling function, its Fourier transform must satisfy a simple linear equation, Eq.(2.113), having a normalized periodic coefficient. The nature of a particular scaling function, and hence the nature of the corresponding MSA, is controlled by the nature of that normalized coefficient function $2\pi$ -periodic on the Fourier domain.

Exercise 26.5 (FUNCTIONAL CONSTRAINT ON THE FILTER FUNCTION)
Verify the validity of the functional constraint, Eq.(2.114).

Exercise 26.6 (THE SCALING EQUATION SOLVED)
Consider a function $\phi(t)$ having the property

$$\left\vert\int\limits_{-\infty}^\infty \phi(t)dt\right\vert\ne 0~.$$

Find the solution to the scaling equation, Eq.(2.113).

Answer:   $$\hat\phi(\omega)=\hat\phi(0)\prod\limits_{k=1}^\infty H\left(\frac{\omega}{2^k}\right)$$

Exercise 26.7 (TWO SOLUTIONS TO THE SCALING EQUATION)
Let $\phi^+(t)$ be a solution to the scaling equation

$$\phi(t)=\sqrt{2}\sum_{-\infty}^\infty h_\ell\phi(2t-\ell)~.$$

a) Point out why

$$\hat\phi^-(\omega)=\left\{ \begin{array}{rl} \hat\phi^+(\omega)&\omega\geq 0\\ -\hat\phi^+(\omega)&\omega< 0 \end{array}\right.$$

is the Fourier transform of a second independent solution to the above scaling equation.

b) Show that $\phi^+(t)$ and $\phi^-(t)$ are orthogonal,

$$\int\limits_{-\infty}^\infty \overline\phi^+(t)\phi^-(t)dt=0~,$$

whenever (i) $\phi(t)$ is a real function or whenever (ii) its Fourier transform is an even function of $\omega$ .