Wavelet transform and taking out of frequencies

Question

We use a scaled wavelet and move it across the signal taking out frequencies so that they need not to be processed with a differently scaled wavelet. How does this show up in the math behind wavelet transform?

Answer 1

Denote the wavelet transform of $f$ by $$ (Wf)(a,b) = \frac{1}{\sqrt{a}} \int_{-\infty}^{\infty} f(x) \psi\left(\frac{x-b}{a}\right)^* dx $$

where $z^*$ indicates complex conjugate of $z$.

Let $g_a(x) = \frac{1}{\sqrt{a}} \psi\left( \frac{-x}{a} \right)^*$. Then

$$ (Wf)(a,b) = (f*g_a)(b) $$

Denote $\hat{u}(w)$ as the Fourier transform of $u(t)$. So for fixed scale, $a$,

$$ (\widehat{Wf})(a,w) = \hat{f}(w) \cdot \hat{g_a}(w) $$

where

$$ \hat{g_a}(w) = \sqrt{a} \hat{\psi}(aw)^* $$

This is the transfer function for the wavelet transform at a fixed scale. Furthermore, a wavelet, $\psi$, is usually defined such that $\int_{\mathbb{R}} \psi(x) dx = 0$. The latter property equivalently says $\hat{\psi}(0) = 0$. So the transfer function is zero at constant frequency and goes to zero as $|w| \rightarrow \infty$. So it is a band-pass filter whose width is controlled by $a$.

Obviously, you can be more specific by specifying $\psi$. Take for instance the Haar wavelet where

$$ \psi(x) = \left\{ \begin{matrix} 1 \;\; \text{if} \;\; 0 \leq x \lt 1/2 \\ -1 \;\; \text{if} \;\; 1/2 \leq x \lt 1 \end{matrix} \right. $$

In this case, $$ \hat{\psi}(w) = \frac{-2e^{-iw/2}\sin^2(w/4)}{iw} $$

so magnitude of the transfer function is

$$ |\hat{g_a}(w)| = \frac{2\sin^2(w/4)}{|w|} $$

I've plotted this below for $a=1,\ldots,10$ (blue->red). Notice that small scales allow higher frequencies to pass and large scales allow lower frequencies.