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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?

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  • $\begingroup$ Loosely speaking (in response to your loosely asked question), a wavelet acts as a band pass filter. Look at the Fourier transform of the wavelet you're using. Use scaling properties of the FT. On the other hand the scaling function associated with wavelet acts as a low pass filter. $\endgroup$ – Chester Jul 3 '15 at 13:06
  • $\begingroup$ Thx for the answer. I try to explain it in more detail. $ [W_\psi,f](a,b) = {1 \over \sqrt a} { \int_{-\infty}^\infty f(x)\overline{\psi{(x-b) \over a}}}dx $ $\endgroup$ – gimba Jul 3 '15 at 13:59
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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. enter image description here

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  • $\begingroup$ $|\hat{g_a}(w)| = \frac{2\sin^2(w/4)}{|w|}$ should be $|\hat{g_a}(w)| = \frac{2\sqrt{a}\sin^2(aw/4)}{|aw|}$? $\endgroup$ – mike Aug 4 '17 at 12:08

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