1
$\begingroup$

Is the refinement equation for the orthonormal Daubechies scaling function $$\phi(x) = \sqrt{2} \sum_n h_n \phi(2x-n) \;?$$ The filter coefficients for Daubechies wavelets have been given e.g. in this page. Consider case D20. Do the indices $n$ for $h_n$ run from 0 to 19? Or do we set $h_{-n} = h_n$?

Suppose we have an arbitrary function $f \in L^2(\mathbb{R})$. Can we approximate $f$ with $$f(x) \approx 2^{j/2} \sum_k c_k \phi(2^j x - k)$$ where $$c_k = 2^{j/2} \int f(x) \phi(2^j x - k) dx$$ when $j$ is sufficiently large? Can we approximate $$f\left(\frac{k}{2^j}\right) \approx c_k \;?$$

$\endgroup$
1
$\begingroup$

The Daubechies wavelets are not symmetric, in fact, there are no continuous, orthogonal and symmetric 2-channel wavelets (but 3-channel and up works). So it is the first way.

Yes, that would be the correct way of sampling a signal. As that is unpractical, some discretized approximation is used. The single-point approximation you mentioned is the worst of these. One fast way is point-sampling with a higher density and then applying the low-pass filter of the wavelet.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.