0
$\begingroup$

I am reading a textbook "Efficient Algorithms for Discrete Wavelet Transform". In that textbook (page 22), the authors are talking about filter banks for the discrete wavelet transform. Specifically, they mention that the impulse response is orthogonal to its even shifts.

Put in mathematical symbols, the authors say that if we have a discrete-time filter, g(n), with finite impulse response, then we have this property

My question is: what does this property mean? Why is it important?

And, most importantly, how did they derive this property?

I am trying to learn wavelets on my own, so I appreciate your help. Thanks

$\endgroup$
  • $\begingroup$ Both the Fourier series and the discrete Fourier transform can be understood as $f = \sum_k \langle f,e_k \rangle e_k$ where $(e_k)$ is an orthonormal basis. In the case of the discrete Fourier transform, $e_k[n] = \frac{1}{\sqrt{N}}e^{2i \pi n k/N}$ and the inner product is $\langle f,g \rangle = \sum_{n=0}^{N-1} f[n] \overline{g[n]}$. It is orthonormal because $\langle e_k,e_m\rangle = \sum_{n=0}^{N-1}\frac{1}{\sqrt{N}}e^{2i \pi n(k-m) k/N}=0$ whenever $k \ne m$ $\endgroup$ – reuns Nov 21 '16 at 18:34
  • $\begingroup$ If you have an orthonormal wavelet basis (for example the (decimated) Haar wavelet transform), then $f = \sum_k \langle f,w_k \rangle w_k$ where $w_k$ is your set of wavelets, and the coefficients $\langle f,w_k \rangle$ is your transformed signal, transformation that is easy to invert (see the properties of orthogonal matrices) $\endgroup$ – reuns Nov 21 '16 at 18:37
  • $\begingroup$ ok I know how to specifically prove this for Haar wavelet because it is easy. I know that g(0) = 1/sqrt(2) and g(1) = 1/sqrt(2) and g(n) = 0 otherwise. But how do I prove this property for discrete (dyadic) wavelet filterbanks in general? $\endgroup$ – blair Nov 21 '16 at 19:06
  • $\begingroup$ There are two types of wavelet transform : the orthogonal wavelets corresponding to what I wrote, and the biorthogonal wavelet where the inverse wavelet transform is more complicated and uses the so called father wavelet (inverting a en.wikipedia.org/wiki/Invertible_matrix can be complicated). There is also the non-invertible WT and the non-decimated WT, both special cases of the filter-bank transform (as the STFT) $\endgroup$ – reuns Nov 21 '16 at 19:10

Your Answer

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

Browse other questions tagged or ask your own question.