1
$\begingroup$

I have an understanding problem with Daubechies wavelets.

When I use a multiresolutional analysis, I want to approximate the given input Signal $f\in L^2(\mathbb{R})$ on the subspaces $V_i$. By calculating the coefficients I split the signal in the detail part and the rest. Now I am wondering, how the Daubechies wavelet is fitting in this picture. The Daubechies wavelet is defined by coefficients, but in my opinion, these are only usable if I try to perform a wavelet decomposition on input data in $l^2(\mathbb{Z})$.

To specify my question, I have no idea how to calculate the first inner products $$c_i = \langle f, \varphi_{i,j} \rangle, \text{ and } d_i = \langle f, \psi_{i,j} \rangle $$ if $\varphi$ and $\psi$ are Daubechies scaling function and wavelet, since they have no expression as functions in $L^2(\mathbb{R})$.

Thanks in advance

Matthias

$\endgroup$
  • $\begingroup$ Can you talk a little bit more about the constrain you perceive with $l^2(\mathbb{Z})$? The inner product operation does not change depending on the decomposition function (whether it's DFT, DWT, etc). $\endgroup$ – A_A Mar 2 at 4:27
  • $\begingroup$ Hi, instead of performing a template matching in $L^2(\mathbb{R})$ I could do the same in $l^2(\mathbb{Z})$. But this was only a side note. What do you mean with "The inner product operation does not change depending on the decomposition function"? It depends on $\varphi$ and $\psi$ since it is a $L^2$ inner product. $\endgroup$ – Matthias Lauber Mar 2 at 15:02
  • $\begingroup$ Ok, I assume I understand what is going on. Theoretically one would need the inner products to calculate the coefficients on the finest grid to start the decomposition procedure, but I read often that one assumes the sampling points $f\in l^2(\mathbb{Z})$ as the result from this inner product and start the procedure with this sample points. Furthermore, the coefficients of the Daubechies wavelets as presented in Wikipedia has per se nothing to do with the scaling function. I added some of them and get always 2. It seems, that these are the filter coefficients for the lowpass and highpass. $\endgroup$ – Matthias Lauber Mar 2 at 15:37
0
$\begingroup$

As I mentioned in my comment, the reason seems to be the so-called "wavelet crime" which is taking the sampling data as the inner product. But I found a question, which is similar to mine

Wavelet transform: How to compute the initial coefficients when only samples are available?*

This question was answered with a lot of papers, many thanks to Laurent Duval. In one of the paper

On the Initialization of the Discrete Wavelet Transform Algorithm

there is a derivation to calculate the initial inner products. One can approximate the inner products as $$ c_k = \int_\mathbb{R} f(t)\varphi(t-k)dt \approx \varphi(-k) $$ And one can read the following remark:

"A very good approximation of the time-shape of $\varphi(t)$ can be obtained as the impulse response of the filter h convolved with itself a large number of times. One then simply needs to collect the samples ${\varphi(p) = \varphi(t = p)}_{p\in\mathbb{Z}}$ to perform the approximate initialization"

Therefore it is sufficient to know the filter coefficients as given in Wikipedia.

$\endgroup$

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.