2
$\begingroup$

I am trying to develop a new type of wavelets and I found out a function that following a particular two scale relation. The function at a scale $t$ say $x(t)$ can be related in a finer scale $x(2t)$ by the following relation:

$$x(t)=0.1x\left(2t\right)+ 0.5x\left(2t-1\right)+0.8x\left(2t-2\right)+0.5x\left(2t-3\right)+0.1x\left(2t-4\right),$$

This is found out by trial and error method by simply changing the values. Now I know that the points $\begin{bmatrix}0.1&0.5&0.8&0.5&0.1\end{bmatrix}$ can be considered as the scaling filter coefficients. Now the problem is finding the wavelet filter coefficients, I am not able to find out a particular relation connecting scaling filter coefficients and wavelet filter coefficients. Is there is a relation connecting the two ??

$\endgroup$
  • $\begingroup$ You have an extra parenthesis after .5 (?) $\endgroup$ – Gilles Apr 5 '16 at 13:56
  • $\begingroup$ typing mistake ..sorry ,edited $\endgroup$ – Abhishek Sadasivan Apr 5 '16 at 14:01
1
$\begingroup$

For perfect reconstruction you need:

$$ g \left[ n \right] * \hat{g} \left[ n \right] + h \left[ n \right] * \hat{h} \left[ n \right] = \delta \left[ n \right] $$

Where $ g \left[ n \right] $ is the Analysis Wavelet Function, $ \hat{g} \left[ n \right] $ is the Synthesis Wavelet Function, $ h \left[ n \right] $ Analysis Scaling Function and $ \hat{h} \left[ n \right] $ is the Synthesis Scaling Function.

Practically, Scaling Function is the LPF and Wavelet Function is the HPF.

$\endgroup$
  • $\begingroup$ so,now can you able to tell me how I can get my wavelet coefficients .... $\endgroup$ – Abhishek Sadasivan Apr 5 '16 at 12:25
1
$\begingroup$

If $\{h(n)\}_{n\in\mathbb{N}}$ is the scaling filter, i.e. the sequence such that $$ \phi(x) = \sum_{n} h(n) 2^{1/2} \phi(2x-n) $$ where with $\phi(\cdot)$ I indicated the scaling function, then the mother wavelet satisfies $$ \psi(x) = \sum_{n} g(n) 2^{1/2} \phi(2x - n) $$ being $$ g(n) = (-1)^n \overline{h(1-n)}$$ being $\bar{z}$ the complex conjugate of $z\in\mathbb{C}$. This according to D. F. Walnut, An Introduction to Wavelet Analysis

$\endgroup$

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.