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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 ??

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  • $\begingroup$ You have an extra parenthesis after .5 (?) $\endgroup$
    – Gilles
    Apr 5, 2016 at 13:56
  • $\begingroup$ typing mistake ..sorry ,edited $\endgroup$
    – Abhishek
    Apr 5, 2016 at 14:01

2 Answers 2

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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

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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.

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  • $\begingroup$ so,now can you able to tell me how I can get my wavelet coefficients .... $\endgroup$
    – Abhishek
    Apr 5, 2016 at 12:25
  • $\begingroup$ @Abhishek, Could you please mark my answer? $\endgroup$
    – Royi
    Feb 20, 2022 at 8:47

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