# 0.7 Generalizations of the basic multiresolution wavelet system  (Page 9/28)

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## Two channel biorthogonal filter banks

In previous chapters for orthogonal wavelets, the analysis filters and synthesis filters are time reversal of each other; i.e., $\stackrel{˜}{h}\left(n\right)=h\left(-n\right)$ , $\stackrel{˜}{g}\left(n\right)=g\left(-n\right)$ . Here, for the biorthogonal case, we relax these restrictions. However, in order to perfectlyreconstruct the input, these four filters still have to satisfy a set of relations.

Let ${c}_{1}\left(n\right),n\in \mathbf{Z}$ be the input to the filter banks in [link] , then the outputs of the analysis filter banks are

${c}_{0}\left(k\right)=\sum _{n}\stackrel{˜}{h}\left(2k-n\right){c}_{1}\left(n\right),\phantom{\rule{2.em}{0ex}}{d}_{0}\left(k\right)=\sum _{n}\stackrel{˜}{g}\left(2k-n\right){c}_{1}\left(n\right).$

The output of the synthesis filter bank is

${\stackrel{˜}{c}}_{1}\left(m\right)=\sum _{k}\left[h,\left(2k-m\right),{c}_{0},\left(k\right),+,g,\left(2k-m\right),{d}_{0},\left(k\right)\right].$

Substituting Equation  [link] into [link] and interchanging the summations gives

${\stackrel{˜}{c}}_{1}\left(m\right)=\sum _{n}\sum _{k}\left[h,\left(2k-m\right),\stackrel{˜}{h},\left(2k-n\right),+,g,\left(2k-m\right),\stackrel{˜}{g},\left(2k-n\right)\right]{c}_{1}\left(n\right).$

For perfect reconstruction, i.e., ${\stackrel{˜}{c}}_{1}\left(m\right)={c}_{1}\left(m\right),\forall m\in \mathbf{Z}$ , we need

$\sum _{k}\left[h,\left(2k-m\right),\stackrel{˜}{h},\left(2k-n\right),+,g,\left(2k-m\right),\stackrel{˜}{g},\left(2k-n\right)\right]=\delta \left(m-n\right).$

Fortunately, this condition can be greatly simplified. In order for it to hold, the four filters have to be related as [link]

$\stackrel{˜}{g}\left(n\right)={\left(-1\right)}^{n}h\left(1-n\right),\phantom{\rule{2.em}{0ex}}g\left(n\right)={\left(-1\right)}^{n}\stackrel{˜}{h}\left(1-n\right),$

up to some constant factors. Thus they are cross-related by time reversal and flipping signs of every other element. Clearly, when $\stackrel{˜}{h}=h$ , we get the familiar relations between the scaling coefficients and the wavelet coefficients for orthogonal wavelets, $g\left(n\right)={\left(-1\right)}^{n}h\left(1-n\right)$ . Substituting [link] back to [link] , we get

$\sum _{n}\stackrel{˜}{h}\left(n\right)h\left(n+2k\right)=\delta \left(k\right).$

In the orthogonal case, we have ${\sum }_{n}h\left(n\right)h\left(n+2k\right)=\delta \left(k\right)$ ; i.e., $h\left(n\right)$ is orthogonal to even translations of itself. Here $\stackrel{˜}{h}$ is orthogonal to $h$ , thus the name biorthogonal .

Equation  [link] is the key to the understanding of the biorthogonal filter banks. Let's assume $\stackrel{˜}{h}\left(n\right)$ is nonzero when ${\stackrel{˜}{N}}_{1}\le n\le {\stackrel{˜}{N}}_{2}$ , and $h\left(n\right)$ is nonzero when ${N}_{1}\le n\le {N}_{2}$ . Equation  [link] implies that [link]

${N}_{2}-{\stackrel{˜}{N}}_{1}=2k+1,\phantom{\rule{1.em}{0ex}}{\stackrel{˜}{N}}_{2}-{N}_{1}=2\stackrel{˜}{k}+1,\phantom{\rule{2.em}{0ex}}k,\stackrel{˜}{k}\in \mathbf{Z}.$

In the orthogonal case, this reduces to the well-known fact that the length of $h$ has to be even.  [link] also imply that the difference between the lengths of $\stackrel{˜}{h}$ and $h$ must be even. Thus their lengths must be both even or both odd.

## Biorthogonal wavelets

We now look at the scaling function and wavelet to see how removing orthogonality and introducing a dual basis changes their characteristics.We start again with the basic multiresolution definition of the scaling function and add to that a similar definition of a dual scaling function.

$\Phi \left(t\right)=\sum _{n}h\left(n\right)\sqrt{2}\Phi \left(2t-n\right),$
$\stackrel{˜}{\Phi }\left(t\right)=\sum _{n}\stackrel{˜}{h}\left(n\right)\sqrt{2}\stackrel{˜}{\Phi }\left(2t-n\right).$

From Theorem  [link] in  Chapter: The Scaling Function and Scaling Coefficients, Wavelet and Wavelet Coefficients , we know that for $\phi$ and $\stackrel{˜}{\phi }$ to exist,

$\sum _{n}h\left(n\right)\phantom{\rule{0.277778em}{0ex}}=\phantom{\rule{0.277778em}{0ex}}\sum _{n}\stackrel{˜}{h}\left(n\right)\phantom{\rule{0.277778em}{0ex}}=\phantom{\rule{0.277778em}{0ex}}\sqrt{2}.$

Continuing to parallel the construction of the orthogonal wavelets, we also define the wavelet and the dual wavelet as

$\psi \left(t\right)=\sum _{n}g\left(n\right)\sqrt{2}\Phi \left(2t-n\right)=\sum _{n}{\left(-1\right)}^{n}\stackrel{˜}{h}\left(1-n\right)\sqrt{2}\Phi \left(2t-n\right),$
$\stackrel{˜}{\psi }\left(t\right)=\sum _{n}\stackrel{˜}{g}\left(n\right)\sqrt{2}\stackrel{˜}{\Phi }\left(2t-n\right)=\sum _{n}{\left(-1\right)}^{n}h\left(1-n\right)\sqrt{2}\stackrel{˜}{\Phi }\left(2t-n\right).$

Now that we have the scaling and wavelet functions and their duals, the question becomes whether we can expand and reconstruct arbitrary functionsusing them. The following theorem [link] answers this important question.

Theorem 37 For $\stackrel{˜}{h}$ and $h$ satisfying [link] , suppose that for some C, $ϵ>0$ ,

$|\text{Φ}\left(\text{ω}\right)|\le C{\left(1+\text{ω}\right)}^{-1/2-ϵ},\phantom{\rule{2.em}{0ex}}|\stackrel{˜}{\text{Φ}}\left(\text{ω}\right)|\le C{\left(1+\text{ω}\right)}^{-1/2-ϵ}.$

If $\Phi$ and $\stackrel{˜}{\Phi }$ defined above have sufficient decay in the frequency domain, then ${\psi }_{j,k}\stackrel{\mathrm{def}}{=}{2}^{j/2}\psi \left({2}^{j}x-k\right),\phantom{\rule{0.277778em}{0ex}}j,k\in \mathbf{Z}$ constitute a frame in ${L}^{2}\left(\mathbf{R}\right)$ . Their dual frame is given by ${\stackrel{˜}{\psi }}_{j,k}\stackrel{\mathrm{def}}{=}{2}^{j/2}\stackrel{˜}{\psi }\left({2}^{j}x-k\right),\phantom{\rule{0.277778em}{0ex}}j,k\in \mathbf{Z}$ ; for any $f\in {L}^{2}\left(\mathbf{R}\right)$ ,

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