As a generalization of orthogonal wavelets, the biorthogonal wavelets have become a fundamental tool in many areas of applied mathematics, from signal processing to numerical analysis [2, 8, 12, 13]. It is well known that 2-band orthogonal wavelet suffers from severe constraints, such as nontrivial symmetric (linear-phase) 2-band orthogonal wavelet does not exist [1]. Fortunately, multiwavelets and multiband wavelets with linear phase are designed as alternatives for more freedom and flexibility [2-5, 9, 10, 12]. (Bi)Orthogonal real-valued wavelets with symmetry and dilations being greater than two have been reported in [3-7, 9-14]. For example, symmetric compactly supported orthogonal real-valued wavelets with dilation factor 3 have been obtained by the method of undetermined coefficient and some special treatments in Chui and Lian [3]. By complicated calculation, only some examples of compactly supported $C^1$ symmetric orthogonal real-valued wavelets with dilation factor 4 have been obtained in [4].

Under the framework of MRA, the construction of multiwavelet systems having some desirable properties say, (bi)orthogonality, symmetry, regularity, and so on can be reduced to two main parts: One firstly constructs dual scaling filters and functions and then the wavelet filters and wavelets, which should be able to inherit certain properties similar to those of their scaling filters. Therefore, in order to obtain a wavelet system, the linchpin is to design the scaling functions with good properties. It is well known that the second part can be formulated as a matrix extension problem, see [7, 17]. Goh and Yap in [15] studied the biorthogonal matrix extension problem and presented a step-by-step algorithm for deriving the extension matrices. Yet neither did they concern about the lengths of the coefficient supports of the extension matrices, nor did they considered any symmetry constrain on the extension matrices. In [16], Chui, Han and Zhuang proposed a dual chain approach for this problem, which first constructs a top-down dual-chain that essentially reduces the lengths of the coefficient supports of the given pair of vectors to zero and then derives a bottom-up dual-chain that produces the desired pair of extension matrices. However, the multiwavelet systems have unsolved questions in many applied areas such as image compression coding. In view of this, further study for multi-band wavelet systems is still valuable.

In this paper, we construct the 3-band wavelets, which is perfect reconstruction, biorthogonal, symmetric/antisymmetric, the high vanishing moments and compactly supported. First, by designating the length of the filters and the vanishing moments, we construct the scaling filters with high vanishing moments and symmetry by solving nonlinear equations with less parameters. Then, we provide a step-by-step algorithm to construct a class of biorthogonal 3-band wavelets with symmetry.

In this section, we shall construct the scaling function with desired properties and propose an algorithm for obtaining symbols of the corresponding 3-band wavelets. First, we discuss several results that are useful in the following.

As the case of dyadic wavelet, a pair $(\phi(x), \widetilde{\phi}(x))$ of dual scaling functions can be expressed as the following dilation equations

$ \begin{equation}\phi(x)=\sum\limits_{k\in Z}h_0(k)\phi(3x-k)\end{equation} $

(2.1)

and

$ \begin{equation}\widetilde{\phi}(x)=\sum\limits_{k\in Z}\widetilde{h}_0(k)\widetilde{\phi}(3x-k), \end{equation} $

(2.2)

respectively, with

$ \begin{equation}\langle \phi(.-k), \widetilde{\phi}(.-j) \rangle=\delta(k-j), \end{equation} $

(2.3)

or equivalently,

$ \begin{eqnarray*}\sum\limits_k h_0(k)\widetilde{h}_0(k+3j)=3\delta(j), \end{eqnarray*} $

(2.4)

where $h, g$ are real number, $\delta(j)$ denotes the Dirac sequence such that $\delta(j)=1 $ for $j=0$ otherwise $\delta(j)=0 $.

The polyphase decomposition of $H_0(z)$ and $\widetilde{H}_0(z)$ are defined by $H_{0, i}(z)=\frac{1}{\sqrt{3}}\sum_{k}h_0(i+3k)z^k, i=0, 1, 2$ and $\widetilde{H}_{0, i}(z)=\frac{1}{\sqrt{3}}\sum_{k}\widetilde{h}_0(i+3k)z^k, i=0, 1, 2$, respectively. It is well known that (2.4) is equivalent to the following equation

$ \begin{equation} H_{0, 0}(z)\widetilde{H}_{0, 0}(z^{-1})+H_{0, 1}(z)\widetilde{H}_{0, 1}(z^{-1})+H_{0, 2}(z)\widetilde{H}_{0, 2}(z^{-1})=1.\end{equation} $

(2.5)

In addition, we assume

$ \begin{eqnarray*} \sum\limits_kh_0(k)=\sum\limits_k\widetilde{h}_0(k)=3. \end{eqnarray*} $

(2.6)

Definition 2.1  The filter $h_i$ is symmetric/antisymmetric if and only if

$ \begin{equation}h_i(k)=\pm h_i(N-1-k)(0\leq k < N), \end{equation} $

(2.7)

where $ N $ is the length of the filter. Equivalently,

$ \begin{equation}H_i(z)=\pm z^{(N-1)}H_i(z^{-1}), \end{equation} $

(2.8)

where $H_i(z)=\sum_{k\in Z}h_i(k)z^k $.

In [18], Zhang provided an effective algorithm to construct compactly supported symmetric biorthogonal multiwavelets by given a pair of biorthogonal matrices. Inspired by this idea, we shall provide a step-by-step algorithm to construct a class of compactly supported biorthogonal 3-band wavelets with symmetry. First, we need the following theorem.

Theorem 2.2  Suppose that $(A(z), \widetilde{A}(z))$ be a pair of $1\times 3$ Laurent polynomial vectors such that

$ \begin{equation}A(z)\widetilde{A}^T(z^{-1})=1, \end{equation} $

(2.9)

and

$ \begin{eqnarray} && A_{1}(z)=A_{1}(z^{-1}), A_{2}(z)=A_{2}(z^{-1}), A_{3}(z)=-A_{3}(z^{-1}), \nonumber\\ && \widetilde{A}_{1}(z)=\widetilde{A}_{1}(z^{-1}), \widetilde{A}_{2}(z)=\widetilde{A}_{2}(z^{-1}), \widetilde{A}_{3}(z)=-\widetilde{A}_{3}(z^{-1}), \end{eqnarray} $

(2.10)

where $A(z)=(A_{1}(z), A_{2}(z), A_{3}(z))$, $\widetilde{A}(z)=(\widetilde{A}_{1}(z), \widetilde{A}_{2}(z), \widetilde{A}_{3}(z))$.

Let their coefficient support be csupp $(A)=[-k, k]$, csupp $(\widetilde{A})=[-\widetilde{k}, \widetilde{k}]$, respectively. Then $A(z)$, and $\widetilde{A}(z)$ can be written as

$ \begin{array}{c} A(z)=(a_0, b_0, c_0)z^{-k}+(a_1, b_1, c_1)z^{1-k}+\cdots+(a_1, b_1, -c_1)z^{k-1}+(a_0, b_0, -c_0)z^{k}, \\ \widetilde{A}(z)=(\widetilde{a}_0, \widetilde{b}_0, \widetilde{c}_0)z^{-\widetilde{k}}+(\widetilde{a}_1, \widetilde{b}_1, \widetilde{c}_1)z^{1-\widetilde{k}}+\cdots +(\widetilde{a}_1, \widetilde{b}_1, -\widetilde{c}_1)z^{\widetilde{k}-1}+(\widetilde{a}_0, \widetilde{b}_0, -\widetilde{c}_0)z^{\widetilde{k}}. \\ \end{array} $

Assume that $a_0, b_0, c_0, \widetilde{a}_0, \widetilde{b}_0, \widetilde{c}_0$ are nonzero. Then there exists a pair $(u(z), \widetilde{u}(z))$ of $3 \times 3$ Laurent polynomial matrices with symmetry such that

(a) $(u(z), \widetilde{u}(z))$ are biorthogonal: $u(z) \widetilde{u}^T(1/z))=I_3$ and the coefficient supports of $u(z), \widetilde{ u}(z)$ satisfy csupp $(u) \subseteq [-1,1]$ and csupp $(\widetilde{u}) \subseteq [-1,1]$, respectively.

(b) $u(z)$, $\widetilde{u}(z)$ reduce the lengths of the coefficient support of $A(z)$ and $\widetilde{A}(z)$, respectively.

(c) $B(z)=A(z)u_(z)$, $\widetilde{B}(z)=\widetilde{A}(z)\widetilde{u}_(z)$ satisfies (2.9) and (2.10).

Proof  Let us recall the theory of symmetric polynomial operations.

Assume that Laurent polynomials $f_1(z)$, $f_2(z)$ are symmetric about the points $l_1$, $l_2$, respectively, and $f_3(z)$, $f_4(z)$ are antisymmetric about the points $l_3$, $l_4$, respectively. Then

(ⅰ) $f_1(z)f_2(z)$, $f_3(z)f_4(z)$ are symmetric about the points $l_1+l_2$, $l_3+l_4$, respectively, and $f_1(z)f_3(z)$, $f_2(z)f_4(z)$ are antisymmetric about the points $l_1+l_3$, $l_2+l_4$, respectively.

(ⅱ) If $l_1=l_2$, then $f_1(z)+f_2(z)$ is symmetric about the point $l_1$; If $l_3=l_4$, then $f_3(z)+f_4(z)$ is antisymmetric about the points $l_3$.

Form (2.9), we have

$ a_0\widetilde{a}_0+b_0\widetilde{b}_0-c_0\widetilde{c}_0=0. $

Define

$ \begin{eqnarray} && q_1(z)=\frac{1}{2\sqrt{|c_0\widetilde{c}_0|}}\left[\begin{array}{ccc} \widetilde{a}_0(z^{-1}+1)& 2b_0& \widetilde{a}_0(z^{-1}-1)\\ \widetilde{b}_0(z^{-1}+1)&-2a_0&\widetilde{b}_0(z^{-1}-1)\\ -\widetilde{c}_0(z^{-1}-1)& 0 &-\widetilde{c}_0(z^{-1}+1) \end{array}\right], \nonumber\\ && \widetilde{q}_1(z)={\rm sgn}(c_0\widetilde{c}_0)\frac{1}{2\sqrt{|c_0\widetilde{c}_0|}}\left[\begin{array}{ccc} a_0(z^{-1}+1)& 2\widetilde{b}_0& a_0(z^{-1}-1)\\ b_0(z^{-1}+1)&-2\widetilde{a}_0&b_0(z^{-1}-1)\\ -c_0(z^{-1}-1)& 0 &-c_0(z^{-1}+1) \end{array}\right], \end{eqnarray} $

(2.11)

where sgn $(x)$ is a sign function such that sgn $(x)=1$ for $x>0$, sgn $(x)=0$ for $x=0$ and sgn $(x)=-1 $ for $x<0$.

By direct calculation, we can see that $q_1(z)$ and $\widetilde{q}_1(z)$ are biorthogonal. It is obvious that the columns of $q_1(z)$ are symmetric/antisymmetric about $-1/2, 0, -1/2$, respectively. Since

$ (a_0, b_0, -c_0)z^{k}q_1(z)=\frac{1}{2\sqrt{|c_0\widetilde{c}_0|}}(a_0\widetilde{a}_0+b_0\widetilde{b}_0+c_0\widetilde{c}_0)z^{k-1}(1, 0, 1) $

and

$ (a_0, b_0, c_0)z^{-k}q_1(z)=\frac{1}{2\sqrt{|c_0\widetilde{c}_0|}}(a_0\widetilde{a}_0+b_0\widetilde{b}_0+c_0\widetilde{c}_0)z^{-k}(1, 0, -1), $

we can see that $q_1(z)$ reduces the length of the coefficient support of $A(z)$ by 1. Note that $A_1(z), A_2(z), \widetilde{A}_1(z)$, $\widetilde{A}_2(z)$ in (2.10) are all symmetric about origin and $A_3(z), \widetilde{A}_3(z)$ are both antisymmetric about origin. It is easy to see that the columns of $A(z)q_1(z)$ are symmetric/antisymmetric about $-1/2, 0, -1/2$, respectively. It is similar to $\widetilde{A}(z)\widetilde{q}_1(z)$.

Therefore, without loss of generality, we suppose that

$ \begin{array}{c} A(z)q_1(z)=(d_0, 0, f_0)z^{-k}+\cdots+(d_0, e_1, -f_0)z^{k-1}, \\ \widetilde{A}(z)\widetilde{q}_1(z)=(f_0, 0, d_0)z^{-\widetilde{k}}+\cdots +(f_0, \widetilde{e}_1, -d_0)z^{\widetilde{k}-1}.\\ \end{array} $

Define

$ \begin{eqnarray} && q_2(z)=\frac{1}{2\sqrt{|d_0f_0|}}\left[\begin{array}{ccc} f_0(z+1)& 0&-f_0(z-1)\\ 0& 2d_0&0\\ d_0(z-1)& 0 &-d_0(z+1) \end{array}\right], \nonumber\\ && \widetilde{q}_2(z)={\rm sgn}(d_0f_0)\frac{1}{2\sqrt{|d_0f_0|}}\left[\begin{array}{ccc} d_0(z+1)& 0&-d_0(z-1)\\ 0& 2f_0&0\\ f_0(z-1)& 0 &-f_0(z+1) \end{array}\right].\end{eqnarray} $

(2.12)

Clearly, $q_2(z)$ and $\widetilde{q}_2(z)$ are biorthogonal, and the columns of $q_2(z)$ are symmetric/antisymmetric about $1/2, 0, 1/2$, respectively.

Since $(d_0, 0, f_0)z^{-k}q_2(z)=\frac{d_0f_0}{\sqrt{|d_0f_0|}}z^{-k+1}(1, 0, -1)$ and

$ (d_0, e_0, -f_0)z^{k-1}q_2(z)=\frac{1}{\sqrt{|d_0f_0|}}z^{k-1}(d_0f_0, d_0e_0, d_0f_0), $

we have $q_2(z)$ reduces the length of the coefficient support of $A(z)q_1(z)$ by 1. Moreover, the columns of $A(z)q_1(z)q_2(z)$ are symmetric/antisymmetric to $0, 0, 0$, respectively. It is similar to $\widetilde{A}(z)\widetilde{q}_1(z)\widetilde{q}_2(z)$.

In summary, let $u(z)=q_1(z)q_2(z) $ and $\widetilde{u}(z)=\widetilde{q}_1(z)\widetilde{q}_2(z) $. Then $(u(z), \widetilde{u}(z))$ satisfy (a), (b) and (c). Moreover, $u(z)$ and $\widetilde{u}(z)$ reduce the lengths of the coefficient support of $A(z)$ and $\widetilde{A}(z)$ by 2, respectively. That is, csupp $(A(z)u(z))= [-k+1, k-1]$ and csupp $(\widetilde{A}(z)\widetilde{u}(z)) \subseteq [-\widetilde{k}+1, \widetilde{k}-1]$, respectively.

According to Theorem 2.2, we have a matrix extension algorithm for the polyphase vectors, see Algorithm 1.

Algorithm 1  (1) Let $(H_0(z), \widetilde{H}_0(z))$ be a pair of dual scaling symbols satisfying (2.4), (2.6).

(2) Define

$ \begin{eqnarray} && A_0(z)=z^{-[N_1/6 ]}\left(H_{0, 0}(z), H_{0, 1}(z), H_{0, 2}(z)\right)\frac{1}{\sqrt{2}}\left[\begin{array}{ccc} 1& 0& 1\\ 0&\sqrt{2}&0\\ 1& 0 &-1 \end{array}\right], \nonumber \\ && \widetilde{A}_0(z)=z^{-[N_2/6 ]}\left(\widetilde{H}_{0, 0}(z), \widetilde{H}_{0, 1}(z), \widetilde{H}_{0, 2}(z)\right)\frac{1}{\sqrt{2}}\left[\begin{array}{ccc} 1& 0& 1\\ 0&\sqrt{2}&0\\ 1& 0 &-1 \end{array}\right], \end{eqnarray} $

(2.13)

where $[x]$ denotes the largest integer less than or equal to $x$ for $x \in R$, $N_1$ and $N_2$ are the lengths of the filters $h_0$ and $\widetilde{h}_0$, respectively. Then $(A_0(z), \widetilde{A}_0(z))$ satisfy (2.9) and (2.10).

(3) Recursively applying (2.11) and (2.12) until

$ \begin{equation} A_K(z)=A_{K-1}(z)u_{{K-1}}(z), \widetilde{A}_K(z)=\widetilde{A}_{K-1}(z)\widetilde{u}_{{K-1}}(z), \end{equation} $

(2.14)

and deg $(A_K(z))=0 $, deg $(\widetilde{A}_K(z))\geq 0 $.

Moreover, $A_K(z)=(1, 0, 0)$, $\widetilde{A}_K(z))= (1, p_1(z), p_2(z))$ for Laurent polynomials $p_1(z)$ and $p_2(z)$ with symmetry. Define

$ \begin{equation} \begin{array}{c} u_K(z)=\left[\begin{array}{ccc} 1& 0& 0\\ p_1(z^{-1})&1&0\\ p_2(z^{-1})& 0 &1 \end{array}\right], \widetilde{u}_K(z)=\left[\begin{array}{ccc} 1&-p_1(z)&-p_2(z)\\ 0&1&0\\ 0& 0 &1 \end{array}\right]. \end{array}\end{equation} $

(2.15)

By direct calculation, it is obvious that $(u_K(z), \widetilde{u}_K(z))$ are biorthogonal and $A_K(z)u_K(z)=\widetilde{A}_K(z)\widetilde{u}_K(z)=(1, 0, 0)$.

(4) Define

$ \begin{array}{c} U(z)=\widetilde{u}_{K}^T(z^{-1})\widetilde{u}_{{K-1}}^T(z^{-1})\cdots \widetilde{u}_{0}^T(z^{-1}), \\ \widetilde{U}(z)=u_{K}^T(z^{-1})u_{{K-1}}^T(z^{-1})\cdots u_{0}^T(z^{-1}). \end{array} $

The wavelet symbols can be derived by

$ \begin{array}{c} \left[\begin{array}{ccc} H_0(z)& \\ H_1(z)\\ H_2(z) \end{array}\right]=z^{[N/6]}U(z^{3})\frac{1}{\sqrt{2}}\left[\begin{array}{ccc} 1& 0& 1\\ 0&\sqrt{2}&0\\ 1& 0 &-1 \end{array}\right] \left[\begin{array}{ccc} 1& \\ z\\ z^2 \end{array}\right], \\ \\ \left[\begin{array}{ccc} \widetilde{H}_0(z)& \\ \widetilde{H}_1(z)\\ \widetilde{H}_2(z) \end{array}\right]=z^{[N/6]}\widetilde{U}(z^{3})\frac{1}{\sqrt{2}}\left[\begin{array}{ccc} 1& 0& 1\\ 0&\sqrt{2}&0\\ 1& 0 &-1 \end{array}\right] \left[\begin{array}{ccc} 1& \\ z\\ z^2 \end{array}\right], \end{array} $

(2.16)

where $N=\max\{N_1, N_2\}$.

In this section, some examples are given to verify this scheme. we only discuss the examples of odd lengths 3-band scaling filters.

Example 3.1  Let the lengths of scaling filters be $(15, 9)$. Assume that the scaling symbols $(H_0(z), \widetilde{H}_0(z)) $ have the following form

$ \begin{array}{c} H_0(z)=\left(\frac{1+z+z^2}{3}\right)^5Q(z), \\ \widetilde{H}_0(z)=\left(\frac{1+z+z^2}{3}\right)^3\widetilde{Q}(z), \\ \end{array} $

(3.1)

where $(Q(z), \widetilde{Q}(z))$ are symmetric Laurent polynomials with degree $(4, 2)$.

It is obvious that $H_0(z)$ is symmetric if and only if $Q(z)$ is symmetric. By (2.4) and (2.6), we can obtain the associated scaling filters as follows:

$ \begin{eqnarray*} h_0& \approx& [ 0.0302708750, 0.0197271260, 0.0109853080, \\ &&-0.12261759, 0.011382944, 0.24928687, 0.83207454, 0.93777986], \\ \widetilde{h}_0& \approx& [-0.20140256, -0.090291448, 0.13193077, 1.0694718, 1.1805829]\end{eqnarray*} $

(whereas the other half is symmetric and so skipped). Thus, the polyphase decomposition of $H_0(z)$ and $\widetilde{H}_0(z)$ are as follows:

$ \left[\begin{array}{ccc} H_{0, 0}& H_{0, 1}&H_{0, 2}\\ \end{array}\right], \left[\begin{array}{ccc} \widetilde{H}_{0, 0}& \widetilde{H}_{0, 1}&\widetilde{H}_{0, 2} \end{array}\right], $

where

$ \begin{eqnarray*}&& H_{0, 0}\approx 0.0063423705z^4 + 0.14392584z^3 + 0.48039846z^2 \\ &&- 0.070793301z + 0.017476898, \\ && H_{0, 1}\approx0.011389462z^4 \\ &&+ 0.0065719459z^3 + 0.54142745z^2 + 0.0065719459z + 0.011389462, \\ && H_{0, 2}\approx0.017476898z^4 - 0.070793301z^3 + 0.48039846z^2 \\ &&+ 0.14392584z + 0.0063423705, \\ && \widetilde{H}_{0, 0}\approx 0.076170268z^2 + 0.61745982z - 0.11627982, \\ &&\widetilde{H}_{0, 1}\approx-0.052129792z^2 + 0.68160985z - 0.052129792, \\ && \widetilde{H}_{0, 2}\approx-0.11627982z^2+0.61745982z+0.076170268.\end{eqnarray*} $

Define

$ \begin{eqnarray*} && q_1(z)=\left[\begin{array}{ccc} 0.43323436(z^{-1}+1)&-0.34795513&0.43323436(z^{-1}-1)\\ 0.79629877(z^{-1}+1)& 0.51455697& 0.79629877(z^{-1}-1)\\ -2.0787065(z^{-1}-1)& 0&-2.0787065(z^{-1}+1) \end{array}\right], \\ && \widetilde{q}_1(z)=\left[\begin{array}{ccc} 0.25727848(z^{-1}+1)&-1.5925975& 0.25727848(z^{-1}-1)\\ 0.17397756/z(z^{-1}+1)&0.86646871&0.17397756(z^{-1}-1)\\ -0.12026710(z^{-1}-1)&0&-0.12026710(z^{-1}+1) \end{array}\right]. \end{eqnarray*} $

We have

$ \begin{eqnarray*}A_0(z)q_1(z)&\approx& [0.37597811z+0.43749957+0.43749957/z+0.37597811/z^2, \\ &&-0.014611995z+0.042199421-0.014611995/z, \\ &&-0.31051300z-1.01344319+1.01344319/z+0.31051300/z^2], \\ \widetilde{A}_0(z)\widetilde{q}_1(z)&\approx&[0.31051300+0.31051300/z, -0.80009451, \\ &&-0.37597811+0.37597811/z].\end{eqnarray*} $

It follows that $d_0=0.31051300, f_0=0.37597811$ in (2.12). Thus,

$ \begin{eqnarray*}A_0(z)q_1(z)q_2(z)&\approx& [1.51276001-0.017108015z-0.017108015/z, \\ &&-0.016078699z+ 0.046435260-0.016078699/z, \\&& 0.70047056z-0.70047056/z], \\ \widetilde{A}_0(z)\widetilde{q}_1(z)\widetilde{q}_2(z)&\approx& [0.68336254, -0.72710963, 0].\end{eqnarray*} $

Thus, there exist

$ \begin{array}{c} v=\left[\begin{array}{ccc} 0.68336254&-0.72710963& 0\\ -0.72710963&-0.68336254& 0\\ 0& 0& 1 \end{array}\right], v^{-1}=\left[\begin{array}{ccc} 0.68633245&-0.73026967& 0\\ -0.73026967&-0.68633245& 0\\ 0& 0& 1 \end{array}\right], \end{array} $

such that $A_0(z)q_1(z)q_2(z)v= (1, p_1(z), p_2(z))$, $\widetilde{A}_0(z)\widetilde{q}_1(z)\widetilde{q}_2(z)v^{-1}=(1, 0, 0)$, where

$ \begin{eqnarray*}&& p_1(z)\approx-1.13167449285+0.0234269828611z+0.0234269828611z^{-1}, \\ && p_2(z)\approx0.700470556308z-0.700470556308z^{-1}. \end{eqnarray*} $

Therefore, we can obtain $(u_0(z), \widetilde{u}_0(z))$ in (2.14) and $(u_1(z), \widetilde{u}_1(z))$ in (2.15).

We derive the polyphase matrices of high-pass filter banks as follows:

$ \left[\begin{array}{ccc} H_{1, 0}& H_{1, 1}&H_{1, 2}\\ H_{2, 0}& H_{2, 1}&H_{2, 2} \end{array}\right], \left[\begin{array}{ccc} \widetilde{H}_{1, 0}& \widetilde{H}_{1, 1}&\widetilde{H}_{1, 2}\\ \widetilde{H}_{2, 0}& \widetilde{H}_{2, 1}&\widetilde{H}_{2, 2} \end{array}\right], $

where

$ \begin{eqnarray*}&& H_{1, 0}\approx-0.006778z^3 + 0.4355z^2 - 0.01868z, \\ && H_{1, 1}\approx-0.01217z^3-0.7957z^2-0.01217z, H_{1, 2}\approx-0.01868z^3+0.4355z^2\\ &&-0.006778z, \\ && H_{2, 0}\approx0.009281z^3 + 0.1709z^2 - 0.02557z, H_{2, 1}\approx0.01667z^3 - 0.01667z, \\ && H_{2, 2}\approx0.02557z^3 - 0.1709z^2 -0.009281z, \\ && \widetilde{H}_{1, 0}\approx-0.001784z^4-0.009312z^3+0.4372z^2-0.02233z+0.002724, \\ && \widetilde{H}_{1, 1}\approx0.001221z^4-0.0195z^3-0.7764z^2-0.0195z+0.001221, \\ && \widetilde{H}_{1, 2}\approx 0.002724z^4-0.02233z^3+0.4372z^2-0.009312z-0.001784, \\ && \widetilde{H}_{2, 0}\approx0.05336z^4+0.544z^3+2.818z^2-0.2624z+0.08145, \\ && \widetilde{H}_{2, 1}\approx -0.03652z^4+0.4012z^3-0.4012z+0.03652, \\ && \widetilde{H}_{2, 2}\approx -0.08145z^4+0.2624z^3-2.818z^2-0.544z-0.05336.\end{eqnarray*} $

From (2.16), the wavelet filters $h_1, h_2$ and $\widetilde{h}_1, \widetilde{h}_2$ can be obtained as follows:

$ \begin{eqnarray*} h_1&\approx&[0.032348717, 0.021081228, 0.011739358, -0.75426796, 1.3781973], \\ h_2&\approx& 4[-0.044296947, -0.028867731, -0.016075373, 0.29594173, 0], \\ \widetilde{h}_1&\approx& [-0.0047182543, -0.0021152562, 0.0030907400, 0.038680809, 0.033766352, \\ && 0.016128444, -0.75722871, 1.3447918], \\ \widetilde{h}_2&\approx& [0.14107656, 0.063246501, -0.092413623, -0.45441064, -0.69483538, \\ &&-0.94219467, 4.8815912, 0]/4\end{eqnarray*} $

(whereas the other half is symmetric/antisymmetric and so skipped). See Fig. 1 for the graphs of the scaling functions and wavelets in this example.

Example 3.2  Let the lengths of scaling filters be $(21, 15)$. Assume that the scaling symbols $(H_0(z), \widetilde{H}_0(z)) $ have the following form

$ \begin{array}{c} H_0(z)=\left(\frac{1+z+z^2}{3}\right)^7Q(z), \\ \widetilde{H}_0(z)=\left(\frac{1+z+z^2}{3}\right)^5\widetilde{Q}(z), \\ \end{array} $

(3.2)

where $(Q(z), \widetilde{Q}(z))$ are symmetric Laurent polynomials with degree $(6, 4)$. Thus,

$ \begin{eqnarray*} h_0& \approx& [-0.0066877269, -0.0049799640, \\ &&-0.0036014177, 0.041559784, 0.024297155, \\ &&-0.0036014177, -0.10184906, 0.0418949921, 0.28283863, \\ &&0 .77648664, 0.87757563], \\ \widetilde{h}_0&\approx&[0.045518221, 0.014468319, -0.035285806, -0.30664390, \\ &&-0.15213721, 0.18829749, \\&& 1.1081140, 1.2753378].\end{eqnarray*} $

We can obtain the polyphase decomposition of $H_0(z)$ and $\widetilde{H}_0(z)$ are as follows:

$ \left[\begin{array}{ccc} H_{0, 0}& H_{0, 1}&H_{0, 2}\\ \end{array}\right], \left[\begin{array}{ccc} \widetilde{H}_{0, 0}& \widetilde{H}_{0, 1}&\widetilde{H}_{0, 2} \end{array}\right], $

where

$ \begin{eqnarray*}&& H_{0, 0}\approx-0.002079z^6+0.006497z^5+0.1633z^4+0.4483z^3-0.0588z^2\\ &&+0.02399z-0.003861, \\ &&H_{0, 1}\approx -0.002875z^6 + 0.01403z^5 + 0.02419z^4 + 0.5067z^3\\ && + 0.02419z^2 + 0.01403z - 0.002875, \\ && H_{0, 2}\approx - 0.003861z^6 + 0.02399z^5 - 0.0588z^4 + 0.4483z^3\\ && + 0.1633z^2 + 0.006497z - 0.002079, \\ && \widetilde{H}_{0, 0}\approx - 0.02037z^4 + 0.1087z^3 + 0.6398z^2 - 0.1770z + 0.02628, \\ &&\widetilde{H}_{0, 1}\approx0.008353z^4 - 0.08784z^3 + 0.7363z^2 - 0.08784z + 0.008353, \\ && \widetilde{H}_{0, 2}\approx 0.02628z^4 - 0.1770z^3 + 0.6398z^2 + 0.1087z - 0.02037.\end{eqnarray*} $

Using the same method, we derive the polyphase matrices of high-pass filter banks as follows:

$ \left[\begin{array}{ccc} H_{1, 0}& H_{1, 1}&H_{1, 2}\\ H_{2, 0}& H_{2, 1}&H_{2, 2} \end{array}\right], \left[\begin{array}{ccc} \widetilde{H}_{1, 0}& \widetilde{H}_{1, 1}&\widetilde{H}_{1, 2}\\ \widetilde{H}_{2, 0}& \widetilde{H}_{2, 1}&\widetilde{H}_{2, 2} \end{array}\right], $

where

$ \begin{eqnarray*} H_{1, 0}&\approx& 0.0008741z^5 - 0.008091z^4 + 0.3279z^3 - 0.02004z^2 + 0.001623z, \\ H_{1, 1}&\approx&0.001209z^5 - 0.01331z^4 - 0.5803z^3 - 0.01331z^2 + 0.001209z, \\ H_{1, 2}& \approx& 0.001623z^5 - 0.02004z^4 + 0.3279z^3 - 0.008091z^2 + 0.0008741z, \\ H_{2, 0}&\approx&- 0.002057z^5 + 0.008632z^4 + 0.1253z^3 - 0.02783z^2 + 0.003821z, \\ H_{2, 1}& \approx&- 0.002845z^5 + 0.01693z^4 - 0.01693z^2 + 0.002845z, \\ H_{2, 2}& \approx&- 0.003821z^5 + 0.02783z^4 - 0.1253z^3 - 0.008632z^2 + 0.002057z, \\ \widetilde{H}_{1, 0}& \approx&0.0005821z^6 - 0.004607z^5 - 0.006087z^4 + 0.592z^3\\ && - 0.03672z^2 + 0.006994z - 0.0007509, \\ \widetilde{H}_{1, 1}& \approx&- 0.0002387z^6 + 0.003125z^5 - 0.02922z^4 - 1.05z^3 \\ &&- 0.02922z^2 + 0.003125z - 0.0002387, \\ \widetilde{H}_{1, 2}& \approx& - 0.0007509z^6 + 0.006994z^5 - 0.03672z^4 + 0.592z^3 \\ && - 0.006087z^2 - 0.004607z + 0.0005821, \\ \widetilde{H}_{2, 0}& \approx&- 0.02084z^6 + 0.09103z^5 + 0.9251z^4 + 3.782z^3 - 0.2686z^2 + 0.1551z \\ && - 0.02688, \\ \widetilde{H}_{2, 1}& \approx& 0.008544z^6 - 0.08157z^5 + 0.5993z^4 - 0.5993z^2 + 0.08157z - 0.008544, \\ \widetilde{H}_{2, 2}& \approx&0.02688z^6 - 0.1551z^5 + 0.2686z^4 - 3.782z^3 - 0.9251z^2 - 0.09103z + 0.02084.\end{eqnarray*} $

From (2.16), the wavelet filters $h_1, h_2$ and $\widetilde{h}_1, \widetilde{h}_2$ can be obtained as follows:

$ \begin{eqnarray*}h_1& \approx& [ -0.0028113764, -0.0020934697, -0.0015139585, 0.034710876, 0.023051661, \\ && 0.014014537, -0.56792580, 1.0051351], \\ h_2& \approx& 4[0.0066173484, 0.0049275573, 0.0035635181, -0.048208391, -0.029317970, \\&&-0.014950592, 0.21698428, 0], \\ \widetilde{h}_1& \approx& [0.0013005736, 0.00041339737, -0.0010082069, -0.012113711, -0.0054124454, \\ && 0.0079787011, 0.063598203, 0.050616897, 0.010543400, -1.0253619, 1.8188901], \\ \widetilde{h}_2& \approx& [-0.046554875, -0.014797828, 0.036089423, 0.26858835, 0.14128600, -0.15767139, \\ && -0.46526530, -1.0379341, -1.6023583, 6.5501164, 0]/4\end{eqnarray*} $

(whereas the other half is antisymmetric and so skipped). The graphs of the scaling functions and wavelets are similar to Example 3.1.