# Symmetry Feature and Construction for the 3-Band Tight Framelets with Prescribed Properties

Authors: Jianjun Sun; Bin Huang; Xiaodong Chen; Lihong Cui
Journal: Journal of Applied Mathematics
 (2012)

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## Abstract

A construction approach for the 3-band tight wavelet frames by factorization of paraunitary matrix is developed. Several necessary constraints on the filter lengths and symmetric features of wavelet frames are investigated starting at the constructed paraunitary matrix. The matrix is a symmetric extension of the polyphase matrix corresponding to 3-band tight wavelet frames. Further, the parameterizations of 3-band tight wavelet frames with3N+1 filter lengths are established. Examples of framelets with symmetry/antisymmetry and Sobolev exponent are computed by appropriately choosing the parameters in the scheme.


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## Body

## 1. Introduction

In the theory and applications of wavelets and wavelet frames, certain properties are always desirable. It is well known that symmetry and high vanishing moments are very important features of all wavelets. Symmetry, which is also called linear phase in the language of engineering, is claimed to improve the rate-distortion performance in image compression [1, 2]. On the one hand, parameterizations of FIR systems are of fundamental importance to the design of filters with the desired properties [1, 3]. On the other hand, the advantages of MRA-based tight wavelet frames and their promising features in applications have attracted a great deal of interest and effort in recent years to extensively study them (e.g., [4–15]). The main tools for construction and characterization of wavelet frames are the unitary extension principle (UEP) [16] and its versions generalized such as OEP and MEP [17]. They give sufficient conditions for constructing MRA-based tight and dual wavelet frames. Many authors have worked on the design of wavelet frames with good properties. Most deals with 2-band wavelet frames systems, and a few authors have studied M-band framelets [8]. M(≥2)-band wavelets have advantages over dyadic wavelets in some aspects. For example, it enables a finer frequency partitioning and can provide a more compact representation of signals [18–21].This paper deals with the construction of 3-band tight wavelet frames filters with prescribed properties using factorization and parameterizations of the paraunitary matrices. The parameter space describing 3-band wavelets is much richer than that in 2-band case; thus, it has greater freedom and flexibility. Concretely, with the describing of unitary extension principle (UEP) in the polyphase representation, we firstly construct a paraunitary matrix based on the polyphase matrix corresponding to compactly supported wavelet frames with the least number generators. Further, we establish necessary constraints on the filter lengths and symmetric features of wavelet frames. Then, we investigate the parameterizations of 3-band tight wavelet frames with3N+1 filter lengths. Finally, examples of 3-band wavelet frames with symmetry/antisymmetry and good smoothness are given by applying the proposed scheme.Throughout this paper, letN and Z denote the sets of all natural numbers and integers, respectively. Let A*, AT, and Tr(A) denote conjugate transpose, transpose and the trace of A, respectively. Let A=[aij]m×n, 1≤i≤m; 1≤j≤n denote the m×n matrix with elements aij, for convenience, we omit the subscript m×n when m=n. Let In denotes the n×n identity matrix and Jn denote the n×n exchange matrix with ones on the antidiagonal. For s≥0, we use Ws(R) denote the Sobolev space consisting of all functions f with (1+|ω|2)s/2f^(ω)∈L2(R). In this paper, we only consider compactly supported wavelet frame and causal sequence with real finite impulse response (FIR), such sequences can be identified with Laurent polynomial defined by h(z)=∑k=0K‍hkz-k, where 0 and K are the smallest and largest indices that hk is nonzero, respectively. Assumpsit: h(ω)=h(z)and z=eiω. We use leng(h)=K+1 to denote the filter length of h.


## 2. UEP of Tight Framelets in Terms of Polyphase Representation

In this section, we introduce some notation and state needed results that will be used later in this paper. Let refinable functionsψ0 with dilation factor 3 generate a multiresolution analysis (MRA) {Vj}j∈Z of L2(R) and Ψ={ψ1,…,ψr}⊂V1. Then for real-value sequence hℓ,k, ψℓ, ℓ=0,1,2,…,r satisfy
(2.1)ψℓ(x)=3∑k∈Z‍hℓ,kψ0(3x-k),
or equivalently
(2.2)ψ^ℓ(3ω)=hℓ,k(ω)ψ^0(ω),
where hℓ(ω)=∑k∈Z‍hℓ,ke-ikω or rewritten hℓ(z)=∑k∈Z‍hℓ,kz-k, z=e-iω. h0 is called a refinement mask or the low filter, and hℓ, ℓ=1,2,…,r are called wavelet masks or the high pass filters of the system, respectively. For notational convenience, refinement mask together with wavelet masks {h0,h1,…,hr} is also called combined MRA masks in [7].For givenΨ={ψ1,…,ψr}⊂L2(R), define the wavelet system as X(Ψ)={ψj,kℓ,ℓ=1,…,r;j,k∈Z}, where ψj,kℓ=3j/2ψℓ(3jx-k),j,k∈Z.Definition 2.1.
The systemX(Ψ) is called an MRA-based 3-band tight wavelet frames of L2(R) if (1)
Ψ⊂V1,(2)
X(Ψ) is tight wavelet frames, that is,
(2.3)∥f∥L2(R)2=∑ℓ=0r‍∑j∈Z‍∑k∈Z‍|〈f,ψj,kℓ〉|holds for all f∈L2(R). Furthermore, elements ψj,kℓ, ℓ=1,…,r are said to be framelets.
If we introduce the following two block matrices: modulation matrix(2.4)M(ω)=[hℓ(ω+2kπ3)]0≤ℓ≤r,0≤k≤2,
and polyphase matrix
(2.5)P(z)=[pℓ,n(z)]0≤ℓ≤r,0≤n≤2,where(2.6)pℓ,n(z)=13∑k∈Z‍hℓ,3k+nz-k,ℓ=0,…,r,n=0,1,2,z=eiω,then we describe the UEP for 3-band tight wavelet frames in terms of modulation matrix and polyphase matrix as follows.Lemma 2.2.
Letψ0∈L2(R) be a compactly supported refinable function generated by finitely supported refinement mask h0 with ψ^0(0)=1. Then for combined MRA masks {h0,h1,…,hr}, the system X(Ψ), where Ψ={ψ1,…,ψr} defined by (2.1) or (2.2) forms an MRA-based 3-band tight wavelet frame of L2(R) provided one of the following hold for ω∈[-π,π]. (a)
Modulation matrixM(ω) is paraunitary, that is M*(ω)M(ω)=I3.(b)
Polyphase matrixP(z) is paraunitary, that is P*(z)P(z)=I3.Symmetry and high vanishing moments are very important features of all wavelets.Definition 2.3.
A casual FIR filterq(z)=∑k=0K‍qkz-k is called symmetric/antisymmetric, if z-Kq(z-1)=tq(z), t=±1, where K/2 is called the center of q(z).Lemma 2.4.
Suppose that the maskhℓ(z)=∑k=0Kℓ‍hℓ,kz-k is symmetric/antisymmetric, then ψℓ defined by (2.1) is symmetric/antisymmetric about Kℓ/2, respectively, that is, ψℓ(Kℓ-x)=tψℓ(x), where ℓ=0,1,2,…,r.The vanishing moments of wavelets is related to the order of sum rules.Definition 2.5.
The refinement maskh0(ω) has the sum rules of order p, if
(2.7)h0(ω)|ω=0=1,h0(m)(ω)|ω=2lπ/3=0,l=1,2,m=0,1,…,p-1,or equivalently(2.8)∑k‍ξkkmh0,k=0,m=0,1,…,p-1,ξ=e2πi/3.Lemma 2.6.
Suppose that the wavelet maskhℓ(ω) satisfies hℓ(m)(ω)|ω=0=0, m=0,1,…,p-1, then ψℓ defined by (2.1) has vanishing moments of order pℓ, respectively, that is, ∫‍xmψℓdx=0,m=0,1,…, pℓ-1. Where ℓ=1,2,…,r.


## 3. Symmetry Transform

In this section, we will firstly construct a paraunitary matrix based on the polyphase matrix corresponding to 3-band compactly supported wavelet frames. Further, we will establish necessary constraints on the filter lengths and symmetric features of wavelet frames.


### 3.1. Construction of Paraunitary Matrix Based on the Polyphase Matrix

From now on, we only consider the least number wavelet frames withr=3 and for the sake of convenience, we rewrit refinement mask and wavelet frame mask associated with ϕ and ψℓ, ℓ=1,2,3, respectively, that is, h(z)=∑k∈Z‍hkz-k, g(z)=∑k∈Z‍gkz-k, f(z)=∑k∈Z‍fkz-k, s(z)=∑k∈Z‍skz-k, where z=eiω.Suppose casual FIR filterh, g, f, s are symmetric/antisymmetric and h3k+n(z), f3k+n(z), g3k+n(z), s3k+n(z), n=0,1,2 defined by (2.4). For L∈Z, define
(3.1)AL(z)=[h3k(z)h3k+1(z)h3k+2(z)hm(z)g3k(z)g3k+1(z)g3k+2(z)gm(z)f3k(z)f3k+1(z)f3k+2(z)fm(z)s3k(z)s3k+1(z)s3k+2(z)sm(z)],
where
(3.2)hm(z)=-z-L|g3k(z-1)g3k+1(z-1)g3k+2(z-1)f3k(z-1)f3k+1(z-1)f3k+2(z-1)s3k(z-1)s3k+1(z-1)s3k+2(z-1)|,(3.3)gm(z)=z-L|h3k(z-1)h3k+1(z-1)h3k+2(z-1)f3k(z-1)f3k+1(z-1)f3k+2(z-1)s3k(z-1)s3k+1(z-1)s3k+2(z-1)|,(3.4)fm(z)=-z-L|h3k(z-1)h3k+1(z-1)h3k+2(z-1)g3k(z-1)g3k+1(z-1)g3k+2(z-1)s3k(z-1)s3k+1(z-1)s3k+2(z-1)|,(3.5)sm(z)=z-L|h3k(z-1)h3k+1(z-1)h3k+2(z-1)g3k(z-1)g3k+1(z-1)g3k+2(z-1)f3k(z-1)f3k+1(z-1)f3k+2(z-1)|.
Then AL(z) can be seen as a symmetry paraunitary extension of polyphase matrices P(z) with r=3, and have the following theorem and further investigation.Lemma 3.1.
LetM(z), P(z), and AL(z) are defined in (2.4), (2.5), and (3.1). Then the following three statements are equivalent.(a)
MT(z)M(z-1)=I3;(b)
PT(z)P(z-1)=I3;(c)
ALT(z)AL(z-1)=I4.From the foregoing discussion(3.6)[h(z)g(z)f(z)s(z)]=33AL(z3)[1z-1z-20].
Further, if h, g, f, s are causal, then for a sufficiently large L,AL is also causal. A paraunitary and causal FIR matrix can be factorized into the products of U0U1(z)⋯Uk(z) [1, 3], where k∈N0. So far, the construction for FIR filters h, g, f, s converted the design for 4×4 paraunitary matrix AL(z).


### 3.2. Investigation on Filter Lengths and Symmetry Features

The main content of this section is to investigate the constraint conditions of filter lengths and symmetric features of wavelet frames in order to discuss the method for construction framelets with desired properties.Theorem 3.2.
Leth, g, f, s are symmetric/antisymmetric nonzero FIR filters with the centers of symmetry N/2, K/2, S/2, T/2, respectively, where N, K, S, and T∈Z. If M is paraunitary, then(1)
leng(h)=leng(g).(2)
leng(f), leng(s) have the same parity.Proof.
(
1
) By the conditions of h,g,f,s, we have leng(h)=2k1+N+1, leng(g)=2k2+K+1, leng(f)=2k3+S+1, leng(s)=2k4+T+1, where k1, k2, k3, k4∈Z. From the paraunitariness of M, we have |h(z)|2+|g(z)|2+|f(z)|2+|s(z)|2=1, leng(h) = leng(g). Thus N-K is even.
(
2
) From the paraunitariness of M, we obtain
(3.7)h(z)h(z1z-1)+g(z)g(z1z-1)+f(z)f(z1z-1)+s(z)s(z1z-1)=0,
or
(3.8)h(z-1)h(z1z)+g(z-1)g(z1z)+f(z-1)f(z1z)+s(z-1)s(z1z)=0,
where z1=-1/2+(3/2)i or z1=-1/2-(3/2)i. From the symmetric of h, g, f, s, we have
(3.9)h(z-1)h(z1z)=(-1)Nh(z)h(z1z-1),g(z-1)g(z1z)=(-1)Kg(z)g(z1z-1),f(z-1)f(z1z)=(-1)Sf(z)f(z1z-1),s(z-1)s(z1z)=(-1)Ts(z)s(z1z-1).
Further,
(3.10)(-1)Nh(z)h(z1z-1)+(-1)Kg(z)g(z1z-1)+(-1)Sf(z)f(z1z-1)+(-1)Ts(z)s(z1z-1)=0.
Thus
(3.11)(1-(-1)K-N)g(z)g(z1z-1)+(1-(-1)S-N)f(z)f(z1z-1)+(1-(-1)T-N)s(z)s(z1z-1)=0.
Since N-K is even, hence
(3.12)(1-(-1)S-N)f(z)f(z1z-1)+(1-(-1)T-N)s(z)s(z1z-1)=0.
Obviously, T-N is even when S-N is even. Assume that T-N is even when S-N is odd, then we have f(z)f(z1z-1)=0, which is contradiction since f is nonzero filter.Without loss of generality, we give two assumptions of the casual filtersh,g,f,s: (1)
leng(h)≡ leng(g)≡ leng(f)≡ leng(s)(mod3);(2)
leng(h)=leng(g)≥leng(f)≥ leng(s).Then we have the following Theorem.Theorem 3.3.
Let(3.13)h(z)=∑k=03γ-1‍hkz-k,g(z)=∑k=03γ-1‍gkz-k,f(z)=∑k=03m1-1‍fkz-k,s(z)=∑k=03m2-1‍skz-k
are the causal filters and satisfy h0≠0 and γ≥m1≥m2, γ∈N. Suppose (3γ-1)/2 is the center of h(z) and g(z)=t0z-(3γ-1)g(z-1), f(z)=t1z-(3m1-1)f(z-1), s(z)=t2z-(3m2-1)s(z-1), tj=±1,j=0,1,2,3. If M is a paraunitary matrix, then (1)
leng(h3k)+γ, leng(g3k)+γ, leng(f3k)+m1, and leng(s3k)+m2 are even;(2)
t0+t1+t2+t0t1t2=0, 1+t0+(-1)γ-m1t1+(-1)γ-m2t2+(-1)γ(t0t1t2-1)=0;(3)
AL(z)=AL0(z) defined by (3.1) with L0=γ+(1/2)(m1+m2)-2 is causal, paraunitary, and satisfies
(3.14)AL(z)=z-(γ-1)diag(1,t0,t1z-(m1-γ),t2z-(m2-γ))AL(z-1)diag(J3,-t0t1t2).Proof.
By the symmetry ofh, g, f, s, we have
(3.15)z-(γ-1)h3k(z-1)=h3k+2(z),z-(γ-1)h3k+1(z-1)=h3k+1(z);z-(γ-1)g3k(z-1)=t0g3k+2(z),z-(γ-1)g3k+1(z-1)=t0g3k+1(z);z-(m1-1)f3k(z-1)=t1f3k+2(z),z-(m1-1)f3k+1(z-1)=t1f3k+1(z);z-(m2-1)s3k(z-1)=t2s3k+2(z),z-(m2-1)s3k+1(z-1)=t2s3k+1(z).
Take them into hm defined by (3.2), we have
(3.16)hm(z)=-z-L[g3k(z-1)g3k+1(z-1)g3k+2(z-1)f3k(z-1)f3k+1(z-1)f3k+2(z-1)s3k(z-1)s3k+1(z-1)s3k+2(z-1)]=-z-L[zγ-1t0g3k+2(z)zγ-1t0g3k+1(z)zγ-1t0g3k(z)zm1-1t1f3k+2(z)zm1-1t1f3k+1(z)zm1-1t1f3k(z)zm2-1t2s3k+2(z)zm2-1t2s3k+1(z)zm2-1t2s3k(z)]=z-L·zγ-1·zm1-1·zm1-1t0t1t2[g3k(z)g3k+1(z)g3k+2(z)f3k(z)f3k+1(z)f3k+2(z)s3k+2(z)s3k+1(z)s3k+2(z)]=-t0t1t2z-(2L-γ-m1-m2+3)hm(z-1).
Similarly, for gm, fm, sm defined by (3.3)–(3.5), we also have
(3.17)gm(z)=-t1t2z-(2L-γ-m1-m2+3)gm(z-1),fm(z)=-t0t2z-(2L-2γ-m2+3)fm(z-1),sm(z)=-t0t1z-(2L-2γ-m1+3)sm(z-1).
Thus, AL(z) defined by (3.1) satisfies with
(3.18)AL(z)=z-(γ-1)diag(1,t0,t1z-(m1-γ),t2z-(m2-γ))AL(z-1)diag(J3,-t0t1t2z-(2L-2γ-m1-m2+4)).
From the paraunitariness ofAL(z), we obtain leng(hm) = leng(h3k). Assume that
(3.19)hm(z)=alz-k+al+1z-(l+1)+⋯+(-t0t1t2)z-(l+leng(h3k)-1),l∈Z,al≠0.
By symmetric of h, we have z-(l+leng(h3k)-1)hm(z-1)=-t0t1t2hm(z), which together with (3.17) leads to 2l+leng(h3k)-1=2L-γ-m1-m2+3; thus, leng(h3k)+γ+m1+m2=2L-2l+4 is even. Note that 3m1 and 3m2 have the same parity by Theorem 3.2, hence m1+m2 is even, leng(h3k)+γ is even. One can prove similarly that leng(g3k)+γ, leng(f3k)+m1 and leng(s3k)+m2 are all even.
LetL0=γ+(1/2)(m1+m2)-2. Then hm,gm,fm,sm defined by (3.2)–(3.5) with L0 satisfy
(3.20)hm(z)=-t0t1t2z-(γ-1)hm(z-1),gm(z)=-t1t2z-(γ-1)gm(z-1),fm(z)=-t0t2z-(m1-1)fm(z-1),sm(z)=-t0t1z-(m2-1)sm(z-1).
Note that
(3.21)leng(hm)=leng(h3k)≤γ=γ-1+1,leng(gm)=leng(g3k)≤γ=γ-1+1,leng(fm)=leng(f3k)≤m1=m1-1+1,leng(sm)=leng(s3k)≤m2.
Thus, hm, gm, fm, sm are casual, A(z)=AL0(z) is casual, paraunitary, and satisfies (3.14).
Setz=1 and z=-1 in (3.14), then we have
(3.22)Tr(diag(1,t0,t1,t2))=Tr(diag(J3,-t0t1t2)),Tr((-1)-γdiag(1,t0,(-1)γ-m1t1,(-1)γ-m2t2))=Tr(diag(J3,-t0t1t2)),
which imply (2.2).With similar arguments we can prove the others case and present the following two theorems as follows.Theorem 3.4.
Let(3.23)h(z)=∑k=03γ‍hkz-k,g(z)=∑k=03γ‍gkz-k,f(z)=∑k=03m1‍fkz-k,s(z)=∑k=03m2‍skz-k,
are the causal filters and satisfy h0≠0 and γ≥m1≥m2, γ∈N. Suppose (3γ)/2 is the center of h(z) and g(z)=t0z-3γg(z-1), f(z)=t1z-3m1f(z-1), s(z)=t2z-3m2s(z-1), tj=±1, j=0,1,2,3. If M is a paraunitary matrix, then (1)
γ+m1 and γ+m2 are even.(2)
t0+t1+t2+t0t1t2=0, 1+t0+(-1)γ-m1t1+(-1)γ-m2t2+(-1)γ(t0t1t2-1)=0.(3)
AL(z)=AL0(z) defined by (3.1) with L0=γ+(1/2)(m1+m2-2) is causal, paraunitary and satisfies
(3.24)AL(z)=z-γdiag(1,t0,t1z-(m1-γ),t2z-(m2-γ))AL(z-1)diag(1,zJ2,-t0t1t2).Theorem 3.5.
Let(3.25)h(z)=∑k=03γ+1‍hkz-k,g(z)=∑k=03γ+1‍gkz-k,f(z)=∑k=03m1+1‍fkz-k,s(z)=∑k=03m2+1‍skz-k
are the causal filters and satisfy h0≠0 and γ≥m1≥m2, γ∈N. Suppose (3γ+1)/2 is the center of h(z) and g(z)=t0z-(3γ+1)g(z-1),f(z)=t1z-(3m1+1)f(z-1), s(z)=t2z-(3m2+1)s(z-1), tj=±1, j=0,1,2,3. If M is a paraunitary matrix, then (1)
leng(h3k)+γ, leng(g3k)+γ, leng(f3k)+m1, and leng(s3k)+m2 are even;(2)
t0+t1+t2+t0t1t2=0, 1+t0+(-1)γ-m1t1+(-1)γ-m2t2+(-1)γ(t0t1t2-1)=0;(3)
AL(z)=AL0(z) defined by (3.1) with L0=γ+(1/2)(m1+m2) is causal, paraunitary and satisfies
(3.26)AL(z)=z-(γ+1)diag(1,t0,t1z-(m1-γ),t2z-(m2-γ))AL(z-1)diag(J2,z,-t0t1t2).So far, we have completed the further extension of the our previous work in [22], that is, we presented the several necessary constraints on the filter lengths and symmetric features of wavelet frames. By using these properties, AL(z) defined by (3.1) can be divided into several categories according to the different length of masks. Then we can improve the condition of AL(z) in the following work and prepare for the parameterizations. This paper is devoted to give the parameterizations of 3-band tight wavelet frames with 3N+1 filter lengths.


## 3.1. Construction of Paraunitary Matrix Based on the Polyphase Matrix

From now on, we only consider the least number wavelet frames withr=3 and for the sake of convenience, we rewrit refinement mask and wavelet frame mask associated with ϕ and ψℓ, ℓ=1,2,3, respectively, that is, h(z)=∑k∈Z‍hkz-k, g(z)=∑k∈Z‍gkz-k, f(z)=∑k∈Z‍fkz-k, s(z)=∑k∈Z‍skz-k, where z=eiω.Suppose casual FIR filterh, g, f, s are symmetric/antisymmetric and h3k+n(z), f3k+n(z), g3k+n(z), s3k+n(z), n=0,1,2 defined by (2.4). For L∈Z, define
(3.1)AL(z)=[h3k(z)h3k+1(z)h3k+2(z)hm(z)g3k(z)g3k+1(z)g3k+2(z)gm(z)f3k(z)f3k+1(z)f3k+2(z)fm(z)s3k(z)s3k+1(z)s3k+2(z)sm(z)],
where
(3.2)hm(z)=-z-L|g3k(z-1)g3k+1(z-1)g3k+2(z-1)f3k(z-1)f3k+1(z-1)f3k+2(z-1)s3k(z-1)s3k+1(z-1)s3k+2(z-1)|,(3.3)gm(z)=z-L|h3k(z-1)h3k+1(z-1)h3k+2(z-1)f3k(z-1)f3k+1(z-1)f3k+2(z-1)s3k(z-1)s3k+1(z-1)s3k+2(z-1)|,(3.4)fm(z)=-z-L|h3k(z-1)h3k+1(z-1)h3k+2(z-1)g3k(z-1)g3k+1(z-1)g3k+2(z-1)s3k(z-1)s3k+1(z-1)s3k+2(z-1)|,(3.5)sm(z)=z-L|h3k(z-1)h3k+1(z-1)h3k+2(z-1)g3k(z-1)g3k+1(z-1)g3k+2(z-1)f3k(z-1)f3k+1(z-1)f3k+2(z-1)|.
Then AL(z) can be seen as a symmetry paraunitary extension of polyphase matrices P(z) with r=3, and have the following theorem and further investigation.Lemma 3.1.
LetM(z), P(z), and AL(z) are defined in (2.4), (2.5), and (3.1). Then the following three statements are equivalent.(a)
MT(z)M(z-1)=I3;(b)
PT(z)P(z-1)=I3;(c)
ALT(z)AL(z-1)=I4.From the foregoing discussion(3.6)[h(z)g(z)f(z)s(z)]=33AL(z3)[1z-1z-20].
Further, if h, g, f, s are causal, then for a sufficiently large L,AL is also causal. A paraunitary and causal FIR matrix can be factorized into the products of U0U1(z)⋯Uk(z) [1, 3], where k∈N0. So far, the construction for FIR filters h, g, f, s converted the design for 4×4 paraunitary matrix AL(z).


## 3.2. Investigation on Filter Lengths and Symmetry Features

The main content of this section is to investigate the constraint conditions of filter lengths and symmetric features of wavelet frames in order to discuss the method for construction framelets with desired properties.Theorem 3.2.
Leth, g, f, s are symmetric/antisymmetric nonzero FIR filters with the centers of symmetry N/2, K/2, S/2, T/2, respectively, where N, K, S, and T∈Z. If M is paraunitary, then(1)
leng(h)=leng(g).(2)
leng(f), leng(s) have the same parity.Proof.
(
1
) By the conditions of h,g,f,s, we have leng(h)=2k1+N+1, leng(g)=2k2+K+1, leng(f)=2k3+S+1, leng(s)=2k4+T+1, where k1, k2, k3, k4∈Z. From the paraunitariness of M, we have |h(z)|2+|g(z)|2+|f(z)|2+|s(z)|2=1, leng(h) = leng(g). Thus N-K is even.
(
2
) From the paraunitariness of M, we obtain
(3.7)h(z)h(z1z-1)+g(z)g(z1z-1)+f(z)f(z1z-1)+s(z)s(z1z-1)=0,
or
(3.8)h(z-1)h(z1z)+g(z-1)g(z1z)+f(z-1)f(z1z)+s(z-1)s(z1z)=0,
where z1=-1/2+(3/2)i or z1=-1/2-(3/2)i. From the symmetric of h, g, f, s, we have
(3.9)h(z-1)h(z1z)=(-1)Nh(z)h(z1z-1),g(z-1)g(z1z)=(-1)Kg(z)g(z1z-1),f(z-1)f(z1z)=(-1)Sf(z)f(z1z-1),s(z-1)s(z1z)=(-1)Ts(z)s(z1z-1).
Further,
(3.10)(-1)Nh(z)h(z1z-1)+(-1)Kg(z)g(z1z-1)+(-1)Sf(z)f(z1z-1)+(-1)Ts(z)s(z1z-1)=0.
Thus
(3.11)(1-(-1)K-N)g(z)g(z1z-1)+(1-(-1)S-N)f(z)f(z1z-1)+(1-(-1)T-N)s(z)s(z1z-1)=0.
Since N-K is even, hence
(3.12)(1-(-1)S-N)f(z)f(z1z-1)+(1-(-1)T-N)s(z)s(z1z-1)=0.
Obviously, T-N is even when S-N is even. Assume that T-N is even when S-N is odd, then we have f(z)f(z1z-1)=0, which is contradiction since f is nonzero filter.Without loss of generality, we give two assumptions of the casual filtersh,g,f,s: (1)
leng(h)≡ leng(g)≡ leng(f)≡ leng(s)(mod3);(2)
leng(h)=leng(g)≥leng(f)≥ leng(s).Then we have the following Theorem.Theorem 3.3.
Let(3.13)h(z)=∑k=03γ-1‍hkz-k,g(z)=∑k=03γ-1‍gkz-k,f(z)=∑k=03m1-1‍fkz-k,s(z)=∑k=03m2-1‍skz-k
are the causal filters and satisfy h0≠0 and γ≥m1≥m2, γ∈N. Suppose (3γ-1)/2 is the center of h(z) and g(z)=t0z-(3γ-1)g(z-1), f(z)=t1z-(3m1-1)f(z-1), s(z)=t2z-(3m2-1)s(z-1), tj=±1,j=0,1,2,3. If M is a paraunitary matrix, then (1)
leng(h3k)+γ, leng(g3k)+γ, leng(f3k)+m1, and leng(s3k)+m2 are even;(2)
t0+t1+t2+t0t1t2=0, 1+t0+(-1)γ-m1t1+(-1)γ-m2t2+(-1)γ(t0t1t2-1)=0;(3)
AL(z)=AL0(z) defined by (3.1) with L0=γ+(1/2)(m1+m2)-2 is causal, paraunitary, and satisfies
(3.14)AL(z)=z-(γ-1)diag(1,t0,t1z-(m1-γ),t2z-(m2-γ))AL(z-1)diag(J3,-t0t1t2).Proof.
By the symmetry ofh, g, f, s, we have
(3.15)z-(γ-1)h3k(z-1)=h3k+2(z),z-(γ-1)h3k+1(z-1)=h3k+1(z);z-(γ-1)g3k(z-1)=t0g3k+2(z),z-(γ-1)g3k+1(z-1)=t0g3k+1(z);z-(m1-1)f3k(z-1)=t1f3k+2(z),z-(m1-1)f3k+1(z-1)=t1f3k+1(z);z-(m2-1)s3k(z-1)=t2s3k+2(z),z-(m2-1)s3k+1(z-1)=t2s3k+1(z).
Take them into hm defined by (3.2), we have
(3.16)hm(z)=-z-L[g3k(z-1)g3k+1(z-1)g3k+2(z-1)f3k(z-1)f3k+1(z-1)f3k+2(z-1)s3k(z-1)s3k+1(z-1)s3k+2(z-1)]=-z-L[zγ-1t0g3k+2(z)zγ-1t0g3k+1(z)zγ-1t0g3k(z)zm1-1t1f3k+2(z)zm1-1t1f3k+1(z)zm1-1t1f3k(z)zm2-1t2s3k+2(z)zm2-1t2s3k+1(z)zm2-1t2s3k(z)]=z-L·zγ-1·zm1-1·zm1-1t0t1t2[g3k(z)g3k+1(z)g3k+2(z)f3k(z)f3k+1(z)f3k+2(z)s3k+2(z)s3k+1(z)s3k+2(z)]=-t0t1t2z-(2L-γ-m1-m2+3)hm(z-1).
Similarly, for gm, fm, sm defined by (3.3)–(3.5), we also have
(3.17)gm(z)=-t1t2z-(2L-γ-m1-m2+3)gm(z-1),fm(z)=-t0t2z-(2L-2γ-m2+3)fm(z-1),sm(z)=-t0t1z-(2L-2γ-m1+3)sm(z-1).
Thus, AL(z) defined by (3.1) satisfies with
(3.18)AL(z)=z-(γ-1)diag(1,t0,t1z-(m1-γ),t2z-(m2-γ))AL(z-1)diag(J3,-t0t1t2z-(2L-2γ-m1-m2+4)).
From the paraunitariness ofAL(z), we obtain leng(hm) = leng(h3k). Assume that
(3.19)hm(z)=alz-k+al+1z-(l+1)+⋯+(-t0t1t2)z-(l+leng(h3k)-1),l∈Z,al≠0.
By symmetric of h, we have z-(l+leng(h3k)-1)hm(z-1)=-t0t1t2hm(z), which together with (3.17) leads to 2l+leng(h3k)-1=2L-γ-m1-m2+3; thus, leng(h3k)+γ+m1+m2=2L-2l+4 is even. Note that 3m1 and 3m2 have the same parity by Theorem 3.2, hence m1+m2 is even, leng(h3k)+γ is even. One can prove similarly that leng(g3k)+γ, leng(f3k)+m1 and leng(s3k)+m2 are all even.
LetL0=γ+(1/2)(m1+m2)-2. Then hm,gm,fm,sm defined by (3.2)–(3.5) with L0 satisfy
(3.20)hm(z)=-t0t1t2z-(γ-1)hm(z-1),gm(z)=-t1t2z-(γ-1)gm(z-1),fm(z)=-t0t2z-(m1-1)fm(z-1),sm(z)=-t0t1z-(m2-1)sm(z-1).
Note that
(3.21)leng(hm)=leng(h3k)≤γ=γ-1+1,leng(gm)=leng(g3k)≤γ=γ-1+1,leng(fm)=leng(f3k)≤m1=m1-1+1,leng(sm)=leng(s3k)≤m2.
Thus, hm, gm, fm, sm are casual, A(z)=AL0(z) is casual, paraunitary, and satisfies (3.14).
Setz=1 and z=-1 in (3.14), then we have
(3.22)Tr(diag(1,t0,t1,t2))=Tr(diag(J3,-t0t1t2)),Tr((-1)-γdiag(1,t0,(-1)γ-m1t1,(-1)γ-m2t2))=Tr(diag(J3,-t0t1t2)),
which imply (2.2).With similar arguments we can prove the others case and present the following two theorems as follows.Theorem 3.4.
Let(3.23)h(z)=∑k=03γ‍hkz-k,g(z)=∑k=03γ‍gkz-k,f(z)=∑k=03m1‍fkz-k,s(z)=∑k=03m2‍skz-k,
are the causal filters and satisfy h0≠0 and γ≥m1≥m2, γ∈N. Suppose (3γ)/2 is the center of h(z) and g(z)=t0z-3γg(z-1), f(z)=t1z-3m1f(z-1), s(z)=t2z-3m2s(z-1), tj=±1, j=0,1,2,3. If M is a paraunitary matrix, then (1)
γ+m1 and γ+m2 are even.(2)
t0+t1+t2+t0t1t2=0, 1+t0+(-1)γ-m1t1+(-1)γ-m2t2+(-1)γ(t0t1t2-1)=0.(3)
AL(z)=AL0(z) defined by (3.1) with L0=γ+(1/2)(m1+m2-2) is causal, paraunitary and satisfies
(3.24)AL(z)=z-γdiag(1,t0,t1z-(m1-γ),t2z-(m2-γ))AL(z-1)diag(1,zJ2,-t0t1t2).Theorem 3.5.
Let(3.25)h(z)=∑k=03γ+1‍hkz-k,g(z)=∑k=03γ+1‍gkz-k,f(z)=∑k=03m1+1‍fkz-k,s(z)=∑k=03m2+1‍skz-k
are the causal filters and satisfy h0≠0 and γ≥m1≥m2, γ∈N. Suppose (3γ+1)/2 is the center of h(z) and g(z)=t0z-(3γ+1)g(z-1),f(z)=t1z-(3m1+1)f(z-1), s(z)=t2z-(3m2+1)s(z-1), tj=±1, j=0,1,2,3. If M is a paraunitary matrix, then (1)
leng(h3k)+γ, leng(g3k)+γ, leng(f3k)+m1, and leng(s3k)+m2 are even;(2)
t0+t1+t2+t0t1t2=0, 1+t0+(-1)γ-m1t1+(-1)γ-m2t2+(-1)γ(t0t1t2-1)=0;(3)
AL(z)=AL0(z) defined by (3.1) with L0=γ+(1/2)(m1+m2) is causal, paraunitary and satisfies
(3.26)AL(z)=z-(γ+1)diag(1,t0,t1z-(m1-γ),t2z-(m2-γ))AL(z-1)diag(J2,z,-t0t1t2).So far, we have completed the further extension of the our previous work in [22], that is, we presented the several necessary constraints on the filter lengths and symmetric features of wavelet frames. By using these properties, AL(z) defined by (3.1) can be divided into several categories according to the different length of masks. Then we can improve the condition of AL(z) in the following work and prepare for the parameterizations. This paper is devoted to give the parameterizations of 3-band tight wavelet frames with 3N+1 filter lengths.


## 4. The Parameterizations for Combined MRA Masks with3N+1 Filter Lengths

For simplicity we give the following two assumptions:(1)
leng(h)=leng(g)=3γ+1, leng(f)=leng(s)=3m+1;(2)
m≡1(mod3), 0≤γ-m≤2.


### 4.1. The Case ofγ=3n+1

ForAL(z) with L=6n+1 in (3.1), denote A(z)=AL(z), then (3.24) can be converted to
(4.1)A(z)=z-(3n+1)diag(1,t0,t1,t2)A(z-1)diag(1,diag(z,z)J2,-t0t1t2).
While
(4.2)t0+t1+t2+t0t1t2=0,ti=±1,i=1,2,3
imply that(i)
The first case: one ofti is 1 and the other two are -1;(ii)
The second case: one ofti is -1 and the other two are 1.Lemma 4.1.
ForA(z) in (4.1), let
(4.3)B(z)=A(z)diag(1,R1,1),
where R1=(2/2)[1-111], then A(z) satisfies if and only if B(z) satisfies
(4.4)B(z)=z-γdiag(1,t0,t1,t2)B(z-1)diag(1,z,-z,-t0t1t2).(i) For the first case, without loss of generality, supposet0=1,t1=t2=-1, (4.4) is converted to
(4.5)z-(3n+1)diag(1,1,-1,-1)B(z-1)diag(1,z,-z,-1)=B(z).
To give the parameterizations of B(z) satisfying (4.5), one defines
(4.6)W0(z)=[-24-24242422-22000022-22-24-242424]+[-24-24-24-240000000024242424]z.
One can check that W0(z) is paraunitary and satisfies
(4.7)W0(z)=zdiag(1,z-1,-z-1,-1)W0(z-1)diag(1,1,-1,-1).Lemma 4.2.
ForB(z) in (4.3), define
(4.8)E(z)=B(z)W0(z).
Then E(z) satisfies
(4.9)E(z)=z-3ndiag(1,1,-1,-1)E(z-1)diag(1,1,-1,-1).Lemma 4.3.
Let a paraunitary matrixE(z) defined by (4.8), then for n=0, E(z) have the form of
(4.10)P0=[cosθ1ρ1sinθ100sinθ1-ρ1cosθ10000cosθ2ρ2sinθ200sinθ2-ρ2cosθ2],
where θ∈[-π,π),ρj=±1,j∈Z.Proof.
Letn=0, (4.9) is converted to
(4.11)E(z)=diag(1,1,-1,-1)E(z)diag(1,1,-1,-1).
If set 4×4 matrix P0=[aij], i,j=1,2,3,4, then we can obtain
(4.12)a112+a122=1,a212+a222=1,a11a21+a12a22=0,a332+a342=1,a432+a442=1,a33a43+a34a44=0,a13=a14=a23=a24=a31=a32=a41=a42=0.
Assume that a11=cosθ1, a33=cosθ2, then (4.10) is obtained immediately.Remark 4.4.
For convenience, we only choose the following form ofP0 in the following:
(4.13)P0=[cosθsinθ00sinθ-cosθ0000cosθsinθ00sinθ-cosθ].Lemma 4.5.
Let a paraunitary matrixE(z) satisfies (4.8), then for n=1, E(z) have the form of
(4.14)V(z)=12[0110][cosα1,sinα1,cosα2,sinα2]+12[1001][sinα1,-cosα1,sinα2,-cosα2]z-1+12[100-1][sinα1,-cosα1,-sinα2,cosα2]z-2+12[01-10][cosα1,sinα1,-cosα2,-sinα2]z-3,
where αj∈[-π,π), j=1,2.Proof.
Whenn=1, (4.9) is converted to
(4.15)E(z)=z-3diag(1,1,-1,-1)E(z-1)diag(1,1,-1,-1).
Suppose E(z)=A+Bz-1+Cz-2+Dz-3, here A, B, C, D are all 4×4 matrices. Let A=[aij] then, from the paraunitariness of E(z), we have
(4.16)A=12[abcd][cosα1,sinα1,cosα2,sinα2],D=12[efgh][cosα1,sinα1,-cosα2,-sinα2],
where b, c, f, g are all not zeros or a, d, e, h are all not zeros. Now suppose b, c, f, g are all not zeros, then b=f=1, c=-g=1.
Similarly,B and C are given as follows:
(4.17)B=12[1001][sinα1,-cosα1,sinα2,-cosα2],C=12[abcd][sinα1,-cosα1,-sinα2,cosα2].
Equation (4.14) is obtained.Theorem 4.6.
A causal paraunitary matrix filterE(z) defined by (4.8) for some n∈N if and only if it can be factorized in the form of
(4.18)E(z)=P0V1(z)⋯Vn(z),
where P0 is defined by (4.13) with θ∈[-π,π) and Vj(z), j=1,2 are defined by (4.14) with αj∈[-π,π), j=1,2.Proof.
IfE(z) can be factorized as (4.18), then it is a causal paraunitary filter and satisfies (4.9). Conversely, we set a paraunitary matrix En(z)=e0+e1z-1+⋯+enz-3n satisfies with (4.18), where n∈N. If n=0, it is easy to get that E(z) is P0. For n∈N∖{0}, define
(4.19)En-1(z)=En(z)VT(z-1),
where V(z) is defined by (4.14). Now we prove that there exists α1, α2 such that En-1(z) defined by (4.19) is causal. From VT(z-1), we only need to prove that there exists α1, α2 such that
(4.20)e0[sinα1,-cosα1,sinα2,-cosα2]T=0,(4.21)e0[sinα1,-cosα1,-sinα2,cosα2]T=0,(4.22)e0[cosα1,sinα1,-cosα2,-sinα2]T=0,(4.23)e1[sinα1,-cosα1,-sinα2,cosα2]T=0,(4.24)e1[cosα1,sinα1,-cosα2,-sinα2]T=0,(4.25)e2[cosα1,sinα1,-cosα2,-sinα2]T=0.
Let us start by discussing the casee0=0. Then (4.20)–(4.22) hold. By the symmetry of En(z),    ej=diag(1,1,-1,-1)e3n-jdiag(1,1,-1,-1), 0≤j≤3n. By the paraunitariness of En(z),  e1e3n-1T=0. Then we have e1diag(1,1,-1,-1)e1T=0, rank(e1)≤2. For simplicity, we only consider the case rank(e1)=1 in the following. Let e1=u1[cosθ1,sinθ1,cosθ2,sinθ2], u1∈R4, θ1,θ2∈[-π,π). Then for α1=θ1, α2=θ2, (4.23) and (4.24) hold. Let Q=e2[cosθ1,sinθ1,-cosθ2,-sinθ2]T, by the paraunitariness of En(z), e1e3n-1T+e2e3nT=0. Then we have u1QT+Qu1T=0. It is clear that Q=0 since u1≠0. Then (4.25) holds for α1=θ1, α2=θ2.
Ife0≠0, we have e0e3nT=0 and e3n=diag(1,1,-1,-1)e0diag(1,1,-1,-1). Let e0=u1[cosθ1,sinθ1,cosθ2,sinθ2], then (4.20)–(4.22) hold for α1=θ1, α2=θ2. Similarly, when e1=u2[-sinθ1,cosθ1,-sinθ2,cosθ2], (4.23), and (4.24) hold for α1=θ1, α2=θ2. Let Q=e2[cosα1,sinα1,-cosα2,-sinα2]T, from the paraunitariness of En(z), we have e1e3n-1T+e2e3nT=0. Thus, u2QT+Qu2T=0. It is clear that Q=0 since u2≠0. So (4.25) holds for α1=θ1, α2=θ2.
SoEn-1(z) is causal and En-1(z) can be factorized in the form of En-1(z)=P0V1(z)⋯Vn-1(z). Thus, En(z) can be factorized in the form of En(z)=P0V1(z)⋯Vn-1(z) by induction assumption. The proof of Theorem 4.6 is complete.Therefore, we have the following corollary.Corollary 4.7.
Leth(z), g(z), f(z), s(z) is symmetric or antisymmetric. If M is a paraunitary matrix, then h(z), g(z), f(z), s(z) can be factorized as
(4.26)(1)[h(z)g(z)f(z)s(z)]=612P0V1(z3)⋯Vn(z3)[-1+2z-1+2z-2-z-3-1-2z-1-2z-2-z-31-2z-1+2z-2-z-31+2z-1-2z-2-z-3],
for the case of t0=1, t1=t2=-1;
(4.27)(2)[h(z)g(z)f(z)s(z)]=612V0V1(z3)⋯Vn(z3)[-1+2z-1+2z-2-z-3-1-2z-1-2z-2-z-31-2z-1+2z-2-z-31+2z-1-2z-2-z-3],
for the case oft0=-1, t1=1, t2=-1, where V0=diag(1,J2,1)P0;
(4.28)(3)[h(z)g(z)f(z)s(z)]=612U0V1(z3)⋯Vn(z3)[-1+2z-1+2z-2-z-3-1-2z-1-2z-2-z-31-2z-1+2z-2-z-31+2z-1-2z-2-z-3],
for the case of t0=t1=-1, t2=1, where U0=diag(1,J3)P0.
WhereP0 is defined by (4.13) with θ∈[-π,π) and Vj(z) are defined by (4.14) with αj∈[-π,π).(ii) For the second case, without loss of generality, supposet0=1, t1=1, t2=-1, then E(z) defined by (4.8) satisfies
(4.29)E(z)=z-3ndiag(1,1,1,-1)E(z-1)diag(1,1,-1,1).
Let n=0 in (4.29), then one can get an orthogonal matrix Q0 as follows:
(4.30)Q0=[cosθ00-ρ1sinθ0ρ20000ρ30sinθ00ρ1cosθ].
Then one has the following results.Theorem 4.8.
LetE(z) is a causal paraunitary matrix filter and satisfies (4.29) for some n∈N, then E(z) cannot be factorized in the form of
(4.31)E(z)=Q0V1(z)⋯Vn(z),
for Q0 defined by (4.30) with θ∈[-π,π) and Vj(z), j=1,2 defined by (4.14) with αj∈[-π,π), j=1,2.Proof.
SupposeE(z) can be factorized in the form of (4.31). Then when n=1, we have
(4.32)V1(z)=z-3diag(1,1,1,-1)V1(z-1)diag(1,1,-1,1);
when n=2, we have
(4.33)V2(z)=z-3diag(1,1,-1,1)V2(z-1)diag(1,1,-1,1).
However, (4.33) is in contradiction with (4.32).


### 4.2. The Case ofγ=3n+2

We have the following results.Theorem 4.9.
E
(
z
) is a causal paraunitary matrix filter defined by (4.8) for some n∈N if and only if it can be factorized in the form of(4.34)
E
(
z
)
=
P
1
V
2
(
z
)
V
3
(
z
)
⋯
V
n
(
z
)
,whereP1 is defined by the following with θj∈[-π,π), j=1,2(4.35)P1(z)=12[cosθ1sinθ1cosθ2sinθ2cosθ1sinθ1cosθ2sinθ2-2sinθ12cosθ10000-2sinθ22cosθ2]+12[cosθ1sinθ1-cosθ2-sinθ2-cosθ1-sinθ1cosθ2sinθ200000000]z-1,andVj(z) are defined by (4.14) with αj∈[-π,π).Corollary 4.10.
Leth(z), g(z), f(z), s(z) is symmetric or antisymmetric. If M is a paraunitary matrix, then h(z), g(z), f(z), s(z) can be factorized as
(4.36)(1)[h(z)g(z)f(z)s(z)]=612P1(z)V2(z3)⋯Vn(z3)[-1+2z-1+2z-2-z-3-1-2z-1-2z-2-z-31-2z-1+2z-2-z-31+2z-1-2z-2-z-3],
for the case t0=-1, t1=1, t2=-1;
(4.37)(2)[h(z)g(z)f(z)s(z)]=612diag(1,1,J2)P1(z)V2(z3)⋯Vn(z3)[-1+2z-1+2z-2-z-3-1-2z-1-2z-2-z-31-2z-1+2z-2-z-31+2z-1-2z-2-z-3],
for the case of t0=t1=-1, t2=1. Where P1(z) is defined by (4.35) with θ∈[-π,π) and Vj(z) are defined by (4.14) with αj∈[-π,π).Remark 4.11.
In the case oft0=1, t1=t2=-1, we cannot construct a paraunitary matrix P1(z). The reason is similarly as Theorem 4.8 when t0=1, t1=t2=-1.


### 4.3. The Case ofγ=3n+3

In this situation, we cannot also construct the corresponding paraunitary matrix.


## 4.1. The Case ofγ=3n+1

ForAL(z) with L=6n+1 in (3.1), denote A(z)=AL(z), then (3.24) can be converted to
(4.1)A(z)=z-(3n+1)diag(1,t0,t1,t2)A(z-1)diag(1,diag(z,z)J2,-t0t1t2).
While
(4.2)t0+t1+t2+t0t1t2=0,ti=±1,i=1,2,3
imply that(i)
The first case: one ofti is 1 and the other two are -1;(ii)
The second case: one ofti is -1 and the other two are 1.Lemma 4.1.
ForA(z) in (4.1), let
(4.3)B(z)=A(z)diag(1,R1,1),
where R1=(2/2)[1-111], then A(z) satisfies if and only if B(z) satisfies
(4.4)B(z)=z-γdiag(1,t0,t1,t2)B(z-1)diag(1,z,-z,-t0t1t2).(i) For the first case, without loss of generality, supposet0=1,t1=t2=-1, (4.4) is converted to
(4.5)z-(3n+1)diag(1,1,-1,-1)B(z-1)diag(1,z,-z,-1)=B(z).
To give the parameterizations of B(z) satisfying (4.5), one defines
(4.6)W0(z)=[-24-24242422-22000022-22-24-242424]+[-24-24-24-240000000024242424]z.
One can check that W0(z) is paraunitary and satisfies
(4.7)W0(z)=zdiag(1,z-1,-z-1,-1)W0(z-1)diag(1,1,-1,-1).Lemma 4.2.
ForB(z) in (4.3), define
(4.8)E(z)=B(z)W0(z).
Then E(z) satisfies
(4.9)E(z)=z-3ndiag(1,1,-1,-1)E(z-1)diag(1,1,-1,-1).Lemma 4.3.
Let a paraunitary matrixE(z) defined by (4.8), then for n=0, E(z) have the form of
(4.10)P0=[cosθ1ρ1sinθ100sinθ1-ρ1cosθ10000cosθ2ρ2sinθ200sinθ2-ρ2cosθ2],
where θ∈[-π,π),ρj=±1,j∈Z.Proof.
Letn=0, (4.9) is converted to
(4.11)E(z)=diag(1,1,-1,-1)E(z)diag(1,1,-1,-1).
If set 4×4 matrix P0=[aij], i,j=1,2,3,4, then we can obtain
(4.12)a112+a122=1,a212+a222=1,a11a21+a12a22=0,a332+a342=1,a432+a442=1,a33a43+a34a44=0,a13=a14=a23=a24=a31=a32=a41=a42=0.
Assume that a11=cosθ1, a33=cosθ2, then (4.10) is obtained immediately.Remark 4.4.
For convenience, we only choose the following form ofP0 in the following:
(4.13)P0=[cosθsinθ00sinθ-cosθ0000cosθsinθ00sinθ-cosθ].Lemma 4.5.
Let a paraunitary matrixE(z) satisfies (4.8), then for n=1, E(z) have the form of
(4.14)V(z)=12[0110][cosα1,sinα1,cosα2,sinα2]+12[1001][sinα1,-cosα1,sinα2,-cosα2]z-1+12[100-1][sinα1,-cosα1,-sinα2,cosα2]z-2+12[01-10][cosα1,sinα1,-cosα2,-sinα2]z-3,
where αj∈[-π,π), j=1,2.Proof.
Whenn=1, (4.9) is converted to
(4.15)E(z)=z-3diag(1,1,-1,-1)E(z-1)diag(1,1,-1,-1).
Suppose E(z)=A+Bz-1+Cz-2+Dz-3, here A, B, C, D are all 4×4 matrices. Let A=[aij] then, from the paraunitariness of E(z), we have
(4.16)A=12[abcd][cosα1,sinα1,cosα2,sinα2],D=12[efgh][cosα1,sinα1,-cosα2,-sinα2],
where b, c, f, g are all not zeros or a, d, e, h are all not zeros. Now suppose b, c, f, g are all not zeros, then b=f=1, c=-g=1.
Similarly,B and C are given as follows:
(4.17)B=12[1001][sinα1,-cosα1,sinα2,-cosα2],C=12[abcd][sinα1,-cosα1,-sinα2,cosα2].
Equation (4.14) is obtained.Theorem 4.6.
A causal paraunitary matrix filterE(z) defined by (4.8) for some n∈N if and only if it can be factorized in the form of
(4.18)E(z)=P0V1(z)⋯Vn(z),
where P0 is defined by (4.13) with θ∈[-π,π) and Vj(z), j=1,2 are defined by (4.14) with αj∈[-π,π), j=1,2.Proof.
IfE(z) can be factorized as (4.18), then it is a causal paraunitary filter and satisfies (4.9). Conversely, we set a paraunitary matrix En(z)=e0+e1z-1+⋯+enz-3n satisfies with (4.18), where n∈N. If n=0, it is easy to get that E(z) is P0. For n∈N∖{0}, define
(4.19)En-1(z)=En(z)VT(z-1),
where V(z) is defined by (4.14). Now we prove that there exists α1, α2 such that En-1(z) defined by (4.19) is causal. From VT(z-1), we only need to prove that there exists α1, α2 such that
(4.20)e0[sinα1,-cosα1,sinα2,-cosα2]T=0,(4.21)e0[sinα1,-cosα1,-sinα2,cosα2]T=0,(4.22)e0[cosα1,sinα1,-cosα2,-sinα2]T=0,(4.23)e1[sinα1,-cosα1,-sinα2,cosα2]T=0,(4.24)e1[cosα1,sinα1,-cosα2,-sinα2]T=0,(4.25)e2[cosα1,sinα1,-cosα2,-sinα2]T=0.
Let us start by discussing the casee0=0. Then (4.20)–(4.22) hold. By the symmetry of En(z),    ej=diag(1,1,-1,-1)e3n-jdiag(1,1,-1,-1), 0≤j≤3n. By the paraunitariness of En(z),  e1e3n-1T=0. Then we have e1diag(1,1,-1,-1)e1T=0, rank(e1)≤2. For simplicity, we only consider the case rank(e1)=1 in the following. Let e1=u1[cosθ1,sinθ1,cosθ2,sinθ2], u1∈R4, θ1,θ2∈[-π,π). Then for α1=θ1, α2=θ2, (4.23) and (4.24) hold. Let Q=e2[cosθ1,sinθ1,-cosθ2,-sinθ2]T, by the paraunitariness of En(z), e1e3n-1T+e2e3nT=0. Then we have u1QT+Qu1T=0. It is clear that Q=0 since u1≠0. Then (4.25) holds for α1=θ1, α2=θ2.
Ife0≠0, we have e0e3nT=0 and e3n=diag(1,1,-1,-1)e0diag(1,1,-1,-1). Let e0=u1[cosθ1,sinθ1,cosθ2,sinθ2], then (4.20)–(4.22) hold for α1=θ1, α2=θ2. Similarly, when e1=u2[-sinθ1,cosθ1,-sinθ2,cosθ2], (4.23), and (4.24) hold for α1=θ1, α2=θ2. Let Q=e2[cosα1,sinα1,-cosα2,-sinα2]T, from the paraunitariness of En(z), we have e1e3n-1T+e2e3nT=0. Thus, u2QT+Qu2T=0. It is clear that Q=0 since u2≠0. So (4.25) holds for α1=θ1, α2=θ2.
SoEn-1(z) is causal and En-1(z) can be factorized in the form of En-1(z)=P0V1(z)⋯Vn-1(z). Thus, En(z) can be factorized in the form of En(z)=P0V1(z)⋯Vn-1(z) by induction assumption. The proof of Theorem 4.6 is complete.Therefore, we have the following corollary.Corollary 4.7.
Leth(z), g(z), f(z), s(z) is symmetric or antisymmetric. If M is a paraunitary matrix, then h(z), g(z), f(z), s(z) can be factorized as
(4.26)(1)[h(z)g(z)f(z)s(z)]=612P0V1(z3)⋯Vn(z3)[-1+2z-1+2z-2-z-3-1-2z-1-2z-2-z-31-2z-1+2z-2-z-31+2z-1-2z-2-z-3],
for the case of t0=1, t1=t2=-1;
(4.27)(2)[h(z)g(z)f(z)s(z)]=612V0V1(z3)⋯Vn(z3)[-1+2z-1+2z-2-z-3-1-2z-1-2z-2-z-31-2z-1+2z-2-z-31+2z-1-2z-2-z-3],
for the case oft0=-1, t1=1, t2=-1, where V0=diag(1,J2,1)P0;
(4.28)(3)[h(z)g(z)f(z)s(z)]=612U0V1(z3)⋯Vn(z3)[-1+2z-1+2z-2-z-3-1-2z-1-2z-2-z-31-2z-1+2z-2-z-31+2z-1-2z-2-z-3],
for the case of t0=t1=-1, t2=1, where U0=diag(1,J3)P0.
WhereP0 is defined by (4.13) with θ∈[-π,π) and Vj(z) are defined by (4.14) with αj∈[-π,π).(ii) For the second case, without loss of generality, supposet0=1, t1=1, t2=-1, then E(z) defined by (4.8) satisfies
(4.29)E(z)=z-3ndiag(1,1,1,-1)E(z-1)diag(1,1,-1,1).
Let n=0 in (4.29), then one can get an orthogonal matrix Q0 as follows:
(4.30)Q0=[cosθ00-ρ1sinθ0ρ20000ρ30sinθ00ρ1cosθ].
Then one has the following results.Theorem 4.8.
LetE(z) is a causal paraunitary matrix filter and satisfies (4.29) for some n∈N, then E(z) cannot be factorized in the form of
(4.31)E(z)=Q0V1(z)⋯Vn(z),
for Q0 defined by (4.30) with θ∈[-π,π) and Vj(z), j=1,2 defined by (4.14) with αj∈[-π,π), j=1,2.Proof.
SupposeE(z) can be factorized in the form of (4.31). Then when n=1, we have
(4.32)V1(z)=z-3diag(1,1,1,-1)V1(z-1)diag(1,1,-1,1);
when n=2, we have
(4.33)V2(z)=z-3diag(1,1,-1,1)V2(z-1)diag(1,1,-1,1).
However, (4.33) is in contradiction with (4.32).


## 4.2. The Case ofγ=3n+2

We have the following results.Theorem 4.9.
E
(
z
) is a causal paraunitary matrix filter defined by (4.8) for some n∈N if and only if it can be factorized in the form of(4.34)
E
(
z
)
=
P
1
V
2
(
z
)
V
3
(
z
)
⋯
V
n
(
z
)
,whereP1 is defined by the following with θj∈[-π,π), j=1,2(4.35)P1(z)=12[cosθ1sinθ1cosθ2sinθ2cosθ1sinθ1cosθ2sinθ2-2sinθ12cosθ10000-2sinθ22cosθ2]+12[cosθ1sinθ1-cosθ2-sinθ2-cosθ1-sinθ1cosθ2sinθ200000000]z-1,andVj(z) are defined by (4.14) with αj∈[-π,π).Corollary 4.10.
Leth(z), g(z), f(z), s(z) is symmetric or antisymmetric. If M is a paraunitary matrix, then h(z), g(z), f(z), s(z) can be factorized as
(4.36)(1)[h(z)g(z)f(z)s(z)]=612P1(z)V2(z3)⋯Vn(z3)[-1+2z-1+2z-2-z-3-1-2z-1-2z-2-z-31-2z-1+2z-2-z-31+2z-1-2z-2-z-3],
for the case t0=-1, t1=1, t2=-1;
(4.37)(2)[h(z)g(z)f(z)s(z)]=612diag(1,1,J2)P1(z)V2(z3)⋯Vn(z3)[-1+2z-1+2z-2-z-3-1-2z-1-2z-2-z-31-2z-1+2z-2-z-31+2z-1-2z-2-z-3],
for the case of t0=t1=-1, t2=1. Where P1(z) is defined by (4.35) with θ∈[-π,π) and Vj(z) are defined by (4.14) with αj∈[-π,π).Remark 4.11.
In the case oft0=1, t1=t2=-1, we cannot construct a paraunitary matrix P1(z). The reason is similarly as Theorem 4.8 when t0=1, t1=t2=-1.


## 4.3. The Case ofγ=3n+3

In this situation, we cannot also construct the corresponding paraunitary matrix.


## 5. Examples

In this section, we will construction some examples of 3-band wavelet frames with symmetry/antisymmetry by applying the parameterizations of masksh, g, f, s provided in Corollarys 4.7 and 4.10.Example 5.1.
Leth, g, f, s be the filters given by (4.36) with n=1. Then
(5.1)h(z)=624[(-cosθ1-sinθ1+cosθ2+sinθ2)+2(cosθ1-sinθ1-cosθ2+sinθ2)z-1+2(cosθ1-sinθ1+cosθ2-sinθ2)z-2-2(cosθ1+sinθ1+cosθ2+sinθ2)z-3+2(cosθ1-sinθ1+cosθ2-sinθ2)z-4+2(cosθ1-sinθ1-cosθ2+sinθ2)z-5+(-cosθ1-sinθ1+cosθ2+sinθ2)z-6],g(z)=624[(-cosθ1-sinθ1+cosθ2+sinθ2)+2(cosθ1-sinθ1-cosθ2+sinθ2)z-1+2(cosθ1-sinθ1+cosθ2-sinθ2)z-2-2(cosθ1-sinθ1+cosθ2-sinθ2)z-4+2(cosθ1-sinθ1-cosθ2+sinθ2)z-5-(-cosθ1-sinθ1+cosθ2+sinθ2)z-6],f(z)=624[(2sinθ1-2cosθ1)+2(-2sinθ1-2cosθ1)z-1+2(-2sinθ1-2cosθ1)z-2]+(2sinθ1-2cosθ1)z-3,s(z)=624[(-2sinθ1+2cosθ1)+2(2sinθ1+2cosθ1)z-1+2(-2sinθ1-2cosθ1)z-2]+(2sinθ1-2cosθ1)z-3.
For h(1)=1, g(1)=f(1)=s(1)=0, h(k)=0, k=-1/2±(3/2)i, we have cosθ1=-3/6(2-2), sinθ1=-3/6(2+2), cosθ2=σ, sinθ2=±1-σ2. Let h has sum rules of order 2, such as h'(k)=0, then we can get that sinθ2=-3/18(2+52), cosθ2=-3/18(-2+52). Thus
(5.2)h0=-118,h1=19,h2=29,h3=49,h4=h2,h5=h1,h6=h0,g0=-118,g1=19,g2=29,g3=0,g4=-g2,g5=-g1,g6=-g0,f0=-26,f1=26,f2=f1,f3=f0,s0=218,s1=-5218,s2=-s1,s3=-s0,
The resulting symmetric refinable function ϕ∈W0.8155(R). See Figure 1 for the graphs of ϕ, ψ1, ψ2, and ψ3.(a) Refinable functionϕ∈W0.8155(R) with symmetry, (b) framelets ψ1 with antisymmetry, and (c) and (d) ψ2 symmetry and ψ3 with antisymmetry, respectively.
(a)
(b)
(c)
(d)Example 5.2.
Leth, g, f, s be the filters given by (4.26) with n=1. For h(1)=1, g(1)=f(1)=s(1)=0,h(k)=0, k=-1/2±(3/2)i, let a1=sinθ, a2=cosθ, b1=sinθ1, b2=cosθ1, c1=sinθ2, c2=cosθ2, then we have
(5.3)a1=-16[-22b22+82b21-b22-4b22+8b21-b22+2+63b2+1-b22+22b2],a2=-66(-b2+1-b22-2b2-2-2b22),b1=1-b22,b2=b2,c1=±1-c22,c2=c2.
Let h has sum rules of order 2, such as h'(k)=0, then we can get that sinθ=-0.8625, cosθ=-0.5061, sinα1=0.7645, cosα1=-0.6447, sinα2=-0.5172, cosα2=-0.8559.
Thus(5.4)h0=0.1314,h1=0.1333,h2=0.2176,h3=-0.0550,h4=-0.1091,h5=0.0915,h6=0.1806,h7=h5,…,h12=h0,g0=-0.0771,g1=-0.0782,g2=-0.1277,g3=0.1590,g4=-0.1858,g5=0.1560,h6=0.3077,g7=g5,…,g12=g0,f0=0.0771,f1=0.0782,f2=0.1277,f3=0.0295,f4=-0.1058,f5=0.1560,f6=0,f7=-f5,…,f12=-f0,s0=0.1314,s1=0.1333,s2=0.2176,s3=-0.1656,s4=0.1091,s5=-0.0915,s6=0,s7=-s5,…,s12=-s0.
The resulting symmetric refinable function ϕ∈W0.3048(R). See Figure 2 for the graphs of ϕ, ψ1, ψ2,   and ψ3.(a) Refinable functionϕ∈W0.3048(R) with symmetry, (b) framelets ψ1 with symmetry, and (c) and (d) ψ2 symmetry and ψ3 with antisymmetry, respectively.
(a)
(b)
(c)
(d)



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*Source: 907175-2012-10-03.xml*