Chapter Introduction. Jianzhong Wang

Transcription

1 Chapter Elementary Matrix Decomposition Algorithm for Symmetric Extension of Laurent Polynomial Matrices and its Application in Construction of Symmetric M-band Filter Banks Jianzhong Wang Abstract In this paper, we develop a novel and effective algorithm for the construction of perfect reconstruction filter banks (PRFBs) with linear phase. In the algorithm, the key step is the symmetric Laurent polynomial matrix extension (SLPME). There are two typical problems in the construction: () For a given symmetric finite low-pass filter a with the polyphase, to construct a PRFBs with linear phase such that its low-pass band of the analysis filter bank is a. (2) For a given dual pair of symmetric finite low-pass filters, to construct a PRFBs with linear phase such that its low-pass band of the analysis filter bank is a, while its low-pass band of the synthesis filter bank is b. In the paper, we first formulate the problems by the SLPME of the Laurent polynomial vector(s) associated to the given filter(s). Then we develop a symmetric elementary matrix decomposition algorithm based on Euclidean division in the ring of Laurent polynomials, which finally induces our SLPME algorithm.. Introduction The main purpose of this paper is to develop a novel and effective algorithm for the construction of perfect reconstruction filter banks (PRFBs) with linear phase. In the algorithm, the key step is the symmetric Laurent polynomial matrix extension (SLPME). PRFBs have been widely used in many areas such as signal and image processing, data mining, feature extraction, and compressive sensing [,, 2,, 4]. A PRFB consists of two sub-filter banks: an analysis filter bank, which decomposes a signal into different bands, and a synthesis filter bank, which composes a signal from its different band components. Either an analysis filter bank or a synthesis one consists of several band-pass filters. Assume that an analysis filter bank con- Jianzhong Wang ( ) Department of Mathematics and Statistics, Sam Houston State University, Huntsville, USA, jzwang@shsu.edu 47

2 48 Jianzhong Wang sists of the filter set {H 0,H,,H M } and a synthesis filter bank consists of the set {B 0,B,,B M }, where H 0 and B 0 are low-pass filters. Then they form an M-band PRFB if and only if the following condition holds: M B j ( M)( M)H j = I, (.) j=0 where M is the M-downsampling operator, M is the M-upsampling operator, I is the identity operator, and H j denotes the conjugate filter of H j. Note that the conjugate of a real filter a = (,a,a 0,a, ) is defined as ā = (,a,a 0,a, ). In signal processing, a filter H having only finite non-zero entries is called a finite impulse response (FIR). Otherwise it is called an infinite impulse response (IIR). Since FIR is much more often used than IIR, in this paper we only study FIR with real entries. Recall that the z-transform of a FIR H is a Laurent polynomial (LP) and the z- transform of the conjugate filter of H is H(z) = [ H(/z). We define the M-polyphase ] form of a signal (or a filter) x by the LP vector a [M,0] (z),,a [M,M ] (z)}, where a [M,k] (z) = a(m j + k)z j, 0 k M. j For convenience, we will simplify a [M,k] to a [k] if it does not cause confusion. The polyphase form of a M-band filter bank {H 0,,H M } is the following LP matrix. H [0] 0 (z) H[] 0 (z) H[M ] 0 (z) H [0] H(z) = (z) H[] (z) H[M ] (z)... H [0] M (z) H[] M (z) H[M ] M (z) Using polyphase form, we represent (.) as a LP matrix identity in the following theorem. Theorem.. The filter bank pair of {H 0,,H M } and {B 0,,B M } realizes a PRFB if and only if the following identity holds: H(z)B (z) = I, (.2) M where both H(z) and B(z) are LP matrices, and B (z) denotes the conjugate transpose matrix of B(z). We denote by L the ring of all Laurent polynomials, and call a LP matrix is L - invertible, if its inverse is a LP matrix too. Since MB (z) = H (z) in a PRFB, the polyphases of its analysis filter bank and its synthesis one are L -invertible. By (.2), we also have M MH [ j] 0 (z) B [ j] 0 (z) =. j=0

3 Symmetric LP Matrix Extension 49 In general, we will call a LP vector a(z) = [a (z),,a M (z)] a prime one if there is a LP vector b(z) = [b (z),,b M (z)] such that a(z)b T (z) =. More details of the theory of PRFBs are referred to [0, 5]. The filters with symmetry (also called with linear phases) are more desirable in application [0]. They are formally defined as follows: Definition.2. Let c be an integer. A filter (or signal) x is called symmetric or antisymmetric about c/2 if x(k) = x(c k) or x(k) = x(c k), k Z, respectively. Later, for simplification, we will use the term symmetric to mention both symmetric and antisymmetric. Thus, x is symmetric if and only if x(k) = εx(c k), where ε (= or ) is the symbol of the symmetry-type. Note that we can always shift a signal/filter x such that the shifted one has the symmetric center at c = 0 or c = ±. Hence, without loss of generality, in this paper we always assume that a symmetric filter has the center at c = 0, c =, or c =, and simply call it 0- symmetric, -symmetric, or ( )-symmetric, respectively. Correspondingly, the set of all 0-symmetric filters (-symmetric, or ( )-symmetric ones) is denoted by V 0 (V or V ). Besides, when we need to stress on the symmetry-type, we denote by V +,V 0 +,V + for ε = + and V,V 0,V for ε =. It is clear that if x V 0, then so is x, and if x V, then x V. We also have the following: x V 0 if and only if x(z) = εx(/z), x V if and only if x(z) = εzx(/z), and x V if and only if x(z) = ε/zx(/z). In addition, if a(z) = εb(/z) (a(z) = εzb(/z),a(z) = ε/zb(/z)), we call [a(z),b(z)] a V 0 (V, V ) pair. For a symmetric filter H, we modify its M-polyphase to the following: H [k] (z) = H(M j + k)z j, m k M m, m = j [ M Later, a LP vector is called a S-LP one if it is a polyphase form of a symmetric filter. Similarly, we will call a LP matrix Sr-LP matrix (Sc-LP matrix) if its rows (columns) are S-LP vectors. They will be simply called S-LP matrices if row and column are not stressed. Similarly,, a PRFB is called symmetric if all of its bandfilters are symmetric. Two fundamental problems in the construction of symmetric PRFBs are the following: Problem. Assume that a given symmetric low-band filter H 0 has a prime polyphase. How to construct a symmetric PRFB, in which the first band of its analysis filter bank is H 0? Problem 2. Assume that a dual pair of symmetric low-band filters H 0 and B 0 are given. How to find other symmetric components H,,H M and B,,B M so that they form a symmetric PRFB? By Theorem., we have the following: Corollary.. The symmetric filter banks {H 0,,H M } and {B 0,,B M } form a symmetric PRFB if and only if the following identity holds: 2 ]. H(z)B (z) = I, (.) M

4 50 Jianzhong Wang where both H(z) and B(z) are Sr-LP matrices. Ignoring the factor M on the right-hand side of (.) in Corollary., we can see that the two fundamental problems are equivalent the following symmetric Laurent polynomial Matrix extension (SLPME) problems: SLPME Problem. Assume that a given S-LP row vector a(z) L M is prime. To find an L -invertible Sr-LP matrix A(z) such that A(,:) = a. SLPME Problem 2. Assume that a given pair of S-LP row vectors [a(z),b(z)] satisfies a(z)b T (z) =. To find an L -invertible Sr-LP matrix A(z) such that A(,: ) = a and A (:,) = b T. Laurent polynomial matrix extension (LPME) has been discussed in [, 4, 9]. Having the aid of LPME technique, several algorithms have been developed for the construction of PRFBs [2, 5, 6, 7, 6, 5]. Unfortunately, the methods for constructing LPME usually do not produce SLPME. The main difficulty in SLPME is how to preserve the symmetry. Recently, Chui, Han, and Zhuang in [2] proposed a bottomup algorithm for solving SPLME Problem 2 based on the properties of dual filters. In this paper, we solve the problem in the framework of the algebra of Laurent polynomials. Our approach to SLPME is based on the decomposition of L - invertible S-LP matrix in the LP ring [5]. To make the paper more readable, we restrict our discussion for M = 2,,4. The readers can find that our algorithms can be extended for any integer M without essential difficulty. The paper is organized as follows. In Section 2, we discuss the properties of S-LP vectors and the symmetric Euclidean division in the LP ring. In Section, we introduce the elementary S-LP matrix decomposition technique and apply it in the development of the SLPME algorithms. Finally, two illustrative examples are presented in Section 4..2 S-LP Vectors and Symmetric Euclidean Division For simplification, in the paper, we only discuss LP with real coefficients. Readers will find that our results can be trivially generalized to LP with coefficients in the complex field or other number fields. Let the ring of all polynomials be denoted by P and write P h = P \ {0}. Similarly, let the ring of all Laurent polynomials be denoted by L and write L h = L \ {0}. If a L h, we can write a(z) = n k=m a kz k, where n m and a m a n 0. We define the highest degree and the lowest degree of a L h by deg + (a) = n and deg (a) = m respectively. When a = 0, we agree that deg + (0) = and deg (0) =. We define the support length of a by supp(a) = deg + (a) deg (a). Particularly, when a(z) L h is 0-symmetric, -symmetric, or ( )-symmetric, we have deg (a) = deg + (a), deg (a) = deg + (a) +, or deg (a) = deg + (a), respectively. Let the semi-group G P h be defined by G = {p P h : p(0) 0}. Then, the power mapping π : L h G, π(a(z)) = z deg (a) a(z), defines an equivalent relation in L h, i.e., a b if and only if π(a) = π(b). For convenience, we agree that

5 Symmetric LP Matrix Extension 5 π(0) = 0. Let L m denote the group of all non-vanished Laurent monomials: L m = {m L h ; m = cz l,c 0,l Z}. Then, we have π(m) = c. For a LP vector a = [a,,a s ], we define π(a) = [π(a ),,π(a s )]. Then the greatest common divisor (gcd) of a nonzero row (or column) LP vector a L s is defined by gcd L (a) = gcd(π(a)) G. A LP a(z) L h is said to be in the subset L d if a(z) = εa(/z) and gcd L (a(z),a(/z)) =. A LP matrix A(z) L s s is said to be L -invertible if A(z) is invertible and A (z) L s s too. It is obvious that A(z) is L -invertible if and only if det(a(z)) L m. We now discuss the properties of S-LP vectors. Recall that an M dimensional S-LP vector is defined as the M-polyphase form of a symmetric filter. Let x(z) be an M-dimensional S-LP vector. We list its symmetric properties for M = 2,,4, in Table.. M = 2 c = 0 x [0] (z) = εx [0] (/z),x [] (z) = ε/zx [] (/z) c = x [0] (z) = εx [] (/z) M = c = 0 x [0] (z) = εx [0] (/z),x [] (z) = εx [ ] (/z) c = x [0] (z) = εx [] (/z),x [ ] (z) = εzx [ ] (/z) M = 4 c = 0 x [0] (z) = εx [0] (/z),x [] (z) = εx [ ] (/z),x [2] (z) = ε/zx [2] (/z) c = x [0] (z) = εx [] (/z),x [ ] (z) = εx [2] (/z) Table.: The symmetry of the components in a S-LP vector. Let m = [ M 2 ]. We can verify that, when M is even and c = 0, x[0] (z) V 0,x [m] (z) V, and [x [i] (z),x [ i] (z)],i =,,M, are V 0 pairs; when M is even and c =, (x [i] (z),x [ i+] (z)),i =,,M, are V 0 pairs; when M is odd and c = 0, x [0] (z) V 0, and [x [i] (z),x [ i] (z)],i =,,M, are V 0 pairs; when M is odd and c =, [x [i] (z),x [ i+] (z)],i =,,M, are V 0 pair and x [ M] (z) V. We need the following L -Euclid s division theorem [5] in our discussion. Theorem.4. Let (a,b) L h L h and supp(a) supp(b). Then there exists a unique pair (q,r) L L such that a(z) = q(z)b(z) + r(z) with supp(r) + deg (a) deg + (r) < supp(b) + deg (a), (.4) which implies that supp(q) supp(a) supp(b) and supp(r) < supp(b). By Theorem.4, it is also clear that if deg (a) = deg (b), then q P h and deg(q) deg + (a) deg + (b). From Theorem.4, we derive the symmetric L - Euclid s division theorem to deal with S-LP vectors. Theorem.5. Let a(z) V 0 with supp(a) = 2m, b(z) V with supp(b) = 2k, c(z) V with supp(c) = 2s, and d(z) L h with supp(d) = l be given. Then we have the following:. If m k, then there is p(z) V + with supp(p) 2(m k) and a (z) V 0 with supp(a ) < supp(b) such that a(z) = b(z)p(z) + a (z). If m < k, then there is q(z) V + with supp(q) 2(k m) and b (z) V with supp(b ) < supp(a) such that b(z) = q(z)a(z) + b (z).

6 52 Jianzhong Wang 2. If m s, then there is q(z) V + with supp(q) 2(m k) and a (z) V 0 with supp(a ) < supp(c) such that a(z) = c(z)q(z) + a (z). If m < s, then there is p(z) V + with supp(p) 2(k m) and c (z) V with supp(c ) < supp(a) such that c(z) = p(z)a(z) + c (z).. If supp(a) > supp(d), there is a p(z) P h with deg(p) m [ l+ 2 ] and a (z) V 0 with supp(a ) l such that a(z) = p(z)d(z) + ε p(/z)d(/z) + a (z). 4. If supp(b) > supp(d), there is a q(z) P h with deg(q) k [ l+ 2 ] and b (z) V with supp(b ) l such that b(z) = q(z)d(z) + ε/zq(/z)d(/z) + b (z). Similarly, if supp(c) > supp(d), there is a p(z) P h with deg(p) c [ l+ 2 ] and c (z) V with supp(c ) l such that c(z) = p(z)d(z) + εzp(/z)d(/z) + c (z). Proof. To prove (), we write a(z) = m j= m a jz j and set a t (z) = m j=k a jz j + 2 k j= k+ a jz j so that a t (z) + εa t (/z) = a(z). By Theorem.4, we can find a ˆp(z) P h with deg( ˆp) m k such that a t (z) = ˆp(z)b(z) + r(z), where r L with deg + (r) < k,deg (r) > k. It leads to a(t) = ˆp(z)b(z) + ε ˆp(/z)b(/z) + r(z) + εr(/z). Since b(z) V, we have b(z) = ε/zb(/z), which yields a(z) = ( ˆp(z) + z ˆp(/z))b(z) + (r(z) + εr(/z)). Write p(z) = ˆp(z) + z ˆp(/z),a (z) = r(z) + εr(/z). It is obvious that p(z) V + with supp(p) 2(m k) and a (z) V 0 with supp(r) < supp(b). The proof of the first statement of () is completed. The proofs of the remains are similar.. SLPME Algorithms Based on Elementary S-LP Matrix Decomposition We now discuss SLPME algorithms for M = 2,,4, respectively... The Case of M = 2 We say a(z) = [a (z),a 2 (z)] V 0,2 if a (z) V 0,a 2 (z) V ; and say a(z) V,2 if it is a V 0 pair. We also say b(z) V 0,2 if b (z) V 0,b 2 (z) V. Define S 0,2 = { S(z) = [s i j (z)] 2 i, j=; s ii (z) V + 0,i =,2,s 2(z) V +,s 2(z) V + }. Thus, if S(z) S 0,2, then S(,:)(z) V 0,2 and S(:,)(z) V 0,2.

7 Symmetric LP Matrix Extension 5... The Case of a V 0,2. To develop our SLPME algorithm, we give the following: Definition.6. Let s(z) V +, t(z) V +, k Z, and r R \ {0}. Then the following matrices [ ] [ ] [ s(z) 0 rz E u (s) = E 0 l (t) =, D(r,k) = k ] 0 (.5) t(z) 0 are called the elementary S 0,2 matrices, and their product is called a S 0,2 - fundamental matrix. It can verify that all of the matrices in (.5) are L -invertible and their inverses are also in S 0,2. Indeed, we have (E u (s)) = E u ( s) (E l (t)) = E l ( t) (D(r,k)) = D(/r, k). (.6) Later, we simply denote by E u,e l,d for the matrices in (.6). We now return the SLPME for a V 0,2. WLOG, we assume supp(a ) > supp(a 2 ). Since gcd L (a) =, By Theorem.5, we can use elementary S 0,2 matrices to make the following: a 0 E l( p ) a E u( q ) a 2 El( p n ) a 2n E u( q n ) 2n D(r,k) a [,0], where a 0 = a, a 2i E l ( p i+ ) = a 2i+,a 2i+ E u ( q i+ ) = a 2i+2,i =,,n. Let E a = E l ( p )E u ( q ) E l ( p n )E u ( q n )D(r,k). (.7) Then E a S 0,2,aE a = [,0], and its inverse A a (z) = E a (z) D(/r, k)e u (q n )E l (p n ) E u (q )E l (p ) (.8) provides a solution for SLPME Problem. We now consider the SPLME Problem 2. WLOG, assuming that the symmetric dual pair [a(z),b(z)] V 0,2 V 0,2 is given. Let E a (z),a a (z) be the matrices given in (.7) and (.8). By a(z)b T (z) = and A a (,:) = a, we have A a (z)b T (z) = [,w(z)] T with w(z) V, which yields E a (z)[,w(z)] T = b T (z). Then the matrices Ã(z) = E l ( w)(z)a a (z) and B(z) = E a (z)e l (w)(z) give the solution....2 The Case of a(z) V,2. If its dual b(z) is not given, then by the extended Euclidean algorithm in [5], we can find LP vector s(z) = [s (z),s 2 (z)], such that as T =. We define b (z) = 2 (s (z) + εs 2 (/z)),b 2 (z) = εb (/z). The vector b(z) is a V 0 pair and ab T =. We now define

8 54 Jianzhong Wang [ ] a (z) a A(z) = 2 (z), b 2 (z) b (z) which is L -invertible and its inverse is B(z) = A (z) = [ ] b (z) a 2 (z). b 2 (z) a (z) Then A(z) provides the solution of SPLME Problem, and the pair [A(z),B(z)] gives the solution of SPLME Problem The Case of M = We say a(z) = [a (z),a 2 (z),a (z)] V 0, if a 2 (z) V 0 and [a (z),a (z)] is a V 0 pair, and say a(z) V, if a 2 (z) V and [a (z),a (z)] is a V 0 pair. We also say b(z) V, if b 2 (z) V and [b (z),b (z)] is a V 0 pair. Note that, in the case of M =,c =, the polyphase form x(z) is not in V,, but [x [0] (z),x [ ] (z),x [] (z)] V,...2. The Case of a V 0, Definition.7. Let q(z) L. The matrices of the following two types are called elementary S 0, matrices: E v (q) = 0 0 q(z) q(/z), E h (q) = q(z) q(/z) In general, we simply denote by E an elementary S 0, matrix, and call their product a Fundamental S 0, one. It is clear that Ev (q) = E v ( q) and Eh (q) = E h( q). By the same argument for M = 2, using the elementary S 0, matrices, we can obtain the following chain: a 0 E a E 2 a 2 En a n where a n has the same symmetry as a and gcd L (a n ) =. Therefore, a n = [p(z),0,ε p(/z)], p(z) L d. Besides, when ε =, it may have another form a n = [0,r,0]. Writing E = E E 2 E n, we have a(z) = a n (z)e (z). Let q(z) L d satisfy p(z)q(z)+ p(/z)q(/z) = and set q(z) = [q(z), 0, εq(/z)]. We define

9 Symmetric LP Matrix Extension 55 0 /2 /2 q(z) 0 ε p(/z) Q (z) = r 0 0 Q 2 (z) = 0 0, (.9) 0 /2 /2 εq(/z) 0 p(z) whose inverses are Q (z) = 0 r p(z) 0 ε p(/z) 2 (z) = 0 0. εq(/z) 0 q(z) Q Finally, we define A(z) = { Q (z)e (z), if a n = [0,r,0], Q 2 (z)e (z), if a n = [p(z),0,ε p(/z)]. It is clear that A(z) is a SLPME of a(z). We now return to SPLME Problem 2. Assume a symmetric dual pair [a(z),b(z)] V 0, V 0, is given so that a(z)b T (z) =. Let E(z) be the LP matrix above. Define w(z) = b(z)(e ) T (z) V 0,. Then a n w T = a n E b T = ab T =. Hence, { [u(z),/r,εu(/z)], if a n = [0,r,0], w(z) = [v(z),v c (z),εv(/z)], if a n = [p(z),0,ε p(/z)], where p(z)v(z) + p(/z)v(/z) =. Write q + (z) = u(z) + εu(/z),q (z) = u(z) εu(/z). Define Q (z) = u(z) v(z) 0 ε p(/z) /r 0 0, Q 2 (z) = v c (z) 0, (.0) εu(/z) εv(/z) 0 p(z) whose inverses are Q (z) = 2 We now define 0 2r 0 rq + (z), Q 2 (z) = rq (z) A(z) = p(z) 0 ε p(/z) v c (z)p(z) εv c (z)p(/z). εv(/z) 0 v(z) { Q (z)e (z), if a n = [0,r,0], Q 2 (z)e (z), if a n = [p(z),0,ε p(/z)]. Then, the pair [A(z),A (z)] is a SLPME of the pair [a(z),b(z)].

10 56 Jianzhong Wang..2.2 The Case of a V, The discussion is very similar to the case of a V 0,. Definition.8. Let q(z) L. The matrices of the following two types are called elementary S, matrices: E v (q) = 0 0 q(z) zq(/z), E h (q) = q(z) zq(/z) The product of elementary S, matrices is called a Fundamental S, one. Using the elementary S, matrices, we can obtain the following chain: a 0 E a E 2 a 2 En a n where a n has the same symmetry as a and gcd L (a n ) =. Therefore, a n = [p(z),0,ε p(/z)], p(z) L d. Note that, because a 2 (z) S,, a n does not have other forms. Writing E = E E 2 E n, we have a(z) = a n (z)e (z). Let Q 2 (z) be the LP matrix in (.9). Then A(z) is a SLPME of a(z). We now consider SPLME Problem 2. In the given dual pair, b(z) V,. Let E(z) be the LP matrix above. Then the LP vector w(z) = b(z)(e ) T (z) V, too. Hence, it has the only form of w(z) = [v(z),v c (z),εv(/z)], v c (z) V, where p(z)v(z) + p(/z)v(/z) =. Let Q 2 (z) be the LP matrix in (.0), and A(z) = Q 2 (z)e (z). Then [(E(z)Q 2 (z)),e(z)q 2 (z)] is a SLPME of the dual pair [a(z),b(z)]... The Case of M = 4 In this case, we say a(z) V 0,4 if a (z) V 0, [a 2 (z),a (z)] is a V 0 pair, and a 4 (z) V ; say a(z) V,4 if both [a (z),a 4 (z)] and [a 2 (z),a (z)] are V 0 pairs. We also say b(z) V 0,4 if b(/z) V 0,4. Note that, if x(z) is the polyphase form of an asymmetry filter in the case of M = 4,c = 0, then [x [0] (z),x [ ] (z),x [] (z),x [2] (z)] V 0,4.

11 Symmetric LP Matrix Extension The Case of a V 0,4 Definition.9. Let q(z) L,s(z) V,t(z) V. The followings are called elementary S 0,4 matrices: q(z) q(/z) 0 Eh 0 (q) = , E0 v (q) = ( 0 0 s(z) Eh 0 (q)) T, Et (s) = , Eh (q) = , E v (q) = ( Eh (q)) T, Eb (t) = q(z) zq(/z) t(z) 0 0 Using the elementary S 0,4 matrices, we can obtain the following chain: a 0 E a E 2 a 2 En a n, where a n has the same symmetry as a and gcd L (a n ) =. Therefore, a n = [0, p(z),ε p(/z),0], p(z) L d. If ε =, it possibly can also have the form (a) n (z) = [r,0,0,0],r 0. Let q(z) L d satisfy p(z)q(z) + p(/z)q(/z) =. Write E = E E 2 E n and define /r Q (z) = 0 /2 /2 0 0 /2 /2 0 Q 2(z) = q(z) 0 0 ε p(/z) εq(/z) 0 0 p(z), whose inverses are r Q (z) = p(z) ε p(z) 0 2 (z) = εq(/z) q(z) 0 Q Set A(z) = { Q (z)e (z), if a n = [r,0,0,0], Q 2 (z)e (z), if a n = [0, p(z),ε p(/z),0]. (.) Then A(z) is a SLPME of a. We now consider SPLME Problem 2. In the given dual pair, b(z) is now in V 0,4. Let E(z) be the LP matrix above. Then the LP vector w = b(z)(e ) T (z) is in V 0,4 too. Hence, if a n = [r,0,0,0], w(z) = [/r,v(z),εv(/z),v (z)], v (z) V,

12 58 Jianzhong Wang else if a n = [0, p(z),ε p(/z),0], w(z) = [v 0 (z),v(z),εv(/z),v (z)], v 0 (z) V 0,v (z) V, where p(z)v(z) + p(/z)v(/z) =. Let /r v 0 (z) 0 0 Q (z) = v(z) /2 /2 0 εv(/z) /2 /2 0 Q 2(z) = v(z) 0 0 ε p(/z) εv(/z) 0 0 p(z), v (z) 0 0 v (z) 0 0 whose inverses are r Q (z) = w + (z) 0 w (z) 0, v (z) p(z) ε p(z) 0 2 (z) = p(z)v 0 (z) ε p(/z)v 0 (z) 0 0 p(z)v (z) ε p(/z)v (z), 0 εv(/z) v(z) 0 Q where w + (z) = s(z) + εs(/z) and w (z) = s(z) εs(/z). Let A(z) be given by (.). Then [A(z),A (z)] is a SLPME of the dual pair [a(z),b(z)]....2 The Case of a V, Let P = be the permutation matrix Definition.0. Let q(z) L. The matrix with the form of q(z) q(/z) 0 E (q) = and E 2 (q) = PE (q)p are called elementary S,4 matrices. Using the elementary S,4 matrices, we can obtain the following chain: a 0 E a E 2 a 2 En a n, where a n (z) = [p(z),0,0,ε p(/z)] with p(z) L d or a n = [0, p(z),ε p(/z),0], p(z) L d. Since [p(z),0,0,ε p(/z)] = [0, p(z),ε p(/z),0]p, we only discuss the first case. Let q(z) L d satisfy p(z)q(z) + p(/z)q(/z) =. Write E = E E 2 E n and define

13 Symmetric LP Matrix Extension 59 q(z) 0 0 ε p(z) Q(z) = 0 /2 /2 0 0 /2 /2 0, εq(z) 0 0 p(z) whose inverse is p(z) 0 0 ε p(/z) Q (z) = εq(/z) 0 0 q(z) The matrix A(z) = Q (z)e (z) is a SLPME of a(z). We now consider SPLME Problem 2. In the given dual pair, b(z) V,4. Let E(z) be the LP matrix above. We still assume that a n = [p(z),0,0,ε p(/z)], p(z) L d. Then the LP vector w(z) = b(z)(e ) T (z) has the form of w(z) = [v(z),s(z),εs(/z),εv(/z)], v(z) L d, where p(z)v(z) + p(/z)v(/z) =. Let w + (z) = s(z) + εs(/z),w (z) = s(z) εs(/z), and v(z) 0 0 ε p(/z) Q(z) = v(z)w + (z) /2 /2 ε p(/z)w + (z) v(z)w (z) /2 /2 ε p(/z)w (z), εv(/z) 0 0 p(z) whose inverse is p(z) 0 0 ε p(/z) Q (z) = s(z) 0 εs(/z) 0. εv(/z) 0 0 p(z) Define A(z) = Q (z)e (z). Then [A(z),A (z)] is a SLPME of the dual pair [a(z),b(z)]..4 Illustrative Examples In this section, we present two examples to the readers for demonstrating the SLPME algorithm we developed in the previous section. Example. (Construction of -band symmetric PRFB). Let H 0 and B 0 be two given low-pass symmetric filters with the z-transforms ( z + + z H 0 (z) = ) 2

14 60 Jianzhong Wang and B 0 (z) = 27 (z + + z) 2 ( + ) We want to construct the -band symmetric PRFB {H 0,H,H 2 },{B 0,B,B 2 }, which satisfies 2 B j ( )( )H j = f raci, j=0 Their polyphase forms are the following: [ H [0] 0 (z), H [] 0 (z), H [2] 0 (z)] = [2 + z,,2 + /z] 9 [ [B [0] 2 0 (z),b[] 0 (z),b[2] 0 (z)] = ,, 2 + z ] 9z 27 9 To normalize them, we set and a = [H [0] 0,H[] 0,H[2] 0 ] = [2 + z,,2 + /z] [ b = [B [0] 2 0,B[] 0,B[2] 0 ] = ; ; 2 + z ] z 9 so that ab T =. We now use elementary S 0, matrix decomposition technique. Let E(z) = /z z We have a (z) = a(z)e(z) = [0,/,0]. To make the SLPME for a(z), we set 0 /2 /2 Q(z) = /2 /2 Then the LP matrix A(z) = Q (z)e (z) = z 9 / 2+/z 9 is a SLPME for a. To obtain the SLPME for the dual pair [a,b], we compute w(z) = b(z)(e (z)) T = [ 2+/z ], which yields,, 2+z 2+/z Q(z) = 0 0, 2+z

15 Symmetric LP Matrix Extension 6 where Q(:,) = w T. Finally, we have and A(z) = (E(z)Q(z)) = B(z) = E(z)Q(z) = 2+z z+2z2 +z 5 2 8z+6z2 +z 5 2+/z 2+/z 9 +z 2 8z +2z+26z2 2z z2 8z +6z 8z2 +2z ) 5 2 4/z+7 /z+z 9 /z+4+z 2+z which are SLPME of (a(z),b(z)). Recovering the filters from their polyphases and applying the normalization factor to B(z), we have the following z-transforms for the -band PRFB: and ( z ) z H 0 (z) =,, H (z) = ( z) ( + 5z + 5z 2 + z + z 4 + 5z 5 + 5z 6 + z 7 ) 5 5, H 2 (z) = (z )2 ( + + 0z z + 6z z 5 + 0z z 8 ) 5 5, B 0 (z) = 27 (z + + z) 2 ( + ), B (z) = 9z + z z z 9, B 2 (z) = 9z + z z z 9. Example.2 (Symmetric LP matrix extension of 4 4 matrix). Let a = 6 [ 2 z + 2 2z, z z, z + 8 z,4 + 4 ] V 0,4 z be a given LP independent vector. We first consider the SLPME Problem. The symmetric Euclidean divisions yields the matrices S = S 2 = ( + z) (7 + z) 4 ( + 7z)

16 62 Jianzhong Wang 2z 2 z z S = S 4 = and a = [,0,0,0]. Let E = S S 2 S S 4 = and Then the SLPME of a is z 2z 2 +z z 2 A(z) = 6z z 2 6z+z z z 2 8z Q = 0 /2 /2 0 0 /2 / z z 2 +8z+z 2 8z 6z +8z z2 6z +z z 2 7+z We now solve the SLPME Problem 2. Let b = 6 +7z 4. [ z + 0 z, 2z + 6,2z + 6,4 + ]. [ ] Then (a,b) is a dual pair. We have w = b(e ) T =, z+ 8z, +z 8, (+z). Set Then, the SLPME for the dual pair is A(z) = (E(z)Q(z)) = z Q(z) = 8 /2 /2 0 +z 8 /2 /2 0. (+z) z z 2 8z +8z+z 2 6z +8z z 2 6z +z 6z+6z z 4 8z+26z 2 +8z +z 4 +8z+26z2 8z +z 4 (z )(z+) z 2 28z 2 2z 2 2z 2 +z 4 +z 4 +80z 2 2z +z z 2 28z 2 +7z+7z2 +z 2z 2 z 0z 2 0z z 4 +z 5 5z+90z 2 0z z 4 z 5 z 0z 2 +90z 5z 4 +z 5 2z (z )2 (+6z+z 2 ) 6z 2 and

17 Symmetric LP Matrix Extension 6 +0z z 2 z 2 +z 2 6z +z z+ B(z) = E(z)Q(z) = 8z z 8 /2 /2 0. +z 5z+5z 2 z 7z 7z 2 +z +6z z 2 4 8z 8z 8z References. C. K. Chui and J.-A. Lian, Construction of compactly supported symmetric and antisymmetric orthonormal wavelets with scale =, Appl. Comput. Harmon. Anal. 2 (995), C. Chui, B. Han, and X. Zhuang, A dual-chain approach for bottom-up construction of wavelet filters with any integer dilation, Appl. Comput. Harmon. Anal. (2) (202), R. Crochiere and L. R. Rabiner, Multirate Digital Signal Processing, Prentice-Hall, Englewood Cliffs, S. Goh and V. Yap, Matrix extension and biorthogonal multiwavelet construction, Linear Algebra Appl. 269 (998), B. Han, Matrix extension with symmetry and applications to symmetric orthonormal complex m-wavelets, Journal of Fourier Analysis and its Applications 5 (2009), B. Han and X. Zhuang, Matrix extension with symmetry and its applications to symmetric orthonormal multiwavelets, SIAM J. Math. Anal. 42 (200), B. Han and Z. Zhuang, Algorithms for matrix extension and orthogonal wavelet filter banks over algebraic number fields, Mathematics of Computation 2 (20), S. Mallat, A wavelet tour of signal processing, Academic Press, San Diego, X. Shi and Q. Sun, A class of m-dilation scaling function with regularity growing proportionally to filter support width, Proc. Amer. Math. Soc. 26 (998), G. Strang and T. Nguyen, Wavelets and Filter Banks, Wellesley-Cambrige, P. Vaidyanathan, Theory and design of M-channel maximally decimated quadrature mirror filters with arbitrary M, having the perfect-reconstruction property, IEEE Trans. Acoust. Speech, Signal Processing 5 (4) (987), P. Vaidyanathan, How to capture all FIR perfect reconstruction QMF banks with unimodular matrices, Proc. IEEE Int l Symp. Circuits Syst., vol., 990, pp P. Vaidyanathan, Multirate Systems and Filter Banks, Prentice-Hall, Englewood Cliffs, M. Vetterli, A theory of multirate filter banks, IEEE Trans. Acoust., Speech, Signal Processing 5 () (987), J. Z. Wang, Euclidean algorithm for Laurent polynomial matrix extension, Appl. Comput. Harmon. Anal. (2004). 6. X. Zhuang, Matrix extension with symmetry and construction of biorthogonal multiwavelets with any integer dilation, Applied and Computational Harmonic Analysis (202), 59 8.