Two-channel sampling in wavelet subspaces

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DOI: 10.1515/auom-2015-0009 An. Şt. Univ. Ovidius Constanţa Vol. 23(1),2015, 115 125 Two-channel sampling in wavelet subspaces J.M. Kim and K.H.

1 DOI: /auom An. Şt. Univ. Ovidius Constanţa Vol. 23(1),2015, Two-channel sampling in wavelet subspaces J.M. Kim and K.H. Kwon Abstract We develop two-channel sampling theory in the wavelet subspace V 1 from the multi resolution analysis {V j} j Z. Extending earlier results by G. G. Walter 11, W. Chen and S. Itoh 2 and Y. M. Hong et al 5 on the sampling theory in the wavelet or shift invariant spaces, we find a necessary and sufficient condition for two-channel sampling expansion formula to hold in V 1. 1 Indroduction The classical Whittaker-Shannon-Kotel nikov(wsk) sampling theorem 4 states that any signal f(t) with finite energy and the bandwidth π can be completely reconstructed from its discrete values by the formula f(t) = n= f(n) sin π(t n). π(t n) WSK sampling theorem has been extended in many directions (see 1, 2, 5, 6, 7, 8, 10, 11, 12 and references therein). G. G. Walter 11 developed a sampling theorem in wavelet subspaces, noting that the sampling function sinct := sin πt/πt in the WSK sampling theorem is a scaling function of a multi-resolution analysis. A. J. E. M. Janssen 6 used the Zak transform to generalize Walter s work to regular shifted sampling. Later, W. Chen and S. Itoh 2 (see also 12) extended Walter s result further by relaxing conditions Key Words: two-channel sampling, wavelet subspaces, Riesz basis 2010 Mathematics Subject Classification: Primary 42C15 Received: 26 April, Revised: 30 May, Accepted: 27 June,

2 TWO-CHANNEL SAMPLING IN WAVELET SUBSPACES 116 on the scaling function φ(t). Recently in 5,8 general sampling expansion are handled on shift invariant spaces(9). In this work, we find a necessary and sufficient condition for two-channel sampling expansion to hold in the wavelet subspace V 1 of a multi resolution analysis {V j } j Z. 2 Preliminaries For a measurable function f(t) on R, we let f(t) 0 := sup E =0 inf f(t) and f(t) := inf R\E sup E =0 R\E f(t) be the essential infimum and the essential supremum of f(t) on R respectively, where E is the Lebesgue measure of E. For any f(t) L 2 (R) L 1 (R), we let F(f)(ξ) = ˆf(ξ) := f(t)e itξ dt be the Fourier transform of f(t) so that 1 2π F( ) becomes a unitary operator on L 2 (R). A sequence {φ n : n Z} in a Hilbert space H is called a Riesz sequence if {φ n : n Z} is a Riesz basis of the closed subspace V := span{φ n : n Z} of H. Definition 1. A function φ(t) L 2 (R) is called a scaling function of a multiresolution analysis (MRA in short) {V j } j Z if the closed subspaces V j of L 2 (R), { } V j := span φ(2 j t k) : k Z, j Z satisfy the following properties; 1. V 1 V 0 V 1 ; 2. V j = L 2 (R) ; 3. V j = {0}; 4. f(t) V j if and only if f(2t) V j+1 ; 5. {φ(t n) : n Z} is a Riesz basis of V 0. The wavelet subspace W j is the orthogonal complement of V j in V j+1 so that V j+1 = V j W j. Then there is a wavelet ψ(t) L 2 (R) that induces a Riesz basis {ψ(2 j t k) : k Z} of W j. Moreover, {φ(2 j t k), ψ(2 j t k) : k Z} forms a Riesz basis of V j+1.

3 TWO-CHANNEL SAMPLING IN WAVELET SUBSPACES 117 For any φ(t) L 2 (R), {φ(t n) : n Z} is a Bessel sequence if there is a constant B > 0 such that f(t), φ(t n) 2 B f 2 L 2 (R), f L2 (R) or equivalently (see Theorem in 3) G φ (ξ) := ˆφ(ξ + 2nπ) 2 B a.e. on 0, 2π. Lemma 1. (Lemma 2.2 in 5 and Lemma in 3) Let φ(t) L 2 (R) be such that {φ(t k) : k Z} is a Bessel sequence. Then, for any {c k } l 2, c kφ(t k) converges in L 2 (R) and ( ) ( F c k φ(t k) = c k e ikξ) ˆφ(ξ). For any c = {c k } l 2, let ĉ(ξ) := c ke ikξ. Then, ĉ(ξ) L 2 0, 2π or C0, 2π if {c k } l 2 or l 1 respectively. Lemma 2. If a = {a k }, b = {b k } l 2 and â(ξ) L 0, 2π, then a b := { j Z a jb k j } l2 and â b(ξ) = â(ξ) b(ξ). Proof. Since â(ξ) L 0, 2π and b(ξ) L 2 0, 2π, â(ξ) b(ξ) L 2 0, 2π. Hence we can expand â(ξ) b(ξ) into its Fourier series n c ne inξ in L 2 0, 2π, where c n = 1 â(ξ) b(ξ), e inξ = 1 ( a k e ikξ, b k e ikξ) e inξ 2π L 2 0,2π 2π L 2 0,2π = 1 a k e ikξ, b n k e ikξ = a k b n k 2π L 2 0,2π by Parseval s identity. Hence the conclusion follows. For any φ(t) L 2 (R), let H φ (ξ) := ˆφ(ξ + 2nπ) and C φ (t) := φ(t + n) 2.

4 TWO-CHANNEL SAMPLING IN WAVELET SUBSPACES Main Result and an example In the following, let φ(t) L 2 (R) be a scaling function of an MRA {V j } j Z and ψ(t) the associated wavelet, of which we always assume that H φ (ξ) and H ψ (ξ) are in L 0, 2π. Then (cf. Proposition 2.4 in 8) φ(t) and ψ(t) L 2 (R) C(R) and sup C φ (t) <, sup C ψ (t) <. Hence for any c = {c n } l 2, ( ) R R c φ (t) := c n φ(t n) converges both in L 2 (R) and uniformly in R so that each V j L 2 (R) C(R), j Z. Let L j be the LTI (linear time invariant) systems with frequency responses M j (ξ) L 2 (R) L (R) for j = 1, 2. Then L j f(t) = F 1 ( ˆfM j )(t) L 2 (R) C(R), f L 2 (R) and j = 1, 2 and lim L jf(t) = 0 by the Riemann-Lebesgue Lemma since ˆf(ξ)M j (ξ) t L 1 (R). Moreover by the Poisson summation formula (cf. Lemma 5.1 in 8), for any fixed t R and j = 1, 2, ˆφ(ξ + 2nπ)M j (ξ + 2nπ)e it(ξ+2nπ) and L j φ(t + n)e inξ = L j ψ(t + n)e inξ = ˆψ(ξ + 2nπ)M j (ξ + 2nπ)e it(ξ+2nπ) are in L 0, 2π as functions in ξ since H φ (ξ), H ψ (ξ) L 0, 2π. In particular, A i,j (ξ) = A i,j (ξ + 2π) L 0, 2π for i, j = 1, 2, where A 1,j (ξ) := L j φ(n)e inξ, A 2,j (ξ) := L j ψ(n)e inξ. Let A(ξ) = A i,j (ξ) 2 i,j=1. Then for any f(t) = c 1,k φ(t k) + c 2,k ψ(t k) V 1, where {c 1,k } and {c 2,k } in l 2, we have for i = 1, 2 and n Z L i (f)(t) = c 1,k L i (φ)(t k) + c 2,k L i (ψ)(t k), which converges both in L 2 (R) and absolutely on R. In particular L i (f)(n) = c 1,k L i (φ)(n k)+ c 2,k L i (ψ)(n k), n Z and i = 1, 2. (1)

5 TWO-CHANNEL SAMPLING IN WAVELET SUBSPACES 119 Lemma 3. Assume that det A(ξ) 0 a.e. in 0, 2π. Let λ 1,B (ξ) λ 2,B (ξ) be eigenvalues of the Hermitian matrix B(ξ) = (A(ξ) A(ξ)) 1. If det A(ξ) 0 > 0, then 0 < λ 1,B (ξ) 0 λ 2,B (ξ) <. Proof. Since B(ξ) is a nonsingular positive semi-definite Hermitian matrix a.e. in 0, 2π, 0 < λ 1,B (ξ) λ 2,B (ξ) a.e. in 0, 2π. Since A i,j (ξ) L 0, 2π and det A(ξ) 0 > 0, all entries of B(ξ) are also in L 0, 2π so that the characteristic equation of B(ξ) is of the form λ(ξ) 2 + f(ξ)λ(ξ) + g(ξ) = 0 where f(ξ) and g(ξ) are real-valued functions in L 0, 2π. Hence 0 < λ 2,B (ξ) <. Since λ 1,B (ξ)λ 2,B (ξ) = det B(ξ) = det A(ξ) 2, det A(ξ) 2 λ 1,B (ξ)λ 2,B (ξ) det A(ξ) 2 0 a.e. in 0, 2π so that 0 < det A(ξ) 2 λ 2,B (ξ) 1 λ 1,B (ξ) a.e. in 0, 2π. Definition 2. For any f 1 (t) and f 2 (t) in L 2 (R), let F (ξ) := F i,j (ξ) 2 i,j=1 be the Gramian of {f 1, f 2 }, where F i,j (ξ) := ˆf i (ξ + 2kπ) ˆf j (ξ + 2kπ). Then as a Hermitian matrix, F (ξ) has real eigenvalues. Proposition 1. (9) Let λ 1,F (ξ) λ 2,F (ξ) be eigenvalues of the Gramian F (ξ) of {f 1, f 2 }. Then {f 1 (t n), f 2 (t n) : n Z} is a Riesz sequence if and only if 0 < λ 1,F (ξ) 0 λ 2,F (ξ) <. Lemma 4. Assume det A(ξ) 0 > 0. Let Ŝ1 (ξ) Ŝ 2 (ξ) := A(ξ) 1 ˆφ(ξ) ˆψ(ξ) and S i (t) := F 1 (Ŝi)(t) for i = 1, 2. Then S i (t) V 1 for i = 1, 2 and {S i (t n) : i = 1, 2 and n Z} is a Riesz sequence. Proof. Let A(ξ) 1 = C(ξ) = C i,j (ξ) 2 i,j=1. Since C i,j(ξ) L (R), Ŝ i (ξ) = C i,1 (ξ) ˆφ(ξ) + C i,2 (ξ) ˆψ(ξ) L 2 (R) for i = 1, 2. Since C i,j (ξ) = C i,j (ξ + 2π) L 0, 2π, we may expand C i,j (ξ) into its Fourier series C i,j (ξ) = c i,j,k e ikξ where {c i,j,k } l 2. Then by Lemma 1, Ŝ i (ξ) = (c i,1,k e ikξ ˆφ(ξ) + ci,2,k e ikξ ˆψ(ξ) )

6 TWO-CHANNEL SAMPLING IN WAVELET SUBSPACES 120 so that S i (t) := F 1 (Ŝi)(t) = ( ) c i,1,k φ(t k) + c i,2,k ψ(t k) V 1. Let G(ξ) and S(ξ) be the Gramians of {φ, ψ} and {S 1, S 2 } respectively and λ 1,G (ξ) λ 2,G (ξ) and λ 1,S (ξ) λ 2,S (ξ) the eigenvalues of G(ξ) and S(ξ) respectively. Then S(ξ) = C(ξ)G(ξ)C(ξ). Let U S (ξ) and U G (ξ) be unitary matrices, which diagonalize S(ξ) and G(ξ) respectively, i.e., and S(ξ) = U S (ξ) Then λ1,s (ξ) 0 0 λ 2,S (ξ) where so that On the other hand, λ1,s (ξ) 0 0 λ 2,S (ξ) λ1,g (ξ) 0 G(ξ) = U G (ξ) 0 λ 2,G (ξ) U S (ξ) U G (ξ). λ1,g (ξ) 0 = R(ξ) 0 λ 2,G (ξ) R(ξ) R(ξ) = U S (ξ) R1,1 (ξ) R C(ξ)U G (ξ) := 1,2 (ξ) R 2,1 (ξ) R 2,2 (ξ) λ 1,S (ξ) = λ 1,G (ξ) R 1,1 (ξ) 2 + λ 2,G (ξ) R 1,2 (ξ) 2 ; (2) λ 2,S (ξ) = λ 1,G (ξ) R 2,1 (ξ) 2 + λ 2,G (ξ) R 2,2 (ξ) 2. (3) R(ξ)R(ξ) = U S (ξ) C(ξ)C(ξ) U S (ξ) = U S (ξ) B(ξ)U S (ξ) (4) = U S (ξ) λ1,b (ξ) 0 U B (ξ) U 0 λ 2,B (ξ) B (ξ) U S (ξ), where U B (ξ) is the unitary matrix such that λ1,b (ξ) 0 B(ξ) = U B (ξ) 0 λ 2,B (ξ) U B (ξ) with λ 1,B (ξ) λ 2,B (ξ). Set U S (ξ) U B (ξ) = D i,j (ξ) 2, which is also a i,j=1 unitary matrix. Then we have from diagonal entries of both sides of (4), R 1,1 (ξ) 2 + R 1,2 (ξ) 2 = λ 1,B (ξ) D 1,1 (ξ) 2 + λ 2,B (ξ) D 1,2 (ξ) 2 ; (5) R 2,1 (ξ) 2 + R 2,2 (ξ) 2 = λ 1,B (ξ) D 2,1 (ξ) 2 + λ 2,B (ξ) D 2,2 (ξ) 2. (6)

7 TWO-CHANNEL SAMPLING IN WAVELET SUBSPACES 121 Then we have a.e. in 0, 2π (from (2), (3), (5), and (6)) ( λ 1,S (ξ) λ 1,G (ξ) R 1,1 (ξ) 2 + R 1,2 (ξ) 2) λ 1,G (ξ)λ 1,B (ξ); ( λ 2,S (ξ) λ 2,G (ξ) R,21 (ξ) 2 + R 2,2 (ξ) 2) λ 2,G (ξ)λ 2,B (ξ) since D 1,1 (ξ) 2 + D 1,2 (ξ) 2 = D 2,1 (ξ) 2 + D 2,2 (ξ) 2 = 1 a.e. in 0, 2π. Hence 0 < λ 1,G (ξ) 0 λ 1,B (ξ) 0 λ 1,S (ξ) 0 λ 2,S (ξ) λ 2,G (ξ) λ 2,B (ξ) < by Lemma 3 and Proposition 4 so that {S i (t n) : i = 1, 2 and n Z} is a Riesz sequence by Proposition 4. Now we are ready to give the main result of this work. Theorem 1. There exist S i (t) V 1 (i = 1, 2) such that {S i (t n) : i = 1, 2 and n Z} is a Riesz basis of V 1 for which two-channel sampling formula f(t) = L 1 (f)(n)s 1 (t n) + L 2 (f)(n)s 2 (t n) (7) holds for f V 1 if and only if det A(ξ) 0 > 0. In this case S i (t) = F 1( C i,1 (ξ) ˆφ(ξ) + C i,2 (ξ) ˆψ(ξ) ) (t) for i = 1, 2. (8) Proof. Assume det A(ξ) 0 > 0 and define S i (t) by (8). Then S i (t) V 1 (i = 1, 2) and {S i (t n) : i = 1, 2 and n Z} is a Riesz sequence by Lemma 5. For any f(t) V 1 f(t) = c 1,k φ(t k) + c 2,k ψ(t k) where {c i,k } l 2 for i = 1, 2, we have by Lemma 1, ( ˆf(ξ) = c 1,k e ikξ) ( ˆφ(ξ) + c 2,k e ikξ) ˆψ(ξ). (9) Since ˆφ(ξ) ˆψ(ξ) ˆf(ξ) = Ŝ1 (ξ) = A(ξ) Ŝ 2 (ξ), we have by (1), (9) and Lemma 1 2 ( c 1,k e ikξ) ( A 1,j (ξ) + c 2,k e ikξ) A 2,j (ξ)ŝj (ξ)(10) j=1 = L 1 f(n)e inξ Ŝ 1 (ξ) + L 2 f(n)e inξ Ŝ 2 (ξ).

8 TWO-CHANNEL SAMPLING IN WAVELET SUBSPACES 122 Taking the inverse Fourier transform on (10) gives (7), which implies V 1 = span {S i (t n) : i = 1, 2 and n Z} so that {S i (t n) : i = 1, 2 and n Z} is a Riesz basis of V 1. Conversely assume that there exist S i (t) V 1 (i = 1, 2) such that {S i (t n) : i = 1, 2 and n Z} is a Riesz basis of V 1 and (7) holds. In particular, φ(t) = L 1 φ(n)s 1 (t n) + L 2 φ(n)s 2 (t n); ψ(t) = L 1 ψ(n)s 1 (t n) + L 2 ψ(n)s 2 (t n). By taking Fourier transform and using Lemma 1, we have ˆφ(ξ) ˆψ(ξ) Ŝ1 (ξ) = A(ξ) Ŝ 2 (ξ). (11) We then have as in the proof of Lemma 5, G(ξ) = A(ξ)S(ξ)A(ξ), where G(ξ) and S(ξ) are Gramians of {φ, ψ} and {S 1, S 2 } respectively. Hence det G(ξ) = det S(ξ) det A(ξ) 2 so that det A(ξ) 2 = det G(ξ) det S(ξ) = λ 1,G(ξ)λ 2,G (ξ) λ 1,S (ξ)λ 2,S (ξ) λ 1,G(ξ) 2 λ 2,S (ξ) 2 a.e. in 0, 2π, where λ 1,G (ξ) λ 2,G (ξ) and λ 1,S (ξ) λ 2,S (ξ) are eigenvalues of G(ξ) and S(ξ) respectively. Therefore, det A(ξ) λ 1,G(ξ) λ 2,S (ξ) λ 1,G(ξ) 0 λ 2,S (ξ) a.e. in 0, 2π so that det A(ξ) 0 > 0 since both {φ(t n), ψ(t n) : n Z} and {S i (t n) : i = 1, 2 and n Z} are Riesz sequences. Finally (8) comes from (11) immediately. Note that if {L i (φ)(n)} and {L i (ψ)(n)} l 1, then A i,j (ξ) C0, 2π for i, j = 1, 2 so that A i,j (ξ) L 0, 2π and det A(ξ) 0 > 0 is equivalent to deta(ξ) 0 on 0, 2π. Example. (2-channel sampling in Paley-Wiener space) Let φ(t) = sinct so that V 0 = span{φ(t n) : n Z} = P W π and V 1 = span{φ(2t n) : n Z} = P W 2π. Then V 1 = V 0 W 0 where W 0 = span{ψ(t n) : n Z} and ψ(t) = (cos 3 2 πt)(sinc 1 2 t).

9 TWO-CHANNEL SAMPLING IN WAVELET SUBSPACES 123 Note that ˆφ(ξ) = 1 2π χ π,π (ξ) and ˆψ(ξ) = 1 2π χ 2π, π π,2π (ξ), where χ E ( ) is the characteristic function of a set E in R. We have by (12) { L i φ(n)e inξ Mi (ξ), ξ 0, π) = M i (ξ 2π), ξ π, 2π and Hence L i ψ(n)e inξ = { Mi (ξ 2π), ξ 0, π) M i (ξ), ξ π, 2π. M 1 (ξ) M 2 (ξ) M 1 (ξ 2π) M 2 (ξ 2π) A(ξ) = M1 (ξ 2π) M 2 (ξ 2π) M 1 (ξ) M 2 (ξ) on 0, π) on π, 2π so that the determinant condition deta(ξ) 0 > 0 is equivalent to detm(ξ) 0 > 0 where M1 (ξ) M M(ξ) = 1 (ξ 2π). M 2 (ξ) M 2 (ξ 2π) Take M 1 (ξ) = 1 and M 2 (ξ) = isgnξ so that L 1 f(t) = f(t) and L 2 f(t) = f(t) where f(t) is the Hilbert transform of f(t). Then 1 1 M(ξ) = i i so that detm(ξ) 0 = 2. As a consequence, the sampling formula holds on V 1 = P W 2π. In fact, we have from (11), on 2π, 0) 2π i so that f(t) = Ŝ1 (ξ) Ŝ 2 (ξ) = π i 0 0 f(n)sinc2(t n) on 0, 2π otherwise f(n) sin π(t n)sinc(t n), f P W 2π.

10 TWO-CHANNEL SAMPLING IN WAVELET SUBSPACES 124 As another example, take M 1 (ξ) = 1 and M 2 (ξ) = iξ so that L 1 f(t) = f(t) and L 2 f(t) = f (t). Then 1 1 M(ξ) = iξ i(ξ 2π) so that detm(ξ) 0 = 2π. By the similar procedure as above, we obtain a sampling formula f(t) = f(n)sinc 2 2(t n) + 1 f (n) sin π(t n)sinc(t n), f P W 2π. π Acknowledgments : This work is partially supported by the National Research Foundation of Korea (NRF) (2012R1A1A ) References 1 A. Aldroubi and K. Gröchenig, Nonuniform sampling and reconstruction in shift-invariant spaces, SIAM Review, vol. 43, no. 4, pp , W. Chen and S. Itoh, A sampling theorem for shift-invariant subspace, IEEE Trans. Signal Processing, vol. 46, pp , O. Christensen, An Introduction to frames and Riesz Bases, Birkhäuser, J. R. Higgins, Sampling theory in Fourier and signal analysis, Oxford University Press, Y. M. Hong, J. M. Kim, K. H. Kwon and E. H. Lee, Channeled sampling in shift invariant spaces, Int. J. Wavelets Multiresolut. Inf. Process., vol. 5, pp , A. J. E. M. Janssen, The Zak-transform and sampling theorem for wavelet subspaces, IEEE Trans. Signal Processing, vol. 41, pp , A. J. Jerri, The Shannon sampling theorem-its various extentions and applications: A tutorial review, Proc. IEEE, vol. 65, pp , J. M. Kim and K. H. Kwon, Sampling expansion in shift invariant spaces, Int. J. Wavelets Multiresolut. Inf. Process., vol. 6, no. 2, pp , 2008.

11 TWO-CHANNEL SAMPLING IN WAVELET SUBSPACES A. Ron and Z. Shen, Frames and stable bases for shift-invariant subspaces of L 2 (R), Canadian J. Math., vol. 47, pp , M. Unser, Sampling-50 years after Shannon, Proc. IEEE, vol. 88, pp , G. G. Walter, A sampling theorem for wavelet subspaces, IEEE Trans. Inform. Theory, vol. 38, pp , P. Zhao, G. Liu and S. Itoh, F rames and sampling theorem for trnaslation-invariant subspaces, IEEE Signal Processing Letters, vol. 11, pp. 8-11, J.M. KIM, Faculty of Mathematics and Computer Sciences, Korea Science Academy of KAIST, Korea frankim@gmail.com K. H. KWON, Faculty of Mathematical Sciences, Korea Advanced Institute of Science and Technology Korea khkwon@kaist.edu

12 TWO-CHANNEL SAMPLING IN WAVELET SUBSPACES 126

MRA Frame Wavelets with Certain Regularities Associated with the Refinable Generators of Shift Invariant Spaces

Chapter 6 MRA Frame Wavelets with Certain Regularities Associated with the Refinable Generators of Shift Invariant Spaces Tian-Xiao He Department of Mathematics and Computer Science Illinois Wesleyan University,