Christensen, Ole

Kim, Hong Oh

Rae Young, Kim

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Downloaded from orbit.dtu.dk on: Dec 17, 2017Gabor frames with reduced redundancyChristensen, Ole; Rae Young, Kim; Kim, Hong OhPublished in:Proceeding of SAMPTA conference (published on CDROM)Publication date:2009Document VersionEarly version, also known as pre-printLink back to DTU OrbitCitation (APA):Christensen, O., Rae Young, K., & Kim, H. O. (2009). Gabor frames with reduced redundancy. In Proceeding ofSAMPTA conference (published on CDROM)Gabor frames with reduced redundancyOle Christensen (1), Hong Oh Kim (2) and Rae Young Kim (3)(1) Department of Mathematics, Technical University of Denmark, Building 303, 2800 Lyngby, Denmark.(2) Department of Mathematical Sciences, KAIST, Daejeon, Korea.(3) Department of Mathematics, Yeungnam University, Gyeongsan-si,Korea.Ole.Christensen@mat.dtu.dk, kimhong@kaist.edu, rykim@ynu.ac.krThis work was supported by the Korea Science and Engineering Foundation (KOSEF) Grant funded by the KoreaGovernment(MOST)(R01-2006-000-10424-0) and by the Korea Research Foundation Grant funded by the KoreanGovernment (MOEHRD) (KRF-2006-331-C00014).Abstract:Considering previous constructions of pairs of dual Gaborframes, we discuss ways to reduce the redundancy. Thefocus is on B-spline type windows.1. IntroductionWe will consider Gabor systems in L2(R), i.e., families offunctions {EmbTng}m,n∈Z, whereEmbTng(x) := e2piimbxg(x− na).If there exists a constant B > 0 such that∑m,n∈Z|〈f,EmbTng〉|2 ≤ B ||f ||2, ∀f ∈ L2(R),then {EmbTng}m,n∈Z is called a Bessel sequence. If thereexist two constants A,B > 0 such thatA ||f ||2 ≤∑m,n∈Z|〈f,EmbTng〉|2 ≤ B ||f ||2, ∀f ∈ L2(R),then {EmbTng}m,n∈Z is called a frame. If{EmbTng}m,n∈Z is a frame with dual frame{EmbTnh}m,n∈Z, thenf =∑m,n∈Z〈f,EmbTnh〉EmbTng, f ∈ L2(R),where the series expansion converges unconditionally inL2(R).Our starting point is the duality condition for Gaborframes, originally due to Ron and Shen [4]. We use theversion due to Janssen [3]:Lemma 1..1 Two Bessel sequences {EmbTng}m,n∈Z and{EmbTnh}m,n∈Z form dual Gabor frames for L2(R) ifand only if∑k∈Zg(x− n/b+ k)h(x+ k) = bδn,0 (1..1)for a.e. x ∈ [0, 1].The Bessel condition in Lemma 1..1 is always satisfiedfor bounded windows with compact support, see [1]. Notethat if g and h have compact support, we only need tocheck a finite number of conditions in (1..1). In this paperwe will usually choose b so small that only the conditionfor n = 0 has to be verified.2. The range 12N−1 < b < 1NWe first cite a result from [2]. It yields an explicit con-struction of dual Gabor frames:Theorem 2..1 Let N ∈ N. Let g ∈ L2(R) be a real-valued bounded function with supp g ⊂ [0, N ], for which∑n∈Zg(x− n) = 1. (2..1)Let b ∈]0, 12N−1 ]. Consider any scalar sequence{an}N−1n=−N+1 for whicha0 = b and an + a−n = 2b, n = 1, 2, · · ·N − 1, (2..2)and define h ∈ L2(R) byh(x) =N−1∑n=−N+1ang(x+ n). (2..3)Then g and h generate dual frames {EmbTng}m,n∈Z and{EmbTnh}m,n∈Z for L2(R).The above result can be extended:Corollary 2..2 Consider any b ≤ 1/N. With g and an asin Theorem 2..1, the functionh(x) =(N−1∑n=−N+1ang(x+ n))χ[0,N ](x) (2..4)is a dual frame generator of g.Proof. Consider the condition (1..1) for n = 0; only thevalues of h(x) for x ∈ [0, N ] play a role, so since thecondition holds for the function in (2..3), it also holds forthe function in (2..4). The cut-off in (2..4) yields a non-smooth function. How-ever, for any b < 1/N, we might modify h slightly andobtain a smooth dual generator:In particular, we obtain the following:Corollary 2..3 Consider any b < 1/N, and take ǫ <1/b −N. With g as in Theorem 2..1, the function h(x) =b, x ∈ [0, N ] has an extension to a function of desiredsmoothness, supported on [−ǫ,N + ǫ], which is a dualframe generator of g.Proof. The choice an = b, n = −N + 1, . . . , N − 1,leads toN−1∑n=−N+1ang(x+ n) = b, x ∈ [0, N ].Given ǫ < 1/b − N and any functions φ1 : [−ǫ, 0[→ Rand φ2 :]N,N + ǫ] → R, the functionh(x) =φ1(x), x ∈ [−ǫ, 0[,∑N−1n=−N+1 ang(x+ n) = b, x ∈ [0, N ],φ2, x ∈]N,N + ǫ],0, x /∈ [−ǫ,N + ǫ],will satisfy (1..1); in fact, for n 6= 0, the support of thefunctions g(· ± n/b) and h are disjoint, and for n = 0we are (for all relevant values of x) back at the function in(2..4). The functions φ1 and φ2 can be chosen such thatthe function h has the desired smoothness. The assumptions in Theorem 2..1 are tailored to B-splines,defined inductively byB1 := χ[0,1], BN+1 := BN ∗B1.Direct calculations shows thatB2(x) =x if x ∈ [0, 1],2− x if x ∈ [1, 2],0 otherwise,andB3(x) =12x2 if x ∈ [0, 1],−x2 + 3x− 32 if x ∈ [1, 2],12x2 − 3x+ 92 if x ∈ [2, 3],0 otherwise.In general, the functions BN are (N − 2)−times differ-entiable piecewise polynomials (explicit expressions areknown). Furthermore, suppBN = [0, N ], and the parti-tion of unity condition (2..1) is satisfied.In case g = BN , the dual generators in Theorem 2..1 aresplines, of the same smoothness as BN itself. By com-pressing the function∑N−1n=−N+1 ang(x+ n) from the in-terval [−N + 1, 0] to [−ǫ, 0] and from [N, 2N − 1] to[N,N + ǫ] we obtain a dual in (2..3) with the same fea-tures:Example 2..4 For the B-spline B3(x) and b = 1/5, The-orem 2..1 yields the symmetric dualh3(x) =151/2 x2 + 2 x+ 2, x ∈ [−2,−1[,−1/2 x2 + 1, x ∈ [−1, 0[,1, x ∈ [0, 3[,−1/2 x2 + 3 x− 7/2, x ∈ [3, 4[,1/2 x2 − 5x+ 25/2, x ∈ [4, 5[,0, x /∈ [0, 5[.(2..5)See Figure 1.Now, for b = 1/4, we can use Corollary 2..3 for ǫ <4 − 3 = 1. Taking ǫ = 1/2, we compress the functionx2 64-2 00.70.60.30.200.50.40.1Figure 1: B3 and the dual generator h3 in (2..5).K1 0 1 2 3 40.050.100.150.200.25Figure 2: The function h in (3..13)..h3 in (2..5) from [−2, 0] to [−1/2, 0] and from [3, 5] to[3, 31/2] and obtain the dualh(x) =141/2 (4x)2 + 2 (4x) + 2, x ∈ [−1/2,−1/4[,−1/2 (4x)2 + 1, x ∈ [−1/4, 0[,1, x ∈ [0, 3[,−1/2 (4(x− 3) + 3)2 + 3 (4(x− 3) + 3)− 7/2,x ∈ [3, 3 + 1/4[,1/2 (4(x− 3) + 3)2 − 5(4(x− 3) + 3) + 25/2,x ∈ [3 + 1/4, 3 + 1/2[,0, x /∈ [−1/2, 3 + 1/2[.=148 x2 + 8 x+ 2, x ∈ [−1/2,−1/4[,−8 x2 + 1, x ∈ [−1/4, 0[,1, x ∈ [0, 3[,−8 x2 + 48 x− 71, x ∈ [3, 3 + 1/4[,8 x2 − 56 x+ 98, x ∈ [3 + 1/4, 3 + 1/2[,0, x /∈ [−1/2, 3 + 1/2[.See Figure 2. 3. B2 and 1/2 < b < 1In the following discussion, we consider dual windows as-sociated with a Gabor frame {EmbTnB2}m,n∈Z generatedby the B-spline B2. The arguments can be extended togeneral functions supported on [0, 2]. Take any functionh with values specified only on [0, 2] and such that∑k∈ZB2(x + k)h(x+ k) = 1, x ∈ [0, 1]. (3..1)In fact, due to the support of B2, only the values for h(x)for x ∈ [0, 2] play a role for that condition. We know thatfor any b ≤ 1/2 the function generates – up to a certainscalar multiple – a dual of g.Now consider any 1/2 < b < 1; that is, we have 1 <1/b < 2.Lemma 3..1 Assume that h(x), x ∈ [0, 2] is chosen suchthat (3..1) is satisfied. The the following hold:(i) If∑k∈ZB2(x− 1/b+ k)h(x+ k) = 0, x ∈ R, (3..2)and∑k∈ZB2(x+ 1/b+ k)h(x+ k) = 0, x ∈ R, (3..3)thenB2(x− 1/b)h(x) +B2(x − 1/b+ 1)h(x+ 1) = 0,x ∈ [1/b, 2], (3..4)B2(x+ 1/b− 1)h(x− 1) +B2(x + 1/b)h(x) = 0x ∈ [0, 2− 1/b]. (3..5)These equations determine h(x) forx ∈ [−1, 1− 1/b] ∪ [1 + 1/b, 3].(ii) If h(x) for x ∈ [−1, 1− 1/b]∪ [1 + 1/b, 3] is chosensuch that (3..4) and (3..5) are satisfied, andh(x) = 0, x /∈ [0, 2] ∪ [−1, 1− 1/b] ∪ [1 + 1/b, 3],then (3..2) and (3..3) hold.Proof. We consider (3..2) for x ∈ [1, 2], and split into twocases:For x ∈ [1, 1/b], (3..2) yields that0 = B2(x− 1/b+ 1)h(x+ 1) +B2(x− 1/b+ 2)h(x+ 2);(3..6)the equation only involve h(x) forx ∈ [2, 1 + 1/b] ∪ [3, 2 + 1/b].For x ∈ [1/b, 2], (3..2) yields that0 = B2(x− 1/b)h(x) +B2(x− 1/b+ 1)h(x+ 1);since h(x) is known, this implies thath(x+ 1) =−B2(x− 1/b)h(x)B2(x− 1/b+ 1), x ∈ [1/b, 2],that is,h(x) =−B2(x− 1/b− 1)h(x− 1)B2(x − 1/b), x ∈ [1/b+ 1, 3].Similarly, considering (3..3) forx ∈ [0, 1] = [0, 2− 1/b] ∪ [2 − 1/b, 1]leads to (3..5) andB2(x+ 1/b− 2)h(x− 2) +B2(x+ 1/b− 1)h(x− 1)= 0, x ∈ [2− 1/b, 1]; (3..7)the equation (3..7) only involves h(x) forx ∈ [−1/b,−1]∪ [1− 1/b, 0],and (3..5) implies thath(x − 1) =−B2(x+ 1/b)h(x)B2(x+ 1/b− 1), x ∈ [0, 2− 1/b],i.e.,h(x) =−B2(x+ 1/b+ 1)h(x+ 1)B2(x+ 1/b), x ∈ [−1, 1− 1/b].For the proof of (ii), the conditionh(x) = 0, x /∈ [0, 2] ∪ [−1, 1− 1/b] ∪ [1 + 1/b, 3],implies that (3..6) and (3..7) are satisfied. By construction,(3..2) and (3..3) are satisfied. Lemma 3..1 shows that if we want that (3..1), (3..2), and(3..3) hold for some b ∈]1/2, 1], then h in general will takevalues outside [0, 2]. However, the proof shows that weunder certain circumstances can find a solution h havingsupport in [0, 2]. In that case, the support will actually bea subset of [0, 2]:Corollary 3..2 Let b ∈]1/2, 1]. Assume that supph ⊆[0, 2] and that (3..1) and (3..2) holds. Thenh(x) = 0, x ∈ [0, 2− 1/b] ∪ [1/b, 2]. (3..8)Proof. According to the proof of Lemma 3..1, we obtainthat h(x) = 0 on [1/b+1, 3]by requiring that h(x) = 0 forx ∈ [1/b, 2]; and we obtain that h(x) = 0 on [−1, 1−1/b]by requiring that h(x) = 0 for x ∈ [0, 2− 1/b]. If supph ⊆ [0, 2], the condition (3..8) implies that h atmost can be nonzero on the interval [2 − 1/b, 1/b] havinglength 2/b − 2. In order for (3..1) to hold, this intervalmust have length at least 1; thus, we need to consider bsuch that 2/b− 2 ≥ 1, i.e., b ≤ 2/3. Note that if b ≤ 2/3,then 2/b ≥ 3 : that is, because B2 and h are supported on[0, 2], Janssen’s duality conditions in (1..1) are automati-cally satisfied for n = ±2,±3, . . . .Corollary 3..3 Consider b ∈]1/2, 2/3]. Then there existsa function h with supph ⊆ [0, 2] such that (3..1) and (3..2)hold; and bh(x) is a dual generator of B2 for these valuesof b.Proof. For x ∈ [0, 2 − 1/b] ∪ [1/b, 2], let h(x) = 0. Forx ∈ [0, 1], the equation (3..1) means thatxh(x) + (1− x)h(x + 1) = 1.This implies thatxh(x) = 1, x ∈ [1/b− 1, 1],(1 − x)h(x+ 1) = 1, x ∈ [0, 2− 1/b];that is,h(x) =1x, x ∈ [1/b− 1, 1], (3..9)andh(x) =12− x, x ∈ [1, 3− 1/b]. (3..10)Finally, for x ∈ [2− 1/b, 1/b− 1] and x ∈ [3− 1/b, 1/b],choose h(x) such thatxh(x) + (1− x)h(x + 1) = 1.By construction, bh(x) is a dual generator. For b = 3/5 we will now explicitly construct a continu-ous dual generator h of B2 with support in [0, 2]. PuttingCorollary 3..2, (3..9), and (3..10) together, we can state aresult about how a dual window supported on [0, 2] mustlook like on parts of [0, 2]:Lemma 3..4 For b = 3/5, every dual generator of B2with support in [0, 2] has the formh(x) =0 if x ≤ 1/3;1xif x ∈ [2/3, 1];12−x if x ∈ [1, 4/3];0 if x ≥ 5/3.That is, we only have freedom on the definition of h on]1/3, 2/3[∪]4/3, 5/3[.Note that on [2/3, 4/3], the function h is symmetricaround x = 1. We will now show that it is possible todefine h on ]1/3, 2/3[∪]4/3, 5/3[ in such a way that h be-comes symmetric around x = 1.First, we note that this form of symmetry means thath(1− x) = h(1 + x), x ∈]1/3, 2/3[. (3..11)Put together with the duality condition, we thus requirethatxh(x) = 1− (1 − x)h(1− x), x ∈]1/3, 2/3[. (3..12)The condition (3..12) shows that must define h(1/2) =1. Now, taking any continuous function h defined on[1/3, 1/2] with the properties that h(1/3) = 0 andh(1/2) = 1, the condition (3..12) shows how to de-fine h(x) on ]1/2, 2/3[; and, finally, the condition (3..11)shows how to define h on ]4/3, 5/3[ such that the resultingfunction is a symmetric dual generator.1.51.250.750.250.51.51.00.50.0x2.01.00.0Figure 3: The function h in (3..13)..Puth(x) = 6x− 2, x ∈ [1/3, 1/2].Then, for x ∈ [1/2, 2/3],h(x) =1− (1 − x)h(1− x)x=−6x2 + 10x− 3x.The condition h(1 + x) = h(1 − x), x ∈]1/3, 2/3[ canalso be expressed as h(x) = h(2 − x), x ∈]4/3, 5/3[.Thus, for x ∈ [4/3, 3/2] we arrive ath(x) = h(2 − x) =−6x2 + 14x− 72− x, x ∈ [4/3, 3/2];while, for x ∈ [3/2, 5/3],h(x) = h(2− x) = 6(2− x)− 2 = 10− 6x.We have arrived at the following conclusion:Lemma 3..5 For b = 3/5, the functionh(x) =0 if x ≤ 1/3;6x− 2 if x ∈ [1/3, 1/2];−6x2+10x−3xif x ∈ [1/2, 2/3];1xif x ∈ [2/3, 1];12−x if x ∈ [1, 4/3];−6x2+14x−72−x if x ∈ [4/3, 3/2];10− 6x if x ∈ [3/2, 5/3];0 if x ≥ 5/3(3..13)is a continuous symmetric dual generator of B2.References:[1] Christensen, O.: Frames and bases. An introductorycourse. Birkha¨user 2007.[2] Christensen, O. and Kim, R. Y.: On dual Gaborframe pairs generated by polynomials. J. FourierAnal. Appl., accepted for publication.[3] Janssen, A.J.E.M.: The duality condition for Weyl-Heisenberg frames. In ”Gabor analysis: theoryand applications” (eds. H.G. Feichtinger and T.Strohmer). Birkha¨user, Boston, 1998.[4] Ron, A. and Shen, Z.: Frames and stable bases forshift-invariant subspaces ofL2(Rd). Canad. J. Math.47 no. 5 (1995), 1051–1094.

Gabor frames with reduced redundancy

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Gabor frames with reduced redundancy

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