Hyperbolic secants yield Gabor frames

12 Pages·0.15 MB·

by  A. J. E. M. Janssen

3 Hyperbolic secants yield Gabor frames
0
0
2
A.J.E.M. Janssen∗and Thomas Strohmer†
n
a
J

3
1
Abstract

]
A
We show that (g ,a,b) is a Gabor frame when a> 0,b > 0,ab < 1
2
F 1
. and g2(t) = (12πγ)2(coshπγt)−1 is a hyperbolic secant with scaling
h
parameter γ > 0. This is accomplished by expressing the Zak trans-
t
a form of g in terms of the Zak transform of the Gaussian g (t) =
m 2 1
1
(2γ)4 exp(−πγt2), together with an appropriate use of the Ron-Shen
[
criterion for being a Gabor frame. As a side result it follows that the

1
windows, generating tight Gabor frames, that are canonically associ-
v
4 ated to g2 and g1 are the same at critical density a= b = 1. Also, we
3 display the “singular” dual function corresponding to the hyperbolic
1
secant at critical density.
1
0
3 AMS Subject Classification: 42C15, 33D10, 94A12.
0
Key words: Gabor frame, Zak transform, hyperbolic secant, theta functions.
/
h
t
a
m 1 Introduction and results
:
v
i Let a > 0,b > 0, and let g ∈ L2(R). A Gabor system (g,a,b) consists of
X
all time- and- frequency shifted functions g with integer n,m, where for
r na,mb
a x,y ∈ R we denote
g (t) = e2πiytg(t−x), t ∈ R. (1)
x,y
∗Philips ResearchLaboratories WY-81, 5656 AA Eindhoven,The Netherlands; Email:
[e-mail]
†Department of Mathematics, University of California, Davis, CA 95616-8633, USA;
Email: [e-mail]. T.S.acknowledgessupportfromNSFgrant9973373.
1
We say that (g,a,b) is a Gabor frame when there are A > 0,B < ∞ (lower,
upper frame bound, respectively) such that for all f ∈ L2(R) we have
Akfk2 ≤ |hf,g i|2 ≤ Bkfk2. (2)
na,mb
Xn,m
Here k·kand h·,·idenote thestandard normandinner product ofL2(R). We
refer to [2, 3, 4] for generalities about frames for a Hilbert space and for both
basic and in-depth information about Gabor frames, frame operators, dual
frames, tight frames, etc. For a very recent and comprehensive treatment of
Gabor frames, we refer to [5], Chs. 5-9, 11-13; many of the more advanced
results of modern Gabor theory are covered there in a unified manner.
It is well known that a triple (g,a,b) cannot be a Gabor frame when
ab > 1, see for instance [5], Corollary 7.5.1. Also such a system cannot
be a Gabor frame when ab = 1 and g′ ∈ L2(R),tg(t) ∈ L2(R) (extended
Balian-Low theorem; see for instance [4], Ch. 2).
In this paper we demonstrate that (g ,a,b) is a Gabor frame when ab < 1
2
and
πγ 1 1
g (t) = 2 , t ∈ R, (3)
2,γ
2 coshπγt
(cid:16) (cid:17)
with γ > 0 (normalization such that kg k = 1). The result is obtained by
2,γ
(i) relating the Zak transform
∞
(Zg)(t,ν) = g(t−l)e2πilν, t,ν ∈ R, (4)
X
l=−∞
of g = g to the Zak transform of g = g , where g is the normalized
2,γ 1,γ 1,γ
Gaussian
g (t) = (2γ)14e−πγt2, t ∈ R, (5)
1,γ
(ii) using the observation that (g ,a,b) is a Gabor frame and (iii) an ap-
1,γ
propriate use of the Ron-Shen time domain criterion [10] for a Gabor system
(g,a,b) to be a Gabor frame. Explicitly, we show that there is a positive
constant E such that
E(Zg )(t,ν)
(Zg )(t,ν) = 1,γ , t,ν ∈ R. (6)
2,γ ϑ (πν;e−πγ)ϑ (πt;e−π/γ)
4 4
Here ϑ (z;q) is the theta function, see [11], Ch. 21,
4
∞
ϑ (z;q) = (−1)nqn2e2inz, z ∈ C. (7)
4
nX=−∞
2
Thisϑ ispositiveandboundedforrealargumentsz andparameterq ∈ (0,1).
4
The rest of the paper is organized as follows. In Section 2 we present
the proof or our main result. In Section 3 we prove formula (6) by using
elementary properties of theta functions and basic complex analysis. These
same methods, combined with formula (6), allow us to compute explicitly
the “singular” dual function gd (at critical density a = b = 1) canonically
2,γ
associated to g according to
2,γ
1
dν
gd (t) = , t ∈ R. (8)
2,γ Z (Zg )∗(t,ν)
2,γ
0
Thisisbrieflypresented inSection4. WealsoshowinSection4thatthetight
Gabor frame generating windows gt ,gt , canonically associated to g ,g
1,γ 2,γ 1,γ 2,γ
and given at critical density a = b = 1 by
1
(Zg)(t,ν)
gt(t) = dν, t ∈ R, (9)
Z |(Zg(t,ν)|
0
for g = g ,g , are the same. In [9] a comprehensive study is made of
1,γ 2,γ
the process, embodied by formula (9), of passing from windows g with few
zeros in the Zak transform domain to tight Gabor frame generating windows
gt at critical density. One of the observations in [9] is that the operation
in (9) seems to diminish distances between positive, even, unimodal windows
enormously. The fact that gt = gt is an absurdly accurate illustration of
1,γ 2,γ
this phenomenon.
2 Proof of the main result
In this section we present a proof for the result that (g ,a,b) is a Gabor
2,γ
frame when a > 0,b > 0,ab < 1 and g is given by (3). According to the
2,γ
Ron-Shen criterion in the time-domain [10] we have that (g,a,b) is a Gabor
frame with frame bounds A > 0,B < ∞ if and only if
1
AI ≤ M (t)M∗(t) ≤ BI, a.e. t ∈ R. (10)
b g g
Here I is the identity operator of ℓ2(Z) and M (t) is the linear operator of
g
ℓ2(Z) (in the notation of [4], Subsec. 1.3.2), whose matrix with respect to the
3
standard basis of ℓ2(Z) is given by
M (t) = g(t−na−l/b) , a.e. t ∈ R (11)
g l∈Z,n∈Z
(cid:0) (cid:1)
(row index l, column index n). Since g is rapidly decaying, the finite
2,γ
frame upper bound condition is easily seen to be satisfied, and we therefore
concentrate on the positive lower frame bound condition. We may restrict
here to the case where a < 1,b = 1, since (g,a,b) is a Gabor frameif andonly
if(D g,a/c,bc)isa Gaborframe. HereD isthedilationoperator(D f)(t) =
c c c
c12f(ct),t ∈ R, defined for f ∈ L2(R) when c > 0. Since Dcg2,γ = g2,γc, we
only need to replace γ > 0 by γb > 0 when b 6= 1. Hence we shall show that
there is an A > 0 such that
∞ ∞
2
clg(t−na−l) ≥ Akck2, c = (cl)l∈Z ∈ ℓ2(Z),t ∈ R, (12)
nX=−∞(cid:12)(cid:12)l=X−∞ (cid:12)(cid:12)
(cid:12) (cid:12)
with g = g .
2,γ
Taking c ∈ ℓ2(Z) it follows from Parseval’s theorem for Fourier series and
the definition of the Zak transform in (4) that for any n ∈ Z
1
∞
c g(t−na−l) = (Zg)(t−na,ν)C∗(ν)dν, (13)
l
Z
X
l=−∞
0
where C(ν) is defined by
∞
C(ν) = c∗e2πilν, a.e. ν ∈ R. (14)
l
X
l=−∞
Now assuming the result (6) (with E > 0) we have that
1
∞
c g (t−na−l) = (Zg )(t−na,ν)C∗(ν)dν
l 2,γ 2,γ
Z
X
l=−∞
0
1
E C∗(ν)
= (Zg )(t−na,ν) dν. (15)
1,γ
ϑ (π(t−na);e−π/γ) Z ϑ (πν;e−πγ)
4 4
0
4
Define the 1-periodic function D(ν) by
∞
C(ν)
D(ν) = d∗e2πilν := , ν ∈ R. (16)
l ϑ (πν;e−πγ)
X 4
l=−∞
It is well known that (g ,a,1) is a Gabor frame, see for instance [5], Theo-
1,γ
rem 7.5.3. Accordingly, see (12), there is an A > 0 such that
1,γ
∞ ∞
2
d g (t−na−l) ≥ A kdk2, d ∈ ℓ2(Z),t ∈ R. (17)
l 1,γ 1,γ
nX=−∞(cid:12)(cid:12)l=X−∞ (cid:12)(cid:12)
(cid:12) (cid:12)
Letting
1
m := min > 0 (18)
δ z∈R ϑ2(z;e−πδ)
4
for δ > 0, we then see from (13)–(18) that
∞ ∞
2
c g (t−na−l) ≥
l 2,γ
nX=−∞(cid:12)(cid:12)l=X−∞ (cid:12)(cid:12)
(cid:12) (cid:12)
1
∞
2
m E2 (Zg )(t−na,ν)D∗(ν)dν =
1/γ 1,γ
nX=−∞(cid:12)(cid:12)Z0 (cid:12)(cid:12)
(cid:12) (cid:12)
∞ ∞
2
m E2 d g (t−na−l) ≥ m E2A kdk2. (19)
1/γ l 1,γ 1/γ 1,γ
nX=−∞(cid:12)(cid:12)l=X−∞ (cid:12)(cid:12)
(cid:12) (cid:12)
Finally, with D and C related as in (16) we have from Parseval’s theorem
for Fourier series that
1 1
1
kdk2 = |D(ν)|2dν = |C(ν)|2dν
Z Z |ϑ (πν;e−πγ)|2
4
0 0
1
≥ m |C(ν)|2dν = m kck2. (20)
γ γ
Z
0
Hence
∞ ∞
| c g (t−na−l)|2 ≥ m m E2A kck2, (21)
l 2,γ γ 1/γ 1,γ
nX=−∞ l=X−∞
as required.
5
3 Expressing Zg in terms of Zg
2,γ 1,γ
In this section we express Zg , with g given in (3), in terms of Zg , with
2,γ 2,γ 1,γ
g given in (5). For this we need some basic facts of the theory of theta
1,γ
functions as they can be found in [11], Ch. 21. The four theta functions are
for q ∈ (0,1) given by
∞
ϑ (z;q) =1 (−1)nq(n+21)2e(2n+1)iz, (22)
1
i
nX=−∞
∞
ϑ (z;q) = q(n+12)2e(2n+1)iz, (23)
2
nX=−∞
∞
ϑ (z;q) = qn2e2niz, (24)
3
nX=−∞
∞
ϑ (z;q) = (−1)nqn2e2niz, (25)
4
nX=−∞
where z ∈ C. One thus easily gets from (4) and (5) that
(Zg )(t,ν) = (2γ)41e−πγt2ϑ (π(ν −iγt);e−πγ), t,ν ∈ R. (26)
1,γ 3
Theorem 3.1 We have for t,ν ∈ R
(Zg )(t,ν) = 2−21π21γϑ′(0;e−πγ)e−πγt2 ϑ3(π(ν −iγt);e−πγ)
2,γ 1 ϑ (πν;e−πγ)ϑ (πt;e−π/γ)
4 4
= π12 γ 34ϑ′(0;e−πγ) (Zg1,γ)(t,ν) . (27)
2 1 ϑ (πν;e−πγ)ϑ (πt;e−π/γ)
(cid:16) (cid:17) 4 4
Proof: For fixed t ∈ R we compute the Fourier coefficients b (t) of the
n
1-periodic function
ϑ (π(ν −iγt);e−πγ) ∞
3 = b (t)e2πinν, ν ∈ R, (28)
ϑ (πν;e−πγ) n
4 nX=−∞
by the method that can be found in [11], Sec. 22.6. For brevity we shall
suppress the expression “;e−πγ” in ϑ (π(ν−iγt);e−πγ), etc. in the remainder
3
of this proof.
6
We thus have
1
2
ϑ (π(ν −iγt))
b (t) = 3 e−2πinνdν, n ∈ Z, (29)
n
Z ϑ (πν)
4
1
−
2
and we consider for n ∈ Z
ϑ (π(ν −iγt))
c (t) = 3 e−2πinνdν, (30)
n
Z ϑ (πν)
4
C
where C is the edge of the rectangle with corner points −1, 1, 1+iγ,−1+iγ,
2 2 2 2
taken with positive orientation. It follows from [11], Sec. 21.12 that the
function ϑ (πν) has a first order zero at ν = 1iγ and no zeros elsewhere on
4 2
or within C. Therefore, by Cauchy’s theorem,
ϑ (π(ν −iγt)) 2iϑ (πi(1 −t)γ)eπγn
c (t) = 2πi Res 3 e−2πinν = 3 2 . (31)
n 1 (cid:20) ϑ (πν) (cid:21) ϑ′(1πiγ)
ν=2iγ 4 4 2
By [11], Ex. 2, first identity, on p. 464 we have (τ = iγ,q = exp(−πγ),z = 0)
1 1
ϑ′ πiγ = ie4πγϑ′(0). (32)
4 2 1
(cid:16) (cid:17)
Next, by the first formula on p. 475 of [11] (z = πi(1 − t)γ,τ′ = −1/τ =
2
i/γ,q′ = exp(πiτ′) = exp(−π/γ))
1 1 1 1
ϑ πi( −t)γ = γ−2 exp πγ( −t)2 ϑ π( −t);e−π/γ . (33)
3 3
2 2 2
(cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1)
and by [11], Ex. 2, fourth identity, on p. 464 (z = −πt, q′ instead of q)
1
ϑ π( −t);e−π/γ = ϑ (−πt;e−π/γ) = ϑ (πt;e−π/γ), (34)
3 4 4
2
(cid:0) (cid:1)
wherewealsohaveusedthatϑ isanevenfunction. Itthusfollowsfrom(31)–
4
(34) that
c (t) = 2 γ−21e−14πγeπγ(12−t)2ϑ (πt;e−π/γ)eπγn. (35)
n ϑ′(0) 4
1
7
On the other hand, we have by 1-periodicity of the integrand in (30) (in
ν) that the two integrals along the vertical edges of C cancel one another.
Hence
1 1
+iγ
2 2
ϑ (π(ν −iγt)) ϑ (π(ν −iγt))
c (t) = 3 e−2πinνdν − 3 e−2πinνdν
n
Z ϑ (πν) Z ϑ (πν)
4 4
1 1
− − +iγ
2 2
1
2
ϑ (π(ν −iγt)+πiγ)
= b (t)−e2πγn 3 e−2πinνdν. (36)
n
Z ϑ (πν +πiγ)
4
1
−
2
Furthermore, by the table in [11], Ex. 3 on p. 465 we have
ϑ π(ν −iγt)+πiγ = eπγe−2iπ(ν−iγt)ϑ π(ν −iγt) , (37)
3 3
(cid:0) (cid:1) (cid:0) (cid:1)
and
ϑ (πν +πiγ) = −eπγe−2iπνϑ (πν). (38)
4 4
It thus follows that
1
2
ϑ (π(ν −iγt))
c (t) = b (t)+e−2πγ(t−n) 3 e−2πinνdν
n n
Z ϑ (πν)
4
1
−
2
= b (t)(1+e−2πγ(t−n)). (39)
n
By (35) we then find that
1
b (t) =2γ−2 e−41πγeπγ(12−t)2ϑ (πt;e−π/γ) eπγn
n ϑ′(0) 4 1+e−2πγ(t−n)
1
1
= γ−2 e−41πγeπγ(12−t)2+πγt ϑ4(πt;e−π/γ)
ϑ′(0) coshπγ(t−n)
1
1
= γ−2 eπγt2 ϑ4(πt;e−π/γ) . (40)
ϑ′(0) coshπγ(t−n)
1
We thus conclude, see (28), that
ϑ3(π(ν −iγt)) = γ−12 eπγt2ϑ (πt;e−π/γ) ∞ e2πinν . (41)
ϑ (πν) ϑ′(0) 4 coshπγ(t−n)
4 1 nX=−∞
8
That is,
πγ 1 ∞ e2πinν
2
(Zg )(t,ν) =
2,γ
2 coshπγ(t−n)
(cid:16) (cid:17) nX=−∞
= 2−21π12γϑ′(0)e−πγt2ϑ3(π(ν −iγt)), (42)
1 ϑ (πν)ϑ (πt;e−π/γ)
4 4
and this is the first line identity in (27). The second line identity in (27)
follows from (26). This completes the proof.
Note: We have that
ϑ′(0;e−πγ),ϑ (πν;e−πγ),ϑ (πt;e−π/γ) > 0 (43)
1 4 4
for ν ∈ R,t ∈ R. Indeed, we have by [11], Sec.21.41 that
ϑ′(0;e−πγ) = ϑ (0;e−πγ)ϑ (0;e−πγ)ϑ (0;e−πγ). (44)
1 2 3 4
One sees directly from (23) that ϑ (0;q) > 0. Also, from the formula on
2
p. 476 of [11] just before Ex. 1, one sees that
1
ϑ (z;q) = ϑ z + π;q > 0, z ∈ R. (45)
3 4
2
(cid:16) (cid:17)
Remark: Theorem 3.1 implies that (Zg )(t,ν)/(Zg )(t,ν) can be factored
2 1
into a function of t and a function of ν. This factorization is crucial in the
proofofthemainresultinSection2. Itseemspossibletoextendtheapproach
in Section 2 to other pairs of windows g ,g where (g ,a,b) is a frame and
1 2 1
Zg /Zg (nearly) factorizes. We do not pursue this extension in this paper.
2 1
4 Canonical dual window and tight window
at critical density
Theorem 3.1 allows us to calculate the Zak transform of the canonical dual
gd of g for rational values of ab < 1 using Zibulski-Zeevi matrices repre-
2,γ 2,γ
senting the frame operator in the Zak transform domain, see for instance [8],
Sec. 1.5, [5], Sec. 8.3, and, of course, [12]. The dual window can then be
obtained as
1
gd (t) = (Zgd )(t,ν)dν, t ∈ R. (46)
2,γ 2,γ
Z
0
9
Fora = b = 1 theZibulski-Zeevi matrices arejust scalars, viz. |(Zg )(t,ν)|2,
2,γ
and we have formally
1
(Zgd )(t,ν) = , t,ν ∈ R, (47)
2,γ (Zg )∗(t,ν)
2,γ
and
1
dν
gd (t) = , t ∈ R. (48)
2,γ Z (Zg )∗(t,ν)
2,γ
0
Theidentities here holdonlyformallysince 1/Zg 6= L2 (R2), and, assaid,
2,γ loc
(g ,1,1) is not a frame. However, nothing prevents us from computing the
2,γ
right-hand side of (48) for t ∈ R not of the form n+ 1,n ∈ Z. For these t
2
we have that (Zg )(t,ν) is a well-behaved zero-free function of ν ∈ R.
2,γ
When we do this calculation for the case that γ = 1, we find, using the
same methods as in the proof of Theorem 3.1, that for t ∈ (−1, 1),n ∈ Z
2 2
221(−1)nϑ2(πt)e2πt2+2πnt
gd (t+n) = 4 . (49)
2,1 1
π2(ϑ′(0))2coshπ(t+n)
1
Here all theta functions are with parameter q = e−π. The same procedure
for the Gaussian g yields for t ∈ R not of the form n+1/2,n ∈ Z
1,1
1
gd (t) = 2−4 eπt2 (−1)ne−π(n−12)2, (50)
1,1 ϑ′(0)
1 X1
n− ≥|t|
2
see for instance [1], [6], Subsec. 2.14, and [7], Subsec. 4.4. Although the two
functions in (49)–(50) seem quite different, one can show that both functions
are bounded but not in Lp(R) when 1 ≤ p < ∞. Moreover, there is for both
functions exponential decay as |t|→ ∞ away from the set of half-integers.
In a similar fashion one can consider the canonical tight windows gt as-
sociated with g = g ,g at critical density a = b = 1. We have that
1,γ 2,γ
these windows are given in the Zak transform domain through the formula
Zgt = Zg/|Zg|, whence
1
(Zg)(t,ν)
gt(t) = dν, t ∈ R. (51)
Z |(Zg)(t,ν)|
0
10