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NONSTATIONARY GABOR FRAMES:
FUNCTIONS AND SEQUENCES
POH WEI SHAN CHARLOTTE
(B.Sc(Hons),NUS)
A THESIS SUBMITTED FOR
THE DEGREE OF MASTER OF SCIENCE
DEPARTMENT OF MATHEMATICS
NATIONAL UNIVERSITY OF SINGAPORE
2012
Acknowledgements
First of all, I would like to express my deepest gratitude to my supervisor Associate
Professor Goh Say Song for his guidance and patience throughout the entire course of
this thesis. He has always been very encouraging and enthusiastic in sharing with me
his views and insights in the study and has motivated me greatly in doing research.
I would like to thank him for the enjoyable and inspiring learning experience that I
had with him.
Also, I would like to thank my boyfriend, Tze Siong, for his encouragement
and support during my course of study in NUS. Having to cope with his studies as
well, he has always been very patient with me whenever I am stressed in studies and
never stopped to help me in my work.
Lastly, I would also like to thank my best friend, Yanjun, whose company has
made the learning journey a fun and enriching one.
i
Contents
Acknowledgements
i
Summary
iii
1 Introduction
1
1.1
Gabor Frames . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1.2
Scope of Thesis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
2 Nonstationary Gabor Frames in L2 (R)
7
2.1
Varying Modulations . . . . . . . . . . . . . . . . . . . . . . . . . . .
7
2.2
Dual Frames . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
25
2.3
Varying Translations . . . . . . . . . . . . . . . . . . . . . . . . . . .
34
3 Nonstationary Gabor Frames in S(2K )
40
3.1
Discrete Frames with Varying Modulations . . . . . . . . . . . . . .
41
3.2
Necessary Condition . . . . . . . . . . . . . . . . . . . . . . . . . . .
48
3.3
Varying Circular Shifts . . . . . . . . . . . . . . . . . . . . . . . . . .
53
4 Time-Frequency Analysis
58
4.1
Dual Frames in S(2K ) . . . . . . . . . . . . . . . . . . . . . . . . . .
58
4.2
Adaptive Time-Frequency Representation . . . . . . . . . . . . . . .
63
4.3
Time-Frequency Representation of Signals . . . . . . . . . . . . . . .
66
Bibliography
75
ii
Summary
Time-frequency analysis was introduced during the 1930s in the early development
of quantum mechanics [23] and in the theoretical foundation of signal analysis by
D. Gabor in 1946. After Gabor’s article “Theory of Communication” [14], timefrequency analysis was mainly focused on the practical applications by engineers
until an independent mathematical field was established in the 1980s. There are
many tools to perform time-frequency analysis for signals, in which the short-time
Fourier transform is one of them.
The short-time Fourier transform has its restrictions when it comes to resolution
issues. This thesis aims to tackle the problem by moving to nonstationary signal
analysis, in particular, by extending the standard Gabor analysis to a nonstationary
setup, aiming at adaptive time-frequency analysis and also reconstruction of the
signals from their time-frequency representations.
We start off the thesis with an introductory chapter by giving a brief review
on the short-time Fourier transform and Gabor frames. This is to highlight the
motivation of the move to nonstationary analysis and also present some standard
results on Gabor frames, allowing the reader to see the difference between results
of the stationary setup and the theory of the nonstationary case developed in the
subsequent chapters.
In Chapter 2, we focus on the space of square-integrable functions, L2 (R). There
are two structures of nonstationary Gabor frames that we look at, namely the one
with varying modulations in Section 2.1 and the one with varying translations in
iii
CHAPTER 0. SUMMARY
Section 2.3. In the first section, we discuss about sufficient conditions to make the
collection with varying modulations a frame in Theorem 2.3, adapting the proofs of
the stationary analogues in [5]. We then propose ways to construct such frames in
the subsequent results. Not forgetting the importance of reconstruction of the signals
from their time-frequency representations, we develop the theory on alternative dual
frames in the second section, identifying a setting such that there is an explicit
expression for an alternative dual. This is achieved in Theorem 2.14, for which the
proof is adapted from techniques developed in [6] and [7] on other situations. In
Section 2.3, we use properties of the Fourier transform to obtain results analogous
to those in the first section for the collection with varying translations.
We then shift our focus to another space in Chapter 3, which is the space of
periodic sequences. This move to a discrete setting is crucial for the applications
to signals in the final chapter. Similar to the previous chapter, we also develop
the theory for the frame collections with varying modulations and varying circular
shifts in Sections 3.1 and 3.3 respectively. To our advantage, as we are dealing with
frames in a finite-dimensional Hilbert space, the conditions for the collections to
form frames become weaker. In Section 3.2, we establish necessary conditions for a
discrete nonstationary Gabor frame in Proposition 3.9 and Theorem 3.10, giving us
more information on the number of frame elements in a frame.
Lastly, in Chapter 4, we investigate some applications of the theory developed
in Chapter 3. Before that, in Section 4.1, we present a theorem that gives an explicit
expression for an alternative dual for discrete frames in a particular setting. This is
then used in the next section where we propose an algorithm to automate the process
of adaptive time-frequency analysis, and the alternative dual generated is crucial in
reconstructing the signals from their time-frequency representations. We then end
with the final section by showing a few examples to illustrate the advantages of
performing adaptive time-frequency analysis with our nonstationary Gabor frames.
iv
List of Figures
4.1
Time-frequency analysis of a signal with high frequencies close to each
other in time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2
69
Time-frequency analysis of a signal with low frequencies close to each
other in frequency . . . . . . . . . . . . . . . . . . . . . . . . . . . .
71
4.3
Time-frequency analysis of a linear chirp . . . . . . . . . . . . . . . .
72
4.4
Time-frequency analysis of a hyperbolic chirp . . . . . . . . . . . . .
74
v
Chapter 1
Introduction
In time-frequency analysis of signals, the concept of frames is important, especially when we are also concerned about reconstructing the signals from their timefrequency representations. Thus, the short-time Fourier transform and Gabor frames
are closely connected in signal processing. In this chapter, we give a brief introduction on these two topics to see the link between them and also observe how it
motivates the development of the theory in this thesis. Interested readers may refer
to [5], [16], [18] for more details and exposition of these topics.
1.1
Gabor Frames
Time-frequency representation of signals is a form of signal processing which comprises many techniques to study a signal in the time and frequency domains. The
short-time Fourier transform, in particular, is one of the methods to do so (see for
instance [8], [16]). Given a signal f ∈ L2 (R), the space of square-integrable functions, and a window function w ∈ L2 (R), the short-time Fourier transform of f ,
defined as
∞
Ww f (a, b) =
f (t)w(t − a)e−2πibt dt,
−∞
1
a, b ∈ R,
CHAPTER 1. INTRODUCTION
2
provides information of the signal in a time-frequency window with the size fixed by
w. However, the Heisenberg uncertainty principle (see for instance [12], [20]) tells
us that there is a limit to the accuracy of the time-frequency localisation of a signal
using the short-time Fourier transform as there is a lower bound to the area of the
time-frequency window. To be more precise, we may choose a window function w
that gives very good time resolution when the time window is small. However, the
frequency window is consequently large, compromising the frequency resolution and
vice versa.
In addition, the use of only one window function fixes the window size across
the entire time domain and has less flexibility in the analysis of signals, especially
when the properties of a signal differ with time, which is mostly the case in real life.
Therefore, it is an advantage to alter the technique to allow nonstationary
analysis of signals. In particular, we are aiming for adaptive time-frequency analysis
of signals so that we can accurately capture the different properties of a signal at
different times.
To perform nonstationary signal analysis, a possible tool is nonstationary Gabor
frames (see [1]). Before we go into that, we need to be familiar with the standard
results in Gabor analysis, especially Gabor frames.
A collection of elements {fk }k∈I in a Hilbert space H is a frame for H if and
only if there exist A, B > 0 such that for every f ∈ H,
A f
2
| f, fk |2 ≤ B f
≤
2
.
k∈I
The constants A and B are called lower and upper frame bounds respectively and
when A = B, we call {fk }k∈I a tight frame. A collection that satisfies the upper
inequality above is called a Bessel sequence and the constant B is a Bessel bound.
A frame has a frame operator S defined by
Sf :=
f, fk fk ,
k∈I
f ∈ H,
CHAPTER 1. INTRODUCTION
3
which is a positive and invertible operator on H. We may then associate with the
frame its canonical dual {S −1 fk }k∈I and canonical tight frame {S −1/2 fk }k∈I which
are both frames for H as well, with the latter having frame bound 1 (see [5]). The
canonical dual is related to the frame by the expansion
f, fk S −1 fk =
f=
k∈I
f, S −1 fk fk .
(1.1)
k∈I
Frames were introduced in 1952 by Duffin and Schaeffer in [11] and further potential
of the topic was noted when nonorthogonal expansion of functions in L2 (R) can be
found using frames in [10]. This is also related to signal processing when it concerns
the reconstruction of signals. To have a more complete picture of the topic, the
reader may refer to [5], [9], [17].
A standard Gabor frame is a frame for L2 (R) of the form {Emb Tna g}m,n∈Z ,
where a, b > 0, g ∈ L2 (R) and the operators Eb and Ta are defined by
Modulation by b ∈ R :
Eb : L2 (R) → L2 (R), Eb f (t) := e2πibt f (t),
Translation by a ∈ R :
Ta : L2 (R) → L2 (R), Ta f (t) := f (t − a),
t ∈ R,
t ∈ R.(1.2)
Observe that the short-time Fourier transform can actually be re-expressed as
Ww f (a, b) = f, Eb Ta w ,
which is the reason why we look particularly at standard Gabor frames and see if we
can extend it to a nonstationary setup. For real-life applications, we probably look
at discrete points f, Emb Tna w , m, n ∈ Z, a, b > 0 as it is not feasible to decompose
a signal into uncountable points f, Eb Ta w for all a, b ∈ R. Hence, in the literature,
we can find necessary and sufficient conditions for a collection {Emb Tna g}m,n∈Z to
form a frame for L2 (R), which then allows us to reconstruct a signal from its frame
coefficients f, Emb Tna g using (1.1). The following are some results on Gabor frames
CHAPTER 1. INTRODUCTION
4
(see for instance [5]).
Theorem 1.1. Consider g ∈ L2 (R), a, b > 0 such that
B :=
1
sup
b t∈[0,a]
g(t − na)g(t − na − k/b) < ∞.
k∈Z n∈Z
Then {Emb Tna g}m,n∈Z is a Bessel sequence with Bessel bound B. If also
1
A :=
inf
b t∈[0,a]
|g(t − na)|2 −
n∈Z
g(t − na)g(t − na − k/b) > 0,
k=0 n∈Z
then {Emb Tna g}m,n∈Z is a frame for L2 (R) with bounds A, B.
Theorem 1.2. Let g ∈ L2 (R) and a, b > 0 such that {Emb Tna g}m,n∈Z is a frame
for L2 (R). Then ab ≤ 1.
In the later chapters, we will obtain some similar results that are nonstationary
analogues of the above.
1.2
Scope of Thesis
It is not the first time here when one wants to extend the standard Gabor analysis
to a nonstationary setup. For example, there is already some work done in [1]. In
this thesis, when we look into nonstationary Gabor frames, namely to consider the
collections {Embn gn }m,n∈Z and {Tnam hm }m,n∈Z , where gn , hm ∈ L2 (R), bn , am > 0
for m, n ∈ Z, we would also like to derive general sufficient conditions to make them
form a frame for L2 (R). In addition, we also look for a setting where there exists an
explicit expression of an alternative dual frame. In fact, after developing the theory
for the collection {Embn gn }m,n∈Z , we can obtain the analogous results for the other
collection using properties of the Fourier transform. The Fourier transform of a
function f ∈ L2 (R) is defined formally by
fˆ(ω) :=
∞
−∞
e−2πiωt f (t) dt,
ω ∈ R.
CHAPTER 1. INTRODUCTION
5
For the Fourier transform, we have the Parseval identity (see for instance [15], [16]):
f, g = fˆ, gˆ
f, g ∈ L2 (R),
(1.3)
which we use mainly in the last section of the next chapter to yield results for the
collection {Tnam hm }m,n∈Z . The use of a collection of functions {gn }n∈Z or {hm }m∈Z
instead of only one function to generate the frames introduce more flexibility in
the time-frequency resolution by allowing different window sizes at different times,
which is what we are aiming for in adaptive time-frequency representation.
After developing the theory on nonstationary Gabor frames for functions in
L2 (R), we will then look at nonstationary frames for sequences, namely the space of
periodic sequences S(2K ), where K is a nonnegative integer. This is to develop the
tools for applications in the last chapter, when we apply our theory to develop algorithms for time-frequency representation of discrete signals and also reconstruction
of the signals.
The theory for sequences is surprisingly much simpler when we deal with finite
number of elements in a collection {Em2K−pn dn }m=0,...,2pn −1 , where dn ∈ S(2K ),
n=0,...,Q−1
pn , Q, K are nonnegative integers, pn ≤ K. This is advantageous in applications to
have a larger collection of frames when the conditions for {Em2K−pn dn }m=0,...,2pn −1
n=0,...,Q−1
to form a frame for S(2K ) are not as strong as those in the case of L2 (R). Similarly,
after developing the theory for {Em2K−pn dn }m=0,...,2pn −1 , we may also easily obtain
n=0,...,Q−1
results for {Tn2K−qm cm }n=0,...,2qm −1 , cm ∈ S(2K ), qm ≤ K, through properties of
m=0,...,Q−1
the discrete Fourier transform.
Not forgetting our aim of adaptive approach in time-frequency analysis, we
then propose algorithms to create the collection {dn }n=0,··· ,Q−1 , where the dn ’s have
different supports, and then use them to analyse signals. After performing timefrequency analysis of the signals, we also use the theory on dual frames to reconstruct
the signals from their time-frequency representations. The theory of dual frames is
developed in the search for an alternative dual to the canonical dual formed using
CHAPTER 1. INTRODUCTION
the frame operator, which, in real-life applications, may not be easily available.
In the next few chapters, we will see the details of the development of the theory
and its applications. In particular, we may observe how the proofs of some results
in [5], [7] are adapted for our use.
6
Chapter 2
Nonstationary Gabor Frames in
L2(R)
Gabor frames in L2 (R) refer to frames that are of the form {Emb Tna g}m,n∈Z , where
g ∈ L2 (R). A Gabor frame was formed using one function g and the translation and
modulation operators, as defined in (1.2). As a result, the frame elements have fixed
window size, same as the original function g. In this chapter, our aim is to introduce
more flexibility in the structure of the frames, so as to allow variable window sizes
within a frame. In addition, we may use more than one function to generate such a
collection.
2.1
Varying Modulations
Given a sequence of functions {gn }n∈Z in L2 (R), we generate a collection of functions {Embn gn }m,n∈Z using the modulation operator and a sequence of positive real
numbers {bn }n∈Z . We observe that if gn are selected to be translates of one function
g and the parameters bn are the same, the collection then has the same structure
as Gabor frames. We want to look for conditions that are sufficient to make the
collection {Embn gn }m,n∈Z a frame for L2 (R).
Most of the results that we see in this section are closely related to results
7
CHAPTER 2. NONSTATIONARY GABOR FRAMES IN L2 (R)
8
proved in [5] regarding standard Gabor frames. In fact, in the original proofs, the
translation operator is of little influence, which allows us to easily generalise the
results to the new collection of frames. However, this generalisation exposes us
to great potential in exploring the possibility of having better methods of signal
analysis, or at least the possibility of more techniques that can be made flexible to
suit different needs.
Take note that in the following results, we may make reference to the following
function
G(t) :=
n∈Z
1
|gn (t)|2 ,
bn
t ∈ R.
(2.1)
In addition, in this chapter, all the interchanges of summations and integrations can
be justified by the Fubini and Tonelli theorems (see for instance [21], [22]).
Lemma 2.1. Let f, gn ∈ L2 (R) and bn > 0 for all n ∈ Z. For a fixed n ∈ Z, let the
function Fn be defined by
f (t −
Fn (t) :=
k∈Z
k
k
)gn (t − ),
bn
bn
t ∈ R.
(2.2)
Then, for any m, n ∈ Z,
1
bn
f, Embn gn =
Fn (t)e−2πimbn t dt.
0
Proof. Firstly, we show that Fn is well defined by showing that the series converges
absolutely a.e. Indeed,
1
bn
0
k∈Z
k
k
|f (t − )gn (t − )| dt =
bn
bn
1
bn
|f (t −
k∈Z 0
1
− bk
bn
n
=
k
k∈Z − bn
∞
k
k
)gn (t − )| dt
bn
bn
|f (y)gn (y)| dy
|f (y)gn (y)| dy.
=
−∞
CHAPTER 2. NONSTATIONARY GABOR FRAMES IN L2 (R)
9
Since f, gn ∈ L2 (R), by the Cauchy-Schwarz inequality,
∞
1
2
∞
2
|f (t)gn (t)| dt ≤
∞
2
|f (t)| dt
−∞
|gn (t)| dt
−∞
1
2
< ∞.
−∞
The above shows that the series defined by Fn is in L1 [0, b1n ], the space of integrable
functions on [0, b1n ], and thus converges absolutely a.e in [0, b1n ]. Since Fn is
1
bn -
periodic, we then have the series to converge absolutely a.e in R.
k
bn
Next, substituting t = y −
∞
f, Embn gn
=
yields
f (t)gn (t)e−2πimbn t dt
−∞
1
− bk
b
n
=
− bk
n
1
bn
k∈Z
=
f (t)gn (t)e−2πimbn t dt
f (y −
k∈Z 0
1
bn
=
k
k
)gn (y − )e−2πimbn y dy
bn
bn
Fn (y)e−2πimbn y dy.
0
To form a frame, we need the collection to satisfy the frame inequality. Hence,
we want to simplify the expression for
m∈Z
n∈Z |
f, Embn gn |2 in the next lemma
for easier manipulation.
Lemma 2.2. Suppose that gn ∈ L2 (R), bn > 0 for all n ∈ Z such that
B := sup
t∈R k∈Z n∈Z
k
1
gn (t)gn (t − ) < ∞.
bn
bn
Then for every f ∈ L2 (R),
∞
| f, Embn gn |2 =
m∈Z n∈Z
|f (t)|2 G(t) dt
−∞
∞
+
f (t −
f (t)
k=0
−∞
n∈Z
k
1
k
) · gn (t)gn (t − ) dt.
bn bn
bn
CHAPTER 2. NONSTATIONARY GABOR FRAMES IN L2 (R)
k
k∈Z f (t − bn )gn (t
Proof. Fix n ∈ Z. As shown in Lemma 2.1, Fn (t) =
10
−
k
bn )
defined and is in L1 [0, b1n ]. In addition, for every m, n ∈ Z,
1
bn
f, Embn gn =
Fn (t)e−2πimbn t dt.
0
Next, we want to show that for every n ∈ Z, Fn ∈ L2 [0, b1n ].
n∈Z
=
n∈Z
≤
n∈Z
Substituting y = t −
1
bn
1
bn
1
bn
1
bn
1
bn
k
bn ,
|Fn (t)|2 dt
0
0
f (t −
k
k
)gn (t − ) · Fn (t) dt
bn
bn
f (t −
k
k
)gn (t − ) · Fn (t) dt =: C.
bn
bn
k∈Z
1
bn
k∈Z 0
we obtain
1
bn
C =
n∈Z
1
− bk
bn
n
− bk
n
k∈Z
1
bn
=
k
n∈Z k∈Z − bn
−
∞
n∈Z −∞
∞
≤
n∈Z −∞
=
n∈Z
1
bn
1
bn -periodic.
1
f (y)gn (y)
bn
1
|f (y)gn (y)|
bn
f (y −
k∈Z
Consequently, we have
k
k
)gn (y − ) dy
bn
bn
f (y −
k∈Z
∞
f (y)gn (y)f (y −
k∈Z −∞
k
) dy
bn
k 1
|f (y)gn (y)Fn (y)| dy,
bn bn
with the last equation due to Fn being
C =
f (y)gn (y) · Fn (y +
k
k
)gn (y − ) dy
bn
bn
k
k
)gn (y − ) dy.
bn
bn
is well
CHAPTER 2. NONSTATIONARY GABOR FRAMES IN L2 (R)
11
By the Cauchy-Schwarz inequality on the integral and summation, for any n ∈ Z,
1
bn
≤
1
bn
∞
|f (y)gn (y)f (y −
k∈Z −∞
k
k
)gn (y − )| dy
bn
bn
∞
k
|f (y)| |gn (y)gn (y − )| dy
bn
−∞
1/2
2
k∈Z
·
∞
k∈Z
Substituting t = y −
∞
|f (y −
k∈Z −∞
k
bn
k
k
|f (y − )|2 |gn (y)gn (y − )| dy
b
b
n
n
−∞
1/2
:= Cn .
and replacing k by −k, we obtain
k 2
k
)| |gn (y)gn (y − )| dy =
bn
bn
∞
|f (t)|2 |gn (t +
k
)gn (t)| dt
bn
|f (t)|2 |gn (t −
k
)gn (t)| dt,
bn
k∈Z −∞
∞
=
k∈Z −∞
hence
Cn =
1
bn
∞
|f (y)|2 |gn (y)gn (y −
k∈Z −∞
k
)| dy.
bn
Therefore,
C ≤
∞
1
bn
|f (y)|2 |gn (y)gn (y −
k∈Z −∞
n∈Z
∞
=
−∞ n∈Z
≤ B f
As a result, for every n ∈ Z,
2
1
bn
|gn (y)gn (y −
k∈Z
k
)| dy
bn
k
)||f (y)|2 dy
bn
< ∞.
1
bn
0
|Fn (t)|2 dt < ∞ and so Fn ∈ L2 [0, b1n ]. Hence, we
may write
f, Embn gn
L2 (R)
= Fn , Embn
L2 [0, b1 ] ,
n
where for simplicity, we also use the operator notation Embn for e2πimbn t . Also, when
there are more than one space involved, we may add a subscript of the space to the
CHAPTER 2. NONSTATIONARY GABOR FRAMES IN L2 (R)
12
inner products and norms to indicate the space they apply to. When there is no
confusion in the spaces, the subscripts will be omitted.
√
Since { bn Embn }m∈Z is an orthonormal basis for L2 [0, b1n ], for each n ∈ Z,
| Fn ,
bn Embn
2
L2 [0, b1 ] |
=
2
,
L2 [0, b1 ]
Fn
n
m∈Z
n
which gives
| Fn , Embn
2
L2 [0, b1 ] |
1
Fn
bn
=
n
m∈Z
2
.
L2 [0, b1 ]
n
This implies that
| f, Embn gn |2 =
n∈Z m∈Z
L2 [0, b1 ] |
| Fn , Embn
n∈Z m∈Z
=
n∈Z
1
Fn
bn
2
L2 [0, b1 ]
1
bn
=
n
Now, by rewriting |Fn (t)|2 = Fn (t)Fn (t) =
2
n
n∈Z 0
l∈Z f (t
−
1
|Fn (t)|2 dt.
bn
l
bn )gn (t
−
l
bn )Fn (t),
have, after integration by substitution,
2
| f, Embn gn |
1
bn
=
n∈Z 0
n∈Z m∈Z
=
n∈Z
Noting that Fn (y +
l
bn )
n∈Z
=
n∈Z
=
n∈Z
1
bn
1
bn
f (t −
l∈Z
− bl + b1
n
l∈Z
− bl
n
l
l
)gn (t − )Fn (t) dt
bn
bn
f (y)gn (y)Fn (y +
n
l
) dy.
bn
= Fn (y) due to the periodicity of Fn , we obtain
1
bn
1
bn
1
bn
− bl + b1
n
l∈Z
∞
− bl
n
n
f (y)gn (y)Fn (y +
l
) dy
bn
f (y)gn (y)Fn (y) dy
−∞
∞
f (y −
f (y)gn (y)
−∞
k∈Z
k
k
)gn (y − ) dy.
bn
bn
we
CHAPTER 2. NONSTATIONARY GABOR FRAMES IN L2 (R)
Lastly, by considering separately k = 0 and k = 0, we have
| f, Embn gn |2
n∈Z m∈Z
∞
|f (y)|2 G(y) dy
=
−∞
∞
+
f (y)
k=0
−∞
n∈Z
1
k
k
f (y − )gn (y)gn (y − ) dy.
bn
bn
bn
This completes the proof.
Using Lemma 2.2, we have the following theorem that gives us a sufficient
condition for {Embn gn }m,n∈Z to be a frame for L2 (R). This theorem generalises
Theorem 1.1 (see [5]) into the nonstationary setting.
Theorem 2.3. Consider gn ∈ L2 (R), bn > 0 for all n ∈ Z such that
B := sup
t∈R k∈Z n∈Z
1
k
gn (t)gn (t − ) < ∞.
bn
bn
Then {Embn gn }m,n∈Z is a Bessel sequence with Bessel bound B. If also
A := inf
t∈R
n∈Z
1
|gn (t)|2 −
bn
k=0 n∈Z
1
k
gn (t)gn (t − ) > 0,
bn
bn
then {Embn gn }m,n∈Z is a frame for L2 (R) with bounds A, B.
Proof. By Lemma 2.2, we have
∞
| f, Embn gn |2 =
m∈Z n∈Z
|f (t)|2 G(t) dt
−∞
∞
f (t −
f (t)
+
k=0
−∞
n∈Z
k
1
k
) · gn (t)gn (t − ) dt.
bn bn
bn
13
CHAPTER 2. NONSTATIONARY GABOR FRAMES IN L2 (R)
14
Looking at the second term above, we have
∞
f (t −
f (t)
k=0
−∞
n∈Z
∞
≤
−∞ n∈Z
k=0
=
n∈Z
1
bn
k
1
k
) · gn (t)gn (t − ) dt
bn bn
bn
1
k
k
f (t)f (t − )gn (t)gn (t − ) dt
bn
bn
bn
∞
f (t)f (t −
k=0
−∞
k
k
)gn (t)gn (t − ) dt.
bn
bn
Following the same steps as a computation in the proof of Lemma 2.2, we apply
Cauchy-Schwarz inequality on the integral and the summation over all k = 0, and
replace k with −k. We then obtain
n∈Z
≤
n∈Z
1
bn
1
bn
∞
f (t)f (t −
k=0
−∞
k
k
)gn (t)gn (t − ) dt
bn
bn
∞
|f (t)|2 gn (t)gn (t −
k=0
−∞
k
) dt.
bn
This means that
∞
| f, Embn gn |2 ≤
m∈Z n∈Z
|f (t)|2
−∞
n∈Z
1
|gn (t)|2
bn
+
k=0 n∈Z
1
k
gn (t)gn (t − ) dt
bn
bn
∞
|f (t)|2
=
−∞
≤ B f
k∈Z n∈Z
2
.
1
k
gn (t)gn (t − )
bn
bn
dt
CHAPTER 2. NONSTATIONARY GABOR FRAMES IN L2 (R)
15
In addition,
∞
| f, Embn gn |2 ≥
m∈Z n∈Z
|f (t)|2
−∞
n∈Z
1
|gn (t)|2
bn
−
k=0 n∈Z
≥ A f
2
1
k
|gn (t)||gn (t − )| dt
bn
bn
.
The constants A and B defined in the theorem above are not easy to check
for any collection of functions since they involved infinite sums. However, if we can
make the sums involved finite, it would be much easier to check. Hence, we aim to
create our collection in certain ways, ensuring that it satisfies the conditions stated.
Corollary 2.4. Let bn > 0, gn ∈ L2 (R) with support of gn in an interval of length
1
bn
for every n ∈ Z such that the function G satisfies A ≤ G(t) ≤ B a.e for some
A, B > 0. Then {Embn gn }m,n∈Z is a frame for L2 (R) with bounds A, B. In addition,
the frame operator and its inverse are
Sf = Gf,
S −1 f =
1
f,
G
f ∈ L2 (R).
Proof. Since for each n ∈ Z, the support of gn lies in an interval of length
k = 0,
n∈Z
1
k
gn (t)gn (t − ) = 0 a.e.
bn
bn
Hence,
sup
t∈R k∈Z n∈Z
1
k
gn (t)gn (t − ) = sup G(t) ≤ B.
bn
bn
t∈R
Also,
inf
t∈R
n∈Z
1
|gn (t)|2 −
bn
k=0 n∈Z
1
k
gn (t)gn (t − ) = inf G(t) ≥ A.
t∈R
bn
bn
1
bn ,
for
CHAPTER 2. NONSTATIONARY GABOR FRAMES IN L2 (R)
16
Hence, by Theorem 2.3, {Embn gn }m,n∈Z forms a frame for L2 (R) with bounds A, B.
To find the frame operator S, we note that by Lemma 2.2, for every f ∈ L2 (R),
Sf, f
=
f, Embn gn Embn gn , f
| f, Embn gn |2
=
m∈Z n∈Z
∞
m∈Z n∈Z
|f (t)|2 G(t) dt = Gf, f .
=
−∞
Thus, for every f ∈ L2 (R), (S − GI)f, f = 0 which gives S = GI.
Corollary 2.4 was proved directly in [1]. Here we first establish general sufficient
conditions for nonstationary Gabor frames and then obtain it as a consequence of
these conditions.
Corollary 2.5. Consider {gn }n∈Z in L2 (R) and bn > 0, n ∈ Z, that satisfy the
following:
(i) There exist C, D > 0 such that for every n ∈ Z, C ≤
|gn (t)|
√
bn
≤ D a.e in the
support of gn ;
(ii) For each n ∈ Z, the support of gn lies in an interval of length less than or
equal to
1
bn ;
(iii) There exists a positive integer K such that for a.e t ∈ R, {n ∈ Z : gn (t) = 0}
has at most K elements and at least one element.
Then {Embn gn }m,n∈Z forms a frame for L2 (R).
Proof. It is easy to see that from (i) and (iii), C 2 ≤
1
2
n∈Z bn |gn (t)|
≤ KD2 a.e.
Hence, by Corollary 2.4, {Embn gn }m,n∈Z forms a frame for L2 (R).
Remark 2.6. The condition (iii) poses some restriction on the intersection of the
supports of gn . It means that while we allow the supports to intersect, there should
only be at most K intersections. On the other hand, it also ensures that the union of
the supports essentially covers the real line, which is not surprising since the union
of the supports of frame elements should do so.
CHAPTER 2. NONSTATIONARY GABOR FRAMES IN L2 (R)
Now, we look at some examples of Corollaries 2.4 and 2.5.
Example 2.7. Firstly, let bn =
g(t) :=
1
3
for every n ∈ Z. Consider
3
9
3
1
1 2
2t + 2t + 8, −2 ≤ t < −2;
−t2 + 3 ,
− 21 ≤ t < 12 ;
4
1 2
3
9
1
3
2t − 2t + 8, 2 ≤ t ≤ 2;
0,
otherwise.
For every n ∈ Z, let gn (t) =
g(t − 2n). We then observe that gn has support in
an interval of length 3. In addition,
G(t − 2) =
n∈Z
1
|gn (t − 2)|2 = 3
bn
|gn (t − 2)|2
n∈Z
|gn+1 (t)|2 = G(t).
= 3
n∈Z
Hence, G is 2-periodic and we have supt∈R G(t) = supt∈[− 3 , 1 ) G(t). Noting that the
2 2
support of gn is
G(t) =
[− 23
3
2
+ 2n, + 2n],
1
2
b−1 |g−1 (t)|
+
1
2
b0 |g0 (t)| ,
− 32 ≤ t < − 12 ;
1 |g0 (t)|2 ,
− 21 ≤ t < 12 ,
b0
1 (t + 2)2 − 3 (t + 2) + 9 + 1 t2 + 3 t + 9 , − 3 ≤ t < − 1 ;
2
2
8
2
2
8
2
2
= 3
3
1
1
2
−t + ,
−2 ≤ t < 2,
4
(t + 1)2 + 1 , − 3 ≤ t < − 1 ;
4
2
2
= 3
−t2 + 3 ,
− 12 ≤ t < 12 .
4
Therefore,
3
4
≤ G(t) ≤
9
4
and by Corollary 2.4, {Embn gn }m,n∈Z forms a frame for
L2 (R).
Example 2.8. Next, we look at a simple example for Corollary 2.5. For n ∈ Z, let
√
1
bn = (|n|+1)
bn χ[n,n+1) (t). Clearly, the support of each gn has length
2 and gn (t) =
17
CHAPTER 2. NONSTATIONARY GABOR FRAMES IN L2 (R)
1 which is less than
1
bn
18
= (|n| + 1)2 . In its support,
|gn (t)|
√
= 1.
bn
Lastly, the supports of the gn do not intersect with each other but the union of them
is the real line. Hence, fulfilling all the conditions of Corollary 2.5, {Embn gn }m,n∈Z
forms a frame for L2 (R).
In the following, we propose ways of forming the sequence of functions {gn }n∈Z
from a continuous compactly supported function so that {Embn gn }m,n∈Z forms a
frame for L2 (R). In the previous two examples, we look at {gn }n∈Z for which the
supports of gn have the same length. This may not be useful enough for our initial
aim of varying the window sizes for signal analysis. Thus, what we can try is to
dilate one function by bn for each n ∈ Z, so that we still have the lengths of the
support to be less than
1
bn
and they may vary with bn .
The first proposition we have below is to dilate a continuous function g, supported on [0, 1], by
1
,
a|n|
for a positive constant a, to first obtain a sequence of
functions with varying support lengths. The next step is to translate each function
accordingly to make sure that the union of the supports covers the entire real line.
Hence, we impose conditions on the translation and yield the following.
Proposition 2.9. Let g ∈ L2 (R) be continuous and compactly supported on [0, 1],
and positive on (0, 1). Let a > 1 and c ∈ R satisfy
tn = can . In addition, for n ∈ Z\{0}, let bn =
1
a|n|
1
a2 −1
1
a2 −1
1
a−1
implies that can+1 ≤ can +an , and
gives can+2 > can + an and so tn+2 > tn + b1n .
1
1
Hence, [tn , b1n + tn ] intersects with [tn+1 , bn+1
+ tn+1 ] but not with [tn+2 , bn+2
+ tn+2 ].
We can then write
t1 +
1
,∞
b1
=
tn−1 +
n≥2
and for each n ≥ 2, G(t) =
1
2
bn |gn (t)|
+
1
2
bn+1 |gn+1 (t)|
1
2
bn |gn (t)|
1
bn−1
, tn+1 ∪ tn+1 , tn +
for t ∈ (tn−1 +
for t ∈ [tn+1 , tn +
1
bn
1
bn−1 , tn+1 ]
,
and G(t) =
1
bn ].
1
We will next show that for every n ≥ 1, G[tn+1 , tn + b1n ] = G[tn+2 , tn+1 + bn+1
],
which means that the set of values of G on the intersection of the supports of each
pair of gn and gn+1 is the same for every nonzero n. Let t ∈ [tn+1 , tn +
t = tn+1 + x for some 0 ≤ x ≤ tn +
G(t) =
1
bn
1
bn ].
Thus
− tn+1 = can + an − can+1 . Then
1
1
|gn (t)|2 +
|gn+1 (t)|2
bn
bn+1
= |g(bn (tn+1 + x − tn ))|2 + |g(bn+1 (tn+1 + x − tn+1 ))|2
=
g ca +
x
−c
an
Similarly, let t ∈ [tn+2 , tn+1 +
0 ≤ y ≤ tn+1 +
1
bn+1
2
1
bn+1 ],
+ g
x
an+1
2
.
which means that t = tn+2 + y for some
− tn+2 = can+1 + an+1 − can+2 . Following the same argument
CHAPTER 2. NONSTATIONARY GABOR FRAMES IN L2 (R)
20
as above yields
G(t ) = g ca +
Therefore, for every t ∈ [tn+1 , tn +
y
an+1
1
bn ],
−c
y
2
+ g
2
.
an+2
writing t = tn+1 + x for some 0 ≤ x ≤
1
can + an − can+1 , there exists t ∈ [tn+2 , tn+1 + bn+1
], where t = tn+2 + ax such that
G(t ) = G(t). Similarly, for every t ∈ [tn+2 , tn+1 +
1
bn+1 ],
where t = tn+2 + y for
some 0 ≤ y ≤ can+1 +an+1 −can+2 , there exists t ∈ [tn+1 , tn + b1n ] where t = tn+1 + ay
such that G(t ) = G(t).
Next, since each gn is continuous, G =
on each [tn+1 , tn +
1
bn ].
1
2
bn |gn |
+
1
2
bn+1 |gn+1 |
is also continuous
Thus it is clearly bounded and B1 = sup G < ∞, where
the supremum is taken over [tn+1 , tn +
1
bn ]
for every n ≥ 1. In addition, since g
is positive on (0, 1), G is also positive on [tn+1 , tn +
1
bn ]
which implies that A1 =
inf G > 0, with the infimum taken over each [tn+1 , tn +
1
bn ].
infimum and supremum of G over the union of all [tn+1 , tn +
Consequently, the
1
bn ]
for n ≥ 1 are A1
and B1 respectively.
With a similar argument, we can also show that for every n ≥ 2, G(tn−1 +
1
bn−1 , tn+1 ]
= G(tn +
n ≥ 1, over [tn +
1
bn , tn+2 ].
1
bn , tn+2 ],
We will also arrive at the conclusion that for every
A2 = inf G > 0 and B2 = sup G < ∞. These will then
show that min{A1 , A2 } = inf G > 0 and max{B1 , B2 } = sup G < ∞ on [t1 +
and due to the definition of gn for negative n, on (−∞, −t1 −
1
b1 , ∞)
1
b1 ).
Lastly, we are left to show that G is bounded on [−t1 − b11 , t1 + b11 ]. Noting that
on this interval, gn = 0 for |n| ≥ 3, we see that G =
1
2
|n|≤2 bn |gn (t)|
is bounded
above by some B3 < ∞ and below by some A3 > 0 by the same argument as
above. Therefore, there exist A, B > 0 such that A ≤ G ≤ B a.e. By Corollary 2.4,
{Embn gn }m,n∈Z forms a frame for L2 (R).
Now, in Proposition 2.9, we dilate a function with support length one by bn in
order to create a sequence of functions with support lengths
1
bn
exactly. However,
CHAPTER 2. NONSTATIONARY GABOR FRAMES IN L2 (R)
21
we notice that what we need is for the support lengths of gn to be smaller than or
equal to
1
bn
for each n ∈ Z. Hence, in the next result, we dilate a function of support
length β by another sequence of parameters {cn }n∈Z and impose conditions on the
parameters to get what we want.
In addition, since the union of the supports of {gn }n∈Z should essentially cover
R, we need to translate the dilated functions accordingly to achieve this. This gives
rise to the sequence {an }n∈Z , for which limn→∞ an = ∞ and limn→−∞ an = −∞.
The key point in the construction of {gn }n∈Z in the following result is to make
use of a function g that has an increasing portion at the left end of its support
and a decreasing one at the other end. Consequently, each of the {gn }n∈Z will
have the same characteristic. We then let the decreasing portion of gn intersects
with sufficient amount of the increasing portion of gn+1 for every n. There will be
conditions on all these features to make sure that the function G is bounded below
by a constant.
Proposition 2.10. Let g be a continuous function, compactly supported on [0, β]
for some β > 0, and positive on (0, β). Assume that there exist constants δ1 , δ2 ,
0 < δ1 ≤ δ2 < β, such that g is increasing on [0, δ1 ] and decreasing on [δ2 , β].
Choose a strictly increasing real sequence, {an }n∈Z , where limn→∞ an = ∞ and
limn→−∞ an = −∞, and a positive sequence {cn }n∈Z satisfying the following:
(i) For every n ∈ Z, an+1 − an <
(ii) There exist constants
1, 2,
β
cn
and an+2 − an >
0 <
1
< δ1 , δ2 <
2
β
cn ;
< β, such that for every
n ∈ Z,
cn+1
2
β
+ an − an+1
cn
≥
1
and
β + cn (an+1 − an )
≤
2
(iii) For every n ∈ Z,
cn+1
β
+ an − an+1
cn
≤ δ1 and cn (an+1 − an ) ≥ δ2 .
2;
CHAPTER 2. NONSTATIONARY GABOR FRAMES IN L2 (R)
In addition, assume that g(t) ≥ min{g( 1 ), g( 2 )} for t ∈ [δ1 , δ2 ]. Lastly, choose
a positive sequence {bn }n∈Z such that for every n ∈ Z, cβn ≤ b1n . Define gn (t) :=
√
bn g(cn (t − an )). Then {Embn gn }m,n∈Z forms a frame for L2 (R).
Proof. To apply Corollary 2.4, we check the conditions in the corollary. Observe
that the length of the support of gn , [an , an +
β
cn ]
is
β
cn
≤
1
bn .
We can also see that
for every n ∈ Z, supp gn intersects supp gn+1 but not supp gn+2 . This is clear from
(i) that an+1 < an +
β
cn
and an+2 > an +
β
cn .
Now, we want to show that there exist A, B > 0 such that A ≤ G(t) ≤ B a.e.
where G is as defined in (2.1). Firstly, we note that for every n ∈ Z, on [an + cβn , an+2 ],
the interval in supp gn+1 not intersected by supp gn or supp gn+2 ,
G(t) =
1
bn+1
Also, on [an+1 , an +
|gn+1 (t)|2 = |g(cn+1 (t − an+1 ))|2 ≤ max |g(x)|2 < ∞.
x∈[0,β]
β
cn ],
G(t) =
the intersection of supp gn and supp gn+1 ,
1
1
|gn (t)|2 +
|gn+1 (t)|2 ≤ 2 max |g(x)|2 .
bn
bn+1
x∈[0,β]
This gives G being bounded above by B = 2 maxx∈[0,β] |g(x)|2 on every [an+1 , an+2 ],
hence on R.
Next, we fix n ∈ Z. We will look at G on [an+1 , an+2 ], showing that there exists
a positive constant A, independent of n, such that G ≥ A on [an+1 , an+2 ]. Consider
t ∈ [an+1 , 21 (an+1 + an +
β
cn )],
G(t) =
the first half-interval of supp gn ∩ supp gn+1 . Then
1
1
|gn (t)|2 +
|gn+1 (t)|2
bn
bn+1
= |g(cn (t − an ))|2 + |g(cn+1 (t − an+1 ))|2 .
Since an+1 ≤ t ≤ 12 (an+1 + an +
β
cn ),
it follows that
cn (an+1 − an ) ≤ cn (t − an ) ≤
β + cn (an+1 − an )
,
2
22
CHAPTER 2. NONSTATIONARY GABOR FRAMES IN L2 (R)
23
and so
δ2 ≤ cn (t − an ) ≤
by (ii) and (iii). Since g is decreasing on [δ2 ,
2 ],
2,
g(cn (t−an )) ≥ g( 2 ) and as a result,
G(t) ≥ |g( 2 )|2 . Next, consider t in the other half interval of supp gn ∩ supp gn+1 ,
i.e. t ∈ [ 21 (an+1 + an +
Since 12 (an+1 + an +
cn+1
2
β
cn ), an
β
cn )
+
β
cn ].
≤ t ≤ an +
β
+ an − an+1
cn
Similarly, G(t) =
β
cn ,
1
2
bn |gn (t)|
+
1
2
bn+1 |gn+1 (t)| .
we have
≤ cn+1 (t − an+1 ) ≤ cn+1
β
+ an − an+1 ,
cn
which gives
1
≤ cn+1 (t − an+1 ) ≤ δ1 .
Since g is increasing on [ 1 , δ1 ], g(cn+1 (t − an+1 )) ≥ g( 1 ). Thus, G(t) ≥ |g( 1 )|2 .
Lastly, we consider t in supp gn+1 not intersected by supp gn or gn+2 ,i.e. [an +
β
cn , an+2 ].
On this interval, G(t) =
1
2
bn+1 |gn+1 (t)|
= |g(cn+1 (t − an+1 ))|2 . We observe
that
cn+1 (t − an+1 ) ∈
cn+1
=
cn+1
By (ii),
cn+1 β
2 ( cn
β
+ an − an+1 , cn+1 (an+2 − an+1 )
cn
β
+ an − an+1 , δ1 ∪ [δ1 , δ2 ] ∪ [δ2 , cn+1 (an+2 − an+1 )].
cn
+ an − an+1 ) ≥
1
implies that cn+1 ( cβn + an − an+1 ) ≥
1.
Hence, for
cn+1 (t − an+1 ) ∈ [cn+1 ( cβn + an − an+1 ), δ1 ], cn+1 (t − an+1 ) ∈ [ 1 , δ1 ] and we obtain
G(t) ≥ |g( 1 )|2 . Similarly, we have for cn+1 (t − an+1 ) ∈ [δ2 , cn+1 (an+2 − an+1 )],
cn+1 (t − an+1 ) ∈ [δ2 ,
2]
and G(t) ≥ |g( 2 )|2 . Lastly, for cn+1 (t − an+1 ) ∈ [δ1 , δ2 ],
G(t) ≥ min{|g( 1 )|2 , |g( 2 )|2 } due to the given assumption of g(t) ≥ min{g( 1 ), g( 2 )}
for t ∈ [δ1 , δ2 ].
Hence, we have shown that on [an+1 , an+2 ], G(t) ≥ min{|g( 1 )|2 , |g( 2 )|2 }. This
CHAPTER 2. NONSTATIONARY GABOR FRAMES IN L2 (R)
24
implies the same inequality on R and completes the proof.
The following result is a corollary of the above construction. We let {an }n∈Z
to be the integers and choose g such that δ1 = δ2 so that we definitely have the
condition of g greater than min{g( 1 ), g( 2 )} satisfied.
Corollary 2.11. Let g be a continuous function, compactly supported on [0, 2k] for
some k > 0, and positive on (0, 2k). Assume that g increases on [0, k] and decreases
on [k, 2k]. Let 0 < σ ≤
1
3
and choose α1 , α2 > 0 such that 1 + σ < α1 < α2 <
1+σ
1−σ .
1
For every n ∈ Z, choose cn , bn > 0 satisfying α1 k ≤ cn ≤ α2 k and 2k
cn ≤ bn . Define
√
the sequence of functions gn (t) := bn g(cn (t − n)). Then {Embn gn }m,n∈Z forms a
frame for L2 (R).
Proof. Since for every n ∈ Z, α1 k ≤ cn ≤ α2 k, we obtain
α1
cn
α2
≤
≤
,
2
2k
2
2
2k
2
≤
≤
.
α2
cn
α1
Given the choice of α1 , α2 , σ,
2(1 − 13 )
2
2(1 − σ)
>
≥
= 1,
α2
1+σ
1 + 31
This gives us 1 <
2k
cn
2
2
<
< 2.
α1
1+σ
< 2, satisfying condition (i) in the previous proposition. Next,
since
cn+1
2
we see that
0<
1
α1 k 2
2 ( α2
≥
α1 k
2
2
−1 ,
α2
− 1) is a probable constant for
< k. Indeed, α2 <
0<
2k
−1
cn
α1 k
2
1+σ
1−σ
≤ 2 implies that
2
−1
α2
<
k
2
1+σ
1−σ
2
α2
1.
We just need to check that
− 1 is positive. As a result,
2
−1
1+σ
Similarly,
2k + cn
2 + α2
≤
k.
2
2
=
k
< k.
2
CHAPTER 2. NONSTATIONARY GABOR FRAMES IN L2 (R)
Letting
2
=
2+α2
2 k,
25
we can verify that
1
2 + α2
k<
k<
2
2
1+σ
2+
1−σ
1
k≤
2
1+
2+
1−
1
3
1
3
k = 2k.
Lastly, we check (iii) in the previous proposition. We have
cn+1
2k
−1
cn
≤ α2
2
−1 k <
α1
1+σ
1−σ
2
− 1 k = k,
1+σ
and
cn ≥ α1 k > k.
Hence, by Proposition 2.10, we have the collection {Embn gn }m,n∈Z forming a frame.
Remark 2.12. We see from the definition of gn in Corollary 2.11 that the window
width of the frame elements are determined by the constants cn . This may allow us
to choose the constants cn carefully in order to adjust the window sizes in different
circumstances during signal analysis.
2.2
Dual Frames
One property of frames in Hilbert space is the existence of dual frames. Given a
frame {fk }k∈I in a Hilbert space H, there exists another frame {gk }k∈I in H such
that for every element f in H,
f=
f, fk gk =
k∈I
f, gk fk .
(2.3)
k∈I
The collections {fk }k∈I and {gk }k∈I are called dual frames for H. In particular, with
S being the frame operator, {gk }k∈I defined by gk = S −1 fk is called the canonical
dual frame for {fk }k∈I . In Corollary 2.4, we have seen a special case of how the
frame operator is like. However, we may not always have an explicit expression for
CHAPTER 2. NONSTATIONARY GABOR FRAMES IN L2 (R)
26
S, let alone the canonical dual frame. Thus, we hope to find, in certain cases, other
dual frames so that we can use them in the recovery of signals via (2.3). Before that,
we will look at a slight generalisation of Lemma 2.2.
Lemma 2.13. Suppose that for all n ∈ Z, gn , hn ∈ L2 (R) and bn > 0 are chosen
such that the following constants are all finite,
C1 := sup
t∈R k∈Z n∈Z
C2 := sup
t∈R k∈Z n∈Z
B1 := sup
t∈R k∈Z n∈Z
B2 := sup
t∈R k∈Z n∈Z
1
k
gn (t)hn (t − ) ;
bn
bn
1
k
gn (t + )hn (t) ;
bn
bn
1
k
gn (t)gn (t − ) ;
bn
bn
k
1
hn (t)hn (t − ) .
bn
bn
Then for every f1 , f2 ∈ L2 (R),
∞
f1 , Embn gn f2 , Embn hn =
n∈Z m∈Z
f1 (t)f2 (t)
−∞
n∈Z
∞
+
f1 (t)
k=0
−∞
n∈Z
1
gn (t)hn (t) dt
bn
1
k
k
f2 (t − )gn (t)hn (t − ) dt.
bn
bn
bn
Proof. We note that as shown in Lemma 2.1, for each n ∈ Z, f ∈ L2 (R), the functions Gn f (t) :=
k∈Z f (t
−
k
bn )gn (t
−
k
bn )
and Hn f (t) :=
k∈Z f (t
−
k
bn )hn (t
−
k
bn )
are well defined and are in L1 [0, b1n ]. In addition, due to B1 , B2 < ∞, we have also
shown in Lemma 2.2 that Gn f, Hn f ∈ L2 [0, b1n ].
Therefore, by Lemma 2.1, for every f1 , f2 ∈ L2 (R) and m, n ∈ Z,
f1 , Embn gn
L2 (R)
1
bn
=
Gn f1 (t)e−2πimbn t dt = Gn f1 , Embn
L2 [0, b1 ] ,
Hn f2 (t)e−2πimbn t dt = Hn f2 , Embn
L2 [0, b1 ] .
0
f2 , Embn hn
L2 (R)
1
bn
=
0
n
n
CHAPTER 2. NONSTATIONARY GABOR FRAMES IN L2 (R)
27
√
As { bn Embn }m∈Z is an orthonormal basis for L2 [0, b1n ], we obtain
f1 , Embn gn
L2 (R)
f2 , Embn hn
1
Gn f1 ,
bn
bn Embn
L2 (R)
n∈Z m∈Z
=
n∈Z m∈Z
=
n∈Z
1
Gn f1 , Hn f2
bn
L2 [0, b1 ]
Hn f2 ,
n
bn Embn
L2 [0, b1 ]
n
L2 [0, b1 ] ,
n
where in the last equality, we are using the well-known result that (see for instance
[5]) that if {ek }k∈I is an orthonormal basis for a Hilbert space H, then for every
f, g ∈ H,
k∈I
f, ek g, ek = f, g . Therefore,
f1 , Embn gn f2 , Embn hn
n∈Z m∈Z
=
n∈Z
=
n∈Z
1
bn
1
bn
1
bn
0
f1 (t −
k
k
)gn (t − )Hn f2 (t) dt
bn
bn
f1 (t −
k
k
)gn (t − )Hn f2 (t) dt.
bn
bn
k∈Z
1
bn
k∈Z 0
We then make a change of variables and obtain
n∈Z
=
n∈Z
=
n∈Z
=
n∈Z
1
bn
1
bn
1
bn
1
bn
1
bn
f1 (t −
k∈Z 0
1
− bk
bn
n
k∈Z
− bk
n
1
− bk
bn
n
k
k∈Z − bn
∞
k
k
)gn (t − )Hn f2 (t) dt
bn
bn
f1 (y)gn (y)Hn f2 (y +
f1 (y)gn (y)Hn f2 (y) dy
f1 (y)gn (y)Hn f2 (y) dy.
−∞
k
) dy
bn
(2.4)
CHAPTER 2. NONSTATIONARY GABOR FRAMES IN L2 (R)
28
Writing out the expression of Hn f2 ,
n∈Z
=
n∈Z
∞
1
bn
f1 (y)gn (y)Hn f2 (y) dy
−∞
∞
1
bn
−∞
k∈Z
∞
=
k∈Z −∞ n∈Z
∞
=
−∞
n∈Z
f1 (y)
−∞
(2.5)
1
gn (y)hn (y) dy
bn
∞
+
k
k
)hn (y − ) dy
bn
bn
k
1
k
f1 (y)gn (y)f2 (y − )hn (y − ) dy
bn
bn
bn
f1 (y)f2 (y)
k=0
f2 (y −
f1 (y)gn (y)
n∈Z
k
k
1
f2 (y − )gn (y)hn (y − ) dy.
bn
bn
bn
This gives the result once we justify the interchanges of summation and integral for
(2.4) and (2.5). In this connection, following the same argument as above, we have
1
bn
n∈Z k∈Z 0
∞
=
n∈Z −∞
1
k
k
f1 (t − )gn (t − )Hn f2 (t) dt
bn
bn
bn
1
|f1 (y)gn (y)Hn f2 (y)| dy.
bn
Further,
∞
n∈Z −∞
∞
=
n∈Z −∞
∞
≤
n∈Z −∞
1
|f1 (y)gn (y)Hn f2 (y)| dy
bn
1
f1 (y)gn (y)
bn
1
bn
f2 (y −
k∈Z
f1 (y)gn (y)f2 (y −
k∈Z
k
k
)hn (y − ) dy
bn
bn
k
k
)hn (y − ) dy := D.
bn
bn
CHAPTER 2. NONSTATIONARY GABOR FRAMES IN L2 (R)
29
Applying the Cauchy-Schwarz inequality yields
D ≤
n∈Z
n∈Z
=
n∈Z
n∈Z
≤
1
bn
1
bn
1
bn
1
bn
1
2
∞
k∈Z
k
|f1 (y)|2 gn (y)hn (y − ) dy
bn
−∞
∞
k∈Z −∞
k
f2 (y − )
bn
2
·
k
gn (y)hn (y − ) dy
bn
1
2
∞
k∈Z
k
|f1 (y)|2 gn (y)hn (y − ) dy
bn
−∞
∞
k∈Z
C1 f1 ·
k
|f2 (t)|2 gn (t + )hn (t) dt
b
n
−∞
1
2
·
1
2
C2 f2 < ∞.
This completes the proof.
For the next result, the proof is similar to those of results in [6] and [7]. The
paper [6] aims to find an explicit expression of dual frames for certain stationary
Gabor frames in L2 (R) while [7] aims to find that for certain nonstationary frames
in L2 [0, 2π]. Our following theorem extends and adapts these proofs for the nonstationary setting in L2 (R). Our aim is also to formulate conditions that {gn }n∈Z
satisfies so that we can find an appropriate alternative dual instead of restricting
ourselves to only the canonical dual. To check whether two collections {fk }k∈Z and
{gk }k∈Z are dual frames for L2 (R) (see for instance [5]), it is sufficient for them to
be Bessel sequences and for every f, f ∈ L2 (R),
f, fk f , gk = f, f .
k∈Z
With Lemma 2.13 on hand, we may derive the following theorem in which for
the type of frame defined, we have an explicit expression for an alternative dual. In
this result, we are concerned with real-valued frame functions only.
Theorem 2.14. Suppose that gn ∈ L2 (R) are real-valued and bn > 0 for all n ∈ Z
CHAPTER 2. NONSTATIONARY GABOR FRAMES IN L2 (R)
such that B := supt∈R
1 2
n∈Z bn gn (t)
30
< ∞. In addition, assume that
(i) There exists P ∈ N such that for every n ∈ Z, supp gn ∩ supp gn+ν = ∅ for
ν ≥ P;
(ii)
n∈Z
√1 gn (x)
bn
= 1 a.e;
(iii) For every n ∈ Z, ρn := sup{|t1 − t2 | : t1 , t2 ∈
P −1
ν=0 supp gn+ν }
<
1
bn .
For each n ∈ Z, define
P −1
hn (t) := gn (t) +
ν=1
√
2 bn
gn+ν (t),
bn+ν
t ∈ R.
Then {Embn gn }m,n∈Z and {Embn hn }m,n∈Z form a pair of dual frames for L2 (R).
Proof. We note that (iii) implies that for each n ∈ Z, supp gn is contained in an interval of length not more than
k
bn )|
1
bn .
Hence, the expression supt∈R
k∈Z
1
n∈Z bn |gn (t)gn (t−
is actually equal to B. We have shown in Lemma 2.2 that for every f ∈ L2 (R),
∞
| f, Embn gn |2 =
m∈Z n∈Z
|f (t)|2 (
−∞
≤ B f
n∈Z
2
1
|gn (t)|2 ) dt
bn
,
which means that {Embn gn }m,n∈Z forms a Bessel sequence for L2 (R).
Next, in order to show that {Embn hn }m,n∈Z also forms a Bessel sequence for
L2 (R), we observe that supp hn =
P −1
ν=0 supp gn+ν .
is contained in an interval of length not more than
D := sup
t∈R k∈Z n∈Z
Hence, (iii) implies that supp hn
1
bn .
This gives us
1
k
1
|hn (t)hn (t − )| = sup h2n (t).
bn
bn
t∈R bn
CHAPTER 2. NONSTATIONARY GABOR FRAMES IN L2 (R)
31
Fixing n ∈ Z,
√
2 bn
gn+ν (t)
bn+ν
P −1
h2n (t)
=
gn (t) +
ν=1
1
√ gn (t) +
bn
= bn
1 2
g (t) +
bn n
= bn
2
P −1
ν=1
P −1
ν=1
2
2
gn+ν (t)
bn+ν
P −1
4
bn+ν
2
gn+ν
(t)
ν=1
8
+
bn+ν1
1≤ν1 0 for all m ∈ Z such that
B := sup
γ∈R k∈Z m∈Z
k
1
hm (γ)hm (γ −
) < ∞.
am
am
35
CHAPTER 2. NONSTATIONARY GABOR FRAMES IN L2 (R)
36
Then {Tnam hm }m,n∈Z is a Bessel sequence with Bessel bound B. If also
A := inf
γ∈R
m∈Z
1
|hm (γ)|2 −
am
k=0 m∈Z
1
k
hm (γ)hm (γ −
) > 0,
am
am
then {Tnam hm }m,n∈Z is a frame for L2 (R) with bounds A, B.
Corollary 2.17. Let am > 0, hm ∈ L2 (R) with support of hm in an interval of
length
A≤
1
am
for every m ∈ Z, such that there exists some constants A, B > 0 for which
1
2
m∈Z am |hm (γ)|
≤ B a.e. Then {Tnam hm }m,n∈Z is a frame for L2 (R) with
bounds A, B.
Unlike Corollary 2.4, the frame operator is not as simple. However, we can still
work out the frame operator in this case. Firstly, we know that when {fk }k∈Z is a
frame for a Hilbert space H and U is a unitary operator on H, then {U fk }k∈Z is
also a frame for H (see for instance [5]). We want to know how the frame operator
T for {U fk }k∈Z is related to that of {fk }k∈Z , S. The frame operators S and T are
defined as follows: for every f ∈ H,
Sf :=
f, fk fk ,
k∈Z
and
T f :=
f, U fk U fk .
k∈Z
Applying the adjoint U ∗ of U , on (2.7), we obtain
U ∗T f =
f, U fk U ∗ U fk =
k∈Z
U ∗ f, fk fk .
k∈Z
Then substituting f for U f gives
U ∗T U f =
U ∗ U f, fk fk = Sf,
k∈Z
(2.7)
CHAPTER 2. NONSTATIONARY GABOR FRAMES IN L2 (R)
37
and therefore U ∗ T U = S, implying T = U SU ∗ . By Corollary 2.4, for hm satisfying
the conditions as stated in Corollary 2.17, the frame {Enam hm }m,n∈Z has frame
operator
Sf =
m∈Z
1
|hm (·)|2
am
f.
Since {Tnam hm }m,n∈Z is the same as the collection {F −1 (Enam hm )}m,n∈Z , where F
is the Fourier transform, letting U = F −1 , the frame operator of {Tnam hm }m,n∈Z
and its inverse are
T f := F −1 SFf,
T −1 f := F −1 S −1 Ff.
Next, we look at the analogous result of Corollary 2.5.
Corollary 2.18. Consider {hm } in L2 (R) and am > 0, m ∈ Z, that satisfy the
following:
(i) There exist C, D > 0 such that for every m ∈ Z, C ≤
|hm (γ)|
√
am
≤ D a.e in the
support of hm ;
(ii) For each m ∈ Z, the support of hm lies in an interval of length less than or
equal to
1
am ;
(iii) There exists a positive integer K such that for a.e γ ∈ R, {m ∈ Z : hm (γ) = 0}
has at most K elements and at least one element.
Then {Tnam hm }m,n∈Z forms a frame for L2 (R).
So far, we have seen the conditions that make {Tnam hm }m,n∈Z a frame for L2 (R).
Using the same technique, we can also derive an analogous result of Theorem 2.14.
Firstly, we know that {Tnam gm }m,n∈Z and {Tnam hm }m,n∈Z are dual frames for L2 (R)
if and only if for every f ∈ L2 (R),
f=
f, Tnam gm Tnam hm =
n∈Z m∈Z
f, Tnam hm Tnam gm .
n∈Z m∈Z
(2.8)
CHAPTER 2. NONSTATIONARY GABOR FRAMES IN L2 (R)
38
Taking the Fourier transform on (2.8), we have
fˆ =
f, Tnam gm Tnam hm =
n∈Z m∈Z
f, Tnam hm Tnam gm .
n∈Z m∈Z
By the Parseval identity and (2.6), this is equivalent to
fˆ =
fˆ, Tnam hm Tnam gm ,
fˆ, Tnam gm Tnam hm =
n∈Z m∈Z
n∈Z m∈Z
fˆ =
fˆ, E−nam hm E−nam gm .
fˆ, E−nam gm E−nam hm =
n∈Z m∈Z
n∈Z m∈Z
Since we are summing over all integers n, we may change −n to n and also fˆ to f ,
yielding
f=
f, Enam gm Enam hm =
n∈Z m∈Z
f, Enam hm Enam gm .
n∈Z m∈Z
This implies that {Tnam gm }m,n∈Z and {Tnam hm }m,n∈Z form dual frames for L2 (R) if
and only if {Enam gm }m,n∈Z and {Enam hm }m,n∈Z do so. This gives us the following.
Corollary 2.19. Suppose that for all m ∈ Z, gm ∈ L2 (R) and am > 0 such that gm
are real-valued and B := supt∈R
2
1
m∈Z am gm (t)
< ∞. In addition, assume that
(i) There exists P ∈ N such that for every m ∈ Z, supp gm ∩ supp gm+ν = ∅ for
ν ≥ P;
(ii)
m∈Z
√1 gn (x)
am
= 1 a.e;
(iii) For every m ∈ Z, ρm := sup{|t1 − t2 | : t1 , t2 ∈
P −1
ν=0 supp gm+ν }
<
1
am .
For each m ∈ Z, define
P −1
hm (t) := gm (t) +
ν=1
√
2 am
gm+ν (t),
√
am+ν
t ∈ R.
Then {Tnam gm }m,n∈Z and {Tnam hm }m,n∈Z form a pair of dual frames for L2 (R).
CHAPTER 2. NONSTATIONARY GABOR FRAMES IN L2 (R)
We have seen in this chapter how we may form some nonstationary Gabor
frames using a sequence of functions {gn }n∈Z or {hm }m∈Z and the modulation and
translation operators and also how in certain circumstances, we can have an alternative dual for the frames. The functions {gn }n∈Z or {hm }m∈Z may be chosen to be
compactly supported in the time or frequency domains but with different support
lengths to cater to different window sizes for signal analysis. Nevertheless, during
our applications in the last chapter, we will focus only on the collection with modulation operator. To be more precise, we will be extending the results to the sequence
space in the next chapter before we use the theory for applications. However, we
may take note that the other collection may also be of use in applications.
39
Chapter 3
Nonstationary Gabor Frames in
S(2K )
In Chapter 2, we have developed sufficient conditions for a collection of functions
in L2 (R), of the form {Embn gn }m,n∈Z or {Tnam hm }m,n∈Z , to form a frame for the
space. We notice that there are several conditions imposed on the collections. These
conditions mainly show their influence in the proofs when we are justifying the
interchanges of integrals or infinite summations. In this chapter, however, we are
able to make simpler the conditions as we deal with a finite-dimensional space, the
space of N -periodic sequences, S(N ), where N is a positive integer. In particular,
we look at 2K -periodic sequences in the space S(2K ), where K is a nonnegative
integer.
The main reason for developing the theory again for this space is for our application to signals in the final chapter. During our applications to real signals, we
sample the signals at discrete points a finite number of times. Hence, we need not
restrict ourselves to the conditions for continuous functions. We may concentrate on
finite summation in this chapter and make use of the advantage of weaker conditions,
which may provide us a larger collection of frames.
For the sequence space S(N ), which one may consider to be the same as CN ,
40
CHAPTER 3. NONSTATIONARY GABOR FRAMES IN S(2K )
41
−1
N −1
the inner product and norm are defined to be, for c = {c(r)}N
r=0 , d = {d(r)}r=0 in
S(N ),
N −1
c, d
S(N )
:=
c(r)d(r),
r=0
1
2
N −1
c
S(N )
2
|c(r)|
:=
.
r=0
If there is no confusion with other sequences spaces, the subscripts will be dropped.
We will encounter two classes of operators on S(N ), namely,
Modulation by m ∈ Z :
Circular shift by n ∈ Z :
(N )
(N )
Em
: S(N ) → S(N ), Em
c(k) := e
2πi
mk
N
c(k), k ∈ Z,
Tn : S(N ) → S(N ), Tn c(k) := c(k − n), k ∈ Z.
(3.1)
Due to the periodicity of the sequences, we see that Tn+kN = Tn for every n, k ∈
Z, hence the name circular shift. We note here that unlike the modulation operator
for L2 (R), the modulation operator for sequence spaces requires a superscript to
indicate which sequence space it is acting on. However, when there is no confusion
in the spaces, the superscript will be dropped. In addition, since applying the
(N )
operator Em
(N )
use Em
on a sequence of ones gives the sequence {e
to denote the sequence {e
2πi
mk
N
2πi
mk
N
−1
}N
k=0 , we will also
−1
}N
k=0 . We also keep in mind that for every
c ∈ S(N ), r ∈ Z with r ∈
/ {0, · · · , N − 1}, c(r) = c(k) where r mod N = k.
3.1
Discrete Frames with Varying Modulations
Finite frames in S(N ) are well studied (see for instance [2], [3], [4]). However, our focus in this thesis would be on nonstationary frames. In this first section, same as the
previous chapter, we would like to propose conditions for {Em2K−pn dn }m=0,...,2pn −1
n=0,...,Q−1
to form a frame for S(2K ), where pn , Q, K are nonnegative integers, pn ≤ K. Before
that, we need to take note of a few observations.
Firstly, by the division algorithm, if we have N ∈ N and a ∈ N is its factor,
CHAPTER 3. NONSTATIONARY GABOR FRAMES IN S(2K )
42
for every n ∈ {0, 1, · · · , N − 1}, n = ja + k where k ∈ {0, 1, · · · , a − 1}, j ∈
{0, 1, · · · , Na − 1}. Hence, we have
N
a
N
a
−1
{0, 1, · · · , N − 1} =
{ja, ja + 1, · · · , ja + a − 1} =
j=0
Therefore, the sum
−1 ja+a−1
N −1
l=0
{l}.
j=0
and the double sum
N
−1
a
j=0
ja+a−1
l=ja
(3.2)
l=ja
are the same and
may be used interchangeably.
−1
We also observe that if A = {An }N
n=0 is a N -periodic sequence, then for any
consecutive N integers, say {k, k + 1, · · · , k + N − 1}, the sums
N −1
r=0 An
k+n−1
An
r=k
and
are the same. This explains why we can rewrite the summation in many
of the computations later. Moreover, in any finite sum, we can always reverse the
order of summation, i.e.
K
K
Hn =
n=1
HK−n+1 .
(3.3)
n=1
In the results that follow, we always fix pn , Q, K ∈ N, and assume pn ≤ K.
Lemma 3.1. For a fixed n = 0, · · · , Q − 1, let c, dn ∈ S(2K ) and define Cn =
pn
{Cn (r)}2r=0−1 by
2K−pn −1
c(r + j2pn )dn (r + j2pn ),
Cn (r) :=
r ∈ Z.
(3.4)
j=0
Then Cn is in S(2pn ).
Proof. Fixing any r ∈ Z, it suffices to show that Cn (r + 2pn ) = Cn (r). We have, by
substituting j = m + 1,
2K−pn −1
pn
c(r + (m + 1)2pn )dn (r + (m + 1)2pn )
Cn (r + 2 ) =
m=0
2K−pn
c(r + j2pn )dn (r + j2pn ).
=
j=1
CHAPTER 3. NONSTATIONARY GABOR FRAMES IN S(2K )
43
Separating the last term in the sum, we obtain
2K−pn −1
c(r + j2pn )dn (r + j2pn ) + c(r + 2K−pn · 2pn )dn (r + 2K−pn · 2pn )
j=1
2K−pn −1
c(r + j2pn )dn (r + j2pn ) + c(r)dn (r)
=
j=1
= Cn (r).
Lemma 3.2. For a fixed n = 0, 1, · · · , Q − 1, let c, dn ∈ S(2K ) and define Cn as
in (3.4). Then for any m = 0, 1, · · · , 2pn − 1,
(2K )
c, Em2K−pn dn
S(2K )
2pn −1
Cn (r)e−
=
2πimr
2p n
(2
= Cn , Em
pn )
r=0
S(2pn )
.
Proof. By Lemma 3.1, we see that Cn ∈ S(2pn ). Writing out the expression, we
have
(2K )
c, Em2K−pn dn
S(2K )
2K −1
c(l)e
=
2πim2K−pn
2K
l
dn (l)
l=0
2K −1
c(l)dn (l)e−
=
2πim
l
2pn
.
l=0
With (3.2), we can rewrite the summation as a double summation,
2K−pn −1 j2pn +2pn −1
2K −1
− 2πim
l
2p n
c(l)dn (l)e
c(l)dn (l)e−
=
j=0
l=0
2πim
l
2pn
l=j2pn
2K−pn −1 2pn −1
=
j=0
2pn −1
2πim
(r+j2pn )
2pn
c(r + j2pn )dn (r + j2pn )e−
2πim
r
2pn
r=0
2K−pn −1
=
r=0
c(r + j2pn )dn (r + j2pn )e−
j=0
2pn −1
Cn (r)e−
=
r=0
2πimr
2p n
.
CHAPTER 3. NONSTATIONARY GABOR FRAMES IN S(2K )
44
This completes the proof.
Remark 3.3. Let us compare the proofs of Lemma 3.2 and Lemma 2.1. Notice
that there is no need for us to worry about whether Cn is well defined because the
definition of Cn only involves finite sums. On the other hand, we have to make sure
that Fn , as defined in (2.2), converges before we can proceed with the proof. This
is an example of how our proofs become simpler when we develop our theory on a
finite-dimensional space.
The following result is analogous to Lemma 2.2. We may observe that the proof
is relatively much shorter. This is partially because we have already proved that Cn
is in the space S(2pn ) in Lemma 3.1. Even so, we can see that the proof of this fact
is much shorter than that of Fn being in L2 [0, b1n ].
Lemma 3.4. Let c, dn ∈ S(2K ) for n = 0, 1, · · · , Q − 1. Then
Q−1 2pn −1
n=0
2K −1
2
(2K )
c, Em2K−pn dn
n=0
l=0
2K −1
2pn |dn (l)|2
|c(l)|
=
m=0
Q−1
2
Q−1 2K−pn −1
2pn c(l)c(l + j2pn )dn (l)dn (l + j2pn ).
+
l=0 n=0
j=1
Proof. From the previous lemma, we know that for each m = 0, 1, · · · , 2pn − 1,
n = 0, 1, · · · , Q − 1,
(2K )
c, Em2K−pn dn
pn )
S(2K )
(2
= Cn , Em
S(2pn )
(2pn ) 2pn −1
}m=0
where Cn is as defined in (3.4). Since { √21pn Em
,
is an orthonormal basis for
S(2pn ) and Cn ∈ S(2pn ),
2pn −1
m=0
(2K )
c, Em2K−pn dn
S(2K )
2pn −1
2
(2pn )
Cn , Em
=
m=0
2
S(2pn )
= 2pn Cn
2
S(2pn ) .
CHAPTER 3. NONSTATIONARY GABOR FRAMES IN S(2K )
45
Therefore, we obtain
Q−1 2pn −1
n=0
2pn −1
Q−1
2
(2K )
c, Em2K−pn dn
2
=
m=0
|Cn (r)|2 .
pn
r=0
n=0
Writing |Cn (r)|2 = Cn (r)Cn (r) and substituting the expression for Cn , we have
2pn −1 2K−pn −1
Q−1
2
pn
c(r + j2pn )dn (r + j2pn )Cn (r)
n=0
r=0
Q−1
2
=
j=0
2K−pn −1
2pn −1
c(r + j2pn )dn (r + j2pn )Cn (r)
pn
n=0
r=0
j=0
2K−pn −1 j2pn +2pn −1
Q−1
2p n
=
c(l)dn (l)Cn (l),
n=0
l=j2pn
j=0
where the last line is due to substituting l = r + j2pn and observing that Cn (l −
j2pn ) = Cn (l). Next, noting that we can merge
2K−pn −1
j=0
j2pn +2pn −1
l=j2pn
into
2K −1
l=0 ,
we obtain
Q−1 2pn −1
(2K )
c, Em2K−pn dn
n=0
c(l)dn (l)Cn (l)
l=0
2K−pn −1
c(l + j2pn )dn (l + j2pn )
c(l)dn (l)
j=0
l=0
2K −1
Q−1
2pn |dn (l)|2
|c(l)|2
=
2
n=0
2K −1
2pn
=
pn
=
n=0 m=0
Q−1
2K −1
Q−1
2
n=0
l=0
2K −1 Q−1
2K−pn −1
pn
+
c(l)c(l + j2pn )dn (l)dn (l + j2pn ).
2
l=0 n=0
j=1
Lastly, we want to use the previous lemma to prove the following theorem.
Observe that the constant B defined in this result is less restrictive than the one
defined in Theorem 2.3. In Theorem 2.3, we need to make sure that the constant is
finite. We do not have the same concern in this case. In fact, the following theorem
CHAPTER 3. NONSTATIONARY GABOR FRAMES IN S(2K )
46
is telling us that any collection {Em2K−pn dn }m=0,...,2pn −1 is a Bessel sequence for
n=0,...,Q−1
S(2K ). This is in line with the well-known fact that a finite collection in a Hilbert
space always forms a Bessel sequence, but the theorem provides additional information of B being a Bessel bound of the collection. Whether the collection forms a
frame depends on whether the constant A defined in this result is positive.
Theorem 3.5. Let dn ∈ S(2K ) for n = 0, 1, · · · , Q − 1. Define
Q−1 2K−pn −1
B :=
2pn |dn (l)dn (l + j2pn )| ,
max
l=0,··· ,2K −1
n=0
j=0
and
A :=
min
2pn |dn (l)dn (l + j2pn )| .
2pn |dn (l)|2 −
l=0,··· ,2K −1
Q−1 2K−pn −1
Q−1
n=0
n=0
j=1
Then {Em2K−pn dn }m=0,...,2pn −1 forms a Bessel sequence for S(2K ) with Bessel bound
n=0,...,Q−1
B. In addition, if A > 0, then {Em2K−pn dn }m=0,...,2pn −1 forms a frame for S(2K )
n=0,...,Q−1
with frame bounds A, B.
Proof. From Lemma 3.4, we have, for every c ∈ S(2K ),
Q−1 2pn −1
n=0
(2K )
c, Em2K−pn dn
2K −1
2
m=0
Q−1
2pn |dn (l)|2
|c(l)|2
=
n=0
l=0
2K −1
Q−1
2K−pn −1
2pn c(l)c(l + j2pn )dn (l)dn (l + j2pn ).
+
l=0 n=0
j=1
Denote the second term above by X. Then by the Cauchy-Schwarz inequality,
1
Q−1 2K−pn −1 2K −1
2
pn
|X| ≤
2
pn
2 |c(l)| |dn (l)dn (l + j2 )| ·
n=0
j=1
l=0
1
Q−1 2K−pn −1 2K −1
2
pn
pn
2
pn
2 |c(l + j2 )| |dn (l)dn (l + j2 )| .
n=0
j=1
l=0
CHAPTER 3. NONSTATIONARY GABOR FRAMES IN S(2K )
Now
Q−1 2K−pn −1 2K −1
2pn |c(l + j2pn )|2 |dn (l)dn (l + j2pn )|
n=0
j=1
Q−1
2K−pn −1
l=0
j2pn +2K −1
2pn |c(r)|2 |dn (r − j2pn )dn (r)|
=
n=0
r=j2pn
j=1
Q−1 2K−pn −1 2·2K −j2pn −1
2pn |c(r)|2 |dn (r − 2K + j2pn )dn (r)|.
=
n=0
j=1
r=2K −j2pn
The last term is obtained by reversing the order of the sum over index j. To be
j2pn +2K −1 pn
2 |c(r)|2 |dn (r
r=j2pn
more precise, by letting Hj =
that
2K−pn −1
2K−pn −1
Hj =
j=1
− j2pn )dn (r)|, (3.3) says
2K−pn −1
H2K−pn −1−j+1 =
j=1
H2K−pn −j .
j=1
Thus, we replace j with 2K−pn − j to give the last result. Next, observe that the
summation over index r is actually summing a 2K -periodic term over 2K consecutive
integers. This is the same as summing over r = 0, 1, · · · , 2K − 1. Hence, the last
expression obtained is the same as
Q−1 2K−pn −1 2K −1
2pn |c(r)|2 |dn (r − 2K + j2pn )dn (r)|.
n=0
j=1
r=0
This yields
Q−1 2K−pn −1 2K −1
2pn |c(l)|2 |dn (l)dn (l + j2pn )|.
|X| ≤
n=0
j=1
l=0
Therefore, we get the Bessel inequality,
Q−1 2pn −1
n=0
m=0
(2K )
c, Em2K−pn dn
2
Q−1 2K−pn −1 2K −1
2pn |c(l)|2 |dn (l)dn (l + j2pn )|
≤
n=0
j=0
2
≤ B c .
l=0
47
CHAPTER 3. NONSTATIONARY GABOR FRAMES IN S(2K )
Similarly, if A > 0, we have the lower bound in the frame inequality,
Q−1 2pn −1
n=0
m=0
(2K )
c, Em2K−pn dn
2
2K −1
Q−1
2pn |dn (l)|2
2
≥
|c(l)|
−
n=0
l=0
Q−1 2K−pn −1 2K −1
2pn |c(l)|2 |dn (l)dn (l + j2pn )|
n=0
2
j=1
l=0
≥ A c .
Here, we have extended a few results in Chapter 2. We will look at more
of them later, in addition to some new results that we have not discussed before.
From the above, it is clear that compared to L2 (R), it is easier to obtain sufficient
conditions for the collection {Em2K−pn dn }m=0,...,2pn −1 to form a frame for S(2K )
n=0,...,Q−1
than for the analogous collection in L2 (R). This is of course more advantageous for
our subsequent application to time-frequency analysis. In the final chapter, we will
see how the theory on the sequence space can be applied in applications.
3.2
Necessary Condition
Up till now, in the previous chapter and this chapter, we have always been looking
for sufficient conditions for a collection of functions to be a frame for either L2 (R)
or S(2K ). In this section, we fix the modulation parameter pn = p for every n =
0, ..., Q − 1 and find necessary conditions for {Em2K−p dn }m=0,...,2p −1 to be a frame
n=0,...,Q−1
for S(2K ).
The results that we are seeing are nonstationary analogues of some well-known
results on stationary Gabor frames (see for instance [5]). The original results are
necessary conditions for standard Gabor frames in L2 (R) based on a single window
function, for which the proofs may be quite long and complicated. Some of our
proofs for S(2K ), on the other hand, may be shorter and simpler, notwithstanding
the nonstationary setup. In the results that follow, we always fix p, Q, K ∈ N,
48
CHAPTER 3. NONSTATIONARY GABOR FRAMES IN S(2K )
49
and assume p ≤ K. For the first result we have below, we want to show that the
modulation operator and the frame operator commute.
Lemma 3.6. Consider dn ∈ S(2K ), n = 0, 1, · · · , Q − 1, such that the collection
{Em2K−p dn }m=0,...,2p −1 forms a frame for S(2K ) with frame operator S. Then for
n=0,...,Q−1
every m = 0, ..., 2p − 1, SEm2K−p = Em2K−p S.
Proof. Consider c ∈ S(2K ) and fix m ∈ {0, ..., 2p − 1}. Then
2p −1 Q−1
SEm2K−p c =
Em2K−p c, Ej2K−p dn Ej2K−p dn
j=0 n=0
2p −1 Q−1
=
c, E(j−m)2K−p dn Ej2K−p dn .
j=0 n=0
Substituting m = j − m, we obtain
2p −1 Q−1
2p −1−m Q−1
c, E(j−m)2K−p dn Ej2K−p dn =
j=0 n=0
c, Em 2K−p dn E(m +m)2K−p dn
m =−m n=0
2p −1−m Q−1
= Em2K−p
c, Em 2K−p dn Em 2K−p dn
m =−m n=0
2p −1 Q−1
= Em2K−p
c, Em 2K−p dn Em 2K−p dn
m =0 n=0
= Em2K−p Sc,
where we rewrite the summation
2p −1−m
m =−m
as
2p −1
m =0
defined by
Q−1
Hm =
p
because the sequence {Hm }2m −1
=0
c, Em 2K−p dn Em 2K−p dn ,
n=0
is 2p -periodic. Hence, we have SEm2K−p c = Em2K−p Sc.
Remark 3.7. Observe that if S commutes with Em2K−p , then so does S −1 as
S −1 Em2K−p = S −1 Em2K−p SS −1 = S −1 SEm2K−p S −1 = Em2K−p S −1 .
CHAPTER 3. NONSTATIONARY GABOR FRAMES IN S(2K )
50
Hence, by a standard functional analysis result on positive operators in a Hilbert
1
space (see for instance [19]), the square root of S −1 , S − 2 also commutes with the
operator Em2K−p . The canonical tight frame can also be expressed as
{S −1/2 Em2K−p dn }m=0,...,2p −1 = {Em2K−p S −1/2 dn }m=0,...,2p −1 .
n=0,...,Q−1
n=0,...,Q−1
This gives us the following.
Theorem 3.8. Consider dn ∈ S(2K ), n = 0, 1, · · · , Q − 1, such that the collection
{Em2K−p dn }m=0,...,2p −1 is a frame. Then the canonical dual is also a discrete nonn=0,...,Q−1
stationary Gabor frame and is given by {Em2K−p S −1 dn }m=0,...,2p −1 . The canonical
n=0,...,Q−1
tight frame associated with {Em2K−p dn }m=0,...,2p −1 is {Em2K−p S −1/2 dn }m=0,...,2p −1 .
n=0,...,Q−1
n=0,...,Q−1
Next, we need another proposition before we have our main result for this
section. This result is also extended from a result for standard Gabor frames in
L2 (R) in [5]. It shows that for a frame {Em2K−p dn }m=0,...,2p −1 in S(2K ), the sum
n=0,...,Q−1
Q−1 p
2
n=0 2 |dn (l)|
is bounded above and below for all l = 0, 1, · · · , 2K − 1. The
corresponding expression of this sum for the nonstationary Gabor frames in L2 (R)
discussed in Chapter 2 is the function G defined in (2.1) when we fix bn = b for
every n ∈ Z. Hence, it is not surprising that we actually also have an analogous
result for this proposition for the collection {Emb gn }m,n∈Z , though it was not stated
in the previous chapter.
Proposition 3.9. Let dn ∈ S(2K ), n = 0, 1, · · · , Q − 1, such that the collection
{Em2K−p dn }m=0,...,2p −1 forms a frame for S(2K ) with bounds A, B. Then for l =
n=0,...,Q−1
0, · · · , 2K − 1,
Q−1
2p |dn (l)|2 ≤ B.
A≤
n=0
Proof. Assume on the contrary that there exists l ∈ {0, · · · , 2K − 1} such that
Q−1
2p |dn (l)|2 > B.
n=0
(3.5)
CHAPTER 3. NONSTATIONARY GABOR FRAMES IN S(2K )
51
Define c ∈ S(2K ) by, for r = 0, · · · , 2K − 1,
c(r) :=
1, if r = l;
0, otherwise.
Then
Q−1 2p −1 2K −1
Q−1 2p −1
2
| c, Em2K−p dn |
n=0 m=0
n=0 m=0
K−p
− 2πim2K
c(r)e
=
2
dn (r)
r=0
Q−1 2p −1
Q−1
|c(l)dn (l)|2 =
=
r
n=0 m=0
2p |dn (l)|2
n=0
2
> B=B c ,
which is a contradiction. The argument works the same way to show the left inequality of (3.5).
Now, with Theorem 3.8 and Proposition 3.9, we have our main result. The
original result for standard Gabor frames, as stated in Theorem 1.2, is that if
{Emb Tna g}m,n∈Z forms a frame for L2 (R), where Emb and Tna are the modulation and translation operators as defined for L2 (R) in (1.2), then ab ≤ 1. This
implies that we do not have absolute freedom in choosing the parameters a and b.
It shows that we cannot choose both modulation and translation parameters to be
arbitrarily large. The following result also poses some contraints on the parameters
in our nonstationary case for S(2K ).
Theorem 3.10. Let dn ∈ S(2K ), n = 0, 1, · · · , Q−1 such that {Em2K−p dn }m=0,...,2p −1
n=0,...,Q−1
forms a frame for S(2K ). Then
2K−p
Q
≤ 1.
Proof. Let {Em2K−p dn }m=0,...,2p −1 be a frame for S(2K ). Then the canonical tight
n=0,...,Q−1
frame associated to it is {Em2K−p S −1/2 dn }m=0,...,2p −1 . Since the canonical tight
n=0,...,Q−1
frame has frame bound 1, i.e A = B = 1, by Proposition 3.9, substituting dn for
CHAPTER 3. NONSTATIONARY GABOR FRAMES IN S(2K )
52
S −1/2 dn , for every l = 0, 1, · · · , 2K − 1,
Q−1
|S −1/2 dn (l)|2 =
n=0
1
.
2p
Hence,
Q−1 2K −1
Q−1
S
−1/2
2
dn
n=0
|S −1/2 dn (l)|2 =
=
n=0 l=0
2K
.
2p
(3.6)
On the other hand, for every c ∈ S(2K ),
2p −1 Q−1
| c, Em2K−p S −1/2 dn |2 = c 2 .
m=0 n=0
Now, given any fixed k ∈ {0, 1, · · · , 2p − 1}, j ∈ {0, · · · , Q − 1},
| Ek2K−p S −1/2 dj , Ek2K−p S −1/2 dj |2
2p −1 Q−1
| Ek2K−p S −1/2 dj , Em2K−p S −1/2 dn |2
≤
m=0 n=0
=
Ek2K−p S −1/2 dj
2
,
which yields
Ek2K−p S −1/2 dj
2
≤ 1.
However,
2K −1
Ek2K−p S
−1/2
dj
2
=
e
2πik2K−p
r
2K
2
S −1/2 dj (r)
r=0
2K −1
|S −1/2 dj (r)|2 = S −1/2 dj
=
r=0
Thus, we have, for every n = 0, 1, · · · , Q − 1,
S −1/2 dn
2
≤ 1.
2
.
CHAPTER 3. NONSTATIONARY GABOR FRAMES IN S(2K )
Combining with (3.6) then gives
Q−1
K−p
2
S −1/2 dn
=
2
≤ Q.
n=0
Remark 3.11. The inequality gives a lower bound to the number Q of sequences dn
that are used to form the frame {Em2K−p dn }m=0,...,2p −1 . We see that when we choose
n=0,...,Q−1
p to be small, then Q will have to be larger; and it may be chosen to be smaller if
p is large. In addition, if we let dn = Tn2q d, q ≤ K being a nonnegative integer, i.e.
circular shifts of a sequence d ∈ S(2K ), and Q = 2K−q , the inequality then becomes
q
p
≤ 1. This stationary case of discrete frames in S(2K ) resembles the standard
Gabor frames in L2 (R). While the translation and modulation parameters in L2 (R)
are restricted by the inequality ab ≤ 1, the parameters involved in modulation and
circular shifts are also constrained.
3.3
Varying Circular Shifts
Analogous to Section 2.3, we would like to use properties of the discrete Fourier
tranform, mainly the Parseval identity (see for instance [13]), to generate results on
frames with varying circular shifts in S(2K ). For c ∈ S(N ), where N is a positive
integer, the discrete Fourier transform of c, cˆ ∈ S(N ) is defined by
N −1
c(k)e−
cˆ(j) :=
2πi
kj
N
,
j = 0, 1, · · · , N − 1.
k=0
Observe that the definition of the discrete Fourier transform is dependent on the
integer N , which hence results in the properties of it also reliant on N . The Parseval
identity states that for every c, d ∈ S(N ),
c, d =
1
cˆ, dˆ ,
N
53
CHAPTER 3. NONSTATIONARY GABOR FRAMES IN S(2K )
54
where cˆ and dˆ are the discrete Fourier transform of c and d respectively.
Next, simple computation shows us that the discrete Fourier transform of Tn c
is E−n cˆ, where Tn and Em are the circular shift and modulation operators for S(2K )
as defined in (3.1).
Now, we shall see how we can use the Parseval identity for the discrete Fourier
transform to establish the relationship between the frames that use only the modulation operator and that which use only the circular shift. By definition, given
K
a collection of sequences {cm }Q−1
m=0 in S(2 ) and nonnegative integers qm ≤ K ,
m = 0, 1, · · · , Q − 1, the collection {Tn2K−qm cm }n=0,...,2qm −1 is a frame for S(2K ) if
m=0,...,Q−1
and only if there exist A, B > 0 such that for every d ∈ S(2K ),
Q−1 2qm −1
A d
2
| d, Tn2K−qm cm |2 ≤ B d 2 .
≤
m=0
n=0
This is equivalent to, by the Parseval identity,
A ˆ
d
2K
Q−1 2qm −1
2
≤
m=0
n=0
1
ˆ T K−qm cm |2 ≤ B dˆ 2 ,
| d,
n2
22K
2K
which gives
Q−1 2qm −1
2 A dˆ
K
2
ˆ E K−qm cm |2 ≤ 2K B dˆ 2 .
| d,
−n2
≤
m=0
n=0
Summing over n = 0, 1, · · · , 2qm −1 is the same as summing over −n = 0, 1, · · · , 2qm −
1, which we shall show later; hence, we replace −n for n and obtain
Q−1 2qm −1
2 A dˆ
K
2
ˆ E K−qm cm |2 ≤ 2K B dˆ 2 .
| d,
n2
≤
m=0
(3.7)
n=0
Next, replacing dˆ with d, the above implies that {En2K−qm cm }n=0,...,2qm −1 forms a
m=0,...,Q−1
frame for S(2K ). Therefore, {Tn2K−qm cm }n=0,...,2qm −1 forms a frame for S(2K ) if
m=0,...,Q−1
and only if {En2K−qm cm }n=0,...,2qm −1 does. We may then use the results in Section
m=0,...,Q−1
CHAPTER 3. NONSTATIONARY GABOR FRAMES IN S(2K )
55
3.1 to formulate conditions for the frame {En2K−qm cm }n=0,...,2qm −1 to satisfy.
m=0,...,Q−1
Thus, it remains to show that the collection {E−n2K−qm cm }n=0,...,2qm −1 is the
m=0,...,Q−1
same as the collection {En2K−qm cm }n=0,...,2qm −1 so that the substitution of −n for
m=0,...,Q−1
n in (3.7) is justified. Firstly, we observe that
E−n2K−qm cm = E2K −n2K−qm cm = E(2qm −n)2K−qm cm .
Hence, substituting n = 2qm − n, we have
{E−n2K−qm cm }n=0,...,2qm −1 = {E(2qm −n)2K−qm cm }n=0,...,2qm −1
m=0,...,Q−1
m=0,...,Q−1
= {En 2K−qm cm }n =1,...,2qm
m=0,...,Q−1
= {En 2K−qm cm }n =0,...,2qm −1 ,
m=0,...,Q−1
since n = 0 and n = 2qm are the same for the operator En 2K−qm . This completes
the proof.
Now, we may look at our first result which is analogous to Theorem 3.5.
Corollary 3.12. Let cm ∈ S(2K ) for m = 0, 1, · · · , Q − 1. Define
Q−1 2K−qm −1
B :=
2qm |cm (l)cm (l + j2qm )| ,
max
l=0,··· ,2K −1
m=0
j=0
and
A :=
min
l=0,··· ,2K −1
Q−1 2K−qm −1
Q−1
2qm |cm (l)|2 −
m=0
2qm |cm (l)cm (l + j2qm )| .
m=0
j=1
Then {Tn2K−qm cm }n=0,...,2qm −1 forms a Bessel sequence for S(2K ) with Bessel bound
m=0,...,Q−1
B
.
2K
In addition, if A > 0, then {Tn2K−qm cm }n=0,...,2qm −1 forms a frame for S(2K )
with frame bounds
m=0,...,Q−1
A
, B.
2K 2K
Remark 3.13. Observe the slight difference between the frame bounds in Theorem
3.5 and Corollary 3.12. In the series of computations earlier on, we would have
CHAPTER 3. NONSTATIONARY GABOR FRAMES IN S(2K )
56
noticed that if {Tn2K−qm cm }n=0,...,2qm −1 forms a frame with bounds A and B, then
m=0,...,Q−1
{En2K−qm cm }n=0,...,2qm −1 will have bounds 2K A and 2K B. Conversely, if A and B
m=0,...,Q−1
are frame bounds for {En2K−qm cm }n=0,...,2qm −1 , then the other collection will form
m=0,...,Q−1
A
, B.
2K 2K
a frame with bounds
In view of Theorem 3.5, the conditions in Corollary 3.12 are for the collection
{En2K−qm cm }n=0,...,2qm −1 to form a frame with bounds A, B, thus the conclusion in
m=0,...,Q−1
A
, B
2K 2K
the corollary gives
as frame bounds for {Tn2K−qm cm }n=0,...,2qm −1 .
m=0,...,Q−1
Next, fixing qm = q for m = 0, 1, · · · , Q − 1, we may also obtain an analogous
result of Proposition 3.9. We will get bounds for the sum
Q−1 q
2
m=0 2 |cm (l)| ,
l =
0, 1, · · · , 2K − 1, when {Tn2K−q cm }n=0,...,2q −1 forms a frame. Due to the explanation
m=0,...,Q−1
given above, we note that the bounds are again slightly different from those in
Proposition 3.9.
Corollary 3.14. Let cm ∈ S(2K ), m = 0, 1, · · · , Q − 1, such that the collection
{Tn2K−q cm }n=0,...,2q −1 forms a frame for S(2K ) with bounds A, B. Then for l =
m=0,...,Q−1
0, 1, · · · , 2K − 1,
Q−1
2K−q A ≤
|cm (l)|2 ≤ 2K−q B.
(3.8)
m=0
Proof. If {Tn2K−q cm }n=0,...,2q −1 forms a frame for S(2K ) with bounds A and B, then
m=0,...,Q−1
the frame {En2K−q cm }n=0,...,2q −1 will have bounds 2K A and 2K B. By Proposition
m=0,...,Q−1
3.9, for every l = 0, 1, · · · , 2K − 1,
Q−1
2K A ≤
2q |cm (l)|2 ≤ 2K B.
m=0
We then obtain (3.8).
Lastly, we will look at the corresponding necessary condition when the collection
{Tn2K−q cm }n=0,...,2q −1 forms a frame. It turns out that the condition is exactly the
m=0,...,Q−1
same as in Theorem 3.10 since {Tn2K−q cm }n=0,...,2q −1 being a frame implies that
m=0,...,Q−1
{En2K−q cm }n=0,...,2q −1 also forms a frame. Hence, we have the following.
m=0,...,Q−1
CHAPTER 3. NONSTATIONARY GABOR FRAMES IN S(2K )
57
Corollary 3.15. Let cm ∈ S(2K ), m = 0, 1, · · · , Q − 1, such that the collection
{Tn2K−q cm }n=0,...,2q −1 forms a frame for S(2K ). Then
m=0,...,Q−1
2K−q
Q
≤ 1.
In this chapter, we have developed the theory for nonstationary Gabor frames
in S(2K ). In the next chapter, we will expand more on {Em2K−pn dn }m=0,...,2pn −1 , in
n=0,...,Q−1
particular an alternative dual for frames with some extra conditions, in order to apply them on discrete signals for time-frequency analysis and also the reconstruction
of the signals from their time-frequency representations.
Chapter 4
Time-Frequency Analysis
After developing the theory on discrete nonstationary Gabor frames, we can apply
them to discrete signals and obtain time-frequency representations of the signals.
With the time-frequency representations, it is of interest to have efficient ways of
reconstructing the signals from them. This is where we are going to use the theory
on dual frames that will be developed in this chapter. We will also make use of
the advantage of having variable window sizes in our approach to allow adaptive
time-frequency analysis.
4.1
Dual Frames in S(2K )
As mentioned in Chapter 2, when we have a frame {fk }k∈I in a Hilbert space with
frame operator S, we have the canonical dual {S −1 fk }k∈I that can reconstruct a
signal f from the sequence { f, fk }k∈I by the formula
f, fk S −1 fk .
f=
k∈I
However, we may also have an alternative dual {gk }k∈I in place of the canonical dual.
Our aim in this section is to formulate conditions for {Em2K−pn dn }m=0,...,2pn −1 to
n=0,...,Q−1
satisfy to be a frame so that we have an explicit expression for an alternative dual.
58
CHAPTER 4. TIME-FREQUENCY ANALYSIS
59
Firstly, we shall look at the discrete analogue of Lemma 2.13, which is also a
generalisation of Lemma 3.4. As before, we always fix in the results, pn , Q, K ∈ N
and assume pn ≤ K.
Lemma 4.1. Consider c, c , dn , dn ∈ S(2K ), n = 0, 1, · · · , Q − 1. Then
Q−1 2pn −1
(2K )
c, Em2K−pn dn
n=0
m=0
2K −1
=
(2K )
c , Em2K−pn dn
Q−1
2pn dn (l)dn (l)
c(l)c (l)
n=0
l=0
2K −1
Q−1 2K−pn −1
2pn c (l + j2pn )dn (l)dn (l + j2pn ).
c(l)
+
n=0
l=0
j=1
Proof. The proof of the equation is similar to that of Lemma 3.4. Firstly, we may
rewrite the inner products
(2K )
c, Em2K−pn dn
S(2K )
(2K )
c , Em2K−pn dn
S(2K )
pn )
=
(2
Cn , Em
=
(2
Cn , Em
S(2pn )
pn )
S(2pn )
,
,
where Cn and Cn , defined by
2K−pn −1
Cn (r) :=
c(r + j2pn )dn (r + j2pn ),
r ∈ Z;
c (r + j2pn )dn (r + j2pn ),
r ∈ Z,
j=0
2K−pn −1
Cn (r) :=
j=0
are in S(2pn ). Therefore,
2pn −1
(2pn )
pn )
(2
Cn , Em
m=0
S(2pn )
Cn , Em
S(2pn )
= 2pn Cn , Cn
S(2pn )
.
CHAPTER 4. TIME-FREQUENCY ANALYSIS
60
Next, we have
Q−1 2pn −1
(2K )
c, Em2K−pn dn
n=0
Q−1
m=0
Q−1 2pn −1
pn
2
=
(2K )
c , Em2K−pn dn
2pn Cn (r)Cn (r)
Cn , Cn =
n=0 r=0
n=0
Q−1 2pn −1 2K−pn −1
2pn c(r + j2pn )dn (r + j2pn )Cn (r)
=
n=0 r=0
Q−1
j=0
2K−pn −1
j2pn +2pn −1
2pn c(l)dn (l)Cn (l − j2pn )
=
n=0
j=0
l=j2pn
Q−1 2K −1
2pn c(l)dn (l)Cn (l).
=
n=0 l=0
Writing out the expression for Cn (l) and splitting the sum into the cases when j = 0
and j = 0, we obtain the result. This completes the proof.
Remark 4.2. Note that there is no need to impose any additional condition on the
sequences in the lemma, unlike its analogue in L2 (R) given by Lemma 2.13, which
thus makes the proof simpler.
Now, we shall look at our main theorem which we will use to construct our
frames later. This is analogous to Theorem 2.14 on constructing explicitly pairs of
dual frames in L2 (R). In contrast to Theorem 2.14, the two collections forming the
dual pairs here, being finite collections, are automatically Bessel sequences. Note
that when we denote dn where n ≥ Q, dn = dm where m = n mod Q.
Theorem 4.3. Let dn ∈ S(2K ), n = 0, 1, · · · , Q − 1, be real sequences satisfying the
following:
(i) There exists P ≤ Q − 1 such that for every n = 0, 1, · · · , Q − 1, supp dn ∩
supp dn+ν = ∅ for P ≤ ν ≤ Q − 1;
(ii)
Q−1 √ pn
2 dn (l)
n=0
= 1 for every l = 0, 1, · · · , 2K − 1;
(iii) For each n = 0, 1, · · · , Q − 1, max{|l1 − l2 | : l1 , l2 ∈
P −1
ν=0 supp dn+ν }
< 2p n .
CHAPTER 4. TIME-FREQUENCY ANALYSIS
61
For each n = 0, 1, · · · , Q − 1, define
P −1
dn (l) := dn (l) + 2
ν=1
√
2pn+ν
√
dn+ν (l),
2p n
l = 0, 1, · · · , 2K − 1.
Then {Em2K−pn dn }m=0,...,2pn −1 and {Em2K−pn dn }m=0,...,2pn −1 form a pair of dual
n=0,...,Q−1
n=0,...,Q−1
frames for S(2K ).
Proof. Firstly, since the two collections have only finite number of elements, they
are clearly Bessel sequences. Thus we only need to show that for any c, c ∈ S(2K ),
Q−1 2pn −1
c, Em2K−pn dn c , Em2K−pn dn = c, c .
n=0
m=0
Fix l ∈ {0, 1, · · · , 2K − 1}. Then there exists a smallest nonnegative integer nl such
that dn (l) is nonzero for at most n = nl , nl + 1, · · · , nl + P − 1. This implies that
√
Q−1 √ pn
l +P −1
2 dn (l) = nn=n
2pn dn (l). Hence,
n=0
l
nl +P −1
1 =
√
2
2pn dn (l)
n=nl
nl +P −1
= 2pnl dnl (l) dnl (l) +
n=nl +1
√
2 2pn
√ p dn (l)
2 nl
nl +P −1
+2pnl +1 dnl +1 (l) dnl +1 (l) +
n=nl +2
+··· + 2
pnl +P −1
√
2 2pn
√ p
dn (l)
2 nl +1
dnl +P −1 (l)[dnl +P −1 (l)].
We note that in each pair of square brackets above, for µ = 0, 1, · · · , P − 1,
√
√
2 2pnl +µ+1
2 2pnl +µ+P −1
√ p
dnl +µ (l) + √ p
dnl +µ+1 (l) + · · · +
dnl +P −1 (l)
2 nl +µ
2 nl +µ
P −1−µ √ pn +µ+ν
2 2 l
√ p
= dnl +µ (l) +
dnl +µ+ν (l)
2 nl +µ
ν=1
P −1 √ pn +µ+ν
2 2 l
√ p
= dnl +µ (l) +
dnl +µ+ν (l).
nl +µ
2
ν=1
CHAPTER 4. TIME-FREQUENCY ANALYSIS
62
The difference between the last two lines are the terms
√p
n +µ+ν
√2 p l
dnl +µ+ν (l) where
n +µ
2
2
l
ν ≥ P − µ. These terms are all zero because, due to the choice of l, dnl +µ+ν (l) = 0
for ν ≥ P − µ. Therefore,
√
2 2pnl +ν
√ p
1 = 2 dnl (l) dnl (l) +
dnl +ν (l)
nl
2
ν=1
P −1 √ pn +1+ν
2 2 l
pnl +1
√ p
dnl +1+ν (l)
+2
dnl +1 (l) dnl +1 (l) +
2 nl +1
ν=1
P −1 √ pn +P −1+ν
2 2 l
pnl +P −1
√ p
+··· + 2
dnl +P −1+ν (l)
dnl +P −1 (l) dnl +P −1 (l) +
nl +P −1
2
ν=1
P −1
pnl
nl +P −1
Q−1
2pn dn (l)dn (l) =
=
n=nl
2pn dn (l)dn (l).
n=0
Next, we note that by (iii), for each n = 0, 1, · · · , Q − 1, supp dn ∪ supp dn has a
maximum ‘length’ of 2pn , which means that
2K−pn −1
2pn c (l + j2pn )dn (l)dn (l + j2pn ) = 0
j=1
for any c ∈ S(2K ). Hence, by Lemma 4.1, for every c, c ∈ S(2K ),
Q−1 2pn −1
2K −1
c, Em2K−pn dn c , Em2K−pn dn
n=0
m=0
Q−1
2pn dn (l)dn (l)
c(l)c (l)
=
l=0
n=0
2K −1
=
c(l)c (l) = c, c .
l=0
This completes the proof.
Thus, we have formulated our main theorem above which we shall use in the
next section, when we fix pn = p for every n = 0, 1, · · · , Q − 1. Observe that after
fixing pn = p, the expression for dn that is used to generate the alternative dual
CHAPTER 4. TIME-FREQUENCY ANALYSIS
{Em2K−pn dn }m=0,...,2pn −1 becomes very simple, which is given by
n=0,...,Q−1
P −1
dn (l) = dn (l) + 2
dn+ν (l),
l = 0, 1, · · · , 2K − 1.
ν=1
4.2
Adaptive Time-Frequency Representation
Recall our aim for developing the theory on frames that are generated by a collection
of functions, instead of just one function, in the previous two chapters. It is to take
advantage of the allowance of more than one fixed window size in a frame. Given
a signal with high frequency in a particular time interval, we hope for high time
resolution in that interval. This can be achieved by having a few dn ’s with very
small support and the union of their supports covers the time interval. Similarly,
for a signal with low frequency in a particular time interval, we want to have high
frequency resolution which can be achieved by having a few dn ’s with large support.
Before devising a method to generate such a collection of sequences {dn }Q−1
n=0 ,
where Q is a positive integer, we want to find a way to be able to recognise the
time intervals in which a signal has high or low frequencies. More precisely, when
a signal is sampled at 2K points per second, where K is a positive integer, we can
form a partition vector such that at time intervals with high frequencies or low
frequencies, the endpoints of these intervals will be reflected in the vector. For
example, a signal sampled at 256 points per second with a high frequency for the
first half second and a low frequency for the second half second may yield a vector
(1, 32, 64, 96, 128, 192, 256). When consecutive entries in the vector are close, it will
imply that the signal has a high frequency between the time-points indicated by the
entries; and it has a low frequency between the time-points if the entries are not
close to each other.
The technique we shall use is to make use of an idea similar to the total variation
of a function defined on an interval. Since a signal is represented by discrete points,
it is easier, when compared to the function case, to calculate the “variation” of the
63
CHAPTER 4. TIME-FREQUENCY ANALYSIS
64
signal in an interval. In more precise terms, given a vector x = (x1 , x2 , · · · , x2K ),
and two integers 1 ≤ a < b ≤ 2K , we define the variation of x in the interval [a, b]
as
b−1
|xi − xi+1 |.
V [x; a, b] :=
(4.1)
i=a
Hence, given a signal x, we can scan from the first value and calculate the
variations between time-point 1 and the following time-points. When the variation
of x in the interval [1, n] exceeds a preset tolerance value, we set (n − 1) to be the
second entry of the partition vector that we want to form. Then, we start calculating
the variations between time-point (n − 1) and the subsequent time-points until the
tolerance level is exceeded again and we get the next entry of the partition vector.
We keep repeating the steps until the variation between the time-point indicated
by the last entry of the partition vector and the time-point 2K is already below the
tolerance level. This process is summarised in the following algorithm.
K
Algorithm 4.4. Let x ∈ R2
and tolerance level t > 0.
1. Set v = [1], left counter i = 1, and right counter j = 2.
2. If j < 2K , obtain V [x; i, j] as defined in (4.1). If j = 2K , set v = [v 2K ] and
stop.
3. If V [x; i, j] ≤ t, set j = j + 1. Goto 2.
4. If V [x; i, j] > t, set v = [v j − 1], i = j − 1. Goto 2.
After obtaining the vector from the above algorithm, we may create a collection
K
of sequences {dn }Q−1
n=0 in S(2 ) that satisfies the conditions in Theorem 4.3. Firstly,
we fix pn = p for all n = 0, 1, · · · , Q − 1. Secondly, the number of entries in the
partition vector determines the constant Q, which is the number of sequences dn
that we are using. Excluding 1 and 2K , we have 1 < x1 < x2 < · · · < xQ−1 < 2K
in the vector. These points are the intersection of the consecutive dn ’s which we are
going to construct. Here is how we proceed to obtain our dn ’s.
CHAPTER 4. TIME-FREQUENCY ANALYSIS
65
Suppose that {1 < x1 < x2 < · · · < xQ−1 < 2K } is given. Let
> 0, satisfying
some constraints that we shall state later, and consider any continuous real-valued
function f defined on the interval [0, 2 ] such that f (0) = 0 and f (2 ) = 1. Let
g = 1 − f . Set x0 = and xQ = 2K + . Then for each n = 0, 1, · · · , Q − 1, we define
dn ∈ S(2K ) by, for l = 0, 1, · · · , 2K − 1,
dn (l) :=
√1 f (l
2p
− (xn − )),
xn − ≤ l < xn + ;
√1 ,
2p
xn + ≤ l < xn+1 − ;
(4.2)
√1 g(l − (xn+1 − )), xn+1 − ≤ l ≤ xn+1 + ;
2p
0,
otherwise.
Thus, the support of each dn are integers in [xn − , xn+1 + ] and it intersects with
dn+1 for integers in [xn+1 − , xn+1 + ]. In this interval, dn is defined by the function
g and dn+1 is defined by the function f , which assures that they add up to
√1 .
2p
In
the intervals where there is no intersection, which is [xn + , xn+1 − ] for each dn , we
also make sure that
Q−1
n =0 dn
(l) = dn (l) =
√1
2p
so that condition (ii) of Theorem
4.3 is satisfied.
Now, all the above assumes that each dn intersects with dn−1 and dn+1 but
dn−1 and dn+1 do not intersect with each other. This would give condition (i) of
Theorem 4.3 with P = 2. The setup poses constraints on the
which we shall now
see. Firstly, for each n = 0, 1, · · · , Q − 1, xn + < xn+1 − . Thus
1. For n = 0, 3 < x1 ;
2. for n = 1, 2, · · · , Q − 2, 2 < xn+1 − xn ;
3. for n = Q − 1,
< 2K − xQ−1 .
Next, we want dn ’s to satisfy (iii) in Theorem 4.3. This gives
4. For n = 0, 1, · · · , Q − 3, xn+2 + − (xn − ) < 2p which implies 2 < 2p −
(xn+2 − xn ).
CHAPTER 4. TIME-FREQUENCY ANALYSIS
5. Considering dQ−2 and dQ−1 , we have xQ +
66
− (xQ−2 − ) < 2p , yielding
3 < xQ−2 + 2p − 2K .
6. Considering dQ−1 and d0 , where we define d0 on [2K , 2K + x1 + ], i.e. on
its second positive period rather than its usual support [0, x1 + ], we have
2K + x1 + − (xQ−1 − ) < 2p , resulting in 2 < xQ−1 − x1 + 2p − 2K .
Therefore, after finding an
> 0 that satisfies all the above constraints, we are able
to define our dn ’s by (4.2). Note that as
is positive, conditions 4, 5 and 6 have
imposed necessary conditions on the constant p. More explicitly, p has to satisfy
2p > xn+2 − xn for n = 0, 1, · · · , Q − 3, 2p > 2K − xQ−2 and 2p > 2K − xQ−1 + x1 .
The resulting collections {Em2K−p dn }m=0,...,2p −1 and {Em2K−p dn }m=0,...,2p −1 , where
n=0,...,Q−1
n=0,...,Q−1
dn , n = 0, 1, · · · , Q − 1, are defined in Theorem 4.3, then form a pair of dual frames
for S(2K ). An additional note here is that with the constant P set to be two in this
setup, we have to make sure that Q ≥ 3. Thus, in the scenario where the signal
given for Algorithm 4.4 has very little activity, resulting in Q being one or two, we
may adjust the tolerance level t to be lower.
In the next section, we shall use this construction to generate time-frequency
representations of discrete signals.
4.3
Time-Frequency Representation of Signals
In the previous section, we have shown how to generate the collection of sequences
K
{dn }Q−1
n=0 in S(2 ) after using Algorithm 4.4 to form a partition, {1 < x1 < x2 <
· · · < xQ−1 < 2K }, of [0, 2K ]. Hence, we shall now use the method to produce the
time-frequency representations of some discrete signals to illustrate the advantages
of the adaptive method. In particular, the continuous function that is used in
(4.2) is f = sin2 ( 4π ·), after the constant
{1 < x1 < x2 < · · · < xQ−1 < 2K } and
> 0 is fixed. This means that given
> 0, the collection {dn }Q−1
n=0 used in the
CHAPTER 4. TIME-FREQUENCY ANALYSIS
67
following representations is defined by
dn (l) :=
√1
2p
sin2 ( 4π (l − (xn − ))),
√1 ,
2p
xn − ≤ l < xn + ;
xn + ≤ l < xn+1 − ;
√1 cos2 ( π (l − (xn+1 − ))), xn+1 − ≤ l ≤ xn+1 + ;
4
2p
0,
otherwise.
p
After obtaining {dn }Q−1
n=0 , we then generate the frames {Em2K−p dn }m=0,...,2 −1
n=0,...,Q−1
and {Em2K−p dn }m=0,...,2p −1 . The first frame collection will be used for decomposing
n=0,...,Q−1
a signal c ∈ S(2K ) into the frame coefficients c, Em2K−p dn , m = 0, 1, · · · , 2p − 1,
n = 0, 1, · · · , Q − 1, and the second frame collection will be used to reconstruct the
signal from the frame coefficients. Recall in Section 4.1 that once we fix pn = p for
all n = 0, 1, · · · , Q − 1, dn , n = 0, 1, · · · , Q − 1, are simply defined by
P −1
dn (l) = dn (l) + 2
dn+ν (l),
l = 0, 1, · · · , 2K − 1.
ν=1
For the signals that we are going to decompose later, the results will be reflected
on a spectrogram. The horizontal axis of the spectrogram is first divided into Q
columns with endpoints at {0, x1 , x2 , · · · , xQ−1 , 2K }, and then into 2p rows with
endpoints at {0, 2K−p , (2)2K−p , · · · , (2p − 1)2K−p , (2p )2K−p }. Thus, each rectangle
is represented by a frame coefficient c, Em2K−p dn . More explicitly, the rectangle
with bottom right corner (xn , m2K−p ) will be represented by c, Em2K−p dn−1 , m =
0, 1, · · · , 2p −1, n = 1, 2, · · · , Q−1. For the rectangles in the last column, which have
bottom right corners (2K , m2K−p ), m = 0, 1, · · · , 2p − 1, they will be represented by
c, Em2K−p dQ−1 .
After generating the time-frequency representation, we will also display the reconstructed signal obtained from the frame coefficients using the dual frame formula
Q−1 2p −1
c=
c, Em2K−p dn Em2K−p dn .
n=0 m=0
CHAPTER 4. TIME-FREQUENCY ANALYSIS
This will illustrate the perfect reconstruction of the signals from their frame coefficients, which is supported by the theory developed.
The first signal we see below is a sinusoidal signal with frequency 50 Hz, which
experiences a sudden increase in the frequency at very short time intervals. To be
more precise, the signal is defined by
sin(2π50t) + 4 sin(2π200t), 0.055 < t < 0.075 and 0.45 < t < 0.5;
c1 (t) :=
sin(2π50t) + 4 sin(2π210t), 0.1 < t < 0.12 and 0.55 < t < 0.565;
sin(2π50t),
otherwise.
Hence, sampled at 512 points per second, i.e. K = 9, we obtain the results in Figure
4.1.
From Figure 4.1, we can see that even though the high frequency periods are
very short and close to each other, it is still reflected quite well in the spectrogram.
In fact, the time resolution is high enough to indicate the intervals when there is a
sharp change in frequencies, which can be observed by comparing the original signal
and the spectrogram. Nevertheless, the frequency resolution is compromised and
it is not clear enough to identify the exact frequency during these time intervals.
On the other hand, during other time intervals when there is only one frequency in
the signal, we can still observe it from the spectrogram. Moreover, we also notice
that the reconstructed signal is the same as the original signal, which shows the
usefulness of the dual frame generated.
Next, we will look at the signal, defined by
0.5 sin(2π50t) + 0.5 sin(2π40t), 0.0 < t < 0.25;
c2 (t) :=
0.5 sin(2π50t) + 0.5 sin(2π60t), 0.55 < t < 0.75;
sin(2π50t),
otherwise.
The original signal has comparable amplitude throughout the one second. Hence, by
direct observation in the time domain, it will not be as easy as the previous signal
68
CHAPTER 4. TIME-FREQUENCY ANALYSIS
Figure 4.1: Time-frequency analysis of a signal with high frequencies close to each
other in time. Top: Original signal of c1 . Centre: Spectrogram of c1 . Bottom:
Reconstructed signal of c1 from the frame coefficients.
69
CHAPTER 4. TIME-FREQUENCY ANALYSIS
70
c1 to see the hidden frequencies at different time intervals. However, as the activity
of the signal is not as strong as c1 , the partition vector obtained from Algorithm
4.4 will have less elements, resulting in dn ’s with wider support. This unavoidably
affects the time resolution but increases the frequency resolution, which allows the
different frequencies to be separated in the spectrogram. The results are shown in
Figure 4.2, where we sampled the signal at 128 points per second, i.e. K = 7.
The next figure shows the results obtained from a linear chirp with an initial
frequency of 20 Hz, increasing linearly to 60 Hz in one second. More precisely, the
linear chirp is defined by
c3 (t) := sin(2π(20 + 20t)t),
0 < t < 1,
and is sampled at 512 points per second, i.e. K = 9. We can see from Figure 4.3 the
changing resolution of both time and frequency. In the beginning, as the signal activity is lower, the dn ’s created for smaller n’s have wider support, resulting in better
frequency resolution. However, as the frequency increases, dn ’s with smaller support
are generated for larger n’s, resulting in better time resolution but the frequency
resolution decreases. Nevertheless, we can still observe the changing frequency from
the spectrogram.
Lastly, we generate the results for a hyperbolic chirp. The signal is defined by
c4 (t) :=
E(t) sin(7.5π(0.8 − t)−1 ), 0.1 < t < 0.68;
0,
otherwise,
with the envelope
E(t) :=
1 + sin π(0.1625 −
2,
1 + sin π(0.1625 −
0,
1 −1
512 ) (t
−
1
512 )
− 0.5π ,
1
512
≤ t < 0.1625;
0.1625 ≤ t < 0.4875 +
1 −1
512 ) (0.65
− t) − 0.5π , 0.4875 +
1
512
otherwise.
1
512 ;
≤ t < 0.65;
CHAPTER 4. TIME-FREQUENCY ANALYSIS
Figure 4.2: Time-frequency analysis of a signal with low frequencies close to each
other in frequency. Top: original signal of c2 . Center: Spectrogram of c2 . Bottom:
Reconstructed signal of c2 from the frame coefficients.
71
CHAPTER 4. TIME-FREQUENCY ANALYSIS
Figure 4.3: Time-frequency analysis of a linear chirp. Top: Original signal of c3 .
Centre: Spectrogram of c3 . Bottom: Reconstructed signal of c3 from the frame
coefficients.
72
CHAPTER 4. TIME-FREQUENCY ANALYSIS
Sampled at 512 points per second, i.e. K = 9, we have the spectrogram in Figure
4.4. From the spectrogram, we can see that the frequency of the signal increases
with time and suddenly there is no more activity in the signal. As there is higher
frequency at around 0.6 second, the time resolution is higher as dn ’s of smaller
supports are generated, allowing us to see clearly when the signal activity stops.
In conclusion, we have seen in this section how the proposed adaptive timefrequency representation is useful in providing information of some signals. In particular, for a signal of a relatively low frequency which encounters sudden sharp
increase in frequency in short periods of time, the adaptive approach allows us to
see clearly when the changes occur because the time resolution is very good at these
periods of time. By decreasing the time resolution during other time intervals, we
also achieve good frequency resolution of the signal at these intervals.
In addition, for a signal that has more than one low frequencies in it, even
when the frequency values are quite near to each other, we are able to decompose
the signal to show the different frequencies in the spectrogram. Nevertheless, the
time resolution is affected in this case.
We have also seen in all the figures how the signals are perfectly reconstructed
from their frame coefficients using the corresponding dual frames. The simple and
explicit formula of the dual frames make the reconstruction process efficient, which
is an important aspect in signal processing.
In general, the applications that we have seen in this section are based on
Theorem 4.3, where we fixed pn = p and P = 2. We note here that other possibilities
may be explored as we allow more freedom in the choices of pn and P , hopefully,
bringing more advantages to this adaptive approach in time-frequency analysis of
signals.
73
CHAPTER 4. TIME-FREQUENCY ANALYSIS
Figure 4.4: Time-frequency analysis of a hyperbolic chirp. Top: Original signal of
c4 . Centre: Spectrogram of c4 . Bottom: Reconstructed signal of c4 from the frame
coefficients.
74
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76
[...]... details of the development of the theory and its applications In particular, we may observe how the proofs of some results in [5], [7] are adapted for our use 6 Chapter 2 Nonstationary Gabor Frames in L2(R) Gabor frames in L2 (R) refer to frames that are of the form {Emb Tna g}m,n∈Z , where g ∈ L2 (R) A Gabor frame was formed using one function g and the translation and modulation operators, as defined... one function g and the parameters bn are the same, the collection then has the same structure as Gabor frames We want to look for conditions that are sufficient to make the collection {Embn gn }m,n∈Z a frame for L2 (R) Most of the results that we see in this section are closely related to results 7 CHAPTER 2 NONSTATIONARY GABOR FRAMES IN L2 (R) 8 proved in [5] regarding standard Gabor frames In fact,... collection of functions {gn }n∈Z or {hm }m∈Z instead of only one function to generate the frames introduce more flexibility in the time-frequency resolution by allowing different window sizes at different times, which is what we are aiming for in adaptive time-frequency representation After developing the theory on nonstationary Gabor frames for functions in L2 (R), we will then look at nonstationary frames. .. proof For the next result, the proof is similar to those of results in [6] and [7] The paper [6] aims to find an explicit expression of dual frames for certain stationary Gabor frames in L2 (R) while [7] aims to find that for certain nonstationary frames in L2 [0, 2π] Our following theorem extends and adapts these proofs for the nonstationary setting in L2 (R) Our aim is also to formulate conditions that... Since an+1 ≤ t ≤ 12 (an+1 + an + β cn ), it follows that cn (an+1 − an ) ≤ cn (t − an ) ≤ β + cn (an+1 − an ) , 2 22 CHAPTER 2 NONSTATIONARY GABOR FRAMES IN L2 (R) 23 and so δ2 ≤ cn (t − an ) ≤ by (ii) and (iii) Since g is decreasing on [δ2 , 2 ], 2, g(cn (t−an )) ≥ g( 2 ) and as a result, G(t) ≥ |g( 2 )|2 Next, consider t in the other half interval of supp gn ∩ supp gn+1 , i.e t ∈ [ 21 (an+1 + an... collections {fk }k∈Z and {gk }k∈Z are dual frames for L2 (R) (see for instance [5]), it is sufficient for them to be Bessel sequences and for every f, f ∈ L2 (R), f, fk f , gk = f, f k∈Z With Lemma 2.13 on hand, we may derive the following theorem in which for the type of frame defined, we have an explicit expression for an alternative dual In this result, we are concerned with real-valued frame functions only... (R), (S − GI)f, f = 0 which gives S = GI Corollary 2.4 was proved directly in [1] Here we first establish general sufficient conditions for nonstationary Gabor frames and then obtain it as a consequence of these conditions Corollary 2.5 Consider {gn }n∈Z in L2 (R) and bn > 0, n ∈ Z, that satisfy the following: (i) There exist C, D > 0 such that for every n ∈ Z, C ≤ |gn (t)| √ bn ≤ D a.e in the support... intersect, there should only be at most K intersections On the other hand, it also ensures that the union of the supports essentially covers the real line, which is not surprising since the union of the supports of frame elements should do so CHAPTER 2 NONSTATIONARY GABOR FRAMES IN L2 (R) Now, we look at some examples of Corollaries 2.4 and 2.5 Example 2.7 Firstly, let bn = g(t) := 1 3 for every n ∈ Z... 2 2 = 3 −t2 + 3 , − 12 ≤ t < 12 4 Therefore, 3 4 ≤ G(t) ≤ 9 4 and by Corollary 2.4, {Embn gn }m,n∈Z forms a frame for L2 (R) Example 2.8 Next, we look at a simple example for Corollary 2.5 For n ∈ Z, let √ 1 bn = (|n|+1) bn χ[n,n+1) (t) Clearly, the support of each gn has length 2 and gn (t) = 17 CHAPTER 2 NONSTATIONARY GABOR FRAMES IN L2 (R) 1 which is less than 1 bn 18 = (|n| + 1)2 In its support,... supports covers the entire real line Hence, we impose conditions on the translation and yield the following Proposition 2.9 Let g ∈ L2 (R) be continuous and compactly supported on [0, 1], and positive on (0, 1) Let a > 1 and c ∈ R satisfy tn = can In addition, for n ∈ Z\{0}, let bn = 1 a|n| 1 a2 −1 ... short-time Fourier transform and Gabor frames This is to highlight the motivation of the move to nonstationary analysis and also present some standard results on Gabor frames, allowing the reader... After developing the theory on nonstationary Gabor frames for functions in L2 (R), we will then look at nonstationary frames for sequences, namely the space of periodic sequences S(2K ), where K is... gm }m,n∈Z and {Tnam hm }m,n∈Z form a pair of dual frames for L2 (R) CHAPTER NONSTATIONARY GABOR FRAMES IN L2 (R) We have seen in this chapter how we may form some nonstationary Gabor frames using