Wavelets and Signal Processing · 2012-02-08 · Wavelets and Signal Processing Reinhold Schneider Sommersemester 2000 Recommended Literature [1] St´ephane Mallat [1998]: A Wavelet

Wavelets and Signal Processing

Reinhold Schneider

Sommersemester 2000

Recommended Literature

[1] Stephane Mallat [1998]: A Wavelet Tour of Signal Processing, Academic Press, Inc.

[2] Ingrid Daubechies [1992]: Ten Lectures on Wavelets, SIAM, Philadelphia

[3] Charles K. Chui [1992]: An Introduction to Wavelets in Wavelet Analyis and its Appli-cations, Vol. 1, Academic Press, Inc.

[4] A. V. Oppenheim and R. W. Schafer [1989]: Discrete-Time Signal Processing, Prentice-Hall, Englewood Cliffs

CONTENTS 2

Contents

1 Fourier Analysis 31.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Uncertainty Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.3 Linear Time-Invariant Operators (Filtering) . . . . . . . . . . . . . . . . . . . 11

2 Discrete-Time Signal Processing 152.1 Sampling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.2 Aliasing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.3 Discrete Time-Invariant Filters . . . . . . . . . . . . . . . . . . . . . . . . . . 202.4 Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222.5 Discrete Fourier Transform (DFT) . . . . . . . . . . . . . . . . . . . . . . . . . 232.6 Fast Fourier Transform (FFT) . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

3 Time meets Frequency 263.1 Heisenberg Boxes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273.2 Windowed Fourier Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . 283.3 Wavelet Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

3.3.1 Real Wavelets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323.3.2 Analytic Wavelets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

4 Time-Frequency Energy 374.1 Wigner-Ville Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 374.2 Interferences and Positivity of the Wigner-Ville distribution . . . . . . . . . . 41

5 Wavelet Bases 435.1 Frames and Riesz Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 435.2 Multi Resolution Analysis (MRA) . . . . . . . . . . . . . . . . . . . . . . . . . 455.3 Orthogonal Wavelet Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 505.4 Construction of Orthogonal Wavelet Bases . . . . . . . . . . . . . . . . . . . . 585.5 Daubechies Compactly supported wavelets . . . . . . . . . . . . . . . . . . . . 635.6 Fast Wavelet Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

1 FOURIER ANALYSIS 3

1 Fourier Analysis

With the observation (1807) that a continuous periodical function can be decomposed in aseries of trigonometrical functions Fourier has put the foundation stone for one of the mostimportant tools of modern mathematics - the Fourier analysis.

Because of its deep practical as well as theoretical impact and fundamental character themost essential results shall be summarized in this chapter.

1.1 Introduction

Definition 1.1 Let f : R → C where f ∈ L1(R), i.e.∫∞−∞ |f(t)| dt < ∞, and ξ ∈ R. Then

the Fourier transform of f at point ξ is defined by the (Lebesgue) integral

f(ξ) =

∫ ∞

−∞e−iξtf(t) dt. (1.1)

The function ξ → f(ξ) is called the Fourier transform of f .

Note that the Fourier transform is well defined, i.e. the integral always exists since

∣∣∣∣∫ ∞

−∞e−iξtf(t) dt

∣∣∣∣ ≤∫ ∞

−∞|f(t)| dt = ‖f‖L1. (1.2)

The auditors who are familiar with measure theory and the Lebesgue measure will observethat in most of the following proofs details related with measure theory are ommitted in favourof a compact and simplified representation. You can regard the completion of the proofs asan exercise for you.

Lemma 1.2 ’Riemann-Lebesgue’Let f ∈ L1(R). Then its Fourier transform f is uniformly continuous on R. Further-

more f(ξ) → 0, as ξ → ∞ or ξ → −∞.

Proof: We already know from (1.2) that f ∈ L∞(R) with ‖f‖∞ ≤ ‖f‖1.

Let |ω − ξ| < δ, then

|f(ξ) − f(ω)| ≤∫ ∞

−∞|f(t)||e−iξt − e−iωt| dt.

Now, since |e−iξt − e−iωt||f(x)| ≤ 2|f(x)| ∈ L1(R) and |e−iξt − e−iωt| → 0 as δ → 0, theLebesgue dominated convergence theorem implies that the quantity above tends to zeroas δ → 0.

To prove the second part, we first assume that f ′ exists and is in L1(R). Integrating(1.1) by parts yields

f(ξ) =

∫ ∞

−∞e−iξtf(t) dt = − 1

iξ

[e−iξtf(t)

]∞−∞ +

1

ξi

∫ ∞

−∞e−iξtf ′(t) dt.


From the continuity of f ∈ L1(R) we know f(t) → 0 as t → ±∞ and hence f ′(ξ) =iξf(ξ):

|f(ξ)| =1

|ξ||f′| ≤ 1

|ξ|‖f‖1 → 0, as ξ → ±∞.

In general, for any given ε > 0 we can find a function g such that g, g′ ∈ L1(R) and‖f − g‖1 < ε. Thus, we have

|f(ξ)| ≤ |f(ξ) − g(ξ)| + |g(ξ)|

≤∫ ∞

−∞|f(t) − g(t)|

∣∣e−iξt∣∣ dt+ |g(ξ)|

= ‖f − g‖1 + |g(ξ)| < ε+ |g(ξ)|

completing the proof.

The following example is of fundamental importance.

Lemma 1.3 The Fourier transform of t→ f(t) = e−t2 (Gaussian function) is

f(ξ) =√πe−

ξ2

4 .

−5 −4 −3 −2 −1 0 1 2 3 4 50

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

f

Fourier Transform

Figure 1.1: The Gaussian function and its Fourier transform

Proof: Consider the function

g(y) :=

∫ ∞

−∞e−x2+xy dx


By completing squares and setting u := x− y2

we have

g(y) =

∫ ∞

−∞e−(x− y

2)2+ y2

4 dx = ey2

4

∫ ∞

−∞e−u2

du =√πe

y2

4 . (1.3)

Now ,since h(y) :=√πe

y2

4 and g(y) can be extended to be entire (analytic) functions,and since they agree on R as shown above, they must agree on the whole complex planeC. In particular, by setting y to be −iω the equality (1.3) becomes

g(−iξ) =

∫ ∞

−∞e−iξte−t2 dt =

√πe−

ξ2

4 .

Definition 1.4 Let f, g ∈ L1(R) then

h(x) = (f ∗ g)(x) :=

∫ ∞

−∞f(x− t)g(t) dt

is the convolution of f and g.

Convolution is an important composition in signal analysis. For example the quite popularlinear time-invariant filters are completely characterized by a convolution with a so calledtransfer function. The following theorem is one of the most fundamental in signal processingand serves as the basis for the construction, analysis and fast implementation of digital filters.

Theorem 1.5 Let f, g ∈ L1(R) then h = (f ∗ g) ∈ L1(R) and furthermore the formula

h(ξ) = (f ∗ g)(ξ) = f(ξ)g(ξ) (1.4)

holds for almost every ξ ∈ R.

Proof: Applying Fubinis theorem yields

‖h‖1 =

∫ ∞

−∞|h(x)| dx =

∫ ∞

−∞

∣∣∣∣∫ ∞

−∞f(x− t)g(t) dt

∣∣∣∣ dx

≤∫ ∞

−∞

∫ ∞

−∞|f(x− t)||g(t)| dt dx

=

∫ ∞

−∞

∫ ∞

−∞|g(t)||f(x− t)| dx dt

(u := x− t) ⇒ =

(∫ ∞

−∞|g(t)| dt

)(∫ ∞

−∞|f(u)| du

)

= ‖g‖1‖f‖1.

Since |f(x− t)||g(t)| ∈ L1(R × R) we may consider

h(ξ) =

∫ ∞

−∞e−ixξ

∫ ∞

−∞f(x− t)g(t) dt dx

=

∫ ∞

−∞

∫ ∞

−∞e−ixξf(x− t)g(t) dt dx.


Again, by applying Fubinis theorem and substituting u := x− t we obtain

h(ξ) =

∫ ∞

−∞

∫ ∞

−∞e−iξ(t+u)f(u)g(t) du dt

=

(∫ ∞

−∞e−iξuf(u) du

)(∫ ∞

−∞e−iξtg(t) dt

)

which proves (1.4).

Theorem 1.6 Inverse Fourier transformLet f ∈ L1(R) and f ∈ L1(R). Then

f(t) =1

2π

∫ ∞

−∞f(ξ)eiξt dξ (1.5)

at each point t on the real axis where f is continuous.

Proof: Let us consider the convolution (f ∗ gα)(t) with the (more general) Gaussianfunction

gα(t) =1

2√πα

e−t2

4α .

For those who are familiar with statistics; gα is the probability density function of anormal distribution with zero mean and variance

√2α. First, we will show that as

α tends to 0+ the convolution term will converge to f(t) at each point t where f iscontinuous. For an arbitrary ε > 0 we select an η > 0 such that

|f(t− τ) − f(t)| < ε

for all τ ∈ R with |τ | < η. Then

|(f ∗ gα)(t) − f(t)| =∣∣∣∫ ∞

−∞f(t− τ)gα dτ − f(t)

=1︷︸︸︷∫ ∞

−∞gα(τ) dτ

∣∣∣

=

∣∣∣∣∫ ∞

−∞[f(t− τ) − f(t)]gα(τ) dτ

∣∣∣∣

≤∫ η

−η

|f(t− τ) − f(t)|gα(τ) dτ +

∫

|τ |≥η

|f(t− τ) − f(t)|gα(τ) dτ

≤ ε

∫ η

−η

gα(τ) dτ + ‖f‖1 max|τ |≥η

gα(τ) + |f(t)|∫

|τ |≥η

gα(τ) dτ

≤ ε

∫ ∞

−∞gα(τ) dτ + ‖f‖1gα(η) + |f(t)|

∫

|τ |≥η/√

α

g1(τ) dτ

= ε+ ‖f‖1gα(η) + |f(t)|∫

|τ |≥η/√

α

g1(τ) dτ


Since both gα(η) and the last term converge to zero as α tends to 0+, this completesthe first part of the proof, f ∗ gα → f for α → 0+.

In the second part we will show that (f ∗ gα)(t) converges to the right hand side ofequation (1.5). Consider the function

h(x) :=1

2πeixte−αx2

.

We compute its Fourier transform using the same technique as in the proof of Lemma1.3

h(ξ) =1

2π

∫ ∞

−∞e−iξxeixte−αx2

dx

=1

2π

∫ ∞

−∞e−i(ξ−t)xe−αx2

dx

=1

2π

√π

αe−

(ξ−t)2

4α = gα(t− ξ).

Hence, we obtain

(f ∗ gα)(t) =

∫ ∞

−∞f(ξ)gα(t− ξ) dξ

=

∫ ∞

−∞f(ξ)h(ξ) dξ

=

∫ ∞

−∞f(ω)h(ω) dω (by applying Fubinis theorem)

=1

2π

∫ ∞

−∞eiωtf(ω)e−αω2

dω.

Since e−αω2tends to 1 as α→ 0+, this completes the proof.

The definition of the Fourier transform is not best suited for the function space L1(R). Notethat in the assumptions of Theorem 1.6 both, f and f , have to be absolutely integrable.Although for the proper definition of f the function f should be in L1(R) there is no goodreason for f .

Example 1.7 Let

χ(t) =

1 for t ∈ [−1, 1]0 else

be the characteristic function on the interval [−1, 1]. The associated Fourier transformis

χ(ξ) =

∫ 1

−1

e−itξ dt =1

iξ(eiξ − e−iξ) = 2

sin ξ

ξ

which is a wildly oscillating function and not absolutely integrable. However, the integral∫∞−∞ |χ(ξ)|2 dξ is finite.


−20 −15 −10 −5 0 5 10 15 20−0.5

0

0.5

1

1.5

2

Figure 1.2: The characteristic function and its Fourier transform

Motivated by the previous example we consider a more convenient function space:

L2(R) :=

u : ‖u‖2 :=

(∫ ∞

−∞|u(t)|2 dt

) 12

<∞

L2(R) is a so called Hilbert space with the ”inner product” (scalar product)

〈u, v〉 :=

∫ ∞

−∞u(x)v(x) dx

which induces an associated norm ‖u‖2 =√

〈u, u〉.

Theorem 1.8 Parseval and PlancherelLet f, g ∈ L1(R) ∩ L2(R). Then the following holds

〈f, g〉 =

∫ ∞

−∞f(t)g(t) dt =

1

2π

∫ ∞

−∞f(ξ)g(ξ) dξ =

1

2π〈f , g〉, (1.6)

and hence

‖f‖22 = 〈f, f〉 =

1

2π〈f , f〉 =

1

2π‖f‖2

2. (1.7)

Proof: Let g(t) = g(−t) and h(t) = (f ∗ g)(t), then the application of the convolutionTheorem 1.5 yields

h(ξ) = f(ξ)ˆg(ξ) = f(ξ)g(ξ).


Using Theorem 1.6 about the inverse Fourier transform we get∫ ∞

−∞f(t)g(t) dt =

∫ ∞

−∞f(−t)g(−t) dt

= h(0) =1

2π

∫ ∞

−∞h(ξ) dξ =

1

2π

∫ ∞

−∞f(ξ)g(ξ) dξ.

As a consequence of this theorem, we observe that F : f → f can be considered as a boundedlinear operator on L1(R)∩L2(R) with range in L2(R). Equation (1.7) tells us that ‖F‖ =

√2π.

Since L2(R) is dense in L1(R) ∩ L2(R), F has a normpreserving extension to all elements ofL2(R). Furthermore the equations (1.6) and (1.7) hold for all f, g ∈ L2(R) so that the domainof F can be extended to the whole L2(R).

The following theorem shows the beautiful connection between F and L2(R).

Theorem 1.9 The Fourier transform F is a linear one-to-one map of L2(R) onto itself. Inother words, to every g ∈ L2(R), there corresponds one and only one f ∈ L2(R) suchthat f = g.

Proof: For the proof and related details see [3].

A major drawback of the Fourier analysis is that f and f cannot simultaneously belocalized on the corresponding domains.

Theorem 1.10 If f 6≡ 0 has compact support then f(ω) cannot vanish on any interval [a, b] ⊂R.

Proof: Let f be zero outside the interval [−M,M ] then

f(ω) =

∫ M

−M

f(t)e−iωt dt.

Assume f(ω) to be zero for all ω ∈ [c, d] then also all its derivatives vanish on theinterior of this interval. Let us consider a w0 ∈ (c, d) and differentiate n-times withrespect to ω

0 = f (n)(ω0) =

∫ M

−M

f(t)(−it)ne−iω0t dt.

Using the equality

f(ω) =

∫ M

−M

f(t)e−it(ω−ω0)e−itω0 dt.

and from the power series expansion of

e−it(ω−ω0) =∞∑

k=0

[−it(ω − ω0)]k

k!

we get

f(ω) =∞∑

k=0

(ω − ω0)k

k!

∫ M

−M

f(t)(−it)ke−itω0 dt = 0.

Hence f ≡ 0 ⇒ f ≡ 0 which contradicts the assumption.


Summary 1.11 Properties of the Fourier transform

Property Function Fourier transform

t→ f(t) ξ → f(ξ)

linearity αf(t) + βg(t) αf(ξ) + βg(ξ)

convolution (f ∗ g)(t) f(ξ)g(ξ)

translation f(t− τ) e−iτξf(ξ)

scaling f( ts) |s|f(sξ)

derivative f ′(t) iξf(ξ)

inverse f(t) 2πf−(ξ)

reflection 1 f−(t) f(ξ)

reflection 2 f−(t) (f)−(ξ)

Gaussian function e−t2√πe−

ξ2

4

characteristic function χ[−1,1] 2 sin ξξ

Dirac distribution δx(t) e−iξx

11+x2 πe−|ξ|

The Dirac distribution δx is a functional C∞ → R defined via δx(ϕ) = ϕ(x), ∀ϕ ∈C∞

0 (R). The function f− is called the reflection of f and is defined as f−(t) := f(−t).

1.2 Uncertainty Principle

Let us define the variances of f in the time and the frequency domains by

σ2t (x) =

1

‖f‖22

∫ ∞

−∞(t− x)2|f(t)|2 dt

and

σ2ω(ξ) =

1

2π‖f‖22

∫ ∞

−∞(ω − ξ)2|f(ω)|2 dω.

Then we have the following result:

Proposition 1.12 Heisenberg Uncertainty

a) σ2t σ

2ω ≥ 1

4,

b) σ2t σ

2ω = 1

4iff there exists (x, ξ, a, z) ∈ R2 × C2 such that f(t) = aeiξte−z(t−x)2.


Proof: The following proof requires the assumption that√tf(t) → 0 for t→ ∞. Indeed

the result is valid for f ∈ L2(R). Obviously for x = ξ = 0 we have

σ2t (0)σ2

ω(0) =1

2π‖f‖4

∫ ∞

−∞t2|f(t)|2 dt ·

∫ ∞

−∞ω2|f(ω)|2 dω,

and obtain from Plancherels formula and the Cauchy-Schwarz inequality

σt(0)2σω(0)2 =1

‖f‖4

∫ ∞

−∞|tf(t)|2 dt ·

∫ ∞

−∞|f ′(t)|2 dt

≥ 1

‖f‖4

(∫ ∞

−∞|tf ′(t)f(t)| dt

)2

≥ 1

‖f‖4

(∫ ∞

−∞

t

2|f ′(t)f(t) + f ′(t)f(t)| dt

)2

=1

4‖f‖4

(∫ ∞

−∞t(|f(t)|2)′ dt

)2

.

Integration by parts together with the assumption limt→∞

√tf(t) = 0 yields

σt(0)2σω(0)2 ≥ 1

4‖f‖4

(∫ ∞

−∞|f(t)|2 dt

)2

=1

4.

For x 6= 0 or ξ 6= 0 we substitute

z := (t− x) and η := (ω − ξ)

and by the same arguments we receive the same result.

The proof of part b) is left to the diligent reader.

This result is somehow disappointing since it guarantees that it is never possible to localizethe energy spreads of f in the time and frequency domains arbitrarily exact.

1.3 Linear Time-Invariant Operators (Filtering)

Definition 1.13 A linear operator L : C∞0 (R) → C(R) (resp. Lp(R)) is called time-

invariant if it commutes with the time translation operator Tτ defined by Tτ (f(·)) =f(· − τ), i.e.,

TτLf(t) = (Lf)(t− τ) = LTτf(t) = (Lf(· − τ))(t).

We denote the Dirac distribution located at τ ∈ R by δτ where

δτ (ϕ) =

∫ ∞

−∞δ(t− τ)ϕ(t) dt = ϕ(τ)

and write for the sake of simplicity

δτ = δ(· − τ) and ϕ(τ) = 〈δ(· − τ), ϕ〉.


Let L⋆ be the dual operator of L and h = L⋆δ0, then

(Lϕ)(τ) = 〈δ(· − τ), Lϕ〉 = 〈δ0, (Lϕ)(· + τ)〉= 〈δ0, (Lϕ(· + τ))(·)〉 = 〈L⋆δ0, ϕ(· + τ)〉= 〈(L⋆δ0)(· − τ), ϕ〉 = 〈h(· − τ), ϕ〉= (h ∗ ϕ)(τ).

Thus, any linear time-invariant operation may be rewritten as a convolution which is aremarkable property of LTI operators.

- -x ∈ C∞0 (R)

Output

h ∗ x

FilterInput

y ∈ C(R)

The function h is called the transfer function or impulse response of L. The followingequalities hold:

1. h ∗ ϕ = ϕ ∗ h,

2. ddt

(h ∗ ϕ)(t) =(

dhdt∗ ϕ)(t) =

(h ∗ dϕ

dt

)(t).

Definition 1.14 A convolution with h = L⋆δ0 is called a linear time-invariant filtering.Furthermore a filter h is called causal if supp h ⊆ R+ and stable if h ∈ L1(R).

Stability implies for every ϕ ∈ C0(R):

|(h ∗ ϕ)(t)| =

∣∣∣∣∫ ∞

−∞h(t− τ)ϕ(τ) dτ

∣∣∣∣ =

∣∣∣∣∫ ∞

−∞ϕ(t− τ)h(τ) dτ

∣∣∣∣

≤ supt∈R

|ϕ(t)|∫ ∞

−∞|h(τ)| dτ = ‖ϕ‖C0 · ‖h‖L1 .

⇒ ‖h ∗ ϕ‖L∞ ≤ ‖ϕ‖C0 · ‖h‖L1

In other words, stable filters do not amplify the input data. Causality is nothing but theindependence of the output data on input data from the future.

Example 1.15 Time averaging over an interval of length T .As a trivial example for LTI filter design consider

h(t) =1

Tχ[−T/2,T/2] ⇒ (h ∗ ϕ)(t) =

1

T

∫ t+T/2

t−T/2

ϕ(τ) dτ.


−10 −8 −6 −4 −2 0 2 4 6 8 10−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

f

h * f

Figure 1.3: Time averaging for f(t) = sin tt

and T = 8

Example 1.16 Gibbs oscillationLet us consider an ideal low pass filter with band width 2ξ described by the Fourier

transform of the transfer function,

hξ(ω) =

1, ω ∈ [−ξ, ξ]0, otherwise.

This is a more useful application of LTI filters, it is often desirable in sound processingto remove all frequencies which are greater than a certain treshold from the input signal.However, the obtained output data is not necessarily the sound you would like to havein the back of your car. The following result demonstrates that (Gibbs) oscillations arecreated by f ∗ hξ:

For any ξ > 0 and the staircase function

u(t) =

1, t ≥ 00, t < 0

one has

(u ∗ hξ)(t) =1

2+

1

π

∫ ξt

0

sin τ

τdτ. (1.8)

Proof: Application of Theorem 1.6 about the inverse Fourier transform yields

hξ(t) =1

2π

∫ ξ

−ξ

eitω dω =1

2π

1

iteitω∣∣∣ξ

ω=−ξ

=1

π

sin ξt

t,


u(t) (u ∗ h4ν)(t) (u ∗ h2ν)(t) (u ∗ hν)(t)

0 500 1000 1500−0.2

−0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0 500 1000 1500−0.1

0

0.1

0.2

0.3

0.4

0.5

0 500 1000 1500−0.1

0

0.1

0.2

0.3

0.4

0.5

0 500 1000 1500−0.1

0

0.1

0.2

0.3

0.4

0.5

Figure 1.4: Gibbs oscillations created by ideal low-pass filters with cut-off frequencies thatdecrease from left to right.

and hence

(u ∗ hξ)(t) =1

π

∫ ∞

0

sin ξ(t− τ)

t− τdτ

x:=ξ(t−τ)=

1

π

∫ ξt

−∞

sin x

xdx =

1

π

∫ 0

−∞

sin x

xdx+

1

π

∫ ξt

0

sin x

xdx

which proves (1.8) .

Example 1.17 Passive Electronic CircuitLet the input voltage f(t) and output voltage g(t) satisfy the ordinary differential

equationK∑

k=0

akf(k)(t) =

M∑

k=0

bkg(k)(t). (1.9)

ThenK∑

k=0

ak(iξ)kf(ξ) =

M∑

k=0

bk(iξ)kg(ξ).

which gives

g(ξ) =

∑Kk=0 ak(iξ)

k

∑Mk=0 bk(iξ)

kf(ξ).

Therefore the impulse response of (1.9) satisfies

h(ξ) =

∑Kk=0 ak(iξ)

k

∑Mk=0 bk(iξ)

k.

The roots of the polynomials∑K

k=0 ak(iξ)k and

∑Kk=0 bk(iξ)

k are called zeros and polesof the linear system (1.9) which are significant values in linear filter design. For moredetails see e.g. [4].

2 DISCRETE-TIME SIGNAL PROCESSING 15

2 Discrete-Time Signal Processing

Digital signal processing plays an important role in many areas like speech processing, televi-sion, tape recording and all other types of information manipulation. Whether sound record-ings or images, most discrete signals are obtained by sampling an analog signal so that wecan expect some parallelism to section 1.3. In this section we will study conditions for recon-structing an analog signal from a uniform sampling. Once more, the Fourier transformation isunavoidable because the eigenvectors of discrete linear time-invariant operators are sinusoidalwaves. The Fourier transform may be discretized for signals of finite size and implementedwith fast computational algorithms.

2.1 Sampling

Let us consider the ’Dirac comb’, for T > 0

C :=

∞∑

k=−∞δkT

which is defined by

C(ϕ) = 〈C, ϕ〉 =∞∑

k=−∞ϕ(kT ), ∀ϕ ∈ C∞

0 (R).

We write δkT := δ(t−kT ) keeping in mind that δ(t−kT ) is not a function value at t−kT .The Fourier transform of C is defined by

C(ϕ) := C(ϕ) =

∞∑

k=−∞ϕ(kT )

which is equal to

C(ϕ) =∞∑

k=−∞

∫ ∞

−∞e−ikT tϕ(t) dt =

⟨ ∞∑

k=−∞e−ikT ·, ϕ(·)

⟩.

Therefore we can write

C(ξ) =∞∑

k=−∞e−ikTξ, ξ ∈ R.

Theorem 2.1 Poisson-Summation FormulasIn the sense of distributions the following two equations hold for ϕ ∈ C∞

0 (R):∞∑

k=−∞ϕ(kT ) =

∞∑

k=−∞

∫ ∞

−∞e−ikTξϕ(ξ) dξ =

∫ ∞

−∞

∞∑

k=−∞e−ikTξϕ(ξ) dξ

=2π

T

∞∑

k=−∞ϕ

(2πk

T

), (2.1)

and∞∑

k=−∞e−ikTξ =

2π

T

∞∑

k=−∞δ 2πk

T(ξ). (2.2)


−3 −2 −1 0 1 2 3−1

−0.5

0

0.5

1

−3 −2 −1 0 1 2 3−1

−0.5

0

0.5

1

Figure 2.1: Sampling f(t) = sin t with the Dirac comb.

Proof: It is sufficient to consider ϕ ∈ C∞0 (R) with support in the interval ϕ ⊆

[− π

T, π

T

].

In this case the Poisson formula (2.1) reduces to

limN→∞

∫ πT

− πT

N∑

k=−N

e−ikTξϕ(ξ) dξ =2π

Tϕ(0). (2.3)

The geometric series gives for ξ 6= 0:

N∑

k=−N

e−ikTξ = e−iT ξN

2N∑

k=0

(e−iT ξ)k =e−i(N+1)Tξ − eiNTξ

e−iT ξ − 1

=e−i(N+ 1

2)Tξ − ei(N+ 1

2)Tξ

e−i T2

ξ − ei T2

ξ=

sin(N + 1

2

)Tξ

sin T2ξ

.

Note that this formula is also consistent for the case ξ = 0, since

limξ→0

sin(N + 1

2

)Tξ

sin T2ξ

= 2N + 1 =

N∑

k=−N

1.

We insert this result into the left hand side of (2.3) and obtain

limN→∞

∫ πT

− πT

sin(N + 1

2

)Tξ

Tξ· Tξ ϕ(ξ)

sin T2ξ

dξ = limN→∞

∫ ∞

−∞2sin(N + 1

2

)Tξ

Tξ· ψ(ξ) dξ


where

ψ(ξ) =

ϕ(ξ) Tξ

2 sin T2

ξ, ξ ∈

[− π

T, π

T

]

0, otherwise.

The latter integral may be rewritten as

(2N + 1)

⟨sin(N + 1

2

)T ·(

N + 12

)T · , ψ

⟩=π

T

⟨χ[−(N+ 1

2)T,(N+ 12)T ], ψ

⟩

where we stressed Plancherels formula of Theorem 1.8 and a scaled version of Example1.7. ψ denotes the Fourier preimage of ψ. Altogether we get

limN→∞

∫ ∞

−∞2sin(N + 1

2

)Tξ

Tξ· ψ(ξ) dξ =

2π

Tlim

N→∞

∫ (N+ 12)T

−(N+ 12)T

ψ(t) dt

=2π

Tψ(0) =

2π

Tϕ(0).

A discrete signal can be represented as a sum of Dirac distributions. For any sample valuef(kT ) at the sample point t = kT, k ∈ Z, T > 0, we associate the distribution f(kT )δkT . Auniform sampling corresponds to the distribution

fd =∞∑

k=−∞f(kT )δkT .

The Fourier transform of fd is represented by the Fourier series

fd(ξ) =

∞∑

k=−∞f(kT )e−ikTξ. (2.4)

Proposition 2.2 The Fourier transform of the distribution fd satisfies

fd(ξ) =1

T

∞∑

k=−∞f

(ξ − 2kπ

T

). (2.5)

Proof: We have f(kT )δkT (ϕ) = f(t)δkT (ϕ) and so that we can write

fd(t) = f(t) ·∞∑

k=−∞δkT = f(t) · C.

Computing the Fourier transform yields

fd(ξ) =1

2π(f ∗ C)(ξ).


Inserting Poissons formula (2.2)

C(ξ) =2π

T

∞∑

k=−∞δ 2πk

T(ξ)

proves that

1

2π(f ∗ C)(ξ) =

1

T

∫ ∞

−∞f(ξ − ω)

∞∑

k=−∞δ

(ω − 2πk

T

)dω =

1

T

∞∑

k=−∞f

(ξ − 2πk

T

).

The following result was primarily proved by Whittaker (1935) but rediscovered by Shan-non (1949).

Theorem 2.3 Shannon, WhittakerIf supp f ⊆

[− π

T, π

T

]and hT (t) = T

πtsin πt

T, then

f(t) =

∞∑

k=−∞f(kT ) · hT (t− kT ) (2.6)

holds.

Proof: From (2.5) in Proposition 2.2 we have

fd(ξ) =1

T

∞∑

k=−∞f

(ξ − 2kπ

T

).

Since the frequency band is bounded, the supports of f(· − 2kπT

), k ∈ Z have no overlapand thus

fd(ξ) =f(ξ)

T, for |ξ| ≤ π

T.

The Fourier transform of hT is Tχ[−π/T,π/T ]. Therefore f(ξ) = fd(ξ)·hT (ξ) and by meansof convolution Theorem 1.5,

f(t) = (hT ∗ fd)(t) =

(hT ∗

∞∑

k=−∞f(kT )δkT

)(t) =

∞∑

k=−∞f(kT ) · hT (t− kT ).

Equality (2.6) allows us a decomposition of functions with bounded frequency band as a sumof sinusoidal waves.


2.2 Aliasing

The length of the sampling interval is restricted - by storage and efficiency requirements.Additionally the condition supp f ⊆

[− π

T, π

T

]is not necessarily fulfilled. This causes an

additional error which can be severe. Have you ever wondered why car tires are spinningbackwards in ancient action movies ? Such effects are called aliasing.

Example 2.4 Let us consider a high frequency signal

f(t) = cosω0t =1

2(eiω0t + e−iω0t)

whose Fourier transform is

f(ξ) = π(δ(ξ − ω0) + δ(ξ + ω0)).

If 2πT> ω0 >

πT

we get

hT (ξ)fd(ξ) = πχ[− πT

, πT

]

∞∑

k=−∞

[δ

(ξ − ω0 −

2kπ

T

)+ δ

(ξ + ω0 −

2kπ

T

)]

= π

[δ

(ξ − ω0 +

2π

T

)+ δ

(ξ + ω0 −

2π

T

)].

Applying the inverse Fourier transform on both sides yields

(fd ∗ hT )(t) = cos t

(ω0 −

2π

T

).

Thus, the high frequency signal is reduced to a low frequency signal with a frequencyω0 − 2π

T∈[− π

T, π

T

].

As a more practical example you might try to compress a picture with sharp edges andlots of tiny details via JPEG.

The effect of aliasing can be minimized through approximating f by a suitable band limitedsignal fb with supp fb ⊆

[− π

T, π

T

]. The error can be measured using Theorem 1.8:

‖f − fb‖2L2(R) =

1

2π‖f − fb‖2

L2(R) =1

2π

∫ ∞

−∞|f(ξ) − fb(ξ)|2 dξ

=1

2π

∫ π/T

−π/T

|f(ξ) − fb(ξ)|2 dξ +

∫

|ξ|>π/T

|f(ξ) − fb(ξ)|2 dξ.

This is minimized iff f(ξ) = fb(ξ) for |ξ| ≤ πT. Setting

fb(ξ) = χ[− πT

, πT ]f(ξ)

is obviously a good way for doing this. Therefore sampling of the filtered signal fb(t) =(f ∗ hT )(t) will give best results.

In practice an analog to digital converter is composed of a low pass filter that limits to[− π

T, π

T

]and uniform sampling at intervals T .


Proposition 2.5 Let hT (t) = Tπt

sin πtT, then

hT (t− kT ) : k ∈ Z (2.7)

forms an orthogonal family of functions in UT ⊂ L2(R). UT is the space of all functionsin L2(R) whose Fourier transform is supported in the interval

[− π

T, π

T

]. Let f ∈ UT be

continuous, then it may be represented as

f(t) =

∞∑

k=−∞f(kT ) · hT (t− kT ) (2.8)

and, conversely, the basis coefficients are given by

f(kT ) =1

T〈f, h(· − kT )〉. (2.9)

Proof: Remind that hT = Tχ[− πT

, πT ]. Parsevals formula (Theorem 1.8) yields

〈hT (· − kT ), hT (· − nT )〉 =1

2π〈e−ikT ·hT (·), e−inT ·hT (·)〉

=T 2

2π

∫ ∞

−∞χ[− π

T, πT ](ξ)e

−i(k−n)Tξ dξ

=T 2

2π

∫ π/T

−π/T

e−i(k−n)Tξ dξ =

0 for k 6= nT for k = n

.

Clearly each hT (· − kT ) is a member of UT .

Shannons sampling Theorem 2.3 gives the first representation (2.8) and, again withTheorem 1.8, we compute

〈f, hT (· − kT )〉 =1

2π

∫ ∞

−∞f(ξ)hT (ξ)eikTξ dξ

=T

2π

∫ π/T

−π/T

f(ξ)eikTξ dξ

(since f ∈ UT ) ⇒ =T

2π

∫ ∞

−∞f(ξ)eikTξ dξ = Tf(kT ),

which proves (2.9).

2.3 Discrete Time-Invariant Filters

To simplify notation, we now assume T = 1 and set

f [k] := f(kT ), k ∈ Z

as the sample values. Now, f : Z → C where we can assume that f ∈ l2(Z). A discretelinear time-invariant operator L satisfies

(Lf [· − n])[k] = Lf [k − n]. (2.10)


The discrete Dirac distribution δ[k] is defined by δ[k] := δk. A discrete signal can berepresented as

f [k] =

∞∑

n=−∞f [n]δ[k − n].

Any time-invariant linear operator Lmay be expressed by its action on the Dirac distributions.Let Lδ[k] =: g[k], the so called discrete impulse response, then

Lf [k] =

∞∑

n=−∞f [n]Lδ[k − n] =

∞∑

n=−∞f [n]g[k − n].

If g has finite support then the above term reduces to a finite sum, such operators are calledfinite impulse response Filters (FIR).

Since

|Lf [n]| ≤ supk∈Z

|f [k]| ·∞∑

k=−∞|g[k]|

we call L analogous to Definition 1.14 stable if g ∈ l1(Z).

Transfer functions

We consider

Leiωn =

∞∑

k=−∞eiω(n−k)g[k] = eiωn

∞∑

k=−∞e−iωkg[k] = g(ω)eiωn,

which shows that L has eigenvectors eiωn with eigenvalues g(ω). The function g(ω) =∑∞k=−∞ e−iωkg[k] is called the discrete filter transfer function.

Example 2.6 Uniform discrete averageThis example corresponds to the continuous version of Example 1.15.

Lf [n] =1

2N + 1

N+n∑

k=n−N

f [k] = (g ∗ f)[n],

where

g[k] =1

2N + 1

1 k = −N, . . . , N,0 otherwise,

which is obviously a stable filter. The associated transfer function is

g(ω) =1

2N + 1

N∑

k=−N

e−iωk =1

2N + 1

sin(N + 12)ω

sin ω2

(see also proof of Theorem 2.1).


2.4 Fourier Series

The Fourier transfrom of the discrete signal

f(t) =

∞∑

k=−∞f [k]δ(t− k)

is given by

f(ω) =

∞∑

k=−∞f [k]e−iωk

where the latter one is a periodic function of period 2π. Therefore we consider the spaceL2[−π, π] of (periodic) square integrable functions where

〈f, g〉 =1

2π

∫ π

−π

f(x)g(x) dx and ‖f‖ =√〈f, f〉

are the scalar product and norm, respectively. L2[−π, π] is again a Hilbert space.

Theorem 2.7 The family of functions

e−ikω : k ∈ Z

forms an orthonormal basis of L2[−π, π].

Proof: Have a look in your favourite Analysis book.

The above result implies that any function f ∈ L2[−π, π] can be approximated by a Fourierseries

f(ω) =∞∑

k=−∞f [k]e−iωk

with Fourier coefficients

f [k] = 〈f(·), e−ik·〉 =1

2π

∫ π

−π

f(ω)eikω dω.

Analogous to Theorem 1.8 we have the Plancherel identity:

∞∑

k=−∞|f [k]|2 = ‖f‖2

l2 = ‖f‖2 =1

2π

∫ π

−π

|f(ω)|2 dω.


2.5 Discrete Fourier Transform (DFT)

In practice we obtain only a finite number of sample points, say k = 0, . . . , N − 1. To get a”complete f” we consider N -periodic signals f with

f [k] = f [k mod N ].

The circular convolution is defined by

(f ⊛ g)[n] :=N−1∑

k=0

f [k]g[n− k]

for two periodic signals f and g. Let ~f,~g ∈ RN then we can rewrite the convolution as amatrix-vector product:

(~f ⊛ ~g)[n] = G~f [n], (2.11)

where

G =

g[0] g[N − 1] . . . . . . g[1]

g[1] g[0] g[N − 1]...

... g[1]. . .

. . ....

.... . .

. . . g[N − 1]g[N − 1] . . . . . . g[1] g[0]

. (2.12)

G is a structured matrix, a so called circulant which is a special form of a Toeplitz matrix.Applying G on the vector

~v =[e2πi kn

N

]N−1

k=0

for some n ∈ 0, . . . , N − 1 yields

G~v =

[N−1∑

k=0

e2πi knN g[m− k]

]N−1

m=0

=

[N−1∑

l=0

e2πi n(m−l)N g[l]

]N−1

m=0

= g[n][e2πi mn

N

]N−1

m=0,

where

g[n] =N−1∑

l=0

e2πi lnN g[l]

Hence, any vector [e2πi knN ]N−1

k=0 is an eigenvector of G associated with the eigenvalue g[n].

Theorem 2.8 The family ek[n] = e2πi knN , k = 0, . . . , N − 1 forms an orthogonal basis in the

space of signals with period N .

Proof: Using the hermitian scalar product for CN we get

〈ek, el〉 =N−1∑

n=0

e2πi knN e−2πi ln

N =N−1∑

n=0

e2πi (k−l)nN .


If k = l then the last term equals to∑N−1

n=0 1 = N . In the case of k 6= l we get 〈ek, el〉 = 0,because using geometric series gives

N−1∑

n=0

e2πi (k−l)nN =

e2πi (k−l)NN − 1

e2πi k−lN − 1

= 0.

Therefore

〈ek, el〉 =

N if k = l0 otherwise

= Nδk,l.

This implies that any signal f of period N can be decomposed with respect to this basis:

f =

N−1∑

k=0

〈f, ek〉ek

‖ek‖2.

Definition 2.9 The discrete Fourier transform (DFT) of a vector f ∈ Cn is defined by

f [k] = 〈f, ek〉 =N−1∑

n=0

f [n]e−2πi knN .

As in the convolution case we can rewrite this formula in matrix language:

f = FN · f, (2.13)

where

FN =[fN

pk

]N−1

p,k=0=[e2πi pk

N

]N−1

p,k=0.

Since ‖ek‖2 = N and (FN)−1 = 1N

(FN)H we obtain the inverse DFT by

f [n] =1

N

N−1∑

k=0

f [k]e2πi knN

and the Plancherel formula

‖f‖2 =

N−1∑

n=0

|f [n]|2 =1

N

N−1∑

n=0

|f [k]|2 =1

N‖f‖2.

Remark 2.10

1. The discrete Fourier transform may be considered as a solution of the followinginterpolation problem:For a given function f ∈ C0[0, 2π] find n coefficients αk ∈ C such that for alln = 0, . . . , N − 1:

f

(2πn

N

)=

N−1∑

k=0

αke2πi kn

N .


The above conclusions tell us that

αk =1

Nf [k] =

1

N

N−1∑

k=0

f

(2πn

N

)e−2πi kn

N .

2. If ak denotes the Fourier coefficient

ak =

∫ 2π

0

f(t)e−2πikt dt,

then αk is an approximation of ak using the trapezoidal rule for computing theintegral.

2.6 Fast Fourier Transform (FFT)

The computation of the vector [f [k]]N−1k=0 by a naive matrix-vector product implementation

requires 2n2 flops since the matrix

FN =[e2πi pk

N

]N−1

p,k=0

is fully populated. This complexity can be drastically reduced by the following observation:Let N be a power of 2. For even coefficients f [2k], k = 0, . . . , N/2 − 1, we observe that

e−2πi 2knN = e−2πi 2k(n+N/2)

N .

Therefore the computation of these coefficients is reduced to

f [2k] =

N/2−1∑

n=0

(f [n] + f [n+N/2])e−2πi knN/2 = FN/2fe, (2.14)

wherefe = [f [n] + f [n+N/2]]N/2−1

n=0 .

For odd indices [2k + 1] we obtain by analogous observations

f [2k + 1] =

N/2−1∑

n=0

e−2πin

N (f [n] − f [n+N/2])e−2πi knN/2 = FN/2fo, (2.15)

where

fo =[e

−2πinN (f [n] − f [n+N/2])

]N/2−1

n=0.

To compute FN/2fe (resp. FN/2fo) we proceed by recursion, applying the same ideas as above.We end up with the following algorithm of complexity O(n logn).

Algorithm 2.11 Fast Fourier Transform

3 TIME MEETS FREQUENCY 26

Purpose Given a vector f of length N = 2t, this routine computes FNf where FN =

[e2πi pkN ]N−1

p,k=0.

function y = FFT (f,N)if n ≤ 1 then

y = felse

m = N/2; ω = e−2πi/N

ye = FFT (f [0 :2 :N ], m); yo = FFT (f [1 :2 :N ], m)d = [1, ω, . . . , ωm−1]T

z = d. ∗ yo

y =

[ye + zye − z

]

end

Remark 2.12 Algorithm 2.11 is one of the standard routines which is included in almost anynumerical package. A brief overview of the related functions implemented in Matlab:

fft - Discrete Fourier transform

fft2 - Two-dimensional discrete Fourier Transform

ifft - Inverse discrete Fourier transform

ifft2 - Two-dimensional inverse discrete Fourier transform

conv - Convolution and polynomial multiplication

conv2 - Two dimensional convolution

deconv - Deconvolution and polynomial division

filter - One-dimensional digital filter

Further recommended toolboxes are Mathworks signal processing toolbox and WaveLabwhich is available via

http://www-stat.stanford.edu/~wavelab/

3 Time meets Frequency

We consider so called time / frequency atoms with small energy spread in the time/frequencydomains. They form a family of functions

Φγ , γ ∈ Γ,

where γ is a multi index parameter. We suppose Φγ ∈ L2(R) and normalize such that‖Φγ‖L2 = 1.

We define a time/frequency operator T : Γ → C/R for functions f ∈ L2(R) by

Tf(γ) := 〈f,Φr〉 =

∫ ∞

−∞f(t)Φγ(t) dt, (3.1)


and reformulate with the help of Parsevals identity (Theorem 1.8)

Tf(γ) =1

2π

∫ ∞

−∞f(ξ)Φ∗

γ(ξ) dξ. (3.2)

If Φγ is nearly zero besides a small interval then (3.1) tell us that Tf(γ) depends only on thevalues of f in this interval. Similarly, if Φγ has an approximately small frequency support

then from (3.2) we know that Tf(γ) reveals the properties of f on a small interval.

Example 3.1

1. ”Windowed Fourier Transform Atom”

Φγ = gx,ξ(t) := g(t− x)eiξt,

where g(t) = g(−t) ∈ R and ξ, x ∈ R.

2. ”Wavelet Atom”

Φγ(t) = Ψa,x(t) :=1√aΨ

(t− x

a

)

Ψ is a so called wavelet with zero mean, i.e.∫∞−∞ Ψ(x) dx = 0. a > 0 is the

scaling parameter.

Both are examples for localization in the time variable.

3.1 Heisenberg Boxes

We interpret the function |Φγ(t)|2 as a probality density function (‖Φγ‖ = 1). The center(resp. expectation) xγ of |Φγ(t)|2 is defined by

xγ =

∫ ∞

−∞t|Φγ(t)|2 dt.

The variance σt(γ) expresses the scattering from the center:

σ2t (γ) =

∫ ∞

−∞(t− xγ)

2|Φγ(t)|2 dt.

Since ‖Φr‖2L2 = 2π we get analog results for the Fourier transform. The frequency center is

ωγ =1

2π

∫ ∞

−∞ξ|Φ(ξ)|2 dξ,

and the variance with respect to frequency

σ2ξ (γ) =

1

2π

∫ ∞

−∞(ξ − ωγ)

2|Φ(ξ)|2 dξ.


Frequency

Time

ξ

γ

u v

σt

σω

σt

σω

gu,ξ(t) g

v,γ(t)

F gv,γ(ω)

F gu,ξ(ω)

Figure 3.1: Heisenberg Box of two windowed Fourier atoms gu,ξ and gv,γ.

Now, the time/frequency localization may be expressed by an rectangle in the t-ξ-time-freqency plane:

[xγ −

σt(γ)

2, xγ +

σt(γ)

2

]×[ωγ −

σξ(γ)

2, ωγ +

σξ(γ)

2

]

which is the so called ”Heisenberg Box”. Our desired property is to have Φγ well localizedin time and frequency. So, optimally σt and σξ should be very small values. But we alreadyknow from Heisenbergs uncertainty (see Proposition 1.12) that

σt(γ) · σξ(γ) ≥1

2.

3.2 Windowed Fourier Transform

We consider as a special time atom the ”time window” g : R → R as an even function whichis modulated by a frequency ξ:

gx,ξ(t) = g(t− x)eitξ, ‖g‖ = 1.

Consequently we have ‖gx,ξ‖ = 1 and the center of gx,ξ is

∫ ∞

−∞tg2(t− x) dt = x


with a variance of

σ2t =

∫ ∞

−∞(t− x)2|g(t− x)|2 dt =

∫ ∞

−∞τ 2g2(τ) dτ =: σ2

which is independent of x and ξ. For gx,ξ(ω) = e−ix(ω−ξ)g(ω − ξ) we get the variance

σ2ξ =

∫ ∞

−∞(ω − ξ)2|gx,ξ(ω − ξ)|2 dω =

∫ ∞

−∞τ 2|g(τ)|2 dτ =: σ2

independent of x and ξ. The center of gx,ξ is ξ.

Definition 3.2 The Windowed-Fourier-Transform (WFT) of f ∈ L2(R) is defined by

Sf(x, ξ) = 〈f, gω,ξ〉 =

∫ ∞

−∞f(t)e−iξtg(t− x) dt, (3.3)

while the energy density is

PSf(x, ξ) = |Sf(x, ξ)|2.

Example 3.3

1. Sinusoidal Wave

f(t) = eiω0t ⇒ f(ξ) = 2πδ(ξ − ω0)

WFT ⇒ Sf(x, ξ) = e−ix(ξ−ω0)g(ξ − ω0)

2. Dirac impulsf(t) = δ(t− x0) ⇒ Sf(x, ξ) = eix0ξ0g(x0 − x)

Theorem 3.4 Reconstruction formulaLet f ∈ L2(R). Then

f(t) =1

2π

∫ ∞

−∞

∫ ∞

−∞Sf(x, ξ)g(t− x)eitξ dx dξ, (3.4)

‖f‖2 =1

2π

∫ ∞

−∞

∫ ∞

−∞|Sf(x, ξ)|2 dx dξ. (3.5)

Proof:

Sf(x, ξ) =

∫ ∞

−∞f(t)g(t− x)eiξ(x−t)e−iξx dt = e−iξx

∫ ∞

−∞f(t)gξ(x− t) dt,

where gξ(t) = g(t)eiξt. Hence

Sf(x, ξ) = e−iξx(f ∗ gξ)(x) =: fξ(x). (3.6)


Using convolution Theorem 1.5 yields

fξ(ω) = f(ω + ξ)gξ(ω + ξ) = f(ω + ξ)g(ω). (3.7)

Recall that the Fourier transform of g(· − t) is g(ω)e−iωt. Applying Parsevals formula(Theorem 1.8) on the equality (3.4) yields

I =1

2π

∫ ∞

−∞

(∫ ∞

−∞Sf(x, ξ)g(t− x) dx

)eitξ dξ

(3.6) ⇒ =1

2π

∫ ∞

−∞

1

2π

∫ ∞

−∞fξ(ω)g(ω)eit(ξ+ω) dω dξ

(3.7) ⇒ =1

4π2

∫ ∞

−∞

∫ ∞

−∞f(ω + ξ)|g(ω)|2eit(ξ+ω) dω dξ

=

(1

2π

∫ ∞

−∞|g(ω)|2 dω

)(1

2π

∫ ∞

−∞f(ω + ξ)eit(ξ+ω) dξ

)

= f(t)1

2π

∫ ∞

−∞|g(ω)|2 dω =

1

2πf(t)‖g‖2 = f(t)‖g‖2 = f(t).

To prove (3.5) we use the same technique as above:

1

2π

∫ ∞

−∞

∫ ∞

−∞|Sf(x, ξ)|2 dx dξ =

1

4π2

∫ ∞

−∞

∫ ∞

−∞|fξ(ω)|2 dω dξ

=1

4π2

∫ ∞

−∞

∫ ∞

−∞|f(ξ + ω)|2|g(ω)|2 dξ dω

=1

2π‖f‖2 1

2π‖g‖2 = ‖f‖2.

The reconstruction formula may be rewritten as

f(t) =1

2π

∫ ∞

−∞

∫ ∞

−∞〈f, gx,ξ〉gx,ξ dξ dx.

Theorem 3.5 Let Φ(·, ·) ∈ L2(R2). Then there exists a function f ∈ L2(R) with Φ(x, ξ) =Sf(x, ξ) if and only if

Φ(x0, ξ0) =1

2π

∫ ∞

−∞

∫ ∞

−∞Φ(x, ξ)K(x, x0, ξ, ξ0) dx dξ, (3.8)

where

K(x, x0, ξ, ξ0) = 〈gx,ξ, gx0,ξ0〉 =

∫ ∞

−∞g(t− x)eiξtg(t− x0)e

−iξ0t dt

holds for all (t0, ξ0) ∈ R2).


Proof: Suppose there exists an f ∈ L2(R) with Φ(x, ξ) = Sf(x, ξ), then

Φ(x0, ξ0) = Sf(x0, ξ0) =

∫ ∞

−∞f(t)g(t− x0)e

−iξ0t dt

(Theorem 3.4) ⇒ =1

2π

∫ ∞

−∞

∫ ∞

−∞

∫ ∞

−∞g(t− x0)e

−iξ0tΦ(x, ξ)g(t− x)eiξt dx dξ dt.

Using Fubinis theorem proves (3.8).

Conversely, define

f(t) =1

2π

∫ ∞

−∞

∫ ∞

−∞g(x− t)eiξtΦ(x, ξ) dx dξ,

and Sf(x0, ξ0) = Φ(x0, ξ0) results from the same considerations as above.

3.3 Wavelet Transform

Definition 3.6 Ψ ∈ L2(R) is called a wavelet if ‖Ψ‖ = 1 and∫∞−∞ Ψ(x) dx = 0. We regard

Ψ to be centered if the expectation∫∞−∞ t|Ψ(t)|2 dt is finite. For x ∈ R and a > 0 a

shifted and scaled wavelet version is defined by

Ψa,x(t) :=1√aΨ

(t− x

a

). (3.9)

Definition 3.7 The continuous wavelet transform (WT) of f ∈ L2(R) at the time xand scale a > 0 is defined by

Wf(x, a) =

∫ ∞

−∞f(t)

1√aΨ∗(t− x

a

)dt = 〈f,Ψa,x〉. (3.10)

We can reformulate (WT) as a convolution

Wf(x, a) =(f ∗ Ψa

)(x), (3.11)

where

Ψa(t) =1√aΨ∗(− t

a

).

SinceWf(·, a)(ξ) = f(ξ) ˆΨa(ξ) =

√af(ξ)Ψ∗(aξ) (3.12)

and Ψ(0) =∫∞−∞ Ψ(x) dx = 0, the (WT) acts as a band filter with scaling a.


3.3.1 Real Wavelets

A real wavelet transform is complete and satsifies an energy conservation, as long as a weakcondition is satisfied, as the following theorem shows.

Theorem 3.8 ’Calderon, Grossmann, Morlet’Let Ψ ∈ L2(R) be real-valued with

∫∞−∞ Ψ(x) dx = 0 and

CΨ =

∫ ∞

0

|Ψ(ω)|2ω

dω <∞ (3.13)

which is the Wavelet admissibility condition. Then we have

f(t) =1

CΨ

∫ ∞

0

∫ ∞

−∞

1√aWf(x, a)Ψ

(x− t

a

)dx

da

a2, (3.14)

‖f‖2 =1

CΨ

∫ ∞

0

∫ ∞

−∞|Wf(x, ξ)|2 dx

da

a2. (3.15)

Proof: We use the convolution formulation (3.11), Wf(x, a) = (f ∗ Ψa)(x). Let

b(t) =1

CΨ

∫ ∞

0

(Wf(·, a) ∗ Ψa)(t)da

a2,

then with (3.12) and Fubinis theorem

b(ξ) =1

CΨ

∫ ∞

0

f(ξ)√aΨ∗(aξ)

√aΨ(aξ)

da

a2

(τ = aξ) ⇒ =f(ξ)

CΨ

∫ ∞

0

|Ψ(τ)|2 dτ

τ= f(ξ).

Hence b(t) = f(t), ∀t ∈ R and (3.14) is proven. Formula (3.15) results from similarconsiderations as they were outlined in the proof of (3.5) in Theorem 3.4.

Example 3.9 The so called Mexican hat wavelet (see also Figure 3.2) is defined as

Ψ(t) =2

π14

√3σ

(t2

σ2− 1

)e−

t2

2σ2 ∈ R,

and its Fourier transform is given by

Ψ(ξ) = −√

8σ52π

14

√3

ξ2e−ξ2σ2

2 ∈ R.

For continuous functions Ψ the condition Ψ(0) = 0 or equivalently∫∞−∞ Ψ(t) dt = 0 is neces-

sary for the Wavelet admissibility condition. In other words, the first moment of Ψ is zero.If furthermore Ψ ∈ C1(R) ∩ L1(R) holds, then the admissibility is guaranteed. The propertyΨ ∈ C1(R) is satisfied e.g. if

∫ ∞

−∞(1 + |x|)|Ψ(x)| dx <∞.


Reproducing Kernels

Applying the Wavelet transformation on the representation formula (3.14) yields

Wf(x, y) =

∫ ∞

−∞

(1

CΨ

∫ ∞

0

∫ ∞

−∞

Wf(u, s)√s

Ψ

(t− u

s

)du

ds

s2

)1√yΨ∗(t− x

y

)dt

=

∫ ∞

0

∫ ∞

−∞

Wf(u, s)

CΨ

(∫ ∞

−∞

1√sy

Ψ

(t− u

s

)Ψ∗(t− x

y

)dt

)du

ds

s2

=

∫ ∞

0

∫ ∞

−∞

1

CΨ√sy

⟨Ψ

( · − u

s

),Ψ

( · − x

y

)⟩Wf(u, s) du

ds

s2.

Setting

K(x, u; y, s) =1√sy

⟨Ψ

( · − u

s

),Ψ

( · − x

y

)⟩

we obtain the reproducing kernel equation

Wf(x, y) =1

CΨ

∫ ∞

0

∫ ∞

−∞K(x, u; y, s)Wf(u, s) du

ds

s2. (3.16)

Note that K(x, u; y, s) can be considered as the wavelet analogon of the WFT kernel (seeTheorem 3.5). The modulus of the reproducing kernel K(x, u; y, s) is a measure for thecorrelation between two wavelets Ψx,y and Ψu,s. The diligent reader may verify that anyfunction Φ(u, s) is the wavelet transform of some f ∈ L2(R) if and only if it satisfies equation(3.16).

Scaling Functions

When Wf(u, s) is only known for s < s0, what can we say about the contained information ?Let us introduce the scaling function ϕ ∈ L2(R) by the modulus of its Fourier transform

|ϕ(ω)|2 =

∫ ∞

1

|Ψ(sω)|2 ds

s=

∫ ∞

ω

|Ψ(u)|2 du

u.

The complex phase of ϕ can be abritrarily chosen. It can be shown that

‖ϕ‖ = 1 and limω→0

|ϕ(ω)|2 = CΨ.

We denote ϕs(t) = 1√sϕ(

ts

)and ϕs(t) = ϕs(−t) so that for f ∈ L2(R) the low-frequency

approximation of f at scale s is

Lf(u, s) =⟨f,

1√sϕ

( · − u

s

)⟩= (f ∗ ϕs)(u).

With a minor modification of the proof of the reconstruction formula in Theorem 3.4 one canshow that

f(t) =1

CΨ

∫ s0

0

(W (·, s) ∗ Ψs)(t)ds

s2+

1

CΨs0(Lf(·, s0) ∗ ϕs0)(t). (3.17)

Example 3.10 If Ψ is again the Mexican hat wavelet from Example 3.9, then

ϕ(ω) =2σ

32π

14

√3

√ω2 +

1

σ2e−

σ2ω2

2 .


−5 0 5

0

0.2

0.4

0.6

0.8

−5 0 5

0

0.5

1

1.5

Figure 3.2: Scaling function associated with a Mexican hat wavelet and its Fourier transformfor σ = 1.

3.3.2 Analytic Wavelets

Definition 3.11 We call a signal fa ∈ L2(R) analytic if its Fourier transform satisfiesfa(ω) = 0 for all ω < 0.

An analytic function fa is in general complex but completely characterized by its real partf(t) = Re fa(t). The Fourier transform of f is given by

f(ω) =fa(ω) + fa(−ω)

2(3.18)

and vice versa

fa(ω) =

2f(ω) for ω ≥ 0,

0 for ω < 0.(3.19)

Example 3.12 The Fourier transform of

f(t) = a cos(ω0t+ ϕ) =a

2

(ei(ω0t+ϕ) + e−i(ω0t+ϕ)

)

isf = π

(eiϕδ(ω − ω0) + e−iϕδ(ω + ω0)

).

Its analytic part satsifies fa(ω) = 2πaeiϕδ(ω − ω0) and hence

fa(t) = aei(ω0t+ϕ).

Definition 3.13 Let Ψ be an analytic Wavelet, then the analytic Wavelet transform isdefined by

Wf(u, s) = 〈f,Ψs,u〉 =

∫ ∞

−∞f(t)

1√sΨ∗(t− u

s

)dt.


Suppose Ψ to be centered at 0, i.e.∫∞−∞ t|Ψ(t)|2 dt = 0, and define

σ2t :=

∫ ∞

−∞t2|Ψ(t)|2 dt.

Then Ψs,u is centered at u ∈ R and

∫ ∞

−∞(t− u)2

∣∣∣∣1√sΨ

(t− u

s

)∣∣∣∣2

dt = s2σ2t =: σ2

t .

Let η = 12π

∫∞−∞ ω|Ψ(ω)|2 dω be the center frequency of Ψ and

σ2ω =

∫ ∞

−∞(ω − η)2|Ψ(ω)|2 dω.

The Fourier transform of Ψu,s is a dilation of Ψ by 1/s:

Ψu,s(ω) =√sΨ(sω)e−iωu. (3.20)

Its center frequency is therefore η/s. The energy spread of Ψu,s around η/s is

∫ ∞

−∞

(ω − η

s

)2

s|Ψ(sω)|2 dω =σ2

ω

s2=: σ2

ω

Hence σ2t σ

2ω = σ2

t σ2ω is constant. Thus, the energy spread of a wavelet-time frequency atom

Ψu,s corresponds to a sσt × σω

sHeisenberg box centered at (u, η

s) with constant area.

PWf(u, ξ) := |Wf(u, s)|2 (3.21)

is the so called scalogram of f and represents the energy density.The following theorem derives a reconstruction formula for the wavelet transform and

proves that energy is preserved for real signals.

Theorem 3.14 Let Ψ be an analytic wavelet, then for any f ∈ L2(R)

Wf(u, s) =1

2Wfa(u, s). (3.22)

If Ψ is admissible (i.e. CΨ <∞, see (3.13) in Theorem 3.8) and f real valued, then

f(t) =2

CΨ

Re

[∫ ∞

0

∫ ∞

−∞Wf(u, s)

1√sΨ

(u− t

s

)du

ds

s2

], (3.23)

and

‖f‖2 =2

CΨ

∫ ∞

0

∫ ∞

−∞|Wf(u, s)|2 du

ds

s2. (3.24)


100 200 300 400 500 600 700 800 900 1000

−2

−1

0

1

2

Fre

quen

cy

0 100 200 300 400 500 600 700 800 900 10000

50

100

150

200

250

300

350

400

450

Figure 3.3: The top signal includes a linear chirp whose frequency increases, a quadratic chirpwhose frequency decreases, and two modulated Gaussian functions located at t = 512 andt = 896. The bottom image is the scalogram.

Proof: We first prove (3.22). The Fourier transform of fs(u) = Wf(u, s) = (f ∗ Ψs)(u)with s > 0 is

fs(ω) = f(ω)Ψ∗s(ω) = f(ω)

√sΨ∗(sω).

Since Ψ(ω) = 0 for negative frequencies, and fa = 2f(ω) we get

fs(ω) =1

2fa(ω)

√sΨ∗(ωs) =

1

2Wfa(·, s)(ω),

which proves (3.22).

With the same derivations as done in the proof of (3.14) in Theorem 3.8 one can showthat

fa(t) =1

CΨ

∫ ∞

0

∫ ∞

−∞Wfa(u, s)

1√sΨ

(u− t

s

)du

ds

s2.

Since f = Refa, inserting (3.22) proves (3.23).

Analogous we claim that

‖fa‖2 =1

CΨ

∫ ∞

0

∫ ∞

−∞|Wfa(u, s)|2 du

ds

s2=

4

CΨ

∫ ∞

0

∫ ∞

−∞|Wf(u, s)|2 du

ds

s2.

4 TIME-FREQUENCY ENERGY 37

Parsevals identity and formula (3.18) give ‖fa‖2 = 12π‖fa‖2 = 1

π‖f‖2 = 2‖f‖2 which

completes the proof.

Wavelet Modulated Windows

An analytic wavelet can be constructed with a frequency modulation of a real and symmetricwindow g. The Fourier transform of

Ψ(t) = eiηtg(t)

is Ψ(ω) = g(ω− η). If g(ω) = 0 for |ω| > η then Ψ(ω) = 0 for all ω < 0. Hence Ψ is analytic.The center frequency of Ψ is η and

|Ψ(η)| = supω∈R

|Ψ(ω)| = g(0).

Example 3.15 ’Gabor Wavelet’Consider the Gaussian window

g(t) =1

(σ2π)1/4e−

t2

2σ2 ⇒ g(ω) = (4πσ2)1/4e−σ2ω2

2 .

If σ2η2 ≫ 1 then g(ω) ≈ 0 for |ω| > η. Sufficiently large η supply approximatelyanalytic wavelets Ψ(t) = g(t)eiηt, the so called Gabor Wavelets.

4 Time-Frequency Energy

The wavelet and windowed Fourier transforms are computed by correlating the signal withfamilies of time-frequency atoms. The time and frequency resolution of these transforms isthus limited by the time-frequency resolution of the corresponding atoms. Ideally, one wouldlike to define a density of energy in a time-frequency plane, which does not spread the signalenergy in time or in frequency.

The Wigner-Ville distribution is a time-frequency energy density computed by correlatingf with a time and frequency translation by itself. This avoids any loss of time-frequencyresolution.

4.1 Wigner-Ville Distribution

Definition 4.1 Let f ∈ L2(R). The Wigner-Ville distribution of f is defined by

PV f(u, ξ) =

∫ ∞

−∞f(u+

τ

2

)f ∗(u− τ

2

)e−iξτ dτ (4.1)

=(f(u+

·2

)f ∗(u− ·

2

))∧(ξ). (4.2)

The Wigner-Ville distribution remains real and using Parsevals formula (4.1) it can be rewrit-ten as

PV f(u, ξ) =1

2π

∫ ∞

−∞f(ξ +

ω

2

)f ∗(ξ − ω

2

)eiωu dω. (4.3)


Proposition 4.2 Let f ∈ L2(R), then

∫ ∞

−∞PV f(u, ξ) du = |f(ξ)|2, (4.4)

and ∫ ∞

−∞PV f(u, ξ) dξ = 2π|f(u)|2. (4.5)

Proof: Let gu(ξ) = PV f(u, ξ). Applying the inverse Fourier transform on (4.2) yields

gu(t) = f

(u+

t

2

)f

(u− t

2

),

and∫ ∞

−∞PV f(u, ξ) dξ =

∫ ∞

−∞gu(ξ) dξ = 2πgu(0)

= 2πf(u)f(u) = 2π|f(u)|2,

which proves (4.5). For the first equality we consider gξ(u) = PV f(u, ξ) where (4.3)supplies the suitable Fourier transform

gξ(ω) = f(ξ +

ω

2

)f ∗(ξ − ω

2

).

Hence, (4.4) is proven by

∫ ∞

−∞gξ(u) du = gξ(0) = |f(ξ)|2.

The Wigner-Ville distribution is a quadratic form, it may be considered as an energydensity in the time-frequency plane. However, it is lacking one fundamental property of anenergy density, namely positivity, which demonstrates the following example.

Example 4.3 The W-V distribution of f(t) = ξ[−T,T ] is

PV f(u, ξ) =

∫ ∞

−∞f(u+

τ

2

)f(u− τ

2

)e−iξτ dτ

= 2sin(2(T − |u|)ξ)

ξ.

This is an oscillating function that takes negative values.

The following proposition shows that the W-V distribution does not alter time and frequencysupports of f and f , respectively.


Proposition 4.4 If f has compact support, i.e. supp f ⊆[u0 − T

2, u0 + T

2

], then

supp PV f(u, ξ) ⊆[u0 −

T

2, u0 +

T

2

]× R, (4.6)

and if supp f ⊆[ω0 − Ω

2, ω0 + Ω

2

], then

supp PV f(u, ξ) ⊆ R ×[ω0 −

Ω

2, ω0 +

Ω

2

]. (4.7)

Proof: Let f(t) = f(−t), then

PV f(u, ξ) =

∫ ∞

−∞f

(τ + 2u

2

)f ∗(τ − 2u

2

)e−iξτ dτ, (4.8)

and for the supports of f and f we have

supp f

( · + 2u

2

)⊆ [2(u0 − u) − T, 2(u0 − u) + T ],

supp f

( · − 2u

2

)⊆ [−2(u0 + u) − T,−2(u0 + u) + T ].

The Wigner-Ville integral (4.8) is zero if these two supports do not overlap, which isthe case only if |u0 − u| < T

2. (4.7) can be proven in a similar way using the following

alternative representation of the W-V distribution

PV f(u, ξ) =1

2π

∫ ∞

−∞f

(ω + 2ξ

2

)f∗(ω − 2ξ

2

)eiωu dω.

Example 4.5 Proposition 4.4 shows that the Wigner-Ville distribution does not spread thetime or frequency support of Diracs or sinusoids, unlike windowed Fourier or wavelettransforms. In fact we get

f(t) = δ(t− u0) ⇒ PV f(u, ξ) = δ(u− u0),

f(t) = eiω0t ⇒ PV f(u, ξ) =1

2πδ(ξ − ω0).

Let

f(t) =1

(σ2π)1/4e−

t2

2σ2

be a Gaussian function, then its Fourier transform f is also a Gaussian and

PV f(u, ξ) =1

πe−

u2

σ−σ2ξ2

=1

πe−

u2

σ e−σ2ξ2

= |f(u)|2|f(ξ)|2.

One can show that PV f(u, ξ) = |f(u)|2|f(ξ)|2 forces f to be a Gaussian function. Fur-thermore Gaussians are the only family of functions whose W-V distribution remainspositive.


One of the most important properties of the W-V distribution is its unitarity. This would notbe possible if the transform remained positive.

Theorem 4.6 ’Moyal’Let f and g in L2(R), then

2π

∣∣∣∣∫ ∞

−∞f(t)g(t) dt

∣∣∣∣2

=

∫∫PV f(u, ξ)PV g(u, ξ) du dξ. (4.9)

Proof: Let us compute the right hand side

I =

∫∫PV f(u, ξ)PV g(u, ξ) du dξ

=

∫∫∫∫f(u+

τ

2

)f ∗(u− τ

2

)g

(u+

τ ′

2

)g∗(u− τ ′

2

)e−iξ(τ+τ ′) dτ dτ ′ du dξ.

The integral ∫ ∞

−∞e−iξ(τ+τ ′) = 2πδ(τ + τ ′)

is the Fourier transform of the function h(t) ≡ 1 at point τ + τ ′. Inserting this resultyields

I = 2π

∫∫∫f(u+

τ

2

)f(u− τ

2

)g

(u+

τ ′

2

)g

(u− τ ′

2

)δ(τ + τ ′) dτ dτ ′ du

= 2π

∫∫f(u+

τ

2

)f(u− τ

2

)g

(u+

τ ′

2

)g

(u− τ ′

2

)dτ du.

A change of variables

t(u, τ) = u+ τ2

s(u, τ) = u− τ2

∣∣∣∣∂(s, t)

∂(u, τ)

∣∣∣∣ = 1

completes the proof.

Summary 4.7 Properties of the Wigner-Ville distribution

Function Wigner-Ville

t→ f(t) (u, ξ) → PV f(u, ξ)

eiφf(t) PV f(u, ξ)

af(t) |a|2PV f(u, ξ)

f(t− u0) PV f(u− u0, ξ)

eiξ0tf(t) PV f(u, ξ − ξ0)

eiat2f(t) PV f(u, ξ − 2au)1√sf(

ts

)PV f

(us, sξ)


Example 4.8 Let g be a symmetric and sufficient smooth window function (i.e. g(t) = 0for all |t| > T ). Its W-V distribution PV g(u, ξ) is centered at u = ξ = 0. The W-Vdistribution of the time-frequency atom

f(t) = aeiφ0g(t− u0)eiξ0t

is derived from PV g(u, ξ) by applying the rules of summary 4.7

PV f(u, ξ) = |a|2PV g(u− u0, ξ − ξ0).

Its energy is concentrated in the neighbourhood of (u0, ξ0).

4.2 Interferences and Positivity of the Wigner-Ville distribution

The Wigner-Ville distribution is lacking in two important aspects. One is the already men-tioned nonpositivity and the other results from so called interference phenomenons createdby cross terms.

Let f = f1 + f2 be a composite signal. Since the Wigner-Ville distribution is a quadraticform,

PV f = PV f1 + PV f2 + PV [f1, f2] + PV [f2, f1],

where PV [h, g] is the cross Wigner-Ville distribution of two signals

PV [h, g](u, ξ) =

∫ ∞

−∞h(u+

τ

2

)g∗(u− τ

2

)e−iτξ dτ.

The interference termI[f1, f2] := PV [f1, f2] + PV [f2, f1]

is a real function but the term may be nonzero at points (u, ξ) where neither f1, f2 nor theirFourier transforms are localized.

Example 4.9 Let us consider the two time-frequency atoms defined by

f1(t) = a1eiφ1g(t− u1)e

iξ1t,

f2(t) = a2eiφ2g(t− u2)e

iξ2t,

where g is a time window centered at t = 0. Their Wigner-Ville distributions are

PV f1(u, ξ) = a21PV g(u− u1, ξ − ξ1),

PV f2(u, ξ) = a22PV g(u− u2, ξ − ξ2).

The interference term is given by

I[f1, f2](u, ξ) = 2a1a2PV g(u− u0, ξ − ξ0) cos((u− u0)ξ − (ξ − ξ0)u+ φ)

with

u0 =u1 + u2

2, ξ0 =

ξ1 + ξ22

,

u = u1 − u2, ξ = ξ1 − ξ2 and φ = φ1 − φ2 + u0ξ.It is an oscillatory waveform centered at (u0, ξ0) but neither f1 nor f2 are concentratedat this point.


200 250 300 350 400 450

−2

−1

0

1

2

Time

Fre

quen

cy

200 250 300 350 400 4500

20

40

60

80

100

120

Figure 4.1: Wigner-Ville distribution PV f(u, ξ) of two Gabor atoms shown at the top. Theoscillating interferences are centered at the middle time-frequency location.

Positive time-frequency distributions totally remove the inteference terms but produce a lossof resolution. The following theorem shows that there exists no positive time-frequency dis-tribution which satisfy (4.4) and (4.5). This is one of the reasons why the Wigner-Villedistribution - despite its disadvantages - is a widely used tool in practice.

Theorem 4.10 There is no positive quadratic energy distribution Pf that satisfies the fol-lowing time and frequency marginal integrals:

∫ ∞

−∞Pf(u, ξ) dξ = 2π|f(u)|2,

∫ ∞

−∞Pf(u, ξ) du = |f(ξ)|2.

Proof: Suppose Pf to be a positive quadratic distribution that satisfies these marginals.Since Pf(u, ξ) ≥ 0, this implies that if the support of f is included in an interval Ithen Pf(u, ξ) = 0 for u 6∈ I. We can associate to the quadratic form Pf a bilineardistribution defined for any f and g by

P [f, g] =1

4(P (f + g) − P (f − g)).

5 WAVELET BASES 43

Let f1 and f2 be two non-zero signals whose supports are two intervals I1 and I2 thatdo not intersect, so that f1f2 = 0. Let f = af1 + bf2:

Pf = |a|2Pf1 + abP [f1, f2] + abP [f2, f1] + |b|2Pf2.

Since I1 does not intersect I2, Pf1(u, ξ) = 0 for u ∈ I2. The positivity of Pf for alla, b ∈ C forces

P [f1, f2](u, ξ) = P [f2, f1](u, ξ) = 0

for all u ∈ I2. Similarly we can show that these cross terms are zero for u ∈ I1 andhence

Pf(u, ξ) = |a|2Pf1(u, ξ) + |b|2Pf2(u, ξ).

Applying∫∞−∞ dξ on both sides and using the marginal integrals yields

|f(ξ)|2 = |a|2|f1(ξ)|2 + |b|2|f2(ξ)|2.

Since f(ξ) = af1(ξ)+ bf2(ξ) it follows that f1(ξ)f2(ξ) = 0. We can assume w.l.o.g. thatf1 is zero on a compact interval. But this is a contradiction to Theorem 1.10.

5 Wavelet Bases

Now, we are looking for wavelets Ψ ∈ L2(R) such that the family of scaled and shiftedwavelets,

Ψj,k := 2j/2Ψ

(t− 2−jk

2−j

)

(j,k)∈Z2

(5.1)

is a basis of L2(R). Of course, the optimal case is obtained if (5.1) forms an orthonormalbasis.

5.1 Frames and Riesz Bases

Let us consider a Hilbert space H with induced norm ‖ · ‖ =√〈·, ·〉 and a family of vectors

φk : k ∈ J ⊆ H

where J is an approbiate countable index set. A natural question is, given an arbitrary vectorf ∈ H , can f be rediscovered from the quantities 〈f, φk〉, k ∈ J ? Or, equivalently, can oneinvert the operator U : H → Im U given by

(Uf)[k] := 〈f, φk〉 ?

Indeed this is possible if φk : k ∈ J forms a frame.

Definition 5.1 A family of vectors φk : k ∈ J ⊆ H is called a frame if there exist frameconstants A,B > 0 such that for all f ∈ H

A∑

k∈J

|〈f, φk〉|2 ≤ ‖f‖2 ≤ B∑

k∈J

|〈f, φk〉|2. (5.2)

A frame is called a tight frame if A = B.

5 WAVELET BASES 44

Note that the vectors φk, k ∈ J are not necessarily linearly independent.

Example 5.2 H = R2 = span~e1, ~e2Consider the three vectors

φ1 = ~e1, φ2 = −~e12

+

√3

2~e2 and φ3 = −~e1

2−

√3

2~e2.

The angle between φi and φj is 2π3

for i 6= j. It can be shown that φ1, φ2, φ3 forms atight frame with A = B = 3

2.

Theorem 5.3 The operator U : H → l2(J) defined by (Uf)[k] := 〈f, φk〉 is a bounded andinvertible operator H → Im U . Im U is a closed subspace of l2(J).

Proof: We only give an outline of the proof. For more details see [1].The adjoint of U maps l2(J) → H and is given by

(U⋆a[·]) =∑

k∈J

a[k]φk.

We get U⋆Uf = U⋆[〈f, φ·〉] =∑

k∈J〈f, φk〉φk, U⋆U is a symmetric operator. One can

show that U⋆U : H → H is bounded and invertible.

Let f ∈ H be arbitrarily,

∑

k∈J

|〈f, φk〉|2 = 〈U⋆Uf, f〉 = 〈Uf, Uf〉 = ‖Uf‖2.

Together with (5.2) we obtain the inequality

A‖Uf‖2 ≤ ‖f‖2 ≤ B‖Uf‖2.

U is bounded since ‖Uf‖2 ≤ 1A‖f‖2. Furthermore from ‖f‖2 ≤ B‖Uf‖2 we get the

injectivity of U and the existence of a bounded inverse U−1 : Im U → H . Finally theclosed graph theorem tells us that Im U is a closed subspace of l2(J).

Definition 5.4 A family φk : k ∈ J ⊆ H is called a Riesz basis if every f posseses aunique representation

f =∑

k∈J

a[k]φk

with approbiate coefficients a[k] and if there exist constants C1, C2 > 0 such that

C1

∑

k∈J

|a[k]|2 ≤ ‖f‖2 ≤ C2

∑

k∈J

|a[k]|2. (5.3)

5 WAVELET BASES 45

5.2 Multi Resolution Analysis (MRA)

Definition 5.5 A family of closed linear subspaces Vj ⊆ L2(R), j ∈ Z is a multiresolutionapproximation if the following 6 properties are satisfied:

1. f ∈ Vj ⇔ f(· − 2−jk) ∈ Vj) for all (j, k) ∈ Z2,

2. Vj ⊂ Vj+1 for all j ∈ Z,

3. f ∈ Vj ⇔ f(2·) ∈ Vj+1 for all j ∈ Z,

4. limj→−∞

Vj =∞⋂

j=−∞Vj = 0,

5. limj→∞

Vj =∞⋃

j=−∞Vj = L2(R),

6. there exists a function ϕ ∈ L2(R) such that ϕ(· − k) : k ∈ Z is a Riesz basis ofV0.

Remark 5.6

(a) The function ϕ in Property 6 is called a scaling function or generating func-tion.

(b) Property 6 can be replaced by the existence of a finite family of scaling functionsϕ1, . . . , ϕN. This leads to the concept of Multiwavelets.

Alternatively we can give a constructive description. Let us consider a scaling functionϕ ∈ L2(R) and

ϕj,k = 2j/2ϕ(2j(· − k)), j, k ∈ Z,

Vj = cl spanϕj,k, k ∈ Z.ϕ should satisfy the following three relations:

1.∑k∈Z

ϕ(x− k) = 1

2. Refinement relation:There exist a[k], k ∈ Z such that

ϕ(x) =√

2∑

k∈Z

a[k]ϕ(2x− k), i.e., ϕ0,0 =∑

k∈Z

a[k]ϕ1,k.

3. ϕ0,k : k ∈ Z forms a Riesz basis of V0.

Note that the third relation also implies that every ϕj,k : k ∈ Z is a Riesz basis of Vj withcertain constants C1, C2 > 0.

5 WAVELET BASES 46

Construction of Wavelet Bases with MRA

If we take an MRA we know that Vj is included in Vj+1. We can find a subspace Wj ⊂ L2(R)which is the complement of Vj in Vj+1 and satisfies

Vj ∩Wj = 0, Vj+1 = Vj ⊕Wj and cos(Vj,Wj) > 0.

The last condition is equivalent to

(a) ”sharpened Cauchy inequality”There exists a γ with 0 < γ < 1 such that

|〈vj, wj〉| ≤ γ‖vj‖‖wj‖, ∀vj ∈ Vj, wj ∈Wj .

(b) ”inverse triangle inequality”There exists a c > 0 such that

‖vj‖2 + ‖wj‖2 ≤ c‖vj + wj‖2, ∀vj ∈ Vj, wj ∈Wj .

If Wj is the orthogonal complement we can replace this by the equality ‖vj‖2 +‖wj‖2 =‖vj + wj‖2.

Again the subspaces Wj are invariant under translation, i.e.,

f ∈Wj ⇔ f(· − 2−jk) ∈Wj, ∀k ∈ Z.

To be more specific: There exists a wavelet (the so called mother wavelet) Ψ ∈ L2(R) suchthat

Ψj,k(x) := 2j/2Ψ(2jt− k)(x), k ∈ Zforms a Riesz basis of Wj. Note that we are now very close to the desired result we drawedat the beginning of this section (see (5.1)). Using the decomposition

Vj+1 = Wj ⊕ Vj = V0 ⊕j⊕

l=0

Wl =

j⊕

l=−∞Wl

we obtain with j → ∞ that∞⊕

l=−∞Wl = L2(R). In particular this means that

Ψj,k := 2j/2Ψ

(t− 2−jk

2−j

)

(j,k)∈Z2

is a basis of L2(R). We call it a wavelet basis if (5.1) is even a Riesz basis. Orthonormalwavelet bases satisfy

〈Ψj,k,Ψj,k′〉 = δk,k′ ∀j ∈ Z.

5 WAVELET BASES 47

Example 5.7 Piecewise constant approximationLet us consider the box function

θ1(x) = χ[0,1)(x) =

1 if x ∈ [0, 1)0 otherwise,

then ϕ(x) := θ1(x) is a scaling function. Indeed ϕ0,k = ϕ(· − k), k ∈ Z formsan orthonormal basis of V0. The space Vj consists of all piecewise constant functions,subordinated to a uniform mesh width hj = 2−j. I.e., all functions fj ∈ Vj are constanton the intervals [2−jk, 2−j(k + 1)).

The refinement relation is

ϕ(x) =√

2

(1√2ϕ(2x) +

1√2ϕ(2x− 1)

)=

1√2ϕ1,0(x) +

1√2ϕ1,1(x).

A matching Wavelet may be found by

Ψ(x) = ϕ(2x) − ϕ(2x− 1) =1√2(ϕ1,0(x) − ϕ1,1(x)).

So the following equalities are satisfied:

〈Ψ, ϕ〉 =

∫ 1

0

Ψ(x) dx =

∫ 1/2

0

dx−∫ 1

1/2

dx = 0

〈Ψ(· − k),Ψ(· − k′)〉 = δk,k′

〈Ψ,Ψ(2·)〉 = 0

〈Ψj,k,Ψj′,k′〉 = δj,j′δk,k′

〈Ψj,k, ϕl,k′〉 = 0, ∀l ≤ j, ∀k, k′ ∈ Z ⇒ Wj ⊥ Vj , Vj−1, . . .

The Wavelet basis defined by those Ψj,k is called the Haar basis. It has the drawbackthat the functions of Vj are not smooth enough, indeed they are not even continuous.This implies low regularity and slow approximation of smooth functions.

Example 5.8 Shannon approximationHere we consider the scaling function

ϕ(x) := 2πθ0(x) =sin πx

πx.

Then ϕ(· − k) : k ∈ Z forms again an orthonormal basis of V0.

Example 5.9 B-spline approximationLet us consider

θd(x) = (θ1 ∗ · · · ∗ θ1︸︷︷︸

d times

)(x),

then

θd(ξ) = (θ0(ξ))m =

(sin ξ/2

ξ/2

)d

e−iξ/2.

These are piecewise polynomials of degree d− 1 on intervals (2−jk, 2−j(k + 1)), k ∈ Z.They are d− 2 times continuously differentiable.

5 WAVELET BASES 48

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

1

2

3

4

5

6

7

8

9

(3, 2)

(3, 5)

(4, 8)

(5,13)

(6,21)

(6,32)

(6,43)

(7,95)

Figure 5.1: Some Haarlets at various scales and locations.

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.2

0.4

0.6

0.8

1

Figure 5.2: B-splines of order 0, . . . , 4.

The following lemma gives an answer to the question whether an L2(R)-function ϕ generateswith ϕ(· − k), k ∈ Z a Riesz basis of V0.

Lemma 5.10 A family ϕ(· − k), k ∈ Z forms a Riesz basis in L2(R) if and only if thereexist constants A,B > 0 such that

A ≤∞∑

k=−∞|ϕ(ξ − 2kπ)| ≤ B <∞ (5.4)

for all ξ ∈ [−π, π].

Proof: Let f ∈ V0 be decomposed as

f(t) =

∞∑

k=−∞a[k]ϕ(t− k) (5.5)

5 WAVELET BASES 49

with approbiate coefficients a[k]. Taking the Fourier transform of (5.5) yields f(ξ) =a(ξ)ϕ(ξ) where

a(ξ) =∞∑

k=−∞a[k]e−ikξ.

Then

‖f‖2 =1

2π

∫ ∞

−∞|f(ξ)|2 dξ

=1

2π

∫ ∞

−∞|a(ξ)|2|ϕ(ξ)|2 dξ

=1

2π

∞∑

k=−∞|a[k]|2

∫ 2π

0

|ϕ(ξ + 2kπ)|2 dξ. (5.6)

On the other hand the family ϕ(· − k), k ∈ Z forms a Riesy basis iff

A

∞∑

k=−∞|a[k]|2 ≤ ‖f‖2 ≤ B

∞∑

k=−∞|a[k]|2. (5.7)

By the discrete Plancherel formula we get

∞∑

k=−∞|a[k]|2 =

1

2π

∫ 2π

0

|a(ξ)|2 dξ.

Together with (5.6) we conclude that

‖f‖2 ≥ ‖a[·]‖l2(Z) infξ∈[0,2π]

∞∑

k=−∞|ϕ(ξ − 2kπ)|2

and

‖f‖2 ≤ ‖a[·]‖l2(Z) supξ∈[0,2π]

∞∑

k=−∞|ϕ(ξ − 2kπ)|2.

which shows (5.7). The functions ϕ(· − k) are linearly independent since f ≡ 0 impliesa[k] = 0 for all k ∈ Z and vice versa.

In order to prove that the condition is necessary let us assume there exists a ξ0 ∈ [0, 2π]such that ∞∑

k=−∞|ϕ(ξ0 + 2kπ)|2 = 0.

We construct a sequence (an(ξ))∞n=1 such that

supp an(ξ) ⊂[ξ0 −

1

n, ξ0 +

1

n

]

and ‖a(ξ)‖L2([0,2π]) = 1. Then we get

an · ϕ(ξ) → 0 for n→ ∞,

which contradicts the Riesz basis property.

5 WAVELET BASES 50

5.3 Orthogonal Wavelet Bases

Theorem 5.11 Let θ ∈ L2(R) ∩L1(R) and θ(· − k) : k ∈ Z be a Riesz basis in V0. Thenϕ defined by

ϕ(ω) =1

(∑∞k=−∞ |θ(ω + 2kπ)|2

)1/2θ(ω) (5.8)

generatesϕj,k(t) : 2j/2ϕ(2jt− k), k ∈ Z

as an orthonormal basis of Vj.

Proof: We represent

ϕ(t) =∞∑

k=−∞a[k]ϕ(t− k)

or, equivalently,ϕ(ω) = a(ω)θ(ω),

where a(ω) =∑∞

k=−∞ a[k]e−ikω is 2π-periodic. The orthonormality condition is

〈ϕ(· − k), ϕ(· − k′)〉 =

∫ ∞

−∞ϕ(t− k)ϕ∗(t− k′) dt

= (ϕ ∗ ϕ)(k′ − k) = δk,k′,

where ϕ(t) = ϕ∗(−t) The Fourier transform of this equality gives

ϕ ∗ ϕ =

∞∑

k=−∞|ϕ(ω + 2kπ)|2 = 1. (5.9)

(5.8) is verified by choosing

a(ω) =1

(∑∞k=−∞ |θ(ω + 2kπ)|2

)1/2.

Example 5.12 For piecewise constant box functions and the Shannon scaling function wehave constructed orthonormal bases in Examples 5.7 and 5.8. However, the B-splinebases are not orthogonal for d > 1. Therefore we insert the known Fourier transform

θ(ω) =

(sin ω

2ω2

)d

ei ω2

ε,

where ε = 1 if d is odd and ε = 0 for even d, in the previous theorem and obtain

ϕ(ω) =e−i ω

2ε

ωd√S2d(ω)

, with S2d(ω) =

∞∑

k=−∞(ω + 2kπ)−2d.

5 WAVELET BASES 51

The sum S2d(ω) can be computed explicitly. At first we observe by differentation that

S2d(ω) = S ′′2(d−1)(ω) = · · · = S2d−2

2 (ω).

Together with the known formula

S2(ω) =

∞∑

k=−∞

1

ω + 2kπ=

1

4 sin2(ω2)

we obtain for linear splines (d = 2)

S4(ω) =1 + 2 cos2 ω

2

48 sin4 ω2

and ϕ(ω) =4√

3 sin2 ω2

ω2√

1 + 2 cos2 ω2

.

and cubic splines (d = 4)

S8(ω) =5 + 30 cos2 ω

2+ 30 sin2 ω

2cos2 ω

2+ 70 cos4 ω

2+ 2 sin4 ω

2cos4 ω

2+ 2

3sin6 ω

2

105 · 28 sin8 ω2

.

The corresponding cubic spline scaling function is displayed in Figure 5.3.

Conjugate Mirror Filters

Let us consider the orthogonal projection Pj : L2(R) → Vj given by

Pjf(t) :=

∞∑

k=−∞〈f, ϕj,k〉ϕj,k(t) =

∞∑

k=−∞2j

∫ ∞

−∞f(x)ϕ(2jx− k) dx ϕ(2jt− k).

In the case of nested subspaces · · · ⊂ Vj ⊂ Vj+1 ⊂ . . . , j ∈ Z we have

⋂

j∈Z

Vj = 0 ⇔ limj→∞

Pjf(t) = 0, ∀f ∈ L2(R), t ∈ R

where the limit is taken pointwise. Furthermore we get the equivalence

⋃

j∈Z

Vj is dense in L2(R) ⇔ limj→∞

‖f − Pjf‖ = 0, ∀f ∈ L2(R).

The inclusion Vj ⊂ Vj+1 implies the existence of coefficients mj,0q,k such that

ϕj,k =

∞∑

q=−∞mj,0

q,k ϕj+1,k.

This can be reduced to the case j = 0 by dilation and translation , i.e. to the relation V0 ⊂ V1:

ϕ(x) = ϕ0,0(x) =

∞∑

q=−∞m0,0

q,0 ϕ1,q(x) =√

2

∞∑

q=−∞h[q]ϕ(2x− q), (5.10)

5 WAVELET BASES 52

or1√2ϕ(x

2

)=

∞∑

q=−∞h[q]ϕ(x− q),

with

h[q] = 〈ϕ, ϕ1,q〉 =√

2

∫ ∞

−∞ϕ(x)ϕ(2x− q) dx.

The matrix M with [M ]q,k = m0,0q,k is called the refinement mask matrix and q → h[q] the

mask or filter. If ϕ has compact support then only finite coefficients h[k] are nonzero and his a trigonometric polynom. In that case we call h a conjugate mirror filter.

Taking the Fourier transform of (5.10) yields

ϕ(2ω) =1√2h(ω)ϕ(ω) (5.11)

for h(ω) =∑∞

k=−∞ h[k]e−ikω. For any j ≥ 0 the equality (5.11) implies

ϕ(2−j+1ω) =1√2h(2−jω)ϕ(2−jω),

and by substitution

ϕ(ω) =

(j∏

k=1

h(2−kω)√2

)ϕ(2−jω).

If ϕ is continuous at ω = 0 then limj→∞

ϕ(2−jω) = ϕ(0) and so

ϕ(ω) =

∞∏

k=1

h(2−kω)√2

ϕ(0). (5.12)

This means that the scaling function is completely determined by the values of h[k] fork ∈ Z which are the filter coefficients. The following theorem gives necessary and sufficientconditions on h to guarantee that the infinite product (5.12) is the Fourier transform of ascaling function.

Theorem 5.13 (Mallat, Meyer 89)Let ϕ ∈ L1(R) ∩ L2(R) be a scaling function such that 〈ϕ(·), ϕ(· − k)〉 = δ0,k. Then the

Fourier series h(ω) =∑∞

k=−∞ h[k]e−ikω, where h[k] = 〈 1√2ϕ( ·

2), ϕ(· − k)〉, satisfies

h(0) =√

2, (5.13)

|h(ω)|2 + |h(ω + π)|2 = 2, for all ω ∈ R. (5.14)

Conversely, if a 2π-periodic function h satisfies (5.13, 5.14) and h ∈ C1([−ε, ε]) forsome ε > 0 and if

infω∈[−π

2, π2]|h(ω)| > 0 (5.15)

5 WAVELET BASES 53

then

ϕ(ω) =

∞∏

j=1

h(2−jω)√2

(5.16)

is the Fourier transform of some scaling function ϕ such that 〈ϕ(·), ϕ(· − k)〉 = δ0,k.

Additionally, ϕ generates an MRA by Vj = spanϕj,k, k ∈ Z.

Proof: This theorem is a central result with a long and technical proof. Therefore it isdivided in several parts.

Proof of the necessary condition (5.14) The orthogonality 〈ϕ(·), ϕ(·−k)〉 = δ0,k impliesthat (see (5.9) in the proof of theorem 5.11)

∞∑

k=−∞|ϕ(ω + 2kπ)|2 = 1, ∀ω ∈ R.

Inserting ϕ(ω) = 1√2h(ω

2)ϕ(ω

2) (see (5.11)) yields

∞∑

k=−∞

∣∣∣h(ω

2+ kπ

)∣∣∣2 ∣∣∣ϕ

(ω2

+ kπ)∣∣∣

2

= 2.

Exploiting the 2π-periodicity of h, we can seperate the even and odd integer terms ofthe above sum

∣∣∣h(ω

2

)∣∣∣2

∞∑

k=−∞

∣∣∣ϕ(ω

2+ 2kπ

)∣∣∣2

+∣∣∣h(ω

2+ π)∣∣∣

2∞∑

k=−∞

∣∣∣ϕ(ω

2+ π + 2kπ

)∣∣∣2

= 2.

Again, we observe that

∞∑

k=−∞

∣∣∣ϕ(ω

2+ 2kπ

)∣∣∣2

=∞∑

k=−∞

∣∣∣ϕ(ω

2+ π + 2kπ

)∣∣∣2

= 1.

Setting ξ = ω2

yields

|h(ξ)|2 + |h(ξ + π)|2 = 2,

which proves (5.14).

Proof of the necessary condition (5.13) Let Pj : L2(R) → Vj be defined as

Pju =

∞∑

k=−∞〈u, ϕj,k〉ϕj,k.

Obviously Pj is an orthogonal projection which means PjPj = Pj and 〈Pju, w〉 =

〈u, Pjw〉 for all u, w ∈ L2(R). From ϕ(0) = 1√2h(0)ϕ(0) we may conclude h(0) =

√2

provided that ϕ(0) 6= 0. Indeed that can be shown by the following arguments.

5 WAVELET BASES 54

The density of∞⋃

j=−∞Vj together with the Plancherel formula imply that

limj→∞

‖u− Pju‖ = limj→∞

1

2π‖u− Pju‖ = 0. (5.17)

To compute the Fourier transform of Pju let us denote ϕj(t) = 2j/2ϕ(2jt). Then

Pju(t) =

∞∑

k=−∞(u ∗ ϕ∗

j)(2−jk) · ϕj(t− 2−jk) =

(ϕj ∗

∞∑

k=−∞(u ∗ ϕ∗

j)(2−jk)δ

)(t− 2−jk).

The Fourier transform of (u ∗ ϕ∗j)(t) is

2−j/2u(ω) · ϕ∗(2−jω).

A uniform sampling has a periodized Fourier transform which we calculated in Propo-sition 2.2, and hence

Pju(ω) = ϕ(2−jω)

∞∑

k=−∞u(ω − 2j+1kπ) · ϕ(2−j[ω − 2j+1kπ]). (5.18)

We choose u with supp u ⊆ [−π, π]. Then for j > 0 and ω ∈ [−π, π] we obtain

Pju(ω) = u(ω)|ϕ(2−jω)|2. (5.19)

The L2 convergence (5.17) implies pointwise convergence almost everywhere:

limj→∞

Pju(ω) = u(ω).

It follows from (5.19) that limj→∞

|ϕ(2jω)| = |ϕ(0)| = 1.

Proof that ϕ(·−k)k∈Z is orthonormal This is the first step to prove that the functionϕ whose Fourier transform is given by (5.16) is a scaling function. Since (5.13) weknow that |h(ω)| ≤

√2 and this implies that the infinite product (5.15) converges and

|ϕ(ω)| ≤ 1. Using Parsevals formula the orthonormality relation reads as follows

〈ϕ(·), ϕ(· − k)〉 =

∫ ∞

−∞ϕ(t)ϕ(t− k) dt =

1

2π

∫ ∞

−∞|ϕ(ω)|2e2kπiω dω.

Verifying orthonormality is thus equivalent to showing that∫ ∞

−∞|ϕ(ω)|2e2kπiω dω = 2πδ[k].

Let us consider the function

ϕn(ω) =

n∏

p=1

h(2−pω)√2

χ[−2nπ,2nπ](ω)

5 WAVELET BASES 55

and the integrals

In[k] =

∫ ∞

−∞|ϕn(ω)|2e2kπiω dω =

∫ 2nπ

−2nπ

n∏

p=1

|h(2−pω)|22

e2kπiω dω.

First, let us show that In[k] = 2πδ[k] for all n ≥ 1. To do this, we decompose In[k] as

In[k] =

∫ 0

−2nπ

n∏

p=1

|h(2−pω)|22

e2kπiω dω +

∫ 2nπ

0

n∏

p=1

|h(2−pω)|22

e2kπiω dω. (5.20)

We change the variable ω′ = ω + 2nπ in the first integral taking into account theperiodicity of h, i.e. when p < n then |h(2−p[ω′ − 2nπ])|2 = |h(2−pω′)|2. When p = nthe hypothesis (5.13) tells us that

|h(2−n[ω′ − 2nπ])|2 + |h(2−nω′)|2 = 2.

For n > 1 we can add both integrals in (5.20) and obtain

In[k] =

∫ 2nπ

0

(1 − |h(2−nω′)|2

2

)n−1∏

p=1

|h(2−pω)|22

e2kπiω +n∏

p=1

|h(2−pω)|22

e2kπiω dω

=

∫ 2nπ

0

n−1∏

p=1

|h(2−pω)|22

e2kπiω dω (5.21)

=

∫ 2n−1π

−2n−1π

n−1∏

p=1

|h(2−pω)|22

e2kπiω dω = In−1[k].

By induction we get In[k] = I1[k]. Writing (5.21) for n = 1 gives

I1[k] =

∫ 2π

0

e2kπiω dω = 2πδ[k],

which proves that In[k] = 2πδ[k], for all n ≥ 1.

Now, we prove that ϕ ∈ L2(R). For ω ∈ R

limn→∞

|ϕn(ω)|2 =

∞∏

p=1

|h(2−pω)|22

= |ϕ(ω)|2.

Fatous lemma shows that∫ ∞

−∞|ϕ(ω)|2 dω ≤ lim

n→∞

∫ ∞

−∞|ϕn(ω)|2 dω = 2π,

because In[0] = 2π for n ≥ 1. Since

|ϕ(ω)|2e2kπiω = limn→∞

|ϕn(ω)|2e2kπiω,

5 WAVELET BASES 56

we will verify that∫ ∞

−∞|ϕ(ω)|2e2kπiω dω = lim

n→∞

∫ ∞

−∞|ϕn(ω)|2e2kπiω dω = In[k] = 2πδ[k]

by applying the dominated convergence theorem. This requires the proof of an upperbound condition. In our case we will prove that there exists a constant C > 0 such that

∣∣|ϕn(ω)|2e2kπiω∣∣ = |ϕn(ω)|2 ≤ C|ϕ(ω)|2. (5.22)

We already showed that |ϕ(ω)|2 is integrable and hence a possible upper bound function.Since supp ϕn ⊆ [−2nπ, 2nπ] we may assume that |ω| ≤ 2nπ. From ϕ(ω) = 1√

2h(ω

2)ϕ(ω

2)

it follows that

|ϕ(ω)|2 = |ϕn(ω)|2∣∣∣ϕ( ω

2n

)∣∣∣2

.

To prove (5.22) it is therefore sufficient to show that |ϕ(ω)|2 ≥ 1C

for ω ∈ [−π, π].

Let us first investigate the ε-neighbourhood of ω = 0. Since h ∈ C[−ε, ε] and |h(ω)|2 ≤2 = |h(0)|2, the derivatives of the functions |h(ω)|2 and ln |h(ω)|2 exist and vanish atω = 0. So there exists a δ > 0 such that for all |ω| ≤ δ:

0 ≥ ln

(|h(ω)|2

2

)≥ −|ω|.

Hence, for |ω| ≤ δ

|ϕ(ω)|2 = exp

[ ∞∑

p=1

ln

(|h(2−pω)|2

2

)]

≥ exp

[−

∞∑

p=1

|2−pω|]

= e−|ω| ≥ e−δ. (5.23)

Next, we analyze the behaviour for ω with π ≥ |ω| > δ. Let l ∈ N such that 2−l < δ.From (5.15) we know that K := infω∈[−π

2, π2] |h(ω)| > 0, so we can estimate that

|ϕ(ω)|2 =l∏

p=1

|h(ω)|22

∣∣∣ϕ(ω

2l

)∣∣∣2

≥ K2l

2le−δ =:

1

C,

which finally proves the upper bound (5.22) for |ϕ(·)|2.Now, we have that ϕ(t − k)k∈Z is orthonormal. A simple change of variable showsthat ϕj,kk∈Z is othonormal for all j ∈ Z.

Proof that Vjj∈Z forms a multiresolution We have to show the 6 properties of anMRA in Definition 5.5. The multiresolution properties 1 and 3 are clearly true. Thedefinition of V0 yields Property 6. We are still lacking Properties 2, 4 and 5. Let usstart with the causality Vj ⊂ Vj+1. From the refinement relation

ϕ(t) =

∞∑

k=−∞

√2h[k]ϕ(2t− k)

5 WAVELET BASES 57

we conclude that

ϕj,n(t) = 2j/2ϕ(2jt− n) =

∞∑

k=−∞h[k]2j/2ϕ(2j+1t− 2n− k) =

∞∑

p=−∞h[p− 2n]ϕj+1,p,

where we substituted p = 2n+ k. Consequently Vj ⊂ Vj+1.

To prove Property 4 we must show that any u ∈ L2(R) satisfies limj→−∞

‖Pju‖ = 0. Since

ϕj,kk∈Z is an orthonormal basis of Vj we get

‖Pju‖ =∞∑

k=−∞|〈u, ϕj,k〉|2.

Suppose first that |u(t)| ≤ A for all t ∈ R and supp u ⊆ [2−J , 2J ]. Then

∞∑

k=−∞|〈u, ϕj,k〉|2 ≤ 2j

[ ∞∑

k=−∞

∫ 2J

−2J

|u(t)||ϕ(2jt− k)| dt

]2

≤ 2jA2

[ ∞∑

k=−∞

∫ 2J

−2J

|ϕ(2jt− k)| dt

]2

≤ 2J+1A2

∞∑

k=−∞

∫ 2J

−2J

|ϕ(2jt− k)|22j dt

= 2J+1A2

∫

Sj

|ϕ(t)|2 dt,

with Sj =⋃

k∈Z[k − 2J+j, k + 2J+j] for j < −J . If j → −∞ then the measure of Sj

tends to zero and by the dominated convergence theorem we have

limj→−∞

∞∑

k=−∞|〈u, ϕj,k〉|2 = 0.

Since the set of bounded functions with compact support is dense in L2(R) we getlim

j→−∞‖Pju‖ = 0 for all u ∈ L2(R) and finally lim

j→−∞Vj = 0.

For the multiresolution property 5 we have to show that for any u ∈ L2(R)

limj→∞

‖u− Pju‖2 = limj→∞

(‖f‖2 − ‖Pjf‖2) = 0. (5.24)

Let us only consider those u with supp u ⊆ [−2Jπ, 2Jπ] for sufficiently large J . We havealready computed in (5.18) that the Fourier transform of Pju is

Pju(ω) = ϕ(2−jω)

∞∑

k=−∞u(ω − 2j+1kπ) · ϕ∗(2−j[ω − 2j+1kπ]).

5 WAVELET BASES 58

If j > J then the supports of u(· − 2j+1kπ) are disjoint for different k ∈ Z and

‖Pju‖2 =1

2π

[∫ ∞

−∞|u(ω)|2|ϕ(2−jω)|4 dω

+

∫ ∞

−∞

∞∑

k=−∞

k 6=0

|u(ω − 2j+1kπ)|2|ϕ(2−jω)|2|ϕ(2−j[ω − 2j+1kπ])|2 dω

].

≥ 1

2π

∫ ∞

−∞|u(ω)|2|ϕ(2−jω)|4 dω

We know that |ϕ(ω)| ≤ 1 and (5.23) supplies for sufficiently small ω the inequality|ϕ(ω)| ≥ e−|ω|/2, hence

limω→0

|ϕ(ω)| = 1.

Since |u(ω)|2|ϕ(2−jω)|4 ≤ |u(ω)|2 and limj→∞

|u(ω)|2|ϕ(2−jω)|4 = |u(ω)|2 one can apply

the dominated convergence theorem, which yields

limj→∞

∫ ∞

−∞|u(ω)|2|ϕ(2−jω)|4 dω =

∫ ∞

−∞|u(ω)|2 dω = ‖u‖2 = 2π‖u‖2.

Therefore ‖Pju‖2 ≥ ‖u‖2 and on the other hand the orthogonality of the projector Pj

implies ‖Pju‖ ≤ ‖u‖ which finally proves

limj→∞

‖Pju‖2 = ‖u‖.

Again, the result may be extended to the complete L2(R) by a density argument.

5.4 Construction of Orthogonal Wavelet Bases

Similar to Section 5.2 we want to construct orthogonal wavelet bases. Since Vj is included inVj+1 we can form the orthogonal complement of Vj in Vj+1

Vj+1 = Vj ⊕Wj , with Wj⊥Vj .

Now, we are looking for a function ψ ∈ L2(R) such that

spanΨj,k = Wj and 〈Ψj,k,Ψj,k′〉 = δk,k′,

where Ψj,k = 2j/2ψ(2jt− k) for j, k ∈ Z. Since Ψ ∈ V1 our desired representation is

Ψ(t) =

∞∑

k=−∞g[k]

√2ϕ(2t− k). (5.25)

Let us define

g(ω) =

∞∑

k=−∞g[k]e−ikω,

then the Fourier transform of (5.25) yields

Ψ(2ω) =1√2g(ω)ϕ(ω). (5.26)

5 WAVELET BASES 59

Lemma 5.14 The family Ψj,kk∈Z is an orthonormal basis of Wj if and only if

|g(ω)|2 + |g(ω + π)|2 = 2 (5.27)

andg(ω)h∗(ω) + g(ω + π)h∗(ω + π) = 0. (5.28)

Proof: The lemma will be proven for j = 0 from which it is extensible to the case j 6= 0via an appropriate scaling. As for (5.9) one can verify that Ψ(t−k)k∈Z is orthonormalif and only if

I(ω) =∞∑

k=−∞|Ψ(ω + 2kπ)|2 = 1, ∀ω ∈ R. (5.29)

Since Ψ(ω) = 2−1/2g(ω/2)ϕ(ω/2) and g(ω) is 2π periodic,

I(ω) =1

2

∞∑

k=−∞

∣∣∣g(ω

2+ kπ

)∣∣∣2 ∣∣∣ϕ

(ω2

+ kπ)∣∣∣

2

=1

2

[∣∣∣g(ω

2

)∣∣∣2

∞∑

p=−∞

∣∣∣ϕ(ω

2+ 2pπ

)∣∣∣2

+∣∣∣g(ω

2+ π)∣∣∣

2∞∑

p=−∞

∣∣∣ϕ(ω

2+ (2p+ 1)π

)∣∣∣2].

Now, from∞∑

p=−∞|ϕ(ω + 2pπ)|2 = 1

we have shown that (5.27) and (5.29) are equivalent.

Next we investigate V0⊥W0 which is equivalent to the condition

〈Ψ(·), ϕ(· − k)〉 = (Ψ ∗ ϕ∗)(k) = 0.

The sampled sequence Ψ ∗ ϕ∗ is zero if its Fourier series (see Proposition 2.2) satisfies

∞∑

k=−∞Ψ(ω + 2kπ)ϕ∗(ω + 2kπ) = 0, ∀ω ∈ R.

With the same argumentation as above we prove that the latter equation is equivalentto (5.28).

Finally we show that V0 ⊕W0 = V1. Since √

2ϕ(2 · −k)k∈Z is an orthonormal basis ofV1 and ϕ(· − k)k∈Z, Ψ(· − k)k∈Z are orthonormal bases of V0 and W0, respectively,we have to show that for any a[·] ∈ l2(Z) there exist b[·], c[·] ∈ l2(Z) such that

∞∑

k=−∞a[k]

√2ϕ(2t− k) =

∞∑

k=−∞b[k]ϕ(t− k) +

∞∑

k=−∞c[k]Ψ(t− k). (5.30)

Taking the Fourier transform of (5.30) yields

1√2a(ω

2

)ϕ(ω

2

)= b(ω)ϕ(ω) + c(ω)Ψ(ω).

5 WAVELET BASES 60

Inserting Ψ(ω) = 2−1/2g(ω/2)ϕ(ω/2) and ϕ(ω) = 2−1/2h(ω/2)ϕ(ω/2) shows that (5.30)is necessarily satisfied if

a(ω

2

)= b(ω)h

(ω2

)+ c(ω)g

(ω2

). (5.31)

Let us set b[·] and c[·] so that

b(2ω) :=1

2[a(ω)h(ω) + a(ω + π)h(ω + π)],

c(2ω) :=1

2[a(ω)g(ω) + a(ω + π)g(ω + π)].

Calculating the right-hand side of (5.31) one easily verifies the equality to the left-handside by inserting (5.27), (5.28) and using

|h(ω)|2 + |h(ω + π)|2 = 2. (5.32)

Since b(ω) and c(ω) are 2π-periodic functions they are the Fourier series of two sequencesb[·] and c[·] that satisfy (5.30) which completes the proof.

The following theorem shows how to construct an orthonormal basis of Wj by scaling andtranslating a wavelet Ψ.

Theorem 5.15 Mallat, MeyerLet ϕ ∈ L2(R) be an orthogonal scaling function and h the corresponding conjugatemirror filter. Then Ψ defined by

Ψ(ω) =1√2g(ω

2

)ϕ(ω

2

), (5.33)

whereg(ω) = e−iωh∗(ω + π), (5.34)

is an orthogonal wavelet, i.e.,

Ψj,k(t) = 2j/2Ψ(2jt− k)k∈Z

form orthonormal bases of Wj for all j ∈ Z and Ψj,kj,k∈Z is an orthonormal basis ofL2(R).

Proof: Since

Ψ(t) =∞∑

k=−∞g[k]

√2ϕ(2t− k) ∈ V1,

and on the other hand

Ψ(t) =

∞∑

k=−∞〈Ψ, ϕ1,k〉ϕ1,k,

we conclude thatg[k] = 〈Ψ, ϕ1,k〉 =

√2〈Ψ, ϕ(2 · −k)〉.

5 WAVELET BASES 61

In order to show that Ψ(· − k) : k ∈ Z forms an orthonormal bases of W0 we involvethe previous lemma. Indeed, the function g(ω) = e−iωh∗(π + ω) satisfies

|g(ω)|2 + |g(ω + π)|2 = |h(ω + π)|2 + |h(ω)|2 = 2,

and

g(ω)h∗(ω) + g(ω + π)h∗(ω + π) = (e−iω + e−i(ω+π))h∗(ω)h∗(ω + π) = 0.

Finally we show that Ψj,k : j, k ∈ Z is an orthonormal basis of L2(R). From W0⊥V0

we conclude that〈Ψj,k, ϕj,k′〉 = 〈Ψ(· − k), ϕ(· − k′)〉 = 0,

i.e. Wj⊥Vj . Therefore, Wj⊥Wl for all j 6= l. Since Vj : j ∈ Z forms a multiresolutionanalysis the proof is completed by observing that spanΨj,k : j, k ∈ Z is dense in L2(R).

Ψ(t) |Ψ(ω)| |Ψ(2−jω)|2

−8 −6 −4 −2 0 2 4 6 8−1

−0.5

0

0.5

1

−25 −20 −15 −10 −5 0 5 10 15 20 250

0.2

0.4

0.6

0.8

1

−3 −2 −1 0 1 2 30

0.2

0.4

0.6

0.8

1

Figure 5.3: Cubic spline wavelet, its Fourier transform and the graphs of |Ψ(2−jω)|2 for1 ≤ j ≤ 5.

The proof points out that

g[k] = 〈Ψ, ϕ1,k〉 =√

2〈Ψ(·), ϕ(2 · −k)〉,

when

Ψ(t) =

∞∑

k=−∞g[k]

√2ϕ(2t− k).

Since g(ω) = e−iωh(ω + π) is the Fourier transform of g we get the following useful equality:

g[k] = (−1)1−kh[1 − k]

Remark 5.16 It can be shown that for all ω ∈ R\0∞∑

j=−∞|Ψ(2jω)|2 = 1.

5 WAVELET BASES 62

This is illustrated in Figure 5.3. According to the previous theorem any f ∈ L2(R) canbe represented as

f =

∞∑

j=−∞

∞∑

k=−∞〈f,Ψj,k〉Ψj,k =

∞∑

j=−∞Qjf,

where Qj = (Pj+1 −Pj)f . The calculations to obtain the coefficients of a signal decom-posed in a certain wavelet basis may can be done by fast wavelet transform algorithms.

50 100 150 200 250 300 350 400 450 500

0

2−1

2−2

2−3

2−4

2−5

2−6

Scaling function coefficients

Figure 5.4: Wavelet coefficients dj[n] = 〈f,Ψj,n〉 calculated at scales 2j, j = −1, . . . ,−6 withthe cubic spline wavelet. At the top is the remaining coarse signal approximation aJ [n] =〈f, ϕ−5,n〉.

Example 5.17 Examples of orthogonal basis functions

1. Box function

h[k] =

1√2, k = 0, 1

0, otherwise.

2. B-splines of order d generate the so called Battle-Lemarie wavelets

h(ω) = eiǫω/2

√S2d(ω)

22d−1S2d(2ω),

5 WAVELET BASES 63

where ǫ = 0 for even d and ǫ = 1 otherwise. For the definition of S2d see example5.12. For the special case of piecewise linear orthogonal scaling functions we obtain

h(ω) =√

2

[1 + 2 cos2 ω

2

1 + 2 cosω

]1/2

cos2 ω

2,

g(ω) = e−iωh(ω + π).

Remark 5.18 Counter exampleLet

h(ω) =√

2 cos

(3

2ω

),

then

|h(ω)|2 + |h(ω + π)|2 = 2 cos2

(3

2ω

)+ 2 cos

(3

2ω +

3

2π

)= 2.

But it can be shown that

ϕ(ω) =

∞∏

j=1

h(2−jω)√2

=1

3χ[− 3

2, 32](ω),

which is obviously not an orthogonal scaling function.

5.5 Daubechies Compactly supported wavelets

The scaling function ϕ has compact support if and only if the associated conjugate mirrorfilter h[·] has compact support, i.e., there exists an N ∈ N such that

h(ω) =N−1∑

k=0

h[k]e−ikω.

Compactly supported filters h[·] have been computed by Ingrid Daubechies. For further detailssee [2] or attend a more advanced course on wavelets.

5.6 Fast Wavelet Transform

Decomposition coefficients in a wavelet orthogonal basis are computed with a fast algorithmthat cascades discrete convolutions with h and g, and subsamples the output. A fast Wavelettransform decomposes successively each approximation Pj+1f into a coarser approximationPjf plus the wavelet coefficients carried by Pj+1f .

Because of Vj+1 = Vj ⊕Wj a function f ∈ Vj+1 may be represented either by an expansionwith respect to a scaling function basis

f =

∞∑

k=−∞〈f, ϕj+1,k〉ϕj+1,k =

∞∑

k=−∞aj+1[k]ϕj+1,k,

5 WAVELET BASES 64

ϕ(t) ϕ(t) ϕ(t)

0 0.5 1 1.5 2 2.5 3−0.5

0

0.5

1

1.5

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5−0.5

0

0.5

1

1.5

0 1 2 3 4 5 6 7−0.5

0

0.5

1

1.5

Ψ(t) Ψ(t) Ψ(t)

−1 −0.5 0 0.5 1 1.5 2−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.5 3−1.5

−1

−0.5

0

0.5

1

1.5

2

−3 −2 −1 0 1 2 3 4−1

−0.5

0

0.5

1

1.5

p = 2 p = 3 p = 4

Figure 5.5: Daubechies scaling function ϕ and wavelets Ψ with p vanishing moments.

or with respect to orthogonal bases of Vj and Wj

f =∞∑

k=−∞〈f, ϕj,k〉ϕj,k +

∞∑

k=−∞〈f,Ψj,k〉Ψj,k

=

∞∑

k=−∞aj [k]ϕj,k +

∞∑

k=−∞dj[k]Ψj,k.

Theorem 5.19 MallatAt the decomposition

aj [n] =∞∑

k=−∞h[k − 2n]aj+1[k] = (aj+1 ∗ h∗)[2n], (5.35)

dj[n] =

∞∑

k=−∞g[k − 2n]aj+1[k] = (aj+1 ∗ g∗)[2n], (5.36)

and at the reconstruction

aj+1[n] =∞∑

k=−∞h[n− 2k]aj[k] +

∞∑

k=−∞g[n− 2k]dj [k] (5.37)

= (aj ∗ h)[n] + (dj ∗ g)[n],

where x[k] =

x[k], if k = 2p, p ∈ Z,0, otherwise,

denotes the oversampling.

Proof:

5 WAVELET BASES 65

Proof of (5.35) Resulting from

aj [n] = 〈f, ϕj,n〉 =

∞∑

k=−∞〈f, ϕj+1,k〉〈ϕj,n, ϕj+1,k〉,

and with x = 2jt− n,

〈ϕj,n, ϕj+1,k〉 =

∫ ∞

−∞2

j2ϕ(2jt− n)2

j+12 ϕ∗(2j+1t− k) dt

=√

2

∫ ∞

−∞ϕ(x)ϕ∗(2x− k + 2n) dx

= h[k − 2n],

we conclude

ϕj,n =

∞∑

k=−∞h[k − 2n]ϕj+1,k.

Taking the inner product with f on both sides of this equality yields (5.35).

Proof of (5.36) Analogously 〈Ψj,n, ϕj+1,n〉 = g[k − 2n] proves

Ψi,n =∞∑

k=−∞g[k − 2n]ϕj+1,k

which results in (5.36).

Proof of (5.37) With the formulas from above we obtain

ϕj+1,n =

∞∑

k=−∞〈ϕj+1,n, ϕj,k〉ϕj,k +

∞∑

k=−∞〈ϕj+1,n,Ψj,k〉Ψj,k

=∞∑

k=−∞h[k − 2n]ϕj,k +

∞∑

k=−∞g[k − 2n]Ψj,k.

Again, taking the inner product with f on both sides completes the proof.

The decomposition and reconstruction,

aj+1[·]aj[·]

րց

dj[·]and

aj [·]ցր

dj[·]aj+1[·]

are nothing but changes of basis functions.Iterating the decomposition yields for given coefficients aj+1[·],

aj+1[·] −→ aj [·] −→ aj−1[·] −→ . . . a1[·] −→ a0[·]ց ց ց . . . ց

dj[·] dj−1[·] . . . d0[·]

5 WAVELET BASES 66

the coefficients(D[l, k])l=−1,...,j := (a0[·], d0[·], d1[·], . . . , dj[·]).

The translation (aj+1[k]) → (D[l, k]) is called the discrete Wavelet transform. Thebackward transform is provided by the reconstruction

d0[·] d1[·] . . . dj−1[·] dj[·]ց ց . . . ց ց

a0[·] −→ a1[·] −→ . . . aj−1[·] −→ aj[·] −→ aj+1[·]

In practice the signal aj+1 is 2j+1 periodic which means that we have N = 2j+1 coefficients.Then the discrete Wavelet transform requires O(N) algebraic operations.

Index

L2[−π, π], 22L2(R), 8

Aliasing, 19minimize, 19

B-spline, 47, 50, 63cubic, 61

Box function, 47, 63

Center, 27, 28Characteristic function, 7Chirp, 36Circulant, 23Conjugate mirror filter, 51Convolution, 5, 12

circular, 23Fourier transform, 5

Cross Wigner-Ville distribution, 41

DFT, 23, 24inverse, 24

Dirac comb, 15, 18Dirac distribution, 10, 11, 17, 21, 39

Energy density, 29, 35Energy spread, 26

Heissenberg uncertainty, 10

FFT, 25Filter, 11

causal, 12conjugate mirror, 51discrete, 20finite impulse response, 21low pass, 13, 19passive electronic circuit, 14stable, 12, 21time averaging, 12, 21

FIR, 21Fourier series, 22Fourier transform, 3, 9, 22

L1(R), 3L2(R), 9characteristic function, 7

continuity, 3convolution, 5discrete, 23, 24fast, 25inverse, 6one-to-one map, 9Parseval and Plancherel identities, 8properties, 10windowed, 27, 28

Frame, 43constants, 43

FWT, 64

Gabor wavelet, 37Gaussian function, 4, 6, 39Gibbs oscillation, 13

Heissenberg boxes, 27Heissenberg uncertainty, 10, 28

Impulse response, 21Interferences, 41

LTI, 11, 20

Mallat theorem, 65Matlab functions, 26Mexcian hat wavelet, 32, 33Moyal formula, 40MRA, 45, 53Multi resolution approximation, 45Multiwavelets, 45

Orthogonal familybanded frequencies, 20periodic functions, 22periodic signals, 23wavelets, 43, 46

Parseval and Plancherel identities, 8, 22, 24Poisson-summation formulas, 15

Reconstruction formula, 29, 35Refinement mask matrix, 52Refinement relation, 45, 52

67

INDEX 68

Reproducing kernel, 30, 33Riemann-Lebesgue lemma, 3Riesz basis, 44

condition, 48

Sampling, 17uniform, 17

Scaling function, 33, 45B-spline, 50construction, 52

Scalogram, 35Shannon approximation, 47Shannon-Whittaker theorem, 18

Time-frequency energy, 37Time-invariant operator, 11, 20Time/frequency atom, 26

wavelet, 27WFT, 27

Time/frequency localization, 28Time/frequency operator, 27Transfer function, 21

Variance, 27, 28, 35

Wavelet, 27, 31, 43, 46admissibility condition, 32analytic, 34centered, 31, 35compactly supported, 63Daubechies, 63modulated window, 37real, 32

Wavelet basis, 43, 46construction, 46, 58, 60Haar, 47orthogonal, 50, 58

Wavelet transform, 31analytic, 34continuous, 31discrete, 66fast, 64reconstruction formula, 35reproducing kernel, 33scaling function, 33

WFT, 27, 28reconstruction, 29

reproducing kernel, 30Wigner-Ville distribution, 37

cross terms, 41positivity, 38, 42properties, 40unitarity, 40

Documents

Wavelets and Signal Processing · 2012-02-08 · Wavelets and Signal Processing Reinhold Schneider Sommersemester 2000 Recommended Literature [1] St´ephane Mallat [1998]: A Wavelet