1. Fourier Transform (Continuous time) A finite energy signal - Lirias

25
Fourier Transform . . . Multi-resolution Properties of the . . . Orthogonality Normalisation Wavelets The wavelet equation The integral of the . . . Normalisation of the . . . Home Page Title Page JJ II J I Page 1 of 100 Go Back Full Screen Close Quit J I 1. Fourier Transform (Continuous time) 1.1. Signals with finite energy A finite energy signal is a signal f (t) for which Z -∞ |f (t)| 2 dt < Scalar product: hf (t),g (t)i = 1 2π Z -∞ f (t)g (t)dt The Hilbert space of all these functions is L 2 (IR) or L 2

Transcript of 1. Fourier Transform (Continuous time) A finite energy signal - Lirias

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Fourier Transform . . .

Multi-resolution

Properties of the . . .

Orthogonality

Normalisation

Wavelets

The wavelet equation

The integral of the . . .

Normalisation of the . . .

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1. Fourier Transform (Continuous time)

1.1. Signals with finite energy

A finite energy signal is a signal f(t) for which∫ ∞−∞|f(t)|2dt <∞

Scalar product:

〈f(t), g(t)〉 =1√2π

∫ ∞−∞

f(t)g(t)dt

The Hilbert space of all these functions is L2(IR) or L2

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Fourier Transform . . .

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Properties of the . . .

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1.2. The Fourier Transform

Suppose f(t) ∈ L2, then its Fourier-transform is

F (ω) = F{f(t)} =1√2π

∫ ∞−∞

f(t)e−iωtdt

This is:F (ω) = 〈eiωt, f(t)〉

Inverse transform

f(t) =1√2π

∫ ∞−∞

F (ω)eiωtdω =⟨e−iωt, F (ω)

⟩Convolution → product

(f ∗ g)(t) =

∫ ∞−∞

f(u)g(t− u)du =

∫ ∞−∞

f(t− u)g(u)du

then:F{f ∗ g} = F{f(t)}F{g(t)}

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1.3. Parseval’s identity

The Fourier transform preserves energy:

‖f(t)‖ = ‖F (ω)‖

or: ∫ ∞−∞|f(t)|2dt =

∫ ∞−∞|F (ω)|2dω

This is Parseval’s identity. It can be stated more generally:

〈f(t), g(t)〉 = 〈F (ω), G(ω)〉

The norm-equivalence follows for g(t) = f(t).

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Fourier Transform . . .

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Properties of the . . .

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Wavelets

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1.4. Heisenberg’s uncertainty

In (t, ω) plane, signal concentrated at (t0, ω0)

t0 =

∫IR|f(t)|2tdt∫

IR|f(t)|2dt

; ω0 =

∫IR|F (ω)|2ωdω∫

IR|F (ω)|2dω

with variances

σ2t =

∫IR|f(t)|2(t− t0)2dt∫

IR|f(t)|2dt

; σ2ω =

∫IR|F (ω)|2(ω − ω0)

2dω∫IR|F (ω)|2dω

then:

σtσω ≥1

2or σ2

t σ2ω ≥

1

4

Higher resolution in the time domain⇔

Lower resolution in the frequency domain

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1.5. Time-frequency plane�

��

��

���������� � ���������

� ��������������������� ���!����"�#%$'&)( *,+.-/�����!���

1. basis functions δ(t− tk) give 1st picture

2. basis functions eiωkt give 2nd picture

how to get pictures like 3 or 4?needs local basis in (t, ω)-plane

3. is more natural for human observation (sound/vision) = wave-lets

4. corresponds to windowed or short time Fourier transform(WFT/STFT)

WFT is older than wavelets

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1.6. Windowed Fourier Transform (WFT)

Use a “windowed” instead of a “pure” sine/cosine:

Fψ(f(t)) =1√2π

∫ ∞−∞

f(t)eiωtψ(x− b)dt

ψ is a (localizing) window function: for example a Gaussian:

ψ(t) = e−t2/2σ0

This leads to the Gabor transform

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

-10 -5 0 5 10

real part

psi(t)*sin(3*t)psi(t)

-psi(t)

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

-10 -5 0 5 10

imaginary part

psi(t)*cos(3*t)psi(t)

-psi(t)

However, this approach has a drawback: the time window isconstant for all frequencies: for high frequencies, we have moreoscillations in the window.

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1.7. The continuous wavelet transform (CWT)

(sneak preview)

Wψ(a, b) =1√a

∫ ∞−∞

f(t)ψ

(t− ba

)dt

This is an overcomplete function representation.(from one dimension, we transform into a 2D space!!).

• the translation factor b→ shifts along t-axis

• the scaling factor a→ shifts along ω-axisshorter time support ↔ higher frequencies

for example mexican hat

� � � � � �� � � � � �� � � � � �� � � � � �� � � � � �� � � � � �� � � � � � �� � � � � � �� � � � � � �� � � � � � �� � � � � � � �� � � � � � �� � � � � � �� � � � � �� � � � � �� � � � � � �� � � � � � � � � �� � � � �� � � � � � �� � � � � � � � � � � �� � �� � � � � � � � � � � � � � � ��� � � � � � � � � � � � � � � � ��� � � � � � � � � � � � � � �� � �� � � � � � � � � � �� � � � � � � � ��� � � � � � �� � � � � � �� � � � � � � � � �� � � � � � � � � � � � � �� � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � �� � � � � � � � � � � �� � � � � � � � � �� � � � � � �� � � � � �� � � � � �� � � � � � �� � � � � � �� � � � � � � �� � � � �

� �� � � � � � �� � � � � � �� � � � � � �� � � � � �� � � � � �� � � � � �� � � � � �� � � � � �� � � � � �

�������

� � � � � � � � � � � � � � �� �� �� �� �� �� � �� �� ��� � �� � � �� ��� ���� ���� ��� ������ �������� ������� ��� �������� ����� �������� ������������ ����������� ������ ������ ���� ����� �� �� ���� ���� ���� ������ ��� ������ ������� ����� �������� ����� �������� ���� ������� �� ������� ������ ����� ���� ���� � �� � � � � � � �� � ��� �� �� �� ��� ��� � �� � � �� � � � � �

� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �

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Fourier Transform . . .

Multi-resolution

Properties of the . . .

Orthogonality

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Wavelets

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2. Multi-resolution

2.1. Objective and Definition

→ approximation of finite energy, continuous time signals.multi-resolution analysis:{

successive (”multi”)approximations (”resolutions”)

}of a function.Successive basis functions all have same shape:→ translations and dilations of one function.

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The precise mathematical definition is as follows:

A multi-resolution analysis (MRA) is a sequence of closed sub-spaces Vj (j ∈ ZZ) of L2(IR) with the following properties

1. Vj ⊂ Vj+1

2. (a)+∞⋃j=−∞

Vj is dense in L2(IR) i.e.:+∞⋃j=−∞

Vj = L2(IR)

(b)+∞⋂j=−∞

Vj = {0} (0 is constant function)

3. (shift invariance)v(t) ∈ V0 ⇔ v(t− k) ∈ V0,∀k ∈ ZZ

4. (scale invariance)v(t) ∈ Vj ⇔ v(2t) ∈ Vj+1,∀j ∈ ZZ

5. (shift invariant basis)∃ϕ(t) ∈ V0 : {ϕ(t− k)}k∈ZZ is a stable basis(Riesz-basis) for V0

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2.2. The dilation equation

(refinement equation, two-scale equation)The function ϕ(t) is called scaling function or father function.

Dilation equation:

∃{ck}k∈ZZ : ϕ(t) =∑k∈ZZ

ckϕ(2t− k)

Immediate consequence:∫ ∞−∞

ϕ(t)dt =∑k∈ZZ

ck

∫ ∞−∞

ϕ(2t− k)dt =∑k∈ZZ

ck1

2

∫ ∞−∞

ϕ(t)dt

Hence, choosing∫ ∞−∞

ϕ(t)dt = θ 6= 0⇔ Φ(0) =1√2π

∫ ∞−∞

ϕ(t)dt =θ√2π

we get ∑k∈ZZ

ck = 2

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2.3. Example

2 31 4

2 31 4

2 31 4

2 31 4

V

V

V

V

1

0

-1

-2

Terminology:j is called the resolution level2j is called the scale

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2.4. Solution of the dilation equation

2.4.1. By iteration

Numerical solution in time domain:

ϕ(t) =∑k∈ZZ

ckϕ(2t− k)

Hence, start with (for instance)

ϕ(0)(t) = χ[0,1](t)

(Box function on [0, 1]) and iterate

ϕ(i)(t) =∑k∈ZZ

ckϕ(i−1)(2t− k)

until convergence.This algorithm is proposed in continuous time. While imple-

menting it (e.g. in MATLAB), be careful with discretisation!!Theoremif cn = 0 for n < n− and for n > n+, then supp(ϕ) ⊂ [n−, n+]

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2.4.2. By recursion

The iteration procedure starts with an initial value in all pointst (in practice only in points of discretisation)Each iteration step computes a new function value in each point.

Recursion starts from the exact function values in some pointsand uses the dilation equation to compute other points, exactly(no function value iteration, the iteration is now on the numberof points)

We have:

ϕ(t

2) =

∑k∈ZZ

ckϕ(t− k)

If we know ϕ(t) for all integer t, then we can compute it for allt = k/2. Repeating this yields ϕ(t) for all dyadic points t = k/2j.

We know ϕ(k) = 0 if k 6∈ {n−, . . . n+}.

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Example: n− = 0, n+ = 5. Then we have

ϕ(0)ϕ(1)ϕ(2)ϕ(3)ϕ(4)ϕ(5)

=

c0

c2 c1 c0

c4 c3 c2 c1 c0

c5 c4 c3 c2 c1

c5 c4 c3

c5

·

ϕ(0)ϕ(1)ϕ(2)ϕ(3)ϕ(4)ϕ(5)

The matrix must have eigenvalue 1 and the function values arethe components of the corresponding eigenvector.

We call this matrix C.From this, it follows: ϕ(n−) = 0 = ϕ(n+)

(unless cn− = 1 or cn+ = 1)

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2.4.3. Cascade algorithm

Recall Vj ⊂ Vj+1.

Suppose f(t) =∑k

sj,kϕ(2jt− k) =∑l

sj+1,lϕ(2j+1t− l)

Use ϕ(x) =∑

i ciϕ(2x− i) to compute sj+1,k from sj,k Result:

sj+1,l =∑k

cl−2ksj,k

Now, start the iteration with s0,k = δk.This corresponds to f(t) = ϕ(t).If j →∞, the support of ϕ(2jt− k) gets smaller and smaller.At a certain moment it can only be represented by one pixel.The sj,k at this level are an approximation for ϕ(k/2j).

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2.4.4. Fourier domain solution

Define the Fourier transform Φ(ω) = 1√2π

∫∞−∞ ϕ(t)e−iωt dt

Then the dilation equation becomes in the Fourier domain

Φ(ω) = C(ω

2)Φ(

ω

2) = · · · = Φ(ω/2k)

k∏j=1

C(ω/2j)

with C(ω) =∑k

cke−iωk.

Since limk→∞ ω/2k = 0, and Φ(0) = θ√

2π, we have:

Φ(ω) =θ√2π

∞∏j=1

C(ω/2j)

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3. Properties of the father function

Theorem: If ϕ(t) is Riemann-integrable, then∑k∈ZZ

ϕ(t− k) = θ

This is called partition of unity.Theorem:∑

k∈ZZ

ϕ(t− k) = θ ⇔∑n∈ZZ

c2n =∑n∈ZZ

c2n+1 ⇔ C(π) = 0

We have

C(ω) =1

2

∑k∈ZZ

cke−ikω

and since∑

k∈ZZ(−1)kck = 0 this becomes: C(π) = 0and since

∑k∈ZZ ck = 2 also C(0) = 1

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4. Orthogonality

L2 is a unitary space with dot/inner/scalar product:

〈f(t), g(t)〉 =1√2π

∫ ∞−∞

f(t)g(t)dt

This allows to define the concept of orthogonality:

f(t) ⊥ g(t)⇔ 〈f(t), g(t)〉 = 0

If ϕ(t− k) ⊥ ϕ(t− l),∀k 6= l, we have an orthogonal basis.If

f(t) =∑k∈ZZ

αkϕ(t− k)

we can easily find αm: multiply both sides by ϕ(t−m) and in-tegrate to get

αm =〈ϕ(t−m), f(t)〉‖ϕ(t−m)‖2 =

〈ϕ(t−m), f(t)〉‖ϕ(t)‖2

since it is easy to see that ‖ϕ(t−m)‖ = ‖ϕ(t)‖.

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5. Normalisationtheorem: ϕ(t−m) is orthogonal basis then∣∣∣∣∫ ∞

−∞ϕ(s)ds

∣∣∣∣2 =

∫ ∞−∞|ϕ(s)|2ds

⇒ If we choose∫∞−∞ ϕ(s)ds = θ, then ‖ϕ(t)‖ = θ

If ϕ(t−m) is an orthonormal basis for V0

⇒ ϕ(2jt−m) is an orthogonal basis for Vj⇒ ϕj,k(t) = 2j/2ϕ(2jt− k) is an orthonormal basis for Vj, j ∈ ZZ

So hk = ck√2

is an normalised sequence:∑k∈ZZ

|hk|2 = 1 and∑k∈ZZ

hk =√

2

The dilation equation ϕ(t) =∑

k∈ZZ ckϕ(2t− k) becomes:

ϕj,m(t) =∑k∈ZZ

hkϕj+1,2m+k(t).

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Orthogonality of the basis, can be expressed as∑n∈ZZ

cn−2kcn = 2δk (∗)

The coefficients have double shift orthogonality .The inverse is also true: coefficients with double shift orthogon-ality lead to orthogonal scaling functions.

In the frequency domain, this is:

|C(ω)|2 + |C(ω + π)|2 = 1

Corollary: Orthogonal scaling functions with compact supportmust have an even number of non-zero coefficients in the dilationequation.Furthermore for Φ(ω) we have:Theorem: If {ϕ0,k} is orthonormal ⇔∑

k∈ZZ

|Φ(ω + 2kπ)|2 =1√2π

⇔ partition of unity property.

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Fourier Transform . . .

Multi-resolution

Properties of the . . .

Orthogonality

Normalisation

Wavelets

The wavelet equation

The integral of the . . .

Normalisation of the . . .

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6. Wavelets

Recall:Vj ⊂ Vj+1

We now try to catch the information presentin Vj+1, but lost in Vj: Suppose we have Wj,such that:

Vj+1 = Vj ⊕Wj

mVj+1 = Vj +Wj and Vj ∩Wj = {0}

Vj+1 is the direct sum of Vj and Wj.

V

W

V

j

j

j+1

Vj +Wj is a linear sum: all elements of Vj +Wj can be writtenas a sum of an element from Vj and an element from Wj. IfVj ∩Wj = {0}, this decomposition is unique.

special case Vj ⊥ Wj: orthogonal sum; Wj is the orthogonalcomplement of Vj in Vj+1.

Page 22: 1. Fourier Transform (Continuous time) A finite energy signal - Lirias

Fourier Transform . . .

Multi-resolution

Properties of the . . .

Orthogonality

Normalisation

Wavelets

The wavelet equation

The integral of the . . .

Normalisation of the . . .

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Theorem: If {φ(t − k)}k∈ZZ constitute an orthogonal basis forV0, then there exists one function ψ(t) such that {ψ(t − k)}k∈ZZ

forms an orthogonal basis for the orthogonal complement W0 ofV0 in V1.

We call this function mother function or wavelet function.We know:

VJ = VJ−1 ⊕WJ−1 = VL ⊕J−1⊕j=L

Wj

A function

f(t) =2J−1∑k=0

skϕ(2Jt− k)

can be written as: (suppose L ≤ 0)

f(t) =J−1∑j=L

2j−1∑k=0

wj,kψ(2jt− k) +2L−1∑k=0

skϕ(2Lt− k)

This decomposition is a discrete wavelet expansion

Page 23: 1. Fourier Transform (Continuous time) A finite energy signal - Lirias

Fourier Transform . . .

Multi-resolution

Properties of the . . .

Orthogonality

Normalisation

Wavelets

The wavelet equation

The integral of the . . .

Normalisation of the . . .

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7. The wavelet equation

Since W0 ⊂ V1, we have ψ(x) =∑k∈ZZ

dkϕ(2x− k)

Orthogonality∫ ∞−∞

ψ(x)ϕ(x− k)dx = 0 &

∫ ∞−∞

ψ(x)ψ(x− k)dx = δk

implies: ∑n∈ZZ

cn−2kdn = 0∑n∈ZZ

dn−2kdn = 2δk

A possible solution is:

dk = (−1)kc1−k

If we put both sequences ck and dk under each other

c . . . c−3 c−2 c−1 c0 c1 c2 c3 c4 . . .

d . . . −c4 c3 −c2 c1 −c0 c−1 −c−2 c−3 . . .

we immediately see that their inner product is zero.

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Fourier Transform . . .

Multi-resolution

Properties of the . . .

Orthogonality

Normalisation

Wavelets

The wavelet equation

The integral of the . . .

Normalisation of the . . .

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8. The integral of the mother function

Sinceψ(x) =

∑k∈ZZ

dkϕ(2x− k)

we have (in the orthogonal case):∫ ∞−∞

ψ(x)dx =∑k∈ZZ

(−1)kc̄1−k

∫ ∞−∞

ϕ(2x− k)dx

Since ∑k∈ZZ

(−1)kc1−k = 0

this is: ∫ ∞−∞

ψ(x)dx = 0

Page 25: 1. Fourier Transform (Continuous time) A finite energy signal - Lirias

Fourier Transform . . .

Multi-resolution

Properties of the . . .

Orthogonality

Normalisation

Wavelets

The wavelet equation

The integral of the . . .

Normalisation of the . . .

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9. Normalisation of the wavelets

Suppose the fully orthogonal case and call:

ψj,m(t) = 2j/2ψ(2jt−m)

Then this is an orthonormal basis for L2. Also:

ψj,m(t) = 2j/2ψ(2jt−m)

= 2j/2∑k∈ZZ

dkϕ(2j+1t− 2m− k)

=∑k∈ZZ

dk√2

2(j+1)/2ϕ(2j+1t− 2m− k)

=∑k∈ZZ

gk ϕj+1,2m+k(t)

where gk = dk√2

is a normalised sequence:∑k∈ZZ

|gk|2 =∑k∈ZZ

|hk|2 = 1 and∑k∈ZZ

gk = 0,∑k∈ZZ

hk =√

2

Part 3