marco de angelis
Post on 17-Nov-2021
11 Views
Preview:
TRANSCRIPT
Introduction Problem statement Interval analysis Main algorithm
The interval (discrete) Fourier transform
Marco De Angelis
Institute for Risk and Uncertainty, University of Liverpool
May 17, 2021
Marco De Angelis The interval Fourier transform May 17, 2021 1 / 22
Introduction Problem statement Interval analysis Main algorithm
Outline
1 Introduction
2 Problem statement
3 Interval analysis
4 Main algorithm
Marco De Angelis The interval Fourier transform May 17, 2021 2 / 22
Introduction Problem statement Interval analysis Main algorithm
Motivations
Imprecise/poor sensormeasurements
Missing data
Relaxed assumptions
=
+
+
Marco De Angelis The interval Fourier transform May 17, 2021 3 / 22
Introduction Problem statement Interval analysis Main algorithm
Motivations
Imprecise/poor sensormeasurements
Missing data
Relaxed assumptions
=
+
+
Marco De Angelis The interval Fourier transform May 17, 2021 3 / 22
Introduction Problem statement Interval analysis Main algorithm
Motivations
Imprecise/poor sensormeasurements
Missing data
Relaxed assumptions
=
+
+
Marco De Angelis The interval Fourier transform May 17, 2021 3 / 22
Introduction Problem statement Interval analysis Main algorithm
Motivations
Imprecise/poor sensormeasurements
Missing data
Relaxed assumptions
=
+
+
Marco De Angelis The interval Fourier transform May 17, 2021 3 / 22
Introduction Problem statement Interval analysis Main algorithm
Motivations
Imprecise/poor sensormeasurements
Missing data
Relaxed assumptions
=
+
+
Marco De Angelis The interval Fourier transform May 17, 2021 3 / 22
Introduction Problem statement Interval analysis Main algorithm
Notation
F : Fourier transform
Fh: Fourier transform for a given harmonic h
f : a function
[f ]: Natural extension
f̂ : United extension
In this presentation we use capital letters to denote interval variables.
Marco De Angelis The interval Fourier transform May 17, 2021 4 / 22
Introduction Problem statement Interval analysis Main algorithm
The discrete Fourier transform
Fh : Rn → C, ∀h ∈ Z+
Fh(x) :=
n−1∑k=0
xk e−i 2π
nhk
Marco De Angelis The interval Fourier transform May 17, 2021 5 / 22
Introduction Problem statement Interval analysis Main algorithm
The discrete Fourier transform
Fh : Rn → C, ∀h ∈ Z+
Fh(x) :=
n−1∑k=0
xk e−i 2π
nhk
or
Fh(x) :=
n−1∑k=0
xk
(cos
2π
nhk − i sin 2π
nhk
)
Marco De Angelis The interval Fourier transform May 17, 2021 5 / 22
Introduction Problem statement Interval analysis Main algorithm
The discrete Fourier transform
0 20 40 60 80 100 120t
10
5
0
5
10
15
20
x 0 10 20 30 40 50 60h
100
50
0
50
100
Rez
0 10 20 30 40 50 60h
100
50
0
50
100
Imz
Marco De Angelis The interval Fourier transform May 17, 2021 6 / 22
Introduction Problem statement Interval analysis Main algorithm
The discrete Fourier transform
0 20 40 60 80 100 120t
10
5
0
5
10
15
20
x 0 10 20 30 40 50 60h
100
50
0
50
100
Rez
0 10 20 30 40 50 60h
100
50
0
50
100
Imz
DFT
iDFT
Marco De Angelis The interval Fourier transform May 17, 2021 6 / 22
Introduction Problem statement Interval analysis Main algorithm
The discrete Fourier transform
0 20 40 60 80 100 120t
10
5
0
5
10
15
20
x 0 10 20 30 40 50 60h
100
50
0
50
100
Rez
0 10 20 30 40 50 60h
100
50
0
50
100
Imz
DFT
iDFTiDFT
Marco De Angelis The interval Fourier transform May 17, 2021 6 / 22
Introduction Problem statement Interval analysis Main algorithm
Amplitude of the Fourier transform.
∣∣∣∣∣n−1∑k=0
xk e−i 2π
nhk
∣∣∣∣∣
0 10 20 30 40 50 60h
100
50
0
50
100
Rez
0 10 20 30 40 50 60h
100
50
0
50
100
Imz
0 10 20 30 40 50 60h
0
20
40
60
80
100
120
|z|
Marco De Angelis The interval Fourier transform May 17, 2021 7 / 22
Introduction Problem statement Interval analysis Main algorithm
United extension F̂
F̂ : S(A)→ S(B)
where, S(A), S(B) is the family of subset of A, B.
F̂(X) =⋃
X∈S(A)
{F(X) | x ∈ X}
United extensions are inclusion monotonic
Marco De Angelis The interval Fourier transform May 17, 2021 8 / 22
Introduction Problem statement Interval analysis Main algorithm
Interval extension [F ]
F : Rn → C
[F ]({x}) = F(x),
where, {x} is a degenerate interval.
Marco De Angelis The interval Fourier transform May 17, 2021 9 / 22
Introduction Problem statement Interval analysis Main algorithm
Fundamental theorems of IA
Theorem 1 (United extension)
If [f ] is an inclusion monotonic extension of f then,
F̂(x) ⊆ [F ](x). (1)
Theorem 2 (Natural extension)
Any real function whose entries have been replaced with intervals is inclusionmonotonic.
f(x) = x− x2, [f ](X) = X −X2. (2)
Marco De Angelis The interval Fourier transform May 17, 2021 10 / 22
Introduction Problem statement Interval analysis Main algorithm
Interval discrete Fourier transform.
n−1∑k=0
Xk e−i 2π
nhk
0 20 40 60 80 100 120t
10
5
0
5
10
15
20
[x] 0 10 20 30 40 50 60
h
150
100
50
0
50
100
150
Re[z
]
0 10 20 30 40 50 60h
100
50
0
50
100
150
Im[z
]Marco De Angelis The interval Fourier transform May 17, 2021 11 / 22
Introduction Problem statement Interval analysis Main algorithm
Computing [F ]
[Fh](x) =
n−1∑k=0
Xk e−i 2π
nhk
X = [X1, X2] = {x ∈ Rn|X1 ≤ x ≤ X2}
[F ] =
[inf
X1≤x≤X2
F(x), supX1≤x≤X2
F(x)
]
For the united extension extra computational power is needed.
Marco De Angelis The interval Fourier transform May 17, 2021 12 / 22
Introduction Problem statement Interval analysis Main algorithm
Example, f : R6 → C, f(x) = b xT
b = ( 1 + i2 −2 + i2 3 + i2 ...
4− i2 −5− i2 6 + i2 )
XT =
[−1, 1]
[−1, 1]
[−1, 1]
[−1, 1]
[−1, 1]
[−1, 1]
20 15 10 5 0 5 10 15 20
Re
10
5
0
5
10
Im
Marco De Angelis The interval Fourier transform May 17, 2021 13 / 22
Introduction Problem statement Interval analysis Main algorithm
n−1∑k=0
Xk
(cos
2π
nhk − i sin 2π
nhk
)
Xk
2 4
rk ik
rk = 0.874
ik = 1.724
1.748 3.496
-6.896 -3.448
1 2 3 4
−6
−5
−4
−3
Re
Im
Marco De Angelis The interval Fourier transform May 17, 2021 14 / 22
Introduction Problem statement Interval analysis Main algorithm
n−1∑k=0
Xk
(cos
2π
nhk − i sin 2π
nhk
)
Xk
2 4
rk ik
rk = 0.874
ik = 1.724
1.748 3.496
-6.896 -3.448
1 2 3 4
−6
−5
−4
−3
Re
Im
Marco De Angelis The interval Fourier transform May 17, 2021 14 / 22
Introduction Problem statement Interval analysis Main algorithm
n−1∑k=0
Xk
(cos
2π
nhk − i sin 2π
nhk
)
Xk
2 4
rk ik
rk = 0.874
ik = 1.724
1.748 3.496
-6.896 -3.448
1 2 3 4
−6
−5
−4
−3
Re
Im
Marco De Angelis The interval Fourier transform May 17, 2021 14 / 22
Introduction Problem statement Interval analysis Main algorithm
n−1∑k=0
Xk
(cos
2π
nhk − i sin 2π
nhk
)
Xk
2 4
rk ik
rk = 0.874
ik = 1.724
1.748 3.496
-6.896 -3.448
1 2 3 4
−6
−5
−4
−3
Re
Im
Marco De Angelis The interval Fourier transform May 17, 2021 14 / 22
Introduction Problem statement Interval analysis Main algorithm
n−1∑k=0
Xk
(cos
2π
nhk − i sin 2π
nhk
)
Xk
2 4
rk ik
rk = 0.874
ik = 1.724
1.748 3.496
-6.896 -3.448
1 2 3 4
−6
−5
−4
−3
Re
Im
Marco De Angelis The interval Fourier transform May 17, 2021 14 / 22
Introduction Problem statement Interval analysis Main algorithm
n−1∑k=0
Xk
(cos
2π
nhk − i sin 2π
nhk
)
Xk
2 4
rk ik
rk = 0.874
ik = 1.724
1.748 3.496
-6.896 -3.448
1 2 3 4
−6
−5
−4
−3
Re
Im
Marco De Angelis The interval Fourier transform May 17, 2021 14 / 22
Introduction Problem statement Interval analysis Main algorithm
n−1∑k=0
Xk
(cos
2π
nhk − i sin 2π
nhk
)
Xk
2 4
rk ik
rk = 0.874
ik = 1.724
1.748 3.496
-6.896 -3.448
1 2 3 4
−6
−5
−4
−3
Re
Im
Marco De Angelis The interval Fourier transform May 17, 2021 14 / 22
Introduction Problem statement Interval analysis Main algorithm
n−1∑k=0
Xk
(cos
2π
nhk − i sin 2π
nhk
)
Xk
2 4
rk ik
rk = 0.874
ik = 1.724
1.748 3.496
-6.896 -3.448
1 2 3 4
−6
−5
−4
−3
Re
Im
Marco De Angelis The interval Fourier transform May 17, 2021 14 / 22
Introduction Problem statement Interval analysis Main algorithm
n−1∑k=0
Xk · rk + i Xk · ik, rk, ik ∈ R
X0 · r0 + X1 · r1 + X2 · r2 + ...
i X0 · i0 + X1 · i1 + X2 · i2 + ...
−3 −2 −1 1 2 3
−3
−2
1
2
3
Re
Im
+−3 −2 −1 1 2 3
−3
−2
1
2
3
Re
Im
+−3 −2 −1 1 2 3
−3
−2
1
2
3
Re
Im
+ ...
Marco De Angelis The interval Fourier transform May 17, 2021 15 / 22
Introduction Problem statement Interval analysis Main algorithm
n−1∑k=0
Xk · rk + i Xk · ik, rk, ik ∈ R
X0 · r0 + X1 · r1 + X2 · r2 + ...
i X0 · i0 + X1 · i1 + X2 · i2 + ...
−3 −2 −1 1 2 3
−3
−2
1
2
3
Re
Im
+−3 −2 −1 1 2 3
−3
−2
1
2
3
Re
Im
+−3 −2 −1 1 2 3
−3
−2
1
2
3
Re
Im
+ ...
Marco De Angelis The interval Fourier transform May 17, 2021 15 / 22
Introduction Problem statement Interval analysis Main algorithm
n−1∑k=0
Xk · rk + i Xk · ik, rk, ik ∈ R
X0 · r0 + X1 · r1 + X2 · r2 + ...
i X0 · i0 + X1 · i1 + X2 · i2 + ...
−3 −2 −1 1 2 3
−3
−2
1
2
3
Re
Im
+−3 −2 −1 1 2 3
−3
−2
1
2
3
Re
Im
+−3 −2 −1 1 2 3
−3
−2
1
2
3
Re
Im
+ ...
Marco De Angelis The interval Fourier transform May 17, 2021 15 / 22
Introduction Problem statement Interval analysis Main algorithm
n−1∑k=0
Xk · rk + i Xk · ik, rk, ik ∈ R
X0 · r0 + X1 · r1 + X2 · r2 + ...
i X0 · i0 + X1 · i1 + X2 · i2 + ...
−3 −2 −1 1 2 3
−3
−2
1
2
3
Re
Im
+−3 −2 −1 1 2 3
−3
−2
1
2
3
Re
Im
+−3 −2 −1 1 2 3
−3
−2
1
2
3
Re
Im
+ ...
Marco De Angelis The interval Fourier transform May 17, 2021 15 / 22
Introduction Problem statement Interval analysis Main algorithm
n−1∑k=0
Xk · rk + i Xk · ik, rk, ik ∈ R
X0 · r0 + X1 · r1 + X2 · r2 + ...
i X0 · i0 + X1 · i1 + X2 · i2 + ...
−3 −2 −1 1 2 3
−3
−2
1
2
3
Re
Im
+
−3 −2 −1 1 2 3
−3
−2
1
2
3
Re
Im
+−3 −2 −1 1 2 3
−3
−2
1
2
3
Re
Im
+ ...
Marco De Angelis The interval Fourier transform May 17, 2021 15 / 22
Introduction Problem statement Interval analysis Main algorithm
n−1∑k=0
Xk · rk + i Xk · ik, rk, ik ∈ R
X0 · r0 + X1 · r1 + X2 · r2 + ...
i X0 · i0 + X1 · i1 + X2 · i2 + ...
−3 −2 −1 1 2 3
−3
−2
1
2
3
Re
Im
+−3 −2 −1 1 2 3
−3
−2
1
2
3
Re
Im
+−3 −2 −1 1 2 3
−3
−2
1
2
3
Re
Im
+ ...
Marco De Angelis The interval Fourier transform May 17, 2021 15 / 22
Introduction Problem statement Interval analysis Main algorithm
n−1∑k=0
Xk · rk + i Xk · ik, rk, ik ∈ R
X0 · r0 + X1 · r1 + X2 · r2 + ...
i X0 · i0 + X1 · i1 + X2 · i2 + ...
−3 −2 −1 1 2 3
−3
−2
1
2
3
Re
Im
+−3 −2 −1 1 2 3
−3
−2
1
2
3
Re
Im
+
−3 −2 −1 1 2 3
−3
−2
1
2
3
Re
Im
+ ...
Marco De Angelis The interval Fourier transform May 17, 2021 15 / 22
Introduction Problem statement Interval analysis Main algorithm
n−1∑k=0
Xk · rk + i Xk · ik, rk, ik ∈ R
X0 · r0 + X1 · r1 + X2 · r2 + ...
i X0 · i0 + X1 · i1 + X2 · i2 + ...
−3 −2 −1 1 2 3
−3
−2
1
2
3
Re
Im
+−3 −2 −1 1 2 3
−3
−2
1
2
3
Re
Im
+−3 −2 −1 1 2 3
−3
−2
1
2
3
Re
Im
+ ...
Marco De Angelis The interval Fourier transform May 17, 2021 15 / 22
Introduction Problem statement Interval analysis Main algorithm
n−1∑k=0
Xk · rk + i Xk · ik, rk, ik ∈ R
X0 · r0 + X1 · r1 + X2 · r2 + ...
i X0 · i0 + X1 · i1 + X2 · i2 + ...
−3 −2 −1 1 2 3
−3
−2
1
2
3
Re
Im
+−3 −2 −1 1 2 3
−3
−2
1
2
3
Re
Im
+−3 −2 −1 1 2 3
−3
−2
1
2
3
Re
Im
+ ...
Marco De Angelis The interval Fourier transform May 17, 2021 15 / 22
Introduction Problem statement Interval analysis Main algorithm
Sum of dependent intervals
−2 2 4
−2
2
4
Re
Im
+−2 2 4
−2
2
4
Re
Im
=−2 2 4
−2
2
4
Re
Im
Marco De Angelis The interval Fourier transform May 17, 2021 16 / 22
Introduction Problem statement Interval analysis Main algorithm
Sum of dependent intervals
−2 2 4
−2
2
4
Re
Im
+−2 2 4
−2
2
4
Re
Im
=
−2 2 4
−2
2
4
Re
Im
Marco De Angelis The interval Fourier transform May 17, 2021 16 / 22
Introduction Problem statement Interval analysis Main algorithm
Sum of dependent intervals
−2 2 4
−2
2
4
Re
Im
+−2 2 4
−2
2
4
Re
Im
=−2 2 4
−2
2
4
Re
Im
Marco De Angelis The interval Fourier transform May 17, 2021 16 / 22
Introduction Problem statement Interval analysis Main algorithm
Sum of dependent intervals
−2 2 4
−2
2
4
Re
Im
+−2 2 4
−2
2
4
Re
Im
=−2 2 4
−2
2
4
Re
Im
The sum is not closed w.r.t. the diagonals.
Marco De Angelis The interval Fourier transform May 17, 2021 16 / 22
Introduction Problem statement Interval analysis Main algorithm
−2 2 4
−2
2
4
Re
Im
+ −3 −2 −1 1 2 3
−3
−2
1
2
3
Re
Im
=−2 2 4 6
−4
−2
2
4
6
Re
Im
Marco De Angelis The interval Fourier transform May 17, 2021 17 / 22
Introduction Problem statement Interval analysis Main algorithm
−2 2 4
−2
2
4
Re
Im
+ −3 −2 −1 1 2 3
−3
−2
1
2
3
Re
Im
=−2 2 4 6
−4
−2
2
4
6
Re
Im
Marco De Angelis The interval Fourier transform May 17, 2021 17 / 22
Introduction Problem statement Interval analysis Main algorithm
−2 2 4
−2
2
4
Re
Im
+ −3 −2 −1 1 2 3
−3
−2
1
2
3
Re
Im
=
−2 2 4 6
−4
−2
2
4
6
Re
Im
Marco De Angelis The interval Fourier transform May 17, 2021 17 / 22
Introduction Problem statement Interval analysis Main algorithm
−2 2 4
−2
2
4
Re
Im
+ −3 −2 −1 1 2 3
−3
−2
1
2
3
Re
Im
=−2 2 4 6
−4
−2
2
4
6
Re
Im
Marco De Angelis The interval Fourier transform May 17, 2021 17 / 22
Introduction Problem statement Interval analysis Main algorithm
Main theorems (Minkowski addition)
Theorem 3 (Linear transformation of convex sets)
Let A ⊆ Rn be a convex set and suppose that f is the linear map of f : Rn → Rm.Then f maps the extreme points of A onto the extreme points of f(A).
Theorem 4 (Sum of a convex set plus a segment)
The maximum distance between a convex polygon and a segment is attained atthe endpoints of the segment.
Marco De Angelis The interval Fourier transform May 17, 2021 18 / 22
Introduction Problem statement Interval analysis Main algorithm
Proof (intuition)
4 2 0 2 4Re
4
2
0
2
4Im
Marco De Angelis The interval Fourier transform May 17, 2021 19 / 22
Introduction Problem statement Interval analysis Main algorithm
Algorithm
0 1 2 3 4Re
3.0
2.5
2.0
1.5
1.0
0.5
0.0
0.5
1.0Im
h=12, k=0,...,10, Vertices: 20
0.6 0.7Re
0.25
0.30
Im
k=10
Marco De Angelis The interval Fourier transform May 17, 2021 20 / 22
Introduction Problem statement Interval analysis Main algorithm
Centered form and MC samples
16.4 16.2 16.0 15.8 15.6 15.4 15.2 15.0Re
7.00
7.25
7.50
7.75
8.00
8.25
8.50
8.75
9.00
Im
n=128, h=27, k=0,...,10Endpoints 1024
Marco De Angelis The interval Fourier transform May 17, 2021 21 / 22
Introduction Problem statement Interval analysis Main algorithm
Centered form and MC samples
17.5 17.0 16.5 16.0 15.5 15.0 14.5 14.0Re
6.0
6.5
7.0
7.5
8.0
8.5
9.0
9.5
Im
n=128, h=27, k=0,...,11Endpoints 2048MC 5000
Marco De Angelis The interval Fourier transform May 17, 2021 21 / 22
Introduction Problem statement Interval analysis Main algorithm
Centered form and MC samples
22 20 18 16 14 12 10Re
2
4
6
8
10
12
14
Im
n=128, h=27, k=0,...,88Endpoints 2048MC 5000Vertices 14466
Marco De Angelis The interval Fourier transform May 17, 2021 21 / 22
Introduction Problem statement Interval analysis Main algorithm
Centered form and MC samples
24 22 20 18 16 14 12 10 8Re
0
2
4
6
8
10
12
14
16
Im
n=128, h=27, k=0,...,127Endpoints 2048MC 5000Vertices 24450
Marco De Angelis The interval Fourier transform May 17, 2021 21 / 22
Introduction Problem statement Interval analysis Main algorithm
Computational cost and summary
Brute Force O(2N )
Combing algorithmO(n2 log n)
Interval extension O(n)
IntervalFourier transform
0 20 40 60 80 100 120t
10
5
0
5
10
15
20
[x]
0 10 20 30 40 50 60h
0
20
40
60
80
100
120
140
160
|Z|
Marco De Angelis The interval Fourier transform May 17, 2021 22 / 22
top related