marco de angelis

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


Outline

1 Introduction

2 Problem statement

3 Interval analysis

4 Main algorithm



Motivations

Imprecise/poor sensormeasurements

Missing data

Relaxed assumptions

=

+

+



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.



The discrete Fourier transform

Fh : Rn → C, ∀h ∈ Z+

Fh(x) :=

n−1∑k=0

xk e−i 2π

nhk




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

)




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




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




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



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|



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



Interval extension [F ]

F : Rn → C

[F ]({x}) = F(x),

where, {x} is a degenerate interval.



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)



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


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.



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



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



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

+ ...



n−1∑k=0


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

+ ...



n−1∑k=0


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

+ ...



n−1∑k=0


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

+ ...



n−1∑k=0


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

+ ...



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




−2 2 4

−2

2

4

Re

Im

+−2 2 4

−2

2

4

Re

Im

=

−2 2 4

−2

2

4

Re

Im




−2 2 4

−2

2

4

Re

Im

+−2 2 4

−2

2

4

Re

Im

=−2 2 4

−2

2

4

Re

Im




−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.



−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



−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



−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



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.



Proof (intuition)

4 2 0 2 4Re

4

2

0

2

4Im



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



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




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




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




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



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|


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