Signal decomposition via SuperEMD Maria van der Walt 1 Joint work - - PowerPoint PPT Presentation

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Signal decomposition via SuperEMD Maria van der Walt 1 Joint work - - PowerPoint PPT Presentation

Signal decomposition via SuperEMD Maria van der Walt 1 Joint work with Charles Chui 2 and Hrushikesh Mhaskar 3 1 Westmont College 2 Stanford University / Hong Kong Baptist University 3 California Institute of Technology / Claremont Graduate


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SLIDE 1

Signal decomposition via SuperEMD

Maria van der Walt1 Joint work with Charles Chui2 and Hrushikesh Mhaskar3

1Westmont College 2Stanford University / Hong Kong Baptist University 3California Institute of Technology / Claremont Graduate University

MWAA October 6-8, 2017

Maria van der Walt | Westmont College 1/ 21

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SLIDE 2

Introduction

Maria van der Walt | Westmont College 2/ 21

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SLIDE 3

Introduction

  • Stationary signal:

f(t) = a0 +

N

  • j=1

aj cos(2π(ωjt + ηj)), t ∈ R, for arbitrary real values aj > 0, ηj and ωj > 0 (frequency).

Maria van der Walt | Westmont College 2/ 21

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SLIDE 4

Introduction

  • Stationary signal:

f(t) = a0 +

N

  • j=1

aj cos(2π(ωjt + ηj)), t ∈ R, for arbitrary real values aj > 0, ηj and ωj > 0 (frequency).

Maria van der Walt | Westmont College 2/ 21

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SLIDE 5

Introduction

  • Stationary signal:

f(t) = a0 +

N

  • j=1

aj cos(2π(ωjt + ηj)), t ∈ R, for arbitrary real values aj > 0, ηj and ωj > 0 (frequency).

  • Non-stationary signal:

Adaptive harmonic model (AHM): f(t) = A0(t) +

N

  • j=1

Aj(t) cos(2πφj(t)), t ∈ R, where Aj(t) > 0 and each φj(t) is a general C1 function such that φ′

j(t) > 0 (instantaneous frequency).

Maria van der Walt | Westmont College 2/ 21

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SLIDE 6

Introduction

  • Stationary signal:

f(t) = a0 +

N

  • j=1

aj cos(2π(ωjt + ηj)), t ∈ R, for arbitrary real values aj > 0, ηj and ωj > 0 (frequency).

  • Non-stationary signal:

Adaptive harmonic model (AHM): f(t) = A0(t) +

N

  • j=1

Aj(t) cos(2πφj(t)), t ∈ R, where Aj(t) > 0 and each φj(t) is a general C1 function such that φ′

j(t) > 0 (instantaneous frequency).

Maria van der Walt | Westmont College 2/ 21

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SLIDE 7

Introduction

  • Stationary signal:

f(t) = a0 +

N

  • j=1

aj cos(2π(ωjt + ηj)), t ∈ R, for arbitrary real values aj > 0, ηj and ωj > 0 (frequency).

  • Non-stationary signal:

Adaptive harmonic model (AHM): f(t) = A0(t) +

N

  • j=1

Aj(t) cos(2πφj(t)), t ∈ R, where Aj(t) > 0 and each φj(t) is a general C1 function such that φ′

j(t) > 0 (instantaneous frequency).

Maria van der Walt | Westmont College 2/ 21

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SLIDE 8

Introduction

  • Stationary signal:

f(t) = a0 +

N

  • j=1

aj cos(2π(ωjt + ηj)), t ∈ R, for arbitrary real values aj > 0, ηj and ωj > 0 (frequency).

  • Non-stationary signal:

Adaptive harmonic model (AHM): f(t) = A0(t) +

N

  • j=1

Aj(t) cos(2πφj(t)), t ∈ R, where Aj(t) > 0 and each φj(t) is a general C1 function such that φ′

j(t) > 0 (instantaneous frequency).

Maria van der Walt | Westmont College 2/ 21

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SLIDE 9

Introduction

  • OBJECTIVE:

Maria van der Walt | Westmont College 3/ 21

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SLIDE 10

Introduction

  • OBJECTIVE:

Given a blind source signal f(t). Find:

Maria van der Walt | Westmont College 3/ 21

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SLIDE 11

Introduction

  • OBJECTIVE:

Given a blind source signal f(t). Find:

  • atoms Aj(t) cos(2πφj(t)), j = 1, . . . , N

Maria van der Walt | Westmont College 3/ 21

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SLIDE 12

Introduction

  • OBJECTIVE:

Given a blind source signal f(t). Find:

  • atoms Aj(t) cos(2πφj(t)), j = 1, . . . , N
  • instantaneous frequency (IF) φ′

j(t) of each atom

Maria van der Walt | Westmont College 3/ 21

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SLIDE 13

Introduction

  • OBJECTIVE:

Given a blind source signal f(t). Find:

  • atoms Aj(t) cos(2πφj(t)), j = 1, . . . , N
  • instantaneous frequency (IF) φ′

j(t) of each atom

  • trend A0(t) (if there is one)

Maria van der Walt | Westmont College 3/ 21

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SLIDE 14

Introduction

  • OBJECTIVE:

Given a blind source signal f(t). Find:

  • atoms Aj(t) cos(2πφj(t)), j = 1, . . . , N
  • instantaneous frequency (IF) φ′

j(t) of each atom

  • trend A0(t) (if there is one)

f(t) = A0(t) +

N

  • j=1

Aj(t) cos(2πφj(t))

Maria van der Walt | Westmont College 3/ 21

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SLIDE 15

EMD

Maria van der Walt | Westmont College 4/ 21

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SLIDE 16

EMD

  • EMD: Empirical mode decomposition
  • Introduced by Norden Huang and others (1998)
  • Data-driven way to analyze non-stationary signals

Maria van der Walt | Westmont College 4/ 21

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SLIDE 17

EMD

  • Given a signal f(t). Set h1,0 := f.

Maria van der Walt | Westmont College 5/ 21

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SLIDE 18

EMD

  • Find its local maxima and minima.

Maria van der Walt | Westmont College 5/ 21

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SLIDE 19

EMD

  • Compute the upper and lower envelopes through (standard)

cubic spline interpolation of the extrema.

Maria van der Walt | Westmont College 5/ 21

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SLIDE 20

EMD

  • Compute the mean envelope m1,1.

Maria van der Walt | Westmont College 5/ 21

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SLIDE 21

EMD

  • Compute the mean envelope m1,1.

Maria van der Walt | Westmont College 5/ 21

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SLIDE 22

EMD

  • Subtract the mean envelope from h1,0.

Maria van der Walt | Westmont College 5/ 21

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SLIDE 23

EMD

  • Repeat this process until h1,ℓ is an intrinsic mode function

(IMF).

Maria van der Walt | Westmont College 5/ 21

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SLIDE 24

EMD

  • Repeat this process until h1,ℓ is an intrinsic mode function

(IMF).

Maria van der Walt | Westmont College 5/ 21

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SLIDE 25

EMD

  • Repeat this process until h1,ℓ is an intrinsic mode function

(IMF).

Maria van der Walt | Westmont College 5/ 21

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SLIDE 26

EMD

  • Repeat this process until h1,ℓ is an intrinsic mode function

(IMF).

Maria van der Walt | Westmont College 5/ 21

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SLIDE 27

EMD

  • Repeat this process until h1,ℓ is an intrinsic mode function

(IMF). – symmetric about the time axis – # local extrema – # zeros = -1,0,1

Maria van der Walt | Westmont College 5/ 21

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SLIDE 28

EMD

  • Subtract the first IMF f1 from h1,0 and repeat the sifting

process to find f2, f3, . . . , fN until the remainder RN is a monotonic function.

Maria van der Walt | Westmont College 5/ 21

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SLIDE 29

EMD

  • Subtract the first IMF f1 from h1,0 and repeat the sifting

process to find f2, f3, . . . , fN until the remainder RN is a monotonic function. f(t) = 3

j=1 fj(t) + R3(t)

Maria van der Walt | Westmont College 5/ 21

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SLIDE 30

Hilbert spectral analysis

Maria van der Walt | Westmont College 6/ 21

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SLIDE 31

Hilbert spectral analysis

  • We note that each IMF fj can be written as

fj(t) = Re

  • f⋆

j (t)

  • = Re (fj(t) + i(Hfj)(t))

Maria van der Walt | Westmont College 6/ 21

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SLIDE 32

Hilbert spectral analysis

  • We note that each IMF fj can be written as

fj(t) = Re

  • f⋆

j (t)

  • = Re (fj(t) + i(Hfj)(t))

Hilbert transform: (Hg)(t) = p.v. 1 π ∞

−∞

g(u) t − u du, g measurable, real-valued

Maria van der Walt | Westmont College 6/ 21

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SLIDE 33

Hilbert spectral analysis

  • We note that each IMF fj can be written as

fj(t) = Re

  • f⋆

j (t)

  • = Re (fj(t) + i(Hfj)(t))

= Re

  • Bj(t)ei2πθj(t)

Maria van der Walt | Westmont College 6/ 21

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SLIDE 34

Hilbert spectral analysis

  • We note that each IMF fj can be written as

fj(t) = Re

  • f⋆

j (t)

  • = Re (fj(t) + i(Hfj)(t))

= Re

  • Bj(t)ei2πθj(t)

= Bj(t) cos(2πθj(t)),

Maria van der Walt | Westmont College 6/ 21

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SLIDE 35

Hilbert spectral analysis

  • We note that each IMF fj can be written as

fj(t) = Re

  • f⋆

j (t)

  • = Re (fj(t) + i(Hfj)(t))

= Re

  • Bj(t)ei2πθj(t)

= Bj(t) cos(2πθj(t)), with Bj(t) = |f⋆

j (t)| and θj(t) = 1

2π tan−1 Hfj(t) fj(t) and IF = θ′

j(t).

Maria van der Walt | Westmont College 6/ 21

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SLIDE 36

EMD + Hilbert spectral analysis

Final result:

Maria van der Walt | Westmont College 7/ 21

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SLIDE 37

EMD + Hilbert spectral analysis

Final result: f(t) = cos 2π(8t) + cos 2π(4t) + cos 2πt

Maria van der Walt | Westmont College 7/ 21

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SLIDE 38

EMD + Hilbert spectral analysis

Limitations of EMD with Hilbert spectral analysis:

Maria van der Walt | Westmont College 8/ 21

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SLIDE 39

EMD + Hilbert spectral analysis

Limitations of EMD with Hilbert spectral analysis:

  • The sifting process cannot distinguish between atoms with

frequencies that are very close together. In other words, a single IMF may actually contain more than one atom.

Maria van der Walt | Westmont College 8/ 21

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SLIDE 40

EMD + Hilbert spectral analysis

Limitations of EMD with Hilbert spectral analysis:

  • The sifting process cannot distinguish between atoms with

frequencies that are very close together. In other words, a single IMF may actually contain more than one atom. f(t) = cos 2π(5t) + cos 2π(4.9t)

Maria van der Walt | Westmont College 8/ 21

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SLIDE 41

EMD + Hilbert spectral analysis

Limitations of EMD with Hilbert spectral analysis:

  • The sifting process cannot distinguish between atoms with

frequencies that are very close together. In other words, a single IMF may actually contain more than one atom. f(t) = cos 2π(5t) + cos 2π(4.9t)

Maria van der Walt | Westmont College 8/ 21

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SLIDE 42

EMD + Hilbert spectral analysis

Limitations of EMD with Hilbert spectral analysis:

  • The sifting process cannot distinguish between atoms with

frequencies that are very close together. In other words, a single IMF may actually contain more than one atom.

  • We can get inaccurate results when computing the Hilbert

transform from discrete samples on a bounded interval.

Maria van der Walt | Westmont College 8/ 21

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SLIDE 43

EMD + Hilbert spectral analysis

Limitations of EMD with Hilbert spectral analysis:

  • The sifting process cannot distinguish between atoms with

frequencies that are very close together. In other words, a single IMF may actually contain more than one atom.

  • We can get inaccurate results when computing the Hilbert

transform from discrete samples on a bounded interval.

  • The standard cubic spline interpolation scheme used to

construct the upper and lower envelopes in the sifting process is not a local method.

Maria van der Walt | Westmont College 8/ 21

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SLIDE 44

Our contribution

Maria van der Walt | Westmont College 9/ 21

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SLIDE 45

Our contribution

Our solution to third limitation:

Maria van der Walt | Westmont College 9/ 21

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SLIDE 46

Our contribution

Our solution to third limitation: Given a function f, we construct an interpolation operator P in terms of cubic B-splines Nt,4,j ...

Maria van der Walt | Westmont College 9/ 21

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SLIDE 47

Our contribution

Our solution to third limitation: Given a function f, we construct an interpolation operator P in terms of cubic B-splines Nt,4,j ...

Maria van der Walt | Westmont College 9/ 21

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SLIDE 48

Our contribution

Our solution to third limitation: Given a function f, we construct an interpolation operator P in terms of cubic B-splines Nt,4,j ...

  • non-decreasing knot sequence t
  • mth order B-splines:

Nt,1,j(t) :=

  • 1,

if tj ≤ t < tj+1 0,

  • therwise

Nt,m,j(t) := wm,jNt,m−1,j(t) + (1 − wm,j+1)Nt,m−1,j+1(t) with wm,j(t) :=

  • t−tj

tj+m−1−tj

if tj = tj+m−1 0,

  • therwise

Maria van der Walt | Westmont College 9/ 21

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SLIDE 49

Our contribution

Our solution to third limitation: Given a function f, we construct an interpolation operator P in terms of cubic B-splines Nt,4,j ...

Maria van der Walt | Westmont College 9/ 21

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SLIDE 50

Our contribution

Our solution to third limitation: Given a function f, we construct an interpolation operator P in terms of cubic B-splines Nt,4,j such that:

Maria van der Walt | Westmont College 9/ 21

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SLIDE 51

Our contribution

Our solution to third limitation: Given a function f, we construct an interpolation operator P in terms of cubic B-splines Nt,4,j such that: (i) P has a local formulation;

Maria van der Walt | Westmont College 9/ 21

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SLIDE 52

Our contribution

Our solution to third limitation: Given a function f, we construct an interpolation operator P in terms of cubic B-splines Nt,4,j such that: (i) P has a local formulation; (ii) P preserves polynomials of degree ≤ 3;

Maria van der Walt | Westmont College 9/ 21

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SLIDE 53

Our contribution

Our solution to third limitation: Given a function f, we construct an interpolation operator P in terms of cubic B-splines Nt,4,j such that: (i) P has a local formulation; (ii) P preserves polynomials of degree ≤ 3; (iii) Pf interpolates f at a = t0 < t1 < · · · < tn = b;

Maria van der Walt | Westmont College 9/ 21

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SLIDE 54

Our contribution

Our solution to third limitation: Given a function f, we construct an interpolation operator P in terms of cubic B-splines Nt,4,j such that: (i) P has a local formulation; (ii) P preserves polynomials of degree ≤ 3; (iii) Pf interpolates f at a = t0 < t1 < · · · < tn = b; (iv) P preserves derivatives (up to order 3) of f at a and b.

Maria van der Walt | Westmont College 9/ 21

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SLIDE 55

Our contribution

Our solution to third limitation: Given a function f, we construct an interpolation operator P in terms of cubic B-splines Nt,4,j such that: (i) P has a local formulation; Q (ii) P preserves polynomials of degree ≤ 3; Q (iii) Pf interpolates f at a = t0 < t1 < · · · < tn = b; (iv) P preserves derivatives (up to order 3) of f at a and b.

Maria van der Walt | Westmont College 9/ 21

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SLIDE 56

Our contribution

Our solution to third limitation: Given a function f, we construct an interpolation operator P in terms of cubic B-splines Nt,4,j such that: (i) P has a local formulation; Q, R (ii) P preserves polynomials of degree ≤ 3; Q (iii) Pf interpolates f at a = t0 < t1 < · · · < tn = b; R (iv) P preserves derivatives (up to order 3) of f at a and b. R

Maria van der Walt | Westmont College 9/ 21

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SLIDE 57

Our contribution

Our solution to third limitation: Given a function f, we construct an interpolation operator P in terms of cubic B-splines Nt,4,j such that: (i) P has a local formulation; Q, R (ii) P preserves polynomials of degree ≤ 3; Q (iii) Pf interpolates f at a = t0 < t1 < · · · < tn = b; R (iv) P preserves derivatives (up to order 3) of f at a and b. R Blending operator [Chui, Diamond, 1990]: P := Q + R − RQ

Maria van der Walt | Westmont College 9/ 21

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SLIDE 58

Blending operator

Quasi-interpolant (Qf)(t) :=

−1

  • j=−3

f(−j)(a)Mj(t) +

n

  • j=0

f(tj)Mj(t) +

n+2

  • j=n+1

f(j−n)(b)Mj(t)

Maria van der Walt | Westmont College 10/ 21

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SLIDE 59

Blending operator

Quasi-interpolant (Qf)(t) :=

−1

  • j=−3

f(−j)(a)Mj(t) +

n

  • j=0

f(tj)Mj(t) +

n+2

  • j=n+1

f(j−n)(b)Mj(t) Spline molecules [Chen, Chui, Lai, 1988; Chui, vdW, 2015]: Mj(t) =                         

j+3

  • k=0

aj,kNt,4,k−3(t), j = −3, . . . , −1;

3

  • k=0

aj,kNt,4,j+k−3(t), j = 0, . . . , n − 1;

2

  • k=j−n

aj,kNt,4,n+k−3(t), j = n, . . . , n + 2.

Maria van der Walt | Westmont College 10/ 21

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SLIDE 60

Blending operator

Quasi-interpolant (Qf)(t) :=

−1

  • j=−3

f(−j)(a)Mj(t) +

n

  • j=0

f(tj)Mj(t) +

n+2

  • j=n+1

f(j−n)(b)Mj(t) Spline molecules [Chen, Chui, Lai, 1988; Chui, vdW, 2015]: Mj(t) =                         

j+3

  • k=0

aj,kNt,4,k−3(t), j = −3, . . . , −1;

3

  • k=0

aj,kNt,4,j+k−3(t), j = 0, . . . , n − 1;

2

  • k=j−n

aj,kNt,4,n+k−3(t), j = n, . . . , n + 2. Find spline coefficients aj,k such that (Qp)(t) = p(t), p ∈ π3.

Maria van der Walt | Westmont College 10/ 21

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SLIDE 61

Blending operator

Local interpolant (Rf)(t) :=

−1

  • j=−3

f(−j)(a)Lj(t) +

n

  • j=0

f(tj)Lj(t) +

n+2

  • j=n+1

f(j−n)(b)Lj(t)

Maria van der Walt | Westmont College 11/ 21

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SLIDE 62

Blending operator

Local interpolant (Rf)(t) :=

−1

  • j=−3

f(−j)(a)Lj(t) +

n

  • j=0

f(tj)Lj(t) +

n+2

  • j=n+1

f(j−n)(b)Lj(t) Spline molecules [Chui, vdW, 2015]: Lj(t) =                       

3

  • k=0

bj,kN˜

tk−3,4(t),

j = −3, . . . , 0; N˜

t,4,2j(t)

t,4,2j(tj),

j = 1, . . . , n − 1;

2

  • k=0

bj,kN˜

tn+k,4(t),

j = n, . . . , n + 2.

Maria van der Walt | Westmont College 11/ 21

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SLIDE 63

Blending operator

Local interpolant (Rf)(t) :=

−1

  • j=−3

f(−j)(a)Lj(t) +

n

  • j=0

f(tj)Lj(t) +

n+2

  • j=n+1

f(j−n)(b)Lj(t) Spline molecules [Chui, vdW, 2015]:

Maria van der Walt | Westmont College 11/ 21

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SLIDE 64

Blending operator

Local interpolant (Rf)(t) :=

−1

  • j=−3

f(−j)(a)Lj(t) +

n

  • j=0

f(tj)Lj(t) +

n+2

  • j=n+1

f(j−n)(b)Lj(t) Spline molecules [Chui, vdW, 2015]: Lj(t) =                       

3

  • k=0

bj,kN˜

tk−3,4(t),

j = −3, . . . , 0; N˜

t,4,2j(t)

t,4,2j(tj),

j = 1, . . . , n − 1; → Lj(tℓ) = δj,ℓ

2

  • k=0

bj,kN˜

tn+k,4(t),

j = n, . . . , n + 2.

Maria van der Walt | Westmont College 11/ 21

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SLIDE 65

Blending operator

Local interpolant (Rf)(t) :=

−1

  • j=−3

f(−j)(a)Lj(t) +

n

  • j=0

f(tj)Lj(t) +

n+2

  • j=n+1

f(j−n)(b)Lj(t) Spline molecules [Chui, vdW, 2015]: Lj(t) =                       

3

  • k=0

bj,kN˜

tk−3,4(t),

j = −3, . . . , 0; → L(ℓ)

j (a) = δ−j,ℓ

t,4,2j(t)

t,4,2j(tj),

j = 1, . . . , n − 1; → Lj(tℓ) = δj,ℓ

2

  • k=0

bj,kN˜

tn+k,4(t),

j = n, . . . , n + 2. → L(ℓ)

j (b) = δj−n,ℓ

Maria van der Walt | Westmont College 11/ 21

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SLIDE 66

Blending operator

Local interpolant (Rf)(t) :=

−1

  • j=−3

f(−j)(a)Lj(t) +

n

  • j=0

f(tj)Lj(t) +

n+2

  • j=n+1

f(j−n)(b)Lj(t) Spline molecules [Chui, vdW, 2015]: Lj(t) =                       

3

  • k=0

bj,kN˜

tk−3,4(t),

j = −3, . . . , 0; → L(ℓ)

j (a) = δ−j,ℓ

t,4,2j(t)

t,4,2j(tj),

j = 1, . . . , n − 1; → Lj(tℓ) = δj,ℓ

2

  • k=0

bj,kN˜

tn+k,4(t),

j = n, . . . , n + 2. → L(ℓ)

j (b) = δj−n,ℓ

Easy to show: (Rf)(tℓ)=f(tℓ) and R preserves derivatives at a, b.

Maria van der Walt | Westmont College 11/ 21

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SLIDE 67

Blending operator

Blending: P := Q + R − RQ Therefore [Chui, vdW, 2015]:

Maria van der Walt | Westmont College 12/ 21

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SLIDE 68

Blending operator

Blending: P := Q + R − RQ Therefore [Chui, vdW, 2015]: (Pp)(t) = (Qp)(t) + (Rp)(t) − (R(Qp))(t) = p(t) + (Rp)(t) − (Rp)(t) = p(t), p ∈ π3;

Maria van der Walt | Westmont College 12/ 21

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SLIDE 69

Blending operator

Blending: P := Q + R − RQ Therefore [Chui, vdW, 2015]: (Pp)(t) = p(t), p ∈ π3; (Pf)(tj) = (Qf)(tj) + (Rf)(tj) − (R(Qf))(tj) = (Qf)(tj) + f(tj) − (Qf)(tj) = f(tj), j = 0, . . . , n;

Maria van der Walt | Westmont College 12/ 21

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SLIDE 70

Blending operator

Blending: P := Q + R − RQ Therefore [Chui, vdW, 2015]: (Pp)(t) = p(t), p ∈ π3; (Pf)(tj) = f(tj), j = 0, . . . , n; (Pf)(j)(a) = f(j)(a), j = 1, 2, 3; (Pf)(j)(b) = f(j)(b), j = 1, 2.

Maria van der Walt | Westmont College 12/ 21

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SLIDE 71

EMD + Hilbert spectral analysis

Limitations of EMD with Hilbert spectral analysis:

  • The sifting process cannot distinguish between atoms with

frequencies that are very close together. In other words, a single IMF may actually contain more than one atom.

  • We can get inaccurate results when computing the Hilbert

transform from discrete samples on a bounded interval.

  • The standard cubic spline interpolation scheme used to

construct the upper and lower envelopes in the sifting process is not a local method.

Maria van der Walt | Westmont College 13/ 21

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SLIDE 72

EMD + Hilbert spectral analysis

Limitations of EMD with Hilbert spectral analysis:

  • The sifting process cannot distinguish between atoms with

frequencies that are very close together. In other words, a single IMF may actually contain more than one atom.

  • We can get inaccurate results when computing the Hilbert

transform from discrete samples on a bounded interval. The standard cubic spline interpolation scheme used to construct the upper and lower envelopes in the sifting process is not a local method – use blending operator.

Maria van der Walt | Westmont College 13/ 21

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SLIDE 73

EMD + Hilbert spectral analysis

Limitations of EMD with Hilbert spectral analysis:

  • The sifting process cannot distinguish between atoms with

frequencies that are very close together. In other words, a single IMF may actually contain more than one atom.

  • We can get inaccurate results when computing the Hilbert

transform from discrete samples on a bounded interval. The standard cubic spline interpolation scheme used to construct the upper and lower envelopes in the sifting process is not a local method – use blending operator.

Maria van der Walt | Westmont College 13/ 21

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SLIDE 74

Our contribution

Our solution to first two limitations:

Maria van der Walt | Westmont College 14/ 21

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SLIDE 75

Our contribution

Our solution to first two limitations:

  • Apply signal separation operator (SSO) [Chui, Mhaskar, 2015]

to each IMF from EMD separately (instead of Hilbert spectral analysis).

Maria van der Walt | Westmont College 14/ 21

slide-76
SLIDE 76

Our contribution

Our solution to first two limitations:

  • Apply signal separation operator (SSO) [Chui, Mhaskar, 2015]

to each IMF from EMD separately (instead of Hilbert spectral analysis).

  • SSO is able to identify very close-by frequencies in a given

IMF and recover/reconstruct the individual atoms associated with these frequencies.

Maria van der Walt | Westmont College 14/ 21

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SLIDE 77

Our contribution

Our solution to first two limitations:

  • Apply signal separation operator (SSO) [Chui, Mhaskar, 2015]

to each IMF from EMD separately (instead of Hilbert spectral analysis).

  • SSO is able to identify very close-by frequencies in a given

IMF and recover/reconstruct the individual atoms associated with these frequencies.

  • SSO is a direct, local method which produces more accurate

results in near real-time.

Maria van der Walt | Westmont College 14/ 21

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SLIDE 78

Our contribution

Our solution to first two limitations:

  • Apply signal separation operator (SSO) [Chui, Mhaskar, 2015]

to each IMF from EMD separately (instead of Hilbert spectral analysis).

  • SSO is able to identify very close-by frequencies in a given

IMF and recover/reconstruct the individual atoms associated with these frequencies.

  • SSO is a direct, local method which produces more accurate

results in near real-time.

  • EMD + SSO = “SuperEMD”

Maria van der Walt | Westmont College 14/ 21

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SLIDE 79

AHM

Adaptive harmonic model (AHM): f(t) = A0(t) +

N

  • j=1

Aj(t) cos(2πφj(t)), t ∈ R

Maria van der Walt | Westmont College 15/ 21

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SLIDE 80

AHM

Adaptive harmonic model (AHM): f(t) = A0(t) +

N

  • j=1

Aj(t) cos(2πφj(t)), t ∈ R Assumptions:

  • Aj ∈ C(R), Aj(t) > 0 and φj ∈ C1(R), φ′

j(t) > 0

  • There exists α = α(t) > 0 s.t. for any u with

|u| ≤ α−1(8πB)−1/2, where B = B(t) := max

j

φ′

j(t):

  • |Aj(t + u) − Aj(t)| ≤ α3|u|Aj(t)
  • |φ′

j(t + u) − φ′ j(t)| ≤ α3|u|φ′ j(t)

  • There exists η = η(t) s.t. min

j=k |φ′ j(t) − φ′ k(t)| =: 2Bη π

> 0 Note: M = M(t) :=

N

  • j=1

Aj(t) and µ = µ(t) := min

1≤j≤N Aj(t)

Maria van der Walt | Westmont College 15/ 21

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SLIDE 81

SuperEMD

Maria van der Walt | Westmont College 16/ 21

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SLIDE 82

SuperEMD

Given IMF fj(t) from (modified) EMD, containing Nj true atoms.

Maria van der Walt | Westmont College 16/ 21

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SLIDE 83

SuperEMD

Given IMF fj(t) from (modified) EMD, containing Nj true atoms.

  • Apply SSO to IMF fj [Chui, Mhaskar, vdW 2016]:

(Ta,δfj)(t, θ) = 1

  • k∈Z h

k

a

  • ℓ∈Z

h ℓ a

  • eiℓθfj(t − ℓδ)

(t ∈ R, θ ∈ [−π, π])

Maria van der Walt | Westmont College 16/ 21

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SLIDE 84

SuperEMD

Given IMF fj(t) from (modified) EMD, containing Nj true atoms.

  • Apply SSO to IMF fj [Chui, Mhaskar, vdW 2016]:

(Ta,δfj)(t, θ) = 1

  • k∈Z h

k

a

  • ℓ∈Z

h ℓ a

  • eiℓθfj(t − ℓδ)

(t ∈ R, θ ∈ [−π, π])

  • h = window function (non-neg, even, supported on [−1, 1])

Maria van der Walt | Westmont College 16/ 21

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SLIDE 85

SuperEMD

Given IMF fj(t) from (modified) EMD, containing Nj true atoms.

  • Apply SSO to IMF fj [Chui, Mhaskar, vdW 2016]:

(Ta,δfj)(t, θ) = 1

  • k∈Z h

k

a

  • ℓ∈Z

h ℓ a

  • eiℓθfj(t − ℓδ)

(t ∈ R, θ ∈ [−π, π])

  • h = window function (non-neg, even, supported on [−1, 1])
  • a = window width (chosen so that

k∈Z h

k

a

  • > 0)

Maria van der Walt | Westmont College 16/ 21

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SLIDE 86

SuperEMD

Given IMF fj(t) from (modified) EMD, containing Nj true atoms.

  • Apply SSO to IMF fj [Chui, Mhaskar, vdW 2016]:

(Ta,δfj)(t, θ) = 1

  • k∈Z h

k

a

  • ℓ∈Z

h ℓ a

  • eiℓθfj(t − ℓδ)

(t ∈ R, θ ∈ [−π, π])

  • h = window function (non-neg, even, supported on [−1, 1])
  • a = window width (chosen so that

k∈Z h

k

a

  • > 0)
  • δ = sample spacing (adjust based on separation of IF’s)

Maria van der Walt | Westmont College 16/ 21

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SLIDE 87

SuperEMD

Given IMF fj(t) from (modified) EMD, containing Nj true atoms.

  • Apply SSO to IMF fj [Chui, Mhaskar, vdW 2016]:

(Ta,δfj)(t, θ) = 1

  • k∈Z h

k

a

  • ℓ∈Z

h ℓ a

  • eiℓθfj(t − ℓδ)

(t ∈ R, θ ∈ [−π, π])

Maria van der Walt | Westmont College 16/ 21

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SLIDE 88

SuperEMD

Given IMF fj(t) from (modified) EMD, containing Nj true atoms.

  • Apply SSO to IMF fj [Chui, Mhaskar, vdW 2016]:

(Ta,δfj)(t, θ) ≈

Nj

  • n=1

Aj,n(t)e2πiφj,n(t)Φa

  • h; θ − 2πδφ′

j,n(t)

  • Maria van der Walt | Westmont College

16/ 21

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SLIDE 89

SuperEMD

Given IMF fj(t) from (modified) EMD, containing Nj true atoms.

  • Apply SSO to IMF fj [Chui, Mhaskar, vdW 2016]:

(Ta,δfj)(t, θ) ≈

Nj

  • n=1

Aj,n(t)e2πiφj,n(t)Φa

  • h; θ − 2πδφ′

j,n(t)

  • Φa
  • h; θ − 2πδφ′

j,n(t)

  • ≈ Dirac delta function at 0

[Mhaskar, Prestin, 2000]

Maria van der Walt | Westmont College 16/ 21

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SLIDE 90

SuperEMD

Given IMF fj(t) from (modified) EMD, containing Nj true atoms.

  • Apply SSO to IMF fj [Chui, Mhaskar, vdW 2016]:

(Ta,δfj)(t, θ) ≈

Nj

  • n=1

Aj,n(t)e2πiφj,n(t)Φa

  • h; θ − 2πδφ′

j,n(t)

  • Φa
  • h; θ − 2πδφ′

j,n(t)

  • ≈ Dirac delta function at 0

[Mhaskar, Prestin, 2000]

  • . . . → true atom fj,n(t) = Aj,n(t) cos(2πφj,n(t))

Maria van der Walt | Westmont College 16/ 21

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SLIDE 91

SuperEMD

Given IMF fj(t) from (modified) EMD, containing Nj true atoms.

  • Apply SSO to IMF fj [Chui, Mhaskar, vdW 2016]:

(Ta,δfj)(t, θ) ≈

Nj

  • n=1

Aj,n(t)e2πiφj,n(t)Φa

  • h; θ − 2πδφ′

j,n(t)

  • It can be shown [Chui, Mhaskar 2015] that
  • θ ∈ [0, π] : |(Ta,δfj) (t, θ)| ≥ µ

2

  • consists of disjoint clusters Gj,1, . . . , Gj,Nj, centered around

2πδφ′

j,1(t), . . . , 2πδφ′ j,Nj(t).

Maria van der Walt | Westmont College 16/ 21

slide-92
SLIDE 92

SuperEMD

Given IMF fj(t) from (modified) EMD, containing Nj true atoms.

  • Apply SSO to IMF fj [Chui, Mhaskar, vdW 2016]:

(Ta,δfj)(t, θ) ≈

Nj

  • n=1

Aj,n(t)e2πiφj,n(t)Φa

  • h; θ − 2πδφ′

j,n(t)

  • Maria van der Walt | Westmont College

16/ 21

slide-93
SLIDE 93

SuperEMD

Given IMF fj(t) from (modified) EMD, containing Nj true atoms.

  • Apply SSO to IMF fj [Chui, Mhaskar, vdW 2016]:

(Ta,δfj)(t, θ) ≈

Nj

  • n=1

Aj,n(t)e2πiφj,n(t)Φa

  • h; θ − 2πδφ′

j,n(t)

  • Compute true IF’s:

φ′

j,n(t) ≈

1 2πδ arg max

θ∈Gj,n |(Ta,δfj) (t, θ)|

Maria van der Walt | Westmont College 16/ 21

slide-94
SLIDE 94

SuperEMD

Given IMF fj(t) from (modified) EMD, containing Nj true atoms.

  • Apply SSO to IMF fj [Chui, Mhaskar, vdW 2016]:

(Ta,δfj)(t, θ) ≈

Nj

  • n=1

Aj,n(t)e2πiφj,n(t)Φa

  • h; θ − 2πδφ′

j,n(t)

  • Compute true IF’s:

φ′

j,n(t) ≈

1 2πδ arg max

θ∈Gj,n |(Ta,δfj) (t, θ)|

  • Recover true atoms:

fj,n(t) ≈ 2 Re (Ta,δfj) (t, 2πδφ′

j,n(t))

Maria van der Walt | Westmont College 16/ 21

slide-95
SLIDE 95

SuperEMD

Given IMF fj(t) from (modified) EMD, containing Nj true atoms.

  • Apply SSO to IMF fj [Chui, Mhaskar, vdW 2016]:

(Ta,δfj)(t, θ) ≈

Nj

  • n=1

Aj,n(t)e2πiφj,n(t)Φa

  • h; θ − 2πδφ′

j,n(t)

  • Compute true IF’s:
  • φ′

j,n(t) −

1 2πδ arg max

θ∈Gj,n |(Ta,δfj) (t, θ)|

  • ≤ K1(α, B, M, µ, h)
  • Recover true atoms:
  • fj,n(t) − 2 Re (Ta,δfj) (t, 2πδφ′

j,n(t))

  • ≤ K2(α, M, µ)
  • 0 < δ ≤ (4B)−1 and a = (αδ

√ 8πB)−1 for sufficiently small α

  • Maria van der Walt | Westmont College

16/ 21

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SLIDE 96

Examples

Maria van der Walt | Westmont College 17/ 21

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SLIDE 97

Examples

f(t) = cos 2π(5t) + cos 2π(4.9t)

Maria van der Walt | Westmont College 17/ 21

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SLIDE 98

Examples

f(t) = cos 2π(5t) + cos 2π(4.9t) (a = 1800, δ = 1

20)

Maria van der Walt | Westmont College 17/ 21

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SLIDE 99

Examples

f(t) = cos 2π(5t) + cos 2π(4.9t) (a = 1800, δ = 1

20)

Maria van der Walt | Westmont College 17/ 21

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SLIDE 100

Examples

f(t) = cos 2π(5t) + cos 2π(4.9t) + noise (a = 1800, δ = 1

20)

Maria van der Walt | Westmont College 17/ 21

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SLIDE 101

Examples

f(t) = cos 2π(5t) + cos 2π(4.9t) + noise (a = 1800, δ = 1

20)

Maria van der Walt | Westmont College 17/ 21

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SLIDE 102

Examples

f(t) = cos 2π(5t) + cos 2π(4.9t) + cos 2πt + 2 cos 2π(0.96t) + 2 cos 2π(0.92t) + cos 2π(0.9t)

Maria van der Walt | Westmont College 18/ 21

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SLIDE 103

Examples

f(t) = cos 2π(5t) + cos 2π(4.9t) + cos 2πt + 2 cos 2π(0.96t) + 2 cos 2π(0.92t) + cos 2π(0.9t)

Maria van der Walt | Westmont College 18/ 21

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SLIDE 104

Examples

f(t) = cos 2π(5t) + cos 2π(4.9t) + cos 2πt + 2 cos 2π(0.96t) + 2 cos 2π(0.92t) + cos 2π(0.9t) (a = 1024, δ = 2

70)

Maria van der Walt | Westmont College 18/ 21

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SLIDE 105

Examples

f(t) = cos 2π(5t) + cos 2π(4.9t) + cos 2πt + 2 cos 2π(0.96t) + 2 cos 2π(0.92t) + cos 2π(0.9t) (a = 1024, δ = 2

70)

Maria van der Walt | Westmont College 18/ 21

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SLIDE 106

Examples

f(t) = cos 2π(5t) + cos 2π(4.9t) + cos 2πt + 2 cos 2π(0.96t) + 2 cos 2π(0.92t) + cos 2π(0.9t) (a = 1024, δ = 2

70)

Maria van der Walt | Westmont College 18/ 21

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SLIDE 107

Examples

f(t) = cos 2π(5t) + cos 2π(4.9t) + cos 2πt + 2 cos 2π(0.96t) + 2 cos 2π(0.92t) + cos 2π(0.9t) (a = 1024, δ = 2

70)

Maria van der Walt | Westmont College 18/ 21

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SLIDE 108

Examples

f(t) = cos 2π(3t+0.02t2)+cos 2π(2t+0.2 cos t)

Maria van der Walt | Westmont College 19/ 21

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SLIDE 109

Examples

f(t) = cos 2π(3t+0.02t2)+cos 2π(2t+0.2 cos t)

Maria van der Walt | Westmont College 19/ 21

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SLIDE 110

Examples

f(t) = cos 2π(3t+0.02t2)+cos 2π(2t+0.2 cos t) (a = 31, δ = 1

20)

Maria van der Walt | Westmont College 19/ 21

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SLIDE 111

Conclusions

Maria van der Walt | Westmont College 20/ 21

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SLIDE 112

Conclusions

  • We apply a modified EMD (using the blending operator) to

separate a (blind) signal into its IMF components. Then we apply the SSO to compute the instantaneous frequency of each component (instead of Hilbert spectral analysis).

Maria van der Walt | Westmont College 20/ 21

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SLIDE 113

Conclusions

  • We apply a modified EMD (using the blending operator) to

separate a (blind) signal into its IMF components. Then we apply the SSO to compute the instantaneous frequency of each component (instead of Hilbert spectral analysis).

  • Advantages of SuperEMD:

Maria van der Walt | Westmont College 20/ 21

slide-114
SLIDE 114

Conclusions

  • We apply a modified EMD (using the blending operator) to

separate a (blind) signal into its IMF components. Then we apply the SSO to compute the instantaneous frequency of each component (instead of Hilbert spectral analysis).

  • Advantages of SuperEMD:
  • capable of extracting very close-by frequencies

Maria van der Walt | Westmont College 20/ 21

slide-115
SLIDE 115

Conclusions

  • We apply a modified EMD (using the blending operator) to

separate a (blind) signal into its IMF components. Then we apply the SSO to compute the instantaneous frequency of each component (instead of Hilbert spectral analysis).

  • Advantages of SuperEMD:
  • capable of extracting very close-by frequencies
  • capable of recovering atoms with very close-by frequencies

Maria van der Walt | Westmont College 20/ 21

slide-116
SLIDE 116

Conclusions

  • We apply a modified EMD (using the blending operator) to

separate a (blind) signal into its IMF components. Then we apply the SSO to compute the instantaneous frequency of each component (instead of Hilbert spectral analysis).

  • Advantages of SuperEMD:
  • capable of extracting very close-by frequencies
  • capable of recovering atoms with very close-by frequencies
  • more accurate results

Maria van der Walt | Westmont College 20/ 21

slide-117
SLIDE 117

Conclusions

  • We apply a modified EMD (using the blending operator) to

separate a (blind) signal into its IMF components. Then we apply the SSO to compute the instantaneous frequency of each component (instead of Hilbert spectral analysis).

  • Advantages of SuperEMD:
  • capable of extracting very close-by frequencies
  • capable of recovering atoms with very close-by frequencies
  • more accurate results
  • local method

Maria van der Walt | Westmont College 20/ 21

slide-118
SLIDE 118

Conclusions

  • We apply a modified EMD (using the blending operator) to

separate a (blind) signal into its IMF components. Then we apply the SSO to compute the instantaneous frequency of each component (instead of Hilbert spectral analysis).

  • Advantages of SuperEMD:
  • capable of extracting very close-by frequencies
  • capable of recovering atoms with very close-by frequencies
  • more accurate results
  • local method
  • simple algorithm

Maria van der Walt | Westmont College 20/ 21

slide-119
SLIDE 119

Thank you for your attention.

Maria van der Walt | Westmont College 21/ 21