How to extract the oscillating components of a signal? A

How to extract the oscillating components of asignal? A wavelet-based approach compared to

the Empirical Mode Decomposition

Adrien DELIÈGE

University of Liège, Belgium

Liège, December 22, 2016

[email protected]

Introduction

Decomposing time series into several modes has become more and morepopular and useful in signal analysis.Methods such as EMD, SSA, STFT, EWT, wavelets,... have beensuccessfully applied in medicine, finance, climatology, ...Old but gold: Fourier transform allows to decompose a signal as

f (t)≈J

∑k=1

ck cos(ωk t +φk).

Problem: often too many components in the decomposition.Idea: Considering the amplitudes and frequencies as functions of t todecrease the number of terms:

f (t) =K

∑k=1

ak(t)cos(φk(t))

with K ≪ J (AM-FM signals).We will focus on the EMD and a wavelet-based method.

Adrien DELIÈGE (University of Liège, Belgium) EMD and WIME Comprehensible seminar, December 22, 2016

1 EMDDescription of the methodIllustration

2 WIMEDescription of the methodIllustration

3 EMD vs WIMECrossings in the TF planeMode-mixing problemResistance to noise

Real-life example: ECGSome conclusions

4 Edge effectsThe problemA possible solution

5 Wavelets and forecasting?ENSO indexAnalysisModel and skillsSome conclusions


EMD








EMD Description of the method

EMD

Empirical Mode Decomposition

Empirical = no strong theoretical background

Decomposes a signal into IMFs (Intrinsic Mode Functions)

Is often used with the Hilbert-Huang transform to represent the IMFs inthe TF plane (not shown here).



EMD

1) For a signal X(t), let

m1,0(t) =u1,0(t)+ l1,0(t)

2

be the mean of its upper and lower envelopes u(t) and l(t) as determinedfrom a cubic-spline interpolation of local maxima and minima.

2) Compute h1,0(t) as:h1,0(t) = X(t)−m1,0(t).

3) Now h1,0(t) is treated as the data, m1,1(t) is the mean of its upper andlower envelopes, and the process is iterated (“sifting process”):

h1,1(t) = h1,0(t)−m1,1(t).

4) The sifting process is repeated k times, i.e.

h1,k(t) = h1,k−1(t)−m1,k(t),

until a stopping criterion is satisfied.



EMD

5) Then h1,k(t) is considered as the component c1(t) of the signal and thewhole process is repeated with the rest

r1(t) = X(t)− c1(t)

instead of X(t). Get c2(t) then repeat with r2(t) = r1(t)− c2(t), ...

By construction, the number of extrema is decreased when going from ri tori+1, and the whole decomposition is guaranteed to be completed with a finitenumber of modes.Stopping criterion for the sifting process: When computing mi,j(t), alsocompute

ai,j(t) =ui,j(t)− li,j(t)

2σi,j(t) =

∣

∣

∣

∣

mi,j(t)

ai,j(t)

∣

∣

∣

∣

.

The sifting is iterated until σ(t)< 0.05 for 95% of the total length of X(t) andσ(t)< 0.5 for the remaining 5%.


EMD Illustration

EMD - Illustration

Show time!


WIME








WIME Description of the method

Continuous wavelet transform

Given a wavelet ψ and a function f , the wavelet transform of f at time t and atscale a > 0 is defined as

Wf (t,a) =∫

f (x)ψ(

x − t

a

)

dx

a

where ψ is the complex conjugate of ψ.We use the wavelet ψ defined by its Fourier transform as

ψ(ν) = sin( πν

2Ω

)

e−(ν−Ω)2

2

with Ω= π√

2/ ln2, which is similar to the Morlet wavelet but with exactly onevanishing moment.



CWT

Real and Imaginary parts of ψ

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CWT

Real and Imaginary parts of ψ compared to a cosine.

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ℜ(Wf (0,a))≈ 0, ℑ(Wf (0,a))≈ 0 thus |Wf (0,a)| ≈ 0.



CWT


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CWT


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CWT

Real and Imaginary parts of ψ compared to a cosine shifted.

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ℜ(Wf (0,a))≈√

2/2, ℑ(Wf (0,a))≈−√

2/2 thus |Wf (0,a)| ≈ 1.



CWT

One has|ψ(ν)|< 10−5 if ν ≤ 0

thus ψ can be considered as a progressive wavelet (i.e. ψ(ν) = 0 if ν ≤ 0).Property: If f (x) = cos(ωx), then

Wf (t,a) =1

2eitωψ(aω).

Consequence: Given t , if a∗ is the scale at which

a 7→ |Wf (t,a)|

reaches its maximum, thena∗ω =Ω.

The value of ω can be obtained (if unknown) and f is recovered as

f (x) = 2ℜ(Wf (x ,a∗)) = 2|Wf (x ,a

∗(x))|cos(argWf (x ,a∗(x))).



CWT

f (x) = cos(

2π32 x)

−256 −192 −128 −64 0

4

8

16

32

64

b

Per

iod

−256 −192 −128 −64 0

4

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32

64

b

Per

iod

Left: |Wf (t,a)| . Right: arg(Wf (t,a)).



CWT

f (x) = cos(

2π32 x)

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Left: initial signal. Right: Reconstructed signal superimposed to initial signal.Difference of order 10−5.



CWT

What to do with this?

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WIME - Wavelet-Induced Mode Extraction

1) Perform the CWT of the signal f : Wf (t,a).

2) Compute the wavelet spectrum Λ associated to f :

a 7→ Λ(a) = Et |Wf (t,a)|

where Et denotes the mean over time.

3) Segment the spectrum to isolate the scale a∗ at which Λ reaches itshighest local maximum between the scales a1 and a2 at which Λ displaysthe left and right local minima that are the closest to a∗. We setA = [a1,a2].

4) Choose a starting point (t0,a(t0)) with a(t0) ∈ A, e.g.

(t0,a(t0)) = argmaxt, a∈A

|Wf (t,a)|.



WIME

5) Compute the ridge t 7→ (t,a(t)) forward and backward that stems from(t0,a(t0)):

a) Compute b1 and b2 such that b2 −b1 = a2 −a1 and a(t0) = (b1 +b2)/2,i.e. center a(t0) in a frequency band of the same length as the initial one.

b) Among the scales at which the function a 7→ |Wf (t0 +1,a)| (a ∈ [b1,b2])reaches a local maximum, define a(t0 +1) as the closest one to a(t0). Ifthere is no local maximum, then a(t0 +1) = b1 if|Wf (t0 +1,b1)|> |Wf (t0 +1,b2)|, and a(t0 +1) = b2 otherwise.

c) Repeat step 5) with (t0 +1,a(t0 +1)) instead of (t0,a(t0)) until the end ofthe signal.

d) Proceed in the same way backward from (t0,a(t0)) until the beginning ofthe signal.

6) Extract the component associated to the ridge:

t 7→ 2ℜ(Wf (t,a(t))) = 2|Wf (t,a(t))|cos(argWf (t,a(t))).



WIME

7) That component is c1. The whole process is repeated with the rest

r1 = f − c1

instead of f . Get c2 then repeat with r2(t) = r1(t)− c2(t), ...

8) Stop the process when the extracted components are not relevantanymore, e.g. at cn if ||cn||< 0.05maxj<n ||cj ||.

Alternative stopping criterion : EMD-like method e.g. |cn|< 0.05maxj<n |cj | for95% of the duration and |cn|< 0.5maxj<n |cj | for the remaining 5%.

Very useful: (t,a) 7→ |Wf (t,a)| can be seen as a TF representation of f .


WIME Illustration

WIME - Illustration

Show time again!


WIME Illustration

WIME - Illustration

We consider the function f = f1 + f2 defined on [0,1] by

f1(t) =

cos(60πt) if t ≤ 0.5cos(80πt −15π) if t > 0.5

f2(t) = cos(10πt +10πt2).

0 0.5 1

−2−1012

f


WIME Illustration

WIME - Illustration

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160

time

frequency

Time-frequency representation of f : (t,a) 7→ |Wf (t,a)|.Adrien DELIÈGE (University of Liège, Belgium) EMD and WIME Comprehensible seminar, December 22, 2016

WIME Illustration

WIME - Illustration

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160

time

frequency

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spectrum

frequency

Time-frequency representation of f and spectrum a 7→ Λ(a) = Et |Wf (t,a)|.


WIME Illustration

WIME - Illustration

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time

frequency

First ridge.Adrien DELIÈGE (University of Liège, Belgium) EMD and WIME Comprehensible seminar, December 22, 2016

WIME Illustration

WIME - Illustration

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1

time

c1

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1

timef1

First component c1 extracted and expected component f1.


WIME Illustration

WIME - Illustration

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frequency

Time-frequency representation of r1 = f − c1.Adrien DELIÈGE (University of Liège, Belgium) EMD and WIME Comprehensible seminar, December 22, 2016

WIME Illustration

WIME - Illustration

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frequency

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frequency

Time-frequency representation of r1 = f − c1 and spectrum.


WIME Illustration

WIME - Illustration

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frequency

First ridge.Adrien DELIÈGE (University of Liège, Belgium) EMD and WIME Comprehensible seminar, December 22, 2016

WIME Illustration

WIME - Illustration

0 0.5 1

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1

time

c2

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1

timef2

Second component c2 extracted and expected component r1 = f − c1.


WIME Illustration

WIME - Illustration

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f

Original and reconstructed signal.


WIME Illustration

WIME - Illustration

With an AM-FM signal

f1(t) = (2+ sin(5πt))cos(100(t −0.5)3 +100t)

f2(t) =

(1.5+ t)cos(0.2e10t +350t) if t ≤ 0.5t−1 cos(−300t2 +1000t) if t > 0.5

0 0.5 1

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time

f


WIME Illustration

WIME - Illustration

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160320

time

freq

uenc

y

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time

c1

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c2


EMD vs WIME








EMD vs WIME Crossings in the TF plane

EMD vs WIME

Round 1Crossings in the TF plane



EMD vs WIME: Crossings in the TF plane

We consider f = f1 + f2 + f3 (on [0,1]) made of three FM-components withconstant amplitudes of 1.25, 1, 0.75:

f1(t) = 1.25cos((10t −7)3 −1800t)

f2(t) = cos(360(0.5)10t −200t)

f3(t) =

0.75cos(125t + cos(30t)) if t ≤ 0.50.75cos(−500t2 +375t) if t > 0.5

0 0.5 1−3−2−10123

time

f



EMD vs WIME: Crossings in the TF plane - WIME

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ency

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c1

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frequ

ency

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c2

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ency

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c3



EMD vs WIME: Crossings in the TF plane - EMD

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IMF1

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EMD vs WIME: Crossings in the TF plane - WIME-EMD

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EMD vs WIME: Crossings in the TF plane

The influence of the crossings between the patterns in the TF planeremains limited for WIME.

The energy-based hierarchy among the components is respected forWIME

The EMD follows an “upper ridge first” scheme and can’t proceedotherwise.


EMD vs WIME Mode-mixing problem

EMD vs WIME

Round 2Mode-mixing problem



EMD vs WIME: Mode-mixing problem

We consider a signal made of AM-FM components that are not “well-separated”with respect to their frequency nor with their amplitudes. Objective: recover theoriginal frequencies used to build the signal. We consider f = ∑i fi with

f1(t) =

(

1+0.5cos

(

2π200

t

))

cos

(

2π47

t

)

f2(t) =ln(t)

14cos

(

2π31

t

)

f3(t) =

√t

60cos

(

2π65

t

)

f4(t) =t

2000cos

(

2π23+ cos

(

2π1600 t

) t

)

.

Target frequencies: 1/47, 1/31, 1/65, and ≈ 1/23 Hz. Note that t takes integervalues from 1 to 800.



EMD vs WIME: Mode-mixing problem

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f

Target frequencies: 1/47, 1/31, 1/65, and ≈ 1/23 Hz.



EMD vs WIME: Mode-mixing problem - WIME

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1/641/321/16

1/81/41/2

time0 0.4 0.81 1.22

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spectrum

frequ

ency

Target:1/47Detect:1/46.3

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1/81/41/2

time0 0.4 0.81 1.22

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frequ

ency





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1/81/41/2

time0 0.4 0.81 1.22

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1/641/321/16

1/81/41/2

spectrum

frequ

ency


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1/81/41/2

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1/81/41/2

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frequ

ency





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0

1

2

time

c1

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1

2

time

f1

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1

2

time

c2

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f2




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c3

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f3

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c4

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f4



EMD vs WIME: Mode-mixing problem - EMD

0 400 800−2

−1

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2

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IMF1

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−1

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1

2

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IMF2

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2

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IMF3

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−1

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1

2

time

IMF4



EMD vs WIME: Mode-mixing problem - WIME - EMD

Target WIME EMD1/23 1/21.6 -1/31 1/30.6 -1/47 1/46.3 1/411/65 1/65.5 1/75- - 1/165- - 1/284

IMF1 is almost the signal itself - correlation of 0.93.

EMD cannot resolve the mode-mixing problem.

WIME provides accurate information.


EMD vs WIME Resistance to noise

EMD vs WIME

Round 3Resistance to noise



Resistance to noise

We consider the chirp f defined on [0,1] by

f (t) = cos(70t +30t2)

and a Gaussian white noise X of zero mean and variance 1 and we run WIMEon f , f +X , f +2X and f +3X .



Resistance to noise: WIME

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WIME with f and f +X .



Resistance to noise: WIME

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f

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WIME with f +2X and f +3X .



Resistance to noise: EMD

Not performed since:

It is known (and obvious) that EMD is not noise-resistant.

It first gives many noisy IMFS.

It is not fair to compare EMD with WIME; improved versions of the EMDshould be used instead, e.g. Ensemble Empirical Mode Decomposition(EEMD) and Complete Ensemble Empirical Mode Decomposition withAdaptive Noise (CEEMDAN).

Improvements of EMD are made to the detriment of computational costs.

WIME is naturally resistant and the scales to use for the reconstructioncan be selected.


EMD vs WIME Real-life example: ECG

Real-life example: ECG

Real-life exampleElectrocardiogram




−0.2

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The Dirac-like impulses make it an approximation of an AM-FM signal.

WIME extracts valuable information.

It decomposes the signal into simpler components easy to analyze.

It could be useful to compare hundreds of patients.

EMD provides 14 IMFS, many of them are noisy.


EMD vs WIME Some conclusions

EMD vs WIME: Some conclusions

EMD is fully data-driven but sensitive to noise and not flexible (black box).

EMD extract components before visualizing them.

EMD follows “upper ridge first” principle, thus have problems withintersecting frequencies and mode mixing.

EMD has codes available on the internet.

...

WIME is flexible but works in the frequency domain. Visualization prior tothe analysis allows more freedom.

WIME respects the hierarchical structure imposed by the energy of thecomponents thus have better skills when EMD is in trouble.

WIME is naturally tolerant to noise.

WIME can provide a finer analysis of the data.

...


Edge effects








Edge effects The problem

Edge effects

What you often see in practice

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C14



Edge effects

In practice: the signal has to be padded at its edges to obtain the CWT.Possibilities:

zero-padding

orthogonal symmetry (mirroring)

central symmetry (inverse mirroring)

periodization

If possible, the padding needs to have the same properties as the signal.Zero-padding: “universality”, independent of the signal.



Zero-padding

Expected: Wf (t,Ω/ω) = 12eitω thus |Wf (t,Ω/ω)|= 0.5.

What happens with a simple cosine:

0 1 000200 400 600 800 1 200

0.4

0.3

0.5

0.25

0.35

0.45

Ridge : straight line.The amplitude decreases at the borders. The instantaneous frequencyincreases at the end (not shown). This confirms intuition.



Zero-padding


What happens with a simple cosine f (x) = cos(2π/100x):

1 000800700 900750 850 950

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0.45




Zero-padding


What happens with a simple cosine f (x) = cos(2π/100x):

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Zero-padding: In theory

This is due to the finite length of the signal. Mathematically, in this case,

f (x) = cos(ωx)χ]−∞,0](x)

and thus for a =Ω/ω,

Wf (t,Ω/ω) =1

2eitωz(t)

with

ℜ(z(t)) =1

2− 2

π

∫ 1

0

ψ(Ωx)(x2 −2x −1)

(x2 −1)(3− x)sin(tω(1− x))dx

and

ℑ(z(t)) =2

π

∫ 1

0

ψ(Ωx)

(x +1)(3− x)cos(tω(1− x))dx .



Zero-padding: In theory

Wf (t,Ω/ω) = 12eitωz(t) , study of z(t):

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1

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0.8

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0.95

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Amplitude and argument of z as function of t . These confirm intuition andexperiments.


Edge effects A possible solution

A possible solution? Iterations!

The theoretical result is difficult to use in practice.

All the energy has not be drained from the TF plane, there is still someenergy left at the borders.

Iterate the extraction process along the same ridge to sharpen thecomponent before getting interested in another ridge.

Stop iterations when the component extracted is not significant anymore,e.g. at iteration J if the extracted component at iteration J has less than95% of the energy of the extracted component at the first extraction.




Iteration 1

0 1 000200 400 600 800 1 200

0.4

0.3

0.5

0.25

0.35

0.45

0 1 000200 400 600 800 1 200

0

−1

1

−0.8

−0.6

−0.4

−0.2

0.2

0.4

0.6

0.8




Iteration 2

0 1 000200 400 600 800 1 2000

0.02

0.04

0.06

0.08

0.01

0.03

0.05

0.07

0 1 000200 400 600 800 1 200

0

−0.1

0.1

−0.15

−0.05

0.05

0.15




Iteration 3

0 1 000200 400 600 800 1 2000

0.02

0.01

0.03

0.005

0.015

0.025

0.035

0 1 000200 400 600 800 1 200

0

−0.08

−0.06

−0.04

−0.02

0.02

0.04

0.06

0.08




Iteration 4

0 1 000200 400 600 800 1 2000

0.02

0.01

0.002

0.004

0.006

0.008

0.012

0.014

0.016

0.018

0 1 000200 400 600 800 1 200

0

−0.04

−0.02

0.02

0.04

−0.03

−0.01

0.01

0.03




Iteration 5

0 1 000200 400 600 800 1 2000

0.01

0.002

0.004

0.006

0.008

0.012

0.014

0 1 000200 400 600 800 1 200

0

−0.02

0.02

−0.01

0.01

0.03

−0.025

−0.015

−0.005

0.005

0.015

0.025




Iteration 50

0 1 000200 400 600 800 1 2000e00

2e−04

4e−04

1e−04

3e−04

5e−05

1.5e−04

2.5e−04

3.5e−04

4.5e−04

0 1 000200 400 600 800 1 200

0e00

−4e−04

−2e−04

2e−04

4e−04

6e−04

8e−04

−3e−04

−1e−04

1e−04

3e−04

5e−04

7e−04




Iteration 1

0 1 000200 400 600 800 1 200

0

−1

1

−1.2

−0.8

−0.6

−0.4

−0.2

0.2

0.4

0.6

0.8

1.2




Iteration 2

0 1 000200 400 600 800 1 200

0

−1

1

−1.2

−0.8

−0.6

−0.4

−0.2

0.2

0.4

0.6

0.8

1.2




Iteration 3

0 1 000200 400 600 800 1 200

0

−1

1

−1.2

−0.8

−0.6

−0.4

−0.2

0.2

0.4

0.6

0.8

1.2




Iteration 4

0 1 000200 400 600 800 1 200

0

−1

1

−1.2

−0.8

−0.6

−0.4

−0.2

0.2

0.4

0.6

0.8

1.2




Iteration 5

0 1 000200 400 600 800 1 200

0

−1

1

−1.2

−0.8

−0.6

−0.4

−0.2

0.2

0.4

0.6

0.8

1.2




Iteration 10

0 1 000200 400 600 800 1 200

0

−1

1

−1.2

−0.8

−0.6

−0.4

−0.2

0.2

0.4

0.6

0.8

1.2




Iteration 20

0 1 000200 400 600 800 1 200

0

−1

1

−1.2

−0.8

−0.6

−0.4

−0.2

0.2

0.4

0.6

0.8

1.2




Iteration 50

0 1 000200 400 600 800 1 200

0

−1

1

−1.2

−0.8

−0.6

−0.4

−0.2

0.2

0.4

0.6

0.8

1.2




Iteration 100

0 1 000200 400 600 800 1 200

0

−1

1

−1.2

−0.8

−0.6

−0.4

−0.2

0.2

0.4

0.6

0.8

1.2


Wavelets and forecasting?








Wavelets and forecasting?

Some ideas

Perfect correction of border effects ⇒ Terrific forecasts!

Idea: perform the CWT, extract dominant components (with corrected bordereffects), extrapolate the components (smooth AM-FM signals), then add thecomponents to reconstruct and predict the signal.

Great idea. Doesn’t work.

Instead: build a model based on the information provided by the CWT.


Wavelets and forecasting? ENSO index

ENSO index

Analyzed data: Niño 3.4 time series, i.e. monthly-sampled sea surfacetemperature anomalies in the Equatorial Pacific Ocean from Jan 1950 toDec 2014 (http://www.cpc.ncep.noaa.gov/).


http://www.cpc.ncep.noaa.gov/


ENSO index

Niño 3.4 index:

1950 1960 1970 1980 1990 2000 2010 2020

−2.5

−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

2.5

time (years)

Nin

o 3.

4 in

dex

17 El Niño events: SST anomaly above +0.5C during 5 consecutivemonths.14 La Niña events: SST anomaly below −0.5C during 5 consecutivemonths.



ENSO index

Flooding in the West coast of South America

Droughts in Asia and Australia

Fish kills or shifts in locations and types of fish, having economic impactsin Peru and Chile

Impact on snowfalls and monsoons, drier/hotter/wetter/cooler than normalconditions

Impact on hurricanes/typhoons occurrences

Links with famines, increase in mosquito-borne diseases (malaria,dengue, ...), civil conflicts

In Los Angeles, increase in the number of some species of mosquitoes (in1997 notably).

...


Wavelets and forecasting? Analysis

Analysis

1950 2014

160

80

40

20

10

5

Per

iod

(mon

th)



Analysis

3 6 12 25 50 100 200 4000

0.2

0.4

0.6

0.1

0.3

0.5

0.7

0.05

0.15

0.25

0.35

0.45

0.55

0.65

Period (months)

Am

plitu

de

17

31 4361

140

340



Analysis

1975 1980 1985 1990 1995 2000 2005 2010 2015

0

−1

1

−0.5

0.5

1975 1980 1985 1990 1995 2000 2005 2010 2015

0

−1

1

−0.5

0.5

1975 1980 1985 1990 1995 2000 2005 2010 2015

0

−1

1

−0.5

0.5



Analysis

1975 1980 1985 1990 1995 2000 2005 2010 2015

0

−1

1

−0.5

0.5

1975 1980 1985 1990 1995 2000 2005 2010 2015

0

−1

1

−0.5

0.5

1975 1980 1985 1990 1995 2000 2005 2010 2015

0

−1

1

−0.5

0.5



Analysis

Periods of ≈ 17,31,43,61,140 months in agreement with previousstudies.

Period of ≈ 340 months can be an artifact; will be neglected.

The low frequency components (corresponding to 31,43,61,140 months)capture ≈ 90% of the variability of the signal.

These components appear relatively stationary thus easier to model.


Wavelets and forecasting? Model and skills

Idea of the model

Model the decadal oscillation and subtract it.

Model a 61-months component phased with warm events and subtract it.

Model a 31-months component phased with cold events and subtract it.

Model a 43-months component phased with remaining warm and coldevents.

Extrapolate these modeled components and add them to obtain aforecast.



Model

Idea: build components that mimic the low-frequency ones and that are easy toextrapolate. Let us assume we have the signal up to time T (between 1995 and2015).

1. Model the decadal oscillation. The amplitude A140 is estimated with theWS of s as 0.35 and we set

y140(t) = A140 cos(2πt/140+2.02).

2. We now work with s1 = s− y140. The WS of s1 gives A61 = 0.435. Phasey61 with the strongest warm events of s1, which occur approximately every5 years: find the position p of the last local maximum of s1 such thats1(p)> 0.5. If s1(p)> 0.9 then we set

y61(t) = A61 cos(2π(t −p)/61);

elsey61(t) =−A61 cos(2π(t −p)/61).



Model

3. We now work with s2 = s1 − y61. The WS of s2 gives A31 = 0.42. Phasey31 with the cold events of s2, which occur approximately every 2.5 years.Find the position p of the last local minimum of s2 such that s2(p)<−0.5and we set

y31(t) =−A31 cos(2π(t −p)/31).

4. We now work with s3 = s2 −y31. The WS of s3 gives A43 = 0.485. y43 hasto explain the remaining warm and cold events of s3. Find the position p ofthe last local maximum of s3 such that s3(p)> 0.5 and we set

y143(t) = A43 cos(2π(t −p)/43).

Then we find the position p of the last local minimum of s3 such thats3(p)<−0.8 and we set

y243(t) =−A43 cos(2π(t −p)/43).

Finally, we definey43 = (y1

43 + y243)/2.



Model

5. Extend the signals (yi)i∈I up to T +N for N large enough (at least thenumber of data to be predicted). Then

y = ∑i∈I

yi

stands for a first reconstruction (for t ≤ T ) and forecast (for t > T ) of s.

6. We set s(t) = y(t) for t > T , perform the CWT of s and extract thecomponents cj at scales j corresponding to 6, 12, 17, 31, 43, 61 and 140months. These are considered as our final AM-FM components andc = ∑j cj both reconstructs (for t ≤ T ) and forecasts (for t > T ) the initialONI signal in a smooth and natural way.



Skills of the model

1995 1996 1997 1998

0

−2

2

−3

−1

1

3

−2.5

−1.5

−0.5

0.5

1.5

2.5

Tem

pera

ture

ano

mal

y

2000 2001 2002 2003

0

−2

2

−3

−1

1

3

−2.5

−1.5

−0.5

0.5

1.5

2.5

Tem

pera

ture

ano

mal

y

2007 2008 2009 2010

0

−2

2

−3

−1

1

3

−2.5

−1.5

−0.5

0.5

1.5

2.5

Tem

pera

ture

ano

mal

y

2013 2014 2015 2016

0

−2

2

−3

−1

1

3

−2.5

−1.5

−0.5

0.5

1.5

2.5

Tem

pera

ture

ano

mal

y



Skills of the model

1993 1994 1995 1996

0

−2

2

−3

−1

1

3

−2.5

−1.5

−0.5

0.5

1.5

2.5

Tem

pera

ture

ano

mal

y

1996 1997 1998 1999

0

−2

2

−3

−1

1

3

−2.5

−1.5

−0.5

0.5

1.5

2.5

Tem

pera

ture

ano

mal

y

1997 1998 1999 2000

0

−2

2

−3

−1

1

3

−2.5

−1.5

−0.5

0.5

1.5

2.5

Tem

pera

ture

ano

mal

y

2009 2010 2011 2012

0

−2

2

−3

−1

1

3

−2.5

−1.5

−0.5

0.5

1.5

2.5

Tem

pera

ture

ano

mal

y



Skills of the model

1975 1980 1985 1990 1995 2000 2005 2010 2015

0

−2

2

−1

1

ON

I

1975 1980 1985 1990 1995 2000 2005 2010 2015

0

−2

2

−1

1

ON

I

1975 1980 1985 1990 1995 2000 2005 2010 2015

0

−2

2

−1

1

ON

I



Skills of the model

0 6 12 18 24 30 36

1

0.4

0.6

0.8

0.3

0.5

0.7

0.9

0.35

0.45

0.55

0.65

0.75

0.85

0.95

Lead time (months)

RM

SE

0 6 12 18 24 30 36

1

0.4

0.6

0.8

0.3

0.5

0.7

0.9

0.35

0.45

0.55

0.65

0.75

0.85

0.95

Lead time (months)C

orre

latio

n

(slightly unfair) comparison with other works.


Wavelets and forecasting? Some conclusions

Some conclusions

The periods detected are in agreement with previous works

The information provided by the CWT allows to build a model forlong-term forecasting

Early signs of major EN and LN events can be detected 2-3 years inadvance

The ideas could be combined with other models that are better forshort-term predictions

We could improve the model with seasonal and annual variations

We could make the amplitudes vary through time

The important feature is the phase-locking of the components



Some references

EMD: [4, 5, 9, 11, 12] andhttp://perso.ens-lyon.fr/patrick.flandrin/emd.html

CWT: [1, 2, 6, 7, 10]

WIME: [3, 8] + coming soon


http://perso.ens-lyon.fr/patrick.flandrin/emd.html


Thank you

Thank youfor your attention



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