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Detecting and estimating the shape of a periodic component in short duration signals Ali Mohammad-Djafari Laboratoire des Signaux et Syst`emes (L2S) UMR8506 CNRS-CentraleSup´elec-UNIV PARIS SUD SUPELEC, 91192 Gif-sur-Yvette, France http://lss.centralesupelec.fr Email:
[email protected] http://djafari.free.fr http://publicationslist.org/djafari MaxEnt 2016 workshop, July 10-15, 2016, Gent, Belgium.
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Contents
1. Description of the problem 2. Putting the things in equations) 3. Classical Regularization approach 4. Bayesian approach 5. Simulation results 6. Conclusions
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Description of the problem
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Detecting a periodic component in a short duration signal, estimating its period and its shape. Low noise example
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High noise example
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Description of the problem
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Detecting a periodic component in a short duration signal, estimating its period and its shape. Low noise example
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High noise example
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Description of the problem I
g = [g 1 , · · · , g N , g N+1 , · · · , · · · , ...g KN+1 , · · · , g M ]0 = [f 1 , · · · , f N , f 1 , · · · , f N , · · · , f 1 , · · · · · · , f r ]0
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where M = KN + r where K is the number of complete repetition of the periodic shape and r is the rest. This relation can be written as a linear relation: g = HN f where H has the following structure HN = [IN |IN |...|IN |I(:, 1 : r )] where IN is the unitary matrix of size N × N and I(:, 1 : r ) is its first r columns. g = HN f + the vector represents those errors.
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Description of the problem I
Also, we require some regularity in the shape of f. This regularity can be modelled as f = Df + ξ → (I − D)f = ξ → Cf = ξ where C can be of the form 1 −1 0 · · · . 0 1 −1 . . . .. .. .. CN = .. . . . 0 ··· 0 1 −1 0 · · · 0
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0 .. . 0 −1 1
A criterion which measures the regularity can be kCfk2
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Deterministic regularization method g = HN f + ,
Cf = ξ
With these two equations, we have at least two possibilities: I Deterministic regularization: bf = arg min {J(f)} with J(f) = kg = HN fk2 + λkCN fk2 2 2 f The solution is given by: bf = [H0 HN + λC0 CN ]−1 H0 g N
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N
N
The above criterion was given for a given value of N. We can now try to define a criterion which depends explicitly on N : J(N, f) = kg = HN fk22 + λkCN fk22 and try to optimize it to find both the seeked period N and the shape f: b bf) = arg min {J(N, f)} (N, (N,f )
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Bayesian approach g = HN f + , I
Cf = ξ
Likelihood p(g|f) = N (g|HN f, v I),
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Prior p(f) = N (f|0, vξ (C0N CN )−1 )
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Poserior b p(f|g) ∝ p(g|f)p(f) = N (f|bf, Σ) with bf = arg max {p(f|g)} = arg min {J(f)} f f which is equivalent to the quadratic regularization as before bf = [H0 HN + λC0 CN ]−1 H0 g N N N with λ = v /vξ .
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Bayesian approach I
Poserior b p(f|g) ∝ p(g|f)p(f) = N (f|bf, Σ) with bf = arg max {p(f|g)} = arg min {J(f)} f f and b = [H0 HN + λC0 CN ]−1 Σ N N which can be used to put error bars on the solution.
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Joint posterior p(N, f|g) and and try to optimize it to find the seeked solution b bf) = arg max {p(N, f|g)} (N, (N,f )
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We can do better.
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Bayesian approach with more appropriate priors I
Forward and prior model equations: g = HN f + ,
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Cf = ξ
i and ξj are Gaussian but with unknown variances that we want to estimate to. p(i |vi ) = N (i |0, vi ),
p(vi |α0 , β0 ) = IG(vi |α0 , β0 )
p(ξi |v ξ i ) = N (ξi |0, v ξ i ),
p(v ξ i |αξ0 , βξ0 ) = IG(vi |αξ0 , βξ0 )
This can also be interpreted as a wish to model them by a heavier tailed probability laws such as Student-t: Z ∞ St(i |αi , βi ) = N (i |, 0, vi ) IG(vi |α0 , β0 ) dvi 0
Z St(ξ j |α, β) = 0
∞
N (ξ j |, 0, v ξ j ) IG(v ξ j |αξ0 , βξ0 ) dv ξ j
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Non stationary noise and sparsity enforcing model – Forward model: g = Hf+, i ∼ N (i |0, vi ) → ∼ N (|0, V ), V = diag [v1 , · · · , vM ]) – Prior model: CN f = ξ, ξj ∼ N (ξi |0, ξ j ) → ξ ∼ N (ξ|0, Vξ ), Vξ = diag v ξ 1 , · · · , v ξ N )
p(g|f, v ) = N (g|Hf, V ), V = diag [v ] p(f|vf ) = N (f|0, Vξ CC0 ), Vξ = diag [vξ ] Q ? ? p(v ) = Qi IG(vi |α0 , β0 ) vξ v p(vξ ) = i IG(v ξ j |αξ0 , βξ0 )
αξ0 , βξ0 α0 , β0
? ?
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p(f, v , vξ |g) ∝ p(g|f, v ) p(f|vξ ) p(v ) p(vξ )
H
?
g
Objective: Infer (f, v , vξ ) – VBA: Approximate p(f, v , vξ |g) by q1 (f) q2 (v ) q3 (vξ )
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Results N=96, period=29, I
By using Fourier Transform technic, no way to find the right value of period.
It is also difficult to estimate the shape of this repeating scheme. Low noise case: I
High noise case:
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Results N=96, period=29, Low noise case:
High noise case:
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Results
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Results
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Conclusions
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the first step in any inference is to write down the relation between what you observe (data g) and the unknowns f.
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The second step is to model and assign priors to account for all uncertainties
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The third step is to use the Bayes rule to find the expression of the joint probability law of all the unknowns given the data and all the hyper parameters.
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Do the Bayesian computation, show the results
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Interpret your results
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Enjoy
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