Filtering random medium effects for imaging Liliana Borcea Computational and Applied Mathematics Rice University borcea@caam.rice.edu Collaborators: alez del Cueto , CAAM, Rice University Fernando Gonz´ George Papanicolaou , Mathematics, Stanford University Chrysoula Tsogka , Mathematics, University of Crete Support: ONR: N00014-05-1-0699 NSF: DMS-0604008; DMS-0354658 1
Topic Problem: Image with arrays of sensors compactly supported re- flectors buried in heterogeneous, strongly backscattering media. Difficulty: The backscattered field can overwhelm the echoes from the reflectors that we wish to find. • Can we differentiate the coherent echoes from the incoherent, backscattered field? Can we design filters that emphasize the coherent field needed in imaging? • We discuss such filters for finely layered media. - These media are a good case study because they give a worse scenario in terms of backscattering effects. For example, wave localization is sure to occur. 2
Formulation of the problem Data: time traces of pressure p ( t,� x r ,� x s ). Source is fixed at � x s = ( x s , z = 0). | h | ≤ a Receivers at � x r = � x s + ( h , 0) , 2 . Image the support S of reflectivity ν ( � x ). 1 x ) ∂ 2 t p ( t,� x ,� x s ) − ∆ p ( t,� x ,� x s ) = − f ( t ) ∂ z δ ( � x − � x s ) , t > 0 , V 2 ( � p ( t,� x ,� x s ) = 0 , t ≤ 0 , where � x = ( x , z ) and 1 1 x ) = v 2 ( z ) + ν ( � x ) V 2 ( � • v ( z ) is rough (fine layering + interfaces) � scattering. 3
Illustration motivated by exploration geophysics x (cross − range) speed x r 10 15 20 25 30 35 40 10 15 20 25 30 35 40 0 5 0 5 35 30 25 20 15 10 0 5 0 0 30 5 100 10 10 133 40 20 20 167 30 30 50 z (range) 200 depth 40 40 time 233 60 50 50 60 60 267 70 70 70 300 80 80 80 333 Frequeny band: 20 − 40Hz, v ( z ) fluctuates about 3km/s. Central wavelength is 100m, the reflectors are at depth 6km and the fine layering is at scale 2m. y s ) = � y s ,� Imaging function: J ( � x r p ( τ ( � x r , � x s ) ,� x r ,� x s ) � y s ,� τ ( � x r , � x s ) = travel time computed with smooth part of v ( z ). 4
Illustration of filtering the layer echoes 5 5 10 10 15 15 20 x r 20 x r 25 25 30 30 35 35 100 133 167 200 233 267 300 333 100 133 167 200 233 267 300 333 time time Recorded traces Filtered traces 40 40 35 35 30 30 x (cross � range) 25 x (cross � range) 25 20 20 15 15 10 10 5 5 0 0 30 40 50 60 70 80 30 40 50 60 70 80 z (range) z (range) Migration image Image with filtered data. 5
Outline • Mathematical model of the data. • Introduce the filters, which use simple ideas from geophysics. • Explain through analysis why they work. • Present numerical results. 6
Model of the medium • The sound speed v ( z ) has a (piecewise) smooth part c ( z ) and a rough part supported in z < 0. For z ≥ 0 we have v ( z ) = c o . � z �� � 1 1 v 2 ( z ) = 1 + σµ − L j < z < − L j − 1 , j = 1 , 2 . . . , c 2 ( z ) ℓ • Interfaces at z = − L j , j = 1 , 2 . . . , due to jump discontinuities of c ( z ) or to isolated blips over depth intervals ∼ wavelength. • Fine layering modeled with random, stationary process µ . The process µ has mean zero, no long range correlations and it is properly normalized. ℓ = correlation length and σ gives strength of fluctuations. 7
Mathematical model of the data • We let: p ( t,� x s + ( h , 0) ,� x s ) � D ( t, h ) . • The layer echoes to be annihilated are � ω � dω � 2 � R ( ω, K ) e − iωt + iω K · h � D lay ( t, h ) ∼ f ( ω ) d K 2 π 2 π where R ( ω, K ) is the reflection coefficient of the medium and K = horizontal slowness vector of plane waves. • The echoes from the compact reflectors to be imaged are modeled with the Born approximation. The transmitted field between the array surface and the compact reflectors is determined by T ( ω, K ) = transmission coefficient. 8
The scattering series ∗ T 1 T 1 e − 2 iω ( τ 1 + τ 2 ) + . . . R ( ω, K ) = R 1 + · · · + T 1 ˜ T 1 T 2 ˜ R 2 T 2 ˜ T 1 R 1 T 1 − L 1 ˜ ˜ T 1 T 1 T 2 T 2 − L 2 ˜ R 2 - We use tilde for the strong interfaces. - Travel times of the plane waves between the interfaces � � − L j 1 c 2 ( z ) − K 2 τ j ( K ) = j = 1 , 2 , . . . dz , − L j − 1 � �� � vertical slowness 9 ∗ Series for transmission T ( ω, K ) between A and compact scatterers is similar.
The coherent and incoherent field N ( t, h ) D ( t, h ) = C ( t, h ) + � �� � � �� � coherent incoherent • The coherent field corresponds to pure transmission through the random medium. It consists of the echoes from the deter- ministic structures (the interfaces and the compact scatterers). C ( t, h ) is modeled with the O’Doherty Anstey (ODA) theory: The fine layering effects on the coherent echoes are: pulse broad- ening, strong attenuation (exponential) and small random arrival time shifts. arrival time � �� � t − [ τ P ( h ) + ǫδ P ( h )] � C ( t, h ) = ϕ P , h . ǫ P • N ( t, h ) is due to reflections in the random slabs. 10
Coherent and incoherent intensity • The coherent echoes from the top scattering interfaces are strong and dominant in the data. • The incoherent backscattered field is weaker, but the coherent echoes from deep deterministic structures are also weak. 5 5 10 10 15 15 20 x r 20 x r 25 25 30 30 35 35 100 133 167 200 233 267 300 333 100 133 167 200 233 267 300 333 time time Recorded traces Filtered traces • The layer annihilator filters are designed to suppress the primary echoes in C ( t, h ) scattered once at an interface. 11
Definition of annihilator: Step 1 normal move-out � �� � • D ( t, h ) → D ( T c ( h, z ) , h ) using the roundtrip time to z < 0 � � 0 1 − c 2 ( s ) K 2 c T c ( h, z ) = 2 ds + hK c where c ( s ) z � 0 c ( s ) K c h K c = d = ds dhT c ( h, z ) � � 2 1 − c 2 ( s ) K 2 z c Example for c ( z ) = c o � h 2 + 4 z 2 T c o = c o and o T c o ( h, z ) = cos θ 1 h K [ T c o ] = z c 2 c o 12
Definition of annihilator • The annihilation step � 1 D ( T c ( h, z ) , h ) − D ( T c ( h + ξ, z ) , h + ξ e h ) dξ | I ( h ) | I ( h ) where I ( h ) is an averaging interval around h and e h = h h . return to ( t, h ) � �� � • z = ζ c ( h, t ) with inverse of travel time map T c ( h, ζ c ( h, t )) = t � 0 t 2 − h 2 If c ( z ) = c o we have ζ c o = − 1 c 2 2 • The filter is � � � 1 � � Q c D ( t, h )= D ( T c ( h, z ) , h ) − D ( T c ( h + ξ, z ) , h + ξ e h ) dξ � | I ( h ) | � I ( h ) z = ζ c ( h, t ) 13
Numerical results 6 5 sound speed 4 3 2 0 10 20 30 40 50 60 70 80 depth 5 10 15 20 x r 25 30 35 100 150 200 250 300 time 5 10 15 20 x r 25 30 35 100 150 200 250 300 time 14
Numerical results 5 4 sound speed 3 2 1 0 10 20 30 40 50 60 70 80 depth 5 10 15 20 x r 25 30 35 100 150 200 250 300 time 5 10 15 20 x r 25 30 35 100 150 200 250 300 time 15
Numerical results 5 4 sound speed (km/s) 3 2 1 0 0 10 20 30 40 50 60 70 80 depth 5 10 15 20 x r 25 30 35 133 167 200 233 267 300 333 time 5 10 15 20 x r 25 30 35 133 167 200 233 267 300 333 time 16
Why is the annihilation better than expected? 50 50 55 55 60 60 65 65 z z 70 70 75 75 80 80 0 5 10 15 20 25 30 35 40 0 5 10 15 20 25 30 35 40 x x • By design, the filter Q c annihilates the coherent echoes from the strong interfaces. The numerics confirm this. • But all the examples show that the incoherent, backscattered field is annihilated too. To explain the surprising effectiveness of Q c we need to go deeper in the fluctuation theory, beyond ODA. 17
Analysis setup for the annihilation of incoherent echoes • Take time t ∈ (0 , t ⋆ ], with t ⋆ < first coherent arrival time � as if we had just the random medium, up to depth L t ⋆ . • The model of the data becomes � ω � dω � � 2 D ( t, h ) = N ( t, h ) = 1 R t ⋆ ( ω, K ) e − iωt + iω K · h , � f ( ω ) d K 2 2 π 2 π where R t ⋆ ( ω, K ) = reflection coefficient of the medium in [ − L t ⋆ , 0]. • It is easy to get that E { R t ⋆ ( ω, K ) } = 0 so that E {N ( t, h ) } = 0. • Our goal is to show � [ Q c N ( t, h )] 2 � � � N 2 ( t, h ) ≪ E E . 18
The moments of the reflection coefficients • To study R t ⋆ ( ω, K ), it is convenient to define it as R t ⋆ ( ω, K ) = lim z ր 0 R t ⋆ ( ω, K, z ) , where R t ⋆ ( ω, K, z ) = reflection coefficient of the medium in [ − L t ⋆ , z ]. • We have the Ricatti equation � ∂ − iωσµ ( z/ℓ ) c ( K, z ) e 2 iωτ ( K,z ) [ R t ⋆ ( ω, K, z )] 2 + ∂z R t ⋆ ( ω, K, z ) = 2 c 2 ( z ) � e − 2 iωτ ( K,z ) − 2 R t ⋆ ( ω, K, z ) + · · · for z > − L t ⋆ and the initial condition R t ⋆ ( ω, K, − L t ⋆ ) = 0 . � z � dz ′ c 2 ( z ) − K 2 and τ ( K, z ) = 1 Here c ( K, z ) = c ( K, z ′ ). − L t⋆ 19
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