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Compensation for geometric mismodelling by anisotropies in optical tomography

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Abstract

We propose an approach for the estimation of the optical absorption coefficient in medical optical tomography in the presence of geometric mismodelling. We focus on cases in which the boundaries of the measurement domain or the optode positions are not accurately known. In general, geometric distortion of the domain produces anisotropic changes for the material parameters in the model. Hence, geometric mismodelling in an isotropic case may correspond to an anisotropic model. We seek to approximate the errors due to geometric mismodelling as extraneous additive noise and to pose a simple anisotropic model for the diffusion coefficient. We show that while geometric mismodelling may deteriorate the estimates of the absorption coefficient significantly, using the proposed model enables the recovery of the main features.

©2005 Optical Society of America

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Figures (5)

Fig. 1.
Fig. 1. Schematic representation of the mapping Φ : Ω'→Ω and its inverse Φ-1 : Ω→Ω′.
Fig. 2.
Fig. 2. Transformation defined in Eq. (16). (a) The domain Ω′ used in the inversion. The 16 optodes are located at equal distances on the boundary. The figure also depicts two inclusions inside the domain. (b) The actual domain Ω. The data used in the numerical inversion example is generated using this model with constant isotropic κ=κ 0=0.05 cm-1 and background absorption µ a,bg=0.1 cm-1. In the inclusions, µ a=0.2 cm-1. The simulated data set was collected by placing the source at each optode location at turn, and using the rest of the optodes for measurement, resulting in 240 measurements of amplitude and phase each.
Fig. 3.
Fig. 3. Diffusion tensor generated by the transformation defined in Eq. (16). The tensor distribution is illustrated by ellipses. The axes of an ellipse correspond to the directions of the diffusion tensor eigenvectors. The diffusion tensor is written as κ=UΛUT , where U=[1 2] contains the eigenvectors i , i=1, 2, and Λ=diag(λ 1,λ 2) the eigenvalues. The axis corresponding to 1 and the larger eigenvalue λ 1 is given a constant length, and the second axis is scaled by λ 2/λ 1. The colour reflects the value of the larger eigenvalue λ 1.
Fig. 4.
Fig. 4. (a) Estimate of the absorption coefficient using the isotropic LM iteration in domain Ω′ used for inversion, and (b) the estimate in (a) mapped onto the actual domain Ω.
Fig. 5.
Fig. 5. Estimates for the optical absorption using the modelling error approach combined with LM iteration. (a) Case 1 with θ∗=π/2 in the domain Ω′, and (b) the estimate in (a) mapped into the real domain Ω. (c) Case 2 with θ∗=0 in Ω′, and (d) the estimate in (c) mapped into Ω.

Tables (1)

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Table 1. The parameter values used in the modelling error approach.

Equations (36)

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div ( κ u ) ( μ a ik ) u = q ,
( γ u + 1 2 n · κ u ) Ω = f .
x = Φ ( x ) Ω , x = Φ 1 ( x ) Ω .
u ( x ) = u ( Φ ( x ) ) u Φ ( x ) u ( x ) .
Ω ψ n · κ u d S Ω ( ψ · κ u + ψ μ ˜ u ) d x = Ω ψ q d x ,
2 γ Ω ψ u d S + Ω ( ψ · κ u + ψ μ ˜ u ) d x = Ω 2 ψ f d S Ω ψ q d x .
= j = 1 n x j e j .
u = [ D Φ 1 ] T u ,
d x = 1 det ( D Φ 1 ) d x .
2 Ω γ ψ u d S + Ω ' ( ψ · κ u + ψ μ ˜ u ) d x = 2 Ω ' ψ f d S Ω ' ψ q d x ,
κ = 1 det ( D Φ 1 ) [ D Φ 1 ] κ [ D Φ 1 ] T ,
μ ˜ = 1 det ( D Φ 1 ) μ ˜ ,
q = 1 det ( D Φ 1 ) q ,
f = 1 det ( D ϕ 1 ) f ,
γ = 1 det ( D ϕ 1 ) γ ,
{ r = r ( 1 + 0.15 ( cos 2 θ 0.3 sin 6 θ ) ) θ = θ + 0.04 r sin 5 θ ,
( D Φ 1 ) j , k = x j x k ,
x j x k = i , x j y y y i y i x k .
y = G ( μ a , κ ) + ν ,
y = G ( x , z ) + ν .
y = G ( x , z * ) + ( G ( x , z ) G ( x , z * ) ) + ν = G ( x , z * ) + ε ( x , z ) + ν .
ε ( x , z ) D z G ( x c , z * ) δ z , δ z = z z * ,
cov ( z ) = E { δ z δ z T } = Γ z ,
Γ ε = E { ε ( x , z ) ε ( x , z ) T } D z G ( x c , z * ) Γ z D z G ( x c , z * ) T .
y = G ( x , z * ) + e , cov ( e ) = Γ ε + Γ ν ,
y = G ( x ) + e ,
π ( y x ) exp ( 1 2 ( y G ( x ) ) T Γ 1 ( y G ( x ) ) ) .
0 ( x ) = 1 2 ( y G ( x ) ) T Γ 1 ( y G ( x ) ) = 1 2 y G ( x ) Γ 1 2 .
G ( x ) = G ( x c + δ x ) G ( x c ) + D G ( x c ) δ x .
Minimize y ( G ( x c ) + D G ( x c ) δ x ) Γ 1 2 subject to δ x r .
y ( G ( x c ) + D G ( x c ) δ x ) Γ 1 2 + λ δ x 2 .
δ x = ( D G ( x c ) T Γ 1 D G ( x c ) + λ I ) 1 D G ( x c ) T Γ 1 ( y G ( x c ) ) .
π ( x y ) π ( y x ) π pr ( x )
exp ( 1 2 y G ( x ) Γ 1 2 V ( x ) ) ,
V ( x ) = log π pr ( x ) .
( x ) = 0 ( x ) + V ( x ) .
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