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Three-dimensional optical tomography of hemodynamics in the human head

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Abstract

We report on the first three-dimensional, volumetric, tomographic localization of vascular reactivity in the brain. To this end we developed a model-based iterative image reconstruction scheme that employs adjoint differentiation methods to minimize the difference between measured and predicted data. The necessary human-head geometry and optode locations were determined with a photogrammetric method. To illustrate the performance of the technique, the three-dimensional distribution of changes in the concentration of oxyhemoglobin, deoxyhemoglobin, and total hemoglobin during a Valsalva maneuver were visualized. The observed results are consistent with previously reported effects concerning optical responses to hemodynamic perturbations.

©2001 Optical Society of America

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Supplementary Material (3)

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Figures (6)

Fig. 1.
Fig. 1. Patient with distribution of markers on forehead with overlaid mesh; optodes (red) surface information (blue/green/red), the sources are located at positions 3,6,10, and 13.
Fig. 2.
Fig. 2. Time series of the experimental protocol at 760nm: 3 epochs - consisting of a Valsalva maneuver followed by a rest period (between dashed black lines).
Figure 3.
Figure 3. Time series for first Valsalva maneuver epoch for source 3 (flat red trace) at 760nm.
Fig. 4.
Fig. 4. (1.6MB) Time-series reconstruction showing change in deoxyhemoglobin in units of mM during the Valsalva maneuver; Coronal view (upper left), Parasagittal view (upper right), Horizontal view (lower left), scale (lower right).
Fig. 5.
Fig. 5. (1.6MB) Time-series reconstruction showing change in oxyhemoglobin in units of mM during the Valsalva maneuver; Coronal view (upper left), Parasagittal view (upper right), Horizontal view (lower left), scale (lower right).
Fig 6.
Fig 6. (1.6MB) Time-series reconstruction showing change in blood volume as reflected by total hemoglobin in units of mM during the Valsalva maneuver; Coronal view (upper left), Parasagittal view (upper right), Horizontal view (lower left), scale (lower right).

Equations (22)

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A i , j e = D φ i φ j + μ a φ i φ j d Ω e
Φ = s d ( M s , d P s , d ) 2 σ s , d 2 ,
Φ = s d [ ( M s , d pert ( t ) M s , d ref P s , d ( ξ 0 ) ) P s , d ( ξ n ) ] 2 ( M s , d pert ( t ) M s , d ref P s , d ( ξ 0 ) ) 2
Φ μ a r = 2 s , d ( ( φ s ( x d ) M s , d ) M s , d 2 · φ s ( x d ) μ a r ) .
φ s ( x r ) μ a r = P r T φ s μ a .
Φ μ a r = 2 s ( ( s ( φ s ( x r ) M s , d ) ( M s , d ) 2 P r T ) φ s μ a ) .
A φ s μ a = A μ a φ s
φ s μ a = A 1 A μ a φ s
Φ μ a r = 2 s ( d ( φ s ( x r ) M s , d ) ( M s , d ) 2 · P r T A 1 ) A μ a φ s .
( φ s * ) T d ( φ s ( x r ) M s , d ) ( M s , d ) 2 · P d T A 1
φ μ a r = 2 s ( φ s * ) T A μ a φ s .
Φ μ a r = 2 s el i , j A i , j e μ a r φ i φ j *
Φ μ a r = el i , j { 2 A i , j e μ a r s φ i φ j * } .
Φ μ a r = el i , j A i , j e μ a r { s 2 φ i φ j * } .
A i , j e μ a r = μ a r μ a r φ i φ j d Ω e
= μ a r k = 1 4 φ i φ j I ( μ a 1 , μ a 2 , μ a 3 , μ a 4 ) x = x k
I ( μ a 1 , μ a 2 , μ a 3 , μ a 4 ) = l = 1 4 μ a l φ l
So I μ a k = φ k and therefore : A i , j e μ a r = k = 1 4 i , j = 1 4 φ i φ j φ k
Δ μ a λ = ε Hb O 2 λ Δ [ Hb O 2 ] + ε Hb λ Δ [ Hb ]
Δ [ Hb ] meas = ε Hb O 2 λ 2 Δ μ a λ 1 ε Hb O 2 λ 1 Δ μ a λ 2 ε Hb λ 1 ε Hb O 2 λ 2 ε Hb λ 2 ε Hb O 2 λ 1 ,
Δ [ HbO 2 ] meas = ε Hb λ 1 Δ μ a λ 2 ε Hb λ 2 Δ μ a λ 1 ε Hb λ 1 ε HbO 2 λ 2 ε Hb λ 2 ε HbO 2 λ 1 .
x i = 1 2 n k = i n k = i + n x k .
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