The Kaczmarz Method for Ultrasound Tomography Frank Natterer University of Münster Department of Mathematics and Computer Science
The model problem ∂ 2 u ∂ t 2 ( x, t ) = c 2 ( x ) ( ∆ u ( x, t ) + q ( t ) p ( x − s )) , 0 < t < T, u = 0 , t < 0 , g s ( x 0 , t ) = u ( x 0 , 0 , t ) = ( R s ( f ))( x 0 , t ) seismogram for source s, c 2 ( x ) = c 2 0 / (1 + f ( x )) .
TechniScan
Coverage in Fourier domain Transmission Reflection
Kaczmarz’ method for nonlinear problems (consecutive time reversal) Solve R s ( f ) = g s for all sources s . ∂ 2 z Update: ∂ t 2 = c 2 ( x ) ∆ z for x 2 > 0 , ∂ z f − α ( R s ʹ″ ( f )) ∗ ( R s ( f ) − g s ) = r on x 2 = 0 , f ← ⎯ ⎯ ∂ x 2 z = 0 for t > T. Compute the adjoint by time reversal: T 2 u ( x , t ) z ( x , t ) ∂ ( R s ʹ″ ( f )) ∗ r )( x ) = ∫ dt 2 ∂ t 0
Kaczmarz‘ method for breast phantom, eight sources 1 sweep 3 sweeps
Kaczmarz‘ method for breast phantom, Rays eight sources 1 sweep 3 sweeps
Questions: How does the algorithm work in the presence of caustics? How does the algorithm work for trapped rays? Answer: The algorithm doesn’t even realize the presence of caustics and trapped rays.
Reconstruction in the presence of caustics 200 kHz Luneberg lens wavelength 5-10 mm 10 cm
Reconstruction in the presence of trapped rays crater 200 kHz
Problems in transmission imaging: 1. Find an initial approximation for the iteration 2. Solution of the wave equation on fine grids
Condition for the initial approximation: − ∆ u 0 − k 2 (1 + f 0 ) u 0 = δ ( x − s ) − ∆ u − k 2 (1 + f 0 ) u = − k 2 ( f − f 0 ) u + δ ( x − s ) . First step of iteration: − ∆ u − k 2 (1 + f 0 ) u = − k 2 ( f − f 0 ) u 0 + δ ( x − s ) . Highly necessary condition for convergence: | phase( u ) − phase( u 0 ) | < π .
WKB-approximation: u ≈ A exp ( ik Φ ) u 0 ≈ A 0 exp ( ik Φ 0 ) Φ ≈ Φ 0 + 1 � ( f − f 0 ) ds 2 phase( u ) − phase( u 0 ) ≈ k � ( f − f 0 ) ds 2 ( f − f 0 ) ds | < 2 π � | k = λ
Condition is plausible: f seismogram of f f 0 seismogram of f 0 traces
THE problem in reflection imaging: Missing of low frequencies in the source pulse
Easy case Nr. 1: Clutter 5 sweeps of Original Kaczmarz 12 cm Diameter of Frequency range 50 to 150 kHz dots 5 mm
Easy case Nr. 2: Source wavelet q is Gaussian peak. Original 6 sweeps
Easy case Nr. 2: Source wavelet q is Gaussian peak. Original 6 sweeps
original reconstruction Difficult case 10 kHz - 150 kHz
Idea: Fill the white circles W by analytic continuation!
Kaczmarz‘ method doesn‘t work! Why? Because Kaczmarz‘ method necessarily works with a finite aperture.
Layered medium: f ( x 1 , x 2 ) = f ( x 2 ). Born approximation, one source at x 1 = 0 , x 2 = 0: e − ix ξ ˆ p k 2 − ξ 2 . g k ( x ) = (2 π ) − 1 / 2 R f ( − 2 κ ( ξ )) d ξ , κ = Finite aperture: Data available for | x | ≤ A only. δ A ( η − ξ ) ˆ f ( − 2 κ ( ξ ) d ξ , δ A ( ξ ) = A R All we can determine: π sinc ( A ξ ).
Determine ˆ f from δ A ( η − ξ ) ˆ f ( − 2 κ ( ξ )) d ξ , δ A ( ξ ) = A p R π sinc ( A ξ ) , κ = k 2 − ξ 2 . bandwidth 2 z | κ 0 ( ξ ) | = 2 z | ξ | / κ ( ξ ) peaks in η , bandwidth A for line object at depth z : ˆ f ( − 2 κ ( ξ )) can be stably f ( x ) = δ ( x − z ) , ˆ f ( ξ ) ∼ e − iz ξ , determined for A > 2 z | ξ | / κ ( ξ ) f ( − 2 κ ( ξ )) ∼ e − 2 iz κ ( ξ ) for | ξ | < k . ˆ 2 k i.e. 1+ A 2 / 4 z 2 < 2 κ < 2 k . √ Sirgue & Pratt 2004
Kaczmarz‘ method, frequencies 5-25 Hz true profile Kaczmarz starting at f=0 Kaczmarz with analytic continuation Aperture A=12000 m!
Reflection Imaging 3D scanner of U-Systems
What can we achieve in medical ultrasound? diameters of 10 cm tumors 2.5 mm. 20 sources on top 5 sweeps of Kaczmarz at 100 kHz (100% bandwidth) 5 sweeps of Kaczmarz at 200 kHz (100% bandwidth)
Conclusions 1. Transmission imaging doable 2. Reflection imaging doable if low frequencies are available 3. True challenge: Reflection imaging without low frequencies. 4. We need low frequency transducers.
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