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The model problem
∂2u ∂t2 (x, t) = c2(x) (∆u(x, t) + q(t)p(x − s)) , 0 < t < T, u = 0, t < 0, gs(x0, t) = u(x0, 0, t) = (Rs(f))(x0, t) seismogram for source s, c2(x) = c2
0/(1 + f(x)).
The Kaczmarz Method for Ultrasound Tomography Frank Natterer - - PowerPoint PPT Presentation
The Kaczmarz Method for Ultrasound Tomography Frank Natterer - - PowerPoint PPT Presentation
The Kaczmarz Method for Ultrasound Tomography Frank Natterer University of Mnster 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 <
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TechniScan
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Coverage in Fourier domain
Transmission Reflection
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Kaczmarz’ method for nonlinear problems (consecutive time reversal)
Solve Rs( f ) = gs for all sources s. Update: f ← ⎯ ⎯ f −α(Rsʹ″( f ))
∗(Rs( f ) − gs)
Compute the adjoint by time reversal: (Rsʹ″( f ))
∗r)(x) =
z(x,t)∂
2u(x,t)
∂t
2
dt
T
∫
∂2z ∂t2 = c2(x)∆z for x2 > 0, ∂z ∂x2 = r on x2 = 0, z = 0 for t > T.
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Kaczmarz‘ method for breast phantom, eight sources 1 sweep 3 sweeps
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Kaczmarz‘ method for breast phantom, eight sources 1 sweep 3 sweeps Rays
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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
- f caustics and trapped rays.
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Reconstruction in the presence of caustics 10 cm Luneberg lens 200 kHz wavelength 5-10 mm
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Reconstruction in the presence of trapped rays 200 kHz crater
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Problems in transmission imaging:
- 1. Find an initial approximation for the iteration
- 2. Solution of the wave equation on fine grids
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Condition for the initial approximation:
−∆u0 − k2(1 + f0)u0 = δ(x − s) −∆u − k2(1 + f0)u = −k2(f − f0)u + δ(x − s). −∆u − k2(1 + f0)u = −k2(f − f0)u0 + δ(x − s). First step of iteration: |phase(u) − phase(u0)| < π. Highly necessary condition for convergence:
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WKB-approximation: u ≈ A exp (ikΦ) u0 ≈ A0 exp (ikΦ0) Φ ≈ Φ0 + 1 2
- (f − f0)ds
phase(u) − phase(u0) ≈ k 2
- (f − f0)ds
|
- (f − f0)ds| < 2π
k = λ
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Condition is plausible:
f f0 seismogram of f0 seismogram of f traces
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THE problem in reflection imaging: Missing of low frequencies in the source pulse
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Easy case Nr. 1: Clutter Frequency range 50 to 150 kHz Original 12 cm Diameter of dots 5 mm 5 sweeps of Kaczmarz
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Easy case Nr. 2: Source wavelet q is Gaussian peak. Original 6 sweeps
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Easy case Nr. 2: Source wavelet q is Gaussian peak. Original 6 sweeps
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Difficult case
10 kHz - 150 kHz
- riginal
reconstruction
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Idea: Fill the white circles W by analytic continuation!
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Kaczmarz‘ method doesn‘t work! Why? Because Kaczmarz‘ method necessarily works with a finite aperture.
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Layered medium:
f(x1, x2) = f(x2). Born approximation, one source at x1 = 0, x2 = 0: gk(x) = (2π)−1/2 R e−ixξ ˆ f(−2κ(ξ))dξ, κ = p k2 − ξ2. Finite aperture: Data available for |x| ≤ A only. All we can determine: R δA(η − ξ) ˆ f(−2κ(ξ)dξ, δA(ξ) = A
π sinc(Aξ).
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peaks in η, bandwidth A bandwidth 2z|κ0(ξ)| = 2z|ξ|/κ(ξ) Determine ˆ f from R δA(η − ξ) ˆ f(−2κ(ξ))dξ, δA(ξ) = A
π sinc(Aξ), κ =
p k2 − ξ2. for line object at depth z: f(x) = δ(x − z), ˆ f(ξ) ∼ e−izξ, ˆ f(−2κ(ξ)) ∼ e−2izκ(ξ) for |ξ| < k. ˆ f(−2κ(ξ)) can be stably determined for A > 2z|ξ|/κ(ξ) i.e.
2k
√
1+A2/4z2 < 2κ < 2k.
Sirgue & Pratt 2004
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Kaczmarz‘ method, frequencies 5-25 Hz true profile Kaczmarz starting at f=0 Kaczmarz with analytic continuation Aperture A=12000 m!
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3D scanner of U-Systems
Reflection Imaging
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10 cm diameters of 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)
What can we achieve in medical ultrasound?
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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.