Limited-Data CT for Underwater Pipeline...
Transcript of Limited-Data CT for Underwater Pipeline...
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Limited-Data CT for Underwater Pipeline Inspection
Nicolai André Brogaard Riis ([email protected])
Tuesday 24th September, 2019
IDA Matematik 2019, Copenhagen, Denmark.
Industrial collaborator:
Joint work with:
Per Christian Hansen, Yiqiu Dong – Technical University of Denmark
Jacob Frøsig, Rasmus D. Kongskov – 3Shape
Torben Klit Pedersen – FORCE Technology
Arvid P. L. Böttiger – Exensor Technology
Jürgen Frikel – OTH Regensburg
Todd Quinto – Tu�s University
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Subsea CT-Scanner by FORCE Technology, Denmark
Alternative to ultrasound: use X-ray scanning to compute cross-sectional images ofoil pipes lying on the seabed, to detect defects, cracks, etc. in the pipe.
Illustration courtesy of FORCE Technology.
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Subsea CT-Scanner by FORCE Technology, Denmark
https://forcetechnology.com/da/innovation/afsluttede-projekter/subsea-inspektion-konstruktion-levetid-reparation-planlaegning
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https://forcetechnology.com/da/innovation/afsluttede-projekter/subsea-inspektion-konstruktion-levetid-reparation-planlaegning
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Subsea CT-Scanner by FORCE Technology, Denmark
https://forcetechnology.com/da/innovation/afsluttede-projekter/subsea-inspektion-konstruktion-levetid-reparation-planlaegning
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https://forcetechnology.com/da/innovation/afsluttede-projekter/subsea-inspektion-konstruktion-levetid-reparation-planlaegning
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Comparing Geometries
Limitations in the scanner deviceData: Narrow high intensity X-Ray beam→
available from a limited view.Goal: Design set-up to capture as much detail
of the pipe as possible.
Full illuminationNot possible
CenteredPossible set-up
O�-centeredPossible set-up
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Comparing Geometries
Limitations in the scanner deviceData: Narrow high intensity X-Ray beam→
available from a limited view.Goal: Design set-up to capture as much detail
of the pipe as possible.
Full illuminationNot possible
CenteredPossible set-up
O�-centeredPossible set-up
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The Computed Tomograhy Forward Model
Continuous formulation, limited data
Themeasured projections g for the object f are described by
g(θ, s) = (R`f)(θ, s) + noise
where (R`f)(θ, s) = (Rf)(θ, s) for those pairs (θ, s) corresponding to the `imitedillumination, andR is the Radon transform.
Corresponding algebraic model
Themeasured data b for a discretized objectx is described by
b = A` x+ e, b ∈ Rm, x ∈ Rn, A` ∈ Rm×n,
where e ∈ Rm with ei ∼ N(0, σ2) andA` is the discretion ofR`.
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Full and Two Di�erent Limited Illuminations
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Characterising What We Can Measure – Centered Beam
Microlocal analysis: a singularity at position χwith direction ξ is visible if and onlyif data from the line through χ perpendicular to ξ is present.
Measured data from oneview/projection.
Visible from oneview/projection.
Visible from allviews/projections.
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Characterising What We Can Measure – O�-Center Beam
Microlocal analysis: a singularity at position χwith direction ξ is visible if and onlyif data from the line through χ perpendicular to ξ is present.
Measured data from oneview/projection.
Visible from oneview/projection.
Visible from allviews/projections.
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Reconstruction: Incorporating Sparsity
We need a robust algorithm that can utilize all information in the data.
Use the algebraic system
b = A` x+ e, b ∈ Rm, x ∈ Rn, A` ∈ Rm×n.
and compute a sparse representation of the solution x in a “basis” that is wellsuited for the problem.If we can represent xwith few non-zero basis functions, we have fewer unknownsto determine→ same data, fewer unknowns, “easier” problem.
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Representing pipe in a sparse “basis”
Frame decomposition Given an object, x, we decompose it into building blocks
x =∑µ
cµϕµ,
whereϕµ are the building blocks and cµ = 〈x,ϕµ〉. are coe�icients.*If the object is fully represented by few building blocks, we have a sparse representation.
Example:
= c1 + c2 + c3
+c4 + c5
Why is this useful?
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Representing pipe in a sparse “basis”
Frame decomposition Given an object, x, we decompose it into building blocks
x =∑µ
cµϕµ,
whereϕµ are the building blocks and cµ = 〈x,ϕµ〉. are coe�icients.*If the object is fully represented by few building blocks, we have a sparse representation.
Example:
= c1 + c2 + c3
+c4 + c5
Why is this useful?
11 DTU Compute Nicolai A. B. Riis: Limited-Data CT for Underwater Pipeline Inspection 24.9.2019
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Representing pipe in a sparse “basis”
Frame decomposition Given an object, x, we decompose it into building blocks
x =∑µ
cµϕµ,
whereϕµ are the building blocks and cµ = 〈x,ϕµ〉. are coe�icients.*If the object is fully represented by few building blocks, we have a sparse representation.
Example:
= c1 + c2 + c3
+c4 + c5
Why is this useful?
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Why enforce sparsity of frame coe�icients?
Noise/artefact reduction:
−−−−−−→Decompose
c1 + c2 + c3
+c4 + c5
=
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Why enforce sparsity of frame coe�icients?
Noise/artefact reduction:
−−−−−−→Decompose
c1 + c2 + c3
+c4 + c5
=
Too strong prior!
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Why enforce sparsity of frame coe�icients?
Noise/artefact reduction:
−−−−−−→Decompose
c1 + c2 + c3
+c4 + c5
=
Too strong prior!Not all images arefully represented bythe decompositon.
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Shearlets: A tight frame
Tight frames,Φ ={ϕµ}, generalises ONB, i.e., for all images xwe have
x =∑µ
〈x,ϕµ〉ϕµ.
Examples of 2D Shearlets:
Shearlets yield a sparse representation of defects, contours, etc.
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Shearlet-based optimization problem
Recall discrete model of scanning process: b = A` x+ e.
Reconstruction with a weighted shearlet-based sparsity penalty
Optimization problem:
argminx≥0
12‖A`x− b‖22 + α‖W c‖1,
s.t. c = Φx
with α > 0 regularization parameter,W = diag(wi) ∈ Rp×p with weightswi > 0,Φ ∈ Rp×n is the shearlet analysis transform and cµ = 〈x,ϕµ〉 are coe�icients.
Shearlet transform contains the basis functions:ΦT =[ϕ1, . . . , ϕp
]. hey
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Shearlet-based optimization problem
Recall discrete model of scanning process: b = A` x+ e.
Reconstruction with a weighted shearlet-based sparsity penalty
Optimization problem:
argminx≥0
12‖A`x− b‖22 + α‖W c‖1,
s.t. c = Φx
with α > 0 regularization parameter,W = diag(wi) ∈ Rp×p with weightswi > 0,Φ ∈ Rp×n is the shearlet analysis transform and cµ = 〈x,ϕµ〉 are coe�icients.
Shearlet transform contains the basis functions:ΦT =[ϕ1, . . . , ϕp
].
Data-fitting
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Shearlet-based optimization problem
Recall discrete model of scanning process: b = A` x+ e.
Reconstruction with a weighted shearlet-based sparsity penalty
Optimization problem:
argminx≥0
12‖A`x− b‖22 + α‖W c‖1,
s.t. c = Φx
with α > 0 regularization parameter,W = diag(wi) ∈ Rp×p with weightswi > 0,Φ ∈ Rp×n is the shearlet analysis transform and cµ = 〈x,ϕµ〉 are coe�icients.
Shearlet transform contains the basis functions:ΦT =[ϕ1, . . . , ϕp
].
Weighted sparsity penalty on shearlet coe�icients
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Final algorithmRecall optimization problem:
arg minx≥0
12‖A
`x− b‖22 + α‖W c‖1,
s.t. c = Φx
ADMM-based algorithm
We solve it using the ADMMmethod [Boyd et. al. 2011]. Auxiliary variable c = Φx. Iterativeupdates:
xk+1 := minx≥0
12‖A` x−b‖22+ ρ2‖Φx−c
k+uk‖22,
ck+1 := minc
α‖W c‖1 + ρ2‖Φxk+1−c+uk‖22,
uk+1 := uk+Φxk+1−ck+1,
where u are the scaled Lagrange multipliers and ρ > 0 the penalty parameter.
The updates are calculated using:xk+1: CGLS + non-negativity projection.ck+1: Element-wise so� thresholding.15 DTU Compute Nicolai A. B. Riis: Limited-Data CT for Underwater Pipeline Inspection 24.9.2019
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Subsea CT-Scanner by FORCE Technology, Denmark
https://forcetechnology.com/da/innovation/afsluttede-projekter/subsea-inspektion-konstruktion-levetid-reparation-planlaegning
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https://forcetechnology.com/da/innovation/afsluttede-projekter/subsea-inspektion-konstruktion-levetid-reparation-planlaegning
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Synthetic andmeasured data from both geometries
Synthetic centered
Projection angle
Detectorpixel
0
0.5
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1.5
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2.5
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3.5
4Measured centered
Projection angle
Detectorpixel
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4Synthetic off-centered
Projection angle
Detectorpixel
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1.5
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2.5
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3.5
4Measured off-centered
Projection angle
Detectorpixel
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Reconstructions from Real Data – Both GeometriesCentered beam: there are many artifacts.
Prev. alg.
Kaczmarz, 5 iterations
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0.2Shearlet, α = 0.010
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Our alg.
O�-center beam: singularities are easy to detect; artifacts are reduced.
Prev. alg.
Kaczmarz, 5 iterations
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0.02
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0.2Ground truth
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Our alg.
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Reconstructions from Real Data – Zoom
O�-center beam: singularities are easy to detect; artifacts are reduced.
Prev. alg. Our alg.
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Reconstructions from Real Data – Zoom
O�-center beam: singularities are easy to detect; artifacts are reduced.
arg minx≥0‖Ax− b‖22 + λ‖∇x‖2,1 Our alg. hey
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Centered Versus O�-Center Beam
Centered beam O�-center beamPros Good reconstruction in the
center domain.Captures singularities out-side the center domain.
Cons Terrible reconstruction outsidethe center domain.
Less good reconstruction inthe center domain.
Comments Requires less projections be-cause the center domain iswellcovered by rays.
Requiresmore projections togive good reconstruction ev-erywhere.Better suited for this applica-tion.
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Conclusions
• For technical reasons the X-ray beam cannot cover the whole pipe.• An o�-centered beam can give a satisfactory reconstruction.• A weighted shearlets-based sparsity penalty gives better reconstructions – especially withfew projections.• It is important to include weights in the sparsity penalty.
Future work:• Optimize the algorithm for performance and robustness.• Design heuristics for choosing the weights and the reg. parameter.• Derive more theory for the continuous model with limited data.• Quantify the uncertainties in the model and the solution.
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Conclusions
• For technical reasons the X-ray beam cannot cover the whole pipe.• An o�-centered beam can give a satisfactory reconstruction.• A weighted shearlets-based sparsity penalty gives better reconstructions – especially withfew projections.• It is important to include weights in the sparsity penalty.
Future work:• Optimize the algorithm for performance and robustness.• Design heuristics for choosing the weights and the reg. parameter.• Derive more theory for the continuous model with limited data.• Quantify the uncertainties in the model and the solution.
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IntroductionFormulation of problem
Uncertain view angle CT
b = A(θ)x + e, θ ∼ πθ(·), e ∼ πe(·). (1)
Measured / fixed:b ∈ Rm: measured noisy sinogram.A ∈ Rm×n: discretized Radon transform.
Known but with uncertainty:θ ∈ Rq : view angles.e ∈ Rm: measurement noise.
Unknown:x ∈ Rn: attenuation coe�icients.
πθ1 (θ
1 )
πθ2(θ2)
b1
=A
(θ1 )x
+e1
b2 =A(θ2
)x +e2
x
ModelMachine
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IntroductionFormulation of problem
Uncertain view angle CT
b = A(θ)x + e, θ ∼ πθ(·), e ∼ πe(·). (1)
Measured / fixed:b ∈ Rm: measured noisy sinogram.A ∈ Rm×n: discretized Radon transform.
Known but with uncertainty:θ ∈ Rq : view angles.e ∈ Rm: measurement noise.
Unknown:x ∈ Rn: attenuation coe�icients.
πθ1 (θ
1 )
πθ2(θ2)
b1
=A
(θ1 )x
+e1
b2 =A(θ2
)x +e2
x
ModelMachine
Actual data comes from b := A(θ̄) x + enoise.
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IntroductionFormulation of problem
Uncertain view angle CT
b = A(θ)x + e, θ ∼ πθ(·), e ∼ πe(·). (1)
Goal:Reconstruct x from bwith uncertainty in θ and e!
Applications:Inaccuracies in rotation, patient motion etc.
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IntroductionEarly results
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IntroductionEarly results 2
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IntroductionOngoing Research - Uncertain View Angles
• Large scale problems (2D→ 3D)• Apply to pipe scanner (with shearlet regularizer)
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Thank you!
Contact:Nicolai André Brogaard [email protected] ComputeBuilding 303B Room 118
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Introduction
anm5: 5.29: 5.28: 5.27: 5.26: 5.25: 5.24: 5.23: 5.22: 5.21: 5.20: 5.19: 5.18: 5.17: 5.16: 5.15: 5.14: 5.13: 5.12: 5.11: 5.10: 5.9: 5.8: 5.7: 5.6: 5.5: 5.4: 5.3: 5.2: 5.1: 5.0: anm4: 4.29: 4.28: 4.27: 4.26: 4.25: 4.24: 4.23: 4.22: 4.21: 4.20: 4.19: 4.18: 4.17: 4.16: 4.15: 4.14: 4.13: 4.12: 4.11: 4.10: 4.9: 4.8: 4.7: 4.6: 4.5: 4.4: 4.3: 4.2: 4.1: 4.0: anm3: 3.29: 3.28: 3.27: 3.26: 3.25: 3.24: 3.23: 3.22: 3.21: 3.20: 3.19: 3.18: 3.17: 3.16: 3.15: 3.14: 3.13: 3.12: 3.11: 3.10: 3.9: 3.8: 3.7: 3.6: 3.5: 3.4: 3.3: 3.2: 3.1: 3.0: anm2: 2.29: 2.28: 2.27: 2.26: 2.25: 2.24: 2.23: 2.22: 2.21: 2.20: 2.19: 2.18: 2.17: 2.16: 2.15: 2.14: 2.13: 2.12: 2.11: 2.10: 2.9: 2.8: 2.7: 2.6: 2.5: 2.4: 2.3: 2.2: 2.1: 2.0: anm1: 1.29: 1.28: 1.27: 1.26: 1.25: 1.24: 1.23: 1.22: 1.21: 1.20: 1.19: 1.18: 1.17: 1.16: 1.15: 1.14: 1.13: 1.12: 1.11: 1.10: 1.9: 1.8: 1.7: 1.6: 1.5: 1.4: 1.3: 1.2: 1.1: 1.0: anm0: 0.29: 0.28: 0.27: 0.26: 0.25: 0.24: 0.23: 0.22: 0.21: 0.20: 0.19: 0.18: 0.17: 0.16: 0.15: 0.14: 0.13: 0.12: 0.11: 0.10: 0.9: 0.8: 0.7: 0.6: 0.5: 0.4: 0.3: 0.2: 0.1: 0.0: