Tomographic Image Reconstruction. - An Introduction.

Tomographic Image Reconstruction.An Introduction.

Milan ZvolskýLecture on Medical Physics

by Erika Garutti & Florian Grüner

28.11.2014

Outline.

1 Introduction2 PET, SPECT, CT3 Basic Idea: Projections4 Analytic Image Reconstruction

BackprojectionFiltered Backprojection

5 Iterative Image ReconstructionBasic IdeaML-EM Algorithm

6 The EndoTOFPET-US Project7 Summary

Milan Zvolský (DESY.) 2 / 48

Introduction.

Cutting into slices is a bad idea to perform on humans


TomographyImaging by sectioning: The word tomography is derived from thegreek tome (cut) and graphein (to write).

Introduction.

Cutting into slices is a bad idea to perform on humans


TomographyImaging by sectioning: The word tomography is derived from thegreek tome (cut) and graphein (to write).

Introduction.

Figure: PET-CT

Figure: PET brain images

Figure: CT: 2D → 3D


Introduction.Reconstruction quality is a crucial point in medical imagingNo use in building accurate detector when image quality is poorVery active field of research


Computer Tomography (CT).

Rotating X-ray tube + detectorX-rays propagate through a x-section of patient


Computer Tomography (CT).

Rotating X-ray tube + detectorX-rays propagate through a x-section of patient

Measure the exit beam intensity integrated alonga line between X-ray source and detector

Id = I0 exp

[−∫ d

0µ(s;E)ds

]with µ linear attenuation coefficient as a function of the

location s and the effective energy E

Basic measurement of CT: Line integral of thelinear attenuation coefficient


Positron Emission Tomography (PET).

Glucose-like 18FDG (β+ emitter) concentrates inthe metabolical active areasWe are interested in the radioactivitydistribution within the bodye+e− → 2γ (back-to-back, 511 keV each)


Positron Emission Tomography (PET).

Glucose-like 18FDG (β+ emitter) concentrates inthe metabolical active areasWe are interested in the radioactivitydistribution within the bodye+e− → 2γ (back-to-back, 511 keV each)

Detect the two γs in coincidence2 detectors fire at the same time → Draw a lineThese are called Lines of Response (LOR)Measure the integrated activity of the LOR


SPECT.Single Photon Emission Computed Tomographyγ-emitting radionuclide → 1 γ → no LORGamma camera with collimators → reconstructlines perpendicular to the detector


SPECT.Single Photon Emission Computed Tomographyγ-emitting radionuclide → 1 γ → no LORGamma camera with collimators → reconstructlines perpendicular to the detector

Similar to PET (morphologic images)Cheaper/simpler than PETWorse spacial resolution, lower sensitivity


Emission & Transmission Tomography.

Transmission Tomography (CT)Radiation source outside the patient: X-ray tubeor long-lived radionuclide rotates around thebodyQuantity to be reconstructed is the photon linearattenuation coefficient of the body

Emission Tomography (PET & SPECT)Radiation source inside the patient: γ- ore+-emitting radionuclideQuantity to be reconstructed is the activityconcentration of the radiopharmacon inside thebody


1 Introduction

2 PET, SPECT, CT

3 Basic Idea: Projections

4 Analytic Image ReconstructionBackprojectionFiltered Backprojection


6 The EndoTOFPET-US Project

7 Summary


Basic Idea.Example: PhotographyTwo trees in a park, make 2 pictures from east and south, try tocreate a map of the park.

A photo is a projection of an object onto a plane


Basic Idea.Example: Another PhotographyOther configuration: If you see two separate trees on both views, canyou uniquely reconstruct the map of trees?Here you cannot reconstruct the position and height of both trees.

Figure: Two trees seen on both viewsFigure: There are two solutions

If we take another picture at 45◦, we are able to solve the ambiguity.


Basic Idea.Example: Another PhotographyOther configuration: If you see two separate trees on both views, canyou uniquely reconstruct the map of trees?Here you cannot reconstruct the position and height of both trees.

Figure: Two trees seen on both viewsFigure: There are two solutions

If we take another picture at 45◦, we are able to solve the ambiguity.


Basic Idea: Projections.Before: photo, now: Projection is a line integralProjection p(s, φ) at angle φ, s is coordinate on detector

Projection p(s) the same for any φ

p(s,ϕ2)

p(s,ϕ1)

Projection p(s, φ) depends on orientation


Projections: Angle dependency.Example: Point source on the y axisLocation s of the spike on the 1D detector: s = r sinφ.The projection p(s, φ) in the s-φ-coordinate system is a sine function.

x

sp(s,Φ)

yΦ

s'

s'

0

0

Point Source

Detector

Φ

r

SinogramA sinogram is a representation of the projections on the s-φ plane.


Projections: Angle dependency.Example: Point source on the y axisLocation s of the spike on the 1D detector: s = r sinφ.The projection p(s, φ) in the s-φ-coordinate system is a sine function.

SinogramA sinogram is a representation of the projections on the s-φ plane.


Projections: Angle dependency.TomographyFind the image f(x, y) from the measured projections p(s, φ)

We measure a sinogram: We want an image:


1 Introduction

2 PET, SPECT, CT





7 Summary


Backprojection Procedure.Backprojection

Placing a value of p(s, φ) back into the position of theappropriate LORBut the knowledge of where the values came from was lost inthe projection stepThe best we can do is place a constant value into all elementsalong the line


Backprojection Example.1st projection

a b

c d

a+b

c+d


Backprojection Example.2nd projection

a b

c d

a+b

c+d

a+d

b

c


Backprojection Example.3rd projection

a b

c d

b+da+c

a+b

c+d

a+d

b

c


Backprojection Example.4th projection

a b

c d

b+da+c

a

c+bd

a+b

c+d

a+d

b

c


Backprojection Example.Backproject

b+da+c

a

c+bd

a+b

c+d

a+d

b

c

? ?

? ?


Backprojection Example.1st backprojection

b+da+c

a

c+bd

a+b

c+d

a+d

b

c

a+ba+b

c+d c+d


Backprojection Example.2nd backprojection

b+da+c

a

c+bd

a+b

c+d

a+d

b

c

a+b+a+b+

c+d+ c+d+

ba+d

c a+d


Backprojection Example.3rd backprojection

b+da+c

a

c+bd

a+b

c+d

a+d

b

c

a+b+a+b+

c+d+ c+d+

b+a+d+

c+ a+d+

a+c

a+c

b+d

b+d


Backprojection Example.4th backprojection

b+da+c

a

c+bd

a+b

c+d

a+d

b

c

a+b+a+b+

c+d+ c+d+

b+a+d+

c+ a+d+

a+c+

a+c+c+b

ab+d+c+b

b+d+d


Backprojection Example.Subtract projection sum from each entry

b+da+c

a

c+bd

a+b

c+d

a+d

b

c

a+ba+b

c+d c+d

ba+d

c a+d

a+c

a+cc+b

ab+dc+b

b+dd


Backprojection Example.Subtract projection sum from each entry

b+da+c

a

c+bd

a+b

c+d

a+d

b

c

c c+d

ba+d

c a+d

a

c

abc+b

b+d


Backprojection Example.Divide by number of projections −1 = 3

b+da+c

a

c+bd

a+b

c+d

a+d

b

c

c c+d

ba+d


Backprojection Procedure.1 view: spike of intensity 1.This is sum of activity alongprojection pathRe-distribute activity back toits original pathGive equal activity everywherealong the lineMany angles → Tall spike atthe location of the point source(d) Ups...


Backprojection: Blurring of the image.

Figure: OriginalShepp-Logan phantom

Figure: Backprojected image for(1,3,7,15,31,63) projections


Backprojection.The Radon transformThe Radon transform of a distribution f(x, y) is given by

p(s, φ) =

∫ ∞−∞

∫ ∞−∞

f(x, y) · δ(x cosφ+ y sinφ− s)dxdy

.Delta function: Integrand is zero everywhere except on the line L(s, φ)

Backprojected image:

b(x, y) =

∫ π

0p(s, φ)|s=x cosφ+y sinφdφ

.Integrate over 180◦, the other half doesn’t give extra information

Reconstructed image is blurred:

b(x, y) = f(x, y)× 1√x2 + y2


Interlude: Fourier Transform.There is a close relationship between Radon and the Fourier trafo!

Fourier transform (FT)

F{p(s)} = P (ω) =1√2π

∫ ∞−∞

p(s)e−iωsds

Summation of lines causes duplication in the centerOversampling in the center of the Fourier space


Central Slice Theorem.Central Slice Theorem (CST)

F1{p(s, φ′)} = F2{f(x, y)}|φ=φ′

The following operations are the equivalent:Take a 2D function f(x, y), project it onto a line, and do a FT of that projection.

Do a 2D FT of f(x, y) first, and then take a slice through the origin, parallel to

the projection line.

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Filtered Backprojection: “Proof”.Re-write the image f(x, y) via the inverse FT:

f(x, y) = F−12 {F (vx, vy)}

=

∫ ∞−∞

dvx∫ ∞−∞

dvyF (vx, vy)e2πi(vxx+vyy)

Polar coordinates: vx = ω cosφ, vy = ω sinφ, dvxdvy = ωdωdφ:

f(x, y) =

∫ 2π

0dφ∫ ∞0

dωω F (ω cosφ, ω sinφ)︸︷︷︸=P (ω) (CST)

e2πiω(

=s︷︸︸︷x cosφ+ y sinφ)

=

∫ π

0dφ∫ ∞−∞

dω|ω|P (ω)e2πiωs (changing the integration limits)

f(x, y) =

∫ π

0dφ

[∫ ∞−∞

dω|ω|P (ω)e2πiωs]

︸︷︷︸=F−1{|ω|P (ω)}≡p′(s,φ)

=

∫ π

0dφp′(s, φ)



f(x, y) = F−12 {F (vx, vy)} =∫ ∞−∞

dvx∫ ∞−∞



f(x, y) =

∫ 2π

0dφ∫ ∞0


e2πiω(


=

∫ π

0dφ∫ ∞−∞


f(x, y) =

∫ π

0dφ

[∫ ∞−∞


︸︷︷︸=F−1{|ω|P (ω)}≡p′(s,φ)

=

∫ π

0dφp′(s, φ)



f(x, y) = F−12 {F (vx, vy)} =∫ ∞−∞

dvx∫ ∞−∞



f(x, y) =

∫ 2π

0dφ∫ ∞0


e2πiω(


=

∫ π

0dφ∫ ∞−∞


f(x, y) =

∫ π

0dφ

[∫ ∞−∞


︸︷︷︸=F−1{|ω|P (ω)}≡p′(s,φ)

=

∫ π

0dφp′(s, φ)



f(x, y) = F−12 {F (vx, vy)} =∫ ∞−∞

dvx∫ ∞−∞



f(x, y) =

∫ 2π

0dφ∫ ∞0


e2πiω(


=

∫ π

0dφ∫ ∞−∞


f(x, y) =

∫ π

0dφ

[∫ ∞−∞


︸︷︷︸=F−1{|ω|P (ω)}≡p′(s,φ)

=

∫ π

0dφp′(s, φ)



f(x, y) = F−12 {F (vx, vy)} =∫ ∞−∞

dvx∫ ∞−∞



f(x, y) =

∫ 2π

0dφ∫ ∞0


e2πiω(


=

∫ π

0dφ∫ ∞−∞


f(x, y) =

∫ π

0dφ

[∫ ∞−∞


︸︷︷︸=F−1{|ω|P (ω)}≡p′(s,φ)

=

∫ π

0dφp′(s, φ)


Filtered Backprojection.

f(x, y) =

∫ π

0dφ[ ∫ ∞−∞

dω|ω|P (ω)e2πiωs]s=x cosφ+y sinφ

FT of projection p(s, φ)Multiply by frequency filter |ω|Inverse-transform this productThis filtered projection is backprojectedThen sum over all filtered projections

h(t)

x

|ω|

ω-ωs ωs

Ramp filter |ω|Blurring (low ω) minimisedContrasts (high ω) accentuated



f(x, y) =

∫ π

0dφ[ ∫ ∞−∞


FT of projection p(s, φ)

Multiply by frequency filter |ω|Inverse-transform this productThis filtered projection is backprojectedThen sum over all filtered projections

h(t)

x

|ω|

ω-ωs ωs




f(x, y) =

∫ π

0dφ[ ∫ ∞−∞


FT of projection p(s, φ)Multiply by frequency filter |ω|

Inverse-transform this productThis filtered projection is backprojectedThen sum over all filtered projections

h(t)

x

|ω|

ω-ωs ωs




f(x, y) =

∫ π

0dφ[ ∫ ∞−∞


FT of projection p(s, φ)Multiply by frequency filter |ω|Inverse-transform this product

This filtered projection is backprojectedThen sum over all filtered projections

h(t)

x

|ω|

ω-ωs ωs




f(x, y) =

∫ π

0dφ[ ∫ ∞−∞


FT of projection p(s, φ)Multiply by frequency filter |ω|Inverse-transform this productThis filtered projection is backprojected

Then sum over all filtered projectionsh(t)

x

|ω|

ω-ωs ωs




f(x, y) =

∫ π

0dφ[ ∫ ∞−∞



h(t)

x

|ω|

ω-ωs ωs



Filtered Backprojection.f(x, y) =

∫ π

0dφ[ ∫ ∞−∞



h(t)

x

|ω|

ω-ωs ωs


A complete set of 1D projections allows the reconstruction of the original 2D

distribution without loss of informationMilan Zvolský (DESY.) 25 / 48

BP vs. FBP.

One single projection for both methods:

Figure: Backprojection Figure: Filtered Backprojection


Filtered Backprojection.p( )ξ,φ

f(x,y)

p ( )ξ,φ’

BP

ramp filterFigure 8. Flow of the FBP algorithm.

Demonstration of the FBP algorithm:

projection

ramp filter

backprojection

original image

sinogram

FBP image

filtered sinogram

1 view 2 views 10 views

30 views 50 views 80 views

Figure 8. Demonstration of the FBP algorithm.

6


BP vs. FBP.

Figure: Backprojection Figure: Filtered Backprojection


Image Quality vs. Number of Events.

102 103 104

105 106

fFBP(x)

f (x)

107

Number of events follows Poisson distribution→ Signal-to-noise ratio ≈

√N

Analytical Reconstruction works exactly for N =∞Milan Zvolský (DESY.) 29 / 48

Effect of Poissonian Noise.

σ=0 σ=0.05 σ=0.1 σ=0.2 σ=0.5

Typically low noise for CT, high noise for PETHigh signal variability but FBP assumes exact distributionAmplification of high-frequency noise (→ use cut-off)

SignalNoiseMeasure

RampfilterFilteredsignalFilterednoise

=+

νc

FBP bad if data is noisy → Iterative Image Reconstruction


1 Introduction

2 PET, SPECT, CT





7 Summary


FBP vs. Reality.Reality is more complex:

Data is discrete: FBP only precise if all angles are availableData is noisy: Measurements follow a probability distributionDetectors are unprecise: mis-positioning of photonsDetector geometry may not provide complete dataNot all photons travel along straight lines: scatter, absorption

Input prior knowledge of the detector & the physics process


Iterative Methods - Basic Idea.Emission tomography described asy = S(x) + noisey: measurement, S: System operator, x: activity

Model system operator S with systemmatrix A → linear approximationAij : Probability that emission in voxel i(xi) results in a detection in detector j (yj)y = Ax or yj =

∑iAijxi

Inverse problem: Know yi, want xiCould do x = A−1yBut: A is huge & cannot be inverted

LOR i

Voxel j

f(x,y)

LOR i

Modeling of detectionimperfections

Iterative Analytical

Iterative image reconstructionSolve inverse problem iterativelyAssume Poissonian noise on yi


Iterative Methods - Basic Idea.Measurements yi: independent random variables (Poisson)Expectation value: µi = Ai · x =

∑j Aijxj

Probability to measure k given λ: p(k|λ) = e−λλk

k!

Likelihood:

L(x) = p(y|x) =∏i

p(yi|x) =∏i

e−Ai·x(Ai · x)yiyi!

Among all possible images x we choose the one that maximisesthe probability of producing the data (find most likely image)Maximise the log-likelihood, i.e. maximise log(p(y|x))

Maximum Likelihood - Expectation Maximisation (ML-EM)Objective function to maximise: log-Likelihood (ML)Maximisation algorithm: Expectation Maximisation (EM)


ML-EM Algorithm.Initial guess for the image (uniform)x(0)i

Simulate measurements from estimate (forward proj.)ysimuj =

∑k Akjx

(0)k

Compare this with actual measurementsRatio Rj =

yjysimuj

Improve image estimate (backward projection)x(1)i = x

(0)i ·

1∑j Aij

·∑

j AijRj

Repeat until convergence

ML-EM

x(n+1)i = x

(n)i ·

1∑j Aij

·∑j

Aijyj∑

k Akjx(n)k




∑k Akjx

(0)k


yjysimuj


(0)i ·

1∑j Aij

·∑

j AijRj


ML-EM

x(n+1)i = x

(n)i ·

1∑j Aij

·∑j

Aijyj∑

k Akjx(n)k




∑k Akjx

(0)k


yjysimuj


(0)i ·

1∑j Aij

·∑

j AijRj


ML-EM

x(n+1)i = x

(n)i ·

1∑j Aij

·∑j

Aijyj∑

k Akjx(n)k




∑k Akjx

(0)k


yjysimuj


(0)i ·

1∑j Aij

·∑

j AijRj


ML-EM

x(n+1)i = x

(n)i ·

1∑j Aij

·∑j

Aijyj∑

k Akjx(n)k




∑k Akjx

(0)k


yjysimuj


(0)i ·

1∑j Aij

·∑

j AijRj


ML-EM

x(n+1)i = x

(n)i ·

1∑j Aij

·∑j

Aijyj∑

k Akjx(n)k


ML-EM Algorithm.ML-EM

x(n+1)i = x

(n)i ·

1∑j Aij

·∑j

Aijyj∑

k Akjx(n)k

Image Domain Projection Domain

ForwardProject

Backproject

Compare with measurement

Initialguess

ith estimate

Correction factorfor each projection

Correction factorfor image estimate

Multiply byimage estimate

Divide by weighting term



x(n+1)i = x

(n)i ·

1∑j Aij

·∑j

Aijyj∑

k Akjx(n)k


ForwardProject

Backproject


Initialguess

ith estimate





??

?? 3

7

4 6

2.52.5

2.52.5 5

5

5 5

x=(3+7)/4=10/4=2.5

AssumeAij = 1 ∀i, j∑

j Aij = 2 ∀i



x(n+1)i = x

(n)i ·

1∑j Aij

·∑j

Aijyj∑

k Akjx(n)k


ForwardProject

Backproject


Initialguess

ith estimate





??

?? 3

7

4 6

2.52.5

2.52.5 5

5

5 5

c11 = (3/5 + 4/5)/2 = 0.7 x11 = 1.75


j Aij = 2 ∀i



x(n+1)i = x

(n)i ·

1∑j Aij

·∑j

Aijyj∑

k Akjx(n)k


ForwardProject

Backproject


Initialguess

ith estimate





??

?? 3

7

4 6

2.52.5

2.51.75 5

5

5 5

c12 = (3/5 + 6/5)/2 = 0.9x12 = 2.25


j Aij = 2 ∀i



x(n+1)i = x

(n)i ·

1∑j Aij

·∑j

Aijyj∑

k Akjx(n)k


ForwardProject

Backproject


Initialguess

ith estimate





??

?? 3

7

4 6

2.52.5

5

5

5 5

c13 = (7/5 + 4/5)/2 = 1.1x13 = 2.75

1.752.25


j Aij = 2 ∀i



x(n+1)i = x

(n)i ·

1∑j Aij

·∑j

Aijyj∑

k Akjx(n)k


ForwardProject

Backproject


Initialguess

ith estimate





??

?? 3

7

4 6

5

5

5 5

c14 = (7/5 + 6/5)/2 = 1.3x14 = 3.25

1.752.25

2.75 2.5


j Aij = 2 ∀i



x(n+1)i = x

(n)i ·

1∑j Aij

·∑j

Aijyj∑

k Akjx(n)k


ForwardProject

Backproject


Initialguess

ith estimate





3

7

4 6

5

5

5 5

1.752.25

2.753.25

1 2

3 4


j Aij = 2 ∀i



x(n+1)i = x

(n)i ·

1∑j Aij

·∑j

Aijyj∑

k Akjx(n)k


ForwardProject

Backproject


Initialguess

ith estimate





1 1.5 2 2.5 3 3.5 4

1

1.5

2

2.5

3

3.5

4

4 projections

No convergence to true image

1 1.5 2 2.5 3 3.5 4

1

1.5

2

2.5

3

3.5

4

5 projections

Deviation < 1% after 50 it.


ML-EM: Number of Iterations.0 1 2 3 4 6 8 16 32

Figure: Number of iterations

With noise: > 50 iterations needed≈ 100 times slower than FBPLow frequencies are reconstructed first. For higher iterations,the alg. will attempt to recover noise → high spacial variance

Early termination (fast, but non-uniform convergence)Post-smoothing with Gaussian kernel

k = 4 k = 16 k = 64 k = 1600


System Model.Reconstruction can only be as good as our model agrees with reality→ Consider all physical effects governing our system:

Positron rangePhoton acollinearityScattering in the bodyAttenuation in the bodyScattering in the detectorAbsorption in the detectorDeadtime of detectorRandom coincidences comingfrom the background

LOR i

Voxel j

f(x,y)

LOR i

Modeling of detectionimperfections

Iterative Analytical


ML-EM vs. FBP.Iterative methods perform much better than analytical methods ifdata is noisy:

FBP

EM-ML withattenuation pre-correction

EM-ML withattenuation

correction build-in


1 Introduction

2 PET, SPECT, CT





7 Summary


The EndoTOFPET-US Project

Pancreas

Cecum

Commonbile duct

Stomach

Pancreaticduct

Liver

Gallbladder

Duodenum

Transverse colon

Ascending colon

Descending colon

Colon


Positron Emission Tomography (PET)

Radiotracer (β+ emitter) concentrates in the metabolical active arease+e− → 2γ (back-to-back, 511 keV each)Detect the two γs in coincidence


Pancreas

Cecum

Commonbile duct

Stomach

Pancreaticduct

Liver

Gallbladder

Duodenum

Transverse colon

Ascending colon

Descending colon

Colon


Objectives

Development of new biomarkersIntra-operative Time-of-Flight (TOF) PET DetectorPrototype for prostate & pancreas cancer


Pancreas


Objectives



Pancreas


Objectives



Pancreas


Objectives


Challenges

Extreme miniaturisationFast crystals & ultra-fast photodetectionAim for our project: Coincidence timeresolution 200 ps FWHM (3 cm)This reduces background noise from otherorgans → better image qualityImage reconstruction for free-handimaging, image resolution of 1 mm

EndoTOFPET Image ReconstructionChallenges

Time of flight (TOF)Limited angle problemFreehand→ undefined volume of interestLow sensitivity (≈ 1%), high noiseReconstruct the image on-line toprovide guidance for the physician

Solution

ML-EM iterative reconstruction: Good performance in case of Poissonian noiseGPU Computation: Solving a massively parallel problem

GPU speedup by factor O(10)

Image reconstruction in O(min.)


Full System Simulation.Simulation of whole detector system → optimise design, test reconstructionGeant4-based toolkit GAMOS with custom extensions (e.g. TOF)Study sensitivity, image resolution etc. on simple phantomsStudy medical procedures on full-body phantoms

Expected resolution in X & Y: ≈ 1 mm(limited by crystal size and tracking precision)Expected resolution in Z: ≈ 5 mm(limited by low angular coverage)

Quantify image qualityHow good can wesuppress backgroundfrom other organs?How much data do weneed to acquire?


Patient Data: PSMA.PET/CT scans performed at TU Munich (Klinikum rechts der Isar)Prostate-specific membrane antigen (PSMA)One specific patient with injected dose of 140 MBqBladder uptake ≈ 20 MBq, prostatic lesion ≈ 0.5 MBqFull-body PET resolution: ≈ 6 mmDownsize datasets: Merge voxels to speed up simulationLoad PET (activity) and CT (scatter) DICOM images into the simulation

Bladder

Prostatic Lesion

BladderProstatic Lesion


Patient Data: PSMA.First reconstructed image of full-body PSMA PET data with the EndoTOFPET system

(a) Transverse (b) Coronal (c) Sagittal

External Plate

Increase lesion uptake ×10:


External Plate

Endoscope

Prostatic lesion can be well separated from bladder backgroundMilan Zvolský (DESY.) 47 / 48

Patient Data: PSMA.First reconstructed image of full-body PSMA PET data with the EndoTOFPET system


External Plate

Increase lesion uptake ×10 and tilt the PET head:

4.2 cm


External Plate

Endoscope

Prostatic lesion can be well separated from bladder backgroundMilan Zvolský (DESY.) 47 / 48

Summary.Tomography

Take projection data,convert it into cross-section imagesA projection is a line integral of an objectBackprojection: sums data from all projection views

Analytic ReconstructionSimple backprojection → BlurringUse the Central Slice Theorem to solve the oversampling(blurring) in the Fourier space → Filtered Backprojection

Iterative ReconstructionStart with guess, compare with measurement, update the guess,iterate until best solution is reachedModel scatter, attenuation and limited detector resolution tobetter reflect reality (e.g. ML-EM)


Thank you for your attention!

Literatur: [1, 2, 3, 4, 2, 4, 5, 6, 7, 8, 9, 10, 11, 12]

Literature I

J.L. Prince and J.M. Links.Medical imaging signals and systems.Pearson Prentice Hall, 2006.

G. Zeng.Medical image reconstruction: A conceptual Tutorial.Springer, 2010.

R. Brinks and T.M. Buzug.Image reconstruction in positron emission tomography (pet): the 90th anniversaryof radon’s solution.Biomedizinische Technik, 52(6):361–364, 2007.

A. Alessio and P. Kinahan.Pet image reconstruction.Nuclear Medicine, 1, 2006.

C. Grupen and I. Buvat.Handbook of particle detection and imaging, volume 1.Springer, 2011.

M. Slaney and A. Kak.Principles of computerized tomographic imaging.SIAM, Philadelphia, 1988.


Literature II

R.M. Leahy and J. Qi.Statistical approaches in quantitative positron emission tomography.Statistics and Computing, 10(2):147–165, 2000.

J.M. Ollinger and J.A. Fessler.Positron-emission tomography.Signal Processing Magazine, IEEE, 14(1):43–55, 1997.

L.A. Shepp and Y. Vardi.Maximum likelihood reconstruction for emission tomography.Medical Imaging, IEEE Transactions on, 1(2):113–122, 1982.

J. Beyerer and F. Puente León.Die Radontransformation in der digitalen Bildverarbeitung.at-Automatisierungstechnik, 50(10/2002):472, 2002.

J. Cui, G. Pratx, S. Prevrhal, and C.S. Levin.Fully 3D list-mode time-of-flight PET image reconstruction on GPUs using CUDA.Medical Physics, 38(12):6775, 2011.

A.J. Reader and H. Zaidi.Advances in PET image reconstruction.PET Clinics, 2(2):173–190, 2007.


Spare Slides


4D NURBS XCAT* Cardiac Torso Phantom.

Realistic model of human anatomy incl.cardiac and respiratory patient motionPlace Tumor in Prostate, injectRadiopharmacon


Frequency cut-off.


Interlude: Fast Fourier Transform (FFT).Discrete FTs are computationally very expensive. Solution: FFTFT: O(N2), FFT: O(N log2N) (2 weeks vs. 30 s CPU time)FT of length N can be written as sum of 2 FTs of length N/2,one for even (e) and one for odd (o) parts:

Fk =

N−1∑j=0

fje− 2πi

Njk =

N/2−1∑j=0

f2je− 2πi

N(2j)k +

N/2−1∑j=0

f2j+1e− 2πi

N(2j+1)k

= F(e)k +W kF

(o)k , W k = exp(2πik/N) (root of unity)

= F(e...e)k + . . .+W ...F

(eoo...o)k + . . .+W pkF

(o...o)k

Divide data all the way down to FTs of length 1.FT of length 1 is function value itselfF

(eoee...o)k = fn for some n→ Each combination (eoeoo...o)

corresponds to an element of the input vector f .Milan Zvolský (DESY.) 56 / 48

Interlude: Fast Fourier Transform (FFT).1 signal of16 points

2 signals of8 points



16 signals of1 point

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

0 2 4 6 8 10 12 14 1 3 5 7 9 11 13 15

0 4 8 12 2 6 10 14 1 5 9 13 3 7 11 15

8 4 12 2 10 6 14 1 9 5 13 3 11 7 150

8 4 12 2 10 6 14 1 9 5 13 3 11 7 150

Figure: An N point signal is decomposed into N signals à 1 point

Reverse the pattern (eoee...), assign binary number (e = 0,o = 1)E.g.: (eoeo)→ (oeoe) = (1010) = 10

Combine adjacent elements (with the Wnk’s) to get 2-pointtrafo, etc


Interlude: Fast Fourier Transform (FFT).1 signal of16 points




16 signals of1 point

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

0 2 4 6 8 10 12 14 1 3 5 7 9 11 13 15

0 4 8 12 2 6 10 14 1 5 9 13 3 7 11 15

8 4 12 2 10 6 14 1 9 5 13 3 11 7 150

8 4 12 2 10 6 14 1 9 5 13 3 11 7 150

e

o

e

o

Figure: An N point signal is decomposed into N signals à 1 point

Reverse the pattern (eoee...), assign binary number (e = 0,o = 1)E.g.: (eoeo)→ (oeoe) = (1010) = 10

Combine adjacent elements (with the Wnk’s) to get 2-pointtrafo, etc


Backprojection Procedure.Sinogram represents theprojection of theradiopharmacon distributiononto the detectorFixed point in the objecttraces a sinusoidal path inprojection spaceSinogram: Superposition of allsinusoids corresponding to eachpoint of activity in the objectLORs described by p(s, φ)f(x, y)→ p(s, φ): RadontransformThe task of tomography is tofind f(x, y) from p(s, φ)


Some examples.

102 103 104

105 106

fFBP(x)

f (x)

107

Figure: The effect of the number of detected photons on the quality of thereconstructed image. A synthetic image was projected and Poisson noisewas added to the projections. Reconstructed with the FBP algorithm

Noise has a high impact on the image quality


Backprojection: Blurring of the image.

Assign equal intensity alongeach LORThe projections are formed oneview at a timeSum up the LORsThe final backprojected imageis the summation of thebackprojections from all viewsBackprojected image 6=original imageSolution? For this we needsome math...


Backprojection.

Discrete caseProjection: P = AXBackprojected image: B = ATP

X = (x1, x2, x3, x4)T

P =(p(1, 0◦), p(2, 0◦), p(1, 90◦), p(2, 90◦)

)T= (7, 2, 5, 4)T

B = ATP =

1 0 1 00 1 0 11 1 0 00 0 1 1

T

7254

=

127116


Interlude: GPU Programming.ML-EM is much slower than FBP. But this is not a real problemnowadays...

Parallelise your work384 cores, 32768 registers/blockEach one can perform simple tasks very quickly


Backprojection Numerical Example.a b cdg

e fh i

a+b+c

d+e+f

g+h+i

a+d+gb+e+h

c+f+i

c

g

a

i

d+b

g+e+c

h+f d+h

a+e+i

b+f

e→ 3e+∑

proj⇒ e→ e

a→ a+ b+ c+ a+ e+ i+ a+ d+ g + a = 3a+∑

proj− f − h

⇒ a→ a− 1/3(f + h)

Blurring of the image!




Backprojection: Blurring.original image after backprojection

image comparison

1/r blurring of the image!


Central Slice Theorem.Central Slice Theorem (CST)

F1{p(s, φ′)} = F2{f(x, y)}|φ=φ′

The CST relates the Fourier transform (FT) of f(x, y) with theFT of p(s;φ)Summation of lines causes duplication in the centerHigh density of central slices in Fourier spaceOversampling in the center of the Fourier space needs to befiltered in order to have equal sampling throughout the Fourierspace

A complete set of 1D projections allows the reconstruction of theoriginal 2D distribution without loss of information


Iterative Methods - Basic Idea.1 Image Model:.Discretisation of image into N distinct voxels fj

2 System Model:.System matrix Hij : probability that an emission

from voxel j is detected in projection i:

pi =∑Nj=1Hijfj

3 Data Model:.Variation of projection measurements around

expected mean values (Poisson). Derived from our

basic understanding of the acquisition process

4 Governing Principle:.Defines the ’best’ image..e.g. Maximum Likelihood cost function

5 Algorithm:.Optimises the cost function: Maximum

Likelihood - Expectation Maximisation

𝑙1

𝑙𝑓𝑢 𝑙𝑙

𝑙𝑣

𝑙2


FBP Numerical Example.black = 1, white = 0

For projections at 0◦ and 90◦ the spacial resolution is one

pixel (i.e. τ = 1), for the other projection it is√2 pixels.

Convolute filter with projection(Product in Fourier space = convolution in position space)

Discrete convolution: (f ∗ g)(t) =∑

k f(k)g(t− k)

0

* =

Filter

2.0

1QP1

0.25 for tau =10.354 for tau =1/sqrt(2)

-0.1013 for tau =1-0.1433 for tau =1/sqrt(2)

1.0

0.39

-0.20

0.05

-0.10-0.1

0.25


FBP Numerical Example.

0.40

-0.10

0.05

-0.20

-0.20

-0.41

-0.20

0.09

0.80

0.09

0.30

-0.10

0.05

-0.20

0.40

2Q

1Q

Q

4Q

3

Push filtered projections back into image

-0.20

0.40

0.05

-0.100.05-0.20

-0.10

-0.20

0.40

0.05

-0.10

-0.20

0.40

0.05

-0.10

-0.100.05-0.20

-0.20

-0.200.09

-0.20

0.05-0.20

-0.100.05-0.20

-0.100.05

-0.10

0.09

-0.410.800.09

-0.20

-0.20

0.80

0.80

0.09

0.09

-0.410.800.09

-0.41

-0.200.40

0.05

-0.10

-0.20

0.40

0.05

-0.20

-0.20

0.40

0.40

0.40

0.40

0.40

-0.10

0.30

-0.20 0.30 0.09

-0.20 0.30 0.09

-0.20

0.09 0.30

0.30 0.09 -0.200.30

0.30

0.30

1

Q backprojectedQ backprojected

Q backprojectedQ backprojected

43

2

.Usually do filtering in Fourier space..FT computationally expensive → use FFT (2 weeks vs. 1 min. CPU time)Milan Zvolský (DESY.) 69 / 48

Basic Idea.Example: 2× 2 MatrixSum of 1st row is 2, sum of 2nd row is 3, sumof 1st column is 0, sum of 2nd column is 5.⇒ x1 = 0, x2 = 2, x3 = 0, x4 = 3

A B

C D

0 5

2

5

Example: Another 2× 2 MatrixSum of 1st row is 5, sum of 2nd row is 4, sumof 1st column is 7, sum of 2nd column is 2.⇒ No unique solution.

Way out: Take more views?

Hypothesis: With many views frommany angles, we are able toperfectly reconstruct the image


Backprojection.

SummaryTomography is a process of taking projection data andconverting the data into cross-section images. Projection datafrom multiple views are requiredA projection is a line integral of an objectBackprojection is a superposition procedure and it sums thedata from all projection views. Backprojection evenlydistributes the projection domain data back along the same linesfrom which the line-integrals were formed.


Filtered Backprojection.Eliminate blurring by introducing negative “wings” around the spikein the projections before backprojecting


BP vs. FBP.

Figure: Reconstruction of a pointsource with simple backprojection (top)and filtered Backprojection (bottom)


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Tomographic Image Reconstruction. - An Introduction.

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