Jakob Sauer J˝rgensen - Technical University of Denmarkpcha/HDtomo/SC/Week1Days1and2.pdf ·...

Introduction to tomographic reconstruction

Jakob Sauer Jørgensen

Postdoc, Technical University of Denmark (DTU)

02946 Scientific Computing for X-Ray Computed TomographyDTU Training School, Week 1, Days 1+2

January 2–6, 2017

SC for CT, week 1, Mon+Tues: Overview

Monday

09.00 – 09.30 Welcome

09.30 – 12.00 Forward problem: Radon transform & Lambert-Beer

12.00 – 13.00 - - - Lunch break - - -

13:00 – 15.30 Reconstruction: Filtered Back-Projection (FBP)

15.30 – 16.30 Intro to micro-CT scan and micro project

Tuesday

09.00 – 12.00 Micro-CT scan at DTU Physics

12.00 – 13.00 - - - Lunch break - - -

13:00 – 15.30 Reconstruction real data set

15.30 – 16.30 Time for micro project

Outline

1 Data model: Lambert-Beer law and Radon transform

2 Reconstruction: Filtered back-projection (FBP) algorithm

3 Introduction to hands-on CT scan and micro project

4 Practical aspects for reconstruction from real data

3 / 73 Introduction to tomographic reconstruction Jakob Sauer Jørgensen, DTU

Tomography

What is tomography?

I From Greek: Tomos a section or slice, Graphos: to describe.

I Imaging of slices of an object – without actually slicing it!

I These days not just slices but 3D maps can be obtained.

I To see the inside, need information obtained from the outside.

I Some applications of tomography...


Medical imaging


Materials scienceDevelopment of advanced materials requires understanding theirproperties at the micro and nano scale.

Example: Maximize strength of glass fibre for wind turbine blades.

Laboratory micro-CT scanners and large-scale synchrotron facilities


Non-destructive testing

Production, security, metrology, etc.

Example: Airport luggage scanner for threat detection.


Tomography: Imaging from projectionsI Projections are measured around an object using X-rays.I Goal is to reconstruct the object from the projections.I Simplest is 2D parallel beam geometry, which we focus on.I Used in early scanners and in large scale synchrotron facilities.

θ

θ

1

2


Cone-beam geometry

I Cone-beam (medical CT scanners, lab-based micro-CT, etc)

I Cone-beam restricted to central slice: Fan-beam

I Move source far away: Parallel-beam (synchrotron)


Contrast mechanism: X-ray attenuation

Heavier matter attenuate X-rays more: Air – tissue – bone – metal.Quantified by so-called linear attenuation coefficient µ.

Wilhelm Conrad Rontgen and the first X-ray image ever takenshowing his wife’s hand (1895).


Lambert-Beer law of attenuationHomogeneous material:

I = I0 exp{− µ0D

}Non-homogeneous (more interesting) material:

I = I0 exp

{−∫Lµ(x)dx

}Rearrange to line integral form:

− logI

I0=

∫Lµ(x)dx

A tomographic scan:I Measure I along many lines to get many line integral values

through the object from which to determine µ(x).I The intensity I is called the transmission, while the

corresponding − log(I/I0) is called absorption or projection.

I0 Iµ0-

D

I0 Iµ(x)

X-ray L

��

-


Transmission vs absorption

I0 = 10000 I I/I0 − log (I/I0)

∫L µ(x)dx =

-

-

-

-

-

10000

5000

2500

1250

625

1.0000

0.5000

0.2500

0.1250

0.0625

0.0

0.7

1.4

2.1

2.8


The origin of tomographic reconstruction

Original reference:

Uber die Bestimmung von Funktionen durchihre Integralwerte langs gewisserMannigfaltigkeiten. (Johann Radon, 1917):

An object can be reconstructed perfectlyfrom a full set of line integrals.

What does a “full set” mean?


Parametrizing lines in the planeHow to describe a line?One way is by slope and y intercept. Vertical lines excluded.Alternative: “normal form”:

Lφ,ρ = {(x, y) | x cosφ+ y sinφ = ρ}ρ is signed orthogonal distance of line to origin.φ is angle between positive x-axis and unit normal vector to Lφ,ρ

φ x

y

(cosφsinφ

) ρ

Lφ,ρ


The Radon transform for parallel-beam geometryObject f(x, y):

I contained in disk of radius R

Line of integration Lφ,ρ given by:

I φ: angle of line to be projected onto

I ρ: position on line

Projection: All line integrals at one φ:

pφ(ρ) =

∫Lφ,ρ

f(x, y)ds for ρ ∈ [−R,R].

The Radon transform is:

[Rf ](φ, ρ) = pφ(ρ) =

∫Lφ,ρ

f(x, y)ds

for φ ∈ [0◦, 180◦[ and ρ ∈ [−R,R].

φ x

y

ρ

ρ

pφ(ρ)


Pen and paper exercise: Radon transform of a disk

Given image of a small centered disk of radius r:

f(x, y) =

{1 for x2 + y2 ≤ r20 otherwise

Derive that the Radon transform is:

[Rf ](φ, ρ) ={

2√r2 − ρ2 for |ρ| ≤ r

0 otherwise

Image Radon transformed image

-�

RadontransformR

R−1 ?


Radon transform of an ellipse (Kak & Slaney)

φ x

y

ρ

pφ(ρ)


Radon transform of an ellipse (Kak & Slaney)

Given ellipse

f(x, y) =

{c for x2

A2 + y2

B2 ≤ 1 (inside ellipse)0 otherwise (outside the ellipse)

If centered (x1 = y1 = 0) and not rotated (α = 0) then

pφ(ρ) =

{2cABa2(φ)

√a2(φ)− ρ2 for |ρ| ≤ a(φ)

0 otherwise

where a2(φ) = A2 cos2 φ+B2 sin2 φ.

If centered at (x1, y1), rotated by α then given from pφ(ρ) above as

pφ(ρ) = pφ−α(ρ− s cos(γ − φ)),

where s =√x21 + y21 and γ = tan−1(y1/x1).


Parametrized form of Radon transformThe Radon transform

pφ(ρ) =

∫Lφ,ρ

f(x, y)ds, (1)

can be written explicitly using a parametrization of the line:

pφ(ρ) =

∫ ∞−∞

f(x(s), y(s))ds, (2)

where at fixed φ and ρ:

x(s) = ρ cosφ− s sinφy(s) = ρ sinφ+ s cosφ

Line is traced as s runsfrom −∞ to ∞.

φ x

y

(cosφsinφ

)(− sinφ

cosφ

)ρ

Lφ,ρ


A convenient explicit form of Radon transform

Useful alternative expression for the Radon transform:

pφ(ρ) =

∫ ∞−∞

∫ ∞−∞

f(x, y)δ(x cosφ+ y cosφ− ρ)dxdy

Interpretation:The line Lφ,ρ consists of exactly those (x, y) at whichx cosφ+ y cosφ− ρ = 0, which is the argument of the Dirac deltafunction δ. Thus, the integrand is restricted to function values off(x, y) on Lφ,ρ, which amounts to the corresponding line integral.


Dirac delta function

Generalized function with heuristic definition:

δ(t) =

{+∞ for t = 00 for t 6= 0

and ∫ ∞−∞

δ(t)dt = 1

Important property of Dirac delta function:∫ ∞−∞

f(t)δ(t− T )dt = f(T )

This is called the sifting property, because the Dirac delta functionacts as a sieve and “sifts out” the value at t = T .


Connection Radon Transform and Lambert-Beer

The Radon transform

pφ(ρ) =

∫Lφ,ρ

f(x, y)ds

and the Lambert-Beer law along the same line Lφ,ρ

Iφ,ρ = I0 exp

(−∫Lφ,ρ

µ(x, y)ds

)

are connected through the identifications

f(x, y) = µ(x, y)

pφ(ρ) = − log

(Iφ,ρI0

)


Projections all around the object

The Radon transform describes the forward problem of how (ideal)X-ray projection data arises in a parallel-beam scan geometry.

6

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@@

@@@

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Image and sinogram

The output of the Radon transform is called a sinogram:

Example image

20 40 60 80 100

10

20

30

40

50

60

70

80

90

100 0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

Sinogram

φ50 100 150

ρ

20

40

60

80

100

120

1400

0.1

0.2

0.3

0.4

0.5

Angle φ

Positionρ

Note that 180◦ captures all necessary projections of the object.The other 180◦ are mirror images of the opposite projections.


A bit about noise in CTLambert-Beer law normally expressed in intensity I, i.e., number ofphotons per time. Equivalently if during an interval the sourceemits N photons, the average transmitted number of photons is

N = N0 · p

where the transmission probability is

p = exp

(−∫Lµ(x, y)ds

)The number of transmitted photons in interval is a Poisson randomvariable, i.e., the probability that k photons are transmitted is

P (N = k) =λk

k!exp(−λ)

where the parameter

λ = N0 · p

is also the expected value of the number of transmitted photons.25 / 73 Introduction to tomographic reconstruction Jakob Sauer Jørgensen, DTU

Gaussian approximation and effect on projection

For large number of photons a Poisson distribution can be closelyapproximated by a Gaussian distribution. In particular theapproximating Gaussian distribution has both mean and standarddeviation of N0p, i.e., approximately the number of photonstransmitted

N ∼ N (N0p,N0p)

How does this affect the reconstruction? It turns out that for theprojection value b = − log(N/N0) to be used in the reconstructionapproximately

b ∼ N (− log p,1

N0p)

Thus, with increasing intensity (larger N0) the expected projectionvalue is constant but the standard deviation decays; in other wordsnoise is reduced.


Some other error sources

In practice, in addition to Poisson noise data can be affected bynumerous other issues:

I Detector noise

I Scatter (some X-rays do not follow straight line)

I X-rays not monochromatic, but full spectrum. Attenuationcoefficient depends on energy, e.g., beam hardening

I Bad detectors, e.g., void measurements

I Too dense features, e.g., metal blocking rays completely

I Object changing during acquisition, e.g., motion


Exercises: Radon transform

Available from:http://www2.compute.dtu.dk/~pcha/HDtomo/SCforCT.html

File:ExWeek1Days1and2.pdf


http://www2.compute.dtu.dk/~pcha/HDtomo/SCforCT.html

Outline






Reconstruction

Overview:

I Now we have seen how the Radon transform describes theforward problem of how projections arise from an object.

I Question: How can we do the inverse operation, namelyreconstruct an image of the object from the projections?

I Class of analytical reconstruction methods construct aninverse transform. For parallel beam case, inverse Radontransform.

I This gives rise to the most common reconstruction method:Filtered Back-Projection (FBP).

I To derive the FBP method, we need an important resultknown as the Fourier slice theorem, and to set some notation.


Back-projection

Mathematically, back-projection is written as integration over all φ:

B[pφ(ρ)](x, y) =∫ π

0bφ(x, y)dφ

of the back-projection images at each angle φ:

bφ(x, y) = pφ(x cosφ+ y sinφ)

Interpretation: “Smearing and summation”Each point (x, y) and each angle φ define a unique linear positionρ = x cosφ+ y sinφ. In the back-projection image the point (x, y)is assigned the value at ρ from the projection pφ(ρ). This is“smearing”. Back-projection then adds up all contributions to each(x, y) by integrating over φ. This is “summation”.


Back-projection: Does it invert projection? (No)

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The Fourier transformFourier transform (“Frequency representation”):

−10 −5 0 5 10−0.5

0

0.5

1

1.5

2

2.5

−10 −5 0 5 10−0.5

0

0.5

1

1.5

-�

1D Fouriertransform F1

1D InverseFouriertransform F−11

ρ ω

Also possible for functions of more than 1 variable:

-�

2D Fouriertransform F2

2D InverseFouriertransform F−12


The Fourier transform pair definitions

1D Fourier transform and inverse:

f(ω) = F1[f(t)](ω) =

∫ ∞−∞

∫ ∞−∞

f(t)e−j2πωtdt (3)

f(t) = F−11 [f(ω)](t) =

∫ ∞−∞

∫ ∞−∞

f(ω)e+j2πωtdω (4)

2D Fourier transform and inverse:

F (u, v) = F2[f(x, y)](u, v) =

∫ ∞−∞

∫ ∞−∞

f(x, y)e−j2π(ux+yv)dxdy (5)

f(x, y) = F−12 [F (u, v)](x, y) =

∫ ∞−∞

∫ ∞−∞

F (u, v)e+j2π(ux+yv)dudv (6)


Fourier Slice Theorem: Key to reconstruction

Projection at angle φ 1D Fourier transform At angle φ in Fourier space

All 1D Fouriertransformed projections

=

2D Fourier transform

All angles φ ∈ [0◦, 180◦[ needed to build complete 2D Fourier transform


Derivation of Fourier Slice Theorem (1)

Strategy: Manipulate 1D Fourier-transformed projection into slicethrough 2D Fourier-transformed image.

pφ(ω) =

∫ ∞−∞

pφ(ρ)e−j2πωρdρ

=

∫ ∞−∞

∫ ∞−∞

∫ ∞−∞

f(x, y)δ(x cosφ+ y sinφ− ρ)e−j2πωρdxdydρ

=

∫ ∞−∞

∫ ∞−∞

f(x, y)

∫ ∞−∞

δ((x cosφ+ y sinφ)− ρ)e−j2πωρdρdxdy

=

∫ ∞−∞

∫ ∞−∞

f(x, y)

∫ ∞−∞

δ(ρ− (x cosφ+ y sinφ))e−j2πωρdρdxdy

=

∫ ∞−∞

∫ ∞−∞

f(x, y)e−j2πω(x cosφ+y sinφ)dxdy

using definition of 1D Fourier transform, Dirac delta expression of pφ(ρ),

reordering, that δ(−t) = δ(t), and sifting property.


Derivation of Fourier Slice Theorem (2)Continuing by reordering and recognizing as 2D Fouirer transform

pφ(ω) =

∫ ∞−∞

∫ ∞−∞

f(x, y)e−j2π(xω cosφ+yω sinφ)dxdy

=

∫ ∞−∞

∫ ∞−∞

f(x, y)e−j2π(xu+yv)dxdy

∣∣∣∣u=ω cosφ,v=ω sinφ

= F (u, v)

∣∣∣∣u=ω cosφ,v=ω sinφ

yields finally the Fourier slice theorem:

pφ(ω) = F (ω cosφ, ω sinφ). (7)

Interpretation: (u, v) = (ω cosφ, ω sinφ) for φ ∈ [0, π) and

ω ∈ (−∞,∞) specifies a line in 2D Fourier space rotated by φ relative to

the positive u axis. This corresponds to the ρ axis in real space. Thus,

the 1D Fourer transform of a projection is equivalent to the

corresponding slice/line through the 2D Fourier transform.


Fourier reconstruction method

−10 −5 0 5 10−0.5

0

0.5

1

1.5

2

2.5

−10 −5 0 5 10−0.5

0

0.5

1

1.5

-

Radontransform R

�

Piece 1Dstogetherto form 2D

?

1D FouriertransformsF1

62D inverseFouriertransformF−12


Problems for Fourier reconstruction methodI In practice we do not have all φ and ρ since we record a finite

number of projections with a detector of fixed size.I Interpolation from polar to Cartesian grid known as

“regridding” required for 2D inverse Fourier transform:

I Accurate interpolation in Fourier domain difficult.I 2D inverse Fourier transform relatively expensive and requires

all data simultaneously.I Result: Fourier method rarely used in practice.I Alternative: Filtered Back-Projection


Derive filtered back-projection, 1

Strategy: Rewrite inverse 2D Fourier transform:

f(x, y) = F−12 [F2f ]

[2D Fourier transform definitions]

=

∫ ∞−∞

∫ ∞−∞

F (u, v)ej2π(ux+vy)dudv

[Change to polar coordinates, including Jacobian]

=

∫ 2π

0

∫ ∞0

F (ω cosφ, ω sinφ)ej2π(ω cosφx+ω sinφy)ωdωdφ

[Split integral over 2π in two:]

=

∫ π

0

∫ ∞0


+

∫ π

0

∫ ∞0

F (ω cos(φ+ π), ω sin(φ+ π))ej2π(ω cos(φ+π)x+ω sin(φ+π)y)ωdωdφ



[Using sin(φ+ π) = − sinφ and cos(φ+ π) = − cosφ :]

=

∫ π

0

∫ ∞0


+

∫ π

0

∫ ∞0

F (−ω cosφ,−ω sinφ)ej2π(−ω cosφx−ω sinφy)ωdωdφ

[Change sign/bounds in second integral:]

=

∫ π

0

∫ ∞0


+

∫ π

0

∫ 0

−∞F (ω cosφ, ω sinφ)ej2π(ω cosφx+ω sinφy)(−ω)dωdφ

Recollect using absolute value:

=

∫ π

0

∫ ∞−∞

F (ω cosφ, ω sinφ)ej2πω(x cosφ+y sinφ)|ω|dωdφ



[Apply Fourier Slice Theorem:]

f(x, y) =

∫ π

0

∫ ∞−∞

pφ(ω)ej2πω(x cosφ+y sinφ)|ω|dωdφ

[Use that x cosφ+ y sinφ is constant wrt ω-integration, say ρ]

=

∫ π

0

[∫ ∞−∞

pφ(ω)ej2πωρ|ω|dω

]ρ=x cosφ+y sinφ

dφ.

Recognizing inner integral as 1D inverse Fourier transform wedefine the filtered projection:

qφ(ρ) =

∫ ∞−∞

pφ(ω)ej2πωρ|ω|dω

= F−11 [pφ(ω) · |ω|](ρ) = F−11 [F1[pφ](ω) · |ω|] (ρ)

Projections filtered by multiplying ramp filter |ω| in Fourier domain.42 / 73 Introduction to tomographic reconstruction Jakob Sauer Jørgensen, DTU


Finally we can write

f(x, y) =

∫ π

0qφ(x cosφ+ y sinφ)dφ = B[qφ](x, y)

by recognizing the back-projection operation.

This is the Filtered Back-Projection inversion formula for theRadon transform.

Interpretation:Given each point (x, y) of the image f to be reconstructed eachangle φ defines a position ρ = x cosφ+ y sinφ. Throughback-projection, the point (x, y) is assigned the value at ρ from thefiltered projection, and contributions at all angles are summed up.


Back-projection: Does it invert projection?

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Filter projections by “ramp” before back-projection

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Back-projection vs. filtered back-projection

Projections must be filteredwith a “ramp” filter beforeback-projection.

In Fourier domain: |ω|

ω

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1

|ω|

0

0.2

0.4

0.6

0.8

1Ramp filter

[Buzug 2008: Computedtomography, Springer.]


Additional filtering needed in practice

Ramp filter is part of the inversion formula but is a high-pass filterwhich is problematic in practice when noise is present.

In practice additional filters can be used:

ω

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1

|ω|

0

0.2

0.4

0.6

0.8

1Ramp filter

[Buzug 2008: Computed tomography, Springer.]

(Ram-Lak)


Filtered back-projection (FBP) step by step

To compute FBP reconstructed image from projections:

1. Fourier-transform projections.

2. Apply ramp filter by multiplication of |ω|.3. Optionally apply additional (low-pass) filters to handle noise.

4. Inverse Fourier-transform to obtain filtered projections.

5. Backproject filtered projections and sum up.

In practice:

I Often done automatically for you by scanner/instrument

I FBP implementations available.

I Easy to use: Main user input is choice of low-pass filter.

I In MATLAB: iradon.


FBP for fan-beam and cone-beam

Parallel-beam FBP:

Projections → Filter → Back-project → Reconstruction

Fan-beam – 2 strategies:

I Rebinning to parallel-beam

I Dedicated filtered back-projection algorithm

Cone-beam:

I Dedicated filtered back-projection algorithm

I Feldkamp-Davis-Kress (FDK) is standard

Projections → Weighting → Filter → Weighting → Back-project → Reconstruction


Strengths and weaknesses of FBP

Strengths:

I Fast: Based on FFT and a single back-projection

I Few parameters to adjust

I Conceptually easy to understand and implement

I Reconstruction behavior well understood

I Typically works very well (for complete/good data)

Weaknesses:

I Large number of projections required.

I Full angular range required (limited angle problem)

I Only modest amount of noise in data can be tolerated.

I Fixed scan geometries – others require own inversion formulas

I Cannot make use of prior knowledge such as non-negativity.


Exercises

Exercises on reconstruction using FBP (exercise 2).


Outline






SC for CT, week 1, Mon+Tues: Overview

Monday

09.00 – 09.30 Welcome

09.30 – 12.00 Forward problem: Radon transform & Lambert-Beer

12.00 – 13.00 - - - Lunch break - - -

13:00 – 15.30 Reconstruction: Filtered Back-Projection (FBP)

15.30 – 16.30 Intro to micro-CT scan and micro project

Tuesday

09.00 – 12.00 Micro-CT scan at DTU Physics

12.00 – 13.00 - - - Lunch break - - -

13:00 – 15.30 Reconstruction real data set

15.30 – 16.30 Time for micro project

Hands-on CT scan session – Meeting time and place

I Micro CT scans take place in building 309, ground floor, from9:00 to 12.00 tomorrow (Tuesday) morning.

I We will meet in the cafe area of building 306, ground floor at8.50 and walk to 309 as a group.


How to find it


Hands-on CT scan session

I To get started there will be a joint introduction given byCarsten Gundlach, DTU Physics.

I There will be 2 groups doing separate scans, one on a Zeissinstrument, one on a Nikon instrument.

I Once the data has been acquired a 3D reconstruction will becomputed automatically.

I After the scans, there will be time to experiment with 3Dvisualization using in-house software until lunch break.


Zeiss micro-CT instrument

I Rotate sample, keep X-ray source and detector fixed.

I Cone-beam scan geometry.

I Reconstruction by FDK.

I Similar Nikon micro-CT instrument.


Micro project – exterior tomography

I In deriving the FBP reconstruction formula, we assume fulldata is available, i.e., 180◦ and sample fully contained withinfield of view.

I In many cases full data is not available, e.g., limited angularrange, few projections or the region-of-interest/interiorproblem all encountered in the exercises.

I In the ROI/interior problem only rays passing through centralpart of the object are available, because object is too large orwe have zoomed in on a small region.

I In the exterior problem, only rays through outer annulus ofobject are measured, not through the center. This can happenfor example in non-destructive testing of rockets, which arelarge and dense so that X-rays do not penetrate sufficiently.


Shepp Logan, full data 180◦

Original phantom and generated sinogram.


Shepp Logan, reconstruction from full data 180◦

Full sinogram and FBP reconstruction.


Shepp Logan, reconstruction region-of-interest data

Region-of-interest/interior sinogram and reconstruction.


Shepp Logan, reconstruction exterior data

Exterior sinogram.

Reconstruction?


Goals of project

Investigate the exterior problem, in particular try to establishwhat can and cannot be reconstructed.

The micro project is quite open-ended. In the lectures andexercises you are introduced to theory, techniques and tools fortomographic reconstruction. In the project it is up to you tochoose from this material and apply to the exterior problem inorder to understand the problem and the limitations ofreconstruction from exterior data.


Some ideas

I Design phantoms with features to illustrate howreconstruction quality depends on size of features, closeness toboundary, radius of missing interior region, etc. (Mon+Tues)

I Apply FBP including practical corrections such as padding asused for ROI data (Mon+Tues)

I Apply SVD analysis and compare singular values and vectorswith full data case (Wed)

I Apply ideas from micro-local analysis to assess which featurescan be reconstructed (Thurs)

You are free to pick from this list or pursue your own ideas withinthe course material and exterior tomography – please do nothesitate to discuss ideas with us.


Time for micro project

I Form groups of 2–3 persons.

I Start experimenting and familiarizing yourselves with theexterior problem.

I For example,I create code to simulate missing data in the sinogram caused

by the exterior problem.I create interesting phantoms and do FBP reconstruction from

simulated exterior data.I ...

I Suggestions:I Take notes/keep a log of the experiments you do and what you

learn from each, etc.I Make separate scripts for each study you do to make it easy

for yourself to remember and reproduce later as well as tocreate figures for the oral presentation on Friday.


Outline






Flat and dark field correction

Typical data acquired:

I I: Transmission images (sample in, source on)

I I0: Flat (or white) field (sample out, source on)

I ID: Dark field (sample out, source off)

FBP needs line integrals (Radon transform):

I Convert using Lambert-Beer:∫Lf(x, y)ds = − log

I

I0

To obtain projections:

Z = − log Y, Y =I − IDI0 − ID

← pixelwise division


Center-of-rotation (COR) correction

I Standard FBP implementationssuch as MATLAB’s iradon assumea perfectly centered object.

I In other words the center ofrotation should be mapped to thecentral detector pixel.

I In practice only approximatecentering physically possible.

I Naively reconstructing yieldsartifacts. Need to docenter-of-rotation correction.

I Can be done by “shifting”projections by padding sinogramwith sufficiently many artificaldetector pixel values.

&%'$s s

---------Centered

&%'$s s

---------

Non-centered


Region-of-interest (ROI) correction

I In some cases the object to be scanned is too large to fit inthe field of view.

I Or we want to focus on some small region-of-interest (ROI).

I Projections are truncated (not covering entire object).

I Can we still reconstruct the object? Or just the ROI?

Image: [Wang & Yu 2009, Med. Phys. 36, 3575–3581]


The region-of-interest problem

I Mathematically, the answer is NO.

I Interior Radon data contaminated by exterior Radon data.

I Interior reconstruction will also be contaminated.

Fig. from [Bilgot et al. 2011, IEEE Nuc. Sci. Symp. Conf. Rec. , 4080–4085]


ROI correction in practice

I Boundaries correct.

I Large artifact apparent.

I Trick of padding of sinogramyields large improvement,but values still off.

I The exercise data set is ROIdata!


Many other possible artifacts and corrections

I Ring artifacts

I Beam hardening

I Zingers

I Misalignment

I Metal artifacts

I Motion

I ...


Exercises

Reconstruction of a real data set (2D parallel-beam). Exercise 3.


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Jakob Sauer J˝rgensen - Technical University of Denmarkpcha/HDtomo/SC/Week1Days1and2.pdf ·...

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