Radon Inversion in the Computed Tomography Problem ...

The CT ProblemDomainRanges

ReconstructionReferences

Details for the Curious

Radon Inversion in the Computed TomographyProblem

Ryan Walker

November 17, 2010

Ryan Walker Radon Inversion in the Computed Tomography Problem




Acknowledgements

Much of the theoretical background and intuition on the CTscan problem is drawn from a series of brilliant lecturesdelivered by Gunther Uhlmann, Peter Kuchment, and LeonidKunyansky at the IPDE Summer School 2010 at theUniversity of Washington.

A comprehensive introduction to the mathematics of CTscans can be found in Charles Epstein’s book, Introduction tothe Mathematics of Medical Imaging [1].

A more advanced treatment can be found in the classic bookof Frank Natter, The Mathematics of Computed Tomography,[3].





The Radon Transform

The Computed Tomography Problem

The Computed Tomography problem: Want to view theinternal structure of something without cutting it open.Physical setup:

1 Send radiation through the object and look at how itattenuates.

2 Determine the object’s density by looking at the attenuationpatterns.

3 Recover detailed image of internal structure from the density

Example: (Medical CT) Tumors and other abnormalities havespecific densities, distinct from healthy tissues.





The Radon Transform

A typical fan beam scan setup. Patient lies between a fixeddetector panel and a rotating source. The source emits x-rayradiation in straight beams and the attenuated signals are collectedby the detector.





The Radon Transform

Direct Problem: A model for X-ray Attenuation

Goal: Image a 2D slice of the patient’s head.

Let I(x) be the flux of radiation at the point x ∈ R2. and letI (x) = ‖I(x)‖ be the intensity of the radiation at x.

Let µ(x) be the attenuation coefficient for X-ray radiation.This coefficient describes the energy loss for radiation passingthrough x and is determined by the density and materialsproperties of the body. µ(x) is a reasonable proxy for thepatient’s density function.





The Radon Transform

A model for X-ray Attenuation

For X-ray radiation traveling along a straight line we haveBeer’s Law:

dI

dx= −µ(x)I

So if l is the line segment on the source and detector and thesource emits initial intensity I0, then the detector on theopposite side feels a radiation intensity

I = I0e−

∫l µ(x) dx





The Radon Transform

Inverse Problem

We want to find µ(x).

We know all the source intensities and we can compute all thedetector intensities by computing all the integrals over all thelines on the sources and detectors.

So solving the inverse problem is equivalent to resolving thefollowing question: If we know all the line integrals of afunction f (x) in a domain Ω, is this enough information torecover the function in Ω?





The Radon Transform

The Inverse Problem Sketch





The Radon Transform

The Radon Transform

Let l be a line in R2, t (affine parameter) denote the perpendiculardistance from l to the origin, ω a unit vector perpendicular to l ,and ω⊥ a unit vector perpendicular to ω. Then the points x ∈ l arethose which satisfy

< x , ω >= t





The Radon Transform

The Radon Transform

The Radon Transform of f ∈ S(R) is the map given by:

Rf (t, ω) =

∫Lt,ω

f (x) dx =

∫ ∞−∞

f (tω + sω⊥) ds.

The major questions:

Domain

Range

Inversion





The Radon Transform

Easy Properties of the Radon Transform

R maps a function on R2 into the set of its line integrals.

Even: Rf (−t,−ω) = Rf (t, ω).

An easy computation shows that translations Ta for a ∈ R2

and rotations O(θ) ∈ SO(R2) commute with the Radontransform in the sense that

R(O(θ)f )(t, ω) = O(θ)Rf (t, ω)

R(Taf )(t, ω) = Ta·ωRf (t, ω)





The Radon Transform

The Radon and Fourier Transform

The translational and rotational invariance of the Radon transformwill suggest to the harmonic analyst that the Radon transformmight be related to the Fourier transform.

Theorem (Projection-Slice)

Let f ∈ L1(R) and the natural domain of R∫ ∞−∞Rf (t, ω)e−itr dt = f (rω)

Proof

The 1D Fourier transform of Rf in the affine parameter t isthe 2D Fourier transform of f expressed in polar coordinates.





The Radon Transform

Radon Tranform Data

Obviously in CT scanning we sample only a finite number ofvalues of R(µ).

The raw output of a CT scan procedure is called a sinogram.

Typically, we visualize this data by a “heat” plot of the valuesof the Radon transform against t and θ.

Darker and lighter areas of the sinogram plot correspond todifferent values of the Radon transform.

We won’t usually have real X-ray data to use, but image filesare a great substitute. Pixel locations correspond to points inspace and gray scale colors correspond to densities.





The Radon Transform

A Phantom:





The Radon Transform

Sinogram





The Radon Transform

Whose sinogram is this?





The Radon Transform






The Radon Transform


Image Source http://images.hollywood.com/site/homer-simpson.jpg





The Radon Transform

Sine Waves in Sinograms

The appearance of sine waves in sinograms is due to the factthat the Radon transform of a delta function has support on atrigonometric curve.

Why? Regard the point x as a vector with angle θ0. Bytrigonometry the only lines containing x will be those withangle θ and affine parameter t satisfying

t = |x | cos (θ − θ0).

Since Rδx(t, θ) = 0 whenever x is not in Lt,θ, the support ofδx is a cosine curve in the θ-t plane.

We can show that the Radon transform is continuous and thusan object made up of many small,sharp-edged features willhave a sinogram that is a combination of blurred sine curves.





The Domain

On first pass, we notice that for the Radon transform of f tobe defined on R× S1, we must have that the restriction of fto lines makes sense and that f decays fast enough alongevery line so that the line integrals converge. This suggest thenatural domain of the Radon transform should be f satisfying

1 f restricted to any line in R2 is locally integrable.2 f decays rapidly enough for the improper integrals in the

definition to converge.

For CT, this is probably much more general than necessary.Tissue structures should be piecewise continuous and peoplecompactly supported!


http://images.hollywood.com/site/homer-simpson.jpg




More on Domains

If f is compactly supported then Rf (t, ω) is compactlysupported in the t parameter. (Why?)

Note that continuous functions of compact support belong tothe natural domain of R, so by density we may extend thedefinition of the Radon transform to f ∈ L1(R). Forf ∈ L1(R2), define Rf to be the limit in L1 norm of thesequence of approximates Rfn, where fn is a sequence ofsmooth, compactly supported functions converging to f in theL1 norm.





Continuity

Continuity: For f ∈ L1(R2) and the natural domain of R,define

‖Rf (t, ω)‖1,∞ = supω∈S1

∫ ∞−∞|Rf (t, ω)| dt.

Then‖Rf ‖1,∞ ≤ ‖f ‖1.

Proof

Continuity implies the scanning procedure is stable even if thepatient can’t keep completely still.





Schwartz Functions

It will be useful to occasionally consider R defined over theset of Schwartz class functions:

S(R2)

= f ∈ C∞(R) | supx∈Rn|xαDβf (x) ≤ ∞ ∀α, β.

We can define an analogous space over T = R× S1, wherethe Schwartz decay happens in the affine parameter.





A Reconstruction

Let f ∈ S(R2). Since Rf is even, we can prove:

FtRf (−r ,−ω) = FtRf (r , ω).

Then by the Fourier inversion theorem:

f (x) =1

(2π)2

∫R2

e i<x,ξ>f (ξ) dξ

=1

(2π)2

∫ 2π

0

∫ ∞0

e ir<x,ω>f (rω)r drdω

=1

(2π)2

∫ 2π

0

∫ ∞0

e ir<x,ω>FtRf (r , ω)r drdω

=1

(2π)2

∫ π

0

∫ ∞−∞

e ir<x,ω>FtRf (r , ω)|r | drdω.





The Range Question

We have the reconstruction formula

f (x) =1

(2π)2

∫ π

0

∫ ∞−∞

e irx ·ωFtRf (t, ω)|r | drdω

By the projection slice theorem, R has trivial kernel over mostreasonable domains.

Though we recover f uniquely by the above formula, withoutknowledge of the range there is no guarantee that this is theonly reconstruction formula.





Range

Geometry: Suppose g is in the range of the radon transform.Then g is a function defined on T = R× S1, but by evenness,we must identify g(t, ω) and g(−t,−ω). So T is in fact aninfinitely wide Mobius strip.

Equip T with the inner product

(f , g)T =

∫Tg(t, ω)f (t, ω) dωds

and define the Hilbert space L2(T ) as the set of functions onT with ‖f ‖2L(T ) =

√(f , f )T <∞.

The natural domain of f is also dense in L2(R2). Thus wemight hope that there is a continuous extension of R to amap from L2(R2) to L2(T ). This does not happen.





A continuous L2 extension of R?

R does not extend to a continuous operator between L2

spaces. More importantly, R−1 is not continuous as anoperator from L2(T )→ L2(R2).

Parseval relation:∫R2

|f (x)|2 dx =

∫ 2π

0

∫ ∞0|Fr (R(r , ω))|2|r | drdω

Consequences:1 R is unbounded on L2(R2).2 R is a smoothing operator





R as a smoothing operator

Recall the Fourier transform identity

F(dn

dxnf )(ξ) = inξnF(f )(ξ).

This suggests a meaningful way to think about fractional orderderivatives. In particular, define the 1/2 derivative operator by

D1/2f = F−1(|ξ|1/2F f )

Parseval:∫R2

|f (x)|2 dx =1

(2π)

∫ π

0

∫ ∞−∞|D1/2Rf (t, ω)|2 dtdω

Thus, a function which is a Radon transform has someadditional smoothness beyond that of an arbitrary L2 function.





Range Conditions

Let us attempt to come up with a description of the range forthe Radon transform.

For suitable choice of domain, the range of R turns out to behighly structured.

Evenness.

For f in the natural domain of R, define

Mn(f )(ω) =

∫RtnRf (t, ω) dt.

Claim: Mn(f )(ω) is a homogeneous polynomial of degree n inω = (ω1, ω2) for any n.





Moments

Use the coarea formula to obtain:

Mn(f )(ω) =

∫RtnRf (t, ω) dt

=

∫R

∫x ·ω=t

tnf (x) dsdt

=

∫R2

(x · ω)nf (x) dx

Thus the moments of the Radon transform of any smooth,compactly supported function are homogeneous polynomials withrespect to components of ω.





The Range Result

Theorem

Let g(t, ω) satisfy

1 g ∈ S(T )

2 g is even

3 Mng(ω) =∫R tng(t, ω) dt is a homogeneous polynomial in

the components of ω.

Then there is an f in S(R2) so that Rf = g. In particular, R is abijection from S(R2) to S(T ).





Sketch of Proof.

Given g satisfying the conditions, the candidate inverse is specifiedby

f (rω) = Ftg(s, ω).

If f is in the S(T ) then by projection-slice

Ftg(s, ω) = FtRf (s, ω).

Thus the proof amounts to obtaining f ∈ S(T ).





Back-projectionHilbert Transform Formulation

First Reconstruction Attempt: Back-Projection

Reconstruction:

f (x) =1

(2π)2

∫ π

0

∫ ∞−∞

e ir<x,ω>FtRf (r , ω)|r | drdω.

For g ∈ S(T ), define the back-projection operator by

R]g(x) =

∫S1

g(x · ω, θ) dω

Intuitively, back projection averages g over every line throughx .






Back-projection

Write the reconstruction formula as:

f (x) =1

2

1

(2π)2R](∫ ∞−∞

e ir<x,ω>FtRf (r , ω)|r | dr).

This reconstruction can be viewed as a two stage process:inner step is filtration and outer step is back-projection.

Radial integral is a filter (|r |) applied to the Radon transform.Back-projection is the angular integral.Formula is often called filtered back-projection formula.Factor |r | suppresses low-frequency components and amplifieshigh frequency components.Numerically useful but not in this form.






Back-projection

Does back-projecting alone recover f from Rf ?

This might seem like a good guess since we know the value ofevery

∫L f dx where L is a line through x .

Answer: no.






A Phantom:






Back-projection: Blurring






Back-projection: Fail






Back-Projection

For functions f (x) from a “good domain” space, recalling thatt = x · ω, compute:

R#Rf (x) =

∫S1

Rf (x · ω, θ) dω

=

∫S1

∫ ∞−∞

f ((x · ω)ω + sω⊥) dsdω

Basically, a polar integral. Rewrite:

R#Rf (x) =

∫2f (y)

|y − x |dy =

2

|x |∗ f (x).

Thus, we should recover not f (x) but a blurred version of f (x)with back-projection.






A Little More on Back-Projection

Formally, we can show that R] : L2(T )→ L2(R2) is up to aconstant the (non-continuous) dual to R : L2(R2)→ L2(T ).

That R] does not recover f from Rf reflects the fact the R isnot a unitary transformation on L2 spaces.






A second look at filtration

It is very interesting to note that if we replace |r | with just r in theformula

f (x) =1

2

1

(2π)2R](∫ ∞−∞

e ir<x,ω>FtRf (r , ω)|r | dr)

we could write

f (x) =1

4πiR](

1

2π

∫ ∞−∞

e ir<x,ω>Ft(∂tRf )(r , ω) dr

)=

1

4πiR] (∂tRf (t, ω))






The Hilbert Transform

This motivates the introduction of the Hilbert transform: forg ∈ L2(R) with g ∈ L1(R): We see that the filtration step in ourreconstruction is

1

2π

∫ ∞−∞

e ir<x,ω>FtRf (r , ω)|r | dr =1

2π

∫ ∞−∞

sgnrrFtRf (r , ω)e irt dr

=1

2πi

∫ ∞−∞

sgnrFt(∂tRf )(r , ω)e itr dr






Hilbert Transform Formulation

We have the filtered back-projection formula:

f (x) =1

4πiR]H d

dt(Rf )

Filtration is itself two steps: derivative and Hilbert transform.

Computationally, this is a superior formulation.






Numerical Methods

Given: real or simulated Radon transform data over a discreterange of affine parameter values and angles.Our samples are approximately 500 pixels x 500 pixels. Wesample 200 evenly spaced angles in [0, 2π]. For each anglesample we approximate the line integral through the angle at130 different values of t. This gives about 26,000 sample datapoints.Use the Hilbert transform back-projection formula:

f =1

4πR#H d

dt(Rf ).

Use differencing to approximate ddtRf .

Possible to approximate H ddt (Rf ) as a convolution

Integrate approximately to back-project.






Reconstruction of “bump” phantom






Reconstruction of Homer






To-Do

Prove the range conditions.

More sophisticated description and analysis for Hilberttransform.

Incomplete data problem: “hole” theorem

Edge detection viewpoint.

Relation to microlocal analysis





References

Charles L. Epstein.Introduction to the mathematics of medical imaging.Society for Industrial and Applied Mathematics (SIAM),Philadelphia, PA, second edition, 2008.

Peter Kuchment.The radon and x-ray transforms.Unpublished lecture notes from IPDE-RTG summer school atthe University of Washington, 2010.

F. Natterer.The mathematics of computerized tomography.B. G. Teubner, Stuttgart, 1986.





Proof of Continuity Estimate

Fix ω and make the orthogonal change of variables(x , y)T = [ω, ω⊥](t, s)T to see∫ ∞−∞|Rf (t, ω)| dt =

∫R2

|f (tω + sω⊥)| dt =

∫R2

|f (x , y)| dxdy .

Back





Proof of the Projection-Slice Theorem

Proof. By definition∫ ∞−∞Rf (t, ω)e−itr dt =

∫R2

f (tω + sω⊥)e−itr dsdt.

Change variables: x = tω + sω⊥ and use the fact thatt =< x, ω > to obtain∫

R2

f (tω + sω⊥)e itr dsdt =

∫R2

f (x)e−i<x,ω>r dx = f (rω).

Back


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