Geometry in CT Reconstruction - Welcome to Computer...

Geometry in CT Reconstruction

Copyright 2011 Benjamin Kimia

March 23, 2011

Contents

1 Introduction to CT Reconstruction 31.1 Linear Attenuation Coefficient . . . . . . . . . . . . . . . . . . 31.2 Linear Tomography . . . . . . . . . . . . . . . . . . . . . . . . 51.3 Radon Transform . . . . . . . . . . . . . . . . . . . . . . . . . 61.4 Understanding Back Projection . . . . . . . . . . . . . . . . . 81.5 The Fourier Approach to CT Reconstruction . . . . . . . . . . 121.6 Filtered Backprojection . . . . . . . . . . . . . . . . . . . . . . 121.7 Watch out for those singularities: A review of Calculus . . . . 171.8 The Spatial Domain Approach to CT Reconstruction . . . . . 181.9 Inverse Radon Transform . . . . . . . . . . . . . . . . . . . . . 20

2 Fan Beam Geometry 212.1 Radon Transform and Fan Beam Geometry . . . . . . . . . . . 242.2 Radon Inversion Formula for Fan Beam Geometry . . . . . . . 262.3 Convolution in Fan Beam Geometry . . . . . . . . . . . . . . . 30

3 Flirting with Filtering 323.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323.2 Filtering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323.3 Matched Filtering . . . . . . . . . . . . . . . . . . . . . . . . . 363.4 Detecting Spots . . . . . . . . . . . . . . . . . . . . . . . . . . 37

3.4.1 Detecting A Gaussian Spot . . . . . . . . . . . . . . . 373.4.2 Detecting A Circular Spot . . . . . . . . . . . . . . . . 38

4 Taylor Expansion in the View Space 41

5 Geometric Tomography 415.1 Fan Beam Geometry . . . . . . . . . . . . . . . . . . . . . . . 46

1

6 Cone Beam Tomography and Volumetric CT 486.1 Notation and Definition . . . . . . . . . . . . . . . . . . . . . 506.2 Relating Cone Beam and Radon Transform . . . . . . . . . . . 546.3 Reconstruction Algorithms for Cone Beam Helical CT . . . . . 576.4 History of Helical Cone Beam CT Development . . . . . . . . 636.5 Material to be worked into this section . . . . . . . . . . . . . 646.6 Numerical algorithms for helical cone-beam reconstruction . . 65

7 Radon’s Proof 66

8 To Do 67

A Integrals Involving Gaussians and Derivatives of Gaussians 71A.1 Integrals involving a Gaussian and its derivatives . . . . . . . 71A.2 Fourier Transform of the Gaussian . . . . . . . . . . . . . . . . 72A.3 Integrals of the Gabor Filter . . . . . . . . . . . . . . . . . . . 73A.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

B Integrals Involving Characteristic Function of Circular Do-mains 77B.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77B.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

C Some Standard Fourier Formulas 79

D A Numerically stable method to find an extremum 80

E Solving A sin θ +B cos θ = C 81

F Π-line membership 83F.1 References not included in the paper yet . . . . . . . . . . . . 88

2

1 Introduction to CT Reconstruction

The key to CT reconstruction is a characterization of how X-rays are ab-sorbed by material. Since X-rays are electromagnetic waves, we considerhow the latter are absorbed as a function of the properties of the materialthrough which the light is traveling, as studied in the field of optics. Thisabsorption is governed by the Beer Lambert law, also alternatively knownas the Beer’s law or the Lambert Beer law or the Beer-Lambert-Bouguer,Figure xx. In short, this law states that the intensity of light passed througha slab of length l drops exponentially with l, i.e.,

I(l) = I0 e−αl, (1)

where I0 is the original intensity of the light, etc...

Figure 1: An illustration of the BeerLambert law in optics

1.1 Linear Attenuation Coefficient

Consider an X-ray beam going through some material. The probability thata photon gets removed from a beam (either absorbed or scattered) dependson (i) the energy of the photon, and (ii) the nature of the material throughwhich the beam passes. Define the linear attenuation coefficient µte of tissuet at energy e as

µte = − ln(pe), (2)

where pe is the probability that a photon of energy e transmits through auniform slab of unit length of tissue t in a direction perpendicular to the faceof the slab. For example, the linear attenuation coefficient of water at 73 kevis 0.19 cm−1.

3

A general model of X-ray absorption is based on an independence as-sumption: The extent of X-ray absorption by the material by any portionof space is independent of that corresponding to the complementary space.This independence is essentially what gives rise to the exponential decay inX-ray intensity. The following discussion is adapted from [11]. Specifically,consider a slab of non-uniform material and denote the linear attenuation co-efficient as a function of the coordinate x along the space as µe(x). Let pe(x)be the probability of absorption in a slab of width dx at x, i.e., the proba-bility that a photon of energy e is transmitted as far as x. Note that pe(x)is a monotonically decreasing function of x. Now, consider an infinitesimalsection dx of the slab from x to x+ dx. Let qe(x, dx) denote the probabilitythat a photon of energy e which has reached x will not be transmitted be-yond x+ dx. It is clear, therefore, that the probability of a photon reachingx+ dx is the probability of it reaching x and being transmitted through thedx slab:

pe(x+ dx) = pe(x)(1− qe(x, dx)),

or

qe(x, dx) = −pe(x+ dx)− pe(x)

pe(x).

Now let

µe(x) = limdx→0

qe(x, dx)

dx= −p

′e(x)

pe(x)= −d(ln(pe(x))

dx.

Observe that∫ A

0

µe(x)dx = −∫ A

0

−d(ln(pe(x))

dxdx = − ln(pe(x))|A0 = − ln(pe(A))+ln(1) = − ln(pe(A)).

Now, for a uniform slab of unit length, i.e. where µe(x) = µe and A = 1, wehave

µe =

∫ 1

0

µe(x)dx = − ln(pe(1)).

Comparing this to Equation(2), it is clear that µe = µte. Since qe as aprobability is dimensionless, µe has dimension of length−1 and so does µte.

In actual practice, since calibration is required to make measurements uni-form across detection, the actual linear attenuation coefficient µte is measuredrelative to µce, the linear attenuation coefficient of the calibration material.Define the relative linear attenuation coefficient as

µt,ae = µte − µae .

4

Note that line integrals of these quantities are simply related:∫ A

0

µt,ae (x)dx =

∫ A

0

(µte(x)− µae(x))dx

=

∫ A

0

µte(x)dx−∫ A

0

µae(x)dx

= − ln(pte(A)) + ln(pae(A))

= − ln(pte(A)

pae(A)).

The CT number is defines as

H = 1000µte − µwatere

µwatere

, (3)

which is scaled so that the CT number of water is 0 and the CT number ofair is -1000. CT numbers are expressed in Hounsfield units1

1.2 Linear Tomography

Linear Tompography is a technique used to emphasize certain depths in pa-tients before computers could be used to do this. Basically, with the patienton the table, the X-ray source and film were moved in opposite directions,each at a fixed rate. Transaxial Tomography essentially accomplishes thesame idea by rotating both the patient and the film at the same speed.

The basic idea is that the images of two points A and B (both at depthh with respect to the source), A and B, have a distance |A B| = H

h|AB|,

1[From Wiki] Sir Godfrey Newbold Hounsfield CBE, FRS, (28 August 1919 12 August2004) was an English electrical engineer who shared the 1979 Nobel Prize for Physiology orMedicine with Allan McLeod Cormack for his part in developing the diagnostic techniqueof X-ray computed tomography (CT).

5

(a) (b)

Figure 2: Linear Tomography was used to focus on a plane at a certain depth bymoving both the source and the film in a predetermined fashion

regardless of the position of S, as long as the depth h is fixed and correspon-dance is maintained. Figure(2) shows the process of emphasizing the planeat depth h from the source. Thus, the film and the source can be movedsimultaneously so that the image of A on the film in the new position of thefilm, A, corresponds exactly to A. Since |A B| = |A B| = H

h|AB|, all points

at the same depth align with their corresponding points. For points at otherdepths, however, there is no correspondence and each point C contributesto points in an extended range. Thus, the effect of these points is to addmore blurring to the film, but at the same time isolating the contribution ofpoints at depth h since they add up in the same spot on the film. Linearand transaxial tomography motivate the idea of backprojection as a way offinding the linear attenuation coefficient at each point. We first develop someformalities.

1.3 Radon Transform

Let µ(x, y) represent the distribution of the X-ray absorption/attenuation co-efficients in Hounsfield units. This is our unknown. We will assume through-out this document that µ is continuous and bounded, and that it vanishesoutside the region of interest (scanning area). The Radon Transform, Rµ isa map from µ(x, y) to a 1-D function ρθ(ξ) where θ is the angle along linesof X-ray projection and ξ is the variable orthogonal to these lines (Figure 3),i.e.,

ρθ(ξ) =

∫ ∞−∞

µ(x(ξ, η), y(ξ, η))dη

=

∫ ∞−∞

µ(ξ cos θ − η sin θ, ξ sin θ + η cos θ)dη.

(4)

6

We also often use Rµ(ξ, θ) or ρ(ξ, θ) to denote the Radon Transform over arange of angles.

(a) (b)

Figure 3: (a) The Radon Transform is the line integral of µ, where each linerepresents the X-ray that goes from source to sensor. The line can be expressedas ξ = ξ0 in a rotated coordinate system in (b).

Ideally, these measurements need to be made only for θ ∈ [0, π) since theentire [0, 2π) range can be covered using:

ρθ−π(ξ) = ρθ(−ξ). (5)

Note that:[xy

]=

[cos θ − sin θsin θ cos θ

] [ξη

]and

[ξη

]=

[cos θ sin θ− sin θ cos θ

] [xy

]. (6)

The set of measurements, ρθ(ξ) where θ ∈ [0, π) and ξ ∈ (−∞,∞) representsthe view data or the sinogram in the 2-D space spanned by ξ and θ,

ρ(ξ, θ) = ρθ(ξ). (7)

7

Our goal is to recover µ(x, y) from ρ(ξ, θ).

Remark: The Radon Transform is integration along a line. We can specifya line by its distance ξ from the origin and the angle it makes with the x-axisθ. A point (x, y) is on the line if and only if (x, y).(cos θ, sin θ) = ξ. Thus ina 2D plane δ(x cos θ+ y sin θ− ξ) is non-zero only when the point (x, y) is onthis line. Thus, we see the Radon Transform.

Alternatively, in polar notation, a point (r, φ) is on the line if and only if(r cosφ, r sinφ)(cos θ, sin θ) = ξ which reduces to r cos(φ−θ) = ξ. This givesan expression for the Radon Transform in polar notation

ρ(ξ, θ) =

∫ π

0

∫ ∞−∞

µ(r, φ)δ(r cos(φ− θ)− ξ)rdrdφ

also written as

ρ(ξ, θ) =

∫ ∞−∞

∫ ∞−∞

µ(x, y)δ(x cos θ + y sin θ − ξ)dxdy.

1.4 Understanding Back Projection

Figure 4: Back projection sums over measurements along rays passing through afixed point as an estimate of µ there.

8

The idea of backprojection goes back to Linear and Transaxial Tomography.Consider the same idea with a single source and a single sensor rotatingaround the object around some fixed point ρ(x0, y0), Figure(4). The idea isto backproject each view data to all points giving rise to it in an accumulativefashion,i.e., the estimated linear attenuation coefficient at each point in thesum of intensities measured for all rays going through the point. The sensorreading for each ray through ρ, indexed by θ, is obtained from Equation (4)

ρ(ξ0, θ) =

∫ ∞−∞

µ(ξ0 cos θ − η sin θ, ξ0 sin θ + η cos θ)dη, (8)

where ξ0 = x0 cos θ + y0 sin θ. Each ray passes through ρ, but any otherpoint participates in the view data in only one ray. Thus, in the cumulativeresponse, the contribution of all points fades in comparison to that of ρ. Theprocess of accumulating all ray measurements through ρ and using it as anestimate of µ(x0, y0) at ρ is called backprojection. Formally,

µBP (x0, y0) =1

π

∫ π

0

ρ(ξ0, θ)dθ, where ξ0 = x0 cos θ + y0 sin θ. (9)

How good an estimate of µ is µBP ? The following theorem relates thebackprojection estimate of µ to µ.

Theorem 1.

µBP (x, y) = µ(x, y) ∗ 1

π

1√x2 + y2

. (10)

Proof. We can relate the two by simplifying Equation (9) using Equation(8):

µBP (x0, y0) =1

π

∫ π

0

[∫ ∞−∞

µ(ξ0(θ) cos θ − η sin θ, ξ0(θ) sin θ + η cos θ) dη

]dθ,

(11)where ξ0(θ) = x0 cos θ+y0 sin θ. We can gain a better insight into this integralby switching to a polar coordinate system centered around (x0, y0),

x = x0 + r cosφ

y = y0 + r sinφ.(12)

Now, given (θ, η) in integral (11), these identify (x, y)x = ξ0(θ) cos θ − η sin θ

y = ξ0(θ) sin θ + η cos θ.(13)

9

Equating (12) and (13) relates (r,φ) to (η, θ):r cosφ = ξ0(θ) cos θ − η sin θ − x0 = (ξ0(θ) cos θ − η sin θ)− (ξ0(θ) cos θ − η0(θ) sin θ)

r sinφ = ξ0(θ) sin θ + η cos θ − y0 = (ξ0(θ) sin θ + η cos θ)− (ξ0(θ) sin θ + η0(θ) cos θ)

or, r cosφ = −(η − η0(θ)) sin θ

r sinφ = (η − η0(θ)) cos θ

where ξ0(θ) = x0 cos θ + y0 sin θ

η0(θ) = −x0 sin θ + y0 cos θ.

When η < η0(θ), we have r = −(η − η0(θ))

φ = θ − π

2

or, solving for (η, θ) η = −r − x0 cosφ− y0 sinφ

θ = φ+ π/2.

When η > η0(θ), we have r = η − η0(θ)

φ = θ +π

2

or, solving for (η, θ) η = r + x0 cosφ+ y0 sinφ

θ = φ− π/2.

In rewriting (9) in terms of r and φ, we will also need an expression for ξ0(θ).When η < η0(θ)

ξ0(θ) = x0 cos(φ+π

2) + y0 sin(φ+

π

2)

= −x0 sinφ+ y0 cosφ.

10

and when η > η0(θ)

ξ0(θ) = x0 cos(φ− π

2) + y0 sin(φ− π

2)

= x0 sinφ− y0 cosφ.

We can now rewrite µBP in (9) as

µBP (x0, y0) =1

π

∫ π

0

∫ η0(θ)

−∞µ(ξ0(θ) cos θ − η sin θ, ξ0(θ) sin θ + η cos θ) dηdθ

+1

π

∫ π

0

∫ ∞η0(θ)

µ(ξ0(θ) cos θ − η sin θ, ξ0(θ) sin θ + η cos θ) dηdθ

=1

π

∫ π2

−π2

∫ 0

−∞µ((−x0 sinφ+ y0 cosφ)(− sinφ)− (−r − x0 cosφ− y0 sinφ)(cosφ),

(−x0 sinφ+ y0 cosφ)(cosφ) + (−r − x0 cosφ− y0 sinφ)(− sinφ)) drdφ

+1

π

∫ 3π2

π2

∫ ∞0

µ((x0 sinφ− y0 cosφ)(sinφ)− (r + x0 cosφ+ y0 sinφ)(− cosφ),

(x0 sinφ− y0 cosφ)(− cosφ) + (r + x0 cosφ+ y0 sinφ)(sinφ)) drdφ

=1

π

∫ π2

−π2

∫ ∞0

µ((x0 sinφ− y0 cosφ) sinφ+ (x0 cosφ+ y0 sinφ) cosφ+ r cosφ,

(−x0 sinφ+ y0 cosφ) cosφ+ (x0 cosφ+ y0 sinφ) sinφ+ r sinφ) drdφ

+1

π

∫ 3π2

π2

∫ ∞0

µ((x0 sinφ− y0 cosφ) sinφ+ (x0 cosφ+ y0 sinφ) cosφ+ r cosφ,

(−x0 sinφ+ y0 cosφ) cosφ+ (x0 cosφ+ y0 sinφ) sinφ+ r sinφ) drdφ

=1

π

∫ 2π

0

∫ ∞0

µ(x0 + r cosφ, y0 + r sinφ)drdφ.

We are now in a position to interpret this as an area integral, i.e.

µBP (x0, y0) =1

π

∫ 2π

0

∫ ∞0

[µ(x0 + r cosφ, y0 + r sinφ)

r

]rdrdφ

=1

π

∫ ∞−∞

∫ ∞−∞

µ(x, y)√(x− x0)2 + (y − y0)2

dxdy

= µ(x, y) ∗ 1

π

1√x2 + y2

x=x0,y=y0

,

which is the 1r

filter well-known from the Fourier approach to the CT recon-struction. In summary, the back projection process gives a blurred estimateof µ(x, y) with a 1

rblurring filter.

11

1.5 The Fourier Approach to CT Reconstruction

In this approach, the CT-reconstruction of µ(x, y) relies on its Fourier Trans-form,

µ(ωx, ωy) = Fµ(x, y) =

∫ ∞−∞

∫ ∞−∞

µ(x, y)e−iωxxe−iωyydxdy.

Consider a polar representation of µ in the Fourier domain, i.e., µ(ω, θ),where ωx = ω cos θ and ωy = ω sin θ. We have

µ(ω, θ) =

∫ ∞−∞

∫ ∞−∞

µ(x, y)e−i(ω cos θ)xe−i(ω sin θ)ydxdy

=

∫ ∞−∞

∫ ∞−∞

µ(x, y)e−i(x cos θ+y sin θ)ωdxdy

=

∫ ∞−∞

∫ ∞−∞

µ(ξ cos θ − η sin θ, ξ sin θ + η cos θ)e−iξωdξdη.

The latter integral can be viewed in the form of a 1-D Fourier Transformwhere θ is fixed, but ω is varying:

µ(ω, θ) =

∫ ∞−∞

[∫ ∞−∞

µ(ξ cos θ − η sin θ, ξ sin θ + η cos θ)dη

]e−iξωdξ

=

∫ ∞−∞

ρ(ξ, θ)e−iξωdξ

= Fξρ(ξ, θ)(ω) = ρ(ω, θ),

where ρ(ω, θ) is the one dimensional Fourier Transform of ρ(ω, θ) for fixedθ along the variable ξ. In other words, when keeping θ fixed in the Fourierdomain, this slice of the Fourier Transform of µ(ω, θ) corresponds exactly tothe 1-D Fourier Transform of the Radon Transform for that θ, i.e., ρθ(ξ).This is the central slice theorem. Alternatively, replacing each column ofthe sinogram by its Fourier Transform gives the Fourier Transform of µ inpolar coordinates.

Need some figures

1.6 Filtered Backprojection

The central slice theorem allows us to recover µ(x, y) from the measurementsusing the inverse Fourier Transform and a change of coordinates from the

12

Cartesian coordinates (ωx, ωy) to the polar coordinates ω, θ.

µ(x, y) =1

4π2

∫ ∞−∞

∫ ∞−∞

µ(ωx, ωy)eiωxxeiωyydωxdωy

=1

4π2

∫ π

0

∫ ∞−∞

µ(ω, θ)eiω(cos θ)xeiω(sin θ)y|ω|dωdθ

=1

2π

∫ π

0

[1

2π

∫ ∞−∞

µ(ω, θ) |ω| eiω(x cos θ+y sin θ)dω

]dθ. (14)

The inner integral in the square brackets is the one-dimensional inverseFourier Transform of µ(ω, θ)|ω| along the variable ω with fixed θ. Define

ρ∗(ξ, θ) = F−1[µ(ω, θ)|ω|] =1

2π

∫ ∞−∞

µ(ω, θ) |ω| eiω(x cos θ+y sin θ)dω. (15)

Then

µ(x, y) =1

2π

∫ π

0

ρ∗θ(x cos θ + y sin θ)dθ. (16)

Equation (16) provided the recipe for recovery of µ:

1)Filtering step: From ρ(ξ, θ), find ρ∗(ξ, θ) = F−1ω Fξρ(ξ, θ)|ω|,

2)Backprojection: Sum over ρ∗(ξ, θ) over all θ.The first step is called the filtering step because it seems to be inter-

pretable as a convolution: Observing that a product in the Fourier domainis a convolution in the spatial domain it is tempting to write

ρ∗(ξ, θ) = ρ(ξ, θ) ∗ F−1|ω|. (17)

However, the function |ω| has infinite energy and such a split into two func-tions is not valid. A different split is possible:

F−1ω Fξρ(ξ, θ)|ω| = F−1

ω Fξρ(ξ, θ)H(w)

.H(w)|ω| (18)

= F−1ω Fξρ(ξ, θ)H(w)

∗ F−1ω H(w)|ω|, (19)

where H(ω) is expected to decrease sufficiently fast to compensate for theincrease by |ω| so that F−1

ω H(w)|ω| exists. If ρ(ξ, θ) is band-limited sothat frequencies beyond Ω are not present,

Fξρ(ξ, θ) = 0 when ω > Ω, (20)

then we can design H(ω) to be 1 up to Ω, and then drop monotonically tozero.

13

H(ω) = 1, |ω| ≤ ΩH(ω) 6= 0, otherwise.

(21)

In this case we have we have

F−1ω Fξρ(ξ, θ)H(w)

= F−1ω Fξρ(ξ, θ)

1 = ρ(ξ, θ), (22)

andρ∗(ξ, θ) = ρ(ξ, θ) ∗ F−1

ω H(w)|ω|. (23)

The filter H(ω) can be designed a number of ways. First, the filter definedby Ramachandran and Lakshiminarayanan [23], known now as the Ram-Lak filter, is simply obtained by limiting the frequency domain response,Figure(5).

H(ω) =

1, |ω ≤ Ω0, otherwise.

(24)

so that the overall filter is HRL(ω) = HRL(ω)|ω|, or,

HRL(ω) =

|ω|, |ω| ≤ Ω0, otherwise.

(25)

Using Appendix(C), Formula 3, the spatial domain filter is

hRL(x) =Ω2

π[sinc(Ωx)− 1

2sinc2(

Ωx

2)], (26)

See Figure(5).Second, the Shepp-Logan filter [25] has a similar design that it also limits

the frequency upper bound, but in contrast to the Ram-Lak filter, it does sogradually using a sinc function 2Ω

πsinc(πω

2Ω), Figure(5), with the overall filter

defined as

HSL(ω) =

∣∣sin (πω2Ω

)∣∣ , |ω| < Ω0, otherwise

(27)

14

(a) (b)

Figure 5: (a)Examples of the band-limited filter function of sampled data. Notethe cyclic repetitiveness of the digital filter. (b) Spatial domain filter kernels cor-responding to the filter functions shown in the Ram-Lak filter is a high-pass filterwith a sharp response but results in some noise enhancement, while the Shepp-Logan and the Hamming window filters are noise-smoothed filters and thereforehave better SNR. (Images taken from [3])

The spatial filter corresponding to the Shepp-Logan filter is

hSL(ω) =1

2π

∫ ∞−∞

HSL(ω)eiωxdω

=1

2π

∫ Ω

−Ω

∣∣∣sin(πω2Ω

)∣∣∣ eiωxdω=

1

2π

∫ Ω

0

2 sin(πω

2Ω

)cosωxdω

=1

2π

∫ Ω

0

sin(( π

2Ω+ x)ω)dω +

1

2π

∫ Ω

0

sin(( π

2Ω− x)ω)dω

=−1

2π

cos((

π2Ω

+ x)ω)(

π2Ω

+ x)

Ω

0

− 1

2π

cos((

π2Ω− x)ω)(

π2Ω− x)

Ω

0

=1

2π

1− cos(π2

+ Ωx)

π2Ω

+ x+

1

2π

1− cos(π2− Ωx

)π

2Ω− x

=1

2π

1 + sin(Ωx)π

2Ω+ x

+1

2π

1− sin(Ωx)π

2Ω− x

=1

2π

1(π

2Ω

)2 − x2

[πΩ− 2x sin(Ωx)

]=

1

π

π2Ω− x sin(Ωx)(π

2Ω

)2 − x2.

15

The Ram-Lak filter has the following formula in the spatial domain whensampled at Nyquist rate which is two times the bandwidth of a bandlimitedsignal so that . ∆x = π

Ω:

H(k) =

π4, if k=0− 1πk2 , if k odd

0, if even

while the Shepp-Logan filter is given by:

H(k) =−8Ω2

π2(4k2 − 1), k = 0,±1,±2, . . .

in the spatial domain.Explain the Hamming Window approach proposed originally by Richard

Hamming. using the raised cosine:

w[n] = 0.54− 0.46 cos(2πn

N − 1). (28)

In a different approach, Rosenfeld and Kak [24] use H(ω) = e−ε|ω| so that theoverall filter is Hε(ω) = |ω|e−ε|ω|. In the spatial domain this filter is written

as hε(x) = ε2−(2πx)2

[ε2+(2πx)2]2. This filter is plotted in Figure(6) for several values of

ε.It is not clear why a filter that balances being 1 in the range up to Omega

and smaller after that has not been used:

HKim =|ω|

1 + (ωΩ

)6. (29)

Note: Add a formal definition of backprojection as “Adjoint Radon Trans-form.”

16

Figure 6: The approximate reconstruction filter hε(x) from [24] plotted for ε = 5(blue), ε = 4 (red) and ε = 3 (green).

1.7 Watch out for those singularities: A review of Cal-culus

What do you do when the integrand has a singularity? Consider the integral

I =

∫ a

−a

1√xdx, (30)

where the integrand blows up at x = 0; what happens to the value of theintegral? A substitution y =

√x gives

I =1

2

∫ √a−√a

dy =√a, (31)

a finite value, even when a→ 0. Now consider the integral

II =

∫ a

−a

1

x2dx. (32)

17

This can be integrated directly, giving

II = − 1

x

a−a

= −2

a, (33)

which goes to infinity as a→ 0. In such cases, the evaluation above does nothold because of the singularity. Instead, we must use the Cauchy PrincipalValue which states that

PV

∫ b

a

f(x) dx = limε→∞

[

∫ c−ε

a

f(x) dx+

∫ c+ε

a

f(x) dx]. (34)

The Hilbert Transform pair are defined as

g(y) = PV1

π

∫ ∞−∞

f(x)

x− ydx, (35)

f(x) = PV1

π

∫ ∞−∞

g(y)

y − xdy. (36)

1.8 The Spatial Domain Approach to CT Reconstruc-tion

Radon[22] showed that

µ(x, y) =1

4π2

∫ 2π

0

∫ ∞−∞

∂ρ∂ξ

x cos θ + y sin θ − ξdξdθ (37)

The following is a summary of his elegant argument. [to summarize]Let ξ = x cos θ+y sin θ be the ξ-coordinate of µ(x, y). Then, x cos θ+y sin θ−ξ = ξ − ξ indicates the distance of the point M(x, y) from the X-ray beamidentified by (ξ, θ). Observe that one can view the inner integral as theconvolution of ρξ with the filter 1

ξ.

ρ(ξ) = (ρξ ∗1

ξ)(ξ)

=

∫ ∞−∞

ρξ(ξ)1

ξ − ξdξ, (38)

so that

µ(x, y) =1

4π2

∫ 2π

0

ρ(ξ)dθ.

18

When the points are represented in polar representation M(r, φ), i.e. (x, y) =(r cosφ, r sinφ), we have

x cos θ + y sin θ − ξ = r cosφ cos θ + r sinφ sin θ

= r cos(θ − φ)− ξ.

Thus, the Radon Inversion formula in polar form is

µ(r, φ) =1

4π2

∫ 2π

0

∫ ∞−∞

∂ρ∂ξ

(ξ, θ)

r cos(θ − φ)− ξdξdθ. (39)

Alternatively, one can reparametrize ξ = r cos(θ−φ)−ξ in the inner integralto get

µ(r, φ) =1

4π2

∫ 2π

0

∫ ∞−∞

∂ρ∂ξ

(r cos(θ − φ) + ξ, θ)

ξdξdθ. (40)

Unfortunately, the denominator for each θ will vanish at ξ = 0 for Equation(40)and ξ = x cos θ + y sin θ for Equation(37). Ben to expand on this

Remark: As Horn [13] noted the Radon’s inversion formula containing ρξdoes not strictly apply when ρξ is not continuous. Instead, one rewrites theinner integral by integration by parts as:∫ ∞−∞

ρξ

ξ − ξdξ = lim

ε→0

[∫ ξ−ε

−∞

ρξ

ξ − ξdξ +

∫ ∞ξ+ε

ρξ

ξ − ξdξ

]

= limε→0

[ρ

ξ − ξ

ξ−ε−∞−∫ ξ−ε

−∞

ρ

(ξ − ξ)2dξ +

ρ

ξ − ξ

∞ξ+ε

−∫ ξ+ε

−∞

ρ

(ξ − ξ)2dξ

]

= limε→0

[ρ(ξ − ε, θ)

ε+ρ(ξ + ε, θ)

ε−∫ ξ−ε

−∞

ρ

(ξ − ξ)2dξ −

∫ ∞ξ+ε

ρ

(ξ − ξ)2dξ

]

= limε→0

[∫ ξ+ε

ξ−ε

ρ(ξ, θ)

ε2dξ −

∫ ξ−ε

−∞

ρ

(ξ − ξ)2dξ −

∫ ∞ξ+ε

ρ

(ξ − ξ)2dξ

]

=

∫ ∞−∞

ρ(ξ, θ)Fε(ξ − ξ)dξ

= ρ ∗ξ Fε, (41)

19

where

Fε(ξ) =

−1

ξ2|ξ| ≥ ε

1

ε2|ξ| < ε,

(42)

as shown in Figure(7).

Figure 7: Plot of Fε(ξ) for ε = (0.5, 0.2, 0.1).

Then,

µ(x, y) = limε→0

1

2π2

∫ π

0

∫ ∞−∞

ρ(ξ, θ)Fε(x cos θ + y sin θ − ξ)dθ

= limε→0

1

2π2

∫ π

0

[ρ(ξ, θ) ∗ξ Fε](x cos θ + y sin θ − ξ)dθ (43)

Figure(8) shows a reconstruction using this method.

1.9 Inverse Radon Transform

Ben to write this using a summary of Radon’s 1917 paper andHorn’s papers

20

Figure 8: (a) The retinal image (b,c,d) Reconstruction of the image for ε =(0.5, 0.2, 0.1). The resulting images were scaled such that [minimum value, maxi-mum value] maps to [0 1].

2 Fan Beam Geometry

In the fan beam geometry, all rays emanate from a single source S located atpolar coordinates (D0, β+ π

2), where D0 = |OS| and β is the angle of a virtual

detector array passing through O. The actual detector array is some distanceD1 away from origin, but the analysis is simplified if the measurements arescaled down to pass through the origin O, Figure(9). An arbitrary point Mis described in polar coordinates (r, φ). The source detector assembly S isassumed to rotate around the origin so that β ∈ [0, 2π]. Thus, the sinogramcollected in this geometry is a 2D set of line integral measurements: for eachβ the line integrals for all A, as specified by ξ, are reported. Let g(ξ, β)denote the line integrals. Note that |ξ| < D0. The sinogram specified byg(ξ, β) is natural when the samples are along a line perpendicular to OS.However, when detectors are arranged in a circular arc centered at S, theangle α = ∠OSA is a better description of the detectors, and the sinogram

21

Figure 9: The fan beam geometry is illustrated for a point source S and a virtuallinear detection array aligned with the ξ-axis. The actual detector array is aconstant distance array.

in the form of g(α, β) is used. The two representations are interchangeable:

g(α, β) = g(D0 tanα, β)

g(ξ, β) = g

(tan−1

(ξ

D0

), β

).

(44)

Proposition 1 (Rebinning from fan beam geometry to parallel beam geom-etry).

ρ(ξ, θ) = g(ξ, β), (45)

where ξ =

D0ξ√D2

0 − ξ2

β = θ − sin−1

(ξ

D0

) and

ξ =

D0ξ√D2

0 + ξ2

θ = β + tan−1

(ξ

D0

).

(46)

22

Figure 10: This figure illustrates the relationship between a parallel beam raythrough a detector array at angle θ and distance ξ from origin to a fan beamthrough S with detector at angle β.

Proof. We have three relationships from the geometry in Figure(10)cosα =

ξ

ξ

sinα =ξ

D0

θ = β + α.

By squaring the first two equations we have

ξ2

ξ2+ξ2

D20

= 1,

which gives ξ in terms of ξ and vice versa. Observing that tanα = ξD0

givesthe remaining equations.

23

Corollary 1. We haveα(ξ, θ) = sin−1

(ξ

D0

)

β(ξ, θ) = θ − sin−1

(ξ

D0

) and

ξ(α, β) = D0 sinα

θ(α, β) = β + α.

Thus, ρ(ξ, θ) = g

(sin−1

(ξ

D0

), θ − sin−1

(ξ

D0

))g(α, β) = ρ(D0 sinα, β + α).

Proof. We have from Proposition(45) that

ρ(ξ, θ) = g(ξ, β)

= g

(tan−1

(ξ

D0

), β

)

= g

tan−1

ξ√D2

0 − ξ2

, θ − sin−1

(ξ

D0

)= g

(sin−1

(ξ

D0

), θ − sin−1

(ξ

D0

)).

Conversely, by Proposition(1)


= ρ

(D0(D0 tanα)√D2

0 + (D0 tanα)2, β + α

)= ρ(D0 sinα, β + α).

2.1 Radon Transform and Fan Beam Geometry

Deriving an expression to describe fan beam measurements requires a care-ful examination. An X-ray measurment in the parallel beam geometry,

24

Figure(11a) indicates the attenuation of X-ray intensity through a narrowbeam with ∆ξ, i.e. ∫ ξ0+ ∆ξ

2

ξ0−∆ξ2

∫ ∞−∞

µ(x(ξ, η), y(ξ, η))dηdξ

which in the limit as ∆ξ → 0 is asymptotically

∆ξ

∫ ∞−∞

µ(x(ξ0, η), y(ξ0, η))dη, (47)

with the latter part corresponding to the idealized Radon Transform. How-ever, an X-ray measurement in the fan beam geometry corresponds fans outso one might infer the measurement is∫ α0+ ∆α

2

α0−∆α2

∫ ∞0

µ(x(η, α), y(η, α))ηdηdα,

where η is the distance from S to the point P along the ray SA. Rather, thebeam density also decreases along the front by the same η factor so that theexpression is asymptotically equivalent to

∆α

∫ ∞0

µ(x(η, α), y(η, α))dη, (48)

giving again the Radon line integral. What this analysis shows, however,is that numerically the fan beam focuses on smaller regions initially and onlarger regions as the ray propogates. When the image is not varying rapidly,this does not present a problem. However, when the variation approachesvoxel size, numerical approximation may not be faithful to the true values.

The expression of the Radon Transform in the fan beam geometry can nowbe easily derived by noting a point along the line SA is specified by

P = ξ(cos β, sin β) + η(cos(β + α + π/2), sin(β + α + π/2))

= (ξ cos β − η sin(β + α), ξ sin β + η cos(β + α)),

where η is length parameter along SA. It is easy to see that η = ηcosα

. Thus,

g(ξ, β) =

∫ ∞−∞

µ(ξ cos β − η

cosαsin(β + α), ξ sin β +

η

cosαcos(β + α))

dη

cosα

=1

cosα

∫ ∞−∞

µ(ξ cos β − η sin β − η cos β tanα, ξ sinα + η cos β − η sin β tanα)dη.

25

Similarly,


=1

cosα

∫ ∞−∞

µ((D0 − η) cos β tanα− η sin β, (D0 − η) sin β tanα + η cos β)dη.

(a) (b)

Figure 11: The relationship between x-ray attenuation measurements and theRadon Transform. (a) Parallel beam (b) Fan beam.

2.2 Radon Inversion Formula for Fan Beam Geometry

Proposition 2.

µ(r, φ) =1

4π2

∫ 2π

0

1

D

[∫ π2

−π2

gα − gβsin(α− α)

dα

]dβ, (49)

where D =√D2

0 + r2 + 2rD0 sin(β − φ) is the distance from M(r, φ) to the

source S(D0, β + π2) and α = tan−1

[r cos(β−φ)

D0+r sin(β−φ)

]is the angle ∠OSM .

Proof. Let M(r, φ) be an arbitrary point, Figure(9). Let D denote its dis-tance to source D = |SM |, so that we have via the triangle inequality in thetriangle SOM

D2 = D20 + r2 − 2rD0 cos

(β +

π

2− φ)

= D20 + r2 + 2rD0 sin(β − φ).

26

Also, let α be the angle of the ray through M , i.e., α = ∠( ~SO, ~SM). Then,

tan(α) =MF

SF=

r sin(β + π

2− φ)

D0 − r cos(β + π

2− φ) =

r cos(β − φ)

D0 + r sin(β − φ). (50)

Note that this gives the polar representation of M in the coordinates of sourceS with respect to the direction ~SO.

Radon’s inversion formula in polar notation was derived in Equation(39).This can be represented in the fan beam geometry as follows. First, we canobtain an expression for ρξ using Corollary(1)

ρ(ξ, θ) = g(α(ξ, θ), β(ξ, θ)),

where α(ξ, θ) = sin−1( ξD0

) and β(ξ, θ) = θ − sin−1( ξD0

). Then

∂ρ

∂ξ(ξ, θ) =

∂g

∂α(α(ξ, θ), β(ξ, θ))

∂α

∂ξ(ξ, θ) +

∂g

∂β(α(ξ, θ), β(ξ, θ))

∂β

∂ξ(ξ, θ). (51)

Since sinα(ξ, θ) = ξD0

, we can differentiate to get cosα.αξ = 1D0

leading to

∂α

∂ξ(ξ, θ) =

1

D0 cosα(ξ, θ)

∂β

∂ξ(ξ, θ) =

−1

D0 cosα(ξ, θ).

Thus,

∂ρ

∂ξ(ξ, θ) =

1

D0 cosα[gα(α, β)− gβ(α, β)].

Second, the coordinates (ξ, θ) have to be changed to (α, β). The Jacobian

requires(∂α

∂ξ, ∂β∂ξ

), which we have already derived and

(∂α∂θ, ∂β∂θ

)which are

∂α

∂θ= 0

∂β

∂θ= 1.

Thus

J =∂(α, β)

∂(ξ, θ)=

∣∣∣∣ 1D0 cosα

0−1

D0 cosα1

∣∣∣∣ =1

D0 cosα.

27

Thus,

µ(r, φ) =1

4π2

∫ 2π

0

∫ ∞−∞

ρξ

r cos(θ − φ)− ξdξdθ

=1

4π2

∫ 2π

0

∫ π2

−π2

1D0 cosα

(gα − gβ)

r cos(β + α− φ)−D0 sinα(D0 cosα)dαdβ

=1

4π2

∫ 2π

0

∫ π2

−π2

gα − gβr cos(β + α− φ)−D0 sinα

dαdβ.

Observe from Figure(9) that

r cos(β + α− φ)−D0 sinα = r cosα cos(φ− β) + r sinα sin(φ− β)−D0 sinα

= FM cosα +OF sinα−D0 sinα

= D sinα cosα + (D0 −D cosα) sinα−D0 sinα

= D(sinα cosα− cosα sinα)

= D sin(α− α).

Thus

µ(r, φ) =1

4π2

∫ 2π

0

∫ π2

−π2

gα − gβD sin(α− α)

dαdβ

=1

4π2

∫ 2π

0

1

D

[∫ π2

−π2


dα

]dβ.

Corollary 2.

µ(r, φ) =1

4π2

∫ 2π

0

G(α, β)√D2

0 + r2 + 2rD0 sin(β − φ)dβ, (52)

where

G(α, β) =

∫ π2

−π2


dα = (gα − gβ) ∗α1

sinα(53)

and α(β) = tan−1[

r cos(β−φ)D0+r sin(β−φ)

].

This corollary gives a reconstruction procedure from the sinogram g(α, β) fora point M(r, φ): Take derivatives in the α and β directions, subtract, andconvolve with 1

sinαfor each column β = constant, separately, to get G(β).

Then, µ can be reconstructed from Formula(52). Figure(2.2) illustrates thisprocedure.

28

Can: 1. Form fan beam Radon Transform 2. Reparametrize in terms of α, β instead of ξ, β 3. Follow the procedure in Corollary(2) 4. Compare to rebinning and reconstruction from ρ(ξ, θ).

Corollary 3.

µ(r, φ) =1

4π2

∫ 2π

0

G(ξ, β)√D2

0 + r2 + 2rD0 sin(β − φ)dβ,

where

G(ξ, β) =

∫ ∞−∞

∫ ∞−∞

D0

(1 + ξ2

D20

)(gξ − gβ)

ξ − ξ

√ξ

2+D2

0√ξ2 +D2

0

dξ,

and ξ = D0 tanα = D0

[r cos(β−φ)

D0+r sin(β−φ)

].

Proof. Let ξ = D0 tanα. Then

sin(α− α) = sinα cosα− cosα sinα

=ξ√

ξ2

+D20

D0√ξ2 +D2

0

− D0√ξ

2+D2

0

ξ√ξ2 +D2

0

= D0ξ − ξ√

ξ2

+D20

√ξ2 +D2

0

.

Also, note that

gα − gβ = gξD0(1 + tan2 α)− gβ

= gξD0

(1 +

ξ2

D20

)− gβ.

Then, using the substitution ξ = D0 tanα and ξ = D0 tanα in the integralof Equation(53) defines

G(ξ, β) =

∫ ∞−∞

D0

(1 + ξ2

D20

)gξ − gβ

ξ − ξ

√ξ

2+D2

0

√ξ2 +D2

0

D0

1

D0

(1 + ξ2

D20

)dξ=

∫ ∞−∞

∫ ∞−∞

D0

(1 + ξ2

D20

)gξ − gβ

ξ − ξ

√ξ

2+D2

0√ξ2 +D2

0

dξ (54)

This clearly demonstrates that g(α, β) is a better representation since Equa-tion (53) is a convolution while Equation (54) is not.

29

2.3 Convolution in Fan Beam Geometry

Proposition 3. Let µ∗(x, y) = µ(x, y) ∗ h(x, y). Then,

gµ∗(α, β) = D0

∫ π2

−π2

gµ(α, β + α− α)gh(α, β + α− α) cosαdα, (55)

where α = sin−1(sinα− sin α).

Proof. We have from Corollary 1, Theorem 2, letting ξ = D0 sinα and ξ =D0 sin α in Figure(10) that

gµ∗(α, β) = ρµ∗(D0 sinα, β + α)

=

∫ ∞−∞

ρµ(ξ, β + α)ρh(D0 sinα− ξ, β + α)dξ

=

∫ ∞−∞

ρµ(D0 sin α, β + α)ρh(D0 sinα−D0 sin α, β + α)D0 cos αdα

=

∫ ∞−∞

gµ(α, β + α− α)ρh(D0 sinα, β + α)D0 cos αdα

=

∫ ∞−∞

gµ(α, β + α− α)gh(α, β + α− α)D0 cos αdα.

Corollary 4. Filtering in the image space is equivalent to separately trans-forming each line α + β = c0 in the fan beam sinogram gµ(α, β) to gµ∗(α,β),Figure(12), via Equation(55).

Proof. Consider a point (α1, β1) on the line α + β = c0. Then gµ∗ requiresgµ(α, β1 + α1 − α) for all α, namely, gµ for all points of the line α + β = c0.It also requires gh(α, β + α − α) for all α, again pointing to the same line.Since no other information is required, the filtering operation transforms aline of the sinogram to the same line.

Remark: The transformation is local due to the local extent of the filterh(x, y) which implies that gh(α, β) has limited extent.

Example: (Gaussian filter in the fan beam geometry) According to Corollary(5),

the Radon Transform of a Gaussian filter h(x, y) = 12πσ2 e

− (x2+y2)

2σ2 is ρh(ξ, θ) =

1√2πσ

e−ξ2

2σ2 . Thus, by Corollary(1)

gh(α, β) = ρh(D0 sinα, β + α)

=1√2πσ

e−D2

0 sin2 α

2σ2 .

30

Figure 12: 2D spatial filtering becomes a 1D transform in the projection data infan beam geometry. The sinogram gµ∗ of the filtered image µ∗ = µ ∗ h can beobtained by transforming each line α+ β = c0 separately from other points.

Similarly, for a shifted filter h(x, y) = 12πσ2 e

− (x−x0)2+(y−y0)2

2σ2 , ρh = 1√2πσ

e−(ξ−x0 cos θ−y0 sin θ)2

2σ2 .Thus,

gh = ρh(D0 sinα, β + α)

=1√2πσ

e−(D0 sinα−x0 cos(β+α)−y0 sin(β+α))2

2σ2

Figure(13) shows the sinogram for the filter h for several selections of (x0, y0) ∈(0, 0), (50, 0), (0, 50), (50, 50).

Figure 13:

31

3 Flirting with Filtering

3.1 Introduction

The idea that structure in the image space can be more effectively detectedwhen working directly with view data is based on a number of intuitions.First, the inherent discretization grid arises from the measurement processitself where a discrete set of views (orientations) are taken using a discrete setof detectors, while the image space discretization is artificially introduced toenable computation and visualization in the physically meaningful domain.Thus, in cases where improved resolution is of crucial importance, it wouldbe wise to re-consider computations directly in the view space. Second, ageneral expectation is that since image space filtering and reconstructionfrom view data are both linear operations, perhaps filtering in the imagespace can be viewed as filtering in the view space. Specifically, matchedfiltering for detecting particular structures can be achieved by consideringa filter projected in the view space and doing the matched filtering there.Thus, the current algorithm projects the matched filter h(x, y) centered at(x0, y0) to the view space and considers the inner product of this view spacerepresentation of the filter, ρh,x0,y0(ξ, θ) and the view space representation ofthe image µ(x, y) , namely ρµ(ξ, θ) in search of products exceeding a pre-determined threshold τ :

structure at(x0, y0) iff

∫ 2π

0

∫ ∞−∞

ρµ(ξ, θ)ρh,x0,y0(ξ, θ)dξdθ > τρ. (56)

In this chapter, we evaluate the validity of this algorithm and examinematched filtering in the image space:

structure at(x0, y0) iff µ∗(x0, y0) =

∫ ∞−∞

∫ ∞−∞

µ(x, y)h(x−x0, y−y0)dxdy > τµ

(57)in regard to an equivalent operation in the view space.

3.2 Filtering

Proposition 4. Let h(x, y) = h(x − x0, y − y0). Then ρh(ξ, θ) = ρh(ξ −x0 cos θ − y0 sin θ, θ), where ρh and ρh are the view space representations of

h and h, respectively. If (r0, φ0) is the polar representation of (x0, y0), thenρh(ξ, θ) = ρh(ξ − r0 cos(θ − φ), θ).

Proof. Let[ξ0

η0

]=


] [x0

y0

],

[x0

y0

]=


] [ξ0

η0

].

32

Then,

ρh(ξ, θ) =

∫ ∞−∞

h(ξ cos θ − η sin θ, ξ sin θ + η cos θ)dη

=

∫ ∞−∞

h(ξ cos θ − η sin θ − x0, ξ sin θ + η cos θ − y0)dη

=

∫ ∞−∞

h(ξ cos θ − η sin θ − ξ0 cos θ + η0 sin θ, ξ sin θ + η cos θ − ξ0 sin θ − η0 cos θ)dη

=

∫ ∞−∞

h((ξ − ξ0) cos θ − (η − η0) sin θ, (ξ − ξ0) sin θ + (η − η0) cos θ)dη

=

∫ ∞−∞

h((ξ − ξ0) cos θ − η sin θ, (ξ − ξ0) sin θ + η cos θ)dη

= ρh(ξ − ξ0, θ)

= ρh(ξ − x0 cos θ − y0 sin θ, θ).

In other words, shifting a filter by (x0, y0) in the image space is equivalentto vertically shifting each column θ in the view space by x0 cos θ + y0 sin θ.

Theorem 2. Let ρµ(ξ, θ) and ρµ∗(ξ, θ) be the view space representation ofµ(x, y) and µ∗(x, y), respectively, where µ∗ = µ ∗ h, for an arbitrary filter h.Then,

ρµ∗(ξ, θ) = ρµ(ξ, θ) ∗ξ ρh(ξ, θ), (58)

where ∗ξ is convolution with respect to the variable ξ.

Proof. In the following algebraic manipulation, we will use auxiliary variables(u, v) and (u, v) related by[

uv

]=


] [uv

]and

[uv

]=


] [ξη

],

and use a substitution η = η − η. By definition of the Radon Transform, we

33

have

ρµ∗(ξ, θ) =

∫ ∞−∞

µ∗(ξ cos θ − η sin θ, ξ sin θ + η cos θ)dη

=

∫ ∞−∞

∫ ∞−∞

∫ ∞−∞

µ(ξ cos θ − η sin θ − u, ξ sin θ + η cos θ − v)h(u, v) dudvdη

=

∫ ∞−∞

∫ ∞−∞

[∫ ∞−∞

µ(ξ cos θ − η sin θ − u, ξ sin θ + η cos θ − v)dη

]h(u, v) dudv

=

∫ ∞−∞

∫ ∞−∞

[∫ ∞−∞

µ(ξ cos θ − η sin θ − ξ cos θ + η sin θ,

ξ sin θ + η cos θ − ξ sin θ − η cos θ)dη]h(u, v) dudv

=

∫ ∞−∞

∫ ∞−∞

[∫ ∞−∞

µ((ξ − ξ) cos θ − (η − η) sin θ, (ξ − ξ) sin θ + (η − η) cos θ)dη

]h(u, v)dudv

=

∫ ∞−∞

∫ ∞−∞

[∫ ∞−∞

µ((ξ − ξ) cos θ − η sin θ, (ξ − ξ) sin θ + η cos θ)dη

]h(u, v)dudv

=

∫ ∞−∞

∫ ∞−∞

ρµ(ξ − ξ, θ)h(ξ cos θ − η sin θ, ξ sin θ + η cos θ) dξdη

=

∫ ∞−∞

ρµ(ξ − ξ, θ)[∫ ∞−∞


]dξ

=

∫ ∞−∞

ρµ(ξ − ξ, θ) ρh(ξ, θ)dξ

= ρµ(ξ, θ) ∗ξ ρh(ξ, θ).

In other words, to convolve a filter h with image µ, we can equivalentlyconvolve each orientation column θ of the sinogram ρµ with the respectivecolumn of the sinogram ρh to reconstruct ρµ∗ at that column.

Remark: This approach can be used to reconstruct the original data. Ob-serve that when the filter h is an impulse, i.e. h(x, y) = δ(

√x2 + y2), we

have

ρh(ξ, θ) =

∫ ∞−∞

h(ξ cos θ−η sin θ, ξ sin θ+η cos θ)dη =

∫ ∞−∞

δ(√ξ2 + η2)dη = δ(ξ).

Since ρh(ω, θ) = 1, ρ∗h(ξ, θ) = F−1(|ω|), thus giving the filtered backpro-jection process. However, while |ω| is not well-behaved in the frequencydomain in the case of h being an impulse, in the general case ρh can be well-behaved!!! For example, when h has a Gaussian drop-off, the rising effectsof |ω| is curtailed. In summary, general purpose reconstruction probes the

34

image with an impulse filter to generate image value at pixels, and only afterreconstruction is the existence of specific structures probed. However, whilethe general reconstruction problem is ill-conditioned, the direct probing ofspecific structures in the view space may not be!!!

Proposition 5 (Radially Symmetric Filters). The Radon Transfom of aradially symmetric filter h(x, y) = H(

√x2 + y2) centered at (x0, y0) is

ρh(ξ, θ) =

∫ ∞−∞

H(√

(ξ − (x0 cos θ − y0 sin θ))2 + η2)dη.

Proof. The Radon Transform of h(x, y) centered at the origin is

ρh(ξ, θ) =

∫ ∞−∞


=

∫ ∞−∞

H(√ξ2 + η2)dη.

Now, for h(x, y) = h(x− x0, y − y0), we have by Proposition(4)

ρh(ξ, θ) = ρh(ξ − (x0 cos θ − y0 sin θ), θ),

and the proposition follows.

Corollary 5 (Radon Transform of a Gaussian Filter). The Radon Transformof a Gaussian filter where centered at (x0, y0), i.e., h(x, y) = h(x−x0, y−y0)

h(x, y) =1

2πσ2e−(x2+y2)

2σ2 ,

is

ρh(ξ, θ) =1√2πσ


2σ2 .

Proof. We have

ρh(ξ, θ) =

∫ ∞−∞


=

∫ ∞−∞

1

2πσ2e−(ξ2+η2)

2σ2 dη

=1√2πσ

e−ξ2

2σ2 . (59)

Now, by Proposition(5)

ρh(ξ, θ) =1√2πσ


2σ2 .

35

This implies that a 2D Gaussian spot in the image space maps to a collectionof 1D Gaussians of the same spread on each column of the view space, butwhose center ξ0 = x0 cos θ + y0 sin θ = r0 cos(θ − φ0) is on a sinusoid withrespect to θ.

3.3 Matched Filtering

Theorem 3 (Matched filtering in the view space). A structure is present atthe image indicated by image space matched filtering

µ∗(x0, y0) =

∫ ∞−∞

∫ ∞−∞

µ(x, y)h(x− x0, y − y0)dxdy > τµ (60)

if and only if∫ 2π

0

[ρµ(ξ, θ) ∗ξ νh(ξ, θ)] (x0 cos θ + y0 sin θ)dθ > 2πτµ, (61)

where νh is the view space matched filter, the projected filter, defined as

νh(ξ, θ) = F−1ω ρh(ω, θ)|ω|, (62)

where ρh(ω, θ) = Fξρh(ξ, θ)(ω). (bar denotes the Fourier Transform,i.e., f(ω) = Ff(x).)

Proof. We have µ∗ = µh. Recall also that by the Central Slice Theorem,µ(ω, θ) is the Fourier Transform of ρµ(ξ, θ) with ξ as the free variable and θas a parameter, and similarly for h,

µ(ω, θ) = Fξρµ(ξ, θ)(ω) = ρµ(ω, θ)

h(ω, θ) = Fξρh(ξ, θ)(ω) = ρh(ω, θ).(63)

36

Then, we can rewrite µ∗ in the Fourier domain

µ∗(x0, y0) =1

4π2

∫ 2π

0

∫ ∞−∞

µ∗(ω, θ)eiω(x0 cos θ+y0 sin θ)|ω|dωdθ (64)

=1

4π2

∫ 2π

0

∫ ∞−∞

µ(ω, θ)h(ω, θ)eiω(x0 cos θ+y0 sin θ)|ω|dωdθ

=1

4π2

∫ 2π

0

∫ ∞−∞

ρµ(ω, θ)ρh(ω, θ)eiω(x0 cos θ+y0 sin θ)|ω|dωdθ

=1

2π

∫ 2π

0

[1

2π

∫ ∞−∞

ρµ(ω, θ)ρh(ω, θ)|ω|eiω(x0 cos θ+y0 sin θ)dω

]dθ

=1

2π

∫ 2π

0

[1

2π

∫ ∞−∞

ρµ(ω, θ)νh(ω, θ)eiω(x0 cos θ+y0 sin θ)dω

]dθ

=1

2π

∫ 2π

0

F−1[ρµ(ω, θ)νh(ω, θ)](x0 cos θ + y0 sin θ, θ)dθ

=1

2π

∫ 2π

0

[ρµ(ξ, θ) ∗ξ ν∗h(ξ, θ)] (x0 cos θ + y0 sin θ, θ)dθ. (65)

Observe that matched filtering thresholds in the view space and image spaceare now related.

3.4 Detecting Spots

3.4.1 Detecting A Gaussian Spot

We now attempt to analytically derive an expression for the view spacematched filter ρ∗h(ξ, θ) for the specific case of the Gaussian spot. Consider amatched filter

h(x, y) =1

2πσ21

e−x

2+y2

2σ21 − 1

2πσ22

e−x

2+y2

2σ22 , where σ2 > σ1, (66)

where σ1 is the size of the Gaussian spot. Then, by additivity,

ρh(ξ, θ) =1√

2πσ1

e−ξ2

2σ21 − 1√

2πσ2

e−ξ2

2σ22 . (67)

Then, as shown in Appendix(A.2)

ρh(ξ, θ) = Fξρh(ξ, θ) = e−ω2σ2

12 − e

−ω2σ22

2 . (68)

37

Then,

ρ∗h(ω, θ) = |ω|e−ω2σ2

12 − |ω|e

−ω2σ22

2 . (69)

or

ρ∗h(ξ, θ) = F−1|ω|e−ω2σ2

12 − |ω|e−

ω2σ22

2 . (70)

Using Equation(110) in Appendix(A.4), we have

ρ∗h(ξ, θ) =1

πσ21

−√

2

πσ31

ξe− ξ2

2σ21

∫ ξ√2σ1

0

eζ2

dζ − 1

πσ22

+

√2

πσ32

ξe− ξ2

2σ22

∫ ξ√2σ2

0

eζ2

dζ.

(71)Figure(14) shows examples of the filter h and the projected filter νh for severalvalues of σ.

(a)

(b)

Figure 14: First row figures show the profile of h(x, y) along a line passing throughthe origin for σ1 = 0.5, 1 and 2 respectively. Second row figures show ρ∗h(ξ, θ) forσ1 = 0.5, 1 and 2 respectively.

3.4.2 Detecting A Circular Spot

A circular spot can be represented by its characteristic function:

µR(x, y) =

1 x2 + y2 ≤ R2

0 x2 + y2 > R2.

38

The Radon Transform of µR can be easily computed using (ξ cos θ−η sin θ)2+(ξ sin θ + η cos θ)2 = ξ2 + η2

ρµR(ξ, θ) =

∫ ∞−∞

µR(ξ cos θ − η sin θ, ξ sin θ + η cos θ)dη

=

∫ √R2−ξ2

−√R2−ξ2

dη = 2√R2 − ξ2 if ξ < R

0 if ξ ≥ R.

Now for detecting this structure using a center-surround matched filter, weuse h(x, y) = µR(x, y)−AµR(x, y), when R > R and A is chosen to give zero

area. Since the area under µR is πR2, we need πR2 −AπR2= 0, or A = R2

R2 ,

so that a matched filter for a circular spot is

h(x, y) = µR(x, y)− R2

R2µR(x, y). (72)

In our previous experiments, the area of the center and surround have also

been matched, i.e. πR2 = πR2 − πR2, or R =

√2R. Then, the Radon

Transform of the matched filter (72) is [Was there a reason for this?]

ρh(ξ, θ) = 2√R2 − ξ2 − 2

R2

R2

√R

2 − ξ2

=

2√R2 − ξ2 − 2

R2

R2

√R

2 − ξ2 if 0 < ξ < R

− 2R2

R2

√R

2 − ξ2 if R < ξ < R

0 if R < ξ

(73)

As shown in Appendix(B.1),

Fx(√A2 − x2) =

πA

ω

(sin(ωA)

(ωA)2− cos(ωA)

ωA

).

Using A = R and x = ξ, we get

Fξ(√R2 − ξ2) =

πR

ω

(sin(ωR)

(ωR)2− cos(ωR)

ωR

).

Hence,

Fξ(ρh(ξ, θ)) = 2

[πR

ω

(sin(ωR)

(ωR)2− cos(ωR)

ωR

)]−2

R2

R2

[πR

ω

(sin(ωR)

(ωR)2− cos(ωR)

ωR

)]39

and

Fξ(ρ∗h(ξ, θ)) = 2

[|ω|πR

ω

(sin(ωR)

(ωR)2− cos(ωR)

ωR

)]−2

R2

R2

[|ω|πR

ω

(sin(ωR)

(ωR)2− cos(ωR)

ωR

)]or

ρ∗h(ξ, θ)) = 2F−1

|ω|πR

ω

(sin(ωR)

(ωR)2− cos(ωR)

ωR

)−2

R2

R2F−1

|ω|πR

ω

(sin(ωR)

(ωR)2− cos(ωR)

ωR

).

(a)

(b)

Figure 15: First row figures show the profile of h(x, y) along a line passing throughthe origin for R = 1, 2 and 4 respectively. Second row figures show νh(ξ, θ) forR = 1, 2 and 4 respectively.

As shown in Appendix(B.2),

F−1

|ω|πR

ω

(sin(ωR)

(ωR)2− cos(ωR)

ωR

)= 1 +

ξ

2Rln|ξ −R||ξ +R|

.

Thus,

ρ∗h(ξ, θ)) = 2

[1 +

ξ


]− 2

R2

R2

[1 +

ξ


]. (74)

Figure(15) shows examples of the filter h and the projected filter νh for severalvalues of R.

40

4 Taylor Expansion in the View Space

5 Geometric Tomography

Assume the image µ(x, y) is made of several smooth regions of smoothlyvarying intensity such as the phantom shown in Figure(16a) , i.e. that µ ispiecewise smooth, with smooth contours of discontinuity. Consider a partic-ular projection angle θ and some ξ along the detector array. In general, thisray goes through a number of regions Ri, but is not tangent to them. In somecases the ray would be tangent to a region Ri. In general, ρ is continuous asa function of ξ, for each θ.

(a) (b)

(c) (d)

Figure 16: (a)The image µ(x, y) is assumed to be piecewise smooth. (b) TheRadon Transform of µ. (c) Edges extracted from the Radon Transform image(shown in green). A hand traced contour is shown in red. (d)The red contour isbackprojected on the original image.

41

When the ray is not tangent to any region, it integrates through the regionsit intersects with. In this case, the continuity of the image and the regionimplies that ρξ is smooth. When the ray is tangent to a region, however,as in the blue ray shown in Figure(16a), the first derivative of the Radontransform is no longer continuous since it is determined on each side by twodistinct regions, one on the left (ξ−) and one on the right side (ξ+) of ξ. Inthe rare case of multiple tangent rays, the same statement holds. We haveinformally shown:

Remark: When the image µ(x, y) is piecewise smooth with smooth dis-continuity curves, the derivative if the Radon Transform ρξ is discontinuousif and only if the projection ray is tangent to at least one image discontinuity.

This implies that edges detected on the projection image point to edgesin the image, although they are not enough to localize these edges in space,Figure(16b). While this localization is not possible from a single view θ, itmay be possible to integrate multiple views to achieve this. Unfortunately,the rays in all other views passing through the edge of interest are transver-sal to the edge and thus no discontinuities can arise! The information abouteach edge is fully focused on a single view. What is worse is that if due todiscretization, this view is missed, so is all the information pertaining to thatedge!

A second constraint pertaining to X-ray images of realistic objects suchas those obtained from anatomy is that not only the images are piecewisesmooth, but also the discontinuity curves themselves are piecewise smooth.What this implies is that while for each image edge information can be ob-tained from only one discontinuity point in the Radon Transform, throughthe continuity of the edge on a curve we will have access to the Radon Trans-form discontinuities of its neighboring points!

Formally, consider a curve Cµ(s) seperating two distinct regions in the µimage plane. Let the tangent and normal to this curve be denoted as Tµ(s)and Nµ(s), respectively. For each point Cµ(s), there is a discontinuity inρξ(ξ, θ) where θ denotes the orientation of the detector array normal to Tµ(s),and ξ id the projection of Cµ(s) on that view, Figure(??). As we traverseCµ(s), this traces out a curve in the sinogram space, Figure(??), denoted byCρ(s) = (ξ(s), θ(s)). Along the curve Cρ(s) in the view (sinogram) space, ρhas discontinuous first derivatives.

Proposition 6. Let ρ(ξ, θ) be the Radon Transform of µ(x, y) containing a

42

(a) (b)

(c)

Figure 17: (a) The discontinuity A at ξ1 detected in the Radon Transform deriva-tive corresponsing to angle θ, predicts a discontinuity A in the image domain,(b) Similarly B at ξ2 points to the existance of an edge B in the projection viewθ+ dθ, (c) The intersection point P of the two rays localizes A and B in the limitas dθ → 0.

smooth curve Cµ(s) seperating two distinct regions, which traces out a curveCρ(s) = (ξ(s), θ(s)) in the view space along which ρξ is discontinuous. Then,Cµ(s) can be obtained for Cρ(s) as

Cµ(s) = ξ(s)(cos θ(s), sin θ(s)) +dξdsdθds

(− sin θ(s), cos θ(s)). (75)

Proof. For each point of Cρ(s) = (ξ(s), θ(s)), the point Cµ(s) is located onthe line orthogonal to the detector array, ξ-axis, in the image domain, atξ, Figure(18). Let η(s) denote the distance between Cµ(s) and the detectorarray. Then

Cµ(s) = ξ(s)(cos θ, sin θ) + η(s)(− sin θ, cos θ). (76)

43

Figure 18:

We also know that the tangent to Cµ(s) is orthogonal to the ξ-axis

d

dsCµ(s) = gµ(s)(− sin θ, cos θ), (77)

where gµ(s) is the metric of parametrization with respect to s. Differentiatingboth sides of Equation(76)

d

dsCµ(s) =

dξ

ds(cos θ, sin θ) + ξ(s)(− sin θ, cos θ)

dθ

ds+

dη

ds(− sin θ, cos θ) + η(s)(− cos θ,− sin θ)

dθ

ds(78)

=

(dξ

ds− η(s)

dθ

ds

)(cos θ, sin θ) +

(ξ(s)

dθ

ds+dη

ds

)(− sin θ, cos θ).

44

Figure 19:

Equating Equation(77) and (78), we havedξ

ds− η(s)

dθ

ds= 0

ξ(s)dθ

ds+dη

ds= gµ(s).

(79)

The first equation leads to a solution for η(s)

η(s) =dξdsdθds

=dξ

dθ, (80)

then leading to

Cµ(s) = ξ(s)(cos θ(s), sin θ(s)) +dξdsdθds

(− sin θ(s), cos θ(s)). (81)

Let ddsCρ(s) = gρ(s)τρ(s), where the tangent vector to Cρ(s) is represented as

τρ(s) = (cosφ(s), sinφ(s)). Then, dξds

= gρ(s) cosφ(s) and dηds

= gρ(s) sinφ(s),so that

dξ

dθ=gρ(s) cosφ(s)

gρ(s) sinφ(s)= cotφ(s). (82)

45

Note that the second equation in Equation(79) gives the relative speed(?) ofparametrization at µ by differentiating the first equation and substituting inthe second.

dη

ds=

d

ds

(dξ

dθ

)=−dφ(s)

ds

sin2 φ= −gρ(s)kρ(s)

sin2 φ,

so that

gµ(s) = ξ(s)dθ

ds− gρ(s)kρ(s)

sin2 φ.

Corollary 6. The edge map of µ(x, y) can be obtained by Equation(75) fromthe edge map of ρξ(ξ, θ).

Figure 20: Can to put the simulation results using Proposition 6 here...

5.1 Fan Beam Geometry

The derivation for the fan beam geometry is similar. Consier Figure(21) andlet the source S be at distances D0 and D away from the real detector arrayand the origin, respectively. Let the sinogram be specified in the normalizeddetector array passing through the origin. Assume S rotates around O andconsider view at angle θ. Let α = ∠(SO, SP ) denote the angle of the raytangent to the curve Cµ at P. Then

Cµ(s) =−→OA+

−→AP,

where −→OA = ξ(cos θ, sin θ),

and

−→AP =

η

cosα

(cos(θ +

π

2+ α

), sin

(θ +

π

2+ α

))=

η

cosα(− sin(θ+α), cos(θ+α)).

46

Figure 21:

We also have ξ = D tanα, so that

Cµ(s) = D tanα(cos θ, sin θ) +η

cosα(− sin(θ + α), cos(θ + α)).

On the other hand, the tangent to the curve Cµ(s) is

Tµ(s) = −−→AP

|AP |= gµ(s)(− sin(θ + α), cos(θ + α))

On the other hand,

dCµ(s)

ds= D(1 + tan2 α)

dα

ds(cos θ, sin θ) +D tanα(− sin θ, cos θ)

dθ

ds

+dη

ds

1

cosα(− sin(θ + α), cos(θ + α)) +

η sinα

cos2 α

dα

ds(− sin(θ + α), cos(θ + α))

+η

cosα(− cos(θ + α),− sin(θ + α))

(dθ

ds+dα

ds

)47

Now, dCµ(s)

ds.Nµ = dCµ(s)

ds(cos(θ + α), sin(θ + α)) = 0 so that after this dot

product

0 = D(1 + tan2 α)dα

ds(cos θ cos(θ + α), sin θ sin(θ + α))

+D tanα(− sin θ cos(θ + α), cos θ sin(θ + α))dθ

ds

+η

cosα(− cos2(θ + α)− sin2(θ + α))

(dθ

ds+dα

ds

)=

D

cos2 α

dα

dscosα +

D sinα

cosαsinα

dθ

ds− η

cosα

(dθ

ds+dα

ds

)which gives

η(s) = Ddαds

+ sin2 αdθds

dαds

+ dθds

. (83)

This can be written in terms of ξ and its derivatives using ξ = D tanα,dξds

= D 1cos2 α

dαds

= DD2+ξ2

D2dαds

= D2+ξ2

Ddαds

, and sinα = ξ2

D2+ξ2 , so that

η(s) = D

DD2+ξ2

dξds

+ ξ2

D2+ξ2dθds

DD2+ξ2

dξds

+ dθds

= DD dξ

ds+ ξ2 dθ

ds

D dξds

+ (D2 + ξ2)dθds

. (84)

6 Cone Beam Tomography and Volumetric

CT

In this section, we consider imaging a volume with an X-ray source and aplanar detector. This is a natural progression from the evolution of singleslice CT to spiral CT and then to multislice CT. As the number of slicesincreased dramatically from the initial 4 to 8, 16, 32, 128, the situationwhere an entire plane can be sensed simultaneously has become realistic,Figure(22). This is the case for µCT and the more experimental volumet-ric CT (e.g., Siemens) machines which use flat panel detectors. At a firstglance, the tomographic reconstruction might appear to be one of collectingall line integrals through the volume and reconstructing the linear attenu-ation coefficient for the volume from these measurements. Note, however,that while the dimension of the unknowns is three, namely, µ(x, y, z), thedimensionality of the measured data collected in this way is four, two for thedetector (u, v) and two for the position of the source, say as described by

48

points on a unit sphere. One can then speculate that not all line integralsmay be needed, but rather a one-dimensional subset, where the source movesalong some curve, is sufficient. This is indeed true: a one-dimensional set ofmeasurements by moving the source along a curve (source orbit, or sourcetrajectory) is shown to be sufficient if the trajectory is chosen appropriately.The proof, however, requires an orginizing measurements of attenuation co-efficients along integration planes, which defines the 3D Radon Transform,as opposed to integration lines which defines the cone beam transform. Wedefine each below.

(a)

(b)

Figure 22: (a) The cone beam geometry: The detection array is some distanceaway from the object space origin O which the source S is a distance D0 away andmoves on a curve Γ.

−→SO is orthogonal to the detection array, which in analogy to

the fan beam case is scaled to pass through origin (b).

49

S X-ray sourceΓ(s) space curve describing the source orbit (trajectory) with s as length parameter

O,~i,~j,~k object coordinate systemD0 the distance from S to origin OD1 the distance from the detection array to origin Ox, y, z coordinates in object coordinate systemµ linear attenuation coefficient of the object

O, ~u,~v, ~w (virtual) detection coordinate system, ~w =−→OS|OS|

ξ, η, ζ coordinates in the detection coordinate systemd distance of the integration plane from object space origin~n unit normal to the integration plane

S, (ν, ν⊥, n) integration plane coordinate systemRµ(d, ~n) 3D Radon Transform: integration along a plane of distance d from origin

and normal ~n

Cµ(s, ~Θ) Cone Beam Transform: integration along a half-line from source S

directed along ~Θ

Table 1: Table of Notations for the Cone Beam Geometry

6.1 Notation and Definition

Let the object coordinate system be described by coordinates (x, y, z) with

O serving as the origin, Figure 22(a), and (~i,~j,~k) as unit vectors along thecoordinates, respectively. The linear attenuation coefficient of the object isthen described as µ(x, y, z).

It is assumed that the source is at a fixed distance D0 from this plane andthat the closest point to S on the virtual detector array is O. We assumethat O = O (Ben, this may need to be relaxed). As in the case of fanbeam geometry, it is easier to consider a virtual detector array, parallel to theactual detector array, that is parametrized along its own intrinsic coordinatesystem described by vectors (~u,~v), Figure 22(b). Define (O, ~u,~v, ~w) as thedetection coordinate system. The source S is assumed to move along a curveΓ(s) referred to as the source orbit or source trajectory.

The notion of 3D Radon Transform requires integration of linear attenuationcoefficients along planes passing through the source, each of which we referto as an integration plane, see Table 1. An integration plane is specified bydistance from origin O to the plane d and a unit normal ~n. Let M0 be theclosest point on this plane to the origin. Since the source S is by definition

50

on this plane, the vector−−→SM0 when normalized defines a unique direction

~ν =−−−→SM0

|−−−→SM0|

which together with ~ν⊥ = ~n × ~ν defines the integration plane co-

ordinate system (~ν, ~ν⊥, ~n), Figure 23.

(a) (b) (c)

Figure 23: An integration plane is specified in the object space by the distance dof the integration plane from the origin and by the unit normal ~n. The integrationplane can be intrinsically described by a coordinate system centered at S and withaxis ~ν, ~ν⊥ where ~ν =

−−−→SM0

|−−−→SM0|

, where M0 is the closest point to the origin.

The vector ~n can be described in the spherical coordinate system of the objectspace as

~n = (cos θ sinφ, sin θ sinφ, cosφ), (85)

where φ = ∠(~n,~k) and θ is the angle the projection of ~n onto the xy-planemakes with~i, Figure 23b. Similarly, in the detection space ~n can be expressedas

~n = (cosα sin β, sinα sin β, cos β). (86)

51

Remark: We can express ν and ν⊥ as follows.

~ν =(−→OS.~n)~n−

−→OS

|(−→OS.~n)~n−

−→OS|

=d0(~w.~n)~n− d0 ~w

|d0(~w.~n)~n− d0 ~w|

=d0 cos β(cosα sin β~u+ sinα sin β~v + cos β ~w)− d0 ~w

|d0 cos β(cosα sin β~u+ sinα sin β~v + cos β ~w)− d0 ~w|

=cosα sin β cos β~u+ sinα sin β cos β~v − sin2 β ~w

| cosα sin β cos β~u+ sinα sin β cos β~v − sin2 β ~w|

=cosα sin β cos β~u+ sinα sin β cos β~v − sin2 β ~w√

sin2 β cos2 β + sin4 β

= (cosα cos β~u+ sinα cos β~v − sin β ~w). (87)

Similarly,

~ν⊥ = ~n× ~ν= (cosα sin β~u+ sinα sin β~v + cos β ~w)× (cosα cos β~u+ sinα cos β~v − sin β ~w)

= (− cosα sin2 β − cosα cos2 β)(−~v)(− sinα sin2 β − sinα cos2 β)(~u)

= − sinα~u+ cosα~v. (88)

We are now ready to define the cone beam transform and the 3D RadonTransform.

Definition 1. The Cone Beam Transform of a scalar function µ(x, y, z) isdefined as

Cµ : R3 → R3

Cµ(S, ~Θ) =

∫ ∞0

µ(−→OS + r~Θ) dr,

where S is the source along its trajectory, and ~Θ is an arbitrary unit vector.

One can extend the definition of the cone-beam transform from R × R2 toR3 × R3 and by an abuse of notation use Cµ to also denote the extendedfunction

Cµ(S, P ) =1

|P |Cµ(S,

P

|P |

). (89)

Then the Fourier Transform of Cµ is

Cµ(S, ω) =

∫ ∫ ∫R3

Cµ(S, P )e−i.P.ωdP.

52

The function Cµ is homogenous of degree -2, i.e.,

Cµ(S, λω) =

∫ ∫ ∫R3

Cµ(S, P )e−iPλωdP

=1

λ3

∫ ∫ ∫Cµ(S,

1

λP

)e−iPωdP

=1

λ2

∫ ∫ ∫Cµ(S, P )e−iPωdP

=1

λ2Cµ(S, ω).

We follow [34] in their reformulation of Tuy’s inversion formula [29].

Theorem 4 (Tuy’s Inversion Formula). Suppose Tuy’s (modified) complete-ness condition is satisfied, i.e., that the set

Λ(P,Θ) = s ∈ R, (P − Γ(s)).Θ = 0,Γ′(s).Θ 6= 0,WP (s,Θ) 6= 0

is non-empty , where P is a point in space, P ∈ R3, Θ is an arbitrarydirection Θ ∈ S2, and WP (s,Θ) is integrable with respect to its second variablesatisfying ∑

s∈Λ(P,Θ)

WP (s,Θ) = 1 almost everywhere in Θ ∈ S2.

Then

µ(P ) =1

(2π)3

∫ ∫S2

∑s∈Λ(P,Θ)

WP (λ,Θ)

iΓ′(s).Θ

∂

∂tCµ(Γ(t),Θ)

t=s

dΘ. (90)

53

Proof.

∂

∂tCµ(Γ(t),Θ) =

∂

∂t

∫ ∫ ∫R3

Cµ(Γ(t), z)e−iz.Θdz

=∂

∂t

∫ ∫ ∫R3

[∫ ∞0

µ(Γ(t) + rz)dr

]e−iz.Θdz

=∂

∂t

∫ ∞0

∫ ∫ ∫R3

µ(W )e−iW−Γ(t)

rΘ 1

r3dWdr

=∂

∂t

∫ ∞0

1

r3

∫ ∫ ∫R3

µ(W )e−i1rW.ΘdW

ei 1r

Γ(t).Θdr

=∂

∂t

∫ ∞0

µ

(Θ

r

)ei

1r

Γ(t)Θ 1

r3dr

=∂

∂t

∫ ∞0

µ(r,Θ)eirΓ(t)Θrdr

= (Γ′(t),Θ)

∫ ∞0

µ(rΘ)eirΓ(t)Θr2dr. (91)

Next, To be continued

Definition 2. The 3D Radon Transform of a scalar function µ(x, y, z) isdefined as

Rµ : R3 → R3

Rµ(d, ~n) =

∫ ∫−−→OM.~n=d

µ(M) dM,

where M is a point on the integration plane defined by−−→OM.~n = d and dM

is the area element.

6.2 Relating Cone Beam and Radon Transform

The following theorem from [20] asserts that the set of all integration planemeasurements is sufficient to invert the transform.

Theorem 5 (Inverse 3D Radon Tranform[20]).

µ(M) =−1

8π2

∫ ∫n∈S2

∂2Rµ

∂d2(−−→OM.~n, ~n) d~n.

54

Remark: The 3D Radon Transform can be expressed as

Rµ(d, ~n) =

∫ ∫OM.n=d0

µ(M)dM

=

∫ 2π

0

∫ ∞0

µ(−→OS + r cos γ~ν + r sin γ~ν⊥)rdrdγ, (92)

where ~ν = (OS.n)~n−−→OS

|(OS.n)~n−−→OS|

and ~ν⊥ = ~n × ~ν. This allows us to explore the

relationship between the cone beam transform and the 3D Radon Transform:When ~Θ is restricted to the integration plane we have

Cµ(S, ~Θ) =

∫ ∞0

µ(−→OS + r~Θ)dr

=

∫ ∞0

µ(−→OS + r cos γ~ν + r sin γ~ν⊥)dr.

Thus, integrating all cone beam measurements along the integration planegives ∫ 2π

0

Cµ(s, ~Θ)dγ =

∫ 2π

0

∫ ∞0

µ(−→OS + r cos γ~ν + r sin γ~ν⊥)drdγ (93)

Observe that this closely corresponds, but is not identical to the 3D RadonTransform in Equation(92). Since the inversion formula of Theorem 5 re-quires Equation(92) and the practical measurements are those of the conebeam transform, a way must be found to relate these two quantities. Grangeat’sinsight [9, 10] was that by differentiating the cone beam transform with re-spect to β, which is implicit in ~ν, the r term can be brought out and theintegrals can be related. The following theorem is based on Grangeat’s proofbut is significantly shorter. This requires a weighted integral of cone beamrays along the integration plane.

Definition 3. For each point S and unit normal ~n define the weighted conebeam integral.

X (S,~n) =

∫ 2π

0

Cµ(S,~n)1

cos γdγ =

∫ 2π

0

∫ ∞0

µ(−→OS + r cos γ~ν + r sin γ~ν⊥)

1

cos γdrdγ(94)

where γ, ~ν, and ~ν⊥ are defined as before.

Theorem 6 (Relating the Cone Beam and 3D Radon Transforms[9]).

∂

∂βX (S,~n(α, β)) =

∂Rµ

∂d(−→OS.~n, ~n).

55

Proof. Observe from Equation 87 and 88 that ~ν and ~ν⊥ only depend on~n(α, β) but not on d. We therefore have

∂~ν

∂β= − cosα sin β~u− sinα sin β~v − cos β ~w = −~n

∂~ν⊥

∂β= 0.

Now, observe that for ~Θ in the integration plane ~Θ = r cos γ~ν + r sin γ~ν⊥

which implies that ~Θ is implicitly a function of ~n(α, β).

∂µ

∂β(−→OS + r cos γ~ν + r sin γ~ν⊥) = ∇µ(

−→OS + r cos γ~ν + r sin γ~ν⊥)r cos γ

∂~ν

∂β

= −r cos γ∇µ(−→OS + r cos γ~ν + r sin γ~ν⊥).~n.

This gives the r term we needed so that as

∂X∂β

(s, ~n) =∂

∂β

∫ 2π

0

Cµ(s, ~m)

(1

cos γ

)dγ

= −∫ 2π

0

∫ ∞0

∇µ(−→OS + r cos γ~ν + r sin γ~ν⊥)~nrdrdγ

On the other hand, differentiating the 3D Radon Transform with respect tod gives

∂Rµ

∂d=

∫ 2π

0

∫ ∞0

∇µ(−→OS + r cos γ~ν + r sin γ~ν⊥).~nrdrdγ,

which proves the result.

This theorem allows for the reconstruction of µ(x, y, z) cone beam measure-ments. It suggests that the cone beam measurements Cµ(s, ~m) be organizedby integrating then along all integration planes passing through the sourceS, by applying Equation(94), i.e. for all ~n(α, β). The result is a three-dimensional set of mesurements, X (s, α, β) corresponding to s along thecurve Γ(s), α and β. Second, differentiating this scalar function X (s, α, β)with respect to β gives ∂X

∂βwhich the theorem equates to ∂Rµ

∂d(d, α, β), where

d =−→OS.~n. Finally, the 3D Radon Inverse Formula states

µ(x, y, z) =−1

8π2

∫ π2

−π2

∫ 2π

0

∂2Rµ

∂d2(x cosα sin β+y sinα sin β+z cos β, α, β)dαdβ

56

where

∂

∂d

(∂Rµ∂d

)=

∂

∂d

(∂X∂β

)=

∂2X∂s∂β

∂d∂s

=

∂2X∂s∂β

∂(Γ(s).~n)∂s

=

∂2X∂s∂β

Γ′(s).~n.

6.3 Reconstruction Algorithms for Cone Beam HelicalCT

Discuss why helix: Tuy’s completeness condition; long object prob-lem

The helix is the most practical and popular source trajectory. Let Γ bethe helix with radius R and pitch h, i.e.,

Γ(s) = (R cos s, R sin s,h

2πs). (95)

The Tam-Danielsson Window [27, 4] for each point of a helix is the regionon the detector plane bounded by the projections of the upper and lowerturns of a helical trajectory, Figure 24(a). Tam et al. [27] using the tri-angular partition procedure showed that exact reconstruction based on the3D Radon Transform only requires cone beam data in the Tam-Danielssonwindows. Then, for any region of interest only the Tam-Danielsson windowscorresponding to this region are required.

A π-line is a straight line connecting two points of a helix which are less thana full 2π turn apart [4], Figure 24(b). The reason for the π-line terminologyis that the data corresponding to the source locus on the helix segment be-tween the two ends of each π-line covers 180 of data for any point on theπ-line [4] conjectured that the cone-beam data on this helix segment (axiallytruncated) was sufficient to reconstruct the object on the π-line and this wasfirst shown by Katsevich [17, 14]. The cylinder containing the helix is calledthe helix cylinder. A π-line segment is the portion of the π-line enclosedby the helix cylinder. The π-helix segment on the π-arc is the segment ofthe helix bounded by the two end points of a π-line segment. The supportcylinder of is the smallest cylinder radius R0 < R through the same centralaxis containing the object of interest.

57

(a) (b)

(c) (d)

Figure 24: (a) The Tam-Danielsson Window (b) π-line segment (c) Tam-Danielsson Window simulation, the red point on the helix is the source, the pro-jection of the helix points (blue) on the sensor array (boundaries shown in green)are shown in purple. (d) The sensor array is shown in 2D.

Observe that the tangent for the helix is

~τ =Γ′(s)

|Γ′(s)|=

(−R sin s, R cos s, h2π

)√R2 + h2

4π2

=(− sin s, cos s, h

2πR)√

1 + h2

4π2R2

while the normal is

~n =dτds

|dτds|

= (− cos s,− sin s, 0).

We see that the normal to the helix is orthogonal to and goes through itscentral axis.

The following proposition states that any region of interest in a helical cylin-der can completely be filled by π-line segments if there is data for a sufficientnumber of turns of the helix.

Theorem 7. [4, 5] Every point P (x, y, z) = P (r cosφ, r sinφ, z) in a helixcylinder belongs to one and only one π-line segment, bounded by s1(P ) ands2(P ), as given by solving Equations 137.

Next, to arrive at Katsevich’s exact reconstruction formula based on a filteredbackprojection algorithm [14], we first need to define the notion of a K-planewhich is a plane that intersects the helix at three equi-spaced points,i.e. at

58

Γ(s−∆s), Γ(s), and Γ(s+ ∆s) for some s and ∆s, and denoted as K(s,∆s).It is clear that the K-planes associated to any source point Γ(s) form a one-parameter family indexed by ∆s. In the limit, as ∆s → 0, the K-plane isthe bi-osculating plane (Ben to check the terminology) of the curve Γ,namely, the plane going through both the tangent and the normal to Γ, i.e.,the plane orthogonal to the helical axis.

Figure 25: For each point P and any source point Γ(s) with s ∈ [s1(p), s2(p)],there exists a K-plane, K(s,∆s) passing through Γ(s) and P . The normal to thisK-plane is ~m.

Proposition 7. The K-plane K(s,∆s) going through any point P (x, y, z) =P (r cosφ, r sinφ, z), and source Γ(s) on a helix is obtained from ∆s satisfying

sinc(∆s) =−hr2πR

sin(s− φ)

z − h2πs.

Ben: Existence of solution needs work.

Proof. The K-plane passing through Γ−1 = Γ(s−∆s),Γ0 = Γ(s),Γ1 = Γ(s+

∆s) can be characterized as the plane through Γ0 with normal along−−−→Γ−1Γ0×−−→

Γ0Γ1. The point P (x, y, z) is on this plane if and only if

−−→Γ0P .(

−−−→Γ−1Γ0 ×

−−→Γ0Γ1) = 0. (96)

59

Since Γ0 (or s) and P are both fixed, only ∆s is unknown leading to oneequation and one unknown. Using the equation for the helix,

Γ−1 = Γ(s−∆s) =

(R cos(s−∆s), R sin(s−∆s),

h

2π(s−∆s)

)Γ0 = Γ(s) =

(R cos s, R sin s,

h

2πs

)Γ1 = Γ(s+ ∆s) =

(R cos(s+ ∆s), R sin(s+ ∆s),

h

2π(s+ ∆s)

),

which gives

−−→Γ0Γ1 =

(R cos(s+ ∆s)−R cos s, R sin(s+ ∆s)−R sin s,

h

2π∆s

)=

(−2R sin

(∆s

2

)sin

(s+

∆s

2

), 2R sin

(∆s

2

)cos

(s+

∆s

2

),h

2π∆s

),

and similarly

−−−→Γ0Γ−1 =

(2R sin

(∆s

2

)sin

(s− ∆s

2

),−2R sin

(∆s

2

)cos

(s− ∆s

2

),−h2π

∆s

).

Thus,

−−→Γ0Γ1 ×

−−−→Γ0Γ−1 =

(Rh

π∆s sin

(∆s

2

)[− cos

(s+

∆s

2

)+ cos

(s− ∆s

2

)],

Rh

π∆s sin

(∆s

2

)[sin

(s− ∆s

2

)− sin

(s+

∆s

2

)],

4R2 sin2

(∆s

2

)[sin

(s+

∆s

2

)cos

(s− ∆s

2

)− cos

(s+

∆s

2

)sin

(s− ∆s

2

)])=

(Rh

π∆s sin

(∆s

2

).2 sin

(∆s

2

)sin s,

−Rh2

∆s sin

(∆s

2

).2 sin

(∆s

2

)cos s,

4R2 sin2

(∆s

2

)sin ∆s

)= 2R sin2

(∆s

2

)(h

π∆s sin s,

−hπ

∆s cos s, 2R sin ∆s

). (97)

Now,−−→Γ0P =

(x−R cos s, y −R sin s, z − h

2πs

), (98)

60

so that

−−→Γ0P .(

−−→Γ0Γ1 ×

−−−→Γ0Γ−1) = 2R sin2

(∆s

2

)[h

π∆s sin s(x−R cos s)+

−hπ

∆s cos s(y −R sin s) +

2R sin ∆s

(z − h

2πs

)]= 2R sin2

(∆s

2

)[h

π∆s(x sin s− y cos s) + 2R sin ∆s

(z − h

2πs

)]Setting this equal to zero gives (Ben: What about ∆s = 0?)

sinc(∆s) =sin ∆s

∆s= −h(x sin s− y cos s)

2πR(z − h

2πs) .

Using the cylindrical coordinates of P simplifies the equation to

sinc(∆s) =−hr2πR

sin(s− φ)

z − h2πs

. (99)

from which ∆s can be obtained. Note that the solution to this equation maynot exist, may be unique or multi-valued.

Corollary 7. The normals defined by the one-parameter family of K-planesK(s,∆s) through a given point Γ(s) forms a curve on the 2D unit sphere ofall possible directions given by

~m(s,∆s) =

(sin s, cos s, −2πR

hsinc(∆s)

)√1 + 4π2R2

h2 sinc2(∆s). (100)

Thus, for a given point P (x, y, z) and source Γ1 = Γ(s) we may not have aK-plane. However, varying s may lead to a solution.

Remark 1. It is clear from Equation 96 that the K-plane with respect tosource Γ0 of a point P lying on the K-plane of P and source Γ0 is the sameK-plane specifically, the K-plane for all points along a given unit direction~Θ passing through Γ0, i.e., P (τ) = Γ0 + τ ~Θ is the same as long as for Γ0 =

Γ(s), s ∈ [s1(P ), s2(P )]. Thus, a unit direction ~Θ defines the K-planes though

Γ0 = Γ(s). We let m(s, ~Θ) define the normal to such a K-plane.

For any point P and any source point Γ(s), s ∈ [s1(P ), s2(P )], where s1(P )and s2(P ) are source parameters corresponding to the end points of π-line of

61

P , there exists a K-plane K(s,∆s) passing through Γ(s) and P solve throughEquation 99. (Ben: This still needs work.) Denote the normal to this

K-plane as ~m(s, ~Θ), where ~Θ is the unit direction from Γ(s) to P . This no-tation is permissible since we showed in Remark 1 that the K-plane is onlydependent on ~Θ not on P . Thus, the significance of a K-plane is in defininga local direction ~m(s, ~Θ), for any given direction ~Θ, Figure 25. Note that

this defines a local coordinate system (~Θ, ~m, ~m⊥) where ~m⊥ = ~Θ× ~m.

The strategy underlying Katsevich’s formulation is to consider K-planesthrough a source point Γ0 = Γ(s) defined by any given direction ~Θ. Foreach K-plane, a kernel C∗µ(s,Θ) is defined and the object is reconstructedfrom it.

Theorem 8 (Katsevich’s inversion formula [14]). The object µ(x, y, z) canbe reconstructed as

µ(P ) =−1

2π2

∫ s2(P )

s1(P )

C∗µ

(s,

P − Γ(s)

|P − Γ(s)|

)1

|P − Γ(s)|ds. (101)

where

C∗µ(s, ~Θ) =1

π

∫ 2π

0

∂Cµ∂s

(s, cos γ~Θ + sin γ ~m⊥(s, ~Θ))s=s

1

sin γdγ. (102)

Note that the differentiation is with respect to the first occurance of s whileleaving the one affecting m⊥ constant. Then, Katsevich showed that theconvolution nature of the computation defining C∗µ(s, ~Θ) in Equation 102is not readily apparent. This aspect is revealed when we fix the K-planeK(s,∆s) through the source point Γ(s), say with normal m0(s,∆s), and

consider all ~Θ in this plane. Consider an arbitrary orthonormal coordinateframe for the K-plane (~e1, ~e2). Then, any unit vector ~Θ in this plane can beexpressed in parametrized form as

~Θ(γ) = cos γ ~e1 + sin γ ~e2. (103)

This gives

~m⊥(s, ~Θ) = ~Θ× ~m0 (104)

= (cos γ ~e1 + sin γ ~e2)× ~m0 (105)

= − cos γ ~e2 + sin γ ~e1

= sin γ ~e1 − cos γ ~e2. (106)

62

Then

C∗µ(s, ~Θ(γ)) =1

π

∫ 2π

0

∂Cµ∂s

(s, cos γ~Θ + sin γ ~m⊥(s, ~Θ))s=s

1

sin γdγ

=1

π

∫ 2π

0

∂Cµ∂s

(s, cos γ cos γ ~e1 + cos γ sin γ ~e2 + sin γ sin γ ~e1 − sin γ cos γ ~e2)

s=s

1

sin γdγ

=1

π

∫ 2π

0

∂Cµ∂s

(s, cos(γ − γ)~e1 + sin(γ − γ)~e2)1

sin γdγ

=1

π

∂Cµ∂s

(s, cos γ ~e1 + sin γ ~e2) ∗γ1

sin γ.

This suggests an algorithm: For each source point Γ(s) and K-plane K(s,∆s)with unit vectorm0, define an arbitrary orthonormal coordinate frame (~e1, ~e2)in the plane and define

F (s,∆s, γ) =∂Cµ∂s

(s, cos γ ~e1 + sin γ ~e2).

Then obtain

F ∗(s,∆s, γ) = F (s,∆s, γ) ∗γ1

sin γ.

(To be completed.)

(Ben: Work out circular and flat panel detector algorithms @ [21])

6.4 History of Helical Cone Beam CT Development

[26] and [19] are extensions of Grangeat’s theory [8].

Exact algorithms have been based on Katsevich’s formula [15], which was arather important breakthrough, in switching from Radon Transform-basedmethods to a filtered backprojection method with 1D shift-invariant filtering.Detector-plane parametrization methods [21] from [38] based on the RadonTransform ??? require more (over-scan) data than those based on Katse-vich’s formula in accurate reconstruction of long objects. On the other hand,Katsevich algorithms require more data with the Tam-Danielsson Window.

Zou and Pan[38] derived an exact formula for helical cone-beam reconstruc-tion and developed a backprojection filtering algorithm where the cone-beamsare first backprojected to π-line segments and then a 1D shift-invariant filter-ing is done over the π-segments. In [39], a filtered-backprojection algortihm

63

is developed where a 1D shift-invariant filtering is performed on the detec-tor plane and then reconstruct the data by backprojecting. This is doneby re-expressing the derivation in Katsevich’s formula [15, 17], Equations 96and 97, in terms of coordinates on the detector plane. Since the number ofsamples along the helix are generaly fewer than the dense samples along thedetector plane, this is claimed to be numerically superior [37]. Katsevich [14]derived a formula with no derivatives that contains five distinct terms. Zouand Pan [35] developed an algorithm with three terms, but which containpartial derivatives with respect to detector plane coordinates.

Generalized Feldkamp-type algorithms were developed for the long objectreconstrcution with longtitudially truncated data obtained in a helical conebeam acquisition [31, 32].

Chen [2] gave an alternative proof of the Katsevich algorithm using Tuy’sformula [29]. Zhao et al. [34] extended this proof to include the Zou-Panalgorithm [38, 36] which is not only valid for helical scan but is generallyvalid for various nonstandard spirals and other curves; see also [33].

Rebinning methods: Heuscher 1999, Chen 2000, Kachelriess 2000

FDK-like algorithms: Feldkamp 1984, Crawford and King 1990, Sourbelle2002, Turbell 1999, lack of theory → cannot predict/detect artefacts

Quasi-exact: Defrise etal 2000, Kudo etal 2000, Tam 2000, zero boundarymethod, some data is processed wrong, Tam 2002

Exact: Tuy SIAM 2000, Schaller 2000 PHI-method, Tam 2000 and Lauritsch2000, Katsevich 2002a, 2003.

Chen Med Phys 2003 ∼ Katsevich:IJMMS:2003

6.5 Material to be worked into this section

(Ben, it was not noted where to add the third hand-written pageof notes, so I added it here. Please mark its correct location.) (Bento work on this) Consider all planes through a point P and a source pointΓ0 = Γ(s) with normal ~m. Then, there are three types of planes intersectingthe π helix segment at one, two, or three points [15]. The planes with twointersections do not contribute to the reconstructed image!

64

Corollary 8. The image µ at the point P (x, y, z) can be fully reconstructedinside the helix cylinder by computing µ along the π-line segment containingP , i.e., the line segment bounded by s1(P ) and s2(P ) as given by Proposition11.

Theorem 9. [38]∫ ∫ ∫K(x, y, z, x, y, z)g(x, y, z) dxdydz,

where

K(x, y, z, x, y, z) =1

2π2

∫ ∫ ∫sgn[−→ν .eπ(x, y, z)]e2πi−→ν (x−x,y−y,z−z)d−→ν ,

g(x, y, z) =

∫ s2

s1

∂∂qCµ(Γ(q),−→n ((x, y, z), s))|q=s|(x, y, z)− Γ(s)|

ds,

and where eπ(x, y, z) is defined based on the unique π-line passing through(x, y, z) which is anchored at Γ1 = Γ(s1) and Γ2 = Γ(s2)

eπ(x, y, z) =Γ2 − Γ1

|Γ2 − Γ1|.

Theorem 10. [17]

µ(x, y, z) =−1

2π2

∫ s2(x,y,z)

s1(x,y,z)

∫ 2π

0

∂∂qCµ(Γ(q),

−→Θ(s, x, y, z, γ))|q=s

|(x, y, z)− Γ(s)| sin γ

where −→Θ(s, x, y, z, γ) = cos γ−→n (s, x, y, z) + sin γ−→e (s, x, y, z)

ande(s, x, y, z) = −→n (s, x, y, z)× u(s, x, y, z).

Ben to work on unifying the notation; u is not explained.

6.6 Numerical algorithms for helical cone-beam recon-struction

Kohler et al. [18] compared two approximate algorithms PI-SLANT [28] andWEDGE [30] with the exact method of Katsevich [15], Figures 27, 28 and29. They showed that a modified WEDGE-PI method performs almost aswell as the Katsevich method for a 64-row detector while PI-SLANT showsartifacts. For a 128-row detector, some artifacts also appear in the WEDGE-PI results, but the Katsevich method is about an order of magnitude slower,Figure 30.

65

Figure 26: Equation taken from [17]

7 Radon’s Proof

66

Figure 27: Illustration of the PI-SLANT method. Left: Original PI acquisition;middle: Part of the focus-centered detector that is used; right: Geometry afterparallel rebinning. Ramp filtering is performed along slanted lines on the virtualdetector.

8 To Do

- Re-examine Radon’s formula in the light of the discontinuity in ρξ. See [13]page 553, equation 3 and resulting Fξ(t).- X-ray lenses.- Summarize ART methods. [6, 12]- Can to get refs 2,5,7,9,10,15 from this paper, and to check INRIA reportcorresponding to the TMI paper. - Put in some description of flat panel x-raydetectors, x-ray tubes

67

Figure 28: Illustration of the WEDGE method. Left: Acquisition with a focus-centered detector; middle: Wedge geometry obtained by parallel rebinning. Rampfiltering is performed along rows of this virtual detector; right: In WEDGE-PI,the filtered data are height rebinned to the PI geometry to ensure that duringback-projection, each object point gets contribution over exactly 180o.

Figure 29: Illustration of the KATSEVICH method. Left: Acquisition with afocus-centered detector; middle: Convolution with 1

sin along K lines defined on aplanar detector; right: Parallel rebinning of the filtered data to allow for back-projection without density correction.

68

Figure 30: Results of the 128 row scanner. From top to bottom: PI-SLANT,WEDGE-PI, KATSEVICH.

Figure 31:

69

Figure 32:

Figure 33:

70

A Integrals Involving Gaussians and Deriva-

tives of Gaussians

A.1 Integrals involving a Gaussian and its derivatives

In the formulations below, all constants (upper case letters) are positive un-less otherwise stated.

1. Integrals of Gaussians∫ ∞−∞

e−x2

dx =√π,

∫ ∞−∞

e−x2σ2 dx =

√2πσ. (107)

∫ ∞−∞

e−Ax2

dx =1√A

∫ ∞−∞

e−y2

dy =

√π

A. (108)

∫ ∞−∞

e−(Ax2+Bx+C)dx =

∫ ∞−∞

e−A(x2+BAx+C

A)dx =

∫ ∞−∞

e−A[(x+ B2A

)2+(CA− B2

4A2 )]dx

= e−A(CA− B2

4A2 )

∫ ∞−∞

e−A(x+ B2A

)2

dx = e(B2

4A−C)

√π

A. (109)

2. Integrals of Gaussians times x∫ ∞−∞

xe−Ax2

dx = 0.

∫ ∞−∞

xe−A(x+D)2

dx =

∫ ∞−∞

(y −D)e−Ay2

dy = −D√π

A.

∫ ∞−∞

xe−(Ax2+Bx+C)dx = e(B2

4A−C)

∫ ∞−∞

xe−A(x+ B2A

)2

dx = e(B2

4A−C)

(−B2A

)√π

A.

3. Integrals of Gaussians times x2∫ ∞−∞

x2e−x2

dx =

[−x2e−x

2

]∞−∞

+1

2

∫ ∞−∞

e−x2

dx =

√π

2.

∫ ∞−∞

x2e−Ax2

dx =

[−x2A

e−Ax2

]∞−∞

+1

2A

∫ ∞−∞

e−Ax2

dx =1

2A

√π

A.

71

∫ ∞−∞

x2e−(Ax2+Bx+C)dx = e(B2

4A−C)

∫ ∞−∞

x2e−A(x+ B2A

)2

dx = e(B2

4A−C)

∫ ∞−∞

(y − B

2A)2e−Ay

2

dy

= e(B2

4A−C)

[∫ ∞−∞

y2e−Ay2

dy −∫ ∞−∞

B

Aye−Ay

2

dy +

∫ ∞−∞

B2

4A2e−Ay

2

dy

]= e(B

2

4A−C)

(1

2A

√π

A+

B2

4A2

√π

A

).

A.2 Fourier Transform of the Gaussian

Fourier Transform of a Gaussian: Let g(x) = 1√2πσ

e−x2

2σ2 . Then, itsFourier Transform is

G(ω) = F 1√2πσ

e−x2

2σ2 =

∫ ∞−∞

1√2πσ

e−x2

2σ2 e−iωxdx

=

∫ ∞−∞

1√2πσ

e−( x√

2σ+ iωσ√

2)2

e−ω2σ2

2 dx

= e−ω2σ2

2

Note that σ is the spatial domain transforms to 1σ

in the spectral domain.

Fourier Transform of a DOG (Difference of Gaussians):

F 1√2πσ1

e− x2

2σ21 − 1√

2πσ2

e− x2

2σ22 = e−

ω2σ21

2 − e−ω2σ2

22

The peak of this distribution can be found by F ′(ω0) = 0, giving

σ21e−ω

20σ

21

2 = σ22e−ω

20σ

22

2 ,

which after simplification gives

ω20 = 2

lnσ22 − lnσ2

1

σ22 − σ2

1

.

If σ2 = cσ1 then

ω20 =

(2 ln c2

(c2 − 1)

)1

σ21

.

72

A.3 Integrals of the Gabor Filter

Proposition 8. For α > 0 and β > 0, we have∫ ∞0

e−αx2

cos(βx)dx =1

2

√π

αe−

β2

4α∫ ∞0

e−αx2

sin(βx)dx = e−β2

4α

∫ β2α

0

eαy2

dy =1√αe−

β2

4α

∫ β2√α

0

ey2

dy =1

2α

∫ β

0

e−(β2−y2)

4α dy.

Proof. The two integrals can be solved by computing∫ ∞0

e−αx2

e−iβxdx = e−β2

4α

∫ ∞0

e−α(x− iβ2α

)2

dx

where we assume α, β > 0. The above integral can be solved by a contour

Figure 34: Rectangle Integral Path in the complex plane, where the x-axis repre-sents the real part and y-axis represents the imaginary part.

integral∮ce−α(z− iβ

2α)2dz along the closed contour C defined in Figure 34. This

contour integral is 0 because there are no poles inside the contour:

0 =

∮c

e−α(z− iβ2α

)2

dz

=

∫ R

0

e−α(x−i β2α

)2

dx+

∫ β2α

0

e−α(R+iy−i β2α

)2

idy +

∫ 0

R

e−αx2

dx+

∫ 0

β2α

e−α(iy−i β2α

)2

idy.

73

Note that when R 7→ ∞,∫ β

2α

0e−α(R+iy−i β

2α)2idy = 0, and we get∫ ∞

0

e−α(x−i β2α

)2

dx =

∫ ∞0

e−αx2

dx+ i

∫ β2α

0

e−α(iy−i β2α

)2

dy

=1

2

√π

α+ i

∫ 0

− β2α

eαy2

dy,

=1

2

√π

α− i√

α

∫ β2√σ

0

ey2

dy.

A.4

Proposition 9.

F−1|ω|e−ω2σ2

2 =1

πσ2

[1− ξ

σ2

∫ ξ

0

e−(ξ2−ζ2)

2σ2 dζ

]=

1

πσ2

[1 +

∫ ξ

0

d

dξe−

(ξ2−ζ2)

2σ2 dζ

].(110)

Proof.

F−1|ω|e−ω2σ2

2 =1

2π

∫ ∞−∞|ω|e−

ω2σ2

2 eiωξdω

=1

2π

∫ ∞−∞|ω|e−

ω2σ2

2 cos(ωξ)dω +i

2π

∫ ∞−∞|ω|e−

ω2σ2

2 sin(ωξ)dω

=1

π

∫ ∞0

ωe−ω2σ2

2 cos(ωξ)dω

=

(−1

πσ2e−

ω2σ2

2 cos(ωξ)

)∣∣∣∞0− ξ

πσ2

∫ ∞0

e−ω2σ2

2 sin(ωξ)dω

=1

πσ2

(1− ξ

∫ ∞0

e−ω2σ2

2 sin(ωξ)dω

). (111)

Now, from Appendix(A.3),∫ ∞0

e−αω2

sin(βω)dω =1√αe−

β2

4α

∫ β2√α

0

eζ2

dζ. (112)

Using α = σ2

2and β = ξ in the above equation, we get∫ ∞

0

e−ω2σ2

2 sin(ωξ)dω =

√2

σe−

ξ2

2σ2

∫ ξ√2σ

0

eζ2

dζ =1

σ2

∫ ξ

0

e−(ξ2−ζ2)

2σ2 dζ. (113)

Substituting back in Equation(111), we have the result.

74

Proposition 10.

F−1iω|ω|e−ω2σ2

2 =1

πσ2

[− ξ

σ2+

∫ ξ

0

d2

dξ2e−

(ξ2−ζ2)

2σ2 dζ

]= − 1

πσ4

[ξ +

(1− ξ2

2σ2

)∫ ξ

0

e−(ξ2−ζ2)

2σ2 dζ

]. (114)

Proof. Approach1:


2 =d

dξF−1|ω|e−

ω2σ2

2

=1

πσ2

d

dξ

[1 +

∫ ξ

0

d

dξe−

(ξ2−ζ2)

2σ2 dζ

]=

1

πσ2

[− ξ

σ2+

∫ ξ

0

d2

dξ2e−

(ξ2−ζ2)

2σ2 dζ

].

Approach2(Direct):


2 =1

2π

∫ ∞−∞

iω|ω|e−ω2σ2

2 eiωξdω

=1

2π

∫ ∞−∞

iω|ω|e−ω2σ2

2 cos(ωξ)dξ − 1

2π

∫ ∞−∞

ω|ω|e−ω2σ2

2 sin(ωξ)dξ

= − 1

π

∫ ∞0

ω2e−ω2σ2

2 sin(ωξ)dξ

= − 1

π

[(−1

σ2e−

ω2σ2

2

)ω sin(ωξ)

∣∣∣∞0

+1

σ2

∫ ∞0

e−ω2σ2

2 (sin(ωξ) + ξω cos(ωξ))dω

]= − 1

πσ2

[∫ ∞0

e−ω2σ2

2 sin(ωξ)dω + ξ

∫ ∞0

ωe−ω2σ2

2 cos(ωξ)dω

]= − 1

πσ2

∫ ∞0

e−ω2σ2

2 sin(ωξ)dω

+ ξ

[(−1

σ2e−

ω2σ2

2

)cos(ωξ)

∣∣∣∞0− ξ

σ2

∫ ∞0

e−ω2σ2

2 sin(ωξ)dω

]= − 1

πσ2

[(1− ξ2

σ2

)∫ ∞0

e−ω2σ2

2 sin(ωξ)dω +ξ

σ2

]= − 1

πσ2

[(1− ξ2

σ2

)1

σ2

∫ ξ

0

e−(ξ2−ζ2)

2σ2 dζ +ξ

σ2

]=

1

πσ2

[− ξ

σ2+

∫ ξ

0

d2

dξ2e−

(ξ2−ζ2)

2σ2 dζ

].

75

Approach3:


2 = − 1

π

∫ ∞0

ω2e−ω2σ2

2 sin(ωξ)dξ

= − 1

π

d

dξ

∫ ∞0

ωe−ω2σ2

2 cos(ωξ)dξ

=1

π

d

dξ

∫ ∞0

ωe−ω2σ2

2 cos(ωξ)dω

=1

π

d2

dξ2

∫ ∞0

e−ω2σ2

2 sin(ωξ)dω

=1

π

d2

dξ2

[1

σ2

∫ ξ

0

e−(ξ2−ζ2)

2σ2 dζ

]=

1

πσ2

d

dξ

[1 +

∫ ξ

0

d

dξ(e−

(ξ2−ζ2)

2σ2 )dζ

]=

1

πσ2

[−ξσ2

+

∫ ξ

0

d2

dξ2(e−

(ξ2−ζ2)

2σ2 )dζ

].

Proposition 11.

F−1ω2|ω|e−ω2σ2

2 =1

πσ2

[2

σ2− ξ2

σ4−∫ ξ

0

d3

dξ3e−

(ξ2−ζ2)

2σ2 dζ

]. (115)

Proof.

F−1ω2|ω|e−ω2σ2

2 = − d

dξF−1iω|ω|e−

ω2σ2

2

= − 1

πσ2

d

dξ

[−ξσ2

+

∫ ξ

0

d2

dξ2e−

(ξ2−ζ2)

2σ2 dζ

]= − 1

πσ2

[−1

σ2+

(−1

σ2+ξ2

σ4

)+

∫ ξ

0

d3

dξ3e−

(ξ2−ζ2)

2σ2 dζ

]=

1

πσ2

[2

σ2− ξ2

σ4−∫ ξ

0

d3

dξ3e−

(ξ2−ζ2)

2σ2 dζ

].

76

B Integrals Involving Characteristic Function

of Circular Domains

B.1

Proposition 12. The Fourier Transform of

f(x) =

√A2 − x2 x ≤ A

0 x > A.

is

F(ω) =πA

ω

(sin(ωA)

(ωA)2− cos(ωA)

ωA

)Proof.

F(ω) =

∫ A

−A

√A2 − x2e−iωxdx = 2

∫ A

0

√A2 − x2 cos(ωx)dx (116)

It is well-known that [7] [1]∫ u

0

(u2 − x2)v− 1

2 cos(ax)dx =

√π

2

(2u

a

)vΓ(v +

1

2)Jv(au)

for[a > 0, u > 0, Re(v) > −1

2

], where Γ(z) is the gamma function

Γ(z) =

∫ ∞0

tz−1e−tdt, Γ(3

2) =

√π

2, Γ(

1

2) =√π

and Jv(z) is the Bessel function of first kind,

Jv(z) =zv

2v

∞∑k=0

(−1)kz2k

22kk!Γ(v + k + 1), J1(z) =

z

2

∞∑k=0

(−1)kz2k

22kk!Γ(k + 2)

for [| arg(z)| < π], or in closed form

Jv(z) = zv(−1

z

d

dz)v sin z

z, J1(z) =

sin z

z2− cos z

z.

Now, returning to Equation(116), we get

2

∫ A

0

(A2 − x2)12 cos(ωx)dx = 2

√π

2

(2A

ω

)Γ(

3

2)J1(ωA)

[ω > 0, A > 0, Re(1) >

−1

2

]=πA

ωJ1(ωA)

=πA

ω

(sin(ωA)

(ωA)2− cos(ωA)

ωA

).

77

B.2

Proposition 13.

f(x) = F−1|ω|πRω

(sin(ωR)

(ωR)2− cos(ωR)

ωR

)

= 1 +x

2Rln|x−R||x+R|

, R > 0

Proof. Using Maple8

f(x) =−1

4R[−xπcsgn((x+R)i)csgn(x+R)i+ xπcsgn((x−R)i)csgn(x−R)i+

x ln(xi+Ri)− x ln(xi−Ri)− 4R + x ln(−xi−Ri)− x ln(−xi+Ri)] ,

where

csgn(x) =

1 if Re(x) > 0 or (Re(x) = 0 and Im(x) > 0)

− 1 if Re(x) < 0 or (Re(x) = 0 and Im(x) < 0).

First, when x < −R, x+R < 0 and x−R < 0. Using ln(i) = πi2

, we have

f(x) =−1

4R

[−xπ(−1)(−1)i+ xπ(−1)(−1)i+ x(ln(−x−R)− πi

2)− x(ln(R− x)− πi

2)

−4R + x(ln(−x−R) +πi

2)− x(ln(R− x) +

πi

2)

]= 1 +

x

2Rln

R− x−(R + x)

.

Second, when −R < x < R we have x+R > 0 and x−R < 0, and

f(x) =−1

4R

[−xπ(1)(1)i+ xπ(−1)(−1)i+ x(ln(x+R) +

πi

2)− x(ln(R− x)− πi

2)

−4R + x(ln(x+R)− πi

2)− x(ln(R− x) +

πi

2)

]= 1 +

x

2RlnR− xR + x

.

Finally, when R < x, x+R > 0 and x−R > 0, which gives

f(x) =−1

4R

[−xπ(1)(1)i+ xπ(1)(1)i+ x(ln(x+R) +

πi

2)− x(ln(x−R) +

πi

2)

−4R + x(ln(x+R)− πi

2)− x(ln(x−R)− πi

2)

]= 1 +

x

2Rlnx−RR + x

.

78

These cases can be combined to give the desired result.

C Some Standard Fourier Formulas

In the following, f(x) and F (ω) are considered as Fourier Transform pairs,i.e., F (ω) = F [f(x)].

1.

F ddxf(x) = iωF (f(x)). (117)

2. Let

F (ω) =

1, |ω| ≤ Ω0, |ω| > Ω

(118)

Then,

f(x) =sin(Ωx)

πx=

Ω

πsinc(Ωx). (119)

3. Let

F (ω) =

|ω|, |ω| ≤ Ω0, |ω| > Ω

(120)

Then,

f(x) =Ω2

π[sinc(Ωx)− 1

2sinc2(

Ωx

2)]. (121)

4. Let2

HBK(ω) =

ω sin

(π

2Ωω), |ω| ≤ Ω

0, |ω| > Ω(122)

2This function is not one near omega 0 which is problematic?

79

Then, the spatial domain Shepp-Logan filter is

hBK(x) =1

2π

∫ ∞−∞

HBK(ω)eiωxdω (123)

=1

2π

∫ Ω

−Ω

ω sin( π

2Ωω)eiωxdω (124)

=1

π

∫ Ω

0

ω sin( π

2Ωω)

cos(ωx)dω (125)

=1

π

d

dx

[∫ Ω

0

sin( π

2Ωω)

sin(ωx)dω

](126)

=1

2π

d

dx

[∫ Ω

0

cos(( π

2Ω− x)ω)dω −

∫ Ω

0

cos(( π

2Ω+ x)ω)dω

](127)

=1

2π

d

dx

sin((

π2Ω− x)ω)(

π2Ω− x)

Ω

0

−sin((

π2Ω

+ x)ω)(

π2Ω

+ x)

Ω

0

(128)

=1

2π

d

dx

[sin(π2− xΩ

)(π

2Ω− x) −

sin(π2

+ xΩ)(

π2Ω

+ x) ]

(129)

=1

2π

d

dx

[cos(Ωx)(π

2Ω− x) − cos(Ωx)(

π2Ω

+ x)] (130)

=1

2π

d

dx

cos(Ωx)

[1(

π2Ω− x) − 1(

π2Ω

+ x)] (131)

=−Ω sin(Ωx)

2π

[1(

π2Ω− x) − 1(

π2Ω

+ x)]+

cos(Ωx)

2π

[1(

π2Ω− x)2 +

1(π

2Ω+ x)2

](132)

=−Ω sin(Ωx)

π

xπ2

4Ω2 − x2+

cos(Ωx)

π

π2

4Ω2 + x2(π2

4Ω2 − x2)2 . (133)

D A Numerically stable method to find an

extremum

Consider the problem of finding the extremum of a function f(x), which byelementary calculus requires

f′(x0) = 0

f′′(x0) > 0

for a minimum, and

f′(x0) = 0

f′′(x0) < 0

for a maximum.

80

(a)

(b)

Figure 35: Plots of (a)HBK(ω) and (b)hBK(x) for Ω = (1, 2, 3).

Since finding zeros of a function is not as stable as finding a zero-crossing,we write this in the equivalent form

f′(x0 + ε) > 0

f′(x0 − ε) < 0

for a minimum, and

f′(x0 + ε) < 0

f′(x0 − ε) > 0

for a maximum,

where ε is a sufficiently small number. This can be further expanded asf(x0 + ε) > f(x0)

f(x0 − ε) > f(x0)for a minimum, and

f(x0 + ε) < f(x0)

f(x0 − ε) < f(x0)for a maximum.

In summary form, let

ρε(x0) =1

2[sgn(f(x0 + ε)− f(x0)) + sgn(f(x0 − ε)− f(x0))].

Then, in the limit as ε → 0, ρε(x0) = 1 if and only if x0 is a minimum andρε(x0) = −1 if and only if x0 is a maximum.

E Solving A sin θ + B cos θ = C

Proposition 14. Solutions of A sin θ +B cos θ + C satisfy

sin θ =AC ±B

√A2 + +B2 − C2

A2 +B2. (134)

81

Proof. Write A sin θ +B cos θ + C as

(A sin θ − C)2 = B2 cos2 θ

= B2(1− sin2 θ).

Thus(A2 +B2) sin2 θ − 2AC sin θ + (C2 −B2) = 0,

a quadratic equation in sin θ giving solutions

sin θ =AC ±

√A2C2 − (A2 +B2)(C2 −B2)

A2 +B2,

or

sin θ =AC ±B

√A2 +B2 − C2

A2 +B2. (135)

Proposition 15. Solution of

A sin θ +B cos θ = C

satisfy

θ = sin−1

(C√

A2 +B2− α

),

when α is defined as

(cosα, sinα) =(A,B)√A2 +B2

. (136)

Proof. We have

√A2 +B2 cosα sin θ +

√A2 +B2 sin θ cos θ = C,

which gives

sin(θ + α) =C√

A2 +B2,

and

θ = sin−1

(C√

A2 +B2

)− α.

82

F Π-line membership

Theorem 11. [4, 5] Every point P (x, y, z) = P (r cosφ, r sinφ, z) in a helixcylinder belongs to one and only one π-line segment, bounded by s1(P ) ands2(P ), as given by solving Equations 137.

Proof. Consider a π-line segment bounded by Γ1 = Γ(s1) and Γ2 = Γ(s2)where by definition |s2 − s1| < 2π. Points on this line segment are therefore

of the form (1− t)−−→OΓ1 + t

−−→OΓ2, t ∈ [0, 1], which can be written as

(1− t)−−→OΓ1 + t

−−→OΓ2 =

(R[(1− t) cos s1 + t cos s2], R[(1− t) sin s1 + t sin s2],

h

2π[(1− t)s1 + ts2]

).

Now, solving for P (x, y, z) = (1− t)−−→OΓ1 + t

−−→OΓ2, we have

x

R= (1− t) cos s1 + t cos s2

y

R= (1− t) sin s1 + t sin s2

2πz

h= (1− t)s1 + ts2,

(137)

which is a system of three equations in three unknowns s1, s2 and t. Thecomputation will be easier if we describe the point P (x, y, z) in cylindricalcoordinates as P (x, y, z) = (r cosφ, r sinφ, z). Then

r

Rcosφ = (1− t) cos s1 + t cos s2

r

Rsinφ = (1− t) sin s1 + t sin s2

2πz

h= (1− t)s1 + ts2.

(138)

The first two equations can be used to write s1 and t in terms of s2. Thethird equation can then be used to solve for s2. First, we can solve for t bysquaring the cos s1 and sin s1 in each term and adding them up to one[ r

Rcosφ− t cos s2

]2

+[ rR

sinφ− t sin s2

]2

= (1− t)2,

orr2

R2+ t2 − 2t

r

Rcos(φ− s2) = t2 − 2t+ 1,

which gives

t(s2) =1− r2

R2

2(1− r

Rcos(φ− s2)

) . (139)

83

Second, s1 can be solved in terms of s2 by substituting t into the first ofEquations 138. However, it is simpler to combine the first two equationsdirectly.

r

R(cosφ sin s2 − sinφ cos s2) = (1− t)(cos s1 sin s2 − sin s2 cos s2)

which givesr

Rsin(s2 − φ) = (1− t) sin(s2 − s1),

so that

sin(s2 − s1) =r

R(1− t)sin(s2 − φ)

=rR

1− 1− r2

R2

2(1− rR

cos(φ−s2))

sin(s2 − φ)

=2 rR

(1− r

Rcos(φ− s2)

)r2

R2 + 1− 2rR

cos(φ− s2)sin(s2 − φ)

= 2rR− r cos(φ− s2)

r2 +R2 − 2rR cos(φ− s2)sin(s2 − φ). (140)

i.e.

s1(s2) = s2 − sin−1

[2r

R− r cos(φ− s2)

r2 +R2 − 2r cos(φ− s2)sin(s2 − φ)

]. (141)

Third, the last equation in 138 can now be used to determine s2 by substi-tuting Equation 139 for t and Equation 141 for s1. However, this tends toa rather complex expression from which the existance and uniqueness of asolution is difficult to infer. Defrise, Noo and Kudo [5] followed a differentapproach. Consider the family of π-segments indexed by s2 that go throughthe line parallel to the z-axis going through P , i.e., when s1 and t satisfy thefirst two equations of 138. In this case, z is determined by s2 via the thirdequation.

z(x, y, s2) =h

2π[(1− t(x, y, s2))s1(x, y, s2) + t(x, y, s2)s2]. (142)

Observe that lims2→±∞ z = ±∞. Then if ∂z∂s2

is positive, z(x, y, s2) = z0 hasone and only one solution. We have

∂z

∂s2

=h

2π

[− ∂t

∂s2

s1 + (1− t)∂s1

∂s2

+∂t

∂s2

s2 + t

]=

h

2π

[t+ (s2 − s1)

∂t

∂s2

+ (1− t)∂s1

∂s2

].

84

Differentiating the first two equations in Equation 138 gives a linear systemof equations in ∂t

∂s2and ∂s1

∂s2.

− ∂t

∂s2

cos s1 − (1− t) sin s1∂s1

∂s2

+∂t

∂s2

cos s2 − t sin s2 = 0

− ∂t

∂s2

sin s1 + (1− t) cos s1∂s1

∂s2

+∂t

∂s2

sin s2 + t cos s2 = 0.

This is simplified as(cos s2 − cos s1)

∂t

∂s2

− (1− t) sin s1∂s1

∂s2

= t sin s2

(sin s2 − sin s1)∂t

∂s2

+ (1− t) cos s1∂s1

∂s2

= −t cos s2.

(143)

By eliminating ∂s1∂s2

, we get

(cos(s2 − s1)− 1)∂t

∂s2

= t sin(s2 − s1),

or∂t

∂s2

=t sin(s2 − s1)

(cos(s2 − s1)− 1).

Observe thatsin θ

cos θ − 1=

2 sin θ2

cos θ2

1− 2 sin2 θ2− 1

= − cotθ

2,

which simplifies the expression for ∂t∂s2

as

∂t

∂s2

= −t cot

(s2 − s1

2

). (144)

Similarly, by eliminating ∂t∂s2

from 143

(1− t)(cos(s2 − s1)− 1)∂s1

∂s2

= −t+ t cos(s2 − s1)

which gives∂s1

∂s2

=t

1− t. (145)

Thus,

∂z

∂s2

=h

2π

[t− (s2 − s1)t cot

(s2 − s1

2

)+ (1− t) t

1− t

]=

ht

π

[1− s2 − s1

2cot

(s2 − s1

2

)](146)

85

Observe that the function f(θ) = 1− θ cot θ is always positive for 0 < θ < π(Figure 36) so that for 0 < s2 − s1 < 2π, the derivative ∂z

∂s2> 0. Since the

function z(x, y, s2) is continuous and monotonic going from (−∞,∞), theequation z(x, y, s2) = z0 has one and only one solution in s2 and then in(t(s2), s1(s2), s2) when Eqautions 139 and 141 are taken into account.

Figure 36: The plot of 1− θ cot θ

Alternatively, we can eliminate t directly from the first two eqautions ofEquation 138 by first rewriting

r

Rcosφ = cos s1 + t(cos s2 − cos s1)

r

Rsinφ = sin s1 + t(sin s2 − sin s1)

and then eliminating t

r

R[cosφ(sin s2 − sin s1)− sinφ(cos s2 − cos s1)] = cos s1(sin s2 − sin s1)− sin s1(cos s2 − cos s1),

which can be organized in two ways. First, we can write

r

R[sin(s2 − φ)− sin(s1 − φ)] = sin(s2 − s1),

which is an implicit equation for solving s1. Second, an explicit formula canbe obtained by writing(

cos s2 −r

Rcosφ

)sin s1 −

(sin s2 −

r

Rsinφ

)cos s1 = − r

Rsin(s2 − φ).

86

This equation is the form of

A sin s1 +B cos s1 = C

or can be solved as indicated in Appendix E, Equation 135, giving

sin s1 =AC ±B

√A2 +B2 − C2

A2 +B2.

In this case

A2 +B2 =(

cos s2 −r

Rcosφ

)2

+(− sin s2 +

r

Rsinφ

)2

= 1 +r2

R2− 2r

Rcos(s2 − φ),

and

A2 +B2 − C2 = 1 +r2

R2− 2r

Rcos(s2 − φ)− r2

R2sin2(s2 − φ)

= 1 +r2

R2cos2(s2 − φ)− 2

r

Rcos(s2 − φ)

=[1− r

Rcos(s2 − φ)

]2

.

Thus,

sin s1 =− rR

(cos s2 − r

Rcosφ

)sin(s2 − φ)∓

(sin s2 − r

Rsinφ

) (1− r

Rcos(s2 − φ)

)1 + r2

R2 − 2 rR

cos(s2 − φ)

=∓ sin s2±

rR

sinφ− rR

[cos s2 sin(s2−φ)∓sin s2 cos(s2−φ)]+ r2

R2 [cosφ sin(s2−φ)∓sinφ cos(s2−φ)]

1+ r2

R2−2 rR

cos(s2−φ)

Taking the plus solution first

sin s1 =− sin s2 + r

Rsinφ− r

Rsin(−φ) + r2

R2 sin(s2 − 2φ)

1 + r2

R2 − 2 rR

cos(s2 − φ)

=r2

R2 sin(s2 − 2φ)− sin s2 + 2rR

sinφ

1 + r2

R2 − 2 rR

cos(s2 − φ).

Taking the minus solution next

sin s1 =sin s2 − r

Rsinφ− r

Rsin(2s2 − φ) + r2

R2 sin s2

1 + r2

R2 − 2 rR

cos(s2 − φ)

=

(r2

R2 + 1)

sin s2 − rR

sin(s2 − φ)− rR

sinφ

1 + r2

R2 − 2 rR

cos(s2 − φ).

87

Alternatively, let

(cosα, sinα) =

(rR

cosφ− cos s2,rR

sinφ− sin s2

)√r2

R2 + 1− 2 rR

cos(s2 − φ). (147)

Then

sin(s1 − α) =− rR

sin(s2 − φ)√1 + r2

R2 − 2 rR

cos(s2 − φ)

so that

s1(s2) = α− sin−1

rR

sin(s2 − φ)√1 + r2

R2 − 2 rR

cos(s2 − φ)

.To be continued.

F.1 References not included in the paper yet

[16]

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