Hindawi Publishing Corporation
International Journal of Mathematics and Mathematical Sciences
Volume 2010, Article ID 915958, 18 pages
doi:10.1155/2010/915958
Research Article
Geometric Sensitivity of a Pinhole Collimator
Howard Jacobowitz1 and Scott D. Metzler2
1
2
Department of Mathematical Sciences, Rutgers University, Camden, NJ 08102, USA
Department of Radiology, University of Pennsylvania, Philadelphia, PA 19104, USA
Correspondence should be addressed to Howard Jacobowitz, jacobowi@camden.rutgers.edu
Received 25 August 2009; Revised 7 December 2009; Accepted 19 February 2010
Academic Editor: Harvinder S. Sidhu
Copyright q 2010 H. Jacobowitz and S. D. Metzler. This is an open access article distributed under
the Creative Commons Attribution License, which permits unrestricted use, distribution, and
reproduction in any medium, provided the original work is properly cited.
Geometric sensitivity for single photon emission computerized tomography SPECT is given by
a double integral over the detection plane. It would be useful to be able to explicitly evaluate this
quantity. This paper shows that the inner integral can be evaluated in the situation where there
is no gamma ray penetration of the material surrounding the pinhole aperture. This is done by
converting the integral to an integral in the complex plane and using Cauchy’s theorem to replace
it by one which can be evaluated in terms of elliptic functions.
1. Introduction
Nuclear-medicine imaging provides images that assess how the body is functioning 1,
2, as opposed to anatomical modalities e.g., X-ray computed tomography, commonly
known as a CT or “CAT” scans that provide little or no information about function,
but great detail of the body’s structure. Nuclear-medicine images the biodistribution of
radiolabeled molecules that are typically intravenously injected into the patient in tracer i.e.,
nonpharmacological quantities. The compounds may have different biochemical properties
that affect the biodistribution and, hence, the choice of pharmaceuticals used to assess the
disease state of a patient.
Two common nuclear-medicine techniques are single photon emission computed
tomography SPECT and positron emission tomography PET. Molecules labeled with a
SPECT tracer emit a single photon. PET tracers emit a positron, which annihilates with a
nearby electron to produce two nearly back-to-back photons at 511 keV, the rest energy of
an electron. The line of these two photons contains the emission point of the positron. Many
photon pairs are detected in coincidence and reconstructed into a three-dimensional 3D
image of the tracer’s distribution.
Since SPECT tracers emit only one photon, the detection of the photon itself gives
little information about the location of the emission since neither the direction nor origin
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International Journal of Mathematics and Mathematical Sciences
Photon source
h
0, 0, 0
α
θ
Z0
θa
ΔL
α
d
ρ cos β, ρ sin β, 0
z
Photon path
y
x
Figure 1: Perspective View of the Knife-edge Pinhole Collimator. A point source at z h and angle θ emits
a photon that intersects the z 0 plane at ρ cos β, ρ sin β, 0 with an incident angle θa . The acceptance
angle of the collimator is α and the diameter of the aperture is d. ΔL is the path length of the photon in the
c 2001 IEEE.
attenuating medium. Reproduced from 3 with permission of the emission is known. Consequently, SPECT uses collimation that allows only photons
along certain lines to be detected. The origin is still unknown, but the 3D image may be
reconstructed when photons from numerous angles are detected 4, 5. Pinhole collimation
is often used when a small organ e.g., the thyroid is imaged by a large detector with a
SPECT tracer. Figure 1 is a diagram of the apparatus. It is not hard to see that the pinhole
collimation’s geometric efficiency, namely, the fraction of emitted photons that pass through
the circular aperture of the collimator, is approximately given by
g
d2 sin3 θ
,
16h2
1.1
where d is the diameter of the pinhole, h is the perpendicular distance, and θ is the incidence
angle of the photon on the aperture plane at the center of aperture. This formula becomes
exact in the limit that d goes to zero. Also, an exact formula is easily obtained when θ π/2
2 −1/2
d2
d
1 d 2
1 1
π 1 1
−
1
1−
≈ −
.
g θ
2
2 2
2h
2 2
2 2h
16h2
1.2
We derive stronger versions of these two results in the next section. But the main goal of this
paper is to simplify the integral representation for the geometric sensitivity without taking
International Journal of Mathematics and Mathematical Sciences
3
d or θ − π/2 to be small. We are concerned with the case where there is no penetration in
the sense that not all gamma rays are stopped by the attenuating material surrounding the
pinhole aperture. We take θ /
π/2. The geometric sensitivity, as defined above, is given by
the integral over the aperture of the flux times the sine of the incidence angle at that point on
the aperture:
g
1
4πh2
d/2
2π
ρ dρ
0
1.3
dβ sin3 θa ,
0
where ρ and β are polar coordinates on the aperture plane and θa is the incidence angle of the
photon at a particular value of ρ and β, and is given by 3
ρ2
ρ
cot2 θa cot2 θ − 2 cotθ cos β − φ 2 ,
h
h
1.4
where φ is the third coordinate of the point source. For integrals over 2π the difference β − φ
can be replaced with only β without loss of generality. This yields
ρ2
ρ
sin θa csc θ − 2 cotθ cos β 2
h
h
2
2
−1
1.5
.
Applying this to 1.3,
1
g
4πh2
d/2
2π
ρ dρ
0
0
ρ2
ρ
dβ csc θ − 2 cotθ cos β 2
h
h
−3/2
2
.
1.6
e−μΔL ,
1.7
When there is penetration the above would be modified to
1
g
4πh2
2π
∞
ρ dρ
0
0
ρ2
ρ
dβ csc θ − 2 cotθ cos β 2
h
h
−3/2
2
where the path length through attenuating material, ΔL, is zero for ρ < d/2 and is otherwise
given by 11 in 3.
Thus, 1.7 can be rewritten as
1
g gNP 4πh2
∞
2π
ρ dρ
d/2
0
ρ2
ρ
dβ csc θ − 2 cotθ cos β 2
h
h
2
−3/2
e−μΔL
1.8
gNP gPEN ,
where gPEN is the efficiency due to penetration and gNP , the efficiency under the assumption
of no penetration, is given by 1.6.
In this paper we use complex variable methods to calculate the β integral in 1.6.
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International Journal of Mathematics and Mathematical Sciences
2. Two Approximations for gNP
The special form of gNP makes it easy to derive approximations for d/h small and for θ − π/2
small which generalize 1.1 and 1.2. Before doing this, we observe that only even powers
of d and of θ − π/2 can occur in the expansion of gNP . We see this by writing the integral as
the sum of integrals from 0 to π and from π to 2π.
2.1. The Expansion for d Small
Let us, in this section, denote gNP by g. Upon substituting r d/2h and η ρ/h into 1.6 we
obtain
gr, θ 1
4π
1
4π
2π r 0
csc2 θ − 2ηcotθ cos β η2
η dη dβ,
0
2π r
0
−3/2
ηφ β, η, θ dη dβ.
2.1
0
Clearly, g0, θ 0. Since
gr r, θ 1
4π
2π
rφ β, r, θ dβ,
2.2
0
we have gr 0, θ 0. Differentiating 2.12 three more times with respect to r we obtain
1 2π rφr β, 0, θ φ β, r, θ dβ,
4π 0
1 2π rφrr β, 0, θ 2φr β, r, θ dβ,
grrr r, θ 4π 0
1 2π rφrrr β, 0, θ 3φrr β, r, θ dβ,
grrrr r, θ 4π 0
grr r, θ 2.3
and then setting r 0 in each of the resulting equations we obtain
grr 0, θ grrr 0, θ 1
4π
1
2π
3
grrrr 0, θ 4π
2π
φ β, 0, θ dβ,
0
2π
φr β, 0, θ dβ,
0
2π
0
φrr β, 0, θ dβ.
2.4
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5
For convenience let us write
2.5
v cotθ,
u csc2 θ,
so φ becomes
−3/2
φ β, r, θ u − 2vr cos β r 2
.
2.6
φ β, 0, θ u−3/2 ,
φr β, 0, θ 3u−5/2 v cos β,
φrr β, 0, θ 15u−7/2 v2 cos2 β − 3u−5/2 .
2.7
Then
We substitute into the Taylor expansion
gr, θ g0, θ gr 0, θr 1
1
1
grr 0, θr 2 grrr 0, θr 3 grrrr 0, θr 4 · · ·
2!
3!
4!
2.8
and obtain
1
gr, θ 2
1
4π
u
−3/2
0
1
4π
1
4!
2π
1
dβ r 3!
2
3
2π
2π
−5/2
6u
v cos β dβ r 3
0
2π −7/2 2
2
−5/2
dβ r 4 · · · .
3 15u
v cos β − 3u
2.9
0
Since
2π
0
cos β dβ 0,
2π
cos2 β dβ π
2.10
0
and u−1 sin2 θ we obtain
gr, θ 3
r2 3
sin θ 5sin5 θcos2 θ − 2sin5 θ r 4 · · · .
4
32
2.11
Finally we replace cos2 θ by 1 − sin2 θ and r by d/2h and obtain
gNP
2
d 4
1
d
3
3
5
2
sin θ
sin θ 3 − 5sin θ
O d6 .
16
h
512
h
2.12
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International Journal of Mathematics and Mathematical Sciences
2.2. The Expansion for θ − π/2 Small
To start we let
−3/2
f θ, η, β csc2 θ − 2ηcotθ cos β η2
2.13
and compute its Taylor expansion at θ π/2,
−3/2
−5/2
π
f θ, η, β 1 η2
− 3 1 η2
η cos β θ −
2
−7/2
−5/2 π 2
1
2
2
2
2
15 1 η
η cos β − 3 1 η
θ−
2!
2
3
π
O θ−
.
2
2.14
Since
gNP 1
4π
2π d/2h
0
f θ, η, β η dη dβ
2.15
0
and since we know that only even powers of θ − π/2 occur, we obtain
⎛
gNP
1⎜
⎝1 − 2
⎞
π 2
π 4
d/2h2
⎟ 3
θ
−
O
θ
−
.
⎠− 5/2
8
2
2
1 d/2h2
1 d/2h2
1
2.16
3. The Contour Integral
We have
I
2π
0
ρ2
ρ
dβ csc θ − 2 cotθ cos β 2
h
h
2
gNP 1
4πh2
−3/2
,
d/2
Iρ dρ.
3.1
3.2
0
I may be recast as a contour integral by setting z eiβ :
−3/2
dz −b 2
a
z −2 z
1
b
Γ iz 2z
3/2 1
2
z − x0 x1 − z −3/2 dz
,
b
i Γ
z
z
I
3.3
International Journal of Mathematics and Mathematical Sciences
7
where
Γ is the unit circle,
a csc2 θ ρ2
,
h2
ρ
b 2 cotθ,
h
⎤
⎡
a⎣
b2 ⎦
x0 1− 1− 2 ,
b
a
3.4
⎤
⎡
a⎣
b2 ⎦
x1 1
1− 2 .
b
a
We note that a > b and so 0 < x0 < 1 < x1 .
In this section we take ρ, θ, and h to be constant. We write
1/2 z
z
1
Fz z
z − x0 x1 − z z − x0 x1 − z
1/2
z
1
,
z − x0 x1 − z z − x0 x1 − z
3/2 2
1
I
Fzdz.
b
i Γ
3.5
We need to be careful because of the square roots. We use the usual convention that
lim± −1 iy ±i.
y→0
3.6
We wish to show that Fz is holomorphic and single valued in
{z : |z| < 1} − {x : 0 < x < x0 }.
3.7
To isolate the square root factor in F we set
gz z
.
z − x0 x1 − z
3.8
We take |z| ≤ 1. If in addition Re z < 0 and | Im z| |z − x0 | then Re z − x0 x1 − z < 0. This
implies that gz never takes on negative real values and so has a single valued square root in
Re z < 0. Similarly, if |z| ≤ 1, and Re z > x0 no condition on Im z now then Re z−x0 x1 −z >
0 and gz has a single valued square root in Re z > x0 .
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International Journal of Mathematics and Mathematical Sciences
ε
2δ
0
z0
Figure 2: The curve γ.
It follows from Cauchy’s theorem that the integral of Fz over the unit circle Γ is equal
to the integral over the “ barbell” γ around 0 and x0 . More explicitly, γ is the boundary of
B B 0 ∪ B x0 ∪ R,
3.9
where B x is the ball of radius and centered at x and R is the rectangle
R z x iy : 0 ≤ x ≤ x0 , −δ ≤ y ≤ δ
3.10
with δ , see Figure 2.
The curve γ decomposes into smooth pieces which we describe using the angle α arcsinδ/
γr z : |z − x0 | , −π α ≤ arg z − x0 ≤ π − α ,
γl z : |z| , α ≤ argz ≤ 2π − α ,
3.11
γ± {z : z x ± iδ, cos α ≤ x ≤ x0 − cos α}.
To bound the integral over γl we set z eiθ
2π−α iθ
iθ
F e e dθ
Fzdz α
γl
O 3/2 .
To evaluate
γr
3.12
Fzdz we set z x0 eiθ . Then
F x0 eiθ C−3/2 e−3/2iθ O −1/2
3.13
with
C
x01/2
x1 − x0 3/2
,
3.14
International Journal of Mathematics and Mathematical Sciences
9
and so
Fzdz γr
π−α −π
α
−3/2 Ce−3/2iθ O −1/2 ieiθ 1 Odθ
π−α
−π
α
Ci
−1/2 −1/2iθ
e
O 1/2
3.15
dθ.
Since α O, we may set α 0 and absorb the error in the O1/2 term. We then do the
integration and obtain
4iC
Fzdz √ O 1/2 .
γr
3.16
We want to compute the integral of F over γ
. We need to be careful about the square root.
Recall that we are assuming
cos α < x < x0 − cos α,
δ .
3.17
Lemma 3.1. On γ
Fz ix0 − xx1 − x−3/2 x1/2 Oδ
3.18
Fz −ix0 − xx1 − x−3/2 x1/2 Oδ.
3.19
and on γ−
Proof. This is a consequence of the branching of the square root function. Let
Hz z
zz − x0 x1 − z
,
z − x0 x1 − z
|z − x0 |2 |x1 − z|2
3.20
hz zz − x0 x1 − z.
We see that
hx iδ xx − x0 x1 − x iδ x2 − 1 O δ2 .
3.21
Recall that δ is replaced by −δ on γ− and that 0 < x < x0 < x1 . So Re Hz < 0 on both γ
and
γ− while Im Hz < 0 on γ
and Im Hz > 0 on γ− . Thus on γ
1/2
x
Oδ
x − x0 x1 − x 1/2
x
−i
Oδ,
x0 − xx1 − x
Hz1/2 −i
3.22
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International Journal of Mathematics and Mathematical Sciences
while on γ−
Hz1/2 i
x
x0 − xx1 − x
1/2
Oδ.
3.23
Since
1
Hz1/2 Oδ
x − x0 x1 − x
Fz Fx Oδ 1
Hz1/2 Oδ
x0 − xx1 − x
1/2 x
1
Oδ,
−i
−
x0 − xx1 − x
x0 − xx1 − x
−
3.24
the lemma follows.
We also need to be careful about the orientation. The curve γ is oriented counterclockwise. This means that the integration over γ
goes in the direction of x0 to 0. We need to
reverse its sign in order to write it in the conventional way
Fzdz −
x0 − cos α
Fx iδdx
cos α
γ
−
x0 − cos α
cos α
3.25
ix1/2 dx
x0 − x3/2 x1 − x3/2
O.
We let x x0 t and Δ x0−1 cos α and use δ to obtain
Fzdz −
1−Δ
γ
ix0 3/2 t1/2 dx
x0 − x0 t3/2 x1 − x0 t3/2
Δ
O.
3.26
Now let
τ
x1
> 1.
x0
3.27
We have
Fzdz −
γ
i
x0
3/2
1−Δ
Δ
t1/2 dt
τ − t3/2 1 − t3/2
O.
3.28
International Journal of Mathematics and Mathematical Sciences
11
We may let Δ → 0 in the lower limit, but not in the upper limit. So we now have
Fzdz −
γ
i
1−Δ
x0 3/2
0
t1/2 dt
τ − t3/2 1 − t3/2
O.
3.29
Here Δ x0 −1 cos α. We may let δ → 0 and redefine Δ to be x0 −1 .
We get the same value for the integral over γ− . This is because the integrand is the
negative of that in 3.29 but the orientation is reversed and so the two negatives cancel.
For any function gt, continuously differentiable on the interval 0, 1, we have
1−Δ
0
gt
1 − t
dt 2
3/2
1
g1
−
2g0
−
2
1 − t−1/2 g tdt O Δ1/2
1/2
Δ
0
3.30
as can easily be seen by an integration by parts.
We apply this to gt t1/2 /τ − t3/2 as in 3.29. This yields
Fzdz γ
−2i
τ −
13/2 x03/2 Δ1/2
1
i
x03/2
t−1/2 τ − t
−3/2
−5/2
3t1/2 τ − t
dt
3.31
O 1/2 .
0
We now take twice this quantity and add to it the results of 3.14 and 3.16,
|z|1
Fzdz Fzdz
γ
√
−2i x0
4iC
√ 2
√
x1 − x0 −3/2 1
i
x0 3/2
3.32
1 − t−1/2 τ − t−3/2 t−1/2 3t1/2 τ − t−5/2 dt
0
O 1/2 .
We substitute the value for C from 3.14 and let → 0
|z|1
Fzdz 2i
x0
3/2
1
0
1 − t−1/2 τ − t−3/2 t−1/2 3t1/2 τ − t−5/2 dt.
3.33
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International Journal of Mathematics and Mathematical Sciences
Finally, we evaluate this integral using Mathematica and obtain
|z|1
Fzdz 2i
x0
3/2
22τE1/τ − −1 τK1/τ
,
√
τ − 12 τ
3.34
where K and E are the complete elliptic functions of the first and second kinds
Km 1
1 − mt2
−1/2 1 − t2
−1/2
dt,
0
Em 1
1 − mt2
1/2 1 − t2
−1/2
3.35
dt.
0
We now have
I
3/2 1
2
Fzdz
b
i Γ
3/2
2
2 22τE1/τ − τ − 1K1/τ
√
b
x0 3/2
τ − 12 τ
3.36
3/2
4τ 1/4
2
2τE1/τ − τ − 1K1/τ,
b
τ − 12
gNP 1
4πh2
d/2
3.37
ρI dρ
0
with
ρ
b 2 cotθ,
h
τ
a csc2 θ ρ2
,
h2
3.38
1
x1
,
x0 x0 2
⎤
⎡
a⎣
b2 ⎦
x0 1− 1− 2 ,
b
a
⎤
⎡
a⎣
b2 ⎦
1
1− 2 .
x1 b
a
4. Limiting Cases
We validate 3.36 by considering two limiting cases.
3.39
International Journal of Mathematics and Mathematical Sciences
13
4.1. Expansion for Small d
We look for an expansion of gNP under the assumption that d/h is small and rederive 2.12.
The dependence of x0 on ρ is given in 3.39. The two-term Taylor expansion of x0 as a
function of ρ is
x0 Aρ Bρ3 O ρ5
4.1
with
A
cos θ sin θ
,
h
B A3 −
A
.
h2 csc2 θ
4.2
Since
τ
1
,
x0 2
4.3
we have
1
A2 ρ2 O ρ4 ,
τ
1
τ − 1
2
1
2
3 O ρ8 .
2
τ
τ
4.4
4.5
It is convenient to work with the quantity ρ2 τ, so we note, using 4.1,
ρ2 τ 1
2B 2
−
ρ
O
ρ4 .
A2 A3
4.6
We start with
I
3/2
4τ 1/4
2
1
1
2τE
−
−
1K
.
τ
2
b
τ
τ
τ − 1
4.7
Substituting the expression for b from 3.38 and for τ − 12 from 4.5 we obtain, after some
manipulation
I
h
cotθ
3/2
4
1
4
2 3/4 τ O ρ
ρ τ
1
1
− τ − 1K
.
2τE
τ
τ
4.8
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International Journal of Mathematics and Mathematical Sciences
We now use the expansions as given by Mathematica
π πt
−
O t2 ,
2
8
π πt
O t2 .
Kt 2
8
Et 4.9
So
I
h
cotθ
3/2
4
2 3/4
ρ τ
π
π
π
2 8τ τ
O ρ4 .
4.10
From the expression for ρ2 τ in 4.6, we derive
1
3B 2
3/2
ρ O ρ4 ,
1
2 3/4 A
2A
ρ τ
4.11
3/2
π
h
π
π 3Bρ2 π
3/2
I4
O ρ4 .
A
cotθ
2 8τ τ
4A
4.12
and this leads to
We now substitute for A and B using 4.2 and also for τ −1 using 4.4
ρ2
3π I sin θ 2π 3sin2 θ − 5sin4 θ 2
2
h
3
O ρ4 .
4.13
Substituting this into 3.37 and integrating, we again derive 2.12.
4.2. An Expansion for θ Near π/2
We show how 4.7 also gives the expansion of Section 2.2. From 3.38 we see that
ρ2 π 2
π 4
θ
−
O
θ
−
,
2
2
h2
2ρ π
π 3
θ−
O θ−
.
b−
h
2
2
a1
4.14
4.15
International Journal of Mathematics and Mathematical Sciences
15
So from 3.39 we have
b
x0 2a
b2
π 3
.
1
2 O θ−
2
4a
4.16
Thus
4a2
π 2
−
2
O
θ
−
,
2
b2
π 4
1
1
2
O
θ
−
1
,
τ
2
τ − 12 τ 2
1
1
π 9π
π 4
1
O θ−
2τE
− τ − 1K
.
τ
τ
τ
2
8τ
2
τ
4.17
4.18
These expansions let us write 4.7 as
I
3/2
2
π 9π
π 4
4
2
O
θ
−
1
b
τ
2
8τ
2
τ 3/4
π 9π
π 4
2
2
O
θ
−
1
.
3/4
τ
2
8τ
2
b2 τ
4.19
7/2
Using 4.15 and 4.17 we obtain a substitute for 4.1
1
b2 τ
−3/4 −3/2
3/4
4
a
3b2
1
2
8a
π 4
O θ−
.
2
4.20
This gives, in light of 4.14,
I 2πa−3/2 15 2 −7/2
π 4
πb a
O θ−
8
2
ρ2
2π 1 2
h
15 ρ2
π 2
2 h
−3/2
ρ2
1
2
h
ρ2
− 3π 1 2
h
−7/2
θ−
−5/2
θ−
π 2
2
π 2
π 4
O θ−
.
2
2
4.21
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International Journal of Mathematics and Mathematical Sciences
Finally, we integrate term by term.
g
1
4πh2
d/2
ρI ρ dρ
0
⎛
⎛
⎞
2 −1/2
d
1 ⎝
2⎝
⎠
2πh 1 − 1 2h
4πh2
⎞
⎛
2 −3/2 2
d
1
⎠ θ− π
− 3πh2 ⎝1 − 1 3
2h
2
⎛
⎞
⎞
2 −5/2 2
π
2 5d/2h2
d
15 2 ⎝ 2
⎠ θ−
⎠
πh
−
1
2
15
15
2h
2
4.22
π 4
O θ−
2
⎞
⎛
2 −1/2
2 −5/2 2 3
π 2
d
d
1⎝
d
⎠−
θ−
1− 1
1
2
2h
8
2h
2h
2
π 4
O θ−
.
2
5. Numerical Confirmation
The result in 3.36 is further validated in this section with numerical integration of 3.1.
In this original formulation, the integral depends on the ratio of ρ/h and on the value of
θ. We analytically considered one limiting case for each of these quantities in the previous
section. We consider these and other values for these quantities in this section using numerical
methods. For fixed values of ρ/h, the numerical integration solid lines is shown with the
results of the residue calculation open circles in Figure 3 as a function of θ. Figure 4 shows
the numerical integration and residue results for fixed values of θ as a function of ρ/h. All
corresponding values of the residue calculation and the numerical integration agree.
6. Conclusions
The accuracy of collimator efficiency is important for design and for quantitative reconstruction programs. Analytic forms obviate the need for numerical models, which do not
readily offer insight into the interplay of collimator parameters. This paper approaches the
problem by recasting one of the integrals in terms of contour integration. This integration
was completed, but the authors were unable to execute the second integral to yield a
complete analytic solution. The result was studied for special cases with known outcomes
for validation. Future efforts may be able to build on this validated result to find a complete
analytic solution.
International Journal of Mathematics and Mathematical Sciences
17
Integrals versus θ
7
Integrals
6
5
4
3
2
π/4
π/2
θ
ρ/h 0
ρ/h 0.2
ρ/h 0.4
ρ/h 0.6
ρ/h 0.8
ρ/h 1
Figure 3: Numerical integration of 3.1 solid lines and residue calculation of 3.36 open circles as a
function of θ for fixed values of ρ/h: 0.0, 0.2, 0.4, 0.6, 0.8, and 1.0.
Integrals versus ρ/h
7
Integrals
6
5
4
3
2
0
0.2
0.4
0.6
0.8
1
ρ/h
θ 50 deg
θ 60 deg
θ 70 deg
θ 80 deg
θ 90 deg
Figure 4: Numerical integration of 3.1 solid lines and residue calculation of 3.36 open circles as a
function of ρ/h for fixed values of θ: 50, 60, 70, 80, and 90 degrees.
Acknowledgment
Scott D. Metzler was supported by the National Institute for Biomedical Imaging and
Bioengineering of the National Institutes of Health under Grant R01-EB-6558.
18
International Journal of Mathematics and Mathematical Sciences
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