4B – Computed Tomography
Projection Profiles
The measured intensity at a detector is Id = I0exp
− d0 μ(s) ds
(1) from which we obtain the projection measurement gd:
gd = − ln Id I0 (2) = d 0 μ(s) ds (3)
- Eq. (3) states that the projection measurement is the integral of μ(s) along the ray path s
Radon transform
Suppose we are scanning a slice (the xy plane):
- the attenuation values at (x, y) are denoted by f (x, y), i.e., f (x, y) ≡ μ(x, y)
- we wish to obtain an expression for the projection profiles g(ρ, θ) in terms of f (x, y)
Consider the projection profile for θ = θk, denoted by g(ρ, θk): - the value at a point (ρj, θk) on this profile is
g(ρj, θk) = gd = − ln(Id/I0) = μ(s) ds (4) = f (x(s), y(s)) ds (5) =
f (x, y)δ(x cos θk+ y sin θk − ρj) dx dy (6) - Eq. 5 follows from the fact that
μ(s) ds is the line integral of μ(x, y) (f (x, y)) along the line L(ρj, θk) with
x(s) = ρjcos θk − s sin θk, y(s) = ρj sin θk + s cos θk
Eq. 6 gives the projection value g(ρj, θk): g(ρj, θk) =
f (x, y)δ(x cos θk + y sin θk − ρj) dx dy - the projection profile for this orientation θ = θk is g(ρ, θk):
g(ρ, θk) =
f (x, y)δ(x cos θk+ y sin θk − ρ) dx dy (7) - generalising to all values of θ to obtain g(ρ, θ):
g(ρ, θ) =
f (x, y)δ(x cos θ + y sin θ− ρ) dx dy (8) - this is the Radon transform, which may be denoted by
Example
General equation: g(ρ, θ) =
f (x, y)δ(x cos θ + y sin θ− ρ) dx dy Projection profile at θ = 30◦:
g(ρ, 30◦) =
f (x, y)δ(x cos 30◦ + y sin 30◦ − ρ) dx dy Projection value at ρ = 0, θ = 30◦
g(0, 30◦) =
f (x, y)δ(x cos 30◦ + y sin 30◦) dx dy Projection value at ρ = 5, θ = 30◦
g(5, 30◦) =
Properties of the Radon transform
g(ρ, θ) =
f (x, y)δ(x cos θ + y sin θ− ρ) dx dy (10)
g(ρ, θ) ≡ R{f(x, y)} or R{f} (11)
or
f (x, y) ←→ g(ρ, θ)R (12)
The Radon transform gives the full set of projection data for an image f (x, y). It can be shown that:
af (x, y) ↔ ag(ρ, θ) (13) f1(x, y) + f2(x, y) ↔ g1(ρ, θ) + g2(ρ, θ) (14) f (x− a, y − b) ↔ g(ρ − a cos θ − b sin θ, θ) (15) f (ax, ay) ↔ 1 a g(aρ, θ), a > 0 (16) δ(x, y) ↔ δ(ρ) (17) δ(x− a, y − b) ↔ δ(ρ − a cos θ − b sin θ) (18) e−(x2+y2) ↔ √πe−ρ2 (19)
Example f (x, y) = δ(x− 3, y − 4) δ(x, y) ←→ δ(ρ)R δ(x− 3, y − 4)} ←→ δ(ρ − 3 cos θ − 4 sin θ)R R ←→ δ(ρ − 5 cos(θ − 53◦)) g(ρ, θ) = δ(ρ − 5 cos(θ − 53◦)) g(ρ, 0◦) = δ(ρ− 5 cos(0◦ − 53◦)) = δ(ρ− 3.0) g(ρ, 20◦) = δ(ρ− 5 cos(20◦ − 53◦)) = δ(ρ− 4.2) Note:
Example
(a) Calculate g(1, 0◦)
g(ρ, θ) =
f (x, y)δ(x cos θ + y sin θ− ρ) dx dy g(1, 0◦) =
f (x, y)δ(x− 1) dx dy = 2
0 f (x(s), y(s)) ds
is the integral of f (x, y) along the line
L1 : x− 1 = 0, or x(s) = 1, y(s) = s Hence, g(1, 0◦) = 2 0 f (1, s) ds = 0.2× 2 = 0.4
(b) Calculate g(2, 30◦)
g(ρ, θ) = f (x, y)δ(x cos θ + y sin θ− ρ) dx dy g(2, 30◦) =
f (x, y)δ(x cos 30◦ + y sin 30◦ − 2) dx dy = 2/
√
3
−2/√3f (x(s)), y(s)) ds
is the integral of f (x, y) along the line
L2 : x cos 30◦ + y sin 30◦ − 2 = 0, or
x(s) = 2 cos 30◦ − s sin 30◦, y(s) = 2 sin 30◦ + s cos 30◦ Hence,
g(2, 30◦) = 0.2× (4/√3) = 0.46
(c) Sketch g(ρ, 30◦) g(ρ, 30◦) =
Example
Obtain the Radon transform of the disk f (x, y) = ⎧ ⎨ ⎩ A x2 + y2 ≤ R2 0 otherwise
The object is circularly symmetric; hence its projections for all θ are the same. We will just obtain the projection for θ = 0◦.
g(ρ, θ) = g(ρ, 0◦) = ∞ −∞ ∞ −∞f (x, y)δ(x cos 0 ◦ + y sin 0◦ − ρ) dx dy = ∞ −∞ ∞ −∞f (x, y)δ(x− ρ) dx dy =
f (x(s), y(s)) ds where x(s) = ρ, y(s) = s = f (ρ, s) ds For |ρ| ≤ R, g(ρ, θ) = y1 −y1A ds = 2Ay1 = 2AR2 − ρ2
In the general case (without restricting ρ), we can write g(ρ, θ) = 2AR2 − ρ2 rect(ρ/2R)
Sinograms
The projection data acquired for one CT slice may be displayed in the form of a sinogram:
- the sinogram is an image of g(ρ, θ) with (ρ, θ) as the coordinate axes
- it shows the data that are necessary to reconstruct f (x, y) - the horizontal axis represents ρ
- the vertical axis represents the projection angle, θ (0◦ ≤ θ ≤ 180◦) - the intensity of a point on the sinogram represents the projection
value
The Fourier-slice Theorem
We take the 1D Fourier transform of a projection with respect to ρ: G(ω, θ1) = g(ρ, θ1)e−j2πωρdρ (20) where ω denotes spatial frequency in an arbitrary direction. We sub-stitute the analytic expression for g(ρ, θ1), the Radon transform, given in Eq. (8):
G(ω, θ1) =
f (x, y)δ(x cos θ1 + y sin θ1 − ρ)dx dy
e−j2πωρdρ = f (x, y)
δ(x cos θ1 + y sin θ1 − ρ)e−j2πωρdρ
dx dy =
f (x, y)e−j2πω(x cos θ1+y sin θ1)dx dy (21)
=
f (x, y)e−j2π[(ω cos θ1)x+(ω sin θ1)y]dx dy (22)
= [F (u, v)]u=ω cos θ1,v=ω sin θ1 (23)
= F (ω cos θ1, ω sin θ1) (24)
F (ω cos θ1, ω sin θ1) are the values of F (u, v) taken along the line that passes through (u, v) = (0, 0) and making an angle θ1 to the x axis.
Generalising,
G(ω, θ) = F (ω cos θ, ω sin θ) (25)
Eq. (25) is known as the Fourier-slice theorem or the projection-slice theorem.
What is F (ω cos θ, ω sin θ)?
F (ω cos θ, ω sin θ) ⇒ u = ω cos θ, v = ω sin θ
⇒ v
u = tan θ
In other words, F (ω cos θ, ω sin θ) gives the values of F (u, v) along the line v = u tan θ:
Ilustration of the Fourier-slice Theorem From above:
g(ρ, θ) ↔ G(ω, θ)F
G(ω, θ) = F(ω cos θ, ω sin θ)
Given an image f (x, y) and its 2D Fourier transform F (u, v), the 1D Fourier transform of a projection taken at an angle θ, g(ρ, θ), is equal to a slice through the origin of F (u, v) at that same angle θ.
Example
Given the function f (x, y) = exp(−x2) exp(−2y2), use the Fourier slice theorem to obtain the projection g(ρ, 30◦).
We have the Fourier transform pair
exp(−x2) exp(−2y2) ↔ (π/√2) exp(−π2u2) exp(−π2v2/2) (26) The Fourier transform of the projection profile g(ρ, 30◦) is
F{g(ρ, 30◦)} = G(ω, 30◦)
= F (ω cos 30◦, ω sin 30◦) (from the Fourier slice theorem) = [F (u, v)]u=ω cos 30◦,v=ω sin 30◦
= (π/√2) exp −π2ω2cos230◦exp
⎛ ⎝−π2 2 ω 2sin230◦ ⎞ ⎠ = 2.22 exp −8.63 ω2 (27)
From Eq. (27), we obtain g(ρ, 30◦) as follows: g(ρ, 30◦) = F−1{G(ω, 30◦)}
= F−1{2.22 exp −8.63 ω2} = 1.34 exp(−1.14ρ2)
