4C – Computed Tomography
Image Reconstruction
BackprojectionConcept
In backprojection, we assign the projection values to each point along the projection path. This backprojection is repeated for each projection ray and for all angles. The reconstructed image value for a particular point in image space is obtained from all the projections passing through that point.
Backprojection value at P : 2 + 2 + 2 = 6 Backprojection value at Q: 1 + 1 + 0 = 2
Backprojection value at R: 0.5 + 1.5 + 1.5 = 3.5
Formulation
Let g(ρ, θ) be the Radon transform of f (x, y) and consider g(ρ, θk), the
projection profile at θ = θk.
- Backprojecting a point g(ρj, θk)is done by assigning the value g(ρj, θk) to every point on the line L(ρj, θk)
- When we repeat this for all ρ, we obtain the backprojection image formed from the profile at angle θk:
bθk(x, y) = g(ρ, θk) (1)
= g(x cos θk + y sin θk, θk) (2) - In the general case,
bθ(x, y) = g(ρ, θ) = g(x cos θ + y sin θ, θ) (3)
- Given a projection profile g(ρ, θ), the corresponding backprojection image is bθ(x, y) = g(x cos θ + y sin θ, θ).
Example
Let g(ρ, θ) be the Radon transform of f (x, y). The projection profile for θ = 0 rad is
g(ρ, 0) = 10 exp[−(ρ/5)2] (4)
- The backprojection image is obtained from Eq. (1): b0(x, y) = g(ρ, 0)
= 10 exp[−(ρ/5)2]
= 10 exp[−(x cos 0 + y sin 0)/5)2]
= 10 exp[−(x/5)2] (5)
- Eq. (5) can be used to calculate the backprojection value b0(x, y) at
any point (x, y), e.g.,
b0(5, 5) = g(5, 0) = 10 exp[−(5/5)2] = 3.7 b0(2, 10) = g(2, 0) = 10 exp[−(2/5)2] = 8.5
Example
The projection profile of an image at θ = 20◦ is
g(ρ, 20◦) = (10− ρ) rect[(ρ − 5)/10], ρ = x cos 20◦ + y sin 20◦ (6) The backprojection image from this projection profile is
b20◦(x, y) = g(ρ, 20◦) (7)
= (10− ρ) rect[(ρ − 5)/10] (8)
= (10− x cos 20◦ − y sin 20◦) rect[x cos 20◦ + y sin 20◦ − 5)/10] (9) - The projection value at ρ = 8 is
g(8, 20◦) = (10− 8)rect[(8 − 5)/10] = 2
This value (2) is assigned to the points lying on line L(8, 20◦), i.e. {(x, y)|x cos 20◦ + y sin 20◦ = 8}
- We can also use Eqs. (7) - (9) to compute the value of any point in the backprojection image. Examples:
(a) At P (0, 0), b20◦(0, 0) = g(0, 20◦) = 10
(b) At Q(3, 3), ρ = 3 cos 20◦ + 3 sin 20◦ = 3.85. Hence b20◦(3, 3) = g(3.85, 20◦)
= (10− 3.85)rect[(3.85 − 5)/10] = 6.15
Example
The figure shows the backprojection image obtained from the profile g(ρ, 30◦):
b30◦(x, y) = g(ρ, 30◦) (10) = ⎧ ⎨ ⎩ 2√52 − ρ2 |ρ| ≤ 5 0 otherwise (11) = 252 − ρ2rect(ρ/10) (12)
where ρ = x cos 30◦+ y sin 30◦. What is b30◦(2, 3)?
(x, y) = (2, 3) ⇒ ρ = 2 cos 30◦ + 3 sin 30◦ = 3.232 b30◦(2, 3) = 2√52 − 3.2322 rect(3.232/10)
= 3.81
The backprojection image obtained from a profile at an angle θ is
bθ(x, y) = g(ρ, θ) = g(x cos θ + y sin θ, θ) (13) - The reconstructed image is obtained by summing (or integrating) the backprojection images from all the projection profiles to yield the back-projection summation image
ˆ
f (x, y) = π
0 bθ(x, y) dθ (14)
= π
0 g(x cos θ + y sin θ, θ) dθ (15)
- The backprojection summation image is also called a laminogram. - For the discrete implementation, we have
ˆ
f (x, y) =
θ bθi (16)
Summary:
Example
For the circle above,
g(ρ, θ) = 225− ρ2rect(ρ/10) bθ(x, y) = g(ρ, θ)
= 225− ρ2rect(ρ/10), ρ = x cos θ + y sin θ
= 225− (x cos θ + y sin θ)2rect [(x cos θ + y sin θ)/10] ˆ
f (x, y) = π
0 bθ(x, y) dθ
Backprojection image for a circular object
The reconstructed image of a circular object shows a “star” pattern, or in the limit of many angles, a blurred image.
- Reconstruction by simple backprojection does not give good results, and more sophisticated methods have to be employed.
- It can be shown that for f (x, y) = δ(x, y) (a delta function at the origin), the image obtained by backprojection is
h(x, y) = √ 1
x2 + y2 = 1 ρ This is the impulse response for backprojection.
- Thus, in general, the reconstructed image from backprojection is ˆ
f (x, y) = f (x, y) h(x, y) (17) = f (x, y) 1
ρ (18)
- This represents a poor reconstruction because of the inherent blurring.
Discrete implementation
- The backprojection process maps the value of a projection sample to a straight line in the reconstructed image.
- This ignores the discrete nature of the backprojection processing i.e., it assumes the projection is a 1D continuous function and the recon-structed image is a 2D continuous function.
- In practice, the projection profile is sampled by the detector into a discrete set of samples. Similarly, the reconstructed image is formed on a discrete matrix. For backprojection, interpolation is required. - For a pixel (P ), we follow the ray path that passes through the center
of the pixel and locate its intersection with the projection (A).
- Some method of interpolation (nearest neighbour, linear, etc) is used to estimate the projection value at location A.
The Fourier method
1. Take the 1D (discrete) Fourier transform of each projection.
2. Insert it with the corresponding correct angular orientation into the correct slice of the 2D Fourier plane.
3. Interpolate from the polar grid onto a Cartesian grid. 4. Take the inverse Fourier transform of the result.
f (x, y) → {g(ρ, θ)} → {G(ω, θ)} → ˆF (u, v) → ˆf (x, y)
The Fourier method (also called direct Fourier reconstruction) is not widely used in CT due to the inherent variable spatial resolution and the practical problem of interpolating from a polar to Cartesian grid.
Filtered backprojection
The inverse Fourier transform of F (u, v) f (x, y) = ∞
−∞
∞
−∞F (u, v)e
j2π(ux+vy)du dv (19)
can be written in polar coordinates as f (x, y) = 2π
0 ∞
0 F (ω cos θ, ω sin θ)e
j2πω(x cos θ+y sin θ)ω dω dθ (20)
Using the Fourier-slice theorem, Eq. (20) becomes f (x, y) = 2π
0
∞
0 G(ω, θ)e
j2πω(x cos θ+y sin θ)ω dω dθ
By splitting this integral into two expressions, one for θ in the range 0 to π, and the other in the range π to 2π, and using the fact that G(ω, θ + π) = G(−ω, θ), we obtain
f (x, y) = π
0 ∞
−∞|ω|G(ω, θ)e
j2πω(x cos θ+y sin θ)dω dθ (21)
= π
0
∞
−∞|ω|G(ω, θ)e
j2πωρdω dθ (22)
where ρ = x cos θ + y sin θ.
Eq. (22) sets out a method for obtaining the original image from the mea-sured projections g(ρ, θ): f (x, y) = π 0 ∞ −∞|ω|G(ω, θ)e j2πωρdω dθ (23) = π 0 ∞ −∞Gh(ω, θ)e j2πωρdω dθ (24) where Gh(ω, θ) = |ω|G(ω, θ) = π 0 gh(ρ, θ) dθ (25) where gh(ρ, θ) = ∞ −∞Gh(ω, θ)e j2πωρdω = π 0 bθ(x, y) dθ (26)
(Note that for the above equations, |ω| = 1 results in simple backprojec-tion.)
The procedure for obtaining the filtered backprojection image is: 1. Compute the 1D Fourier transform of each projection → G(ω, θ) 2. Multiply each Fourier transform by the filter function |ω| → Gh(ω, θ) 3. Obtain the inverse 1D Fourer transform of each resulting filtered
trans-form → gh(ρ, θ)
4. Backproject for each θ (by replacing ρ with x cos θ + y sin θ) → bθ(x, y)
5. Integrate (sum) all the backprojected values → ˆf (x, y)
Filtered backprojection: f (x, y) = π 0 ∞ −∞|ω|G(ω, θ)e j2πωρdω dθ
The function |ω| is a ramp filter which extends to ω = ±∞: - unrealizable in practice
- the range has to be limited to reduce susceptibility to image noise Image noise generally has high-frequency characteristics, i.e., the noise spectrum extends signficantly into the high-frequency region. A filter that amplifies the high-frequency portion of a noisy signal will boost the noise relative to the clean signal.
We use a window, W (ω) to bandlimit the ramp filter, resulting in the filter function
H(ω) = |ω| × W(ω)
A simple implementation of W (ω) is the rectangle function, i.e., W (ω) = rect ω 2ωc (27) where ωc is the cut-off frequency. This gives the Ram-Lak filter:
HRL(ω) = |ω|rect ω 2ωc (28) - undesirable ringing due to the sharp cut-off
- susceptible to noise due to amplification of high frequencies
Ringing due to sharp frequency cut-off
A better approach is to use a smooth window function. Filter Window W (ω) Ram-Lak rect ω 2ωc Shepp-Logan sinc ω 2ωc rect ω 2ωc Cosine cos πω 2ωc rect ω 2ωc Hamming 0.54 + 0.46 cos πω ωc rect ω 2ωc Hann 0.5 + 0.5 cos πω ωc rect ω 2ωc 21
Choice of filter bandwith, ωc:
- a compromise between a cleaner (less noisy) image and better spatial resolution
- greater bandwidth ⇒ more noise
- greater bandwidth ⇒ more high frequency signal content, and hence a sharper image
Example of filtering with different bandwidths
The reconstructed image is obtained as follows:
1. Compute the 1D Fourier transform of each projection. 2. Multiply each Fourier transform by |ω| × W(ω).
3. Obtain the inverse 1D Fourer transform of each resulting filtered trans-form.
4. Backproject for each θ (by replacing ρ with x cos θ + y sin θ). 5. Integrate (sum) all the backprojected values.
Note that
- The filtering step
∞
−∞|ω| W(ω) G(ω, θ)e
j2πωρdω
may be implemented as a convolution.
- Implementation of the reconstruction algorithm will be on a computer operating on digitized data to give a 3D digital image as the output.
Image quality
The filtered backprojection equation (22) is an exact formula for the inverse Radon transform. However, in practice we cannot recover the original function f (x, y) exactly. Some of the reasons for these are:
- the ideal ramp filter |ω| is not realizable - finite size of detectors
- limited number of detectors
- limited number of projection profiles - measurement errors and noise
Compared with projection radiography, CT has worse spatial resolution but better contrast resolution (the ability to reliably show subtle differences in contrast).
Fan beam reconstruction
Fan-beam source-to detector configurations are common in modern com-mercial scanners. The typical fan-beam geometry is shown in the figure. The source and detector rotate together around the rotational isocenter (the image origin). The reconstruction formulation will be different com-pared to parallel-ray projections.
Interative reconstruction
Another approach to image reconstruction is interative reconstruction. - This is a highly computationally intensive method.
- The unknowns are the pixel values in the image array: 512× 512 = 262, 144
- Known information is the entire set of measured projection data. - Basically solving a set of linear equations iteratively to obtain these
values.
- Widely used in nuclear medicine (PET, SPECT) because of smaller data sets but only recently introduced in commercial CT scanners. - Copes well with noisy data, which is more prominent in low-dose CT.
Image Display
One of the great advantages of CT is its ability to measure and distinguish quite small differences in attenuation coefficient. It is convenient to nor-malise to the attenuation coefficient of water, μw, giving a CT number, or Hounsfield unit (HU):
H = 1000(μ− μw)
μw (29)
Tissues with attenuation coefficients less than μw have H between 0 and
−1000. Tissues with higher attenuation coefficients have H between 0 and +3000.
The use of this standardized scale facilitates the comparison of CT values obtained from different CT scanners and with different X-ray beam energy spectra.
• Bone tissues have high attenuation coefficients, with H varying from 1000 to 3000.
• Soft tissues vary from about H = 20 to 80,
• Fat is less attenuating and varies from about H = −70 to H = −100. • Lungs are of low density because of the air, and H varies from about
−750 to −900.
Computer monitors usually display a range of 256 gray levels, with black at 0 and white at 255. The range of interest of the CT numbers is linearly mapped to the gray level range [0, 255]. This process of establishing a variable-width “display window” that be raised or lowered enhances the imaging of specific organs. For example, if an image of heart tissue is desired, this window is raised above water density, but if an image of the detail in the lung is desired, it is lowered.
Artifacts
AliasingThe projections will be aliased if they are undersampled. These artifacts will continue through the process of reconstruction and appear as artifacts in the CT image.
An insufficient number of projections also causes aliasing. These artifacts appear as streaks.
A general rule of thumb for designing CT systems is that the number of detectors should be approximately equal to the number of projections that should be approximately equal to the number of points on the side of a reconstructed image. A typical 3G system has around 700 detectors, acquires 1000 projections, and reconstructs a 512× 512 image.
Ring artifacts
A defective or miscalibrated detector element will result in a bright or dark ring centred at the rotation centre.
Beam hardening
- CT uses a polyenergetic x-ray beam and x-ray attenuation coefficients are energy dependent, with lower-energy x-rays more highly attenuated than higher-energy ones.
- As the x-ray bam propagates through the human body, the average energy of the x-ray beam becomes greater (“harder”), thus decreasing the effective attenuation.
- Because the attenuation of bone is greater than that of soft tissue, bone causes more beam hardening than an equivalent thickness of soft tissue.
Motion artifacts
Motion artifacts occur when the patient moves during the acquisition pro-cess. Small motions cause blurring, while larger displacements produce double images or image ghosting.
Partial volume effect
The partial volume effect (PVE) arises when more than one tissue type occurs in a voxel. In such cases, the voxel intensity depends on the propor-tions of each tissue type present in the voxel. Voxels, in general, contain a mixture of different tissue types, e.g., bone and soft tissue, and this leads to ill-defined edges in the CT image.
