1. Central Section Theorem
If F{f(x,y)} = F(u,v) or F(,)
then F(,) = F{ g (R) }
The Fourier Transform of a projection at angle  is a line in the Fourier transform of the image at the same angle.

2. Filtered Back Projection
π ∞
∫ d ∫ F-1{F{g (R) } • |p| }  ( x cos  + y sin  - R) dR ∞
Symbols in equation

3. Convolution Back Projection
π ∞
∫ d ∫ [ g (R) * • c (R) ]  ( x cos  + y sin  - R) dR ∞
Each projection is convolved with c (R) and then back projected.
Describe c (R)
C (p) = |p|
c (R) = lim 2 (2 - 4π2R2) / (2 + 4π2R2)2
 0

4. Beam Hardening I0 I = I0 e - ∫  dl ln (I0 / I) = ∫  dl (x,y)
Assumptions - zero width pencil beam
- monoenergetic
(x,y)
Bone •

5. Should get same answer for each projection.
Relative
Intensity
Normal emitted energy spectrum
How does beam look after moving
through bone tissue? E
Bone attenuates
Cupping artifact
Soft tissue cannot demonstrate its attenuation  no photons to show it

6. Krestel-
Imaging Systems for Medical Diagnosis

7. First Generation CT Scanner
From Webb, Physics of Medical Imaging

8. Physics of Medical Physics, EditorWebb

9. Interpolation during Back Projection

10. Krestel-
Imaging Systems for Medical Diagnosis

11. Krestel-
Imaging Systems for Medical Diagnosis

12. Sampling Requirements in CT
How many angles must we acquire?
Samples acquired by each projection are shown below as they fill the F(u,v) space. v u
F(u,v)

13. Impulse Response
Let’s assume that the CT scanner acquires exact projections. Then
the impulse response is the inverse 2D FT of the sampled pattern. v u
Inverse 2D FT
F(u,v)
Impulse response h(x,y)

14. Impulse response Bright white line points to correctly imaged impulse.
Inside of White circle shows region that is correctly imaged
Region outside of white circle suffers streak artifacts.
How big is the white circle?
How does it relate to the number of angles?
Easier to think of in frequency domain
Impulse response h(x,y)

15. Recall: Band limiting and Aliasing
F{g(x)} = G(u)
G(u) is band limited to uc, (cutoff frequency)
Thus G(u) = 0 for |u| > uc. uc
To avoid overlap (aliasing) with a sampling interval X, uc
Nyquist Condition:
Sampling rate must be greater than twice the frequency component.

16. Apply to Imaging F{g(x,y)}= G(u,v) G(u,v) g(x,y)
If we sample g(x,y) at intervals of X in x and y, then G(u,v) replicates v
1/X ^
G(u,v) u
1/X

17. Sampling in Frequency Domain
If we sample in the frequency domain, then the image will replicate
at 1/frequency sampling interval.
G(u,v) DF
Intersections depict sampling points. Let sampling interval in
frequency domain be equal in each direction and be DF. Then we
expect the images to replicate every 1/ DF

18. 1/ DF = FOV
FOV

19. How many angles do we need?Dq …
DFr
Three angles acquired from a CT exam are shown, each
acquired Dq apart. Data acquisition is shown in the Fourier
space using the Central Section Theorem. The radial spacing, DFr ,
is the separation between the vertical hash marks on the
horizontal projection. If we consider what the horizontal
projection will look like as it is back projected, we can appreciate
that 1/ DFr = Field of View.

20. How many angles do we need?Dq …
DFr
To avoid aliasing of any spatial frequencies, the spacing of the
points in all directions must be no larger than DFr . The largest
spacing between points will come along the circumference.
If we scale the radius of the circle so it has radius 1, the
circumference will be 2p. However, we only need projections
to sample a distance of p since one projection will provide
two points on the circle.

21. How many angles do we need?Dq …
DFr
To avoid aliasing of any spatial frequencies, the spacing of the
points in all directions must be no larger than DFr . The largest
spacing between points will come along the circumference.
If we scale the radius of the circle so it has radius 1, the
circumference will be 2p. However, we only need projections
to sample a distance of p since one projection will provide
two points on the circle.

22. How many angles do we need?Dq …
For a radius of 1, DFr = 2/N where N is the number of detectors.
That is, there are N hash marks ( samples) on the projection above.
The spacing of points on the circumference will then be p/M where
M is the number of projections. So the azimuthal spacing
on the circumference, p/M,must be equal or smaller to the radial
spacing 2/N. Solving for M gives M= p/2 * N

23. Azimuthal Sampling requires M angles = π/2 Ndetector elements
Azimuthal Undersampling
Sampling Pattern - frequency space or projections acquired
One line of 2D impulse response
2D impulse response
Image of a Square
What causes artifacts? Where do artifacts appear?

24. Krestel-
Imaging Systems for Medical Diagnosis

25. Second Generation CT Scanner
From Webb, Physics of Medical Imaging

26. 3rd and 4th Generation Scanners
From Webb, Physics of Medical Imaging

27. Electron Beam CT