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.

Filtered Back Projection π ∞ ∫ d ∫ F-1{F{g (R) } • |p| }  ( x cos  + y sin  - R) dR 0 -∞ Symbols in equation

Convolution Back Projection π ∞ ∫ d ∫ [ g (R) * • c (R) ]  ( x cos  + y sin  - R) dR 0 -∞ 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

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

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

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Interpolation during Back Projection

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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)

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)

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)

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.

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

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

1/ DF = FOV FOV

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.

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.

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.

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

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?

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