If F {f(x,y)} = F(u,v) or F( , ) then F( , ) = F { g (R) } The Fourier Transform of a projection...

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.

The Central Section Theorem is also referred to as the projection-slice theorem.

Central Section or Projection Slice Theorem

π ∞

∫ d ∫ F-1{F{g (R) } • |p| } ( x cos + y sin - R) dR0 -∞

CT Reconstruction Methods- Filtered Back Projection

Each projection is filtered to account for oversampling of lower spatial frequencies before back projection. Here filtering is done in the frequency domain

π ∞

∫ d ∫ [ g (R) * • c (R) ] ( x cos + y sin - R) dR0 -∞

Same idea as filtered back projection, but filtering is done in projection domain. Here 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

I0

(x,y)

I = I0 e - ∫ dl

ln (I0 / I) = ∫ dl

Assumptions - zero width pencil beam

- monoenergetic

Look at horizontal and vertical projection measure of attenuation at point P. How will bone affect the vertical projection?

Bone

•

CT Artifacts: Beam Hardening

P

Should get same answer for each projection.

Bone attenuates

Cupping artifact

Soft tissue cannot demonstrate its attenuation no photons to show it

Normal emitted energy spectrumHow does beam look after moving through bone tissue?

E

RelativeIntensity

First Generation CT Scanner

From Webb, Physics of Medical Imaging

Physics of Medical Physics, EditorWebb

Interpolation during Back Projection

Second Generation CT Scanner

From Webb, Physics of Medical Imaging

3rd and 4th Generation Scanners

From Webb, Physics of Medical Imaging

Electron Beam CT

Krestel-Imaging Systems for Medical Diagnosis

Krestel-Imaging Systems for Medical Diagnosis

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.

F(u,v)

u

v

Impulse Response

Impulse response h(x,y)F(u,v)

u

v

Let’s assume that the CT scanner acquires exact projections. Thenthe impulse response is the inverse 2D FT of the sampled pattern.

Inverse 2D FT

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

domainImpulse 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.

)(G u

uc

To avoid overlap (aliasing) with a sampling interval X,

cc uuX

1

cu2X

1

)(G u

uc

Nyquist Condition:Sampling rate must be greater than twice the frequency component.

X1

cu1

X

Apply to Imaging

F{g(x,y)}= G(u,v)

g(x,y) G(u,v)

If we sample g(x,y) at intervals of X in x and y, then G(u,v) replicates

u

v

1/X

1/X

G(u,v)^

Sampling in Frequency DomainIf we sample in the frequency domain, then the image will replicateat 1/frequency sampling interval.

Intersections depict sampling points. Let sampling interval infrequency domain be equal in each direction and be F. Then weexpect the images to replicate every 1/ F

G(u,v)F

1/ F = FOV

FOV

How many angles do we need?

…

F

Three angles acquired from a CT exam are shown, each acquired apart. Data acquisition is shown in the Fourierspace using the Central Section Theorem. The radial spacing, F, is the separation between the vertical hash marks on the horizontal projection. If we consider what the horizontalprojection will look like as it is back projected, we can appreciate that 1/ F = Field of View.

How many angles do we need?

…

F

To avoid aliasing of any spatial frequencies, the spacing of thepoints in all directions must be no larger than FThe largestspacing between points will come along the circumference. If we scale the radius of the circle so it has radius 1, thecircumference will be 2However, we only need projectionsto sample a distance of since one projection will providetwo points on the circle.

How many angles do we need?

…

F

To avoid aliasing of any spatial frequencies, the spacing of thepoints in all directions must be no larger than FThe largestspacing between points will come along the circumference. If we scale the radius of the circle so it has radius 1, thecircumference will be 2However, we only need projectionsto sample a distance of since one projection will providetwo points on the circle.

How many angles do we need?

…

Fora radius of 1 or diameter of 2, F 2/N where N is the numberof detectors. That is, there are N hash marks ( samples) on the projection above.The spacing of points on the circumference will then be /M where M is the number of projections. So the azimuthal spacing on the circumference, /M,must be equal or smaller to the radial spacing 2/N. Solving for M gives M= /2 * N

Azimuthal Sampling requires M angles = π/2 Ndetector elements

Azimuthal Undersamplinga) Sampling Pattern - frequency space or projections acquired

b) Plot of one line of 2D impulse responsec) 2D impulse response with undersampling

d) Image of a Square

What causes artifacts? Where do artifacts appear?