BMME 560 & BME 590I Medical Imaging: X-ray, CT, and Nuclear Methods Tomography Part 2.

BMME 560 & BME 590IMedical Imaging: X-ray, CT, and

Nuclear Methods

Tomography Part 2

Today

• Tomography– Radon Transform– Fourier Slice Theorem– Direct Fourier Reconstruction– Simple Backprojection

Tomographic Reconstruction

• Define terms

f(x,y)p(t,q)

t

t

y

q

x

s

The object

The projection

“Object space” or “image space”

“Projection space”

The “rotated frame”


• The problem

f(x,y)p(t,q)

t

t

y

q

x

s

Given p(t,q) for 0<q<pFind f(x,y)

The Radon Transform

• Specifies the 1D projection of the 2D function f(x,y) at any angle q

( , ) ( , ) ( cos sin )p t f x y x y t dxdy

Nonzero only along the line of projection at t

Radon Transform Example

( , ) ( , ) ( cos sin )p t x a y x y t dxdy

( , ) ( , )f x y x a y

Both delta functions have to be nonzero for the integral to be one. This occurs when the following conditions are met:

0

0

cos sin

x a x a

y

t x y

Therefore, the transform is nonzero only when

cos So, ( , ) ( cos )t a p t t a

Radon Transform Example2 2 2( , ) 1 for f x y x y r

Note that you can make use of the circular symmetry here.

The delta function selects x = t since it is nonzero and integrates to one only if this is true.

Now set the limits of integration because the function is one only if t2 + y2 < r2

We could have also reached this by reasoning that the projection for any t would be the length of that chord in the circle.

Because of the circular symmetry, all projections will be the same. We can arbitrarily choose q = 0.

Sinograms

image sinogram

t

0

2p

q

Sinograms

image sinogram

t

0

2p

q

Sinograms

image sinogram

t

0

2p

q

Sinogram

• Here is a sinogram. What does the object look like?


f(x,y)p(t,q)

t

t

y

q

x

s

Recall that

( , ) ( , )p t f s t ds


f(x,y)

p(t,q)t

ty

qx

s

It turns out that there is a relationship between the Fourier Transforms of f(x,y) and p(t,q)

( , ) ( , )p t f s t ds


f(x,y)

p(t,q)t

ty

qx

s

Take the FT of p with respect to a frequency variable wt

( , ) ( , ) ( , ) tj ttP p t f s t ds e dt

Reorder the integrals

( , ) ( , ) tj ttP f s t e ds dt


f(x,y)

p(t,q)t

ty

qx

s

Look at the 2DFT of f(x,y)

( cos sin )( , ) ( , ) tj x ytP f x y e dx dy

( )( , ) ( , ) j ux vyF u v f x y e dx dy

These are the same if the following is true:

cos

sint

t

u

v

The 1DFT of the projection equals the 2DFT of the image at certain spatial frequencies.


f(x,y)p(y)

yy

x

These are equal if .

Consider if we project along x:The 2D FT of our object:

The projection of our object:

The 1D FT of the projection of our object:


f(x,y)

p(t,q)t

ty

qx

s

Therefore

( , ) ( cos , sin )t t tP F u v

This is called the Fourier Slice Theorem or the Projection Slice Theorem


Look at this in the Fourier Space


u

v

q

This line is described by the parametric equations on wt

wt

So, each projection angle gives us information from one line in the 2D FT space.


• Does this suggest to you a method for reconstruction from projections?


u

v

q

wt


• Direct Fourier Reconstruction works, but is generally not done. What is its drawback?


u

v

q

wt


• This does tell us which angles we need for a full reconstruction. What is the rule?


u

v

q

wt


• Also, this proves that the Radon transform is an invertible transform (in the limit).


u

v

q

wt

Simple Backprojection

• Let’s try something else. What if we just cast the projections back across the image field?

1 2

3 4

4 6

3

7

4 6

3

7


• A mathematical expression for simple backprojection

0

ˆ ( , ) ( cos sin , )sbpf x y p t x y d

Sum over the set of angles

The projection that intersects the given location of the image space

The estimate of the true image


• Replace p by its 1D Fourier transform expression

• Which gives

0

ˆ ( , ) ( , ) tj tsbp t tf x y P e d d

( , ) ( , ) tj tt tp t P e d

With t related to x, y, and q as before


• Rewrite in terms of x and y since t = ?

• This tells something about the property of our simple backprojection estimate in the frequency space. But what?

( cos sin )

0

ˆ ( , ) ( , ) tj x ysbp t tf x y P e d d


• Any 2D function can be written in terms of its Fourier transform in polar coordinates

• This looks similar to our expression for the simple backprojection estimate.

( cos sin )

0

( , ) ( , ) rj x yr r rg x y G e d d


• Compare these two– 2D FT of any function

– Simple backprojection estimate

( cos sin )

0

( , ) ( , ) rj x yr r rg x y G e d d

( cos sin )

0



• According to the Fourier slice theorem,

• BUT,

1ˆ ( , ) ( , )sbp tf x y P

1( , ) ( , )tf x y P


• So, what is the simple backprojection estimate?

• Put it in the form of a 2D FT in polar coordinates

( cos sin )

0


( cos sin )

0

( , )ˆ ( , ) tj x ytsbp t t

t

Pf x y e d d


• An LSI system model for projection followed by simple backprojection:

FourierTransform

1

t

filter

InverseFourier

Transform

ˆ ( , )sbpf x y( , )f x y

ProjectionSimple

backprojectionˆ ( , )sbpf x y( , )f x y ( , )p t


• The transfer function of that filter

, spatial frequencyt

resp

onse

2 2

1( )

1( , )

tt

H

H u vu v

Key Point

• Simple backprojection causes a loss of high-frequency details.– Drops off as the inverse of the spatial frequency


• Example

True image Simple backprojection


• Example



• Example


BMME 560 & BME 590I Medical Imaging: X-ray, CT, and Nuclear Methods Tomography Part 2.

Documents

y slide

y py y y x

t t y x s look

t t y x s recall

tomographic reconstruction

t t y x s given pt

projection slice theorem

sinograms imagesinogram