1. Sample CT Image

2. X-ray CT 2 % attenuation change detectable in film
0.2 % change in attenuation coefficient y x
f(x,y)
CT, like X-ray measure line integrals
∑ ∑ ∑ ∑ ∑
g(R) = CT detector array output
g(R)

3. Direct Methods Radon Transform 1917
Central Section Theorem - Bracewell
The transform of each projection forms a line, at that angle, in the 2D FT of f(x,y)
g(R) = ∫ f(x,y) dy
f(x,y)
We will skip the Algebraic Reconstruction Technique
g(R)

4. First, let’s think of our experience on the meaning
of F(u,0) in the Fourier transform. v
F(u,0) u
{f(x,y)} => F (u,v) = ∫∫ f(x,y) e -i 2π (ux +vy) dx dy
F(u,0) = ∫ [ ∫ f(x,y) dy] e -i 2π ux dx
So F(u,0) is the Fourier transform of the projection formed
from line integrals along the y direction.

5. Computed Tomography Uses a collimator to keep exposure to a slice
Builds image from multiple projections
We will assume parallel rays for now.
Actually how first generation scanners worked.
Translate source and single detector across body for one angle
Then rotate source and detector to get next angle

6. Central Section Theorem
The 1D Fourier Transform of a projection at angle  forms a line in the 2D Fourier space of the image at the same angle.
Incident
X-rays y R  ø x R r
f(x,y)
Detected g (R)

7. Incident x-rays pass through the object f(x,y) from upper left to lower right at the angle 90 + . For each point R, a different line integral describes the result on the function g (R). g (R) is measured by an array of detectors or a moving detector.
The thick line is described by
x cos  + y sin  = R
I have just drawn one thick line to show one line integral, but the diagram is general and pertains to any value of R.

8. The projection g (R) can thus be calculated as a set of line integrals, each at a unique R.
g (R) = ∫ ∫ f(x,y)  (x cos  + y sin  - R) dx dy
2π ∞
g (R) = ∫ ∫ f(r, f)  (r cos ( - f ) - R) r dr df
In the second equation, we have translated to polar coordinates.
Again g (R) is a 1D function of R.

9. Let’s consider the 2D FFT
F(u,v) = ∫ ∫ f(x,y) e -i 2π (ux + vy) dx dy
In polar coordinates,
u =  cos  v =  sin 
F(,) = ∫ ∫ f(x,y) exp [ -i 2π  (x cos  + y sin )] dx dy
When x cos  + y sin  = constant, exp [ -i 2π  (x cos  + y sin )] is a linear phase shift. This is the Fourier transform of a shifted delta function. Let’s let the constant = R and write the complex exponential as the FT of a  function.
F(,) = ∫y ∫ xf(x,y) F[( x cos  + y sin  - R)] dx dy
F(,) = ∫y ∫x f(x,y) [∫ [( x cos  + y sin  - R)] e-i 2π rR dR] dx dy

10. F(,) = ∫ ∫ f(x,y) [∫ [( x cos  + y sin  - R)] e-i 2π rR dR] dx dy
Recall how we wrote the projection as a double integral of f(x,y)
where a delta function performs the line integral,
g (R) = ∫ ∫ f(x,y)  (x cos  + y sin  - R) dx dy
We take the Fourier Transform of g (R):
F{g (R)}= ∫R [ ∫y ∫x f(x,y) ( x cos  + y sin  - R) dx dy] e-i 2πrR dR
dxdy
which is exactly what we wrote for F(r, ) above!
Thus, F{g (R)} = F(r,  )

11. Central Section Or Projection Slice Theorem
F{g (R)} = F(r, )
So in words, the Fourier transform of a projection at angle  gives us
a line in the polar Fourier space at the same angle .

12. BATTLESHIP North/South Data East - West Data Battleship: New Rules
Old way
EW 2/ NS 3 Hit!
EW 5/ NS 6 Miss!
North/South Data
EW1
EW2
EW3
EW4
EW5
EW6
EW7
EW8
EW9
NS1
NS2
NS3
NS4
NS5
NS6
NS7
NS8
NS9
We won’t ask the other player if there is a battleship in a certain peg.
Write down how many squares you see a battleship along each column.
Then write down how many times you see a battleship along each row.
I started it for you.
East - West Data

13. BATTLESHIP 1 4 North/South Data East - West Data Battleship: New Rules
Tell the other player how
many times you
see a battleship in
a square along
each column.
I started it for you. Finish up the other gray squares on the bottom row.
North/South Data
EW1
EW2
EW3
EW4
EW5
EW6
EW7
EW8
EW9
NS1
NS2
NS3
NS4
NS5
NS6
NS7
NS8
NS9 1 4
We won’t ask the other player if there is a battleship in a certain peg.
Write down how many squares you see a battleship along each column.
Then write down how many times you see a battleship along each row.
I started it for you.
East - West Data

14. BATTLESHIP 1 4 North/South Data East - West Data Battleship: New Rules
Here it is finished.
North/South Data
EW1
EW2
EW3
EW4
EW5
EW6
EW7
EW8
EW9
NS1
NS2
NS3
NS4
NS5
NS6
NS7
NS8
NS9 1 4
We won’t ask the other player if there is a battleship in a certain peg.
Write down how many squares you see a battleship along each column.
Then write down how many times you see a battleship along each row.
I started it for you.
East - West Data

15. BATTLESHIP 1 4 North/South Data East - West Data If this is all your
opponent told you,
could you find where the battleships were?
North/South Data
EW1
EW2
EW3
EW4
EW5
EW6
EW7
EW8
EW9
NS1
NS2
NS3
NS4
NS5
NS6
NS7
NS8
NS9 1 4
We won’t ask the other player if there is a battleship in a certain peg.
Write down how many squares you see a battleship along each column.
Then write down how many times you see a battleship along each row.
I started it for you.
East - West Data

16. BATTLESHIP 1 4 North/South Data East - West Data Battleship: New Rules
What if your opponent also had to tell you the number of ships in squares along each row?
Complete the gray squares on the right side. I started it for you.
North/South Data
EW1
EW2
EW3
EW4
EW5
EW6
EW7
EW8
EW9
NS1
NS2 1
NS3 4
NS4
NS5
NS6
NS7
NS8
NS9
We won’t ask the other player if there is a battleship in a certain peg.
Write down how many squares you see a battleship along each column.
Then write down how many times you see a battleship along each row.
I started it for you.
East - West Data

17. BATTLESHIP 1 4 3 North/South Data East - West Data Here are all the
EW1
EW2
EW3
EW4
EW5
EW6
EW7
EW8
EW9
NS1 1
NS2
NS3 4
NS4
NS5
NS6
NS7 3
NS8
NS9
Here are all the
answers.
We won’t ask the other player if there is a battleship in a certain peg.
Write down how many squares you see a battleship along each column.
Then write down how many times you see a battleship along each row.
I started it for you.
East - West Data

18. BATTLESHIP 1 4 3 North/South Data East - West Data Can you find the
EW1
EW2
EW3
EW4
EW5
EW6
EW7
EW8
EW9
NS1 1 4
NS2
NS3
NS4
NS5
NS6
NS7 3
NS8
NS9
North/South Data
Can you find the
ships now if you just
knew the gray squares?
One idea:
Smear the East West
numbers all the way
up the columns.
This tells us that we
should not spend much
time looking along
EW4 or EW9.
But there is something
probably up in EW8
We won’t ask the other player if there is a battleship in a certain peg.
Write down how many squares you see a battleship along each column.
Then write down how many times you see a battleship along each row.
I started it for you.
East - West Data

19. BATTLESHIP 1 4 3 North/South Data East - West Data
Next, smear the North/ South numbers to the left and add them
to what was in the
grid before.
Where do you think the ships are?
By the biggest numbers?
Is this always true?
Let’s see.
EW1
EW2
EW3
EW4
EW5
EW6
EW7
EW8
EW9
NS1
1+1=2
4+1=5 1
NS2
NS3 4
NS4
NS5
NS6
NS7
3+1=4 3
NS8
NS9
We won’t ask the other player if there is a battleship in a certain peg.
Write down how many squares you see a battleship along each column.
Then write down how many times you see a battleship along each row.
I started it for you.
East - West Data

20. BATTLESHIP 2 1 5 4 8 3 7 North/South Data East - West Data
EW1
EW2
EW3
EW4
EW5
EW6
EW7
EW8
EW9
NS1 2 1 5
NS2
NS3 4 8
NS4
NS5
NS6
NS7 3 7
NS8
NS9
Here I finished smearing the north/south numbers to the left and adding them to the east/west numbers.
Where do you think the ships are?
By the biggest numbers?
Is this always true?
Let’s see.
We won’t ask the other player if there is a battleship in a certain peg.
Write down how many squares you see a battleship along each column.
Then write down how many times you see a battleship along each row.
I started it for you.
East - West Data

21. BATTLESHIP 2 1 5 4 8 3 7 North/South Data East - West Data
EW1
EW2
EW3
EW4
EW5
EW6
EW7
EW8
EW9
NS1 2 1 5
NS2
NS3 4 8
NS4
NS5
NS6
NS7 3 7
NS8
NS9
Next, smear the North/ South numbers to the left and add them
to what was in the
grid before.
Are the battleships
where the biggest numbers are?
All of the time?
Some of the time?
We won’t ask the other player if there is a battleship in a certain peg.
Write down how many squares you see a battleship along each column.
Then write down how many times you see a battleship along each row.
I started it for you.
East - West Data

22. BATTLESHIP What if we can measure along the diagonals? 1 1 1 1 1 1 2 13 4 5 6 7 8 9 A B C D E F G H I
What if
we can measure
along the
diagonals? 1 1 1
We won’t ask the other player if there is a battleship in a certain peg.
Write down how many squares you see a battleship along each column.
Then write down how many times you see a battleship along each row.
I started it for you. 1 1 1 2 1 1

23. BATTLESHIP Now add the diagonal information to our totals.
Are we doing any
better?
Are the battleships
where the biggest numbers are more
often? 1 2 3 4 5 6 7 8 9 A B C D E F G H I
We won’t ask the other player if there is a battleship in a certain peg.
Write down how many squares you see a battleship along each column.
Then write down how many times you see a battleship along each row.
I started it for you.

24. Reconstruction from Projections
Crude Idea 1: Take each projection and smear it back along the lines of integration it was calculated over.
Result from a back projection:
b (x,y) = ∫ g (R)  (x cos  + y sin  - R) dR
Adding up all the back projections from all the angles gives,
fback-projected (x,y) = ∫ b (x,y) d
π ∞
fb (x,y) = ∫ d ∫ g (R)  (x cos  + y sin  - R) dR
0 -∞

25. Let’s calculate an impulse response to see how the reconstruction does.
g (R) =  (R)
That is,  (x,y) causes a  (R) projection. By calculating the back-
projected image, fb (x,y), we will be calculating the impulse response.
hb (x,y) = fb (x,y) for this delta object
Recall x cos  + y sin  - R = r cos ( - f) - R
hb (r,) = ∫ d ∫  (R)  (r cos ( - f ) - R) dR
We can simplify the integration over R by realizing that the
first delta function will be non-zero only when R=0. Then we will
only have one integral across  π
= ∫  (r cos ( - f)) d

26. Now  [f(x)] = ∑  (x - xn) / |f’(xn)|
Only one root (zero) in range of integral
cos ( - f) = 0  - f = π/2
 = π/2 + f
Since denominator has to be evaluated at  = π/2 + f

27. Back-projected impulse response
hb(r) = 1/r
fb (x,y) = f(x,y) ** 1/r
Fb (,) = F (,) /  since F{1/r}= 1/ 

28. Back projected image is blurred by convolution with 1/r
Where intuitively does the 1/r come from?
We must account for this blurring to properly reconstruct the image.
How does 1/r convolution look in image space?
In frequency space?

29. Low spatial frequency data is overweighted
Low spatial frequency data is overweighted. Filter to compensate for this. Weighted by 1/.
Solution - filter each projection by || to account for the uneven sampling density
Steps:
Calculate projection
Transform projection
Weight with ||
Inverse transform
Back project
Add all angles
Mathematically,
∫ d ∫ F -1 { F{g (R)} || }  (x cos  + y sin  - R) dR
Solution

30. Filtered Back-Projection
The reconstruction described is known as filtered back-projection.
What is downside of the described filter, || ? 

31. Filtered Back-Projection
What is downside of this filter? 
Band limit filter to:
 0
Filter  rect ( / 20)
c(R) = ?

32. Convolution Back Projection
Instead of converting to frequency domain, use convolution
in the image domain
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) = |  |
c (R) = lim 2 (2 - 4π2R2) / (2 + 4π2R2)2
 0

33. Better Design
1.0
-  0
 0
c (R) = 0 [ 2 sinc (2 0 R) - sinc 2 (0 R)]
Actual Filters
: Ram-Lak
: Shepp-Logan

34. a) Ramachandran-Lakshminarayanan kernel
b) Shepp-Logan kernel.
Solid lines represent h() and circles hk
c) Representation of the filter functions corresponding to hR and hS