Projection-Slice Theorem · 2015. 10. 21. · TT Liu, BE280A, UCSD Fall 2015! Fourier...

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TT Liu, BE280A, UCSD Fall 2015

Bioengineering 280A ��Principles of Biomedical Imaging��

��Fall Quarter 2015��

CT/Fourier Lecture 5


Projection-Slice Theorem

Modified from Prince&Links 2006

kx

ky

kU kx ,ky( )µ x, y( )

€

kx = k cosθky = k sinθ

k = kx2 + ky

2

€

G(k,θ) = g(l,θ)e− j2πkl−∞

∞

∫ dl€

U(kx,ky ) =G(k,θ)g(l,θ )


Example (sinc/rect) Exampleµ(x, y) =Π(x)Π(y)U(kx ,ky ) = sinc(kx )sinc(ky )

-1/2 1/2

1/2

-1/2

x

y



Projection at θ = 0;g(l,0) = rect(l)→ F(g(l,0)) = sinc(k)kx = k cosθ = k; ky = k sinθ = 0

U(kx ,ky ) k cosθ ,k sinθ= sinc(k)sinc(0) = sinc(k)

Projection at θ = 90;g(l,90) = rect(l)→ F(g(l,90)) = sinc(k)kx = k cosθ = 0; ky = k sinθ = k

U(kx ,ky ) k cosθ ,k sinθ= sinc(0)sinc(k) = sinc(k)

2



Projection at θ = 45;

g(l,45) = 2Λ l2 / 2

#

$%

&

'(= 2rect l

2 / 2#

$%

&

'(* rect l

2 / 2#

$%

&

'(

→ F(g(l,45)) = 2 22

sinc k 22

#

$%

&

'(

22

sinc k 22

#

$%

&

'(= sinc2 k 2

2

#

$%

&

'(

kx = k cosθ = k 22

; ky = k sinθ = k 22

U(kx ,ky ) k cosθ ,k sinθ= sinc k 2

2

#

$%

&

'(sinc k 2

2

#

$%

&

'(= sinc2 k 2

2

#

$%

&

'(

2

− 2 / 2 2 / 2



Projection at 0 <θ < 90kx = k cosθ; ky = k sinθ

U(kx ,ky ) k cosθ ,k sinθ= sinc k cosθ( )sinc k sinθ( )

g(l,θ ) = F−1 sinc k cosθ( )sinc k sinθ( )( )= F−1 sinc k cosθ( )( )*F−1 sinc k sinθ( )( )

=1

cosθrect l

cosθ#

$%

&

'(* 1

sinθrect l

sinθ#

$%

&

'(

=1

cosθ sinθrect l

cosθ#

$%

&

'(* rect l

sinθ#

$%

&

'(

TT Liu, BE280A, UCSD Fall 2015 TT Liu, BE280A, UCSD Fall 2015

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µ(x, y) = rect(x, y)cos 2π (x + y)( )

In-class Exercise

Sketch this object. What are the projections at theta = 0 and 90 degrees? For what angle is the projection maximized?

PollEv.com/be280a


g(l) = cos 2π l + y( )( )−1/2

1/2∫ dy

=12πsin 2π l + y( )( )

−1/2

1/2

=12π

sin 2π l + 12

#

$%

&

'(

#

$%

&

'(− sin 2π l − 1

2#

$%

&

'(

#

$%

&

'(

#

$%

&

'(

= 0

µ(x, y) = rect(x, y)cos 2π (x + y)( )Projection at 0 degrees

Peaks occur along the lines y = n− x where n is an integer.

Valleys occur along the lines

y = 2n+12

− x

g(l) = cos 2π x + l( )( )−1/2

1/2∫ dx

= 0

Projection at 90 degrees x

y


U(kx ,ky ) =12sinc(kx −1,ky −1)+ sinc(kx +1,ky +1)( )

F g(l)[ ] =U(kx ,ky ) k ,0

=12sinc(k −1,0−1)+ sinc(k +1,0+1)( )

=12sinc(k −1)sinc(−1)+ sinc(k +1)sinc(1)( )

= 0

Fourier transform of projection at 0 degrees:


F g(l)[ ] =U(kx ,ky ) 22k , 22k

=12sinc 2

2k −1, 2

2k −1

"

#$

%

&'+ sinc 2

2k +1, 2

2k +1

"

#$

%

&'

"

#$$

%

&''

=12sinc2 2

2k − 2( )

"

#$

%

&'+ sinc2 2

2k + 2( )

"

#$

%

&'

"

#$$

%

&''

= sinc2 22k

"

#$

%

&'∗12δ k − 2( )+δ k + 2( )( )

at 45 degrees

g(l) = 22rect 2

2l

!

"#

$

%&∗

22rect 2

2l

!

"#

$

%&

!

"#

$

%&⋅cos 2π 2l( )

= 2Λ 22l

!

"#

$

%&cos 2π 2l( )

4


g(l) = 2Λ 22l

"

#$

%

&'cos 2π 2l( )

22

222

2rect 2

2l

!

"#

$

%&∗

22rect 2

2l

!

"#

$

%&

!

"#

$

%&= 2Λ 2

2l

!

"#

$

%&

* = 22

−22

2

24

−24

24

−24

2 2


dx = 0.01;xax = -1:dx:1; [x,y] = meshgrid(xax,xax); z = (abs(x)<=0.5 & abs(y)<=0.5); z0 = z.*cos(2*pi*(x+y)); figure(1); % The object imagesc(xax,xax,z0);colorbar; axis equal; set(gca,’YDir’,’normal’); figure(2);% Image of Fourier Transform kax = -4:.1:4; [kx,ky] = meshgrid(kax,kax); z1 = 0.5*(sinc(kx-1).*sinc(ky-1) + sinc(kx+1).*sinc(ky+1)); imagesc(kax,kax,z1);colorbar;axis equal; set(gca,’YDir’,’normal’); figure(3);% Mesh plot of Fourier Transform mesh(kax,kax,z1); axis([-4 4 -4 4 -0.2 0.5]); view(16,48); figure(4);% Fourier Transform along 45 degree line. plot(diag(kx),diag(z1),’LineWidth’,2);grid; figure(5);% approximate the projection at 45 degrees. % and compare to exact formulation [nr,nc] = size(z0) ; ind = toeplitz(nr:-1:1,[1:nc]+nr-1) ; z0t = flipud(z0);; % approximation proj = accumarray(ind(:),z0t(:))*sqrt(2)*dx; pax = (dx*sqrt(2)/2)*(-(length(proj)-1)/2:(length(proj)-1)/2); % exact formulation. proj2 = sqrt(2).*cos(2*pi*sqrt(2)*pax).*tripuls(pax,sqrt(2)); plot(pax,proj,’b’,pax,proj2,’g--’,’LineWidth’,2);grid; figure(6); % plot consituent parts of the projection at 45 degrees. plot(pax,sqrt(2)*tripuls(pax,sqrt(2)),pax,cos(2*pi*sqrt(2)*pax),pax,proj2,’LineWidth’,2); grid;


Fourier Reconstruction

Suetens 2002

F

Interpolate onto Cartesian grid then take inverse transform


Fourier Interpretation

Kak and Slaney; Suetens 2002

€

Density ≈ Ncircumference

≈N

2π k

Low frequencies are oversampled. So to compensate for this, multiply the k-space data by |k| before inverse transforming.

5


Polar Version of Inverse FT

Suetens 2002

€

µ(x,y) = G(kx,ky−∞

∞

∫−∞

∞

∫ )e j 2π (kxx+kyy )dkxdky

= G(k,θ0

∞

∫0

2π∫ )e j2π (xk cosθ +yk sinθ )kdkdθ

= G(k,θ−∞

∞

∫0

π

∫ )e j 2πk(x cosθ +y sinθ ) k dkdθ

€

Note :g(l,θ + π ) = g(−l,θ)SoG(k,θ + π ) =G(−k,θ)


Filtered Backprojection

Suetens 2002

€

µ(x,y) = G(k,θ−∞

∞

∫0

π

∫ )e j 2π (xk cosθ +yk sinθ ) k dkdθ

= kG(k,θ−∞

∞

∫0

π

∫ )e j2πkldkdθ

= g∗(l,θ)dθ0

π

∫

€

g∗(l,θ) = kG(k,θ−∞

∞

∫ )e j2πkldk

= g(l,θ)∗F −1 k[ ]= g(l,θ)∗q(l)€

where l = x cosθ + y sinθ

Backproject a filtered projection


Ram-Lak Filter

Suetens 2002

kmax=1/Δs


Reconstruction Path

Suetens 2002

F

x F-1

Projection

Filtered Projection

Back- Project

6


Reconstruction Path

Suetens 2002

Projection

Filtered Projection

Back- Project

*


Example

Kak and Slaney


Example

Prince and Links 2005 TT Liu, BE280A, UCSD Fall 2015

Example

Prince and Links 2005

7


Filters

Filtered Projections



(a) (b)

PollEv.com/be280a

Which image used the Hanning filter?


CT Sampling Requirements

Suetens 2002

What should the size of the detectors be? How many detectors (or lines) do we need? How many views do we need?

8


View Aliasing

Kak and Slaney TT Liu, BE280A, UCSD Fall 2015

Aliasing

Kak and Slaney


Artifacts

Suetens 2002

Object Effect of Noise

Aliasing due to insufficient number of detectors

Aliasing due to insufficient number of views


Analog vs. Digital The Analog World: Continuous time/space, continuous valued signals or images, e.g. vinyl records, photographs, x-ray films. The Digital World: Discrete time/space, discrete-valued signals or images, e.g. CD-Roms, DVDs, digital photos, digital x-rays, CT, MRI, ultrasound.

9


The Process of Sampling g(x)

g[n]=g(n Δx)

Δx sample


Questions

How finely do we need to sample? What happens if we don’t sample finely enough?

Can we reconstruct the original signal or image from its samples?


Sampling in the Time Domain


Comb Function

€

comb(x) = δ(x − n)n=−∞

∞

∑

Other names: Impulse train, bed of nails, shah function.

-5 -4 -3 -2 -1 0 1 2 3 4 5 x

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Scaled Comb Function

€

comb xΔx#

$ %

&

' ( = δ( x

Δx− n)

n=−∞

∞

∑

= δ( x − nΔxΔx

)n=−∞

∞

∑

= Δx δ(x − nΔx)n=−∞

∞

∑

x

Δx


1D spatial sampling

€

gS (x) = g(x) 1Δx

comb xΔx#

$ %

&

' (

= g(x) δ(x − nΔx)n=−∞

∞

∑

= g(nΔx)δ(x − nΔx)n=−∞

∞

∑

€

Recall the sifting property g(x)δ(x − a) = g(a)−∞

∞

∫

But we can also write g(a)δ(x − a) = g(a)−∞

∞

∫ δ(x − a)−∞

∞

∫ = g(a)

So, g(x)δ(x − a) = g(a)δ(x − a)


1D spatial sampling g(x)

x

Δx

x

comb(x/Δx)/ Δx

gS(x)


Fourier Transform of comb(x)

€

F comb(x)[ ] = comb(kx )

= δ(kx − n)n=−∞

∞

∑

€

F 1Δx

comb( xΔx)

#

$ % &

' ( =1Δx

Δxcomb(kxΔx)

= δ(kxΔx − n)n=−∞

∞

∑

=1Δx

δ(kx −nΔx)

n=−∞

∞

∑

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Fourier Transform of comb(x/ Δx)

x

Δx

comb(x/ Δx)/ Δx

kx

comb(kx Δx)

1/Δx

1/Δx

F


Fourier Transform of gS(x)

€

F gS (x)[ ] = F g(x) 1Δx

comb xΔx#

$ %

&

' (

)

* +

,

- .

=G(kx )∗F1Δx

comb xΔx#

$ %

&

' (

)

* +

,

- .

=G(kx )∗1Δx

δ kx −nΔx

#

$ %

&

' (

n=−∞

∞

∑

=1Δx

G(kx )∗δ kx −nΔx

#

$ %

&

' (

n=−∞

∞

∑

=1Δx

G kx −nΔx

#

$ %

&

' (

n=−∞

∞

∑


Fourier Transform of gS(x)

G(kx)

GS(kx)

1/Δx

kx

kx

1/Δx


Nyquist Condition

G(kx)

GS(kx)

KS=1/Δx

kx

kx

B -B

To avoid overlap, we require that 1/Δx>2B or KS > 2B where KS=1/ Δx is the sampling frequency

KS

12


Example

Assume that the highest spatial frequency in an object is B = 2 cm-1. Thus, smallest spatial period is 0.5 cm. .

Nyquist theorem says we need to sample with Δx < 1/2B = 0.25 cm This corresponds to 2 samples per spatial period.


Example

PollEv.com/be280a

Assume that the Nyquist sampling periods of f (x) and g(x)are Δf and Δg, respectively. Determine the Nyquist sampling periods fora) f (x − x0 )b) f (x)+ g(x)c) f (x)* f (x)

from Prince and Links 2006

Projection-Slice Theorem · 2015. 10. 21. · TT Liu, BE280A, UCSD Fall 2015! Fourier...

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Transcript of Projection-Slice Theorem · 2015. 10. 21. · TT Liu, BE280A, UCSD Fall 2015! Fourier...