chapter6 image reconstruction -...

Chapter6 Image Reconstruction

• Preview6 1 I d i• 6.1 Introduction

• 6.2 Reconstruction by Fourier Inversioni b l i d b k j i• 6.3 Reconstruction by convolution and backprojection

• 6.4 Finite series-expansion

Digital Image Processing Prof.Zhengkai Liu Dr.Rong Zhang

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PreviewReconstruction by Stereoscopic image pair

measured depth map

left image right image

segmentationdense depth mapi f fDigital Image Processing

Prof.Zhengkai Liu Dr.Rong Zhang2

segmentationdense depth mapwire-frame surface

PreviewReconstruction by range data and CCD image

#5 #4#6#3

#2

INS/GPS

#1#3#1

#2


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PreviewReconstruction by range data and CCD image


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Preview

Reconstruction by range data and CCD image


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PreviewPreviewCT reconstruction


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PreviewPreviewCT reconstruction

MRIMRITCTCT machine


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6.1 Introduction

6.1.1 Overview of Computed Tomography (CT)

parallel-beam fan-beam


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6.1 Introduction

6.1.2 Classification

( ) i i d h(1) Transmission Computed Tomography TCT(2) Emission Computed Tomography, ECT(3) Reflection Computed Tomography, RCT(4) Magnetic resonance imaging, MRI


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6.1 Introduction6.1.3 Physical basis of projection

L t id th i l t i l bl k fLet us consider the simplest case, a single block of homogeneous tissue and a monochromatic beam of X-rays. The linear attenuation coefficientμ is defined byThe linear attenuation coefficientμ is defined by

0LI I e μ−= 0

where

L is the length of the blockL is the length of the block

I and I0 are incident and attenuated intensities of the X-ray, respectivelyrespectively


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6.1 Introduction6.1.3 Physical basis of projection

Let μ(x y) denote the sectional attenuation variation For anLet μ(x,y) denote the sectional attenuation variation. For an infinitely thin beam of monochromatic X-rays, the detected intensity of the X-ray along a straight line Lis expressed as y y g g p

( , )L

x y dl

I I eμ−∫

= 0I I e=

0

ln ( , )L

I x y dlI

μ− = ∫0 L


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6.1 Introduction

6.1.4 Purpose

Reconstruction μ(x,y) from its projections

6.1.5 Methods

Fourier Inversion ( frequency domain)

convolution and backprojection ( spatial domain)

Finite series-expansion ( spatial domain)


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6.2 Reconstruction by Fourier Inversion6.2.1 Mathematical expression of projection

Projection in x- and y-axial

∞

∫( ) ( , )p x f x y dy−∞

= ∫

( ) ( , )p y f x y dx∞

−∞= ∫

Projection in arbitrary direction

( ) ( , )p t f x y dsθ

∞

−∞= ∫


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6.2 Reconstruction by Fourier Inversion6.2.1 Mathematical expression of projection

where sin coscos sin

t y xs y x

θ θθ θ

= += −

yscos sins y xθ θ

cos sinx t sθ θ= −t

cos sinsin cos

x t sy t s

θ θθ θ

= −= +

x

( ) ( )p t f x y ds∞

= ∫( ) ( , )p t f x y dsθ −∞= ∫

( cos sin , sin cos )f t s t s dsθ θ θ θ∞

−∞= − +∫


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6.2 Reconstruction by Fourier Inversion6.2.2 Fourier Slice Theorem

If F(u,v) is the Fourier Transform of f(x,y), pθ(t) is the projectionof f(x,y) in θ –direction, and the Sθ(ω) is the Fourier transformof p (t) then S (ω) is a slice of F(u v) in the θ directionof pθ(t) then, Sθ(ω) is a slice of F(u,v) in the θ –direction

( ) ( )S Fω ω θ( ) ( , )S Fθ ω ω θ=

where F(ω,θ) is the polar coordinate expression of F(u,v)


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6.2 Reconstruction by Fourier Inversion6.2.2 Fourier Slice Theorem: proof

When θ=0∞

∫( ) ( , )p x f x y dy∞

−∞= ∫ ( ) ( ) exp( 2 )P u p x jux dxπ

−∞= −∫

( , ) exp( 2 )f x y dy jux dxπ∞ ∞

= −∫ ∫ ( , ) exp( 2 )f x y dy jux dxπ−∞ −∞∫ ∫

( , ) ( , ) exp[ 2 ( )]F u v f x y j ux vy dxdyπ∞ ∞

−∞ −∞= − +∫ ∫

Let v=0 ( ,0) ( , ) exp( 2 )F u f x y jux dxdyπ∞ ∞

−∞ −∞= −∫ ∫

( ,0) ( )F u P u=∴In other words the Fourier transform of the vertical projection of anIn other words, the Fourier transform of the vertical projection of an image is the horizontal radial profile of the 2D Fourier transform of the image.


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g

6.2 Reconstruction by Fourier Inversion6.2.2 Fourier Slice Theorem :proof

In generalIn generalBy the nature of the Fourier transform, if an image f(x,y) is rotated by an angle with respect to the x axis theis rotated by an angle with respect to the x axis, the Fourier transform F(u,v) will be correspondingly rotated by the same angle with respect to the u axis.

( ) ( , )S Fθ ω ω θ=


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6.2 Reconstruction by Fourier Inversion6.2.2 Fourier Slice Theorem

2D-DFT

projection( ,0) ( )F u P u=

1D-DFT


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6.2 Reconstruction by Fourier Inversion6.2.3 Arithmetic of Fourier inversion

Step1: calculate the projection’s DFT Sθ(ω) in θm –direction, m=0 1 M-1m 0,1, M 1

Step2: combine Sθ(ω) into F(ω,θ)

Step3: interpolation

Step4: 2D-IDFTStep4: 2D IDFT


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6 3 Reconstruction by convolution and backprojection6.3 Reconstruction by convolution and backprojection6.3.1 Parallel-Beam Reconstruction

With the inverse Fourier transform, an image f(x,y)can be expressed as

[ ]( ) ( ) 2 ( )f F j d d∞ ∞

∫ ∫ [ ]( , ) ( , ) exp 2 ( )f x y F u v j ux vy dudvπ−∞ −∞

= +∫ ∫L t iθ θLet cos sinu vω θ ω θ= =

we have[ ]

2

0 0( , ) ( , ) exp 2 ( cos sin )f x y F j x y d d

πθ ω π θ θ ω ω ω θ

∞= +∫ ∫

B ( ) ( )F Fθ θBecause ( , ) ( , )F Fθ π ω θ ω+ = −

[ ]( ) ( ) 2 ( i )f F j d dπ

θ θ θ θ∞

∫ ∫we have


20[ ]

0( , ) ( , ) exp 2 ( cos sin )f x y F j x y d dθ ω ω π θ θ ω ω θ

−∞= +∫ ∫

6.3 Reconstruction by convolution and backprojection6.3 Reconstruction by convolution and backprojection6.3.1 Parallel-Beam Reconstruction

Using the Fourier slice theorem, we have

[ ]( ) ( ) exp 2 ( cos sin )f x y S w j x y d dπ

ω π θ θ ω ω θ∞

+∫ ∫ [ ]0

( , ) ( ) exp 2 ( cos sin )f x y S w j x y d dθ ω π θ θ ω ω θ−∞

= +∫ ∫let cos sint x yθ θ= +cos sint x yθ θ+

[ ]0

( , ) ( ) exp 2f x y S w j t d dπ

θ ω π ω ω θ∞

= ∫ ∫ [ ]0

( ) ( ) pf y jθ−∞∫ ∫( )Sθ ω ω ( ) ( )p t h tθ ∗

where1( ) ( )h t F ω−=

0( , ) ( ) ( )f x y d p t h t d

π

θθ τ τ∞

∞= −∫ ∫


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0 −∞∫ ∫

6.3 Reconstruction by convolution and backprojection6.3 Reconstruction by convolution and backprojection6.3.1 Parallel-Beam Reconstruction

Note that h(t) does not exist in an ordinary sense, but Sθ(ω) is essentially bandlimited, h(t) can be accurately evaluated within th i b d idth f S ( )the maximum bandwidth of Sθ(ω) .

1( ) ( )h t F ω−= exp(2 )c

j t dωω π ω ω= ∫( ) ( )h t F ω= exp(2 )

cj t d

ωω π ω ω

−∫If 1cω = 2If

2cω

τ 2

2 2

1 sin(2 / 2 ) 1 sin(2 / 2 )( )2 2 / 2 4 2 / 2

t th tt tπ τ π τ

τ π τ τ π τ⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

1−

1 ω


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2τ 2τω

6.3 Reconstruction by convolution and backprojection6.3.2 Arithmetic of Fourier inversion

y p j

Step1: applying the filter h(t) to pθ(t), get p’θ(t) ,

Step2: calculate the integration

'

0( , ) ( )f x y p t d

π

θ θ= ∫


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6.3 Reconstruction by convolution and backprojection6.3.3 Experimental results

original

y p j

t t d ioriginal reconstructed image

j iprojection

rampbackprojection

ramp filter

sinogram Filtered sinogram


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6.3 Reconstruction by convolution and backprojection6.3.4 Practical problems:Aliasing - Insufficient angular sampling

y p j

•reducing scanning time

•reducing patient dose


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6.3 Reconstruction by convolution and backprojection6.3.4 Practical problems:Aliasing - Insufficient radial sampling

y p j

occurs when there is a sharp intensity change caused byoccurs when there is a sharp intensity change caused by, for example, bones.


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6.3 Reconstruction by convolution and backprojection6.3.4 Practical problems:Motion artifact

y p j

caused by patient motion, such as respiration and heart beat, during data acquistiong q


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6.4 Finite series-expansion6.4.1 Expression of Discrete projection

A 2-D array can be expressed as a vector

{ }0 1 1( , ) , Nf x y f f f −=A 2 D array can be expressed as a vector

The projection of i-th ray is1N

f−

∑ ,0

0,1, 1i i j jj

p w f i M=

= = −∑

h 1 if th th th th i ti j⎧where,

1 if the -th ray cross the -th point0 else i j

i jw ⎧

= ⎨⎩


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6.4 Finite series-expansion6.4.1 Expression of Discrete projection

1 2 3

接受器

第i条射线

N 第j个象素

第条射线

发射源N 第j个象素

w f w f w f p+ + + =⎧ 00 0 01 1 0( 1) 1 0

10 0 11 1 1( 1) 1 1

N N

N N

w f w f w f pw f w f w f p

− −

− −

+ + + =⎧⎪ + + + =⎪⎨⎪

( 1)0 0 ( 1)1 1 ( 1)( 1) 1 1

M M M N N Mw f w f w f p− − − − − −

⎨⎪⎪ + + + =⎩


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6.4 Finite series-expansion6.4.2 Solution by iteration

f( ) b i t i N D h j tif(x,y) can be seen as a point in a N-D space, each projectionequation is a super-plane of this N-D space. If that equation sethas a unique solution, then all the super-planes intersect in a pointhas a unique solution, then all the super planes intersect in a point

For example N=2p1

f1 p0

p1

p0

p1

00 0 01 1 0w f w f p+ =⎧⎨

f(2)p0

10 0 11 1 1w f w f p⎨ + =⎩

ff(0)f(1)


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f0

6.4 Finite series-expansion6.4.2 Solution by iteration

(0)w f p−(1) (0) 0 00

0 0

w f pf f ww w

−= −

•( )kf ( )

( 1) ( )k

k k k kk

k k

w f pf f ww w

+ −= −

•

6.4.3 Arithmetic of finite series-expansion Step1: given an initial estimation f(0), k=0p g f ,Step2: adjust f(k) by a projection equation to get f(k+1) ,k=k+1

St 3 t t 2 til th dj t l l th δStep3: repeat step2 until the adjust value less than δ

DEMODigital Image Processing

Prof.Zhengkai Liu Dr.Rong Zhang31

DEMO

5.1. 简述CT 发明过程。

5 14 试证明投影定理5.14. 试证明投影定理


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END

2003 Digital Image Processing Prof.Zhengkai Liu Dr.Rong Zhang

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