chapter6 image reconstruction -...
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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
Digital Image Processing Prof.Zhengkai Liu Dr.Rong Zhang
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Preview
Reconstruction by range data and CCD image
Digital Image Processing Prof.Zhengkai Liu Dr.Rong Zhang
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PreviewPreviewCT reconstruction
Digital Image Processing Prof.Zhengkai Liu Dr.Rong Zhang
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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
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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
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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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