1. 2 unknown 3 4 5 6 7 8 9 10 backprojection usually produce a blurred version of the image
TRANSCRIPT
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Tutorial 3CT Image Reconstruction
Part II
Alexandre Kassel
Introduction to Medical Imaging
046831
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Tutorial Overview
Backprojection Filtered Backprojection Other Methods
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Recall : Projection
𝑝𝜃 (𝑟 )=− ln ¿
Unknown𝑝𝜃 (𝑟 )
[𝑟𝑠]=[ cos𝜃 sin𝜃−sin 𝜃 cos𝜃 ][ 𝑥𝑦 ]
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What’s Backprojection ?
Example : 2 projections
(projecting)
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What’s Backprojection ?
Example : 2 projections
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What’s Backprojection ?
Example : 2 projections
(backprojecting)
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What’s Backprojection ?
Example : 2 projections
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What’s Backprojection ?
Example : 2 projections
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What’s Backprojection ?
Example : 2 projections
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Backprojection From 2 Projections
From 10 Projections
From 90 Projections :
From 4 Projections
Backprojection usually produce a blurred version of the image.
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BP: Numerical Example
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1
3
1
3
0 5 3 3 0
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BP : Numerical Example
0.6 0.6 0.6 0.6 0.6
0.2 0.2 0.2 0.2 0.2
0.6 0.6 0.6 0.6 0.6
0.2 0.2 0.2 0.2 0.2
0.6 0.6 0.6 0.6 0.6
3
1
3
1
3
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BP: Numerical Example
0 1 0.6 0.6 0
0 1 0.6 0.6 0
0 1 0.6 0.6 0
0 1 0.6 0.6 0
0 1 0.6 0.6 0
0 5 3 3 0
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0.6 0.6 0.6 0.6 0.6
0.2 0.2 0.2 0.2 0.2
0.6 0.6 0.6 0.6 0.6
0.2 0.2 0.2 0.2 0.2
0.6 0.6 0.6 0.6 0.6
0 1 0.6 0.6 0
0 1 0.6 0.6 0
0 1 0.6 0.6 0
0 1 0.6 0.6 0
0 1 0.6 0.6 0
0.6 1.6 1.2 1.2 0.6
0.2 1.2 0.8 0.8 0.2
0.6 1.6 1.2 1.2 0.6
0.2 1.2 0.8 0.8 0.2
0.6 1.6 1.2 1.2 0.6
BP: Numerical Example
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BP : Mathematical DefinitionThe Backprojection is given by :
And the discrete version:
𝑏(𝑥 𝑖 , 𝑦 𝑗)=𝐵 ¿
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Reminder : Central Slice Theorem
¿ } 1D-FT{}
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Remainder : Central Slice Theorem
2D-FT(I) 1D-FT(Radon(I))
0°
10°
90°
120°
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Remainder : Direct Fourier Reconstruction
We discussed the problematic of interpolating into the Fourier Domain. Can we find a way to avoid doing this ?
Fundamentals of Medical ImagingPaul Suentes
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Let’s do some calculus
𝑓 (𝑥 , 𝑦)=∬𝐹 (𝑘𝑥 ,𝑘𝑦)𝑒+2 𝜋 𝑗𝑘𝑥 𝑥𝑒+ 2𝜋 𝑗 𝑘𝑦 𝑦𝑑𝑘𝑥𝑑𝑘𝑦
2D Inverse Fourier Transform
Function we want to reconstruct
Let’s change F from cartesian coordinates to polar coordinates
¿ ¿
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From Cartesian to Polar
{𝑘𝑥=𝑘cos𝜃𝑘𝑦=𝑘 sin𝜃
¿{ 𝑘=√𝑘𝑥
2+𝑘𝑦2
𝜃=tan− 1(𝑘𝑦
𝑘𝑥)
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With
Form half lines to full lines :
𝑓 (𝑥 , 𝑦)=∫0
𝜋
∫−∞
∞
𝐹 (𝑘 ,𝜃)∙|𝑘|∙𝑒𝑖2𝜋 𝑘𝑟𝑑𝑘𝑑𝜃
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Now the Central Slice Theorem become simply :
=P
And therefore:
𝑓 (𝑥 , 𝑦)=∫0
𝜋
∫∞
∞
𝑃 (𝑘 ,𝜃) ∙|𝑘|∙𝑒𝑖2 𝜋𝑘𝑟 𝑑𝑘𝑑𝜃
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Note that is a filter in the K-space. Let’s define the filtered projection in K-space :
𝑝∗ (𝑟 ,𝜃 )≜∫−∞
∞
𝑃∗ (𝑘 , 𝜃 )𝑒𝑖 2𝜋𝑘𝑟 𝑑𝑘
)
And its 1D inverse Fourier transform from k to r.
In the Radon domain it’s a convolution over r :
)
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𝑓 (𝑥 , 𝑦)=∫0
𝜋
∫∞
∞
𝑃 (𝑘 ,𝜃) ∙|𝑘|∙𝑒𝑖2 𝜋𝑘𝑟 𝑑𝑘𝑑𝜃
𝑓 (𝑥 , 𝑦)=∫0
𝜋
∫∞
∞
𝑃∗(𝑘 ,𝜃)𝑒𝑖 2𝜋𝑘𝑟 𝑑𝑘𝑑𝜃
𝑓 (𝑥 , 𝑦)=∫0
𝜋
𝑝∗ (𝑟 , 𝜃 )𝑑𝜃
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Filtered Backprojection
𝑓 (𝑥 , 𝑦)=∫0
𝜋
𝑝∗ (𝑟 , 𝜃 )𝑑𝜃
Note that it’s a backprojection ! 𝑓 (𝑥 , 𝑦 )=B {𝑝∗ (𝑟 ,𝜃 ) }=B {𝑝 (𝑟 , 𝜃 )∗𝑞 (𝑟 )}
This is called Filtered Backprojection
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FBP : Ramp Filter (Ram-Lak)
In Frequency domain
|𝑘|
Fundamentals of Medical ImagingPaul Suentes
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FBP : Ramp Filter (Ram-Lak) In space domain :
𝑞 (𝑟 )=𝑘𝑚𝑎𝑥
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4𝜋 2 (𝑠𝑖𝑛𝑐 (𝑘𝑚𝑎𝑥 ∙𝑟 )− 12𝑠𝑖𝑛𝑐2(𝑘𝑚𝑎𝑥 ∙𝑟
2 )) A sample at discrete value of gives this simple filter :
𝑞 (𝑛)={14𝑛=0
−1𝑛2𝜋 2 𝑛𝑖𝑠𝑜𝑑𝑑
0𝑛𝑖𝑠𝑒𝑣𝑒𝑛
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FBP : Ramp Filter (Ram-Lak)
-5 -4 -3 -2 -1 0 1 2 3 4 5-0.15
-0.1
-0.05
0
0.05
0.1
0.15
0.2
0.25Ram-Lak filter in space domain
n
q(n)
𝑞 (𝑛)={14𝑛=0
−1𝑛2𝜋 2 𝑛𝑖𝑠𝑜𝑑𝑑
0𝑛𝑖𝑠𝑒𝑣𝑒𝑛
Discrete filter in space domain :
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FBP : Ramp Filter (Ram-Lak)
-5 -4 -3 -2 -1 0 1 2 3 4 5-0.15
-0.1
-0.05
0
0.05
0.1
0.15
0.2
0.25Ram-Lak filter in space domain
n
q(n)
The Ramp Filter is also called the Ram-Lak filter after Ramachandran and Lakshiminarayanan
Problem : High frequencies are unreliable because of noise and aliasing. And Ram-Lak filter enhances them.
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FBP : Smoothed window (Hamming, Hann…)(in frequency domain)
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FBP: A two steps algorithm
Ram-Lak Filter
(or smoothed version of
it)
Projections Backproject
Reconstructed image
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Filtered backprojection : Results Examples(from 360 projections)
No filtered Ram-Lak
Ram-Lak Hamming
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A.R.TAlgebraic Reconstruction Technique
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1
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0 5 3 3 0
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A.R.T(Rectification by difference)
0.6 0.6 0.6 0.6 0.6
0.2 0.2 0.2 0.2 0.2
0.6 0.6 0.6 0.6 0.6
0.2 0.2 0.2 0.2 0.2
0.6 0.6 0.6 0.6 0.6
3
1
3
1
3
0 5 3 3 0
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A.R.T(Rectification By difference)
0.6 0.6 0.6 0.6 0.6
0.2 0.2 0.2 0.2 0.2
0.6 0.6 0.6 0.6 0.6
0.2 0.2 0.2 0.2 0.2
0.6 0.6 0.6 0.6 0.6
3
1
3
1
3
0 5 3 3 0
2.2 2.2 2.2 2.2 2.2
Σ
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A.R.T(Rectification By difference)
0.6 0.6 0.6 0.6 0.6
0.2 0.2 0.2 0.2 0.2
0.6 0.6 0.6 0.6 0.6
0.2 0.2 0.2 0.2 0.2
0.6 0.6 0.6 0.6 0.6
3
1
3
1
3
0 5 3 3 0
-2.2 2.8 0.8 0.8 -2.2 Rectification
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A.R.T(Rectification By difference)
0.16 1.16 0.76 0.76 0.16
-0.22
0.76 0.36 0.36-
0.22
0.16 1.16 0.76 0.76 0.16
-0.22
0.76 0.36 0.36-
0.22
0.16 1.16 0.76 0.76 0.16
3
1
3
1
3
0 5 3 3 0
-2.2 2.8 0.8 0.8 -2.2 Rectification
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And we continue until convergence … We can prove A.R.T is converging. For an
unique solution we need N projections for a NxN matrix.
A.R.T is accurate but very slow. Some elaborate techniques were developed with improved efficiency.
Current CT devices are using FBP anyway.
A.R.TAlgebraic Reconstruction Technique
Next week in Introduction to Medical imaging :
Magnetic Resonance Image Reconstruction