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
3
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
2
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
3
1
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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
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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
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Next week in Introduction to Medical imaging :
Magnetic Resonance Image Reconstruction