computed tomography - university of california, san diego · 2010. 10. 5. · 1! tt liu, be280a,...

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1 TT Liu, BE280A, UCSD Fall 2010 Bioengineering 280A Principles of Biomedical Imaging Fall Quarter 2010 CT Lecture 1 TT Liu, BE280A, UCSD Fall 2010 Computed Tomography Suetens 2002 TT Liu, BE280A, UCSD Fall 2010 Computed Tomography Suetens 2002 Parallel Beam Fan Beam TT Liu, BE280A, UCSD Fall 2010 Scanner Generations Suetens 2002

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  • 1

    TT Liu, BE280A, UCSD Fall 2010

    Bioengineering 280A ���Principles of Biomedical Imaging���

    Fall Quarter 2010���CT Lecture 1���

    TT Liu, BE280A, UCSD Fall 2010

    Computed Tomography

    Suetens 2002

    TT Liu, BE280A, UCSD Fall 2010

    Computed Tomography

    Suetens 2002

    Parallel

    Beam

    Fan

    Beam

    TT Liu, BE280A, UCSD Fall 2010

    Scanner Generations

    Suetens 2002

  • 2

    TT Liu, BE280A, UCSD Fall 2010

    From http://www.sprawls.org/resources/CTIMG/classroom.htm

    TT Liu, BE280A, UCSD Fall 2010

    Single vs. Multi-slice

    Suetens 2002

    TT Liu, BE280A, UCSD Fall 2010

    Scanner Generations

    Prince and Links 2005

    TT Liu, BE280A, UCSD Fall 2010

    1G vs. 2G scanner

    Example 6.1 from Prince and Links. Compare 1G vs. 2G scanner whose source - detector apparatus can move linearlyat speed of 1 m/sec; FOV 0.5m; 360 projections over 180 degrees; 0.5 s for apparatusto rotate one angular increment, regardless of angle.Question : Scan time for 1 G scanner? Scan time for 2G scanner with 9 detectors space 0.5 degrees apart?

    Answer : 1G scanner : 0.5m/(1m/s) = 0.5s per projection. 360 * 0.5 = 180s scan time 360 * 0.5 =180s for rotation of apparatus. Total time = 360 s or 6 minutes.

    2G scanner : Required angular resolution is 180/360 = 0.5 degrees - - agrees with spacing. 360/9 = 40 rotations required. 40*0.5 = 20s for scanning 40*0.5 = 20s for rotations. Total time = 40s.

  • 3

    TT Liu, BE280A, UCSD Fall 2010

    3G, 6G, and 7G scanners

    3G scanner : Typical scanner acquires 1000 projections with fanbeam angleof 30 to 60 degrees; 500 to 700 detectors; 1 to 20 seconds.

    6G : Spiral/Helical CT 60 cm torso scan : 30s. 24 cm lung scan : 12s 15 cm angio : 30s

    7G : Multislice CT 64 or more parallel 1D projections.

    TT Liu, BE280A, UCSD Fall 2010

    Suetens 2002

    TT Liu, BE280A, UCSD Fall 2010

    Detectors

    Prince and Links 2005

    TT Liu, BE280A, UCSD Fall 2010

    CT Line Integral

    Id = S0 E( )E0Emax∫ exp − µ s; ʹ′ E ( )0

    d∫ ds( )dE

    Monoenergetic Approximation

    Id = I0 exp − µ s;E ( )0d∫ ds( )

    gd = −logIdI0

    ⎝ ⎜

    ⎠ ⎟

    = µ s;E ( )0d∫ ds

  • 4

    TT Liu, BE280A, UCSD Fall 2010

    CT Number

    CT_number = µ −µwaterµwater

    ×1000

    Measured in Hounsfield Units (HU)

    Air: -1000 HU

    Soft Tissue: -100 to 60 HU

    Cortical Bones: 250 to 1000 HU

    Metal and Contrast Agents: > 2000 HU

    TT Liu, BE280A, UCSD Fall 2010

    CT Display

    Suetens 2002

    TT Liu, BE280A, UCSD Fall 2010

    Direct Inverse Approach

    µ1

    µ2

    µ3

    µ4

    p1

    p2

    p3

    p4

    p1= µ1+ µ2 p2= µ3+ µ4 p3= µ1+ µ3 p4= µ2+ µ4

    4 equations, 4 unknowns.

    Are these the correct equations to use?

    p1p2p3p4

    ⎢ ⎢ ⎢ ⎢

    ⎥ ⎥ ⎥ ⎥

    =

    1 1 0 00 0 1 11 0 1 00 1 0 1

    ⎢ ⎢ ⎢ ⎢

    ⎥ ⎥ ⎥ ⎥

    µ1µ2µ3µ4

    ⎢ ⎢ ⎢ ⎢

    ⎥ ⎥ ⎥ ⎥

    No, equations are not linearly independent.

    p4= p1+ p2- p3

    Matrix is not full rank.

    TT Liu, BE280A, UCSD Fall 2010

    Direct Inverse Approach

    µ1

    µ2

    µ3

    µ4

    p1

    p2

    p3

    p4

    p1= µ1+ µ2 p2= µ3+ µ4 p3= µ1+ µ3 p4= µ2+ µ4

    4 equations, 4 unknowns.

    Are these the correct equations to use?

    p1p2p3p4

    ⎢ ⎢ ⎢ ⎢

    ⎥ ⎥ ⎥ ⎥

    =

    1 1 0 00 0 1 11 0 1 00 1 0 1

    ⎢ ⎢ ⎢ ⎢

    ⎥ ⎥ ⎥ ⎥

    µ1µ2µ3µ4

    ⎢ ⎢ ⎢ ⎢

    ⎥ ⎥ ⎥ ⎥

    No, equations are not linearly independent.

    p4= p1+ p2- p3

    Matrix is not full rank.

  • 5

    TT Liu, BE280A, UCSD Fall 2010

    Direct Inverse Approach

    µ1

    µ2

    µ3

    µ4

    p1

    p2

    p3

    p4

    p1= µ1+ µ2 p2= µ3+ µ4 p3= µ1+ µ3 p5= µ1+ µ4

    4 equations, 4 unknowns. These are linearly independent now.

    In general for a NxN image, N2 unknowns, N2 equations.

    This requires the inversion of a N2xN2 matrix

    For a high-resolution 512x512 image, N2=262144 equations.

    Requires inversion of a 262144x262144 matrix!

    Inversion process sensitive to measurement errors.

    p1p2p3p5

    ⎢ ⎢ ⎢ ⎢

    ⎥ ⎥ ⎥ ⎥

    =

    1 1 0 00 0 1 11 0 1 01 0 0 1

    ⎢ ⎢ ⎢ ⎢

    ⎥ ⎥ ⎥ ⎥

    µ1µ2µ3µ4

    ⎢ ⎢ ⎢ ⎢

    ⎥ ⎥ ⎥ ⎥

    p5

    TT Liu, BE280A, UCSD Fall 2010

    Iterative Inverse Approach ���Algebraic Reconstruction Technique (ART)

    1

    2

    3

    4

    3

    7

    4

    6

    5

    2.5

    2.5

    2.5

    2.5

    5

    5

    1.5

    1.5

    3.5

    3.5

    3

    7

    5

    5

    1

    2

    3

    4

    3

    7

    5

    5

    TT Liu, BE280A, UCSD Fall 2010

    Backprojection

    Suetens 2002

    0

    0

    0

    0

    3

    0

    0

    0

    0

    0

    3

    0

    3

    0

    3

    0

    3

    0

    0

    0

    1

    1

    1

    0

    0

    0

    1

    0

    0

    1

    2

    1

    0

    0

    1

    1

    1

    0

    1

    3

    1

    0

    1

    1

    1

    1

    1

    1

    4

    1

    1

    1

    1

    TT Liu, BE280A, UCSD Fall 2010

    In-Class Exercise

    Suetens 2002

    µ1

    µ2

    µ3

    µ4

    5.7

    11.3

    8.2

    8.8

    10.1

  • 6

    TT Liu, BE280A, UCSD Fall 2010

    Projections

    Suetens 2002

    rs⎡

    ⎣ ⎢ ⎤

    ⎦ ⎥ =

    cosθ sinθ−sinθ cosθ⎡

    ⎣ ⎢

    ⎦ ⎥ xy⎡

    ⎣ ⎢ ⎤

    ⎦ ⎥

    xy⎡

    ⎣ ⎢ ⎤

    ⎦ ⎥ =

    cosθ −sinθsinθ cosθ⎡

    ⎣ ⎢

    ⎦ ⎥ rs⎡

    ⎣ ⎢ ⎤

    ⎦ ⎥

    TT Liu, BE280A, UCSD Fall 2010

    Projections

    Suetens 2002

    I(r,θ) = I0 exp − µ(x,y)dsLr ,θ∫⎛ ⎝ ⎜ ⎞

    ⎠ ⎟

    = I0 exp − µ(rcosθ − ssinθ,rsinθ + scosθ)dsLr ,θ∫⎛ ⎝ ⎜ ⎞

    ⎠ ⎟

    TT Liu, BE280A, UCSD Fall 2010

    Projections

    Suetens 2002

    I(r,θ) = I0 exp − µ(rcosθ − ssinθ,rsinθ + scosθ)dsLr ,θ∫⎛ ⎝ ⎜ ⎞

    ⎠ ⎟

    p(r,θ) = −ln Iθ (r)I0

    = µ(rcosθ − ssinθ,rsinθ + scosθ)dsLr ,θ∫

    TT Liu, BE280A, UCSD Fall 2010

    Radon Transform

    g(r,θ) = µ(x(s),y(s))ds−∞

    ∫= µ(rcosθ − ssinθ,rsinθ + scosθ)ds

    −∞

    ∫= µ(x,y)δ x cosθ + y sinθ − r( )

    −∞

    ∫−∞

    ∫ dxdy

    z • r r

    = r

    xˆ x + yˆ y ( ) • cosθˆ x + sinθˆ y ( ) = rx cosθ + y sinθ = r€

    θ

    z

    r

  • 7

    TT Liu, BE280A, UCSD Fall 2010

    Example

    f (x,y) =1 x 2 + y 2 ≤10 otherwise

    ⎧ ⎨ ⎩

    g(l,θ = 0) = f (l,y)dy−∞

    = dy− 1− l 21− l 2

    =2 1− l2 l ≤10 otherwise

    ⎧ ⎨ ⎪

    ⎩ ⎪

    TT Liu, BE280A, UCSD Fall 2010

    Sinogram

    Suetens 2002

    TT Liu, BE280A, UCSD Fall 2010

    Backprojection

    Suetens 2002

    b(x,y) = B g l,θ( ){ }

    = g(x cosθ + y sinθ,θ)dθ0

    π

    x

    y

    b(x0,y) = p l,θ = 0( )Δθ= p(x0,0)Δθ

    x0

    l

    bθ (x,y) = g(x cosθ + y sinθ,θ)Δθ

    TT Liu, BE280A, UCSD Fall 2010

    Backprojection

    Suetens 2002

    b(x,y) = B p l,θ( ){ }

    = p(x cosθ + y sinθ,θ)dθ0

    π

  • 8

    TT Liu, BE280A, UCSD Fall 2010

    Backprojection

    Suetens 2002

    b(x,y) = B p l,θ( ){ } = p(x cosθ + y sinθ,θ)dθ0

    π

    TT Liu, BE280A, UCSD Fall 2010

    Example

    Prince & Links 2006