Bioengineering 28A Principles of Biomedical Imaging Fall Quarter 25 X-Rays/CT Lecture Topics X-Rays Computed Tomography Direct Inverse and Iterative Inverse Backprojection Projection Theorem Filtered Backprojection EM spectrum
X-Ray Tube Usually tungsten is used for anode Molybdenum for mammography Tungsten filament heated to about 22 C leading to thermionic emission of electrons. X-Ray Production Characteristic Radiation Bremsstrahlung (braking radiation) http://www.scienceofspectroscopy.info/theory/advanced/x_ray.htm X-Ray Spectrum bremsstrahlung 2
Interaction with Matter Typical energy range for diagnostic x-rays is below 2 kev. The two most important types of interaction are photoeletric absorption and Compton scattering. Photoelectric effect dominates at low x-ray energies and high atomic numbers. Compton scattering dominates at high x-ray energies and low atomic numbers, not much contrast http://www.eee. ntu.ac.uk/research/vision/asobania Interaction with Matter Pair Production Photoelectric absorption Compton Scattering Attenuation d For single-energy x-rays passing through a homogenous object: I out = I in exp( µd) Linear attenuation coefficient 3
Attenuation Photoelectric effect dominates Compton Scattering dominates Attenuation Coefficient 5 Bone Muscle Fat. 5 5 5 Photon Energy (kev) Adapted from www.cis.rit.edu/class/simg25/xrays.ppt X-ray energy (kev) 3 5 5 Half Value Layer HVL, muscle (cm).8 3. 3.9 4.5 HVL Bone (cm).4.2 2.3 2.8 In chest radiography, about 9% of x-rays are absorbed by body. Average energy from a tungsten source is 68 kev. However, many lower energy beams are absorbed by tissue, so average energy is higher. This is referred to as beam-hardening, and reduces the contrast. Values from Webb 23 For an inhomogenous object: Attenuation x I out = I in exp( out µ(x)dx x in ) Integrating over energies I out = x σ(e) exp out µ(e, x)dx ( x in ) de Intensity distribution of beam 4
X-Ray Imaging Chain Reduces effects of Compton scattering X-ray film Emulsion with silver halide crystals Each layer ~ µm Flexible base ~ 5 µm Silver halide crystals absorb optical energy. After development, crystals that have absorbed enough energy are converted to metallic silver and look dark on the screen. Thus, parts in the object that attenuate the x-rays will look brighter. Intensifying Screen http://learntech.uwe.ac.uk/radiography/rscience/imaging_principles_d/diagimage.htm http://www.sunnybrook.utoronto.ca:88/~selenium/xray.html#film 5
X-Ray Examples X-Ray w/ Contrast Agents Angiogram using an iodine-based contrast agent. K-edge of iodine is 33.2 kev Barium Sulfate K-edge of Barium is 37.4 kev Computed Tomography 6
Computed Tomography Parallel Beam Fan Beam CT Number CT_number = µ µ water µ water Measured in Hounsfield Units (HU) Air: - HU Soft Tissue: - to 6 HU Cortical Bones: 25 to HU CT Display 7
Projections r = cosθ s sinθ sinθ cosθ x y x y = cosθ sinθ sinθ cosθ r s Projections I θ (r) = I exp µ(x, y)ds Lr,θ = I exp µ(rcosθ ssinθ,rsinθ + scosθ)ds Lr,θ Projections I θ (r) = I exp Lr,θ µ(rcosθ ssinθ,rsinθ + scosθ)ds p θ (r) = ln I (r) θ I = µ(rcosθ ssinθ,rsinθ + scosθ)ds Lr,θ Sinogram 8
Sinogram µ 3 µ 4 p 2 µ µ 2 p p 3 p 4 Direct Inverse Approach p = µ + µ 2 p 2 = µ 3 + µ 4 p 3 = µ + µ 3 p 4 = µ 2 + µ 4 4 equations, 4 unknowns. Are these the correct equations to use? p µ p 2 = µ 2 p 3 µ 3 p 4 µ 4 No, equations are not linearly independent. p 4 = p + p 2 - p 3 Matrix is not full rank. µ 3 µ 4 p 2 µ µ 2 p p 3 p 4 Direct Inverse Approach p 5 p = µ + µ 2 p 2 = µ 3 + µ 4 p 3 = µ + µ 3 p 5 = µ + µ 4 p µ p 2 = µ 2 p 3 µ 3 p 4 µ 4 4 equations, 4 unknowns. These are linearly independent now. In general for a NxN image, N 2 unknowns, N 2 equations. This requires the inversion of a N 2 xn 2 matrix For a high-resolution 52x52 image, N 2 =26244 equations. Requires inversion of a 26244x26244 matrix! Inversion process sensitive to measurement errors. 9
Iterative Inverse Approach Algebraic Reconstruction Technique (ART) 2 3 3 4 7 2.5 2.5 5 2.5 2.5 5 4 6 5 2 3 3 4 5 5 7.5.5 3 3.5 3.5 5 5 7 Backprojection 3 3 3 3 3 2 3 4 y Backprojection b(x, y) = B{ p( r,θ) } π = p(x cosθ + y sinθ,θ)dθ x x b(x, y) = p( r,θ = )Δθ = p(x )Δθ r
Backprojection b(x, y) = B{ p( r,θ) } π = p(x cosθ + y sinθ,θ)dθ Backprojection π b(x, y) = B{ p( r,θ) } = p(x cosθ + y sinθ,θ)dθ Projection Theorem U(k x,) = µ(x, y)e j 2π (k xx +k yy) dxdy [ ] = µ(x,y)dy e j 2πkxx dx = p (x) e j 2πkxx dx = p (r) e j 2πkr dr
Projection Theorem [ ] U(k x,k y ) = µ(x, y)e j 2π (k x x +k y y) dxdy = F 2D µ(x, y) U(k x,k y ) = P(k,θ) k x = k cosθ k y = k sinθ k = k 2 2 x + k y F P(k,θ) = p θ (r)e j 2πkr dr Fourier Reconstruction F Interpolate onto Cartesian grid then take inverse transform Polar Version of Inverse FT µ(x, y) = 2π π U(k x,k y )e j 2π (k xx +k y y ) dk x dk y = U(k,θ)e j 2π (k cosθx +k sinθy) kdkdθ = U(k,θ)e j 2π (xk cosθ +yk sinθ ) k dkdθ 2
µ(x, y) = Filtered Backprojection π π π U(k,θ)e j 2π (xk cosθ +yk sinθ ) k dkdθ = k U(k,θ)e j 2πkr dkdθ = u (r,θ)dθ where r = x cosθ + y sinθ u (r,θ) = k U(k,θ)e j 2πkr dk = u(r,θ) F k = u(r,θ) q(r) [ ] Backproject a filtered projection Ram-Lak Filter k max =/Δs Projection Reconstruction Path F x F - Filtered Projection Back- Project 3
Projection Reconstruction Path * Filtered Projection Back- Project Example Kak and Slaney Example Prince and Links 25 4
Example Prince and Links 25 Fourier Interpretation Density N N circumference 2π k Low frequencies are oversampled. So to compensate for this, multiply the k-space data by k before inverse transforming. Kak and Slaney; Additional Filtering kmax =/Δs 5
Sampling Requirements Projection Beam Width 2/(Δs) W= 2/(Δs) δ=/w= Δs/2 Smoothed Projection Sampling Requirements Smoothed Projection Detectors Δr Δs/2 Sampled Smooth Projection Sampling Requirements Size of detector Δr = δ=/w= Δs/2 Number of Detectors N = FOV/ Δr where Δr Δs/2 Angular Sampling -- how many views? Want Circumference/(views in 36 degrees) = Δr πfov/(views)=δr=fov/n Number of views in 36 degrees = πn 6