1TT Liu, BE280A, UCSD Fall 2004Bioengineering 280APrinciples of Biomedical ImagingFall Quarter 2004X-Rays/CT Lecture 1TT Liu, BE280A, UCSD Fall 2004Topics• X-Rays• Computed Tomography• Direct Inverse and Iterative Inverse• Backprojection• Projection Theorem• Filtered BackprojectionTT Liu, BE280A, UCSD Fall 2004EM spectrumSuetens 20022TT Liu, BE280A, UCSD Fall 2004X-Ray TubeSuetens 2002Tungsten filament heated to about 2200 C leading to thermionicemission of electrons.Usually tungsten is used for anodeMolybdenum for mammographyTT Liu, BE280A, UCSD Fall 2004X-Ray Productionhttp://www.scienceofspectroscopy.info/theory/ADVANCED/x_ray.htmCharacteristic RadiationBremsstrahlung(braking radiation)TT Liu, BE280A, UCSD Fall 2004X-Ray SpectrumSuetens 2002bremsstrahlung3TT Liu, BE280A, UCSD Fall 2004Interaction with MatterPhotoelectric effectdominates at low x-rayenergies and high atomicnumbers.Typical energy range for diagnostic x-rays is below 200 keV.The two most important types of interaction are photoeletricabsorption and Compton scattering.Compton scatteringdominates at high x-rayenergies and low atomicnumbers, not much contrasthttp://www.eee. ntu.ac.uk/research/vision/asobaniaTT Liu, BE280A, UCSD Fall 2004Interaction with MatterPhotoelectric absorptionCompton ScatteringPair ProductionTT Liu, BE280A, UCSD Fall 2004Attenuation€ Iout= Iinexp(−µd)dFor single-energy x-rays passing through a homogenous object:Linear attenuation coefficient4TT Liu, BE280A, UCSD Fall 2004Attenuation510 50 100 15010.1AttenuationCoefficient500BoneMuscleFatAdapted from www.cis.rit.edu/class/simg215/xrays.ppt Photon Energy (keV)Photoelectric effectdominatesCompton ScatteringdominatesTT Liu, BE280A, UCSD Fall 2004Half Value LayerValues from Webb 20032.84.51502.33.91001.23.0500.41.830HVLBone (cm)HVL,muscle(cm)X-rayenergy(keV)In chest radiography, about 90% 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 averageenergy is higher. This is referred to as beam-hardening, andreduces the contrast.TT Liu, BE280A, UCSD Fall 2004Attenuation€ Iout= Iinexp −µ(x)dxxinxout∫( )For an inhomogenous object:Integrating over energies€ Iout=σ(E )0∞∫exp −µ(E,x)dxxinxout∫( )dEIntensity distribution of beam5TT Liu, BE280A, UCSD Fall 2004X-Ray Imaging ChainSuetens 2002Reduces effects of Compton scatteringTT Liu, BE280A, UCSD Fall 2004X-ray filmFlexible base~ 150 µmEmulsion withsilver halide crystalsEach layer~ 10 µmSilver halide crystals absorb optical energy. After development,crystals that have absorbed enough energy are converted tometallic silver and look dark on the screen. Thus, parts in theobject that attenuate the x-rays will look brighter.TT Liu, BE280A, UCSD Fall 2004Intensifying Screenhttp://learntech.uwe.ac.uk/radiography/RScience/imaging_principles_d/diagimage11.htmhttp://www.sunnybrook.utoronto.ca:8080/~selenium/xray .html#Film6TT Liu, BE280A, UCSD Fall 2004X-Ray ExamplesSuetens 2002TT Liu, BE280A, UCSD Fall 2004X-Ray w/ Contrast AgentsSuetens 2002Angiogram using an iodine-basedcontrast agent.K-edge of iodine is 33.2 keVBarium SulfateK-edge of Barium is 37.4 keVTT Liu, BE280A, UCSD Fall 2004Computed TomographySuetens 20027TT Liu, BE280A, UCSD Fall 2004Computed TomographySuetens 2002ParallelBeamFan BeamTT Liu, BE280A, UCSD Fall 2004CT Number€ CT_number = µ−µwaterµwater×1000Measured in Hounsfield Units (HU)Air: -1000 HUSoft Tissue: -100 to 60 HUCortical Bones: 250 to 1000 HUTT Liu, BE280A, UCSD Fall 2004CT DisplaySuetens 20028TT Liu, BE280A, UCSD Fall 2004ProjectionsSuetens 2002€ rs =cosθsinθ−sinθcosθ xy xy =cosθ−sinθsinθcosθ rs TT Liu, BE280A, UCSD Fall 2004ProjectionsSuetens 2002€ Iθ(r) = I0exp −µ(x, y)dsLr ,θ∫ = I0exp −µ(r cosθ− ssinθ,rsinθ+ scosθ)dsLr ,θ∫ TT Liu, BE280A, UCSD Fall 2004ProjectionsSuetens 2002€ Iθ(r) = I0exp −µ(r cosθ− ssinθ,r sinθ+ scosθ)dsLr,θ∫ € pθ(r) = − lnIθ(r)I0=µ(r cosθ− ssinθ,r sinθ+ scosθ)dsLr,θ∫Sinogram9TT Liu, BE280A, UCSD Fall 2004SinogramSuetens 2002TT Liu, BE280A, UCSD Fall 2004Direct Inverse Approachµ4µ3µ2µ1p1p2p3p4p1= µ1+ µ2p2= µ3+ µ4p3= µ1+ µ3p4= µ2+ µ44 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- p3Matrix is not full rank.TT Liu, BE280A, UCSD Fall 2004Direct Inverse Approachµ4µ3µ2µ1p1p2p3p4p1= µ1+ µ2p2= µ3+ µ4p3= µ1+ µ3p5= µ1+ µ44 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 matrixFor a high-resolution 512x512 image, N2=262144 equations.Requires inversion of a 262144x262144 matrix! Inversion process sensitive to measurement errors. € p1p2p3p4 =1 1 0 00 0 1 11 0 1 01 0 0 1 µ1µ2µ3µ4 p510TT Liu, BE280A, UCSD Fall 2004Iterative Inverse ApproachAlgebraic Reconstruction Technique (ART)4321374 652.52.52.52.5553.53.51.51.5375 54321375 5TT Liu, BE280A, UCSD Fall 2004BackprojectionSuetens 2002000030000 03030303000111000100121001110131011111141111TT Liu, BE280A, UCSD Fall 2004BackprojectionSuetens 2002€ b(x, y) = B p r,θ( ){ }= p(x cosθ+ y sinθ,θ)dθ0π∫11TT Liu, BE280A, UCSD Fall 2004BackprojectionSuetens 2002€ b(x, y) = B p r,θ( ){ }= p(x cosθ+ y sinθ,θ)dθ0π∫TT Liu, BE280A, UCSD Fall 2004Projection TheoremSuetens 2002€ U(kx,0) =µ(x, y)e− j 2π(kxx + kyy)−∞∞∫−∞∞∫dxdy=µ(x, y)dy−∞−∞∫[ ]−∞−∞∫e− j 2πkxxdx= p0(x)−∞−∞∫e− j 2πkxxdx= p0(r)−∞−∞∫e− j 2πkrdrTT Liu, BE280A, UCSD Fall 2004Projection TheoremSuetens 2002€ U(kx,ky) =µ(x, y)e− j 2π(kxx + kyy)−∞∞∫−∞∞∫dxdy= F2Dµ(x, y)[ ]€ P(k,θ) = pθ(r)e− j 2πkr−∞∞∫drF€ U(kx,ky) = P(k,θ)€ kx= k cosθky= k sinθk = kx2+ ky212TT Liu, BE280A, UCSD Fall 2004Fourier ReconstructionSuetens
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