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UCSD BENG 280A - X-Rays/CT Lecture 2

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1TT Liu, BE280A, UCSD Fall 2005Bioengineering 280APrinciples of Biomedical ImagingFall Quarter 2005X-Rays/CT Lecture 2TT Liu, BE280A, UCSD Fall 2005AliasingKak and Slaney2TT Liu, BE280A, UCSD Fall 2005View AliasingKak and SlaneyTT Liu, BE280A, UCSD Fall 2005CT System GenerationsSuetens 20025 minutes/slice 20 seconds /slice0.5 seconds /slice3TT Liu, BE280A, UCSD Fall 2005CT SystemSuetens 2002TT Liu, BE280A, UCSD Fall 2005Fan BeamSuetens 2002€ θ=α+β€ r = R sinαr4TT Liu, BE280A, UCSD Fall 2005Fan BeamSuetens 2002€ θ=α+β€ r = R sinαr0r€ θ€ π€ β= 0€ αmax€ αmaxTT Liu, BE280A, UCSD Fall 2005Spiral CTSuetens 2002From http://www.sprawls.org/resources/CTIMG/classroom.htmNearest NeighborInterpolationLinearInterpolation5TT Liu, BE280A, UCSD Fall 2005Longitudinal Aliasing in Spiral CTSuetens 2002From http://www.sprawls.org/resources/CTIMG/classroom.htmTT Liu, BE280A, UCSD Fall 2005Multislice CTSuetens 20026TT Liu, BE280A, UCSD Fall 2005Poisson ProcessEvents occur at random instants of time at an average rateof λ events per second.Examples: arrival of customers to an ATM, emission ofphotons from an x-ray source, lightning strikes in athunderstorm.Assumptions:1) Probability of more than 1 event in an small timeinterval is small.2) Probability of event occurring in a given small timeinterval is independent of another event occuring inother small time intervals.TT Liu, BE280A, UCSD Fall 2005Poisson Process€ P N (t) = k[ ]=λt( )kk!exp(−λt)λ= Average rate of events per secondλt = Average number of events at time tλt = Variance in number of eventsProbability of interarrival timesP T > t[ ]= e−λt7TT Liu, BE280A, UCSD Fall 2005Example€ A service center receives an average of 15 inquiriesper minute. Find the probability that 3 inquiries arrivein the first 10 seconds. λ=15 /60 = 0.25λt = 0.25(10) = 2.5P[N (t =10) = 3) =(2.5)33!exp(−2.5) = .2138TT Liu, BE280A, UCSD Fall 2005Quantum NoiseFluctuation in the number of photons emitted by the x-raysource and recorded by the detector.€ Pk=N0kexp(−N0)k!Pk: Probability of emitting k photons in a given time interval.N0: Average number of photons emitted in that time interval = λt8TT Liu, BE280A, UCSD Fall 2005Transmitted Photons€ Qk=pN0( )kexp(− pN0)k!Qk: Probability of k photons making it through object N0: Average number of photons emitted in that time interval = λtp = exp(−µdz) = probability of proton being transmitted∫TT Liu, BE280A, UCSD Fall 2005Example€ Over the diagnostic energy range, the photondensity is approximately 2.5×1010 photons/cm2/ Rwhere R stands for roentgen (unit for X- ray exposure).A typical chest x - ray has an exposure of 50 mR.For transmission in regions devoid of bone, p = exp(−µdz) ≈ 0.05∫ What are the mean and standard deviation of the number ofphotons that make it it to a 1 mm2 detector?pN0= 0.05 ⋅ 2.5 ×1010⋅.050 ⋅ (.1)2= 6.25 ×105 photonsmean = 6.25 ×105 photonsstandard deviation = 6.25 ×105= 790 photons9TT Liu, BE280A, UCSD Fall 2005Contrast and SNR for X-Rays€ Contrast = C =ΔII SNR =ΔIσI=Mean difference in # of photonsStandard Deviation of # photons=CpN0pN0= C pN0TT Liu, BE280A, UCSD Fall 2005€ C =ΔII =N0exp(−µ1L) − exp −µ1(L − W ) +µ2W( )( )( )N0exp(−µ1L)SNR =CN0A exp(−µ1L)N0A exp(−µ1L)= C N0A exp(−µ1L)µ1µ2LWArea A10TT Liu, BE280A, UCSD Fall 2005Signal to Noise Ratio for CT€ SNR =Cµ σµ=Cµ TMN 2π2ρ033≈ 0.4kCµ d3 / 2MN TC = contrastµ = mean attenuationN = mean number of transmitted photonT = spacing between detectors M = number of viewsρ0= bandwidth of Ram - Lak filter ≈ k/d where d = width of detectork = scaling constant, order unityTT Liu, BE280A, UCSD Fall 2005CT ApplicationsSuetens 200211TT Liu, BE280A, UCSD Fall 2005Virtual ColonoscopySuetens


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UCSD BENG 280A - X-Rays/CT Lecture 2

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