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## Review

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**Review**Poisson Random Variable P [X = i ] = e- i / i! i.e. the probability that the number of events is i E [ X ] = for Poisson Random Variable X 2 = **Signal to Noise Ratio**Object we are trying to detect ∆I I Background Definitions:**To find I , find the probability density function that**describes the # photons / pixel • ( ) Source Body Detector • X-ray emission is a Poisson process N0 is the average number of emitted X-ray photons, or in the Poisson process.**2) Transmission -- Binomial Process**transmitted p = e - ∫ u(z) dz interacting q = 1 - p 3) Cascade of a Poisson and Binary Process still has a Poisson Probability Density Function - Q(k) represents transmission process Still Poisson, with = p N0 Average Transmission =pN0 Variance = pN0**Recall,**Let the number of transmitted photons = N then describes the signal , Another way,**What limits SNR?**Units of Exposure (X) = Roentgen (R) is defined as a number of ion pairs created in air 1 Roentgen = 2.58 x 10 -4 coulombs/kg of air Ionizing energy creates energy in the body. Dose refers to energy deposition in the body. Units of Dose (D): ergs/ gram (CGS) or J/kg (SI) Section 4.6 of textbook**What limits SNR?**1 Rad - absorbed dose unit: expenditure of 100 ergs/gram 1 R produces 0.87 Rad in air How do Rads and Roentgen relate? -depends on tissue and energy -Rads/Roentgen > 1 for bone at lower energies -Rads/ Roentgen approximately 1 for soft tissue -independent of energy Section 4.6 of textbook**Photons per Pixel**N = AR exp[ - ∫ dz ] R = incident Roentgens A = pixel area (cm2) 3.0 10 10 photons / cm2 / Roentgen 0.5 160 20 Photon Energy**Dose Equivalent:**H(dose equivalent) = D(dose) * Q(Quality Factor) Q is approximately 1 in medical imaging Q is approximately 10 for neutrons and protons Units of H = 1 siever (Sv)= 1 Gray .01 mSv= .001 Rad = 1 mrem Background radiation: 280-360 mrem/yr Typical Exams: Chest X-ray = 10 mRad = 10 mrems CT Cardiac Exam = Several Rad Quantitative Feeling For Dose Fermilab Federal Limits : 5 Rads/year No one over 2.5**Let t = exp [ - ∫ dz ]**Add a recorder with quantum efficiency Example chest x-ray: 50 mRad = 0.25 Res = 1 mm t = 0.05 What is the SNR as a function of C?**Have we made an image yet?**Consider the detector M X light photons / capture Y light photons Transmitted And captured Photons Poisson What are the zeroth order statistics on Y? M Y = Xm m=1 Y depends on the number of x-ray photons M that hit the screen, a Poisson process. Every photon that hits the screen creates a random number of light photons, also a Poisson process.**What is the mean of Y? ( This will give us the signal level**in terms of light photons) Mean Expectation of a Sum is Sum of Expectations (Always) Each Random Variable X has same mean. There will be M terms in sum. There will be M terms in the sum E [Y] = E [M] E [X] Sum of random variables E [M] = N captured x-ray photons / element E [X] = g1 mean # light photons / single x-ray capture so the mean number of light photons is E[Y] = N g1.**What is the variance of Y? ( This will give us the std**deviation) We will not prove this but we will consider the variance in Y as a sum of two variances. The first will be due to the uncertainty in the number of light photons generated per each X-ray photon, Xm. The second will be an uncertainty in M, the number of incident X-ray photons. To prove this, we would have to look at E[Y2]. The square of the summation would be complicated, but all the cross terms would greatly simplify since each process X in the summation is independent of each other.**What is the variance of Y? ( This will give us the std**deviation) If M was the only random variable and X was a constant, then the summation would simply be Y = MX. The variance of Y, s2y=X2 s2m Recall multiplying a random variable by a constant increases its variance by the square of the constant. X is actually a Random variable, so we will write X as E[X] and the uncertainy due to M as s2y=[E[X]]2 s2m If M were considered fixed and each X in the sum was considered a random variable, then the variance of the sum of M random variables would simply be M * s2x . We can make this simplification since each process that makes light photons upon being hit by a x-ray photon is independent of each other.**M2 = N Recall M is a Poisson Process**X2 = g1 Generating light photons is also Poisson Y2 = Ng1 +Ng12 Uncertainty due to X Uncertainty due to M Dividing numerator and denominator by g1**What can we expect for the limit of g1, the generation rate**of light photons? Actually, half of photons escape and energy efficiency rate of screen is only 5%. This gives us a g1 = 500 Since g1 >> 1,**2nd Stage**We still must generate pixel grains Y W = ∑ Zm where W is the number of silver grains developed m=1 Y Z W grains / pixel Light Photons / pixel Z = developed Silver grains / light photons Let E[Z] = g2 , the number of light photons to develop one grain of film. Then, z2 = g2 also since this is a Poisson process, i.e. the mean is the variance. E[W] = E[Y] E[Z] W2 = E[Y] z2 + Y2 E2[Z] uncertainty in gain factor z uncertainty in light photons**Let E [ Z ] = g2, , the mean number of light photons needed**to develop a grain of film**Recall g1 = 500 ( light photons per X-ray)**g2 = 1/200 light photon to develop a grain of film That is one grain of film requires 200 light photons. Is 1/g1 g2 <<1? Is 1/g1 << 1? What is the lesson of cascaded gains we have learned?