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1. 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 = 

2. Signal to Noise Ratio Object we are trying to detect ∆I I Background Definitions:

3. 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.

4. 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

5. Recall, Let the number of transmitted photons = N then describes the signal , Another way,

6. 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

7. 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

8. 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

9. 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

10. 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?

11. 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.

12. 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.

13. 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.

14. 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.

15. 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

16. 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,

17. 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

18. Let E [ Z ] = g2, , the mean number of light photons needed to develop a grain of film

19. 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?