1. In electromagnetic systems, the energy per photon = h. In communication systems, noise can be either quantum or additive from the measurement system ( receiver, etc). The noise power in a communication system is 4kTB, where k is the Boltzman constant,T is the absolute temperature, and B is the bandwidth of the system. When making a measurement (e.g. measuring voltage in a receiver), noise energy per unit time 1/B can be written as 4kT. The in the denominator comes from the standard deviation of the number of photons per time element. Motivation:

2. When the frequency > h In the X-ray region where frequencies are on the order of 10 19, hv >> 4kT So X-ray is quantum limited due to the discrete number of photons per pixel. We need to know the mean and variance of the random process that generate x-ray photons, absorb them, and record them. Motivation: Recall: h = 6.63x10 -34 Js k = 1.38x10 -23 J/K

3. Motivation: We will be working towards describing the SNR of medical systems with the model above. We will consider our ability to detect some object ( here shown in blue)that has a different property, in this case attenuation, from the background ( shown here in green). To do so, we have to be able to describe the random processes that will cause the x-ray intensity to vary across the background. I Contrast = ∆I / I SNR = ∆I /  I = CI /  I ∆I Object we are trying to detect Background

4. The value of a rolled die is a random process. The outcome of rolling the die is a random variable of discrete values. Let’s call the random variable X. We write then that the probability of X being value n is p x (n) = 1/6 1/6 123456 Note: Because the probability of all events is equal, we refer to this event as having a uniform probability distribution

5. 1/6 123456 1 123456 Cumulative Probability Distribution Probability Density Function (pdf)

6. Continuous Random Variables pdf is derivative of cumulative density function 1. cdf is integral of pdf 2. cdf must be between 0 and 1 3. p x (x) > 0 p[x 1 ≤ X ≤ x 2 ] = F(x 2 ) - F(x 1 ) =

7. Zeroth Order Statistics Not concerned with relationship between events along a random process Just looks at one point in time or space Mean of X,  X  or Expected Value of X, E[X] –Measures first moment of p X (x) Variance of X,   X, or E[(X-  ) 2 ] –Measure second moment of p X (x) Standard deviation

8. Zeroth Order Statistics Recall E[X] Variance of X or E[(X-  ) 2 ]

9. p(n) for throwing 2 die 23456 7 891011 12 6/36 1/36 2/36 3/36 4/36 5/36

10. Fair die Each die is independent Let die 1 experiment result be x and called Random Variable X Let die 2 experiment result be y and called Random Variable Y With independence, p XY (x,y) = p X (x) p Y (y) and E [xy] = ∫ ∫ xy p XY (x,y) dx dy = ∫ x p X (x) dx ∫ y p Y (y) dy = E[X] E[Y] if x,y independent

11. 1. E [X+Y] = E[X] + E[Y] Always 2. E[aX] = aE[X] Always 3.  2 x = E[X 2 ] – E 2 [X] Always 4.  2 (aX)= a 2  2 x Always 5. E[X + c] = E[X] + c 6. Var(X + Y) = Var(X) + Var(Y) only if the X and Y are statistically independent. _

12. If experiment has only 2 possible outcomes for each trial, we call it a Bernouli random variable. Success: Probability of one is p Failure: Probability of the other is 1 - p For n trials, P[X = i] is the probability of i successes in the n trials X is said to be a binomial variable with parameters (n,p)

13. Roll a die 10 times. In this game, you win if you roll a 6. Anything else - you lose What is P[X = 2], the probability you win twice?

14. Roll a die 10 times. 6 you win Anything else - you lose P[X = 2] i.e. you win twice = (10! / 8! 2!) (1/6) 2 (5/6) 8 = (90 / 2) (1/36) (5/6) 8 = 0.2907

15. If p is small and n large so that np is moderate, then an approximate (very good) probability is: P[X=i] = e - i / i! Where np = With Poisson random variables, their mean is equal to their variance! E[X] =  x 2 = p[X=i] is the probability exactly i events happen

16. Let the probability that a letter on a page is misprinted is 1/1600. Let’s assume 800 characters per page. Find the probability of 1 error on the page. Binomial Random Variable Calculation. P [ X = 1] = (800! / 799!) (1/1600) (1599/1600) 799 Very difficult to calculate some of the above terms.

17. Let the probability that a letter on a page is misprinted is 1/1600. Let’s assume 800 characters per page. Find the probability of 1 error on the page. P [ X = i] = e - i / i! Here i = 1, p = 1/1600 and n =800, so =np = ½ So P[X=1] = 1/2 e –0.5 =.30 What is the probability there is more than one error per page? Hint: Can you determine the probability that no errors exist on the page?

18.   +   - 

19. 1)Number of Supreme Court vacancies in a year 2) Number of dog biscuits sold in a store each day 3) Number of x-rays discharged off an anode

20. Signal Power Noise Power If X represents power, SNR = E[X]/  x 2 If X represents an amplitude or a voltage, then X 2 represents power. SNR = E[X]/  x

21. X-ray photon d x Light photons Scintillating material High density material stop photons through photoelectric absorption Screen creates light fluorescent photons. These get captured or trapped by silver bromide particles on film. Film

22.  x r h (r) = h(0) cos 3  = h(0) x 3 / (x 2 + r 2 ) 3/2 Since h(0) = K/x 2 K constant x 2 inverse falloff h (r) = Kx / (x 2 + r 2 ) 3/2 Analysis: First calculate spray of light photons from an event at depth x.

23. H 1  = F {h(r)} ∞ = 2π ∫ h(r) J 0 (  r) r dr where h(r) given on previous page 0 H 1  = 2πKe - 2πx  (from a table Hankel transforms) H  = H 1  = e - 2πx  ( Normalize to DC Value) H 1 (0)

24. H  = H 1  = e - 2πx  ( Normalize to DC Value) H 1 (0) Notice this depends on x, depth of event into screen. Let’s come up with based on the likelihood of where events will occur in the scintillating material. F(x) = 1 - e -  x for an infinite screen Probability an x-ray photon will interact in distance x x F(x)

25. p(x) = dF(x) / dx =  e -  x For a screen of thickness d F(x) = 1 - e -  x / 1 - e -  d, then since we are only concerned with captured photons H(p) = ∫ H( ,x) p(x) dx d = (1/ 1 - e -  d )∫ e -2π x   e -  x dx Typical d =.25 mm,  =15/cm for calcium tungstate screen

26. We would like to describe a figure of merit that would describe a cutoff spatial frequency, akin to the bandwidth of a lowpass filter. For a typical screen with d approximately.25 mm and  =15/cm for a calcium tungstate screen, the bracketed term above can be approximated as 1 for spatial frequencies near the cutoff.

27. For moderate k, (i.e. a cutoff frequency) Let (1 - e -  d ) =  the capture efficiency of the screen Then k ≈  / (2πp k +  )  2πk  p k = (1 -  k)  For  k << 1  : As the efficiency increases,  k decreases. This is because  increases as d increases. k Cycles/mm kk Figure of Merit 1.00.110

28. d1d1 d2d2 phosphor Double Emulsion film Intuitively, we would believe this system would work better. Let’s analyze its performance. Let d 1 +d 2 = d so we can compare performance. x

29. H( ,x) = e -2π  (d 1 -x) for 0 < x < d 1 H( ,x) = e -2π  (x –d 1 ) for d 1 < x < d 1 + d 2 d,d 2 H(  ) = (  / 1- e -  d ) { ∫ e -2π  (d 1 - x) e -  x dx + ∫ e -2π  (x –d 1 ) e -  x dx} 0 d 1 = (  / 1- e -  d ) [ ((e -  d 1 - e -2π  d 1 ) / (2π  -  )) + ((e -  d 1 - e -(2π   d 2 +  d) / (2π  + u) )] Again lets determine a cutoff frequency of  k for a low pass filter that has a response of H(  k ) = k, If we assume d 1 ≈ d 2 = d/2, than we can neglect e -2π  d, e -2π  d 1, e -2π  d 2 because they will be small even for relatively small spatial frequencies. e –ud is also small compared to e –ud 1 Since (2π  ) 2 >> u 2 is true for all but lowest frequency, then

30. Compare this cutoff frequency to the single screen cutoff. Concentrate on the factor 2e -  d 1 ( the new factor from the single screen film) Since With  ≈ 0.3, improvement is 1.7 Use improvement to lower dose, quicken exam, improve contrast, or some combination. Improvement is

31. Assuming a circularly symmetric source, I d (x d, y d ) = Kt (x d /M, y d /M) ** (1/m 2 ) s (r d /m) ** h (r d ) Detector response is also circularly symmetric.

32. I d (u,v) = KM 2 T (Mu,Mv) S(mp) · H(p) H 0 (p) Spatial frequencies at detector Object is of interest though I d (u/M,v/M) = KM 2 T (u,v) S((m/M)p) · H(p/M) Product of 2 Low Pass Filters. H 0 (p) = S [(1-z/d) p ] H ((z/d)p)

33. As z  d H(p) S(0) As z  0 S(p) H(0)