Probability and Screen Systems

8/13/2019 Probability and Screen Systems

1/33

In electromagnetic systems, the energy per photon = hn.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 thebandwidth 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.

kThvNNhv

iseAdditiveNohvNNhvSNR

4

Motivation:

N


2/33

When the frequency n> hn

In the X-ray region where frequencies are on the order of 1019, 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.

kThvN

Nhv

iseAdditiveNohvN

NhvSNR4

Motivation:

Recall: h = 6.63x10-34Js

k = 1.38x10-23J/K


3/33

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


4/33

The value of a rolled die is a random process.

The outcome of rolling the die is a random variable of discrete

values. Lets call the random variable X. We write then that

the probability of X being value n is px(n) = 1/6

1/6

1 2 3 4 5 6

Note: Because the probability of all events is equal,we refer to this event as having a uniformprobability

distribution


5/33

1/6

1 2 3 4 5 6

1

1 2 3 4 5 6

Cumulative Probability

Distribution

mn

n

X

X

np

mnpmF

stemDiscreteSy

1

)(

))(()(

)(npX

Probability Density Function (pdf)


6/33

Continuous Random Variables

pdf is derivative of cumulative density function

1)(0

)()(

/)()(

xF

dxxpxF

dxxdFxp

X

XX

XX

1. cdf is integral of pdf

2. cdf must be between 0 and 1

3. px(x) > 0

p[x1 X x2] = F(x2) - F(x1) = 2

1

)(

x

x

X dxxp


7/33

Zeroth Order Statistics

Not concerned with relationship between eventsalong a random process

Just looks at one point in time or space

Mean of X, mX,or Expected Value of X, E[X]

Measures first moment of pX(x)

Variance of X, 2X, or E[(X-m)2]

Measure second moment ofpX(x)

dxxxpXX )(

m

std

dxxpx

X

XX

m )()( 22

Standard deviation


8/33

Zeroth Order Statistics

Recall E[X]

Variance of X or E[(X-m)2]

dxxxpX )(

22222

222

222

22

][][][

][2][

)()(2)(

)()(

m

mm

mm

m

XEXEXE

XEXE

dxxpdxxpxdxxpx

dxxpx

X

X

XXXX

XX


9/33

p(n) for throwing 2 die

2 3 4 5 6 7 8 9 10 11 12

6/36

1/36

2/36

3/36

4/36

5/36


10/33

Fair die

Each die is independent

Let die 1 experiment result be x and called Random Variable XLet die 2 experiment result be y and called Random Variable Y

With independence,

pXY(x,y) = pX(x) pY(y)

and E [xy] = xy pXY(x,y) dx dy= x pX(x) dx y pY(y) dy= E[X] E[Y] if x,y independent


11/33

1. E [X+Y] = E[X] + E[Y] Always2. E[aX] = aE[X] Always

3. 2x= E[X2]E2[X] Always

4. 2(aX) = a22x Always

5. E[X + c] = E[X] + c

6. Var(X + Y) = Var(X) + Var(Y) only if the X and Yare statistically independent.

_


12/33

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)

ini ppiin

niXp

)1(

!)!(

!][


13/33

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?


14/33

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


15/33

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] = x2=

p[X=i] is the probability exactly i events happen


16/33

Let the probability that a letter on a page is misprinted

is 1/1600. Lets assume 800 characters per page. Findthe 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.


17/33

Let the probability that a letter on a page is misprinted

is 1/1600. Lets assume 800 characters per page. Findthe probability of 1 error on the page.

P [ X = i] = e -i / i!

Here i = 1, p = 1/1600 and n =800, so =np =

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?


18/33

m m+ m-

2

2

2

)(

exp2

1

)(

m

x

xpX


19/33

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


20/33

Signal Power

Noise Power

If X represents power,

SNR = E[X]/ x2

If X represents an amplitude or a voltage, then X2represents power.

SNR = E[X]/ x


21/33

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


22/33

x

r

h (r) = h(0) cos3= h(0) x3 / (x2+ r2)3/2

Since

h(0) = K/x2 K constant

x2 inverse falloff

h (r) = Kx / (x2+ r2)3/2

)()cos(

22 rx

x

Analysis: First calculate spray of light photons

from an event at depth x.


23/33

H1()= F{h(r)}

= 2 h(r) J0(2r) r dr where h(r) given on previous page0

H1()= 2Ke-

2x

(from a table Hankel transforms)

H()= H1()= e-2x ( Normalize to DC Value)

H1(0)


24/33

H()= H1()= e-2x ( Normalize to DC Value)

H1(0)

Notice this depends on x, depth of event into screen.Lets come up with based on the likelihood ofwhere events will occur in the scintillating material.

F(x) = 1 - e - mx for an infinite screen

Probability an x-ray photon will interact in distance x

x

F(x)

)(H


25/33

p(x) = dF(x) / dx = me -mx

For a screen of thickness d

F(x) = 1 - e -mx / 1 - e -md , then

since we are only concerned

with captured photons

H(p) = H(,x) p(x) dx

d

= (1/ 1 - e -md ) e-2 x me-mxdx

]1[)1)(2(

)( )2( mmm

m

dd

ee

H

Typical d = .25 mm, m=15/cm for calcium tungstate screen

d

x

e

exp

m

mm

1

)(


26/33

]1[)1)(2(

)( )2( mmm

m

d

d e

eH

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 m=15/cm for a calcium tungstate screen, the bracketed term above

can be approximated as 1 for spatial frequencies near the cutoff.


27/33

For moderate k, (i.e. a cutoff frequency)

Let (1 - e-md) = the capture efficiency of the screen

Then k m/ (2pk + m)2k pk = (1 - k) m

For k


28/33

d1 d2

phosphor phosphor

Double

Emulsion

filmIntuitively, we would believe this system would work better . Letsanalyze its performance. Let d1+d2= d so we can compare

performance.

x


29/33

H(,x) = e -2(d1-x) for 0 < x < d1

H(,x) = e -2(xd1) for d1< x < d1+ d2

d, d2

H() = (m/ 1- e-md

) { e-2(d

1

- x)

e- mx

dx + e-2(xd

1

)

e- mx

dx}0 d1

= (m/ 1- e -md) [ ((e -md1 - e -2d1) / (2- m))+ ((e -md1 - e -(2d2+ md) / (2+ 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 d1 d2= d/2, than we can neglecte -2d, e -2d1, e -2d2 because they will be small even for relatively

small spatial frequencies. e ud is also small compared toe ud1

Since (2)2>> u2 istrue for all but lowest frequency, then

12

2

d

k e

k

m

m

k

d

k ekpH

m

mm

m

2

2

)(

2

2

2

1

2

1

1


30/33

Compare this cutoff frequency to the single screen cutoff.

Concentrate on the factor 2e-md1 ( 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.

)2(2

1ud

k ek

m

mm

m

m

12

1222

1

1

12/

2/

dd

d

d

eeSo

e

e

Improvement is


31/33

Assuming a circularly symmetric source,

Id

(xd

, yd

) = Kt (xd

/M, yd

/M) ** (1/m2) s (rd

/m) ** h (rd

)

Detector response is also circularly symmetric.


32/33

Id(u,v) = KM2T (Mu,Mv) S(mp) H(p)

H0(p)

Spatial frequencies at detectorObject is of interest though

Id(u/M,v/M) = KM2T (u,v) S((m/M)p) H(p/M)

Product of 2 Low Pass Filters.

H0(p) = S [(1-z/d) p ] H ((z/d)p)


33/33

As z d

H(p)

S(0)

As z 0

S(p)

H(0)

Documents

Probability and Screen Systems