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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:.

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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 = 


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Signal to Noise Ratio

Object we are trying to detect

∆I

I

Background

Definitions:


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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.


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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


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Recall,

Let the number of transmitted photons = N

then describes the signal ,

Another way,


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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


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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


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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


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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


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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?


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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.


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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.


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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.


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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.


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deviation)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


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What can we expect for the limit of g deviation)1, 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,


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2nd Stage deviation)

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


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Let E [ Z ] = g deviation)2, , the mean number of light photons needed to

develop a grain of film


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Recall g deviation)1 = 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?


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