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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 =

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

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

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

N = AR exp[ - ∫ dz ]

R = incident Roentgens

A = pixel area (cm2)

3.0

10 10 photons / cm2 / Roentgen

0.5

160

20

Photon Energy

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

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?

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,

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?

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