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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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 =
Object we are trying to detect
To find I , find the probability density function that describes the # photons / pixel
• ( )
Source Body Detector
N0 is the average number of emitted X-ray photons, or in the
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
Let the number of transmitted photons = N
then describes the signal ,
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)
10 10 photons / cm2 / Roentgen
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
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
Let t = exp [ - ∫ dz ]
Add a recorder with quantum efficiency
Example chest x-ray:
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
What are the zeroth order statistics on Y?
Y = Xm
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)
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.
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.
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.
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
What can we expect for the limit of g deviation)1, the generation rate of light
Actually, half of photons escape and energy efficiency rate of screen is
only 5%. This gives us a g1 = 500
Since g1 >> 1,
2nd Stage deviation)
We still must generate pixel grains
W = ∑ Zm where W is the number of silver grains developed
Y Z W grains / 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 ] = g deviation)2, , the mean number of light photons needed to
develop a grain of film
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?