Transcript Document

Poisson Random Variable
P [X = i ] = e- i / i!
i.e. the probability that the number of events is i
E[X]=
X 2 = 
for Poisson Random Variable
Signal to Noise Ratio
Object we are trying to detect
∆I
I
Background
Definitions:
I
Contrast 
I
I
SNR 
I
SNR 
CI
I
To find I , find the probability density function that describes the #
photons / pixel
•
(
Source
)
Body
Detector
1) X-ray emission is a Poisson process
 N0
kemissions
k e
P(
)  N0
unit  tim e
k!
N0 is the average number of emitted X-ray photons, or  in the
Poisson process.
2) Transmission -- Binomial Process
transmitted
interacting
p = e - ∫ u(z) dz
q=1-p
3) Cascade of a Poisson and Binary Process still has a Poisson
Probability Density Function
- Q(k) represents transmission process
Q(k )  ( pN0 )
k
e
 pN 0
k!
Still Poisson, with  = p N0
Average Transmission =pN0
Variance = pN0
Recall,
I
Contrast 
I
I
SNR 
I
SNR 
CI
I
Let the number of transmitted photons = N
then  N describes the signal ,
SNR 
Another way,
CN
N
SNR 

 C N whereC 
N
N
N
CI
N
N
N

N
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 in100the body. Dose refers to energy
deposition in the body.
Units of Dose (D): ergs/ gram (CGS) or J/kg (SI)
ergs
1 Rad  100
gram
ergs 103 g 10
100
(
)(
g
kg erg
1 Gray  100 Rads
7
J )  .01Gray (SI unit)
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
10 10 photons / cm2 / Roentgen
N = AR exp[ - ∫ dz ]
R = incident Roentgens
A = pixel area (cm2)
P hotonFluence/Roentgen
3.0
2
10 Photons/ cm

 2.5  10
R
R
0.5
20
160
Photon Energy
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
Let t = exp [ - ∫ dz ]
SNR  C N  C ARt
Add a recorder with quantum efficiency 
SNR  C N  C ARt
Example chest x-ray:
50 mRad
 = 0.25
Res = 1 mm
t = 0.05
What is the SNR as a function of C?
SNR  C N  C N
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 = m=1
 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)
M
Y   Xm
m 1
Mean
M
E[Y ]  E[ X m ]
m 1
Expectation of a Sum is Sum of Expectations (Always)
E[Y ]  E[ X 1 ]  E[ X 2 ]  ....E[ X m ]
Each Random Variable X has same mean. There will be M terms in
sum.
E[Y ]  E[ X ]  E[ X ]  ....E[ X ]
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)
M
Y   Xm
m 1
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)
M
Y  m1 X m
If M was the only random variable and X was a constant,
then the summation would simply be Y = MX.
The variance of Y, 2y=X2 2m 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 2y=[E[X]]2 2m
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 * 2x . 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 M
Uncertainty due to X
SNR 
CE[Y ]
y

CNg1

Ng1 1  g1
Dividing numerator and denominator by g1

C N
1
1
g1
CNg1
Ng1  Ng12
What can we expect for the limit of g1, the generation rate of light
photons?
g1 
h xray
h light
light 5000A


 20,000

xray
.25A

Actually, half of photons escape and energy efficiency rate of screen is
only 5%. This gives us a g1 = 500
Since g1 >> 1,
SNR  C N
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
+
uncertainty in gain factor z
Y2 E2[Z]
uncertainty in light photons
Let E [ Z ] = g2, , the mean number of light photons needed to
develop a grain of film
 w2  E[Y ] Z2   Y2 E 2 [ Z ]
  g1Ng 2  (Ng1  Ng ) g
2
w
2
1
2
2
 w  g1 g 2 N 1  1 / g1  1 / g1 g 2
SNR 
CE[W ]
W

C N
1  1 / g1  1 / g1 g 2
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?
SNR 
C N
1
1  1 / 500 
500
200
SNR  0.85C N
What is the lesson of cascaded gains we have learned?