Transcript Document

What are inside the gantry?
Schematic Representation o f the Scanning Geometry
of a CT System
Scanner without covers
Scanner with covers
SourceSource
Detector
Generations source
detector
Advantages Disadvantages
collimation
collimation Detector
movement
1st Gen.
2nd Gen.
single
single
Pencil
beam
Fanbeamlet
Fanbeam
single
no
multiple
yes
3rd Gen.
single
4th Gen.
single
Fanbeam
Stationary
ring
no
multiple
Fanbeam
Stationary
ring
no
5th Gen.
6th Gen.
single
7th Gen. single
8th Gen.
single
Fanbeam
Narrow
many
many
Multiple
cone- beam arrays
wide
FPD
cone- beam
no
yes
yes
Trans.+
Rotates
Trans.+
Rotates
Faster
than 1G
Faster
Rotates
than 2G
together
Source
Higher
Rotates
efficiency
only
than 3G
No
Ultrafast
movement for cardiac
3rdGen.+
bed trans.
3rdGen.+
bed trans.
no
No
scatter
3rd Gen.
faster 3D
imaging
faster 3D
imaging
Large 3D
slow
Low
efficiency
High cost
and Low
efficiency
high
scatter
high
cost
higher
cost
higher
cost
Relatively
slow
What is displayed in CT images?
T  water
CT #
 1000HU
water
Water: 0HU
Air: -1000HU
Typical medical scanner display:
[-1024HU,+3071HU],
Range:
12
4096 2
12 bit per pixel is required in display.
Hounsfield scales for typical tissues
For most of the display device, we can only display 8 bit gray scale.
This can only cover a range of 2^8=256 CT number range. Therefore,
for a target organ, we need to map the CT numbers into [0,255] gray
scale range for observation purpose. A window level and window
width are utilized to specify a display.
+3071
255
W

0
 CT #( L  W / 2)
DisplayedGray scale  
I max
W

I max

L
0
-1024
CT #  L  W / 2
L  W / 2  CT #  L  W / 2
CT #  L  W / 2

   ( E , Z )  

Mass density
Mass attenuation coefficient: attenuation per electron or per gram
Reminder:

3
E


3~ 4
Z

Windowing in CT image display
Multi-row CT detector (I) GE Light Speed
Multi-row CT detector (II) Siemens Sensation
Future 256-slice cone-beam CT detector
SNR is dependent on dose, as in X-ray.
Notice how images become grainier and our
ability to see small objects decreases as dose
decreases. Next slides discuss analysis of
SNR in CT. We will see some similarities
with X-ray. But we also see some important
differences.
KrestelImaging Systems for Medical Diagnosis
In CT, the recon algorithm calculates the  of each pixel.
x-ray = No e -∫  dz
recorder intensity
For each point along a projection g(R), the detector calculates a
line integral.
n0=Incoming photon density
X-ray Source
of area A
(x,y)
ith line integral
Ni
Detector
Ni = n0 A exp ∫i - dl = N0 exp ∫i - dl where A is area of detector and
N0 = n0 A
The calculated line integral is ∫i  dl = ln (N0/Ni)
Mean = ≈ ln (N0/Ni)
2(measured variance) ≈ 1/Ni
Now we use these line integrals to form the projections g(R). These
projections are processed with convolution back projection to make
the image.
SNR = C  / 
Discrete Backprojection over M projections
M
(x,y) = ∑
i=1
g(R) * c(R) * (R-R’) ∆
add projections
convolution back projection
where R’ = r cos ( - f )
π
Since ∆ = π/M   = M/π ∫ g(R) * c(R)* (R-R’) d
0
We can view this as:
û
=
estimate
h(r, f )
**
Entire system
and recon process
(r, f)
input image or
desired image
Recall  = M/π ∫ g(R) * c(R)* (R-R') d
H(p) = (M/π) (C() / ||) is system impulse response of CT system
C() is the convolution filter that compensates for the
1/|| weighting from the back projection operation
Let’s get a gain (DC) of 1. Find a C() to do this.
We can consider C() = || a rect(/ 20). Find constant a
C(p)
H(0) = (M/π) a  a = π/M
If we set H(0) = 1, DC gain is 1.
p0
Therefore, C(p) = (π/M) || rect(p/ 20).
This makes sense – if we increase the number of angles M,
we should attenuate the filter gain to get the same gain.
At this point, we have selected a filter for the convolution-back
projection algorithm. It will not change the mean value of the CT
image. So we just have to study the noise now.
The noise in each line integral is due to differing numbers of photons.
The processes creating the difference are independent.
- different section of the tube, body paths, detector
What does this imply about the noise properties along the projection?
The set of projections?
What does this say for a plan of attack?
What effect does the convolution have on the noise?
Recall
π
 = M/π ∫ g(R) * c(R)* (R-R’) d
0
Then the variance at any pixel
convolution
π
2 = M/π ∫ g2(R)(R) * [c2(R)] d
0
variance of any one detector measurement
Theorem described further at end of these notes.
1
Assume  
with n = average number of transmitted
nh
photons per unit beam width
2
g
and h = width of beam
π
2 = (M/π) (1/ (nh)) ∫ d ∫ c2 (R) dR = M/nh ∫ c2 (R) dR
0
Easier to evaluate in frequency domain. Using Parseval’s Rule
2
M
p
2

p
dp
2

n h  p0 M
p0
∞
2 = M/(nh) ∫ |C()|2 d
-∞
p/M
C(p)
p0

2


p 2 2 p03
n hM 3
 C 3 / 2n hM 3 / 2
SNR  C

p0

p
The cutoff for our filter C() will be matched to the detector width w.
Let’s let p0 = K/w where K is a constant
3/ 2

SNR  K C n hM w
Combine all the constants
n was defined over a continuous projection
Let N = nA = nwh = average number of photons per detector element.
SNR  K C NM w
In X-ray, SNR  √N
For CT, there is an additional penalty. To see this, cut w in ½.
What happens to SNR?
Why Due to convolution operation
Another way of looking at it, there is a penalty for oversampling
the center or the Fourier space.
Supplementary Random Process Material
The following slides may be interesting to someone who
has had some background in random processes. It will
show how power spectral density analysis is useful in
understanding imaging systems. No exams in the class will
cover this material. This material is the foundation for the
CT noise derivation.
First Order Statistics ( What we have studied)
m = E[X] = x
2 = E[(x - x)2]
Second Order statistic ( Important if we can’t assume
independence)
RN (x1 , x2) = E [N(x1) N(x2)]
Given an example random process where
N = cos (2π fx + )
f is constant, and  is uniformly distributed
0  2π
RN (x1 ,x2) = ∫ cos (2π fx1 + ) cos (2π fx2 + ) p() d
Use cos (a) cos (b) = 1/2 (cos (a - b) + cos (a + b))
p() = 1/2π
RN (x1 ,x2) = 1/2π∫1/2[( cos (2π (fx1 + fx2) + 2  )
+ (cos(2π(fx1 – fx2)] d
First term integrates to 0 across all 
= ½ cos(2π(fx1 – fx2))
Autocorrelation Statistic: RN (t )
If mN (x) = m for all x ( i.e. mean stays constant) and the random
process is said to be wide sense stationary, then the autocorrelation
statistic, RN(t), depends only on the relative distance between two
points ( time points, voxels, etc). RN (t ) is a measure of the
information one can deduce about a random process if we know the
value of the random process at another location.
RN (x1 ,x2) = RN (t )
RN (t ) = E [ N(x) N(x + t )]
The value of the autocorrelation function at 0 represents average
power of the random process. This is helpful in measuring noise
power.
RN (0) = E [ N2 (x)] Measure of average power of random process
Power spectral density of a Random Process N
We can’t take a meaningful Fourier transform of a random process. But a
Fourier transform of RN(t) gives us its power spectrum. This is an indication of
where the random processes power resides as a function of frequency.
SN (f) = ∫ RN (t ) e -i 2π f t dt
RN (t ) = F-1{SN (f)}
= ∫ SN (f) e i 2π xf df
∞
E [N2 (x)] = Rx (0) = ∫ SN (f) df
-∞
How do statistics change after random process is operated by a
linear system?
N
H
Y
RY,N (t ) = E [Y(x + t ) N(x)]
= E [N(x) ∫ N(x + t - ) h() d]
∞
= ∫ E [N(x) N(x + t - )] h() d
-∞
∞
= ∫ RN (t - ) h() d
-∞
= RN (t ) * h (t )
Cross-Correlation
What about the autocorrelation of the output Y? That is RY (t ) .
∞
∞
-∞
-∞
E [Y(x) Y(x + t )] = E [ ∫ h() N(x - ) d • ∫ h() N(x + t - ) d ]
But h(), h() are deterministic.
∞
= ∫ h() h() E[N(x - ) N(x + t - )] d d
-∞
∞
∫ h() h() RN (t +  - ) d d
-∞
∞
∫ h()• [h(t) * RN (t + )] d
-∞
RN (t ) = h(-t) * h(t ) * RN (t )
h(-t)  H (-f)
H(-f) = H*(f)
Sy(f) = H*(f)• H(f) • Sx(f)
if real h(t )
Sy(f) = |H(f)|2 • Sx(f)
Average power
Ry(0) = ∫ |H(f)|2 • Sx(f) df