Schematic Representation o f the Scanning Geometry of a CT System What are inside the gantry?
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Transcript of Schematic Representation o f the Scanning Geometry of a CT System What are inside the gantry?
Schematic Representation o f the Scanning Geometry
of a CT System
What are inside the gantry?
Scanner without covers
Scanner with covers
DisadvantagesGenerations sourceSource
collimation detectorDetector
collimation
Source- Detector movement
Advantages
single
single
single
single
single
single
multiple
Pencil beam
Fan- beamlet
Fan- beam
Fan- beam
Fan- beam
Fan- beam
Narrow
cone- beam
single
multiple
many
Stationary ring
many
Stationary ring
Multiple arrays
1st Gen.
2nd Gen.
3rd Gen.
4th Gen.
5th Gen.
6th Gen.
7th Gen.
no
yes
no
no
no
yes
yes
Trans.+Rotates Trans.+Rotates Rotates together
Source Rotates only
No movement
3rdGen.+
bed trans.3rdGen.+
bed trans.
No scatter
Faster than 1G
Faster than 2G
Higher efficiency than 3G
Ultrafast for cardiac
faster 3D imaging
faster 3D imaging
slow
Low efficiency
High cost and Low
efficiency
high scatter
high cost
higher cost
higher cost
8th Gen. singlewide
cone- beamFPD no 3rd Gen. Large 3D Relatively
slow
What is displayed in CT images?
HU1000CT# T
water
water
Water: 0HU Air: -1000HU
Typical medical scanner display:
[-1024HU,+3071HU],
Range: 1224096 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.
WL
-1024
+3071
0
255
2/#
2/#2/
2/#
)2/(#
0
scaleGray Displayed
max
max
WLCT
WLCTWL
WLCT
I
IW
WLCT
),( ZE Mass density
Mass attenuation coefficient: attenuation per electron or per gram
Reminder:
4~3
3
Z
E
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
Krestel-Imaging Systems for Medical Diagnosis
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.
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,y)
Detector
ith line integral
Ni
X-ray Sourceof area A
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 /
M
(x,y) = ∑ g(R) * c(R) * (R-R’) ∆ i = 1
add projections convolution back projection where R’ = r cos ( - )
π
Since ∆ = π/M = M/π ∫ g(R) * c(R)* (R-R’) d 0
We can view this as:
û = h(r, ) ** (r, )estimate Entire system input image or
and recon process desired image
Discrete Backprojection over M projections
Recall = M/π ∫ g(R) * c(R)* (R-R') dH(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(/ 20). Find constant a
H(0) = (M/π) a a = π/M
If we set H(0) = 1, DC gain is 1.
Therefore, C(p) = (π/M) || rect(/ 2).
C(p)
p0
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 π
2 = M/π ∫ g
2(R)(R) * [c2(R)] d 0
variance of any one detector measurementTheorem described further at end of these notes.
Assume with n = average number of transmittedphotons per unit beam widthand h = width of beam
hng
12
convolution
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/(nh) ∫ |C()|2 d
-∞
C(p)
p0
2/30
30
22
2/3
3
2
phMnC
CSNR
p
hMn
/M
dpM
phn
Mp
p2
22
0
0
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
Combine all the constants
n was defined over a continuous projection
Let N = nA = nwh = average number of photons per detector element.
2/3whMnCKSNR
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 operationAnother way of looking at it, there is a penalty for oversampling the center or the Fourier space.
wMNCKSNR
First Order Statistics ( What we have studied) m = E[X] = x2 = E[(x - x)2]
Second Order statistic ( Important if we can’t assume independence)
RN (x1 , x2) = E [N(x1) N(x2)]
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.
Given an example random process whereN = 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))
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(), depends only on the relative distance between two points ( time points, voxels, etc). RN ( ) 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 ( ) RN ( ) = E [ N(x) N(x + )]
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
Autocorrelation Statistic: RN ( )
Power spectral density of a Random Process NWe can’t take a meaningful Fourier transform of a random process. But a Fourier transform of RN() gives us its power spectrum. This is an indication of where the random processes power resides as a function of frequency.
SN (f) = ∫ RN ( ) e -i 2π f d
RN ( ) = 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?
HN Y
RY,N ( ) = E [Y(x + ) N(x)] = E [N(x) ∫ N(x + - ) h() d] ∞
= ∫ E [N(x) N(x + - )] h() d -∞
∞
= ∫ RN ( - ) h() d -∞
= RN ( ) * h ( )Cross-CorrelationWhat about the autocorrelation of the output Y? That is RY ( ) .
∞ ∞
E [Y(x) Y(x + )] = E [ ∫ h() N(x - ) d • ∫ h() N(x + - ) d ] -∞ -∞
But h(), h() are deterministic. ∞
= ∫ h() h() E[N(x - ) N(x + - )] d d -∞
∞
∫ h() h() RN ( + - ) d d-∞
∞
∫ h()• [h() * RN ( + )] d-∞
RN ( ) = h(-) * h( ) * RN ( )
h(-) H (-f) if real h( )H(-f) = H*(f)
Sy(f) = H*(f)• H(f) • Sx(f)
Sy(f) = |H(f)|2 • Sx(f)Average power
Ry(0) = ∫ |H(f)|2 • Sx(f) df