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What are inside the gantry?. Schematic Representation o f the Scanning Geometry of a CT System. Scanner without covers. Scanner with covers. Source- Detector movement. Source collimation. Detector collimation. Generations. source. detector. Advantages. Disadvantages. No scatter .
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Schematic Representation o f the Scanning Geometry of a CT System
Faster than 1G
High cost and Low efficiency
Faster than 2G
Source Rotates only
Higher efficiency than 3G
Ultrafast for cardiac
faster 3D imaging
faster 3D imaging
Water: 0HU Air: -1000HU
Typical medical scanner display:
12 bit per pixel is required in display.
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.
Mass attenuation coefficient: attenuation per electron or per gram
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.
Imaging Systems for Medical Diagnosis
In CT, the recon algorithm calculates the of each pixel.
x-ray = No e -∫ dz
For each point along a projection g(R), the detector calculates a line integral.
n0=Incoming photon density
of area A
ith line integral
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 /
i = 1
add projections convolution back projection
where R’ = r cos ( - f )
Since ∆ = π/M = M/π ∫ g(R) * c(R)* (R-R’) d
We can view this as:
û = h(r, f ) ** (r, f)
estimate Entire system input image or
and recon process desired image
Recall = M/π ∫ g(R) * c(R)* (R-R') d
H(p) = (M/π) (C(r) / |r|) is system impulse response of CT system
C(r) is the convolution filter that compensates for the
1/|r| weighting from the back projection operation
Let’s get a gain (DC) of 1. Find a C(r) to do this.
We can consider C(r) = |r| a rect(r/ 2r0). 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) |r| rect(p/ 2r0).
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?
= M/π ∫ g(R) * c(R)* (R-R’) d
Then the variance at any pixel
2 = M/π ∫ g2(R)d(R) * [c2(R)]d
variance of any one detector measurement
Theorem described further at end of these notes.
Assume with n = average number of transmitted
photons per unit beam width
and h = width of beam
2 = (M/π) (1/ (nh)) ∫ d ∫ c2 (R) dR = M/nh ∫ c2 (R) dR
Easier to evaluate in frequency domain. Using Parseval’s Rule
2 = M/(nh) ∫ |C(r)|2 dr
The cutoff for our filter C(r) 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.
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.
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)]
N = cos (2π fx + )
f is constant, and is uniformly distributed
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(t), 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
= E [N(x) ∫ N(x + - ) h() d]
= ∫ E [N(x) N(x + - )] h() d
= ∫ RN ( - ) h() d
= RN ( ) * h ( )
What 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)
Ry(0) = ∫ |H(f)|2 •Sx(f) df