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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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slide1

What are inside the gantry?

Schematic Representation o f the Scanning Geometry of a CT System

slide4

Source- Detector movement

Source collimation

Detector collimation

Generations

source

detector

Advantages

Disadvantages

No scatter

Pencil beam

Trans.+Rotates

1st Gen.

single

single

no

slow

Fan- beamlet

Trans.+Rotates

Faster than 1G

Low efficiency

2nd Gen.

yes

multiple

single

High cost and Low efficiency

Faster than 2G

Fan- beam

Rotates together

3rd Gen.

many

single

no

Source Rotates only

Higher efficiency than 3G

Fan- beam

high scatter

Stationary ring

4th Gen.

single

no

No movement

Stationary ring

Fan- beam

Ultrafast for cardiac

high cost

5th Gen.

multiple

no

3rdGen.+

bed trans.

Fan- beam

faster 3D imaging

higher cost

6th Gen.

single

many

yes

Narrow

cone- beam

3rdGen.+

bed trans.

faster 3D imaging

7th Gen.

single

Multiple arrays

higher cost

yes

wide

cone- beam

Relatively slow

3rd Gen.

8th Gen.

single

no

Large 3D

FPD

slide5

What is displayed in CT images?

Water: 0HU Air: -1000HU

Typical medical scanner display:

[-1024HU,+3071HU],

Range:

12 bit per pixel is required in display.

slide7

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

L

W

0

-1024

slide8

Mass density

Mass attenuation coefficient: attenuation per electron or per gram

Reminder:

slide13

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.

Krestel-

Imaging Systems for Medical Diagnosis

slide14

CT SNR

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

Detector

Ni

slide15

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  / 

End Goal

slide16

Discrete Backprojection over M projections

M

  • (x,y) = ∑ g(R) * c(R) * (R-R’) ∆

i = 1

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 = h(r, f ) ** (r, f)

estimate Entire system input image or

and recon process desired image

slide17

C(p)

p0

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.

slide18

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?

slide19

Recall

π

 = M/π ∫ g(R) * c(R)* (R-R’) d

0

Then the variance at any pixel

π

2 = M/π ∫ g2(R)d(R) * [c2(R)]d

0

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

convolution

slide20

C(p)

p0

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(r)|2 dr

-∞

p/M

slide21

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.

slide22

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.

slide23

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)]

slide24

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))

Sample Problem

slide25

Autocorrelation Statistic: RN ( )

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

slide26

Frequency Domain Expression for 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 ( ) 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?

N

Y

H

slide27

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-Correlation

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

-∞

slide28

∫ 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