Tecniche Diagnostiche 3 – TC

Tecniche Diagnostiche 3 – TC

Corso di laurea in Fisica A.A. 2002-2003

Computed Tomography Principles

• 1. Projection measurement

• 2. Scanner systems

Basic Tomographic Principle

The internal structure of an object can be reconstructed from multiple projections of the object.

Exponential Attenuation of X-ray

Ray-Sum of X-ray Attenuation

Computed Tomography Principles

• 1. Projection measurement

• 2. Scanning modes

Projection & Sinogram

Computed tomography (CT): image reconstruction from projections P(θ,t) Î f(x, y)

Computed Tomography Geometry

reconstruction matrix

focal spot detector

Matrix Representation of a Tissue Slice in CT

•

•

CT Display Scale

linear attenuation coefficient, µ(x,y,z⁾ reconstructed image

displayed image

Linear Attenuation Coefficients (60 keV)

Tissue

• Brain - Grey

• Brain - White

• Cerebro-Spinal Fluid (CSF)

• Pancreas

• Liver

• Water

• Fat

µ ⁽ χµ

⁾

• .213

• .215

• .208 - .213

• .215

• .221

• .205

• .190

Grey - White Matter Contrast

C = (.215 - .213)cm^-1 * 1.0 cm

= .2 %!

CT Number µ(cm^-1) 39

49

.213 .215

CT number allows the computer to present the information with a larger grey scale

Variation of Linear Attenuation Coefficients with Energy

0.14 0.16 0.18 0.2 0.22

60 70 80

Energy (keV) µ

Water Fat

Variation of H.U. with Energy

Energy (keV) 60 70 80

µ(water) .205 .193 .184

µ(fat) .190 .179 .171

H.U.(fat) -73 -73 -73

Image Display

CT Number

- Hounsfield unit

• Air: -1024

• Water: 0

• Bone: +175 to +3071 Viewing Parameters

• Window level (L)

• Window width (W)

• Zoom factor

water water

HU µ

µ µ ) = 1000 µ ⁻ (

-1024 +3071

0 255

W L

Computed Tomography Principles

• 1. Projection measurement

• 2. Scanning modes

• 3. Scanner systems

• 4. Image reconstruction

CT Scanner

Data Acquisition System (DAS)

Data Acquisition System (DAS)

First Generation

One detector Translation-rotation

Parallel-beam

160 rays x 180 views 5 minutes/per slice

First Generation CT Scanner

From Webb, Physics of Medical Imaging

Second Generation

Multiple detectors Translation-rotation Small fan-beam

Second Generation CT Scanner

From Webb, Physics of Medical Imaging

Third Generation

Multiple detectors Translation-rotation Large fan-beam

800 rays x 1000 views

<1 seconds/per slice

Ring Artifact In 3 ^rd Generation

Fourth Generation

Detector ring Source-rotation Large fan-beam

Fourth Generation

Detector fan eliminates ring artifacts.

3 ^rd and 4 ^th Generation

Scanners

Krestel-

Imaging Systems for Medical Diagnosis

Third & Fourth Generations

Fifth Generations

Electron-beam CT for cardiac imaging.

Electron Beam CT

Sixth Generation:

Spiral/Helical/Volumetric CT

Continuous &

Simultaneous :

• Source rotation

• Patient translation

• Data acquisition

Volume Scanning

Scan-Translate Patient, Scan-Translate Patient,

Important years in helical CT history

Single-slice 1989

Dual-slice

1992 Quad-slice

1998

Quad- Slice Single- Slice

8 times faster than single-slice

One rotation / sec

Two rotations / sec +

4 slices / rotation

Why is faster better?

•Improved temporal resolution

•Faster scanning causes less motion artifacts

•Breath holding time is reduced

•Improved spatial resolution

•Narrower collimation leads to higher resolution in the z-axis (MPR)

•Narrower collimation reduces partial volume effect

•Improved contrast media concentration

•Higher contrast media concentration due to faster infusion

•Better separation of arterial and venous phases

•Increased power (mAs)

•The widened x-ray beam and sampling of multiple slices for each rotation

allows for raised mAs

•Decreased image noise

•A direct effect of raised mAs

•Efficient x-ray tube utilization

•Faster scanning causes markedly less waiting for tube cooling

•More images from x-ray tube during tube life cycle

Pitch

Pitch = table travel (mm) per gantry rotation ⁽¹⁾ beam collimation (mm)

Information about table travel relative to beam collimation

> 1

= 1

Pitch

Volume

Volumetric scanning Viewing

Reformatting

x y

z

Reformating multiple slices into a volume produces a volume “image” with unequal

spatial resolution in x, y, and z.

Reformatting with Interpolation

Scan a volume – view a volume

The image data from the volume can be merged into single images.

Coronal Reformat 25, 50, 100,

400, 1000 images can be reconstructed from volume.

Computed Tomography Principles

• 1. Projection measurement

• 2. Scanning modes

• 3. Scanner systems

• 4. Image reconstruction

• Uses a collimator to keep exposure to a slice

• Builds image from multiple projections

• We will assume parallel rays for now.

• Actually how first scanners worked.

• Translate and then rotate as shown in diagram

Physics of Medical Physics, EditorWebb

The 1D Fourier Transform of a projection at angle θforms a line in the 2D Fourier space of the image at the same angle.

Incident y X-rays

θ R

Incident x-rays pass through the object f(x,y) from upper left to lower right at the angle 90 + θ. For each point R, a different line integral describes the result on the function g_θ(R). g_θ(R) is measured by an array of detectors or a moving detector.

The thick line is described by x cos θ+ y sin θ= R

I have just drawn one thick line to show one line integral, but the diagram is general and pertains to any R.

The projection g_θ(R) can thus be calculated as a set of line integrals, each at a unique R.

g_θ(R) = ∫ ∫ f(x,y) δ(x cosθ+ y sin θ- R) dx dy g_θ(R) = ∫ ∫ f(r, φ) δ2π ∞ (r cos (θ-φ) - R) r dr dφ

0 0

In the second equation, we have translated to polar coordinates.

Again g_θ(R) is a 1D function of R.

Let’s consider the 2D FFT

F(u,v) = ∫ ∫ f(x,y) e -i 2π (ux + vy) dx dy In polar coordinates,

µ= ρcosβ ν= ρsin β

F(ρ,β) = ∫ ∫ f(x,y) exp [ -i 2π ρ(x cosβ+ y sin β)] dx dy

When x cosβ+ y sin β= constant, exp [ -i 2π ρ(x cosβ+ y sin β) is a linear phase shift. This is the Fourier transform of a shifted delta function. Let’s let the constant = R and write the complex exponential as the FT of a δfunction.

F(ρ,β) = ∫ ∫ f(x,y) F[δ( x cos B + y sin B - R)] dx dy

F(ρ,β) = ∫_R∫ ∫ f(x,y) [δ( x cos B + y sin B - R)] e^{-i 2πρR} dx dy dR

F(ρ,β) = ∫ ∫ f(x,y) ∫ [δ( x cos B + y sin B - R)] e^{-i 2πρR} dx dy dR Recall how we wrote the projection as a double integral of f(x,y) where a delta function performs the line integral,

g_θ(R) = ∫ ∫ f(x,y) δ(x cosθ+ y sin θ- R) dx dy We take the Fourier Transform of g_θ(r):

F{g_θ(r)}= ∫_R[ ∫_y∫_x f(x,y) δ( x cos Β+ y sin B - R) dx dy] e^{-i 2πρR}dR

Central Section Theorem

Again, F{g_θ(r)} = F(ρ,B) and since B = θin our proof, then the F{g_θ(r)} = F(ρ, θ)

So in words, the Fourier transform of a projection at angle θgives us a line in the polar Fourier space at the same angle θ.

Crude Idea 1: Take each projection and smear it back along the lines of integration it was calculated over.

Result from a back projection:

b_θ(x,y) = ∫ g_θ(R) δ(x cosθ+ y sin θ- R) dR

Adding up all the back projections from all the angles gives, fback-projected(x,y) = ∫ b_θ(x,y) dθ

π ∞

f_b(x,y) = ∫ dθ∫ g_θ(R) δ(x cosθ+ y sin θ- R) dR

0 -∞

Let’s calculate an impulse response to see how the reconstruction does.

g_θ(R) = δ(R)

That is, δ(x,y) causes a δ(R) projection. By calculating the back- projected image, f_b(x,y), we will be calculating the impulse response.

h_b(x,y) = f_b(x,y) for this delta object

Recall x cos θ+ y sin θ- R = r cos (θ-φ) - R h_b(r,θ) = ∫ dθ∫δ(R) δ(r cos (θ-φ) - R) dR

We can simplify the integration over R by realizing that the first delta function will be non-zero only when R=0. Then we will only have one integral across θ

= ∫πδ(r cos (θ-φ)) dθ

0

Now δ[f(x)] = ∑δ(x - x_n) / |f’(x_n)|

n

Only one root (zero) in range of integral

cos (θ-φ) = 0 θ-φ = π/2

θ= π/2 + φ

d r r d

h

1 1

)] | cos(

| [

)) 2 / ( ) (

, (

0

− − +

= ∫

θ θ φ φ π

θ

φ

δ

Back-projected impulse response

h_b(r) = 1/r

f_b(x,y) = f(x,y) ** 1/r

F_b(ρ,θ) = F (ρ,θ) / ρ since F{1/r}= 1/ ρ

Back projected image is blurred by convolution with 1/r

Where intuitively does the 1/r come from?

We must account for this blurring to properly reconstruct the image.

How does 1/r convolution look in image space?

In frequency space?

- Low spatial frequency data is overweighted. Filter to compensate for this. Weighted by 1/ρ.

- Solution - filter each projection by |ρ| to account for the uneven sampling density

Steps:

1) Calculate projection 2) Transform projection 3) Weight with |ρ|

4) Inverse transform 5) Back project 6) Add all angles Mathematically,

∫ dθ ∫ F ^-1{ F{g_θ(R)} |ρ| } δ(x cosθ+ y sin θ- R) dR

The reconstruction described is known as filtered back-projection.

What is downside of the described filter, |ρ| ?

ρ

What is downside of this filter?

ρ Band limit filter to:

p₀ Filter |ρ|rect (ρ/ 2ρ₀)

c(R) = ?

Instead of converting to frequency domain,, use convolution in the image domain

π ∞

∫ dθ∫ [ g_θ(R) * c (R) ] δ( x cosθ+ y sin θ- R) dR

0 -∞

Each projection is convolved with c (R) and then back projected.

Describe c (R) c (p) = |p|

c (R) = lim 2 (ε²- 4π²R²) / (ε²+ 4π²R²)²

ε →0

p₀ - p₀

c (R) = p₀ [ 2 sinc ²(p₀ R) - sinc ²(p₀ R)]

Actual Filters : Ram-Lak : Shepp-Logan

a) Ramachandran-

Lakshminarayanan kernel

b) Shepp-Logan kernel.

Solid lines represent h(η) and circles h_k