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Fund BioImag 2013 5-1 5: Emission (Computed) Tomography 1.What is a tracer ? 2.Why is collimation necessary and what are its consequences ? 3.How are the effects of attenuation taken into account ? 4.What is the principle of x-ray detection ? scintillation followed by amplification 5.How can scattered photons be eliminated ? After this course you 1.Understand the reason for collimation in imaging -emitting tracers and its implication on resolution/sensitivity 2.Understand the implications of x-ray absorption on emission tomography 3.Understand the basic principle of radiation measurement using scintillation 4.Are familiar with the principle/limitations of photomultiplier tube amplification 5.Understand the use of energy discrimination for scatter correction Slide 2 Fund BioImag 2013 5-2 5-1. Emission computed tomography Human brain examples uncontrolled complex partial seizures left temporal lobe has less blood flow than right indicates nonfunctioning brain areas causing the seizures 99m Tc-HMPAO SPECT of Brain (Sagittal Views) Alzheimers Disease Dementia with Lewy Bodies Control patient with AIDS contrast enhanced CT scan (left) ring enhancing lesion with surrounding edema toxoplasma abscess or CNS lymphoma. thallium SPECT scan (right) intense uptake of radiotracer by the lesion lymphoma Thallium SPECT - CNS Lymphoma Slide 3 Fund BioImag 2013 5-3 Emission computed tomography Heart and rodent examples perfusion of heart muscle orange, yellow: good perfusion blue, purple: poor perfusion SPECT Image of a small mouse (25 gr) injected with 10 millicuries of 99mTc-MDP. Slide 4 Fund BioImag 2013 5-4 What is Emission Computed Tomography ? Until now: CT and x-ray imaging measure attenuation of incident x-ray Emission tomography: X-rays emitted by exogenous substance (tracer) in body are measured What is a tracer ? Exogenously administered substance (infused into blood vessel) that (a) alters image contrast (CT, MRI) (b) has a unique signal ( emitting) -> Emission computed tomography Typical tracers for emission tomography half-life and photon energies - Two issues: 1.How to determine directionality of x-rays ? 2.Absorption is undesirable Injection of radioactive tracer [h][keV] 99m Tc6140 201 Tl7370 123 I13159 133 Xe0.0881 Slide 5 Fund BioImag 2013 5-5 What are the basic elements needed for -emitter imaging ? Collimator Photomultipliers Light guide Scintillating crystal Septa - camera Position logic electronics w. collimator boardThyroid scannerWhole body scintigraphy Slide 6 Fund BioImag 2013 5-6 5-2. How can directionality of x-rays be established ? Collimation Problem: Photon detection alone does not give directionality Solution: Collimation establishes direction of x-ray Collimator Thick (lead or tungsten) with thin holes Select rays orthogonal to crystal Signal S(y) Consider one detector, assuming perfect collimation (and neglecting attenuation, see later) x y C T (x,y): tissue radioactivity Radon transform Reconstruction as in CT septa Impact of collimation on resolution Line of incidence (LOI) Slide 7 Fund BioImag 2013 5-7 Crystal scintillant (NaI) D How does collimation affect resolution ? Its never perfect t d a w penetrating scattering Perfect collimation, i.e. resolution ? geometric d/a 0 collimator t Impossible to achieve (Why?) septa D Source Collimator resolution: Two objects have to be separated by distance >R b Septa penetration < 5% occurs when t=t 5% : Price of collimation (resolution) ? Sensitivity ! (a e : imperfect septal absorption) Slide 8 Fund BioImag 2013 5-8 5-3. How to deal with attenuation of the emitted x-rays ? result of x-ray absorption in tissue Signal measured from a homogeneous sphere (C T (x,y)=constant) D Intensity distortion: Cause ? Attenuation T 1.depends on object dimension and source location (D=f(object)) 2.Photon energy =f(E ) 5 cm 10 cm 15 cm 99m Tc in H 2 O T=0.46 T=0.21 T=0.10 Consider point source: Attenuation depends on location of source in tissue Slide 9 Fund BioImag 2013 5-9 What are the basic steps in attenuation correction ? Corrected image Attenuation correction procedure A.Estimated object geometry and estimated (x,y) or measured (x,y) B.Transmission loss : T(projection)=f((object), projection) C.Attenuation correction A(x,y)=1/T(x,y) D.Corrected C corr (x,y)=A(x,y) C(x,y) lung heart A(x,y) of thorax Problem is prior knowledge needed for A (i.e. (x,y)) Measured A B C D (x,y): Attenuation correction Attenuation correction rarely applied! Slide 10 Fund BioImag 2013 5-10 How to simplify attenuation correction ? by measuring at 180 0 using geometric mean Problem: Spatial dependence of correction Signal (S 1 ) x y C T (x,y): tissue radioactivity Signal (S 2 ) x1x1 x2x2 D = x 1 + x 2 Solution: Geometric mean of the two 180 0 opposite signals: Measure at 180 0 simultaneously and take the geometric mean attenuation correction depends only on dimension of object along the measured Radon transform (D = x 1 + x 2 ) Signal=sum of all activity along x: Consider point source: Slide 11 Fund BioImag 2013 5-11 Photomultiplier Tube 5-4. What is the principle of x-ray detection ? Collimation, followed by scintillation and amplification electronsLight El. Signal Scintillator crystal e.g. Tl-doped SodiumIodide (NaI) -energy # scintillation photons Signal Slide 12 Fund BioImag 2013 5-12 What is Scintillation ? T = energy transfer efficiency from excited ion to luminescence centre q A = quantum efficiency of luminescence centre W e-h = energy required to create one electron-hole pair Conduction band Activation band (Doped) Valence band Activator ground state Gap Sequence of events in scintillation crystal 1.Atom ionized by Compton interaction Electron-hole pair 2.Hole ionizes activator, electron falls into activator 3.Activator is deactivated by emission of Photons (10 -7 sec) Efficiency of scintillators Slide 13 Fund BioImag 2013 5-13 What elements characterize scintillation materials ? Overview of some crystals Requirements for scintillator High yield Good linearity Small time constant Transparent for scintillation light good mechanical properties Refraction index close to 1.5 Z eff Refr. Index Yield 732.1513% 511.85100% 59 6679% Most of the energy of the x-ray is lost as heat (to lattice), see e.g. NaI(140keV)=40140 =5600 photons at 400nm E 400nm [keV]=hc/ = 1.2/ [nm] =1.2/400 keV=3eV