ELEG 479 Lecture #6. Mark Mirotznik, Ph.D. Associate Professor The University of Delaware

Transcription

1 ELEG 479 Lecture #6 Mark Mirotznik, Ph.D. Associate Professor The University of Delaware

2 Summary of Last Lecture X-ray Physics What are X-rays and when are they useful for medical imaging? How are X-rays generated? How do X-rays interact with matter (tissue)? How do we detect X-rays How do we mathematically model X-ray projection radiography

3 X-rays Wavelength: nm Frequency: 30 petahertz to 30 exahertz (3x10 16 to 3x10 19 ) Energies: 120 ev to 120 kev Soft X-rays: Hard X-rays: 0.12 kev to 30 kev 30 kev to 120 kev Soft X-rays do not penetrate matter where hard X-rays do

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5 Interaction of Photons with Matter

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7 Interaction of Photons with Matter: Attenuation x I o µ I

8 Interaction of Photons with Matter: Attenuation Lower energies can distinguish different material better than higher energies

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11 Interaction of Photons with Matter: Attenuation x I o µ(x) I(x)

12 Interaction of Photons with Matter: Attenuation x S o (E) µ(x,e) I(x)

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15 * FDA Study

16 & Inevitable problems of not fully understanding the effects of ionizing radiation

17 Today s Lecture Projection Radiography Overview of different systems for projection radiography Instrumentation Overall system layout X-ray sources grids and filters detectors Imaging Equations Basic equations Geometrical distortions More complicated imaging equations

18 Projection Radiography Traditional chest x-ray 2-3 mrem ~100 kev Static X-ray Systems (snap shots) Traditional dental x-ray 2-3 mrem kev Mammography 70 mrem ~30 kev

19 Digital Fluoroscopy 10 mrem/min ~50 kev Projection Radiography Dynamic X-ray Systems (Fluoroscopic) Angiogram 460-1,580 mrem ~50 kev Barium Colon Image 700 mrem ~30 kev

20 Projection Radiography Some new stuff Digital Subtraction Angiography (DSA)

21 Projection Radiography Generic System Description

22 xrays

23 Amount of radiation for a fixed voltage is proportional to the current in the tube. Radiation exposure is often given in units of mas (milliampere-seconds). This gives an idea of how much current was used in the tube for how long. While mas is usually proportional to dose (rems or rads) it is not easy to equate the two however

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30 X-Ray Detectors Screen or Film Image Intensifier Storage Phosphors Direct Flat Panel Radiography

31 X-Ray Detectors

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37 X-Ray Detectors What do we do in dynamic systems that can t use film?

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40 10 x12 85 micron pixel pitch 3600 x 3000 pixels (10.8 Mpixels)

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42 Quiz #1: March 18 th, 2014 (1) Determine the transfer function of an imaging system defined by the following PSF x y h ( x, y) = rect(, ) 2 2 Recall: rect( p, q) = 1 p < 0.5, q < 0 else 0.5 (2) Calculate the value of the transfer function at u=0.25 rad/cm and v=0.25 rad/cm. In words tell me what the value you just calculated means physically. Equations: H ( u, v) = h( x, y) = h( x, y) e H ( u, v) e j2π j2π ( ux+ vy ) ( ux+ vy ) dxdy dudv

43 Quiz #1: March 18 th, 2014 (1) Determine the transfer function of an imaging system defined by the following PSF Recall: rect( p, q) H ( u, v) = = = = = h ( x, y) = 1 p < 0.5, q < 0.5 = 0 else πju e e j2π h( x, y) e j2π ( ux+ vy) x rect(, 2 y ) 2 ( ux+ vy) j2π ( ux) 1 j2π ( vy) [ e dx] [ e dy] j2π ( u ) j2π ( u ) 1 j2π ( v) j2π ( v) [ e e ] e e ( u ) j2π ( u ) j2π ( v) j2π ( v) e 2πju j2π dxdy e dxdy 2πjv e 2πjv 2 j sin(2πu) 2 j sin(2πv) 2πju 2πjv = [ ] = 4 sin c(2π u) sin c(2π v)

44 Quiz #1: March 18 th, 2014 (2) Calculate the value of the transfer function at u=0.25 rad/cm and v=0.25 rad/cm. In words tell me what the value you just calculated means physically. H ( u, v) = 4 sin c(2π u) sin c(2π v) H (0.25,0.25) = 4 sin c(2π 0.25) sin c(2π 0.25) sin( π / 2) sin( π / 2) 16 = 4 = 2 π / 2 π / 2 π If a sinusoidal input of spatial frequencies u=0.25 rad/cm and v=0.25 rad/cm is inputted into the system the output s magnitude will be 16/pi^2 less and in phase.

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46 Monochromatic = ), ( 0 ), ; ( ), ( y x r ds y x s I o e y x I µ Polychromatic ' ),, ; ( ' ' ), ( 0 ' max min ) ( ), ( de e E E S y x I y x r ds E y x s E E o = µ

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48 I = I e min ( ) o I o I = I e max Local contrast: Global contrast: ( ) o I o C l = C = I I I I I max min = max max I + I max min = min

49 I = I e min ( ) o I o I = I e max Local contrast: Global contrast: ( ) o I o C l = C = I I I I I max min = max max I + I max min = min

50 Geometric Effects X-rays are diverging from source Undesirable effects cos 3 (θ) fall off in intensity across detector magnification

51 Geometric Effects

52 I s = photon flux at source I o = photon flux at detector origin I r = photon flux at (x,y) point on detector

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54 Inverse Square Law +Obliquity How big of a deal is this?

55 Inverse Square Law +Obliquity How big of a deal is this? Depends on (1) how far away is the detector from the source and (2) how large of a field of view are we looking at.

56 Inverse Square Law +Obliquity Example How big of a deal is this? Depends on (1) how far away is the detector from the source and (2) how large of a field of view are we looking at. Chest x-ray system has a detector of 18 x 18 The source is placed a distance of 6 (72 ) away What is the maximum change in intensity across the detector? 1 9 θ = tan ( ) = cos( θ ) = I = I min o o Usually not that big of a problem

57 Path Length Variation Uniform (or homogenous) slab of material. Since the rays are going at different angles the amount of material they go through varies depending on the location through the tissue. This difference in path length means more attenuation for points away from the origin.

58 Path Length Variation This results in a prism effect if you are imaging a cube. How big of a deal is this?

59 Inverse Square Law +Obliquity+Path Length I d ( x, y) = I o cos 3 ( θ ) e µ L / cosθ

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61 x d = x z ' o x ' = x z d o = x M ( z)

62 ) ( ) ( ' ' z M y d z y y z M x d z x x o o = = = = ) ) (, ) ( ( ) ( cos ), ( 3 z M y z M x t I y x I z o d = θ ), ( ' ' y x t z Let transmittvity given as Then at the detector we get ), ( ) ( ), ( d yz d xz t y x d d I y x I z o d + + = After substitution we get: be transmission through an infinitely thin object (e.g. thin piece of paper with a lead pattern printed on it)

63 Not done yet! There are additional factors (1) X-rays source are not really a point source but an extended source. This causes blurring (2)X-ray films are not perfect! A single x-ray photon causes a blurry spot on the film. This is equivalent to the impulse response or PSF of the film h(x,y)

64 When we add all that together ), ( ), (, 4 1 ) ( cos ), ( y x h M y M x t m y m x s m d I y x I z o d = π θ Inverse square +obliquity Extended source m=source magnification s=source intensity distribution. Transmittivity of object PSF of film M m z d M o = = 1 where