X-Ray Notes, Part II

Transcription

1 Noll X-ra Notes : Page X-Ra Notes Part Source ssues The Parallel X-ra aging Sste Earlier we consiere a parallel ra sste with an incient intensit that passes through a 3D object having a istribution of attenuation coefficients µ an projects to an iage : ep µ There are essentiall no practical eical project -ra sstes where the source has parallel ras. There are soe scanning sstes that ight be appropriate for inustrial inspection operations for eaple: but these kins of sstes are too slow for eical applications. Practical X-ra Sources There are two ain issues associate with practical -ra sources:. Geoetric istortions ue to point geoetr epth epenent agnification.

2 Noll X-ra Notes : Page. Resolution loss blurring ue to finite large source sies Point Source Geoetr First we will fin epressions for the iage intensit for a point source geoetr: i ep µ r Coents:. is the coorinate sste in the output etector plane.. is the coorinate sste of the object.

3 Noll X-ra Notes : Page 3 3. Notice that i a spatiall variant incient intensit replaces. 4. Notice that the integration is along soe path r with variable of integration r. ntensit Variations The incient intensit is aial at the center of the coorinate sste an falls off towars the eges. This has two coponents an increases in istance fro the source an the ras obliquel striking the etector. ntensit has reall power/unit area. We can write an epression for the intensit i as: i photons ean photon E unit areaepo sure tie where k is a scaling coefficient N is the nuber of photon that are eitte uring the observation tie we assue here that photons are eitte isotropicall over a sphere an Ω/4π is fraction of the surface of a sphere that is subtene b piel area a. [Ω is known as the soli angle an has units of steraians of which there are 4π over the surface of a sphere. This is siilar to there being π raians over circuference of a circle.] For a piel of area a at soe position angle θ awa fro the origin the part of a sphere covere will be acos θ. Thus: Ω a cosθ or 4π 4πr r kn a a cosθ Ω We now efine the intensit at the origin to be. At the origin θ an the istance fro the source to the etector is r thus Ω a/ an: kn i 4π Ω 4π

4 Noll X-ra Notes : Page 4 Note that the intensit falls off with / as the etector oves awa fro the source. The constant k can now be foun in ters of : Substituting: Observing that cos θ r i we get: k 4π N kn Ω cosθ a 4π r 3 cos 3 i θ r we can put this epression in the coorinate sste of the etector using r + an r + : r i + r 3 r + The cos 3 θ ter or its other representations is calle the incient intensit obliquit ter an this has two coponents: the cos θ ter for an increase in istance fro the source to the etector an the cosθ ter for ras obliquel striking the etector. The cos θ ter is reall a /r ter the inverse square law for fallout of intensit. The cosθ ter can be easil visualie if ou think of a flashlight bea hitting a wall obliquel the oblique bea spreas the photons over a larger area of the wall. 3/ Oblique Path ntegration f we look at soe point in the object at epth we see that it will strike the : etector at a position

5 Noll X-ra Notes : Page 5 where is the agnification factor for an object at epth. We can now write the attenuation coefficient at location in ters of the output coorinate sste: µ µ Also instea of integrating along the path r we can rewrite the epression to integrate in : r + + r This epression sas that if with integrate in instea of r the integral will nee to be increase b r + in orer to account of the longer path length in r than. This ter is soeties known as the pathlength obliquit ter.

6 Noll X-ra Notes : Page 6 Finall we put it all together an we get an epression for the output intensit fro a point source: + + r r ep 3/ µ Eaple For the eaple we will reuce the iensions of the proble to an an thus r. Now let s look at a rectangular object at epth : W L rect rect µ µ The epression for the iage intensit will be: + + W L 3/ rect rect ep µ The use of the agnification factor allowe the function of to be converte to a function of for each location in the etector plane. The first rect in the above epression has with L/ an is centere at. The secon rect has with W an is centere at. The integral is the area uner the overlap of these two rect functions.

7 Noll X-ra Notes : Page 7 The integral is: for l W < or > L W / W for l W > + or L < + W / L W + otherwise f we ignore all obliquit ters we get the following: ncluing the pathlength an incient intensit obliquit ters we get:

8 Noll X-ra Notes : Page 8 Uner a parallel ra geoetr we get the following: As we can see the epth epenent agnification has significantl istorte the appearance of the object in the iage. We can efine a fractional transition with be: 4 L W / L + W / 4 L + W / L + W / Thus we can iniie the geoetric istortions b placing the object as far fro the source as possible ake large. Finite Large Sources To gain an unerstaning of this issue we will first consier a thin object. Specificall we will let the attenuation coefficient be: an then: µ τ δ W

9 Noll X-ra Notes : Page ep ep r r i i τ δ τ We let / the object agnification factor an we will ignore the pathlength obliquit ter to get: t i i ep τ where t ep-τ is the transission function. gnoring all obliquit ters we get: t Now we consier a finite source function s an a ver sall pinhole transission function: The iage will now be an iage of the source with the source agnification factor : ks

10 Noll X-ra Notes : Page where k is a scaling factor that is proportional to the area of the pinhole / etc. f we want the above to represent the ipulse response of the sste we nee to ake the pinhole equal to δ an account for all of the scaling ters [t δ is not a realiable transission function since t can never ecee nevertheless we will allow it for atheatical convenience.] The area of the pinhole is δ. The capture efficienc of the pinhole is the fraction of all photons eitte fro the source that pass through the pinhole. This will be equal to: pinhole area η 4 Letting the total nuber of photon eitte be: π 4π N s an the total nuber of photons to get through the pinhole will be: N Nη. 4π This ust be the sae nuber at the etector: ks The scaling coefficient will therefore be: so: k 4π kn N 4π s 4π Now we let the pinhole be at position that is t δ- - :

11 Noll X-ra Notes : Page The iage of the source is not locate at where is the object agnification factor. Thus the ipulse response function is: s h 4 ; π Now we can calculate the iage for an arbitrar transission function using the superposition integral: t s s t s t h t ** 4 4 sub an 4 ; π π π Thus the final iage is equal to the convolution of the agnifie source an the agnifie object. The object is blurre b the source function. The frequenc oain equivalent is: { } 4 v u T v u S F D π Consier / which iels an. The object is agnifie b a factor of an is blurre b the unagnifie source.

12 Noll X-ra Notes : Page Coents:. The least blurring coe when is ae sall. Thus it is esirable to ake the epth plane as far fro the source as possible:. Then -/ an. As we was above aking also reuces geoetric istortions. The coon practice for -ra iaging then is to position the subject ieiatel net to or on top of the etector.. f the thickness of the bo is a liiting factor then let. This will ake the sste close to a parallel ra geoetr with an. The ain proble with this approach is. / an SNR 3. We woul also like the ake s as sall as possible to reuce blurring but s an aking it sall ight reuce the nuber of photons create an thus reuce SNR. 4. For a cople object we can ake µ τ δ an each plane i i will have its own agnification factors. This is not particularl useful but it can give ou soe iea of how blurring an agnification ight affect ifferent parts of a real object ifferentl. Detector ssues Earlier we iscusse the effect of source sie an location on spatial resolution an agnification istortions in -ra iaging. Now we will iscuss etector issues. n selecting etector characteristics we will have a resolution/snr trae-off this coe priaril fro the fact the thicker etectors have better SNR but a larger ipulse response. Conversion of -Ras to Fil Photographic fils are generall not ver sensitive to -ras so -ras ust first be converte to visible light b a scintillating screen:

13 Noll X-ra Notes : Page 3 We will now evelop epressions to represent the ipulse response of the etector. Suppose we have a -ra photon enter the scintillating screen an it interacts at soe epth which we ll call an generates a shower of light photons isotropicall fro a point of which soe eventuall strike the etector. The geoetr is essentiall the sae as a point -ra source striking the etector. Notabl: 3 h r h cos θ but h h b the inverse square law thus: + r 3 h r k + r 3/ The corresponing frequenc oain equivalent is: H ρ πk ep πρ Without loss of generalit we will select k to noralie this epression to have a peak frequenc response of. H ρ ep πρ Notice that right net to the fil : H ρ h r δ Finall we can calculate an average frequenc response b taking:

14 Noll X-ra Notes : Page 4 H ρ H ρ p where p is the probabilit ensit function for an interaction occurring at epth. To eterine this we first recognie that the scintillating screen has its own linear attenuation coefficient µ. The nuber of photons that pass through at an epth is: an the nuber absorbe will be: N Nep µ ep N abs N µ The total fraction absorbe in the etector is: η ep µ where η is etector efficienc which increases with. We can efine the cuulative istribution function as: an thus the probabilit ensit function is: The average frequenc response is then: ep µ P η P µ p ep µ η µ H ρ ep π ρep µ η µ ηπρ+ µ For large ρ this epression looks like: ep πρ+ µ H ρ µ πηρ The high spatial frequencies pla a large role in ictating the shape of the ipulse response close to the peak e.g. h r near r an the low spatial frequencies will ictate the appearance of the tails of h r. Thus near r the average ipulse response will take on the shape:

15 Noll X-ra Notes : Page 5 µ h r πηr recall the inverse Fourier-Bessel transfor of /ρ is /r. The average ipulse response then is ver peake infinite in aplitue. One consequence of this is the coon easures of resolution or blurring e.g. like FWH Full With at Half aiu have no eaning. One wa to evaluate the perforance of the etector sste is to efine a cutoff frequenc ρ k as the frequenc at which the response falls to k H. For saller values of k this is: ρ k µ πηk This in essence give the aiu spatial frequenc that can be etecte where k represents the level of etectabilit. For eaple k. is a coon value an having a higher cutoff frequenc ρ k is esire to iprove spatial resolution. We can now begin to see the SNR resolution trae-off. As increase the etector efficienc η increases which leas to ore -ra photons being etecte an thus the SNR iprove. This however causes ρ k to be saller resulting in lower spatial resolution. Recall that the SNR is proportion to the square root of the nuber -ra photons an in orer to see the the ust be etecte so the SNR is proportional to the root of the nuber -ra photons that are etecte. SNR is therefore proportional to η. Eaple Let s look at a etector with the µ.5 - an.5 an we will use k..

16 Noll X-ra Notes : Page 6 η.3 ρ k 8 an the liiting spatial resolution is approiatel: 5µ ρ Now if we ouble the thickness to.5 : η.53 ρ k k 4.5 an the liiting spatial resolution is approiatel: µ ρ k Coents: n general increasing iproves both η an ρ k. What happens if we put the fil on the back of the scintillator? s the response better or worse? Two Screen Detectors with Double Eulsion Fils To ease the traeoff between resolution an SNR we can use a ouble eulsion fil with a two screen scintillator: We assign a coorinate sste here to ease our analsis:

17 Noll X-ra Notes : Page 7 Since no interaction occur in the fil we can neglect its thickness: For interactions occurring in the first screen : which iels a frequenc response of: h r k ep πρ + r 3/ H ρ ep πρ For interactions occurring in the secon screen : Finall: H ρ ep πρ ep πρ ep πρ for < H ρ ep πρ for < where +. The etector efficienc is again: η ep µ.

18 Noll X-ra Notes : Page 8 We can now fin the average frequenc response in a siilar anner as before: µ ep µ ep πρ ep µ ep πρ + µ H ρ + η πρ µ πρ µ For large ρ this epression looks like: f we take / then: an the cutoff frequenc will be: where [ η ] / Eaple µ H ρ µ πηρ [ ep µ + ep ] µ H ρ η πηρ πηρ ρ k [ ] / µ ep µ [ ] / µ πηk [ η ] / is the iproveent factor over the single eulsion fil sste. Let s look at the previous eaple with a etector with the µ / then an we will use k.. η.3 [ η ] / ρ k 3.7 an the liiting spatial resolution is approiatel: k 76µ ρ Alternativel we can hol ρ k constant b setting. 4 : η.45 ρ k ρ k 8 5µ Overall Sste Response Now we can a the etector response to the other sste eleents:

19 Noll X-ra Notes : Page 9 4 π The ipulse response function will then be: h 4 π or for a circularl setric source function: s ** t ** h r h 4 π s * * h r r s * * h r Object Blurring One issue is how uch oes the etector response blur the object. t is iportant to realie that the etector blurs the agnifie object. Our intuition woul be to ake the object as large as possible b aking / ver large. This woul ictate oving the object as close to the source as possible which is eactl opposite as what we woul like to o to iniie source blurring. Consier also that the agnifie source also blurs the agnifie object source an object have ifferent agnification factors. One wa to look at this is to eaine the response in the coorinate sste of the object rather than the etector : the effective agnification of the source is: ks ** t * * h r an the effective agnification of the etector response is: These are in copetition: to ake the source blurring sall ake to ake the etector response sall ake Coents:

20 Noll X-ra Notes : Page. For ost fil sstes the etector response is ver sall an the source is alost alwas bigger. Therefore we woul like to ake.. For other kins of sstes e.g. igital fluoroscop sstes the etector resolution is uch larger e.g..5 an for these sstes an intereiate a be appropriate.