ProcIEEE_Kak_computerized_tomography_with_xray_emission_ultr(11)

(e, k7)as

s

J-

A2

=

ray path

J-

a 2 ( x, Y ) ds.

(38)

A 1 and A2 are, clearly, ray integrals for the functions ( x, y ) a1 and a2 ( x, y ) . Now if we could somehow determineA 1 and A2 for each ray, fromthis information then the functions ( x, y ) a1 and a2 (x, y ) could be separately reconstructed. And, once we know a l ( x, y ) and a 2 ( x, y ), using ( 3 3 ) an attenuation coefficient tomogram could be presented at any energy free from beam hardening artifacts. A few words about the determination of A 1 and A2: Note that it is the intensity Nd that is measured by the detector. Now suppose instead of making one measurement we make two measurements for each ray path for two differentsource spectra. Let us call these measurements I I and Z2, then

where p{ *} denotes the _probability, Ne(k7)! denotes the factorial of Ne(k7), and Ne(k7) the expected value of the measuremen

t

N (k7)= E{Ne(k7)) e

(43)

where E denotesstatisticalexpectation.Notethatthe ance of each measurement is given by variance (Ne (kr)}= Y (k7). o

vari(44)

Because of the randomness in Ne(k7) thetrue value of Pe(k7) will differ from its measured valuewhich will be denoted by P r ( k 7 ) . To bring out this distinction we reexpress

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1258

PROCEEDINGS OF THE IEEE, VOL. 67, NO. 9, SEPTEMBER 1979

and(45)

We will now define a relative-uncertainty image as follows:’ variance{?(x, y ) ) relative-uncertainty at (x, y )= Ni,[?(x, r ) l‘

By interpreting exp[-Pe(kT)] as the probability that (along a ray such as the one shown in Fig. 3) a photon entering the object from side A will emerge (without scattering or absorption) at side B, one can show thatN e ( k T )= N i, exp[ - P e ( k~ )] .-

(55)

(46)

We will now assume that all fluctuations (departures from the mean) in Ne(k7) that have a significant probability of occurrence are much less than the mean. With this assumption and using (41) and (42) it is easily shown that E and

{PF(k~)) PO ( k~ )=

(47)

In computer simulation studies with this definition the relative-uncertainty image becomes independent of the number of incident photons used for measurements, and is completely determined by the choice of the phantom. Fig. 13(c) shows the relative-uncertainty image for the Shepp and Logan phantom (Fig. 13(a)) for Mpr0j= 120 and T= 2/101 and for h ( t ) given by (7). Fig. 13(d) shows graphically the middle horizontal line through Fig. 13(c). Relative-uncertainty at (x, y ) gives us a relative measure of how much confidence an observer might place in the reconstructed value at the point (x, y ) vis-$-vis those elsewhere. We will now show that the results that have been obtained by Shepp and Logan[ 1071 and Chesler[35] are special cases of our (54) or, equivalently (51). Suppose we want to determine the variance of noise at the origin. From (51) we

From the statistical properties of the measured projections P r ( k T ), we will now derive those of the reconstructed image. By combining (8) and (IO), the relationship between the reconstruction at a point (x, y ) and the measured projections is given bynTMproj

can writevariance{P(~,= o))

z Mproj

(56)

?(x, ),

P;(kT) h(x cos BiMmoji=1

+,

sin Bi - kT).(49)

k

where we haveused the fact that h ( t ) is an even function. Chesler et al.[ 3 5] have argued that since h(k7) drops rapidly with k (see (7)), it is safe to make the following approximation for objects that are approximately homogeneous:

Using (47), (48), and (49), we get

which, when T is small enough, may also be written as+ y sin Bi

- kr) ( 5 0 )

and(58)

. h Z ( x cos B i+ y sin Bi

-k

~ ) (51)

where we have used the assumption that fluctuations in P& (~ T )areuncorrelatedfordifferent rays. Equa

tion ( 5 0 ) shows that the expected value of the reconstructed image is equal to that made from the ideal projection data. Before we interpret (51) we will rewrite it as follows. In terms of the ideal projections Pe(kT), we define new projections as

v ( k~= exp[PO(k7)I e )and a new filter function h, ( t ) ash”(t)= h2(t).

(52)

(53)

Note again that i O i ( O ) are the mean number of exiting photons measured forthecenter ray in each projection. Using (58) Chesler et al.[ 3 5] have arrived at the very interesting result that(forthe same uncertaintyinmeasurement)the total number of photons per resolution element required for X-ray CT (using the filtered-backprojection algorithm) is the same as inthe measurement of attenuation of an isolated (excised) piece of tissue with dimensions equal to those of the resolutionelement. Now consider the case where the cross section forwhich the image is being reconstructed is circularly symmetric. The Nei(0)’s for all i’s will be equal, let’s call t i value No. That hs is,

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