X-ray Development Assume parallel x-rays and ignore scatter Point source with no object a)quantified obliquity fall off b)largely ignore 3) Point source with object - saw object was Magnified 4) Finite source - Ray optics - Source magnification m = 1 - M - Output seen as convolution of Magnified object and magnified source 5) Finite source with an arbitrary planar object

View as convolution of a Magnified object and the magnified source Linear if you view it as a transmission 6) Volumetric Object - see body as a series of slabs or slices in z. Consider attenuation as composed of µ (x,y,z) = µw (x,y,z) + µ∆ (x,y,z) Id (xd, yd) = Tw [Ii - ∑ 1/(4πd2mi2) s ((xd/mi), (yd/mi)) ** ∆i (xd/Mi, yd/Mi) where Tw = exp [ - ∫ µ (xd/m, yd/m, z) dz] 7) Next: Tilted source briefly 8) Motion

Effects of Tilted Source Source magnification in xd function of z Source magnification in y dependent on yd and z We will skip the mathematical development on this section

Consider a 3 x 3 array of pinholes t(x,y) = d(0,y – c) + d(0,y + c) s (xs, ys, zs) Consider a 3 x 3 array of pinholes t(x,y) = d(0,y – c) + d(0,y + c) Id (yd) ys y c anode t -c z d t(x,y) above can be considered a lead plate with two pinholes punched into it. 3 x 3 array of circular pinholes to left shows how source is contracted in y for positive y detector positions and enlarged for negative detector locations. y x

Let’s allow the magnification to be different on each axis Let’s allow the magnification to be different on each axis. We will consider the impulse response using a pinhole at x’, y’. The pinhole models an object, d(x’, y’,z). M’x’ + m’ xs, M’y’ + m’ ys Here we use a general M’ to allow for magnification of the pinhole, and m’ to allow for magnification of the source. The diagram above merely shows how one point in the source, the pinhole, and the detector are related by geometry. d (x’, y’,z)  ys  z d Detector Plane

We desire an expression where  is a collection efficiency for the pinhole

If tan () = a, then the source position z’ = a ys By geometry, object magnification is M’ = (d - a ys)/(z - a ys) Source magnification is m’ = - ((d - a ys) - (z - a ys)) / (z - a ys) m’ = - (d - z)/ (z - a ys) d (x’, y’) (xd, yd) ys  ays z d

Recall xd = M’x’ + m’xs yd = M’y’ + m’ ys Mx = xd /x’ = M’ = (d - a ys)/(z - a ys) ≈ d/z since for practical arrangements d, z >> ys Typical dimensions: z, d ~ 1 m, ys ~ 1mm Similarly My = yd /y’ ≈ d/z mx = xd /xs = m’ ≈ - (d-z)/z = m my = yd /ys This is more interesting derivative since both M’ and m’ are functions of ys yd /ys = (M’ y’)/ ys +  (m’ ys)/ ys

From previous slide, M’ = (d - a ys)/(z - a ys) and m’ = - (d - z)/ (z - a ys) Find yd /ys = (M’ y’)/ ys +  (m’ ys)/ ys my = [((z- a ys)(- a) - (d - a ys)(- a)] y’)/(z - a ys)2 + -(d-z) • [(z- a ys) - ys(- a)]/ (z- ays)2 = (- a[z-d] y’ - (d -z) z)/ (z- ays)2 = - (d - z) (z - ay’)/ ((z- ays)2 my = - (d-z) (z - ay’)/ z2 = m (1- (ay’/z)) Using this relationship and ignoring obliquity, How does magnification change with object position?

Since system is linear, we can write a superposition integral. Id (xd, yd) = ∫ ∫ h ((xd, yd, x’, y’) t(x’,y’) dx’ dy’ =

For developing space-invariance, let’s consider a magnified object Not a space-invariant system since 1- (ay’’/Mz) varies slowly with y’’ or y’ But it doesn’t vary much in a region of an object.

Object Detector y’ Let y’d = My’ Consider a horizontal strip across the detector centered at For this region, ay’’ = aMy’ = ayd where Mz = d Here is a constant over a region in the detector during the convolution.

At ay’= z, source width goes to 0 in y. We call this the “heel effect” In the frequency domain, Anode Detector plane Id Electron beam

Motion Considerations Why don’t we simply use point source, and cut down on current to avoid over heating? Let’s consider object motion with constant velocity in the x direction over the imaging time T at velocity v Over time T, the object size will change position in the detector plane by MvT vT MvT

The impulse response due to movement is Notice that there is no degradation in y, as we expected. The complete impulse response is given below( a planar source parallel to the detector is used here for simplicity). Check equation symbols Blur in x direction gets minimized as T decreases to 0

Assume a L x L source parallel to detector Write s(x,y) energy density as a power integrated over time Assume beam is on for time T p (xs, ys) regional source power density. p is limited by tungsten melting point. Operate tube at maximum power available. Set p (xs, ys) = Pmax. Es = ∫∫∫ p (xs, ys) dxs dys dt=PmaxTL2 , then As L increases, source grows, but T decreases

Extent of the impulse response in the x direction, due to source blurring and motion is, by the convolution of two rectangles, X = |m|L + (MvEs/ Pmax L2) We could choose to minimize several criteria. Area, for example. We will simply minimize X today with respect to L. If v = 0, then L = 0 T = ∞ no source blurring If |m| = 0, object is on detector L ∞ T = 0 Corresponding Exposure Time at Optimal L

Measuring Source Size Pinhole Source Detector will show size of source Detector plane d Place pinhole halfway between source and detector. |m| = 1 M = 2 z = d/2 Pinhole is essentially  (x,y) Image of source s(xd, yd) is measure of s(xs, ys)