PET Image Reconstruction - Correction Techniques in Emission Tomography - page 35

Essentially, the theorem says that when we measure Rf(;s) for all and s,
a 1D{Fourier transform with respect to the second variable will provide us
with the Fourier transform of f, and a further nD{inverse Fourier transform
will produce our function f. In particular, the Radon inversion problem is
uniquely solvable.
While the theorem is easily proved and written, it is hard to implement.
Suppose that Rf is measured on an equidistant grid in and s. Then, f is given
on a polar rather than rectangular grid, which makes it impossible to use plain
FFT for the inversion. While very ecient algorithms for the implementation
of Fourier slice exist (see, e.g., [6]), usually the filtered backprojection approach
is employed in PET.

3.2.2 Filtered backprojection
For simplicity, we assume that f is a 2D function. As a motivation for the
filter algorithms, let us look at a simple interpretation of one of its variants
rst.
Assume that the distribution function f z is the characteristic function of an
arbitrarily small neighborhood of a point z, meaning that all radioactivity
is enclosed in a very small region around z. Thus, only line integrals passing
through will have a nonzero value.
The simplest idea for inversion of the Radon transform is that the value of
f(x) has an impact on Rf(;s) only if the corresponding line passes through
x. Thus, we expect to get an approximation f 0 (x) to f(x) by simply averag-
ing over all line integrals going through x. Mathematically, that amounts to
evaluating (R Rf)(x).
While f 0 is in fact an approximation to f, it is very blurred (see Figure
3.1). The reason for that is easily seen when we refer to f z . f 0 z will have its
maximal value at z, but it will be not be zero outside of since for every
point x we nd some lines going through x and , so Rf > 0 for these lines,
so the average value will not be zero. Rather, a simple geometrical argument
shows that f 0 z decays like c=jjxzjj.
When we move z, f 0 z is moved appropriately, so the mapping from f to f 0
is translation invariant, which implies that it is a convolution; this is already
obvious from Figure 3.1. The convolution function is the response to a single
peak at the origin, which we identied as (x) = c=jjxjj. So, we have

(R Rf)(x) = (f)(x):
According to the corollary of the convolution theorem and using that in
2D, the Fourier transform of 1=r is 1=r, we nally nd that

1
4 jjjj

f() =

· \ R Rf():

Basically, this implies that R Rf is a smooth version of f, and all we have
to do to regain f is to apply an appropriate edge{enhancing lter, the lter is