Image Processing

As in the (1, 3) case, we consider the shifted set of the rays, l(t) → l(t – 1), as the set of 25 geometrical rays

$\tilde{l}\left(t\right)=l_{1,-3}\left(t-1\right)=\left\{\left(x,y\right);x-3y=\frac{t}{7}-\frac{2}{7}\right\},t=6:-1:-18,$

which are shown in Figure 5.22.

For the line-integrals, we obtain the system of equations, which is similar to the system described for the (1, 3)-projection,

$w\left(t\right)=w\left(\tilde{l}\left(t\right)\right)=\frac{7\sqrt{10}}{3}\left[v\left(t\right)+v\left(t-1\right)+v\left(t-2\right)\right],t=6:-1:-18,$

(5.14)

FIGURE 5.22The set of geometrical rays for the (1,−3)-projection.

where v(−19) = v(−20) = 0. The required sums v(t) of the discrete image f_n,m can therefore be defined as $v\left(t\right)=3b_t/\left(7\sqrt{10}\right)$, where the components b_t are calculated recursively by