Page 302 - Introduction to Information Optics

5.10. Geometric Transform 287

The solution for this equation is 9 = rc/4 and a = v/2/2. Therefore, the
mappings in the geometric transformation are

u — (Pi(x, y) = ^—- (x cos 0 + y sin 6)

3 ; cos

Example 5.2
The log-polar transformation is a combination of the logarithmic trans-
formation and the polar coordinate transformation with the mapping as

u =

y
tan J
~~
It is a transformation from the Cartesian coordinate system to the polar
coordinate system; then the logarithmic transformation is applied to the radial
coordinate in the polar coordinate.
Derive the log-polar transformation of the two functions in Example 5, 1 .

Solution
We first process the polar transform by using inverse mapping, x = r cos 0
and v = r sin (9; that leads to
2
2
2
2
2
F p(r, 0) = /(r cos 6, r sin 6) = (r cos #) + 4(r sin 0) = r + 3r sin 0
2
2
G D (r, 9) = g(r cos 9, r sin 9) = 5(r cos Of + 5(r sin 9) - 6r cos 9 sin 0
2
2
2r + 6r sin 2
4
We then execute the transformation by using r = exp u and y = 9
2
2
2
2
F lp(u, v) = F p(exp u, v) — (exp u) + 3(exp u) sin v — exp(2w) + 3 exp(2w) sin i;
2
2
2
G lp(u, v) = G p(exp u, v) = 2(exp u) + 6(exp u) sin I v — -
2
- exp[2(w - ln^/2)] 4 3 exp[2(w - ln^/2)] sin ( u - ? ) •