In order to investigate each term by itself, we notice that (11) is
satisfied when y is chosen in such a way that f = aD' à. The k-th
component becomes
fsaAg ; Km 1,2,...,M (16)
from which we obtain
log. (f, /a$,)
k^ ^y M
UK) = —————— = - log (Ë, /ad, ) (17)
100 Ac 2kri k k
which should be compared to (15). Here y has been given as a function
of k, as we generally obtain a different value of y for every k. The
complex logarithms in these expressions are easily expressed in the
more familiar real form:
Imag (f£,/8,)
M M. abre Eee
Real p(k) = k Luce oT arctg Sei VNIN (18)
Imag u(k) - —- Au log T [Rea (f, /8,) * ing 0,8 3] 7 (19)
Ls:0, 1,:2,. |. 5 K-1
M is the period corresponding to a given spatial frequency k. L is
fne number of complete periods involved in the displacement. Only one
of the k possible integers L is correct. In the case when the image f
is a pure cyclic translation of the image g, we obtain the same value
of u for all values of k:
| = M N=
Real p(k) = k Lot kr Pk Oo € 94 « 27 (20)
SUUM.
Imag p(k) = T log a (21)
where vw = 2kwu/M is the arctg-function in (18). Several procedures
for finding L can be conceived. Here we choose a method which can also
be used in the case where there are discrepancies between the images:
For each k, y is determined for all possible values of L. A vector of
M components (the image size) 1s constructed where the components
indexed [u] and [p+1] (k different cases) are set equal to 1 and the
rest are set equal to O. All the vectors are added and y chosen as the
index of the largest component. This method is used in the following
examples, which are typical for a large number of investigated cases
with translations ranging from zero to half the image size.
Example: Cyclically displaced Sine Curve:
E(x) = sin, 21x/38; x = 1,2, ...8
G(x) = sin 27(x121)/38; x = 1,2, canl3.®= M=2] (22)
g(x) = sin.27(x-43)/38; X =.44 45 ....M
M = 64
This choice of f and g corresponds to y - 21. The vectors defined in
the above section are given in figure 2a, one row for each vector (for
each value k). As is immediately seen, the maximum component of the
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