Image Processing Reference
In-Depth Information
T
Fig. 14.5. The two-to-one mapping of the spectrum, Eq. (14.5), creating an equivalence of
directions differing with
2
solve. After presenting the solution to this problem, we will generalize these results
by means of a theorem which includes the cross-fitting (and the line-fitting) process
as a special case. The proof of the theorem will not explicitly be carried out since it
is a straightforward extension of the method described here. We will show that the
complex moment
I
pq
of
|
2
, with
p
|
F
−
q
=4, fits a cross (4-folded symmetry) to the
|
2
, and thereby to
F
in the TLS sense.
Assuming that the spectrum is expressed in polar coordinates,
F
(
r, ϕ
), the inte-
gral defining the complex moment
I
40
of the power spectrum yields:
I
40
=
2
π
0
function
|
F
∞
r
4
exp(
i
4
ϕ
)
2
r
d
r
d
ϕ
|
F
(
r, ϕ
)
|
(14.4)
0
Using the transformation
ϕ
1
=2
ϕ
(14.5)
the integral can be transformed once more, so that it can be identified as
I
20
of a
remapped spectrum. The latter mapping is a two-to-one mapping, as illustrated in
Fig. 14.5. Consequently, points of the spectrum that have angular coordinates dif-
fering by
π
2
|
2
originating from
are treated as equivalent, i.e. the spectral power
|
F
F
1
|
2
. The complex mo-
such points is added, to produce the folded spectral power
|
|
2
, respectively. Using
theorem 10.2, one can then conclude that
I
40
,I
22
fit a cross to the power spectrum
in the TLS sense. The pairs
I
4
,
0
,I
2
,
2
also constitute a tensor because they together
represent the structure tensor of a (remapped) power spectrum. The cross-fitting pro-
cess, a detailed discussion of which is found in [26], leads naturally to the following
generalization:
Theorem 14.1 (Group direction tensor I).
A pair of complex moments
I
n,
0
and
I
n
2
F
1
|
2
are such that they equal
I
40
,I
22
ments
I
20
,I
11
of
|
of
|
F
of
|F |
2
, with
n
=0
, determines an optimal fit of a set of lines possessing
n
-folded symmetry to a function
F
in the TLS sense. The real quantities
,
n
2
|I
n
0
|,I
n
2
,
n
2
depend on the minimum and the maximum errors of the fit,
e
(
k
n
max
)
,
e
(
k
n
min
)
, respec-
tively,
|I
n
0
|
=
e
(
k
n
)
− e
(
k
n
min
)
(14.6)
max
=
e
(
k
n
max
)+
e
(
k
n
min
I
n
2
)
(14.7)
,
n
2
