Image Processing Reference
In-Depth Information
where
d
equal to
dω
x
dω
y
, and the integral is a double integral over
E
2
. The ex-
pression defines the TLS error function for a continuous data set. Interpreting the
integral as a summation, Eq. (10.13) is a special case of Eq. (10.17). By construc-
tion, the contribution of
F
(
ω
ω
) to the error is zero for points
ω
along the line
t
k
. With
increased distance of
ω
to the latter axis, the contribution of
F
(
ω
) will be amplified.
|
2
, which is the spectral energy, is interpreted as the mass density, then
the error
e
(
k
) corresponds to the inertia of a mass with respect to the axis
k
in
mechanics [83]. Besides this function being continuous in
k
, it has also a screening
effect; that is,
F
is concentrated to a line if and only if
e
(
k
) vanishes for some
k
.
This is to be expected because the resulting quantity when substituting Eq. (10.16)
in Eq. (10.17) is the square norm of a vector-valued function, i.e.,
e
(
k
)=
ω
−
(
ω
If
|
F
(
ω
)
2
=
T
k
)
k
F
ω
−
T
k
)
k
2
|
|
2
d
(
ω
F
(
ω
)
ω
(10.18)
Because
e
(
k
) is the square norm of a function in a Hilbert space,
ω
−
T
k
)
k
F
ω
−
T
k
)
k
F
=
0
2
=0
e
(
k
)=
(
ω
⇒
(
ω
(10.19)
Consequently, when
e
(
k
)=0, the expression
ω
−
T
k
)
k
F
equals (the vector-
(
ω
=0, this will happen if and only if
F
equals zero
outside the line
t
k
. A nontrivial
F
can thus be nonzero only on the line
t
k
, which,
according to lemma 10.1, means that
f
must be linearly symmetric. Conversely, if
F
is zero except on the line
t
k
, the corresponding
e
(
k
) vanishes. To summarize, if and
only if
e
(
k
)=0, all spectral energy,
valued function) zero. Assuming
F
|F |
2
, is concentrated to a line and
f
is linearly
symmetric with the direction
±
k
always
translates to an unambiguous direction aside (we will discuss this further below), we
proceed to the more urgent question of how to calculate
k
.
Noting that
±
k
. Leaving the question whether or not
T
k
is a scalar that is indistinguishable from the scalar
k
T
, and
k
2
=
k
T
k
=1, the square magnitude distance can be written in quadratic form:
ω
ω
,
k
)=
k
T
I
T
k
d
2
(
T
ω
ω
ω
−
ωω
(10.20)
=
k
T
ω
x
+
ω
y
ω
x
k
0
ω
x
ω
y
−
(10.21)
ω
x
+
ω
y
ω
y
0
ω
x
ω
y
=(
ω
x
,ω
y
)
T
Defining the components of
ω
as
ω
x
=
ω
1
,
and
ω
y
=
ω
2
,
(10.22)
for notational convenience, Eq. (10.17) is expressed as
e
(
k
)=
k
T
Jk
=
k
T
(
I
·
Trace(
S
)
−
S
)
k
(10.23)
where
I
is the identity matrix,
J
=
I
·
Trace(
S
)
−
S
(10.24)
and
