Biomedical Engineering Reference
In-Depth Information
φ
(4)
yx
(
k
,
l
)
=
(
yx
)
(
x
,
y
)
φ
(
yx
)
k
,
l
(2)
(
k
)
(2)
(
l
)
,
(
x
,
y
)
dx dy
=
φ
(
x
)
k
,
l
(
x
,
y
)
dx dy
=
D
(
l
)
(3)
(
xx
)
(
x
,
y
)
φ
(3)
(
k
)
,
x
(
k
,
l
)
=
φ
(
yy
)
k
,
l
(3)
(
y
)
(
x
,
y
)
φ
(3)
(
l
)
,
y
(
k
,
l
)
=
(
x
,
y
)
dx dy
=
D
(
k
)
φ
(
x
)
k
,
l
(
x
,
y
)
dx dy
=
D
(
l
)
(2)
(
x
)
(
x
,
y
)
φ
(2)
(
k
)
,
x
(
k
,
l
)
=
φ
(
y
)
k
,
l
(
x
,
y
)
dx dy
=
D
(
k
)
(2)
(
y
)
(
x
,
y
)
φ
(2)
(
l
)
,
y
(
k
,
l
)
=
φ
(1)
x
(
k
,
l
)
=
(
x
)
(
x
,
y
)
φ
k
,
l
(
x
,
y
)
dx dy
=
D
(
l
)
(1)
(
k
)
,
φ
(1)
(
y
)
(
x
,
y
)
φ
k
,
l
(
x
,
y
)
dx dy
=
D
(
k
)
(1)
(
l
)
,
y
(
k
,
l
)
=
where
(1)
(
k
)
=
(
x
)
(
x
)
φ
(
x
−
k
)
dx
,
(2)
(
k
)
=
(
x
)
(
x
)
φ
(
x
)
(
x
−
k
)
dx
,
φ
φ
(3)
(
k
)
=
(
xx
)
(
x
)
φ
(
x
)
(
x
−
k
)
dx
,
(4)
(
k
)
=
(
xx
)
(
x
)
φ
(
xx
)
(
x
−
k
)
dx
φ
φ
are 1D connection coefficients and
D
(0)
=
1
,
D
(
n
)
=
0
,
n
=
1. Note that since
the 2D basis here is constructed from the tensor product of 1D basis, these
2D connection coefficients can be computed by using 1D coefficients. We also
notice that these connection coefficients are independent of the input images;
therefore, they only need to be computed once.
The energy function is then linearized by taking the linear term in its Taylor
expansion at (
p
,
q
). The next step is to solve the optimization problem associated
with the linearized energy function by iterations. Let
δ
p
i
,
j
,δ
q
i
,
j
, and
δ
z
i
,
j
be the
small variation of
p
i
,
j
,
q
i
,
j
, and
z
i
,
j
,
respectively, and set
∂δ
p
i
,
j
=
∂δ
W
∂δ
W
∂δ
q
i
,
j
=
∂δ
W
∂δ
z
i
,
j
=
0
.
We obtain
δ
p
i
,
j
=
[
C
1
D
22
−
C
2
∂
R
∂
p
(
i
,
j
)
∂
R
∂
q
(
i
,
j
)]
/
D
,
(5.73)
δ
q
i
,
j
=
[
C
2
D
11
−
C
1
∂
R
∂
p
(
i
,
j
)
∂
R
∂
q
(
i
,
j
)]
/
D
,
δ
z
i
,
j
=
C
3
/
D
33
,
