Image Processing Reference
In-Depth Information
If one attempts to solve
v
=(
v
x
,v
y
)
T
from the BCC when the partial derivatives
of
f
are known at a given space-time
x, y, t
, one fails because
v
contains two real
variables, whereas we have only one equation. However, if the equation holds at
several image points
s
=(
x
k
,y
k
)
T
at a given time
t
0
, then we can write several
such equations. This happens typically for a local image pattern
g
(
x, y
) wherein all
points translate with the same velocity
v
. This situation is similar to the example in
Fig. 12.2 (right), where the coherent translation of points is depicted. Accordingly,
the BCC for a discrete 2D neighborhood
f
(
x
k
,y
k
,t
0
) yields
10
⎛
⎝
⎞
⎠
⎛
⎝
⎞
⎠
∂f
(
x
1
,y
1
,t
0
)
∂x
∂f
(
x
1
,y
1
,t
0
)
∂y
∂f
(
x
1
,y
1
,t
0
)
∂t
∂f
(
x
2
,y
2
,t
0
)
∂t
.
∂f
(
x
N
,y
N
,t
0
)
∂t
v
x
v
y
∂f
(
x
2
,y
2
,t
0
)
∂x
∂f
(
x
2
,y
2
,t
0
)
∂y
=
−
(12.98)
.
.
∂f
(
x
N
,y
N
,t
0
)
∂x
∂f
(
x
N
,y
N
,t
0
)
∂y
Here,
f
(
x
k
,y
k
,t
0
) with
k
represents
f
(
x, y, t
) evaluated at the
k
th point
of an image neighborhood. This is an overdetermined system of linear equations of
the form
∈{
1
···
N
}
d
=
−
Dv
(12.99)
with
v
being unknown and
⎛
⎞
⎛
⎞
∂f
(
x
1
,y
1
,t
0
)
∂x
∂f
(
x
1
,y
1
,t
0
)
∂y
∂f
(
x
1
,y
1
,t
0
)
∂t
∂f
(
x
2
,y
2
,t
0
)
∂t
.
∂f
(
x
N
,y
N
,t
0
)
∂t
⎝
⎠
⎝
⎠
∂f
(
x
2
,y
2
,t
0
)
∂x
∂f
(
x
2
,y
2
,t
0
)
∂y
d
=
,
D
=
(12.100)
.
.
∂f
(
x
N
,y
N
,t
0
)
∂x
∂f
(
x
N
,y
N
,t
0
)
∂y
are to be assumed known. Suggested by Lucas and Kanade [157], this is a linear re-
gression problem for optical flow estimation. The standard solution of such a system
of equations is given by the MS estimate, obtained by multiplying the equation with
D
T
and solving for the 2
×
2 system of equations for the unknown
v
:
D
T
d
=
−
D
T
Dv
(12.101)
The solution exists if the matrix
S
=
D
T
D
=
k
T
s
k
f
)
(
∇
s
k
f
)
·
(
∇
(12.102)
where
∂f
(
x
k
,y
k
,t
0
)
∂x
∂f
(
x
k
,y
k
,t
0
)
∂y
∇
s
k
f
=
(12.103)
However,
S
is the structure tensor for the 2D discrete image
f
(
x
k
,y
k
,t
0
). A unique
solution exists if
S
is nonsingular, a situation that occurs if the image lacks linear
10
For a fixed
t
0
,
f
(
x, k, y
k
,t
0
) is a 2D function.
