Image Processing Reference
In-Depth Information
U
are two vectors
2
,
This calls for the following important comment. If
U
and
R
then:
U
∗
J
U
=
u
x
u
y
−
u
x
u
y
,
[6.26]
=det
U
U
,
,
and as a result:
=det
R
V
,
W
k
X
+
k
V
,
R
+
k
[6.27]
det
R
k
,
R
k
R
k
R
k
=
r
k
,
r
k
sin
R
k
,
R
k
=
r
k
[6.28]
r
k
,
R
k
R
+
k
V
where:
R
k
R
+
k
V
.
We now simply have to point out that sin(
R
k
,
R
k
)=sin(
β
k
−
β
k
).
This calculation can be applied, not without some difficulties, to the case of a target
and an observer that are maneuvering. We then have the following convergence result
[LEC 99].
Convergence of the iterative methods
Let us assume that the maneuvering times of the target are known, then the deriva-
tive with respect to time
L
(
ˆ
) of the Liapunov function
L
(
ˆ
X
X
) is:
p
L
ˆ
X
=2
ˆ
X
−
X
∗
G
L
ˆ
X
=
r
k
β
k
−
β
k
sin
β
k
−
β
k
.
r
k
−
2
[6.29]
k
=1
G
L
(
ˆ
) of the likelihood
functional that cannot be equal to zero if all of the
β
k
and
β
k
coincide, i.e. if the model
is perfectly estimated. This analysis can be generalized without difficulty to the case
of multiple sensors and most importantly to multi-leg cases
2
.
As you can see, there is a functional of the gradient vector
X
This type of calculation can actually easily be applied to the case of a constantly
accelerating target and, more generally, to an observation model of the type tan(
β
t
)=
A
t
X
B
t
X
A
t
X
B
t
X
where
f
is a monotonic and continuously
differentiable function. Thus, the non-linearity has a relatively simple structure. This is
/
and even
f
(
β
t
)=
/
2. A leg is a section of the trajectory over which the target has a uniform rectilinear motion.
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