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.

Search WWH ::

Custom Search