Global Positioning System Reference
In-Depth Information
A
1
x
−
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
M
−
1
1
A
2
A
1
M
−
1
1
A
1
−
w
1
=
(4.106)
A
2
−
M
2
k
2
−
w
2
Eq
uation (4.106) shows how the normal matrix of the first group must be augmented
in
order to find the solution of both groups. The whole matrix can be inverted in one
ste
p to give the solution for
x
and
k
2
. Alternatively, one can compute the inverse using
th
e matrix partitioning techniques of Section A.3.5, giving
Q
11
A
1
M
−
1
x
=−
w
1
−
Q
12
w
2
(4.107)
1
Q
21
A
1
M
−
1
k
2
=
w
1
+
Q
22
w
2
(4.108)
1
Setting
[11
A
1
M
−
1
N
1
=
A
1
(4.109)
1
Lin
—
0.5
——
Nor
PgE
A
2
M
−
1
N
2
=
A
2
(4.110)
2
then
Q
11
=
N
1
+
N
2
−
1
A
2
M
2
+
A
2
−
1
A
2
N
−
1
N
−
1
1
N
−
1
1
A
2
N
−
1
Q
x
≡
=
−
(4.111)
1
1
N
−
1
A
2
M
2
+
A
2
−
1
Q
21
=
A
2
N
−
1
Q
12
=
(4.112)
1
[11
Q
22
=−
M
2
+
A
2
−
1
A
2
N
−
1
(4.113)
1
Substituting
Q
11
and
Q
12
into (4.107) gives the sequential solution for the param-
eters. We denote the solution of the first group by an asterisk and the contribution of
the second group by
. In that notation, the estimated parameters of the first group
are denoted by
x
∗
, which is simplified to
x
∗
. Thus,
∆
x
∗
+ ∆
x
=
x
(4.114)
Comparing (4.107) and (4.52) the sequential solution becomes
x
∗
=−
N
−
1
1
A
1
M
−
1
w
1
(4.115)
1
and
A
2
M
2
+
A
2
−
1
A
2
x
∗
+
w
2
N
−
1
1
A
2
N
−
1
∆
x
=−
(4.116)
1
Si
milarly, the expression for the Lagrange multiplier
k
2
is
k
2
=−
M
2
+
A
2
−
1
A
2
x
∗
+
w
2
A
2
N
−
1
(4.117)
1
