Global Positioning System Reference
In-Depth Information
1
2
φ
∂
v
1
=
∂
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
B
1
P
1
v
1
−
k
1
=
o
(4.93)
φ
∂
v
2
=
1
2
∂
B
2
k
2
=
P
2
v
2
−
o
(4.94)
1
2
∂
φ
∂
x
=−
A
1
A
2
k
1
−
k
2
=
o
(4.95)
1
2
φ
∂
k
1
=
∂
B
1
v
1
+
A
1
x
+
w
1
=
o
(4.96)
1
2
φ
∂
k
2
=
∂
B
2
v
2
+
A
2
x
+
w
2
=
o
(4.97)
[11
an
d solving for
v
1
,
v
2
,
k
1
,
k
2
and
x
. Equations (4.93) and (4.94) give the residuals
Lin
—
7.3
——
No
*PgE
P
−
1
1
B
1
k
1
v
1
=
(4.98)
P
−
1
2
B
2
v
2
=
k
2
(4.99)
Combining (4.98) and (4.96) yields
M
1
k
1
+
A
1
x
+
w
1
=
o
(4.100)
[11
where
B
1
P
−
1
B
1
M
1
=
(4.101)
1
is
an
r
1
×
r
1
symmetric matrix. The Lagrange multiplier becomes
M
−
1
1
M
−
1
1
k
1
=−
A
1
x
−
w
1
(4.102)
Eq
uations (4.95) and (4.97) become, after combination with (4.102) and (4.99),
A
1
M
−
1
A
1
M
−
1
A
2
k
2
=
A
1
x
+
w
1
−
o
(4.103)
1
1
B
2
P
−
1
B
2
k
2
+
A
2
x
+
w
2
=
o
(4.104)
2
By
using
B
2
P
−
1
B
2
M
2
=
(4.105)
2
we can write both Equations (4.103) and (4.104) in matrix form:
