Global Positioning System Reference
In-Depth Information
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TABLE 4.4
Sequential Solution without Inverting the Normal Matrix
P
1
O
OP
2
1
a
=
x
a
P
=
2
a
=
f
2
a
(
x
a
)
v
1
=
x
+
1
1
=
x
0
−
x
b
v
2
=
A
2
x
+
2
N
2
=
A
2
P
2
A
2
N
1
=
P
1
u
2
=
A
2
P
2
2
u
1
=
P
1
1
[12
(
x
0
−
x
b
)
Q
1
=
P
−
1
x
1
=−
1
Lin
—
0.5
——
Nor
PgE
v
T
Pv
1
=
0
x
i
=
x
i
−
1
+ ∆
x
i
−
1
v
T
Pv
i
=
v
T
Pv
i
−
1
+ ∆
v
T
Pv
i
−
1
Q
i
=
Q
i
−
1
+ ∆
Q
i
−
1
P
−
1
i
+
A
i
Q
i
−
1
A
i
−
1
T
=
[12
Q
i
−
1
A
i
T
A
i
x
i
−
1
+
i
∆
x
i
−
1
=−
∆
v
T
Pv
i
−
1
=
(
A
i
x
i
−
1
+
i
)
T
T
(
A
i
x
i
−
1
+
i
)
∆
Q
i
−
1
=−
Q
i
−
1
A
i
TA
i
Q
i
−
1
Note:
Case of observation equation model.
with
v
1
=
+
1
A
1
x
(4.165)
A
2
x
+
2
=
o
(4.166)
Table 4.5 contains the expression of the sequential solution with conditions between
parameters. If (4.158) is used to impose the conditions, the largest numbers that can
still be represented in the computer should be used. In most situations, it will be
readily clear what constitutes a large weight; the weight must simply be large enough
so that the respective observations or parameters do not change during the adjustment.
For sequential solution, the solution of the first group must exist. Conditions cannot
