Global Positioning System Reference
In-Depth Information
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The symbol
I
indicates the values for
z
j
have already been selected, i.e., are known.
Note that the subscript
j
goes from
i
|
+
1to
n
. Since
m
i,i
=
1 and using (7.188) and
(7.186), we can write the
i
th component as
w
i
=ˆ
ˆ
z
i
−
z
i
+ˆ
w
i
|
I
i
=
1
,n
−
1
(7.189)
The bounds of the
z
parameters follow from (7.187). We begin with the
n
th level to
determine the bounds for the
n
th ambiguity and then proceed to level 1, establishing
the bound for the other ambiguities. Using the term with
w
n
s
n,n
in (7.187), and
ˆ
knowing that the matrix element
m
n,n
=
1 in (7.186), we find
w
n
s
n,n
=
z
n
−ˆ
z
n
2
s
n,n
≤ χ
2
ˆ
(7.190)
[28
The bounds are
−
χ
2
/s
n,n
1
/
2
+
χ
2
/s
n,n
1
/
2
z
n
ˆ
≤
z
n
≤ˆ
z
n
(7.191)
Lin
—
0.0
——
Nor
PgE
U
sing the terms from
i
to
n
in (7.187) and (7.189), we obtain for level
i
,
n
w
i
s
i,i
=
ˆ
w
i
|
I
2
s
i,i
≤
χ
2
w
j
s
j,j
ˆ
z
i
−
z
i
+ˆ
−
1
ˆ
(7.192)
j
=
i
+
1
/
2
n
1
√
s
i,i
χ
2
w
j
s
j,j
[28
z
i
+ˆ
ˆ
w
i
|
I
−
−
1
ˆ
≤
z
i
j
=
i
+
(7.193)
1
/
2
n
1
√
s
i,i
χ
2
w
j
s
j,j
≤ˆ
z
i
+ˆ
w
i
|
I
+
−
1
ˆ
j
=
i
+
The bounds (7.191) and (7.193) can contain one or several integer values
z
n
or
z
i
. All values must be used when locating the bounds and integer values at the next
lower level. The process stops when level 1 is reached. For certain combinations, the
process stops earlier if the square root in (7.193) becomes negative.
Figure 7.20 demonstrates how one can proceed systematically, trying to reach the
first level. At a given level, one proceeds from the left to the right while reaching a
lower level. This example deals with
n
=
4 ambiguities
z
1
,z
2
,z
3
, and
z
4
. The fourth
level produced the qualifying values
z
4
={−
1
,
0
,
1
}
. Using
z
4
=−
1or
z
4
=
1
does not produce a solution at level 3 and the branch terminates. Using
z
4
=
0
gives
z
3
={−
1
,
0
}
at level 3. Using
z
3
=−
1 and
z
4
=
0, or in short notation
z
1
,
0
)
does not
produce a solution at level 1; the branch terminates. Returning to level 3, we try the
combination
z
=
(
−
1
,
0
)
, one gets
z
2
=
0 at level 2. The combination
z
=
(
0
,
−
=
(
0
,
0
)
, giving
z
2
={−
1
,
0
,
1
}
at level 2. Trying the left branch with
z
=
(
−
1
,
0
,
0
)
gives no solution and the branch terminates. Using
z
=
(
0
,
0
,
0
)
gives
