Global Positioning System Reference
In-Depth Information
The equation 6 can be iteratively solved using Newton's method. However, the iterative
approach will be computationally expensive. Moreover, the positioning accuracy will be poor
as there is no proper formalism to identify and mitigate the error components.
3.1 Ordinary trilateration for positioning
x
i
2
,
y
i
2
,
z
i
2
represent satellite positions of the
i
th
pair. Analogous to 2-D linear form LOP of equation 3, a
3-D planar form LOP is found as follows.
x
i
2
−
p
2
be an arbitrary satellite pair, where
x
i
1
,
y
i
1
,
z
i
1
and
i
i
i
i
Let
P
i
=
1
,
p
p
1
=
p
2
=
x
i
1
x
y
i
2
−
y
i
1
y
z
i
2
−
z
i
1
z
+
+
=
2
r
i
2
r
i
1
2
r
i
2
2
r
i
1
−
(8)
i
2
2
i
1
2
p
−
p
+
−
+
2
ξ
Where it is assumed that the noise are equal and constant for a particular satellite pair
i.e.
,
ξ
1
=
ξ
2
=
ξ
.
The equation becomes linear in terms of
x
,
y
,
z
and
ξ
if the noise is represented by a single
parameter
for all pairs. In that case there are four unknowns in this equation and therefore
four equations will be required to solve them. In practicality, the assumption is susceptible
for large positioning error and hence iterative refinement approach of the following is rather
adopted for real implementations.
ξ
3.2 Iterative least squares estimate
0
ρ
=
The iterative approach works by having a preliminary estimate of the receiver position (
x
0
y
0
z
0
T
). Let the rotation rate of the earth be
[
. The position vectors in the earth centered
earth fixed (ECEF) system of the receiver be donated by
]
ω
ρ
(
t
)
ECEF
and geo-stationary position
vector for satellite
i
be denoted by
geo
where the argument
t
denotes the dependence on
time. The range equation can be written as:
p
i
(
t
)
r
i
=
(ωτ
i
)
p
i
(
−τ
i
)
−ρ(
)
ECEF
R
3
t
t
(9)
geo
Where
R
3
is the earth's rotation matrix as defined below.
⎣
⎦
cos
(
ωτ
i
)
sin
(
ωτ
i
)
0
R
3
(
ωτ
i
)=
−
sin
(
ωτ
i
)
cos
(
ωτ
i
)
0
0
0
1
Let
⎣
⎦
=
−τ
i
)
geo
and
⎣
⎦
=
ρ
(
x
i
y
i
z
i
x
y
z
R
3
(
ωτ
i
)
p
i
(
t
t
)
ECEF
(10)
Now, omitting the refraction terms
I
i
and
T
i
and linearizing the equation 6, we get
y
0
x
0
z
0
x
i
−
y
i
−
z
k
−
0
−
0
δ
x
−
0
δ
y
−
0
δ
z
+(
cdt
)=
r
i
−
(
r
i
)
−
i
=
b
i
−
i
(11)
(
r
i
)
(
r
i
)
(
r
i
)
where
b
i
denotes the correction to the preliminary range estimate.
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