Global Positioning System Reference
In-Depth Information
These nonlinear equations can be solved for the unknowns by employing either
(1) closed-form solutions [19-22], (2) iterative techniques based on linearization, or
(3) Kalman filtering. (Kalman filtering provides a means for improving PVT esti-
mates based on optimal processing of time sequence measurements and is described
in Sections 7.3.5 and 9.1.3.) Linearization is illustrated in the following paragraphs.
(The following development regarding linearization is based on a similar develop-
ment in [23].) If we know approximately where the receiver is, then we can denote
the offset of the true position (
x
u
,
y
u
,
z
u
) from the approximate position (
,
$
xyz
uuu
$
,
$
)
by a displacement
, , ). By expanding (2.20) to (2.23) in a Taylor series
about the approximate position, we can obtain the position offset (
(∆
xyz
u
∆
∆
u
u
z
u
)as
linear functions of the known coordinates and pseudorange measurements. This
process is described next.
Let a single pseudorange be represented by
∆
x
u
,
∆
y
u
,
∆
(
)
(
)
(
)
2
2
2
ρ
j
=
xx
−
+−
yy
+−
zz
+
t
(2.24)
j
u
j
u
j
u
u
(
)
=
fx y z t
,
,
,
uuuu
uuu
and time bias estimate
t
u
,
an approximate pseudorange can be calculated:
$
$
,
$
Using the approximate position location (
xyz
,
)
(
)
(
)
(
)
2
2
2
$
$
$
$
$
ρ
j
=
xx
−
+−
yy
+−
zz
+
t
j
u
j
u
j
u
u
(2.25)
(
)
,
$
$
$
$
=
fx y z t
,
,
uuuu
As stated earlier, the unknown user position and receiver clock offset is consid-
ered to consist of an approximate component and an incremental component:
$
xx x
yy y
z
=+
=+
=+
∆
∆
∆
u
u
u
$
$
$
u
u
u
(2.26)
z
z
u
u
u
t
=+
t
∆
t
u
u
u
Therefore, we can write
(
)
(
)
,
$
$
,
$
,
$
fx y z t
,
,
,
=
fx
+
∆
x y
+
∆
y z
+
∆
z t
+
∆
t
uuuu
u
uu
uu
uu
u
This latter function can be expanded about the approximate point and associ-
ated predicted receiver clock offset (
,
$
,
$
$
$
xyzt
uuuu
using a Taylor series:
,
)
(
)
(
)
,
$
,
$
$
$
$
$
$
$
fx
+
∆
xy
,
+
∆
yz
,
+
∆
zt
+
∆
t
=
fxyzt
,
,
u
u
u
u
u
u
u
u
u
u
u
u
(
)
(
)
,
$
,
$
$
$
$
$
$
$
∂
fx y z t
,
,
∂
fx y z t
,
,
uuuu
uuuu
+
∆
x
+
∆
y
(2.27)
u
u
$
$
∂
x
∂
y
u
u
(
)
(
)
,
$
,
$
$
$
$
$
$
$
∂
fx y z t
,
,
∂
fx y z t
,
,
uuuu
uuuu
+
∆
z
+
∆
t
+K
u
$
u
$
∂
z
∂
t
u
u
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