Global Positioning System Reference
In-Depth Information
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x
T
Ax
φ =
(A.68)
is,
according to (A.67),
2
x
T
A
d
x
d
φ =
(A.69)
Th
e gradient of
φ
with respect to
x
is
∂
T
∂
φ
∂
x
≡
φ
∂x
1
φ
∂x
u
∂
···
=
2
Ax
(A.70)
Eq
uation (A.70) can be readily verified by computing the partial derivatives
∂
φ
/∂x
t
at
the
t
th component,
#
&
[35
k
k
∂
x
T
Ax
∂x
t
∂
∂x
i
%
(
=
x
i
x
j
a
ij
j
=
1
i
=
1
Lin
—
0.7
——
No
*PgE
(A.71)
k
k
k
=
x
j
a
tj
+
x
i
a
it
=
2
x
j
a
tj
j
=
1
i
=
1
j
=
1
=
[2
Ax
]
t
because
A
is symmetric.
Equation (A.70) is the foundation for deriving least-squares solutions, which re-
quires locating the stationary point (minimum) for a quadratic function. The proce-
dure is to take the partial derivatives with respect to all variables and equate them
to zero. While the details of the least-squares derivations are given in Chapter 4, the
following example serves to demonstrate the principle of minimization using matrix
notation.
Let
B
denote an
n
[35
×
u
rectangular matrix with
n>u,
is an
n
×
1 vector, and
P
an
n
×
n
symmetric weight matrix that can include the special case
P
=
I
. The
elements of
B
,
, and
P
are constants. The least-squares solution of
v
=
Bx
+
(A.72)
v
T
Pv
requires
φ
(
x
)
≡
=
min. First, we compute the gradient (column vector)
∂
v
T
Pv
∂
x
∂
x
(
Bx
)
∂
)
T
P
(
Bx
=
+
+
∂
x
2
∂
(A.73)
T
PBx
x
T
B
T
PBx
T
P
=
+
+
2
B
T
PBx
2
B
T
P
=
+
and equate it to zero,
