Fundamentals of Satellite Navigation - Understanding GPS: Principles and Applications - page 56

The expansion has been truncated after the first-order partial derivatives to
eliminate nonlinear terms. The partials derivatives evaluate as follows:

( )

, $

$ $ $

∂ fx y z t , , $

xx
r

−

uuuu

j u

=−

$ $

∂ x

u

j

( )

, $

$ $ $

∂ fx y z t , , $

yy
r

−

uuuu

u

j

=−

$ $

∂ y

u j

(2.28)

( )

, $

$ $ $

∂ fx y z t , , $

zz
r

− uuuu

j u

=−

$ $

∂ z

u

j

( )

, $

$

, $ , $ ∂ fx yzt

uuuu

= c

$

∂ t

u

where

( ( (

)

2

)

2

)

2

$ $ $ $

r = x − x +− y y +− z z

j j u j u j u

Substituting (2.25) and (2.28) into (2.27) yields

$ $ $

=− −

xx
r

yy
r

− zz

r

−

j u j u j u

$

ρρ
j

∆ x − ∆ y − ∆ z t + (2.29)

j u u u u

$ $ $

j j j

We have now completed the linearization of (2.24) with respect to the unknowns

∆ t u . (It is important to remember that we are neglecting secondary

error sources such as Earth rotation compensation, measurement noise, propagation
delays, and relativistic effects, which are treated in detail in Section 7.2.)
Rearranging this expression with the known quantities on the left and
unknowns on right yields

x u , ∆ y u , ∆ z u ,and ∆

$ $ $

−= −

xx
r

yy
r

− zz

r

−

u u u

$

j j j

ρρ
j

∆ x + ∆ y − ∆ z t − (2.30)

j

$

u

$

u

$

u u

j j j

For convenience, we will simplify the previous equation by introducing new
variables where

$

∆ρρ ρ =−

j j

$

xx
r

−

u

j

a =

xj

$

j

(2.31)

$

yy
r

−

j u

a =

yj

$

j

$

zz
r

−

j u

a =

zy

$

j