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Antti Lehtinen Doppler Positioning with GPS - Matematiikan laitos

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Fortunately, findingthe matrix G by algebraic differentiation of the function<br />

p(ˆx) is quite straightforward. Let us consider the differentiation in detail:<br />

G = ∂p<br />

∂ ˆx =<br />

⎡<br />

⎢<br />

∂p ⎢<br />

= ⎢<br />

ˆru<br />

∂<br />

⎢<br />

ˆd<br />

⎢<br />

⎣<br />

∂p1<br />

∂ ˆru<br />

∂p1<br />

∂ ˆ ∂p2<br />

∂ ˆru<br />

d<br />

∂p2<br />

∂ ˆ .<br />

∂pn<br />

d<br />

.<br />

∂pn<br />

∂ ˆru ∂ ˆ ⎤<br />

⎥ ∈ R<br />

⎥<br />

⎦<br />

d<br />

n×4<br />

(4.23)<br />

where pi are the components of the vector function p. Usingthe Definition 3,<br />

one obtains an expression for a single component of the vector function p. This<br />

can be written as<br />

pi(ˆx) =vi • ri − ˆru<br />

ri − ˆru + ˆ d − ˙ρi<br />

(4.24)<br />

Differentiatingthe ith component of the delta range residuals function p(ˆx) partially<br />

<strong>with</strong> respect to the estimated receiver position ˆru yields<br />

∂pi<br />

∂ ˆru<br />

= ∂<br />

∂ ˆru<br />

<br />

vi • ri − ˆru<br />

ri − ˆru + ˆ d − ˙ρi<br />

<br />

(4.25)<br />

In equation 4.25 vi, ri, ˆ d and ρi are constants <strong>with</strong> respect to the estimated<br />

receiver position ˆru. Applyingthe formula (fg) ′ = f ′ g + fg ′ for differentiating<br />

products gives us<br />

∂pi<br />

∂ ˆru<br />

= vi • (ri − ˆru) ∂ 1<br />

∂ ˆru ri − ˆru +<br />

vT i ∂<br />

(ri − ˆru)<br />

ri − ˆru ∂ ˆru<br />

1<br />

= vi • (ri − ˆru)<br />

ri − ˆru 2<br />

(ri − ˆru) T<br />

ri − ˆru −<br />

vT i<br />

ri − ˆru I3×3<br />

<br />

vi<br />

=<br />

ri − ˆru • ri<br />

<br />

− ˆru ri − ˆru<br />

ri − ˆru ri − ˆru<br />

<br />

ri − ˆru<br />

−<br />

ri − ˆru • ri<br />

T − ˆru vi<br />

ri − ˆru ri − ˆru<br />

where I3×3 is 3 × 3 identity matrix.<br />

(4.26)<br />

The equation 4.26 gives the partial derivative of the delta range residuals function<br />

<strong>with</strong> respect to the receiver position estimate. The form of the equation is<br />

satisfactory for numerical computation. However, the form is not very intuitive.<br />

The right hand side of the equation 4.26 can be transformed into a geometrically<br />

more illustrative form involvingcross products. Makinguse of the vector triple<br />

28

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