New characteristics of weighted GDOP in multi-GNSS positioning

Abstract

In positioning, navigation and timing applications of multi-GNSS (global navigation satellite system) constellations, the geometric dilution of precision (GDOP) offers an important index for selecting satellites and evaluating positioning accuracy. However, GDOP assumes that the measurement errors of all the tracked satellites are independent and have the same accuracy level, which is impossible in practice, especially when the tracked satellites are from various constellations. Through introducing a weighted matrix describing the measurement errors of different satellites into a common GDOP, we focus on new characteristics of weighted GDOP (WGDOP) in two aspects. First, we compare the sizes of WGDOP and the common GDOP based on the range of the weights of different satellites, i.e., the diagonal elements of the weighted matrix. In addition, when the weights of different satellites increase, the change of WGDOP with the weights is also derived. Moreover, a closed-form formula for calculating WGDOP is also presented. The theoretical derivations demonstrate that the closed-form can reduce the computation burden effectively. Furthermore, numerical tests verify these analyses.

Appendix

The proof of (9) will be clarified in this section. Let ${\varvec{A}},\;{\varvec{B}} \in {R^{K \times K}}$ be the two symmetric positive matrices and let the eigenvalues ${\lambda _i}\left( {\varvec{A}} \right),\;{\lambda _i}\left( {\varvec{B}} \right)$, ${\lambda _i}\left( {{\varvec{A}}+{\varvec{B}}} \right)$ be arranged in increasing order. For any nonzero vector ${\varvec{v}} \in {R^K}$, we have the bound

$${\lambda _1}\left( {\varvec{B}} \right) \leq \frac{{{{\varvec{x}}^T}{\varvec{B}}{\varvec{x}}}}{{{{\varvec{x}}^T}{\varvec{x}}}} \leq {\lambda _K}\left( {\varvec{B}} \right)$$

(41)

and hence for any $1 \leq i \leq \;K$ we have

$$\begin{gathered} {\lambda _i}\left( {{\varvec{A}}+{\varvec{B}}} \right)=\mathop {\hbox{min} }\limits_{{{{\varvec{w}}_1},\; \cdots ,{{\varvec{w}}_{K - i}} \in {R^K}\;}} \mathop {\hbox{max} }\limits_{{\begin{array}{*{20}{c}} {{\varvec{x}} \ne 0} \\ {{\varvec{x}} \bot {{\varvec{w}}_1},\; \cdots ,{{\varvec{w}}_{K - i}}} \end{array}\;}} \frac{{{{\varvec{x}}^T}\left( {{\varvec{A}}+{\varvec{B}}} \right){\varvec{x}}}}{{{{\varvec{x}}^T}{\varvec{x}}}} \\ =\mathop {\hbox{min} }\limits_{{{{\varvec{w}}_1},\; \cdots ,{{\varvec{w}}_{K - i}} \in {R^K}\;}} \mathop {\hbox{max} }\limits_{{\begin{array}{*{20}{c}} {{\varvec{x}} \ne 0} \\ {{\varvec{x}} \bot {{\varvec{w}}_1},\; \cdots ,{{\varvec{w}}_{K - i}}} \end{array}\;}} \left[ {\frac{{{{\varvec{x}}^T}{\varvec{A}}{\varvec{x}}}}{{{{\varvec{x}}^T}{\varvec{x}}}}+\frac{{{{\varvec{x}}^T}{\varvec{B}}{\varvec{x}}}}{{{{\varvec{x}}^T}{\varvec{x}}}}} \right] \\ \end{gathered}$$

(42)

which can be further transformed as

$${\lambda _i}\left( {{\varvec{A}}+{\varvec{B}}} \right) \geq \mathop {\hbox{min} }\limits_{{{{\varvec{w}}_1},\; \cdots ,{{\varvec{w}}_{K - i}} \in {R^K}\;}} \mathop {\hbox{max} }\limits_{{\begin{array}{*{20}{c}} {{\varvec{x}} \ne 0} \\ {{\varvec{x}} \bot {{\varvec{w}}_1},\; \cdots ,{{\varvec{w}}_{K - i}}} \end{array}\;}} \left[ {\frac{{{{\varvec{x}}^T}{\varvec{A}}{\varvec{x}}}}{{{{\varvec{x}}^T}{\varvec{x}}}}+{\lambda _1}\left( {\varvec{B}} \right)} \right]={\lambda _i}\left( {\varvec{A}} \right)+{\lambda _1}\left( {\varvec{B}} \right)$$

(43)

Considering the fact that the matrix ${\varvec{B}}$ is a symmetric positive one, namely, its eigenvalues are all positive, then we can obtain

$${\lambda _i}\left( {{\varvec{A}}+{\varvec{B}}} \right) \geq {\lambda _i}\left( {\varvec{A}} \right)$$

(44)

Assuming that ${\varvec{A}}={\varvec{H}}_{N}^{T}{\vec {{\varvec{W}}}_N}{{\varvec{H}}_N}$and ${\varvec{B}}={\varvec{H}}_{N}^{T}\left( {{{\varvec{I}}_N} - {{\vec {{\varvec{W}}}}_N}} \right){{\varvec{H}}_N}$ in (44), then we can obtain the desired statement in (9).

$${\lambda _i}\left( {{\varvec{H}}_{N}^{T}{{\varvec{H}}_N}} \right) \geq {\lambda _i}\left( {{\varvec{H}}_{N}^{T}{{\vec {{\varvec{W}}}}_N}{{\varvec{H}}_N}} \right)$$

(45)

End of proof.