From 52338e77b16d4590c265f1a48074eeca52161ef8 Mon Sep 17 00:00:00 2001 From: Refik Hadzialic Date: Sat, 30 Jun 2012 19:13:17 +0200 Subject: Position estimation --- vorlagen/thesis/maindoc.pdf | Bin 4779599 -> 4801134 bytes vorlagen/thesis/src/kapitel_x.tex | 41 ++++++++++++++++++++++++++++++++++++++ 2 files changed, 41 insertions(+) diff --git a/vorlagen/thesis/maindoc.pdf b/vorlagen/thesis/maindoc.pdf index 6a5d1e9..cfc207e 100644 Binary files a/vorlagen/thesis/maindoc.pdf and b/vorlagen/thesis/maindoc.pdf differ diff --git a/vorlagen/thesis/src/kapitel_x.tex b/vorlagen/thesis/src/kapitel_x.tex index ecac181..4da0984 100644 --- a/vorlagen/thesis/src/kapitel_x.tex +++ b/vorlagen/thesis/src/kapitel_x.tex @@ -741,7 +741,48 @@ the equation \eqref{eq:MultitaylorDerivAfterRearange} resembles the one given in \label{eq:userPosition} \Delta\rho_i = \alpha_{xi}\Delta x_u + \alpha_{yi}\Delta y_u + \alpha_{zi}\Delta z_u - c\Delta t_u \end{equation} +There are four unknowns, $\Delta x_u$, $\Delta y_u$, $\Delta z_u$ and $\Delta t_u$, in equation \eqref{eq:userPosition}. +By solving this set of linear equations, which will result in finding $\Delta x_u$, $\Delta y_u$, $\Delta z_u$ and $\Delta t_u$, +the GPS receiver position $(x_u, y_u, z_u)$ and clock offset $t_u$ is computed by replacing the +same into equations in \eqref{eq:userCoordinates}. Equation \eqref{eq:userPosition} can be rewritten for four satellites +in the matrix form as in \eqref{eq:userPositionMatrix}. +\begin{equation} +\label{eq:userPositionMatrix} +\Delta\boldsymbol{\rho} = \boldsymbol{\alpha} \Delta \boldsymbol{x} +\end{equation} +\begin{equation} +\Delta\boldsymbol{\rho}= +\begin{bmatrix} +\Delta \rho_1 \\ +\Delta \rho_2 \\ +\Delta \rho_3 \\ +\Delta \rho_4 +\end{bmatrix} +\hspace{1.5em} +\boldsymbol{\alpha}= +\begin{bmatrix} +\alpha_{x1} & \alpha_{y1} & \alpha_{z1} & 1 \\ +\alpha_{x2} & \alpha_{y2} & \alpha_{z2} & 1 \\ +\alpha_{x3} & \alpha_{y3} & \alpha_{z3} & 1 \\ +\alpha_{x4} & \alpha_{y4} & \alpha_{z4} & 1 +\end{bmatrix} +\hspace{1.5em} +\Delta \boldsymbol{x}= +\begin{bmatrix} +\Delta x_u \\ +\Delta y_u \\ +\Delta z_u \\ +-\Delta ct_u +\end{bmatrix} +\end{equation} +Finally, by multiplying both left sides\footnote{Matrix multiplication is not communitative, $\mathbf{AB\neq BA}$.} of the equation \eqref{eq:userPositionMatrix} +with the inverse term of $\boldsymbol{\alpha}$, it yields the result of the unknown terms, as given in equation \eqref{eq:userPositionMatrixFinal}. +\begin{equation} +\label{eq:userPositionMatrixFinal} +\Delta \boldsymbol{x} = \boldsymbol{\alpha}^{-1} \Delta\boldsymbol{\rho} +\end{equation} +The -- cgit v1.2.3-55-g7522