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authorRefik Hadzialic2012-06-30 19:13:17 +0200
committerRefik Hadzialic2012-06-30 19:13:17 +0200
commit52338e77b16d4590c265f1a48074eeca52161ef8 (patch)
tree24160d795a454f3dfc20a3ea8302660ff1684f66
parentExplaining the derivation of the distance (diff)
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Position estimation
-rw-r--r--vorlagen/thesis/maindoc.pdfbin4779599 -> 4801134 bytes
-rw-r--r--vorlagen/thesis/src/kapitel_x.tex41
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diff --git a/vorlagen/thesis/maindoc.pdf b/vorlagen/thesis/maindoc.pdf
index 6a5d1e9..cfc207e 100644
--- a/vorlagen/thesis/maindoc.pdf
+++ b/vorlagen/thesis/maindoc.pdf
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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