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\begin{center}
{\Large\bf Assignment #1}
{(Hand-in date #3)}
\end{center}
}
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%Kommando für Aufgaben
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\textbf{Exercise \arabic{n}: #1} (#2 Punkte)\\}
\begin{document}
%\header{BlattNr}{Tutor}{Abgabedatum}{Vorlesungsname}{Namen}{Semester}{Anzahl Aufgaben}
\header{1}{}{2015-05-05}{Mobile Robots}{
\textit{Jan-Peter Hohloch}\\ \textit{Maximus Mutschler}
}{SS 15}{4}
\vspace{1cm}
\Aufgabe{}{6}
\begin{enumerate}[(a)]
\item Forward kinematic velocity: calculation of the robots' current pose with knowledge about the velocity and position of the last pose \\
inverse kinematic velocity: given a pose the velocity and position needed to reach this pose are calulated
\item $^R\mathbf{v}=\begin{pmatrix}
0.2\frac{m}{s} \\ 1 \frac{1}{s}
\end{pmatrix}$, $l=0.2m$\\
\begin{align}
v_l &= \frac{2\cdot 0.2-0.2}{2}&=0.1\\
v_r &= \frac{2\cdot 0.2+0.2}{2}&=0.3
\end{align}
$\Rightarrow \mathbf{u}_t=\left(0.1\frac{m}{s}, 0.3\frac{m}{s}\right)^T$
\item \begin{align}
v &= \frac{0.3+0.5}{2}\frac{m}{s} &=0.4\frac{m}{s}\\
\omega &= \frac{0.5-0.3}{0.2}\frac{1}{s}&=1\frac{1}{s}
\end{align}
$\Rightarrow \mathbf{v}_t=\left(0.4\frac{m}{s},1\frac{1}{s}\right)$
\item \begin{align}
\frac{1}{2}v_r+\frac{1}{2}v_l &=v\\
\frac{1}{l}v_r -\frac{1}{l}v_l &=\omega
\end{align}$\Leftrightarrow$\begin{align}
v_r+v_l &= 2v\\
v_r-v_l &=l\omega\\
\end{align}
$\Rightarrow 2v_r =2v+l\omega\Leftrightarrow v_r=\frac{2v+l\omega}{2}$\\
$\Rightarrow \frac{2v+l\omega}{2}+v_l =2v \Leftrightarrow v_l=2v-\frac{2v+l\omega}{2}=\frac{2v-l\omega}{2}$\qed
\end{enumerate}
\Aufgabe{}{8}
\begin{enumerate}[(a)]
\item $s_l=s_r=s,\ \theta_{t-1}=\theta_t=\theta$\\
$\mathbf{x}_t=\mathbf{x}_{t-1}+\mathbf{Rot}(z,\theta)\begin{pmatrix}
s\\0\\0
\end{pmatrix}=\mathbf{x}_{t-1}+\begin{pmatrix}
\cos\theta & -\sin\theta & 0\\
\sin\theta & \cos\theta & 0\\
0&0&1
\end{pmatrix}\begin{pmatrix}
s\\0\\0
\end{pmatrix}=\mathbf{x}_{t-1}+\begin{pmatrix}
s\cos\theta\\s\sin\theta\\0
\end{pmatrix}$
\item $r\left(\begin{pmatrix}
2l\\l
\end{pmatrix}\right)=-\frac{3}{2} l$\\
$\frac{\partial r}{\partial s_r}=\frac{(s_l+s_r)'(s_r-s_l)-(s_l+s_r)(s_r-s_l)'}{(s_r-s_l)^2}\cdot\frac{l}{2}=\frac{-2s_l}{(s_r-s_l)^2}\cdot\frac{l}{2}\overset{s=\overline{s}}{=}\frac{-4}{l}\cdot\frac{l}{2}=-2$\\
$\frac{\partial r}{\partial s_l}=\frac{(s_l+s_r)'(s_r-s_l)-(s_l+s_r)(s_r-s_l)'}{(s_r-s_l)^2}\cdot\frac{l}{2}=\frac{2s_r}{(s_r-s_l)^2}\cdot\frac{l}{2}\overset{s=\overline{s}}{=}\frac{2}{l}\cdot\frac{l}{2}=1$\\
$\Rightarrow r(\mathbf{s})\approx -\frac{3}{2}-2\left(s_l-2l\right)+\left(s_r-l\right)$
\end{enumerate}
\Aufgabe{}{6}
\end{document}