ETH - Technical University Homework
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ianmcculloch
Last Updated
9 years ago
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Other (as stated in the work)
Resumé
Comparison of three data providers.
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{
ETH Zürich\\%
D-BAUG\\%
Institute of Environmental Engineering (IfU)\hfill
Master Studies%
}
\subject{\vspace{-1ex} \horrule{2pt}\\[0.15cm] {\textsc{\texttt{Basics and Principles of Radar Remote Sensing for Environmental Applications}}}}
\title{Homework \#1 - SAR System\\[0.5cm]}
\subtitle{\textsc{\texttt{Comparison Of Three Data Providers}}\\\horrule{2pt}\\[0.5cm]}
\author{\bfseries{Peter Zweifel}\vspace{-2ex}}
\date{\begin{tabular}{cc}
\textsc{Date:}& \textsc{\emph{\today}}\\
\textsc{Due :}& \textsc{\emph{20th January 2014}}\vspace{3ex}
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\section{Problem definition} %add a * after \section to get rid of the numbering
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\paragraph{goal}The following task was given in the assignment:
\begin{quotation}"\emph{Task:} You are in charge of developing a new radar remote sensing software for monitoring vessels and sailboats on Lake Zurich. Before the project officially starts, you’ll be required to find a suitable data provider, based on the following goals:
\begin{itemize}
\item Minimum distinguishable separation between targets of 8 m.
\item Complete coverage of Lake Zurich."
\end{itemize}
\end{quotation}
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\paragraph{System Nr. 1}Table~\ref{tab:solutions} clearly shows that System Nr. 1 cannot be used, as its azimuthal resolution exceeds the prerequisites of the tasks (8m).
\begin{table}[h]
\centering
\captionabove{\label{tab:solutions}Shows the computed values computed with the formulas given in the lecture.}
\begin{tabular}{l c c c}
\toprule[1pt] \textbf{Parameter} & \textbf{Nr. 1} & \textbf{Nr. 2} & \textbf{Nr. 3 (near-/far-range)}\\\midrule[0.5pt]
${\rho}_{az}$ & 8.83 m &2.65 m & 1.08 m / 1.71 m \\
${\rho}_{ra}$ & 6 m & 1.87 m & 5 m\\
${r}_{0}$ & 923'760 m & 567'136 m & 3310 / 5230 m \\
$\lambda$ & 0.0566 m & 0.0312 m & 0.2306 m\\
\bottomrule[1pt]
\end{tabular}
\end{table}
\paragraph{Computation}Computed with the following formulas:
\begin{align}
\begin{split}
{\rho}_{az}&={r}_{0} \frac{\lambda}{2{L}_{sa}} \\[0.15cm]
{r}_{0}&=\frac{altitude}{cos{\alpha}}\\[0.15cm]
\lambda&=\frac{{c}_{light}}{f}\\
\end{split}
\end{align}
\subsection{Cost calculation and comparison}
\paragraph{Lake Zürich}
Lake of Zürich has about the following dimensions (see Wikipedia.ch)
\begin{align}
\begin{split}
Length &= 42 km\\
Width &= 4 km
\end{split}
\end{align}
\paragraph{System Nr. 3}With System Nr. 3 it would take 10 pictures (5 à 2 rows) à 1200 CHF.
\begin{center}
\emph{= 12'000 CHF}
\end{center}
\paragraph{System Nr. 2}System Nr. 2 would cost 4500 CHF in this operation, about one third of System Nr. 3.
\begin{center}
\emph{= 4500 CHF}
\end{center}
\paragraph{Final Decision}System Nr. 3 may be precise as well and also more flexible in task operation planning; thrice the price of system Nr. 2, however, seems to be too much of an expense.
System Nr. 2 would be chosen in this configuration as the data provider for this task, as it is cheaper than the others and still very precise as well.
%$\mathsf{({m}^{3}/s)}$
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