main.tex

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\title{Modelling Assignment\\Module CY2A9}
\author{James E. Cleveland\\\textit{Computer Science \& Cybernetics (BSc)}}

\date{December 7th, 2009}
\begin{document}
\begin{titlingpage}

\setlength{\droptitle}{180pt}
\maketitle

\centering
\includegraphics[width=0.4\linewidth]{runi.eps}
\\~\\~\\~
\begin{abstract}
This report has the results and implementation of finding the transfer function for a Wien Bridge Oscillator circuit (used for converting DC current to sinusoidal or AC current).

It also includes the workings and method behind three other systems which support the methods of analysing the Wien Bridge Oscillator circuit.

The complete workings for the first three systems are available in Appendix C. The object of this is so that if any details are missed in the main report, they can be located here to avoid confusion.

\end{abstract}

\end{titlingpage}

%\textit{NOTE: Although the total page count may be higher than the specified maximum, this is likely due to formatting (blank pages, etc), and contents pages necessary in a double-sided report.}

\tableofcontents
%\clearpage
%\listoffigures
\chapter{Introduction}

\input{introduction}

\chapter{Implementation}
This chapter includes the block diagrams and transfer functions and the algebra used to produce the transfer functions for each system.
%\clearpage
\section{System 1: Mass Spring System}
\input{sys1_imp}
%\clearpage
\section{System 2: Electronic Circuit}
\input{sys2_imp}
%\clearpage
\section{System 3a: Feedback Control System}
\input{sys3a_imp}
%\clearpage
\section{System 3b: Feedback Control System}
\input{sys3b_imp}
%\clearpage
\section{System 4: Wien Bridge Oscillator}
\input{sys4_imp}
%\clearpage

\chapter{Results}
This chapter includes the transfer functions found by hand, those found by MATLAB, the simulated and theoretical output response, and the bode plots for each system.
%\clearpage
\section{System 1: Mass Spring System}
\input{sys1_res}
%\clearpage
\section{System 2: Electronic Circuit}
\input{sys2_res}
%\clearpage
\section{System 3a: Feedback Control System}
\input{sys3a_res}
%\clearpage
\section{System 3b: Feedback Control System}
\input{sys3b_res}
%\clearpage
\section{System 4: Wien Bridge Oscillator}
\input{sys4_res}
%\clearpage


\chapter{Discussion}
\input{discussion}
\chapter{Conclusion}
\input{conclusion}

\chapter*{Acknowledgements}
\begin{itemize}
	\item David J. Salisbury: Help with LaTeX, Diagraph
	\item Carlos Sainz-Martinez: Help with Math
\end{itemize}

\bibliography{partb}{}
\bibliographystyle{plainnat}

\appendix
\chapter{Code Listing}
\input{code_listing}

\chapter{Analysis For Theoretical Step Response}
\input{analysis_step_response}
\section{System 1: Mass Spring System}
\input{sys1}
\section{System 2: Electronic Circuit}
\input{sys2}
\section{System 3a \& 3b: Modelling feedback control systems}
\input{sys3}
\section{System 4: Wien Bridge Oscillator}
\input{sys4}

\end{document}

sys3b.tex

\subsection{Task 1}

\begin{figure}[H]
	\centering
	\input{sys3b_1}
	\caption{Block diagram for System 3b with values inserted}
	\label{fig:sys3bbd1}
\end{figure}

\begin{align*}
\frac{O\left(s\right)}{I\left(s\right)} & = \frac{\frac{12.5}{\left(1+s3\right)\left(1+s4\right)}}{1+\frac{0.08s\times12.5}{\left(1+s3\right)\left(1+s4\right)}}\\
	& = \frac{12.5}{\left(1+s3\right)\left(1+s4\right)+s}\\
\end{align*}

\begin{figure}[H]
	\centering
	\input{sys3b_2}
	\caption{Initial block diagram reduction}
	\label{fig:sys3bbd2}
\end{figure}

\begin{align*}
\frac{O\left(s\right)}{I\left(s\right)} & = \frac{0.5\left(\frac{1+s6}{1+s}\right)\left(\frac{12.5}{\left(1+s3\right)\left(1+s4\right)+s}\right)}{1+0.5\left(\frac{1+s6}{1+s}\right)\left(\frac{12.5}{\left(1+s3\right)\left(1+s4\right)+s}\right)}\times\frac{1+s}{1+s}\\
	& = \frac{0.5\left(1+s6\right)\left(\frac{12.5}{\left(1+s3\right)\left(1+s4\right)+s}\right)}{\left(1+s\right)+0.5\left(1+s6\right)\left(\frac{12.5}{\left(1+s3\right)\left(1+s4\right)+s}\right)}\times\frac{\left(1+s3\right)\left(1+s4\right)+s}{\left(1+s3\right)\left(1+s4\right)+s}\\
	& = \frac{6.25\left(1+s6\right)}{\left(1+s\right)\left(\left(1+s3\right)\left(1+s4\right)+s\right)+6.25\left(1+s6\right)}\\
	& = \frac{6.25\left(1+s6\right)}{\left(1+s3\right)\left(1+s4\right)\left(1+s\right)+\left(s^2+s\right)+6.25\left(1+s6\right)}\\
	\\
	\left(1+s3\right)\left(1+s4\right) & = 12s^2+7s+1\\
	\left(12s^2+7s+1\right)\left(1+s\right) & = 12s^3 + 19s^2 + 8s + 1\\
	\\
	\frac{O\left(s\right)}{I\left(s\right)} & = \frac{6.25\left(6s+1\right)}{12s^3+20s^2+9s+1+6.25\left(6s+1\right)}\times\frac{4}{4}\\
	& = \frac{25\left(6s+1\right)}{25\left(6s+1\right)+48s^3+80s^2+36s+4}\\
\end{align*}

Factorising $48s^3+80s^2+36s+4$, (Figure~\ref{fig:sys3bldiv})

\setlength{\tabcolsep}{1pt}
\begin{figure}[H]
\centering
\begin{tabular}{lllll}
& $8s^2+$ & $12s+$ & $4$\\
\cline{2-5}
$6s+1 \div$ & $48s^3+$ & $80s^2+$ & $36s+$ & $4$ \\
       & $48s^3$   & $8s^2$\\
       &           & $72s^2+$ & $36s+$ & $4$ \\
       &           & $72s^2+$ & $12s$\\
       &           &          & $24s+$ & $4$\\
       &           &          & $24s+$ & $4$\\
\end{tabular}
\caption{Long division to factorise $48s^3+80s^2+36s+4$ }
\label{fig:sys3bldiv}
\end{figure}

\begin{align*}
\frac{O\left(s\right)}{I\left(s\right)} & = \frac{25\left(6s+1\right)}{25\left(6s+1\right)+\left(6s+1\right)\left(8s^2+12s+4\right)}\\
	& = \frac{25}{8s^2+12s+29}
\end{align*}

This is the transfer function.

\subsection{Task 2}

\begin{figure}[H]
\begin{alltt}
	% System 3b: Feedback Control Systems
	S=1;
	C= 0.5 * tf( [6 1], [1 1] );
	L = tf( [0.08 0], 1 );
	P = tf( 12.5, [12 7 1] );
	M = 1;
	sys3b_tf = S * feedback ( C * feedback( P, L ), M );
	minreal( sys3b_tf )
	
	Transfer function:
	       3.125
	-------------------
	s^2 + 1.5 s + 3.625
\end{alltt}
\end{figure}

This is the same as the transfer function that was derived manually.

\subsection{Task 3}

To get the function $o(t)$, output in the time domain.

\begin{align*}
\frac{O\left(s\right)}{I\left(s\right)}\times\frac{1}{s} &  = \frac{1}{s}\times\frac{25}{8s^2+12s+29}\\
	& = \frac{25}{s\left(8s^2+12s+29\right)}\\
	& = \frac{A}{s} + \frac{Bs+C}{8s^2+12s+29}\\
     25 & = A\left(8s^2+12s+29\right) + Bs^2 + C\\
\end{align*}
Let $s = 0$:
\begin{align*}
25 & = 29A\\
A & = \frac{25}{29}
\end{align*}
Let $s = 1$:
\begin{align*}
25 & = \frac{1225}{29} + B + C\\
\end{align*}
Let $s = 2$:
\begin{align*}
25 & = \frac{2124}{29} + 4B + 2C\\
\frac{-500}{29} & = B + C\\
\frac{-1400}{29} & = 4B + 2C\\
 B & = \frac{-200}{29}\\
 C & = \frac{-300}{29}\\
\end{align*}

These values can then be put back into the origin output function.
\begin{figure}[H]
	\begin{align*}
	 O\left(s\right) & = \frac{25}{29s} + \frac{\frac{-200}{29}s + \frac{-300}{29}}{8s^2+12s+29}\\
	  & = \frac{\frac{-200}{29}s + \frac{-300}{29}}{8s^2+12s+29}\times\frac{\frac{1}{8}}{\frac{1}{8}}\\
	  & = \frac{\frac{-25}{29}s + \frac{-37.5}{29}}{s^2+\frac{3}{2}s+\frac{29}{8}}\\
	  & = \frac{-\frac{25}{29}\left(s+\frac{3}{2}\right)}{\left(s+\frac{3}{4}\right)^2+\frac{49}{16}}\\
	  & = \frac{-\frac{25}{29}\left(s+\frac{3}{4}\right)}{\left(s+\frac{3}{4}\right)^2+\frac{49}{16}}+\frac{-\frac{25}{29}\times\frac{3}{4}}{\left(s+\frac{3}{4}\right)^2+\frac{49}{16}}\\
	  \frac{-\frac{25}{29}\times\frac{3}{4}}{\left(s+\frac{3}{4}\right)^2+\frac{49}{16}} & = \frac{-\frac{75}{203}\times\frac{7}{4}}{\left(s+\frac{3}{4}\right)^2+\frac{49}{16}}\\
	  & = \frac{\frac{7}{4}\left(\frac{3}{7}\times-\frac{25}{29}\right)}{\left(s+\frac{3}{4}\right)^2+\left(\frac{7}{4}\right)^2}\\
	 \\
	O\left(s\right) & = \frac{25}{29s} + \frac{-\frac{25}{29}\left(s+\frac{3}{4}\right)}{\left(s+\frac{3}{4}\right)^2 + \left(\frac{7}{4}\right)^2} + \frac{\frac{7}{4}\left(\frac{3}{7}\times-\frac{25}{29}\right)}{\left(s+\frac{3}{4}\right)^2 + \left(\frac{7}{4}\right)^2}\\
	\\
	o\left(t\right) & = \frac{25}{29} - \frac{25}{29}e^{-\frac{3}{4}t}\cos{\frac{7}{4}t} - \frac{3}{7}\frac{25}{29}e^{-\frac{3}{4}t}\sin{\frac{7}{4}t}\\
	  & = \frac{25}{29}\left(1-e^{-\frac{3}{4}t}\cos{\frac{7}{4}t}-\frac{3}{7}e^{-\frac{3}{4}t}\sin{\frac{7}{4}t}\right)\\
	  \\
	  & = \frac{25}{29}\left(1-e^{-\frac{3}{4}t}\left(\cos{\frac{7}{4}t}+\frac{3}{7}\sin{\frac{7}{4}t}\right)\right)\\   
	\end{align*}
	\caption{Obtaining the output function in the time domain for System 3b}
	\label{fig:sys4outfunc}
\end{figure}
\subsection{Task 4}

\begin{figure}[H]
	\centering
	\input{sys3b_mat}
	\caption{MATLAB code for System 3b}
	\label{fig:sys3mlab}
\end{figure}

\begin{figure}[H]
	\centering
	\includegraphics[height=100mm]{plots_sys3b.eps}
	\caption{Graph and Bode Plot for System 3b}
	\label{fig:sys3bplots}
\end{figure}