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\begin{document}
\null\vspace{-1.75cm}
{\Large
{\bf ECE 515 / ME 540 \hfill Assignment \# 1}
\vspace{.15in}
Issued: Aug 24 \hfill Due: Aug 31, 2023}
\rule{6in}{.01in}
\vspace*{.1cm}
\head{Reading Assignment:}
\textbf{BMP}, Ch. 1.
\head{Problems:}
\begin{arabnum}
\setcounter{arabnum}{0}
%#1
\item (Exercise 1.7.1 in BMP) Consider a nonlinear scalar input-output system whose input $u(t)$ and output $y(t)$ are related through the differential equation
$$
\ddot{y} = 2y - (y^2+1)(\dot{y}+1) + u.
$$
\vspace{-15pt}
\begin{romannum}
\item Obtain a nonlinear state-space representation.
\item Linearize this system of equations around the equilibrium trajectory when $u(\cdot) \equiv 0$ and write it down in state-space form.
\end{romannum}
\item
Convert each of the following high-order differential equations into the input/state/output form:
\begin{romannum}
\item $\ddot{y} - 2\dot{y} = 4u$
\item $y^{(3)} + 2\dot{y} - 2y = -u$ ($y^{(3)}$ is the $3$rd derivative of $y$ with respect to time)
\end{romannum}
\item A nonlinear state-space model with no controls is a system of first-order ODEs that has the form
$$
\dot{x}(t) = f(x(t)),
$$
where $x(t) \in{\mathbb R}^n$ is the state vector. The term ``no controls'' designates the fact that the external input $u$ is absent, so the system evolves on its own. We say that a point $x_0 \in {\mathbb R}^n$ is an \textit{equilibrium point} of the system if $f(x_0) = 0$.
Consider the following circuit that contains linear components (an inductor and a capacitor) and a nonlinear resistive element:
\Ebox{.4}{resistive.pdf}
The voltage $V$ across the resistive element and the current $I$ flowing into it are related via a nonlinear voltage-current characteristic $I = g(V)$.
\begin{romannum}
\item Derive a second-order ODE for $V$. You may (and should) assume that $g$ is differentiable.
\item Write down nonlinear state-space model with no controls for the ODE you have obtained in part (i).
\item Consider the following voltage-current characteristic:
$$
g(V) = - V + \tfrac{1}{3}V^3.
$$
Show that the zero state is the only eqiulibrium point of the state-space model from part (ii) and linearize the system around this equilibrium point.
\end{romannum}
\item (Exercise 1.7.9 in BMP) Consider the SISO LTI system with the transfer function
$$
G(s) = \frac{s+4}{(s+1)(s+2)(s+3)}.
$$
\vspace{-15pt}
\begin{romannum}
\item Obtain a state-space representation in the controllable canonical form.
\item Now obtain one in the observable canonical form.
\item Use partial fraction expansion to obtain a representation with a diagonal state matrix $A$ (modal canonical form).
\end{romannum}
\item (Exercise 1.7.10 in BMP) Repeat the above steps for the system with the transfer function
$$
G(s) = \frac{s^3+2}{(s+1)(s+3)(s+4)}.
$$
\end{arabnum}
\end{document}