\documentclass[12pt]{article}
\usepackage{latexsym}
\usepackage{graphicx}
\usepackage[hypertex]{hyperref}
\usepackage{epsfig}
\usepackage{subfigure}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{graphicx}
\usepackage{mathrsfs}
\usepackage{cases}
\usepackage{array}
\renewcommand{\baselinestretch}{1.3}
\setlength{\oddsidemargin}{0in}
\setlength{\evensidemargin}{0in}
\setlength{\textwidth}{6.5in}
\setlength{\topmargin}{0in}
\setlength{\textheight}{8.5in}
%\input setbmp
%\input{psfig.sty}
\usepackage{epsfig}
\newtheorem{defn}{Definition}
\newtheorem{thm}{Theorem}
\newtheorem{lemma}{Lemma}
\newtheorem{example}{Example}
\newtheorem{corollary}{Corollary}
\newtheorem{proposition}{Proposition}
\newtheorem{algorithm}{Algorithm}
\newtheorem{exercise}{Exercise}
%\usepackage[centertags]{amsmath}
\newcommand{\markex}{\nopagebreak\begin{center}\rule{5in}{.01in}\end{center}}
\newcommand{\markdefn}{\nopagebreak\hfill$\Diamond$}
\newcommand{\mb}{\mbox}
\newcommand{\bm}{\boldmath}
\def\b#1{\mb{\bm$#1$}}
\newfont{\Bb}{msbm10 scaled\magstep1}
\newfont{\Fr}{eufm10 scaled\magstep1}
\newfont{\Sc}{eusm10 scaled\magstep1}
%\pagestyle{empty}
\begin{document}
\noindent{\large ECE 602}\hfill{\Large Fall 2010}
\begin{center}
{\Large \bf Constructing Robust Controllers Using Lyapunov's Theorem}\\
by\\
{\large Stanislaw H. \.{Z}ak}
\end{center}
\begin{center}
October 20, 2010
\end{center}
\medskip
%\bigskip
\begin{flushright}
%\parbox{5in}{Early amplifiers were made of crude materials that tended to disintegrate over use, causing the amp to ``run away.'' Not only would an aging relay amplify the phone signal, it would mistakenly compound any tiny deviation from the range it expected until the mushrooming error filled and killed the system. What was needed was Heron's {\it regula}, a counter signal to rein in the chief signal, to dampen the effect of the perpetual recycling. Black came up with a {\it negative} feedback loop, which was designated negative in contrast to the snowballing positive loop of the amplifier. Conceptually, the electrical negative feedback loop is a toilet flusher or thermostat. This braking circuit keeps the amplifier honed in on a steady amplification in the same way a thermostat hones in on a steady temperature. But instead of metallic levers, a weak train of electrons talks to itself.~\cite[page~116]{kelly94}}
\end{flushright}


One of the applications of the Lyapunov theorem is to construct robust state-feedback controllers. A basic issue in the control of dynamic systems is the effect of uncertainties, or neglected nonlinearities, on the stability of the closed-loop system.

A controller is viewed robust if it maintains stability for all uncertainties in an expected range. We consider a class of dynamic systems modeled by
\begin{equation}
\dot{\b x}(t) = \b A\b x(t)+\b B\b u(t) +\b f(t, \b x(t))+\b B\b h(t,\b x(t),\b u(t)),
\label{eq:uncersysmolic}
\end{equation}
where $\b x(t) \in \mb{\Bb R}^n$, $\b u(t) \in \mb{\Bb R}^m$, and the functions $\b h$ and $\b f$ model uncertainties, or nonlinearities, in the system. We refer to $\b h$ as the {\em matched uncertainty\/} because it affects the system dynamics via the input matrix $\b B$ in the same fashion as the input $\b u$ does. In other words, the uncertainty $\b h$ matches the system input $\b u$. The vector $\b f$ models the {\em unmatched uncertainty\/}. We assume that the uncertain elements $\b h$ and $\b f$ satisfy the following norm bounds:
\begin{enumerate}
\item $\|\b h(t, \b x, \b u)\| \le \gamma_h\|\b u\|+\alpha_h\|\b x\|$,
\item $\|\b f(t, \b x)\| \le \alpha_f\|\b x\|$,
\end{enumerate}
where $\gamma_h$, $\alpha_h$, and $\alpha_f$ are known nonnegative constants. We further assume that the matrix $\b A$ is asymptotically stable. If this is not the case, we apply a preliminary state-feedback controller $\b u=-\b K\b x$ such that $\b A-\b B\b K$ is asymptotically stable. Such a feedback exists provided that the pair $(\b A,\b B)$ is stabilizable. Our goal is to construct a linear state-feedback controller that would make the closed-loop system asymptotically stable for arbitrary $\b f$ and $\b h$ that satisfy the above norm bounds.
\begin{thm}\cite[p.~168]{zak03}\label{thm:layap_stabcon}
Suppose that $\b A$ is asymptotically stable, and $\b P=\b P^{\top} >0$ is the solution to the Lyapunov matrix equation $\b A^{\top}\b P+\b P\b A=-2\b Q$ for some $\b Q=\b Q^{\top} >0$. Suppose also that
\[
\alpha_f < \frac{\lambda_{\min}(\b Q)}{\lambda_{\max}(\b P)}\quad\mb{and}\quad \gamma_h < 1.
\]
Then, the state-feedback controller $\b u=-\gamma\b B^{\top}\b P\b x$, where
\[
\gamma > \frac{\alpha_h^2}{4\left(\lambda_{\min}(\b Q)-\alpha_f \lambda_{\max}(\b P) \right)(1-\gamma_h)}
\]
stabilizes the uncertain system model~(\ref{eq:uncersysmolic}) for arbitrary $\b f$ and $\b h$ that satisfy the norm bounds.
\end{thm}
{\bf Proof}\quad The time derivative of the positive definite function $V=\b x^{\top}\b P\b x$ evaluated on any trajectory of the closed-loop system is
\begin{eqnarray*}
\dot{V} &=& 2\b x^{\top}\b P\dot{\b x}\\
&=& -2\b x^{\top}\b Q\b x -2\gamma\b x^{\top}\b P\b B\b B^{\top}\b P\b x +2\b x^{\top} \b P\b f+2\b x^{\top} \b P\b B\b h.
\end{eqnarray*}
Our goal is to determine $\tilde{\gamma} >0$ such that if $\gamma > \tilde{\gamma}$, then $\dot{V}<0$, which in turn implies the asymptotic stability of the closed-loop system. To proceed further, recall that if $\b Q$ is a symmetric matrix, then
\[
\lambda_{\min}(\b Q)\|\b x\|^2 \le \b x^{\top}\b Q\b x \le \lambda_{\max}(\b Q)\|\b x\|^2,
\]
and therefore
\[
-\b x^{\top}\b Q \b x \le -\lambda_{\min}(\b Q)\|\b x\|^2.
\]
Furthermore,
\[
\b x^{\top}\b P\b B\b B^{\top}\b P\b x =\b x^{\top}\b P\b B\left( \b x^{\top}\b P\b B\right)^{\top}=\|\b x^{\top}\b P\b B\|^2.
\]
For a symmetric positive definite matrix $\b P$, its induced 2-norm is
\[
\|\b P\|=\lambda_{\max}(\b P).
\]
Taking the above relations into account, using the norm bounds on the uncertain elements, and performing some manipulations yields
\begin{eqnarray}
\dot{V} &\le& -2\lambda_{\min}(\b Q)\|\b x\|^2-2\gamma\|\b x^{\top}\b P\b B\|^2+2\alpha_f\lambda_{\max}(\b P)\|\b x\|^2\nonumber\\
&{}&\quad+2\|\b x^{\top}\b P\b B\|\left( \alpha_h\|\b x\|+\gamma_h \gamma\|\b x^{\top}\b P\b B\|\right).
\label{eq:vdotless}
\end{eqnarray}
Let
\begin{equation}
\tilde{\b Q}=\left[\begin{array}{cc}
\lambda_{\min}(\b Q)-\alpha_f\lambda_{\max}(\b P) & -\frac{\alpha_h}{2}\\
-\frac{\alpha_h}{2} & \gamma(1-\gamma_h)
\end{array}\right]\in \mb{\Bb R}^{2 \times 2}.
\label{eq:big_matrix}
\end{equation}
Then, we can represent~(\ref{eq:vdotless}) as
\begin{equation}
\dot{V} \le -2\left[\begin{array}{cc}
\|\b x\| & \|\b x^{\top}\b P\b B\|
\end{array}\right]\tilde{\b Q}\left[\begin{array}{c}
\|\b x\|\\
\|\b x^{\top}\b P\b B\|
\end{array}\right].
\label{eq:vdotq}
\end{equation}
For $\dot{V}$ to be negative it is enough that the leading principal minors of the matrix $\tilde{\b Q}$ given by~(\ref{eq:big_matrix}) be positive. The first-order leading principal minor is positive by assumption. For the second-order leading principal minor to be positive it is sufficient that $\gamma_h < 1$ and
\[
\gamma > \frac{\alpha_h^2}{4\left(\lambda_{\min}(\b Q)-\alpha_f \lambda_{\max}(\b P) \right)(1-\gamma_h)}.
\]
The proof is now complete.\\
\hspace*{5.5in} $\Box$

Note that if $\gamma_h=0$, then the closed-loop system is asymptotically stable if
\[
\alpha_f < \frac{\lambda_{\min}(\b Q)}{\lambda_{\max}(\b P)}\quad\mb{and}\quad \gamma > \frac{\alpha_h^2}{4\left(\lambda_{\min}(\b Q)-\alpha_f \lambda_{\max}(\b P) \right)}.
\]
\begin{example}\rm
Consider a dynamic system model
\begin{eqnarray*}
\dot{\b x} &=& \b A\b x+\b b(u+h(\b x,u))\\ &=&\left[\begin{array}{cc}
-2 & 0\\
0 & -1
\end{array}\right]\b x+\left[\begin{array}{c}
1\\
1
\end{array}\right](u+h(\b x,u)).
\end{eqnarray*}
Let $\b P$ be the solution of the matrix Lyapunov equation $\b A^{\top}\b P+\b P\b A=-2\b I_2$. We will design a linear state-feedback controller $u=-\gamma \b b^{\top}\b P\b x$ so that the origin of the closed-loop system is uniformly asymptotically stable for any $h(\b x,u)$ such that
\begin{equation}
|h(\b x,u)|\le 4\|\b x\|+\frac{1}{2}|u|.
\label{eq:nobonhinex}
\end{equation}
In particular, we will find a bound $\tilde{\gamma}>0$ so that for any $\gamma>\tilde{\gamma}$ the closed-loop system is uniformly asymptotically stable in the face of uncertainties satisfying the norm bound~(\ref{eq:nobonhinex}).

Solving the Lyapunov equation $\b A^{\top}\b P+\b P\b A=-2\b I_2$ yields
\[
\b P=\left[\begin{array}{cc}
1/2 & 0\\
0 & 1
\end{array}\right].
\]
Hence,
\[
u = -\gamma\b b^{\top}\b P\b x= -\gamma\left(\frac{1}{2}x_1+x_2\right).
\]
We obtain a bound on $\gamma$ by evaluating $\frac{d}{dt}\left( \b x^{\top}\b P\b x\right)$ on the trajectories of the closed-loop system. First, we evaluate
\begin{eqnarray*}
\frac{d}{dt}\left( \b x^{\top}\b P\b x\right)&=& 2\b x^{\top}\b P\dot{\b x}\\
&=& 2\b x^{\top}\b P\left( \b A\b x+\b b(u+h(\b x,u))\right)\\
&=& -2\b x^{\top}\b x-2\gamma\b x^{\top}\b P\b b\b b^{\top}\b P\b x+2\b x^{\top}\b P\b b h.
\end{eqnarray*}
Taking the norms and performing manipulations yields
\begin{eqnarray*}
\frac{d}{dt}\left( \b x^{\top}\b P\b x\right)&\le& -2\b x^{\top}\b x-2\gamma\left|\b x^{\top}\b P\b b\right|^2+2\left| \b x^{\top}\b P\b b\right|\left( 4\|\b x\|+\frac{1}{2}\gamma\left| \b x^{\top}\b P\b b\right|\right)\\
&=&-2\|\b x\|^2-\gamma\left|\b x^{\top}\b P\b b\right|^2+8\left|\b x^{\top}\b P\b b\right|\|\b x\|\\
&=&-2\left[\begin{array}{cc}
\|\b x\| & \left|\b x^{\top}\b P\b b\right|
\end{array}\right]\left[\begin{array}{cc}
1 & -2\\
-2 & \gamma/2
\end{array}\right]\left[\begin{array}{c}
\|\b x\|\\
\left|\b x^{\top}\b P\b b\right|
\end{array}\right].
\end{eqnarray*}
For $\frac{d}{dt}\left( \b x^{\top}\b P\b x\right)$ to be negative definite it is enough that the matrix
\[
\left[\begin{array}{cc}
1 & -2\\
-2 & \gamma/2
\end{array}\right]
\]
be positive definite. This is the case whenever
\[
\gamma > \tilde{\gamma} = 8.
\]
\end{example}\markex

%\section{Summary and Notes}



%\section{Exercises}

%\begin{exercise}\label{ex:toiletsysex}\rm
%An example of a closed-loop feedback system is a toilet flushing device connected to a water tank. A toilet system is shown in Figure~\ref{fig:toiletd}. The control objective is to maintain the level of water in the tank at a constant level. Draw a block diagram of the system.
%\begin{figure}[h]
%\centering
%\includegraphics[scale=0.6]{TOILETD.eps}
%%\vspace{3.5in}
%%\hskip1in\special{bmp:c:/mybook/toiletd.bmp x=4.67in y=3.5in}
%\caption{Toilet-flushing system for Exercise~\ref{ex:toiletsysex}.}
%\label{fig:toiletd}
%\end{figure}
%\end{exercise}



\bibliography{Mybook,root2}
\bibliographystyle{plain}
\end{document} 