\chapter{Unknown Input Observer (UIO) Design}
\label{chap:UIO_chapter}

\begin{flushright}
\parbox{5.5in}{The best piece of advice I got before I wrote this book was from a colleague of mine, David Ignatius, who is editor of the ``Outlook'' section at {\it The Washington Post}. He had written a novel about the Middle East [SIRO], a very good novel. He said, ``When you're writing this book always imagine whether the reader is going to want to turn the page. Whether what you're writing, how you're writing, and the anecdotes you're telling will compel that reader to want to turn the page.'' That's what I tried to keep in mind. Hence the book is a really a combination of analysis of history, of autobiography, of anecdote, put into my own kind of Cuisinart.'' Thomas Friedman in~\cite[p.~288]{lamb97}}
\end{flushright}


\bigskip
%Design procedures are proposed for two different classes of observers for uncertain
%dynamical systems. In the first approach, the state of the system, whose state is to
%be estimated, is decomposed into known and unknown components. The unknown component
%is a projection, not necessarily orthogonal, of the whole state along the subspace
%in which the available state component resides. Then a dynamical system to estimate
%the unknown component is constructed. Combining the output of the dynamical system,
%which estimates the unknown state component, with the available state information
%results in the observer that estimates the whole state. It is shown that some
%previously proposed observer architectures can be obtained using the projection
%operator approach presented in this paper. The second approach combines sliding
%modes and the second method of Lyapunov.

In this Chapter, we present design procedures for full- and reduced-order observers\index{full-order observer}\index{reduced-order observer}
for a class of dynamical systems modeled by a triple $(\b A, \b B,\b C)$ where some
or all of the inputs are unknown.\index{unknown input} The unknown input can be a combination of
unmeasurable or unmeasured disturbances,\index{disturbances!as unknown inputs} unknown control action, and unmodeled
system dynamics.\index{unmodeled
system dynamics} The first design method uses a projection operator\index{projection operator} approach to the
state estimation where the state of the system, whose state is to be estimated, is
decomposed into known and unknown components. The unknown component is, in general,
a skew projection,\index{skew projection} that is, not necessarily orthogonal,\index{orthogonal projection}\index{projection!orthogonal}\index{projection!skew} of the whole state along the
subspace in which the available state component resides. We then construct a
dynamical system to estimate the unknown component. Finally, we combine the output
of the dynamical system, which estimates the unknown state component, with the
available state information to obtain the observer\index{observer} that estimates the whole state.
In the second design method we employ a sliding mode\index{sliding mode} approach combined with the
second method of Lyapunov\index{second method of Lyapunov}\index{Lyapunov's second method}. We include design algorithms and illustrate the results
with numerical examples. Portions of this Chapter are based on our paper~\cite{hui05}.



\section{Introduction}
%Observers use the plant input and output signals to generate an estimate of the
%plant's state, which is then employed to close the control loop. Observers are
%utilized to augment or replace sensors in a control system. The observer was first
%proposed and developed by Luenberger in the early sixties of the last
%century~\cite{luenberger66,luenberger71,luenberger79}. Franklin and
%Powell~\cite[page~139]{franklin80} suggest using the term ``estimator'' rather than
%the commonly used term ``observer'' because ``observer'' seems to imply a direct
%measurement while, in fact, we estimate the state. On the other hand, the term
%estimator seems to suggest the use of statistical methods when estimating the
%uncertain system state. In this Chapter, we utilize only deterministic methods to
%estimate the state of an uncertain dynamical system. In view of the above, we will
%be using the term observer for the deterministic dynamical system whose goal is to
%estimate the state of an uncertain dynamical system employing only the known system
%input and output signals.
Since the early developments of
observers for systems without uncertainties, observers for plants with both known
and unknown inputs have been developed resulting in the so-called unknown input
observer~(UIO)\index{unknown input
observer (UIO)}\index{UIO|see{unknown input
observer}} architectures, such as, for example,
in~\cite{hostetter73,wang75,sundareswaran77,bhattacharyya78,kudva80,
kurek83,yang88,hou92,hui_uio93,darouach94,chen96,corless98,chen99,hou99}. Barmish
and Galimidi~\cite{barmish86} showed that robust stabilization of uncertain systems
via the Luenberger observer\index{Luenberger observer} is possible provided the uncertain parameters do not
exceed a certain computable threshold. More recently, observer architectures
utilizing the concept of sliding modes were proposed for uncertain systems,\index{uncertain systems} see, for
example,~\cite{walcott87,walcott87b,walcott88,hui90,zak90,zak93b,zak93c,edwards98,
utkin99,zak03,ha03,koshkouei04}. Other methods of the observer design for linear
systems developed prior 1983 are reported by O'Reilly in~\cite{oreilly83}.



Observers for uncertain systems\index{observers!for uncertain systems} play an essential role in robust model-based fault
detection~\cite{edwards98,chen99,edwards00,saif03,jiang04}.\index{model-based fault
detection} The basic idea behind
the use of the observers for fault detection is to form residuals\index{residuals} from the
difference between the actual system outputs and the estimated outputs using an
observer. Once a fault\index{fault} occurs, the residuals are expected to react by becoming
greater than a pre-specified threshold. When the system under consideration is
subject to unknown disturbances or unknown inputs, their effect has to be decoupled
from the residuals to avoid false alarms.\index{false alarms!avoidance of}



\section{Uncertain System Model}
\label{sec:uncertain_model}

The class of uncertain dynamical systems\index{uncertain systems} that we consider is modeled by
\begin{eqnarray}
\dot{\b x} &=& \b A\b x + \b B\b u\label{eq:uncert_sys}\\
\b y &=& \b C\b x, \label{eq_lin_system}
\end{eqnarray}
where the input matrix $\b B\in \mb{\Bb R}^{n\times m}$ and the output matrix $\b C
\in \mb{\Bb R}^{p\times n}$. We assume that the model parameters $(\b A, \b B, \b
C)$ are known. We further assume that some or all of the inputs are unknown and that
the last $m_2$ inputs are unknown and the remaining $m_1=m-m_2$ inputs are known. We
partition the input matrix $\b B$ corresponding to the known and unknown inputs as
\[
\b B=\left[\begin{array}{cc} \b B_1 & \b B_2
\end{array}\right],
\]
where $\b B_1\in \mb{\Bb R}^{n\times m_1}$ and $\b B_2\in \mb{\Bb R}^{n\times m_2}$.
Let
\[
\b u=\left[\begin{array}{c}
\b u_1\\
\b u_2
\end{array}\right].
\]
Then, the system model~(\ref{eq:uncert_sys}) can be represented as
\begin{equation}
\fbox{$\displaystyle \dot{\b x} = \b A\b x + \b B_1\b u_1 + \b B_2\b u_2 $}
\label{eq:two_inputs}
\end{equation}
The vector function $\b u_2$ may also model
lumped uncertainties\index{lumped uncertainties!as unknown inputs} or nonlinearities\index{nonlinearities!as unknown inputs} in the plant. The output matrix $\b C \in
\mb{\Bb R}^{p\times n}$. We assume that the pair $\left( \b A,\b C \right)$ is
detectable.


\section{A Projection Operator Approach}

\subsection{Background Results}

In this section we employ a projection operator approach to the problem of state
observation\index{projection operator approach!to state observation} of uncertain systems. In our further discussion, we assume that the
matrix $\b B_2$ has full column rank. We begin our presentation by noticing that
because the system output $\b y$ is known, it would seem reasonable to decompose the
state $\b x$ as
\begin{eqnarray}
\b x &=&(\b I-\b M\b C)\b x+\b M\b C\b x\nonumber\\
&=&(\b I-\b M\b C)\b x+\b M\b y, \label{eq:state_decompose}
\end{eqnarray}
where $\b M$ is an $n\times p$ real matrix, and the unknown part of the
decomposition is $(\b I-\b M\b C)\b x$. Let $\b q=(\b I-\b M\b C)\b x$, then $\b
x=\b q+\b M\b y$, and we have
\begin{eqnarray*}
\dot{\b q}&=& (\b I-\b M\b C)\dot{\b x}\\
&=& (\b I-\b M\b C)(\b A\b x+\b B_1\b u_1+\b B_2\b u_2)\\
&=& (\b I-\b M\b C)(\b A\b x+\b B_1\b u_1)+(\b I-\b M\b C)\b B_2\b u_2\\
&=&(\b I-\b M\b C)(\b A\b q+\b A\b M\b y+\b B_1\b u_1)+(\b I-\b M\b C)\b B_2\b u_2.
\end{eqnarray*}
If $\b M$ is chosen so that $(\b I-\b M\b C)\b B_2=\b O$, then the dynamics of $\b
q$ depend only on the known quantities $\b u_1$ and $\b y$:
\begin{equation} \dot{\b q}=(\b I-\b M\b C)(\b A\b q+\b A\b M\b y+\b B_1\b u_1).
\label{eq:pre_obser}
\end{equation}
Note that if we start the above dynamical system with the initial condition $\b
q(0)=(\b I-\b M\b C)\b x(0)$, then $\b x=\b q+\b M\b C\b x=\b q+\b M\b y$ for all
$t\geq 0$. But since $\b x(0)$ is assumed to be unknown, then $\tilde{\b x}=\b q+\b
M\b y$ is only an approximation of $\b x$. To improve convergence rate\index{convergence rate!of observer}\index{observer!convergence rate of} or to ensure
convergence, we add an extra term to the right-hand side of~(\ref{eq:pre_obser}) to
obtain
\begin{eqnarray*} \dot{\b q}&=&(\b I-\b M\b C)(\b A\b q+\b A\b M\b y+\b B_1\b u_1 +
\b L(\b y-\b C\b q-\b C\b M\b y))\\%\label{eq:pre_obser}
&=&(\b I-\b M\b C)(\b A\b q+\b A\b M\b y+\b B_1\b u_1+\b L\b C(\b x-\b q-\b M\b
y)).\end{eqnarray*} Let $\b e=\b x-\tilde{\b x}$. We will show that
\[ \dot{\b e}=(\b I-\b M\b C)(\b A-\b L\b C)\b e\] and $\b e(t) \rightarrow{\b 0}$ as
$t\rightarrow \infty$ under mild conditions.


Since
\[ \mbox{rank }(\b M\b C\b B_2)\leq \mbox{rank }(\b C\b B_2)\leq \mbox{rank }(\b B_2),\]
the equality $(\b I-\b M\b C)\b B_2=\b O$ makes it necessary that $\mbox{rank }(\b
C\b B_2)=\mbox{rank}(\b B_2)$, which we assume throughout the Chapter. This rank
condition also implies that there must be at least as many independent outputs as
unknown inputs for the method to work.


%We will show that, in order to arrive at a reduced-order observer using the above
%presented approach, it is critical that the term $\b L(\b y-\b C\b q-\b C\b M\b y)$
%is premultiplied by $(\b I-\b M\b C)$, or equivalently, $\b L$ has $(\b I-\b M\b C)$
%as a left factor. Indeed, let $\tilde{\b P}=\b I-\b M\b C$. Then, if $\tilde{\b P}$
%is a projection, that is, $\tilde{\b P}^2=\tilde{\b P}$, then the subspace
%${\cal{V}} = {\tilde{\b P}}{\mb{\Bb R}^n}$ is invariant under $\tilde{\b P}$. It
%follows that $\dot{\b q}$ in equation~(\ref{eq:pre_obser}) lies in $\cal{V}$. If the
%initial condition $\b q(0)$ is also in $\cal{V}$, then the trajectories of the
%system will reside in $\cal{V}$ for $t\geq 0$. If the term $\b L(\b y-\b C\b q-\b
%C\b M\b y)$ is not premultiplied by $(\b I-\b M\b C)$ or $\b L$ does not have $(\b
%I-\b M\b C)$ as a left factor, then the trajectory will not stay in $\cal{V}$ and,
%in general, it would not be possible to transform the full order-observer into a
%reduced order-observer.
%
%
%The condition $\b q(0)\in \cal{V}$ alone is not sufficient to guarantee that the
%observation error $\b e$ tends to $\b 0$. The reason is that we do not know $\b
%x(0)$ and it is not obvious how to choose $\b q(0)$ so that the $\b e(t)$ converges
%to $\b 0$. Unless $\b q(0)$ is chosen appropriately, the observation error $\b e(t)$
%stays in a hyperplane not containing $\b 0$ and thus $\b e(t)$ cannot converge to
%$\b 0$.
%
%
%
%Another difficulty that must be overcome is that,
For the error dynamics  matrix\index{error dynamics  matrix}
$(\b I-\b M\b C)(\b A-\b L\b C)$ to be asymptotically stable, it is not sufficient
for $(\b A-\b L\b C)$ to be asymptotically stable. It is possible for a product of a
projection matrix and an asymptotically stable matrix to be unstable as the
following simple example shows.
\begin{example}\rm
Let
\[\b{\Pi} =\left[\begin{array}{cc}1 & 0\\0& 0\end{array}\right]\mbox{   and   }
\b A=\left[\begin{array}{cc}1 & -3\\3& -2\end{array}\right].\] It is easy to check
that $\b A$ is asymptotically stable while $\b{\Pi}\b A$ is unstable. Furthermore, the
system $\dot{\b x}=\b A\b x$ restricted to the range of $\b{\Pi}$ is governed by
$\dot{z}=z$, which is also unstable in this case.
\end{example}
\markex


We analyze convergence properties of the proposed full-order observer and then
we use the results of our analysis to propose a new type of reduced-order observer
for uncertain systems.\index{reduced-order observer!for uncertain systems} Consider the following dynamical system model,
\begin{eqnarray}
\dot{\b q}&=&(\b I-\b M\b C)(\b A\b q+\b A\b M\b y+\b B_1\b u_1 + \b L(\b y-\b C\b
q-\b C\b M\b y))\\\label{eq:p_obser} \tilde{\b x} &=& \b q + \b M\b
y.\label{eq:q_express}
\end{eqnarray}
We will now show that $\tilde{\b x} \rightarrow{\b x}$ as $t\rightarrow \infty$. To
this end let
\[
\b e(t)=\b x(t) - \tilde{\b x}(t)
\]
denote the estimation error. Then, using $(\b I -\b M \b C)\b B_2=\b O$ and $\b y=\b
C\b x$, we obtain
\begin{eqnarray}
\frac{d \b e}{dt} &=& \frac{d}{dt}\left( \b x-\tilde{\b x}\right)\nonumber\\
 &=& \frac{d}{dt}\left( \b x-\b q - \b
M\b C\b x\right)\nonumber\\&=& \frac{d}{dt}\left( (\b I -\b M \b C)\b x-\b q\right)
\nonumber\\ &=& (\b I -\b M \b C)(\b A\b x+\b B_1\b u_1+\b B_2\b u_2)\nonumber\\
&{}&\quad -(\b I-\b M\b C)(\b A\b q+\b A\b M\b y+\b B_1\b u_1 + \b L(\b y-\b C\b
q-\b C\b M\b y))\nonumber\\
&=& (\b I -\b M \b C)(\b A\b x + \b B_1\b u_1) +
(\b I -\b M \b C)\b B_2\b u_2\nonumber\\
&{}& \quad -(\b I-\b M\b C)(\b A\b q+\b A\b M \b C\b x+\b B_1\b u_1 + \b L(\b C\b x
- \b C\b q-\b C\b M\b C\b x))\nonumber\\
&=& (\b I-\b M\b C)(\b A-\b L\b C)(\b x - \b q-\b M\b C\b x)\nonumber\\
&=& (\b I-\b M\b C)\left(\b A-\b L\b C\right)\b e.\label{der_estierror}
\end{eqnarray}
Our objective is to specify $\b M$ and $\b L$ and a set of initial conditions so
that $\b e (t) \rightarrow \b 0$ as $t\rightarrow\infty$. A particular class of
solutions to $(\b I -\b M \b C)\b B_2=\b O$ is given by
\[ \b M = \b B_2 \left((\b C\b B_2)^{\dagger}+\b H_0\left(\b I_p - (\b C\b B_2)(\b C\b
B_2)^{\dagger}\right)\right),
\]
where the superscript $\dagger$ denotes the Moore-Penrose pseudo-inverse\index{Moore-Penrose pseudo-inverse} operation
and $\b H_0\in \mb{\Bb R}^{m_2\times p}$ is a design parameter matrix. (See, for
example, Kaczorek~\cite[Section~1.5]{kaczorek98} or~\cite[Section~12.5]{chong08} for more information on the subject
of pseudo-inverse matrices.) Because, by assumption, $\mbox{rank }(\b C\b
B_2)=\mbox{rank }\b B_2$ and $\b B_2$ has full rank, we have $(\b C\b
B_2)^{\dagger}(\b C\b B_2)=\b I_{m_2}$. If $\b C \b B_2$ is a square matrix, then
$\b C\b B_2$ is invertible by assumption and the above $\b M$ reduces to $\b B_2(\b
C\b B_2)^{-1}$. Furthermore, it is easy to check that for the above class of $\b M$,
the product $\b M\b C$ is a projection\index{projection} (not necessarily orthogonal):
\[ (\b M\b C)^2 =\b M\b C.\]
It follows that
\[ \tilde{\b{\Pi}} =\b I-\b M\b C\]
is also a projection.




To proceed further, we need the following lemma.
\begin{lemma}\label{projector}
Let $\tilde{\b{\Pi}}:\mb{\Bb R}^n \to \mb{\Bb R}^n$ be a projection, that is,
$\tilde{\b{\Pi}}^2=\tilde{\b{\Pi}}$, and let $\mb{rank}\,\tilde{\b{\Pi}}=n-m_2$. Then
$\tilde{\b{\Pi}}$ has $(n-m_2)$ eigenvalues equal to $1$ while the remaining $m_2$
eigenvalues are located at $0$ and there is a basis of $\mb{\Bb R}^n$ in which the
matrix $\tilde{\b{\Pi}}$ relative to this basis has the form
\[
\b \Pi=\left[ \begin{array}{cc} \b I_{n-m_2} & \b O\\\b O & \b O\end{array}\right],
\]
that is, there is an invertible matrix $\b T$ whose columns are eigenvectors of
$\tilde{\b \Pi}$ such that
\begin{equation}
\b T^{-1} \tilde{\b \Pi}\b T =\b \Pi=\left[ \begin{array}{cc} \b I_{n-m_2} & \b O\\\b O &
\b O\end{array}\right].
\label{eq:sim_trans_for_can_proj}
\end{equation}
\end{lemma}
{\bf Proof}\quad See Smith~\cite[pp.~156--158 and pp.~194--195]{smith84}.\\
\hspace*{5.5in} $\Box$

In the next Section, we provide a numerically efficient method for constructing $\b T$ such that~\eqref{eq:sim_trans_for_can_proj} holds; see~\eqref{eq:Proj_transfor}.

\subsection{Constructing Full-Order UIO}\label{fo_uio}

We begin this section by introducing
the following coordinate transformation,
\begin{equation}
\tilde{\b e}=\b T^{-1}\b e, \label{error_p} \end{equation} where the transformation
matrix $\b T$ is obtained, using Lemma~\ref{projector}, from the representation of
the projection operator $\tilde{\b \Pi}$ in the form
\begin{equation}\label{defQ}
\tilde{\b \Pi}=\b T\b \Pi\b T^{-1}.
\end{equation}
Applying the coordinate transformation~(\ref{error_p}) to the error
equation~(\ref{der_estierror}) gives
\begin{eqnarray}
\dot{\tilde{\b e}}&=& \b \Pi\b T^{-1}(\b A-\b L\b C)\b T\tilde{\b e}\nonumber\\
&=& \b \Pi\left(\b T^{-1}\b A\b T - \left(\b T^{-1}\b L\right)\left(\b C\b T \right)
\right)\tilde{\b e}.\label{err_innew}
\end{eqnarray}
Let
\begin{equation}
\tilde{\b A}=\b T^{-1}\b A\b T =\left[\begin{array}{cc}
\tilde{\b A}_{11} & \tilde{\b A}_{12}\\
\tilde{\b A}_{21} & \tilde{\b A}_{21}
\end{array}\right],\;\; \tilde{\b L}=\b T^{-1}\b L=\left[\begin{array}{c}
\tilde{\b L}_1\\
\tilde{\b L}_2
\end{array}\right],\;\; \tilde{\b C}= \b
C\b T =\left[\begin{array}{cc} \tilde{\b C}_1 & \tilde{\b C}_2
\end{array}\right],
\label{matrices}
\end{equation}
where $\tilde{\b A}_{11}\in \mb{\Bb R}^{(n-m_2)\times (n-m_2)}$, $\tilde{\b
L}_{1}\in \mb{\Bb R}^{(n-m_2)\times p}$, $\tilde{\b C}_{1}\in \mb{\Bb R}^{p\times
(n-m_2)}$, and the remaining block submatrices are of appropriate dimensions. Using
the above notation, we represent~(\ref{err_innew}) in the form,
\begin{eqnarray}
\dot{\tilde{\b e}}&=& \b \Pi\left(\tilde{\b A} -
\tilde{\b L}\tilde{\b C} \right)\tilde{\b e}\nonumber\\
&=& \left[\begin{array}{cc} \b I_{n-m_2} & \b O\\
\b O & \b O
\end{array}\right]\left(\left[\begin{array}{cc}
\tilde{\b A}_{11} & \tilde{\b A}_{12}\\
\tilde{\b A}_{21} & \tilde{\b A}_{21}
\end{array}\right]-\left[\begin{array}{c}
\tilde{\b L}_1\\
\tilde{\b L}_2
\end{array}\right]\left[\begin{array}{cc}
\tilde{\b C}_1 & \tilde{\b C}_2
\end{array}\right]\right)\tilde{\b e}\nonumber\\
&=&  \left[\begin{array}{cc} \tilde{\b A}_{11}-\tilde{\b L}_1\tilde{\b C}_1 &
\tilde{\b A}_{12} -
\tilde{\b L}_1\tilde{\b C}_2\\
\b O & \b O
\end{array}\right]\tilde{\b e}\label{errindiffco}
\end{eqnarray}
Let
\begin{equation}
\tilde{\b e}=\left[\begin{array}{c}
\tilde{\b e}_1\\
\tilde{\b e}_2
\end{array}\right],
\end{equation}
where $\tilde{\b e}_1 \in \mb{\Bb R}^{n-m_2}$. Note that $\dot{\tilde{\b e}}_2 =\b
0$. Hence if $\tilde{\b e}_2(0)=\b 0$, then $\tilde{\b e}_2 = \b 0$ for all $t\ge
0$. Thus if $\tilde{\b e}_2=\b 0$ then
\[
\dot{\tilde{\b e}}_1=\left(\tilde{\b A}_{11}-\tilde{\b L}_1\tilde{\b
C}_1\right)\tilde{\b e}_1,
\]
and so if $\tilde{\b e}_2=\b 0$ and $\tilde{\b e}_1\to\b 0$ then $\tilde{\b e} \to
\b 0$. Obviously $\tilde{\b e}_1\to\b 0$ for arbitrary $\tilde{\b e}_1(0)$ if and
only if the matrix $\left(\tilde{\b A}_{11}-\tilde{\b L}_1\tilde{\b C}_1\right)$ is
asymptotically stable.



We give a condition on $\b q(0)$ that guarantees that $\tilde{\b e}_2=\b 0$. We
have,
\begin{eqnarray*}
\b M\b C &=& \b I-\tilde{\b \Pi}\\
&=& \b T(\b I-\b \Pi)\b T^{-1}.
\end{eqnarray*}
Hence
\begin{eqnarray}
\left[\begin{array}{c}
\b 0\\
\tilde{\b e}_2
\end{array}\right] &=& (\b I_n - \b \Pi)\b T^{-1}\b e\nonumber\\
&=& \b T^{-1}\b M\b C\b e\nonumber\\
&=& \b T^{-1}\b M\b C(\b x-\b q-\b M\b C\b x)\nonumber\\
&=& \b T^{-1}(\b M\b C\b x -\b M\b C\b q - (\b M\b C)^2\b x)\nonumber\\
&=& - \b T^{-1}\b M\b C\b q.\label{e2toq}
\end{eqnarray}
Therefore $\tilde{\b e}_2(0)=\b 0$ if and only if $\b M\b C\b q(0)=\b 0$, which is
equivalent to
\[
\b q(0)=(\b I-\b M\b C)\b v
\]
for arbitrary $\b v\in \mb{\Bb R}^n$. In particular $\b q(0)=\b 0$ satisfies the
above condition.



In summary, we proved the following theorem,
\begin{thm}\label{thm:UIO_full_order}
If the following conditions are satisfied:
\begin{enumerate}
\item $\mb{rank}\,\left(\b C\b B_2\right)= \mb{rank}\,\b B_2$; \item the pair
$(\tilde{\b A}_{11}, \tilde{\b C}_{1})$ defined in~(\ref{matrices}) is detectable;
\item $\b q(0)=(\b I-\b M\b C)\b v$ for arbitrary $\b v\in \mb{\Bb R}^n$,
\end{enumerate}
then there exists a gain matrix $\b L$ such that the estimation error, $\b e=\b
x-\tilde{\b x}$, of the full-order observer given by
\begin{eqnarray*}
\dot{\b q}&=&(\b I-\b M\b C)(\b A\b q+\b A\b M\b y+\b B_1\b u_1 + \b L(\b y-\b C\b
q-\b C\b M\b y))\\
\tilde{\b x} &=& \b q + \b M\b y
\end{eqnarray*}
converges to $\b 0$ as $t\to \infty$. \label{thm:fo_uio}
\end{thm}

In Figure~\ref{fig:full_order_UIO}, we show a block diagram of the full-order UIO\index{full-order UIO!projection operator approach}\index{full-order UIO|see{full-order unknown input observer}} of Theorem~\ref{thm:UIO_full_order}.
\begin{figure}
  \centering
  \includegraphics[width=5in]{zak_f_order}
  \caption{Full-order UIO described in Theorem~\ref{thm:UIO_full_order}.}
  \label{fig:full_order_UIO}
\end{figure}

\begin{thm}\label{thm:detect_condition}
The second condition of Theorem~\ref{thm:fo_uio} that states that the pair
$(\tilde{\b A}_{11}, \tilde{\b C}_{1})$ defined in~(\ref{matrices}) is detectable,
is equivalent to
\[
\mb{rank}\left[\begin{array}{cc}
s\b I_n - \b A & \b B_2\\
\b C & \b O
\end{array}\right]=n+m_2 \mb{  for all  } s \mb{  such that  } \mb{Re}(s)
\ge 0.
\]
\end{thm}
{\bf Proof}\quad We begin the proof by considering the projection matrix,\index{projection matrix} $\tilde{\b
\Pi}=\b I-\b M\b C$, where
\[
\b M=\b B_2 \left((\b C\b B_2)^{\dagger}+\b H_0\left(\b I_p - (\b C\b B_2)(\b C\b
B_2)^{\dagger}\right)\right).
\]
To simplify further analysis, we let $\b F=(\b C\b B_2)^{\dagger}+\b H_0\left(\b I_p
- (\b C\b B_2)(\b C\b B_2)^{\dagger}\right)$ and $\b S=\b F\b C$. Note that
\begin{equation}
\b S\b B_2=\b I_{m_2}. \label{eq:sbtwoone}
\end{equation}
Thus, $\mb{rank}\,\b S=m_2$ and so we can find a full rank $n\times (n-m_2)$ matrix
$\b W$ such that
\begin{equation}\b S\b W=\b O.
\label{eq:swzero}
\end{equation}
Combining~(\ref{eq:sbtwoone}) and~(\ref{eq:swzero}) we conclude that $[\b W \,\, \b
B_2]$ is invertible. Let
\[
\left[\begin{array}{cc} \b W & \b B_2
\end{array}\right]^{-1}=\left[\begin{array}{c}
\b W^g\\
\b N
\end{array}\right],
\]
where $\b W^g$ is $(n-m_2)\times n$ and $\b N$ is $m_2\times n$. Then
\begin{eqnarray*}
\b N \left[\begin{array}{cc} \b W & \b B_2\end{array}\right]
=\left[\begin{array}{cc} \b O & \b I_{m_2} \end{array}\right] =\b S
\left[\begin{array}{cc} \b W & \b B_2\end{array}\right].
\end{eqnarray*}
Since $[\b W\,\,\b B_2]$ is full rank, we conclude that $\b N=\b S$. Since
\[
\left[\begin{array}{c} \b W^g\\ \b S \end{array}\right] \left[\begin{array}{cc} \b W
& \b B_2 \end{array}\right] =\b I_n,
\]
we have
\[
\b W^g\b W = \b I_{n-m_2}\quad\mb{and}\quad \b W^g\b B_2=\b O.
\]
Let
\begin{equation}
\b T = \left[\begin{array}{cc} \b W & \b B_2 \end{array}\right].
\label{eq:Proj_transfor}
\end{equation}
Then
\begin{eqnarray}
\b T^{-1}\tilde{\b \Pi}\b T &=& \left[\begin{array}{c}
\b W^g\\
\b S
\end{array}\right]\left(\b I_n-\b B\b S\right)\left[\begin{array}{cc} \b W & \b B_2
\end{array}\right]\nonumber\\
&=& \b I_n-\left[\begin{array}{c}
\b W^g\b B_2\b S\\
\b S\b B_2\b S
\end{array}\right]\left[\begin{array}{cc} \b W & \b B_2
\end{array}\right]\nonumber\\
&=& \left[\begin{array}{cc} \b I_{n-m_2} & \b O\\
\b O & \b I_{m_2}
\end{array}\right]-\left[\begin{array}{cc}
\b W^g\b B_2\b S\b W & \b W^g\b B_2\b S\b B_2\\
\b S\b B_2\b S\b W & \b S\b B_2\b S\b B_2
\end{array}\right]\nonumber\\
&=& \left[\begin{array}{cc} \b I_{n-m_2} & \b O\\
\b O & \b I_{m_2}
\end{array}\right]-\left[\begin{array}{cc} \b O & \b O\\
\b O & \b I_{m_2}
\end{array}\right]\nonumber\\
&=& \left[\begin{array}{cc} \b I_{n-m_2} & \b O\\
\b O & \b O
\end{array}\right].
\label{eq:qinptq}
\end{eqnarray}
We apply the following coordinate transformation to the system modeled
by~(\ref{eq:uncert_sys}) and~(\ref{eq_lin_system}):
\[
\left[\begin{array}{c} \b z\\
\b{\sigma}
\end{array}\right]=\left[\begin{array}{c}
\b W^g\\
\b S
\end{array}\right]\b x=\b T^{-1}\b x.
\]
Then
\begin{eqnarray*}
\left[\begin{array}{c} \dot{\b z}\\
\dot{\b{\sigma}}
\end{array}\right]&=&\left[\begin{array}{cc}
\b W^g\b A\b W & \b W^g\b A\b B_2\\
\b S\b A\b W & \b S\b A\b B_2
\end{array}\right]\left[\begin{array}{c} \b z\\
\b{\sigma}
\end{array}\right]+\b T^{-1}\b B_1\b u_1 + \left[\begin{array}{c}
\b O\\
\b I_{m_2} \end{array}\right]\b u_2\\
&=& \left[\begin{array}{cc}
\tilde{\b A}_{11} & \tilde{\b A}_{12}\\
\tilde{\b A}_{21} & \tilde{\b A}_{22}
\end{array}\right]\left[\begin{array}{c} \b z\\
\b{\sigma}
\end{array}\right]+\b T^{-1}\b B_1\b u_1 + \left[\begin{array}{c}
\b O\\
\b I_{m_2} \end{array}\right]\b u_2\\
\b y &=& \left[\begin{array}{cc} \b C\b W & \b C\b B_2
\end{array}\right]\left[\begin{array}{c} \b z\\
\b{\sigma}
\end{array}\right]\\
&=&  \left[\begin{array}{cc} \tilde{\b C}_{1} & \tilde{\b C}_{2}
\end{array}\right]\left[\begin{array}{c} \b z\\
\b{\sigma}
\end{array}\right].
\end{eqnarray*}
If the trajectory of the system described by the triple $(\b A, \b B_2, \b F\b C)$
resides in the null space of $\b S$, then such a motion is described by
\[
\dot{\b z}= \tilde{\b A}_{11}\b z.
\]
It follows from~\ref{sec:system_zeros} that the poles of the above system are
the zeros of the system\index{system zeros} described by the triple $(\b A, \b B_2, \b F\b C)$, which
are the complex numbers $s$ for which the system matrix\index{system matrix}
\[
\left[\begin{array}{cc}
s\b I_n - \b A & \b B_2\\
\b F\b C & \b O
\end{array}\right]
\]
loses its normal-rank---see Definition~\ref{defn:normal_rank} on page~\pageref{defn:normal_rank}\index{normal-rank}. On the other hand, the zeros of the triple $(\b A, \b B_2, \b
C)$ are also the zeros of the squared-down system $(\b A, \b B_2, \b F\b C)$, that
is, the zeros of $(\b A, \b B_2, \b C)$ form a subset of the set of the eigenvalues
of $\tilde{\b A}_{11}$. It is well known that zeros are invariant with respect to
similarity transformations. Therefore,
\[
\mb{rank}\left[\begin{array}{ccc}
s\b I_{n-m_2} - \tilde{\b A}_{11} & -\tilde{\b A}_{12} & \b O\\
-\tilde{\b A}_{21} & s\b I_{m_2}-\tilde{\b A}_{22}& \b I_{m_2}\\
\tilde{\b C}_{1} & \tilde{\b C}_{2} & \b O
\end{array}\right] = \mb{rank}\left[\begin{array}{c}
s\b I_{n-m_2} - \tilde{\b A}_{11}\\
\tilde{\b C}_{1}
\end{array}\right]+2m_2\quad\mb{for}\quad s\in\mb{\Bb C},
\]
which means that zeros of the system $(\b A, \b B_2, \b C)$ are in the open
left-half plane if and only if the pair $(\tilde{\b A}_{11},\tilde{\b C}_{1})$ is
detectable.\\
\hspace*{5.5in} $\Box$
%{\bf Proof}\quad See Hui and \.{Z}ak~\cite{hui04_uio}.\\
%\hspace*{5.5in} $\Box$



\subsection{Reduced-Order UIO}
\label{sec:reducedUIO}


The error dynamics of the full-order observer\index{error dynamics!of full-order observer}\index{full-order observer!error dynamics of} that we analyzed above are given by
equation~(\ref{errindiffco}):
\[
\dot{\tilde{\b e}}=  \left[\begin{array}{cc} \tilde{\b A}_{11}-\tilde{\b
L}_1\tilde{\b C}_1 & \tilde{\b A}_{12} -
\tilde{\b L}_1\tilde{\b C}_2\\
\b O & \b O
\end{array}\right]\tilde{\b e}
\]
The reader may have noticed that since we choose the initial condition for $\b q$ to
force $\tilde{\b e}_2(t)=\b 0$ for $t\geq 0$, the dynamics of the error are
completely determined by the dynamics of $\tilde{\b e}_1$, which are given by
\begin{equation}\label{eq:ro_uio_error}
\dot{\tilde{\b e}}_1=\left(\tilde{\b A}_{11}-\tilde{\b L}_1\tilde{\b
C}_1\right)\tilde{\b e}_1,
\end{equation}
an $(n-m_2)$-dimensional system. This motivates us to apply the transformation from
$\b e$ to $\tilde{\b e}$ to $\b q$:
\[ \tilde{\b q}=\b T^{-1} \b q.
\]
From equation~(\ref{eq:p_obser}),
\[
\dot{\b q}=(\b I-\b M\b C)(\b A\b q+\b A\b M\b y+\b B_1\b u_1 + \b L(\b y-\b C\b
q-\b C\b M\b y))
\]
and $\b I-\b M\b C=\b T\b \Pi\b T^{-1}$, we obtain
\begin{eqnarray}
\dot{\tilde{\b q}}&=&\b \Pi(\b T^{-1}\b A\b T\tilde{\b q}+\b T^{-1}\b A\b M\b y+\b
T^{-1}\b B_1\b u_1 + \b T^{-1}\b L(\b y-\b C\b T\tilde{\b q}-\b C\b M\b y))\nonumber\\
&=& \b \Pi\b T^{-1}\b A\b T\tilde{\b q}+ \b \Pi\b T^{-1}\b A\b M\b y + \b \Pi\b T^{-1}\b
B_1\b u_1 + \b \Pi\b T^{-1}\b L(\b y-\b C\b T\tilde{\b q} - \b C\b M\b
y)\nonumber\\
&=& \b \Pi\left(\b T^{-1}\b A\b T -\b T^{-1}\b L\b C\b T \right)\tilde{\b q} + \b
\Pi\left(\b T^{-1}\b A\b M  + \b T^{-1}\b L -\b T^{-1}\b L\b C\b M\right)\b y\nonumber\\
&{}&\quad +  \b \Pi\b T^{-1}\b B_1\b u_1.\label{red_ord_dyn}
\end{eqnarray}
Using the notation defined in~(\ref{matrices}), we have
\[
\dot{\tilde{\b q}}=\b \Pi\left(\tilde{\b A} -\tilde{\b L}\tilde{\b C} \right)\tilde{\b
q} + \b \Pi\b T^{-1}\left[\left(\b A\b M  + \b T\tilde{\b L}(\b I_p - \b C\b M)
\right)\b y + \b B_1\b u_1\right].
\]
Let
\[
\tilde{\b q}=\left[\begin{array}{c} \tilde{\b q}_1\\
\tilde{\b q}_2
\end{array}\right],
\]
where $\tilde{\b q}_1\in \mb{\Bb R}^{n-m_2}$ and $\tilde{\b q}_2\in \mb{\Bb
R}^{m_2}$. Since
\[
\b \Pi=\left[\begin{array}{cc}
\b I_{n-m_2} & \b O\\
\b O & \b O
\end{array}\right],
\]
we have $\dot{\tilde{\b q}}_2(t)=\b 0$. Therefore, setting $\tilde{\b q}_2(0)=\b 0$
ensures that $\tilde{\b q}_2(t)=\b 0$ for $t\geq 0$. We thus can remove $m_2$
observer states from the observer dynamics. Let $\tilde{\b G}=\b A\b M+\b T\tilde{\b
L}(\b I_p-\b C\b M)$. Then the resulting reduced-order observer takes the form,
\begin{eqnarray*}
\dot{\tilde{\b q}}_1&=&(\tilde{\b A}_{11}- \tilde{\b L}_{1}\tilde{\b
C}_{1})\tilde{\b q}_1 + \left[\begin{array}{cc} \b I_{n-m_2} & \b O_{m_2}
\end{array}\right]\b T^{-1}\left(\tilde{\b G}\b y
+ \b B_1\b u_1\right),\quad \tilde{\b q}_1(0)=\b 0\\
\tilde{\b x}&=&\b T\left[\begin{array}{c} \b I_{n-{m_2}}\\
\b O_{m_2\times(n-{m_2})}
\end{array}\right]\tilde{\b q}_1 + \b M\b y,
\end{eqnarray*}
where the vector $\tilde{\b x}$ is the estimate of the plant state $\b x$. In Figure~\ref{fig:reduced_order_UIO}, we show a block diagram of the above reduced-order UIO.\index{reduced-order UIO}
\begin{figure}
  \centering
  \includegraphics[width=5in]{zak_red_ord}
  \caption{Reduced-order UIO.}
  \label{fig:reduced_order_UIO}
\end{figure}

We now summarize the above considerations in the form of the following design
algorithm.
\begin{algorithm}\label{algo:red_order_UIO_design}\rm
\begin{center}
{\bf Reduced-Order Unknown Input Observer Design}
\end{center}
For a given a quadruple of matrices $(\b A, \b B,\b C)$ modeling the plant,
\begin{enumerate}
\item Check that $\mb{rank}\,\left(\b C\b B_2\right)= \mb{rank}\,\b B_2$\\
If $\mb{rank}\,\left(\b C\b B_2\right) < \mb{rank}\,\b B_2$, STOP. The observer does
not exist. \item Compute
\[
\b M=\b B_2\left( (\b C\b B_2)^{\dagger}+\b H_0\left(\b I_p - (\b C\b B_2)(\b C\b
B_2)^{\dagger}\right)  \right),
\]
where the superscript $\dagger$ denotes the Moore-Penrose pseudo-inverse\index{Moore-Penrose pseudo-inverse} operation
and $\b H_0\in \mb{\Bb R}^{m_2\times p}$ is a design parameter matrix.

\item Compute the projector
\[
\tilde{\b \Pi}=\b I_n-\b M\b C.
\]
\item Compute $\b T = \left[\begin{array}{cc} \b W & \b B_2 \end{array}\right]$---see\ref{eq:Proj_transfor}. Note
that
\[
\tilde{\b \Pi}=\b T\b \Pi\b T^{-1},
\]
where
\[
\b \Pi=\left[\begin{array}{cc}
\b I_{n-m_2} & \b O\\
\b O & \b O
\end{array}\right].
\]
\item Compute
\[
\tilde{\b A}=\b T^{-1}\b A\b T=\left[\begin{array}{cc}
\tilde{\b A}_{11} & \tilde{\b A}_{12}\\
\tilde{\b A}_{21} & \tilde{\b A}_{22}
\end{array}\right]\;\;\mb{and}\;\;\tilde{\b C}=\b C\b T =\left[\begin{array}{cc}
\tilde{\b C}_{1} & \tilde{\b C}_{2}
\end{array}\right],
\]
where $\tilde{\b A}_{11} \in \mb{\Bb R}^{(n-m_2)\times (n-m_2)}$ and $\tilde{\b
C}_{1} \in \mb{\Bb R}^{p\times (n-m_2)}$
\item Check detectability\index{detectability} of the pair $(\tilde{\b A}_{11}, \tilde{\b C}_{1})$\\
If the pair $(\tilde{\b A}_{11}, \tilde{\b C}_{1})$ is not detectable, STOP. The
observer does not exist.\\
Note that if the matrix $\tilde{\b A}_{11}$ is asymptotically stable, then the pair
$(\tilde{\b A}_{11}, \tilde{\b C}_{1})$ is detectable for an arbitrary matrix
$\tilde{\b C}_{1}$.



\item If there are eigenvalues of $\tilde{\b A}_{11}$ that are not asymptotically
stable, construct $\tilde{\b L}_1$ so that the matrix $(\tilde{\b A}_{11}-\tilde{\b
L}_1\tilde{\b C}_1)$ has its eigenvalues in locations as close to the desired
eigenvalues as possible. \item Form
\[
\tilde{\b L}=\left[\begin{array}{c}
\tilde{\b L}_{1}\\
\b O_{m_2\times p}
\end{array}\right],
\]
where $\b O_{m_2\times p}$ is an $m_2\times p$ zero matrix. \item Compute matrix
\[
\tilde{\b G}=\b A\b M+\b T\tilde{\b L}(\b I_p-\b C\b M).
\]
\item Construct the observer,
\begin{eqnarray*}
\dot{\tilde{\b q}}_1&=&(\tilde{\b A}_{11}- \tilde{\b L}_{1}\tilde{\b
C}_{1})\tilde{\b q}_1 + \left[\begin{array}{cc} \b I_{n-m_2} & \b O_{m_2}
\end{array}\right]\b T^{-1}\left(\tilde{\b G}\b y
+ \b B_1\b u_1\right),\quad \tilde{\b q}_1(0)=\b 0\\
\tilde{\b x}&=&\b T\left[\begin{array}{c} \b I_{n-{m_2}}\\
\b O_{m_2\times(n-{m_2})}
\end{array}\right]\tilde{\b q}_1 + \b M\b y.
\end{eqnarray*}
The vector $\tilde{\b x}$ is the estimate of the state $\b x$.
%\item\label{WZstop} STOP
\end{enumerate}
\end{algorithm}\markex



\begin{example}\label{ex:eg2}\rm We consider the fifth-order lateral axis model of an
L-1011 fixed-wing aircraft\index{fixed-wing aircraft!model of}, with the actuator dynamics neglected\index{fixed-wing aircraft!actuator dynamics neglected}, at cruise flight
conditions\index{fixed-wing aircraft!cruise flight}. This model can be found in the book by Edwards and Spurgeon~\cite[pages
122, 123 and pages 179, 180]{edwards98}. We assume that the inputs to the system are
unknown and there are no known inputs. We have,
\[
\b A= \left[ \begin{array}{rrrrr}
  0.0000 & 0.0000 & 1.0000 & 0.0000 & 0.0000 \\
  0.0000 & -0.1540 & -0.0042 & 1.5400 & 0.0000 \\
  0.0000 & 0.2490 & -1.0000 & -5.2000 & 0.0000 \\
  0.0386 & -0.9960 & -0.0003 & -0.1170 & 0.0000 \\
  0.0000 & 0.5000 & 0.0000 & 0.0000 & -0.5000
\end{array} \right]
\]
and
\[\b B_2=
\left[ \begin{array}{rr}
  0.0000 & 0.0000 \\
  -0.7440 & -0.0320 \\
  0.3370 & -1.1200 \\
  0.0200 & 0.0000 \\
  0.0000 & 0.0000
\end{array} \right],\quad \b C = \left[ \begin{array}{rrrrr}
  0 & 1 & 0 & 0 & -1 \\
  0 & 0 & 1 & 0 & 0 \\
  0 & 0 & 0 & 1 & 0 \\
  1 & 0 & 0 & 0 & 0
\end{array} \right],
\]
%\end{eqnarray*}
where $\b u_2=\left[\begin{array}{cc} \cos(t) & \sin(t)
\end{array}\right]^{\top}$. We
first check that $\mb{rank}\,\left(\b C\b B_2\right)= \mb{rank}\,\b B_2$. We then
compute the matrix $\b M$, where in this example we set $\b H_0=\b O$,
\[
\b M=\b B_2(\b C\b B_2)^{\dagger}= \left[ \begin{array}{rrrr}
0.0000 & 0.0000 & 0.0000 & 0.0000 \\
0.9993 & 0.0000 & -0.0265 & 0.0000 \\
0.0000 & 1.0000 & 0.0008 & 0.0000 \\
-0.0265 & 0.0008 & 0.0007 & 0.0000 \\
0.0000 & 0.0000 & 0.0000 & 0.0000
\end{array} \right].
\]
Then, the projector $\tilde{\b \Pi}$ is
\[
\tilde{\b \Pi}=\b I_5-\b M\b C=\left[\begin {array}{rrrrr}
  1.0000 & 0.0000 & 0.0000 & 0.0000 & 0.0000 \\
  0.0000 & 0.0007 & -0.0000 & 0.0265 & 0.9993 \\
  0.0000 & -0.0000 & 0.0000 & -0.0008 & 0.0000 \\
  0.0000 & 0.0265 & -0.0008 & 0.9993 & -0.0265 \\
  0.0000 & 0.0000 & 0.0000 & 0.0000 & 1.0000
\end{array}\right].
\]
We next use~(\ref{eq:Proj_transfor}) to compute $\b T$ such that
\[
\tilde{\b \Pi}=\b T\b \Pi\b T^{-1}=\b T\left[\begin{array}{cc} \b I_3 & \b O_{3\times 2}\\
\b O^{\top}_{3\times 2} & \b O_{2\times 2}
\end{array}\right]\b T^{-1}.
\]
We have
\[
\b T =\left[\begin {array}{rrrrr}
-0.7169  & -0.0190 & -0.6969  &      0  &      0\\
-0.4930  &  0.0131  &  0.5069 &  -0.7440 &  -0.0320\\
0.0000 &  -0.0008  &  0.0000  &  0.3370  & -1.1200\\
-0.0138 &   0.9996 &  -0.0131  &  0.0200    &  0\\
-0.4927 &  -0.0135  &  0.5072   &      0    &     0
\end{array}\right].
\]
Hence,
\[
\tilde{\b A}_{11}=\left[\begin {array}{rrr}
-0.0124 &  -0.0102  &  0.0145\\
0.4662 & -0.0863 &  -0.5326\\
-0.0000  &  0.0139 &  -0.0004
\end{array}\right]\mb{ and } \tilde{\b C}_{1}=
\left[\begin {array}{rrr}
-0.0004  &  0.0265 &  -0.0003\\
0.0000 &  -0.0008  &  0.0000\\
-0.0138 &  0.9996 & -0.0131\\
-0.7169 &  -0.0190 &  -0.6969
\end{array}\right].
\]
Note that the pair $(\tilde{\b A}_{11}, \tilde{\b C}_1)$ is detectable and that the
matrix $\tilde{\b A}_{11}$ is unstable with its eigenvalues at $\{0.0000,-0.0495 +
0.1041j, -0.0495 - 0.1041j\}$. We proceed to stabilize $\tilde{\b A}_{11}$. We
select the desired eigenvalues to be located at $-3, -4, -5$. The gain matrix
$\tilde{\b L}_{1}$ such that $\mb{eig}(\tilde{\b A}_{11}-\tilde{\b L}_{1}\tilde{\b
C}_{1})=\{-3, -4, -5\}$ is
\[
\tilde{\b L}_{1}=\left[\begin{array}{rrrr}
0.3866 & -0.0110 &  14.5695 &  -3.5473\\
0.2097 &  -0.0060 &  7.9007 &  -0.2190\\
-0.4028 &  0.0115 & -15.1794 &  -2.0853
\end{array}\right].
\]
We obtained the above gain matrix using MATLAB's command \verb!place!.\index{MATLAB's command!place} We next form
the matrix $\tilde{\b L}$ by adding two zero rows to $\tilde{\b L}_{1}$ and compute
\[
\left[\begin{array}{cc} \b I_{3} & \b O_{3\times 2}
\end{array}\right]\b T^{-1}\tilde{\b G}=\left[\begin{array}{rrrr}
-0.0787 &  -0.7280 &  14.5813 &  -3.5473\\
-0.8007 &  -0.0248 &   7.9275 &  -0.2190\\
0.1035 &  -0.6854 & -15.1934 &  -2.0853
\end{array}\right].
\]
\begin{figure}
\centering\epsfig{file=eg2.eps,width=6.5in} \caption{Plots of $x_i$s and their
estimates versus time for Example~\ref{ex:eg2}.} \label{fig:eg2}
\end{figure}
The reduced-order UIO has the form\index{fixed-wing aircraft!reduced-order UIO}
\begin{eqnarray*}
\dot{\tilde{\b q}}_1 &=& (\tilde{\b A}_{11}- \tilde{\b L}_{1}\tilde{\b
C}_{1})\tilde{\b q}_1 + \left[\begin{array}{cc} \b I_{3} & \b 0
\end{array}\right]\b T^{-1}\tilde{\b G}\b y\\
&=&\left[\begin{array}{rrr}
-2.3538 & -14.6523 &  -2.2668\\
0.4186 &  -7.9939 &  -0.5818\\
-1.7052 &  15.1589  & -1.6523
\end{array}\right]\tilde{\b q}_1 + \left[\begin{array}{rrrr}
-0.0787 &  -0.7280 &  14.5813 &  -3.5473\\
-0.8007 &  -0.0248 &   7.9275 &  -0.2190\\
0.1035 &  -0.6854 & -15.1934 &  -2.0853
\end{array}\right]\b y\\
\tilde{\b x} &=& \b T\left[\begin{array}{c} \b I_{3}\\
\b O_{2\times{3}}
\end{array}\right]\tilde{\b q}_1 + \b M\b y\\
&=& \left[\begin{array}{rrr}
-0.7169  & -0.0190 & -0.6969\\
-0.4930  &  0.0131  &  0.5069\\
0.0000 &  -0.0008  &  0.0000\\
-0.0138 &   0.9996 &  -0.0131\\
-0.4927 &  -0.0135  &  0.5072
\end{array}\right]\tilde{\b q}_1   + \left [\begin {array}{rrrr}
0.0000 & 0.0000 & 0.0000 & 0.0000 \\
0.9993 & 0.0000 & -0.0265 & 0.0000 \\
0.0000 & 1.0000 & 0.0008 & 0.0000 \\
-0.0265 & 0.0008 & 0.0007 & 0.0000 \\
0.0000 & 0.0000 & 0.0000 & 0.0000
\end{array}\right]\b y.
\end{eqnarray*}
In Figure~\ref{fig:eg2}, we show plots of the system state variables and their
estimates versus time. The initial conditions of the plant were selected randomly to
be equal to
\[
\b x(0)=\left[\begin{array}{ccccc}
 0.0342 & 0.3200 & 0.0178 & -0.0287 & -0.2497
\end{array}\right]^{\top}.
\]
The initial conditions of the observer were set to zero.
\end{example}
\markex







\subsection{Relations With Other UIO Architectures}\label{sec:otherUIOarch}
Our analysis, from the previous sections, can be extended to cover the case
\begin{equation}
\dot{\b q}=(\b I-\b M\b C)(\b A\b q+\b A\b M\b y+\b B_1\b u_1) + \b L(\b y-\b C\b
q-\b C\b M\b y), \label{eq:full_order_out}
\end{equation}
where the term $\b L(\b y-\b C\b q-\b C\b M\b y)$ is not premultiplied by $(\b I-\b
M\b C)$. This case leads to the observer analyzed in \cite{chen96, chen99}
even though the approach there is quite different. Indeed, we can equivalently
represent the dynamics of the proposed full-order observer as follows:
\begin{eqnarray*}
\dot{\b q}&=&\left(\left(\b I-\b M\b C\right)\b A - \b L\b C\right)\b q +
\left(\left[\left(\b I-\b M\b C\right)\b A -\b L\b C\right]\b M + \b L \right)\b
y+\left(\b I-\b M\b C\right)\b B_1\b u_1\\
&=& (\b T\b A -\b L\b C)\b q + \b K\b y + \b T\b B_1\b u_1,\\
\tilde{\b x} &=& \b q + \b M\b y,
\end{eqnarray*}
where, using the notation similar to that in~\cite{chen96, chen99},
\begin{eqnarray*}
\b T &=& \b I-\b M\b C,\quad \b K_1 = \b L,\\
\b K_2 &=& \left[\b T\b A -\b L\b C\right]\b M, \quad \b K =\b K_1+\b K_2.
\end{eqnarray*}
In addition, the conditions for the existence of  the full-order observer presented
in~\cite{chen96, chen99} and our observers are equivalent.

The observer given by~(\ref{eq:full_order_out}) is also the same as the one proposed
by Yang and Wilde~\cite{yang88} and further analyzed by Darouach, Zasadzinski, and
Xu~\cite{darouach94}. The connections are as follows: (i) $\b M$ is called $-\b E$
in~\cite{yang88,darouach94}, (ii) $(\b I-\b M\b C)$ corresponds to $\b P$ there,
(iii) $\b B_1$ is $\b B$ and $\b B_2$ is $\b D$ in~\cite{yang88,darouach94}, (iv)
$(\b I-\b M\b C)(\b A-\b L\b C)$ corresponds to $\b N$.

A different approach is proposed by Hou and M\"{u}ller~\cite{hou92} who first transform the
system~(\ref{eq:two_inputs}) into the form
\begin{eqnarray*}
\dot{\b x}&=&\b A\b x+\b B_1\b u_1+\left[\begin{array}{c} \b O\\
\b I_{m_2}
\end{array}\right]\b u_2\\
&=&\left[\begin{array}{cc} \b A_{11} & \b A_{12}\\
\b A_{21} & \b A_{22}
\end{array}\right]\left[\begin{array}{c}
\b x_1\\
\b x_2
\end{array}\right]+\left[\begin{array}{c}
\b B_{11}\\
\b B_{12}
\end{array}\right]\b u_1\\
&{}&\quad +\left[\begin{array}{c} \b O\\
\b I_{m_2}
\end{array}\right]\b u_2.
\end{eqnarray*}
Note that $\b x_1$ in the new coordinates is independent of $\b u_2$ and we have
\begin{equation}
\dot{\b x}_1 = \b A_{11}\b x_1+\b A_{12}\b x_2 + \b B_{11}\b u_1. \label{eq:onlyx1}
\end{equation}
Let
\[
\b y=\b C\b x=\left[\begin{array}{cc} \b C_1 & \b C_2
\end{array}\right]\left[\begin{array}{c}
\b x_1\\
\b x_2 \end{array}\right],
\]
where $\b C_2\in \mb{\Bb R}^{p\times m_2}$. Because by assumption, $\mb{rank} (\b
C\b B_2)=\mb{rank} \b B_2 = m$, the submatrix $\b C_2$ has a left inverse, $\b
C_2^{\dag}$. Hence we can compute
\begin{equation}
\b x_2=-\b C_2^{\dag}\b C_1\b x_1 + \b C_2^{\dag}\b y. \label{eq:x2usingx1andy}
\end{equation}
Substituting the above into~(\ref{eq:onlyx1}) gives
\[
\dot{\b x}_1 = \left(\b A_{11}-\b A_{12}\b C_2^{\dag}\b C_1\right)\b x_1 + \b
A_{12}\b C_2^{\dag}\b y +  \b B_{11}\b u_1.
\]
Hou and M\"{u}ller~\cite{hou92}\index{Hou and M\"{u}ller UIO} propose now to construct an observer for $\b x_1$
using only known signals and then substitute the estimate of $\b x_1$
into~(\ref{eq:x2usingx1andy}) to obtain an estimate of $\b x_2$.
%Thus, the resulting
%architecture of the reduced-order UIO proposed by Hou and M\"{u}ller~\cite{hou92},
%as well as their approach, differs from our design. Yet another approach to
%constructing reduced-order UIOs can be found in~\cite{hui_uio93}.

\section{Sliding Mode UIO}\label{sec:uncertain}

\subsection{Sliding Mode UIO Architecture}
We present a sliding mode UIO\index{sliding mode UIO} following the exposition in our paper~\cite{hui05}. We assume that $\b u_2$ is bounded, that is, there exists a
nonnegative real number, $\rho$, such that
\[
\|\b u_2\| \le \rho.
\]
Let $\hat{\b x}$ be an estimate of $\b x$. Let $\b e$ denote the estimation error,
that is,
\[
\b e(t)=\b x(t) - \hat{\b x}(t).
\]
The observability of $\left( \b A,\b C \right)$ implies the existence of a matrix
$\b L \in \mb{\Bb R}^{n \times p}$ such that the matrix $\left(\b A-\b L\b C\right)$
has prescribed (symmetric with respect to the real axis) eigenvalues in the open
left-half plane. Because $\left(\b A-\b L\b C\right)$ is asymptotically stable, for
any $\b Q=\b Q^{\top}>0$, there is a unique $\b P=\b P^{\top}>0$ such that
\begin{equation}
\left(\b A - \b L\b C \right)^{\top}\b P+\b P\left(\b A - \b L\b C \right)=-\b Q.
\label{eq:lyap_esti}
\end{equation}
We choose $\b Q$, if possible, so that for some $\b F \in \mb{\Bb R}^{m_2 \times
p}$,
\begin{equation}
\b F\b C=\b B_2^{\top}\b P. \label{eq:tech_cond}
\end{equation}
We need this technical condition to ensure the realizablity of the observer.\index{technical condition!to ensure realizablity of observer}

To proceed, we define the vector function
\[
\b E\left(\b e, \eta \right)=\left\{\begin{array}{lcc}
\eta\frac{\b F\b C\b e}{\|\b F\b C\b e\|_2} & \mb{for} & \b F\b C\b e\ne \b 0\\
\b 0  & \mb{for} & \b F\b C\b e=\b 0,
\end{array}\right.
\]
where $\eta \ge \rho$ is a design parameter. In the case of single-input
single-output plant, we can write
\[
\b E(\b e,\eta)=\eta \,\mb{sign}\left(\b F\b C\b e\right).
\]
We note that
\[
\b C\b e=\b C(\b x - \hat{\b x})=\b y - \hat{\b y}.
\]
The vector function $\b E(\b e,\eta)$ is an essential ingredient of the sliding mode
observer that we present next. When implementing the function $\b E$, we use the
output measurements $\b y$ and $\hat{\b y}$, that is, instead of using $E(\b
e,\eta)$, we utilize
\[
\b E(\b y, \hat{\b y}, \eta) = \left\{\begin{array}{lcc}
\eta\frac{\b F(\b y - \hat{\b y} )}{\|\b F(\b y - \hat{\b y})\|_2} & \mb{for} & \b F( \b y - \hat{\b y})\ne \b 0\\
\b 0  & \mb{for} & \b F(\b y - \hat{\b y})=\b 0.
\end{array}\right.
\]
Hence for the case of single-input single-output plant, we have
\[
\b E(\hat{y}, \b e, y)=\eta \,\mb{sign}\left(F(y - \hat{y})\right).
\]

Using arguments similar to those found in~\cite{walcott87}, we can show that the
state $\hat{\b x}$ of the dynamical system
\begin{equation}
\fbox{$\displaystyle \dot{\hat{\b x}}=\b A \hat{\b x}+\b B_1\b u_1 + \b L( \b y - \hat{\b y}) + \b B_2\b E (\b y,\hat{\b y}, \eta)$} \label{eq:WZsm_observer}
\end{equation}
for $\eta \ge \rho$, is an asymptotic estimate of the state $\b x$ of the system
described by~(\ref{eq:two_inputs}) and (\ref{eq_lin_system}), that is,
\[
\lim_{t \to \infty}\b e(t)=\lim_{t \to \infty} \left(\b x(t) - \hat{\b x}(t)\right)=\b 0.
\]
In Figure~\ref{fig:wzsm_obs}, we show a block diagram of the sliding mode observer given by~(\ref{eq:WZsm_observer}).
\begin{figure}
  \centering
  \includegraphics[width=4.5in]{zak_11_20091}
  \caption{Sliding mode observer~(\ref{eq:WZsm_observer}) can also be used as an unknown input estimator---see~\eqref{eq:WZu2recontruct}.}
  \label{fig:wzsm_obs}
\end{figure}
To prove the above statement using Lyapunov's type of arguments, first
represent~(\ref{eq:WZsm_observer}) as
\[
\dot{\hat{\b x}} = \left(\b A-\b L\b C \right)\hat{\b x} + \b L\b y+\b B_1\b u_1 + \b
B_2\b E (\b e,\eta).
\]
Then construct the differential equation describing the dynamics of the estimation
error $\b e$,
\begin{equation}
\dot{\b e} = \dot{\b x} - \dot{\hat{\b x}}= \left(\b A-\b L\b C \right)\b e +\b B_2\b
u_2-\b B_2\b E(\b e,\eta). \label{eq:error_diffeqn}
\end{equation}
Then, show that
\[
\frac{d}{dt}\left(\b e^{\top}\b P\b e \right)=-\b e^{\top}\b Q\b e<0,
\]
which implies
\[
\lim_{t \to \infty}\b e(t)=\b 0.
\]
It follows from the above that the estimation error is insensitive\index{estimation error!insensitive to uncertainty} to the
uncertainty modeled by the term $\b B_2\b u_2$. In summary, the design of the
sliding mode UIO  for a system modeled by
the quadruple $(\b A,\b B_1, \b B_2, \b C)$ can be thought of as finding a triple of
matrices $(\b L, \b F, \b P)$ satisfying~(\ref{eq:lyap_esti}) and~(\ref{eq:tech_cond}) for
some $\b Q=\b Q^{\top}>0$. Edwards and Spurgeon~\cite{edwards98} (see also Saif and
Xiong~\cite{saif03}) present necessary and sufficient conditions for the existence
of the above observer, which are:
\begin{itemize}
\item[{\bf (i)}] $\mb{rank}\, \b B_2=\mb{rank}\,\b C\b B_2=r$; \item[{\bf (ii)}] the
system zeros of the triple $(\b A,\b B_2, \b C)$ are in the open left-hand complex
plane, that is,
\[
\mb{rank}\left[\begin{array}{cc}
s\b I_n - \b A & \b B_2\\
\b C & \b O
\end{array}\right]=n+r\;\;\mb{for all}\;\; s\;\;\mb{such that}\;\; \mb{Re}(s)
\ge 0.
\]
\end{itemize}

It is interesting to note that the above conditions are also necessary and
sufficient for the existence of the observers with unknown inputs analyzed by
us in the previous sections as well for the Edwards and Spurgeon UIO discussed later in Section~\ref{sec:Ed_SpurUIO}.

The rank condition,
\[
\mb{rank}\, \b B_2=\mb{rank}\,\b C\b B_2
\]
is also referred to as the observer matching condition.\index{observer matching condition} This condition can be restrictive in some applications. To construct a sliding mode observer for systems that do not satisfy the observer matching condition, Kalsi et al.~\cite{kalsi10}  propose a method where auxiliary outputs\index{auxiliary outputs} are generated using high-gain approximate differentiators\index{high-gain!approximate differentiators} and then employ the auxiliary outputs in the design of a sliding mode observer.

\subsection{Reconstructing the Unknown Input}

We show here how the UIO given by~\eqref{eq:WZsm_observer} can be used to reconstruct the unknown input\index{unknown input reconstruction}\index{Reconstructing unknown input} $\b u_2$. Consider the manifold,
\begin{equation}\label{eqn:sigma_WZ}
\left\{\b e:\b\sigma (\b e)=\b F\b C\b e=\b 0\right\}.
\end{equation}
We can always choose $\eta$ such that the above manifold is attractive and it will be reached in a finite time.  Thus, after the estimation error trajectory, $\b e$, enters a sliding mode along the sliding surface, we have $\b\sigma (\b e)=\b 0$ and
\setlength{\arraycolsep}{0.0em}
\begin{equation}\label{eqn:WZmanifold_der}
\dot{\b{\sigma}}=\b F\b C\left(\b A-\b L\b C\right)\b e + \b F\b C\b B_2\b u_2 - \b F\b C\b B_2\b E(\b e,\eta) = \b 0.
\end{equation}
\setlength{\arraycolsep}{5pt}Because $\b e\to\b 0$ as $t\to\infty$, we obtain
\begin{equation}
\b F\b C\b B_2\b u_2 = \b F\b C\b B_2\b E(\b e,\eta).
\label{eq:tech_cond_sec}
\end{equation}
Recall that $\b F\b C=\b B_2^{\top}\b P$. Assuming that $\b B_2$ is of full column rank implies that the square matrix $\b B_2^{\top}\b P\b B_2$ is invertible. Hence, the steady-state condition~\eqref{eq:tech_cond_sec} can be re-written as
\[
\b B_2^{\top}\b P\b B_2\b u_2 = \b B_2^{\top}\b P\b B_2\b E(\b e,\rho).
\]
Therefore, the unknown input estimate is the output of the nonlinear term, $\b E(\b e,\rho)$, that is,
\begin{equation}
\fbox{$\displaystyle \hat{\b u}_2 = \b E(\b e,\rho) $}
\label{eq:WZu2recontruct}
\end{equation}
See Figure~\ref{fig:wzsm_obs} for an illustration of~\eqref{eq:WZu2recontruct}. For a rigorous proof of the above result, see~\cite{kalsi10}.

\subsection{Sliding Mode UIO Analysis and Construction}
\label{sec:slidingUIO}

We first present a lemma that will serve us as a design platform of the
sliding mode observer for uncertain systems. The lemma is a minor modification of
Lemma~1 of Corless and Tu~\cite{corless98} who proved it constructively using a
singular value decomposition\index{singular value decomposition} approach. We offer a different constructive proof using
the Q-R decomposition.\index{Q-R decomposition}
\begin{lemma}\label{lem_transform} For a triple $(\b A, \b B_2, \b C)
 \in \mb{\Bb R}^{n\times n}\times
\mb{\Bb R}^{n\times m_2}\times \mb{\Bb R}^{p\times n}$,
\begin{equation}
\mb{rank}\, \b B_2 = \mb{rank}\, (\b C\b B_2) = r, \label{eq:distribute_con}
\end{equation}
if and only if there exist nonsingular matrices $\b T$ and $\b S$ such that
\begin{equation}\label{eq:transAGC}
\b T\b A\b T^{-1}=\left[\begin{array}{cc} \b A_{11} & \b A_{12}\\
\b A_{21} & \b A_{22}
\end{array}\right],\quad \b T\b B_2=\left[\begin{array}{c}
\b B_{2_1}\\
\b O
\end{array}\right], \quad \b S\b C\b T^{-1}=\left[\begin{array}{cc} \b I_r & \b O\\
\b O & \b C_{22}
\end{array}\right],
\end{equation}
where $\b A_{11}\in \mb{\Bb R}^{r\times r}$, $\b A_{22}\in \mb{\Bb R}^{(n-r)\times
(n-r)}$, $\b B_{2_1} \in \mb{\Bb R}^{r\times m_2}$, $\mb{rank}\, \b B_{2_1} = r$,
and $\b C_{22}\in \mb{\Bb R}^{(p-r)\times (n-r)}$.
\end{lemma}
{\bf Proof}\quad ($\Longrightarrow$) The proof is constructive. Using the Q-R
decomposition applied to $\b B_2$, we obtain
\[
\b B_2=\b Q_{B_2}\b R_{B_2},
\]
where $\b Q_{B_2}\in \mb{\Bb R}^{n\times n}$ is an unitary matrix and the matrix $\b
R_{B_2} \in \mb{\Bb R}^{n\times m_2}$ is upper triangular where
\[
\mb{rank}\, \b R_{B_2} =  r.
\]
Let $\b T_1=\b Q_{B_2}^{-1}$. Then, we obtain
\[
\b T_1\b B_2=\left[\begin{array}{c} \tilde{\b B}_{2_1}\\
\b O
\end{array}\right],
\]
where $\tilde{\b G}_1 \in \mb{\Bb R}^{r\times m_2}$. We next partition the matrix
$\b C\b T_1^{-1}$ as follows,
\[
\b C\b T_1^{-1}=\left[\begin{array}{cc} \tilde{\b C}_1 & \tilde{\b C}_2
\end{array}\right],
\]
where $\tilde{\b C}_1 \in \mb{\Bb R}^{p\times r}$. Note that \[ \b C\b B_2 =
\left(\b C\b T_1^{-1} \right)\left(\b T_1\b B_2 \right)=\tilde{\b C}_1\tilde{\b
B}_{2_1}.
\]
By the lemma hypothesis, $\mb{rank}\, \b B_2 = \mb{rank}\, (\b C\b B_2) = r$. Hence
\[
\mb{rank}\, \tilde{\b C}_1=  r.
\]
Applying the Q-R decomposition\index{Q-R decomposition} to $\tilde{\b C}_1$ yields
\[
\tilde{\b C}_1=\b Q_{\tilde{C}_1}\b R_{\tilde{C}_1},
\]
where
\[
\b R_{\tilde{C}_1}=\left[\begin{array}{c} \b C_{11}\\
\b O
\end{array}\right]\quad\mb{and}\quad \det \b C_{11} \ne 0.
\]
Note that $\b C_{11} \in \mb{\Bb R}^{r\times r}$. Let $\b S=\b
Q_{\tilde{C}_1}^{-1}$, then,
\[
\b S\b C\b T_1^{-1}=\left[\begin{array}{cc}
\b C_{11} & \b C_{12}\\
\b O & \b C_{22}
\end{array}\right].
\]
Postmultiplying $\b S\b C\b T_1^{-1}$ by
\[
\b T_2^{-1}=\left[\begin{array}{cc}
\b C_{11}^{-1} & -\b C_{11}^{-1}\b C_{12}\\
\b O & \b I_{n-r}
\end{array}\right]
\]
gives
\[
\b S\b C\b T_1^{-1}\b T_2^{-1}=\left[\begin{array}{cc}
\b C_{11} & \b C_{12}\\
\b O & \b C_{22}
\end{array}\right]\left[\begin{array}{cc}
\b C_{11}^{-1} & -\b C_{11}^{-1}\b C_{12}\\
\b O & \b I_{n-r}
\end{array}\right]=\left[\begin{array}{cc}
\b I_r & \b O\\
\b O & \b C_{22}
\end{array}\right].
\]
We then have $\b T=\b T_2\b T_1$.

($\Longleftarrow$) By inspection.\\
\hspace*{6in}$\Box$

We next show that the systems
\[
\left.\begin{array}{lll}
\dot{\b x} &=& \b A\b x + \b B_2\b u_2\\
\b y &=& \b C\b x
\end{array}\right\}
\]
and
\[
\left.\begin{array}{lll}
\dot{\tilde{\b x}} &=& \b T\b A\b T^{-1}\tilde{\b x} + \b T\b B_2\b u_2\\
\tilde{\b y} &=& \b S\b C\b T^{-1}\b x
\end{array}\right\}
\]
have the same system zeros,\index{system zeros} that is, their system matrices\index{system matrix} have the same rank for
all $s\in \mb{\Bb C}$, where $\mb{\Bb C}$ is the set of complex numbers. To prove
the above statement, we note the following relation between the system matrices of
the above models:
\begin{equation}
\left[\begin{array}{cc}
s\b I_n - \b T\b A\b T^{-1} & \b T\b B_2\\
\b S\b C\b T^{-1} & \b O
\end{array}\right]=\left[\begin{array}{cc}
\b T & \b O\\
\b O & \b S
\end{array}\right]\left[\begin{array}{cc}
s\b I_n - \b A & \b B_2\\
\b C & \b O
\end{array}\right]\left[\begin{array}{cc}
\b T^{-1} & \b O\\
\b O & \b I_q
\end{array}\right].
\label{eq:invariant_z}
\end{equation}
Suppose now that we have two matrices, $\b X\in \mb{\Bb R}^{r\times s}$ and $\b Y\in
\mb{\Bb R}^{s\times t}$. Then, we have the following bounds on the rank of the
product of these matrices:
\[
\mb{rank}\,\b X + \mb{rank}\, \b Y - s \le \mb{rank}\, (\b X\b Y) \le \min (
\mb{rank}\, \b X, \mb{rank}\, \b Y ).
\]
The above are referred to as the Sylvester's inequalities---see,\index{Sylvester's inequalities} for example,
Gantmacher~\cite[pages 65, 66]{gantmacher59v1} for their proof or Section~\ref{sec:vector_spaces}  and Exercise~\ref{ex:sylvesters} in the Appendix. Applying the
Sylvester's inequalities to~(\ref{eq:invariant_z}), we conclude that
\begin{equation}
\mb{rank}\,\left[\begin{array}{cc}
s\b I_n - \b T\b A\b T^{-1} & \b T\b B_2\\
\b S\b C\b T^{-1} & \b O
\end{array}\right]=\mb{rank}\,\left[\begin{array}{cc}
s\b I_n - \b A & \b B_2\\
\b C & \b O
\end{array}\right]\;\;\mb{for all}\;\; s\in \mb{\Bb C}.
\label{eq:same_rank}
\end{equation}

\begin{lemma}\label{lem_detect}
Assume that $\mb{rank}\,\b B_2 = \mb{rank}\,(\b C\b B_2)=r$. Then, the pair $(\b
A_{22}, \b C_{22})$ is detectable if and only if
\begin{equation}
\mb{rank}\left[\begin{array}{cc}
s\b I_n - \b A & \b B_2\\
\b C & \b O
\end{array}\right]=n+r\;\;\mb{for all}\;\; s\;\;\mb{such that}\;\; \mb{Re}(s)
\ge 0. \label{eq:rank_condition}
\end{equation}
\end{lemma}
{\bf Proof}\quad By assumption, $\mb{rank}\,\b B_2 = \mb{rank}\,(\b C\b B_2)=r$. By
Lemma~\ref{lem_transform}, the above condition is equivalent to the existence of
nonsingular matrices $\b T$ and $\b S$ such that
\[
\b T\b A\b T^{-1}=\left[\begin{array}{cc} \b A_{11} & \b A_{12}\\
\b A_{21} & \b A_{22}
\end{array}\right],\quad \b T\b B_2=\left[\begin{array}{c}
\b B_{2_1}\\
\b O
\end{array}\right], \quad \b S\b C\b T^{-1}=\left[\begin{array}{cc} \b I_r & \b O\\
\b O & \b C_{22}
\end{array}\right],
\]
where $\b B_{2_1} \in \mb{\Bb R}^{r\times m_2}$ and $\mb{rank}\, \b B_{2_1} = r$.
Then, by~(\ref{eq:same_rank}) for any $s\in \mb{\Bb C}$,
\begin{eqnarray*}
\mb{rank}\left[\begin{array}{ccc}
s\b I_r - \b A_{11} &-\b A_{12} & \b B_{2_1}\\
-\b A_{21} & s\b I_{n-r} - \b A_{22} & \b O\\
\b I_r & \b O & \b O\\
\b O &\b C_{22} & \b O
\end{array}\right] &=& \mb{rank}\left[\begin{array}{cc}
-\b A_{12} & \b B_{2_1}\\
s\b I_{n-r} - \b A_{22} & \b O\\
\b C_{22} & \b O
\end{array}\right] + r\\
&=& \mb{rank}\left[\begin{array}{c}
s\b I_{n-r} - \b A_{22}\\
\b C_{22}
\end{array}\right]+2r.
\end{eqnarray*}
It follows from the above that the pair $(\b A_{22},\b C_{22})$ is detectable if and
only if the rank condition~(\ref{eq:rank_condition}) holds.\\
\hspace*{6in}$\Box$

Note that if $m_2=p=r$, then $\mb{rank}\left[\begin{array}{cc}
s\b I_n - \b A & \b B_2\\
\b C & \b O
\end{array}\right]=n+r$ for all $s$ such that $\mb{Re}(s)
\ge 0$ if and only if the matrix $\b A_{22}$ is asymptotically stable.


To proceed, we need the following technical lemma.
\begin{lemma}\label{lemma:same_kers}
For a given matrix $\b V \in \mb{\Bb R}^{n\times m}$,
\[
\ker(\b V)=\ker(\b V^{\top}\b V).
\]
\end{lemma}
%{\bf Proof}\quad If $\b x\in \ker(\b V)$, that is, if $\b x$ is such that $\b V\b
%x=\b 0$, then obviously $\b V^{\top}\b V\b x=\b 0$, that is, $\b x\in \ker(\b V^{\top}\b V)$.
%Thus, we showed that if $\b x\in \ker(\b V)$ then $\b x\in \ker(\b V^{\top}\b V)$.
%
%Suppose now that $\b x\in \ker(\b V^{\top}\b V)$, that is, $\b V^{\top}\b V\b x=\b 0$. Then,
%$\b y= \b V\b x \in \ker(\b V^{\top})$. Thus, $\b y\in \mb{range}\,(\b V)\cap\ker(\b
%V^{\top})$. However, it is well know from linear algebra that
%\[
%\mb{range}\,(\b V)\cap\ker(\b V^{\top})=\{\b 0\},
%\]
%that is, $\b y=\b V\b x=\b 0$ which means that $\b x\in \ker(\b V)$. Thus, we showed
%that if $\b x\in \ker(\b V^{\top}\b V)$ then $\b x\in \ker(\b V)$.
%
%In summary, $\ker(\b V)=\ker(\b V^{\top}\b V)$.\\
%\hspace*{6in}$\Box$
{\bf Proof}\quad See Exercise~\ref{ex:same_kers}.\\
\hspace*{5.5in} $\Box$

The following theorem appears in Corless and Tu~\cite[Lemma 3, page 760]{corless98}.
A related result was obtained by Edwards and Spurgeon~\cite[Proposition 6.2, page
138]{edwards98}. An algorithm for constructing matrices $\b L$, $\b F$, and $\b P$
that are essential ingredients of the sliding mode observer\index{sliding mode observer design!essential ingredients of} for the uncertain
systems is contained in the proof of the theorem.
\begin{thm}\label{thm:slidingUIO}
There exists a triple of matrices $(\b L, \b F, \b P) \in \mb{\Bb R}^{n\times
p}\times \mb{\Bb R}^{m_2\times p}\times \mb{\Bb R}^{n\times n}$ such that
\begin{equation}
(\b A-\b L\b C)^{\top}\b P + \b P(\b A-\b L\b C) < 0 \label{eq:pexistence}
\end{equation}
and
\begin{equation}
\b F\b C=\b B_2^{\top}\b P \label{eq:tech_condit}
\end{equation}
if and only if
\begin{itemize}
\item[{\bf (i)}] $\mb{rank}\, \b B_2=\mb{rank}\,(\b C\b B_2)=r$; \item[{\bf (ii)}]
the system zeros\index{system zeros} of the triple $(\b A,\b B_2, \b C)$ are in the open left-hand
complex plane, that is,
\[
\mb{rank}\left[\begin{array}{cc}
s\b I_n - \b A & \b B_2\\
\b C & \b O
\end{array}\right]=n+r\;\;\mb{for all}\;\; s\;\;\mb{such that}\;\; \mb{Re}(s)
\ge 0.
\]
\end{itemize}
\end{thm}
{\bf Proof}\quad $(\Longleftarrow)$ We follow the arguments of Corless and
Tu~\cite{corless98}. By Lemma~\ref{lem_transform},  the condition, $\mb{rank}\,\b
B_2 = \mb{rank}\,(\b C\b B_2)=r$, is equivalent to the existence of nonsingular
matrices $\b T$ and $\b S$ such that
\[
\hat{\b A}=\b T\b A\b T^{-1}=\left[\begin{array}{cc} \b A_{11} & \b A_{12}\\
\b A_{21} & \b A_{22}
\end{array}\right],\hat{\b B}_2=\b T\b {\b B}_2=\left[\begin{array}{c}
\b B_{2_1}\\
\b O
\end{array}\right], \quad \hat{\b C}=\b S\b C\b T^{-1}=\left[\begin{array}{cc} \b I_r & \b O\\
\b O & \b C_{22}
\end{array}\right],
\]
where $\b B_{2_1} \in \mb{\Bb R}^{r\times m_2}$ and $\mb{rank}\, \b B_{2_1} = r$.
Let
\begin{equation}\label{eq:PLFinnewcor}
\hat{\b P}=\b T^{-{\top}}\b P\b T^{-1},\;\;  \hat{\b L}=\b T\b L\b
S^{-1},\;\;\mb{and}\;\;\hat{\b F}=\b F\b S^{-1}.
\end{equation}
To proceed, note that condition {\bf (ii)} is equivalent to the existence of a
matrix $\b L_{22}$ such that the eigenvalues of $(\b A_{22} - \b L_{22}\b C_{22})$
are all in the open left-half complex plane. Then, for any symmetric positive
positive definite $\b Q_{22}$, the symmetric solution $\b P_{22}$ to the Lyapunov
matrix equation,\index{Lyapunov matrix equation}
\[
(\b A_{22} - \b L_{22}\b C_{22})^{\top}\b P_{22}+\b P_{22}(\b A_{22} - \b L_{22}\b
C_{22})=-\b Q_{22},
\]
is also positive definite. Let
\[
\hat{\b L}=\left[\begin{array}{cc}
\kappa\b I_r & \b O\\
\b O & \b L_{22}
\end{array}\right],
\]
where $\kappa >0$ is a design parameter whose lower bound is determined in the
following considerations. We have
\begin{eqnarray*}
\hat{\b A}-\hat{\b L}\hat{\b C} &=& \left[\begin{array}{cc} \b A_{11} & \b A_{12}\\
\b A_{21} & \b A_{22}
\end{array}\right]-\left[\begin{array}{cc}
\kappa\b I_r & \b O\\
\b O & \b L_{22}
\end{array}\right]\left[\begin{array}{cc}
\b I_r & \b O\\
\b O & \b C_{22}
\end{array}\right]\\
&=& \left[\begin{array}{cc} \b A_{11}-\kappa\b I_r & \b A_{12}\\
\b A_{21} & \b A_{22}-\b L_{22}\b C_{22}
\end{array}\right].
\end{eqnarray*}
Let
\[
\hat{\b P}=\left[\begin{array}{cc} \b I_r & \b O\\
\b O &\b P_{22}
\end{array}\right]\quad\mb{and}\quad\hat{\b Q}=\left[\begin{array}{cc} \b Q_{11} & \b Q_{12}\\
\b Q_{12}^{\top} & \b Q_{22}
\end{array}\right].
\]
Using the above, we obtain
\begin{eqnarray}
-\hat{\b Q} &=& (\hat{\b A}-\hat{\b L}\hat{\b C})^{\top}\hat{\b P} + \hat{\b P}(\hat{\b
A}-\hat{\b L}\hat{\b C})\nonumber\\
&=& \left[\begin{array}{cc} \b A_{11}^{\top}-\kappa\b I_r & \b A_{21}^{\top}\\
\b A_{12}^{\top} & (\b A_{22}-\b L_{22}\b C_{22})^{\top}
\end{array}\right]\left[\begin{array}{cc} \b I_r & \b O\\
\b O &\b P_{22}
\end{array}\right]\nonumber\\
&{}&\quad\quad+\left[\begin{array}{cc} \b I_r & \b O\\
\b O &\b P_{22}
\end{array}\right]\left[\begin{array}{cc} \b A_{11}-\kappa\b I_r & \b A_{12}\\
\b A_{21} & \b A_{22}-\b L_{22}\b C_{22}
\end{array}\right]\nonumber\\
&=& \left[\begin{array}{cc} \b A_{11}^{\top}-\kappa\b I_r & \b A_{21}^{\top}\b P_{22}\\
\b A_{12}^{\top} & (\b A_{22}-\b L_{22}\b C_{22})^{\top}\b P_{22}
\end{array}\right]+\left[\begin{array}{cc} \b A_{11}-\kappa\b I_r & \b A_{12}\\
\b P_{22}\b A_{21} & \b P_{22}(\b A_{22}-\b L_{22}\b C_{22})
\end{array}\right]\nonumber\\
&=& \left[\begin{array}{cc} \b A_{11}^{\top}+\b A_{11}-2\kappa\b I_r & \b A_{21}^{\top}\b
P_{22}
+ \b A_{12}\\
\b A_{12}^{\top} + \b P_{22}\b A_{21} & (\b A_{22}-\b L_{22}\b C_{22})^{\top}\b P_{22}+\b
P_{22}(\b A_{22}-\b L_{22}\b C_{22})
\end{array}\right]\nonumber\\
&=&\left[\begin{array}{cc} -\b Q_{11} & -\b Q_{12}\\
-\b Q_{12}^{\top} & -\b Q_{22} \end{array}\right], \label{eq:exp_forqhat}
\end{eqnarray}
where $\b Q_{22}=-\left((\b A_{22}-\b L_{22}\b C_{22})^{\top}\b P_{22}+\b P_{22}(\b
A_{22}-\b L_{22}\b C_{22}) \right)$ is positive definite by construction. Our goal
is to obtain a lower bound on the parameter $\kappa$ that would yield a positive
definite $\hat{\b Q}$. Using the Schur complement\index{Schur complement} of the positive definite $\b
Q_{22}$, we have that $\hat{\b Q}$ is positive definite if and only if
\[
\b Q_{11} > \b Q_{12}\b Q_{22}^{-1}\b Q_{12}^{\top}.
\]
Employing the above in~(\ref{eq:exp_forqhat}), we obtain
\[
2\kappa\b I_r -\left(\b A_{11}^{\top} + \b A_{11}\right) > \left(\b A_{21}^{\top}\b P_{22} +
\b A_{12}\right)\b Q^{-1}_{22}\left( \b A_{12}^{\top} + \b P_{22}\b A_{21}\right).
\]
Hence, $\hat{\b Q}$ is positive definite if
\[
\kappa >\frac{1}{2}\lambda_{\max}\left(\b A_{11}^{\top}+\b A_{11} + \left(\b A_{21}^{\top}\b
P_{22} + \b A_{12}\right)\b Q^{-1}_{22}\left( \b A_{12}^{\top} + \b P_{22}\b
A_{21}\right) \right).
\]
Let
\[
\hat{\b F}=\left[\begin{array}{cc} \b B_{2_1}^{\top} & \b O
\end{array}\right].
\]
Then, it is easy to see that
\begin{equation}\label{eq:layaptrans}
(\hat{\b A}-\hat{\b L}\hat{\b C})^{\top}\hat{\b P} + \hat{\b P}(\hat{\b A}-\hat{\b
L}\hat{\b C}) < 0,
\end{equation}
and
\begin{equation}\label{eq:techtransfor}
\hat{\b F}\hat{\b C} = \hat{\b B}_2^{\top}\hat{\b P},
\end{equation}
which are conditions (\ref{eq:pexistence}) and ({\ref{eq:tech_condit}) in the new
basis. Hence, the proof of the sufficiency conditions for the existence of the
desired triple of matrices $(\b L, \b F, \b P)$ is complete.

\noindent$(\Longrightarrow)$ We assume that there exists a triple of matrices $(\b
L, \b F, \b P)$ such that the conditions
\[
(\b A-\b L\b C)^{\top}\b P + \b P(\b A-\b L\b C) < 0\quad\mb{and}\quad \b F\b C = \b
B_2^{\top}\b P
\]
are satisfied. Because $\b P$ is symmetric positive definite, it can be represented
as
\[
\b P=\b P^{\frac{1}{2}}\b P^{\frac{1}{2}},
\]
where $\b P^{\frac{1}{2}}\in \mb{\Bb R}^{n \times n}$ is also symmetric positive
definite and hence nonsingular. We refer to $\b P^{\frac{1}{2}}$ as a square root of
$\b P$---see, for example, Luenberger~\cite[page 459]{luenberger84} or Section~\ref{sec:quadratform} in the Appendix. We then have
\[
\b B_2^{\top}\b P\b B_2=\b B_2^{\top}\b P^{\frac{1}{2}}\b P^{\frac{1}{2}}\b B_2=\left(\b
P^{\frac{1}{2}}\b B_2 \right)^{\top}\left( \b P^{\frac{1}{2}}\b B_2\right).
\]
By Lemma~\ref{lemma:same_kers} and the fact that $\b P^{\frac{1}{2}}$ is
nonsingular,
\[
\ker\left(\b B_2^{\top}\b P\b B_2\right)=\ker\left(\b P^{\frac{1}{2}}\b B_2
\right)=\ker(\b B_2).
\]
Therefore,
\[
\mb{rank}\, \left(\b B_2^{\top}\b P\b B_2\right)=\mb{rank}\,\b B_2.
\]
Using the above, the condition, $\b B_2^{\top}\b P=\b F\b C$, and the Sylvester's
inequality, we obtain
\[
\mb{rank}\,\b B_2=\mb{rank}\,\left(\b B_2^{\top}\b P\b B_2\right)=\mb{rank}\,\left( \b
F\b C\b B_2\right)\le \mb{rank}\,\left(\b C\b B_2\right)\le \mb{rank}\,\b B_2.
\]
Hence,
\[
\mb{rank}\,\left(\b C\b B_2\right)= \mb{rank}\,\b B_2.
\]

Our goal now is to show that conditions~(\ref{eq:pexistence})
and~(\ref{eq:tech_condit}) imply that the system zeros of the triple $(\b A,\b B_2,
\b C)$ are in the open left-hand complex plane. Because $\mb{rank}\,\left(\b C\b
B_2\right)= \mb{rank}\,\b B_2$, by Lemma~\ref{lem_transform}, we can construct
transformation matrices $\b T$ and $\b S$ so that the transformed matrices $\b A$,
$\b B_2$, and $\b C$ take the form given by~(\ref{eq:transAGC}). By
Lemma~\ref{lem_detect}, the system zeros of the triple $(\b A,\b B_2, \b C)$ are in
the open left-hand complex plane if and only the pair $(\b A_{22}, \b C_{22})$ is
detectable. Let matrices $\hat{\b P}$, $\hat{\b L}$, and $\hat{\b F}$ be defined
by~(\ref{eq:PLFinnewcor}), where
\[
\hat{\b P}=\left[\begin{array}{cc}
\b P_{11} & \b P_{12}\\
\b P_{21} & \b P_{22}
\end{array}\right],\quad \hat{\b L}=\left[\begin{array}{cc}
\b L_{11} & \b L_{12}\\
\b L_{21} & \b L_{22}
\end{array}\right],\quad\mb{and}\quad \hat{\b F}=\left[\begin{array}{cc}
\b F_1 & \b F_2 \end{array}\right].
\]
Because $\hat{\b P}$ is symmetric positive definite, the submatrix $\b P_{22}\in
\mb{\Bb R}^{(n-r)\times (n-r)}$ is symmetric positive definite and hence invertible.
We write down the corresponding 2-by-2 block equation of
inequality~(\ref{eq:layaptrans}),
\begin{equation}\label{eq:22eqn}
(\b A_{12} - \b L_{12}\b C_{22})^{\top}\b P_{12} + (\b A_{22} - \b L_{22}\b C_{22})^{\top}\b
P_{22} + \b P_{21}(\b A_{12} - \b L_{12}\b C_{22}) + \b P_{22}(\b A_{22} - \b L_{22}
\b C_{22})<0.
\end{equation}
Equation~(\ref{eq:techtransfor}) yields
\[
\b F_2\b C_{22}=\b B_{2_1}^{\top}\b P_{12},
\]
where $\b B_{2_1}$ has a full row rank. Therefore, $\b B_{2_1}\b B_{2_1}^{\top}\in
\mb{\Bb R}^{r\times r}$ is invertible and we have
\begin{equation}\label{eq:submatp12}
\b P_{12}=\left( \b B_{2_1}\b B_{2_1}^{\top}\right)^{-1}\b B_{2_1}\b F_2\b C_{22}.
\end{equation}
Note that
\[
\b P_{21}=\b P_{12}^{\top}=\b C_{22}^{\top}\b F_2^{\top}\b B_{2_1}^{\top}\left( \b B_{2_1}\b
B_{2_1}^{\top}\right)^{-1}.
\]
Substituting the above into~(\ref{eq:22eqn}), performing manipulations and denoting
\[
\tilde{\b L}_{22}= \b L_{22}-\b P_{22}^{-1}(\b A_{12}-\b L_{12}\b C_{22})^{\top}\left(\b
B_{2_1}\b B_{2_1}^{\top} \right)^{-1}\b B_{2_1}\b F_2,
\]
we obtain
\[
(\b A_{22}-\tilde{\b L}_{22}\b C_{22})^{\top}\b P_{22}+\b P_{22}(\b A_{22}-\tilde{\b
L}_{22}\b C_{22})<0.
\]
Because $\b P_{22}$ is symmetric and positive definite, the above implies that $(\b
A_{22}-\tilde{\b L}_{22}\b C_{22})$ is asymptotically stable, which in turn implies
that the pair $(\b A_{22}, \b C_{22})$ is detectable. The proof of the
theorem is now complete.\\
\hspace*{6in}$\Box$

We now summarize the above considerations in the form of the following design
algorithm.\index{UIO design!finding a triple of matrices}
\begin{algorithm}\label{slid_mo_UIO_des}\rm
\begin{center}
{\bf Sliding-Mode UIO Design}
\end{center}
Given a quadruple of matrices $(\b A, \b B_1, \b B_2, \b C)$ modeling the plant.
\begin{enumerate}
\item Check that the rank condition, $\mb{rank}\,\left(\b C\b B_2\right)
= \mb{rank}\,\b B_2$, is satisfied.\\
If $\mb{rank}\,\left(\b C\b B_2\right)\ne \mb{rank}\,\b B_2$, the sliding-mode
observer cannot be constructed, STOP.
\item Transform the triple $(\b A, \b C, \b B_2)$.\\
Use the method of the proof of Lemma~\ref{lem_transform} to construct nonsingular
matrices $\b T$ and $\b S$ and compute
\[
\b T\b A\b T^{-1}=\left[\begin{array}{cc} \b A_{11} & \b A_{12}\\
\b A_{21} & \b A_{22}
\end{array}\right],\;\; \b T\b B_2=\left[\begin{array}{c}
\b B_{2_1}\\
\b O
\end{array}\right],\;\;  \b S\b C\b T^{-1}=\left[\begin{array}{cc} \b I_r & \b O\\
\b O & \b C_{22}
\end{array}\right],
\]
where $\b B_{2_1} \in \mb{\Bb R}^{r\times m_2}$ and $\mb{rank}\, \b B_{2_1} = r$.
\item Check detectability of $(\b A_{22}, \b C_{22})$.\\
If the pair $(\b A_{22}, \b C_{22})$ is not detectable, the sliding-mode observer
cannot be constructed, STOP. \item Construct a matrix $\b L_{22}$ so that the
eigenvalues of $(\b A_{22}-\b L_{22}\b C_{22})$ are in the open left-half plane.
\item Choose a positive definite $\b Q_{22} \in \mb{\Bb R}^{(n-r)\times (n-r)}$ and
solve for positive definite $\b P_{22}$ the Lyapunov matrix equation,
\[
(\b A_{22} - \b L_{22}\b C_{22})^{\top}\b P_{22}+\b P_{22}(\b A_{22} - \b L_{22}\b
C_{22})=-\b Q_{22}.
\]
\item Choose $\kappa$ that satisfies the condition,
\[
\kappa >\frac{1}{2}\lambda_{\max}\left(\b A_{11}^{\top}+\b A_{11} + \left(\b A_{21}^{\top}\b
P_{22} + \b A_{12}\right)\b Q^{-1}_{22}\left( \b A_{12}^{\top} + \b P_{22}\b
A_{21}\right) \right).
\]
\item Construct,
\[
\hat{\b L}=\left[\begin{array}{cc}
\kappa\b I_r & \b O\\
\b O & \b L_{22}
\end{array}\right],\quad \hat{\b F}=\left[\begin{array}{cc} \b B_{2_1}^{\top} & \b O
\end{array}\right].
\]
\item Compute
\[
\b L=\b T^{-1}\hat{\b L}\b S,\quad \b F=\hat{\b F}\b S.
\]
\item Construct the observer,
\[
\dot{\hat{\b x}}=\b A \hat{\b x}+\b B_1\b u_1 + \b L(\b y - \hat{\b y})+\b B_2\b E
(\b y, \hat{\b y}, \eta),
\]
where
\[
\b E(\b y, \hat{\b y}, \eta) = \left\{\begin{array}{lcc}
\eta\frac{\b F(\b y - \hat{\b y})}{\|\b F(\b y - \hat{\b y})\|_2} & \mb{for} & \b F(\b y - \hat{\b y})\ne \b 0\\
\b 0  & \mb{for} & \b F(\b y - \hat{\b y}) =
\b 0.
\end{array}\right.
\]
Note that $\b C\b e = \b C(\b x - \hat{\b x})=\b y - \hat{\b y}$.
\end{enumerate}
\end{algorithm}\markex

\begin{example}\label{ex:smo_L-1011}\rm We consider the same fifth-order lateral axis model of an
L-1011 fixed-wing aircraft that we considered in Example~\ref{ex:eg2}.
We have $\mb{rank}\,\left(\b C\b B_2\right)= \mb{rank}\,\b B_2$. We then
use Lemma~\ref{lem_transform} to compute the transformation matrices $T$ and $S$,
\[
\b T=\left[ \begin{array}{rrrrr}
-0.0000 & -0.9106 & 0.4125 & 0.0245 & 0.9106\\
0.0000 & -0.4124 & -0.9110 & 0.0103 & 0.4124\\
-0.9998 & -0.0005 & 0.0000 & -0.0179 & 0\\
-0.0179 & 0.0265 & -0.0008 & 0.9995 & 0\\
0 & 0 & 0 & 0 & 1.0000
\end{array} \right]
\]
and
\[
\b S=\left[\begin {array}{rrrr}
-0.9106  &  0.4125  &  0.0245  & -0.0000\\
-0.4124 &  -0.9110  &  0.0103  &  0.0000\\
0.0265  & -0.0008  &  0.9996 &  -0.0000\\
0.0000  &  0.0000  & -0.0000 &   1.0000
\end{array}\right].
\]
We next transform the given model into the new coordinates and find the pair$(\b
A_{22}, \b C_{22})$, where
\[
\b A_{22}=\left[\begin {array}{rrr}
0.0006  &  0.0025  &  0.0178\\
-0.0368  &  -0.0993 &  -0.9971\\
-0.0002  &  0.0133 &  -0.0004
\end{array}\right]\;\;\;\mb{and}\;\;\;\b C_{22}=\left[\begin {array}{rrr}
-0.0179  &  0.9998  & -0.0265\\
-0.9998 &  -0.0179  & -0.0000
\end{array}\right].
\]
The pair $(\b A_{22}, \b C_{22})$ is detectable and we use MATLAB's \verb!place!
function to construct the gain matrix $\b L_{22}$ so that the eigenvalues of the
matrix $(\b A_{22}-\b L_{22}\b C_{22})$ are located at $-3, -4, -5$, where
\[
\b L_{22}=\left[\begin {array}{rr}
-0.1441 & -4.0022\\
7.5018  &  0.0744\\
-14.9883 &  -0.4586
\end{array}\right].
\]
We then solve the Lyapunov matrix equation,\index{Lyapunov matrix equation!constructing UIO}
\[
(\b A_{22} - \b L_{22}\b C_{22})^{\top}\b P_{22}+\b P_{22}(\b A_{22} - \b L_{22}\b
C_{22})=-\b I_3,
\]
to obtain
\[
\b P_{22}=\left[\begin {array}{rrr}
0.1273 &  -0.0445 &  -0.0236\\
-0.0445  &  0.9995  &  0.4735\\
-0.0236  &  0.4735  &  0.3059
\end{array}\right].
\]
After that we compute
\[
\frac{1}{2}\lambda_{\max}\left(\b A_{11}^{\top}+\b A_{11} + \left(\b A_{21}^{\top}\b P_{22} +
\b A_{12}\right)\b Q^{-1}_{22}\left( \b A_{12}^{\top} + \b P_{22}\b A_{21}\right)
\right)=12.6790
\]
and select
\[
\kappa=13.6790.
\]
Finally we construct
\[
\b L=\b T^{-1}\hat{\b L}\b S=\left[\begin {array}{rrrr}
0.0003 & -0.0000 & 0.0100 & 4.0002\\
13.2773  &  0.0115 & -15.1363  & -0.4544\\
0.0001 &  13.6790  &  0.0044  & -0.0001\\
-0.1532  &  0.0044  &  7.9048  &  0.1580\\
-0.3976 &   0.0114 & -14.9831  & -0.4586
\end{array}\right]
\]
and
\[
\b F=\hat{\b F}\b S=\left[\begin {array}{rrrr}
-0.7440  &  0.3370 &   0.0200 & -0.0000\\
-0.0320 & -1.1200  & -0.0000  &  0.0000
\end{array}\right].
\]
Let $\eta=7$. We then construct the UIO,
\[
\dot{\hat{\b x}}=\b A \hat{\b x} + \b L(\b y - \hat{\b y}) + \b B_2\tilde{\b E} (\b y, \hat{\b y},
\eta),
\]
where
\[
\tilde{\b E}( \b y, \hat{\b y}, \eta) = \left\{\begin{array}{lcc}
\eta\frac{\b F(\b y - \hat{\b y})}{\|\b F(\b y - \hat{\b y})\|+\mu} & \mb{for} & \b F(\b y - \hat{\b y})\ne \b 0\\
\b 0  & \mb{for} & \b F(\b y - \hat{\b y}) =
\b 0,
\end{array}\right.
\]
where $\mu=0.0005$. The parameter $\mu$ was introduced to smooth out the
discontinuity\index{smoothing out discontinuity}\index{discontinuity!smoothing out} to facilitate simulations. In Figure~\ref{fig:smo_L-1011}, we show
plots of the system state variables and their estimates versus time.
\begin{figure}
\centering\epsfig{file=smo_L-1011.eps,width=6.5in} \caption{Plots of $x_i$s and
their estimates versus time for Example~\ref{ex:smo_L-1011}.} \label{fig:smo_L-1011}
\end{figure}
The initial conditions of the plant were selected randomly to be equal to
\[
\b x(0)=\left[\begin{array}{ccccc}
 0.0342 & 0.3200 & 0.0178 & -0.0287 & -0.2497
\end{array}\right]^{\top}.
\]
The initial conditions of the observer were set to zero.
\end{example}
\markex

In the next example, we illustrate the use of the sliding mode UIO as a part of the combined controller-observer compensator\index{combined controller-observer compensator} and an unknown input estimator as shown in Figure~\ref{fig:smo_w_feedbk}.
\begin{figure}
  \centering
  \includegraphics[width=4.5in]{smo_w_feedbk.eps}
  \caption{Sliding mode observer~(\ref{eq:WZsm_observer}) as a part of the combined controller-observer compensator and an unknown input estimator.}
  \label{fig:smo_w_feedbk}
\end{figure}
\begin{example}\label{ex:smo_with_b1andb2}\rm We consider a plant with unknown input modeled by~\eqref{eq:two_inputs}, where
\begin{eqnarray*}
\b A &=&\left[\begin{array}{cccc}
0.6733 & -0.9068 & -0.3089 & 1.6703\\
1.2736 & -0.4747 & -0.8369 & -1.5417\\
-1.0398 & -0.6661 & -0.6283 & -0.2720\\
0.4880 & -0.5020 &  0.5009 & 0.3416
\end{array}\right],\quad\b B_1=\left[\begin{array}{cc}
0.4844 & 0.6678\\
0.9527 & 0.8982\\
0.1314 & -0.5493\\
-1.7419 & -0.3175
\end{array}\right],\\
\b B_2 &=&\left[\begin{array}{c}
-0.5827\\
-0.1642\\
-0.9351\\
1.1881
\end{array}\right],\quad \b C = \left[\begin{array}{cccc}
-1.2128 & 1.3227 & -0.3618 & 0.3871\\
-0.3812 & 0.3853 & -0.4264 & 0.9028
\end{array}\right].
\end{eqnarray*}
Let the unknown input be
\[
u_2=3\cos(5t).
\]
We use~\ref{slid_mo_UIO_des} to design the a sliding-mode UIO to reconstruct the unknown input $u_2$. Let the eigenvalues of $(\b A_{22}-\b L_{22}\b C_{22})$ be placed at $\{-3, -4, -5 \}$. Using MATLAB's function \verb!place! we obtain
\[
\b L_{22}=\left[\begin{array}{c}
-37.5312\\
31.9351\\
-13.5910
\end{array}\right].
\]
We select $\kappa=3.7417$.

The desired closed-loop poles of the system are to be located at $\{-4, -5, -6, -7\}$. Using MATLAB's function \verb!place! we obtain
\[
\b K=\left[\begin{array}{cccc}
-3.6021 & 8.0249 &  8.1459 & -2.3902\\
-9.3810 & -9.7238 & -38.9690 & -13.7649
\end{array}\right]
\]
such that $\mb{eig}(\b A-\b K\b B_1)=\{-4, -5, -6, -7\}$.

In our simulations, the vector of the initial conditions of the plant is
\[
\b x(0)=\left[\begin{array}{cccc}
0.6470 & -3.4975 & -3.4439 & 0.3146
\end{array}\right]^{\top}
\]
and we set zero initial conditions for the observer. In Figure~\ref{fig:state_esti_v_t}, we show plots of the plant's states and their estimates.
\begin{figure}
  \centering
  \includegraphics[width=4.5in]{state_esti_v_t.eps}
  \caption{Plots of the states and their estimates.}
  \label{fig:state_esti_v_t}
\end{figure}
In Figure~\ref{fig:u2_esti_v_t}, we show an estimate of the unknown\index{estimate of unknown input} input $u_2$,\ which is the output of the nonlinear element $E$.
\begin{figure}
  \centering
  \includegraphics[width=3.5in]{u2_esti_v_t.eps}
  \caption{Plots of the unknown input and its estimate.}
  \label{fig:u2_esti_v_t}
\end{figure}
\end{example}\markex


\subsection{Sliding Mode UIO Design Using LMIs}\label{sec:lmis_for_uios}

%Finding a triple $(\b L,\b F, \b P)$ such that \eqref{eq:lyap_esti} and \eqref{eq:tech_cond} are satisfied for some symmetric positive definite $\b P$ and $\b Q$ can be accomplished using LMIs.

We can use LMIs to design a sliding mode UIO.\index{sliding mode UIO design!using LMIs}\index{LMIs!to design sliding mode UIO} Xiang, Su, and Chu~\cite{xiang05} represent the problem of finding a triple $(\b L,\b F, \b P)$ subject to \eqref{eq:lyap_esti} and \eqref{eq:tech_cond} in the form of an LMI feasibility problem.\index{LMI feasibility problem}\index{UIO design!as LMI feasibility problem}\index{UIO design!finding a triple of matrices}

To proceed, let $\b G$ be the orthogonal complement\index{orthogonal complement} of $\b B_2$, which means that $\b B_2\b G=\b O$, and we write
\[
\b G=\b B_2^{\perp},
\]
where we assume that the matrix $\mb{rank}\,\b B_2 = m_2$, that is, the matrix $\b B_2$ has full column rank. Then the problem of finding a triple $(\b L,\b F, \b P)$, $\b P=\b P^{\top} > 0$ such that
\begin{equation}
(\b A-\b L\b C)^{\top}\b P + \b P(\b A-\b L\b C)< 0\quad\mb{and}\quad \b B_2^{\top}\b P = \b F\b C
\label{eq:spr_as_ineq}
\end{equation}
%\eqref{eq:lyap_esti} and \eqref{eq:tech_cond} are satisfied for some symmetric positive definite $\b P$ and $\b Q$
is equivalent to finding matrices
\begin{eqnarray*}
\b W_1 &=& \b W_1^{\top}\in\mb{\Bb R}^{(n-m_2)\times (n-m_2)},\\
\b W_2 &=& \b W_2^{\top}\in\mb{\Bb R}^{p\times p},\\
\b Y &\in & \mb{\Bb R}^{n\times p},
\end{eqnarray*}
such that
\begin{equation}
\b G\b W_1\b G^{\top} + \b C^{\top}\b W_2\b C > 0
\label{eq:first_lmi_for_smo}
\end{equation}
and
\begin{equation}
\b G\b W_1 \b G^{\top}\b A + \b C^{\top}\b W_2\b C \b A - \b Y\b C + \left(\b G\b W_1 \b G^{\top}\b A + \b C^{\top}\b W_2\b C \b A - \b Y\b C  \right)^{\top} < 0.
\label{eq:second_lmi_for_smo}
\end{equation}

If a feasible solution to~\eqref{eq:first_lmi_for_smo} and \eqref{eq:second_lmi_for_smo} exists, then
\begin{eqnarray}
\b P &=& \b G\b W_1\b G^{\top} + \b C^{\top}\b W_2\b C, \label{eq:xiangetalP}\\
\b L &=& \b P^{-1}\b Y, \label{eq:xiangetalL}\\
\b F &=& \b G^{\top}\b C^{\top}\b W_2. \label{eq:xiangetalF}
\end{eqnarray}
\begin{example}\rm We use LMIs to find a triple $(\b L,\b F, \b P)$ satisfying~\eqref{eq:spr_as_ineq} for the same system as in Example~\ref{ex:smo_with_b1andb2}. Recall that
\begin{eqnarray*}
\b A &=&\left[\begin{array}{cccc}
0.6733 & -0.9068 & -0.3089 & 1.6703\\
1.2736 & -0.4747 & -0.8369 & -1.5417\\
-1.0398 & -0.6661 & -0.6283 & -0.2720\\
0.4880 & -0.5020 &  0.5009 & 0.3416
\end{array}\right],\quad\b B_2=\left[\begin{array}{c}
-0.5827\\
-0.1642\\
-0.9351\\
1.1881
\end{array}\right],\\
\b C &=& \left[\begin{array}{cccc}
-1.2128 & 1.3227 & -0.3618 & 0.3871\\
-0.3812 & 0.3853 & -0.4264 & 0.9028
\end{array}\right].
\end{eqnarray*}
We mention that the above matrices were generated using MATLAB's function \verb!randn!.\index{MATLAB's function!randn}

We use MATLAB's LMI toolbox to solve the LMIs given by~\eqref{eq:second_lmi_for_smo}. Then, using~\eqref{eq:xiangetalP}, \eqref{eq:xiangetalL}, and \eqref{eq:xiangetalF}, we compute the matrices
\begin{eqnarray*}
\b P &=& \left[\begin{array}{cccc}
178.2434 & -54.7774  & 11.9292 &  66.0388\\
-54.7774 &  67.3161  & 16.2845  & 16.3716\\
11.9292 &  16.2845 &  61.8596 &  6.8944\\
66.0388 &  16.3716 & 6.8944 & 154.8361
\end{array}\right], \quad \b L= \left[\begin{array}{cc}
 -80.7624 & -30.7668\\
 79.1421 & -578.7866\\
 -194.5307 & 302.2489\\
 174.7920  &  2.1652
\end{array}\right],\\
\b F &=&\left[\begin{array}{cc}
-28.5969 & 163.2860
\end{array}\right].
\end{eqnarray*}
\end{example}\markex



\section{The Edwards and Spurgeon Sliding Mode UIO}\label{sec:Ed_SpurUIO}

In this Section, we present an alternative approach to the sliding mode UIO design proposed by Edwards and Spurgeon~\cite{edwards98}.\index{Edwards and Spurgeon UIO}\index{UIO of Edwards and Spurgeon} The construction of this observer involves a number of state variable transformations that bring the plant model into a special canonical form that is then used to construct the observer. The inverse transformations are performed to obtain the final form of the observer. The necessary and sufficient conditions for the existence of the Edwards and Spurgeon observer are the same as for the previously discussed UIO architectures.

\subsection{The Canonical Form---The Design Platform}

We begin by assuming that
\[
\mb{rank}\, \b B_2 = m_2\quad\mb{and}\quad \mb{rank}\, \b C = p.
\]
We then construct
\[
\b T_c=\left[\begin{array}{c}
\b N_c^{\top}\\
\b C
\end{array}\right],
\]
where $\b N_c \in \mb{\Bb R}^{n\times (n-p)}$ is the matrix whose columns span the null space of $\b C$. The matrix $\b T_c$ is invertible.



We use $\b T_c$ to transform the triple $(\b A,\b B_2,\b C)$ into the form,
    \setlength{\arraycolsep}{0.0em}
    \begin{eqnarray*}
%    \b A_c&{}={}&\setlength{\arraycolsep}{5pt}\b T_c\b A\b T_c^{-1}
%    {}={}\left[\begin{array}{cc}
%            \b A_{11} & \b A_{12} \\
%            \b A_{21} & \b A_{22}
%           \end{array}\right], \\
\b A_c&{}={}&\setlength{\arraycolsep}{5pt}\b T_c\b A\b T_c^{-1},\quad
    \b B_{2_c}=\setlength{\arraycolsep}{5pt}\b T_c\b B_2{}={}\left[\begin{array}{c}
            \b B_{2_{c_1}} \\
            \b B_{2_{c_2}}
           \end{array}\right], \\
    \b C_c&{}={}&\setlength{\arraycolsep}{5pt}\b C\b T_c^{-1}
    {}={}\left[\begin{array}{cc}
              \b O_{p\times (n-p)} & \b I_p \\
              \end{array}\right],
    \end{eqnarray*}
where $\b B_{2_{c_1}}\in\mb{\Bb R}^{(n-p)\times m_2}$ and $\b B_{2_{c_2}}\in\mb{\Bb R}^{p\times m_2}$.


The next state coordinate transformation, $\b T_b$, contains two submatrices that require their careful description. First note that by assumption
\[
\mb{rank}\, \b B_2 = \mb{rank} (\b C\b B_2) = m_2
\]
and
\[
\b C\b B_2 =  \b C_c\b B_{2_c} = \b B_{2_{c_2}}.
\]
Hence
\[
\mb{rank}\, \b B_{2_{c_2}} = m_2,
\]
where recall $\b B_{2_{c_2}}\in\mb{\Bb R}^{p\times m_2}$ and $p\ge m_2$. Therefore
\[
\b B_{2_{c_2}}^{\dagger}=\left(\b B_{2_{c_2}}^{\top} \b B_{2_{c_2}} \right)^{-1}\b B_{2_{c_2}}^{\top}
\]
is well defined. Next, there exists an orthogonal matrix\index{orthogonal matrix} $\b T_{b_2}\in \mb{\Bb R}^{p\times p}$ such that
\[
\b T_{b_2}\b B_{2_{c_2}} = \left[
                             \begin{array}{c}
                               \b O \\
                               \b B_{2_{b_2}}
                             \end{array}
                           \right],
\]
where $\b B_{2b}\in\mb{\Bb R}^{m_2\times m_2}$ and $\mb{rank}\b B_{2b}  = m_2$. We use $\b B_{2_{c_2}}^{\dagger}$ and $\b T_{b_2}$ to construct the coordinate transformation\index{coordinate transformation}
\[
\b T_b=\left[
           \begin{array}{cc}
             \b I_{n-p} & -\b B_{2_{c_1}}\b B_{2_{c_2}}^{\dagger} \\
             \b O_{p\times (n-p)} & \b T_{b_2}  \\
           \end{array}
         \right].
\]
Applying the transformation $\b T_b$ to the triple $(\b A_c,\b B_{2c},\b C_c)$ gives
\setlength{\arraycolsep}{0.0em}
\[
\b B_{2_b} = \b T_b\b B_{2c},\quad \b C_b = \b C_c\b T_b^{-1}
\]
\setlength{\arraycolsep}{5pt}and
\[
\b A_b = \b T_b\b A_c\b T_b^{-1} = \left[
                                     \begin{array}{cc}
                                       \b A_{b11} & \b A_{b12} \\
                                       \b A_{b21} & \b A_{b22}
                                     \end{array}
                                   \right],
\]
where $\b A_{b11}\in\mb{\Bb R}^{(n-m_2)\times (n-m_2)}$ and it has the form
\[
\b A_{b11}=\left[\begin{array}{cc}
\b A_{1111} & \b A_{1112}\\
\b A_{1121} & \b A_{1122}
\end{array}\right],
\]
where $\b A_{1111} \in \mb{\Bb R}^{(n-p)\times (n-p)}$ and $\b A_{1121} \in \mb{\Bb R}^{(p-m_2)\times (n-p)}$. If the pair $(\b A_{1111}, \b A_{1121})$ is not observable, we construct a transformation $\b T_{obs} \in \mb{\Bb R}^{(n-p)\times (n-p)}$ such that
\[
\b T_{obs}\b A_{1111}\b T_{obs}^{-1}=\left[\begin{array}{cc}
\b A_{11}^o & \b A_{12}^o\\
\b O & \b A_{22}^o
\end{array}\right]\quad\mb{and}\quad \b A_{1121}\b T_{obs}^{-1}=\left[\begin{array}{cc}
\b O & \b A_{21}^o
\end{array}\right],
\]
where $\b A_{11}^o \in \mb{\Bb R}^{r\times r}$, $\b A_{21}^o \in \mb{\Bb R}^{(p-m_2)\times (n-p-r)}$, $\b A_{22}^o \in \mb{\Bb R}^{(n-p-r)\times (n-p-r)}$, the pair $(\b A_{22}^o, \b A_{21}^o)$ is observable, and $r$ is number of unobservable states of the pair $(\b A_{1111}, \b A_{1121})$. We embed the transformation $\b T_{obs}$ into
\[
\b T_a=\left[\begin{array}{cc}
\b T_{obs} & \b O\\
\b O & \b I_p
\end{array}\right].
\]
Applying $\b T_a$ to the triple $(\b A_b,\b B_{2_b},\b C_b)$ yields
\[
\b A_a=\b T_a\b A_b\b T_a^{-1}=\left[
                                     \begin{array}{cc}
                                       \b A_{a11} & \b A_{a12} \\
                                       \b A_{a21} & \b A_{a22}
                                     \end{array}
                                   \right],
\]
where
\[
\b A_{a11}=\left[\begin{array}{cc|c}
\b A_{11}^o & \b A_{12}^o & {}\\
{}& {}& \b A^{m}_{12}\\
\b O & \b A_{22}^o & {}\\
\hline
\b O & \b A_{21}^o & \b A^{m}_{22}
\end{array}\right]\in \mb{\Bb R}^{(n-m_2)\times (n-m_2)},
\]
\[
\b B_{2_a}=\b T_a\b B_{2_b}=\left[\begin{array}{c}
\b O\\
\b B_{2_{b_2}}
\end{array}\right],\quad \b B_{2_{b_2}}\in \mb{\Bb R}^{m_2\times m_2},
\]
and
\[
\b C_a=\b C_b\b T_a^{-1}=\left[\begin{array}{cc}
\b O & \b T_{b_2}^{\top}
\end{array}\right],\quad \b T_{b_2}\in \mb{\Bb R}^{p\times p}.
\]

\begin{remark}
The eigenvalues of $\b A_{11}^o$ are the system zeros of the system represented by the triple $(\b A_a, \b B_{2_a}, \b C_a)$. Because system zeros are invariant under similarity transformations, the eigenvalues of $\b A_{11}^o$ are the system zeros of the system represented by the triple $(\b A, \b B_{2}, \b C)$.
\end{remark}\markdefn


By assumption, the system zeros are in the open left-half complex plane. The pair $(\b A_{22}^o, \b A_{21}^o)$ is observable; therefore we can construct a gain matrix $\b L_a\in \mb{\Bb R}^{(n-p-r)\times (p-m_2)}$ such that the matrix
\[
\b A_{22}^o-\b L_a\b A_{21}^o
\]
has its eigenvalues in pre-specified locations in the left-half plane. Let
\[
\b L=\left[\begin{array}{c}
\b O_{r\times (p-m_2)}\\
\b L_a
\end{array}\right]
\]
Then because the eigenvalues of the matrix $\b A_{11}^o$ are asymptotically stable, the matrix
\[
{\cal A}_{11}=\left[\begin{array}{cc}
\b A_{11}^o & \b A_{12}\\
\b O & \b A_{22}^o
\end{array}\right]-\left[\begin{array}{c}
\b O_{r\times (p-m_2)}\\
\b L_a
\end{array}\right]\left[\begin{array}{cc}
\b O_{(p-m_2)\times r} & \b A_{21}^o
\end{array}\right]
\]
has its all eigenvalues in the left-half plane, where ${\cal A}_{11}\in \mb{\Bb R}^{(n-p)\times (n-p)}$.


Let
\begin{equation*}
\bar{\b L} = \left[
               \begin{array}{cc}
                 \b L & \b O_{(n-p)\times m_2} \\
               \end{array}
             \right].
\end{equation*}
We construct the transformation
\begin{equation*}
\b T_L = \left[
           \begin{array}{cc}
             \b I_{n-p} & -\bar{\b L} \\
             \b O_{p\times (n-p)} & \b T_{b2} \\
           \end{array}
         \right]
\end{equation*}
and apply it to the triple $(\b A_a,\b B_{2_a},\b C_a)$ to obtain
\[
{\cal B}_{2} = \b T_L\b B_{2_a}=\left[\begin{array}{c}
\b O\\
{\cal B}_{2_2}
\end{array}\right],\quad
{\cal C} = \b C_a\b T_L^{-1}=\left[\begin{array}{cc}
\b O & \b I_p
\end{array}\right],
\]
and
\[
{\cal A} = \b T_L\b A_a\b T_L^{-1} =\left[
                                    \begin{array}{cc}
                                      {\cal A}_{11} & {\cal A}_{12}  \\
                                      {\cal A}_{21} & {\cal A}_{22}\\
                                    \end{array}
                                  \right],
\]
where ${\cal A}_{11}\in\mb{\Bb R}^{(n-p)\times (n-p)}, {\cal A}_{12}\in\mb{\Bb R}^{(n-p)\times p}, {\cal A}_{21}\in\mb{\Bb R}^{p\times (n-p)}$ and ${\cal A}_{22}\in\mb{\Bb R}^{p\times p}$. The matrix ${\cal A}_{11}$ has asymptotically stable eigenvalues.

The plant model represented by the quadruple $(\b A, \b B_1, \b B_2, \b C)$ takes the following form in  the new coordinates
\begin{equation}
\left.\begin{array}{lcl}
\dot{\b x}_1 &=& {\cal A}_{11}\b x_1 + {\cal A}_{12}\b y + {\cal B}_{1_1}\b u_1\\
\dot{\b y} &=& {\cal A}_{21}\b x_1 + {\cal A}_{22}\b y + {\cal B}_{1_2}\b u_1 + {\cal B}_{2_2}\b u_2
\end{array}\right\}
\label{eq:ed_spur_canon}
\end{equation}
We refer to the above representation as the {\em Edwards-Spurgeon canonical form}.\index{Edwards-Spurgeon canonical form}


\subsection{Observer Architecture}
We are now ready to present the Edwards-Spurgeon UIO.\index{Edwards-Spurgeon UIO} Let $\hat{\b x}_1$ denote an estimate of $\b x_1$, and let $\hat{\b y}$ denote an estimate of $\b y$. We also let
\[
\b e_y=\b y-\hat{\b y}
\]
to denote the estimation error of $\b y$. Edwards and Spurgeon~\cite[p.~134]{edwards98} propose their UIO for the plant model in the canonical form~\eqref{eq:ed_spur_canon},
\begin{equation}
\left.\begin{array}{lcl}
\dot{\hat{\b x}}_1 &=& {\cal A}_{11}\hat{\b x}_1 + {\cal A}_{12}\hat{\b y} + {\cal B}_{1_1}\b u_1 + {\cal A}_{12}\b e_y\\
\dot{\hat{\b y}} &=& {\cal A}_{21}\hat{\b x}_1 + {\cal A}_{22}\hat{\b y} + {\cal B}_{1_2}\b u_1 + \left({\cal A}_{22} - {\cal A}_{22}^s\right)\b e_y + \b{\nu},
\end{array}\right\}
\label{eq:ed_spur_canon_UIO}
\end{equation}
where ${\cal A}_{22}^s$ is a stable design matrix. The discontinuous injection term $\b{\nu}$ is
\[
\b{\nu}=\left\{\begin{array}{lcl}
\rho\|{\cal B}_{2_2}\|\frac{\b P_2\b e_y}{\|\b P_2\b e_y\|} &{}& \mb{if }  \b e_y\ne \b 0\\
\b 0 &{}& \mb{otherwise}
\end{array} \right.
\]
where $\b P_2=\b P_2^{\top}>0$ is the solution to the matrix Lyapunov equation
\begin{equation}
{\cal A}_{22}^{s{\top}}\b P_2 + \b P_2{\cal A}_{22}^s = -\b Q_2
\label{eq:sol_redu_layap}
\end{equation}
for some symmetric positive definite $\b Q_2$, and the gain $\rho \ge \|\b u_2\|$.

Let $\b e_1=\b x_1-\hat{\b x}_1$. The dynamics of the observation error are
\begin{eqnarray*}
\dot{\b e}_1 &=& {\cal A}_{11}\hat{\b e}_1\\
\dot{\b e}_y &=& {\cal A}_{21}\hat{\b e}_1 + {\cal A}_{22}^s\hat{\b e}_y + {\cal B}_{2_2}\b u_2 -\b{\nu}.
\end{eqnarray*}
One can show using the Lyapunov second method that the above error system is asymptotically stable.

Let the linear term coefficient\index{linear term coefficient!in Edwards-Spurgeon UIO} $\b G_l$ be defined as
\begin{equation*}
\b G_l = \b T_c^{-1}\b T_b^{-1}\b T_a^{-1}\b T_L^{-1}\left[
               \begin{array}{c}
                 {\cal A}_{12} \\
                 {\cal A}_{22} - {\cal A}_{22}^s  \\
               \end{array}
             \right],
\end{equation*}
and the nonlinear term coefficient\index{nonlinear term coefficient!in Edwards-Spurgeon UIO} ${\cal G}_n$ be defined as
\begin{eqnarray*}
\b G_n = \b T_c^{-1}\b T_b^{-1}\b T_a^{-1}\b T_L^{-1}\left[
              \begin{array}{c}
                \b O_{(n-p)\times p} \\
                \b I_p \\
              \end{array}
            \right].
\end{eqnarray*}
Recall that
\[
\b e_y=\b y-\hat{\b y}=\b C\b x - \b C\hat{\b x}=\b C\b e.
\]
Then the Edwards-Spurgeon UIO in the original coordinates takes the form
\begin{equation}
\fbox{$\displaystyle \dot{\hat{\b x}}=\b A\hat{\b {x}}+\b B_1\b u_1 + \b G_l\b e_y + \b G_n\b{\nu} $}
\label{eqn:smo_edsp_or_coor}
\end{equation}

\subsection{Reconstructing the Unknown Input}

We show here how the Edwards-Spurgeon observer~\eqref{eqn:smo_edsp_or_coor} can be used to reconstruct the unknown input $\b u_2$.\index{reconstructing unknown input!by Edwards-Spurgeon UIO} Consider the manifold,
\begin{equation}\label{eqn:sigma_edsp}
\left\{\b e:\b\sigma (\b e)=\b C\b e=\b 0\right\}.
\end{equation}
We can always choose $\rho$ such that the above manifold is attractive and it will be reached in a finite time.  Thus, after the estimation error trajectory, $\b e$, enters a sliding mode along the sliding surface, we have $\b\sigma (\b e)=\b 0$, and
\setlength{\arraycolsep}{0.0em}
\begin{equation}\label{eqn:manifold_der}
\dot{\b\sigma}=\b C\left(\b A-\b G_l\b C\right)\b e+\b C\b B_2\b u_2-\b C\b G_n\b{\nu}(\b C\b e)=\b 0.
\end{equation}
\setlength{\arraycolsep}{5pt}Because $\b e\to\b 0$ as $t\to\infty$, we obtain
\[
\b C\b B_2\b u_2 = \b C\b G_n\b{\nu} .
\]
From the above, we obtain a least-squares estimate of the unknown input\index{least-squares estimate of unknown input!by Edwards-Spurgeon UIO} of the form
\begin{equation}
\fbox{$\displaystyle \hat{\b u}_2 = (\b C\b B_2)^{\dag}\b C\b G_n\b{\nu} $}
\label{eq:ESu2recontruct}
\end{equation}


As in the previous case, the Edwards-Spurgeon sliding mode UIO can also be used in the combined controller-observer compensator\index{combined controller-observer compensator} architecture and as unknown input estimator as depicted in Figure~\ref{fig:smo_ES_w_feedbk}.
\begin{figure}
  \centering
  \includegraphics[width=4.5in]{smo_ES_w_feedb.eps}
  \caption{The Edwards-Spurgeon sliding mode UIO given by~\eqref{eqn:smo_edsp_or_coor} as a part of the combined observer-controller compensator and an unknown input estimator.}
  \label{fig:smo_ES_w_feedbk}
\end{figure}

\begin{example}\label{ex:ES_simu_illustr}\rm We construct the Edwards-Spurgeon sliding UIO given by~\eqref{eqn:smo_edsp_or_coor} for the plant in Example~\ref{ex:smo_with_b1andb2} and use it in the architecture depicted in Figure~\ref{fig:smo_ES_w_feedbk}.

We have $m_2=1$. We start with computing $\b T_c$, where we employ the command, \verb!Nc=null(C)!, to find $\b N_c$. We use
\[
\rho \|{\cal B}_{2_2}\| = 10\|\left[\begin{array}{cc}
1.2877 & 1.6302
\end{array}\right]^{\top}\|.
\]
The gain matrix $\b L$ was selected using the \verb!place! command so that the eigenvalues of ${\cal A}_{11}$ are placed at $\{-5, -6\}$. The design parameter matrix, $\b A_{22}^s$, was selected as
\[
\b A_{22}^s=\left[\begin{array}{cc}
-6& 0\\
0 & -7
\end{array}\right].
\]
For $\b Q_2=2\b I_2$, the solution to~\eqref{eq:sol_redu_layap} is
\[
\b P_2=\left[\begin{array}{cc}
0.1667 &  0\\
0  &  0.1429
\end{array}\right].
\]

The nonlinear discontinuous term $\b{\nu}$ was approximated\index{approximating discontinuous term}\index{discontinuous term!approximation} with
\[
\tilde{\b{\nu}}=\rho\|{\cal B}_{2_2}\|\frac{\b P_2\b e_y}{\|\b P_2\b e_y\| + \mu},
\]
where $\mu=0.01$.

The feedback matrix $\b K$ was computed using the \verb!place! function so that the eigenvalues of $(\b A-\b B_1\b K)$ are located at $\{-4,  -5,  -6,  -7\}$.

In Figure~\ref{fig:ES_state_estimation}, we show plots of the plant's states and their estimates.
\begin{figure}
  \centering
  \includegraphics[width=4.5in]{ES_state_estiamtion.eps}
  \caption{Plots of states and their estimates using the Edwards-Spurgeon UIO.}
  \label{fig:ES_state_estimation}
\end{figure}
In Figure~\ref{fig:ES_u2_estimation}, we show plots of the unknown input, $u_2=3\cos(5t)$, and its estimate.
\begin{figure}
  \centering
  \includegraphics[width=3.75in]{ES_u2_estiamtion.eps}
  \caption{Plots of the unknown input and its estimate using the Edwards-Spurgeon UIO.}
  \label{fig:ES_u2_estimation}
\end{figure}
\end{example}\markex

\section{Notes}
In his practical guide for the selection and installation of observers in control
systems, Ellis~\cite[page~3]{ellis02} writes: ``Observers add complexity to the
system and require computational resources. They may be less robust than physical
sensors,\index{physical sensors} especially when plant parameters change substantially during operation.
Still, an observer applied with skill can bring substantial performance benefits and
do so, in many cases, while reducing cost or increasing reliability.'' Examples of
impressive applications of nonlinear observers to the control of electric machinery\index{observers!to control electric machinery}
can be found in~\cite{dawson98, utkin99,solsona03}. R\"{o}benack
and Lynch~\cite{robenack04} presented a method to the observer design for a class of
nonlinear plants that yields almost linear observation error dynamics.\index{almost linear observation error dynamics} An application of the sliding mode observer presented in Subsection~\ref{sec:uncertain} to decentralized dynamic output feedback control of nonlinear interconnected systems is reported in~\cite{kalsi10b}.

Sliding-mode observers for dynamical systems with unknown inputs have found applications in robust detection and reconstruction of actuator and sensor faults~\cite{tan03,yan07,tan08,yan08,yan08IJC,alwi08,alwi09,tan10}\index{reconstructing!actuator and sensor faults} and fault detection and isolation~(FDI)~\cite{saif03}.\index{fault detection and isolation (FDI)}\index{FDI|see{fault detection and isolation}} In particular, the purpose of the FDI scheme is to generate an alarm when a fault, such as a component malfunction, develops in the process being monitored and to identify the location of the fault.
The fault signal can often be modeled as an unknown input\index{fault model!as an unknown input} to the plant. Thus, unknown input observers can be used for FDI purposes. For example, Edwards and Spurgeon (see \cite[Section~6.5]{edwards98}, Edwards, Spurgeon, and Patton~\cite{edwards00}), and Saif and Xiong~\cite{saif03} use sliding mode observers to develop FDI schemes for reconstructing the actuator and sensor fault signals.


\section{Exercises}
\begin{exercise}\rm Design and test the full-order UIO\index{full-order UIO} described in Theorem~\ref{thm:UIO_full_order}  for the system model of Example~\ref{ex:eg2}.
\end{exercise}

\begin{exercise}\rm Design and test the UIO of Hou and M\"{u}ller,\index{Hou and M\"{u}ller UIO} described in Section~\ref{sec:otherUIOarch}, for the system model of Example~\ref{ex:eg2}.
\end{exercise}


\begin{exercise}\label{ex:two systmod4UIO}\rm Consider two system models that appear in~\cite{yang88} and~\cite{darouach94}. The first system, using our notation, is modeled as
\begin{eqnarray*}
\b A &=&\left[\begin{array}{ccc}
-2 & -2 & 0\\
0 & 0 & 1\\
0 & -3 & -4
\end{array}\right], \quad \b B_1=\b O,\quad \b B_2=\left[\begin{array}{cc}
1 & 0\\
0 & 1\\
0 & 0
\end{array}\right]\\
\b C&=& \left[\begin{array}{ccc}
1 & 0 & 1\\
0 & 1 & 0
\end{array}\right].
\end{eqnarray*}
The second system is modeled by
\begin{eqnarray*}
\b A &=&\left[\begin{array}{ccc}
-1 & -1 & 0\\
-1 & 0 & 0\\
0 & -1 & -1
\end{array}\right], \quad \b B_1=\b O,\quad \b B_2=\left[\begin{array}{c}
-1\\
0\\
0
\end{array}\right]\\
\b C&=& \left[\begin{array}{ccc}
1 & 0 & 0\\
0 & 0 & 1
\end{array}\right].
\end{eqnarray*}
Construct UIOs developed in Section~\ref{fo_uio},~\ref{sec:reducedUIO}, and~\ref{sec:slidingUIO} for the above systems.
\end{exercise}

\begin{exercise}\rm Construct the Hou and M\"{u}ller UIO, described in Section~\ref{sec:otherUIOarch}, for each system of Exercise~\ref{ex:two systmod4UIO}.
\end{exercise}

\begin{exercise}\label{ex:same_kers}\rm
Prove Lemma~\ref{lemma:same_kers}.
%{\bf Proof}\quad If $\b x\in \ker(\b V)$, that is, if $\b x$ is such that $\b V\b
%x=\b 0$, then obviously $\b V^{\top}\b V\b x=\b 0$, that is, $\b x\in \ker(\b V^{\top}\b V)$.
%Thus, we showed that if $\b x\in \ker(\b V)$ then $\b x\in \ker(\b V^{\top}\b V)$.
%
%Suppose now that $\b x\in \ker(\b V^{\top}\b V)$, that is, $\b V^{\top}\b V\b x=\b 0$. Then,
%$\b y= \b V\b x \in \ker(\b V^{\top})$. Thus, $\b y\in \mb{range}\,(\b V)\cap\ker(\b
%V^{\top})$. However, it is well know from linear algebra that
%\[
%\mb{range}\,(\b V)\cap\ker(\b V^{\top})=\{\b 0\},
%\]
%that is, $\b y=\b V\b x=\b 0$ which means that $\b x\in \ker(\b V)$. Thus, we showed
%that if $\b x\in \ker(\b V^{\top}\b V)$ then $\b x\in \ker(\b V)$.
%
%In summary, $\ker(\b V)=\ker(\b V^{\top}\b V)$.\\
%\hspace*{6in}$\Box$
\end{exercise}


\begin{exercise}\label{exer:equvalence_w_lmis}\rm
Show that the conditions in~\eqref{eq:spr_as_ineq} are equivalent to the LMIs given by~\eqref{eq:first_lmi_for_smo} and~\eqref{eq:second_lmi_for_smo}.
\end{exercise}

\begin{exercise}\rm Use an LMI toolbox to find matrices $\b L$, $\b F$, and $\b P$\index{find matrices $\b L$, $\b F$, and $\b P$!using LMIs} of Theorem~\ref{thm:slidingUIO} for the system of Example~\ref{ex:smo_L-1011}.
\end{exercise}

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