\chapter{A Glimpse at Linear Time-Varying (LTV) Systems}
\begin{flushright}
\parbox{5.5in}{''As we shall see, the concept of time\index{time!concept of} has no meaning before the beginning of the universe. This was first pointed out by St. Augustine. When asked: What did God do before he created the universe? Augustine didn't reply: He was preparing Hell for people who asked such questions. Instead, he said that time was a property of the universe that God created, and that time did not exist before the beginning of the universe.'' Stephen W. Hawking~\cite[page~8]{hawking88}}
\end{flushright}

\bigskip

\section{Motivating Example}
Of considerable importance in many applications are linear systems in which the elements of $\b A$ and $\b B$ are continuous functions of time\index{continuous functions!of time}. We present a linear time-varying (LTV)\index{linear time-varying (LTV)}\index{LTV|see{linear time-varying}} model\index{time-varying model} of a battery as a motivation for our discussion on linear time-varying systems. Our derivation follows that of  Chiasson and Vairamohan~\cite{chiasson03,chiasson05}.


The simplest battery model\index{battery model} is shown in Figure~\ref{fig:SOC_mod1}, which is called a practical voltage source\index{practical voltage source!as battery model}.
\begin{figure}[h]
\centering \epsfig{file=SOC_mod1.eps,height=5.75cm}
\caption{Practical voltage source as a simple battery model.}
\label{fig:SOC_mod1}
\end{figure}
This model consists of an ideal voltage source that produces voltage $V_{oc}$. We refer to $V_{oc}$ as the open-circuit voltage\index{open-circuit!voltage}\index{open-circuit}. Thus the open-circuit voltage is obtained by measuring the voltage across the open-circuit, that is, when $I_b=0$. The internal resistance of the practical voltage source is denoted  $R_{int}$. The battery terminal voltage is labeled as $V_b$. This model, however, does not take into account the fact that the internal resistance\index{internal resistance}\index{resistance!internal} is different under discharge\index{discharge} and charge\index{charge} conditions. To account for this, the circuit of Figure~\ref{fig:SOC_mod1} is modified by introducing two different internal resistances; $R_c$---the charging resistance\index{charging resistance}\index{resistance!charging} and $R_d$---the discharging resistance\index{discharging resistance}\index{resistance!discharging}. These resistances are connected to the diodes\index{diode!to model charging} to model charging and discharging\index{diode!to model discharging} modes of the battery operation. The modified circuit is shown in Figure~\ref{fig:SOC_mod2}.
\begin{figure}
\centering \epsfig{file=SOC_mod2.eps,height=5.75cm}
\caption{Battery model accounting for the different charging, $R_c$, and discharging, $R_d$, resistances.}
\label{fig:SOC_mod2}
\end{figure}
This model, on the other hand, does not take into account the diffusion of the electrolytic through the battery. To model this effect, a polarization capacitor\index{capacitor!polarization} is added to model electro-chemical reaction\index{electro-chemical!reaction}\index{reaction!electro-chemical} taking place in the battery. The resulting model is depicted in Figure~\ref{fig:SOC_mod3}.
\begin{figure}
\centering \epsfig{file=SOC_mod3.eps,height=5.75cm}
\caption{Battery model accounting for the different charging, $R_c$, and discharging, $R_d$, resistances and polarization capacitance, $C$.}
\label{fig:SOC_mod3}
\end{figure}


In electric vehicles, the state of charge (SOC)\index{state of charge (SOC)}\index{SOC|see{state of charge}} of the battery corresponds to a fuel gauge in a conventional, internal combustion, cars. To define the state of charge, consider a completely charged battery. Let $I_b(t)$ be the discharging current\index{discharging current}\index{current!discharging}. Then, the charge delivered by the battery is
\[
Q(t)=\int_{t_0}^t I_b(\tau)d \tau .
\]
Let $Q_0=\int_{t_0}^{\infty} I_b(\tau)d \tau $ be the total charge that battery can hold. Then the state of charge (SOC) of the battery is
\begin{eqnarray}
\mb{SOC}(t) &=& \frac{Q_0-Q(t)}{Q_0}\times 100\nonumber\\
&=& \frac{Q_0 - \int_{t_0}^t I_b(\tau)d \tau}{Q_0}\times 100 .
\label{eq:soc_defn_q}
\end{eqnarray}
Direct application of~\eqref{eq:soc_defn_q} to estimate the battery SOC\index{battery SOC!estimate} is not accurate. An alternative approach is to use an estimate of the open circuit voltage\index{open circuit!voltage}, $V_{oc}$ of the battery. Then, the SOC can be evaluated\index{battery SOC!evaluate} from the relation,
\begin{equation}
\mb{SOC}(t) = \frac{V_{oc}(t) - a_0}{a_1},
\label{eq:scoviavoc}
\end{equation}
where $a_0$ is the battery terminal voltage\index{battery!terminal voltage} when $\mb{SOC}(t)=0\%$ and $a_1$ is obtained using $a_0$ and $V_{oc}$ at $\mb{SOC}(t)=100\%$. Thus estimating the SOC\index{estimating SOC!estimating open-circuit voltage} is equivalent to estimating the battery open-circuit voltage $V_{oc}(t)$. Now, in order to measure $V_{oc}$, the battery must be disconnected from its load, which is not possible during the electric vehicle\index{electric vehicle} operation that uses the battery whose SOC we wish to estimate. The approach proposed by Chiasson and Vairamohan~\cite{chiasson03,chiasson05} is to construct the open-circuit voltage estimator\index{open-circuit voltage estimator}. This estimator construction is based on the battery model\index{battery model}\index{open-circuit voltage estimator!model based} that we develop next.


We analyze the condition when the battery is in the discharging mode\index{battery!in discharging mode}, that is, when $I_b\ge 0$. The charging mode is analyzed in a similar way. The battery model in the discharging mode is depicted in Figure~\ref{fig:SOC_mod4}.
\begin{figure}
\centering \epsfig{file=SOC_mod4.eps,height=5.75cm}
\caption{A battery model in the discharging mode of operation.}
\label{fig:SOC_mod4}
\end{figure}
To proceed, we apply Kirchhoff's current law (KCL)\index{Kirchhoff's current law (KCL)}\index{KCL|see{Kirchhoff's current law}} at the node between the resistances $R_d$ and $R_b$ to obtain
\[
I_c+I_b-I_d=0,
\]
that is,
\[
I_c=I_d-I_b.
\]
Applying Ohm's law\index{Ohm's law} gives
\begin{equation}
I_d=\frac{V_{oc}-V_p}{R_d}\quad\mb{and}\quad I_b=\frac{V_p-V_b}{R_b}.
\label{eq:two_eqns}
\end{equation}
We also have
\[
I_c=C\dot{V_p}.
\]
Combining the above equations yields
\begin{eqnarray}
C\dot{V_p} &=& I_d-I_b\nonumber\\
&=& \frac{V_{oc}-V_p}{R_d}-I_b\nonumber\\
&=& -\frac{V_p}{R_d}+\frac{V_{oc}}{R_d} -I_b.
\label{eq:combo_v_dot}
\end{eqnarray}
Dividing both sides by $C$, we obtain
\begin{equation}
\dot{V_p}=-\frac{V_p}{R_d C}+\frac{V_{oc}}{R_d C} -\frac{1}{C}I_b.
\label{eq:v_dot_afterC}
\end{equation}
The output equation is obtained from the second of the equations in~\eqref{eq:two_eqns},
\begin{equation}
V_b=V_p-R_bI_b
\label{eq:out_vb_as_f_of_ib}
\end{equation}
We define the state variables,
\[
x_1=V_p,\quad x_2=\frac{1}{R_d C},\quad x_3 = \frac{V_{oc}}{R_d C},\quad x_4=\frac{1}{C},\quad\mb{and}\quad x_5=R_b.
\]
Applying the state variable definitions to~\eqref{eq:v_dot_afterC} and~\eqref{eq:out_vb_as_f_of_ib}, and taking into account that $x_2$, $x_3$, $x_4$, and $x_5$ are constant, we obtain a nonlinear time-varying\index{nonlinear time-varying!battery model} state-space model,
\begin{equation}
\left.
\begin{array}{lll}
\dot{x}_1 &=& -x_1x_2 + x_3 - I_b(t) x_4\\
\dot{x}_2 &=& 0\\
\dot{x}_3 &=& 0\\
\dot{x}_4 &=& 0\\
\dot{x}_5 &=& 0\\
y&=& V_b=x_1 - I_b(t) x_5
\end{array}\right\}
\label{eq:non_tv_mod_of_batt}
\end{equation}
The initial state vector is unknown to us. The state variable $x_3$ depends on $V_{oc}$. Recall that the state variables, $x_2$, $x_3$, $x_4$ and $x_5$ are constant. Our objective is to compute the open-circuit voltage\index{open-circuit voltage!compute} $V_{oc}$, which is a constant. We note that
\[
x_3=\frac{V_{oc}}{R_d C}=V_{oc}x_2.
\]
Therefore
\[
V_{oc}=\frac{x_3}{x_2}=\frac{x_{30}}{x_{20}}
\]
because $V_{oc}$ is constant. Hence, we can determine $V_{oc}$ if we can estimate the initial values of the state variables\index{initial values of state variables!estimate} $x_{20}$ and $x_{30}$. We proceed by redefining the state variables as
\[
z_1=x_1,\quad z_2=x_3,\quad z_3=x_4,\quad z_4=x_5.
\]
With the above in mind, and using the fact that $x_2=\mb{constant}=x_{20}$, we re-write~\eqref{eq:non_tv_mod_of_batt} as
\begin{eqnarray}
\left[\begin{array}{c}
\dot{z}_1\\
\dot{z}_2\\
\dot{z}_3\\
\dot{z}_4
\end{array}\right] &=& \left[\begin{array}{cccc}
-x_{20} & 1 & -I_b(t) & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0
\end{array}\right]\left[\begin{array}{c}
{z}_1\\
{z}_2\\
{z}_3\\
{z}_4
\end{array}\right]\label{eqstate_eqnabatt}\\
y&=& \left[\begin{array}{cccc}
1 & 0 & 0 & -I_b(t)
\end{array}\right]\left[\begin{array}{c}
{z}_1\\
{z}_2\\
{z}_3\\
{z}_4
\end{array}\right].
\label{eq:out_batt_mod}
\end{eqnarray}
The above is a LTV system\index{LTV!system} because the current $I_b(t)$ is time-varying. In summary, we have a homogenous, or free, or unforced, LTV system\index{unforced LTV system}\index{homogenous LTV system|see{unforced LTV system}}\index{free LTV system|see{unforced LTV system}} model
\begin{eqnarray*}
\dot{\b z} &=& \b A(t,x_{20})\b z, \quad \b z(t_0)\\
y &=& \b c(t)\b z.
\end{eqnarray*}
Our objective is to estimate $\b z(t_0)$ from knowledge of the output $y(t)$ and the current\index{estimate!from output and current} $I_b(t)$ on some time interval $[t_0, \; t_f]$.
%To compute $\b z(t_0)$, we need the state-transition matrix of this LTV system.

We now outline a solution to the problem of estimating $\b z(t_0)$. The first step is to compute the state-transition matrix\index{state-transition matrix}, $\b{\Phi}(t,t_0)$, which relates the state at time $t$ to the initial state,
\[
\b z(t) = \b{\Phi}(t,t_0)\b z(t_0).
\]
We then use the output equation\index{output!equation} to obtain
\begin{eqnarray*}
y(t) &=& \b c \b z(t)\\
&=& \b c \b{\Phi}(t,t_0)\b z(t_0).
\end{eqnarray*}
Premultiply both sides of the above by $\b{\Phi}(t,t_0)^{\top}\b c^{\top}$ and then integrate both sides to get
\[
\int_{t_0}^{t_f} \b{\Phi}(t,t_0)^{\top}\b c^{\top} y(t) dt = \int_{t_0}^{t_f} \b{\Phi}(t,t_0)^{\top}\b c^{\top}\b c \b{\Phi}(t,t_0) dt \;\b z(t_0).
\]
Let
\[
\b V(t_0,t_f) = \int_{t_0}^{t_f} \b{\Phi}(t,t_0)^{\top}\b c^{\top}\b c \b{\Phi}(t,t_0) dt ,
\]
which, as in the LTI case, we refer to as the observability gramian\index{observability!gramian}. If $\b V(t_0,t_f)$ is invertible then
\[
\b z(t_0)=\b V^{-1}(t_0,t_f)\int_{t_0}^{t_f} \b{\Phi}(t,t_0)^{\top}\b c^{\top} y(t) dt .
\]
Having obtained $\b z(t_0)$ we can compute $V_{oc}$ and then the SOC. We refer to Chiasson and Vairamohan~\cite{chiasson03,chiasson05} for a complete solution to this problem; see also Pang et al.~\cite{pang01}, and McIntyre et al.~\cite{mcintyre06} for alternative solutions.

To be able to solve this problem, we need to acquaint ourselves with the state transition matrix for LTV systems. We start with the LTV uncontrolled system\index{LTV!uncontrolled system}.



\section{Solving the Uncontrolled LTV  System}\label{sec:sol_LTV_uncon}

In general, it is not possible to give explicit expressions for solutions of linear time-varying systems. We obtain some general properties that for some cases will be used to obtain explicit solutions. We first consider the uncontrolled system
\begin{equation}
\dot{\b x}(t) = \b A(t) \b x(t), \quad \b x(t_0) = \b x_0.
\label{eq:LTV_uncontrolled}
\end{equation}
\begin{thm}\label{thm:ltv_sol_unique}
If $\b A(t)$ is continuous for $t\ge t_0$ then the uncontrolled system modeled by~\eqref{eq:LTV_uncontrolled} has a unique solution for $t\ge t_0$ given by
\[
\b x(t) = \b X(t)\b x_0,
\]
where $\b X(t)$ is the unique $n\times n$ matrix satisfying
\[
\frac{d \b X}{d t} = \b A(t) \b X, \quad \b X(t_0)=\b I_n .
\]
\end{thm}
{\bf Proof}\quad See Rugh~\cite[Chp.~3]{rugh93} or Antsaklis and Michel~\cite[Section~1.8]{antsaklis07}. See also Exercise~\ref{ex:using contraction}.\\
\hspace*{5.5in} $\Box$

Note that it follows from the above theorem that for every initial state $\b x^{(i)}(t_0)$, $i=1,2,\ldots, n$, there exists a unique solution $\b x^{(i)}(t)$, $i=1,2,\ldots, n$. We arrange these $n$ solutions as
\[
\b X(t)=\left[\begin{array}{cccc}
\b x^{(1)}(t) & \b x^{(2)}(t) & \cdots & \b x^{(n)}(t)
\end{array}\right].
\]
Because every $\b x^{(i)}(t)$ satisfies~\eqref{eq:LTV_uncontrolled}, we have
\[
\dot{\b X}(t) = \b A(t) \b X(t) .
\]
If $\b X(t_0)$ is nonsingular, then $\b X(t)$ is called a {\em fundamental matrix}\index{fundamental matrix} of $\dot{\b x}=\b A(t)\b x$. Because the initial states are linearly independent, the fundamental matrix is nonsingular---see Exercise~\ref{ex:non_sing_fundm}. The fundamental matrix is not unique\index{fundamental matrix!not unique} because we can choose initial states arbitrarily.

We now analyze a scalar LTV case\index{LTV!scalar case},
\begin{equation}
\dot{x}=a(t)x,\quad x(0)=x_0 .
\label{eq:scalarltvdiffeq}
\end{equation}
Recall that when $a(t)=a=$constant, then the solution has the form
\[
x(t)=e^{at}x(0)=e^{\int_0^t a d\tau}x(0).
\]
The above formula works for the scalar LTV case, that is,
\begin{equation}
x(t)= e^{\int_0^t a(\tau)  d\tau}x(0).
\label{eq:sol_formula_scalltv}
\end{equation}
It is easy to verify that the above is the solution of~\eqref{eq:scalarltvdiffeq} because
\begin{eqnarray*}
\frac{d}{dt} e^{\int_0^t a(\tau) d \tau} &=& e^{\int_0^t a(\tau) d \tau } a(t)\\
&=& a(t) e^{\int_0^t a(\tau) d \tau } .
\end{eqnarray*}
That is, $a(t)$ and $\int_0^t a(\tau) d \tau$ commute\index{commute}.
\begin{example}\rm We use the solution formula~\eqref{eq:sol_formula_scalltv} to solve the equation,
\[
\dot{x} = (\sin t) x,\quad x(0)=x_0.
\]
We have
\begin{eqnarray*}
x(t) &=& e^{\int_0^t a(\tau)  d\tau}x_0\\
&=& x_0  e^{\int_0^t \sin (\tau)  d\tau}\\
&=& x_0\left. e^{-\cos t} \right\arrowvert^t_0\\
&=&  e^{-\cos t +1 } x_0\\
&=&  e \times e ^{-\cos t} x_0 .
\end{eqnarray*}
\end{example}\markex


The solution formula~\eqref{eq:sol_formula_scalltv} does not generalize to the multi-dimensional\index{multi-dimensional case} case because, in general,
\[
\frac{d}{dt} e^{\int_0^t \b A(\tau) d \tau} \ne \b A(t) e^{\int_0^t \b A(\tau) d \tau}.
\]
If, however, $\b A(t)$ and $\int_0^t \b A(\tau) d \tau$ commute, then the solution to~\eqref{eq:LTV_uncontrolled} can be written as
\[
\b x(t) = e^{\int_0^t \b A(\tau) d \tau} \b x(0) .
\]
We note that $\b A(t)$ and $\int_0^t \b A(\tau) d \tau$ commute if
\[
\b A(t)\b A(s) = \b A(s) \b A(t) .
\]

We now analyze a general case when $t_0$ is not necessarily equal to $0$. If $\b A(t)=\b A$ in~\eqref{eq:LTV_uncontrolled}, then the solution is
\[
\b x(t) = e^{\b A (t - t_0)}\b x(t_0).
\]
We re-write the above as
\[
\b x(t) = e^{\b A t} e^{- \b A t_0}\b x(t_0).
\]
Let $\b X(t)=e^{\b A t}$, then
\[
\b X(t_0)^{-1} = e^{- \b A t_0}
\]
and we represent the solution as
\begin{eqnarray}
\b x(t)&=& \b X(t)\b X(t_0)^{-1} \b x(t_0)\nonumber\\
&=& \b{\Phi} (t, t_0) \b x(t_0),
\label{eq:stae_tran_matr_general}
\end{eqnarray}
where $\b{\Phi} (t, t_0)$ is the state transition matrix\index{state transition matrix}.
\begin{defn}
Let $\b X(t)$ be a fundamental matrix of~\eqref{eq:LTV_uncontrolled}. Then
\[
\b{\Phi} (t, t_0)=\b X(t)\b X(t_0)^{-1}
\]
is the state transition matrix of~\eqref{eq:LTV_uncontrolled}. The state transition matrix is the unique solution\index{state transition matrix!unique solution} of
\[
\frac{\partial}{\partial t}\b{\Phi} (t, t_0) = \b A(t) \b{\Phi} (t, t_0),\quad \b{\Phi} (t_0, t_0)=\b I_n .
\]
\end{defn}\markdefn

As in the LTI case, the state transition matrix in the LTV case has the following properties (see Exercise~\ref{ex:state_trns_propos}):
\begin{eqnarray}
\b{\Phi} (\tau, \tau) &=& \b I_n \label{eq:state_trans_prop1}\\
\b{\Phi} (t, \tau)^{-1} &=& \b{\Phi} (\tau, t) \label{eq:state_trans_prop2}\\
\b{\Phi} (t, t_0)&=& \b{\Phi} (t, t_1)\b{\Phi} (t_1, t_0) \label{eq:state_trans_prop3}\\
\frac{\partial}{\partial \tau}\b{\Phi} (t, \tau) &=& - \b{\Phi} (t, \tau) \b A(\tau), \mb{  for all   } t, \tau \label{eq:state_trans_prop4}
\end{eqnarray}
The third of the above properties is referred to as the {\em semigroup property}\index{semigroup property!state transition matrix}.


\begin{example}\rm The state transition matrix of
\[
\dot{x}=(\sin t) x, \quad x(t_0)=x_0,
\]
has the form
\begin{eqnarray*}
\Phi(t, t_0) &=& X(t)X(t_0)^{-1}\\
&=& \frac{e^{-\cos t + 1}}{e^{-\cos t_0 + 1}}.
\end{eqnarray*}
\end{example}
\markex
\begin{example}\rm We construct the state transition matrix for the LTV system,
\[
\dot{\b x} = \left[\begin{array}{cc}
0 & -t\\
0 & 0
\end{array}\right]\b x,\quad \b x(0)=\b x(0).
\]
The above system consists of two first-order linear differential equations,
\[
\dot{x}_1 = -t x_2,\quad\mb{and}\quad \dot{x}_2=0.
\]
We can easily obtain the solution to the second of the above equations,
\begin{equation}
x_2(t) = x_2(0).
\label{eq:sol_to_sec_eq}
\end{equation}
We next obtain the solution to $\dot{x}_1 = -t x_2$ by integrating its both sides,
\begin{eqnarray}
x_1(t) &=& x_1(0) - \int_0^t \tau x_2(0) d \tau\nonumber\\
&=& x_1(0) - \frac{t^2 x_2(0)}{2}.
\label{eq:sol_to_first_eq}
\end{eqnarray}
We can  now construct a fundamental matrix\index{fundamental matrix}. Let
\[
\b x^{(1)}(0)=\left[\begin{array}{c}
1\\
1
\end{array}\right]\quad\mb{and}\quad \b x^{(2)}(0)=\left[\begin{array}{c}
0\\
1
\end{array}\right].
\]
Then using~\eqref{eq:sol_to_first_eq} and~\eqref{eq:sol_to_sec_eq}, we obtain the fundamental matrix corresponding to the above initial conditions is
\begin{eqnarray*}
\b X(t) &=& \left[\begin{array}{cc}
\b x^{(1)}(t) & \b x^{(2)}(t)
\end{array}\right]\\
&=& \left[\begin{array}{cc}
1-\frac{t^2}{2} & -\frac{t^2}{2}\\
1 & 1
\end{array}\right].
\end{eqnarray*}

If we select the initial conditions
\[
\b x^{(1)}(0)=\left[\begin{array}{c}
1\\
0
\end{array}\right]\quad\mb{and}\quad \b x^{(2)}(0)=\left[\begin{array}{c}
0\\
1
\end{array}\right]
\]
then we obtain a different fundamental matrix\index{fundamental matrix!another one},
\[
\b X(t)= \left[\begin{array}{cc}
1 & -\frac{t^2}{2}\\
0 & 1
\end{array}\right].
\]
The state transition matrix, however, is unique\index{state transition matrix!unique} and has the form
\begin{eqnarray*}
\b{\Phi}(t, t_0) &=& \b X(t) \b X (t_0)^{-1}\\
&=& \left[\begin{array}{cc}
1-\frac{t^2}{2} & -\frac{t^2}{2}\\
1 & 1
\end{array}\right]\left[\begin{array}{cc}
1-\frac{t_0^2}{2} & -\frac{t_0^2}{2}\\
1 & 1
\end{array}\right]^{-1}\\
&=& \left[\begin{array}{cc}
1-\frac{t^2}{2} & -\frac{t^2}{2}\\
1 & 1
\end{array}\right]\left[\begin{array}{cc}
1 & \frac{t_0^2}{2}\\
-1 & 1-\frac{t_0^2}{2}
\end{array}\right]\\
&=& \left[\begin{array}{cc}
1 & \frac{1}{2}(t_0^2-t^2)\\
0 & 1
\end{array}\right].
\end{eqnarray*}
\end{example}
\markex

\begin{example}\rm We construct the state-transition matrix for the LTV system model found in Luenberger~\cite[p.~115]{luenberger79},
\[
\dot{\b x}=\left[\begin{array}{cc}
0 & 0\\
t & \frac{1}{t}
\end{array}\right]\b x, \quad t \ge 1 .
\]
Then, for the given $\b x(2)=\left[\begin{array}{cc}
2 & -1
\end{array}\right]^{\top}$, we compute $\b x(1)$.

First, we solve for $x_1$ to obtain,
\[
x_1(t)=x_1(1), \quad t\ge 1
\]
Substituting the above into the equation for $\dot{x}_2$ gives
\[
\dot{x}_2 = t x_1(t) + \frac{x_2(t)}{t}.
\]
Next, we make the substitution
\[
z(t)=\frac{x_2(t)}{t}.
\]
Computing the derivative\index{derivative!computing} of $z(t)$ yields
\begin{eqnarray*}
\dot{z} &=& \frac{\dot{x}_2 t -x_2}{t^2}\\
&=& \frac{\dot{x}_2}{t}-\frac{x_2}{t^2}.
\end{eqnarray*}
Multiplying both sides of the above by $t$, we obtain
\begin{eqnarray}
t\dot{z} &=& \dot{x}_2 - \frac{x_2}{t^2}\nonumber\\
&=& \dot{x}_2 -z.
\label{eq:expr_for_tzdot}
\end{eqnarray}
We substitute $\dot{x}_2 = t x_1(1) + z$ into~\ref{eq:expr_for_tzdot}  to obtain
\[
t\dot{z}+z=t x_1(1) + z,
\]
that is,
\[
\dot{z}=x_1(1).
\]
Integrating the above from $1$ to $t$ gives
\[
z(t)=z(1)+x_1(1)(t-1).
\]
Therefore,
\begin{eqnarray*}
z &=& \frac{x_2}{t}\\
&=&z(1)+x_1(1)(t-1),
\end{eqnarray*}
and multiplying both sides by $t$, we obtain
\[
x_2(t)=t x_2(1) + t x_1(1) (t-1).
\]
We are now ready to generate a fundamental matrix. Let
\[
\b x^{(1)}(1)=\left[\begin{array}{c}
1\\
0
\end{array}\right]\quad\mb{and}\quad \b x^{(2)}(1)=\left[\begin{array}{c}
0\\
1
\end{array}\right],
\]
then we have
\[
\b x^{(1)}(t)=\left[\begin{array}{c}
1\\
t^2-t
\end{array}\right]\quad\mb{and}\quad \b x^{(2)}(t)=\left[\begin{array}{c}
0\\
t
\end{array}\right].
\]
The fundamental matrix corresponding to the above initial conditions is
\[
\b X(t) = \left[\begin{array}{cc}
\b x^{(1)}(t) & \b x^{(2)}(t)
\end{array}\right]
= \left[\begin{array}{cc}
1 & 0\\
t^2-t & t
\end{array}\right].
\]
The state transition matrix is
\begin{eqnarray*}
\b{\Phi}(t,\tau) &=& \b X(t)\b X(\tau)^{-1}\\
&=& \left[\begin{array}{cc}
1 & 0\\
t^2-t & t
\end{array}\right]\left[\begin{array}{cc}
1 & 0\\
\tau^2-\tau & \tau
\end{array}\right]^{-1}\\
&=& \left[\begin{array}{cc}
1 & 0\\
t^2-t & t
\end{array}\right]\left[\begin{array}{cc}
1 & 0\\
-\tau + 1& \frac{1}{\tau}
\end{array}\right]\\
&=& \left[\begin{array}{cc}
1 & 0\\
t(t-\tau) & \frac{t}{\tau}
\end{array}\right].
\end{eqnarray*}
Note that
\[
\b x(t)=\b{\Phi}(t,\tau)\b x(\tau).
\]
Thus, for example, if we are given $\b x(2)=\left[\begin{array}{cc}
2 & -1
\end{array}\right]^{\top}$ and we wish to find $\b x(1)$, we have,
\begin{eqnarray*}
\b x(1) &=& \b{\Phi}(1,2)\b x(2)\\
&=& \left[\begin{array}{cc}
1 & 0\\
-1 & \frac{1}{2}
\end{array}\right]\left[\begin{array}{c}
2\\
-1
\end{array}\right]\\
&=& \left[\begin{array}{c}
2\\
-\frac{5}{2}
\end{array}\right].
\end{eqnarray*}
\end{example}
\markex





\section{Solving the LTV Controlled System}

We now discuss the problem of obtaining the solution to the LTV controlled system\index{LTV controlled system}
\begin{equation}
\dot{\b x}=\b A(t)\b x + \b B(t)\b u,\quad \b x(t_0)=\b x_0 .
\label{eq:ltv_control_syst}
\end{equation}
We claim that, by analogy with the LTI controlled system\index{LTI controlled system}, the solution to~\eqref{eq:ltv_control_syst} is
\begin{equation}
\b x(t) = \b{\Phi}(t,t_0)\b x_0 + \int_{t_0}^t \b{\Phi}(t,\tau)\b B(\tau) \b u(\tau) d \tau .
\label{eq:sol_lvi_co_sys}
\end{equation}
We can prove our claim by verification utilizing Leibniz's rule\index{Leibniz's rule} (see Exercise~\ref{exer:leibnitzs_rule}),
\begin{equation}
\frac{\partial}{\partial t} \int_{t_0}^t f(t, \tau) d \tau = \int_{t_0}^t \left(\frac{\partial}{\partial t} f(t,\tau)  \right) d \tau + \left. f(t, \tau)  \right\arrowvert_{\tau = t}
\label{eq:leibnitzs_rule}
\end{equation}
\begin{example}[R. A. DeCarlo]\rm The LTV state model,
\[
\dot{\b x}(t) = \b A(t)\b x(t) + \left[\begin{array}{c}
\pi e^{-\alpha(t)^2}\\
0
\end{array}\right] u(t),
\]
has associated state transition matrix
\[
\b{\Phi}(t,\tau)=e^{\alpha(\tau)^2 - \alpha(t)^2}\left[\begin{array}{cc}
1 & 0\\
\cos (\pi t) - \cos (\pi \tau) & 1
\end{array}\right],
\]
where $\alpha(\cdot)$ is a continuous function\index{continuous function}\index{function!continuous}.
\begin{enumerate}
\item Compute $\b A(t)$ using the given state transition matrix.
\item Find the zero-state response\index{zero-state response} when the input $u(t)$ is the unit step\index{unit step!input}.
\end{enumerate}

We first compute $\b A(t)$ using the given state transition matrix. We use the state transition matrix properties to accomplish this. Recall that the state transition matrix satisfies the equation
\[
\frac{\partial}{\partial t}\b{\Phi} (t, \tau) = \b A(t) \b{\Phi} (t, \tau),\quad \b{\Phi} (\tau, \tau)=\b I_n .
\]
Note that, we can also write.
\[
\left.\b{\Phi}(t,\tau)\right\arrowvert_{\tau = t}=\b I_n
\]
Taking this into account, we obtain
\[
\b A(t) = \left. \frac{\partial}{\partial t}\b{\Phi} (t, \tau)\right\arrowvert_{\tau = t}.
\]
To proceed, we first evaluate
\begin{eqnarray*}
\left.\frac{\partial}{\partial t} e^{\alpha(\tau)^2 - \alpha(t)^2}\right\arrowvert_{\tau = t} &=& \left. e^{\alpha(\tau)^2 - \alpha(t)^2}(-2\alpha(t)\dot{\alpha}(t))\right\arrowvert_{\tau = t}\\
&=& -2\alpha(t)\dot{\alpha}(t).
\end{eqnarray*}
We next evaluate the derivative
\begin{eqnarray*}
\left.\frac{\partial}{\partial t} \left(e^{\alpha(\tau)^2 - \alpha(t)^2}\left(\cos (\pi t) - \cos (\pi \tau)\right)\right)\right\arrowvert_{\tau = t} &=& -2\alpha(t)\dot{\alpha}(t)(\cos (\pi t) - \cos (\pi t)) - \pi \sin (\pi t)\\
&=& - \pi \sin (\pi t).
\end{eqnarray*}
Hence
\begin{eqnarray*}
\b A(t) &=& \left. \frac{\partial}{\partial t}\b{\Phi} (t, \tau)\right\arrowvert_{\tau = t}\\
&=& \left[\begin{array}{cc}
-2\alpha(t)\dot{\alpha}(t) & 0\\
-\pi \sin (\pi t) & -2\alpha(t)\dot{\alpha}(t)
\end{array}\right].
\end{eqnarray*}

We now compute the zero-state response. The formula for the zero-state response\index{zero-state response} is obtained from~\eqref{eq:sol_lvi_co_sys} by setting $\b x_0 = \b 0$,
\[
\b x(t) = \int_{t_0}^t \b{\Phi}(t,\tau)\b B(\tau) \b u(\tau) d \tau .
\]
We first compute the integrand
\begin{eqnarray*}
\b{\Phi}(t,\tau)\b B(\tau) &=& e^{\alpha(\tau)^2 - \alpha(t)^2}\left[\begin{array}{cc}
1 & 0\\
\cos (\pi t) - \cos (\pi \tau) & 1
\end{array}\right]\left[\begin{array}{c}
\pi e^{-\alpha(\tau)^2}\\
0
\end{array}\right]\\
&=& \pi e^{-\alpha(t)^2}\left[\begin{array}{c}
1\\
\cos (\pi t) - \cos (\pi \tau)
\end{array}\right].
\end{eqnarray*}
Therefore, the zero-state response when $u(t) = 1(t)$ is
\begin{eqnarray*}
\b x(t) &=& \int_{t_0}^t \b{\Phi}(t,\tau)\b B(\tau) \b u(\tau) d \tau\\
&=& \pi e^{-\alpha(t)^2}\int_0^t\left[\begin{array}{c}
1\\
\cos (\pi t) - \cos (\pi \tau)
\end{array}\right] d \tau\\
&=& \pi e^{-\alpha(t)^2}\left[\begin{array}{c}
t\\
t\cos (\pi t) - \frac{\sin (\pi \tau)}{\pi}
\end{array}\right].
\end{eqnarray*}
\end{example}\markex

\section{Notes} For additional information on the fundamentals of the LTV systems, we recommend Kamen~\cite{kamen11}. An in-depth treatment of the LTV systems\index{LTV systems} can be found in Rugh~\cite{rugh96}, and in Antsaklis and Michel~\cite{antsaklis07}. For results on observers for LTV systems\index{LTV systems!observers for}, see O'Reilly~\cite{oreilly83}. A novel approach to the analysis and control of LTV systems\index{LTV systems!analysis and control} is proposed by Bourl{\`{e}}s and Marinescu~\cite{bourles11}.

\section{Exercises}


\begin{exercise}\label{ex:non_sing_fundm}\rm Prove that a fundamental matrix is nonsingular\index{fundamental matrix!nonsingular} for any t.
\end{exercise}


\begin{exercise}\label{ex:using contraction}\rm Use the contraction map theorem~\ref{thm:contramap}\index{contraction map theorem} to prove that~\eqref{eq:LTV_uncontrolled} has a unique solution going through $\b x_0$ at $t=t_0$, where $t_0$  is an element of a finite time interval $[a, \;b]$. (Start with the Lipschitz condition.)
\end{exercise}

\begin{exercise}\rm\begin{itemize}\item[{\bf (a)}] Find the state-transition matrix $\b{\Phi}(t, \tau)$ associated with a homogenous LTI system\index{homogenous LTI system}\index{LTI system!homogenous} with
    \[
    \b A=\b T\left[\begin{array}{cc}
    -2 & 0\\
    0 & 0
    \end{array}\right]\b T^{-1},
    \]
where
\[
\b T=\left[\begin{array}{cc}
1 & 0\\
-3 & 1
\end{array}\right].
\]
\item[{\bf (b)}] Find the state-transition matrix\index{state-transition matrix} $\b{\Phi}(t, \tau)$ associated with a homogenous LTV system with
    \[
    \b A(t)=\b T\left[\begin{array}{cc}
    -2  & 0\\
    0 & \cos t
    \end{array}\right]\b T^{-1},
    \]
where
\[
\b T=\left[\begin{array}{cc}
1 & 0\\
-3 & 1
\end{array}\right].
\]
\end{itemize}

\end{exercise}
\begin{exercise}\rm Compute the state transition matrix $\b{\Phi}(t,\tau)$ for the linear time-varying (LTV) system\index{LTV system} model,
\[
\dot{\b x}(t)=\left[\begin{array}{cc}
0 & 0\\
1 & \frac{1}{t}
\end{array}\right]\b x(t)
\]
defined for $t\ge 1$.
Then, for the given $\b x(2)=\left[\begin{array}{cc}
    0 & 4
    \end{array}\right]^{\top}$, compute $\b x(1)$.
%We have two linear time-varying differential equations,
%\begin{eqnarray*}
%\dot{x}_1 &=& 0\\
%\dot{x}_2 &=& x_1 +\frac{x_2}{t}.
%\end{eqnarray*}
%The solution to the first of the above equations is
%\[
%x_1(t)=x_1(1).
%\]
%To solve the second equation, we make a substitution,
%\[
%z=\frac{x_2}{t}.
%\]
%Differentiating both sides of the above gives
%\[
%\dot{z}=\frac{\dot{x}_2}{t}-\frac{x_2}{t^2}.
%\]
%Rearranging the terms, then multiplying both sides by $t$ and finally taking into account the substitution, we obtain
%\[
%\dot{x}_2=t\dot{z}+z.
%\]
%Hence,
%\[
%\dot{x}_2=t\dot{z}+z=x_1(1)+z
%\]
%and therefore,
%\[
%\dot{z}=\frac{x_1(1)}{t}.
%\]
%Integrating the above gives
%\[
%z(t)=z(1)+x_1(1)\ln t.
%\]
%Hence
%\[
%x_2(t)=tx_2(1) + x_1(1) t \ln t.
%\]
%Let
%\[
%\b x^{(1)}(1)=\left[\begin{array}{c}
%1\\
%0
%\end{array}\right]\quad\mb{and}\quad \b x^{(2)}(1)=\left[\begin{array}{c}
%0\\
%1
%\end{array}\right].
%\]
%Then we obtain a fundamental matrix
%\[
%\b X(t)=\left[\begin{array}{cc}
%1 & 0\\
%t\ln t & t
%\end{array}\right].
%\]
%The state transition matrix is
%\begin{eqnarray*}
%\b{\Phi}(t,\tau)&=&\b X(t)\b X(\tau)^{-1}\\
%&=& \left[\begin{array}{cc}
%1 & 0\\
%t\ln t & t
%\end{array}\right]\left[\begin{array}{cc}
%1 & 0\\
%-\ln \tau & \frac{1}{\tau}
%\end{array}\right]\\
%&=& \left[\begin{array}{cc}
%1 & 0\\
%t\ln t-t \ln \tau & \frac{t}{\tau}
%\end{array}\right].
%\end{eqnarray*}
%
%
%For the given $\b x(2)=\left[\begin{array}{cc}
%    0 & 4
%    \end{array}\right]^{\top}$, we compute $\b x(1)$.
%
%We have
%\begin{eqnarray*}
%\b x(1) &=& \b{\Phi}(1,2)\b x(2)\\
%&=& \left[\begin{array}{cc}
%1 & 0\\
%-\ln 2 & \frac{1}{2}
%\end{array}\right]\left[\begin{array}{c}
%0\\
%4
%\end{array}\right]\\
%&=& \left[\begin{array}{c}
%0\\
%2
%\end{array}\right].
%\end{eqnarray*}
\end{exercise}

\begin{exercise}\label{ex:state_trns_propos}\rm Use the definition of the state transition matrix to prove its properties~\eqref{eq:state_trans_prop1}, \eqref{eq:state_trans_prop2}, \eqref{eq:state_trans_prop3}, and \eqref{eq:state_trans_prop4}.
\end{exercise}

\begin{exercise}\rm \begin{itemize}\item[{\bf (a)}] The value of the state-transition matrix for a homogeneous linear system is
    \[
    \b{\Phi}(1, 3)=\left[\begin{array}{cc}
    1 & 1\\
    1 & 2
    \end{array}\right].
    \]
Find $\b x(3)$ when
\[
\b x(1)=\left[\begin{array}{c}
1\\
3
\end{array}\right].
\]
\item[{\bf (b)}] The values of the state-transition matrix for a homogeneous linear system is
    \[
    \b{\Phi}(1, 3)=\left[\begin{array}{cc}
    1 & 1\\
    1 & 2
    \end{array}\right]\quad\mb{and}\quad \b{\Phi}(2, 3)=\left[\begin{array}{cc}
    2 & 3\\
    4 & 5
    \end{array}\right].
    \]
If
\[
\b x(1)=\left[\begin{array}{c}
2\\
-3
\end{array}\right],
\]
find $\b x(2)$.

\end{itemize}
\end{exercise}

\begin{exercise}\label{exer:leibnitzs_rule}\rm Derive Leibniz's rule\index{Leibniz's rule!derive} given by~\eqref{eq:leibnitzs_rule}.
\end{exercise}

\begin{exercise}\rm \begin{itemize}
\item[{\bf (a)}] The state-transition matrix of a homogenous LTV system is
    \[
    \b{\Phi}(t,\tau)=\left[\begin{array}{cc}
    e^{\cos t - \cos \tau} & 0\\
    0 & e^{-\sin t +\sin \tau}
    \end{array}\right].
    \]
    Find the associated $\b A(t)$-matrix.

\item[{\bf (b)}] A fundamental matrix associated with a homogenous continuous-time LTV state model has the form,
    \[
    \b X(t)=\left[\begin{array}{cc}
    2 & -1\\
    -1 & 1
    \end{array}\right]\left[\begin{array}{cc}
    1 & 0\\
    t^2-t & t
    \end{array}\right].
    \]
    Compute the state-transition matrix $\b{\Phi}(t, \tau)$.
\end{itemize}
\end{exercise}

\begin{exercise}[R. Balakrishnan]\label{fw5_ltv_F2010}\rm
%    \begin{figure}[h]
%\centering
%\includegraphics[scale=0.75]{Fwk4p1.eps}
%\caption{Block diagram of a DC motor system for Exercise~\ref{fw4p1dc}.}
%\label{fig:Fwk4p1}
%\end{figure}
For the linear time-varying (LTV) system model\index{LTV system},
\[
\dot{\b x}(t)=\left[\begin{array}{cc}
0 & -2t\\
0 & -t
\end{array}\right]\b x(t) + \left[\begin{array}{c}
0\\
t
\end{array}\right]u(t),
\]
\begin{itemize}
\item Find the state transition matrix $\b{\Phi}(t,\tau)$;
    \item Let $\b x(1)=\left[\begin{array}{cc}
    1 & 1
    \end{array}\right]^{\top}$ and $u(t)=0$ for $t\ge 0$. Find $\b x(0)$;
\item Let $\b x(0)=\b 0$ and $u(t)=1$ for $t\ge 0$. Find $\b x(t)$ for $t\ge 0$.
\end{itemize}

\end{exercise}

