\section{The Modeling of an Automotive Suspension}
The role of an automotive suspension is to support the vehicle body on the axles. A simple linear model of the suspenssion assembly associated with a single wheel, called quarter-car model, is shown in Figure~\ref{fig:actsus}. We represent the vehicle body by the ``sprung'' mass, while the tire and axle by the ``unsprung'' mass.
\begin{figure}[h]
\vspace{4in}
\hskip1in\special{bmp:c:/mybook/actsus.bmp x=5.33in y=4in}
\caption{Linear quarter-car model}
\label{fig:actsus}
\end{figure}
We model the tire as a spring with spring coefficient $k_t$, and assume its damping to be negligible. The suspension consists of the spring, shock absorber, and the variable force element. The role of the variable force element is to generate a force that compensates for the uneven road which is the source of the system disturbances. We assume that the variable force element $F_a$ can instantaneously provide any desired force. Thus we have a linear active suspension. The equations modeling the dynamics of the two degree of freedom system depicted in Figure~\ref{fig:actsus} are derived by applying Newton's law to each mass of the system, and taking into account the positive direction of the $z$ coordintes. We obtain
\begin{eqnarray*}
m_s\ddot{z}_s&=& -k_s(z_s-z_u)-b_s(\dot{z}_s-\dot{z}_u)+F_a\\
m_u\ddot{z}_u &=& k_s(z_s-z_u)+b_s(\dot{z}_s-\dot{z}_u)-F_a-k_t(z_u-z_r).
\end{eqnarray*}
We define the following state variables:
\begin{eqnarray*}
x_1 &=& z_s-z_u=\mb{suspension stroke},\\
x_2 &=& \dot{z}_s =\mb{sprung mass velocity},\\
x_3 &=& z_u-z_r =\mb{tire deflection},\\
x_4 &=& \dot{z}_u=\mb{unsprung mass velocity}.
\end{eqnarray*}
Let $u=F_a$ be the system input, and let $d(t)=\dot{z}_r(t)$ be the disturbance input due to the road roughness. Then the linear model active suspension can be represented in state-space format as
\[
\dot{\b {x}}(t)=\left[\begin{array}{cccc}
0 & 1 & 0 & -1\\
-\frac{k_s}{m_s}& -\frac{b_s}{m_s}& 0 & \frac{b_s}{m_s}\\
0 & 0 & 0 & 1\\
\frac{k_s}{m_u} & \frac{b_s}{m_u} & -\frac{k_t}{m_u} & -\frac{b_s}{m_u}
\end{array}\right]\b x(t)+\left[\begin{array}{c}
0\\
\frac{1}{m_s}\\
0\\
-\frac{1}{m_u}
\end{array}\right]u(t)+\left[\begin{array}{c}
0\\
0\\
-1\\
0
\end{array}\right]d(t).
\]
For the simulation purpose we use tha data of a commercial vehicle that is given in Table~\ref{tab:actsus}. The data comes from Sunwoo and Cheok~\cite{sunwoo90}.
\begin{table}
\begin{center}
\begin{tabular}{cc}
$m_s$ & $355\,kg$\\
$m_u$ & $46\,kg$\\
$k_s$ & $14384\,N/m$\\
$b_s$ & $1860\,N/m/sec$\\
$k_t$ & $300000\,N/m$
\end{tabular}
\end{center}
\caption{Data for a vehicle suspension.}
\label{tab:actsus}
\end{table}
Substituting the vehicle data into the modeling equations yields
\[
\dot{\b {x}}(t)=\left[\begin{array}{cccc}
0 & 1 & 0 & -1\\
-40.5 & -5.2 & 0 & 5.2\\
0 & 0 & 0 &1\\
312.7 & 40.4 & -6521.7 & -40.4
\end{array}\right]\b x(t)+\left[\begin{array}{c}
0\\
0.0028\\
0\\
-0.0217
\end{array}\right]u(t)+\left[\begin{array}{c}
0\\
0\\
-1\\
0
\end{array}\right]d(t).
\]
Our goal is to isolate the sprung mass from road disturbances. In other words, we wish to design a control law such that the vertical acceleration of the sprung mass $\ddot{z}_s(t)=\dot{x}_2$ is as close to $0$ as possible in spite of road disturbances. Thus, we assume that we can measure the vertical acceleration of the sprung mass, which means that the output of the plant is
\begin{eqnarray*}
y(t) &=& \left[\begin{array}{cccc}
-\frac{k_s}{m_s}& -\frac{b_s}{m_s}& 0 & \frac{b_s}{m_s}
\end{array}\right]+\frac{1}{m_s}u(t)\\
&=& \left[\begin{array}{cccc}
-40.5 & -5.2 & 0 & 5.2
\end{array}\right]\b x(t) + 0.0028u(t).
\end{eqnarray*}