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\begin{document}
\section{Antilock Braking System Modeling}
The main components of an antilock brake system~(ABS) are: a control unit, a brake force actuator, and wheel speed sensors. The controller unit monitors the wheel speed, performs calculations based on the wheel speed, and sends the control signal to the brake force actuator. A diagram of an ABS unit is shown in Figure~\ref{fig:abs_p2}.
\begin{figure}
%\par
%\centerline{\psfig{figure=Absc-3.eps,height=8cm}}
%\par
\vspace{2in}
%\hskip.7in\special{bmp:c:/tony/absc.bmp x=5.33in y=4in}
\hskip.7in\special{ps:c:/mybook/Absc.ps y=2in}
\caption{Antilock brake system components.}
\label{fig:abs_p2}
\end{figure}
The ABS system is considered a safety feature in that the main purpose of an ABS is to minimize stopping distance under emergency braking conditions and yet maintain vehicle stability. Antilock brake systems minimize the stopping distance by preventing the wheels from locking up during braking. The vehicle brake force is proportional to the normal force on the tires. The proportionality factor that relates the vehicle brake force to the tire's normal force is the road coefficient of adhesion. The road coefficient of adhesion is a function of wheel slip as can be seen in Figure~\ref{fig:four_p2}.
\begin{figure}
%\par
%\centerline{\psfig{figure=Lmu-3.ps,height=8cm}}
%\par
\vspace{3.5in}
\hskip.75in\special{bmp:c:/tony/lmu.bmp y=3.5in}
%\hskip1in\special{ps:c:/mybook/Lmu.ps y=4in}
\caption{Road coefficient of adhesion versus wheel slip curves for different road surfaces.}
\label{fig:four_p2}
\end{figure}
The wheel slip is defined as the ratio of the slip velocity in the contact patch (forward velocity minus tire circumferential speed) to forward velocity. Let $\lambda$ denote the wheel slip, then
\begin{equation}
\lambda  = \frac{v-\omega R_w}{v},
\label{eq:wheeslipde}
\end{equation}
where
\begin{eqnarray*}
v  &=& \mbox{vehicle forward velocity}\\
\omega &=& \mbox{tire angular velocity}\\
R_w &=& \mb{tire rolling radius}.
\end{eqnarray*}
The above parameters are shown in Figure~\ref{fig:five_p2}.
\begin{figure}
%\par
%\centerline{\psfig{figure=Abs2-3.ps,height=8cm}}
%\par
\vspace{3in}
%\hskip.2in\special{bmp:c:/tony/abs2.bmp x=5.33in y=4in}
\hskip1.7in\special{ps:c:/mybook/abs2.ps y=3in}
\caption{Front wheel free body diagram; for rear wheel $F_{bf}$ becomes $F_{br}$.}
\label{fig:five_p2}
\end{figure}
Note that $0\le \lambda \le 1$. Following the convention, we specify the value of $\lambda$ using \%. Brake lock-up corresponds to 100\% wheel slip, and Figure~\ref{fig:four_p2} shows that at 100\% slip the road coefficient of adhesion is not maximized. The maximum road coefficient of adhesion is attained when the wheels are slipping. Therefore, the maximum braking force acting on the vehicle occurs when the wheels are slipping. Hence, the vehicle stopping distance is minimized by controlling the wheels to slip in the range where the road coefficient of adhesion is maximal. The maximum coefficient of adhesion is in the 5\% to 30\% slip range. Thus, by allowing the wheels to slip in the 5\% to 30\% range, we maximize brake force. The operation of an ABS systems is shown in Figure~\ref{fig:two_p2} that was adapted from Gillespie~\cite[page 68]{gillespie92}.
\begin{figure}
%\par
%\centerline{\psfig{figure=Abs4-3.ps,height=8cm}}
%\par
\vspace{3.5in}
\hskip1in\special{bmp:c:/tony/abs4.bmp y=3.5in}
\caption{ABS operating modes.}
\label{fig:two_p2}
\end{figure}
The application region is the range of slip (point A) where the driver applies the brakes and where the slip level is below impending wheel lock-up. When the wheel slip reaches the range where the coefficient of adhesion is maximized, the ABS system takes control from the driver (point B).  The ABS system releases the brake torque before the wheels lock-up (point C) and the wheels begin to pick up speed again.  As the wheel speed increases, the wheel slip increases until its slip falls in the maximum brake force region (point B). The ABS system then repeatedly applies the brake torque and releases the brake to keep the wheel slip within the optimal slip region (point B) near the peak of the $\mu$-slip curve.

An ABS model must include the wheel dynamics and road surface for accurate results. Here, we develop a simple model of an ABS for straight line braking with zero steering wheel input. In our model, we neglect pitching motion and consider motion only along the longitudinal axis. Our development follows that of Will and \.{Z}ak~\cite{will00}. A straight line braking vehicle free body diagram is shown in Figure~\ref{fig:three_p2}. The masses $m_s$, $m_{uf}$, and $m_{ur}$ are the vehicle's sprung, front unsprung, and rear unsprung masses. The front and rear tractive forces are denoted $F_{tf}$ and $F_{tr}$, respectively. Tractive force is the force developed at the road-to-tire interface that provides traction for either accelerating or braking. Tractive force is a function of the tire normal load and the road coefficient of adhesion.
\begin{figure}
\vspace{3.5in}
\hskip.75in\special{bmp:c:/tony/abs1.bmp y=3.5in}
\caption{A vehicle free body diagram.}
\label{fig:three_p2}
\end{figure}

The road coefficient of adhesion model we use was adapted from the data obtained by Layne, Passino, and Yurkovich~\cite{layne93} in their fuzzy learning control for antilock brake systems. The coefficient of adhesion $\mu$ versus wheel slip $\lambda$ curves for asphalt, wet asphalt, and ice surfaces are shown in Figure~\ref{fig:four_p2}. This is the surface model we use in later simulation studies. We denote the adhesion coefficient of front wheels by $\mu(\lambda_f)$ and that of the rear wheels by $\mu(\lambda_r)$

We now derive the equations for the total tractive force.  From the vehicle free body diagram, shown in Figure~\ref{fig:three_p2}, we write the equation of the total tractive force $F_{tot}$ as
\begin{equation}
F_{tot} = F_{tf} + F_{tr},
\label{eq:totracfo}
\end{equation}
where
\begin{eqnarray*}
F_{tf} &=& \mu(\lambda_f)F_{zf},\\ F_{tr} &=&
\mu(\lambda_r)F_{zr}.
\end{eqnarray*}
Upon substituting the above into~(\ref{eq:totracfo}), we obtain the expression for the total tractive force,
\begin{equation}
F_{tot} = \mu(\lambda_f)F_{zr} + \mu(\lambda_r) F_{zf}.
\label{eq:tofoaftsubs}
\end{equation}
Let
\begin{eqnarray*}
a &=& \mb{distance from sprung mass to front wheels}\\
b &=& \mb{distance from sprung mass to rear wheels}\\
h_{cg} &=& \mb{sprung mass height}\\
h_f &=& \mb{front unsprung mass height}\\
h_r &=& \mb{rear unsprung mass height}\\
m_s &=& \mb{sprung mass}\\
m_{uf} &=& \mb{front unsprung mass}\\
m_{ur} &=& \mb{rear unsprung mass}\\
m_{tot} &=& m_s + m_{uf} + m_{ur}.
\end{eqnarray*}
The normal forces acting on both the front and rear tires have two components. One component is due to the longitudinal weight transfer due to braking or acceleration. The other component is due to the static distribution of the masses. In our derivations of the expressions for the above mentioned forces, we follow arguments of Nalecz and Bindemann~\cite{nalecz88b,nalecz88c}, and Will and \.{Z}ak~\cite{will97}. We first concern ourselves with the components due to the longitudinal weight transfer. Consider the diagram shown in Figure~\ref{fig:three_p2}. Summing the moments about the line passing through the points of contact of the rear tires, and dividing the resulting equation by the wheel base $a+b$ yields
\begin{equation}
F_{zf1} = \left(m_{uf}h_{f}+ m_{s}h_{cg} +  m_{ur} h_{r}\right)\frac{\ddot{x}}{a+b}.
\end{equation}
Similarly, summing the moments about the line passing through the points of contact of the front tires and performing simple manipulations we obtain
\begin{equation}
F_{zr1} = -\left(m_{uf}h_{f}+ m_{s}h_{cg} +  m_{ur} h_{r}\right)\frac{\ddot{x}}{a+b}.
\end{equation}

We denote the components of normal forces due to the static distribution of the masses by $F_{zf2}$ and $F_{zr2}$. The static force acting on the front tires is
\begin{equation}
F_{zf2} = \frac{b}{a+b}m_{tot}g.
\end{equation}
The static force acting on the rear tires is
\begin{equation}
F_{zr2} = \frac{a}{a+b}m_{tot}g.
\end{equation}

The total normal force acting on the front tires is
\begin{eqnarray}
F_{zf}&=&-F_{zf1}+F_{zf2} \nonumber\\
&=&-\left(m_{uf}h_{f}+ m_{s}h_{cg} +  m_{ur} h_{r}\right)\frac{\ddot{x}}{a+b} + \frac{b}{a+b}m_{tot}g.
\label{eq:force_zf}
\end{eqnarray}
Similarly, the total normal force acting on the rear tires is
\begin{eqnarray}
F_{zr}&=&-F_{zr1}+F_{zr2} \nonumber\\
&=&\left(m_{uf}h_{f}+ m_{s}h_{cg} +  m_{ur} h_{r}\right)\frac{\ddot{x}}{a+b} + \frac{a}{a+b}m_{tot}g.
\label{eq:force_zr}
\end{eqnarray}
Let
\begin{eqnarray*}
m_{eq1} &=& \frac{b m_{tot}}{a+b},\\
m_{eq2} &=& \frac{h_s m_s + h_r m_{ur} + h_f m_{uf}}{a+b},\\
m_{eq3} &=& \frac{a m_{tot}}{a+b}.
\end{eqnarray*}
Then, equations~(\ref{eq:force_zf}) and~(\ref{eq:force_zr}) can be represented as
\begin{eqnarray*}
F_{zf} &=&  g m_{eq1} - m_{eq2}\ddot{x},\\
F_{zr} &=&  g m_{eq3} + m_{eq2}\ddot{x}.
\end{eqnarray*}
Hence, the total tractive force is
\begin{eqnarray}
F_{tot} &=& F_{zf}+F_{zr} \nonumber\\  &=&\mu(\lambda_f)(m_{eq1}g-m_{eq2}\ddot{x})+ \mu(\lambda_r)(m_{eq3}g+m_{eq2}\ddot{x}).
\label{eq:tractive}
\end{eqnarray}
Then, the equation of motion of the vehicle undergoing a braking maneuver is
\[
m_{tot}\ddot{x} =-F_{tot}.
\]
Hence,
\begin{equation}
\ddot{x} = -g\frac{\mu(\lambda_f)m_{eq1}+ \mu(\lambda_r)m_{eq3}}{m_{tot} - \mu(\lambda_f)m_{eq2}+ \mu(\lambda_r)m_{eq2}}.
\label{eq:long_a}
\end{equation}


Next, we model the front and rear wheel dynamics using the free body diagram shown in Figure~\ref{fig:five_p2}. Let $T_e$ be the engine torque, $F_{bf}$ and $F_{br}$ denote the front and rear brake forces, and $R_b$ be the mean radius of the brake lining contact. It is assumed that the vehicle is front wheel driven, and hence $T_e$ will only appear in the front wheel dynamics equation. We model the front and rear wheel dynamics as:
\begin{eqnarray}
2J_f\dot{\omega}_f &=& -F_{bf}R_b + \mu(\lambda_f )F_{zf}R_w + T_e \label{eq:omega_f}\\
2J_r\dot{\omega}_r &=& -F_{br}R_b + \mu(\lambda_r )F_{zr}R_w.\label{eq:omega_r}
\end{eqnarray}
Substituting equations~(\ref{eq:force_zf}) and~(\ref{eq:force_zr}) into (\ref{eq:omega_f}) and (\ref{eq:omega_r}),  we get
\begin{eqnarray}
\dot{\omega}_f &=& \frac{1}{2J_f}\left(-F_{bf}R_b +\mu(\lambda_f )m_{eq1}R_wg -\mu(\lambda_f)m_{eq2}R_w\ddot{x}  + T_e\right)\label{eq:omfdot}\\
\dot{\omega}_r &=& \frac{1}{2J_r}\left(-F_{br}R_b + \mu(\lambda_r)m_{eq3}R_wg +\mu(\lambda_r)m_{eq2}R_w\ddot{x} \right).\label{eq:omrdot}
\end{eqnarray}
We assume that a single brake force actuator supplies the braking forces $F_{bf}$ and $F_{br}$. This actuator powers both the front and rear brakes. Let the brake stiffness coefficient for the $i$-th wheel be denoted by $C_i$ where $i$=1, 2 corresponds to the left and right front wheels, while i=3, 4 corresponds to the left and right rear wheels. Then, we model the front brake force actuator as
\begin{equation}
F_{bf} = \mu_b(C_1 + C_2)u,
\label{eq:fbf}
\end{equation}
and we model the rear force actuator as
\begin{equation}
F_{br} = \mu_b(C_3 + C_4) k_{br} u ,
\label{eq:fbr}
\end{equation}
where $u$ is the brake actuator displacement, $\mu_b$ is the brake pad coefficient of friction, and $k_{br}$ is the brake displacement proportionality constant. In a manual four-wheel brake system, the rear brake force is a fraction of the front wheel brake force. Typical values for $k_{br}$ are in the range $[0.2,\, 0.6]$. The reason for proportioning the rear brake force in a brake system is to prevent the rear wheels from lock-up, even when front brake lock-up occurs. Rear wheel lock-up destabilizes the vehicle so that no steering input can stabilize it. However, when the front brakes lock-up, the vehicle still can be controlled to a certain degree. We now write the expressions for the front and rear wheel dynamics. Substituting~(\ref{eq:fbf}) into~(\ref{eq:omfdot}), we obtain an equation that models the front wheel dynamics. We have,
\begin{equation}
\dot{\omega}_f = \frac{1}{2J_f}\left(-\mu_b (C_1+C_2) R_bu + \mu(\lambda_f )m_{eq1}R_wg -\mu(\lambda_f)m_{eq2}R_w\ddot{x}  + T_e\right).
\label{eq:omff}
\end{equation}
Similarly, substituting~(\ref{eq:fbr}) into (\ref{eq:omrdot}) yields an equation that models the rear wheel dynamics. We have,
\begin{equation}
\dot{\omega}_r = \frac{1}{2J_r}\left(-\mu_b (C_3+C_4) R_b k_{br} u + \mu(\lambda_r)m_{eq3}R_wg +\mu(\lambda_r)m_{eq2}R_w\ddot{x}\right).
\label{eq:omrr}
\end{equation}


Let $x_1 = x$, $x_2 = \dot{x}_1$,  $x_3 = \omega_f$ , and $x_4 = \omega_r$. Then, we can represent equations~(\ref{eq:long_a}), (\ref{eq:omff}), and (\ref{eq:omrr}), that model a braking vehicle, in the state space format as
\begin{equation}
\left.\begin{array}{lll} \dot{x}_1 &=& x_2\\ \dot{x}_2  &=&
-g\frac{\mu(\lambda_f) m_{eq1}+
\mu(\lambda_r)m_{eq3}}{m_{tot}-\mu(\lambda_f)m_{eq2}+\mu(\lambda_r)m_{eq2}}
\\ \dot{x}_3 &=& \frac{1}{J_f}\left(-\mu_b (C_1+C_2) R_bu +
\mu(\lambda_f )m_{eq1}R_wg -\mu(\lambda_f)m_{eq2}R_w\ddot{x} +
T_e\right) \\ \dot{x}_4 &=& \frac{1}{J_r}\left(-\mu_b (C_3+C_4)
R_b k_{br} u + \mu(\lambda_r)m_{eq3}R_wg
+\mu(\lambda_r)m_{eq2}R_w\ddot{x}\right).
\end{array}\right\}
\label{eq:state}
\end{equation}
The output variable is
\begin{equation}
\b y =\left[\begin{array}{c}
x_1\\
x_2 \\
\lambda_f\\
\lambda_r
\end{array}\right]=\left[\begin{array}{c}
x_1\\
x_2 \\
\frac{x_2 - R_wx_3}{x_2}\\
\frac{x_2 - R_wx_4}{x_2}
\end{array}\right].
\label{eq:output}
\end{equation}

The vehicle data that we use in the simulation experiment is shown in Table~\ref{tab:veh_p2}. Upon substituting the data into equations~(\ref{eq:state}) and (\ref{eq:output}), we get
\begin{equation}
\left.
\begin{array}{lll}
\dot{x}_1 &=& x_2 \\ \dot{x}_2  &=& \frac{-7574.0\mu(\lambda_r)-
7140.7\mu(\lambda_f)}{1500 + 341.86(\mu(\lambda_r) -
\mu(\lambda_f))}
\\ \dot{x}_3 &=&
726.237\mu(\lambda_f)-32.778\mu(\lambda_r)\ddot{x}-31264.0u\\
\dot{x_4}&=&683.974\mu(\lambda_f)
+32.778\mu(\lambda_r)\ddot{x}-15632.0u,
\end{array}\right\}
\label{eq:final_s}
\end{equation}
and
\begin{equation}
\b y =\left[\begin{array}{c}
x_1\\
x_2 \\
\frac{x_2 - 0.326x_3}{x_2}\\
\frac{x_2 - 0.326x_4}{x_2}
\end{array}\right].
\label{eq:final_o}
\end{equation}
\begin{table}
\caption{Antilock brake system vehicle data.}
\begin{center}
\begin{tabular}{cc}\hline
$g$ & $9.81\,\mb{m/s}^2$\\
$a$ & $1.186\,\mb{m}$\\
$b$ & $1.258\,\mb{m}$\\
$h_{s}$ & $0.6\,\mb{m}$\\
$h_f$ & $0.30\,\mb{m}$\\
$h_r$ & $0.30\,\mb{m}$\\
$m_{tot}$ & $1500\,\mb{kg}$\\
$m_{s}$& $1285\,\mb{kg}$\\
$m_{uf}$ & $96\,\mb{kg}$\\
$m_{ur}$ & $119\,\mb{kg}$\\
$\mu_b$ & $0.90$\\
$J_f$ & $1.7\,\mb{kg\,m}^2$\\
$J_r$ & $1.7\,\mb{kg\,m}^2$\\
$R_b$ & $0.163\,\mb{m}$\\
$R_w$ & $0.326\,\mb{m}$\\
$T_e$ & $0.0\,\mb{Nm}$\\
$C_1=C_2$  &$362.3\,\mb{kN/m}$\\
$C_3=C_4$  &$362.3\,\mb{kN/m}$ \\
$k_{br}$  &$0.50\,$
\\ \hline
\end{tabular}
\end{center}
\label{tab:veh_p2}
\end{table}

We now test the above brake system model by simulating an emergency brake situation. The vehicle is traveling with constant speed, $\dot{x} = 20\,\mb{m/s}$, on a dry asphalt road before the brakes are applied. We first validate our brake model by simulating the situation when the brakes lock the wheels. Specifically, we apply a step brake force input of $1500\,\mb{Nm}$ and this input is held for 50~milliseconds and then stepped back to zero brake torque. The step brake input simulation result is shown in Figure~\ref{fig:six_p2}.
\begin{figure}
\vspace{4in}
\hskip.5in\special{bmp:c:/tony/abs8.bmp x=5.33in y=4in}
\caption{Response of the proposed brake system model to an emergency brake input.}
\label{fig:six_p2}
\end{figure}
The figure shows that the front wheels lock-up and the wheel skids at zero velocity until brake force is released. When the brake force is released, the wheel accelerates to the velocity of the vehicle. The vehicle decelerates during the brake application. The response of the proposed brake system model, shown in Figure~\ref{fig:six_p2}, is in agreement with the dynamic behavior of an actual brake system.


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