Question 5.5: A Two-Degree-of-Freedom Quarter-Car Model Consider a quarter...
A Two-Degree-of-Freedom Quarter-Car Model
Consider a quarter-car model shown in Figure 5.34a, where m_{1} is the mass of one-fourth of the car body and m_{2} is the mass of the wheel-tire-axle assembly. The spring k_{1} represents the elasticity of the suspension and the spring k_{2} represents the elasticity of the tire. z(t) is the displacement input due to the surface of the road. The actuator force, f, applied between the car body and the wheel-tire-axle assembly, is controlled by feedback and represents the active components of the suspension system.
a. Draw the necessary free-body diagrams and derive the differential equations of motion.
b. Determine the state-space representation. Assume that the displacements of the two masses, x_{1} and x_{2}, are the outputs and the state variables are x_{1}=x_{1}, x_{2}=x_{2}, x_{3}=\dot{x}_{1}, and x_{4}=\dot{x}_{2}.
c. The parameter values are m_{1}=290 \mathrm{~kg}, m_{2}=59 \mathrm{~kg}, b_{1}=1000 \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}, k_{1}=16,182 \mathrm{~N} / \mathrm{m}, and k_{2}=19,000 \mathrm{~N} / \mathrm{m}. Use MATLAB commands to define the system in the state-space form and then convert it to the transfer function form. Assume that all the initial conditions are zero.
a. We choose the displacements of the two masses x_{1} and x_{2} as the generalized coordinates.
The static equilibrium positions of m_{1} and m_{2} are set as the coordinate origins. Assume
x_{1}>x_{2}>z>0,
which implies that the springs are in tension and
\dot{x}_{1}>\dot{x}_{2}>\dot{z}>0.
The free-body diagrams of m_{1} and m_{2} are shown in Figure 5.34b. According to the assumption, the mass m_{1} moves faster than the mass m_{2}, and the elongation of the spring k_{1} is x_{1}-x_{2}. The force exerted by the spring k_{1} on the mass m_{1} is downward as it tends to restore to the undeformed position. Because of Newton’s third law, the force exerted by the spring k_{1} on the mass m_{2} has the same magnitude, but opposite in direction. Other spring forces and damping forces can be determined using the same logic. Note that the gravitational forces, m_{1} g and m_{2} g, are not included in the free-body diagrams.
Applying Newton’s second law to the masses m_{1} and m_{2}, respectively, gives
\begin{gathered} +\uparrow x: \sum F_{x}=m a_{x \prime} \\ f-k_{1}\left(x_{1}-x_{2}\right)-b_{1}\left(\dot{x}_{1}-\dot{x}_{2}\right)=m_{1} \ddot{x}_{1} \\ -f+k_{1}\left(x_{1}-x_{2}\right)+b_{1}\left(\dot{x}_{1}-\dot{x}_{2}\right)-k_{2}\left(x_{2}-z\right)=m_{2} \ddot{x}_{2} . \end{gathered}
Rearranging the equations into the standard input-output form,
\begin{gathered} m_{1} \ddot{x}_{1}+b_{1} \dot{x}_{1}-b_{1} \dot{x}_{2}+k_{1} x_{1}-k_{1} x_{2}=f \\ m_{2} \ddot{x}_{2}-b_{1} \dot{x}_{1}+b_{1} \dot{x}_{2}-k_{1} x_{1}+\left(k_{1}+k_{2}\right) x_{2}=-f+k_{2} z_{1} \end{gathered}
which can be expressed in second-order matrix form (Section 4.1) as
\left[\begin{array}{cc} m_{1} & 0 \\ 0 & m_{2} \end{array}\right]\left\{\begin{array}{l} \ddot{x}_{1} \\ \ddot{x}_{2} \end{array}\right\}+\left[\begin{array}{cc} b_{1} & -b_{1} \\ -b_{1} & b_{1} \end{array}\right]\left\{\begin{array}{l} \dot{x}_{1} \\ \dot{x}_{2} \end{array}\right\}+\left[\begin{array}{cc} k_{1} & -k_{1} \\ -k_{1} & k_{1}+k_{2} \end{array}\right]\left\{\begin{array}{l} x_{1} \\ x_{2} \end{array}\right\}=\left[\begin{array}{cc} 1 & 0 \\ -1 & k_{2} \end{array}\right]\left\{\begin{array}{l} f \\ z \end{array}\right\} .
b. Note that the inputs to the system are the actuator force f and the road surface irregularity z. The state, the input, and the output vectors are
\mathbf{x}=\left\{\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array}\right\}=\left\{\begin{array}{l} x_{1} \\ x_{2} \\ \dot{x}_{1} \\ \dot{x}_{2} \end{array}\right\}, \quad \mathbf{u}=\left\{\begin{array}{l} f \\ z \end{array}\right\}, \quad \mathbf{y}=\left\{\begin{array}{l} x_{1} \\ x_{2} \end{array}\right\}
The state-variable equations are then obtained as
\begin{aligned} \dot{x}_{1} & =x_{3}, \\ \dot{x}_{2} & =x_{4} \\ \dot{x}_{3} & =\ddot{x}_{1}=-\frac{k_{1}}{m_{1}} x_{1}+\frac{k_{1}}{m_{1}} x_{2}-\frac{b_{1}}{m_{1}} \dot{x}_{1}+\frac{b_{1}}{m_{1}} \dot{x}_{2}+\frac{f}{m_{1}} \\ & =-\frac{k_{1}}{m_{1}} x_{1}+\frac{k_{1}}{m_{1}} x_{2}-\frac{b_{1}}{m_{1}} x_{3}+\frac{b_{1}}{m_{1}} x_{4}+\frac{1}{m_{1}} u_{1} \\ \dot{x}_{4} & =\ddot{x}_{2}=\frac{k_{1}}{m_{2}} x_{1}-\frac{k_{1}+k_{2}}{m_{2}} x_{2}+\frac{b_{1}}{m_{2}} \dot{x}_{1}-\frac{b_{1}}{m_{2}} \dot{x}_{2}-\frac{f}{m_{2}}+\frac{k_{2}}{m_{2}} z \\ & =\frac{k_{1}}{m_{2}} x_{1}-\frac{k_{1}+k_{2}}{m_{2}} x_{2}+\frac{b_{1}}{m_{2}} x_{3}-\frac{b_{1}}{m_{2}} x_{4}-\frac{1}{m_{2}} u_{1}+\frac{k_{2}}{m_{2}} u_{2} . \end{aligned}
The output equation is
\mathbf{y}=\left\{\begin{array}{l} x_{1} \\ x_{2} \end{array}\right\}
Thus, the state-space representation is
\begin{aligned} \left\{\begin{array}{l} \dot{x}_{1} \\ \dot{x}_{2} \\ \dot{x}_{3} \\ \dot{x}_{4} \end{array}\right\} & =\left[\begin{array}{cccc} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -\frac{k_{1}}{m_{1}} & \frac{k_{1}}{m_{1}} & -\frac{b_{1}}{m_{1}} & \frac{b_{1}}{m_{1}} \\ \frac{k_{1}}{m_{2}} & -\frac{k_{1}+k_{2}}{m_{2}} & \frac{b_{1}}{m_{2}} & -\frac{b_{1}}{m_{2}} \end{array}\right]\left\{\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array}\right\}+\left[\begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \frac{1}{m_{1}} & 0 \\ -\frac{1}{m_{2}} & \frac{k_{2}}{m_{2}} \end{array}\right]\left\{\begin{array}{l} u_{1} \\ u_{2} \end{array}\right\}, \\ \mathbf{y} & =\left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{array}\right]\left\{\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array}\right\}+\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]\left\{\begin{array}{l} u_{1} \\ u_{2} \end{array}\right\} . \end{aligned}
c. The following is the MATLAB session:
>> m1 = 290;
>> m2 = 59;
>> b1 = 1000;
>> k1 = 16182;
>> k2 = 19000;
>> A = [0 0 1 0;
0 0 0 1;
-k1/m1 k1/m1 -b1/m1 b1/m1;
k1/m2 -(k1+k2)/m2 b1/m2 -b1/m2];
>> B = [0 0; 0 0; 1/m1 0; -1/m2 k2/m2];
>> C = [1 0 0 0; 0 1 0 0];
>> D = zeros(2,2);
>> sys_ss = ss(A,B,C,D);
>> sys_tf = tf(sys_ss);
The command tf returns two transfer functions from the input #1 (i.e., f ) to the two outputs x_{1} and x_{2} :
\begin{aligned} & \frac{X_{1}(s)}{F(s)}=\frac{0.003448 s^{2}+1.11}{s^{4}+20.4 s^{3}+652.1 s^{2}+1110 s+17,970}, \\ & \frac{X_{2}(s)}{F(s)}=\frac{-0.01695 s^{2}+0.04655 s+0.7533}{s^{4}+20.4 s^{3}+652.1 s^{2}+1110 s+17,970}, \end{aligned}
and another two transfer functions from the input #2 (i.e., z ) to the outputs x_{1} and x_{2} :
\begin{aligned} & \frac{X_{1}(s)}{Z(s)}=\frac{1110 s+17,970}{s^{4}+20.4 s^{3}+652 \cdot 1 s^{2}+1110 s+17,970}, \\ & \frac{X_{2}(s)}{Z(s)}=\frac{322 s^{2}+1110 s+17,970}{s^{4}+20.4 s^{3}+652 \cdot 1 s^{2}+1110 s+17,970} . \end{aligned}
Note that the system has two inputs and two outputs. Thus, there are a total of four transfer functions, which can be formed as a 2 \times 2 transfer matrix
\mathbf{G}(s)=\left[\begin{array}{ll} G_{11}(s) & G_{12}(s) \\ G_{21}(s) & G_{22}(s) \end{array}\right]=\left[\begin{array}{cc} \frac{X_{1}(s)}{F(s)} & \frac{X_{1}(s)}{Z(s)} \\ \frac{X_{2}(s)}{F(s)} & \frac{X_{2}(s)}{Z(s)} \end{array}\right] .
Note that the gravity terms in Example 5.5 do not appear in the equations of motion because the static equilibrium positions are chosen as the coordinate origins. Two independent coordinates, x_{1} and x_{2}, are required to specify the system dynamics. Such a system is called a two-degree-offreedom system, which is a special case of multiple-degree-of-freedom systems.