Question 6.11: A Single-Link Robot Arm Driven by a DC Motor Consider the dy...
A Single-Link Robot Arm Driven by a DC Motor
Consider the dynamic system shown in Figure 6.46, in which a single-link robot arm is driven by a DC motor. The differential equation of the robot arm in terms of the motor variable \theta_{\mathrm{m}} was determined in Example 5.17 to be
\left(I_{\mathrm{m}}+N^{2}I\right) \ddot{\theta}_{\mathrm{m}}+\left(B_{\mathrm{m}}+N^{2} B\right) \dot{\theta}_{\mathrm{m}}=\tau_{\mathrm{m}^{\prime}}
where I_{\mathrm{m}} and I are the mass moments of inertia of the motor and the load, respectively, B_{\mathrm{m}} and B are the coefficients of the torsional viscous damping of the motor and the load, respectively, \tau_{\mathrm{m}} is the torque generated by the motor, and N is the gear ratio. Assume that the armature inductance is negligibly small, that is, L_{\mathrm{a}} \approx 0. The torque and the back emf constants of the motor are K_{\mathrm{t}} and K_{\mathrm{e}}, respectively.
a. Derive the differential equation relating the applied voltage v_{\mathrm{a}} and the link angular displacement \theta.
b. Determine the transfer function \Theta(s) / V_{\mathrm{a}}(s) using the differential equation obtained in Part (a). Assume that all initial conditions are zero.
a. For the electrical circuit, applying Kirchhoff’s voltage law gives
R_{\mathrm{a}} i_{\mathrm{a}}+\mathrm{e}_{\mathrm{b}}-v_{\mathrm{a}}=0,
where e_{\mathrm{b}}=K_{\mathrm{e}} \dot{\theta}_{\mathrm{m}}. With the given gear ratio,
\theta=N \theta_{m},
we have
R_{\mathrm{a}} i_{\mathrm{a}}=v_{\mathrm{a}}-K_{\mathrm{e}} \frac{1}{N} \dot{\theta}.
Thus, the current i_{\mathrm{a}} can be expressed as
i_{\mathrm{a}}=\frac{1}{R_{\mathrm{a}}} v_{\mathrm{a}}-\frac{K_{\mathrm{e}}}{R_{\mathrm{a}}} \frac{1}{N} \dot{\theta}.
The model of the mechanical part in terms of \theta is given by
\left(I_{\mathrm{m}}+N^{2} I\right) \frac{1}{N} \ddot{\theta}+\left(B_{\mathrm{m}}+N^{2} B\right) \frac{1}{N} \dot{\theta}=\tau_{\mathrm{m}^{\prime}}
where \tau_{\mathrm{m}}=K_{\mathrm{t}} i_{\mathrm{a}}. Substituting the expression of the current i_{\mathrm{a}^{\prime}} we obtain
\left(I_{\mathrm{m}}+N^{2} I\right) \frac{1}{N} \ddot{\theta}+\left(B_{\mathrm{m}}+N^{2} B\right) \frac{1}{N} \dot{\theta}=K_{\mathrm{t}}\left(\frac{1}{R_{\mathrm{a}}} V_{\mathrm{a}}-\frac{K_{\mathrm{e}}}{R_{\mathrm{a}}} \frac{1}{N} \dot{\theta}\right) .
Rearranging the equation yields
\left(I_{\mathrm{m}}+N^{2} I\right) \ddot{\theta}+\left(B_{\mathrm{m}}+N^{2} B+\frac{K_{\mathrm{t}} K_{\mathrm{e}}}{R_{\mathrm{a}}}\right) \dot{\theta}=\frac{N K_{\mathrm{t}}}{R_{\mathrm{a}}} V_{\mathrm{a}} .
b. Taking the Laplace transform results in
\frac{\Theta(s)}{V_{\mathrm{a}}(s)}=\frac{N K_{\mathrm{t}} / R_{\mathrm{a}}}{\left(I_{\mathrm{m}}+N^{2} I\right) s^{2}+\left(B_{\mathrm{m}}+N^{2} B+K_{\mathrm{t}} K_{\mathrm{e}} / R_{\mathrm{a}}\right) s} .
This electromechanical system can be modeled using Simulink and Simscape. The details will be discussed in Section 6.6.