Solutions of Exercise 1.2

Fig. 4 Schematic of the motor with a load (courtesy of Exercise Manual for Automatic Control)

Assume that the DC-motor in Exercise 1.1 is running a load attached to an elastic axis, as in the figure above. Let \(k\) be the axis’ spring constant, and let \(J_L\) be the moment of inertia of the load.

1. Write the transfer function between the input voltage \(u\) and the angle \(\theta_L\).

Hint

To retrieve the governing equations, you can write the balance of momentum of both the load:

\[ J_L \ddot{\theta}_L = - f_L \dot{\theta}_L - k (\theta_L - \theta_m) \]

where, \(J_L \ddot{\theta}_L\) is the motor inertia equaling the sum of the friction (\(-f_L \dot{\theta}_L\)) and of the elastic force \( k (\theta_L - \theta_m)\), and of the motor

\[ J_m \ddot{\theta}_m = -f_m \dot{\theta}_m + k (\theta_L - \theta_m) + k_a i \]

where the motor inertia (\(J_m \ddot{\theta}_m\)) equals the sum of the friction term (\(-f_m \dot{\theta}_m\)), the elastic force (\( k (\theta_L - \theta_m)\)), and the motor torque (\(k_a i\)).

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Solution

To solve the system’s ODE in the Laplace domain, we replace the time derivative \(\frac{d}{dt}\) with the Laplace operator \(s\):

\[\begin{split} \begin{aligned} J_L s^2 {\Theta}_L(s) =& - f_L s {\Theta}_L(s) - k (\Theta_L(s) - \Theta_m(s)) \\ J_m s^2 {\Theta}_m(s) =& -f_m s {\Theta}_m(s) + k (\Theta_L(s) - \Theta_m(s)) + k_a I(s) \end{aligned} \end{split}\]

We can isolate \(\Theta_m(s)\) from the second equation:

\[ \Theta_m(s) = \frac{k}{J_m s^2 + f_m s + k} \Theta_L(s) + \frac{k_a}{J_m s^2+ f_m s + k} I(s) \]

Replacing \(\Theta_m(s)\) into the first equation, we get

\[ (J_L s^2 + f_L s + k) \Theta_L(s) = \frac{k^2}{J_m s^2 + f_m s + k} \Theta_L(s) + \frac{k \cdot k_a}{J_m s^2+ f_m s + k} I(s) \]

Isolating \(\Theta_L(s)\), and recalling that \(I(s) = \frac{U(s)}{R_a}\), we finally get our transfer function

\[ \Theta_L(s) = \frac{\frac{k \cdot k_a}{R_a}}{J_m J_L s^4 + (J_m f_L + J_L f_m)s^3 + (k J_m + k J_L + f_M f_L) s^2 + k(f_L + f_m) s} U(s) \]