Exercise 1.1

Analysis ∙ Linear dynamical systems

Fig. 1 Schematic of the motor (courtesy of Exercise Manual for Automatic Control)

The scheme above is a DC motor characterized by the following physical relationships. The rotating axis is described by

\[J\ddot{\theta} = -f\dot{\theta} + M,\]

where \(\theta\) is the angle of rotation, \(M\) is the torque, \(J\) is the angular momentum, and \(f\) is the frictional coefficient.

The torque is related to the stator current by the following relationship

\[M = k_a i\]

where \(i\) is the stator current and \(k_a\) is a proportional constant characteristic for the motor. In addition,

\[v = k_v\dot{\theta}\]

where \(v\) is the induced voltage of the motor and \(k_v\) is a known proportional constant.

Let the input voltage \(u\) be the control signal and \(\theta\) be the output.

1. Write a differential equation that relates \(u\) and \(\theta\) (the inductance \(L_a\) can be neglected) in the Laplace domain.

2. Determine the transfer function between the input and the output.

3. Discuss the behavior of the system by calculating step response of the system. That is, \(\theta(t)\) when \(u = \textrm{step}(t)\).

Hint 1

To get the relationship between the voltage \(u\) and the current flowing through the motor, we can apply Ohm’s law:

\[ u(t) = R_a i(t) + L_a \frac{d}{dt} i(t) + v(t) \]

Note that \(L_a \frac{d}{dt} i(t)\) is the voltage drop at the pins of the inductance, see Equation (1) of this page.

Hint 2

To get the transfer function between the input and the output, the balance of momentum needs to be applied:

\[ J \ddot{\theta} = -f_s \dot{\theta} + M = -f_s \dot{\theta} + k_a i(t) \]

where:

• \(J \ddot{\theta}\) is the angular intertia (\(\ddot{\theta} = \frac{d^2}{dt^2}\theta\) is the angular acceleation)

• \(-f_s \dot{\theta}\) is the friction term, proportional to the angular velocity

• \(M = k_a i(t)\) is the force applied by the motor, proportional to the current \(i(t)\) flowing throught the motor.