PID Controller

The general open loop transfer function which models the angular velocity of $\omega(t)$ of a motor is:

where $A$ and $\tau$ are positive constants.

1

What is the transfer function of the angular position of a motor $\theta(t)$?

Since angular velocity is the derivative of angular position, we have: $\omega(t) = \frac{d \theta(t)}{dt}$

From Theorem 7 of Table 2.2: $\mathcal{L} \{ \frac{d f(t)}{dt} \} = s F(s) - f(0^-)$.

Assuming the initial angular position is zero, or $\theta(0^-) = 0$: $\mathcal{L}(\omega(t)) = \omega(s) = s \Theta(s)$

Transfer function of the angular position of a motor is $\Theta(s) = \frac{A}{s(\tau s + 1)} \cdot U(s)$

2

What, if any, is the steady state value of $\omega(t)$ in open loop response to a step input:

$$u(t) = \begin{cases} U_{\mathcal{o}}, & t \geq 0 \\\ 0, & t < 0 \end{cases}$$

Using the final value theorem from Laplace transform, we have:

Laplace transform of $u(t)$ is $U(s) = \frac{U_{\mathcal{o}}}{s}$

the steady-state value of $\omega(t)$ is: