Part 11 - Design of controllers (2)

PD-controller

Given that $u(t) = K_p(e(t) + T_d \dot{e}(t))$

In the $s$-domain, the PD controller is: $$ F_{PD}(s) = \dfrac{U(s)}{E(s)} = K_p(1 + T_d s) $$

PD-controller are often paired with low-pass filters (LP), so that $\dot{e}(t)$ doesn’t become to large.

$$ \begin{align*} F_{PD}(s) & = K_p(1 + \dfrac{T_d s}{1 + T_f s}) \\ & = K_p(\dfrac{1 + (T_d + T_f) s}{1 + T_f s}) \\ & = K_p(\dfrac{1 + \tau_d s}{1 + \frac{\tau_d}{b} s}) \ | \ \tau_d = T_d + T_f, b = \frac{T_d + T_f}{T_f} \end{align*} $$

This means that: $$ F(j\omega) = K_p(\dfrac{1 + \tau_d j \omega}{1 + \frac{\tau_d}{b} j\omega}) $$

$$ \omega_{max} = \dfrac{\sqrt{b}}{\tau_d} \ | \ \text{if we choose } \omega_c = \dfrac{\sqrt{b}}{\tau_d} $$

$$ arg(L(j\omega)) = arg(F_{PD}(j\omega)) + arg(G(j\omega)) $$

$$ \begin{align*} \phi_{max} & = arg(F_{PD}(j\omega)) \\ & = arctan(\tau_d \cdot \frac{\sqrt{b}}{\tau_d}) - arctan(\frac{\tau_d}{b} \cdot \frac{\sqrt{b}}{\tau_d}) \\ & = arctan(\sqrt{b}) - arctan(\frac{1}{\sqrt{b}}) \end{align*} $$

$$ b = \ldots = \dfrac{1 + sin(\phi_{max})}{1 - sin(\phi_{max})} $$

Example

$$ G(s) = \dfrac{2 - s}{s(1 + s)^2}, \phi_m = 50^\circ, \omega_c = 0.5 $$

Design a PD-controller $$ F_{PD}(s) = K_p \dfrac{1 + \tau_d s}{1 + \frac{\tau_d}{b} s} $$

$$ \begin{align*} \phi_{max} & = 50^\circ - 180^\circ - arg(G(j\omega_c)) \\ & = 50^\circ - 180^\circ - (arctan(\frac{\omega_c}{2}) - 90^\circ - 2 arctan(\omega_c)) \\ & = 27^\circ \end{align*} $$

$$ b = \dfrac{1 + sin(\phi_{max})}{1 - sin(\phi_{max})} = 2.66 $$

$$ \tau_d = \dfrac{\sqrt{b}}{\omega_c} = 3.26 $$

$$ |F(j\omega_c)||G(j\omega_c)| = 1 $$

$$ K_p \cdot \dfrac{\sqrt{1^2 + (\tau_d \cdot \omega_c)^2}}{\sqrt{1^2 + (\frac{\tau_d}{b} \cdot \omega_c)^2}} \cdot \dfrac{\sqrt{2^2 + (-\omega_c)^2}}{\omega_c (\sqrt{1^2 + \omega_c^2})^2} = 1 $$

$$ K_p = 0.186 $$

Ziegler-Nichols method

1. Set up the controller as a pure P-controller ($T_i = \infty, T_d = 0$).

2. Increase $K_p$ until self-oscillation occurs, let $K_p = K_0$ and the period be $T_0$.

Then, approximately: