Controller design for peak time 1.2 s (ζ = 0.55), Kv

A lead-type digital controller $G_D(z)=K\dfrac{z-\alpha}{z-\beta}$ is designed by the root-locus method in the z-plane. Its zero cancels the plant pole at $e^{-T}$.

Plant pulse transfer function ($T = 0.2$ s)

$$G(z) = (1-z^{-1})\,\mathcal{Z}\left[\frac{1}{s^2(s+1)}\right] = \frac{0.01873(z+0.9355)}{(z-1)(z-0.8187)}$$

Desired dominant poles

$$\begin{aligned} \omega_d &= \frac{\pi}{t_p} = \frac{\pi}{1.2} = 2.618\ \text{rad/s}\\ \omega_n &= \frac{2.618}{\sqrt{1-0.55^2}} = 3.135\ \text{rad/s},\quad \zeta\omega_n = 1.724\\ |z| &= e^{-1.724\times0.2} = 0.7083,\quad \angle z = 2.618\times0.2 = 0.5236\ \text{rad} = 30^\circ\\ z_1 &= 0.6134 + j0.3542 \end{aligned}$$

Angle condition

The angle deficiency is $64.72^\circ$, so lead compensation is needed.

• Place the controller zero at $\alpha = 0.8187$ to cancel the plant pole.
• The controller pole angle must be $\theta_\beta = 12.88^\circ - 137.50^\circ + 180^\circ = 55.38^\circ$:

$$\beta = 0.6134 - \frac{0.3542}{\tan 55.38^\circ} = 0.3689$$

Gain from the magnitude condition

$$K = \left|\frac{(z_1-1)(z_1-0.3689)}{0.01873(z_1+0.9355)}\right| = 7.582$$ $$G_D(z) = 7.582\,\frac{z-0.8187}{z-0.3689}$$

Check

• Closed-loop characteristic equation: $z^2 - 1.2269z + 0.5018 = 0$, roots $0.6134 \pm j0.3542$.
• Step response peak $c = 1.126$ at $k = 6$, i.e. $t_p = 1.2$ s.

Velocity error constant

$$\begin{aligned} K_v &= \frac{1}{T}\lim_{z\to1}(z-1)G_D(z)G(z) = \frac{1}{0.2}\cdot\frac{7.582\times0.01873(1.9355)}{1-0.3689}\\ &= 2.18\ \text{s}^{-1} \end{aligned}$$

So the steady-state error to a unit ramp is $1/K_v = 0.459$.

Answer: $G_D(z) = 7.582\,\dfrac{z-0.8187}{z-0.3689}$; $K_v = 2.18\ \text{s}^{-1}$.