Controller design for peak time ≤ 1.2 s (ζ = 0.5)

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

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

With $e^{-0.2} = 0.8187$:

$$\begin{aligned} G(z) &= (1-z^{-1})\,\mathcal{Z}\left[\frac{1}{s^2(s+1)}\right] = \frac{(T-1+e^{-T})z + (1-e^{-T}-Te^{-T})}{(z-1)(z-e^{-T})}\\ &= \frac{0.01873(z+0.9355)}{(z-1)(z-0.8187)} \end{aligned}$$

Desired dominant poles

$$\begin{aligned} t_p &= \frac{\pi}{\omega_d} = 1.2 \Rightarrow \omega_d = 2.618\ \text{rad/s}\\ \omega_n &= \frac{\omega_d}{\sqrt{1-0.5^2}} = 3.023\ \text{rad/s},\quad \zeta\omega_n = 1.511\\ |z| &= e^{-\zeta\omega_nT} = e^{-0.3023} = 0.7391, \quad \angle z = \omega_dT = 0.5236\ \text{rad} = 30^\circ\\ z_1 &= 0.6401 + j0.3696 \end{aligned}$$

That is 12 samples per cycle of damped oscillation, which is adequate.

Angle deficiency

The controller must add $+56.84^\circ$.

Controller zero and pole

• Zero at $\alpha = 0.8187$ cancels the plant pole and removes its $-115.80^\circ$.
• Then the controller pole must give $13.20^\circ - 134.24^\circ - \theta_\beta = -180^\circ$, so $\theta_\beta = 58.96^\circ$:

$$\beta = 0.6401 - \frac{0.3696}{\tan 58.96^\circ} = 0.4177$$

Gain

$$K = \left|\frac{(z_1-1)(z_1-0.4177)}{0.01873(z_1+0.9355)}\right| = 7.340$$ $$G_D(z) = 7.340\,\frac{z-0.8187}{z-0.4177}$$

Check

• Characteristic equation: $z^2 - 1.2802z + 0.5463 = 0$, roots $0.6401 \pm j0.3696$, which is the design point.
• Step response: $c(kT) = 0,\ 0.137,\ 0.442,\ 0.757,\ 0.994,\ 1.125,\ 1.163,\ 1.141,\ \ldots$. The peak is 1.163 at $t = 6T = 1.2$ s, so $t_p \le 1.2$ s is met.
• Velocity error constant:

$$K_v = \frac{1}{T}\lim_{z\to1}(z-1)G_D(z)G(z) = \frac{7.340\times0.03625}{0.2\times(1-0.4177)} = 2.28\ \text{s}^{-1}$$

Answer: $G_D(z) = 7.340\,\dfrac{z-0.8187}{z-0.4177}$, giving closed-loop poles $0.6401\pm j0.3696$ ($\zeta=0.5$, $t_p=1.2$ s).