Root-locus design: ζ = 0.5, ts = 2 s, plant 1/(s(s+2))

Step 1: Desired closed-loop poles

From the specifications $\zeta = 0.5$ and $t_s = 2$ s (2% criterion, $t_s = 4/\zeta\omega_n$):

$$\begin{aligned} \zeta\omega_n &= \frac{4}{t_s} = 2, \qquad \omega_n = 4\ \text{rad/s} \\ \omega_d &= \omega_n\sqrt{1-\zeta^2} = 3.464\ \text{rad/s} \end{aligned}$$

With $T = 0.2$ s, the samples per cycle are $\omega_s/\omega_d = 31.42/3.464 \approx 9$.

$$\begin{aligned} |z| &= e^{-\zeta\omega_nT} = e^{-0.4} = 0.6703 \\ \angle z &= \omega_dT = 0.6928\ \text{rad} = 39.69^\circ \\ z_{1,2} &= 0.6703\angle \pm 39.69^\circ = 0.5158 \pm j0.4281 \end{aligned}$$

Step 2: Plant pulse transfer function

$$\begin{aligned} G(z) &= (1-z^{-1})\,\mathcal{Z}\left[\frac{1}{s^2(s+2)}\right] = \frac{(2T-1+e^{-2T})z + (1-e^{-2T}-2Te^{-2T})}{4(z-1)(z-e^{-2T})} \\ &= \frac{0.01758(z + 0.8753)}{(z-1)(z-0.6703)} \end{aligned}$$

Step 3: Angle deficiency at $z_1 = 0.5158 + j0.4281$

The angle condition requires $-180^\circ$, so the controller must add $+51.26^\circ$ of phase lead.

Step 4: Lead controller

Take $G_D(z) = K\dfrac{z-\alpha}{z-\beta}$. Put the zero on the plant pole, $\alpha = 0.6703$, to cancel it. This zero adds $+109.85^\circ$. The pole must then contribute $109.85^\circ - 51.26^\circ = 58.59^\circ$:

$$\beta = 0.5158 - \frac{0.4281}{\tan 58.59^\circ} = 0.2543$$

Step 5: Gain from the magnitude condition

After cancellation, the open-loop function is $\dfrac{0.01758K(z+0.8753)}{(z-1)(z-0.2543)}$, and $|G_DG| = 1$ at $z_1$:

$$K = \frac{|z_1-1|\,|z_1-0.2543|}{0.01758\,|z_1+0.8753|} = \frac{0.6464 \times 0.5017}{0.01758 \times 1.4555} = 12.67$$ $$G_D(z) = 12.67\,\frac{z - 0.6703}{z - 0.2543}$$

Step 6: Check

$$\begin{aligned} G_D(z)G(z) &= \frac{0.2228(z + 0.8753)}{(z-1)(z-0.2543)} \\ \text{Characteristic eq.:}\quad z^2 &- 1.0316z + 0.4493 = 0 \end{aligned}$$

The roots are $z = 0.5158 \pm j0.4281$, the desired poles. ✔ This is a second-order closed loop, so the poles are exactly dominant.

Static velocity error constant:

$$K_v = \lim_{z\to1}\frac{(1-z^{-1})G_DG}{T} = \frac{1}{0.2}\cdot\frac{0.2228 \times 1.8753}{0.7457} = 2.80\ \text{s}^{-1}$$

Answer: $G_D(z) = 12.67\dfrac{z-0.6703}{z-0.2543}$, a lead compensator, with $K_v = 2.80\ \text{s}^{-1}$.