Compensator for 10/[s(s+1)], ζ=0.5, ωn=3

The specification is on the closed-loop pole positions, so a lead compensator is designed by the root-locus method.

Step 1: Present system

$$\frac{C}{R}=\frac{10}{s^2+s+10}\Rightarrow\omega_n=\sqrt{10}=3.16\ \text{rad/s},\ \zeta=\frac{1}{2\times3.16}=0.158$$

The response is very oscillatory ($M_p\approx60\%$).

Step 2: Desired dominant poles

$$s_d=-\zeta\omega_n\pm j\omega_n\sqrt{1-\zeta^2}=-1.5\pm j2.598$$

Step 3: Angle deficiency

Angle of $G(s)=\dfrac{10}{s(s+1)}$ at $s_d=-1.5+j2.598$:

$$\begin{aligned} \angle s_d&=180^\circ-\tan^{-1}\frac{2.598}{1.5}=120^\circ\\ \angle(s_d+1)&=180^\circ-\tan^{-1}\frac{2.598}{0.5}=100.9^\circ\\ \angle G(s_d)&=-(120^\circ+100.9^\circ)=-220.9^\circ \end{aligned}$$

To satisfy the angle condition ($-180^\circ$), the compensator must add

$$\phi=-180^\circ-(-220.9^\circ)=40.9^\circ\ \text{(lead)}$$

Step 4: Locate zero and pole

Choose the compensator zero at $s=-1$ to cancel the plant pole at $-1$. Then the pole $-p$ must satisfy

$$-\angle s_d-\angle(s_d+p)=-180^\circ\Rightarrow-120^\circ-\angle(s_d+p)=-180^\circ\Rightarrow\angle(s_d+p)=60^\circ$$ $$\tan^{-1}\frac{2.598}{p-1.5}=60^\circ\Rightarrow p-1.5=\frac{2.598}{1.732}=1.5\Rightarrow p=3$$

Check of angle added: zero $+100.9^\circ$, pole $-60^\circ$, net $+40.9^\circ$ ✓.

Step 5: Gain $K_c$ (magnitude condition)

$$G_c(s)G(s)=K_c\frac{s+1}{s+3}\cdot\frac{10}{s(s+1)}=\frac{10K_c}{s(s+3)}$$ $$\left|\frac{10K_c}{s_d(s_d+3)}\right|=1\Rightarrow10K_c=|s_d|\,|s_d+3|=3\times3=9\Rightarrow K_c=0.9$$ $$G_c(s)=0.9\,\frac{s+1}{s+3}=0.3\,\frac{1+s}{1+0.333s}$$

Step 6: Verification

$$\frac{C(s)}{R(s)}=\frac{9}{s^2+3s+9}$$

$\omega_n=3$ rad/s, $2\zeta\omega_n=3\Rightarrow\zeta=0.5$ ✓. Closed-loop poles: $-1.5\pm j2.598$.

         jw
   x sd  |  j2.598
    \    |
 ---x--o-x------- sigma
   -3 -1 0   (zero cancels pole at -1)

Note: $K_v$ falls to $10\times0.9/3=3$; if a higher $K_v$ is needed, the bisector method or an extra lag section can be used.

Answer: $G_c(s)=0.9\,\dfrac{s+1}{s+3}$, giving closed-loop poles at −1.5 ± j2.6 (ζ = 0.5, ωn = 3 rad/s).