Lead compensator for 10/[s(s+1)], ts=2 s, Mp=5%

Time-domain specifications are met by placing the dominant closed-loop poles with a lead compensator designed on the root locus.

Step 1: Present system

$$\frac{C}{R}=\frac{10}{s^2+s+10}:\ \omega_n=3.16\ \text{rad/s},\ \zeta=0.158,\ M_p=60.5\%,\ t_s=\frac{4}{0.5}=8\ \text{s}$$

Step 2: Desired poles

From $M_p=5\%$:

$$\zeta=\frac{-\ln0.05}{\sqrt{\pi^2+(\ln0.05)^2}}=\frac{2.996}{4.341}=0.690$$

From $t_s=2$ s (2% criterion):

$$\zeta\omega_n=\frac{4}{2}=2\Rightarrow\omega_n=\frac{2}{0.690}=2.898\ \text{rad/s},\quad\omega_d=2.898\sqrt{1-0.690^2}=2.097$$ $$s_d=-2\pm j2.097$$

Step 3: Angle deficiency

$$\begin{aligned} \angle s_d&=180^\circ-\tan^{-1}\frac{2.097}{2}=133.6^\circ\\ \angle(s_d+1)&=180^\circ-\tan^{-1}\frac{2.097}{1}=115.5^\circ\\ \angle G(s_d)&=-249.1^\circ\Rightarrow\phi=249.1^\circ-180^\circ=69.1^\circ \end{aligned}$$

Step 4: Zero and pole

Place the zero at $s=-1$ to cancel the plant pole. Then the open loop becomes $\dfrac{10K_c}{s(s+p)}$ and the closed loop $s^2+ps+10K_c=0$. Matching with $s^2+2\zeta\omega_ns+\omega_n^2=s^2+4s+8.4$:

$$p=2\zeta\omega_n=4$$

Angle check: zero adds $+115.5^\circ$, pole at $-4$ adds $-\tan^{-1}\frac{2.097}{2}=-46.4^\circ$, net $+69.1^\circ$ ✓.

Step 5: Gain

$$10K_c=\omega_n^2=8.4\Rightarrow K_c=0.84$$

(same as magnitude condition $K_c=|s_d||s_d+4|/10=2.898\times2.898/10$).

$$G_c(s)=0.84\,\frac{s+1}{s+4}=0.21\,\frac{1+s}{1+0.25s}$$

Step 6: Verification

$$\frac{C(s)}{R(s)}=\frac{8.4}{s^2+4s+8.4}$$

Poles $-2\pm j2.10$; $\zeta=\frac{4}{2\sqrt{8.4}}=0.690$, $M_p=5.0\%$, $t_s=\frac{4}{2}=2$ s ✓.

          jw
  sd x    | j2.1
      \   |
 --x---o--x------ sigma
  -4  -1  0
 (zero at -1 cancels plant pole)

If a higher $K_v$ is needed, a lag section can be added in cascade (lag–lead).

Answer: $G_c(s)=0.84\,\dfrac{s+1}{s+4}$; closed loop $\dfrac{8.4}{s^2+4s+8.4}$ with ts = 2 s and Mp = 5%.