Lead compensator for 4/[s(s+2)], ts=2 s

The specification is in the time domain, so the lead compensator is designed by the root-locus method.

Step 1: Present system

$$\frac{C(s)}{R(s)}=\frac{4}{s^2+2s+4}\Rightarrow\omega_n=2\ \text{rad/s},\ \zeta=\frac{2}{2\times2}=0.5$$ $$t_s=\frac{4}{\zeta\omega_n}=\frac{4}{1}=4\ \text{s (2\% criterion)},\qquad M_p=e^{-\pi\zeta/\sqrt{1-\zeta^2}}=16.3\%$$

Step 2: Desired poles

Same overshoot means same $\zeta=0.5$. New settling time 2 s:

$$\zeta\omega_n=\frac{4}{2}=2\Rightarrow\omega_n=4\ \text{rad/s},\quad\omega_d=4\sqrt{1-0.25}=3.464$$ $$s_d=-2\pm j3.464$$

Step 3: Angle deficiency

$$\begin{aligned} \angle s_d&=180^\circ-\tan^{-1}\frac{3.464}{2}=120^\circ\\ \angle(s_d+2)&=90^\circ\\ \angle G(s_d)&=-(120^\circ+90^\circ)=-210^\circ\\ \phi&=-180^\circ-(-210^\circ)=30^\circ \end{aligned}$$

Step 4: Pole and zero by the bisector method

1. Draw a horizontal line PA from $P=s_d$ to the left, and line PO to the origin. Angle APO $=180^\circ-60^\circ=120^\circ$.
2. Bisect it: the bisector PB makes $60^\circ$ with PA (direction $240^\circ$).
3. Draw PC and PD at $\pm\phi/2=\pm15^\circ$ from PB; they meet the real axis at the zero and pole.

$$\begin{aligned} \text{zero: }z&=-2-\frac{3.464}{\tan75^\circ}=-2-0.928=-2.93\\ \text{pole: }p&=-2-\frac{3.464}{\tan45^\circ}=-2-3.464=-5.46 \end{aligned}$$

Check: $\angle(s_d+2.93)=75^\circ$, $\angle(s_d+5.46)=45^\circ$, net lead $=30^\circ$ ✓.

Step 5: Gain (magnitude condition)

$$K_c=\frac{|s_d|\,|s_d+2|\,|s_d+5.46|}{4\,|s_d+2.93|}=\frac{4\times3.464\times4.899}{4\times3.586}=4.73$$ $$G_c(s)=4.73\,\frac{s+2.93}{s+5.46}$$

Step 6: Compensated system

$$G_c(s)G(s)=\frac{18.93(s+2.93)}{s(s+2)(s+5.46)}$$

Closed-loop poles: $-2\pm j3.464$ (desired) and a third pole at $-3.46$, which lies close to the zero at $-2.93$, so its effect is small and the response is dominated by the desired pair.

          jw
    sd x  | j3.46
       \  |
 --x---o--x---x--- sigma
 -5.46 -2.93 -2  0

Alternative: placing the zero at $-2$ (cancelling the plant pole) needs a pole at $-4$ and gives $G_c=4\frac{s+2}{s+4}$ with closed loop exactly $\frac{16}{s^2+4s+16}$.

Answer: $G_c(s)=4.73\,\dfrac{s+2.93}{s+5.46}$; dominant poles at −2 ± j3.46 give ts = 2 s with the same ζ = 0.5 (Mp ≈ 16%).