Compensator for 4/[s(s+2)]: Mp 16.3%, ts, Kv

The settling time must be halved while keeping $\zeta$, so the dominant poles must move left; a lead compensator designed by the root locus method is used.

Step 1: Desired closed-loop poles

Peak overshoot:

$$\begin{aligned} M_p &= e^{-\pi\zeta/\sqrt{1-\zeta^2}} = 0.163 \\ \zeta &= \frac{-\ln 0.163}{\sqrt{\pi^2 + (\ln 0.163)^2}} = \frac{1.814}{3.628} = 0.5 \end{aligned}$$

Settling time (2% criterion):

$$t_s = \frac{4}{\zeta\omega_n} = 2 \;\Rightarrow\; \zeta\omega_n = 2,\quad \omega_n = \frac{2}{0.5} = 4\ \text{rad/s}$$ $$s_d = -\zeta\omega_n \pm j\omega_n\sqrt{1-\zeta^2} = -2 \pm j3.464$$

Step 2: Uncompensated system

Closed-loop characteristic equation: $s^2 + 2s + 4 = 0$, so $\omega_n = 2$, $\zeta = 0.5$.

• $M_p = 16.3\%$ (already correct), $t_s = \dfrac{4}{0.5\times2} = 4$ s (too slow).
• $K_v = \lim_{s\to0} s\,\dfrac{4}{s(s+2)} = 2\ \text{s}^{-1}$.

So the poles must move further left along the same $\zeta = 0.5$ line, which needs a lead compensator.

Step 3: Angle deficiency

At $s_d = -2 + j3.464$:

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

The compensator must add $+30^\circ$ at $s_d$.

Step 4: Place the zero and pole

Let $G_c(s) = K_c\dfrac{s+z}{s+p}$. Put the zero at $s = -2$ to cancel the plant pole at $-2$, so the zero gives $90^\circ$. The pole must then give $90^\circ - 30^\circ = 60^\circ$:

$$\tan 60^\circ = \frac{3.464}{p - 2} \;\Rightarrow\; p - 2 = 2 \;\Rightarrow\; p = 4$$

Step 5: Gain from the magnitude condition

The compensated open loop is $\dfrac{4K_c}{s(s+4)}$:

$$\begin{aligned} 4K_c &= |s_d|\,|s_d+4| = |-2+j3.464|\,|2+j3.464| = 4 \times 4 = 16 \\ K_c &= 4 \end{aligned}$$

Check: characteristic equation $s^2 + 4s + 16 = 0$ gives $\omega_n = 4$, $\zeta = 4/(2\times4) = 0.5$. Correct.

$$G_c(s) = \frac{4(s+2)}{s+4} = \frac{2(0.5s+1)}{0.25s+1}$$

Step 6: Check velocity error constant

$$K_v = \lim_{s\to0} s\,G_c(s)G(s) = \lim_{s\to0} s\cdot\frac{16}{s(s+4)} = 4\ \text{s}^{-1} \;\geq\; 2\ \text{s}^{-1}$$

Result

Answer: $G_c(s) = \dfrac{4(s+2)}{s+4}$ (lead compensator, zero at $-2$, pole at $-4$, $\alpha = 0.5$). All three specifications are met.