Lag compensator for 1.06/[s(s+1)(s+2)], ramp error < 0.2

Since the transient must stay almost the same while $K_v$ is raised, a lag compensator by the root-locus method is used: a pole–zero pair very close to the origin raises the low-frequency gain but hardly moves the dominant poles.

$$G_c(s)=\hat K_c\,\frac{s+\frac1T}{s+\frac{1}{\beta T}},\qquad\beta>1$$

Step 1: Present system

$$K_v=\lim_{s\to0}sG(s)=\frac{1.06}{1\times2}=0.53,\qquad e_{ss}=\frac{1}{K_v}=1.89$$

Characteristic equation: $s^3+3s^2+2s+1.06=0$; roots $-2.339$ and $-0.3307\pm j0.5864$.

Dominant poles: $\omega_n=0.673$ rad/s, $\zeta=0.491$.

Step 2: Required $K_v$ and β

$$e_{ss}<0.2\Rightarrow K_v>5\Rightarrow\beta\ge\frac{5}{0.53}=9.4;\ \text{take}\ \beta=10$$

Step 3: Place the pole–zero pair

Put the zero and pole close to the origin, with ratio 10:

$$\text{zero at }-0.05,\quad\text{pole at }-0.005\Rightarrow T=20\ \text{s},\ \beta T=200\ \text{s}$$

Angle added at $s_1=-0.3307+j0.5864$:

$$\angle(s_1+0.05)-\angle(s_1+0.005)=115.6^\circ-119.0^\circ\approx-3.5^\circ$$

This is less than about 5°, so the root locus near the dominant poles changes very little.

Step 4: Gain $\hat K_c$

The new dominant poles lie on the compensated root locus with almost the same ζ (≈0.49): $s_1'=-0.312\pm j0.551$. Magnitude condition:

$$\hat K_c=\frac{|s_1'|\,|s_1'+1|\,|s_1'+2|\,|s_1'+0.005|}{1.06\,|s_1'+0.05|}=0.966$$ $$G_c(s)=0.966\,\frac{s+0.05}{s+0.005}=9.66\,\frac{1+20s}{1+200s}$$

Step 5: Check

$$K_v=\lim_{s\to0}sG_cG=0.966\times10\times0.53=5.12\Rightarrow e_{ss}=\frac{1}{5.12}=0.195<0.2$$

The requirement is met.

Closed-loop poles of the compensated system: $-0.312\pm j0.551$, $-2.326$, $-0.055$. The pole at $-0.055$ is almost cancelled by the zero at $-0.05$, so it adds only a small, slow tail.

Transient response is nearly unchanged (slightly slower, since $\omega_n$ dropped about 6%).

Answer: $G_c(s)=0.966\,\dfrac{s+0.05}{s+0.005}$ gives Kv = 5.12 s⁻¹, ramp error ≈ 0.195 (< 0.2), with almost the same damping.