Lag compensator for 0.53/[s(1+s)(1+0.5s)], Kv = 5

"Without significant change in transient properties" means the dominant closed-loop poles should stay almost where they are, so a lag compensator is designed by the root locus method, with its pole and zero close to the origin.

$$G(s) = \frac{0.53}{s(1+s)(1+0.5s)} = \frac{1.06}{s(s+1)(s+2)}$$

Step 1: Uncompensated system

Characteristic equation: $s^3 + 3s^2 + 2s + 1.06 = 0$.

Roots: $s = -0.331 \pm j0.586$ (dominant) and $s = -2.34$.

$$\omega_n = \sqrt{0.331^2 + 0.586^2} = 0.673\ \text{rad/s},\qquad \zeta = \frac{0.331}{0.673} = 0.491$$ $$K_v = \lim_{s\to0} s\,G(s) = 0.53\ \text{s}^{-1}$$

Step 2: Required increase in $K_v$

$$\beta \ge \frac{5}{0.53} = 9.43 \quad\Rightarrow\quad \text{choose } \beta = 10$$

Step 3: Place the pole and zero

$G_c(s) = \hat{K}_c\,\dfrac{s + 1/T}{s + 1/(\beta T)}$. Put the zero close to the origin, well to the right of the dominant poles:

$$\text{zero at } s = -0.05, \qquad \text{pole at } s = -\frac{0.05}{10} = -0.005$$

Angle added at $s_1 = -0.331 + j0.586$:

$$\angle(s_1 + 0.05) - \angle(s_1 + 0.005) = 115.6^\circ - 119.1^\circ = -3.5^\circ$$

This is less than $5^\circ$, so the root locus near the dominant poles hardly changes.

Step 4: New dominant poles and gain

Keeping $\zeta = 0.491$, the compensated locus $\dfrac{1.06\hat{K}_c(s+0.05)}{s(s+1)(s+2)(s+0.005)}$ crosses the $\zeta$ line at

$$s_1' = -0.312 \pm j0.553$$

Magnitude condition at $s_1'$:

$$\hat{K}_c = \frac{|s_1'|\,|s_1'+1|\,|s_1'+2|\,|s_1'+0.005|}{1.06\,|s_1'+0.05|} = 0.970$$

Step 5: Compensator

$$G_c(s) = 0.970\,\frac{s+0.05}{s+0.005} = 9.70\,\frac{20s+1}{200s+1}$$ $$G_c(s)G(s) = \frac{1.028(s+0.05)}{s(s+0.005)(s+1)(s+2)}$$

Step 6: Check

$$K_v = \lim_{s\to0} s\,G_cG = \frac{1.028\times0.05}{0.005\times2} = 5.14\ \text{s}^{-1} \;(\ge 5)$$

The dominant poles keep $\zeta = 0.491$ and $\omega_n$ falls by only about 6%, so the dominant-pole transient response is practically the same while $K_v$ rises nearly ten times. The extra closed-loop pole at $-0.055$ lies close to, but not exactly on, the zero at $-0.05$, so it leaves a small, slowly decaying tail (amplitude about 0.11, time constant about 18 s). In a step simulation this tail raises the peak overshoot from about 16% to about 26% and lengthens the 2% settling time. This slow tail is the usual price of lag compensation.

Answer: $G_c(s) = 0.970\dfrac{s+0.05}{s+0.005}$, giving $K_v = 5.14\ \text{s}^{-1}$ with nearly unchanged transient response.