Compensator for 4/[s(s+2)], ts=2 s, Kv=2

"Without change in overshoot" means the damping ratio stays the same, while the settling time is halved. The dominant poles must move left along the same $\zeta$ line, so a lead compensator designed by root locus is used.

Step 1: Present (uncompensated) performance

Closed loop: $\dfrac{4}{s^2 + 2s + 4}$, so $\omega_n = 2$ rad/s and $\zeta = \dfrac{2}{2\times2} = 0.5$.

• $M_p = e^{-\pi(0.5)/\sqrt{0.75}} = 16.3\%$
• $t_s = \dfrac{4}{\zeta\omega_n} = \dfrac{4}{1} = 4$ s (2% criterion)
• $K_v = \lim_{s\to0} s\dfrac{4}{s(s+2)} = 2\ \text{s}^{-1}$

Step 2: Desired dominant poles

Keep $\zeta = 0.5$ and make $t_s = 2$ s:

$$\zeta\omega_n = \frac{4}{2} = 2,\qquad \omega_n = 4\ \text{rad/s},\qquad s_d = -2 \pm j4\sqrt{1-0.25} = -2 \pm j3.464$$

Step 3: Angle deficiency

$$\angle G(s_d) = -\left[\angle s_d + \angle(s_d+2)\right] = -(120^\circ + 90^\circ) = -210^\circ, \qquad \phi = 30^\circ$$

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: Velocity error constant

$$K_v = \lim_{s\to0} s\cdot\frac{16}{s(s+4)} = 4\ \text{s}^{-1}$$

This is not less than the required $2\ \text{s}^{-1}$, so steady-state accuracy is also better.

(If $K_v$ must be exactly 2, the zero can be moved to $-0.8$ with the pole at $-2.67$ and $K_c = 3.33$. The angle condition still holds and $K_v = 2$, but a third closed-loop pole appears at $-0.67$, close to the origin. A step simulation then gives only about 3% overshoot and a settling time of about 3.4 s, so the overshoot and settling-time specifications are lost. The design above, which meets $K_v \ge 2$, is therefore preferred.)

Result

Answer: $G_c(s) = \dfrac{4(s+2)}{s+4}$; closed-loop poles $-2 \pm j3.464$, overshoot 16.3%, settling time 2 s, $K_v = 4\ \text{s}^{-1}$.