Lead compensator for 4/[s(s+2)], Mp≤16.3%, ts≤2 s

A lead compensator is designed by the root locus method: find the desired dominant poles, find the angle the compensator must add, place its zero and pole, then find its gain.

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: Locate zero and pole (bisector method)

Let $G_c(s) = K_c\dfrac{s+z}{s+p}$. At point P $(-2, j3.464)$ draw a horizontal line PA to the left and the line PO to the origin. Angle APO $= 120^\circ$. Draw its bisector PB, then two lines at $\pm\phi/2 = \pm15^\circ$ from PB; they cut the real axis at the zero and the pole.

            P (-2, j3.46)
           /|\ 
          / | \  bisector +-15 deg
         /  |  \
 -------x---o--x------x-----o---> Re
      -5.46 -2.93  -2        0
       pole  zero  plant   plant

The lines from P make angles $255^\circ$ (zero) and $225^\circ$ (pole) with the positive real axis:

$$\begin{aligned} z &= 2 + \frac{3.464}{\tan 75^\circ} = 2 + 0.928 = 2.93 \\ p &= 2 + \frac{3.464}{\tan 45^\circ} = 2 + 3.464 = 5.46 \end{aligned}$$

Check: zero angle $75^\circ$, pole angle $45^\circ$, net lead $= 30^\circ$.

Step 5: Gain

$$\begin{aligned} K_c &= \frac{|s_d|\,|s_d+2|\,|s_d+5.46|}{4\,|s_d+2.93|} = \frac{4 \times 3.464 \times 4.899}{4 \times 3.586} = 4.73 \\ G_c(s) &= 4.73\,\frac{s+2.93}{s+5.46} \end{aligned}$$

$K_v = \lim_{s\to0} s\,G_cG = \dfrac{4.73\times4\times2.93}{2\times5.46} = 5.07\ \text{s}^{-1}$.

Closed-loop poles: $-2 \pm j3.464$ and $-3.46$. The closed-loop zero at $-2.93$ is not close enough to the third pole to cancel it, so a step simulation gives $M_p \approx 21\%$, $t_s \approx 2.04$ s, slightly above the limits.

Step 6: Final design (pole–zero cancellation)

To meet $M_p \le 16.3\%$ exactly, place the zero on the plant pole at $-2$. The pole must give $90^\circ - 30^\circ = 60^\circ$:

$$p = 2 + \frac{3.464}{\tan 60^\circ} = 4, \qquad 4K_c = |s_d|\,|s_d+4| = 4 \times 4 = 16 \Rightarrow K_c = 4$$ $$G_c(s) = \frac{4(s+2)}{s+4}, \qquad G_cG = \frac{16}{s(s+4)}$$

Closed loop: $s^2 + 4s + 16 = 0$, so $\omega_n = 4$, $\zeta = 0.5$.

Answer: lead compensator $G_c(s) = \dfrac{4(s+2)}{s+4}$ gives $M_p = 16.3\%$ and $t_s = 2$ s with $K_v = 4\ \text{s}^{-1}$.