--- title: "Lead compensator for 10/[s(s+1)], ts=2 s, Mp=5%" subject: "Control Systems" chapter: 7 key: "semester_3/control_systems:lead-10-s-s1-ts2-mp5" source: https://www.bctengineer.com/q/semester_3/control_systems/lead-10-s-s1-ts2-mp5 site: BCT Engineering --- # Lead compensator for 10/[s(s+1)], ts=2 s, Mp=5% Control Systems · Chapter 7: Performance Specifications and Compensation Design - Asked in: 2076 Chaitra, 12 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/control_systems/lead-10-s-s1-ts2-mp5 - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-7 ## Question For a unity feedback system with feed forward transfer function G(s) = 10/[s(s+1)], design a lead compensator such that the settling time of the system will become 2 sec and maximum percent overshoot 5%. ## Short answer A lead compensator Gc(s)=0.84(s+1)/(s+4) places the closed-loop poles at -2±j2.1, giving damping ratio 0.69. This yields settling time 2 s and maximum overshoot 5% as required. ## Answer Time-domain specifications are met by placing the dominant closed-loop poles with a **lead compensator designed on the root locus**. #### Step 1: Present system $$ \frac{C}{R}=\frac{10}{s^2+s+10}:\ \omega_n=3.16\ \text{rad/s},\ \zeta=0.158,\ M_p=60.5\%,\ t_s=\frac{4}{0.5}=8\ \text{s} $$ #### Step 2: Desired poles From $M_p=5\%$: $$ \zeta=\frac{-\ln0.05}{\sqrt{\pi^2+(\ln0.05)^2}}=\frac{2.996}{4.341}=0.690 $$ From $t_s=2$ s (2% criterion): $$ \zeta\omega_n=\frac{4}{2}=2\Rightarrow\omega_n=\frac{2}{0.690}=2.898\ \text{rad/s},\quad\omega_d=2.898\sqrt{1-0.690^2}=2.097 $$ $$ s_d=-2\pm j2.097 $$ #### Step 3: Angle deficiency $$ \begin{aligned} \angle s_d&=180^\circ-\tan^{-1}\frac{2.097}{2}=133.6^\circ\\ \angle(s_d+1)&=180^\circ-\tan^{-1}\frac{2.097}{1}=115.5^\circ\\ \angle G(s_d)&=-249.1^\circ\Rightarrow\phi=249.1^\circ-180^\circ=69.1^\circ \end{aligned} $$ #### Step 4: Zero and pole Place the zero at $s=-1$ to cancel the plant pole. Then the open loop becomes $\dfrac{10K_c}{s(s+p)}$ and the closed loop $s^2+ps+10K_c=0$. Matching with $s^2+2\zeta\omega_ns+\omega_n^2=s^2+4s+8.4$: $$ p=2\zeta\omega_n=4 $$ Angle check: zero adds $+115.5^\circ$, pole at $-4$ adds $-\tan^{-1}\frac{2.097}{2}=-46.4^\circ$, net $+69.1^\circ$ ✓. #### Step 5: Gain $$ 10K_c=\omega_n^2=8.4\Rightarrow K_c=0.84 $$ (same as magnitude condition $K_c=|s_d||s_d+4|/10=2.898\times2.898/10$). $$ G_c(s)=0.84\,\frac{s+1}{s+4}=0.21\,\frac{1+s}{1+0.25s} $$ #### Step 6: Verification $$ \frac{C(s)}{R(s)}=\frac{8.4}{s^2+4s+8.4} $$ Poles $-2\pm j2.10$; $\zeta=\frac{4}{2\sqrt{8.4}}=0.690$, $M_p=5.0\%$, $t_s=\frac{4}{2}=2$ s ✓. ```text jw sd x | j2.1 \ | --x---o--x------ sigma -4 -1 0 (zero at -1 cancels plant pole) ``` | Item | Before | After | |---|---|---| | $\zeta$ | 0.158 | 0.69 | | $\omega_n$ | 3.16 rad/s | 2.90 rad/s | | $M_p$ | 60.5% | 5% | | $t_s$ | 8 s | 2 s | | $K_v$ | 10 | 2.1 | If a higher $K_v$ is needed, a lag section can be added in cascade (lag–lead). **Answer: $G_c(s)=0.84\,\dfrac{s+1}{s+4}$; closed loop $\dfrac{8.4}{s^2+4s+8.4}$ with ts = 2 s and Mp = 5%.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What damping ratio does Mp = 5% require? A. About 0.69 B. About 0.05 C. About 0.5 D. About 0.9 Answer: A. zeta = -ln(0.05)/sqrt(pi^2 + ln(0.05)^2) = 2.996/4.341 = 0.690. (Key point: Damping from overshoot) 2. With ts = 2 s (2% criterion) and zeta = 0.69, what are the desired poles? A. -4 +/- j2.097 B. -2 +/- j2.097 C. -2 +/- j2.9 D. -0.5 +/- j3.16 Answer: B. zeta*wn = 4/2 = 2 so wn = 2.898 and wd = 2.898 sqrt(1 - 0.69^2) = 2.097. (Key point: Desired poles) 3. Where is the compensator zero placed, and why? A. At the origin, to add an integrator B. At -4, to cancel the compensator pole C. At s = -1, cancelling the plant pole, so the open loop becomes 10Kc/[s(s+p)] D. Far left at -10, to add phase lead cheaply Answer: C. Cancelling the pole at -1 leaves a second-order loop; the pole p is then set by matching s^2 + p s + 10Kc. (Key point: Zero placement) 4. Matching s^2 + p s + 10Kc with s^2 + 4s + 8.4 gives what compensator? A. Gc = 8.4 (s+1)/(s+4) B. Gc = 0.84 (s+4)/(s+1) C. Gc = 0.4 (s+1)/(s+2) D. Gc = 0.84 (s+1)/(s+4) Answer: D. p = 2 zeta wn = 4 and 10Kc = wn^2 = 8.4, so Kc = 0.84. 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