--- title: "Compensator for 4/[s(s+2)], ts=2 s, Kv=2" subject: "Control Systems" chapter: 7 key: "semester_3/control_systems:comp-4-s-s2-ts2-kv2" source: https://www.bctengineer.com/q/semester_3/control_systems/comp-4-s-s2-ts2-kv2 site: BCT Engineering --- # Compensator for 4/[s(s+2)], ts=2 s, Kv=2 Control Systems · Chapter 7: Performance Specifications and Compensation Design - Asked in: 2071 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/comp-4-s-s2-ts2-kv2 - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-7 ## Question Design a suitable compensator for a unity feedback system with open loop transfer function G(s) = 4/[s(s+2)] such that the settling time will become 2 seconds without change in overshoot and velocity error constant will be 2 s⁻¹. ## Short answer Lead compensator Gc(s)=4(s+2)/(s+4) cancels the plant pole at -2, placing closed-loop poles at -2±j3.464. Settling time reduces to 2 s, overshoot remains 16.3%, and velocity error constant becomes 4 s⁻¹. ## Answer "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 | Quantity | Before | After | |---|---|---| | $\zeta$ | 0.5 | 0.5 | | $M_p$ | 16.3% | 16.3% (unchanged) | | $\omega_n$ | 2 rad/s | 4 rad/s | | $t_s$ | 4 s | 2 s | | $K_v$ | 2 s$^{-1}$ | 4 s$^{-1}$ | **Answer:** $G_c(s) = \dfrac{4(s+2)}{s+4}$; closed-loop poles $-2 \pm j3.464$, overshoot 16.3%, settling time 2 s, $K_v = 4\ \text{s}^{-1}$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does 'without change in overshoot' mean here? A. zeta stays 0.5 while ts is halved B. wn stays at 2 rad/s C. Poles stay at -1 +/- j1.732 D. Kv stays at 2 Answer: A. Same overshoot means the same damping ratio. (Key point: Meaning of spec) 2. Why is a lead compensator used? A. Kv must be reduced B. Poles must move left along the same zeta line, needing +30 deg angle at s_d C. Phase margin must be lowered D. Gain must be cut by half Answer: B. Angle deficiency is 30 deg. (Key point: Compensator type) 3. What compensator results? A. Gc = 2(s+4)/(s+2) B. Gc = 4(s+4)/(s+2) C. Gc = 4(s+2)/(s+4) D. Gc = 16/(s+4) Answer: C. Zero cancels the plant pole; closed loop becomes s^2 + 4s + 16. (Key point: Design) 4. What Kv does the design give, and is it acceptable? A. Kv = 2 exactly B. Kv = 1, too low C. Kv = 16 D. Kv = 4 s^-1, not less than the required 2 Answer: D. Kv = 16/4 = 4 which is not less than 2. (Key point: Kv) ## Close variants set in other exams - [Lead compensator for 4/[s(s+2)], ts=2 s](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Alead-4-s-s2-ts2) — Control Systems, chapter 7, asked 2× ## Related in Control Systems - [Overshoot, settling time and error for 50/[s(s+2)]](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Amp-ts-50-s-s2-sse) — Control Systems, chapter 3 - [Compensator for 4/[s(s+2)]: Mp 16.3%, ts, Kv](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Acomp-4-s-s2-mp163-ts2) — Control Systems, chapter 7 - [Transfer function from overshoot and settling time](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Atf-from-mp25-ts6) — Control Systems, chapter 3 - [Lag compensator for 0.53/[s(1+s)(1+0.5s)], Kv = 5](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Alag-053-s-1s-05s-kv5) — Control Systems, chapter 7 - [Lead compensator for 10/[s(s+1)], ts=2 s, Mp=5%](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Alead-10-s-s1-ts2-mp5) — Control Systems, chapter 7 - [Lag compensator for 1.06/[s(s+1)(s+2)], ramp error < 0.2](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Alag-106-ramp-ess02) — Control Systems, chapter 7 ## Related in other subjects - [z-plane design with Kv, plant 1/(s(s+2))](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Az-plane-design-kv-s2) — Digital Control System, chapter 4 - [Root-locus design: ζ = 0.5, ts = 2 s, plant 1/(s(s+1))](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Aroot-locus-design-plant-s1) — Digital Control System, chapter 4, asked 5× - [Controller design for peak time 1.2 s (ζ = 0.55), Kv](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Adesign-peak-time-zeta-0-55) — Digital Control System, chapter 4, asked 2× - [Root-locus design: ζ = 0.5, ts = 2 s, plant 1/(s(s+2))](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Aroot-locus-design-plant-s2) — Digital Control System, chapter 4, asked 5×