--- title: "Lead compensator for 4/[s(s+2)], Kv=20, PM≥50°" subject: "Control Systems" chapter: 7 key: "semester_3/control_systems:lead-4-s-s2-kv20-pm50" source: https://www.bctengineer.com/q/semester_3/control_systems/lead-4-s-s2-kv20-pm50 site: BCT Engineering --- # Lead compensator for 4/[s(s+2)], Kv=20, PM≥50° Control Systems · Chapter 7: Performance Specifications and Compensation Design - Asked in: 2078 Bhadra, 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-4-s-s2-kv20-pm50 - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-7 ## Question Design a suitable phase lead compensating network for G(s) = 4/[s(s+2)] to meet the following specification: Kv = 20 sec⁻¹, P.M ≥ 50°. ## Short answer A phase lead network Gc(s)=10(1+0.233s)/(1+0.0513s) with α≈0.22 provides the required velocity error constant Kv=20 s⁻¹. The compensated phase margin is approximately 52°, meeting the specification of at least 50°. ## Answer Lead network $G_c(s)=K_c\dfrac{1+Ts}{1+\alpha Ts}$ ($\alpha<1$), maximum phase lead $\sin\phi_m=\frac{1-\alpha}{1+\alpha}$ at $\omega_m=\frac{1}{T\sqrt\alpha}$. #### Step 1: Gain for $K_v$ $$ K_v=\lim_{s\to0}s\,K_c\frac{4}{s(s+2)}=2K_c=20\Rightarrow K_c=10 $$ $$ G_1(s)=\frac{40}{s(s+2)}=\frac{20}{s(1+0.5s)} $$ #### Step 2: Bode plot of $G_1$ 20log20 = 26 dB at $\omega=1$, slope $-20$ dB/dec, corner 2 rad/s, then $-40$ dB/dec. $$ \frac{40}{\omega\sqrt{\omega^2+4}}=1\Rightarrow\omega^4+4\omega^2-1600=0\Rightarrow\omega_{gc}=6.17\ \text{rad/s} $$ $$ PM=180^\circ-90^\circ-\tan^{-1}\frac{6.17}{2}=90^\circ-72.0^\circ=18.0^\circ $$ #### Step 3: Required lead $$ \phi_m=50^\circ-18^\circ+8^\circ=40^\circ $$ #### Step 4: α $$ \alpha=\frac{1-\sin40^\circ}{1+\sin40^\circ}=\frac{1-0.643}{1+0.643}=0.217\approx0.22 $$ #### Step 5: New crossover Gain added at $\omega_m$: $10\log(1/0.22)=6.58$ dB. Solve $|G_1(j\omega)|=-6.58$ dB $=\sqrt{0.22}=0.469$: $$ \frac{40}{\omega\sqrt{\omega^2+4}}=0.469\Rightarrow\omega_m=9.13\ \text{rad/s} $$ #### Step 6: T $$ T=\frac{1}{\omega_m\sqrt\alpha}=\frac{1}{9.13\times0.469}=0.233\ \text{s},\qquad\alpha T=0.0513\ \text{s} $$ Zero at $1/T=4.28$ rad/s, pole at $1/\alpha T=19.5$ rad/s. $$ G_c(s)=10\,\frac{1+0.233s}{1+0.0513s}=45.4\,\frac{s+4.28}{s+19.5} $$ #### Step 7: Check $$ G_c(s)G(s)=\frac{40(1+0.233s)}{s(s+2)(1+0.0513s)} $$ At $\omega_{gc}'=9.11$ rad/s: $$ \begin{aligned} \phi&=-90^\circ-\tan^{-1}\frac{9.11}{2}+\tan^{-1}(2.12)-\tan^{-1}(0.467)\\ &=-90^\circ-77.62^\circ+64.78^\circ-25.05^\circ=-127.89^\circ\\ PM&=180^\circ-127.9^\circ\approx52^\circ\ (\ge50^\circ) \end{aligned} $$ GM = ∞. ```text dB 26 |\ -20 uncompensated | \__ 2 / 0 |------\--x---x------ w | 6.17 9.11 compensated | \__ -40 ``` | Item | Before | After | |---|---|---| | $K_v$ | 2 | 20 | | $\omega_{gc}$ | 6.17 rad/s (with gain 10) | 9.11 rad/s | | PM | 18° | ≈52° | **Answer: $G_c(s)=10\,\dfrac{1+0.233s}{1+0.0513s}$ (α ≈ 0.22) gives Kv = 20 s⁻¹ and PM ≈ 52°.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What gain Kc is used first to meet Kv = 20 for G = 4/[s(s+2)]? A. Kc = 10 B. Kc = 5 C. Kc = 20 D. Kc = 2 Answer: A. Kv = Kc x 4/2 = 2Kc = 20, so Kc = 10. (Key point: Gain for Kv) 2. Why is phi_m = 40 deg chosen when PM of the gain-adjusted system is 18 deg and 50 deg is needed? A. 50 + 18 = 68 deg B. 50 - 18 = 32 deg, plus about 8 deg extra because the crossover shifts right and phase falls C. 50 - 18 = 32 deg exactly, no extra D. 18 deg, the existing PM Answer: B. An allowance (8 deg) covers the extra phase lost when the gain crossover moves to a higher frequency. (Key point: Required phase lead) 3. What alpha results from phi_m = 40 deg? A. About 0.64 B. About 4.6 C. About 0.22 D. About 0.5 Answer: C. alpha = (1 - sin40)/(1 + sin40) = 0.357/1.643 = 0.217. (Key point: Alpha) 4. How is the new crossover wm found? A. Where the phase of G1 is -180 deg B. At the old crossover 6.17 rad/s C. At 1/T where the zero sits D. Where the gain of G1 is -10log(1/alpha) = -6.58 dB, because the lead adds that gain at wm Answer: D. The lead boosts gain by 6.58 dB at wm, so wm is placed where |G1| = -6.58 dB (9.13 rad/s). (Key point: New crossover) ## Close variants set in other exams - [Lag compensator for k/[s(s+2)(s+20)], Kv = 20](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Alag-k-s-s2-s20-kv20) — Control Systems, chapter 7, asked 4× ## Related in Control Systems - [Lead compensator for 2/[s(1+0.5s)], Kv > 20, PM 50°](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Alead-2-s-1-05s-kv20) — Control Systems, chapter 7 - [Lead compensator for K/[s(1+0.2s)], Kv≥20, PM≥44°](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Alead-k-s-1-02s-kv20) — Control Systems, chapter 7, asked 2× - [Lead compensator for K/[s(1+0.2s)], Kv=10, PM=50°](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Alead-k-s-1-02s-kv10) — Control Systems, chapter 7, asked 2× - [Lead compensator for k/[s²(1+0.25s)], Ka = 10, PM 35°](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Alead-k-s2-025s-ka10-pm35) — Control Systems, chapter 7, asked 5× - [Lead compensator for 5/[(s+2)(s+20)], Kv ≥ 20](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Alead-5-s2-s20-kv20) — Control Systems, chapter 7 - [Lead compensator for K/[s²(0.2s+1)], Ka = 10, PM 35°](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Alead-k-s2-02s-ka10-pm35) — Control Systems, chapter 7, asked 5× ## Related in other subjects - [Short note: Phase-lead/phase-lag compensator](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Ashort-note-lead-lag-compensator) — Digital Control System, chapter 4 - [Bode design: PM 55°, GM 10 dB, Kv = 5](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Abode-design-phase-margin) — Digital Control System, chapter 4 - [Lead compensator for satellite position control](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Alead-compensator-satellite) — Digital Control System, chapter 4