--- title: "Lag compensator for k/[s(s+2)(s+20)], Kv = 20" subject: "Control Systems" chapter: 7 key: "semester_3/control_systems:lag-k-s-s2-s20-kv20" source: https://www.bctengineer.com/q/semester_3/control_systems/lag-k-s-s2-s20-kv20 site: BCT Engineering --- # Lag compensator for k/[s(s+2)(s+20)], Kv = 20 Control Systems · Chapter 7: Performance Specifications and Compensation Design - Asked in: 2080 Baisakh, 12 marks - Video: [Lag Compensator design/Using Bode Plot/Problem/ Control System](https://www.youtube.com/watch?v=Z5CF8btPHR4) — Dr.Jayaudhaya ,Simple and Easy Way - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/control_systems/lag-k-s-s2-s20-kv20 - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-7 ## Question Design a suitable lag compensating network for G(s) = k/[s(s+2)(s+20)] to meet the following specification: Kv = 20 sec⁻¹, P.M ≥ 35°. ## Short answer Set k=800 to meet Kv=20 sec^-1. The lag network Gc(s)=(1+5.92s)/(1+53.4s) with zero at -0.169 rad/s and pole at -0.0187 rad/s (β=9.02) shifts the gain crossover to 1.69 rad/s, yielding a phase margin of 39.9°≥35° while maintaining Kv=20 sec^-1. ## Answer A **lag network** $G_c(s) = \dfrac{1+Ts}{1+\beta Ts}$ ($\beta > 1$) is designed by the Bode plot method. #### Step 1: Gain for $K_v$ $$ K_v = \lim_{s\to0} s\,\frac{k}{s(s+2)(s+20)} = \frac{k}{40} = 20 \;\Rightarrow\; k = 800 $$ $$ G(s) = \frac{800}{s(s+2)(s+20)} = \frac{20}{s(0.5s+1)(0.05s+1)} $$ #### Step 2: Uncompensated margins - $|G(j\omega)| = 1$ at $\omega_{gc} = 6.03$ rad/s; $\angle G = -90^\circ - \tan^{-1}\frac{6.03}{2} - \tan^{-1}\frac{6.03}{20} = -178.4^\circ$, so $PM = 1.6^\circ$. - $\omega_{pc} = \sqrt{40} = 6.32$ rad/s, $GM = 0.8$ dB. The system is on the verge of instability. #### Step 3: New gain crossover frequency Allow $10^\circ$ for the lag of the network: $$ \angle G(j\omega_{gc}') = -180^\circ + 35^\circ + 10^\circ = -135^\circ $$ $$ -90^\circ - \tan^{-1}\frac{\omega}{2} - \tan^{-1}\frac{\omega}{20} = -135^\circ \;\Rightarrow\; \omega_{gc}' = 1.69\ \text{rad/s} $$ #### Step 4: Value of $\beta$ $$ |G(j1.69)| = \frac{800}{1.69\times\sqrt{1.69^2+4}\times\sqrt{1.69^2+400}} = 9.02\;(19.1\ \text{dB}) \;\Rightarrow\; \beta = 9.02 $$ #### Step 5: Corner frequencies $$ \begin{aligned} \frac{1}{T} &= \frac{1.69}{10} = 0.169\ \text{rad/s},\quad T = 5.92\ \text{s} \\ \frac{1}{\beta T} &= \frac{0.169}{9.02} = 0.0187\ \text{rad/s},\quad \beta T = 53.4\ \text{s} \end{aligned} $$ #### Step 6: Compensator and check $$ G_c(s) = \frac{1+5.92s}{1+53.4s} = 0.111\,\frac{s+0.169}{s+0.0187} $$ $$ G_c(s)G(s) = \frac{800(1+5.92s)}{s(s+2)(s+20)(1+53.4s)} $$ | Quantity | Uncompensated | Compensated | |---|---|---| | Gain crossover | 6.03 rad/s | 1.69 rad/s | | Phase margin | 1.6° | 39.9° | | Gain margin | 0.8 dB | 19.2 dB | | $K_v$ | 20 s$^{-1}$ | 20 s$^{-1}$ | (With only $5^\circ$ allowance the PM comes out at $35.0^\circ$, just on the limit; the $10^\circ$ allowance gives a safe margin.) **Answer:** $k = 800$, $G_c(s) = \dfrac{1+5.92s}{1+53.4s}$ (zero at $-0.169$, pole at $-0.0187$); $K_v = 20\ \text{s}^{-1}$, $PM = 39.9^\circ \ge 35^\circ$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For G = k/[s(s+2)(s+20)], what k gives Kv = 20? A. k = 40 B. k = 800 C. k = 400 D. k = 20 Answer: B. Kv = k/(2*20) = k/40 = 20, so k = 800. (Key point: Gain from Kv) 2. What are the uncompensated margins? A. PM = 40 degrees, GM infinite B. PM negative, unstable C. PM = 35 degrees, GM = 10 dB D. PM = 1.6 degrees and GM = 0.8 dB, nearly unstable Answer: D. At crossover 6.03 rad/s the phase is -178.4 degrees, so the system is on the verge of instability. (Key point: Uncompensated state) 3. How is the new crossover 1.69 rad/s chosen? A. At the phase crossover 6.32 rad/s B. Where plant phase = -90 degrees C. Where plant phase = -180 + 35 + 10 = -135 degrees D. Where plant gain is 20 Answer: C. Required PM plus a 10 degree allowance for the lag network gives -135 degrees. (Key point: New crossover) 4. With plant gain 9.02 at 1.69 rad/s, which choices follow? A. beta = 9.02 and zero at 16.9 rad/s B. beta = 20 and zero at 6.03 C. beta = 1/9.02 and zero at 1.69 D. beta = 9.02 and zero at 0.169 rad/s (one decade below crossover) Answer: D. The lag must cancel the plant gain at the new crossover, and its zero is placed one decade lower to keep phase lag small. 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