--- title: "Cascade lag compensator for 50K/[s(s+5)(s+10)]" subject: "Control Systems" chapter: 7 key: "semester_3/control_systems:lag-50k-s-s5-s10-kv30" source: https://www.bctengineer.com/q/semester_3/control_systems/lag-50k-s-s5-s10-kv30 site: BCT Engineering --- # Cascade lag compensator for 50K/[s(s+5)(s+10)] Control Systems · Chapter 7: Performance Specifications and Compensation Design - Asked in: 2069 Chaitra, 16 marks - Video: [Designing a Lag Compensator with Bode Plot](https://www.youtube.com/watch?v=-4bY4W0hvFA) — Brian Douglas - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/control_systems/lag-50k-s-s5-s10-kv30 - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-7 ## Question Design a suitable cascade lag compensator network for the given system G(s) = 50K/[s(s+5)(s+10)] such that the requirement of velocity error constant of 30 sec⁻¹ and phase margin of ≥ 45° are met. ## Short answer K=30; lag compensator Gc(s)=(1+4.71s)/(1+59.9s) lowers the gain crossover to 2.14 rad/s. This yields Kv=30 s⁻¹ and a phase margin of 49.6°, stabilizing the originally unstable system. ## Answer A **cascade lag network** $G_c(s) = \dfrac{1+Ts}{1+\beta Ts}$ ($\beta > 1$, unity DC gain) is designed with the Bode plot. A lag network is suitable because the plant has enough phase at a lower frequency; we only need to lower the gain there. #### Step 1: Gain for $K_v$ $$ K_v = \lim_{s\to0} s\,\frac{50K}{s(s+5)(s+10)} = \frac{50K}{50} = K = 30 $$ $$ G(s) = \frac{1500}{s(s+5)(s+10)} = \frac{30}{s(0.2s+1)(0.1s+1)} $$ #### Step 2: Uncompensated margins - $|G(j\omega)| = 1$ at $\omega_{gc} = 9.77$ rad/s; $\angle G = -90^\circ - \tan^{-1}\frac{9.77}{5} - \tan^{-1}\frac{9.77}{10} = -197.2^\circ$, so $PM = -17.2^\circ$. - $\omega_{pc} = \sqrt{50} = 7.07$ rad/s, $|G(j7.07)| = 2$, $GM = -6.0$ dB. The uncompensated system is **unstable**. #### Step 3: New gain crossover frequency Allow $10^\circ$ for the lag of the network (needed for this large shift): $$ \angle G(j\omega_{gc}') = -180^\circ + 45^\circ + 10^\circ = -125^\circ $$ $$ -90^\circ - \tan^{-1}\frac{\omega}{5} - \tan^{-1}\frac{\omega}{10} = -125^\circ \;\Rightarrow\; \omega_{gc}' = 2.12\ \text{rad/s} $$ #### Step 4: Value of $\beta$ $$ |G(j2.12)| = \frac{1500}{2.12\times\sqrt{2.12^2+25}\times\sqrt{2.12^2+100}} = 12.72 \;(22.1\ \text{dB}) \;\Rightarrow\; \beta = 12.72 $$ #### Step 5: Corner frequencies $$ \begin{aligned} \frac{1}{T} &= \frac{\omega_{gc}'}{10} = 0.212\ \text{rad/s},\quad T = 4.71\ \text{s} \\ \frac{1}{\beta T} &= \frac{0.212}{12.72} = 0.0167\ \text{rad/s},\quad \beta T = 59.9\ \text{s} \end{aligned} $$ #### Step 6: Compensator and check $$ G_c(s) = \frac{1 + 4.71s}{1 + 59.9s} = 0.0786\,\frac{s+0.212}{s+0.0167} $$ $$ G_c(s)G(s) = \frac{1500(1+4.71s)}{s(s+5)(s+10)(1+59.9s)} $$ | Quantity | Uncompensated | Compensated | |---|---|---| | Gain crossover | 9.77 rad/s | 2.14 rad/s | | Phase margin | $-17.2^\circ$ (unstable) | $49.6^\circ$ | | Phase crossover | 7.07 rad/s | 6.86 rad/s | | Gain margin | $-6.0$ dB | 15.5 dB | | $K_v$ | 30 s$^{-1}$ | 30 s$^{-1}$ | A practical RC lag network ($R_1$ in series, $R_2$–$C$ in shunt) gives $T = R_2C$ and $\beta = (R_1+R_2)/R_2$; e.g. $C = 10\ \mu$F gives $R_2 = 471$ k$\Omega$ and $R_1 = 5.52$ M$\Omega$. **Answer:** $K = 30$, $G_c(s) = \dfrac{1+4.71s}{1+59.9s}$; $K_v = 30\ \text{s}^{-1}$, $PM = 49.6^\circ \ (\ge 45^\circ)$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For G(s) = 50K/[s(s+5)(s+10)], what is the relation between K and Kv? A. Kv = 50K/15, so K = 9 B. Kv = 5K, so K = 6 C. Kv = 50K/50 = K, so K = 30 D. Kv = K/50, so K = 1500 Answer: C. Kv = 50K/(5*10) = K = 30, giving G = 1500/[s(s+5)(s+10)]. (Key point: Gain from Kv) 2. Why is a lag network suitable for this plant? A. The plant is stable with infinite gain margin B. The plant has enough phase at a lower frequency, so only the gain there needs reducing C. The plant has enough phase at crossover already D. The plant has too little gain at high frequency Answer: B. The uncompensated system is unstable, but at lower frequencies the phase is far from -180 degrees; the lag network brings the crossover there. (Key point: Why lag works) 3. How is the new gain crossover 2.12 rad/s located? A. Where plant magnitude is 30 B. Where plant phase = -180 + 45 + 10 = -125 degrees C. Where plant phase = -180 + 45 = -135 degrees exactly D. Where plant phase = -45 degrees Answer: B. The target PM of 45 degrees plus a 10 degree allowance for the network lag gives -125 degrees for the plant. (Key point: New crossover) 4. How is beta = 12.72 obtained? A. It equals the uncompensated crossover divided by the new crossover B. It equals the plant gain |G(j2.12)| at the new crossover, which the lag must cancel C. It equals Kv/ 2.36 D. It equals the ratio of the two plant poles Answer: B. The lag network gives 1/beta attenuation at high frequency, so beta must equal the plant gain (22.1 dB) at the new crossover. 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