--- title: "Lag compensator for 1/[s(s+1)(0.5s+1)], Kv = 5" subject: "Control Systems" chapter: 7 key: "semester_3/control_systems:lag-1-s-s1-05s1-kv5" source: https://www.bctengineer.com/q/semester_3/control_systems/lag-1-s-s1-05s1-kv5 site: BCT Engineering --- # Lag compensator for 1/[s(s+1)(0.5s+1)], Kv = 5 Control Systems · Chapter 7: Performance Specifications and Compensation Design - Asked in: 2075 Asoj, 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/lag-1-s-s1-05s1-kv5 - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-7 ## Question The open loop transfer function of a system is given by G(s) = 1/[s(s+1)(0.5s+1)]. Compensate the system such that Kv = 5 sec⁻¹ and phase margin is at least 40° and the gain margin is at least 10 dB with a lag compensator. ## Short answer Use a lag compensator with β=10 and T=10 s. Attenuate high frequencies by 20 dB to shift gain crossover to 0.45 rad/s, giving phase margin ≈42° and gain margin ≈14 dB. Compensator: Gc(s)=5(10s+1)/(100s+1). ## Answer A **lag compensator** $G_c(s)=K_c\beta\dfrac{Ts+1}{\beta Ts+1}$ ($\beta>1$) is used: it supplies the low-frequency gain for $K_v$ and attenuates by $20\log\beta$ at higher frequencies, moving the gain crossover down to where the plant phase gives the required margins. #### Step 1: Gain for $K_v$ $$ K_v=\lim_{s\to0}s\,G_c(s)G(s)=K_c\beta=5 $$ Let $K=K_c\beta=5$: $$ G_1(s)=\frac{5}{s(s+1)(0.5s+1)} $$ #### Step 2: Bode plot of $G_1$ Corners at 1 and 2 rad/s; slopes $-20$, $-40$, $-60$ dB/dec; 14 dB at $\omega=1$ on the first asymptote. $$ \phi=-90^\circ-\tan^{-1}\omega-\tan^{-1}0.5\omega $$ | $\omega$ | 0.01 | 0.1 | 0.2 | 0.5 | 1 | 1.41 | 2 | |---|---|---|---|---|---|---|---| | $|G_1|$ (dB) | 54.0 | 33.9 | 27.8 | 18.8 | 10.0 | 4.5 | $-2.0$ | | $\phi$ | $-90.9^\circ$ | $-98.6^\circ$ | $-107^\circ$ | $-130.6^\circ$ | $-161.6^\circ$ | $-180^\circ$ | $-198.4^\circ$ | $\omega_{gc}=1.80$ rad/s, PM $=-13^\circ$; $\omega_{pc}=1.414$ rad/s, GM $=-4.4$ dB: **unstable**. #### Step 3: New gain crossover Required phase $=-180^\circ+40^\circ+12^\circ=-128^\circ$ (12° allowed for the lag network). Solving gives $\omega\approx0.47$ rad/s; it is rounded to $\omega_{gc}'=0.5$ rad/s ($\phi=-130.6^\circ$ there). #### Step 4: Zero and β - Corner of the zero well below $\omega_{gc}'$: $\frac1T=0.1$ rad/s (one-fifth of 0.5) $\Rightarrow T=10$ s. - Attenuation needed at 0.5 rad/s is about 19–20 dB (18.8 dB from the table), so take $20\log\beta=20$ dB: $\beta=10$. - Pole at $\frac{1}{\beta T}=0.01$ rad/s. $$ K_c=\frac{K}{\beta}=\frac{5}{10}=0.5 $$ $$ G_c(s)=0.5\times10\,\frac{10s+1}{100s+1}=5\,\frac{10s+1}{100s+1}=0.5\,\frac{s+0.1}{s+0.01} $$ #### Step 5: Compensated system $$ G_c(s)G(s)=\frac{5(10s+1)}{s(100s+1)(s+1)(0.5s+1)} $$ | $\omega$ | 0.01 | 0.1 | 0.2 | 0.45 | 1 | 1.32 | |---|---|---|---|---|---|---| | $|G_cG|$ (dB) | 51.0 | 16.9 | 8.7 | 0 | $-10.0$ | $-14.3$ | | $\phi$ | $-130^\circ$ | $-138^\circ$ | $-131^\circ$ | $-138.4^\circ$ | $-167^\circ$ | $-180^\circ$ | $$ PM=180^\circ-138.4^\circ=41.6^\circ\ (\ge40^\circ),\qquad GM=14.3\ \text{dB}\ (\ge10\ \text{dB}) $$ ```text dB 54 |\___ uncompensated | \___ 0 |--x-----\----------- w | 0.45 \ 1.8 | compensated: 20 dB lower | above 0.1 rad/s ``` | Item | Before | After | |---|---|---| | $K_v$ | 5 | 5 | | $\omega_{gc}$ | 1.80 rad/s | 0.45 rad/s | | PM | $-13^\circ$ | $41.6^\circ$ | | GM | $-4.4$ dB | 14.3 dB | Closed-loop poles: $-0.286\pm j0.520$, $-0.123$, $-2.32$ (all in LHP). **Answer: $G_c(s)=5\,\dfrac{10s+1}{100s+1}$ (β = 10, T = 10 s): Kv = 5 s⁻¹, PM ≈ 42°, GM ≈ 14 dB.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is K = Kc*beta = 5? A. Kv = lim s Gc(s)G(s) = Kc*beta, and Kv = 5 is required B. Kv = Kc/beta C. Kv = beta/Kc D. Kv = 20log(beta) Answer: A. The lag network has DC gain beta times Kc, so the product must give Kv = 5. (Key point: Gain for Kv) 2. What is the state of G1 = 5/[s(s+1)(0.5s+1)]? A. Stable with PM = 40 deg B. Unstable, PM = -13 deg C. Marginally stable, PM = 0 D. Stable with GM = 10 dB Answer: B. Crossover at 1.8 rad/s has phase below -180 deg, so it is unstable. (Key point: Uncompensated state) 3. How is the new gain crossover selected? A. Where the plant phase is -180 deg B. At the old crossover 1.8 rad/s C. Where the plant phase is about -180 + 40 + 12 = -128 deg (12 deg allowed for the lag), about 0.5 rad/s D. Where the gain is +20 dB Answer: C. The lag is not allowed to add much phase lag, so 12 deg is reserved for it. (Key point: New crossover) 4. How are beta and T chosen? A. beta = 5 from Kv; zero at the new crossover B. beta = 0.1; zero at 1.8 rad/s C. beta from the gain margin of 10 dB; zero above the crossover D. beta from the attenuation needed at the new crossover (about 20 dB, so beta = 10); zero 1/T about one-fifth of crossover (0.1 rad/s) Answer: D. The network must attenuate about 19-20 dB at 0.5 rad/s, so beta = 10, and the zero is placed well below crossover so little phase lag appears there. 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