--- title: "Lag compensator for 2/[s(s+1)(s+2)], Kv = 5, PM 40°" subject: "Control Systems" chapter: 7 key: "semester_3/control_systems:lag-2-s-s1-s2-kv5" source: https://www.bctengineer.com/q/semester_3/control_systems/lag-2-s-s1-s2-kv5 site: BCT Engineering --- # Lag compensator for 2/[s(s+1)(s+2)], Kv = 5, PM 40° Control Systems · Chapter 7: Performance Specifications and Compensation Design - Asked in: 2075 Chaitra, 12 marks; 2080 Bhadra, 12 marks (asked 2 times) - 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-2-s-s1-s2-kv5 - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-7 ## Question Consider a system with open loop transfer function G(s)H(s) = 2/[s(s+1)(s+2)]. It is desired to compensate the system so that the static velocity error constant Kv is 5 per second, the phase margin is at least 40° and gain margin is at least 10 dB. Determine transfer function of appropriate lag compensator. ## Short answer Lag compensator: Gc(s) = 5(1+20s)/(1+177s) with β ≈ 8.9, T = 20 s. This yields Kv = 5 s⁻¹, phase margin ≈ 45°, and gain margin ≈ 14 dB. ## Answer A **lag compensator** $G_c(s)=K_c\dfrac{1+Ts}{1+\beta Ts}$ ($\beta>1$) is used here: it gives high gain at low frequency (for $K_v$) while its attenuation of $20\log\beta$ at high frequency lowers the gain crossover to a frequency where the plant already has enough phase margin. #### Step 1: Gain for the required $K_v$ $$ G(s)H(s)=\frac{2}{s(s+1)(s+2)}=\frac{1}{s(1+s)(1+0.5s)}\Rightarrow K_v=\lim_{s\to0}sGH=1 $$ To get $K_v=5$ the compensator must give a DC gain $K_c=5$: $$ G_1(s)=\frac{5}{s(1+s)(1+0.5s)} $$ #### Step 2: Uncompensated (gain-adjusted) system $$ |G_1|=\frac{5}{\omega\sqrt{1+\omega^2}\sqrt{1+0.25\omega^2}},\qquad\phi=-90^\circ-\tan^{-1}\omega-\tan^{-1}0.5\omega $$ | $\omega$ | 0.05 | 0.1 | 0.49 | 1 | 1.41 | 1.80 | |---|---|---|---|---|---|---| | $|G_1|$ (dB) | 40.0 | 33.9 | 19.0 | 10.0 | 4.5 | 0 | | $\phi$ | $-94.3^\circ$ | $-98.6^\circ$ | $-129.9^\circ$ | $-161.6^\circ$ | $-180^\circ$ | $-193^\circ$ | - $\omega_{pc}=\sqrt2=1.414$ rad/s, GM $=-4.4$ dB; $\omega_{gc}=1.80$ rad/s, PM $=-13^\circ$: **unstable**. #### Step 3: New gain crossover frequency Required phase at $\omega_{gc}'$: $-180^\circ+40^\circ+\epsilon$ with $\epsilon=10^\circ$ (for the lag of the compensator), i.e. $-130^\circ$: $$ -90^\circ-\tan^{-1}\omega-\tan^{-1}0.5\omega=-130^\circ\Rightarrow\omega_{gc}'=0.49\ \text{rad/s} $$ #### Step 4: β At 0.49 rad/s, $|G_1|=18.95$ dB $=8.86$. The lag network must cut this to 0 dB: $$ 20\log\beta=18.95\ \text{dB}\Rightarrow\beta=8.86 $$ #### Step 5: T Place the zero one decade below the new crossover so the compensator lag there is small: $$ \frac1T=\frac{\omega_{gc}'}{10}\approx0.05\ \text{rad/s}\Rightarrow T=20\ \text{s},\quad\beta T=177\ \text{s} $$ Pole at $1/\beta T=0.00565$ rad/s. $$ G_c(s)=5\,\frac{1+20s}{1+177s}=0.565\,\frac{s+0.05}{s+0.00565} $$ #### Step 6: Check $$ G_c(s)G(s)H(s)=\frac{5(1+20s)}{s(1+s)(1+0.5s)(1+177s)} $$ | Quantity | Before | After | |---|---|---| | $K_v$ | 1 (5 with gain) | 5 s⁻¹ | | $\omega_{gc}$ | 1.80 rad/s | 0.494 rad/s | | PM | $-13^\circ$ | $44.7^\circ$ | | $\omega_{pc}$ | 1.414 rad/s | 1.37 rad/s | | GM | $-4.4$ dB | 13.9 dB | ```text dB 40 |\__ uncompensated | \___ 0 |---x---\------------ w | 0.49 \ 1.8 | compensated curve is | 19 dB lower above 0.05 ``` All specifications are met (PM ≥ 40°, GM ≥ 10 dB, $K_v=5$). The price is a lower bandwidth (slower response). **Answer: $G_c(s)=5\,\dfrac{1+20s}{1+177s}$ (β ≈ 8.9, T = 20 s): Kv = 5 s⁻¹, PM ≈ 45°, GM ≈ 14 dB.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. G(s)H(s) = 2/[s(s+1)(s+2)] has Kv = 1. How is Kv = 5 achieved? A. Include DC gain Kc = 5 in the compensator B. Use beta = 5 C. Move the pole at -2 to the origin D. Reduce the system gain by 5 Answer: A. Multiplying by Kc = 5 gives G1 = 5/[s(1+s)(1+0.5s)] with Kv = 5. (Key point: Gain for Kv) 2. How is the new gain crossover (0.49 rad/s) chosen? A. Where |G1| = 1 for the uncompensated plant B. At the frequency where the plant phase is -130 degrees (= -180 + 40 + 10 allowance) C. At the phase crossover 1.414 rad/s D. At the corner frequency 2 rad/s Answer: B. Choose w where phase gives PM 40 degrees plus a 10 degree allowance for the lag network, i.e. -130 degrees. (Key point: New crossover choice) 3. At 0.49 rad/s |G1| = 18.95 dB. How is beta found? A. 20 log beta = 18.95 dB, so beta = 8.86 B. beta = 18.95 C. beta = 1/8.86 D. 20 log beta = 40 dB Answer: A. The lag must attenuate by the plant gain at the new crossover so the loop gain is 0 dB there: beta = 8.86. (Key point: Beta) 4. Why is the compensator zero placed about a decade below the new crossover (T = 20 s)? A. So the lag network adds little phase lag at the new crossover B. So that Kv is increased tenfold C. To raise the gain crossover D. To make the system type 2 Answer: A. With 1/T = w_gc'/10, the lag's phase contribution at crossover is small, so PM is preserved (final PM 44.7 degrees). 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