--- title: "Lag compensator for 1.06/[s(s+1)(s+2)], ramp error < 0.2" subject: "Control Systems" chapter: 7 key: "semester_3/control_systems:lag-106-ramp-ess02" source: https://www.bctengineer.com/q/semester_3/control_systems/lag-106-ramp-ess02 site: BCT Engineering --- # Lag compensator for 1.06/[s(s+1)(s+2)], ramp error < 0.2 Control Systems · Chapter 7: Performance Specifications and Compensation Design - Asked in: 2079 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/lag-106-ramp-ess02 - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-7 ## Question For a unity feedback system with open loop transfer function G(s) = 1.06/[s(s+1)(s+2)], design a lag compensator such that steady state error for ramp input would be less than 0.2 without significant change in transients. ## Short answer A lag compensator Gc(s)=0.966(s+0.05)/(s+0.005) raises the velocity error constant to Kv=5.12 s⁻¹, giving ramp steady-state error 0.195 (<0.2). Dominant poles remain near -0.312±j0.551, so damping and transient response are nearly unchanged. ## Answer Since the transient must stay almost the same while $K_v$ is raised, a **lag compensator by the root-locus method** is used: a pole–zero pair very close to the origin raises the low-frequency gain but hardly moves the dominant poles. $$ G_c(s)=\hat K_c\,\frac{s+\frac1T}{s+\frac{1}{\beta T}},\qquad\beta>1 $$ #### Step 1: Present system $$ K_v=\lim_{s\to0}sG(s)=\frac{1.06}{1\times2}=0.53,\qquad e_{ss}=\frac{1}{K_v}=1.89 $$ Characteristic equation: $s^3+3s^2+2s+1.06=0$; roots $-2.339$ and $-0.3307\pm j0.5864$. Dominant poles: $\omega_n=0.673$ rad/s, $\zeta=0.491$. #### Step 2: Required $K_v$ and β $$ e_{ss}<0.2\Rightarrow K_v>5\Rightarrow\beta\ge\frac{5}{0.53}=9.4;\ \text{take}\ \beta=10 $$ #### Step 3: Place the pole–zero pair Put the zero and pole close to the origin, with ratio 10: $$ \text{zero at }-0.05,\quad\text{pole at }-0.005\Rightarrow T=20\ \text{s},\ \beta T=200\ \text{s} $$ Angle added at $s_1=-0.3307+j0.5864$: $$ \angle(s_1+0.05)-\angle(s_1+0.005)=115.6^\circ-119.0^\circ\approx-3.5^\circ $$ This is less than about 5°, so the root locus near the dominant poles changes very little. #### Step 4: Gain $\hat K_c$ The new dominant poles lie on the compensated root locus with almost the same ζ (≈0.49): $s_1'=-0.312\pm j0.551$. Magnitude condition: $$ \hat K_c=\frac{|s_1'|\,|s_1'+1|\,|s_1'+2|\,|s_1'+0.005|}{1.06\,|s_1'+0.05|}=0.966 $$ $$ G_c(s)=0.966\,\frac{s+0.05}{s+0.005}=9.66\,\frac{1+20s}{1+200s} $$ #### Step 5: Check $$ K_v=\lim_{s\to0}sG_cG=0.966\times10\times0.53=5.12\Rightarrow e_{ss}=\frac{1}{5.12}=0.195<0.2 $$ The requirement is met. Closed-loop poles of the compensated system: $-0.312\pm j0.551$, $-2.326$, $-0.055$. The pole at $-0.055$ is almost cancelled by the zero at $-0.05$, so it adds only a small, slow tail. | Item | Before | After | |---|---|---| | Dominant poles | $-0.331\pm j0.586$ | $-0.312\pm j0.551$ | | $\zeta$ | 0.491 | 0.493 | | $\omega_n$ | 0.673 rad/s | 0.633 rad/s | | $K_v$ | 0.53 | 5.12 | | Ramp $e_{ss}$ | 1.89 | 0.195 | Transient response is nearly unchanged (slightly slower, since $\omega_n$ dropped about 6%). **Answer: $G_c(s)=0.966\,\dfrac{s+0.05}{s+0.005}$ gives Kv = 5.12 s⁻¹, ramp error ≈ 0.195 (< 0.2), with almost the same damping.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is a lag compensator chosen here? A. Ramp error must fall (higher Kv) without significantly changing the transients B. The transient response must be made much faster C. Phase margin must be raised by adding phase lead D. The plant needs a higher bandwidth Answer: A. A lag pole-zero pair near the origin raises low-frequency gain but hardly moves the dominant poles. (Key point: Why lag) 2. With Kv = 0.53 and the requirement ess < 0.2, why is beta = 10 taken? A. beta = 0.53 x 0.2 B. Kv must exceed 5, so beta >= 5/0.53 = 9.4; 10 is chosen C. beta = 1/0.2 = 5 D. beta = 1.06 x 10 Answer: B. ess < 0.2 requires Kv > 5; the gain increase needed is 5/0.53 = 9.4, rounded to 10. (Key point: Value of beta) 3. Where is the lag zero and pole placed? A. Near the dominant poles at -0.33 +/- j0.59 B. Far left of the plant poles C. Very close to the origin (zero at -0.05, pole at -0.005), so the angle contribution at the dominant poles is under about 5 deg D. At -1 and -2 to cancel the plant poles Answer: C. Keeping the pair near the origin makes the angle added at the dominant poles tiny, so the root locus barely changes. (Key point: Placement) 4. What Kv and ramp error does the final design give? A. Kv = 0.53, ess = 1.89 B. Kv = 10, ess = 0.1 C. Kv = 5.12, ess = 5.12 D. Kv = 5.12, ess = 0.195 Answer: D. Kv = 0.966 x 10 x 0.53 = 5.12 and ess = 1/Kv = 0.195 < 0.2. 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