--- title: "Lag compensator for 0.53/[s(1+s)(1+0.5s)], Kv = 5" subject: "Control Systems" chapter: 7 key: "semester_3/control_systems:lag-053-s-1s-05s-kv5" source: https://www.bctengineer.com/q/semester_3/control_systems/lag-053-s-1s-05s-kv5 site: BCT Engineering --- # Lag compensator for 0.53/[s(1+s)(1+0.5s)], Kv = 5 Control Systems · Chapter 7: Performance Specifications and Compensation Design - Asked in: 2081 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-053-s-1s-05s-kv5 - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-7 ## Question Design a lag compensator for a unity feedback system with open loop transfer function G(s) = 0.53/[s(1+s)(1+0.5s)] such that velocity error constant would become 5 s⁻¹ without significant change in transient properties. ## Short answer The lag compensator is Gc(s) = 0.970(s+0.05)/(s+0.005). By placing zero at -0.05 and pole at -0.005 (β=10) near the origin, the velocity error constant becomes Kv=5.14 sec^-1 while the dominant closed-loop poles remain near -0.33±j0.59 (ζ≈0.49), so the transient response is essentially unchanged. ## Answer "Without significant change in transient properties" means the dominant closed-loop poles should stay almost where they are, so a **lag compensator** is designed by the **root locus method**, with its pole and zero close to the origin. $$ G(s) = \frac{0.53}{s(1+s)(1+0.5s)} = \frac{1.06}{s(s+1)(s+2)} $$ #### Step 1: Uncompensated system Characteristic equation: $s^3 + 3s^2 + 2s + 1.06 = 0$. Roots: $s = -0.331 \pm j0.586$ (dominant) and $s = -2.34$. $$ \omega_n = \sqrt{0.331^2 + 0.586^2} = 0.673\ \text{rad/s},\qquad \zeta = \frac{0.331}{0.673} = 0.491 $$ $$ K_v = \lim_{s\to0} s\,G(s) = 0.53\ \text{s}^{-1} $$ #### Step 2: Required increase in $K_v$ $$ \beta \ge \frac{5}{0.53} = 9.43 \quad\Rightarrow\quad \text{choose } \beta = 10 $$ #### Step 3: Place the pole and zero $G_c(s) = \hat{K}_c\,\dfrac{s + 1/T}{s + 1/(\beta T)}$. Put the zero close to the origin, well to the right of the dominant poles: $$ \text{zero at } s = -0.05, \qquad \text{pole at } s = -\frac{0.05}{10} = -0.005 $$ Angle added at $s_1 = -0.331 + j0.586$: $$ \angle(s_1 + 0.05) - \angle(s_1 + 0.005) = 115.6^\circ - 119.1^\circ = -3.5^\circ $$ This is less than $5^\circ$, so the root locus near the dominant poles hardly changes. #### Step 4: New dominant poles and gain Keeping $\zeta = 0.491$, the compensated locus $\dfrac{1.06\hat{K}_c(s+0.05)}{s(s+1)(s+2)(s+0.005)}$ crosses the $\zeta$ line at $$ s_1' = -0.312 \pm j0.553 $$ Magnitude condition at $s_1'$: $$ \hat{K}_c = \frac{|s_1'|\,|s_1'+1|\,|s_1'+2|\,|s_1'+0.005|}{1.06\,|s_1'+0.05|} = 0.970 $$ #### Step 5: Compensator $$ G_c(s) = 0.970\,\frac{s+0.05}{s+0.005} = 9.70\,\frac{20s+1}{200s+1} $$ $$ G_c(s)G(s) = \frac{1.028(s+0.05)}{s(s+0.005)(s+1)(s+2)} $$ #### Step 6: Check $$ K_v = \lim_{s\to0} s\,G_cG = \frac{1.028\times0.05}{0.005\times2} = 5.14\ \text{s}^{-1} \;(\ge 5) $$ | Quantity | Uncompensated | Compensated | |---|---|---| | Dominant poles | $-0.331 \pm j0.586$ | $-0.312 \pm j0.553$ | | $\zeta$ | 0.491 | 0.491 | | $\omega_n$ | 0.673 rad/s | 0.635 rad/s | | Other poles | $-2.34$ | $-2.33$, $-0.055$ | | $K_v$ | 0.53 s$^{-1}$ | 5.14 s$^{-1}$ | The dominant poles keep $\zeta = 0.491$ and $\omega_n$ falls by only about 6%, so the dominant-pole transient response is practically the same while $K_v$ rises nearly ten times. The extra closed-loop pole at $-0.055$ lies close to, but not exactly on, the zero at $-0.05$, so it leaves a small, slowly decaying tail (amplitude about 0.11, time constant about 18 s). In a step simulation this tail raises the peak overshoot from about 16% to about 26% and lengthens the 2% settling time. This slow tail is the usual price of lag compensation. **Answer:** $G_c(s) = 0.970\dfrac{s+0.05}{s+0.005}$, giving $K_v = 5.14\ \text{s}^{-1}$ with nearly unchanged transient response. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the root locus method used for this lag design? A. The Bode plot cannot handle type 1 systems B. Lag compensators only work with root locus C. The dominant closed-loop poles must stay almost where they are D. A lead is needed Answer: C. Keeping transient properties unchanged means the dominant poles should hardly move, so the lag pole-zero pair is placed near the origin. (Key point: Method) 2. Uncompensated Kv = 0.53 and Kv needed = 5. How is beta chosen? A. beta = 5 - 0.53 B. beta >= 5/0.53 = 9.43, choose 10 C. beta = 0.53/5 D. beta = 2 Answer: B. The lag network raises Kv by the factor beta, so beta must be at least 9.43. (Key point: Beta) 3. Why are the zero at -0.05 and the pole at -0.005 chosen? A. They shift the dominant poles far left B. They give 60 degrees of lead C. They cancel the plant poles D. They are close to the origin so the angle they add at the dominant poles is under 5 degrees Answer: D. The net angle is about -3.5 degrees, so the root locus near the dominant poles barely changes. (Key point: Placing pole and zero) 4. What is the result of the design? A. Kv = 5.14 with zeta doubled B. Kv = 5.14 with practically the same dominant poles and zeta C. Kv = 10 with unstable response D. Kv = 0.53 with faster response Answer: B. Dominant poles move only slightly (-0.312 +/- j0.553) at the same zeta, so the transient is nearly unchanged while Kv rises about ten times. 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