--- title: "Lag compensator for K/[s(s+4)(s+80)], Kv = 30" subject: "Control Systems" chapter: 7 key: "semester_3/control_systems:lag-k-s-s4-s80-kv30" source: https://www.bctengineer.com/q/semester_3/control_systems/lag-k-s-s4-s80-kv30 site: BCT Engineering --- # Lag compensator for K/[s(s+4)(s+80)], Kv = 30 Control Systems · Chapter 7: Performance Specifications and Compensation Design - Asked in: 2070 Chaitra, 12 marks - 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-k-s-s4-s80-kv30 - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-7 ## Question Design series lag compensator for the unity feedback system with feedforward transfer function G(s) = K/[s(s+4)(s+80)]. The velocity error constant is 30 s⁻¹ and phase margin at least 33°. ## Short answer K=9600; lag compensator Gc(s)=(1+2.19s)/(1+9.50s) shifts the gain crossover to 4.56 rad/s. This achieves Kv=30 s⁻¹ and a phase margin of 33.6°. ## Answer A **lag compensator** $G_c(s) = \dfrac{1+Ts}{1+\beta Ts}$ ($\beta > 1$) is designed with the Bode plot method: the gain is fixed by $K_v$, and the lag network lowers the high-frequency gain so the crossover moves to a frequency with enough phase margin. #### Step 1: Gain for $K_v$ $$ K_v = \lim_{s\to0} s\,\frac{K}{s(s+4)(s+80)} = \frac{K}{320} = 30 \;\Rightarrow\; K = 9600 $$ $$ G(s) = \frac{9600}{s(s+4)(s+80)} = \frac{30}{s(0.25s+1)(0.0125s+1)} $$ #### Step 2: Uncompensated margins - Gain crossover: $|G(j\omega)| = 1$ at $\omega_{gc} = 10.55$ rad/s; $\angle G = -90^\circ - \tan^{-1}\frac{10.55}{4} - \tan^{-1}\frac{10.55}{80} = -166.7^\circ$, so $PM = 13.3^\circ$. - Phase crossover: $\omega_{pc} = \sqrt{4\times80} = 17.9$ rad/s, $GM = 8.9$ dB. The PM is far below $33^\circ$. #### Step 3: New gain crossover frequency Allow $5^\circ$ for the lag of the compensator at the new crossover. Required plant phase: $$ \angle G(j\omega_{gc}') = -180^\circ + 33^\circ + 5^\circ = -142^\circ $$ $$ -90^\circ - \tan^{-1}\frac{\omega}{4} - \tan^{-1}\frac{\omega}{80} = -142^\circ \;\Rightarrow\; \omega_{gc}' = 4.56\ \text{rad/s} $$ #### Step 4: Value of $\beta$ At 4.56 rad/s: $$ |G(j4.56)| = \frac{9600}{4.56\times\sqrt{4.56^2+16}\times\sqrt{4.56^2+6400}} = 4.33 \;(12.7\ \text{dB}) $$ The lag network must reduce the gain by this amount: $\beta = 4.33$. #### Step 5: Corner frequencies Place the zero one decade below the new crossover: $$ \begin{aligned} \frac{1}{T} &= \frac{4.56}{10} = 0.456\ \text{rad/s},\quad T = 2.19\ \text{s} \\ \frac{1}{\beta T} &= \frac{0.456}{4.33} = 0.105\ \text{rad/s},\quad \beta T = 9.50\ \text{s} \end{aligned} $$ #### Step 6: Compensator and check $$ G_c(s) = \frac{1 + 2.19s}{1 + 9.50s} = 0.231\,\frac{s+0.456}{s+0.105} $$ $$ G_c(s)G(s) = \frac{9600(1+2.19s)}{s(s+4)(s+80)(1+9.50s)} $$ | Quantity | Uncompensated | Compensated | |---|---|---| | Gain crossover | 10.55 rad/s | 4.57 rad/s | | Phase margin | 13.3° | 33.6° | | Gain margin | 8.9 dB | 20.9 dB | | $K_v$ | 30 s$^{-1}$ | 30 s$^{-1}$ | The lag network adds only $-4.4^\circ$ at the new crossover, within the $5^\circ$ allowed. **Answer:** $K = 9600$, $G_c(s) = \dfrac{1+2.19s}{1+9.50s}$; $K_v = 30\ \text{s}^{-1}$ and $PM = 33.6^\circ$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For G(s) = K/[s(s+4)(s+80)], what K gives Kv = 30? A. K = 9600 B. K = 320 C. K = 2400 D. K = 120 Answer: A. Kv = K/(4*80) = K/320 = 30, so K = 9600. (Key point: Gain from Kv) 2. How does a lag compensator achieve the required phase margin? A. It adds a pole at the origin to raise Kv B. It increases the gain at crossover C. It lowers the high-frequency gain so the crossover moves to a frequency where the plant already has enough phase D. It adds positive phase near crossover Answer: C. Lag design attenuates the gain so the new crossover occurs where the plant phase gives the needed margin; the compensator phase lag is only allowed for. (Key point: Lag design idea) 3. Why is the new crossover found where plant phase = -180 + 33 + 5 = -142 degrees? A. The 5 degrees is the gain margin B. The 5 degrees compensates for Kv C. The 5 degrees is the lead from the plant zero D. The extra 5 degrees allows for the phase lag the lag network adds at the new crossover Answer: D. The lag network contributes a small negative phase at crossover, so the plant must have 5 degrees extra margin. (Key point: New crossover) 4. At the new crossover the plant gain is 4.33. How are beta and the zero chosen? A. beta = 4.33 and the pole is one decade above the crossover B. beta = 33 and the zero is at 80 rad/s C. beta = 4.33 and the zero 1/T is one decade below the crossover D. beta = 1/4.33 and the zero is at the crossover Answer: C. The network must cut the gain by 4.33 at crossover, so beta = 4.33; the zero at 4.56/10 = 0.456 rad/s keeps the lag small at crossover. 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