--- title: "Lead compensator for k/[s²(s+5)], Ka = 5, PM 30°" subject: "Control Systems" chapter: 7 key: "semester_3/control_systems:lead-k-s2-s5-ka5-pm30" source: https://www.bctengineer.com/q/semester_3/control_systems/lead-k-s2-s5-ka5-pm30 site: BCT Engineering --- # Lead compensator for k/[s²(s+5)], Ka = 5, PM 30° Control Systems · Chapter 7: Performance Specifications and Compensation Design - Asked in: 2065 Shrawan (old course), 16 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/control_systems/lead-k-s2-s5-ka5-pm30 - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-7 ## Question Design a lead compensator for a system with open loop transfer function G(s) = k/[s²(s+5)] for the specifications of phase margin = 30° and acceleration error constant Ka = 5 sec⁻². Also draw the bode magnitude and phase plots after compensation. ## Short answer k=25 from Ka=5. Uncompensated PM is –23.2°. Two identical lead stages with α=0.271 each provide net lead. Gc(s)=((1+0.502s)/(1+0.136s))^2 yields Ka=5 s⁻² and PM=32.6° (≥30°). ## Answer The plant is type 2 with an extra pole, so it is unstable for every $k$. A **phase-lead** compensator $G_c(s) = \dfrac{1+Ts}{1+\alpha Ts}$ (unity DC gain) is designed on the Bode plot. #### Step 1: Gain for $K_a$ $$ K_a = \lim_{s\to0} s^2\,\frac{k}{s^2(s+5)} = \frac{k}{5} = 5 \;\Rightarrow\; k = 25 $$ $$ G(s) = \frac{25}{s^2(s+5)} = \frac{5}{s^2(0.2s+1)} $$ #### Step 2: Uncompensated phase margin $$ \begin{aligned} \frac{25}{\omega^2\sqrt{\omega^2+25}} &= 1 \;\Rightarrow\; \omega_{gc} = 2.14\ \text{rad/s} \\ \angle G &= -180^\circ - \tan^{-1}(2.14/5) = -203.2^\circ,\qquad PM = -23.2^\circ \end{aligned} $$ #### Step 3: Phase lead required $$ \phi_m = 30^\circ - (-23.2^\circ) + 5^\circ = 58.2^\circ $$ A single stage ($\alpha = 0.081$) moves the crossover to 3.75 rad/s, where the plant phase is already $-216.9^\circ$, so the PM is only $21.3^\circ$. Hence two identical lead stages of $35^\circ$ each are used: $$ \alpha = \frac{1-\sin35^\circ}{1+\sin35^\circ} = 0.271 \text{ per stage} $$ #### Step 4: New gain crossover Two stages add $2\times10\log(1/\alpha) = 11.34$ dB at $\omega_m$, so $\omega_m$ is where $|G| = \alpha = 0.271$: $$ \frac{25}{\omega^2\sqrt{\omega^2+25}} = 0.271 \;\Rightarrow\; \omega_m = 3.83\ \text{rad/s} $$ #### Step 5: Corner frequencies $$ T = \frac{1}{3.83\sqrt{0.271}} = 0.502\ \text{s},\quad \frac{1}{T} = 1.99\ \text{rad/s},\quad \frac{1}{\alpha T} = 7.35\ \text{rad/s},\quad \alpha T = 0.136\ \text{s} $$ $$ G_c(s) = \left(\frac{1+0.502s}{1+0.136s}\right)^2 = 13.6\left(\frac{s+1.99}{s+7.35}\right)^2 $$ $$ G_c(s)G(s) = \frac{25(1+0.502s)^2}{s^2(s+5)(1+0.136s)^2} $$ PM check at 3.83 rad/s: $180^\circ - 180^\circ - \tan^{-1}(3.83/5) + 2\times35^\circ = -37.45^\circ + 70^\circ = 32.55^\circ \approx 32.6^\circ$. #### Step 6: Bode plots after compensation | $\omega$ (rad/s) | Magnitude (dB) | Phase | |---|---|---| | 0.1 | 54.0 | $-177.0^\circ$ | | 0.5 | 26.5 | $-165.3^\circ$ | | 1 | 15.6 | $-153.4^\circ$ | | 1.99 | 6.8 | $-142.0^\circ$ | | 3.83 | 0.0 | $-147.4^\circ$ | | 7.35 | $-8.4$ | $-176.1^\circ$ | | 10 | $-13.7$ | $-193.3^\circ$ | | 20 | $-28.7$ | $-227.0^\circ$ | | 50 | $-51.5$ | $-252.1^\circ$ | Asymptotic magnitude slopes: $-40$ dB/dec up to 1.99 rad/s (14 dB at $\omega = 1$), $0$ dB/dec from 1.99 to 5 rad/s (double zero; the flat level is about 2 dB), $-20$ dB/dec from 5 to 7.35 rad/s (plant pole), and $-60$ dB/dec beyond 7.35 rad/s (double pole). ```text dB 54 |\ | \ -40 2 | \______ 0 0 |---------\--------- 0 dB | \ -20 -20 | \ | \ -60 +---+-----+--+------> w (log) 1.99 5 7.35 deg -140| .--. -160| / \ -180|----'------\------- -180 -220| \__ -250| \___ +---+-----+--+------> w (log) 1.99 3.83 7.89 ``` Phase rises from $-180^\circ$ to a peak of about $-142^\circ$ near 2 rad/s, is $-147.4^\circ$ at the crossover 3.83 rad/s, and crosses $-180^\circ$ at 7.89 rad/s, where the magnitude is $-9.5$ dB. **Answer:** $k = 25$, $G_c(s) = \left(\dfrac{1+0.502s}{1+0.136s}\right)^2$; $K_a = 5\ \text{s}^{-2}$, $PM = 32.6^\circ\ (\ge 30^\circ)$, $GM = 9.5$ dB. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For G(s) = k/[s^2(s+5)], what k gives Ka = 5? A. k = 25 B. k = 125 C. k = 1 D. k = 5 Answer: A. Ka = k/5 = 5, so k = 25. (Key point: Gain from Ka) 2. Why does a single lead stage fail here? A. It reduces Ka B. It cannot add any phase C. It moves the crossover up to where plant phase is already -216.9 degrees, giving PM of only 21.3 degrees D. It makes the gain margin infinite Answer: C. Plant phase falls quickly as crossover shifts, so the required lead is not realised; two 35-degree stages are used. (Key point: Single stage fails) 3. What is alpha for each 35-degree lead stage? A. cos35 = 0.819 B. (1+sin35)/(1-sin35) = 3.69 C. sin35 = 0.574 D. (1-sin35)/(1+sin35) = 0.271 Answer: D. alpha = (1-sin phi)/(1+sin phi) = 0.271. (Key point: Alpha per stage) 4. How are the zero and pole of each stage located from omega_m = 3.83? A. Zero at 5, pole at 25 B. Zero at omega_m, pole at 2*omega_m C. Zero at 7.35, pole at 1.99 D. T = 1/(omega_m*sqrt(alpha)): zero at 1/T = 1.99, pole at 1/(alpha T) = 7.35 Answer: D. The geometric mean of the zero and pole is omega_m, so 1/T = omega_m*sqrt(alpha) and pole = 1/(alpha*T). 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