--- title: "Lead compensator for K/[s²(0.2s+1)], Ka = 10, PM 35°" subject: "Control Systems" chapter: 7 key: "semester_3/control_systems:lead-k-s2-02s-ka10-pm35" source: https://www.bctengineer.com/q/semester_3/control_systems/lead-k-s2-02s-ka10-pm35 site: BCT Engineering --- # Lead compensator for K/[s²(0.2s+1)], Ka = 10, PM 35° Control Systems · Chapter 7: Performance Specifications and Compensation Design - Asked in: 2066 Jestha (old course), 10 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-s2-02s-ka10-pm35 - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-7 ## Question The open loop transfer function of a unity feedback system is given by G(s) = K/[s²(0.2s+1)]. Design a lead compensator to meet the following specifications: Acceleration error constant = 10, Phase margin = 35°. ## Short answer Set K=10 for Ka=10. Uncompensated PM is –30.4°. Two identical lead stages with α=0.1716 are needed. Gc(s)=((1+0.397s)/(1+0.0681s))^2 gives Ka=10 s⁻² and PM=39.4° (≥35°). ## Answer A type-2 plant has $-180^\circ$ phase at low frequency, so the plant pole makes it unstable for any $K$. A **phase-lead** network is designed on the Bode plot: $G_c(s) = \dfrac{1+Ts}{1+\alpha Ts}$, $\alpha < 1$ (unity DC gain, so $K_a$ is set by $K$). #### Step 1: Gain for $K_a$ $$ K_a = \lim_{s\to0} s^2\,\frac{K}{s^2(0.2s+1)} = K = 10, \qquad G(s) = \frac{10}{s^2(0.2s+1)} $$ #### Step 2: Uncompensated phase margin $$ \begin{aligned} |G(j\omega)| &= \frac{10}{\omega^2\sqrt{1+0.04\omega^2}} = 1 \;\Rightarrow\; \omega_{gc} = 2.94\ \text{rad/s} \\ \angle G &= -180^\circ - \tan^{-1}(0.2\times2.94) = -210.4^\circ \\ PM &= -30.4^\circ \quad (\text{unstable}) \end{aligned} $$ #### Step 3: Phase lead required $$ \phi_m = 35^\circ - (-30.4^\circ) + 5^\circ = 70.4^\circ $$ This is more than about $60^\circ$, which a single lead network cannot give with a sensible $\alpha$. Use **two identical lead stages** in cascade. Also, the plant magnitude falls at $-40$ to $-60$ dB/decade, so the crossover moves a lot and its phase falls quickly; a first trial with $2 \times 35^\circ$ gives only $PM = 24.5^\circ$. So take $2 \times 45^\circ$: $$ \alpha = \frac{1-\sin45^\circ}{1+\sin45^\circ} = 0.1716 \text{ per stage} $$ #### Step 4: New gain crossover The two stages together add $2 \times 10\log(1/\alpha) = 15.3$ dB at $\omega_m$. So $\omega_m$ is where $|G(j\omega)| = \alpha = 0.1716$ ($-15.3$ dB): $$ \frac{10}{\omega^2\sqrt{1+0.04\omega^2}} = 0.1716 \;\Rightarrow\; \omega_m = 6.08\ \text{rad/s} $$ #### Step 5: Corner frequencies $$ \begin{aligned} T &= \frac{1}{\omega_m\sqrt{\alpha}} = \frac{1}{6.08\times0.4142} = 0.397\ \text{s} \\ \text{zero: } 1/T &= 2.52\ \text{rad/s},\qquad \text{pole: } 1/(\alpha T) = 14.69\ \text{rad/s} \end{aligned} $$ #### Step 6: Compensator and check $$ G_c(s) = \left(\frac{1+0.397s}{1+0.0681s}\right)^2 = 34.0\left(\frac{s+2.52}{s+14.69}\right)^2 $$ $$ G_c(s)G(s) = \frac{10(1+0.397s)^2}{s^2(0.2s+1)(1+0.0681s)^2} $$ At $\omega = 6.08$: plant phase $= -180^\circ - \tan^{-1}(1.216) = -230.6^\circ$, compensator phase $= +90^\circ$, so $PM = 180^\circ - 230.6^\circ + 90^\circ = 39.4^\circ$. | Quantity | Uncompensated | Compensated | |---|---|---| | Gain crossover | 2.94 rad/s | 6.08 rad/s | | Phase margin | $-30.4^\circ$ | $39.4^\circ$ | | Gain margin | negative (unstable) | 11.5 dB at 14.5 rad/s | | $K_a$ | 10 s$^{-2}$ | 10 s$^{-2}$ | **Answer:** two cascaded lead stages, $G_c(s) = \left(\dfrac{1+0.397s}{1+0.0681s}\right)^2$, give $K_a = 10$ and $PM = 39.4^\circ \ (\ge 35^\circ)$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For G(s) = K/[s^2(0.2s+1)], what K gives Ka = 10? A. K = 10 B. K = 2 C. K = 0.2 D. K = 50 Answer: A. Ka = lim s^2 G = K, so K = 10. (Key point: Gain from Ka) 2. What is the uncompensated system's condition? A. PM = -30.4 degrees, unstable B. PM = 0, marginally stable C. PM = 35 degrees D. PM = 30.4 degrees, stable Answer: A. Crossover is at 2.94 rad/s where phase is -210.4 degrees, so PM is negative. (Key point: Uncompensated PM) 3. Why are two cascaded lead stages used? A. Each stage adds 10 dB of gain margin B. One stage would increase steady-state error C. Two stages are needed to set Ka D. The required lead (about 70 degrees) is more than one lead network can sensibly give Answer: D. A single network is limited to about 60 degrees; splitting into two stages gives the needed lead with reasonable alpha. (Key point: Why two stages) 4. With two stages of alpha = 0.1716, how is the new crossover omega_m found? A. Where |G| = 1 B. Where the plant phase is -180 degrees C. Where |G| = alpha (-15.3 dB), since two stages add 15.3 dB at omega_m D. Where |G| = sqrt(alpha) Answer: C. Each stage adds 10log(1/alpha) = 7.66 dB at omega_m, so the plant must be at -15.3 dB there for the loop to cross 0 dB. (Key point: Crossover for lead) ## Close variants set in other exams - [Lead compensator for k/[s²(1+0.25s)], Ka = 10, PM 35°](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Alead-k-s2-025s-ka10-pm35) — Control Systems, chapter 7, asked 5× ## Related in Control Systems - [Lead compensator for K/[s(1+0.2s)], Kv=10, PM=50°](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Alead-k-s-1-02s-kv10) — Control Systems, chapter 7, asked 2× - [Lead compensator for k/[s²(s+5)], Ka = 5, PM 30°](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Alead-k-s2-s5-ka5-pm30) — Control Systems, chapter 7 - [Lead compensator for K/[s(1+0.2s)], Kv≥20, PM≥44°](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Alead-k-s-1-02s-kv20) — Control Systems, chapter 7, asked 2× - [Lead compensator for 2/[s(1+0.5s)], Kv > 20, PM 50°](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Alead-2-s-1-05s-kv20) — Control Systems, chapter 7 - [Lead compensator for 5/[(s+2)(s+20)], Kv ≥ 20](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Alead-5-s2-s20-kv20) — Control Systems, chapter 7 - [Lead compensator for K/[s(s+10)(s+2)], Kv ≥ 800](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Alead-k-s-s10-s2-kv800) — Control Systems, chapter 7 ## Related in other subjects - [Bode design: PM 55°, GM 10 dB, Kv = 5](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Abode-design-phase-margin) — Digital Control System, chapter 4