--- title: "Lead compensator for 2/[s(1+0.5s)], Kv > 20, PM 50°" subject: "Control Systems" chapter: 7 key: "semester_3/control_systems:lead-2-s-1-05s-kv20" source: https://www.bctengineer.com/q/semester_3/control_systems/lead-2-s-1-05s-kv20 site: BCT Engineering --- # Lead compensator for 2/[s(1+0.5s)], Kv > 20, PM 50° Control Systems · Chapter 7: Performance Specifications and Compensation Design - Asked in: 2082 Baisakh, 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/lead-2-s-1-05s-kv20 - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-7 ## Question Design a lead compensator for unity feedback system with open loop transfer function G(s) = 2/[s(1+0.5s)] such that velocity error constant becomes greater than 20 sec⁻¹ and phase margin would be at least 50°. ## Short answer Set Kc=10 to achieve Kv=20 s⁻¹. The required phase lead is about 42°, so choose α=0.2. The new gain crossover is 9.35 rad/s, giving T=0.24 s. The compensator is Gc(s)=10(1+0.24s)/(1+0.048s), yielding phase margin ≈54° at 9.4 rad/s. ## Answer Lead compensator: $G_c(s)=K_c\dfrac{1+Ts}{1+\alpha Ts}$, $\alpha<1$, with maximum lead $\sin\phi_m=\frac{1-\alpha}{1+\alpha}$ at $\omega_m=\frac{1}{T\sqrt\alpha}$, where it raises the magnitude by $10\log(1/\alpha)$ dB. #### Step 1: Gain for $K_v$ $$ K_v=\lim_{s\to0}s\,G_c(s)G(s)=K_c\times2=20\Rightarrow K_c=10 $$ Gain-adjusted plant: $$ G_1(s)=\frac{20}{s(1+0.5s)} $$ #### Step 2: Uncompensated (gain-adjusted) system Asymptotes: $-20$ dB/dec (26 dB at $\omega=1$), corner at 2 rad/s, then $-40$ dB/dec. $$ \frac{20}{\omega\sqrt{1+0.25\omega^2}}=1\Rightarrow0.25\omega^4+\omega^2-400=0\Rightarrow\omega_{gc}=6.17\ \text{rad/s} $$ $$ PM=90^\circ-\tan^{-1}(0.5\times6.17)=90^\circ-72.0^\circ=18.0^\circ $$ #### Step 3: Phase lead needed $$ \phi_m=50^\circ-18^\circ+\epsilon $$ With $\epsilon=5^\circ$ the check gives PM = 49.8° (just short), so take $\epsilon\approx10^\circ$: $\phi_m\approx42^\circ$. #### Step 4: α $$ \alpha=\frac{1-\sin42^\circ}{1+\sin42^\circ}=0.198\approx0.2\quad(\phi_m=41.8^\circ) $$ #### Step 5: New gain crossover Lead gain at $\omega_m$: $10\log(1/0.2)=6.99$ dB. Find where $|G_1|=-6.99$ dB $(=\sqrt{0.2}=0.447)$: $$ \frac{20}{\omega\sqrt{1+0.25\omega^2}}=0.447\Rightarrow\omega_m=9.35\ \text{rad/s} $$ #### Step 6: T $$ T=\frac{1}{\omega_m\sqrt\alpha}=\frac{1}{9.35\times0.447}=0.239\approx0.24\ \text{s},\qquad\alpha T=0.048\ \text{s} $$ Zero at 4.17 rad/s, pole at 20.8 rad/s. $$ G_c(s)=10\,\frac{1+0.24s}{1+0.048s}=50\,\frac{s+4.17}{s+20.8} $$ #### Step 7: Check $$ G_c(s)G(s)=\frac{20(1+0.24s)}{s(1+0.5s)(1+0.048s)} $$ | $\omega$ | 1 | 2 | 4.17 | 6.17 | 9.37 | 20.8 | |---|---|---|---|---|---|---| | $|G_1|$ (dB) | 25.1 | 17.0 | 6.3 | 0 | $-7.0$ | $-20.7$ | | $|G_cG|$ (dB) | 25.3 | 17.9 | 9.2 | 4.7 | 0 | $-9.6$ | | $\phi_{comp}$ | $-106^\circ$ | $-115^\circ$ | $-121^\circ$ | $-123^\circ$ | $-126.1^\circ$ | $-141^\circ$ | $$ PM=180^\circ-126.1^\circ=53.9^\circ\ (\ge50^\circ),\quad GM=\infty $$ ```text dB 26 |\ -20 | \__ 2 (corner) 0 |------\----x------- w | 6.17\ 9.37 (new wgc) | \__ -40 lead adds phase between 4.17 and 20.8 ``` **Answer: $G_c(s)=10\,\dfrac{1+0.24s}{1+0.048s}$ (α = 0.2): Kv = 20 s⁻¹, PM ≈ 54° at ω ≈ 9.4 rad/s.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Plant 2/[s(1+0.5s)] has Kv = 2. To get Kv = 20, what is done? A. Kc = 10 so that Kv = Kc x 2 = 20 B. Kc = 20 C. Kc = 0.1 D. alpha is chosen as 10 Answer: A. Kv = Kc x 2 = 20 gives Kc = 10 and G1 = 20/[s(1+0.5s)]. (Key point: Gain Kc) 2. For G1 = 20/[s(1+0.5s)], gain crossover is 6.17 rad/s. The PM is A. 72 degrees B. 50 degrees C. 90 degrees D. 18 degrees Answer: D. PM = 90 - tan^-1(0.5 x 6.17) = 90 - 72 = 18 degrees. (Key point: Gain-adjusted PM) 3. Why is epsilon raised from 5 to about 10 degrees? A. With 5 degrees the check gives PM 49.8 degrees, just short of 50 B. Because Kv would otherwise fail C. To lower alpha to zero D. Because the plant is unstable Answer: A. Higher crossover means more plant lag, so extra margin is needed; epsilon = 10 gives phi_m about 42 degrees and alpha = 0.2. (Key point: Margin epsilon) 4. With alpha = 0.2, lead gain at w_m is 6.99 dB. The next steps are A. w_m where |G1| = -6.99 dB (9.35 rad/s), T = 1/(w_m sqrt(alpha)) = 0.24 s B. w_m = 6.17 rad/s, T = 0.5 s C. w_m where |G1| = +6.99 dB, T = 0.048 s D. w_m = 20.8 rad/s, T = 0.24 s Answer: A. w_m is where plant gain is -6.99 dB, then T = 0.24 s, alpha T = 0.048 s, giving Gc = 10(1+0.24s)/(1+0.048s) and PM about 54 degrees. 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