--- title: "Lead compensator for K/[s(1+0.2s)], Kv=10, PM=50°" subject: "Control Systems" chapter: 7 key: "semester_3/control_systems:lead-k-s-1-02s-kv10" source: https://www.bctengineer.com/q/semester_3/control_systems/lead-k-s-1-02s-kv10 site: BCT Engineering --- # Lead compensator for K/[s(1+0.2s)], Kv=10, PM=50° Control Systems · Chapter 7: Performance Specifications and Compensation Design - Asked in: 2078 Kartik, 12 marks; 2066 Bhadra (old course), 16 marks (asked 2 times) - 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-s-1-02s-kv10 - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-7 ## Question The open loop transfer function of a unity feedback control system is given by G(s) = K/[s(1+0.2s)]. Design a lead compensator such that the velocity error constant Kv = 10 sec⁻¹ and phase margin = 50°. Also draw the bode diagram for compensated system. ## Short answer Lead compensator: Gc(s) = (1+0.18s)/(1+0.1s) with α ≈ 0.56, K=10. This achieves velocity error constant Kv = 10 s⁻¹ and phase margin ≈ 50° at ω ≈ 7.5 rad/s. ## Answer Lead compensator: $G_c(s)=\dfrac{1+Ts}{1+\alpha Ts}$, $\alpha<1$, maximum phase lead $\phi_m$ with $\sin\phi_m=\frac{1-\alpha}{1+\alpha}$ at $\omega_m=\frac{1}{T\sqrt\alpha}$, where it adds $10\log(1/\alpha)$ dB. #### Step 1: K from $K_v$ $$ K_v=\lim_{s\to0}sG(s)=K=10\ \text{s}^{-1}\Rightarrow G(s)=\frac{10}{s(1+0.2s)} $$ #### Step 2: Uncompensated system Asymptotic magnitude: 20 dB at $\omega=1$ with $-20$ dB/dec, corner at $1/0.2=5$ rad/s, then $-40$ dB/dec. Phase $\phi=-90^\circ-\tan^{-1}0.2\omega$. Gain crossover: $\dfrac{10}{\omega\sqrt{1+0.04\omega^2}}=1\Rightarrow0.04\omega^4+\omega^2-100=0$ $$ \begin{aligned} \omega^2&=\frac{-1+\sqrt{1+16}}{0.08}=39.04\Rightarrow\omega_{gc}=6.25\ \text{rad/s}\\ PM&=180^\circ-90^\circ-\tan^{-1}(1.25)=90^\circ-51.3^\circ=38.7^\circ \end{aligned} $$ #### Step 3: Required phase lead $$ \phi_m=50^\circ-38.7^\circ+5^\circ=16.3^\circ $$ (5° added because the crossover moves right, where the plant has more lag.) #### Step 4: α $$ \alpha=\frac{1-\sin16.3^\circ}{1+\sin16.3^\circ}=\frac{1-0.281}{1+0.281}=0.561 $$ #### Step 5: New gain crossover Lead gain at $\omega_m$: $10\log(1/0.561)=2.51$ dB. Find where $|G|=-2.51$ dB: $$ \frac{10}{\omega\sqrt{1+0.04\omega^2}}=\sqrt{0.561}=0.749\Rightarrow\omega_m=7.44\ \text{rad/s} $$ #### Step 6: T $$ \begin{aligned} T&=\frac{1}{\omega_m\sqrt\alpha}=\frac{1}{7.44\times0.749}=0.179\approx0.18\ \text{s}\\ \alpha T&=0.561\times0.179=0.10\ \text{s} \end{aligned} $$ $$ G_c(s)=\frac{1+0.18s}{1+0.1s} $$ Zero at 5.56 rad/s, pole at 10 rad/s. An amplifier of gain $1/\alpha\approx1.8$ makes up the network's attenuation, keeping $K_v=10$. #### Step 7: Compensated system $$ G_c(s)G(s)=\frac{10(1+0.18s)}{s(1+0.2s)(1+0.1s)} $$ Check at the new crossover $\omega_{gc}'=7.47$ rad/s: $$ \begin{aligned} \phi&=-90^\circ+\tan^{-1}1.345-\tan^{-1}1.494-\tan^{-1}0.747\\ &=-90^\circ+53.4^\circ-56.2^\circ-36.8^\circ=-129.6^\circ\\ PM&=180^\circ-129.6^\circ=50.4^\circ \end{aligned} $$ GM = ∞ (phase never reaches −180°). #### Bode diagram of the compensated system Asymptotes: $-20$ dB/dec from 20 dB at $\omega=1$; corner 5 (pole) → $-40$; corner 5.56 (zero) → $-20$; corner 10 (pole) → $-40$ dB/dec. | $\omega$ | 0.1 | 1 | 2 | 5 | 7.47 | 10 | 20 | 100 | |---|---|---|---|---|---|---|---|---| | $|G|$ uncomp. (dB) | 40.0 | 19.8 | 13.3 | 3.0 | $-2.6$ | $-7.0$ | $-18.3$ | $-46.0$ | | $|G_cG|$ (dB) | 40.0 | 19.9 | 13.7 | 4.6 | 0 | $-3.7$ | $-13.9$ | $-41.0$ | | $\phi$ uncomp. | $-91^\circ$ | $-101^\circ$ | $-112^\circ$ | $-135^\circ$ | $-146^\circ$ | $-153^\circ$ | $-166^\circ$ | $-177^\circ$ | | $\phi$ comp. | $-91^\circ$ | $-97^\circ$ | $-103^\circ$ | $-120^\circ$ | $-130^\circ$ | $-138^\circ$ | $-155^\circ$ | $-175^\circ$ | ```text dB 40 |\ -20 dB/dec 20 | \ (w=1) | \ compensated 0 |------\---x------------- w | 5 \ 7.47 -20 between | 5.56 \__ 5.56 and 10 | \__ -40 after 10 deg -90|__ -130| \___o PM = 50.4 deg -180|-------------\_______ ``` **Answer: $G_c(s)=\dfrac{1+0.18s}{1+0.1s}$ (α ≈ 0.56) with K = 10 gives Kv = 10 s⁻¹ and PM ≈ 50° at ω ≈ 7.5 rad/s.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For G(s) = K/[s(1+0.2s)] with Kv = 10, what is K? A. K = 2 B. K = 10 C. K = 50 D. K = 0.2 Answer: B. Kv = lim s G(s) = K = 10. (Key point: K from Kv) 2. The gain crossover of 10/[s(1+0.2s)] is 6.25 rad/s. What is the PM? A. About 51.3 degrees B. About 90 degrees C. About 38.7 degrees D. About 50 degrees Answer: C. PM = 180 - 90 - tan^-1(0.2 x 6.25) = 90 - 51.3 = 38.7 degrees. (Key point: Uncompensated PM) 3. The needed lead is phi_m = 50 - 38.7 + 5 = 16.3 degrees. Why add 5 degrees, and what alpha results? A. To offset the lead's DC attenuation; alpha = 0.56 B. Because the crossover moves right where the plant has more lag; alpha = 0.561 C. To raise Kv; alpha = 0.18 D. To reduce overshoot; alpha = 1.78 Answer: B. alpha = (1 - sin16.3)/(1 + sin16.3) = 0.561, with the 5 degrees covering the extra plant lag at the higher crossover. (Key point: Lead required and alpha) 4. Which compensator and result does the design give? A. Gc = (1+0.1s)/(1+0.18s), PM = 50 degrees B. Gc = (1+0.18s)/(1+0.1s) with gain K=10; Kv = 10 and PM about 50 degrees C. Gc = (1+0.18s)/(1+0.1s); PM = 38.7 degrees D. Gc = (1+0.2s)/(1+0.56s); Kv = 5 Answer: B. T = 0.18 s and alpha T = 0.10 s; the compensated system has PM about 50.4 degrees and Kv = 10. 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