--- title: "Lead compensator for k/[s(1+0.1s)(1+0.001s)], Kv = 1000" subject: "Control Systems" chapter: 7 key: "semester_3/control_systems:lead-k-s-01s-0001s-kv1000" source: https://www.bctengineer.com/q/semester_3/control_systems/lead-k-s-01s-0001s-kv1000 site: BCT Engineering --- # Lead compensator for k/[s(1+0.1s)(1+0.001s)], Kv = 1000 Control Systems · Chapter 7: Performance Specifications and Compensation Design - Asked in: 2074 Asoj, 12 marks; 2068 Chaitra, 16 marks; 2081 Baisakh, 12 marks (asked 3 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-01s-0001s-kv1000 - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-7 ## Question Design a suitable lead compensator for a system having open loop TF G(s) = K/[s(1+0.1s)(1+0.001s)] such that the compensated system should have phase margin of at least 45° and static velocity error constant of at least 1000. ## Short answer Set K = 1000. Using Bode design, the required phase lead is 55°, α = 0.1, and ω_m is placed at 176 rad/s where uncompensated gain is −10 dB. The lead compensator is (1+0.018s)/(1+0.0018s) with amplifier gain 10, yielding PM ≈ 48° and Kv = 1000 s⁻¹. ## Answer A lead compensator $G_c(s)=\dfrac{1+Ts}{1+\alpha Ts}$ ($\alpha<1$) is designed on the Bode plot: it supplies its maximum lead $\phi_m$ ($\sin\phi_m=\frac{1-\alpha}{1+\alpha}$) at the new gain crossover $\omega_m=\frac{1}{T\sqrt\alpha}$, where it adds $10\log(1/\alpha)$ dB of gain. #### Step 1: K from the velocity error constant $$ K_v=\lim_{s\to0}sG(s)=K=1000\ \text{s}^{-1} $$ $$ G(s)=\frac{1000}{s(1+0.1s)(1+0.001s)} $$ #### Step 2: Uncompensated Bode plot Corners at 10 and 1000 rad/s. Asymptotic magnitude: 60 dB at $\omega=1$, $-20$ dB/dec to 10 (40 dB), $-40$ dB/dec to 1000 ($-40$ dB), then $-60$ dB/dec. $$ \phi(\omega)=-90^\circ-\tan^{-1}0.1\omega-\tan^{-1}0.001\omega $$ | $\omega$ | 1 | 10 | 55.6 | 100 | 176 | 556 | 1000 | |---|---|---|---|---|---|---|---| | $|G|$ (dB) | 60.0 | 37.0 | 10.1 | $-0.1$ | $-10.0$ | $-31.0$ | $-43.0$ | | $\phi$ | $-95.8^\circ$ | $-135.6^\circ$ | $-173^\circ$ | $-180^\circ$ | $-186.7^\circ$ | $-208^\circ$ | $-224.4^\circ$ | From $|G|=1$: $\omega_{gc}=99.5$ rad/s, $\phi=-179.94^\circ$, so $$ PM_{unc}\approx0^\circ $$ (and $\omega_{pc}=100$ rad/s, GM ≈ 0 dB). The system is on the verge of instability. #### Step 3: Required phase lead $$ \phi_m=PM_d-PM_{unc}+\epsilon=45^\circ-0^\circ+10^\circ=55^\circ $$ A margin of $\epsilon=10^\circ$ is taken because the crossover will move to a higher frequency where the plant phase is lower (a 5° margin gives only 44° and fails the check). #### Step 4: α $$ \alpha=\frac{1-\sin55^\circ}{1+\sin55^\circ}=\frac{1-0.819}{1+0.819}=0.0994\approx0.1 $$ With $\alpha=0.1$: $\phi_m=\sin^{-1}(0.9/1.1)=54.9^\circ$. #### Step 5: New gain crossover frequency The lead adds $10\log(1/0.1)=10$ dB at $\omega_m$, so place $\omega_m$ where $|G(j\omega)|=-10$ dB: $$ \frac{1000}{\omega\sqrt{1+0.01\omega^2}\sqrt{1+10^{-6}\omega^2}}=0.316\Rightarrow\omega_m\approx176\ \text{rad/s} $$ #### Step 6: Time constants $$ \begin{aligned} T&=\frac{1}{\omega_m\sqrt\alpha}=\frac{1}{176\times0.316}=0.018\ \text{s}\\ \alpha T&=0.0018\ \text{s} \end{aligned} $$ Zero at $1/T=55.6$ rad/s, pole at $1/\alpha T=556$ rad/s. $$ G_c(s)=\frac{1+0.018s}{1+0.0018s} $$ An amplifier of gain $1/\alpha=10$ makes up the lead network's attenuation, so $K_v$ stays 1000. #### Step 7: Compensated system and check $$ G_c(s)G(s)=\frac{1000(1+0.018s)}{s(1+0.1s)(1+0.001s)(1+0.0018s)} $$ | $\omega$ | 10 | 55.6 | 100 | 176.8 | 556 | 696 | |---|---|---|---|---|---|---| | $|G_cG|$ (dB) | 37.1 | 13.0 | 6.1 | 0 | $-13.9$ | $-17.5$ | | $\phi$ | $-126^\circ$ | $-134^\circ$ | $-129^\circ$ | $-131.9^\circ$ | $-169^\circ$ | $-180^\circ$ | At $\omega_{gc}'=176.8$ rad/s: $$ \begin{aligned} \phi&=-90^\circ-\tan^{-1}17.68-\tan^{-1}0.177+\tan^{-1}3.18-\tan^{-1}0.318\\ &=-90^\circ-86.8^\circ-10.0^\circ+72.5^\circ-17.6^\circ=-131.9^\circ\\ PM&=180^\circ-131.9^\circ=48.1^\circ\ (\ge45^\circ) \end{aligned} $$ GM ≈ 17.5 dB at $\omega_{pc}\approx696$ rad/s. ```text dB 60 |\ uncompensated -20 37 | \__ 10 compensated 0 |------\----x----------- w | 100 \ 176.8 | \__ -40/-60 lead adds phase between 55.6 and 556 rad/s ``` | Item | Before | After | |---|---|---| | $K_v$ | 1000 | 1000 | | $\omega_{gc}$ | 99.5 rad/s | 176.8 rad/s | | PM | ≈ 0° | 48.1° | | GM | ≈ 0 dB | 17.5 dB | **Answer: $G_c(s)=\dfrac{1+0.018s}{1+0.0018s}$ (α = 0.1, T = 0.018 s) with K = 1000 gives Kv = 1000 s⁻¹ and PM ≈ 48° (≥ 45°).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For G(s) = K/[s(1+0.1s)(1+0.001s)], Kv = 1000 requires what K? A. K = 100 B. K = 1000 C. K = 10 D. K = 1 Answer: B. Kv = lim s G(s) = K, so K = 1000. (Key point: Gain from Kv) 2. What is the phase margin of the uncompensated system (gain crossover about 99.5 rad/s)? A. About 45 degrees B. About 90 degrees C. About 0 degrees (verge of instability) D. About -45 degrees Answer: C. At w_gc = 99.5 rad/s the phase is about -179.94 degrees, so PM is about 0. (Key point: Uncompensated margin) 3. Why is 10 degrees extra added to the required lead, giving phi_m = 55 degrees and alpha about 0.1? A. To compensate for the DC gain loss B. Because the crossover moves higher where the plant phase is lower, so a small margin is needed C. Because lead networks have a 10 degree lag D. To make Kv larger Answer: B. The new crossover is at a higher frequency where the plant has more lag; a 5 degree allowance gives only 44 degrees. (Key point: Phase lead and alpha) 4. The lead adds 10 dB at w_m (alpha = 0.1). How are w_m and T obtained? A. w_m where plant gain is -10 dB (176 rad/s); T = 1/(w_m sqrt(alpha)) = 0.018 s B. w_m where plant gain is +10 dB; T = 0.1 s C. w_m at 100 rad/s; T = 0.01 s D. w_m where phase is -180 degrees; T = 0.0018 s Answer: A. w_m is at -10 dB, about 176 rad/s, and T = 1/(176 x 0.316) = 0.018 s, giving Gc = (1+0.018s)/(1+0.0018s). 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