--- title: "GM and PM from Bode plot of 1/[s(0.1s+1)(0.2s+1)]" subject: "Control Systems" chapter: 6 key: "semester_3/control_systems:bode-gm-pm-1-s-01s" source: https://www.bctengineer.com/q/semester_3/control_systems/bode-gm-pm-1-s-01s site: BCT Engineering --- # GM and PM from Bode plot of 1/[s(0.1s+1)(0.2s+1)] Control Systems · Chapter 6: Frequency Response Techniques - Asked in: 2070 Chaitra (old course), 8 marks - Video: [Bode plot in control system](https://www.youtube.com/watch?v=qnsznGf70eU) — Smart Engineer - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/control_systems/bode-gm-pm-1-s-01s - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-6 ## Question Find the Gain Margin and Phase Margin using Bode plots for the following transfer function: G(s) = 1/[s(0.1s+1)(0.2s+1)]. ## Short answer Phase crossover at 7.07 rad/s gives GM = 23.5 dB. Gain crossover near 0.98 rad/s gives PM ≈ 73.4°. Both margins positive; closed-loop system is stable. ## Answer $$ G(s)=\frac{1}{s(1+0.1s)(1+0.2s)} $$ Type-1 system, $K=1$, corner frequencies $\omega_1 = 1/0.2 = 5$ rad/s and $\omega_2 = 1/0.1 = 10$ rad/s. #### Magnitude plot (asymptotic) | Range of $\omega$ | Slope | |---|---| | $\omega < 5$ | $-20$ dB/dec (pole at origin) | | $5 < \omega < 10$ | $-40$ dB/dec | | $\omega > 10$ | $-60$ dB/dec | - At $\omega=0.1$: $20\log(1/0.1) = 20$ dB; at $\omega=1$: 0 dB. - At $\omega=5$: $20\log(1/5) = -14$ dB. - At $\omega=10$: $-14 - 40\log 2 = -26$ dB. - At $\omega=100$: $-26 - 60 = -86$ dB. #### Phase plot $$ \phi(\omega) = -90^\circ - \tan^{-1}0.1\omega - \tan^{-1}0.2\omega $$ | $\omega$ (rad/s) | 0.1 | 1 | 2 | 5 | 7.07 | 10 | 20 | 100 | |---|---|---|---|---|---|---|---|---| | $\phi$ | $-91.7^\circ$ | $-107^\circ$ | $-123^\circ$ | $-161.6^\circ$ | $-180^\circ$ | $-198.4^\circ$ | $-229.4^\circ$ | $-261.4^\circ$ | ```text dB 20 |\ -20dB/dec 0 |-\---------------------- w | \ wgc~1 -14 | \___ w=5 -40 -26 | \__ w=10 | \ -60 deg -90 |--__ -180 |-------x---------- wpc=7.07 -270 | ----___ ``` #### Phase crossover frequency and gain margin At $\omega_{pc}$, $\phi=-180^\circ$: $\tan^{-1}0.1\omega+\tan^{-1}0.2\omega = 90^\circ \Rightarrow 0.02\omega^2 = 1$. $$ \begin{aligned} \omega_{pc} &= \sqrt{50} = 7.07\ \text{rad/s}\\ |G(j\omega_{pc})| &= \frac{1}{7.07\sqrt{1+0.5}\sqrt{1+2}} = \frac{1}{15} = 0.0667\\ GM &= 20\log\frac{1}{0.0667} = 20\log 15 = 23.52\ \text{dB} \end{aligned} $$ #### Gain crossover frequency and phase margin From the asymptotic plot, the magnitude is 0 dB at $\omega\approx1$ rad/s. Solving $|G(j\omega)|=1$ exactly gives $\omega_{gc}=0.977$ rad/s. $$ \begin{aligned} \phi(\omega_{gc}) &= -90^\circ - \tan^{-1}(0.0977) - \tan^{-1}(0.1954)\\ &= -90^\circ - 5.58^\circ - 11.05^\circ = -106.6^\circ\\ PM &= 180^\circ + \phi(\omega_{gc}) = 73.4^\circ \end{aligned} $$ (Using the asymptotic value $\omega_{gc}=1$: $PM = 180^\circ-107.0^\circ = 73^\circ$.) **Answer: GM = 23.5 dB at $\omega_{pc}=7.07$ rad/s; PM ≈ 73.4° at $\omega_{gc}\approx0.98$ rad/s. Both margins are positive, so the closed-loop system is stable.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For G(s) = 1/[s(0.1s+1)(0.2s+1)], what is the phase crossover frequency? A. 5 rad/s B. 10 rad/s C. 1 rad/s D. 7.07 rad/s Answer: D. Setting tan^-1 0.1w + tan^-1 0.2w = 90 degrees gives 0.02 w^2 = 1, so w = sqrt(50) = 7.07 rad/s. (Key point: Phase crossover frequency) 2. The magnitude at w = 7.07 rad/s is 1/15. What is the gain margin? A. 15 dB B. 23.5 dB C. -23.5 dB D. 3.9 dB Answer: B. GM = -20 log|G| = 20 log 15 = 23.5 dB. (Key point: Gain margin) 3. At the gain crossover frequency (about 0.98 rad/s) the phase is about -106.6 degrees. What is PM? A. 106.6 degrees B. 73.4 degrees C. -73.4 degrees D. 286.6 degrees Answer: B. PM = 180 degrees + phase = 180 - 106.6 = 73.4 degrees. (Key point: Phase margin) 4. What is the asymptotic slope between w = 5 and w = 10 rad/s? A. -20 dB/dec B. -40 dB/dec C. -60 dB/dec D. 0 dB/dec Answer: B. The pole at the origin and the pole at 5 rad/s together give -40 dB/dec until the pole at 10 rad/s. 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