--- title: "Bode plot of 60/[s(s+2)(s+6)] with gain margin" subject: "Control Systems" chapter: 6 key: "semester_3/control_systems:bode-60-s-s2-s6" source: https://www.bctengineer.com/q/semester_3/control_systems/bode-60-s-s2-s6 site: BCT Engineering --- # Bode plot of 60/[s(s+2)(s+6)] with gain margin Control Systems · Chapter 6: Frequency Response Techniques - Asked in: 2079 Bhadra, 8 marks - Video: [1. Bode Plot Solved Example: Steps, Procedure, Parameters, and Stability](https://www.youtube.com/watch?v=lTnqZVNReF8) — Engineering Funda - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/control_systems/bode-60-s-s2-s6 - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-6 ## Question Draw Bode plot for system with open loop transfer function G(s)H(s) = 60/[s(s+2)(s+6)]. Also find GM and comment on stability. ## Short answer Phase crossover occurs at ω=3.46 rad/s with gain margin 4.08 dB. Gain crossover at 2.71 rad/s gives PM ≈ 12.2°. Both margins are positive, so the closed-loop system is stable. ## Answer #### Time-constant form $$ G(s)H(s) = \frac{60}{s(s+2)(s+6)} = \frac{5}{s\left(1 + \dfrac{s}{2}\right)\left(1 + \dfrac{s}{6}\right)} $$ $K = 5$ ($20\log5 = 13.98$ dB), type 1. Corner frequencies: 2 and 6 rad/s. #### Magnitude asymptotes | Range | Slope | Values | |---|---|---| | ω < 2 | −20 dB/dec | 33.98 dB at 0.1; 13.98 dB at 1; 7.96 dB at 2 | | 2 – 6 | −40 dB/dec | 7.96 → $7.96 - 40\log3 = -11.12$ dB | | ω > 6 | −60 dB/dec | −24.4 dB at 10; −42.5 dB at 20 | #### Phase $$ \phi = -90^\circ - \tan^{-1}\frac{\omega}{2} - \tan^{-1}\frac{\omega}{6} $$ | ω | 0.1 | 0.5 | 1 | 2 | 2.71 | 3.46 | 4 | 6 | 10 | 20 | |---|---|---|---|---|---|---|---|---|---|---| | φ (deg) | −93.8 | −108.8 | −126.0 | −153.4 | −167.8 | −180.0 | −187.1 | −206.6 | −227.7 | −247.6 | #### Bode plot ```text 40 | |******** | ******** 10 | ******** |---------------------****--------------------- | **** -20 | *** | **** | *** -50 | **** | *** | **** -80 | *** +---------------------------------------------- 0.1 1 10 100 Magnitude asymptotes (dB) vs w (rad/s) ``` ```text -90 |***** | ********* | **** -135 | **** | *** | *** -180 |----------------------***--------------------- | *** | *** -225 | **** | **** | ******** -270 | **** +---------------------------------------------- 0.1 1 10 100 Phase (deg); crosses -180 at 3.46 rad/s ``` #### Gain margin Phase crossover: $$ \tan^{-1}\frac{\omega}{2} + \tan^{-1}\frac{\omega}{6} = 90^\circ \;\Rightarrow\; \frac{\omega^2}{12} = 1 \;\Rightarrow\; \omega_{pc} = \sqrt{12} = 3.464\ \text{rad/s} $$ $$ |G(j3.464)| = \frac{60}{3.464 \times \sqrt{12 + 4} \times \sqrt{12 + 36}} = \frac{60}{3.464 \times 4 \times 6.928} = 0.625 $$ $$ GM = \frac{1}{0.625} = 1.6 = 20\log1.6 = 4.08\ \text{dB} $$ (The asymptotic plot gives about −1.6 dB at 3.46 rad/s, i.e. GM ≈ 1.6 dB; the exact value is 4.08 dB.) #### Phase margin Exact gain crossover: $\omega_{gc} = 2.71$ rad/s, where φ = −167.8°, so $PM = 12.2^\circ$. #### Stability GM = +4.08 dB > 0 and PM = +12.2° > 0, and $\omega_{gc} < \omega_{pc}$. The **closed-loop system is stable**, but with small margins, so the response will be quite oscillatory. Check: $s^3 + 8s^2 + 12s + 60$ gives $8 \times 12 = 96 > 60$. ✓ The gain could rise by a factor of 1.6 (to 96) before instability. **Answer:** **GM = 4.08 dB** at $\omega_{pc}$ = 3.46 rad/s. PM ≈ 12.2°. The system is **stable**. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. G(s)H(s) = 60/[s(s+2)(s+6)] in time-constant form has which gain K? A. 60 B. 10 C. 5 D. 12 Answer: C. K = 60/(2 x 6) = 5, i.e. 13.98 dB. (Key point: Time-constant gain) 2. What are the asymptote slopes below 2, between 2 and 6, and above 6? A. -20, -40, -60 dB/dec B. -20, -20, -20 dB/dec C. 0, -20, -40 dB/dec D. -40, -60, -80 dB/dec Answer: A. The integrator gives -20 and each simple pole at 2 and 6 adds a further -20 dB/dec. (Key point: Slope after each corner) 3. What is w_pc and the exact gain margin? A. w_pc = 2 rad/s, GM = 8 dB B. w_pc = 3.46 rad/s (sqrt 12), |G| = 0.625, GM = 4.08 dB C. w_pc = 2.71 rad/s, GM = 12.2 dB D. w_pc = 6 rad/s, GM = 1.6 dB Answer: B. tan^-1(w/2) + tan^-1(w/6) = 90 deg gives w^2 = 12; |G| = 0.625 so GM = 1/0.625 = 1.6 = 4.08 dB. (Key point: Phase crossover and GM) 4. Why is this system stable? A. GM and PM are both positive (w_gc = 2.71 < w_pc = 3.46) B. Because it is type 1 C. Because the phase is always above -180 deg D. Because GM is negative in dB Answer: A. Gain crosses 0 dB before the phase reaches -180 deg, giving PM = 12.2 deg and GM = 4.08 dB, though margins are small; Routh (8 x 12 > 60) agrees. 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