--- title: "Bode plot of 40(s+2)/[(2+s+25s²)(1+2s)s]" subject: "Control Systems" chapter: 6 key: "semester_3/control_systems:bode-40-s2-2-s-25s2" source: https://www.bctengineer.com/q/semester_3/control_systems/bode-40-s2-2-s-25s2 site: BCT Engineering --- # Bode plot of 40(s+2)/[(2+s+25s²)(1+2s)s] Control Systems · Chapter 6: Frequency Response Techniques - Asked in: 2076 Chaitra, 10 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-40-s2-2-s-25s2 - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-6 ## Question Draw Bode plot for unity feedback system with open loop transfer function G(s) = 40(s+2)/[(2 + s + 25s²)(1+2s)s]. Find GM and comment on stability. ## Short answer Phase crosses -180° at 0.275 rad/s with magnitude 58.8 dB, giving GM = -58.8 dB. Phase margin is -124.5°. Both margins are negative, so the closed-loop system is unstable. ## Answer #### Time-constant form $$ G(s) = \frac{40(s+2)}{s(1+2s)(25s^2 + s + 2)} = \frac{40\left(1 + \dfrac{s}{2}\right)}{s(1+2s)\left(1 + 0.5s + 12.5s^2\right)} $$ Quadratic factor: $\omega_n^2 = 2/25$, so $\omega_n = 0.283$ rad/s; $2\zeta\omega_n = 1/25$, so $\zeta = 0.0707$ (very lightly damped). #### Factors | Factor | Corner (rad/s) | Slope change | Phase | |---|---|---|---| | $K = 40$ | — | 32.04 dB | 0° | | $1/s$ | — | −20 dB/dec | −90° | | quadratic, ζ = 0.071 | 0.283 | −40 dB/dec | 0 to −180° (sharp) | | $1/(1+2s)$ | 0.5 | −20 dB/dec | 0 to −90° | | $(1 + s/2)$ | 2 | +20 dB/dec | 0 to +90° | #### Magnitude asymptotes | Range | Slope | Values | |---|---|---| | ω < 0.283 | −20 dB/dec | 72.0 dB at 0.01; 52.0 dB at 0.1; 43.0 dB at 0.283 | | 0.283 – 0.5 | −60 dB/dec | 43.0 → 28.2 dB | | 0.5 – 2 | −80 dB/dec | 28.2 → −20.0 dB | | ω > 2 | −60 dB/dec | −61.9 dB at 10 | Resonant correction at $\omega_n$: $-20\log(2\zeta) = +17$ dB, so the true peak is about 59 dB near 0.28 rad/s. #### Phase $$ \phi = -90^\circ - \tan^{-1}2\omega + \tan^{-1}\frac{\omega}{2} - \tan^{-1}\frac{0.5\omega}{1 - 12.5\omega^2} $$ | ω | 0.01 | 0.1 | 0.2 | 0.25 | 0.275 | 0.283 | 0.3 | 0.5 | 1 | 2 | 10 | |---|---|---|---|---|---|---|---|---|---|---|---| | φ (deg) | −91 | −102 | −117 | −139 | −180 | −201 | −242 | −294 | −304 | −300 | −278 | #### Bode plot ```text 90 | |**** | ************* 45 | ********* | ***** | *** 0 |-----------------------------****------------- | **** | ***** -45 | ***** | *** | -90 | +---------------------------------------------- 0.01 0.1 1 10 Magnitude asymptotes (dB); true peak ~59 dB at 0.28 ``` ```text -90 |*************** | ******* | ** -150 | * |---------------------*------------------------ | * -210 | * | * | * -270 | * ** | ***** ********** | ******** -330 | +---------------------------------------------- 0.01 0.1 1 10 Phase (deg); crosses -180 at 0.275 rad/s ``` #### Gain margin The phase crosses −180° at $\omega_{pc} = 0.275$ rad/s, just below $\omega_n$. Because of the resonance, the magnitude there is very large: $|G| = 871.6$ (58.8 dB). $$ GM = -58.8\ \text{dB} \quad (\text{i.e. } 1/871.6) $$ #### Phase margin Gain crossover is at $\omega_{gc} = 1.16$ rad/s (asymptotic estimate 1.12), where φ = −304.5°. $$ PM = 180^\circ - 304.5^\circ = -124.5^\circ $$ #### Stability GM is negative (in dB), PM is negative, and $\omega_{pc} < \omega_{gc}$. The **closed-loop system is unstable**. Check: $50s^4 + 27s^3 + 5s^2 + 42s + 80 = 0$ gives Routh first column $50, 27, -72.8, \dots$. There are two sign changes, so there are 2 RHP poles. ✓ **Answer:** GM ≈ **−58.8 dB**, PM ≈ −124.5°. The system is **unstable**. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For the factor 25s^2 + s + 2, what are wn and zeta, and what does this imply? A. wn = 0.283, zeta = 0.0707: very lightly damped with a large resonant peak B. wn = 2, zeta = 0.5: moderate peak C. wn = 0.283, zeta = 0.707: no peak D. wn = 0.08, zeta = 1: critically damped Answer: A. wn^2 = 2/25 gives wn = 0.283; 2 zeta wn = 1/25 gives zeta = 0.0707, so the resonant correction is large. (Key point: Lightly damped quadratic) 2. What is the resonant correction at wn and the true peak? A. 0 dB; peak 43 dB B. -17 dB; peak 26 dB C. +17 dB (= -20 log 2 zeta); true peak about 59 dB near 0.28 rad/s D. +3 dB; peak 46 dB Answer: C. -20 log(2 x 0.0707) = +17 dB added to the 43 dB asymptote value gives about 59 dB. (Key point: Resonant peak size) 3. The phase hits -180 deg at 0.275 rad/s where |G| = 871.6. What is the GM? A. +58.8 dB B. -58.8 dB C. 0 dB D. Infinite Answer: B. |G| is much greater than 1 at the phase crossover, so GM = 1/871.6, i.e. -58.8 dB. (Key point: Gain margin sign) 4. Why is the closed loop unstable? A. Because K is large B. Because the phase never crosses -180 deg C. Because it is type 1 D. Because GM and PM are negative; w_pc (0.275) is less than w_gc (1.16); Routh shows 2 RHP poles Answer: D. The magnitude is still above 0 dB when the phase reaches -180 deg, so both margins are negative (PM = -124.5 deg). 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