--- title: "Bode plot of (4s+40)/[(s²+4s+25)(s²+50s)]" subject: "Control Systems" chapter: 6 key: "semester_3/control_systems:bode-4s40-s2-4s-25-s2" source: https://www.bctengineer.com/q/semester_3/control_systems/bode-4s40-s2-4s-25-s2 site: BCT Engineering --- # Bode plot of (4s+40)/[(s²+4s+25)(s²+50s)] Control Systems · Chapter 6: Frequency Response Techniques - Asked in: 2072 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-4s40-s2-4s-25-s2 - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-6 ## Question Draw Bode plot for the system with transfer function G(s) = (4s + 40)/[(s² + 4s + 25)(s² + 50s)]. Determine gain margin, phase margin and comment on stability of the system according to your plot. ## Short answer Gain crossover at 0.032 rad/s yields PM = 89.9°. Phase crossover at 5.97 rad/s yields GM = 44.5 dB. Both margins are positive, so the closed-loop system is stable. ## Answer #### Time-constant form $$ G(s) = \frac{4(s+10)}{s(s+50)(s^2 + 4s + 25)} = \frac{0.032\left(1 + \dfrac{s}{10}\right)}{s\left(1 + \dfrac{s}{50}\right)\left(1 + 0.16s + 0.04s^2\right)} $$ - $K = \dfrac{4 \times 10}{50 \times 25} = 0.032$, so $20\log K = -29.9$ dB. Type 1. - Quadratic: $\omega_n = 5$ rad/s, $\zeta = 4/(2 \times 5) = 0.4$. Peak correction $-20\log(0.8) = +1.9$ dB. #### Factors | Factor | Corner (rad/s) | Slope change | Phase | |---|---|---|---| | $0.032/s$ | — | −20 dB/dec | −90° | | quadratic, ζ = 0.4 | 5 | −40 dB/dec | 0 to −180° | | $(1 + s/10)$ | 10 | +20 dB/dec | 0 to +90° | | $1/(1 + s/50)$ | 50 | −20 dB/dec | 0 to −90° | #### Magnitude asymptotes | Range | Slope | Values | |---|---|---| | ω < 5 | −20 dB/dec | 30.1 dB at 0.001; 10.1 dB at 0.01; 0 dB at 0.032; −29.9 dB at 1; −43.9 dB at 5 | | 5 – 10 | −60 dB/dec | −43.9 → −61.9 dB | | 10 – 50 | −40 dB/dec | −61.9 → −89.9 dB | | ω > 50 | −60 dB/dec | −108.0 dB at 100 | #### Phase $$ \phi = -90^\circ + \tan^{-1}\frac{\omega}{10} - \tan^{-1}\frac{\omega}{50} - \tan^{-1}\frac{0.16\omega}{1 - 0.04\omega^2} $$ | ω | 0.01 | 0.032 | 0.1 | 1 | 5 | 5.97 | 10 | 50 | 100 | |---|---|---|---|---|---|---|---|---|---| | φ (deg) | −90.0 | −90.1 | −90.5 | −94.9 | −159.2 | −180.0 | −208.2 | −231.7 | −246.9 | #### Bode plot ```text 45 | |**** | ******** 0 |----------********---------------------------- | ******** | ******* -45 | ****** | *** | ***** -90 | *** | ** | -135 | +---------------------------------------------- 0.001 0.01 0.1 1 10 100 Magnitude asymptotes (dB) vs w (rad/s) ``` ```text -90 |***************************** | **** | ** -135 | ** | * | ** -180 |----------------------------------*----------- | ** | ***** -225 | **** | *** | -270 | +---------------------------------------------- 0.001 0.01 0.1 1 10 100 Phase (deg); crosses -180 at 5.97 rad/s ``` #### Gain and phase margin - **Gain crossover:** on the −20 dB/dec segment, $|G| = 1$ at $\omega_{gc} = K = 0.032$ rad/s. There φ = −90.1°, so $$ PM = 180^\circ - 90.1^\circ = 89.9^\circ $$ - **Phase crossover:** φ = −180° at $\omega_{pc} = 5.97$ rad/s, where the magnitude is −44.5 dB (asymptote ≈ −48.5 dB, plus the resonance lift). So $$ GM = +44.5\ \text{dB} $$ #### Stability GM ≈ +44.5 dB and PM ≈ +89.9°, both positive, with $\omega_{gc} \ll \omega_{pc}$. The **closed-loop system is stable**. The very low gain crossover (0.032 rad/s) means the response is very slow; the gain could be raised by about 44 dB (×168) before instability. **Answer:** **GM ≈ 44.5 dB**, **PM ≈ 89.9°**. The system is **stable**. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the time-constant gain found for G(s) = 4(s+10)/[s(s+50)(s^2+4s+25)]? A. K = 4 B. K = 4 x 10/(50 x 25) = 0.032, i.e. -29.9 dB C. K = 40/50 = 0.8 D. K = 10/25 = 0.4 Answer: B. Dividing out the constants of each factor gives 40/1250 = 0.032, and 20 log 0.032 = -29.9 dB. (Key point: Computing K) 2. What is the asymptote slope between 5 and 10 rad/s? A. -40 dB/dec B. -20 dB/dec C. -80 dB/dec D. -60 dB/dec (integrator -20 plus quadratic -40) Answer: D. At 5 the quadratic adds -40 to the integrator's -20; the zero at 10 later adds +20. (Key point: Slope between 5 and 10) 3. Where is the gain crossover and what is the PM? A. w = 0.032 rad/s (equal to K), PM = 89.9 deg B. w = 5.97 rad/s, PM = 0 C. w = 1 rad/s, PM = 85 deg D. w = 50 rad/s, PM = -52 deg Answer: A. The -20 dB/dec line reaches 0 dB at w = K = 0.032, where phase is -90.1 deg, so PM = 89.9 deg. (Key point: Gain crossover and PM) 4. Phase crosses -180 deg at 5.97 rad/s with magnitude -44.5 dB. What follows? A. GM = -44.5 dB, unstable B. GM = +44.5 dB; the gain could rise about 168 times before instability, so it is stable C. GM = 1.9 dB D. GM = 89.9 dB Answer: B. GM = 44.5 dB (x168); with both margins positive the system is stable, though the low crossover makes it slow. 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