--- title: "Bode plot of (20s+200)/[(s²+?s+25)(s²+40s)]" subject: "Control Systems" chapter: 6 key: "semester_3/control_systems:bode-20s200-s2-25-s2-40s" source: https://www.bctengineer.com/q/semester_3/control_systems/bode-20s200-s2-25-s2-40s site: BCT Engineering --- # Bode plot of (20s+200)/[(s²+?s+25)(s²+40s)] Control Systems · Chapter 6: Frequency Response Techniques - Asked in: 2074 Asoj, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/control_systems/bode-20s200-s2-25-s2-40s - 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) = (20s + 200)/[(s² + [?]s + 25)(s² + 40s)]. Determine gain margin, phase margin and comment on stability of the system according to your plot. ## Short answer Gain crossover at 0.2 rad/s yields PM ≈ 89°. Phase crosses -180° at 5.88 rad/s with magnitude -28.3 dB, giving GM ≈ 28.3 dB. Both margins positive, so closed-loop stable. ## Answer One coefficient in the quadratic is unreadable in the paper ("s² + [?]s + 25"). It is taken here as **4**, giving $s^2 + 4s + 25$ (as in the similar 2072 question). Then $\omega_n = 5$ rad/s and $\zeta = 4/(2 \times 5) = 0.4$. #### Time-constant form $$ G(s) = \frac{20(s+10)}{s(s+40)(s^2 + 4s + 25)} = \frac{0.2\left(1 + \dfrac{s}{10}\right)}{s\left(1 + \dfrac{s}{40}\right)\left(1 + 0.16s + 0.04s^2\right)} $$ $K = 0.2$ ($20\log0.2 = -13.98$ dB), type 1. #### Corner frequencies and slopes | Range | Factor added | Slope | |---|---|---| | ω < 5 | $0.2/s$ | −20 dB/dec | | 5 – 10 | quadratic (ω_n = 5) | −60 dB/dec | | 10 – 40 | zero at 10 | −40 dB/dec | | ω > 40 | pole at 40 | −60 dB/dec | #### Magnitude asymptotes | ω | 0.01 | 0.1 | 0.2 | 1 | 5 | 10 | 40 | 100 | |---|---|---|---|---|---|---|---|---| | dB | 26.0 | 6.0 | 0 | −14.0 | −28.0 | −46.0 | −70.1 | −94.0 | The quadratic peak correction at $\omega_n$ is $-20\log(2\zeta) = +1.9$ dB. #### Phase $$ \phi = -90^\circ + \tan^{-1}\frac{\omega}{10} - \tan^{-1}\frac{\omega}{40} - \tan^{-1}\frac{0.16\omega}{1 - 0.04\omega^2} $$ | ω | 0.1 | 0.2 | 1 | 2 | 5 | 5.88 | 10 | 20 | 40 | 100 | |---|---|---|---|---|---|---|---|---|---|---| | φ (deg) | −90.5 | −91.0 | −95.2 | −102.4 | −160.6 | −180.0 | −211.0 | −221.1 | −233.2 | −251.6 | #### Bode plot ```text 30 |* | ****** | ******* 0 |------------******---------------------------- | ****** | ******* -30 | **** | *** | *** -60 | **** | *** | ** -90 | *** +---------------------------------------------- 0.01 0.1 1 10 100 Magnitude asymptotes (dB) vs w (rad/s) ``` ```text -90 |************************* | **** | ** -135 | ** | * | ** -180 |-------------------------------*-------------- | ** | **** -225 | ***** | ***** | ** -270 | +---------------------------------------------- 0.01 0.1 1 10 100 Phase (deg); crosses -180 at 5.88 rad/s ``` #### Margins - **Gain crossover:** the −20 dB/dec line cuts 0 dB at ω = K = 0.2 rad/s (exact 0.200). There φ = −91.0°, so **PM = 89.0°**. - **Phase crossover:** φ = −180° at $\omega_{pc} = 5.88$ rad/s, where the exact magnitude is −28.3 dB. So **GM = +28.3 dB**. #### Stability GM > 0 and PM > 0, and $\omega_{gc}$ (0.2) < $\omega_{pc}$ (5.88). The **closed-loop system is stable** with large margins, though it will be slow because the bandwidth is small. **Answer (with $s^2 + 4s + 25$):** GM ≈ **28.3 dB**, PM ≈ **89°**. The system is **stable**. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Taking s^2 + 4s + 25, what are wn, zeta and the peak correction? A. wn = 25, zeta = 0.4, +1.9 dB B. wn = 5, zeta = 0.4, +1.9 dB C. wn = 5, zeta = 0.8, -1.9 dB D. wn = 5, zeta = 0.4, 0 dB Answer: B. wn = sqrt(25) = 5, zeta = 4/(2 x 5) = 0.4, and -20 log(0.8) = +1.9 dB. (Key point: Damping from quadratic) 2. What is K in time-constant form, and what is the type? A. K = 0.2 (-13.98 dB), type 1 B. K = 20, type 0 C. K = 0.8, type 1 D. K = 200, type 2 Answer: A. K = 20 x 10 / (40 x 25) = 0.2; the pole at the origin makes it type 1. (Key point: Gain in time-constant form) 3. Where is the gain crossover and what is the PM? A. At 5.88 rad/s with PM = 0 B. At w = K = 0.2 rad/s on the -20 dB/dec line, PM about 89 deg C. At 10 rad/s with PM = 31 deg D. At 40 rad/s with PM = 53 deg Answer: B. A -20 dB/dec line with K = 0.2 hits 0 dB at w = 0.2, where phase is about -91 deg, giving PM = 89 deg. (Key point: Phase margin) 4. Phase is -180 deg at 5.88 rad/s where |G| is -28.3 dB. What is GM and the conclusion? A. GM = -28.3 dB, unstable B. GM = +28.3 dB, stable since w_gc < w_pc C. GM = 1.9 dB, marginal D. GM = infinite, stable Answer: B. GM is the negative of the magnitude in dB at phase crossover: +28.3 dB, with PM also positive, so the system is stable. 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