--- title: "Bode plot of 50(s+10)/[(s+1)(s+100)]" subject: "Control Systems" chapter: 6 key: "semester_3/control_systems:bode-50-s10-s1-s100" source: https://www.bctengineer.com/q/semester_3/control_systems/bode-50-s10-s1-s100 site: BCT Engineering --- # Bode plot of 50(s+10)/[(s+1)(s+100)] Control Systems · Chapter 6: Frequency Response Techniques - Asked in: 2080 Baisakh, 10 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-50-s10-s1-s100 - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-6 ## Question Draw Bode plot for given open loop transfer function G(s)H(s) = 50(s+10)/[(s+1)(s+100)]. Also comment on stability from the plot. ## Short answer Phase never reaches –180°, so GM = ∞. At gain crossover ω=5.64 rad/s phase is –53.7°, giving PM ≈ 126°. The closed-loop system is stable. ## Answer #### Time-constant form $$ G(s)H(s) = \frac{50(s+10)}{(s+1)(s+100)} = \frac{5\left(1 + \dfrac{s}{10}\right)}{(1+s)\left(1 + \dfrac{s}{100}\right)} $$ It is a type-0 system with $K = 5$, so $20\log5 = 13.98$ dB. #### Corner frequencies | Factor | Corner (rad/s) | Slope change | |---|---|---| | $1/(1+s)$ | 1 | −20 dB/dec | | $(1 + s/10)$ | 10 | +20 dB/dec | | $1/(1 + s/100)$ | 100 | −20 dB/dec | #### Magnitude asymptotes | Range | Slope | Value at ends | |---|---|---| | ω < 1 | 0 | 13.98 dB | | 1 – 10 | −20 dB/dec | 13.98 → −6.02 dB | | 10 – 100 | 0 | −6.02 dB | | ω > 100 | −20 dB/dec | −26.02 dB at 1000 | The asymptotic gain crossover is where $13.98 - 20\log\omega = 0$, i.e. ω = 5 rad/s. The exact value is 5.64 rad/s. #### Phase $$ \phi = \tan^{-1}\frac{\omega}{10} - \tan^{-1}\omega - \tan^{-1}\frac{\omega}{100} $$ | ω | 0.1 | 1 | 2 | 5 | 5.64 | 10 | 20 | 50 | 100 | 1000 | |---|---|---|---|---|---|---|---|---|---|---| | φ (deg) | −5.2 | −39.9 | −53.3 | −55.0 | −53.7 | −45.0 | −35.0 | −36.7 | −50.1 | −84.8 | The phase dips to a minimum of about −56.8° near ω ≈ 3.4 rad/s and tends to −90° at high frequency. It **never reaches −180°**. #### Bode plot ```text 15 |************* | **** | **** 0 |------------------***------------------------- | *************** | *** -15 | **** | **** | *** -30 | | | -45 | +---------------------------------------------- 0.1 1 10 100 1000 Magnitude asymptotes (dB) vs w (rad/s) ``` ```text 0 |*** | ****** | **** ******* -45 | **** ****** **** | ******* **** | ****** -90 | *** | | -135 | | | -180 |---------------------------------------------- +---------------------------------------------- 0.1 1 10 100 1000 Phase (deg); never reaches -180 ``` #### Stability from the plot - **Phase margin:** at $\omega_{gc} = 5.64$ rad/s, φ = −53.7°, so $PM = 180^\circ - 53.7^\circ = 126.3^\circ$. - **Gain margin:** the phase never crosses −180°, so $\omega_{pc} = \infty$ and **GM = ∞**. Both margins are positive and large, so the **closed-loop system is stable** (and stays stable for any increase in gain). Check: the closed loop $s^2 + 151s + 600 = 0$ has both roots in the LHP. ✓ **Answer:** PM ≈ **126°**, GM = **∞**. The system is **stable**. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For G(s)H(s) = 50(s+10)/[(s+1)(s+100)], what are the type and the low-frequency magnitude? A. Type 1, magnitude falls at -20 dB/dec B. Type 0, K = 5, so 13.98 dB flat at low frequency C. Type 0, K = 50, so 33.98 dB D. Type 2, K = 5 Answer: B. No pole at the origin makes it type 0; K = 50 x 10/(1 x 100) = 5, so 20 log 5 = 13.98 dB. (Key point: System type and DC gain) 2. Which sequence of asymptote slopes is correct as w increases past corners 1, 10 and 100? A. 0, -20, 0, -20 dB/dec B. 0, -20, -40, -20 dB/dec C. 0, +20, 0, -20 dB/dec D. -20, 0, -20, 0 dB/dec Answer: A. Pole at 1 gives -20, zero at 10 cancels it to 0, pole at 100 gives -20 again. (Key point: Slopes from corner frequencies) 3. What is the gain margin and why? A. Zero, because the phase reaches -180 deg at 5.64 rad/s B. About 126 dB, read at w = 100 C. Infinite, because phase dips to only about -55 deg and tends to -90 deg, never crossing -180 deg D. 13.98 dB, equal to the DC gain Answer: C. With no phase crossover, w_pc is infinite and GM is infinite. (Key point: Phase never reaches -180) 4. At the gain crossover 5.64 rad/s the phase is -53.7 deg. What is the PM and conclusion? A. PM = 53.7 deg, marginally stable B. PM = 126.3 deg, stable C. PM = -126.3 deg, unstable D. PM = 233.7 deg, unstable Answer: B. PM = 180 - 53.7 = 126.3 deg, positive and large, so the closed loop is stable (confirmed by s^2 + 151s + 600 with LHP roots). 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