--- title: "Bode plot of 512(s+3)/[s(s²+16s+256)]" subject: "Control Systems" chapter: 6 key: "semester_3/control_systems:bode-512-s3-s2-16s-256" source: https://www.bctengineer.com/q/semester_3/control_systems/bode-512-s3-s2-16s-256 site: BCT Engineering --- # Bode plot of 512(s+3)/[s(s²+16s+256)] Control Systems · Chapter 6: Frequency Response Techniques - Asked in: 2081 Bhadra, 8 marks - Video: [Control System Lectures - Bode Plots, Introduction](https://www.youtube.com/watch?v=_eh1conN6YM) — Brian Douglas - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/control_systems/bode-512-s3-s2-16s-256 - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-6 ## Question Draw bode plot for open loop transfer function of closed loop system as G(s) = 512(s+3)/[s(s²+16s+256)]. Also comment on stability. ## Short answer Bode plot yields gain crossover at 24.4 rad/s with phase -137.9°, giving phase margin 42.1°. Phase never reaches -180°, so gain margin is infinite. Closed-loop system is stable. ## Answer #### Time-constant form $$ G(s) = \frac{512(s+3)}{s(s^2 + 16s + 256)} = \frac{6\left(1 + \dfrac{s}{3}\right)}{s\left(1 + \dfrac{s}{16} + \dfrac{s^2}{256}\right)} $$ Quadratic factor: $\omega_n = 16$ rad/s and $2\zeta\omega_n = 16$, so $\zeta = 0.5$. #### Factors and corner frequencies | Factor | Corner (rad/s) | Slope change | Phase | |---|---|---|---| | $K = 6$ | — | $20\log6 = 15.56$ dB | 0° | | $1/s$ | — | −20 dB/dec | −90° | | $(1 + s/3)$ | 3 | +20 dB/dec | $+\tan^{-1}(\omega/3)$ | | quadratic, ζ = 0.5 | 16 | −40 dB/dec | 0 to −180° | #### Magnitude asymptotes | Range | Slope | Value | |---|---|---| | ω < 3 | −20 dB/dec | 35.56 dB at 0.1; 15.56 dB at 1; 6.02 dB at 3 | | 3 < ω < 16 | 0 dB/dec | 6.02 dB | | ω > 16 | −40 dB/dec | 0 dB at $16 \times 10^{6.02/40} = 22.6$; −25.8 dB at 100 | With ζ = 0.5 the resonant correction at $\omega_n$ is $-20\log(2\zeta) = 0$ dB, so the asymptotes are close to the true curve. #### Phase $$ \phi(\omega) = -90^\circ + \tan^{-1}\frac{\omega}{3} - \tan^{-1}\frac{\omega/16}{1 - (\omega/16)^2} $$ | ω | 0.1 | 1 | 3 | 5 | 10 | 16 | 24.4 | 50 | 100 | ∞ | |---|---|---|---|---|---|---|---|---|---|---| | φ (deg) | −88.5 | −75.2 | −56.0 | −50.1 | −62.4 | −100.6 | −137.9 | −163.8 | −172.4 | −180 | #### Bode plot ```text 40 |* |******* | ****** 10 | ************** |-------------------------****----------------- | **** -20 | *** | *** | **** -50 | **** | **** | * -80 | +---------------------------------------------- 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); -180 line shown ``` #### Margins and stability - **Gain crossover:** the exact $|G| = 1$ at $\omega_{gc} = 24.4$ rad/s (asymptotic estimate 22.6 rad/s). There, φ = −137.9°, so $$ PM = 180^\circ - 137.9^\circ = 42.1^\circ $$ - **Phase crossover:** the phase tends to −180° only as ω → ∞, so it never actually crosses −180°. Hence $\omega_{pc} = \infty$ and **GM = ∞**. Both margins are positive, so the **closed-loop system is stable**, with PM ≈ 42° (reasonably damped). Check: $s^3 + 16s^2 + 768s + 1536$ gives $16 \times 768 > 1536$. ✓ **Answer:** GM = ∞, PM ≈ 42.1° (at 24.4 rad/s). The system is **stable**. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Converting G(s) = 512(s+3)/[s(s^2+16s+256)] to time-constant form, what is the constant K and its dB value? A. K = 512, 54.2 dB B. K = 6, 15.56 dB C. K = 3, 9.54 dB D. K = 2, 6.02 dB Answer: B. K = 512 x 3 / 256 = 6 once each factor is written as (1 + s/a); 20 log 6 = 15.56 dB. (Key point: Time-constant gain K) 2. For the quadratic s^2 + 16s + 256, what are wn and zeta, and what is the resonant peak correction at wn? A. wn = 16, zeta = 1, correction 6 dB B. wn = 256, zeta = 0.03, correction +30 dB C. wn = 16, zeta = 0.5, correction 0 dB D. wn = 16, zeta = 0.25, correction +6 dB Answer: C. wn^2 = 256 gives wn = 16 and 2*zeta*wn = 16 gives zeta = 0.5; -20 log(2 zeta) = 0 dB, so asymptotes are close to the true curve. (Key point: Damping ratio of quadratic) 3. Between 3 rad/s and 16 rad/s, what is the slope of the magnitude asymptote and why? A. -20 dB/dec, because only the integrator acts B. -40 dB/dec, because the quadratic acts C. 0 dB/dec, because the -20 of the integrator is cancelled by the +20 of the zero D. +20 dB/dec, because the zero acts alone Answer: C. The 1/s gives -20 dB/dec and the zero at 3 adds +20 dB/dec, so the slope is 0 until the quadratic at 16 adds -40. (Key point: Slope between corners 3 and 16) 4. Why is the gain margin infinite for this system, and what is the conclusion? A. The magnitude never reaches 0 dB, so the system is unstable B. The phase only approaches -180 deg as w tends to infinity, so there is no phase crossover; with PM about 42 deg the system is stable C. The phase crosses -180 deg at 24.4 rad/s, so GM = 0 dB D. PM is negative, so GM is infinite and the system is unstable Answer: B. Phase reaches -180 deg only at infinity, so w_pc is infinite and GM is infinite; PM = 180 - 137.9 = 42.1 deg > 0 means stable. (Key point: Margins and stability) ## Close variants set in other exams - [Bode plot of 60/[s(s+2)(s+6)] with gain margin](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Abode-60-s-s2-s6) — Control Systems, chapter 6, asked 4× - [Bode plot of 1000/[s(1+0.1s)(1+0.001s)]](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Abode-1000-s-01s-0001s) — Control Systems, chapter 6, asked 3× - [Bode plot of 50(s+10)/[(s+1)(s+100)]](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Abode-50-s10-s1-s100) — Control Systems, chapter 6, asked 3× - [Bode plot of 40(s+2)/[(2+s+25s²)(1+2s)s]](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Abode-40-s2-2-s-25s2) — Control Systems, chapter 6, asked 3× ## Related in Control Systems - [Transfer function from asymptotic Bode plot](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Atf-from-bode-asymptote) — Control Systems, chapter 6, asked 2× - [Determining transfer function from Bode plot](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Abode-to-tf) — Control Systems, chapter 6, asked 2× - [Transfer function from Bode plot with example](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Abode-to-tf-example) — Control Systems, chapter 6, asked 2× - [Transfer function from field asymptotic Bode plot](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Atf-from-bode-engineer) — Control Systems, chapter 6 - [Traffic light system of Kathmandu: open or closed loop](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Atraffic-light-justify) — Control Systems, chapter 1 - [Bode plot of 4(s+4)/[(s+2)(s²+2s+4)]](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Abode-4-s4-s2-s2-2s) — Control Systems, chapter 6 ## Related in other subjects - [DT transfer function; drawing a Bode plot](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Adt-transfer-function-bode) — Signals and Systems, chapter 6 - [Open loop versus closed loop control](https://www.bctengineer.com/question.md?key=semester_6%2Fembedded_system%3Aopen-vs-closed-loop) — Embedded System, chapter 7, asked 6× - [Open loop vs closed loop instrumentation](https://www.bctengineer.com/question.md?key=semester_5%2Finstrumentation_ii%3Aopen-closed-loop) — Instrumentation II, chapter 1, asked 2× - [MBI features; open vs closed loop compared](https://www.bctengineer.com/question.md?key=semester_5%2Finstrumentation_ii%3Ambi-features-open-closed) — Instrumentation II, chapter 1