--- title: "Bode plot: 20(S+5)/(S(S²+2S+10)(S²+21S+20))" subject: "Electrical Circuit Theory" chapter: 5 key: "semester_3/electrical_circuit_theory:bode-20s5-s-s2-2s-10" source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/bode-20s5-s-s2-2s-10 site: BCT Engineering --- # Bode plot: 20(S+5)/(S(S²+2S+10)(S²+21S+20)) Electrical Circuit Theory · Chapter 5: Frequency Response of Network - Asked in: 2068 Shrawan, 8 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/electrical_circuit_theory/bode-20s5-s-s2-2s-10 - Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-5 ## Question Sketch the asymptotic Bode plots for the transfer function given by G(S) = 20(S + 5) / [S(S² + 2S + 10)(S² + 21S + 20)]. ## Short answer Time-constant form: 0.5(1+s/5)/[s(1+s)(1+s/20)(1+0.2s+s²/10)], 20log0.5=−6 dB. Corners: 1 (pole, −20 dB/dec), 3.16 rad/s (quad pole, ζ=0.316, −40 dB/dec), 5 (zero, +20), 20 rad/s (pole). Initial −20 dB/dec through −6 dB at ω=1, phase −90° to −360°, ~4 dB resonant peak. ## Answer A Bode plot has two curves against log ω: the gain 20 log|H(jω)| in dB and the phase ∠H(jω). The asymptotic (straight-line) magnitude plot is drawn by writing H(s) in time-constant form, finding the corner frequencies and changing the slope by ±20 dB/decade for each simple factor (±40 for each quadratic factor). #### Step 1: Standard form ```text G(s) = 20(s + 5) / [ s (s² + 2s + 10)(s² + 21s + 20) ] = 20(s + 5) / [ s (s + 1)(s + 20)(s² + 2s + 10) ] since s² + 21s + 20 = (s + 1)(s + 20) ``` ```text H(s) = 0.5 × (1 + s/5) / [ s (1 + s/1) (1 + s/20) (1 + 0.2 s + s²/10) ] ``` Constant term K₀ = 0.5, so 20 log K₀ = −6.02 dB. #### Step 2: Corner frequencies | Corner ω (rad/s) | Factor | Slope change | |---|---|---| | 1 | pole 1/(1 + s/1) | −20 dB/dec | | 3.162 | quadratic pole, ωn = 3.162, ζ = 0.316 | −40 dB/dec | | 5 | zero (1 + s/5) | +20 dB/dec | | 20 | pole 1/(1 + s/20) | −20 dB/dec | The quadratic s² + 2s + 10 has ωn = √10 = 3.162 rad/s and ζ = 0.316. #### Step 3: Magnitude plot The net number of poles at the origin is 1, so the low-frequency asymptote has slope −20 dB/decade and passes through −6 dB at ω = 1 rad/s. The slope then changes at each corner as below. | ω (rad/s) | Slope after corner | Asymptotic gain (dB) | Phase (°) | |---|---|---|---| | 0.1 | −20 | 14 | −96 | | 1 | −40 | −6 | −139 | | 3.162 | −80 | −26 | −229 | | 5 | −60 | −41.9 | −284 | | 20 | −80 | −78.1 | −320 | Asymptotic magnitude plot (each * is the asymptote; 1 row = 10 dB): ```text 20 | |****** 0 | ****** | *** −20 | *** | ** −40 | ** | * −60 | ** | ** −80 | ** | ** −100 | * | ** −120 | * | ** −140 | ++-----------+-----------+-----------+ 0.1 1 10 100 ω (rad/s) ``` At ωn = 3.162 the quadratic pole has ζ = 0.316, so the true curve rises about 4 dB above the asymptote there (resonant peak, −20 log 2ζ); the actual peak is near ωn. #### Step 4: Phase plot Low-frequency phase = 90° × (zeros at origin − poles at origin) = −90°. Each simple factor changes the phase by 90° (zero: +, pole: −), spread over the range 0.1ωc to 10ωc with 45° at ωc; each quadratic factor changes it by 180°, with ±90° at ωn. High-frequency phase = 90° × (m − n) = −360°. Exact phase values at the corners are in the table above; join them with a smooth curve starting at −90° and ending at −360°. Accuracy check: at the first corner ω = 1 rad/s the exact gain is −8.2 dB, the asymptote gives −6 dB. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How does s^2 + 21s + 20 factor? A. (s+1)(s+20) B. (s+4)(s+5) C. (s+2)(s+10) D. (s+3)(s+7) Answer: A. 1 x 20 = 20 and 1 + 20 = 21. (Key point: Factoring) 2. What is K0? A. 2 B. 0.5 (-6.02 dB) C. 20 D. 0.05 Answer: B. K0 = 20 x 5 /(1 x 20 x 10) = 0.5. (Key point: Constant term) 3. What is the slope just after omega_n = 3.162 (before 5)? A. -60 dB/dec B. -40 dB/dec C. -80 dB/dec D. -100 dB/dec Answer: C. -20 at start, -40 after the pole at 1, and -80 after the quadratic pole. (Key point: Slope after omega_n) 4. What is the high-frequency phase? A. -270 deg B. -180 deg C. -450 deg D. -360 deg Answer: D. 90 x (1 - 5) = -360 deg. (Key point: High-frequency phase) ## Close variants set in other exams - [Bode plot: 2s²(S+5)/((S²+22S+40)(S+10))](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Abode-2s2-s5-s2-22s-40) — Electrical Circuit Theory, chapter 5, asked 2× ## Related in Electrical Circuit Theory - [Bode plot: 20(s+1)/(s(s+5)(s²+2s+10))](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Abode-20s1-s-s5-s2-2s) — Electrical Circuit Theory, chapter 5 - [Bode plot: 10 S(S+3)/((S+1)(S²+2S+16))](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Abode-10s-s3-s1-s2-2s) — Electrical Circuit Theory, chapter 5, asked 4× - [Bode plot: 40(s+1)/((2s²+10s)(s²+2s+10))](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Abode-40s1-2s2-10s-s2-2s) — Electrical Circuit Theory, chapter 5 - [Bode plot: 64(S+2)/(S(S²+0.5S)(S²+3.2S+64))](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Abode-64s2-s-s2-05s-s2) — Electrical Circuit Theory, chapter 5, asked 4× - [Bode plot: 200(S+1)/(S(S+5)(S²+2S+100))](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Abode-200s1-s-s5-s2-2s) — Electrical Circuit Theory, chapter 5, asked 4× - [Bode plot: 20(s+5)/(S(S+20)(s²+80s+200))](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Abode-20s5-s-s20-s2-80s) — Electrical Circuit Theory, chapter 5, asked 4× ## Related in other subjects - [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× - [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 - [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×