--- title: "Bode plot: 40(s+1)/((2s²+10s)(s²+2s+10))" subject: "Electrical Circuit Theory" chapter: 5 key: "semester_3/electrical_circuit_theory:bode-40s1-2s2-10s-s2-2s" source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/bode-40s1-2s2-10s-s2-2s site: BCT Engineering --- # Bode plot: 40(s+1)/((2s²+10s)(s²+2s+10)) Electrical Circuit Theory · Chapter 5: Frequency Response of Network - Asked in: 2075 Asoj, 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-40s1-2s2-10s-s2-2s - Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-5 ## Question Sketch Bode Plot for the following transfer function. H(s) = 40(s + 1) / [(2s² + 10s)(s² + 2s + 10)] ## Short answer Factor as 0.4(1+s/1)/[ s (1+s/5)(1+0.2s+s²/10) ]; ωn=3.162 rad/s (ζ=0.316). Initial slope –20 dB/dec, asymptote at ω=1: –7.96 dB. Corner frequencies: 1 (zero, +20 dB/dec), 3.162 (quadratic pole, –40 dB/dec), 5 (pole, –20 dB/dec). Phase: –90° to –270°, with resonant peak near ωn. ## 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 H(s) = 40(s + 1) / [ (2s² + 10s)(s² + 2s + 10) ] = 20(s + 1) / [ s (s + 5)(s² + 2s + 10) ] since 2s² + 10s = 2s(s + 5) ``` ```text H(s) = 0.4 × (1 + s/1) / [ s (1 + s/5) (1 + 0.2 s + s²/10) ] ``` Constant term K₀ = 0.4, so 20 log K₀ = −7.96 dB. #### Step 2: Corner frequencies | Corner ω (rad/s) | Factor | Slope change | |---|---|---| | 1 | zero (1 + s/1) | +20 dB/dec | | 3.162 | quadratic pole, ωn = 3.162, ζ = 0.316 | −40 dB/dec | | 5 | pole 1/(1 + s/5) | −20 dB/dec | The quadratic s² + 2s + 10 has ωn = √10 = 3.162 rad/s and 2ζωn = 2, so ζ = 0.316 (underdamped). #### 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 −8 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 | 12 | −87 | | 1 | 0 | −8 | −69 | | 3.162 | −40 | −8 | −140 | | 5 | −60 | −15.9 | −203 | Asymptotic magnitude plot (each * is the asymptote; 1 row = 5 dB): ```text | 10 |*** | *** 0 | *** | *** −10 | ******** | * −20 | * | * −30 | * | * ++-----------+-----------+ 0.1 1 10 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) = −270°. Exact phase values at the corners are in the table above; join them with a smooth curve starting at −90° and ending at −270°. Accuracy check: at the first corner ω = 1 rad/s the exact gain is −4.4 dB, the asymptote gives −8 dB. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How does 2s^2 + 10s simplify, giving the standard form? A. 2s(s+5), so H = 20(s+1)/[s(s+5)(s^2+2s+10)] B. 2s(s+10), so H = 20(s+1)/[s(s+10)(s^2+2s+10)] C. s(s+5), so H = 40(s+1)/[s(s+5)(s^2+2s+10)] D. 2s^2(s+5) Answer: A. 2s^2+10s = 2s(s+5), which cancels the 2 against 40. (Key point: Simplified form) 2. What is K0? A. 20 (26 dB) B. 0.4 (-7.96 dB) C. 2 (6 dB) D. 4 (12 dB) Answer: B. K0 = 20 x 1/(5 x 10) = 0.4, which is -7.96 dB. (Key point: Constant term) 3. For s^2 + 2s + 10, what are omega_n and zeta? A. omega_n = 10, zeta = 0.1 B. omega_n = 3.162, zeta = 1 C. omega_n = 3.162, zeta = 0.316 D. omega_n = 2, zeta = 5 Answer: C. omega_n = sqrt(10) = 3.162 and 2 zeta omega_n = 2 give zeta = 0.316. (Key point: Quadratic factor) 4. About how far does the true curve rise above the asymptote at omega_n, and why? A. About 14 dB B. About 0 dB since zeta < 1 C. About 8 dB D. About 4 dB, from -20 log(2 zeta) Answer: D. -20 log(2 x 0.316) = about 4 dB for an underdamped quadratic pole. 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