--- title: "Bode plot: 2s²(S+5)/((S²+22S+40)(S+10))" subject: "Electrical Circuit Theory" chapter: 5 key: "semester_3/electrical_circuit_theory:bode-2s2-s5-s2-22s-40" source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/bode-2s2-s5-s2-22s-40 site: BCT Engineering --- # Bode plot: 2s²(S+5)/((S²+22S+40)(S+10)) Electrical Circuit Theory · Chapter 5: Frequency Response of Network - Asked in: 2068 Chaitra, 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-2s2-s5-s2-22s-40 - 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: N(S) = 2s²(S + 5) / [(S² + 22S + 40)(S + 10)]. ## Short answer Standard form: 0.025s²(1+s/5)/[(1+s/2)(1+s/10)(1+s/20)], 20log|0.025|=−32 dB. Double zero at origin → initial slope +40 dB/dec through −32 dB at ω=1. Corners: 2 (pole), 5 (zero), 10, 20 rad/s (poles). Phase 180° to 0°. ## 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 N(s) = 2s²(s + 5) / [ (s² + 22s + 40)(s + 10) ] = 2s²(s + 5) / [ (s + 2)(s + 20)(s + 10) ] since s² + 22s + 40 = (s + 2)(s + 20) ``` ```text H(s) = 0.025 × s² (1 + s/5) / [ (1 + s/2) (1 + s/20) (1 + s/10) ] ``` Constant term K₀ = 0.025, so 20 log K₀ = −32.04 dB. #### Step 2: Corner frequencies | Corner ω (rad/s) | Factor | Slope change | |---|---|---| | 2 | pole 1/(1 + s/2) | −20 dB/dec | | 5 | zero (1 + s/5) | +20 dB/dec | | 10 | pole 1/(1 + s/10) | −20 dB/dec | | 20 | pole 1/(1 + s/20) | −20 dB/dec | There is a double zero at the origin, giving +40 dB/decade at low frequency. #### Step 3: Magnitude plot The net number of zeros at the origin is 2, so the low-frequency asymptote has slope +40 dB/decade and passes through −32 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 (°) | |---|---|---|---| | 0.2 | +40 | −60 | 175 | | 2 | +20 | −20 | 140 | | 5 | +40 | −12 | 116 | | 10 | +20 | 0 | 93 | | 20 | 0 | 6 | 63 | Asymptotic magnitude plot (each * is the asymptote; 1 row = 10 dB): ```text | ********** 0 | **** | **** −20 | **** | *** −40 | *** | *** −60 | *** |*** −80 | ++-----------+-----------+-----------+ 0.1 1 10 100 ω (rad/s) ``` #### Step 4: Phase plot Low-frequency phase = 90° × (zeros at origin − poles at origin) = 180°. 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) = 0°. Exact phase values at the corners are in the table above; join them with a smooth curve starting at 180° and ending at 0°. Accuracy check: at the first corner ω = 2 rad/s the exact gain is −22.6 dB, the asymptote gives −20 dB. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How does s^2 + 22s + 40 factor? A. (s+2)(s+20) B. (s+4)(s+10) C. (s+1)(s+40) D. (s+5)(s+8) Answer: A. 2 x 20 = 40 and 2 + 20 = 22. (Key point: Factoring) 2. What is the starting slope and why? A. +20 dB/dec due to one zero B. +40 dB/dec due to a double zero at the origin C. -40 dB/dec due to poles D. 0 dB/dec Answer: B. s^2 in the numerator gives two zeros at the origin. (Key point: Low-frequency slope) 3. What is K0? A. 2 (6 dB) B. 0.25 C. 0.025 (-32 dB) D. 0.1 Answer: C. K0 = 2 x 5 /(2 x 20 x 10) = 0.025. (Key point: Constant term) 4. What are the low- and high-frequency phases? A. 90 deg and -90 deg B. 0 deg and 180 deg C. 180 deg and -180 deg D. 180 deg at low frequency, 0 deg at high Answer: D. Low = 90 x 2 = 180; high = 90 x (3 - 3) = 0, so the final slope is 0 (flat 6 dB). 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