--- title: "Bode plot: 20(S+1)/((S²+4S+2)(S²+5S))" subject: "Electrical Circuit Theory" chapter: 5 key: "semester_3/electrical_circuit_theory:bode-20s1-s2-4s-2-s2" source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/bode-20s1-s2-4s-2-s2 site: BCT Engineering --- # Bode plot: 20(S+1)/((S²+4S+2)(S²+5S)) Electrical Circuit Theory · Chapter 5: Frequency Response of Network - Asked in: 2072 Chaitra, 6 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-20s1-s2-4s-2-s2 - Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-5 ## Question Sketch the bode plot for the transfer function given by G(S) = 20(S + 1) / [(S² + 4S + 2)(S² + 5S)] ## Short answer Factor into simple poles: H(s)=2(1+s)/[s(1+s/0.586)(1+s/3.414)(1+s/5)]; K₀=2 gives 6.02 dB. Corner frequencies: 0.586, 1, 3.414, 5 rad/s. Initial −20 dB/dec passes 6 dB at ω=1, phase −90° to −270°. ## 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 + 1) / [ (s² + 4s + 2)(s² + 5s) ] = 20(s + 1) / [ s (s + 5)(s + 0.586)(s + 3.414) ] ``` ```text H(s) = 2 × (1 + s/1) / [ s (1 + s/5) (1 + s/0.586) (1 + s/3.414) ] ``` Constant term K₀ = 2, so 20 log K₀ = 6.02 dB. #### Step 2: Corner frequencies | Corner ω (rad/s) | Factor | Slope change | |---|---|---| | 0.586 | pole 1/(1 + s/0.586) | −20 dB/dec | | 1 | zero (1 + s/1) | +20 dB/dec | | 3.414 | pole 1/(1 + s/3.414) | −20 dB/dec | | 5 | pole 1/(1 + s/5) | −20 dB/dec | s² + 4s + 2 has real roots −2 ± √2 = −0.586 and −3.414 (discriminant 16 − 8 > 0), so it splits into two simple factors. Also s² + 5s = s(s + 5). #### 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 (°) | |---|---|---|---| | 0.059 | −20 | 30.7 | −94 | | 0.586 | −40 | 10.7 | −121 | | 1 | −20 | 1.4 | −132 | | 3.414 | −40 | −9.3 | −176 | | 5 | −60 | −15.9 | −195 | Asymptotic magnitude plot (each * is the asymptote; 1 row = 10 dB): ```text |* 40 | ****** | ****** 20 | ****** | **** 0 | ***** | ***** −20 | ** | ** −40 | ++-----------+-----------+-----------+ 0.01 0.1 1 10 ω (rad/s) ``` #### 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 = 0.586 rad/s the exact gain is 8.8 dB, the asymptote gives 10.7 dB. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why does s^2 + 4s + 2 give two simple factors instead of a quadratic pole? A. Its discriminant 16 - 8 > 0, giving real roots -0.586 and -3.414 B. It is underdamped with zeta < 1 C. Its roots are complex at -2 +/- j1.4 D. It has a double root at -2 Answer: A. Real roots mean zeta > 1, so it is split into two first-order factors. (Key point: Quadratic with real roots) 2. What is K0? A. 20 (26 dB) B. 2 (6.02 dB) C. 0.2 D. 10 Answer: B. The roots multiply to 2, so K0 = 20 x 1 /(0.586 x 3.414 x 5) = 2. (Key point: Constant term) 3. The slope is -40 after 0.586. What is it after the zero at 1? A. 0 dB/dec B. -60 dB/dec C. -20 dB/dec D. -40 dB/dec Answer: C. A zero adds +20, from -40 to -20. (Key point: Slope after zero at 1) 4. What is the final slope after 5 rad/s? A. -80 dB/dec B. -40 dB/dec C. -20 dB/dec D. -60 dB/dec Answer: D. Poles minus zeros = 4 - 1 = 3, so -60 dB/dec. 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