--- title: "Bode plot: 200(S+1)/(S(S+5)(S²+2S+100))" subject: "Electrical Circuit Theory" chapter: 5 key: "semester_3/electrical_circuit_theory:bode-200s1-s-s5-s2-2s" source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/bode-200s1-s-s5-s2-2s site: BCT Engineering --- # Bode plot: 200(S+1)/(S(S+5)(S²+2S+100)) Electrical Circuit Theory · Chapter 5: Frequency Response of Network - Asked in: 2074 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-200s1-s-s5-s2-2s - 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 H(S) = 200(S + 1) / [S(S + 5)(S² + 2S + 100)] ## Short answer Factor as 0.4(1+s/1)/[ s (1+s/5)(1+0.02s+s²/100) ]; ωn=10 rad/s, ζ=0.1. Initial slope –20 dB/dec, asymptote at ω=1: –7.96 dB. Corner frequencies: 1 (zero, +20 dB/dec), 5 (pole, –20 dB/dec), 10 (quadratic pole, –40 dB/dec). Slopes become 0, –20, –60 dB/dec. Phase: –90° to –270°, sharp 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) = 200(s + 1) / [ s (s + 5)(s² + 2s + 100) ] ``` ```text H(s) = 0.4 × (1 + s/1) / [ s (1 + s/5) (1 + 0.02 s + s²/100) ] ``` 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 | | 5 | pole 1/(1 + s/5) | −20 dB/dec | | 10 | quadratic pole, ωn = 10, ζ = 0.1 | −40 dB/dec | The quadratic s² + 2s + 100 has ωn = 10 rad/s and 2ζωn = 2, so ζ = 0.1 (very lightly damped). #### 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 | −86 | | 1 | 0 | −8 | −57 | | 5 | −20 | −8 | −64 | | 10 | −60 | −14 | −159 | Asymptotic magnitude plot (each * is the asymptote; 1 row = 10 dB): ```text 20 | |***** 0 | ****** | ************** −20 | ** | ** −40 | ** | ** −60 | ** | ** −80 | ++-----------+-----------+-----------+ 0.1 1 10 100 ω (rad/s) ``` At ωn = 10 the quadratic pole has ζ = 0.1, so the true curve rises about 14 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 −5 dB, the asymptote gives −8 dB. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is K0 for H(s) = 200(s+1)/[s(s+5)(s^2+2s+100)]? A. 0.4 (-7.96 dB) B. 2 (6 dB) C. 40 (32 dB) D. 0.2 (-14 dB) Answer: A. K0 = 200/(5 x 100) = 0.4. (Key point: Constant term) 2. What is the slope between 1 rad/s and 5 rad/s? A. -20 dB/dec B. 0 dB/dec C. +20 dB/dec D. -40 dB/dec Answer: B. The start is -20; the zero at 1 adds +20, giving 0 until the pole at 5. (Key point: Slope between corners 1 and 5) 3. What is zeta for s^2 + 2s + 100? A. 1 B. 0.02 C. 0.1 (very lightly damped) D. 0.5 Answer: C. omega_n = 10 and 2 zeta (10) = 2, so zeta = 0.1. (Key point: Damping ratio) 4. About how far above the asymptote is the peak at omega_n = 10? A. About 4 dB B. About 6 dB C. About 20 dB D. About 14 dB Answer: D. -20 log(2 x 0.1) = 14 dB. 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