--- title: "Bode plot: 20(s+5)/(S(S+20)(s²+80s+200))" subject: "Electrical Circuit Theory" chapter: 5 key: "semester_3/electrical_circuit_theory:bode-20s5-s-s20-s2-80s" source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/bode-20s5-s-s20-s2-80s site: BCT Engineering --- # Bode plot: 20(s+5)/(S(S+20)(s²+80s+200)) Electrical Circuit Theory · Chapter 5: Frequency Response of Network - Asked in: 2071 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-20s5-s-s20-s2-80s - Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-5 ## Question For the transfer function below, draw the asymptotic Bode plot. G(s) = 20(s + 5) / [S(S + 20)(s² + 80s + 200)] ## Short answer Time-constant form: H(s)=0.025(1+s/5)/[s(1+s/2.583)(1+s/20)(1+s/77.4)]; 20log|0.025|=−32 dB. Corners at 2.58, 5, 20, 77.4 rad/s. Asymptote starts −20 dB/dec through −32 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 + 5) / [ s (s + 20)(s² + 80s + 200) ] = 20(s + 5) / [ s (s + 20)(s + 2.583)(s + 77.42) ] ``` ```text H(s) = 0.025 × (1 + s/5) / [ s (1 + s/20) (1 + s/2.583) (1 + s/77.417) ] ``` Constant term K₀ = 0.025, so 20 log K₀ = −32.04 dB. #### Step 2: Corner frequencies | Corner ω (rad/s) | Factor | Slope change | |---|---|---| | 2.583 | pole 1/(1 + s/2.583) | −20 dB/dec | | 5 | zero (1 + s/5) | +20 dB/dec | | 20 | pole 1/(1 + s/20) | −20 dB/dec | | 77.417 | pole 1/(1 + s/77.417) | −20 dB/dec | s² + 80s + 200 has real roots −40 ± √1400 = −2.583 and −77.42 (ζ = 2.83 > 1), so it splits into two simple factors. #### 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 −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 (°) | |---|---|---|---| | 0.258 | −20 | −20.3 | −94 | | 2.583 | −40 | −40.3 | −117 | | 5 | −20 | −51.8 | −125 | | 20 | −40 | −63.8 | −156 | | 77.417 | −60 | −87.3 | −212 | Asymptotic magnitude plot (each * is the asymptote; 1 row = 10 dB): ```text |** −20 | ****** | ****** −40 | ***** | **** −60 | ***** | *** −80 | *** | *** −100 | ++-----------+-----------+-----------+ 0.1 1 10 100 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 = 2.583 rad/s the exact gain is −42.3 dB, the asymptote gives −40.3 dB. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is s^2 + 80s + 200 split into two simple factors? A. zeta = 2.83 > 1, so it has real roots -2.583 and -77.42 B. zeta = 0.35, so it is underdamped C. zeta = 1, a double root D. It has a zero at the origin Answer: A. Real roots result when zeta > 1. (Key point: Real-root quadratic) 2. What is the order of corners? A. 5 zero, 2.583 pole, 20 pole, 77.417 pole B. 2.583 pole, 5 zero, 20 pole, 77.417 pole C. 2.583 pole, 20 pole, 5 zero, 77.417 pole D. 5 zero, 20 pole, 2.583 pole, 77.417 pole Answer: B. Sorted by frequency, the types are as given. (Key point: Corner order) 3. What is K0 and 20 log K0? A. 0.5, -6 dB B. 0.25, -12 dB C. 0.025, -32 dB D. 5, 14 dB Answer: C. K0 = 20 x 5 /(20 x 200) = 0.025, -32.04 dB. (Key point: Constant term) 4. What is the slope between 5 and 20 rad/s? A. -40 dB/dec B. 0 dB/dec C. -60 dB/dec D. -20 dB/dec Answer: D. Slope -20 at start, -40 after 2.583, then the zero at 5 raises it to -20. 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