--- title: "Bode plot: 64(S+2)/(S(S²+0.5S)(S²+3.2S+64))" subject: "Electrical Circuit Theory" chapter: 5 key: "semester_3/electrical_circuit_theory:bode-64s2-s-s2-05s-s2" source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/bode-64s2-s-s2-05s-s2 site: BCT Engineering --- # Bode plot: 64(S+2)/(S(S²+0.5S)(S²+3.2S+64)) Electrical Circuit Theory · Chapter 5: Frequency Response of Network - Asked in: 2074 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-64s2-s-s2-05s-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 H(S) = 64(S + 2) / [S(S² + 0.5S)(S² + 3.2S + 64)] ## Short answer Factor as 4(1+s/2)/[ s² (1+s/0.5)(1+0.05s+s²/64) ]; ωn=8 rad/s (ζ=0.2). Low-frequency slope –40 dB/dec, asymptote at ω=1: 12.04 dB. Corner frequencies: 0.5 (pole, –20), 2 (zero, +20), 8 (quadratic pole, –40). Slopes: –60, –40, –80 dB/dec. Phase: –180° to –360°. ## 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) = 64(s + 2) / [ s (s² + 0.5s)(s² + 3.2s + 64) ] = 64(s + 2) / [ s² (s + 0.5)(s² + 3.2s + 64) ] ``` ```text H(s) = 4 × (1 + s/2) / [ s² (1 + s/0.5) (1 + 0.05 s + s²/64) ] ``` Constant term K₀ = 4, so 20 log K₀ = 12.04 dB. #### Step 2: Corner frequencies | Corner ω (rad/s) | Factor | Slope change | |---|---|---| | 0.5 | pole 1/(1 + s/0.5) | −20 dB/dec | | 2 | zero (1 + s/2) | +20 dB/dec | | 8 | quadratic pole, ωn = 8, ζ = 0.2 | −40 dB/dec | Here s² + 3.2s + 64 has ωn = 8 rad/s and 2ζωn = 3.2, so ζ = 0.2 (a sharp resonance). There is a double pole at the origin. #### Step 3: Magnitude plot The net number of poles at the origin is 2, so the low-frequency asymptote has slope −40 dB/decade and passes through 12 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 (°) | |---|---|---|---| | 0.05 | −40 | 64.1 | −184 | | 0.5 | −60 | 24.1 | −212 | | 2 | −40 | −12 | −217 | | 8 | −80 | −36.1 | −280 | Asymptotic magnitude plot (each * is the asymptote; 1 row = 10 dB): ```text 100 | |*** 80 | *** | *** 60 | *** | *** 40 | *** | *** 20 | ** | ** 0 | ** | ** −20 | *** | *** −40 | ** | ++-----------+-----------+-----------+ 0.01 0.1 1 10 ω (rad/s) ``` At ωn = 8 the quadratic pole has ζ = 0.2, so the true curve rises about 8 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) = −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) = −360°. Exact phase values at the corners are in the table above; join them with a smooth curve starting at −180° and ending at −360°. Accuracy check: at the first corner ω = 0.5 rad/s the exact gain is 21.4 dB, the asymptote gives 24.1 dB. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How many poles are at the origin and what is the starting slope? A. Two; -40 dB/dec B. One; -20 dB/dec C. Two; -20 dB/dec D. Three; -60 dB/dec Answer: A. s(s^2+0.5s) = s^2(s+0.5), a double pole at the origin gives -40 dB/dec. (Key point: Poles at origin) 2. What is K0? A. 128 (42 dB) B. 4 (12.04 dB) C. 0.5 (-6 dB) D. 64 (36 dB) Answer: B. K0 = 64 x 2 /(0.5 x 64) = 4, which is 12.04 dB. (Key point: Constant term) 3. s^2 + 3.2s + 64 gives which values and consequence? A. omega_n = 64, zeta = 0.2, resonance about 8 dB B. omega_n = 8, zeta = 0.4, no peak C. omega_n = 8, zeta = 0.2, sharp resonance about 8 dB D. omega_n = 8, zeta = 2, overdamped Answer: C. 2 zeta x 8 = 3.2 gives zeta = 0.2; the peak is -20 log 0.4 = 8 dB. (Key point: Quadratic damping) 4. What are the low- and high-frequency phase values? A. -90 deg and -270 deg B. -180 deg and -180 deg C. 0 deg and -360 deg D. -180 deg and -360 deg Answer: D. Low phase = -90 x 2 = -180; high = 90 x (m - n) = 90 x (1 - 5) = -360. 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