--- title: "Bode plot: 10(S+10)/(S(S²+5S+4)(S+40))" subject: "Electrical Circuit Theory" chapter: 5 key: "semester_3/electrical_circuit_theory:bode-10s10-s-s2-5s-4" source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/bode-10s10-s-s2-5s-4 site: BCT Engineering --- # Bode plot: 10(S+10)/(S(S²+5S+4)(S+40)) Electrical Circuit Theory · Chapter 5: Frequency Response of Network - Asked in: 2069 Chaitra, 8 marks; 2068 Baisakh, 8 marks; 2070 Asar (old course), 10 marks (asked 3 times) - 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-10s10-s-s2-5s-4 - Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-5 ## Question Sketch the asymptotic Bode plot for the transfer function N(S) = 10(S + 10) / [S(S² + 5S + 4)(S + 40)]. ## Short answer Factor as 0.625(1+s/10)/[ s (1+s/1)(1+s/4)(1+s/40) ]. Low-frequency slope –20 dB/dec, asymptotic gain at ω=1 rad/s is –4.08 dB. Corner frequencies: 1 (pole), 4 (pole), 10 (zero), 40 (pole). The slope becomes –40, –60, –40, –60 dB/dec respectively. Phase starts at –90° and ends at –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 N(s) = 10(s + 10) / [ s (s² + 5s + 4)(s + 40) ] = 10(s + 10) / [ s (s + 1)(s + 4)(s + 40) ] since s² + 5s + 4 = (s + 1)(s + 4) ``` ```text H(s) = 0.625 × (1 + s/10) / [ s (1 + s/1) (1 + s/4) (1 + s/40) ] ``` Constant term K₀ = 0.625, so 20 log K₀ = −4.08 dB. #### Step 2: Corner frequencies | Corner ω (rad/s) | Factor | Slope change | |---|---|---| | 1 | pole 1/(1 + s/1) | −20 dB/dec | | 4 | pole 1/(1 + s/4) | −20 dB/dec | | 10 | zero (1 + s/10) | +20 dB/dec | | 40 | pole 1/(1 + s/40) | −20 dB/dec | #### 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 −4.1 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 | 15.9 | −97 | | 1 | −40 | −4.1 | −145 | | 4 | −60 | −28.2 | −195 | | 10 | −40 | −52 | −212 | | 40 | −60 | −76.1 | −232 | Asymptotic magnitude plot (each * is the asymptote; 1 row = 10 dB): ```text 20 |* | ****** 0 | ****** | *** −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 = 1 rad/s the exact gain is −7.3 dB, the asymptote gives −4.1 dB. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How does the quadratic s^2 + 5s + 4 factor, and what corner frequencies result for the whole function? A. (s+1)(s+4); corners at 1, 4, 10, 40 rad/s B. (s+2)(s+2); corners at 2, 10, 40 rad/s C. (s+1)(s+5); corners at 1, 5, 10, 40 rad/s D. Quadratic pole at 2 rad/s; corners at 2, 10, 40 rad/s Answer: A. Roots of s^2+5s+4 are -1 and -4 (real), so it splits into two simple poles. (Key point: Factorisation) 2. What is K0 in the time-constant form and 20 log K0? A. K0 = 10, 20 dB B. K0 = 0.625, -4.08 dB C. K0 = 0.0625, -24 dB D. K0 = 1.6, 4.08 dB Answer: B. K0 = 10 x 10 /(1 x 4 x 40) = 0.625, i.e. -4.08 dB. (Key point: Constant term) 3. Why is the starting slope -20 dB/decade? A. There is one zero at the origin B. The zero at 10 is first C. There is one pole at the origin D. There are two poles at the origin Answer: C. The factor s in the denominator is a single pole at the origin, giving -20 dB/dec. (Key point: Low-frequency slope) 4. What is the slope just after the zero at 10 rad/s? A. -60 dB/dec B. -20 dB/dec C. 0 dB/dec D. -40 dB/dec Answer: D. Slopes go -20, -40 (at 1), -60 (at 4), then the zero adds 20 to give -40. 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