--- title: "Asymptotic Bode magnitude of 125/[s(s²+10s+25)]" subject: "Control Systems" chapter: 6 key: "semester_3/control_systems:bode-mag-125-s-s2-10s" source: https://www.bctengineer.com/q/semester_3/control_systems/bode-mag-125-s-s2-10s site: BCT Engineering --- # Asymptotic Bode magnitude of 125/[s(s²+10s+25)] Control Systems · Chapter 6: Frequency Response Techniques - Asked in: 2066 Jestha (old course), 6 marks - Video: [Intro to Control - 15.2 Bode Plot Approximation Example](https://www.youtube.com/watch?v=ANRrwj_JHhE) — Engineering with Prof. Kim - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/control_systems/bode-mag-125-s-s2-10s - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-6 ## Question Draw the asymptotic Bode magnitude plot of the unity feedback system whose open loop transfer function is given by G(s) = 125/[s(s²+10s+25)]. ## Short answer Initial slope –20 dB/dec, passing 13.98 dB at ω=1 rad/s, 0 dB at corner ω=5 rad/s, then –60 dB/dec. ## Answer #### Time-constant form $$ G(s)=\frac{125}{s(s^2+10s+25)}=\frac{125}{s(s+5)^2}=\frac{5}{s\left(1+\frac{s}{5}\right)^2} $$ - $K=5$, type-1: initial slope $-20$ dB/dec. - $s^2+10s+25$: $\omega_n=5$ rad/s, $2\zeta\omega_n=10\Rightarrow\zeta=1$ (critically damped, i.e. two equal real poles at $s=-5$). - Corner frequency $\omega_c=5$ rad/s; after it the slope falls by 40 dB/dec. #### Asymptotic magnitude | Factor | Corner | Contribution | |---|---|---| | $K=5$ | — | $+13.98$ dB constant | | $1/s$ | — | $-20$ dB/dec (passes 0 dB at $\omega=1$) | | $1/(1+s/5)^2$ | 5 rad/s | $-40$ dB/dec after 5 | | Range | Net slope | |---|---| | $\omega<5$ | $-20$ dB/dec | | $\omega>5$ | $-60$ dB/dec | Points on the asymptotic plot: $$ \begin{aligned} \omega=0.1:&\ 20\log\frac{5}{0.1} = 33.98\ \text{dB}\\ \omega=1:&\ 20\log5 = 13.98\ \text{dB}\\ \omega=5:&\ 20\log\frac{5}{5}=0\ \text{dB}\\ \omega=10:&\ 0-60\log2 = -18.06\ \text{dB}\\ \omega=50:&\ 0-60\log10 = -60\ \text{dB} \end{aligned} $$ ```text dB 34 |\ | \ -20 dB/dec 14 | \ (w=1) 0 |------\-------------- w (log) | 5 \ -18 | \ (w=10) | \ -60 dB/dec -60 | \ (w=50) 0.1 1 5 10 50 ``` #### Correction (optional) For $\zeta=1$ the correction at the corner is $20\log\frac{1}{2\zeta}=-6.02$ dB, i.e. the exact curve lies 6 dB below the asymptote there. So the actual magnitude at $\omega=5$ is $-6.02$ dB (exact values: 13.64 dB at $\omega=1$, $-20$ dB at $\omega=10$). The asymptotic gain crossover frequency is $\omega_{gc}=5$ rad/s. **Answer: a straight line of −20 dB/dec through 13.98 dB at ω = 1, meeting 0 dB at the corner ω = 5 rad/s, then −60 dB/dec.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is 125/[s(s^2+10s+25)] in Bode form? A. 125 / [s(1 + s/5)^2] B. 5 / [s(1 + s/5)^2] C. 25 / [s(1 + s/5)^2] D. 5 / [s(1 + s/10)] Answer: B. s^2+10s+25 = (s+5)^2 = 25(1+s/5)^2, so K = 125/25 = 5. (Key point: Time-constant form) 2. What does zeta = 1 for s^2 + 10s + 25 mean? A. Two equal real poles at -5 B. Complex poles with a resonant peak C. Poles at the origin D. Two poles at +5 Answer: A. zeta = 1 is critical damping: a repeated real pole at s = -5. (Key point: Damping ratio) 3. What is the slope for w > 5 rad/s? A. -40 dB/dec B. -20 dB/dec C. -80 dB/dec D. -60 dB/dec Answer: D. The integrator gives -20 and the double pole at 5 adds -40, giving -60 dB/dec. (Key point: Slope after corner) 4. What is the asymptotic magnitude at w = 1 rad/s? A. 20 dB B. 0 dB C. 13.98 dB D. 33.98 dB Answer: C. Below the corner the magnitude is 20 log(5/w); at w = 1 this is 20 log 5 = 13.98 dB. 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