--- title: "Transfer function from field asymptotic Bode plot" subject: "Control Systems" chapter: 6 key: "semester_3/control_systems:tf-from-bode-engineer" source: https://www.bctengineer.com/q/semester_3/control_systems/tf-from-bode-engineer site: BCT Engineering --- # Transfer function from field asymptotic Bode plot Control Systems · Chapter 6: Frequency Response Techniques - Asked in: 2070 Chaitra (old course), 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/control_systems/tf-from-bode-engineer - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-6 ## Question An engineer is called in to consult on a control system in a piece of equipment in the field. No one can find the design report or test results from the original design of control system. The engineer therefore decided to take a frequency response of the system. The resulting asymptotic frequency response is obtained as below. Determine the transfer function. [Figure: asymptotic magnitude plot; 40 dB at ω = 0.1 with slope −20 dB/dec; at ω = √2 there is a resonant peak 4 dB above the asymptote, after which the slope is −60 dB/dec; the curve crosses 0 dB at ω = 2 where the slope changes to −40 dB/dec; at ω = 3 the slope changes to −60 dB/dec] ## Short answer From the Bode plot: initial −20 dB/dec slope indicates a pole at origin; 40 dB at ω=0.1 gives K=10. Complex poles at ωn=√2 rad/s with ζ=0.315 from a 4 dB resonant peak. A zero at ω=2 and a pole at ω=3 yield G(s)=30(s+2)/[s(s+3)(s²+0.892s+2)]. ## Answer We read the asymptotic plot from left to right: each change of slope is a corner frequency, and the low-frequency line fixes the gain. #### Step 1: Initial slope and gain The initial slope is $-20$ dB/dec, so there is **one pole at the origin** (type-1). Its line is $20\log(K/\omega)$: $$ 20\log\frac{K}{0.1} = 40\ \text{dB} \Rightarrow \frac{K}{0.1}=100 \Rightarrow K=10 $$ #### Step 2: Corner at $\omega=\sqrt2$ (complex poles) The slope changes from $-20$ to $-60$ dB/dec, a change of $-40$ dB/dec, and the curve shows a resonant peak. So there is a **pair of complex poles** with $\omega_n=\sqrt2$ rad/s. The exact curve at the corner lies $20\log\frac{1}{2\zeta}$ above the asymptote, so $$ \begin{aligned} 20\log\frac{1}{2\zeta} &= 4\ \text{dB}\\ \frac{1}{2\zeta} &= 10^{0.2} = 1.585\\ \zeta &= 0.315 \end{aligned} $$ Quadratic factor: $\dfrac{1}{1+\frac{2\zeta}{\omega_n}s+\frac{s^2}{\omega_n^2}} = \dfrac{1}{1+0.446s+0.5s^2} = \dfrac{2}{s^2+0.892s+2}$ #### Step 3: Corner at $\omega=2$ (zero) Slope changes from $-60$ to $-40$ dB/dec (+20 dB/dec), so there is a **simple zero** at $\omega=2$: factor $(1+s/2)$. #### Step 4: Corner at $\omega=3$ (pole) Slope changes from $-40$ to $-60$ dB/dec (−20 dB/dec), so there is a **simple pole** at $\omega=3$: factor $\frac{1}{1+s/3}$. #### Transfer function $$ \begin{aligned} G(s) &= \frac{10\,(1+s/2)}{s\,(1+0.446s+0.5s^2)(1+s/3)}\\ &= \frac{10\times\frac12\times2\times3\,(s+2)}{s(s^2+0.892s+2)(s+3)}\\ &= \frac{30(s+2)}{s(s+3)(s^2+0.892s+2)} \end{aligned} $$ Check: as $s\to0$, $G(s)\to \frac{30\times2}{s\times3\times2} = \frac{10}{s}$, which gives 40 dB at $\omega=0.1$ as required. | Corner $\omega$ | Factor | Slope after | |---|---|---| | — | $10/s$ | $-20$ dB/dec | | $\sqrt2$ | complex poles, $\zeta=0.315$ | $-60$ dB/dec | | 2 | zero $(1+s/2)$ | $-40$ dB/dec | | 3 | pole $1/(1+s/3)$ | $-60$ dB/dec | Note: with $K=10$ the straight-line asymptote is about 8 dB at $\omega=2$, not exactly 0 dB as labelled in the sketch; the gain is taken from the clearly marked low-frequency point (40 dB at $\omega=0.1$), which is the usual method. **Answer: $G(s)=\dfrac{30(s+2)}{s(s+3)(s^2+0.892s+2)}$, with $\zeta\approx0.315$ and $\omega_n=\sqrt2$ rad/s for the complex poles.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The initial slope is -20 dB/dec and the magnitude is 40 dB at w = 0.1. What is K in K/s? A. K = 4 B. K = 100 C. K = 10 D. K = 40 Answer: C. 20 log(K/0.1) = 40 dB gives K/0.1 = 100, so K = 10. (Key point: Gain from the low-frequency line) 2. At w = sqrt(2) the slope changes by -40 dB/dec and a resonant peak appears. What does this indicate? A. Two simple real poles at different frequencies B. A pair of complex poles with wn = sqrt(2) C. A double zero D. A pole at the origin Answer: B. A -40 dB/dec change together with a resonant peak means an underdamped quadratic pole pair. (Key point: Complex poles from slope change and peak) 3. The peak is 4 dB above the asymptote, so 20 log(1/2 zeta) = 4. What is zeta? A. 0.63 B. 0.158 C. 0.5 D. 0.315 Answer: D. 1/(2 zeta) = 10^0.2 = 1.585, so zeta = 0.315. (Key point: Damping from peak height) 4. At w = 2 the slope changes from -60 to -40 dB/dec, and at w = 3 from -40 to -60 dB/dec. What are these? A. Pole at 2, zero at 3 B. Zero at 2, pole at 3 C. Pole at 2, pole at 3 D. Zero at 2, zero at 3 Answer: B. A +20 dB/dec change is a zero and a -20 dB/dec change is a pole, giving a zero at 2 and a pole at 3. 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