--- title: "Transfer function from Bode plot with example" subject: "Control Systems" chapter: 6 key: "semester_3/control_systems:bode-to-tf-example" source: https://www.bctengineer.com/q/semester_3/control_systems/bode-to-tf-example site: BCT Engineering --- # Transfer function from Bode plot with example Control Systems · Chapter 6: Frequency Response Techniques - Asked in: 2071 Shrawan, 6 marks - Video: [Bode Plot Example: Identifying Transfer Function from Gain Plot in Control Systems](https://www.youtube.com/watch?v=dgsYZXTlxjg) — Engineering Funda - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/control_systems/bode-to-tf-example - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-6 ## Question Discuss how a Bode plot can be used to determine transfer function of the system. Explain with an example. ## Short answer Fit asymptotes: initial flat slope at 20 dB gives type 0, K = 10. Breaks at 2 and 10 rad/s (each −20 dB/dec) indicate two poles. Phase goes 0° to −180°, confirming poles. Transfer function: G(s) = 10/[(1+s/2)(1+s/10)] = 200/[(s+2)(s+10)]. ## Answer A Bode plot obtained from a frequency-response test can be used to identify the transfer function. The asymptotic magnitude curve gives the gain, type and corner frequencies, and the phase curve confirms them (for a minimum-phase system). #### Steps 1. **Fit asymptotes** with slopes of 0, ±20, ±40 … dB/decade to the magnitude curve. 2. **Low-frequency slope gives the system type:** 0, −20 or −40 dB/dec means 0, 1 or 2 integrators. 3. **Gain K from the low-frequency segment:** - Type 0: $M_{low} = 20\log K$. - Type 1: the extended −20 dB/dec line crosses 0 dB at ω = K. - Type 2: the −40 dB/dec line crosses 0 dB at $\omega = \sqrt K$. 4. **Each change of slope is a corner frequency $\omega_c$:** - −20 dB/dec change: pole $1/(1 + s/\omega_c)$; - +20 dB/dec change: zero $(1 + s/\omega_c)$; - −40 dB/dec change: a quadratic (complex) pole pair or a double real pole. For a quadratic, find ζ from the height of the resonant peak: peak above asymptote ≈ $-20\log 2\zeta$. 5. **Verify with phase:** at very low ω the phase should be $-90^\circ \times$ (number of integrators), and at very high ω it should be $-90^\circ \times$ (poles − zeros). Unexplained extra lag means a time delay or a non-minimum-phase zero. #### Worked example Suppose the asymptotic magnitude plot is: - a horizontal line at 20 dB up to ω = 2 rad/s; - then −20 dB/dec up to ω = 10 rad/s; - then −40 dB/dec beyond 10 rad/s. The phase goes from 0° to −180°. ```text 30 | |****************** | ****** 0 |-----------------------****------------------- | *** | **** -30 | **** | **** | **** -60 | *** | *** | -90 | +---------------------------------------------- 0.1 1 10 100 1000 Magnitude asymptotes (dB) vs w (rad/s) ``` **Reading it:** 1. Initial slope 0 dB/dec, so the system is **type 0**. 2. $20\log K = 20$ dB, so **K = 10**. 3. At ω = 2 the slope changes by −20 dB/dec: a pole $1/(1 + s/2)$. 4. At ω = 10 the slope changes by a further −20 dB/dec: a pole $1/(1 + s/10)$. 5. Phase check: two poles give 0° → −180°, which is consistent. $$ G(s) = \frac{10}{\left(1 + \dfrac{s}{2}\right)\left(1 + \dfrac{s}{10}\right)} = \frac{200}{(s+2)(s+10)} $$ **Check of one point:** at ω = 10 the asymptote gives $20 - 20\log(10/2) = 6.0$ dB. From the factors of the TF: $20\log10 - 20\log5 = 6.0$ dB. ✓ (The true curve is about 3 dB lower at each corner.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. On an asymptotic Bode magnitude plot, the low-frequency slope is -40 dB/dec. What does this tell you about the system? A. Type 0, with no integrators B. Type 1, with one integrator C. Type 2, with two integrators D. Two zeros at the origin Answer: C. A slope of 0, -20 or -40 dB/dec at low frequency means 0, 1 or 2 integrators, so -40 dB/dec means type 2. (Key point: Type from low-frequency slope) 2. For a type-0 system the low-frequency magnitude is a flat line at 20 dB. What is K? A. K = 20 B. K = 10 C. K = 2 D. K = 100 Answer: B. 20 log K = 20 dB gives K = 10. (Key point: Gain of a type-0 system) 3. At a corner frequency the asymptotic slope changes by -20 dB/dec. What factor does this correspond to? A. A simple zero (1 + s/wc) B. A simple pole 1/(1 + s/wc) C. A pair of complex poles D. A pole at the origin Answer: B. A -20 dB/dec change means a simple pole at that corner; +20 dB/dec would mean a zero. (Key point: Corner frequency and slope change) 4. The plot is flat at 20 dB to 2 rad/s, then -20 dB/dec to 10 rad/s, then -40 dB/dec. Which transfer function fits? A. 10 / [(1 + s/2)(1 + s/10)] B. 20 / [(1 + s/2)(1 + s/10)] C. 10(1 + s/2) / (1 + s/10) D. 10 / [s(1 + s/2)(1 + s/10)] Answer: A. Type 0 with K = 10 and poles at 2 and 10 rad/s gives G(s) = 10/[(1 + s/2)(1 + s/10)]. 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