---
title: "Transfer function from asymptotic Bode plot"
subject: "Control Systems"
chapter: 6
key: "semester_3/control_systems:tf-from-bode-asymptote"
source: https://www.bctengineer.com/q/semester_3/control_systems/tf-from-bode-asymptote
site: BCT Engineering
---

# Transfer function from asymptotic Bode plot

Control Systems · Chapter 6: Frequency Response Techniques

- Asked in: 2078 Kartik, 3 marks; 2068 Chaitra, 6 marks (asked 2 times)
- Video: [Find Transfer Function from Bode (Magnitude) Plot: Lect no.1](https://www.youtube.com/watch?v=zlLgyw3bniU) — EasyMechLearn
- 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-asymptote
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-6

## Question

Find the open loop transfer function with the help of following Bode plot. [Figure: asymptotic magnitude plot; 20 dB at ω = 0.1, falling at −20 dB/dec through 0 dB at ω = 1, then changing at ω = 2 to −40 dB/dec (axis marks at 0.1, 1, 2, 5, 10)]

## Short answer

The transfer function is G(s) = 2/[s(s+2)], i.e. 1/[s(1+0.5s)], derived from an initial -20 dB/dec slope crossing 0 dB at 1 rad/s and a further -20 dB/dec change at 2 rad/s.

## Answer

A Bode magnitude asymptote is read segment by segment. The initial slope gives the type and gain, and each change of slope gives a corner frequency.

#### Reading the plot
1. **Initial slope −20 dB/dec:** one pole at the origin (type 1), i.e. a factor $K/s$.
2. **Gain K:** the −20 dB/dec line (or its extension) cuts 0 dB at $\omega = K$. It crosses 0 dB at ω = 1, so **K = 1**.
   Check at ω = 0.1: $20\log(1/0.1) = 20$ dB, which matches the plot.
3. **At ω = 2 the slope changes from −20 to −40 dB/dec:** the change is −20 dB/dec, so there is a simple pole with corner frequency 2 rad/s, i.e. a factor $1/(1 + s/2)$.
4. There are no further changes, so there are no more poles or zeros.

#### Transfer function
$$
G(s) = \frac{1}{s\left(1 + \dfrac{s}{2}\right)} = \frac{2}{s(s+2)}
$$

#### Check
| ω | Asymptote value | Plot |
|---|---|---|
| 0.1 | $20\log\frac{1}{0.1} = 20$ dB | 20 dB ✓ |
| 1 | $20\log1 = 0$ dB | 0 dB ✓ |
| 2 | $20\log\frac12 = -6.02$ dB | corner ✓ |
| 10 | $-6.02 - 40\log5 = -34.0$ dB | on −40 line ✓ |

**Answer:** $G(s) = \dfrac{2}{s(s+2)}$, i.e. $\dfrac{1}{s(1+0.5s)}$.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The Bode magnitude starts with slope -20 dB/dec. What does this tell you?
   A. The system is type 0
   B. There is one pole at the origin (type 1), giving a factor K/s
   C. There is a zero at the origin
   D. There is a pole at 1 rad/s
   Answer: B. A starting slope of -20 dB/dec corresponds to a single integrator, so the TF contains K/s. (Key point: Initial slope gives type)

2. The -20 dB/dec line crosses 0 dB at omega = 1. What is K?
   A. K = 0.1
   B. K = 2
   C. K = 20
   D. K = 1
   Answer: D. For K/s the asymptote crosses 0 dB at omega = K, so K = 1; at omega = 0.1, 20 log(1/0.1) = 20 dB matches the plot. (Key point: Finding the gain K)

3. At omega = 2 the slope changes from -20 to -40 dB/dec. What factor does this add?
   A. A zero (1 + s/2) in the numerator
   B. A simple pole 1/(1 + s/2)
   C. A double pole at 2
   D. A pole at 0.5
   Answer: B. A slope change of -20 dB/dec is one simple pole at the corner frequency 2 rad/s, giving G(s) = 1/[s(1 + s/2)] = 2/[s(s+2)]. (Key point: Corner frequency)

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