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Transfer function from asymptotic Bode plot

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.

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)]

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/sK/s.
  2. Gain K: the −20 dB/dec line (or its extension) cuts 0 dB at ω=K\omega = K. It crosses 0 dB at ω = 1, so K = 1. Check at ω = 0.1: 20log⁡(1/0.1)=2020\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)1/(1 + s/2).
  4. There are no further changes, so there are no more poles or zeros.

Transfer function

G(s)=1s(1+s2)=2s(s+2)G(s) = \frac{1}{s\left(1 + \dfrac{s}{2}\right)} = \frac{2}{s(s+2)}

Check

ωAsymptote valuePlot
0.120log⁡10.1=2020\log\frac{1}{0.1} = 20 dB20 dB ✓
120log⁡1=020\log1 = 0 dB0 dB ✓
220log⁡12=−6.0220\log\frac12 = -6.02 dBcorner ✓
10−6.02−40log⁡5=−34.0-6.02 - 40\log5 = -34.0 dBon −40 line ✓

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

✓ Answer checked by Claude Opus 5.5 (high effort)

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