Transfer function from field asymptotic Bode plot

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.

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.