Question 10.18: System Identification in Frequency Domain The Bode plot of a...

As shown in Figure 10.73, the Bode magnitude plot starts from −20  \text{dB}, the slope changes at 10  \text{rad/s} from 0  \text{dB/decade} to −40  \text{dB/decade}, and there is a lightly damped resonant peak at 10  \text{rad/s}. This implies that the frequency response function of the system consists of one constant term and one second-order term in the denominator,

G(\text{j}ω) = \frac{K}{(\text{j}ω/ω_n)^2 + 2 ζ(\text{j}ω/ω_n) + 1}

Note that the Bode phase plot starts from 0° and changes at 10  \text{rad/s} to −180°. This also indicates a second-order term in the denominator.
The initial magnitude −20  \text{dB} = 20  \text{log}_{10}K, which gives

K= 10^{\frac{-20}{20}} = 0.1

The slope changes at 10  \text{rad/s}, indicating the corner frequency, also the natural frequency, of the second-order term is ω_n = 10  \text{rad/s}. The value of the peak is −12  \text{dB}, or 8  \text{dB} above the asymptotes, indicating

20\text{log}_{10}\frac{1}{2ζ} = 8  \text{dB}

which gives

ζ = \frac{1}{2\left( 10 ^{\frac{8}{20}}\right) } = 0.20

Thus, the frequency response function of the system is

G(\text{j}ω) = \frac{0.1}{(\text{j}ω/10)^2 + 2 (0.2)(\text{j}ω/10) + 1}

and replacing \text{j}ω with s gives the transfer function of the system as

G(s) = \frac{10}{s^2 +4s + 100}