Question 8.5.2: Identifying a Second-Order System Measured response data are...

After drawing the asymptotes shown by the dashed lines, we first note that the data have a lowfrequency asymptote of zero slope, and a high-frequency asymptote of slope −40 dB/decade.
This suggests a second-order model without numerator dynamics, either of the form having real roots:
T (s) = \frac{K}{(τ_{1}s  +  1)(τ_{2}s  +  1)}
or the form having complex roots:
T (s) = \frac{K}{s^{2}  +  2ζω_{n}s + ω^{2}_{n}}
However, the peak in the data eliminates the form having real roots.
At low frequencies, m \approx 20 \log K . From the plot, at low frequencies, m = 75 dB. Thus 75 = 20 log K, which gives
K = 10^{75/20} = 5623
The peak is estimated to be 83 dB. From Table 8.2.1 the peak when K = 1 is given by m_{r} = −20 \log(2ζ \sqrt{1  −  ζ^{2})} . Thus with K = 5623 the formula for the peak becomes

or
83 = 75  −  20 \log \left(2ζ \sqrt{1  −  ζ^{2}} \right)

Thus,
\log \left(2ζ \sqrt{1  −  ζ^{2}}\right) = \frac{75  −  83}{20} = −0.4
and
2ζ \sqrt{1  −  ζ^{2}} = 10^{−0.4}
Solve for ζ by squaring both sides.
4ζ^{2} (1  −  ζ^{2}) = 10^{−0.8}
4ζ^{4}  −  4ζ^{2} + 10^{−0.8} = 0
This gives ζ² = 0.9587 and 0.0413. The positive solutions are ζ = 0.98 and 0.2. Because there is a resonance peak in the data, the first solution is not valid, and we obtain ζ = 0.2.
Knowing ζ , we can now estimate ω_{n} from the peak frequency, which is estimated to be ω_{r} = 70  rad/s. Thus from Table 8.2.1, ω_{r} = ω_{n} \sqrt{1  −  2ζ^{2}} , or
70 = ω_{n} \sqrt{1  −  2(0.2)^{2}}
This gives ω_{n} = 73  rad/s.
Thus, the estimated model is
T (s) = \frac{5623}{s^{2}  +  29.2s  +  5329}