Question 8.5.1: Identifying a First-Order System An input vs(t) = 5 sin ωt V...

First divide the output amplitude |v_{o}| by the amplitude of the input. The result is given in the third column. This is the amplitude ratio M. Next convert this data to decibels using the conversion

m = 20 \log M. This is the fourth column. The plot of m versus ω is shown by the small circles in Figure 8.5.1.
After drawing the asymptotes shown by the dashed lines, we first note that the data has a low-frequency asymptote of zero slope, and a high-frequency asymptote of slope−20 dB/decade.
This suggests a model of the form
T (s) = \frac{K}{τ  s  +  1}
In decibel units,
m = 20 \log K  −  10 \log(τ^{2} ω^{2} + 1)
The corner frequency ω = 1/τ occurs where m is 3 dB below the peak value of 6.81. From the plot or the data we can see that the corner frequency is ω = 8 rad/s. Thus τ = 1/8  s.
At low frequencies, ω \ll 1/τ and m \approx 20 \log K. From the plot, at low frequency, m = 6.81 dB. Thus 6.81 = 20 log K, which gives
K = 10^{6.81/20} = 2.19
Thus the estimated model is
\frac{V_{o}(s)}{V_{s}(s)} = \frac{2.19}{\frac{1}{8} s  +  1}
Another first-order model form is
\frac{V_{o}(s)}{V_{s}(s)} = K \frac{τ_{1} s  +  1}{τ_{2}s  +  1}
Because the high-frequency asymptote of this model has zero slope, it cannot describe the given data.