Question 8.5.3: Application of the Phase Plot Consider the experimentally de...
Application of the Phase Plot
Consider the experimentally determined plots shown in Figure 8.5.3. Determine the forms of the transfer function.
At low frequencies both magnitude curves start at 0 dB and have zero slope. Therefore, the numerator of both transfer functions is 1 at low frequencies. Both magnitudes drop by 3 dB near ω = 8 rad/s, and both phase curves pass near −45° near ω = 8 rad/s. Thus, we conclude that each transfer function has a denominator term \tau s + 1 where \tau = 1/8 s, approximately.
The high-frequency slope of curve B is −20 dB/decade, and thus we conclude that its transfer function is
T_{B}(s) = \frac{1}{\frac{1}{8} s + 1}
The high-frequency slope of curve A is steeper, and thus we conclude that its transfer function has another denominator term that contributes to the slope at high frequencies. This is more apparent in the phase plot, where \phi_{B} becomes horizontal but \phi_{A} appears to be heading toward −180°. Thus we suspect that its transfer function has the form
T_{A}(s) = \frac{1}{(\frac{1}{8} s + 1) (\tau_{2}s + 1)}
but we cannot confirm this or determine the value of \tau_{2} without data at frequencies higher than 10² rad/s