Discussion
Find the corner frequencies if the transfer function is given as:
\(G(s) = \dfrac{10}{s(1 + 0.4s)(1 + 0.1s)}\).
Answer: A
The corner (break) frequency of each factor of the form \((1 + sT)\) in a Bode plot is \(1/T\).
\[ \begin{aligned} \omega_{c1} = \frac{1}{0.4} \\ &= 2.5\ \text{rad/s},\qquad \omega_{c2} \\ &= \frac{1}{0.1} = 10\ \text{rad/s} \end{aligned} \]
The pole at the origin, \(1/s\), has no corner frequency; it gives a line of slope \(-20\) dB per decade through the whole plot. The values 0.4 and 0.1 are the time constants, not frequencies, and 4, 1, 25 and 100 are not reciprocals of the time constants. The answer is option A.
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