Correct Answer - Option 2 : 0.33 rad/s
Concept:
In a bode magnitude plot, a pole adds a slope of -20 dB/decade whereas a zero adds a slope of 20 dB/decade.
The initial slope of the magnitude plot is ±20n dB/decade
+ for n zeros at origin
- for n poles at the origin
The standard transfer function of a Bode magnitude plot is:
\(TF = \frac{{K\left( {1 + \frac{s}{{{\omega _1}}}} \right)\left( {1 + \frac{s}{{{\omega _2}}}} \right) \ldots }}{{{s^n}\left( {1 + \frac{s}{{{\omega _3}}}} \right)\left( {1 + \frac{s}{{{\omega _4}}}} \right) \ldots }}\)
Here, ω1, ω2, ω3, ω4, … are the corner frequencies.
n is the number poles at the origin.
Calculation:
\(G(s) = \frac{{10}}{{(3s + 1)}}\)
Comparing with standard transfer function we get,
cross-over frequency = 1/3 = 0.33 rad/sec