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