Correct Answer - Option 4 :
\(\sqrt {\frac{a}{b}} \)
The transfer function of lead compensation is given by:
\({K_{lead}}\left( s \right) = \frac{{{K_c}\left( {s + a} \right)}}{{\left( {s + b} \right)}},\;a < b\)
Here, the zero is dominating over the pole.
If a pole or zero is close to the origin, then it becomes more dominating.
The magnitude of lead compensation is given by:
\({\left| M \right|_{s = j\omega }} = \sqrt {\frac{{{\omega ^2} + {a^2}}}{{{\omega ^2} + {b^2}}}} \)
The maximum lead attenuation at a frequency of:
\({\omega _m} = \sqrt {ab} \) will be:
\({\left| M \right|_{{\omega _1} = {\omega _m} = \sqrt {ab} }} = {\left. {\sqrt {\frac{{{\omega ^2} + {a^2}}}{{{\omega ^2} + {b^2}}}} } \right|_{\omega = \sqrt {ab} }}\)
\(\sqrt {\frac{{{{\left( {\sqrt {ab} } \right)}^2} + {a^2}}}{{{{\left( {\sqrt {ab} } \right)}^2} + {b^2}}}} = \sqrt {\frac{{ab + {a^2}}}{{ab + {b^2}}}} \)
\(= \sqrt {\frac{{a\left( {a + b} \right)}}{{b\left( {a + b} \right)}}} \)
\( = \sqrt {\frac{a}{b}} \)
