Correct Answer - Option 1 :
\(\frac{1}{ \propto }\frac{{\left( {s + \frac{1}{\tau }} \right)}}{{\left( {s + \frac{1}{{ \propto \tau }}} \right)}}\)
Concept:
The transfer function of the phase lag compensator is given by
\(G\left( s \right) = \frac{{1 + \tau s}}{{1 + a\tau s}}\)
∝ > 1 and τ > 0
\(G\left( s \right) = \frac{\tau ({s + \frac{1}{\tau}})}{\alpha \tau ({s + \frac{1}{\alpha \tau}})}\)
\(G\left( s \right) = \frac{ ({s + \frac{1}{\tau}})}{\alpha ({s + \frac{1}{\alpha \tau}})}\)
For a lag compensator poles are more closer to the origin than zeros.
The transfer function of phase lag compensator is given by
\(G\left( s \right) = \frac{{1 + Ts}}{{1 + aTs}}\)
\(G\left( {j\omega } \right) = \frac{{1 + j\omega T}}{{1 + j\omega aT}}\)
Phase angle, \(\angle G\;\left( {j\omega } \right) = {\tan ^{ - 1}}\omega T - {\tan ^{ - 1}}a\omega T\)
ϕ = tan-1 ωT – tan-1 aωT
The maximum phase lag occurs at ωm such that \(\frac{{d\phi }}{{d\omega }} = 0\)
\(\Rightarrow {\omega _m} = \frac{1}{{T\sqrt a }}\)
It is a Geometric mean of its two corner frequencies
\({\omega _m} = \sqrt {\frac{1}{T} \times \frac{1}{{aT}}} = \frac{1}{{T\sqrt a }}\)
The maximum phase lag,
\({\phi _m} = {\tan ^{ - 1}}\omega T - {\tan ^{ - 1}}a\omega T\)
\(= {\tan ^{ - 1}}\left( {\frac{1}{{T\sqrt \alpha }}} \right)T - {\tan ^{ - 1}}\left( {\frac{1}{{T\sqrt \alpha }}T} \right)\)
\(= {\tan ^{ - 1}}\frac{1}{{\sqrt a }} - {\tan ^{ - 1}}\sqrt a\)
\({\phi _m} = {\tan ^{ - 1}}\left( {\frac{{\frac{1}{{\sqrt a }} - \sqrt a }}{{1 + \sqrt a \cdot \frac{1}{{\sqrt a }}}}} \right)\)
\({\phi _m} = {\tan ^{ - 1}}\left( {\frac{{1 - a}}{{2\sqrt a }}} \right)\)
\(= {\sin ^{ - 1}}\left( {\frac{{1 - a}}{{a + 1}}} \right) = {\cos ^{ - 1}}\left( {\frac{{2\sqrt a }}{{a + 1}}} \right)\)
