Correct Answer - Option 4 :
∞
Concept:
The gain margin is always calculated at the phase Crossover frequency (ωpc)
The phase Crossover frequency (ωpc) is calculated as:
\(\angle G\left( {j\omega } \right)H\left( {j\omega } \right){\left. \right|_{\omega = {\omega _{pc}}}} = \pm 180^\circ \)
Gain margin \( = \frac{1}{{{{\left| {G\left( {j\omega } \right)H\left( {j\omega } \right)} \right|}_{\omega = {\omega _{pc}}}}}}\)
Calculation:
Given:
\(G\left( s \right)H\left( s \right) = \frac{1}{{s\left( {s + 1} \right)}}\)
\(G\left( {j\omega } \right)H\left( {j\omega } \right) = \frac{1}{{j\omega \left( {j\omega + 1} \right)}}\)
The phase Crossover frequency (ωpc) is calculated as:
\(\angle G\left( {j\omega } \right)H\left( {j\omega } \right){\left. \right|_{\omega = {\omega _{pc}}}} = \pm 180^\circ \)
\(- 90^\circ - {\tan ^{ - 1}}{{{\omega _{pc}}}} = 180^\circ \)
\({\tan ^{ - 1}}{{{\omega _{pc}}}} = 90^\circ \)
\({\omega _{pc}} = \infty\)
Now,
\(\left| {G\left( {j\omega } \right)H\left( {j\omega } \right)} \right| = \frac{1}{{\omega \sqrt {4 + {\omega ^2}} }}\)
\({\left| {G\left( {j\omega } \right)H\left( {j\omega } \right)} \right|_{\omega = {\omega _{pc}}}} = 0\)
Thus,
Gain margin \(= \frac{1}{{{{\left| {G\left( {j\omega } \right)H\left( {j\omega } \right)} \right|}_{\omega = {\omega _{pc}}}}}}\)
\(Gain\;margin = \frac{1}{0} = \infty\)
