Correct Answer - Option 4 : ∞
Gain margin (GM): The gain margin of the system defines by how much the system gain can be increased so that the system moves on the edge of stability.
It is determined from the gain at the phase cross-over frequency.
\(GM = \frac{1}{{{{\left| {G\left( {j\omega } \right)H\left( {j\omega } \right)} \right|}_{\omega = {\omega _{pc}}}}}}\)
Phase crossover frequency (ωpc): It is the frequency at which the phase angle of G(s) H(s) is -180°.
\(\angle G\left( {j\omega } \right)H\left( {j\omega } \right){|_{\omega = {\omega _{pc}}}} = - 180^\circ \)
Phase margin (PM): The phase margin of the system defines by how much the phase of the system can increase to make the system unstable.
\(PM = 180^\circ + \angle G\left( {j\omega } \right)H\left( {j\omega } \right){|_{\omega = {\omega _{gc}}}} = - 180^\circ \)
It is determined from the phase at the gain cross over frequency.
Gain crossover frequency (ωgc): It is the frequency at which the magnitude of G(s) H(s) is unity.
\({\left| {G\left( {j\omega } \right)H\left( {j\omega } \right)} \right|_{\omega = {\omega _{gc}}}} = 1\)
Calculation:
\(G\left( s \right)H\left( s \right) = \frac{2}{{s\left( {s + 1} \right)}}\)
\(G\left( {j\omega } \right)H\left( {j\omega } \right) = \frac{2}{{j\omega \left( {j\omega + 1} \right)}}\)
∠G(jω)H(jω) = - 90° - tan-1 (ω)
At phase crossover frequency, - 90° - tan-1 (ωpc) = -180°
⇒ ωpc = ∞
Now,
\(\left| {G\left( {j\omega } \right)H\left( {j\omega } \right)} \right| = \frac{2}{{\omega \sqrt {1 + {\omega ^2}} }}\)
\({\left| {G\left( {j\omega } \right)H\left( {j\omega } \right)} \right|_{\omega = {\omega _{pc}}}} = 0\)
Thus,
Gain margin \(= \frac{1}{{{{\left| {G\left( {j\omega } \right)H\left( {j\omega } \right)} \right|}_{\omega = {\omega _{pc}}}}}}\)
\(Gain\;margin = \frac{1}{0} = \infty\)
Important Points:
- If both GM and PM are positive, the system is stable (ωgc < ωpc)
- If both GM and PM are negative, the system is unstable (ωgc > ωpc)
- If both GM and PM are zero, the system is just stable (ωgc = ωpc)
