Question
Download Solution PDFThe phase margin of a system having the loop transfer function \(G(s) H(s)=\frac{2 \sqrt{3}}{s(s+1)}\) is
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Let 's' be \(\omega_{gc}\)
\(\frac{2 \sqrt{3}}{\sqrt\omega_{gc}^2(\sqrt{\omega_{gc}^2+1)}} =1\)
\(\frac{2 \sqrt{3}}{\omega_{gc}(\sqrt{\omega_{gc}^2+1)}} =1\)
\({2 \sqrt{3}}={\omega_{gc}(\sqrt{\omega_{gc}^2+1)}} \)
Square on both sides
\(12=\omega_{gc}^4 + \omega_{gc}^2\)
\(\omega_{gc}^4 + \omega_{gc}^2-12=0\)
Let \(w_{gc}^2 = x\)
\(x^2+x-12 = 0\)
Solving the equation, we get x = 3, - 4.
So, \(\omega_{gc} = \sqrt3\)
Phase margin = \(180 + \angle G(jw)H(jw)|_{w_{gc}=\sqrt3}\)
\(\angle G(jw)H(jw) = -90 - tan^{-1}(\frac{w_{gc}}{1}) \)
\(\angle G(jw)H(jw) = -90 - tan^{-1}(\frac{\sqrt3}{1}) \)
\(\angle G(jw)H(jw) = -90 - 60 = -150 \)
Phase margin = 180 - 150 = 30°
Last updated on May 28, 2025
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