The phase margin of a system having the loop transfer function $G (s) H (s) = \frac{2 \sqrt{3}}{s (s + 1)}$ is

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Question

This question was previously asked in

UPSC ESE (Prelims) Electronics and Telecommunication Engineering 19 Feb 2023 Official Paper

Answer 1

Concept:

Let 's' be $ω_{g c}$

$\frac{2 \sqrt{3}}{{\sqrt{ω}}_{g c}^{2} (\sqrt{ω_{g c}^{2} + 1)}} = 1$

$\frac{2 \sqrt{3}}{ω_{g c} (\sqrt{ω_{g c}^{2} + 1)}} = 1$

$2 \sqrt{3} = ω_{g c} (\sqrt{ω_{g c}^{2} + 1)}$

Square on both sides

$12 = ω_{g c}^{4} + ω_{g c}^{2}$

$ω_{g c}^{4} + ω_{g c}^{2} - 12 = 0$

Let $w_{g c}^{2} = x$

$x^{2} + x - 12 = 0$

Solving the equation, we get x = 3, - 4.

So, $ω_{g c} = \sqrt{3}$

Phase margin = $180 + ∠ G (j w) H (j w) |_{w_{g c} = \sqrt{3}}$

$∠ G (j w) H (j w) = - 90 - t a n^{- 1} (\frac{w_{g c}}{1})$

$∠ G (j w) H (j w) = - 90 - t a n^{- 1} (\frac{\sqrt{3}}{1})$

$∠ G (j w) H (j w) = - 90 - 60 = - 150$

Phase margin = 180 - 150 = 30°