Polar plot of transfer function is

G(jω) = \(\frac{1}{\text{jω}\left(1 + \text{jω T}_{1}\right)}\)

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Question

This question was previously asked in

UPPSC AE Electrical 29 May 2022 Official Paper I

Answer 1

Polar plot:

The polar plot is a plot, that is drawn between the magnitude and the phase angle of G(jω) by varying ω from zero 0 to ∞.

Given, G(jω) = \(\frac{1}{\text{jω}\left(1 + \text{jω T}_{1}\right)}\)

|G(jω)| = \(\frac{1}{\text T_1{ω}\sqrt {\left(({1\over T_1})^2 + \text{ω}^2\right)}}\)

∠G(jω) = -90° - tan^-1(ωT₁)

ω	G(jω)\| = \(\frac{1}{\text T_1{ω}\sqrt {\left(({1\over T_1})^2 + \text{ω}^2\right)}}\)	∠G(jω) = -90° - tan^-1 (ωT₁)
0	∞	-90°
∞	0	-180°

F1 Savita Engineering 2-8-22 D6