The phase cross-over frequency of the transfer function Gs-100s-13 in rad-s is

G(s)=100(s+1)^3 G(s)=100(1+jω)^3 =1001+(jω)^3+3.(jω)^2+3jω) =100(1 3ω^2)+j(3ω ω^3) =100[(1 3ω^2) jω(3 ω^2)^2][(1 3ω^2)+ω^2(3 ω^2)^2] For phase crossover frequen ...

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Question

The phase cross-over frequency of the transfer function $G (s) = \frac{100}{(s + 1)^{3}}$ in rad/s is

\sqrt{3}

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\frac{1}{\sqrt{3}}

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3 \sqrt{3}

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Solution

The correct option is A $\sqrt{3}$
$G (s) = \frac{100}{(s + 1)^{3}}$
$G (s) = \frac{100}{{(1 + j ω)}^{3}}$
$= \frac{100}{1 + (j ω)^{3} + 3. (j ω)^{2} + 3 j ω)}$
$= \frac{100}{(1 - 3 ω^{2}) + j (3 ω - ω^{3})}$
$= \frac{100 [(1 - 3 ω^{2}) - j ω (3 - ω^{2})^{2}]}{[(1 - 3 ω^{2}) + ω^{2} (3 - ω^{2})^{2}]}$
For phase crossover frequency $ω_{p h}, I m g [G (j ω)] = 0,$
$H e n c e, ω (3 - ω^{2}) = 0$ or $ω = 0; \pm \sqrt{3}$
$∴ ω_{p h} = \sqrt{3} r a d / s e c$

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The phase cross-over frequency of the transfer function G(s)=100(s+1)3 in rad/s is

The correct option is A √3G(s)=100(s+1)3 G(s)=100(1+jω)3 =1001+(jω)3+3.(jω)2+3jω) =100(1−3ω2)+j(3ω−ω3) =100[(1−3ω2)−jω(3−ω2)2][(1−3ω2)+ω2(3−ω2)2] For phase crossover frequencyωph, Img[G(jω)]=0, Hence,ω(3−ω2)=0 or ω=0;±√3 ∴ωph=√3rad/sec

The phase cross-over frequency of the transfer function $G (s) = \frac{100}{(s + 1)^{3}}$ in rad/s is