An open loop transfer funtion Gs of a system is Gs - Kss-1s-2- For a unity feedback system- the breakway point of the root loci on the real axis occurs at-

G_1(s) = Ks(s+1)(s+2) ⇒ nbsp;1+ Ks(s+1)(s+2) = 0 ⇒ nbsp;s(s^2+3s+2)+k = 0 ⇒ s^3+3s^2+2s+k = 0 ⇒ nbsp; nbsp;k = s^3 3s^2 2s nbsp; nbsp; nbs ...

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Standard VIII

Physics

Normal Force

An open loop ...

Question

An open loop transfer funtion G(s) of a system is $G (s) = \frac{K}{s (s + 1) (s + 2)} .$ For a unity feedback system, the breakway point of the root loci on the real axis occurs at,

-0.42

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-1.58

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-0.42 and -1.58

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None of the above

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Solution

The correct option is A -0.42
$G_{1} (s) = \frac{K}{s (s + 1) (s + 2)}$

$\Rightarrow 1 + \frac{K}{s (s + 1) (s + 2)}$ = 0

$\Rightarrow s (s^{2} + 3 s + 2) + k$ = 0

$\Rightarrow s^{3} + 3 s^{2} + 2 s + k$ = 0

$\Rightarrow k = - s^{3} - 3 s^{2} - 2 s$ ......(i)

$\Rightarrow \frac{d k}{d s}$ = 0

$\frac{d k}{d s} = - 3 s^{2} - 6 s - 2 = 0$

$s = - 0.422, - 1.577$

Only s = -0.422 lie on root locus. Therefore breakway point is s = -0.42.

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An open loop transfer funtion G(s) of a system is G(s)=Ks(s+1)(s+2). For a unity feedback system, the breakway point of the root loci on the real axis occurs at,

The correct option is A -0.42G1(s)=Ks(s+1)(s+2) ⇒1+Ks(s+1)(s+2) = 0 ⇒s(s2+3s+2)+k = 0 ⇒s3+3s2+2s+k = 0 ⇒k=−s3−3s2−2s ......(i) ⇒dkds = 0 dkds=−3s2−6s−2=0 s=−0.422,−1.577 Only s = -0.422 lie on root locus. Therefore breakway point is s = -0.42.

An open loop transfer funtion G(s) of a system is $G (s) = \frac{K}{s (s + 1) (s + 2)} .$ For a unity feedback system, the breakway point of the root loci on the real axis occurs at,