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 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Other

Control Systems

Break Points

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

-1.58

No worries! We‘ve got your back. Try BYJU‘S free classes today!

-0.42 and -1.58

No worries! We‘ve got your back. Try BYJU‘S free classes today!

None of the above

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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.

Suggest Corrections

Similar questions

Q. 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,

A unity negative feedback system has the open-loop transfer function $G (s) = \frac{K}{s (s + 1) (s + 3)}$
The value of the gain $K (> 0)$ at which the root locus crosses the imaginary axis is

Q. Consider open loop transfer function of unity negative feedback system as

G (s) = \frac{K (s^{2} + 16)}{s (s + 4)}

breakaway point is /are :

Q. A unity feedback system has the open loop transfer function,

G (s) = \frac{1}{(s - 1) (s + 2) (s + 3)}

The Nyquist plot of G(s) encircle the origin.

Q. The gain at the breakaway point of the root locus of a unity feedback system with open loop transfer function

G (s) = \frac{K s}{(s - 1) (s - 4)}

Join BYJU'S Learning Program

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,