The transfer function is K(s + 1)(s + 2)(s + 3). The break

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The transfer function is
K
(s + 1)(s + 2)(s + 3)
.
The break point lies between—

1. 0 and – 1
2. – 1 and – 2
3. – 2 and –3
4. beyond –3

Correct Option: B

The C.E. is
(s + 1) (s + 2) (s + 3) + K = 0
K = – (s² + 3s + 2) (s + 3)
or
K = – (s³ + 3s² + 2s + 3s² + 9s + 6)
K = – (s³ + 6s² + 11s + 6)
dK/ds = – (3s² + 12s + 11)
put
dK/ds = 0
3s² + 12s + 11 = 0

s =	– 12 ±√12² – 4 × 3 × 11	=	– 12 ± √12
	2 × 3		6

which is lies between – 1 and – 2.