If the characteristic equation of a system is 3 + 14s2 +

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If the characteristic equation of a system is ³ + 14s² + 56s + K = 0, then it will be stable only if—

1. 0 < K < 784
2. 1 < K < 64
3. 10 > K > 660
4. 4 < K < 784

Correct Option: A

Given C.E., s³ + 14s² + 56s + K = 0
Using R.H array
s³ 1 56
s² 14 K
s¹ 784 – K /14
s⁰ K
Therefore, for the system to be stable
K > 0 and 784 – K 14 > 0 or 784 – K > 0 or K < 784
Therefore 0 < K < 784 is the required condition for the system to be stable.