Correct Answer - Option 3 : 6 < K < 10
Concept:
The characteristic equation for a given open-loop transfer function G(s) is
1 + G(s) H(s) = 0
According to the Routh tabulation method,
The system is said to be stable if there are no sign changes in the first column of the Routh array
The number of poles lies on the right half of s plane = number of sign changes
Calculation:
s3 + 4s2 + s – 6 + K = 0
By applying the Routh tabulation method,
\(\begin{array}{*{20}{c}} {{s^3}\\{s^2}}\\ {{s^1}}\\ {{s^0}} \end{array}\left| {\begin{array}{*{20}{c}} {1}&{1}\\ { 4}&{K-6}&{}\\ {\frac{10-K}{4}}&{}&{}\\{K-6} \end{array}} \right.\)
The system to become stable, the sign changes in the first column of Routh table must be zero.
K - 6 > 0
K > 6 ----(1)
\(\frac{10-K}{4}>0\)
10 > K ----(2)
From equation (1) and (2)
6 < K < 10
