Correct Answer - Option 2 : The system is unstable for K > 0
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 Routh array
The number of poles lie on the right half of s plane = number of sign changes
Calculation:
Characteristic equation: q(s) = s4 + s3 + s2 + s + K = 0
By applying Routh tabulation method,
\(\begin{array}{*{20}{c}} {{s^4}}\\ {{s^3}}\\ {{s^2}}\\ {{s^1}}\\ {{s^0}} \end{array}\left| {\begin{array}{*{20}{c}} 1&1&{K}\\ 1&1&0\\ { 0}&{K}&{}\\ {-K/0}&0&{}\\ {K}&{}&{} \end{array}} \right.\)
The system is stable for all the values k < 0 i.e. the system is unstable for all the values K > 0.
