Which of the following is the correct comment on stability based on unknown k for the feedback system with characteristic s⁴ + 2ks³ + s² + 5s + 5 = 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: s⁴ + 2ks³ + s² + 5s + 5 = 0

By applying the Routh tabulation method,

\(\begin{array}{*{20}{c}} {{s^4}}\\ {{s^3}}\\ {{s^2}}\\ {{s^1}}\\ {{s^0}} \end{array}\left| {\begin{array}{*{20}{c}} 1&1&5\\ {2k}&5&0\\ {1 - \frac{5}{{2k}}}&5&{}\\ {5 - \frac{{20{k^2}}}{{2k - 5}}}&0&{}\\ 5&{}&{} \end{array}} \right.\)

The system to become stable, the sign changes in the first column of Routh table must be zero.

k > 0, \(1 - \frac{5}{{2k}} > 0\)

⇒ k > 2.5

For all the values of k > 2.5, \(5 - \frac{{20{k^2}}}{{2k - 5}}\) gives negative values.

Therefore, the given system is unstable for all the values of k.