Stability and Routh Hurwitz Criterion Questions and Answers

Answer: B

Apply the Routh array to \(s^5 + 6s^4 + 3s^3 + 2s^2 + s + 1\).

\[ \begin{array}{c|ccc} s^5 & 1 & 3 & 1 \\ s^4 & 6 & 2 & 1 \\ s^3 & 2.667 & 0.833 & \\ s^2 & 0.125 & 1 & \\ s^1 & -20.5 & & \\ s^0 & 1 & & \end{array} \]

The first column is \(1, 6, 2.667, 0.125, -20.5, 1\). The sign changes twice: from \(0.125\) to \(-20.5\) and from \(-20.5\) to \(1\). The number of sign changes equals the number of roots in the right half plane, so there are 2 roots with a positive real part, and the system is unstable. The answer is option B.

Answer: B

Apply the Routh array to \(s^4 + 2s^3 + 3s^2 + 4s + 5 = 0\).

\[ \begin{array}{c|ccc} s^4 & 1 & 3 & 5 \\ s^3 & 2 & 4 & \\ s^2 & 1 & 5 & \\ s^1 & -6 & & \\ s^0 & 5 & & \end{array} \]

The \(s^2\) row is \((2 \times 3 - 4)/2 = 1\) and \((2 \times 5)/2 = 5\), and the \(s^1\) entry is \((1 \times 4 - 2 \times 5)/1 = -6\). The first column is \(1, 2, 1, -6, 5\), which has two sign changes, so two roots are in the right half plane and the system is unstable. The options stable, partially stable and stable at origin all need no sign change in the first column. The answer is Unstable, option B.