3. Solved Problems

Example 1: Checking for Stability The characteristic equation of a closed loop control system is \[s^4 + 4 s^3 + 6 s^2 + 2 s + 3 = 0\] Check whether the system is stable or not.

1994-28

Solution: Characteristic equation: \[s^4 + 4 s^3 + 6 s^2 + 2 s + 3 = 0\] Routh array: \[\begin{array}{rlll} \text{Row } 1 \ \ & 1 & 6 & 3\\ 2 \ \ & 4 & 2 & \ \\ 3 \ \ & \dfrac{4\times 6 - 1 \times 2}{4} = 5.5 & \dfrac{4\times 3 - 1 \times 0}{4} = 3 & \ \\\\ 4 \ \ & \dfrac{5.5 \times 2 - 4 \times 3}{5.5} = -\dfrac{1}{5.5} & \ & \ \end{array}\] Since the first column of Routh array is having negative element (i.e., \(-1/5.5\)), the given system is unstable. 0◻