Control Engineering Basics
Which of the following closed-loop systems, given by their characteristic equations, is stable?
- As² + 3s + 2 = 0
- Bs² − 3s + 2 = 0
- Cs² + 4 = 0
Show answer and explanation
Answer: A. s² + 3s + 2 = 0
Since s² + 3s + 2 = (s + 1)(s + 2), the poles are at s = −1 and −2, both with negative real parts, so the system is stable. The poles of s² − 3s + 2 = (s − 1)(s − 2) are at s = 1 and 2, and the response grows exponentially, so it is unstable. The poles of s² + 4 = 0 are at s = ±j2, and an oscillation at an angular frequency of 2 rad/s continues without decaying. For a second-order characteristic equation, all coefficients having the same sign with no missing term is a necessary and sufficient condition for stability.
- A. Correct. The poles are −1 and −2.
- B. The poles are at s = 1 and 2, in the right half of the s-plane, so it is unstable.
- C. The poles lie on the imaginary axis (±j2), giving a sustained oscillation.