Discussion
Find the number of roots with positive real part for the system whose characteristic equation is \(s^5 + 6s^4 + 3s^3 + 2s^2 + s + 1 = 0\).
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.
No comment is present. Be the first to comment.
Loading…
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development