Discussion
The final value of \(2S + 1 / S^{4} + 8S^{3} + 16S^{2} + S\) is
Answer: C
The final value is 1, option (C). By the final value theorem, \(\lim_{t\to\infty} f(t) = \lim_{s\to 0} sF(s)\), provided that all the poles of \(sF(s)\) lie in the left half plane.
\[ sF(s) = \frac{s(2s+1)}{s\,(s^{3}+8s^{2}+16s+1)} \]
\[ = \frac{2s+1}{s^{3}+8s^{2}+16s+1} \]
\[ \lim_{s\to0} sF(s) = \frac{1}{1} = 1 \]
The cubic in the denominator is stable, since all its coefficients are positive and \(8\times16 > 1\), so the theorem applies. The answers 0, \(\infty\) and 2 in options (A), (B) and (D) do not follow from this limit.
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