Discussion
Laplace transform of f(t)=sinht cost is
Answer: B
Write the hyperbolic sine in exponential form and use the shift theorem.
\[ \sinh t \cos t = \frac{1}{2}\left(e^{t} - e^{-t}\right)\cos t \]
With \( \mathcal{L}\{e^{at}\cos t\} = \dfrac{s-a}{(s-a)^{2}+1} \):
\[ \mathcal{L} = \frac{1}{2}\left[\frac{s-1}{s^{2}-2s+2} - \frac{s+1}{s^{2}+2s+2}\right] \]
Over the common denominator \( (s^{2}-2s+2)(s^{2}+2s+2) = s^{4}+4 \):
\[ \text{numerator} = (s-1)(s^{2}+2s+2) - (s+1)(s^{2}-2s+2) = 2s^{2} - 4 \]
\[ \mathcal{L}\{\sinh t \cos t\} = \frac{s^{2}-2}{(s^{2}+2s+2)(s^{2}-2s+2)} \]
As a check, \( \sinh t \cos t \approx t \) for small \( t \), and the expression behaves like \( \tfrac{1}{s^{2}} \) for large \( s \) - which is \( \mathcal{L}\{t\} \).
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