Question

Find the \(L^{-1} \frac{s}{(s^2+4)^2}\).

(a) \(\frac{1}{4} tcos(2t)\)

(b) \(\frac{1}{4} tsin(t)\)

(c) \(\frac{1}{4} tsin(2t)\)

(d) \(\frac{1}{2} tsin(2t)\)

This question was posed to me during an online interview.

I want to ask this question from Convolution in portion Laplace Transform of Engineering Mathematics

Answer 1

The correct choice is (c) \(\frac{1}{4} tsin(2t)\)

For explanation: In the given question,

Let \(p_1(s) = \frac{1}{s^2+4}\) and \(p_2(s) = \frac{1}{s}\)

\(f_1 (t) = L^{-1} (\frac{1}{s^2+4}) = \frac{sin⁡(2t)}{2}\)

\(f_2 (t) = L^{-1} (\frac{s}{s^2+4}) = cos⁡(2t)\)

By Convolution Theorem,

\(L^{-1} (p_1(s)×p_2(s)) = \int_{0}^{t} f_1(u) f_2(t-u) dt\)

\(L^{-1} \left (\frac{s}{(s^2+4^2)^2}\right )=\int_{0}^{t} sin⁡(2u)×\frac{1}{2}×cos⁡(2(t-u))du\)

=\(\frac{1}{4}[tsin(2t)-\frac{cos⁡(2t)}{4}+\frac{cos⁡(2t)}{4}]\)

=\(\frac{1}{4} tsin(2t)\)

Thus, the correct answer is \(\frac{1}{4} tsin(2t)\) .