Question

using convolution theorem find the inverse Laplace transform of 1/(s^2+4)^2.​

Answer 1

Answer:

The Inverse Transform of g′(s) is easy to find:

L−1{g′(s)}=L−1{−1/(s^2+4)}=−sin2t

Answer 2

Answer:

The required inverse Laplace transform is given by $& \frac{1}{4} t \sin 2 t .$

Step-by-step explanation:

Here

$F(s) & G(s)=\frac{s}{\left(s^{2}+a^{2}\right)^{2}} \\+\frac{s}{\left(s^{2}+a^{2}\right)} \cdot \frac{1}{\left(s^{2}+a^{2}\right)}\end{aligned}$

Let

$\begin{aligned}&F(s)=\frac{s}{s^{2}+a^{2}}, \\&G(s)=\frac{1}{s^{2}+a^{2}}\end{aligned}$

Then

$\begin{aligned}&L^{-1}\{F(s)\}=L^{-1}\left\{\frac{s}{s^{2}+a^{2}}\right\} \\&=\cos a t=f(t) \\&L^{-1}\{G(s)\}=L^{-1}\left\{\frac{1}{s^{2}+a^{2}}\right\} \\&=\frac{1}{a} \sin a t=g(t) \\&t=u, \quad f(u)=\cos a u,\end{aligned}$

$g(u)=\frac{1}{a} \sin a u .$

Using Convolution theorem, we have

$\begin{aligned}& L^{-1}\{F(s) G(s)\} \\=& \int_{0}^{t} f(u) g(t-u) d u \\\text { i.e., } L^{-1}\left\{\frac{s}{\left(s^{2}+a^{2}\right)^{2}}\right\} \\=& \int_{0}^{t} \cos a u \frac{1}{a} \sin a(t-u) d u \\=& \frac{1}{a} \int_{0}^{t} \cos a u \sin (a t-a u) d u \\=& \frac{1}{2 a} \int_{0}^{t}[\sin a t+\sin (a t-2 a u)] d u \\=& \frac{1}{2 a}\left[u \sin a t-\frac{1}{-2 a} \cos (a t-2 a u)\right]_{0}^{t} \\=& \frac{1}{2 a} t \sin a t .\end{aligned}$

Substituting $a=2$, in then get the required inverse Laplace transform.

$=& \frac{1}{2 a} t \sin a t .\\=& \frac{1}{4} t \sin 2 t .$