Question

The inverse Laplace transform of the signal $X\left(s\right)=1n[1+\frac{{\omega }^2}{s^2}]$ is

• A. $\frac{2(1-cos\omega t)}{t}u\left(t\right)$
• B. None of the above
• C. $\frac{2(1+cos\omega t)}{t}u\left(t\right)$
• D. $2(1-cos\omega t)u\left(t\right)$

Answer 1

The correct option is A $\frac{2(1-cos\omega t)}{t}u\left(t\right)$

Given, $X\left(s\right)=1n[1+\frac{{\omega }^2}{s^2}]$

Let $x\left(t\right)=L^{-1}\left[X\right(s\left)\right]=L^{-1}\left[ln(1+\frac{{\omega }^2}{s^2})\right]$

$\therefore L\left[x\right(t\left)\right]=ln[1+\frac{{\omega }^2}{s^2}]=ln\left[\frac{s^2+{\omega }^2}{s^2}\right]$

$=ln[s^2+{\omega }^2]-ln{s}^2$

$L\left[tx\right(t\left)\right]=\frac{-d}{ds}\left[ln\right(s^2+{\omega }^2)-ln{s}^2]$

$=\frac{-1}{s^2+{\omega }^2}.2s+\frac{1}{s^2}2s=\frac{2}{s}-\frac{2s}{s^2+{\omega }^2}$

$\therefore tx\left(t\right)=L^{-1}[\frac{2}{s}-\frac{2s}{s^2+{\omega }^2}]$

$=(2-2cos\omega t)u\left(t\right)=2(1-cos\omega t)u\left(t\right)$

$\therefore x\left(t\right)=\frac{2(1-cos\omega t)}{t}u\left(t\right)$