Question

The Laplace transform of a signal x(t) is

$X\left(s\right)=s^{-1}\left(\frac{e^{-3s}}{s}\right),$ ROC : Re(s) > 0

The signal x(t) is

• A. $-0.1[t-4.5]u(t-3)$
• B. $0.25[t^3-9]u(t+3)$
• C. $0.5[t-2]u(t-3)$
• D. $-0.5[t^2-9]u(t-3)$

Answer 1

The correct option is D $-0.5[t^2-9]u(t-3)$

We have,

$\frac{1}{s}\stackrel{L^{-1}}{⟷}u\left(t\right)$

$\frac{e^{-3s}}{s}\stackrel{L^{-1}}{⟷}u(t-3)$ (Time shifting)

$\frac{d}{ds}\left(\frac{e^{-3s}}{s}\right)⟷-tu(t-3)$ (Differentiation in s-domain)

$\frac{1}{s}\frac{d}{ds}\left(\frac{e^{-3s}}{s}\right)\stackrel{L^{-1}}{⟷}{\int }_{-\infty }^t-\tau u(\tau -3)d\tau$

$x\left(t\right)=-{\int }_3^t\tau d\tau =-{\left(\frac{t^2}{2}\right)}_3^t;t>3$

$x\left(t\right)=-\frac{1}{2}(t^2-9)u(t-3)$