Question

The Laplace transform and the ROC for the signal $x\left(t\right)=e^{at}u(t-k)$is

• A. $X\left(s\right)=e^{ka}\frac{1}{s-a};ROC;Re\left(s\right)<a$
• B. $X\left(s\right)=e^{a-s}\frac{1}{s-a};ROC:Re\left(s\right)>a$
• C. $X\left(s\right)=e^{k(a-s)}\frac{1}{s-a};ROC;Re\left(s\right)<a$
• D. $X\left(s\right)=e^{k(a-s)}\frac{1}{s-a};ROC:Re\left(s\right)>a$

Answer 1

The correct option is D $X\left(s\right)=e^{k(a-s)}\frac{1}{s-a};ROC:Re\left(s\right)>a$

Given signals, $x\left(t\right)=e^{at}u(t-k)$

$x\left(t\right)=e^{a(t-k+k)}.u(t-k)$

$=e^{ak}[e^{a(t-k)}.u(t-k\left)\right]$

By using time shifting property,

$x(t-t_0)=X\left(s\right).e^{-s{t}_0}$ (No change in ROC)

$X\left(s\right)=e^{ak}\left[\frac{1}{s-a}{e}^{-ks}\right]=e^{k(n-s)}\frac{1}{s-a}$

ROC of the signal is Re {s} > a
Right side of the right most pole.