Question

Which of the following is the Laplace transform of $e^{-2(t-t_0)}u\left(t\right)$ ?

• A. $\frac{e^{-2s{t}_0}}{s+2},\sigma >-2$
• B. $\frac{e^{-2t_0}}{s+2},\sigma >-2$
• C. $\frac{e^{2s{t}_0}}{s+2},\sigma >-2$
• D. $\frac{e^{2t_0}}{s+2},\sigma >-2$

Answer 1

The correct option is D $\frac{e^{2t_0}}{s+2},\sigma >-2$


$e^{2t_0}\cdot e^{-2t}u\left(t\right)$ $\stackrel{L.T.}{\to }$ $e^{2t_0}.\frac{1}{s+2}$