Question

Laplace transform of $\cos \left(\omega t\right)is\frac{s}{s^2+{\omega }^2}$. The Laplace transform of $e^{-2t}\cos \left(4t\right)$ is

• A. $\frac{s-2}{(s-2{)}^2+16}$
• B. $\frac{s+2}{(s-2{)}^2+16}$
• C. $\frac{s-2}{(s+2{)}^2+16}$
• D. $\frac{s+2}{(s+2{)}^2+16}$

Answer 1

The correct option is D $\frac{s+2}{(s+2{)}^2+16}$

$L\left[e^{at}f\right(t\left)\right]=F(s-a)$
$\therefore L\left[e^{-2t}\cos 4t\right]=\frac{(s+2)}{(s+2{)}^2+4^2}$
$\therefore L\left[e^{-2t}\cos 4t\right]=\frac{(s+2)}{(s+2{)}^2+16}$