Question

Let $x\left(t\right)=\alpha s\left(t\right)+s(-t)withs\left(t\right)=\beta e^{-4t}u\left(t\right),$ where u(t) is unit step function. If the bilateral laplace transform of x(t) is $X\left(s\right)=\frac{16}{s^2-16}$ $-4<Res<4;$
then the value of $\beta$ is ______

• A. -2

Answer 1

The correct option is A -2
$x\left(t\right)=\alpha \beta e^{-4t}u\left(t\right)+\beta e^{4t}u(-t)$

$\text{Given ,}X\left(s\right)=\frac{16}{s^2-16}\text{;}-4<Re\left[s\right]<4$

$=\frac{16}{(s+4)(s-4)}=\frac{-2}{s+4}+\frac{2}{s-4}$

$x\left(t\right)=-2e^{-4t}u\left(t\right)-2e^{4t}u(-t)$

$\text{On comparison,}\beta =-2\text{.}$