retagged by
185 views
0 votes
0 votes
Let $x(t)  = \alpha s(t) + s(-t)$ with $s(t)  =  \beta e^{-4t}u(t),$ where $u(t)$ is unit step function. If the bilateral Laplace transform of $x(t)$ is $$X(s) = \dfrac{16}{s^{2} – 16}\:\: -4 < Re\{s\}<4;$$

then the value of $\beta$ is ______.
retagged by

Please log in or register to answer this question.

Answer: