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 ______. Network Solution Methods gate2015-ec-2 numerical-answers network-solution-methods laplace-transform + – Milicevic3306 asked Mar 27, 2018 • retagged Nov 16, 2020 by soujanyareddy13 Milicevic3306 16.0k points 185 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.