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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 ______.