Topic: Signals and Systems
Laplace transform of $e^{-\alpha t}f(t)$ is
- A. $F(s)e^{\alpha t}$
- B. $F(s-\alpha)$
- C. $F(s+\alpha)$
- D. $\frac{F(s)}{s}+\alpha$
Answer: C. $F(s+\alpha)$
Explanation: $x(t)=e^{-\alpha t}f(t)$.
$X(s)=\int_{-\infty}^{\infty} e^{-\alpha t}f(t)e^{-st} dt=\int_{-\infty}^{\infty} f(t)e^{-(s+\alpha)t} dt=F(s+\alpha)$.