The impulse response of an LTI system is $h(t)=e^{-3t} u(t)$. If the output of the system is $y(t)=e^{-3t} u(t)-e^{-5t} u(t)$, then the input $x(t)$ is given by

GS&EI ElectEngg 14 2024 1601-2025 The relationship between input $x(t)$, output $y(t) $, and impulse response $h(t)$ of an LTI system is $y(t) = x(t) h(t)$. In the Laplace domain, this is $Y(s) = X(s)H(s)$. If the intended output was $y(t) = te^{-3t}u(t)$, its Laplace Transform is $Y(s) = \frac{1}{(s+3)^2}$. Given impulse response $h(t) = e^{-3t}u(t)$, its Laplace Transform is $H(s) = \frac{1}{s+3}$. Then, $X(s) = \frac{Y(s)}{H(s)} = \frac{1/(s+3)^2}{1/(s+3)} = \frac{1}{s+3}$. The inverse Laplace transform of $X(s) = \frac{1} {s+3}$ is $x(t) = e^{-3t}u(t)$, which is Option A.