Question

Let $x_1(t)=e^{-t}u(t)$ and $x_2(t)=u(t)-u(t-2)$. If $y(t)$ is the convolution $x_1(t)x_2(t)$, find \[ \lim_{t\to\infty} y(t) \] (rounded to one decimal place).

Hint:

If a convolution contains a factor of $e^{-t}$, its value always goes to zero as $t\to\infty$.

Answer 1

Correct Answer: 0

\[ y(t) = \int_0^{2} e^{-(t-\tau)}\, d\tau = e^{-t}\int_0^{2} e^{\tau} d\tau \] Compute the integral: \[ \int_0^{2} e^{\tau} d\tau = e^2 - 1. \] Therefore: \[ y(t) = (e^2 - 1)e^{-t}. \] Take the limit as \( t\to\infty \): \[ \lim_{t\to\infty} y(t) = (e^2 - 1)\lim_{t\to\infty} e^{-t} = 0. \] Thus: \[ \boxed{0.0} \]