# Practice Question on Computing the Output of an LTI system by Convolution

The unit impulse response h(t) of a CT LTI system is

$h(t) = e^{-t} u(t+3). \$

Use convolution to compute the system's response to the input

$x(t)=u(t). \$

$y(t)=x(t)*h(t)=\int_{-\infty}^{\infty} u(t)e^{\tau-t}u(t-\tau+3) d\tau$

but

$u(\tau)= \begin{cases} 1, & \mbox{if }\tau \ge 0 \\ 0, & \mbox{if }\tau < 0 \end{cases}$

so

$y(t)=\int_{0}^{\infty} e^{-t} e^{\tau}u(t-\tau+3) d\tau=e^{-t} \int_{0}^{\infty} e^{\tau}u(t-\tau+3) d\tau$

but

$u(t-\tau+3)= \begin{cases} 1, & \mbox{if }(t-\tau+3) \ge 0 \\ 0, & \mbox{if }(t-\tau+3) < 0 \end{cases} = \begin{cases} 1, & \mbox{if }\tau \le t+3 \\ 0, & \mbox{if }\tau > t+3 \end{cases}$

so

$y(t)=\begin{cases} e^{-t} \int_{0}^{t+3} e^{\tau}d\tau, & \mbox{if }t > -3 \\ 0, & \mbox{if }t \le -3 \end{cases} = e^{-t}(e^{t+3}-1)u(t+3)$

$y(t)=(e^3-e^{-t})u(t+3)\,$

--Cmcmican 19:56, 28 January 2011 (UTC)

--Ahmadi 22:59, 29 January 2011 (UTC)
Instructor's comments: The above answer shows a lot of detail, as I did when presenting the class example. However, once you are comfortable with all the steps, it is ok to skip a few details. In particular, the following answer would get full credit on the exam.
\begin{align} y(t)&=x(t)*h(t),\\ & =\int_{-\infty}^{\infty} u(t)e^{\tau-t}u(t-\tau+3) d\tau,\\ & =\int_{0}^{\infty} e^{-t} e^{\tau}u(t-\tau+3) d\tau,\\ & =e^{-t} \int_{0}^{\infty} e^{\tau}u(t-\tau+3) d\tau, \\ &=\begin{cases} e^{-t} \int_{0}^{t+3} e^{\tau}d\tau, & \mbox{if }t > -3 \\ 0, & \mbox{if }t \le -3 \end{cases}, \\ &= e^{-t}(e^{t+3}-1)u(t+3). \end{align}