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

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

$h(t)= u( -t+1 ) \$

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

$x(t)= e^{-2 t }u(t). \$

You will receive feedback from your instructor and TA directly on this page. Other students are welcome to comment/discuss/point out mistakes/ask questions too!

$y(t)=h(t)*x(t)=\int_{-\infty}^\infty u(-\tau+1)e^{-2(t-\tau)}u(t-\tau)d\tau=e^{-2t}\int_{-\infty}^1 e^{2\tau}u(t-\tau)d\tau$ $= \begin{cases} e^{-2t}\int_{-\infty}^t e^{2\tau}d\tau, & \mbox{if }t \le 1 \\ e^{-2t}\int_{-\infty}^1 e^{2\tau}d\tau, & \mbox{if }t > 1 \end{cases} =\begin{cases} e^{-2t}\frac{e^{2t}}{2}, & \mbox{if }t \le 1 \\ e^{-2t}\frac{e^{2}}{2}, & \mbox{if }t > 1 \end{cases}$

$y(t)=\frac{1}{2}\Bigg(u(1-t)+e^{-2(t-1)}u(t-1)\Bigg)$

--Cmcmican 21:19, 4 February 2011 (UTC)

• Instructor's comments: Great! That one was harder then the previous practice problem, and you still got it right. You should do very well on the test! -pm 