(New page: Category:ECE301Spring2011Boutin Category:problem solving = Practice Question on Computing the Output of an LTI system by Convolution= The unit impulse response h(t) of a DT LTI sys...) |
|||
| Line 14: | Line 14: | ||
---- | ---- | ||
===Answer 1=== | ===Answer 1=== | ||
| − | + | ||
| + | <math>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</math> | ||
| + | <math> = \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}</math> | ||
| + | |||
| + | <math>y(t)=\frac{1}{2}\Bigg(u(1-t)+e^{-2(t-1)}u(t-1)\Bigg)</math> | ||
| + | |||
| + | --[[User:Cmcmican|Cmcmican]] 21:19, 4 February 2011 (UTC) | ||
| + | |||
===Answer 2=== | ===Answer 2=== | ||
Write it here. | Write it here. | ||
Revision as of 17:19, 4 February 2011
Contents
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!
Answer 1
$ 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)
Answer 2
Write it here.
Answer 3
Write it here.
