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| − | = Practice Question on Computing the Output of an LTI system by Convolution = | + | = [[: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 system is | The unit impulse response h(t) of a DT LTI system is | ||
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| − | == Share your answers below == | + | == Share your answers below == |
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! | 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! | ||
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| − | === 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} | <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} | ||
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:*<span style="color: green;"> Instructor's comments: Great! That one was harder then the [[Output of LTI CT system by convolution no2 ECE301S11|previous practice problem]], and you still got it right. You should do very well on the test! -pm </span> | :*<span style="color: green;"> Instructor's comments: Great! That one was harder then the [[Output of LTI CT system by convolution no2 ECE301S11|previous practice problem]], and you still got it right. You should do very well on the test! -pm </span> | ||
| − | === Answer 2 === | + | === Answer 2 === |
Write it here. | Write it here. | ||
| − | === Answer 3 === | + | === Answer 3 === |
Write it here. | Write it here. | ||
Latest revision as of 10:24, 11 November 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)
- 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
Answer 2
Write it here.
Answer 3
Write it here.
