(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...)
 
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===Answer 1===
 
===Answer 1===
Write it here.
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<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>
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<math> = \begin{cases}
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e^{-2t}\int_{-\infty}^t e^{2\tau}d\tau,  & \mbox{if }t \le 1 \\
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e^{-2t}\int_{-\infty}^1 e^{2\tau}d\tau, & \mbox{if }t > 1
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\end{cases}
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=\begin{cases}
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e^{-2t}\frac{e^{2t}}{2},  & \mbox{if }t \le 1 \\
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e^{-2t}\frac{e^{2}}{2}, & \mbox{if }t > 1
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\end{cases}</math>
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<math>y(t)=\frac{1}{2}\Bigg(u(1-t)+e^{-2(t-1)}u(t-1)\Bigg)</math>
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--[[User:Cmcmican|Cmcmican]] 21:19, 4 February 2011 (UTC)
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===Answer 2===
 
===Answer 2===
 
Write it here.
 
Write it here.

Revision as of 17:19, 4 February 2011

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). \ $


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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.


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