(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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= [[:Category:Problem_solving|Practice Question]] on Computing the Output of an LTI system by Convolution =
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= Practice Question on Computing the Output of an LTI system by Convolution=
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The unit impulse response h(t) of a DT LTI system is  
 
The unit impulse response h(t) of a DT LTI system is  
  
<math>h(t)= u( -t+1 ) \ </math>
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<math>h(t)= u( -t+1 ) \ </math>  
  
Use convolution to compute the system's response to the input
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Use convolution to compute the system's response to the input  
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<math>x(t)= e^{-2 t }u(t). \ </math>
  
<math>x(t)= e^{-2 t }u(t). \ </math>
 
 
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==Share your answers below==
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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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== Share your answers below ==
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 +
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===
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Write it here.
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=== Answer 1 ===
===Answer 2===
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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> <math> = \begin{cases}
===Answer 3===
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e^{-2t}\int_{-\infty}^t e^{2\tau}d\tau,  & \mbox{if }t \le 1 \\
Write it here.
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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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:*<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>
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=== Answer 2 ===
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Write it here.  
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=== Answer 3 ===
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Write it here.  
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[[2011_Spring_ECE_301_Boutin|Back to ECE301 Spring 2011 Prof. Boutin]]
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[[2011 Spring ECE 301 Boutin|Back to ECE301 Spring 2011 Prof. Boutin]]
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[[Category:ECE301Spring2011Boutin]] [[Category:Problem_solving]]

Latest revision as of 10:24, 11 November 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). \ $


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!


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.


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