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[[Category:ECE301 S11 Exam 3 more practice]]
 
[[Category:ECE301 S11 Exam 3 more practice]]
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[[Category:ECE301Spring2011Boutin]]
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[[Category:Problem_solving]]
  
 
= Problem =
 
= Problem =
<math>\begin{align} x(t) &= u(t) - u(t-1) \\ y(t) &= u(t+2) - u(t-2) \\ z(t) &= x(t) * y(t) \end{align}</math>
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Compute the convolution
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<math>z(t)=x(t)*y(t) \ </math>
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between
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<math> x(t) = u(t) - u(t-1) \ </math>
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and
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<math>y(t) = u(t+2) - u(t-2) \ </math>.
  
 
= Solution =
 
= Solution =
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\end{align}
 
\end{align}
 
</math>
 
</math>
 
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---
 
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==Comments==
 
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Write them here.
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----
 
[[ ECE301 S11 Exam 3 more practice|Back to ECE301 S11 Exam 3 more practice]]
 
[[ ECE301 S11 Exam 3 more practice|Back to ECE301 S11 Exam 3 more practice]]

Revision as of 11:40, 29 April 2011


Problem

Compute the convolution

$ z(t)=x(t)*y(t) \ $

between

$ x(t) = u(t) - u(t-1) \ $

and

$ y(t) = u(t+2) - u(t-2) \ $.

Solution

$ \begin{align} z(t) &= y(t) * x(t) \\ &= \int_{-\infty}^{\infty}x(\tau) y(t-\tau)\mathrm{d}\tau \\ &= \int_{-\infty}^{\infty}\left( u(\tau) - u(\tau-1) \right) \left( u(t-\tau+2) - u(t-\tau-2) \right)\mathrm{d}\tau \\ &= \int_{0}^{1}\left( u(t-\tau+2) - u(t-\tau-2) \right)\mathrm{d}\tau \\ &= \left[ (t-\tau+2)u(t-\tau+2)\right]\big|_0^1 - \left[ (t-\tau-2)u(t-\tau-2)\right]\big|_0^1 \\ &= \left[ (t+1)u(t+1) - (t+2)u(t+2)\right] + \left[ -(t-3)u(t-3) + (t-2)u(t-2)\right] \end{align} $ ---

Comments

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Back to ECE301 S11 Exam 3 more practice

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