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[[Category:ECE301 S11 Exam 3 more practice]] | [[Category:ECE301 S11 Exam 3 more practice]] | ||
+ | [[Category:ECE301Spring2011Boutin]] | ||
+ | [[Category:Problem_solving]] | ||
= Problem = | = Problem = | ||
− | <math> | + | Compute the convolution |
+ | |||
+ | <math>z(t)=x(t)*y(t) \ </math> | ||
+ | |||
+ | between | ||
+ | |||
+ | <math> x(t) = u(t) - u(t-1) \ </math> | ||
+ | |||
+ | and | ||
+ | |||
+ | <math>y(t) = u(t+2) - u(t-2) \ </math>. | ||
= Solution = | = Solution = | ||
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\end{align} | \end{align} | ||
</math> | </math> | ||
− | + | --- | |
− | + | ==Comments== | |
− | + | Write them here. | |
+ | ---- | ||
[[ 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 10:40, 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
Write them here.