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| − | [[Category:ECE301 S11 Exam 3 more practice]][[Category: | + | [[Category:ECE301 S11 Exam 3 more practice]] |
| + | [[Category:ECE301Spring2011Boutin]] | ||
| + | [[Category:Problem_solving]] | ||
| − | = | + | = Problem = |
| + | Compute the convolution | ||
| + | <math>z[n]=x[n]*y[n] \ </math> | ||
| + | between | ||
| − | + | <math> x[t] = u[n] - u[n-1] \ </math> | |
| − | + | ||
| − | + | and | |
| + | <math> y[t] = u[n+2] - u[n-2] \ </math>. | ||
| + | = My Solution= | ||
| + | <math> | ||
| + | \begin{align} | ||
| + | z[n] &= x[n]*y[n]\\ | ||
| + | &= \sum_{k=-\infty}^{\infty} x[k]y[n-k]\\ | ||
| + | &= \sum_{k=-\infty}^{\infty} (u[k] - u[k-1])(u[n-k+2] - u[n-k-2]) \\ | ||
| + | &= \sum_{k=0}^{1} (u[n-k+2] - u[n-k-2]) \\ | ||
| + | &= \begin{cases} | ||
| + | u[n+2] - u[n-2] + u[n+1] - u[n-3] ,& 0 \leq n \leq 1 \\ | ||
| + | 0 ,& \text{else} | ||
| + | \end{cases}\\ | ||
| + | &= \begin{cases} | ||
| + | 2 ,& n = 0 \\ | ||
| + | 2 ,& n = 1 \\ | ||
| + | 0 ,& \text{else} | ||
| + | \end{cases} | ||
| + | \end{align} | ||
| + | </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]] | ||
Latest revision as of 07:53, 6 May 2011
Problem
Compute the convolution
$ z[n]=x[n]*y[n] \ $
between
$ x[t] = u[n] - u[n-1] \ $
and
$ y[t] = u[n+2] - u[n-2] \ $.
My Solution
$ \begin{align} z[n] &= x[n]*y[n]\\ &= \sum_{k=-\infty}^{\infty} x[k]y[n-k]\\ &= \sum_{k=-\infty}^{\infty} (u[k] - u[k-1])(u[n-k+2] - u[n-k-2]) \\ &= \sum_{k=0}^{1} (u[n-k+2] - u[n-k-2]) \\ &= \begin{cases} u[n+2] - u[n-2] + u[n+1] - u[n-3] ,& 0 \leq n \leq 1 \\ 0 ,& \text{else} \end{cases}\\ &= \begin{cases} 2 ,& n = 0 \\ 2 ,& n = 1 \\ 0 ,& \text{else} \end{cases} \end{align} $
Comments
Write them here.
