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| − | a. | + | a.<math>x[n]=cos(\frac{\pi n}{2})\text{ ,n=0,1,2,3}</math> |
<math> | <math> | ||
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</math> | </math> | ||
| − | b. | + | b. <math>h[n]=2^n\text{ ,n=0,1,2,3}</math> |
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| + | <math> | ||
| + | \begin{align} | ||
| + | H[k]&=\sum_{n=0}^3 h[n]e^{-j\frac{2\pi n}{N}k} \\ | ||
| + | &=\sum_{n=0}^3 2^ne^{-j\frac{2\pi n}{N}k} \\ | ||
| + | &=1+2e^{-j\frac{\pi k}{4}}+4e^{-j\pi k}+8e^{-j\frac{3\pi k}{4}}\text{ , }0\le k\le 3 | ||
| + | \end{align} | ||
| + | </math> | ||
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[[ECE438_Week12_Quiz|Back to Quiz Pool]] | [[ECE438_Week12_Quiz|Back to Quiz Pool]] | ||
[[ECE438_Lab_Fall_2010|Back to Lab wiki]] | [[ECE438_Lab_Fall_2010|Back to Lab wiki]] | ||
Revision as of 13:49, 10 November 2010
Solution of Week12 Quiz Question 5
a.$ x[n]=cos(\frac{\pi n}{2})\text{ ,n=0,1,2,3} $
$ \begin{align} X[k]&=\sum_{n=0}^3 x[n]e^{-j\frac{2\pi n}{N}k} \\ &=\sum_{n=0}^3 cos(\frac{n\pi}{2})e^{-j\frac{2\pi n}{N}k} \\ &=1-e^{-j\pi k}\text{ , }0\le k\le 3 \end{align} $
b. $ h[n]=2^n\text{ ,n=0,1,2,3} $
$ \begin{align} H[k]&=\sum_{n=0}^3 h[n]e^{-j\frac{2\pi n}{N}k} \\ &=\sum_{n=0}^3 2^ne^{-j\frac{2\pi n}{N}k} \\ &=1+2e^{-j\frac{\pi k}{4}}+4e^{-j\pi k}+8e^{-j\frac{3\pi k}{4}}\text{ , }0\le k\le 3 \end{align} $
