Line 2: | Line 2: | ||
---- | ---- | ||
− | a. | + | a.<math>x[n]=cos(\frac{\pi n}{2})\text{ ,n=0,1,2,3}</math> |
<math> | <math> | ||
Line 12: | Line 12: | ||
</math> | </math> | ||
− | b. | + | b. <math>h[n]=2^n\text{ ,n=0,1,2,3}</math> |
− | + | ||
+ | <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> | ||
---- | ---- | ||
[[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} $