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a.
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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.  
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b. <math>h[n]=2^n\text{ ,n=0,1,2,3}</math>
 
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<math>
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\begin{align}
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H[k]&=\sum_{n=0}^3 h[n]e^{-j\frac{2\pi n}{N}k} \\
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&=\sum_{n=0}^3 2^ne^{-j\frac{2\pi n}{N}k} \\
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&=1+2e^{-j\frac{\pi k}{4}}+4e^{-j\pi k}+8e^{-j\frac{3\pi k}{4}}\text{ , }0\le k\le 3
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\end{align}
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</math>
 
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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} $


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Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva