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2.  <math>\sum_{k = 0}^{3}x[n] = (2 + j)</math>
 
2.  <math>\sum_{k = 0}^{3}x[n] = (2 + j)</math>
  
3.  for the given value of k, <math>e^{jk2\pi} = 1\,</math>, then that <math>a_{k} = \frac{1}{2}\,</math>  
+
3.  for the given value of k, <math>e^{jk\frac{2\pi}{N}} = 1\,</math>, then that <math>a_{k} = \frac{1}{2}\,</math>  
  
 
4.  All other <math>a_{k} = 0\,</math>  
 
4.  All other <math>a_{k} = 0\,</math>  
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<math>a_{0} = \frac{2 + j}{4}</math>
 
<math>a_{0} = \frac{2 + j}{4}</math>
  
<math>a_{0} = </math>
+
<math>a_{2} = \frac{1}{2}</math>

Revision as of 15:23, 26 September 2008

DT Signal:

1. Signal is periodic with N = 4

2. $ \sum_{k = 0}^{3}x[n] = (2 + j) $

3. for the given value of k, $ e^{jk\frac{2\pi}{N}} = 1\, $, then that $ a_{k} = \frac{1}{2}\, $

4. All other $ a_{k} = 0\, $


Solution

$ a_{0} = \frac{2 + j}{4} $

$ a_{2} = \frac{1}{2} $

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang