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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.  <math> </math>
+
3.  for the given value of k, <math>e^{jk2\pi} = 1</math>, then that <math>a_{k}</math>  
  
  

Revision as of 16:16, 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^{jk2\pi} = 1 $, then that $ a_{k} $


Solution

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

Alumni Liaison

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

Buyue Zhang