(New page: We are given the following information about a signal x[n]: 1. x[n] is periodic with period 3 2. <math>\sum^{2}_{n = 0} x[n] = 5</math> 3. <math>\sum^{2}_{n = 0} (-1)^{n} x[n] = 15</mat...)
 
 
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3. <math>\sum^{2}_{n = 0} (-1)^{n} x[n] = 15</math>
 
3. <math>\sum^{2}_{n = 0} (-1)^{n} x[n] = 15</math>
  
4. x[n] has minimum power among all signals that satisfy 1,2,3.
 
  
 
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Latest revision as of 10:58, 26 September 2008

We are given the following information about a signal x[n]:

1. x[n] is periodic with period 3

2. $ \sum^{2}_{n = 0} x[n] = 5 $

3. $ \sum^{2}_{n = 0} (-1)^{n} x[n] = 15 $



1. => $ x[n] = \frac{1}{3} \sum^{2}_{n = 0} a_k e^{j k \frac{2\pi}{3} n} $

2. => $ x[n] = \frac{5}{3} = a_0 $

3. => $ a_1 = \frac{1}{3}(15) = 5 $

$ a_2 = \frac{-1}{3}(15) = -5 $


Solution = $ \frac{5}{3} + 5e^{j \frac{2\pi}{3} t} - 5e^{j \frac{2\pi}{3} t} $

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