(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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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} $
