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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^{ | + | 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_{ | + | <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} $