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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. <math>a_{1} = a{2}\,</math> |
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| + | 4. for the given value of k, <math>e^{jk\frac{2\pi}{N}} = -1\,</math>, then that <math>a_{k} = \frac{1}{2}\,</math> | ||
| + | |||
| + | 5. 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_{1} = \frac{1}{2}</math> | ||
| + | |||
| + | <math>a_{2} = \frac{1}{2}</math> | ||
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| + | <math>a_{3} = 0</math> | ||
| + | |||
| + | <math>x[n] = \frac{2 + j}{4} + \frac{1}{2}e^{j\frac{\pi}{2}n} + \frac{1}{2}e^{j\pi n}</math> | ||
Latest revision as of 15:33, 26 September 2008
DT Signal:
1. Signal is periodic with N = 4
2. $ \sum_{k = 0}^{3}x[n] = (2 + j) $
3. $ a_{1} = a{2}\, $
4. for the given value of k, $ e^{jk\frac{2\pi}{N}} = -1\, $, then that $ a_{k} = \frac{1}{2}\, $
5. All other $ a_{k} = 0\, $
Solution
$ a_{0} = \frac{2 + j}{4} $
$ a_{1} = \frac{1}{2} $
$ a_{2} = \frac{1}{2} $
$ a_{3} = 0 $
$ x[n] = \frac{2 + j}{4} + \frac{1}{2}e^{j\frac{\pi}{2}n} + \frac{1}{2}e^{j\pi n} $
