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<math>x[n]=\sum_{n=0}^{3}a_ke^{jk\pi n}\,</math> | <math>x[n]=\sum_{n=0}^{3}a_ke^{jk\pi n}\,</math> | ||
| + | |||
| + | From third information, we can find <math>a_0\,</math>: | ||
| + | |||
| + | <math>a_0=\frac{1}{4}\sum_{n=0}^{3}x[n]=\frac{1}{4}2 = \frac{1}{2}</math> | ||
Revision as of 09:59, 26 September 2008
Informations
1. $ N = 2\, $
2. $ a_k = 0\, $ for all |k|>1
3. $ \sum_{n=0}^{3}x[n]=2 $
4. $ \sum_{n=0}^{3}(-1)^nx[n]=5 $
Inspections
From first information, we can directly subtitute N into:
$ x[n]=\sum_{n=0}^{3}a_ke^{jk(2\pi/2)n}\, $
$ x[n]=\sum_{n=0}^{3}a_ke^{jk\pi n}\, $
From third information, we can find $ a_0\, $:
$ a_0=\frac{1}{4}\sum_{n=0}^{3}x[n]=\frac{1}{4}2 = \frac{1}{2} $
