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<math>a_0=\frac{1}{4}\sum_{n=0}^{3}(-1)^nx[n]\,</math> | <math>a_0=\frac{1}{4}\sum_{n=0}^{3}(-1)^nx[n]\,</math> | ||
| − | <math>a_0=\frac{1}{ | + | <math>a_0=\frac{1}{4}\sum_{n=0}^{3}x[n]e^{-jn\pi}\,</math> |
| + | |||
| + | <math>a_0=\frac{5}{4}</math> | ||
Revision as of 10:05, 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 the 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 the 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} $
From the fourth information, we can find $ a_1\, $:
$ a_0=\frac{1}{4}\sum_{n=0}^{3}(-1)^nx[n]\, $
$ a_0=\frac{1}{4}\sum_{n=0}^{3}x[n]e^{-jn\pi}\, $
$ a_0=\frac{5}{4} $
