| Line 23: | Line 23: | ||
From the fourth information, we can find <math>a_1\,</math>: | From the fourth information, we can find <math>a_1\,</math>: | ||
| − | <math>a_0=\frac{1}{4}\sum_{n=0}^{3}(-1)^nx[n] | + | <math>a_0=\frac{1}{4}\sum_{n=0}^{3}(-1)^nx[n]\,</math> |
| − | <math>a_0=\frac{1}{4} 2 = \frac{1}{2}</math> | + | <math>a_0=\frac{1}{4} 2 = \frac{1}{2}\,</math> |
Revision as of 11:02, 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} 2 = \frac{1}{2}\, $
