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| − | =Informations= | + | ==Informations== |
1. <math>N = 2\,</math> | 1. <math>N = 2\,</math> | ||
| − | 2. | + | 2. <math>a_k = 0\,</math> for all |k|>1 |
| + | |||
| + | 3. <math>\sum_{n=0}^{3}x[n]=2</math> | ||
| + | |||
| + | 4. <math>\sum_{n=0}^{3}(-1)^nx[n]=5</math> | ||
| + | |||
| + | ==Inspections== | ||
| + | |||
| + | From the first information, we can directly subtitute N into: | ||
| + | |||
| + | <math>x[n]=\sum_{n=0}^{3}a_ke^{jk(2\pi/2)n}\,</math> | ||
| + | |||
| + | <math>x[n]=\sum_{n=0}^{3}a_ke^{jk\pi n}\,</math> | ||
| + | |||
| + | From the 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> | ||
| + | |||
| + | 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> | ||
| + | |||
| + | <math>a_0=\frac{1}{4}\sum_{n=0}^{3}x[n]e^{-jn\pi}\,</math> | ||
| + | |||
| + | <math>a_0=\frac{5}{4}</math> | ||
| + | |||
| + | Which says that the function is: | ||
| + | |||
| + | |||
| + | <math>x[n]=\frac{1}{2}+\frac{5}{4}e^{jn\pi}\,</math> | ||
Latest revision as of 10:06, 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} $
Which says that the function is:
$ x[n]=\frac{1}{2}+\frac{5}{4}e^{jn\pi}\, $
