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}\, $

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva