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<big><math>x[n] = 1e^{j\pi n}</math></big> | <big><math>x[n] = 1e^{j\pi n}</math></big> | ||
| + | |||
| + | Compare to <math>x[n]=\displaystyle\sum_{k=0}^{n-1}a_ke^{jk\frac{2\pi}{N}n}</math> | ||
| + | |||
| + | <big><math>k\frac{2\pi}{N} = \pi n</math></big> | ||
| + | |||
| + | <math>N = 2k</math> where k is the smallest integer that N is an integer. | ||
| + | |||
| + | So k = 1, N = 2. | ||
Revision as of 18:39, 23 September 2008
For periodic DT signal, x[n] with fundamental period N:
$ x[n]=\displaystyle\sum_{k=0}^{n-1}a_ke^{jk\frac{2\pi}{N}n} $
The Fourier series coefficients can be calculated with:
$ a_k = \frac{1}{N}\displaystyle\sum_{n=0}^{N-1}x[n]e^{-jk\frac{2\pi}{N}n} $
Let us look for the Fourier series coefficients for the DT signal $ x[n] = cos(3\pi n) $
$ x[n] = cos(5\pi n) = \frac{e^{j5\pi n}+e^{-j5\pi n}}{2} = \frac{1}{2}e^{j4\pi n}e^{j\pi n} + \frac{1}{2}e^{-j4\pi n}e^{-j\pi n} = \frac{1}{2}e^{j\pi n} + \frac{1}{2}e^{-j\pi n} = \frac{1}{2}e^{j\pi n} + \frac{1}{2}e^{j\pi n} $
Finally,
$ x[n] = 1e^{j\pi n} $
Compare to $ x[n]=\displaystyle\sum_{k=0}^{n-1}a_ke^{jk\frac{2\pi}{N}n} $
$ k\frac{2\pi}{N} = \pi n $
$ N = 2k $ where k is the smallest integer that N is an integer.
So k = 1, N = 2.
