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<big><math>x[n] = 1e^{j\pi n}</math></big> | <big><math>x[n] = 1e^{j\pi n}</math></big> | ||
| − | + | <math>a_1 = \frac{1}{2}\displaystyle\sum_{n=0}^{2}e^{j\pi n}e^{-j\pi n} = 1 </math> | |
| − | + | <math>a_0 = \frac{1}{2}</math> | |
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Revision as of 17:49, 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} $
$ a_1 = \frac{1}{2}\displaystyle\sum_{n=0}^{2}e^{j\pi n}e^{-j\pi n} = 1 $
$ a_0 = \frac{1}{2} $
