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Let us look for the Fourier series coefficients for the DT signal <math>x[n] = cos(2\pi n)</math> | Let us look for the Fourier series coefficients for the DT signal <math>x[n] = cos(2\pi n)</math> | ||
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| + | <math>x[n] = cos(2\pi n) = \frac{e^{j2\pin}+e^{-j2\pin}}{2}</math> | ||
Revision as of 18:07, 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(2\pi n) $
$ x[n] = cos(2\pi n) = \frac{e^{j2\pin}+e^{-j2\pin}}{2} $
