Revision as of 16:14, 26 September 2008

Fourier series coefficients for DT signal

DT Signal

$ x[n]=sin(3 \pi n)\, $

Fourier series coefficients

To find N:

$ N=\frac{2\pi}{3\pi} \times k $ where k is the smallest integer that makes this be an integer.

$ \therefore k=3\ and\ N=2 $

$ \, x[0]= 1 $

$ \, x[1]=-1 $

$ \, x[2]=1 $

$ \, x[3]=-1 $

$ \, a_k=\frac{1}{N}\sum_{n=0}^{N-1} x[n] e^{-jk\frac{2\pi}{N} n} \rightarrow a_k=\frac{1}{2}\sum_{n=0}^{1} x[n] e^{-jk\pi n} $

$ \, a_0=\frac{1+(-1)}{2}=0 $

$ \, a_1=\frac{1}{2}\sum_{n=0}^{1} x[n] e^{-jk\pi n} $

$ \, a_1=\frac{1}{2}(x[0]e^{0}+x[1]e^{-j\pi})=\frac{1}{2}(1 \cdot 1+(-1)(-1)=1 $

$ \, \therefore \ a_o=0, a_1 =1 $