Defining equation for the DFT and assume that N is even, so we have
<math>X(k) = \sum_{n=0}^{N-1} x(n)e^{-j{\frac{2{\pi}kn}{N}}}<\math>
Defining equation for the DFT and assume that N is even, so we have
<math>X(k) = \sum_{n=0}^{N-1} x(n)e^{-j{\frac{2{\pi}kn}{N}}}<\math>