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To start, we will define the DFT as, | To start, we will define the DFT as, | ||
| − | <math>X[k] = \sum_{n=0}^{N-1} x[n] e^{- | + | <math>X[k] = \sum_{n=0}^{N-1} x[n] e^{-j2\pikn/N} </math> |
Revision as of 08:37, 26 October 2013
Comparison of the DFT and FFT via Matrices
The purpose of this article is to illustrate the differences of the Discrete Fourier Transform (DFT) versus the Fast Fourier Transform (FFT). Please note, the following explanation of the FFT will use the "divide and conquer" method.
To start, we will define the DFT as,
$ X[k] = \sum_{n=0}^{N-1} x[n] e^{-j2\pikn/N} $
x[n] = n2(u[n + 3] − u[n − 1])
x[n] = n2(δ(n + 3) + δ(n + 2) + δ(n + 1) + δ(n))
$ X(z) = \sum_{n=-\infty}^{+\infty} x[n] z^{-n} $
$ X(z) = \sum_{n=-\infty}^{+\infty} n^2(\delta(n+3)+\delta(n+2)+\delta(n+1)+\delta(n)) z^{-n} $
X(z) = 9z3 + 4z2 + z + 1 for all z in complex plane
