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'''Comparison of the DFT and FFT via Matrices'''
 
'''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.  
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<text> The purpose of this article is to illustrate the differences of the Discrete Fourier Transform (DFT) versus the Fast Fourier Transform (FFT) as well as provide an alternative . Please note, the following explanation of the FFT will use the "divide and conquer" method. </text>
  
 
     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^{-j2{\pi}kn/N}  </math>
 
<math>X[k] = \sum_{n=0}^{N-1} x[n] e^{-j2{\pi}kn/N}  </math>

Revision as of 09:54, 26 October 2013

Comparison of the DFT and FFT via Matrices


<text> The purpose of this article is to illustrate the differences of the Discrete Fourier Transform (DFT) versus the Fast Fourier Transform (FFT) as well as provide an alternative . Please note, the following explanation of the FFT will use the "divide and conquer" method. </text>

   To start, we will define the DFT as, 

$ X[k] = \sum_{n=0}^{N-1} x[n] e^{-j2{\pi}kn/N} $

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Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood