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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). In addition, I will provide an alternative view of the FFT calculation path as described in Week 2 of Lab 6 in ECE 438. A link can be found here [https://engineering.purdue.edu/VISE/ee438L/lab6/pdf/lab6b.pdf]. Please note the following explanation of the FFT will use the "divide and conquer" method." | |
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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^{-j2{\pi}kn/N} </math> | <math>X[k] = \sum_{n=0}^{N-1} x[n] e^{-j2{\pi}kn/N} </math> | ||
Revision as of 09:00, 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). In addition, I will provide an alternative view of the FFT calculation path as described in Week 2 of Lab 6 in ECE 438. A link can be found here [1]. 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{\pi}kn/N} $
