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To derive the FFT we will use the "divide and conquer" method as stated before. This means that we will divide Eq. 1 into two separate equations. The first equation processes the even components of x[n] while the second equation processes the odd components of x[n]. This produces, | To derive the FFT we will use the "divide and conquer" method as stated before. This means that we will divide Eq. 1 into two separate equations. The first equation processes the even components of x[n] while the second equation processes the odd components of x[n]. This produces, | ||
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e^{-j2{\pi}k/N}\sum_{m=0}^{{N/2}-1} x[2m+1] e^{-j2{\pi}k(m)/(N/2)} | e^{-j2{\pi}k/N}\sum_{m=0}^{{N/2}-1} x[2m+1] e^{-j2{\pi}k(m)/(N/2)} | ||
| − | {\;} {\;} (Eq. | + | {\;} {\;} (Eq. 3) |
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| − | Our DFT has now been successfully split into two N/2 pt DFT's. For simplification purposes, the first summation will be defined as X0[k] and the second summation as X1[k]. We can now simplify Eq. | + | Our DFT has now been successfully split into two N/2 pt DFT's. For simplification purposes, the first summation will be defined as X0[k] and the second summation as X1[k]. We can now simplify Eq. 3 to the following form. |
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| − | By definition of the discrete fourier transform, X0[k] and X1[k] are periodic with period N/2. | + | By definition of the discrete fourier transform, X0[k] and X1[k] are periodic with period N/2. Therefore we can split Eq. 3 into two separate equations. |
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Revision as of 16:35, 26 October 2013
Comparison of the DFT and FFT via Matrices
BY: Cary A. Wood
The purpose of this article is to illustrate the differences of the Discrete Fourier Transform (DFT) versus the Fast Fourier Transform (FFT). In addition, it will provide an alternative view of the FFT calculation path as described in Lab 6 (Week 2, pg. 5) of ECE 438. The link can be found here [1]. Please note the following explanation of the FFT will utilize the "divide and conquer" method and assume the user understands its basic concepts.
To start, we will define the DFT as,
$ X_N[k] = \sum_{n=0}^{N-1} x[n] e^{-j2{\pi}kn/N} {\;} {\;} (Eq. 1) $
It is fairly easy to visualize this 1 point DFT, but how does it look when x[n] has 8 points, 1024 points, etc. That's where matrices come in. For an N point DFT, we will define our input as x[n] where n = 0, 1, 2, ... N-1. Similarly, the output will be defined as X[k] where k = 0, 1, 2, ... N-1. Referring to our definition of the Discrete Fourier Transform above, to compute an N point DFT, all we need to do is simply repeat Eq. 1 N times. For every value of x[n] in the discrete time domain, there is a corresponding value, X[k].
Input x[n]
$ \begin{bmatrix} x[0] & x[1] & {\;}{\dotsb} & x[N-1] \end{bmatrix} $
Output X[k]
$ \begin{bmatrix} X[0] \\ X[1] \\ {\vdots} \\ X[N-1] \end{bmatrix} $
To solve for X[K], means simply repeating Eq. 1, N times, where x[n] is a real scalar value for each entry. We represent this in the matrices below.
$ \begin{bmatrix} X[0] \\ X[1] \\ X[2] \\ {\vdots} \\ X[N-1] \end{bmatrix} = x[0]\begin{bmatrix} e^{-j2{\pi}(0)n/N} \\ e^{-j2{\pi}(0)n/N} \\ e^{-j2{\pi}(0)n/N} \\ {\vdots} \\ e^{-j2{\pi}(0)n/N} \end{bmatrix} + x[1]\begin{bmatrix} e^{-j2{\pi}(0)/N} \\ e^{-j2{\pi}(1)/N} \\ e^{-j2{\pi}(2)/N} \\ {\vdots} \\ e^{-j2{\pi}(N-1)/N} \end{bmatrix} + x[2]\begin{bmatrix} e^{-j2{\pi}(0)/N} \\ e^{-j2{\pi}(2)/N} \\ e^{-j2{\pi}(4)/N} \\ {\vdots} \\ e^{-j2{\pi}2(N-1)/N} \end{bmatrix} + {\dotsb} + x[N-1]\begin{bmatrix} e^{-j2{\pi}(0)/N} \\ e^{-j2{\pi}(N-1)/N} \\ e^{-j2{\pi}2(N-1)/N} \\ {\vdots} \\ e^{-j2{\pi}{(N-1)^2}/N} \end{bmatrix} {\;} {\;} (Eq. 2) $
From this matrix representation of the DFT, you can see that N^2 complex multiplications and N^2 - N complex additions are necessary to fully compute the discrete fourier transform. There are a few simplifications that can be made right away. The first column vector of complex exponentials can be reduced to a vector of 1's. This is possible because, e^(0*anything) will always equal 1. This also applies for the first entry in each vector of complex exponentials because n always begins at 0.
Whats happens when we want to perform the DFT on 3 minute audio signal recorded at a 44.1 kHz sampling. Our handy-dandy DFT suddenly becomes extremely lengthy since computation time increases with a rate of N^2. Due to the work of Cooley and Turkey and previously developed by Gauss, the development of the Fast Fourier Transform has reduced computation time by orders of magnitude.
To derive the FFT we will use the "divide and conquer" method as stated before. This means that we will divide Eq. 1 into two separate equations. The first equation processes the even components of x[n] while the second equation processes the odd components of x[n]. This produces,
$ X_N[k] = \sum_{m=0}^{{N/2}-1} x[2m] e^{-j2{\pi}k(m)/(N/2)} + e^{-j2{\pi}k/N}\sum_{m=0}^{{N/2}-1} x[2m+1] e^{-j2{\pi}k(m)/(N/2)} {\;} {\;} (Eq. 3) $
Our DFT has now been successfully split into two N/2 pt DFT's. For simplification purposes, the first summation will be defined as X0[k] and the second summation as X1[k]. We can now simplify Eq. 3 to the following form.
$ X[k] = X_0[k] + e^{-j2{\pi}k/n}X_1[k] $
The complex exponential preceding X1[k] is also known as the "twiddle factor" and represented by
$ {W_N}^k = e^{-j2{\pi}k/N} $
By definition of the discrete fourier transform, X0[k] and X1[k] are periodic with period N/2. Therefore we can split Eq. 3 into two separate equations.
$ X[k] = X_0[k] + {{W_N}^k}X_1[k] $
$ X[k] = X_0[k] - {{W_N}^k}X_1[k] $
