# Implementation of the Divide and Conquer DFT via Matrices

By Cary Wood

The purpose of this article is to illustrate the differences of the Discrete Fourier Transform (DFT) versus the Divide and Conquer Discrete Fourier Transform (dcDFT). In addition, it will provide a further explanation of the dcDFT calculation path as described in Lab 6 (Week 2, pg. 5) of ECE 438. Please note the following explanation assumes the user understands the basic concepts of the Discrete Fourier Transform.

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, 256 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], in the frequency domain.

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.

$\begin{bmatrix} X[0] \\ X[1] \\ X[2] \\ {\vdots} \\ X[N-1] \end{bmatrix} = x[0]\begin{bmatrix} e^{-j2{\pi}(0)/N} \\ e^{-j2{\pi}(0)/N} \\ e^{-j2{\pi}(0)/N} \\ {\vdots} \\ e^{-j2{\pi}(0)/N} \end{bmatrix} + x[1]\begin{bmatrix} e^{-j2{\pi}(0)/N} \\ e^{-j2{\pi}(1)/N} \\ e^{-j2{\pi}(2)/N} \\ {\vdots} \\ e^{-j2{\pi}(N-1)/N} \end{bmatrix} + x[2]\begin{bmatrix} e^{-j2{\pi}(0)/N} \\ e^{-j2{\pi}(2)/N} \\ e^{-j2{\pi}(4)/N} \\ {\vdots} \\ e^{-j2{\pi}2(N-1)/N} \end{bmatrix} + {\;} {\dotsb} {\;} + x[N-1]\begin{bmatrix} e^{-j2{\pi}(0)/N} \\ e^{-j2{\pi}(N-1)/N} \\ e^{-j2{\pi}2(N-1)/N} \\ {\vdots} \\ e^{-j2{\pi}{(N-1)^2}/N} \end{bmatrix} {\;} {\;} (Eq. 2)$

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.

$\begin{bmatrix} X[0] \\ X[1] \\ X[2] \\ {\vdots} \\ X[N-1] \end{bmatrix} = x[0]\begin{bmatrix} 1 \\ 1 \\ 1 \\ {\vdots} \\ 1 \end{bmatrix} + x[1]\begin{bmatrix} 1 \\ e^{-j2{\pi}(1)/N} \\ e^{-j2{\pi}(2)/N} \\ {\vdots} \\ e^{-j2{\pi}(N-1)/N} \end{bmatrix} + x[2]\begin{bmatrix} 1 \\ e^{-j2{\pi}(2)/N} \\ e^{-j2{\pi}(4)/N} \\ {\vdots} \\ e^{-j2{\pi}2(N-1)/N} \end{bmatrix} + {\;} {\dotsb} {\;} + x[N-1]\begin{bmatrix} 1 \\ e^{-j2{\pi}(N-1)/N} \\ e^{-j2{\pi}2(N-1)/N} \\ {\vdots} \\ e^{-j2{\pi}{(N-1)^2}/N} \end{bmatrix} {\;} {\;} (Eq. 3)$

Now whats happens when we want to perform the DFT on 3 minute audio signal recorded at a 44.1 kHz sampling rate? Our handy-dandy DFT suddenly becomes extremely lengthy since computation time increases with a rate of N^2 + (N^2-N). Due to the work of Cooley and Turkey (although previously discovered by Gauss), the development of the Fast Fourier Transform has reduced computation time by orders of magnitude.

Introduction of the Divide and Conquer Method

One of the key building blocks of the Fast Fourier Transform, is the Divide and Conquer DFT. As the name implies, we will divide Eq. 1 into two separate summations. The first summation processes the even components of x[n] while the second summation processes the odd components of x[n]. This produces,

$X_N[k] = \sum_{m=0}^{{N/2}-1} x[2m] e^{-j2{\pi}k(m)/(N/2)} + e^{-j2{\pi}k/N}\sum_{m=0}^{{N/2}-1} x[2m+1] e^{-j2{\pi}k(m)/(N/2)} {\;} {\;} (Eq. 4)$

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. 4 to the following form.

$X[k] = X_0[k] + e^{-j2{\pi}k/n}X_1[k] {\;} {\;} {\;} (Eq. 5)$

where

$X_0[k] = \sum_{m=0}^{{N/2}-1} x[2m] e^{-j2{\pi}k(m)/(N/2)}$

and

$X_1[k] = \sum_{m=0}^{{N/2}-1} x[2m+1] e^{-j2{\pi}k(m)/(N/2)}$

The complex exponential preceding X1[k] in Eq. 4 is generally called 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. 4 into two separate equations.

$X[k] = X_0[k] + {{W_N}^k}X_1[k]$

$X[k+(N/2)] = X_0[k] - {{W_N}^k}X_1[k]$

Once again were are left with a number of 1 pt. DFT's. So how do represent this in matrix form? First, note that we have two separate equations and therefore need two separate equations of matrices. Similar to Eq. 2, we will repeat the DFT for the entire length of the input signal. However, since we split x[n] into even and odd components, we will only repeat the DFT (N/2) times for X0 and X1. The first equation solves for the first half of X[k].

Eq. 6 and 7,

$\begin{bmatrix} X_0[0] \\ X_0[1] \\ X_0[2] \\ {\vdots} \\ X_0[{N/2}-1] \\ \end{bmatrix} + \begin{bmatrix} {W_N}^0 & 0 & {\dotsb} & 0 \\ 0 & {W_N}^1 & & {\vdots} \\ {\vdots} & & {W_N}^2 & \\ & & {\ddots} & 0 \\ 0 & {\dotsb} & & {W_N}^{(N/2)-1} \end{bmatrix} \begin{bmatrix} X_1[0] \\ X_1[1] \\ X_1[2] \\ {\vdots} \\ X_1[{N/2}-1] \\ \end{bmatrix} = \begin{bmatrix} X[0] \\ X[1] \\ X[2] \\ {\vdots} \\ X[{N/2}-1] \end{bmatrix}$

The second equation solves for the second half of X[k].

$\begin{bmatrix} X_0[0] \\ X_0[1] \\ X_0[2] \\ {\vdots} \\ X_0[{N/2}-1] \\ \end{bmatrix} - \begin{bmatrix} {W_N}^0 & 0 & {\dotsb} & 0 \\ 0 & {W_N}^1 & & {\vdots} \\ {\vdots} & & {W_N}^2 & \\ & & {\ddots} & 0 \\ 0 & {\dotsb} & & {W_N}^{(N/2)-1} \end{bmatrix} \begin{bmatrix} X_1[0] \\ X_1[1] \\ X_1[2] \\ {\vdots} \\ X_1[{N/2}-1] \\ \end{bmatrix} = \begin{bmatrix} X[N/2] \\ X[(N/2)+1] \\ X[(N/2)+2] \\ {\vdots} \\ X[N-1] \end{bmatrix}$

where

$k = 0, 1, 2, ... (N/2) - 1$

After analyzing the two matrices above, there are a few concepts you should understand.

1) The matrix X0 in each equation is just the condensed form of Eq. 2 and then cut in half. It is simply the DFT repeated (N/2) times where the input is the even indices of x[n].

2) The matrix X1 in each equation, is also the condensed form of Eq. 2 cut in half. However, the input is now the odd indices of x[n].

Now compare these two equations to the dcDFT signal path below, as shown in Week 2 of Lab 6.

As you can see from image above, the signal path image from Lab 6 is derived directly from the two matrix equations above (Eq. 6 and 7). Although matrices X0 and X1 are used in both equations, they only need to be calculated once. X0 and X1 are calculated separately, and then combined to form the full vector X[k]. In conclusion, we have four key steps to implement the dcDFT. 1) Separate x[n] into its even and odd components. 2) Calculate the DFT of x[even] and x[odd]. 3) Multiply X1 by the twiddle factor. 4) Perform operations X0 + W*X1 and X0 - W*X1.

Hopefully this gives you a better idea of how to visualize the map given above. Below is example Matlab code of how to implement the DFT and dcDFT.

## DFT '

function [X] = DFT(x)

%DFT This function implements the Discrete Fourier Transform.

% input = x of length N (time)

% output = X of same length N (frequency)

% Authors: Cary Wood, Bill Gu

N = length(x); %find length of input vector

X(N) = 0; %initialize zero vector X(k)

for k = 1:N
for n = 1:N
X(k) = X(k) + x(n)*exp(-1j*2*pi*(k-1)*(n-1)/N);
end %end for
end %end for

end %end DFT

## dcDFT

function [X] = dcDFT(x)

%dcDFT This function implements the Divide and Conquer Discrete Fourier Transform

% input = x of length N (time)

% output = X of same length N (frequency)

% Authors: Cary Wood, Bill Gu

N = length(x); %find length of input vector

%STEP 1

x0 = x(1:2:N); %even components

x1 = x(2:2:N); %odd components

%STEP 2

X0 = DFT(x0); %calculate X0 once

X1 = DFT(x1); %calculate X1 once

for k = 0:N/2-1
%STEP 3
W(k+1) = exp(-1j*2*pi*k/N);  %calculate twiddle factor "W"
% STEP 4 (Eq. 6 and 7)
X(k+1) = X0(k+1) + W(k+1)*X1(k+1);  %X0 + W*X1
X(k+1+N/2) = X0(k+1) - W(k+1)*X1(k+1);  %X0 - W*X1
end %end for

end %end dcDFT

• It's nice to see a student doing the project early! -pm
• Can you explain why you chose to call the second accproach the "Divide and Conquer Discrete Fourier Transform (dcDFT)" instead of simply calling it a "Fast Fourier Transform (FFT)"?
• The FFT works because of two processes, 1) divide and conquer 2) recursion. I feel it would be misleading to call it an FFT because I did not explain recursion nor its implementation. If you assume the function, DFT, works by recursion, then yes, it would be a Fast Fourier Transform implementation. - CW
• I see. Thanks for the clarification.
• As Prof. Mimi mentioned, a few of these equations can be further simplified, specifically Eq. 2,5 and 6. I will try to simplify these over Thanksgiving break.
• write a comment here.