【数字信号处理】【傅里叶分析】【FFT】快速傅里叶变换的完整公式推导
给出离散傅里叶变换DFT的公式：
D F T [ x ( n ) ] = X ( m ) = ∑ n = 0 N − 1 x ( n ) e x p ( − j 2 π m n N ) , m ∈ [ 0 , N − 1 ] = x ( 0 ) e x p ( − j 2 π m 0 N ) + x ( 1 ) e x p ( − j 2 π m 1 N ) + x ( 2 ) e x p ( − j 2 π m 2 N ) + x ( 3 ) e x p ( − j 2 π m 3 N ) + . . . \begin{aligned} DFT[x(n)]=X(m)&=\sum_{n=0}^{N-1}x(n)exp(-j\frac{2\pi mn}{N}),m\in[0,N-1]\\ &=x(0)exp(-j\frac{2\pi m0}{N})+x(1)exp(-j\frac{2\pi m1}{N})+x(2)exp(-j\frac{2\pi m2}{N})+x(3)exp(-j\frac{2\pi m3}{N})+... \end{aligned} DFT[x(n)]=X(m)=n=0∑N−1x(n)exp(−jN2πmn),m∈[0,N−1]=x(0)exp(−jN2πm0)+x(1)exp(−jN2πm1)+x(2)exp(−jN2πm2)+x(3)exp(−jN2πm3)+...
FFT将DFT公式分成奇和偶两部分，即，将含 x ( 0 ) , x ( 2 ) , . . . x(0),x(2),... x(0),x(2),...的项取出，作为偶数项，将含 x ( 1 ) , x ( 3 ) , . . . x(1),x(3),... x(1),x(3),...的项取出，作为奇数项，有
X ( m ) = ∑ n = 0 N 2 − 1 x ( 2 n ) e x p ( − j 2 π m 2 n N ) + ∑ n = 0 N 2 − 1 x ( 2 n + 1 ) e x p ( − j 2 π m ( 2 n + 1 ) N ) , m ∈ [ 0 , N − 1 ] = ∑ n = 0 N 2 − 1 x ( 2 n ) e x p ( − j 2 π m n N 2 ) + exp ( − j 2 π m N ) ∑ n = 0 N 2 − 1 x ( 2 n + 1 ) exp ( − j 2 π m n N 2 ) , m ∈ [ 0 , N 2 − 1 ] \begin{aligned} X(m)&=\sum_{n=0}^{\frac{N}{2}-1}x(2n)exp(-j\frac{2\pi m2n}{N})+\sum_{n=0}^{\frac{N}{2}-1}x(2n+1)exp(-j\frac{2\pi m(2n+1)}{N}),m\in[0,N-1]\\ &=\sum_{n=0}^{\frac{N}{2}-1}x(2n)exp(-j\frac{2\pi mn}{\frac{N}{2}})+\exp(-j\frac{2\pi m}{N})\sum_{n=0}^{\frac{N}{2}-1}x(2n+1)\exp(-j\frac{2\pi mn}{\frac{N}{2}}),m\in[0,\frac{N}{2}-1]\\ \end{aligned} X(m)=n=0∑2N−1x(2n)exp(−jN2πm2n)+n=0∑2N−1x(2n+1)exp(−jN2πm(2n+1)),m∈[0,N−1]=n=0∑2N−1x(2n)exp(−j2N2πmn)+exp(−jN2πm)n=0∑2N−1x(2n+1)exp(−j2N2πmn),m∈[0,2N−1]
可以看到经过变换后， m m m的取值范围变成了一半。现在考虑另一半的m，有
X ( m + N 2 ) = ∑ n = 0 N 2 − 1 x ( 2 n ) exp ( − j 2 π n ( m + N 2 ) N 2 ) + exp ( − j 2 π ( m + N 2 ) N ) ∑ n = 0 N 2 − 1 x ( 2 n + 1 ) exp ( − j 2 π n ( m + N 2 ) N 2 ) , m ∈ [ 0 , N 2 − 1 ] = ∑ n = 0 N 2 − 1 x ( 2 n ) exp ( − j 2 π m n N 2 ) exp ( − j 2 π n ) + exp ( − j 2 π m N ) exp ( − j π ) ∑ n = 0 N 2 − 1 x ( 2 n + 1 ) exp ( − j 2 π n m N 2 ) exp ( − j 2 π n ) , m ∈ [ 0 , N 2 − 1 ] = ∑ n = 0 N 2 − 1 x ( 2 n ) e x p ( − j 2 π m n N 2 ) − exp ( − j 2 π m N ) ∑ n = 0 N 2 − 1 x ( 2 n + 1 ) exp ( − j 2 π m n N 2 ) , m ∈ [ 0 , N 2 − 1 ] \begin{aligned} X(m+\frac{N}2)&=\sum_{n=0}^{\frac{N}{2}-1}x(2n)\exp(-j\frac{2\pi n(m+\frac{N}2)}{\frac{N}{2}})+\exp(-j\frac{2\pi (m+\frac{N}2)}{N})\sum_{n=0}^{\frac{N}{2}-1}x(2n+1)\exp(-j\frac{2\pi n(m+\frac{N}2)}{\frac{N}{2}}),m\in[0,\frac{N}{2}-1]\\ &=\sum_{n=0}^{\frac{N}{2}-1}x(2n)\exp(-j\frac{2\pi mn}{\frac{N}{2}})\exp(-j2\pi n)+\exp(-j\frac{2\pi m}{N})\exp(-j\pi)\sum_{n=0}^{\frac{N}{2}-1}x(2n+1)\exp(-j\frac{2\pi nm}{\frac{N}{2}})\exp(-j2\pi n),m\in[0,\frac{N}{2}-1]\\ &=\sum_{n=0}^{\frac{N}{2}-1}x(2n)exp(-j\frac{2\pi mn}{\frac{N}{2}})-\exp(-j\frac{2\pi m}{N})\sum_{n=0}^{\frac{N}{2}-1}x(2n+1)\exp(-j\frac{2\pi mn}{\frac{N}{2}}),m\in[0,\frac{N}{2}-1] \end{aligned} X(m+2N)=n=0∑2N−1x(2n)exp(−j2N2πn(m+2N))+exp(−jN2π(m+2N))n=0∑2N−1x(2n+1)exp(−j2N2πn(m+2N)),m∈[0,2N−1]=n=0∑2N−1x(2n)exp(−j2N2πmn)exp(−j2πn)+exp(−jN2πm)exp(−jπ)n=0∑2N−1x(2n+1)exp(−j2N2πnm)exp(−j2πn),m∈[0,2N−1]=n=0∑2N−1x(2n)exp(−j2N2πmn)−exp(−jN2πm)n=0∑2N−1x(2n+1)exp(−j2N2πmn),m∈[0,2N−1]
令
W N θ = exp ( − j 2 π θ N ) W^\theta_N=\exp(-j\frac{2\pi \theta}{N}) WNθ=exp(−jN2πθ)则
X ( m ) = ∑ n = 0 N 2 − 1 x ( 2 n ) W N 2 m n + W N m ∑ n = 0 N 2 − 1 x ( 2 n + 1 ) W N 2 m n , m ∈ [ 0 , N 2 − 1 ] X ( m + N 2 ) = ∑ n = 0 N 2 − 1 x ( 2 n ) W N 2 m n − W N m ∑ n = 0 N 2 − 1 x ( 2 n + 1 ) W N 2 m n , m ∈ [ 0 , N 2 − 1 ] X(m)=\sum_{n=0}^{\frac{N}{2}-1}x(2n)W_{\frac{N}2}^{mn}+W_N^m\sum_{n=0}^{\frac{N}{2}-1}x(2n+1)W_{\frac{N}2}^{mn},m\in[0,\frac{N}{2}-1]\\ X(m+\frac{N}2)=\sum_{n=0}^{\frac{N}{2}-1}x(2n)W_{\frac{N}2}^{mn}-W_N^m\sum_{n=0}^{\frac{N}{2}-1}x(2n+1)W_{\frac{N}2}^{mn},m\in[0,\frac{N}{2}-1]\\ X(m)=n=0∑2N−1x(2n)W2Nmn+WNmn=0∑2N−1x(2n+1)W2Nmn,m∈[0,2N−1]X(m+2N)=n=0∑2N−1x(2n)W2Nmn−WNmn=0∑2N−1x(2n+1)W2Nmn,m∈[0,2N−1]
注意到
∑ n = 0 N 2 − 1 x ( 2 n ) W N 2 m n ① ∑ n = 0 N 2 − 1 x ( 2 n + 1 ) W N 2 m n ② \sum_{n=0}^{\frac{N}{2}-1}x(2n)W_{\frac{N}2}^{mn} ①\\ \sum_{n=0}^{\frac{N}{2}-1}x(2n+1)W_{\frac{N}2}^{mn} ② n=0∑2N−1x(2n)W2Nmn①n=0∑2N−1x(2n+1)W2Nmn②
①和②分别是 N N N点序列 x ( n ) x(n) x(n)中对偶数项与奇数项的 N 2 \frac N 2 2N点的DFT。
上述两个式子 X ( m ) X(m) X(m)和 X ( m + N 2 ) X(m+\frac N2) X(m+2N)为FFT的核心公式，即一个 N N N点DFT可以分成2个 N 2 \frac N2 2N点的DFT，并且前 N 2 \frac N2 2N点DFT与后 N 2 \frac N2 2N点DFT之间，只有正负号的不同，所以可以使用前 N 2 \frac N2 2N点DFT公式中的元素，计算后 N 2 \frac N2 2N点DFT中的元素，使计算量降低。
根据该思路，可以将①和②继续分解，分别求①中奇数项与偶数项的 N 4 \frac N 4 4N点DFT，和②中奇数项和偶数项的 N 4 \frac N 4 4N点DFT。这样一直分解下去，直到分解成最简单的2点DFT为止。注意到，在FFT中，N点DFT会一分为二，二分为四，四分为八，最后会分解成多个2点DFT，所以N必须是2的幂次。
接下来，本文用一个 N = 8 N=8 N=8点FFT进行说明。
首先写出8点DFT原始公式
D F T [ x ( n ) ] = X ( m ) = ∑ n = 0 7 x ( n ) e x p ( − j 2 π m n N ) , m ∈ [ 0 , 7 ] DFT[x(n)]=X(m)=\sum_{n=0}^{7}x(n)exp(-j\frac{2\pi mn}{N}),m\in[0,7] DFT[x(n)]=X(m)=n=0∑7x(n)exp(−jN2πmn),m∈[0,7]
根据上文的FFT核心公式，8点DFT可以分为两个4点DFT，即
X ( m ) = ∑ n = 0 3 x ( 2 n ) W 4 m n + W 8 m ∑ n = 0 3 x ( 2 n + 1 ) W 4 m n , m ∈ [ 0 , 3 ] X ( m + 4 ) = ∑ n = 0 3 x ( 2 n ) W 4 m n − W 8 m ∑ n = 0 3 ( 2 n + 1 ) W 4 m n , m ∈ [ 0 , 3 ] X(m)=\sum_{n=0}^{3}x(2n)W_4^{mn}+W_8^m\sum_{n=0}^3x(2n+1)W_4^{mn},m\in[0,3]\\ X(m+4)=\sum_{n=0}^3x(2n)W_4^{mn}-W_8^m\sum_{n=0}^3(2n+1)W_4^{mn},m\in[0,3]\\ X(m)=n=0∑3x(2n)W4mn+W8mn=0∑3x(2n+1)W4mn,m∈[0,3]X(m+4)=n=0∑3x(2n)W4mn−W8mn=0∑3(2n+1)W4mn,m∈[0,3]
令
X 1 ( m ) = ∑ n = 0 3 x ( 2 n ) W 4 m n X 2 ( m ) = ∑ n = 0 3 x ( 2 n + 1 ) W 4 m n X_1(m)=\sum_{n=0}^{3}x(2n)W_4^{mn}\\ X_2(m)=\sum_{n=0}^{3}x(2n+1)W_4^{mn}\\ X1(m)=n=0∑3x(2n)W4mnX2(m)=n=0∑3x(2n+1)W4mn
则
X ( m ) = X 1 ( m ) + W 8 m X 2 ( m ) , m ∈ [ 0 , 3 ] X ( m + 4 ) = X 1 ( m ) − W 8 m X 2 ( m ) , m ∈ [ 0 , 3 ] X(m)=X_1(m)+W_8^mX_2(m),m\in[0,3]\\ X(m+4)=X_1(m)-W_8^mX_2(m),m\in[0,3]\\ X(m)=X1(m)+W8mX2(m),m∈[0,3]X(m+4)=X1(m)−W8mX2(m),m∈[0,3]
1、对 X 1 ( m ) X_1(m) X1(m)进行奇偶分解，有
X 1 ( m ) = ∑ n = 0 1 x ( 2 ( 2 n ) ) W 4 m 2 n + ∑ n = 0 1 x ( 2 ( 2 n + 1 ) ) W 4 m ( 2 n + 1 ) , m ∈ [ 0 , 3 ] = ∑ n = 0 1 x ( 4 n ) exp ( − 2 π m 2 n 4 ) + W 4 m ∑ n = 0 1 x ( 4 n + 2 ) exp ( − j 2 π m 2 n 4 ) , m ∈ [ 0 , 3 ] = ∑ n = 0 1 x ( 4 n ) W 2 m n + W 4 m ∑ n = 0 1 x ( 4 n + 2 ) W 2 m n , m ∈ [ 0 , 1 ] = x ( 0 ) W 2 0 + x ( 4 ) W 2 m + W 4 m [ x ( 2 ) W 2 0 + x ( 6 ) W 2 m ] , m ∈ [ 0 , 1 ] = x ( 0 ) + x ( 4 ) W 2 m + W 4 m [ x ( 2 ) + x ( 6 ) W 2 m ] , m ∈ [ 0 , 1 ] \begin{aligned} X_1(m)&=\sum_{n=0}^{1}x(2(2n))W_4^{m2n}+\sum_{n=0}^1x(2(2n+1))W_4^{m(2n+1)},m\in[0,3]\\ &=\sum_{n=0}^{1}x(4n)\exp(-\frac{2\pi m2n}{4})+W_4^m\sum_{n=0}^1x(4n+2)\exp(-j\frac{2\pi m2n}4),m\in[0,3]\\ &=\sum_{n=0}^{1}x(4n)W_2^{mn}+W_4^m\sum_{n=0}^1x(4n+2)W_2^{mn},m\in[0,1]\\ &=x(0)W_2^0+x(4)W_2^m+W_4^m[x(2)W_2^0+x(6)W_2^m],m\in[0,1]\\ &=x(0)+x(4)W_2^m+W_4^m[x(2)+x(6)W_2^m],m\in[0,1]\\ \end{aligned} X1(m)=n=0∑1x(2(2n))W4m2n+n=0∑1x(2(2n+1))W4m(2n+1),m∈[0,3]=n=0∑1x(4n)exp(−42πm2n)+W4mn=0∑1x(4n+2)exp(−j42πm2n),m∈[0,3]=n=0∑1x(4n)W2mn+W4mn=0∑1x(4n+2)W2mn,m∈[0,1]=x(0)W20+x(4)W2m+W4m[x(2)W20+x(6)W2m],m∈[0,1]=x(0)+x(4)W2m+W4m[x(2)+x(6)W2m],m∈[0,1]
即
X 1 ( 0 ) = x ( 0 ) + x ( 4 ) + x ( 2 ) + x ( 6 ) X 1 ( 1 ) = x ( 0 ) − x ( 4 ) + W 4 1 [ x ( 2 ) − x ( 6 ) ] X_1(0)=x(0)+x(4)+x(2)+x(6)\\ X_1(1)=x(0)-x(4)+W_4^1[x(2)-x(6)] X1(0)=x(0)+x(4)+x(2)+x(6)X1(1)=x(0)−x(4)+W41[x(2)−x(6)]
考虑 m + 2 ∈ [ 2 , 3 ] m+2\in[2,3] m+2∈[2,3]的情况
X 1 ( m + 2 ) = ∑ n = 0 1 x ( 4 n ) W 2 ( m + 2 ) n + W 4 m + 2 ∑ n = 0 1 x ( 4 n + 2 ) W 2 ( m + 2 ) n , m ∈ [ 0 , 1 ] = ∑ n = 0 1 x ( 4 n ) W 2 m n exp ( − j 2 π n ) + W 4 m exp ( − j π ) ∑ n = 0 1 x ( 4 n + 2 ) W 2 m n exp ( − j 2 π n ) , m ∈ [ 0 , 1 ] = ∑ n = 0 1 x ( 4 n ) W 2 m n − W 4 m ∑ n = 0 1 x ( 4 n + 2 ) W 2 ( m + 2 ) n , m ∈ [ 0 , 1 ] = x ( 0 ) W 2 0 + x ( 4 ) W 2 m − W 4 m [ x ( 2 ) W 2 0 + x ( 6 ) W 2 m ] , m ∈ [ 0 , 1 ] = x ( 0 ) + x ( 4 ) W 2 m − W 4 m [ x ( 2 ) + x ( 6 ) W 2 m ] , m ∈ [ 0 , 1 ] \begin{aligned} X_1(m+2)&=\sum_{n=0}^{1}x(4n)W_2^{(m+2)n}+W_4^{m+2}\sum_{n=0}^1x(4n+2)W_2^{(m+2)n},m\in[0,1]\\ &=\sum_{n=0}^{1}x(4n)W_2^{mn}\exp(-j2\pi n)+W_4^{m}\exp(-j\pi)\sum_{n=0}^1x(4n+2)W_2^{mn}\exp(-j2\pi n),m\in[0,1]\\ &=\sum_{n=0}^{1}x(4n)W_2^{mn}-W_4^{m}\sum_{n=0}^1x(4n+2)W_2^{(m+2)n},m\in[0,1]\\ &=x(0)W_2^0+x(4)W_2^m-W_4^m[x(2)W_2^0+x(6)W_2^m],m\in[0,1]\\ &=x(0)+x(4)W_2^m-W_4^m[x(2)+x(6)W_2^m],m\in[0,1]\\ \end{aligned} X1(m+2)=n=0∑1x(4n)W2(m+2)n+W4m+2n=0∑1x(4n+2)W2(m+2)n,m∈[0,1]=n=0∑1x(4n)W2mnexp(−j2πn)+W4mexp(−jπ)n=0∑1x(4n+2)W2mnexp(−j2πn),m∈[0,1]=n=0∑1x(4n)W2mn−W4mn=0∑1x(4n+2)W2(m+2)n,m∈[0,1]=x(0)W20+x(4)W2m−W4m[x(2)W20+x(6)W2m],m∈[0,1]=x(0)+x(4)W2m−W4m[x(2)+x(6)W2m],m∈[0,1]
即
X 1 ( 2 ) = x ( 0 ) + x ( 4 ) − [ x ( 2 ) + x ( 6 ) ] X 1 ( 3 ) = x ( 0 ) − x ( 4 ) − W 4 1 [ x ( 2 ) − x ( 6 ) ] X_1(2)=x(0)+x(4)-[x(2)+x(6)]\\ X_1(3)=x(0)-x(4)-W_4^1[x(2)-x(6)] X1(2)=x(0)+x(4)−[x(2)+x(6)]X1(3)=x(0)−x(4)−W41[x(2)−x(6)]
2、对 X 2 ( m ) X_2(m) X2(m)进行奇偶分解，有
X 2 ( m ) = ∑ n = 0 1 x ( 2 ( 2 n ) + 1 ) W 4 m 2 n + ∑ n = 0 1 x ( 2 ( 2 n + 1 ) + 1 ) W 4 m ( 2 n + 1 ) , m ∈ [ 0 , 3 ] = ∑ n = 0 1 x ( 4 n + 1 ) exp ( − 2 π m 2 n 4 ) + W 4 m ∑ n = 0 1 x ( 4 n + 3 ) exp ( − j 2 π m 2 n 4 ) , m ∈ [ 0 , 3 ] = ∑ n = 0 1 x ( 4 n + 1 ) W 2 m n + W 4 m ∑ n = 0 1 x ( 4 n + 3 ) W 2 m n , m ∈ [ 0 , 1 ] = x ( 1 ) W 2 0 + x ( 5 ) W 2 m + W 4 m [ x ( 3 ) W 2 0 + x ( 7 ) W 2 m ] , m ∈ [ 0 , 1 ] = x ( 1 ) + x ( 5 ) W 2 m + W 4 m [ x ( 3 ) + x ( 7 ) W 2 m ] , m ∈ [ 0 , 1 ] \begin{aligned} X_2(m)&=\sum_{n=0}^{1}x(2(2n)+1)W_4^{m2n}+\sum_{n=0}^1x(2(2n+1)+1)W_4^{m(2n+1)},m\in[0,3]\\ &=\sum_{n=0}^{1}x(4n+1)\exp(-\frac{2\pi m2n}{4})+W_4^m\sum_{n=0}^1x(4n+3)\exp(-j\frac{2\pi m2n}4),m\in[0,3]\\ &=\sum_{n=0}^{1}x(4n+1)W_2^{mn}+W_4^m\sum_{n=0}^1x(4n+3)W_2^{mn},m\in[0,1]\\ &=x(1)W_2^0+x(5)W_2^m+W_4^m[x(3)W_2^0+x(7)W_2^m],m\in[0,1]\\ &=x(1)+x(5)W_2^m+W_4^m[x(3)+x(7)W_2^m],m\in[0,1]\\ \end{aligned} X2(m)=n=0∑1x(2(2n)+1)W4m2n+n=0∑1x(2(2n+1)+1)W4m(2n+1),m∈[0,3]=n=0∑1x(4n+1)exp(−42πm2n)+W4mn=0∑1x(4n+3)exp(−j42πm2n),m∈[0,3]=n=0∑1x(4n+1)W2mn+W4mn=0∑1x(4n+3)W2mn,m∈[0,1]=x(1)W20+x(5)W2m+W4m[x(3)W20+x(7)W2m],m∈[0,1]=x(1)+x(5)W2m+W4m[x(3)+x(7)W2m],m∈[0,1]
即
X 2 ( 0 ) = x ( 1 ) + x ( 5 ) + x ( 3 ) + x ( 7 ) X 2 ( 1 ) = x ( 1 ) − x ( 5 ) + W 4 1 [ x ( 3 ) − x ( 7 ) ] X_2(0)=x(1)+x(5)+x(3)+x(7)\\ X_2(1)=x(1)-x(5)+W_4^1[x(3)-x(7)] X2(0)=x(1)+x(5)+x(3)+x(7)X2(1)=x(1)−x(5)+W41[x(3)−x(7)]
考虑 m + 2 ∈ [ 2 , 3 ] m+2\in[2,3] m+2∈[2,3]的情况
X 2 ( m + 2 ) = ∑ n = 0 1 x ( 4 n + 1 ) W 2 ( m + 2 ) n + W 4 m + 2 ∑ n = 0 1 x ( 4 n + 3 ) W 2 ( m + 2 ) n , m ∈ [ 0 , 1 ] = ∑ n = 0 1 x ( 4 n + 1 ) W 2 m n exp ( − j 2 π n ) + W 4 m exp ( − j π ) ∑ n = 0 1 x ( 4 n + 3 ) W 2 m n exp ( − j 2 π n ) , m ∈ [ 0 , 1 ] = ∑ n = 0 1 x ( 4 n + 1 ) W 2 m n − W 4 m ∑ n = 0 1 x ( 4 n + 3 ) W 2 m n , m ∈ [ 0 , 1 ] = x ( 1 ) W 2 0 + x ( 5 ) W 2 m − W 4 m [ x ( 3 ) W 2 0 + x ( 7 ) W 2 m ] , m ∈ [ 0 , 1 ] = x ( 1 ) + x ( 5 ) W 2 m − W 4 m [ x ( 3 ) + x ( 7 ) W 2 m ] , m ∈ [ 0 , 1 ] \begin{aligned} X_2(m+2)&=\sum_{n=0}^{1}x(4n+1)W_2^{(m+2)n}+W_4^{m+2}\sum_{n=0}^1x(4n+3)W_2^{(m+2)n},m\in[0,1]\\ &=\sum_{n=0}^{1}x(4n+1)W_2^{mn}\exp(-j2\pi n)+W_4^{m}\exp(-j\pi)\sum_{n=0}^1x(4n+3)W_2^{mn}\exp(-j2\pi n),m\in[0,1]\\ &=\sum_{n=0}^{1}x(4n+1)W_2^{mn}-W_4^{m}\sum_{n=0}^1x(4n+3)W_2^{mn},m\in[0,1]\\ &=x(1)W_2^0+x(5)W_2^m-W_4^m[x(3)W_2^0+x(7)W_2^m],m\in[0,1]\\ &=x(1)+x(5)W_2^m-W_4^m[x(3)+x(7)W_2^m],m\in[0,1]\\ \end{aligned} X2(m+2)=n=0∑1x(4n+1)W2(m+2)n+W4m+2n=0∑1x(4n+3)W2(m+2)n,m∈[0,1]=n=0∑1x(4n+1)W2mnexp(−j2πn)+W4mexp(−jπ)n=0∑1x(4n+3)W2mnexp(−j2πn),m∈[0,1]=n=0∑1x(4n+1)W2mn−W4mn=0∑1x(4n+3)W2mn,m∈[0,1]=x(1)W20+x(5)W2m−W4m[x(3)W20+x(7)W2m],m∈[0,1]=x(1)+x(5)W2m−W4m[x(3)+x(7)W2m],m∈[0,1]
即
X 2 ( 2 ) = x ( 1 ) + x ( 5 ) − [ x ( 3 ) + x ( 7 ) ] X 2 ( 3 ) = x ( 1 ) − x ( 5 ) − W 4 1 [ x ( 3 ) − x ( 7 ) ] X_2(2)=x(1)+x(5)-[x(3)+x(7)]\\ X_2(3)=x(1)-x(5)-W_4^1[x(3)-x(7)] X2(2)=x(1)+x(5)−[x(3)+x(7)]X2(3)=x(1)−x(5)−W41[x(3)−x(7)]
将 X 1 ( m ) X_1(m) X1(m)和 X 2 ( m ) X_2(m) X2(m)分解后的表达式代回 X ( m ) X(m) X(m)和 X ( m + 4 ) X(m+4) X(m+4)中，有
X ( m ) = X 1 ( m ) + W 8 m X 2 ( m ) X ( m + 2 ) = X 1 ( m + 2 ) + W 8 m + 2 X 2 ( m + 2 ) X ( m + 4 ) = X 1 ( m ) − W 8 m X 2 ( m ) X ( ( m + 2 ) + 4 ) = X 1 ( m + 2 ) − W 8 m + 2 X 2 ( m + 2 ) m ∈ [ 0 , 1 ] \begin{aligned} X(m)&=X_1(m)+W_8^mX_2(m)\\ X(m+2)&=X_1(m+2)+W_8^{m+2}X_2(m+2)\\ X(m+4)&=X_1(m)-W_8^mX_2(m)\\ X((m+2)+4)&=X_1(m+2)-W_8^{m+2}X_2(m+2)\\ m&\in[0,1] \end{aligned} X(m)X(m+2)X(m+4)X((m+2)+4)m=X1(m)+W8mX2(m)=X1(m+2)+W8m+2X2(m+2)=X1(m)−W8mX2(m)=X1(m+2)−W8m+2X2(m+2)∈[0,1]
将 m ∈ [ 0 , 1 ] m\in[0,1] m∈[0,1]代入，得到 X ( 0 ) , X ( 1 ) , . . . , X ( 7 ) X(0),X(1),...,X(7) X(0),X(1),...,X(7)的表达式
X ( 0 ) = x ( 0 ) + x ( 4 ) + x ( 2 ) + x ( 6 ) + x ( 1 ) + x ( 5 ) + x ( 3 ) + x ( 7 ) X ( 1 ) = x ( 0 ) − x ( 4 ) + W 4 1 [ x ( 2 ) − x ( 6 ) ] + W 8 1 { x ( 1 ) − x ( 5 ) + W 4 1 [ x ( 3 ) − x ( 7 ) ] } X ( 2 ) = x ( 0 ) + x ( 4 ) − [ x ( 2 ) + x ( 6 ) ] + W 8 2 { x ( 1 ) + x ( 5 ) − [ x ( 3 ) + x ( 7 ) ] } X ( 3 ) = x ( 0 ) − x ( 4 ) − W 4 1 [ x ( 2 ) − x ( 6 ) ] + W 8 3 { x ( 1 ) − x ( 5 ) − W 4 1 [ x ( 3 ) − x ( 7 ) ] } X ( 4 ) = x ( 0 ) + x ( 4 ) + x ( 2 ) + x ( 6 ) − [ x ( 1 ) + x ( 5 ) + x ( 3 ) + x ( 7 ) ] X ( 5 ) = x ( 0 ) − x ( 4 ) + W 4 1 [ x ( 2 ) − x ( 6 ) ] − W 8 1 { x ( 1 ) − x ( 5 ) + W 4 1 [ x ( 3 ) − x ( 7 ) ] } X ( 6 ) = x ( 0 ) + x ( 4 ) − [ x ( 2 ) + x ( 6 ) ] − W 8 2 { x ( 1 ) + x ( 5 ) − [ x ( 3 ) + x ( 7 ) ] } X ( 7 ) = x ( 0 ) − x ( 4 ) − W 4 1 [ x ( 2 ) − x ( 6 ) ] − W 8 3 { x ( 1 ) − x ( 5 ) − W 4 1 [ x ( 3 ) − x ( 7 ) ] } \begin{aligned} X(0)&=x(0)+x(4)+x(2)+x(6)+x(1)+x(5)+x(3)+x(7)\\ X(1)&=x(0)-x(4)+W_4^1[x(2)-x(6)]+W_8^1\{x(1)-x(5)+W_4^1[x(3)-x(7)]\}\\ X(2)&=x(0)+x(4)-[x(2)+x(6)]+W_8^2\{x(1)+x(5)-[x(3)+x(7)]\}\\ X(3)&=x(0)-x(4)-W_4^1[x(2)-x(6)]+W_8^3\{x(1)-x(5)-W_4^1[x(3)-x(7)]\}\\ X(4)&=x(0)+x(4)+x(2)+x(6)-[x(1)+x(5)+x(3)+x(7)]\\ X(5)&=x(0)-x(4)+W_4^1[x(2)-x(6)]-W_8^1\{x(1)-x(5)+W_4^1[x(3)-x(7)]\}\\ X(6)&=x(0)+x(4)-[x(2)+x(6)]-W_8^2\{x(1)+x(5)-[x(3)+x(7)]\}\\ X(7)&=x(0)-x(4)-W_4^1[x(2)-x(6)]-W_8^3\{x(1)-x(5)-W_4^1[x(3)-x(7)]\}\\ \end{aligned} X(0)X(1)X(2)X(3)X(4)X(5)X(6)X(7)=x(0)+x(4)+x(2)+x(6)+x(1)+x(5)+x(3)+x(7)=x(0)−x(4)+W41[x(2)−x(6)]+W81{x(1)−x(5)+W41[x(3)−x(7)]}=x(0)+x(4)−[x(2)+x(6)]+W82{x(1)+x(5)−[x(3)+x(7)]}=x(0)−x(4)−W41[x(2)−x(6)]+W83{x(1)−x(5)−W41[x(3)−x(7)]}=x(0)+x(4)+x(2)+x(6)−[x(1)+x(5)+x(3)+x(7)]=x(0)−x(4)+W41[x(2)−x(6)]−W81{x(1)−x(5)+W41[x(3)−x(7)]}=x(0)+x(4)−[x(2)+x(6)]−W82{x(1)+x(5)−[x(3)+x(7)]}=x(0)−x(4)−W41[x(2)−x(6)]−W83{x(1)−x(5)−W41[x(3)−x(7)]}
即
X ( 0 ) = x ( 0 ) + x ( 4 ) + x ( 2 ) + x ( 6 ) + x ( 1 ) + x ( 5 ) + x ( 3 ) + x ( 7 ) X ( 1 ) = x ( 0 ) − x ( 4 ) − j [ x ( 2 ) − x ( 6 ) ] + exp ( − j π 4 ) { x ( 1 ) − x ( 5 ) − j [ x ( 3 ) − x ( 7 ) ] } X ( 2 ) = x ( 0 ) + x ( 4 ) − [ x ( 2 ) + x ( 6 ) ] − j { x ( 1 ) + x ( 5 ) − [ x ( 3 ) + x ( 7 ) ] } X ( 3 ) = x ( 0 ) − x ( 4 ) + j [ x ( 2 ) − x ( 6 ) ] + exp ( − j 3 π 4 ) { x ( 1 ) − x ( 5 ) + j [ x ( 3 ) − x ( 7 ) ] } X ( 4 ) = x ( 0 ) + x ( 4 ) + x ( 2 ) + x ( 6 ) − [ x ( 1 ) + x ( 5 ) + x ( 3 ) + x ( 7 ) ] X ( 5 ) = x ( 0 ) − x ( 4 ) − j [ x ( 2 ) − x ( 6 ) ] − exp ( − j π 4 ) { x ( 1 ) − x ( 5 ) − j [ x ( 3 ) − x ( 7 ) ] } X ( 6 ) = x ( 0 ) + x ( 4 ) − [ x ( 2 ) + x ( 6 ) ] + j { x ( 1 ) + x ( 5 ) − [ x ( 3 ) + x ( 7 ) ] } X ( 7 ) = x ( 0 ) − x ( 4 ) + j [ x ( 2 ) − x ( 6 ) ] − exp ( − j 3 π 4 ) { x ( 1 ) − x ( 5 ) + j [ x ( 3 ) − x ( 7 ) ] } \begin{aligned} X(0)&=x(0)+x(4)+x(2)+x(6)+x(1)+x(5)+x(3)+x(7)\\ X(1)&=x(0)-x(4)-j[x(2)-x(6)]+\exp(-j\frac{\pi}4)\{x(1)-x(5)-j[x(3)-x(7)]\}\\ X(2)&=x(0)+x(4)-[x(2)+x(6)]-j\{x(1)+x(5)-[x(3)+x(7)]\}\\ X(3)&=x(0)-x(4)+j[x(2)-x(6)]+\exp(-j\frac{3\pi}4)\{x(1)-x(5)+j[x(3)-x(7)]\}\\ X(4)&=x(0)+x(4)+x(2)+x(6)-[x(1)+x(5)+x(3)+x(7)]\\ X(5)&=x(0)-x(4)-j[x(2)-x(6)]-\exp(-j\frac{\pi}4)\{x(1)-x(5)-j[x(3)-x(7)]\}\\ X(6)&=x(0)+x(4)-[x(2)+x(6)]+j\{x(1)+x(5)-[x(3)+x(7)]\}\\ X(7)&=x(0)-x(4)+j[x(2)-x(6)]-\exp(-j\frac{3\pi}4)\{x(1)-x(5)+j[x(3)-x(7)]\}\\ \end{aligned} X(0)X(1)X(2)X(3)X(4)X(5)X(6)X(7)=x(0)+x(4)+x(2)+x(6)+x(1)+x(5)+x(3)+x(7)=x(0)−x(4)−j[x(2)−x(6)]+exp(−j4π){x(1)−x(5)−j[x(3)−x(7)]}=x(0)+x(4)−[x(2)+x(6)]−j{x(1)+x(5)−[x(3)+x(7)]}=x(0)−x(4)+j[x(2)−x(6)]+exp(−j43π){x(1)−x(5)+j[x(3)−x(7)]}=x(0)+x(4)+x(2)+x(6)−[x(1)+x(5)+x(3)+x(7)]=x(0)−x(4)−j[x(2)−x(6)]−exp(−j4π){x(1)−x(5)−j[x(3)−x(7)]}=x(0)+x(4)−[x(2)+x(6)]+j{x(1)+x(5)−[x(3)+x(7)]}=x(0)−x(4)+j[x(2)−x(6)]−exp(−j43π){x(1)−x(5)+j[x(3)−x(7)]}