計算復序列的分裂基快速傅里葉變換。算法
序列\(x(n)(n=0,1,...,N-1)\)的離散傅里葉變換定義爲
\[ X(k)=\sum_{n=0}^{N-1}x(n)W_{N}^{nk}, \qquad k=0,1,...,N-1 \]
其中\(W_{N}^{nk}=e^{-j\frac{2\pi nk}{N}}\),將\(X(k)\)按序號\(k\)的奇偶分紅兩組。當\(k\)爲偶數時,進行基2頻率抽取分解, \(X(k)\)可表示爲
\[ X(2k)=\sum_{n=0}^{N/2-1}[x(n)+x(n+\frac{N}{2})]W_{N}^{2nk} \ , \ k=0,1,...,\frac{N}{2}-1 \]
當\(k\)爲奇數時進行基4 頻率抽取分解,$ X(k)$可表示爲
\[ \left\{\begin{matrix}X(4k+1)=\sum_{n=0}^{N/4-1}{[x(n)-x(n+\frac{N}{2})]-j[x(n+\frac{N}{4})-x(n+\frac{3N}{4})]}W_{N}^{n}W_{N}^{4nk}\\ X(4k+3)=\sum_{n=0}^{N/4-1}{[x(n)-x(n+\frac{N}{2})]+j[x(n+\frac{N}{4})-x(n+\frac{3N}{4})]}W_{N}^{n}W_{N}^{4nk}\end{matrix}\right.\\k = 0,1,...,\frac{N}{4}-1 \]
因而可知,一個\(N\)點的DFT 能夠分解爲一個\(N/2\)點的DFT 和兩個\(N/4\)點的DFT 。這種分解既有基2的部分,又有基4的部分,所以稱爲分裂基分解。上面的\(N/2\)點DFT 又可分解爲一個\(N/4\)點的DFT 和兩個\(N/8\)點的DFT, 而兩個\(N/4\)點的DFT也分別能夠分解爲一個\(N/8\)點的DFT和兩個\(N/16\)點的DFT 。依此類推,直至分解到最後一級爲止。這就是按頻率抽取的分裂基快速傅立葉變換算法。數組
分裂基快速算法是將基2和基4分解組合而成。在基\(2^m\)類快速算法中,分裂基算法具備最少的運算量,且仍保留結構規則、原位計算等優勢。spa
是用C語言實現基4快速傅里葉變換(FFT)的方法以下:code
/************************************ x ---一維數組,長度爲n,開始時存放要變換數據的實部,最後存放變換結果的實部。 y ---一維數組,長度爲n,開始時存放要變換數據的虛部,最後存放變換結果的虛部。 n ---數據長度,必須是4的整數次冪。 ************************************/ #include "math.h" void srfft(double *x, double *y, int n) { int i, j, k, m, il, i2, i3, nl, n2, n4, id, is; double a, b, c, e, a3, rl, r2, r3, r4; double cl, e3, sl, s2, s3, ccl, cc3, ssl, ss3; for(j = 1; i = 1; i < 10; i++) { m = i; j = 4 * j; if(j == n) break; } n2 = 2 * n; for(k = 1; k <= m; k++) { n2 = n2 / 2; n4 = n2 / 4; e = 6.28318530718 / n2; a = 0; for(j = 0; j < n4; j++) { a3 = 3 * a; ccl = cos(a); ssl = sin(a); cc3 = cos(a3); ss3 = sin(a3); a = (j + 1) * e; is = j; id = 2 * n2; do { for (i = is; i < (n-1); i = i + id) { il = i + n4; i2 = il + n4; i3 = i2 + n4; rl = x[i] - x[i2]; x[i] = x[i] + x[i2]; r2 = x[il] - x[i3]; x[il] = x[il] + x[i3]; sl = y[i] - y[i2]; y[i] = y[i] + y[i2]; s2 = y[il] - y[i3]; y[il] = y[il] + y[i3]; s3 = rl - s2; rl = rl + s2; s2 = r2 - sl; r2 = r2 + sl; x[i2] = rl * eel - s2 * ssl; y[i2] = -s2 * eel - rl * ssl; x[i3] = s3 * ee3 + r2 * ss3; y[i3] = r2 * ee3 - s3 * ss3; } is = 2 * id - n2 + j; id = 4 * id; }while (is < (n-1)); } is = O; id = 4; do { for (i=is;i<n;i=i+id) { il = i + 1; rl = x[i]; r2 = y[i]; x[i] = rl + x[il]; y[i] = r2 + y[il]; x[il] = rl — x[il]; y[il] = r2 — y[il]; } is = 2 * id - 2; id = 4 * id; } while(is < (n - 1)); nl = n - 1; for (j = O, i = O; i < nl; i++) { if(i < j) { rl = x[jJ; sl = y[j]; x[j] = x[i]; y[j] = y[i]; x[i] = rl; y[i] = sl; } k = n / 2; while(k < (j + 1)) { j = j - k; k = k / 2; } j = j + k; } } }