共轭对称序列的快速傅里叶反变换

偶尔善良 提交于 2019-12-04 16:34:12

一、功能

计算共轭对称复序列的快速傅里叶反变换,其变换结果是实数。

二、方法简介

序列\(x(n)\)的离散傅里叶变换为
\[ X(k)=\sum_{n=0}^{N-1}x(n)W_{N}^{nk}, \ k=0,1,...,N-1 \]

序列\(X(k)\)的离散傅里叶反变换为
\[ x(n)=\frac{1}{N}\sum_{n=0}^{N-1}X(k)W_{N}^{-nk}, \ n=0,1,...,N-1 \]
共轭对称复序列的傅里叶反变换,可用复序列快速傅里叶反变换算法进行计算。但考虑到\(X(k)\)是共轭对称的,其傅里叶反变换\(x(n)\)是实数,因此,为进一步提高计算效率,需要对一般的复序列IFFT算法进行一定的修改。

共轭对称序列\(X(k)\)具有如下性质:\(X(0)\)\(X(\frac{N}{2})\)都是实数,且有
\[ X(k)=X^{*}(N-k), \ 1\leqslant k\leqslant \frac{N}{2}-1 \]
\(X(k)\)的实部是偶对称,虚部是奇对称。在计算傅里叶反变换时,利用这种共轭对称性,我们就可以不必计算和存储\(X(k)(\frac{N}{2}+1\leqslant k\leqslant N-1)\)以及\(X(0)\)\(X(\frac{N}{2})\)的虚部,这比一般形式的快速傅里叶反变换算法大约可减少一半的运算量和存储量。具体计算时采用的是分裂基算法。

三、使用说明

C语言实现方式如下

/************************************
    x       ----长度为n。开始时存放要具有共轭对称性的复序列X(k)的前n/2+1个值,
                其存储顺序为[Re(0),Re(1),...,Re(n/2),Im(n/2-1),...,Im(1)],
                其中Re(0)=X(0),Re(n/2)=X(n/2)。最后存放变换结果x(i)(i=0,0,1,...,n-1),
                这里x(i)是实数。
    n       ----数据长度,必须是2的整数次幂,即n=2^m。
************************************/
#include "math.h"
void irfft(double *x, int n)
{
    int i, j, k, m, i1, i2, i3, i4, i5, i6, i7, i8, n2, n4, n8, id, is;
    double a, e, a3, t1, t2, t3, t4, t5, cc1, cc3, ss1, ss3;
    for (j = 1, i = 1; i < 16; i++) {
        m = i;
        j = 2 * j;
        if(j == n)
            break;
    }
    n2 = 2 * n;
    for(k = 1; k < m; k++){
        is = 0;
        id = n2;
        n2 = n2 / 2;
        n4 = n2 / 4;
        n8 = n4 / 2;
        e = 6. 28318530718 / n2;
        do {
            for(i = is; i < n; i += id){
                i1 = i;
                i2 = i1 + n4;
                i3 = i2 + n4;
                i4 = i3 + n4;
                t1 = x[i1] - x[i3];
                x[i1] = x[i1] + x[i3];
                x[i2] = 2 * x[i2];
                x[i3] = t1 - 2 * x[i4];
                x[i4] = t1 + 2 * x[i4];
                if(n4 == 1)
                    continue;
                i1 += n8;
                i2 += n8;
                i3 += n8;
                i4 += n8;
                t1 = (x[i2] - x[i1]) / sqrt(2.0);
                t2 = (x[i4] + x[i3]) / sqrt(2.0);
                x[i1] = x[i1] + x[i2];
                x[i2] = x[i4] - x[i3];
                x[i3] = 2 * (-t2 - t1);
                x[i4] = 2 * (-t2 + t1);
            }
            is = 2 * id - n2;
            id = 4 * id;
        }while(is < (n - 1));
        a = e;
        for(j = 1; j < n8; j++) {
            a3 = 3 * a;
            cc1 = cos(a);
            ss1 = sin(a);
            cc3 = cos(a3);
            ss3 = sin(a3);
            a = (j + 1) * e;
            is = 0;
            id = 2 * n2;
            do {
                for(i = is; i <= (n - 1); i = i + id) {
                    i1 = i - j;
                    i2 = i1 + n4;
                    i3 = i2 + n4;
                    i4 = i3 + n4;
                    i5 = i + n4 - j;
                    i6 = i5 + n4;
                    i7 = i6 + n4;
                    i8 = i7 + n4;
                    t1 = x[i1] - x[i6];
                    x[i1] = x[i1] + x[i6];
                    t2 = x[i5] - x[i2];
                    x[i5] = x[i2] + x[i5];
                    t3 = x[i8] + x[i3];
                    x[i6] = x[i8] - x[i3];
                    t4 = x[i4] + x[i7];
                    x[i2] = x[i4] - [i7];
                    t5 = t1 - t4;
                    t1 = t1 + t4;
                    t4 = t2 - t3;
                    t2 = t2 + t3;
                    x[i3] = t5 * cc1 + t4 * ss1;
                    x[i7] = -t4 * cc1 + t5 * ss1;
                    x[i4] = t1 * cc3 - t2 * ss3;
                    x[i8] = t2 * cc3 + t1 * ss3;
                }
                is = 2 * id - n2;
                id = 4 * id;
            }
            while(is < (n - 1));
        }
    }
    is = 0;
    id = 0;
    do {
        for((i = is; i < n; i = i + id)){
            i1 = i + 1;
            t1 = x[i];
            x[i] = t1 + x[i1];
            x[i1] = t1 - x[i1];
        }
        is = 2 * id - 2;
        id = 4 * id;
    }while(is < (n - 1));
    for(j = 0, i = 0; i < (n - 1); i++){
        if(i < j){
            t1 = x[j];
            x[j] = x[i];
            x[i] = t1;
        }
        k = n / 2;
        while(k < (j + 1)){
            j = j - k;
            k = k / 2;
        }
        j = j + k;
    }
    for(i = 0; i < n; i++)
        x[i] = x[j] / n;
}
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