DFT离散傅立叶变换C++实现【转】

来源：互联网发布：nba球队季后赛数据编辑：程序博客网时间：2024/06/07 00:02

来自：http://blog.csdn.net/calcular/article/details/46804779

DFT的执行效率是O(n^2)，FFT为O(log2n)，但是它对点数没有限制。

/*     离散傅立叶算法V1.0   含有：DFT,IDFT         made by xyt  2015/7/5*/#ifndef _DFT_H#define _DFT_H#include<iostream>#include<math.h>using namespace std;#define PI 3.14159265354struct complex{double r,i;};complex multi(complex a,complex b){complex tmp;tmp.r=a.r*b.r-a.i*b.i;tmp.i=a.r*b.i+a.i*b.r;return tmp;}int fi(double in){if((in-(int)in)>0.5) return (int)in+1;else return (int)in;}/* 离散傅立叶正变换,输出[][2]数组实部在前,采样容量n可以任意正整数 */void DFT(int *in,double **out,const int &n){int i,j;complex **W=new complex*[n];for(i=0;i<n;i++){W[i]=new complex[n];}complex *lis=new complex[(n-1)*(n-1)+1];lis[0].r=1;lis[0].i=0;lis[1].r=cos(2.0*PI/n);lis[1].i=-1.0*sin(2.0*PI/n);for(i=2;i<=(n-1)*(n-1);i++){lis[i]=multi(lis[1],lis[i-1]);}for(i=0;i<n;i++){for(j=0;j<n;j++){W[i][j]=lis[i*j];}}complex sum;for(i=0;i<n;i++){sum.r=0;sum.i=0;for(j=0;j<n;j++){sum.r+=in[j]*W[i][j].r;sum.i+=in[j]*W[i][j].i;}out[i][0]=sum.r;out[i][1]=sum.i;}for(i=0;i<n;i++) delete []W[i];delete []W;delete []lis;}/* 离散傅立叶逆变换 */void IDFT(double **in,int *out,const int &n){int i,j;complex **W=new complex*[n];for(i=0;i<n;i++){W[i]=new complex[n];}complex *lis=new complex[(n-1)*(n-1)+1];lis[0].r=1;lis[0].i=0;lis[1].r=cos(2.0*PI/n);lis[1].i=sin(2.0*PI/n);for(i=2;i<=(n-1)*(n-1);i++){lis[i]=multi(lis[1],lis[i-1]);}for(i=0;i<n;i++){for(j=0;j<n;j++){W[i][j]=lis[i*j];}}complex sum;for(i=0;i<n;i++){sum.r=0;sum.i=0;for(j=0;j<n;j++){sum.r+=W[i][j].r*in[j][0]-W[i][j].i*in[j][1];sum.i+=W[i][j].i*in[j][0]+W[i][j].r*in[j][1];}out[i]=fi(sum.r/n);}for(i=0;i<n;i++) delete []W[i];delete []W;delete []lis;}#endif

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