c語言fft
float ar[1024],ai[1024];/* 原始數據實部，虛部 */
float a[2050];
void fft(int nn) /* nn數據長度 */
{
int n1,n2,i,j,k,l,m,s,l1;
float t1,t2,x,y;
float w1,w2,u1,u2,z;
float fsin[10]={0.000000,1.000000,0.707107,0.3826834,0.1950903,0.09801713,0.04906767,0.02454123,0.01227154,0.00613588,};
float fcos[10]={-1.000000,0.000000,0.7071068,0.9238796,0.9807853,0.99518472,0.99879545,0.9996988,0.9999247,0.9999812,};
switch(nn)
{
case 1024: s=10; break;
case 512: s=9; break;
case 256: s=8; break;
}
n1=nn/2; n2=nn-1;
j=1;
for(i=1;i<=nn;i++)
{
a[2*i]=ar[i-1];
a[2*i+1]=ai[i-1];
}
for(l=1;l<n2;l++)
{
if(l<j)
{
t1=a[2*j];
t2=a[2*j+1];
a[2*j]=a[2*l];
a[2*j+1]=a[2*l+1];
a[2*l]=t1;
a[2*l+1]=t2;
}
k=n1;
while (k<j)
{
j=j-k;
k=k/2;
}
j=j+k;
}
for(i=1;i<=s;i++)
{
u1=1;
u2=0;
m=(1<<i);
k=m>>1;
w1=fcos[i-1];
w2=-fsin[i-1];
for(j=1;j<=k;j++)
{
for(l=j;l<nn;l=l+m)
{
l1=l+k;
t1=a[2*l1]*u1-a[2*l1+1]*u2;
t2=a[2*l1]*u2+a[2*l1+1]*u1;
a[2*l1]=a[2*l]-t1;
a[2*l1+1]=a[2*l+1]-t2;
a[2*l]=a[2*l]+t1;
a[2*l+1]=a[2*l+1]+t2;
}
z=u1*w1-u2*w2;
u2=u1*w2+u2*w1;
u1=z;
}
}
for(i=1;i<=nn/2;i++)
{
ar[i]=4*a[2*i+2]/nn; /* 實部 */
ai[i]=-4*a[2*i+3]/nn; /* 虛部 */
a[i]=4*sqrt(ar[i]*ar[i]+ai[i]*ai[i]); /* 幅值 */
}
}
(http://..com/question/284943905.html?an=0&si=2)
❷ 怎樣用C語言實現FFT演算法啊
1、二維FFT相當於對行和列分別進行一維FFT運算。具體的實現辦法如下：
先對各行逐一進行一維FFT，然後再對變換後的新矩陣的各列逐一進行一維FFT。相應的偽代碼如下所示：
for (int i=0; i<M; i++)
FFT_1D(ROW[i]，N);
for (int j=0; j<N; j++)
FFT_1D(COL[j]，M);
其中，ROW[i]表示矩陣的第i行。注意這只是一個簡單的記法，並不能完全照抄。還需要通過一些語句來生成各行的數據。同理，COL[i]是對矩陣的第i列的一種簡單表示方法。
所以，關鍵是一維FFT演算法的實現。
2、常式：
#include<stdio.h>
#include<math.h>
#include<stdlib.h>
#defineN1000
/*定義復數類型*/
typedefstruct{
doublereal;
doubleimg;
}complex;
complexx[N],*W;/*輸入序列,變換核*/
intsize_x=0;/*輸入序列的大小，在本程序中僅限2的次冪*/
doublePI;/*圓周率*/
voidfft();/*快速傅里葉變換*/
voidinitW();/*初始化變換核*/
voidchange();/*變址*/
voidadd(complex,complex,complex*);/*復數加法*/
voidmul(complex,complex,complex*);/*復數乘法*/
voidsub(complex,complex,complex*);/*復數減法*/
voidoutput();
intmain(){
inti;/*輸出結果*/
system("cls");
PI=atan(1)*4;
printf("Pleaseinputthesizeofx: ");
scanf("%d",&size_x);
printf("Pleaseinputthedatainx[N]: ");
for(i=0;i<size_x;i++)
scanf("%lf%lf",&x[i].real,&x[i].img);
initW();
fft();
output();
return0;
}
/*快速傅里葉變換*/
voidfft(){
inti=0,j=0,k=0,l=0;
complexup,down,proct;
change();
for(i=0;i<log(size_x)/log(2);i++){/*一級蝶形運算*/
l=1<<i;
for(j=0;j<size_x;j+=2*l){/*一組蝶形運算*/
for(k=0;k<l;k++){/*一個蝶形運算*/
mul(x[j+k+l],W[size_x*k/2/l],&proct);
add(x[j+k],proct,&up);
sub(x[j+k],proct,&down);
x[j+k]=up;
x[j+k+l]=down;
}
}
}
}
/*初始化變換核*/
voidinitW(){
inti;
W=(complex*)malloc(sizeof(complex)*size_x);
for(i=0;i<size_x;i++){
W[i].real=cos(2*PI/size_x*i);
W[i].img=-1*sin(2*PI/size_x*i);
}
}
/*變址計算，將x(n)碼位倒置*/
voidchange(){
complextemp;
unsignedshorti=0,j=0,k=0;
doublet;
for(i=0;i<size_x;i++){
k=i;j=0;
t=(log(size_x)/log(2));
while((t--)>0){
j=j<<1;
j|=(k&1);
k=k>>1;
}
if(j>i){
temp=x[i];
x[i]=x[j];
x[j]=temp;
}
}
}
/*輸出傅里葉變換的結果*/
voidoutput(){
inti;
printf("Theresultareasfollows ");
for(i=0;i<size_x;i++){
printf("%.4f",x[i].real);
if(x[i].img>=0.0001)printf("+%.4fj ",x[i].img);
elseif(fabs(x[i].img)<0.0001)printf(" ");
elseprintf("%.4fj ",x[i].img);
}
}
voidadd(complexa,complexb,complex*c){
c->real=a.real+b.real;
c->img=a.img+b.img;
}
voidmul(complexa,complexb,complex*c){
c->real=a.real*b.real-a.img*b.img;
c->img=a.real*b.img+a.img*b.real;
}
voidsub(complexa,complexb,complex*c){
c->real=a.real-b.real;
c->img=a.img-b.img;
}
❸ C語言 1024點快速傅里葉變換（FFT）程序，最好經過優化，執行速度快
void fft()
{
int nn,n1,n2,i,j,k,l,m,s,l1;
float ar[1024],ai[1024]; // 實部 虛部
float a[2050];
float t1,t2,x,y;
float w1,w2,u1,u2,z;
float fsin[10]={0.000000,1.000000,0.707107,0.3826834,0.1950903,0.09801713,0.04906767,0.02454123,0.01227154,0.00613588,};// 優化
float fcos[10]={-1.000000,0.000000,0.7071068,0.9238796,0.9807853,0.99518472,0.99879545,0.9996988,0.9999247,0.9999812,};
nn=1024;
s=10;
n1=nn/2; n2=nn-1;
j=1;
for(i=1;i<=nn;i++)
{
a[2*i]=ar[i-1];
a[2*i+1]=ai[i-1];
}
for(l=1;l<n2;l++)
{
if(l<j)
{
t1=a[2*j];
t2=a[2*j+1];
a[2*j]=a[2*l];
a[2*j+1]=a[2*l+1];
a[2*l]=t1;
a[2*l+1]=t2;
}
k=n1;
while (k<j)
{
j=j-k;
k=k/2;
}
j=j+k;
}
for(i=1;i<=s;i++)
{
u1=1;
u2=0;
m=(1<<i);
k=m>>1;
w1=fcos[i-1];
w2=-fsin[i-1];
for(j=1;j<=k;j++)
{
for(l=j;l<nn;l=l+m)
{
l1=l+k;
t1=a[2*l1]*u1-a[2*l1+1]*u2;
t2=a[2*l1]*u2+a[2*l1+1]*u1;
a[2*l1]=a[2*l]-t1;
a[2*l1+1]=a[2*l+1]-t2;
a[2*l]=a[2*l]+t1;
a[2*l+1]=a[2*l+1]+t2;
}
z=u1*w1-u2*w2;
u2=u1*w2+u2*w1;
u1=z;
}
}
for(i=1;i<=nn/2;i++)
{
ar[i]=a[2*i+2]/nn;
ai[i]=-a[2*i+3]/nn;
a[i]=4*sqrt(ar[i]*ar[i]+ai[i]*ai[i]); // 幅值
}
}
❹ 求FFT的c語言程序
快速傅里葉變換 要用C++ 才行吧 你可以用MATLAB來實現更方便點啊
此FFT 是用VC6.0編寫，由FFT.CPP；STDAFX.H和STDAFX.CPP三個文件組成，編譯成功。程序可以用文件輸入和輸出為文件。文件格式為TXT文件。測試結果如下：
輸入文件：8.TXT 或手動輸入
8 //N
1
2
3
4
5
6
7
8
輸出結果為：或保存為TXT文件。（8OUT.TXT）
8
(36,0)
(-4,9.65685)
(-4,4)
(-4,1.65685)
(-4,0)
(-4,-1.65685)
(-4,-4)
(-4,-9.65685)
下面為FFT.CPP文件：
// FFT.cpp : 定義控制台應用程序的入口點。
#include "stdafx.h"
#include <iostream>
#include <complex>
#include <bitset>
#include <vector>
#include <conio.h>
#include <string>
#include <fstream>
using namespace std;
bool inputData(unsigned long &, vector<complex<double> >&); //手工輸入數據
void FFT(unsigned long &, vector<complex<double> >&); //FFT變換
void display(unsigned long &, vector<complex<double> >&); //顯示結果
bool readDataFromFile(unsigned long &, vector<complex<double> >&); //從文件中讀取數據
bool saveResultToFile(unsigned long &, vector<complex<double> >&); //保存結果至文件中
const double PI = 3.1415926;
int _tmain(int argc, _TCHAR* argv[])
{
vector<complex<double> > vecList; //有限長序列
unsigned long ulN = 0; //N
char chChoose = ' '; //功能選擇
//功能循環
while(chChoose != 'Q' && chChoose != 'q')
{
//顯示選擇項
cout << "\nPlease chose a function" << endl;
cout << "\t1.Input data manually, press 'M':" << endl;
cout << "\t2.Read data from file, press 'F':" << endl;
cout << "\t3.Quit, press 'Q'" << endl;
cout << "Please chose:";
//輸入選擇
chChoose = getch();
//判斷
switch(chChoose)
{
case 'm': //手工輸入數據
case 'M':
if(inputData(ulN, vecList))
{
FFT(ulN, vecList);
display(ulN, vecList);
saveResultToFile(ulN, vecList);
}
break;
case 'f': //從文檔讀取數據
case 'F':
if(readDataFromFile(ulN, vecList))
{
FFT(ulN, vecList);
display(ulN, vecList);
saveResultToFile(ulN, vecList);
}
break;
}
}
return 0;
}
bool Is2Power(unsigned long ul) //判斷是否是2的整數次冪
{
if(ul < 2)
return false;
while( ul > 1 )
{
if( ul % 2 )
return false;
ul /= 2;
}
return true;
}
bool inputData(unsigned long & ulN, vector<complex<double> >& vecList)
{
//題目
cout<< "\n\n\n==============================Input Data===============================" << endl;
//輸入N
cout<< "\nInput N:";
cin>>ulN;
if(!Is2Power(ulN)) //驗證N的有效性
{
cout<< "N is invalid (N must like 2, 4, 8, .....), please retry." << endl;
return false;
}
//輸入各元素
vecList.clear(); //清空原有序列
complex<double> c;
for(unsigned long i = 0; i < ulN; i++)
{
cout << "Input x(" << i << "):";
cin >> c;
vecList.push_back(c);
}
return true;
}
bool readDataFromFile(unsigned long & ulN, vector<complex<double> >& vecList) //從文件中讀取數據
{
//題目
cout<< "\n\n\n===============Read Data From File==============" << endl;
//輸入文件名
string strfilename;
cout << "Input filename:" ;
cin >> strfilename;
//打開文件
cout << "open file " << strfilename << "......." <<endl;
ifstream loadfile;
loadfile.open(strfilename.c_str());
if(!loadfile)
{
cout << "\tfailed" << endl;
return false;
}
else
{
cout << "\tsucceed" << endl;
}
vecList.clear();
//讀取N
loadfile >> ulN;
if(!loadfile)
{
cout << "can't get N" << endl;
return false;
}
else
{
cout << "N = " << ulN << endl;
}
//讀取元素
complex<double> c;
for(unsigned long i = 0; i < ulN; i++)
{
loadfile >> c;
if(!loadfile)
{
cout << "can't get enough infomation" << endl;
return false;
}
else
cout << "x(" << i << ") = " << c << endl;
vecList.push_back(c);
}
//關閉文件
loadfile.close();
return true;
}
bool saveResultToFile(unsigned long & ulN, vector<complex<double> >& vecList) //保存結果至文件中
{
//詢問是否需要將結果保存至文件
char chChoose = ' ';
cout << "Do you want to save the result to file? (y/n):";
chChoose = _getch();
if(chChoose != 'y' && chChoose != 'Y')
{
return true;
}
//輸入文件名
string strfilename;
cout << "\nInput file name:" ;
cin >> strfilename;
cout << "Save result to file " << strfilename << "......" << endl;
//打開文件
ofstream savefile(strfilename.c_str());
if(!savefile)
{
cout << "can't open file" << endl;
return false;
}
//寫入N
savefile << ulN << endl;
//寫入元素
for(vector<complex<double> >::iterator i = vecList.begin(); i < vecList.end(); i++)
{
savefile << *i << endl;
}
//寫入完畢
cout << "save succeed." << endl;
//關閉文件
savefile.close();
return true;
}
void FFT(unsigned long & ulN, vector<complex<double> >& vecList)
{
//得到冪數
unsigned long ulPower = 0; //冪數
unsigned long ulN1 = ulN - 1;
while(ulN1 > 0)
{
ulPower++;
ulN1 /= 2;
}
//反序
bitset<sizeof(unsigned long) * 8> bsIndex; //二進制容器
unsigned long ulIndex; //反轉後的序號
unsigned long ulK;
for(unsigned long p = 0; p < ulN; p++)
{
ulIndex = 0;
ulK = 1;
bsIndex = bitset<sizeof(unsigned long) * 8>(p);
for(unsigned long j = 0; j < ulPower; j++)
{
ulIndex += bsIndex.test(ulPower - j - 1) ? ulK : 0;
ulK *= 2;
}
if(ulIndex > p)
{
complex<double> c = vecList[p];
vecList[p] = vecList[ulIndex];
vecList[ulIndex] = c;
}
}
//計算旋轉因子
vector<complex<double> > vecW;
for(unsigned long i = 0; i < ulN / 2; i++)
{
vecW.push_back(complex<double>(cos(2 * i * PI / ulN) , -1 * sin(2 * i * PI / ulN)));
}
for(unsigned long m = 0; m < ulN / 2; m++)
{
cout<< "\nvW[" << m << "]=" << vecW[m];
}
//計算FFT
unsigned long ulGroupLength = 1; //段的長度
unsigned long ulHalfLength = 0; //段長度的一半
unsigned long ulGroupCount = 0; //段的數量
complex<double> cw; //WH(x)
complex<double> c1; //G(x) + WH(x)
complex<double> c2; //G(x) - WH(x)
for(unsigned long b = 0; b < ulPower; b++)
{
ulHalfLength = ulGroupLength;
ulGroupLength *= 2;
for(unsigned long j = 0; j < ulN; j += ulGroupLength)
{
for(unsigned long k = 0; k < ulHalfLength; k++)
{
cw = vecW[k * ulN / ulGroupLength] * vecList[j + k + ulHalfLength];
c1 = vecList[j + k] + cw;
c2 = vecList[j + k] - cw;
vecList[j + k] = c1;
vecList[j + k + ulHalfLength] = c2;
}
}
}
}
void display(unsigned long & ulN, vector<complex<double> >& vecList)
{
cout << "\n\n===========================Display The Result=========================" << endl;
for(unsigned long d = 0; d < ulN;d++)
{
cout << "X(" << d << ")\t\t\t = " << vecList[d] << endl;
}
}
下面為STDAFX.H文件：
// stdafx.h : 標准系統包含文件的包含文件，
// 或是常用但不常更改的項目特定的包含文件
#pragma once
#include <iostream>
#include <tchar.h>
// TODO: 在此處引用程序要求的附加頭文件
下面為STDAFX.CPP文件：
// stdafx.cpp : 只包括標准包含文件的源文件
// FFT.pch 將成為預編譯頭
// stdafx.obj 將包含預編譯類型信息
#include "stdafx.h"
// TODO: 在 STDAFX.H 中
//引用任何所需的附加頭文件，而不是在此文件中引用
❺ 求FFT的C語言程序
float ar[1024],ai[1024];/* 原始數據實部，虛部 */
float a[2050];
void fft(int nn) /* nn數據長度 */
{
int n1,n2,i,j,k,l,m,s,l1;
float t1,t2,x,y;
float w1,w2,u1,u2,z;
float fsin[10]={0.000000,1.000000,0.707107,0.3826834,0.1950903,0.09801713,0.04906767,0.02454123,0.01227154,0.00613588,};
float fcos[10]={-1.000000,0.000000,0.7071068,0.9238796,0.9807853,0.99518472,0.99879545,0.9996988,0.9999247,0.9999812,};
switch(nn)
{
case 1024: s=10; break;
case 512: s=9; break;
case 256: s=8; break;
}
n1=nn/2; n2=nn-1;
j=1;
for(i=1;i<=nn;i++)
{
a[2*i]=ar[i-1];
a[2*i+1]=ai[i-1];
}
for(l=1;l<n2;l++)
{
if(l<j)
{
t1=a[2*j];
t2=a[2*j+1];
a[2*j]=a[2*l];
a[2*j+1]=a[2*l+1];
a[2*l]=t1;
a[2*l+1]=t2;
}
k=n1;
while (k<j)
{
j=j-k;
k=k/2;
}
j=j+k;
}
for(i=1;i<=s;i++)
{
u1=1;
u2=0;
m=(1<<i);
k=m>>1;
w1=fcos[i-1];
w2=-fsin[i-1];
for(j=1;j<=k;j++)
{
for(l=j;l<nn;l=l+m)
{
l1=l+k;
t1=a[2*l1]*u1-a[2*l1+1]*u2;
t2=a[2*l1]*u2+a[2*l1+1]*u1;
a[2*l1]=a[2*l]-t1;
a[2*l1+1]=a[2*l+1]-t2;
a[2*l]=a[2*l]+t1;
a[2*l+1]=a[2*l+1]+t2;
}
z=u1*w1-u2*w2;
u2=u1*w2+u2*w1;
u1=z;
}
}
for(i=1;i<=nn/2;i++)
{
ar[i]=4*a[2*i+2]/nn; /* 實部 */
ai[i]=-4*a[2*i+3]/nn; /* 虛部 */
a[i]=4*sqrt(ar[i]*ar[i]+ai[i]*ai[i]); /* 幅值 */
}
}
❻ 求用C語言實現FFT變換的程序（見下面）
你好，這是我的回答，希望可以幫到你。
1）結果討論
一，如果對信號進行同樣點數N的FFT變換，采樣頻率fs越高，則可以分析越高頻的信號；與此同時，采樣頻率越低，對於低頻信號的頻譜解析度則越好。
二，假設采樣點不在正弦信號的波峰、波谷、以及0電壓處，頻譜則會產生泄露（leakage）。
三，對於同樣的采樣率fs，提高FFT的點數N，則可提高頻譜的解析度。
四，如果采樣頻率fs小於2倍信號頻率2*fs（奈圭斯特定理），則頻譜分析結果會出錯。
五，對於（二）中泄露現象，可以通過在信號後面補零點解決。
2）程序及註解如下
%清除命令窗口及變數
clc;
clear all;
%輸入f、N、T、是否補零（補幾個零）
f=input('Input frequency of the signal: f\n');
N=input('Input number of pointsl: N\n');
T=input('Input sampling time: T\n');
flag=input('Add zero too sampling signal or not? yes=1 no=0\n');
if(flag)
ZeroNum=input('Input nmber of zeros\n');
else
ZeroNum=0;
end
%生成信號，signal是原信號。signal為采樣信號。
fs=1/T;
t=0:0.00001:T*(N+ZeroNum-1);
signal=sin(2*pi*f*t);
t2=0:T:T*(N+ZeroNum-1);
signal2=sin(2*pi*f*t2);
if (flag)
signal2=[signal2 zeros(1, ZeroNum)];
end
%畫出原信號及采樣信號。
figure;
subplot(2,1,1);
plot(t,signal);
xlabel('Time(s)');
ylabel('Amplitude(volt)');
title('Singnal');
hold on;
subplot(2,1,1);
stem(t2,signal2,'r');
axis([0 T*(N+ZeroNum) -1 1]);
%作FFT變換，計算其幅值，歸一化處理，並畫出頻譜。
Y = fft(signal2,N);
Pyy = Y.* conj(Y) ;
Pyy=(Pyy/sum(Pyy))*2;
f=0:fs/(N-1):fs/2;4
subplot(2,1,2);
bar(f,Pyy(1:N/2));
xlabel('Frequency(Hz)');
ylabel('Amplitude');
title('Frequency compnents of signal');
axis([0 fs/2 0 ceil(max(Pyy))])
grid on;
祝你好運！
我可以幫助你，你先設置我最佳答案後，我網路Hii教你。
❼ 用c語言實現FFT
float ar[1024],ai[1024];/* 原始數據實部，虛部 */
float a[2050];
void fft(int nn) /* nn數據長度 */
{
int n1,n2,i,j,k,l,m,s,l1;
float t1,t2,x,y;
float w1,w2,u1,u2,z;
float fsin[10]={0.000000,1.000000,0.707107,0.3826834,0.1950903,0.09801713,0.04906767,0.02454123,0.01227154,0.00613588,};
float fcos[10]={-1.000000,0.000000,0.7071068,0.9238796,0.9807853,0.99518472,0.99879545,0.9996988,0.9999247,0.9999812,};
switch(nn)
{
case 1024: s=10; break;
case 512: s=9; break;
case 256: s=8; break;
}
n1=nn/2; n2=nn-1;
j=1;
for(i=1;i<=nn;i++)
{
a[2*i]=ar[i-1];
a[2*i+1]=ai[i-1];
}
for(l=1;l<n2;l++)
{
if(l<j)
{
t1=a[2*j];
t2=a[2*j+1];
a[2*j]=a[2*l];
a[2*j+1]=a[2*l+1];
a[2*l]=t1;
a[2*l+1]=t2;
}
k=n1;
while (k<j)
{
j=j-k;
k=k/2;
}
j=j+k;
}
for(i=1;i<=s;i++)
{
u1=1;
u2=0;
m=(1<<i);
k=m>>1;
w1=fcos[i-1];
w2=-fsin[i-1];
for(j=1;j<=k;j++)
{
for(l=j;l<nn;l=l+m)
{
l1=l+k;
t1=a[2*l1]*u1-a[2*l1+1]*u2;
t2=a[2*l1]*u2+a[2*l1+1]*u1;
a[2*l1]=a[2*l]-t1;
a[2*l1+1]=a[2*l+1]-t2;
a[2*l]=a[2*l]+t1;
a[2*l+1]=a[2*l+1]+t2;
}
z=u1*w1-u2*w2;
u2=u1*w2+u2*w1;
u1=z;
}
}
for(i=1;i<=nn/2;i++)
{
ar[i]=4*a[2*i+2]/nn; /* 實部 */
ai[i]=-4*a[2*i+3]/nn; /* 虛部 */
a[i]=4*sqrt(ar[i]*ar[i]+ai[i]*ai[i]); /* 幅值 */
}
}
(http://..com/question/284943905.html?an=0&si=2)
打字不易，如滿意，望採納。
❽ C語言輸出到文件吶，就是一個fft的結果輸出來求指點啊
freopen("in.txt","r",stdin);//文件輸入
freopen("out.txt","w",stdout);//文件輸出
頭文件#include<stdlib.h>
using namespace std;
❾ 求基2、基4、基8FFT（快速傅里葉變換）的c語言程序，要能運行得出來的
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
typedef struct{
double r;
double i;
}my_complex
;
//檢查a是否為2的整數次方數
#define NOT2POW(a) (((a)-1)&(a)||(a)<=0)
//pi
#define MYPI 3.14159265358979323846
my_complex* fft(const my_complex* x, unsigned int len){
unsigned int ex=0,t=len;
unsigned int i,j,k;
my_complex *y;
double tr,ti,rr,ri,yr,yi;
if(NOT2POW(len)) return NULL; //如果失敗，返回空指針
for(;!(t&1);t>>=1) ex++; //len應該等於2的ex次方
y=(my_complex*)malloc(len*sizeof(my_complex));
if(!y) return NULL;
//變址計算，庫里-圖基演算法
for(i=0;i<len;i++){
k=i;
j=0;
t=ex;
while((t--)>0){
j<<=1;
j|=k&1;
k>>=1;
}
if(j>=i){
y[i]=x[j];
y[j]=x[i];
}
}
//用變址後的y向量進行計算
for(i=0;i<ex;i++){
t=1<<i;
for(j=0;j<len;j+=t<<1){
for(k=0;k<t;k++){
ti=-MYPI*k/t;
rr=cos(ti);
ri=sin(ti);
tr=y[j+k+t].r;
ti=y[j+k+t].i;
yr=rr*tr-ri*ti;
yi=rr*ti+ri*tr;
tr=y[j+k].r;
ti=y[j+k].i;
y[j+k].r=tr+yr;
y[j+k].i=ti+yi;
y[j+k+t].r=tr-yr;
y[j+k+t].i=ti-yi;
}
}
}
return y;
}
//以下為測試
int main()
{
int i,DATA_LEN;
my_complex *x,*y;
printf("基二FFT測試\n輸入生成序列長度:");
scanf("%d",&DATA_LEN);
x=(my_complex*)malloc(DATA_LEN*sizeof(my_complex));
for(i=0;i<DATA_LEN;i++){
x[i].r=i;
x[i].i=i-1;
}
printf("處理前...\n實部\t\t虛部\n");
for(i=0;i<DATA_LEN;i++)
printf("%lf\t%lf\n",x[i].r,x[i].i);
y=fft(x,DATA_LEN);
if(!y){
printf("序列長度不為2的整數次方!\n");
return 0;
}
printf("處理後...\n實部\t\t虛部\n");
for(i=0;i<DATA_LEN;i++)
printf("%lf\t%lf\n",y[i].r,y[i].i);
free(y);
free(x);
return 0;
}
❿ 誰能給一個C語言的FFT演算法程序啊，急用
typedef struct complex_struct
{
float real;
float img;
} complex;
void FFT(complex *TD,complex *FD,int r)
{
int count;
int i,j,k;
int bfsize,p;
float angle;
complex *W,*X1,*X2,*X;
W=w; X1=x1; X2=x2;
count=1<<r;
for ( i=0;i<count/2;i++ )
{
angle=-i*PI*2/count;
(W+i)->real=cos(angle);
(W+i)->img=sin(angle);
}
for ( i=0;i<count;i++ )
{
(X1+i)->real=(TD+i)->real;
(X1+i)->img=(TD+i)->img;
}
for ( k=0;k<r;k++ )
{
for ( j=0;j<1<<k;j++ )
{
bfsize=1<<(r-k);
for ( i=0;i<bfsize/2;i++ )
{
p=j*bfsize;
(X2+i+p)->real=(X1+i+p)->real+(X1+i+p+bfsize/2)->real;
(X2+i+p)->img=(X1+i+p)->img+(X1+i+p+bfsize/2)->img;
(X2+i+p+bfsize/2)->real=((X1+i+p)->real-(X1+i+p+bfsize/2)->real)
*(W+i*(1<<k))->real-((X1+i+p)->img -(X1+i+p+bfsize/2)->img)*(W+i*(1<<k))->img;
(X2+i+p+bfsize/2)->img=((X1+i+p)->real-(X1+i+p+bfsize/2)->real)
*(W+i*(1<<k))->img+((X1+i+p)->img -(X1+i+p+bfsize/2)->img)*(W+i*(1<<k))->real; }
}
X=X1; X1=X2; X2=X;
}
for ( j=0;j<count;j++ )
{
p = 0;
for ( i=0;i<r;i++ )
if ( j&(1<<i) )
p+=1<<(r-i-1);
(FD+j)->real=(X1+p)->real;
(FD+j)->img=(X1+p)->img;
}
}