求FFT的c语言程序

求FFT的c语言程序,第1张

快速傅里叶变换 要用C++ 才行吧 你可以用MATLAB来实现更方便点啊

此FFT 是用VC6.0编写,由FFT.CPP;STDAFX.H和STDAFX.CPP三个文件组成,编译成功。程序可以用文件输入和输出为文件。文件格式为TXT文件。测试结果如下:

输入文件:8.TXT 或手动输入

8 //N

1

2

3

4

5

6

7

8

输出结果为:或保存为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)

下面为FFT.CPP文件:

// FFT.cpp : 定义控制台应用程序的入口点。

#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>>&) //手工输入数据

void FFT(unsigned long &, vector<complex<double>>&) //FFT变换

void display(unsigned long &, vector<complex<double>>&) //显示结果

bool readDataFromFile(unsigned long &, vector<complex<double>>&) //从文件中读取数据

bool saveResultToFile(unsigned long &, vector<complex<double>>&) //保存结果至文件中

const double PI = 3.1415926

int _tmain(int argc, _TCHAR* argv[])

{

vector<complex<double>>vecList //有限长序列

unsigned long ulN = 0 //N

char chChoose = ' ' //功能选择

//功能循环

while(chChoose != 'Q' &&chChoose != 'q')

{

//显示选择项

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:"

//输入选择

chChoose = getch()

//判断

switch(chChoose)

{

case 'm': //手工输入数据

case 'M':

if(inputData(ulN, vecList))

{

FFT(ulN, vecList)

display(ulN, vecList)

saveResultToFile(ulN, vecList)

}

break

case 'f'://从文档读取数据

case 'F':

if(readDataFromFile(ulN, vecList))

{

FFT(ulN, vecList)

display(ulN, vecList)

saveResultToFile(ulN, vecList)

}

break

}

}

return 0

}

bool Is2Power(unsigned long ul) //判断是否是2的整数次幂

{

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)

{

//题目

cout<<"\n\n\n==============================Input Data===============================" <<endl

//输入N

cout<<"\nInput N:"

cin>>ulN

if(!Is2Power(ulN)) //验证N的有效性

{

cout<<"N is invalid (N must like 2, 4, 8, .....), please retry." <<endl

return false

}

//输入各元素

vecList.clear() //清空原有序列

complex<double>c

for(unsigned long i = 0i <ulNi++)

{

cout <<"Input x(" <<i <<"):"

cin >> c

vecList.push_back(c)

}

return true

}

bool readDataFromFile(unsigned long &ulN, vector<complex<double>>&vecList) //从文件中读取数据

{

//题目

cout<<"\n\n\n===============Read Data From File==============" <<endl

//输入文件名

string strfilename

cout <<"Input filename:"

cin >>strfilename

//打开文件

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()

//读取N

loadfile >>ulN

if(!loadfile)

{

cout <<"can't get N" <<endl

return false

}

else

{

cout <<"N = " <<ulN <<endl

}

//读取元素

complex<double>c

for(unsigned long i = 0i <ulNi++)

{

loadfile >>c

if(!loadfile)

{

cout <<"can't get enough infomation" <<endl

return false

}

else

cout <<"x(" <<i <<") = " <<c <<endl

vecList.push_back(c)

}

//关闭文件

loadfile.close()

return true

}

bool saveResultToFile(unsigned long &ulN, vector<complex<double>>&vecList) //保存结果至文件中

{

//询问是否需要将结果保存至文件

char chChoose = ' '

cout <<"Do you want to save the result to file? (y/n):"

chChoose = _getch()

if(chChoose != 'y' &&chChoose != 'Y')

{

return true

}

//输入文件名

string strfilename

cout <<"\nInput file name:"

cin >>strfilename

cout <<"Save result to file " <<strfilename <<"......" <<endl

//打开文件

ofstream savefile(strfilename.c_str())

if(!savefile)

{

cout <<"can't open file" <<endl

return false

}

//写入N

savefile <<ulN <<endl

//写入元素

for(vector<complex<double>>::iterator i = vecList.begin()i <vecList.end()i++)

{

savefile <<*i <<endl

}

//写入完毕

cout <<"save succeed." <<endl

//关闭文件

savefile.close()

return true

}

void FFT(unsigned long &ulN, vector<complex<double>>&vecList)

{

//得到幂数

unsigned long ulPower = 0 //幂数

unsigned long ulN1 = ulN - 1

while(ulN1 >0)

{

ulPower++

ulN1 /= 2

}

//反序

bitset<sizeof(unsigned long) * 8>bsIndex //二进制容器

unsigned long ulIndex//反转后的序号

unsigned long ulK

for(unsigned long p = 0p <ulNp++)

{

ulIndex = 0

ulK = 1

bsIndex = bitset<sizeof(unsigned long) * 8>(p)

for(unsigned long j = 0j <ulPowerj++)

{

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

}

}

//计算旋转因子

vector<complex<double>>vecW

for(unsigned long i = 0i <ulN / 2i++)

{

vecW.push_back(complex<double>(cos(2 * i * PI / ulN) , -1 * sin(2 * i * PI / ulN)))

}

for(unsigned long m = 0m <ulN / 2m++)

{

cout<<"\nvW[" <<m <<"]=" <<vecW[m]

}

//计算FFT

unsigned long ulGroupLength = 1 //段的长度

unsigned long ulHalfLength = 0 //段长度的一半

unsigned long ulGroupCount = 0//段的数量

complex<double>cw //WH(x)

complex<double>c1 //G(x) + WH(x)

complex<double>c2 //G(x) - WH(x)

for(unsigned long b = 0b <ulPowerb++)

{

ulHalfLength = ulGroupLength

ulGroupLength *= 2

for(unsigned long j = 0j <ulNj += ulGroupLength)

{

for(unsigned long k = 0k <ulHalfLengthk++)

{

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 = 0d <ulNd++)

{

cout <<"X(" <<d <<")\t\t\t = " <<vecList[d] <<endl

}

}

下面为STDAFX.H文件:

// stdafx.h : 标准系统包含文件的包含文件,

// 或是常用但不常更改的项目特定的包含文件

#pragma once

#include <iostream>

#include <tchar.h>

// TODO: 在此处引用程序要求的附加头文件

下面为STDAFX.CPP文件:

// stdafx.cpp : 只包括标准包含文件的源文件

// FFT.pch 将成为预编译头

// stdafx.obj 将包含预编译类型信息

#include "stdafx.h"

// TODO: 在 STDAFX.H 中

//引用任何所需的附加头文件,而不是在此文件中引用

#include <stdio.h>

#include <math.h>

#include <stdlib.h>

#define N 1000

/*定义复数类型*/

typedef struct{

double real

double img

}complex

complex x[N], *W/*输入序列,变换核*/

int size_x=0 /*输入序列的大小,在本程序中仅限2的次幂*/

double PI/*圆周率*/

void fft()/*快速傅里叶变换*/

void initW() /*初始化变换核*/

void change()/*变址*/

void add(complex ,complex ,complex *)/*复数加法*/

void mul(complex ,complex ,complex *)/*复数乘法*/

void sub(complex ,complex ,complex *)/*复数减法*/

void output()

int main(){

int i/*输出结果*/

system("cls")

PI=atan(1)*4

printf("Please input the size of x:\n")

scanf("%d",&size_x)

printf("Please input the data in x[N]:\n")

for(i=0i<size_xi++)

scanf("%lf%lf",&x[i].real,&x[i].img)

initW()

fft()

output()

return 0

}

/*快速傅里叶变换*/

void fft(){

int i=0,j=0,k=0,l=0

complex up,down,product

change()

for(i=0i<log(size_x)/log(2) i++){ /*一级蝶形运算*/

l=1<<i

for(j=0j<size_xj+= 2*l ){ /*一组蝶形运算*/

for(k=0k<lk++){/*一个蝶形运算*/

mul(x[j+k+l],W[size_x*k/2/l],&product)

add(x[j+k],product,&up)

sub(x[j+k],product,&down)

x[j+k]=up

x[j+k+l]=down

}

}

}

}

/*初始化变换核*/

void initW(){

int i

W=(complex *)malloc(sizeof(complex) * size_x)

for(i=0i<size_xi++){

W[i].real=cos(2*PI/size_x*i)

W[i].img=-1*sin(2*PI/size_x*i)

}

}

/*变址计算,将x(n)码位倒置*/

void change(){

complex temp

unsigned short i=0,j=0,k=0

double t

for(i=0i<size_xi++){

k=ij=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

}

}

}

/*输出傅里叶变换的结果*/

void output(){

int i

printf("The result are as follows\n")

for(i=0i<size_xi++){

printf("%.4f",x[i].real)

if(x[i].img>=0.0001)printf("+%.4fj\n",x[i].img)

else if(fabs(x[i].img)<0.0001)printf("\n")

else printf("%.4fj\n",x[i].img)

}

}

void add(complex a,complex b,complex *c){

c->real=a.real+b.real

c->img=a.img+b.img

}

void mul(complex a,complex b,complex *c){

c->real=a.real*b.real - a.img*b.img

c->img=a.real*b.img + a.img*b.real

}

void sub(complex a,complex b,complex *c){

c->real=a.real-b.real

c->img=a.img-b.img

}

function y=myditfft(x)

%本程序对输入序列实现DIT-FFT基2算法,点数取大于等于长度的2的幂次

%------------------------------------

%

myditfft.c

%------------------------------------

m=nextpow2(x)

%求的x长度对应的2的最低幂次m

N=2^m

if length(x)<N

x=[x,zeros(1,N-length(x))]

%若的长度不是2的幂,补0到2的整数幂

end

nxd=bin2dec(fliplr(dec2bin([1:N]-1,m)))+1

%求1:2^m数列的倒序

y=x(nxd)

%将倒序排列作为的初始值

for mm=1:m

%将DFT做m次基2分解,从左到右,对每次分解作DFT运算

Nmr=2^mm

u=1

%旋转因子u初始化

WN=exp(-i*2*pi/Nmr)

%本次分解的基本DFT因子WN=exp(-i*2*pi/Nmr)

for j=1:Nmr/2

%本次跨越间隔内的各次碟形运算

for k=j:Nmr:N

%本次碟形运算的跨越间隔为Nmr=2^mm

kp=k+Nmr/2

%确定碟形运算的对应单元下标

t=y(kp)*u

%碟形运算的乘积项

y(kp)=y(k)-t

%碟形运算的加法项

y(k)=y(k)+t

end

u=u*WN

%修改旋转因子,多乘一个基本DFT因子WN

end

end


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