求FFT的c语言程序

分类: 教育/科学 >>学习帮助

问题描述:

追20分

解析:

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 <plex>

#include <bitset>

#include <vector>

#include <conio.h>

#include <string>

#include <fstream>

using namespace std

bool inputData(unsigned long &, vector<plex<double>>&)手工输入数据

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

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

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

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

const double PI = 3.1415926

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

{

vector<plex<double>>vecList有限长序列

unsigned long ulN = 0N

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<plex<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

plex<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<plex<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

plex<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<plex<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<plex<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<plex<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)

{

plex<double>c = vecList[p]

vecList[p] = vecList[ulIndex]

vecList[ulIndex] = c

}

}

计算旋转因子

vector<plex<double>>vecW

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

{

vecW.push_back(plex<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

plex<double>cwWH(x)

plex<double>c1G(x) + WH(x)

plex<double>c2G(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<plex<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

float ar[1024],ai[1024]/* 原始数据实部，虚部 */

float a[2050]

void fft(int nn) /* nn数据长度 */

{

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=10break

case 512: s=9 break

case 256: s=8 break

}

n1=nn/2 n2=nn-1

j=1

for(i=1i<=nni++)

{

a[2*i]=ar[i-1]

a[2*i+1]=ai[i-1]

}

for(l=1l<n2l++)

{

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=1i<=si++)

{

u1=1

u2=0

m=(1<<i)

k=m>>1

w1=fcos[i-1]

w2=-fsin[i-1]

for(j=1j<=kj++)

{

for(l=jl<nnl=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=1i<=nn/2i++)

{

ar[i]=4*a[2*i+2]/nn/* 实部 */

ai[i]=-4*a[2*i+3]/nn/* 虚部 */

a[i]=4*sqrt(ar[i]*ar[i]+ai[i]*ai[i])/* 幅值 */

}

}

(http://zhidao.baidu.com/question/284943905.html?an=0&si=2)

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