double
A[2049]={0}
double
B[1100]={0}
double
powerA[1025]={0}
改成
A[256]={0}
B[130]={0}
power[129]={0}就行了，
void
FFT(double
data[],
int
nn,
int
isign)
的程序可以针对任何点数，只要是2的n次方
具体程序如下：
#include
<iostream.h>
#include
"math.h"
#include<stdio.h>
#include<string.h>
#include
<stdlib.h>
#include
<fstream.h>
#include
<afx.h>
void
FFT(double
data[],
int
nn,
int
isign)
{
//复数的快速傅里叶变换
int
n,j,i,m,mmax,istep
double
tempr,tempi,theta,wpr,wpi,wr,wi,wtemp
n
=
2
*
nn
j
=
1
for
(i
=
1
i<=n
i=i+2)
//这个循环进行的是码位倒置。
{
if(
j
>
i)
{
tempr
=
data[j]
tempi
=
data[j
+
1]
data[j]
=
data[i]
data[j
+
1]
=
data[i
+
1]
data[i]
=
tempr
data[i
+
1]
=
tempi
}
m
=
n
/
2
while
(m
>=
2
&&
j
>
m)
{
j
=
j
-
m
m
=
m
/
2
}
j
=
j
+
m
}
mmax
=
2
while(
n
>
mmax
)
{
istep
=
2
*
mmax
//这里表示一次的数字的变化。也体现了级数，若第一级时，也就是书是的第0级，其为两个虚数，所以对应数组应该增加4，这样就可以进入下一组运算
theta
=
-6.28318530717959
/
(isign
*
mmax)
wpr
=
-2.0
*
sin(0.5
*
theta)*sin(0.5
*
theta)
wpi
=
sin(theta)
wr
=
1.0
wi
=
0.0
for(
m
=
1
m<=mmax
m=m+2)
{
for
(i
=
m
i<=n
i=i+istep)
{
j
=
i
+
mmax
tempr=double(wr)*data[j]-double(wi)*data[j+1]//这两句表示蝶形因子的下一个数乘以W因子所得的实部和虚部。
tempi=double(wr)*data[j+1]+double(wi)*data[j]
data[j]
=
data[i]
-
tempr
//蝶形单元计算后下面单元的实部，下面为虚部，注意其变换之后的数组序号与书上蝶形单元是一致的
data[j
+
1]
=
data[i
+
1]
-
tempi
data[i]
=
data[i]
+
tempr
data[i
+
1]
=
data[i
+
1]
+
tempi
}
wtemp
=
wr
wr
=
wr
*
wpr
-
wi
*
wpi
+
wr
wi
=
wi
*
wpr
+
wtemp
*
wpi
+
wi
}
mmax
=
istep
}
}
void
main()
{
//本程序已经和MATLAB运算结果对比，准确无误，需要注意的的是，计算中数组都是从1开始取得，丢弃了A[0]等数据
double
A[2049]={0}
double
B[1100]={0}
double
powerA[1025]={0}
char
line[50]
char
dataA[20],
dataB[20]
int
ij
char
ch1[3]="\t"
char
ch2[3]="\n"
int
strl1,strl2
CString
str1,str2
ij=1
//********************************读入文件data1024.txt中的数据,
其中的数据格式见该文件
FILE
*fp
=
fopen("data1024.txt","r")
if(!fp)
{
cout<<"Open
file
is
failing!"<<endl
return
}
while(!feof(fp))
//feof(fp)有两个返回值:如果遇到文件结束，函数feof（fp）的值为1，否则为0。
{
memset(line,0,50)
//清空为0
memset(dataA,0,20)
memset(dataB,0,20)
fgets(line,50,fp)
//函数的功能是从fp所指文件中读入n-1个字符放入line为起始地址的空间内
sscanf(line,
"%s%s",
dataA,
dataB)
//我同时读入了两列值，但你要求1024个，那么我就只用了第一列的1024个值
//dataA读入第一列，dataB读入第二列
B[ij]=atof(dataA)
//将字符型的dataA值转化为float型
ij++
}
for
(int
mm=1mm<1025mm++)//A[2*mm-1]是实部，A[2*mm]是虚部，当只要输入实数时，那么保证虚部A[mm*2]为零即可
{
A[2*mm-1]=B[mm]
A[2*mm]=0
}
//*******************************************正式计算FFT
FFT(A,1024,1)
//********************************************写入数据到workout.txt文件中
for
(int
k=1k<2049k=k+2)
{
powerA[(k+1)/2]=sqrt(pow(A[k],2.0)+pow(A[k+1],2.0))//求功率谱
FILE
*pFile=fopen("workout.txt","a+")
//?a+只能在文件最后补充，光标在结尾。没有则创建
memset(ch1,0,15)
str1.Format("%.4f",powerA[(k+1)/2])
if
(A[k+1]>=0)
str2.Format("%d\t%6.4f%s%6.4f
%s",(k+1)/2,A[k],"+",A[k+1],"i")//保存fft计算的频谱，是复数频谱
else
str2.Format("%d\t%6.4f%6.4f
%s",(k+1)/2,A[k],A[k+1],"i")
strl1=strlen(str1)
strl2=strlen(str2)
//
用
法:fwrite(buffer,size,count,fp)
//
buffer：是一个指针，对fwrite来说，是要输出数据的地址。
//
size：要写入的字节数；
//
count:要进行写入size字节的数据项的个数；
//
fp:目标文件指针。
fwrite(str2,1,strl2,pFile)
fwrite(ch1,1,3,pFile)
fwrite(ch1,1,3,pFile)
fwrite(str1,1,strl1,pFile)
fwrite(ch2,1,3,pFile)
fclose(pFile)
}
cout<<"计算完毕，到fft_test\workout.txt查看结果"<<endl
}
#include <math.h>#include <stdio.h>
#define N 8
void kkfft(double pr[], double pi[], int n, int k, double fr[], double fi[], int l, int il)
void main()
{
double xr[N],xi[N],Yr[N],Yi[N],l=0,il=0
int i,j,n=N,k=3
for(i=0i<Ni++)
{
xr[i]=i
xi[i]=0
}
printf("------FFT------\n")
l=0
kkfft(xr,xi,n,k,Yr,Yi,l,il)
for(i=0i<Ni++)
{
printf("%-11lf + j* %-11lf\n",Yr[i],Yi[i])
}
printf("-----DFFT-------\n")
l=1
kkfft(Yr,Yi,n,k,xr,xi,l,il)
for(i=0i<Ni++)
{
printf("%-11lf + j* %-11lf\n",xr[i],xi[i])
}
getch()
}
void kkfft(double pr[], double pi[], int n, int k, double fr[], double fi[], int l, int il)
{
int it,m,is,i,j,nv,l0
double p,q,s,vr,vi,poddr,poddi
for (it=0it<=n-1it++)
{
m = it
is = 0
for(i=0i<=k-1i++)
{
j = m/2
is = 2*is+(m-2*j)
m = j
}
fr[it] = pr[is]
fi[it] = pi[is]
}
pr[0] = 1.0
pi[0] = 0.0
p = 6.283185306/(1.0*n)
pr[1] = cos(p)
pi[1] = -sin(p)
if (l!=0)
pi[1]=-pi[1]
for (i=2i<=n-1i++)
{
p = pr[i-1]*pr[1]
q = pi[i-1]*pi[1]
s = (pr[i-1]+pi[i-1])*(pr[1]+pi[1])
pr[i] = p-q
pi[i] = s-p-q
}
for (it=0it<=n-2it=it+2)
{
vr = fr[it]
vi = fi[it]
fr[it] = vr+fr[it+1]
fi[it] = vi+fi[it+1]
fr[it+1] = vr-fr[it+1]
fi[it+1] = vi-fi[it+1]
}
m = n/2
nv = 2
for (l0=k-2l0>=0l0--)
{
m = m/2
nv = 2*nv
for(it=0it<=(m-1)*nvit=it+nv)
for (j=0j<=(nv/2)-1j++)
{
p = pr[m*j]*fr[it+j+nv/2]
q = pi[m*j]*fi[it+j+nv/2]
s = pr[m*j]+pi[m*j]
s = s*(fr[it+j+nv/2]+fi[it+j+nv/2])
poddr = p-q
poddi = s-p-q
fr[it+j+nv/2] = fr[it+j]-poddr
fi[it+j+nv/2] = fi[it+j]-poddi
fr[it+j] = fr[it+j]+poddr
fi[it+j] = fi[it+j]+poddi
}
}
/*逆傅立叶变换*/
if(l!=0)
{
for(i=0i<=n-1i++)
{
fr[i] = fr[i]/(1.0*n)
fi[i] = fi[i]/(1.0*n)
}
}
/*是否计算模和相角*/
if(il!=0)
{
for(i=0i<=n-1i++)
{
pr[i] = sqrt(fr[i]*fr[i]+fi[i]*fi[i])
if(fabs(fr[i])<0.000001*fabs(fi[i]))
{
if ((fi[i]*fr[i])>0)
pi[i] = 90.0
else
pi[i] = -90.0
}
else
pi[i] = atan(fi[i]/fr[i])*360.0/6.283185306
}
}
return
}
void fft(){
int nn,n1,n2,i,j,k,l,m,s,l1
float ar[1024],ai[1024]// 实部 虚部
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,}// 优化
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=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]=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
}
}
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