1、二维FFT相当于对行和列分别进行一维FFT运算。具体的实现办法如下:
先对各行逐一进行一维FFT,然后再对变换后的新矩阵的各列逐一进行一维FFT。相应的伪代码如下所示:
for (int i=0; i
for (int j=0; j
其中,ROW[i]表示矩阵的第i行。注意这只是一个简单的记法,并不能完全照抄。还需要通过一些语句来生成各行的数据。同理,COL[i]是对矩阵的第i列的一种简单表示方法。
所以,关键是一维FFT算法的实现。
2、例程:
#include
#include
#include
#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=0;iscanf("%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=0;i< log(size_x)/log(2) ;i++){ /*一级蝶形运算*/
l=1< for(j=0;jfor(k=0;k 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=0;iW[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=0;ik=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;
}
}
}
/*输出傅里叶变换的结果*/
void output(){
int i;
printf("The result are as follows\n");
for(i=0;iprintf("%.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;
}
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=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
if(l
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-k;
k=k/2;
}
j=j+k;
}
for(i=1;i<=s;i++)
{
u1=1;
u2=0;
m=(1< k=m>>1;
w1=fcos[i-1];
w2=-fsin[i-1];
for(j=1;j<=k;j++)
{
for(l=j;l
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; /* 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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