怎么用C语言实现FFT算法 呀

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怎么用C语言实现FFT算法 呀,第1张

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)

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 */

}

}

#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

}