如何用C语言或汇编语言实现FFT(快速傅里叶)变换,并写出C语言或汇编代码,万分感谢。 如何用单片机实现文字滚动,汇编语言或C语言都可以

\u6c42\u4e2a\u5feb\u901f\u5085\u91cc\u53f6\u53d8\u6362\u7684C\u8bed\u8a00\u7a0b\u5e8f

void fft(){ int nn,n1,n2,i,j,k,l,m,s,l1; float ar[1024],ai[1024]; // \u5b9e\u90e8 \u865a\u90e8 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,};// \u4f18\u5316 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=1;i>1; w1=fcos[i-1]; w2=-fsin[i-1]; for(j=1;j<=k;j++) { for(l=j;l<nn;l=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=1;i<=nn/2;i++) { 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]); // \u5e45\u503c }}

\u5982\u4f55\u7528\u5355\u7247\u673a\u5b9e\u73b0\u6587\u5b57\u6eda\u52a8\uff0c
----
\u5148\u5b9e\u73b0\u663e\u793a\u4e00\u5e45\u7a33\u5b9a\u7684\u6587\u5b57\uff0c\u518d\u95f4\u9694\u51e0\u767e\u6beb\u79d2\uff0c\u663e\u793a\u53e6\u4e00\u5e45\uff0c...
\u9010\u5e45\u7684\u66f4\u6362\uff0c\u5373\u53ef\u5b9e\u73b0\u6587\u5b57\u7684\u6eda\u52a8\u3002

\u663e\u793a\u4e00\u5e45\u7a33\u5b9a\u7684\u6587\u5b57\uff0c\u697c\u4e3b\u5df2\u7ecf\u5b9e\u73b0\u4e86\u5417\uff1f
\u663e\u793a\u51fa\u6765\u770b\u770b\uff0c\u6c47\u7f16\u8bed\u8a00\u6216C\u8bed\u8a00\u90fd\u53ef\u4ee5\u3002

\u53e6\u5916\uff0c\u697c\u4e3b\u4e5f\u6ca1\u6709\u7ed9\u51fa\u663e\u793a\u5c4f\u7684\u7535\u8def\uff0c\u600e\u4e48\u597d\u7f16\u5199\u7a0b\u5e8f\uff1f
\u6709\u4e2a\u73b0\u6210\u7684\uff0c\u53ef\u53c2\u8003\u4e00\u4e0b\uff1a
http://hi.baidu.com/%D7%F6%B6%F8%C2%DB%B5%C0/blog/item/7bfd810fd4f54bd07acbe167.html

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<n2;l++)
{
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=1;i<=s;i++)
{
u1=1;
u2=0;
m=(1<<i);
k=m>>1;
w1=fcos[i-1];
w2=-fsin[i-1];
for(j=1;j<=k;j++)
{
for(l=j;l<nn;l=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=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 */
}
}

去c++吧看看,我记得有人发了一个。

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