📄 fft和ifft的c语言程序_声韵.htm
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<DIV class=tit>fft和ifft的C语言程序</DIV>
<DIV class=date>2008-04-16 19:03</DIV>
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最近要做FFT和IFFT，在网上搜来搜去，总是不大好理解，于是就根据大学里学的数字信号处理（西电.刘顺兰等）中的图3-9和图3-10自己写了一下，有用的朋友可以拿去，希望能对大家有帮助。</P>
<P>注：其中的M为11，这是因为我做了N=2048个点的FFT，2^11=2048，M值可根据需要更改，若N不是2的整数次幂，应补零。其中xin为输入信号，并且也应该是COMPX类型量。</P>
<P>1、按时间抽取（DIT）的基-2FFT算法（C）</P>
<P>#include "math.h"<BR>typedef
struct
//定义表示复数的结构体<BR>{<BR> float real;<BR> float
imag;<BR>}COMPX;<BR>COMPX EE(COMPX b1,COMPX b2)
//复数相乘函数<BR>{<BR> COMPX b3;<BR>
b3.real=b1.real*b2.real-b1.imag*b2.imag;<BR>
b3.imag=b1.real*b2.imag+b1.imag*b2.real;<BR>
return(b3);<BR>}</P>
<P>void FFT(COMPX *xin,int N)
//FFT运算<BR>{<BR> int
m,LH,nm,I,k,J,M,K;<BR> float p,ps
;<BR> int B,N1;<BR> COMPX
w,T;<BR>
M=11;
//对应N值为2048</P>
<P>//下面是倒序的程序 <BR>
LH=N/2;<BR> J=LH;<BR>
N1=N-2; <BR>
/*变址运算*/<BR> for(I=1;I<=N1;I++)<BR>
{<BR>
if(I<J)<BR>
{<BR>
T=xin[I];xin[I]=xin[J];xin[J]=T;<BR>
}<BR>
K=LH;<BR>
while(J>=K)<BR>
{<BR>
J=J-K;<BR>
K=K/2;<BR>
}<BR>
J=J+K;<BR> }<BR>
<BR> <BR>//下面是DIT-FFT运算程序</P>
<P> for(m=1;m<=M;m++)<BR> {
<BR> B=pow(2,m-1);
<BR>
for(J=0;J<=B-1;J++)<BR>
{<BR>
p=pow(2,M-m)*J;<BR>
ps=2*pi/N*p;<BR>
w.real=cos(ps);<BR>
w.imag=-sin(ps);<BR>
for(k=J;k<=N-1;k=k+pow(2,m))<BR>
{
<BR>
T=EE(xin[k+B],w);<BR>
xin[k+B].real=xin[k].real-T.real;<BR>
xin[k+B].imag=xin[k].imag-T.imag;<BR>
xin[k].real=xin[k].real+T.real;<BR>
xin[k].imag=xin[k].imag+T.imag;<BR>
}<BR>
}<BR> }<BR><BR>}</P>
<P>2、IFFT</P>
<P>根据定义可知，IFFT与FFT算法没有差别，只需将上面的w.imag=-sin(ps);改写为w.imag=sin(ps);</P>
<P>并在最后一个大括号前加上以下语句就可以了：</P>
<P>for(k=0;k<N;k++)<BR> {<BR>
xin[k].real=xin[k].real/N;<BR>
xin[k].imag=xin[k].imag/N;<BR> }<BR>也就是乘了一个常数因子1/N。</P>
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水林声 作于 嘉兴 JRC大楼 208室 小雨后</P>
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