好久之前玩的音乐频谱，标记一下。
实现代码
int n = 64,k=6;double pr[64],pi[64],fr[64],fi[64];void kfft(double *pr,double *pi,int n,int k,double *fr,double *fi){int it,m,is,i,j,nv,l0;double p,q,s,vr,vi,poddr,poddi;for (it=0; it<=n-1; it++) //将pr[0]和pi[0]循环赋值给fr[]和fi[]{m=it; is=0;for(i=0; i<=k-1; i++){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); //将w=e^-j2pi/n用欧拉公式表示pi[1]=-sin(p);for (i=2; i<=n-1; i++) //计算pr[]{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=0; it<=n-2; it=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-2; l0>=0; l0--) //蝴蝶操作{m=m/2; nv=2*nv;for (it=0; it<=(m-1)*nv; it=it+nv)for (j=0; j<=(nv/2)-1; j++){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;}}for (i=0; i<=n-1; i++){pr[i]=sqrt(fr[i]*fr[i]+fi[i]*fi[i])/100; //幅值计算}return;}/*** @brief TIM callback functions* @param htim: pointer to a TIM_HandleTypeDef structure that contains*the configuration information for TIM module.* @retval None*/void HAL_TIM_PeriodElapsedCallback(TIM_HandleTypeDef *htim){static int tim2Cnt = 0;if(htim==(&TIM2_Handler)){if(tim2Cnt >= 64){kfft(pr,pi,64,6,fr,fi);pr[0] = 0;tim2Cnt = 0;/*修改显示数据*/}else{pr[tim2Cnt] = ADC_GetValue();pi[tim2Cnt] = 0;tim2Cnt++;}}}
