?? 高通.cpp
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#include<iostream>
#include<cmath>
#include<fstream>
using namespace std;
const double pi=4*atan(1);
void butterworthhighpass(double,double,double,double,double);
int main()
{
//數字高通濾波器性能要求
double f,fc,fst;//f:抽樣頻率;fc:通帶截止頻率;fst:阻帶截止頻率
double q1,q2;//q1:通帶允許的最大衰減;q2:阻帶允許的最小衰減
cout<<"please input the parameter:"<<endl;
cin>>f>>fc>>fst>>q1>>q2;
butterworthhighpass(f,fc,fst,q1,q2);
return 0;}
void butterworthhighpass(double f,double fc,double fst,double q1,double q2)
{double Wc,Wst;//數字頻率
Wc=2*pi*fc/f;
Wst=2*pi*fst/f;
double Uc=1;
double C1=Uc*tan(Wc/2);
double Ust=C1*(1/tan(Wst/2));
double log(double num);
double v;
v=log10(pow(10,q2/10)-1)/(2*log10(Ust));//計算階數
double ceil(double x);
int N=ceil(v);//階數圓整
int k;
double b[20];
for(k=1;k<=(N/2);k++)
b[k]=-2*cos(pi/2+(2*k-1)*pi/(2*N));
double A0[20],A1[50],A2[50],B0[50],B1[50],B2[50];//數字低通濾波器系統函數參數
ofstream outfile("f1.dat",ios::out);//定義文件流對象,打開磁盤文件“f1.dat"
if(!outfile)
{
cerr<<"open error!"<<endl;
exit(1);
}
for (k=1;k<=N/2;k++)
{
A0[k]=1;
A1[k]=-2;
A2[k]=1;
B0[k]=C1*C1+b[k]*C1+1;
B1[k]=2*C1*C1-2;
B2[k]=C1*C1+1-b[k]*C1;
}
cout<<N<<endl;
if (N%2==0)
{
for (k=1;k<=N/2;k++)
outfile<<endl<<A0[k]/B0[k]<<" "<<A1[k]/B0[k]<<" "<<A2[k]/B0[k]<<" "<<1<<" "<<B1[k]/B0[k]<<" "<<B2[k]/B0[k]<<endl;
}
else
{
outfile<<endl<<1<<" "<<-1<<" "<<1+C1<<" "<<C1-1<<endl;
for (k=1;k<=N/2;k++)
outfile<<endl<<A0[k]/B0[k]<<" "<<A1[k]/B0[k]<<" "<<A2[k]/B0[k]<<" "<<1<<" "<<B1[k]/B0[k]<<" "<<B2[k]/B0[k]<<endl;
}
outfile.close();}
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