低通滤波器差分方程用C语言实现的问题-fylwzh的回答

低通滤波器差分方程用C语言实现的问题

2019-07-15 18:56发布

本帖最后由 hanl 于 2011-7-21 10:07 编辑

我用C语言实现一个butterworth低通滤波器，现在遇到问题，please help me！
实现过程：
1、采样频率10KHz，通带边界频率2KHz，通带波纹小于1dB，阻带边界频率3KHz，阻带衰减大于40dB。
2、用matlab求出滤波器系数b，a

>> Fp = 2000;
>> Fs = 3000;
>> F = 10000;
>> T = 1/F;
>> Ap = 1;
>> As = 40;
>> Wp = Fp*T*2;
>> Ws = Fs*T*2;
>> [n,Wc] = buttord(Wp,Ws,Ap,As);
>> [b,a] = butter(n,Wc)
b =
0.0021 0.0186 0.0745 0.1739 0.2609 0.2609 0.1739 0.0745 0.0186 0.0021
a =
1.0000 -1.0893 1.6925 -1.0804 0.7329 -0.2722 0.0916 -0.0174 0.0024 -0.0001
>>

复制代码3、差分方程： y(0) = b(0)*x(0) + b(1)*x(1) +...+ b(9)*x(9) - {a(1)*y(1) + a(2)*y(2) +...+ a(9)*y(9)}
4、程序：

double a[L] = {1.0000, -1.0893, 1.6925, -1.0804, 0.7329, -0.2722, 0.0916, -0.0174, 0.0024, -0.0001};
double b[L] = {0.0021, 0.0186, 0.0745, 0.1739, 0.2609, 0.2609, 0.1739, 0.0745, 0.0186, 0.0021};
read(&input); //输入范围：-10V～+10V ，量化范围 0～65535
x[0] = input-32767;
for (i=0; i<=L-1; i++)
{
b_sum = b[i]*x[i] + b_sum;
a_sum = a[i+1]*y[i+1] + a_sum;
}
y[0] = b_sum - a_sum;
output = y[0];
for (i=L-1; i>=1; i--)
{
x[i] = x[i-1];
y[i] = y[i-1];
}
write(output);