如何用C#实现FIR低通滤波器
这个我刚好做过一个滤波器，事实上对时域信号做FFT，截取一定点数再做逆FFT相当于理想滤波。设计滤波器代码如下：
f1=100;f2=200;%待滤波正弦信号频率
fs=2000;%采样频率
m=(0.3*f1)/(fs/2);%定义过度带宽
M=round(8/m);%定义窗函数的长度
N=M-1;%定义滤波器的阶数
b=fir1(N,f2/fs);%使用fir1函数设计滤波器
%输入的参数分别是滤波器的阶数和截止频率
figure(1)
[h,f]=freqz(b,1,512);%滤波器的幅频特性图
%[H,W]=freqz(B,A,N)当N是一个整数时函数返回N点的频率向量和幅频响应向量
plot(f*fs/(2*pi),20*log10(abs(h)))%参数分别是频率与幅值
xlabel(‘频率/赫兹’);ylabel(‘增益/分贝’);title(‘滤波器的增益响应’);
figure(2)
subplot(211)
t=0:1/fs:0.5;%定义时间范围和步长
s=sin(2*pi*f1*t)+sin(2*pi*f2*t);%滤波前信号
plot(t,s);%滤波前的信号图像
xlabel(‘时间/秒’);ylabel(‘幅度’);title(‘信号滤波前时域图’);
subplot(212)
Fs=fft(s,512);%将信号变换到频域
AFs=abs(Fs);%信号频域图的幅值
f=(0:255)*fs/512;%频率采样
plot(f,AFs(1:256));%滤波前的信号频域图
xlabel(‘频率/赫兹’);ylabel(‘幅度’);title(‘信号滤波前频域图’);
figure(3)
sf=filter(b,1,s);%使用filter函数对信号进行滤波
%参数分别为滤波器系统函数的分子和分母多项式系数向量和待滤波信号输入
subplot(211)
plot(t,sf)%滤波后的信号图像
xlabel(‘时间/秒’);ylabel(‘幅度’);title(‘信号滤波后时域图’);
axis([0.2 0.5 -2 2]);%限定图像坐标范围
subplot(212)
Fsf=fft(sf,512);%滤波后的信号频域图
AFsf=abs(Fsf);%信号频域图的幅值
f=(0:255)*fs/512;%频率采样
plot(f,AFsf(1:256))%滤波后的信号频域图
xlabel(‘频率/赫兹’);ylabel(‘幅度’);title(‘信号滤波后频域图’);
如何用C语言实现低通滤波器
float middle_filter(float middle_value [] , intcount)
{
float sample_value, data;
int i, j;
for (i=1; i for(j=count-1; j=i,–j){
if(middle_value[j-1]=middle_value[j]{
data=middle_value[j-1];
middle_value[j-1]=middle_value[j]
middle_value[j]=data;
}
}
sample_value=middle_value(count-1)/2];
return(sample_value);
}
用C语言(！！！不移位方式)实现FIR滤波器 程序尽量简单，还有正确性
short h[], short y[])
{
int i, j, sum; for (j = 0; j 100; j++) {
sum = 0;
for (i = 0; i 32; i++)
sum += x[i+j] * h[i];
y[j] = sum 15;
}
}
2
void fir(short x[], short h[], short y[])
{
int i, j, sum0, sum1;
short x0,x1,h0,h1; for (j = 0; j 100; j+=2) {
sum0 = 0;
sum1 = 0;
x0 = x[j];
for (i = 0; i 32; i+=2){
x1 = x[j+i+1];
h0 = h[i];
sum0 += x0 * h0;
sum1 += x1 * h0;
x0 = x[j+i+2];
h1 = h[i+1];
sum0 += x1 * h1;
sum1 += x0 * h1;
}
y[j] = sum0 15;
y[j+1] = sum1 15;
}
}
3
void fir(short x[], short h[], short y[])
{
int i, j, sum0, sum1;
short x0,x1,x2,x3,x4,x5,x6,x7,h0,h1,h2,h3,h4,h5,h6,h7; for (j = 0; j 100; j+=2) {
sum0 = 0;
sum1 = 0;
x0 = x[j];
for (i = 0; i 32; i+=8){
x1 = x[j+i+1];
h0 = h[i];
sum0 += x0 * h0;
sum1 += x1 * h0;
x2 = x[j+i+2];
h1 = h[i+1];
sum0 += x1 * h1;
sum1 += x2 * h1;
x3 = x[j+i+3];
h2 = h[i+2];
sum0 += x2 * h2;
sum1 += x3 * h2;
x4 = x[j+i+4];
h3 = h[i+3];
sum0 += x3 * h3;
sum1 += x4 * h3;
x5 = x[j+i+5];
h4 = h[i+4];
sum0 += x4 * h4;
sum1 += x5 * h4;
x6 = x[j+i+6];
h5 = h[i+5];
sum0 += x5 * h5;
sum1 += x6 * h5;
x7 = x[j+i+7];
h6 = h[i+6];
sum0 += x6 * h6;
sum1 += x7 * h6;
x0 = x[j+i+8];
h7 = h[i+7];
sum0 += x7 * h7;
sum1 += x0 * h7;
}
y[j] = sum0 15;
y[j+1] = sum1 15;
}
}
大家好，求助c语言怎么编写设计fir数字滤波器。要求:用一个矩形窗设计
用四种窗函数设计线形相位低通滤波器。要求在两种窗口长度下，绘制相应的幅频和相频特性曲线，观察3dB和20dB带宽以及阻带最小衰减，比较四种窗函数对滤波器特性的影响。
求用C语言实现一个FIR数字低通滤波
没有定义这个函数，此函数为
function hd=ideal_lp(wc,M);
%Ideal Lowpass filter computation
%————————————
%[hd]=ideal_lp(wc,M)
% hd=ideal impulse response between 0 to M-1
% wc=cutoff frequency in radians
% M=length of the ideal filter
%
alpha=(M-1)/2;
n=[0:1:(M-1)];
m=n-alpha+eps;
hd=sin(wc*m)./(pi*m);
点击file中的new中M-file，新建上面的函数，保存后就可以运行了
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各位好心人帮帮忙，谁能给我段基于c语言fir滤波器设计的程序代码啊。急急急!!!!!
1 void fir(short x[], short h[], short y[]) { int i, j, sum; for (j = 0; j 100; j++) { sum = 0; for (i = 0; i 32; i++) sum += x[i+j] * h[i]; y[j] = sum 15; } } 2 void fir(short x[], short h[], short y[]) { int i, j, sum0, sum1; short x0,x1,h0,h1; for (j = 0; j 100; j+=2) { sum0 = 0; sum1 = 0; x0 = x[j]; for (i = 0; i 32; i+=2){ x1 = x[j+i+1]; h0 = h[i]; sum0 += x0 * h0; sum1 += x1 * h0; x0 = x[j+i+2]; h1 = h[i+1]; sum0 += x1 * h1; sum1 += x0 * h1; } y[j] = sum0 15; y[j+1] = sum1 15; } } 3 void fir(short x[], short h[], short y[]) { int i, j, sum0, sum1; short x0,x1,x2,x3,x4,x5,x6,x7,h0,h1,h2,h3,h4,h5,h6,h7; for (j = 0; j 100; j+=2) { sum0 = 0; sum1 = 0; x0 = x[j]; for (i = 0; i 32; i+=8){ x1 = x[j+i+1]; h0 = h[i]; sum0 += x0 * h0; sum1 += x1 * h0; x2 = x[j+i+2]; h1 = h[i+1]; sum0 += x1 * h1; sum1 += x2 * h1; x3 = x[j+i+3]; h2 = h[i+2]; sum0 += x2 * h2; sum1 += x3 * h2; x4 = x[j+i+4]; h3 = h[i+3]; sum0 += x3 * h3; sum1 += x4 * h3; x5 = x[j+i+5]; h4 = h[i+4]; sum0 += x4 * h4; sum1 += x5 * h4; x6 = x[j+i+6]; h5 = h[i+5]; sum0 += x5 * h5; sum1 += x6 * h5; x7 = x[j+i+7]; h6 = h[i+6]; sum0 += x6 * h6; sum1 += x7 * h6; x0 = x[j+i+8]; h7 = h[i+7]; sum0 += x7 * h7; sum1 += x0 * h7; } y[j] = sum0 15; y[j+1] = sum1 15; } }
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