C语言 低通滤波器带通滤波器高通滤波器
C语言?低通滤波器\带通滤波器\高通滤波器
(-07-08 23:07:17)
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分类： [知识交流]当信号和干扰信号同时存在时，不会发生叠加吗？要如何滤掉干扰信号？
文章发表于：-10-21 18:16
这是对滤波的理解有误，滤波不是指滤掉干扰信号的幅值，而是滤掉它的频率。有用信号和干扰信号的频率不同，如果干扰信号的频率高，则用一个低通滤波器滤掉干扰信号，如果信号频率高，则用一个高通滤波器滤掉干扰信号。
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MATLAB:
?? y(m)=-(a(2)*y(m-1)+a(3)*y(m-2)+a(4)*y(m-3)+a(5)*y(m-4)+a(6)*y(m-5)+a(7)*y(m-6)+a(8)*y(m-7)+a(9)*y(m-8))+(b(1)*x(m)+b(2)*x(m-1)+b(3)*x(m-2)+b(4)*x(m-3)+b(5)*x(m-4)+b(6)*x(m-5)+b(7)*x(m-6)+b(8)*x(m-7)+b(9)*x(m-8));
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C语言以下一个结果：通带 5Hz，1db， 阻带8Hz，40db
b=[0.0117?? -0.0329??? 0.0451?? -0.0329??? 0.0117]a=[1.0000?? -3.6048??? 4.9787?? -3.1156??? 0.7447]
int??????? filterBegin=5;double xBuf[5];double yBuf[5];double filter(double x){??????? //把历史数据保存，算出yBuf[]并返回??????? for(int i=4;i>0;i--)??????? {??????????? yBuf[i] = yBuf[i-1]; xBuf[i] = xBuf[i-1];??????? }?????? xBuf[0] = x;
??? if(filterBegin>0)??? {??????? filterBegin =0;?????? yBuf[0] = x;?????? return x;??? }??? yBuf[0] = 0.0117* ( xBuf[0] +xBuf[4])? -0.0329 * (xBuf[1 ]+xBuf[3]) +0.0451*xBuf[2 ]??????????????? -( -3.6048* yBuf[1] + 4.9787*yBuf[2]? -3.1156*yBuf[3] +??? 0.7447*yBuf[4]);??? return yBuf[0];}
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2、关于FIR带通滤波器的C语言设计程序 代码
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;?}}
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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;?}}
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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;??
