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这是一个对语音信号(0.3kHz~3.4kHz)进行低通滤波的C语言程序，低通滤波的截止频率为800Hz，滤波器采用19点的有限冲击响应FIR滤波。语音信号的采样频率为8kHz，每个语音样值按16位整型数存放在insp.dat文件中。<br />例3.7 语音信号800Hz 19点FIR低通滤波C语言浮点程序<br />#include <stdio.h><br />const int length = 180 /*语音帧长为180点＝22.5ms@8kHz采样*/<br />//22.5ms是怎么得来的 ？？？<br />void filter(int xin[ ],int xout[ ],int n,float h[ ]); /*滤波子程序说明*/<br />/*19点滤波器系数*/<br />static float h[19]=<br /> {0.01218354,-0.009012882,-0.02881839,-0.04743239,-0.04584568,<br />-0.008692503,0.06446265,0.1544655,0.2289794,0.257883,<br />0.2289794,0.1544655,0.06446265,-0.008692503,-0.04584568,<br />-0.04743239,-0.02881839,-0.009012882,0.01218354};<br />static int x1[length+20];// x1是做什么用的？？？？<br />/*低通滤波浮点子程序*/<br />void filter(int xin[ ],int xout[ ],int n,float h[ ])<br />{ //哪位大虾能告诉我一下滤波器的公式？？？？<br />int i,j;<br />float sum;<br />for(i=0;i<length;i++) x1[n+i-1]=xin;<br />for (i=0;i<length;i++)<br />{<br />sum=0.0;<br />for(j=0;j<n;j++) sum+=h[j]*x1[i-j+n-1];<br />xout=(int)sum;<br /> }<br />for(i=0;i<(n-1);i++) x1[n-i-2]=xin[length-1-i]; //这句话是 ？？<br />}<br /> |
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