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 = 0j <100j++) {

sum = 0

for (i = 0i <32i++)

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 = 0j <100j+=2) {

sum0 = 0

sum1 = 0

x0 = x[j]

for (i = 0i <32i+=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 = 0j <100j+=2) {

sum0 = 0

sum1 = 0

x0 = x[j]

for (i = 0i <32i+=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

}

}

例题：

设计一个低通数字滤波器，给定抽样频率为fs=12000Hz，通带截止频率wp=0Hz，阻带起始频率ws=5000Hz（假设阻带衰减不小于-50dB）。

解答：

由于阻带衰减为50dB，查表，可选海明窗，其阻带最小衰减为53dB，过渡带宽度为6.6π/N。

MATLAB程序如下：

wp=0% 频率归一化

ws=5000*2/12000

wdel=ws-wp% 过渡带宽

wn=0.5*(wp+ws)% 近似计算截止频率

N=ceil(6.6*pi/wdel)% 根据过渡带宽度求滤波器阶数

window=hamming(N+1)% 海明窗

b=fir1(N,wn,window)% FIR滤波器设计

freqz(b,1,512)% 查看滤波器幅频及相频特性

参考：

http://zhidao.baidu.com/question/350250283.html

你的问题没有说明衰减，我直接在百度上找的答案，使用海明窗的低通滤波器。我运行了下，语句是对的。

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