主要是獲取濾波器參數(shù)和卷積算法，參數(shù)根據(jù)自己的濾波器特性用matlab上面的一個(gè)fdatool工具配置生成，然后將這些參數(shù)寫入程序里面即可(參數(shù)可適當(dāng)?shù)某艘砸粋€(gè)系數(shù)，對(duì)濾波沒任何影響)，我這里Fs = 2k，Fpass= 1k,Fstop = 2k，至于輸入數(shù)據(jù)你可以自己產(chǎn)生幾個(gè)不同頻率正弦波的疊加。我這里隨便給的一個(gè)輸入例子：

f1=1k,f2=6k,fs=20k

采樣數(shù)據(jù):200

3012-35-2 -10911-14-18-10-19-1213 7-10-14-3 153010-35 -4-9119-16 -17-10 -20-9 144-1113 -317297-2 5-5-8127 -17-16 -11-20 -7152-102 2-220285 -25-7-614 4-19-15-12-20-415 0-1031-1 22263-15 -8-4142-20 -13-12 -20-1 15-2-940 1252510 4-9-215-1 -20-12 -13-20 214-4-85 -132622-1 13-10015 -4-20 -12-15 -19414-6-7 5-252820 -222-102 15-7-20-11-16-177 12-8-55-2 72917-33 1-11414-9 -20-10 -17-16 911-9-45 -3103015-3 4-1-10713 -12-19 -10-18 -1411 9-10-25-3 1230

39階參數(shù):

110-2-5 -9-13 -18-22 -23-21 -130173964 88109 124133 133124 10988 6439170-13 -21-23 -22-18 -13-9 -5-2011

#include

#include

#define samples 200

#define taps 40

int input[samples] = {30,12,-3,5,-2,-10,9,11,-14,-18,-10,-19,-12,13,7,-10,-1,4,-3,15,30,10,-3,5,-4,-9,11,9,-16,-17,-10,-20,-9,14,4,-11,1,3,-3,17,29,7,-2,5,-5,-8,12,7,-17,-16,-11,-20,-7,15,2,-10,2,2,-2,20,28,5,-2,5,-7,-6,14,4,-19,-15,-12,-20,-4,15,0,-10,3,1,-1,22,26,3,-1,5,-8,-4,14,2,-20,-13,-12,-20,-1,15,-2,-9,4,0,1,25,25,1,0,4,-9,-2,15,-1,-20,-12,-13,-20,2,14,-4,-8,5,-1,3,26,22,-1,1,3,-10,0,15,-4,-20,-12,-15,-19,4,14,-6,-7,5,-2,5,28,20,-2,2,2,-10,2,15,-7,-20,-11,-16,-17,7,12,-8,-5,5,-2,7,29,17,-3,3,1,-11,4,14,-9,-20,-10,-17,-16,9,11,-9,-4,5,-3,10,30,15,-3,4,-1,-10,7,13,-12,-19,-10,-18,-14,11,9,-10,-2,5,-3,12,30};

int output[samples];

int coefficients[taps] = {1,1,0,-2,-5,-9,-13,-18,-22,-23,-21,-13,0,17,39,64,88,109,124,133,133,124,109,88,64,39,17,0,-13,-21,-23,-22,-18,-13,-9,-5,-2,0,1,1};

int state[taps+1]={0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0};

main()

{

int temp;

int i;

int j;

int k;

for (k = 0; k < samples; k++)

{

state[0] = input[k];

for (i = 0, temp = 0; i < taps; i++)

temp += coefficients[i] * state[i];

output[k] = temp;

for (j = taps-1; j > -1 ; j--)

state[j+1] = state[j];

}

for( i = 0; i<200;i++){

printf("%d,",&output[i]);

}

exit(0);

}

下面使用matlab仿真的結(jié)果

fs =20000; f1 = 1000; f2 = 3000; f3 = 6000; ? T=10*10^(-3); t=linspace(0,T,fs*T); y =round(10*cos(2*pi*f1*t)+10*cos(2*pi*f2.*t)+10*cos(2*pi*f3.*t)); lpf=[11 0-2-5-9-13 -18-22 -23-21 -130173964 88109 124133 133124 10988 6439170-13 -21-23 -22-18 -13-9 -5-2011] ? y1 =conv(y,lpf); ? ? figure; plot(y); figure; plot(y1);

總結(jié)

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