Question

Goodevening，我在C中编写一个简单的软件，使用libsndfile库（http://www.mega-nerd.com/libsndfile/api.html）读取一个wav音频文件，而不是样本转到一个过程函数，在那里我对信号应用滤波器（二阶巴特沃思）低通滤波器，适用于转置直接形式II）。之后，我将结果写入新的wav文件。如果我代替过滤器，我应用简单的操作（比如多次采样常数），它可以正常工作，但是当我应用滤波器时会产生很多噪音。我尝试在过滤器之后以及在新文件中写入之前打印样本的值，并且我获得了与Matlab相同的值（我获得的输出是完美的），但它们与值I不同如果我读了库写的输出，那就得到了。

static void processAudio (double *buffer, int length)
{   
    //arrays a and b are the coefficients
    double a[] = {   1,
                    -1.799096409484668,
                     0.817512403384758};


    double b[] = {  0.004603998475022,
                    0.009207996950045,
                    0.004603998475022};

    //arrays a_ and b_ are the coefficients normalized 
    double a_[] = {1,a[1]/b[0],a[2]/b[0]};
    double b_[] = {1,b[1]/b[0],b[2]/b[0]};
    double gain = b[0]/a[0];

    double reg[] ={0,0}; //memory registers

    for(int i = 0; i<length; i++)
    {   
        if(i%2==0) //just left channel is changed
        {  
            //TRANSPOSED DIRECT FORM II
            double input = *(buffer+i);
            double output =(input + reg[0])*gain;

            reg[0] = reg[1]+b_[1]*input-a_[1]*output;
            reg[1] = b_[2]*input - a_[2]*output;

            *(buffer+i) = output;

        }
    }   
} 



//that function is called in the main method inside this cycle
int main(void)
{
    ...

    while ((framesRead = sf_read_double(inputFile, buffer, BUFFER_LENGTH)))
    {   

            processAudio(buffer, framesRead);

            sf_write_double(outputFile, buffer, framesRead);
    }

    ...
}

如果我在过滤器之后打印结果：

0.00000000000000
0.00000000000000
0.00000000000000
0.00000000000000
0.00000000000000
0.00000000000000
0.00000000000000
0.00000000000000
0.00000000000000
0.00000000000000
0.00000000000000
0.00000000000000
0.00000000000000
0.00000000000000
0.00000000000000
-0.00000014050288
-0.00000025277823
0.00000050310774
0.00000209530885
0.00000420138311
0.00000725078112
0.00001115570320
0.00001554761090
0.00002053775981
0.00002550357112
0.00002796948679
0.00000727086200
-0.00006401853354
...

如果我读取输出文件，结果是

0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
3.05175781250000e-05
3.05175781250000e-05
3.05175781250000e-05
3.05175781250000e-05
0
-6.10351562500000e-05
-0.000366210937500000
-0.00183105468750000
-0.00601196289062500
-0.0140075683593750
-0.0261840820312500
-0.0421142578125000
-0.0610656738281250
-0.0820617675781250
-0.104156494140625
-0.126342773437500
-0.147766113281250
-0.167663574218750
-0.185363769531250
-0.200378417968750
-0.212280273437500
...

他们非常不同。我真的不知道发生了什么。如果您有任何想法，请告诉我。提前谢谢！

Answer 1

你得到的输出不同，因为你没有保持过滤器的状态。在帧循环的每次迭代中，您都将过滤器状态重置为初始状态{0.0, 0.0}。另外，不要对系数进行标准化，因为MATLAB会为您执行此操作。只需从MATLAB中取出系数并将其应用于滤波器。

我会形成一个结构来保持过滤器的状态，如下所示：

struct biquad {
    double b0, b1, b2, a1, a2; // Note: MATLAB will always normalize a0 to 1.0, so no need to process that.
    double r0, r1;
};

然后，我会添加像biquad_process这样的函数，如下所示：

void process_biquad(struct biquad *self, double *buffer, int length)
{
    int i;

    for (i = 0; i < length; i++) {
        double x = buffer[i];
        double y = (x * self->b0) + self->r0;

        self->r0 = self->r1 + (x * self->b1) - (self->a1 * y);
        self->r1 = (self->b2 * x) - (self->a2 * y);

        buffer[i] = y;
    }
}

然后，在您的main()函数中，您可以在缓冲区中加载音频并像这样处理每个块：

int main(void)
{
    struct biquad *bq = calloc(1, sizeof(struct biquad));

    // Take coefficients from MATLAB and put them here
    bq->b0 = ...
    bq->b1 = ...

    // Set initial state to 0.0
    bq->r0 = bq->r1 = 0.0;
    ...

    while ((framesRead = sf_read_double(inputFile, buffer, BUFFER_LENGTH)))
    {   

            process_biquad(bq, buffer, framesRead);

            sf_write_double(outputFile, buffer, framesRead);
    }

    ...
}

希望这有帮助。

将滤波器应用于音频信号时的不同输出

1 个答案: