C 编程中的 FIR 滤波器实现

Question

谁能告诉我如何使用 c 编程语言实现 FIR 滤波器。

Answer 1

设计一个 FIR 滤波器不是一个简单的话题，但是实现一个已经设计好的滤波器（假设你已经有了 FIR 系数）也不错。该算法称为convolution。这是一个幼稚的实现......

void convolve (double *p_coeffs, int p_coeffs_n,
               double *p_in, double *p_out, int n)
{
  int i, j, k;
  double tmp;

  for (k = 0; k < n; k++)  //  position in output
  {
    tmp = 0;

    for (i = 0; i < p_coeffs_n; i++)  //  position in coefficients array
    {
      j = k - i;  //  position in input

      if (j >= 0)  //  bounds check for input buffer
      {
        tmp += p_coeffs [i] * p_in [j];
      }
    }

    p_out [k] = tmp;
  }
}

基本上，卷积对输入信号进行移动加权平均。权重是滤波器系数，假设总和为 1.0。如果权重总和不是 1.0，则会得到一些放大/衰减以及滤波。

顺便说一句 - 这个函数有可能将系数数组向后 - 我没有仔细检查过，我已经有一段时间没有考虑这些事情了。

对于如何计算特定滤波器的 FIR 系数，这背后有相当多的数学知识 - 您确实需要一本关于数字信号处理的好书。 This one 可免费获得 PDF，但我不确定它有多好。我有Rorabaugh 和Orfandis，它们都在九十年代中期发布，但这些东西并没有真正过时。

Answer 2

组合多个过滤器：

以单位脉冲开始（第一个位置为 1，其他位置为 0 的信号）。应用第一个过滤器。应用第二个过滤器。继续，直到应用所有过滤器。结果显示了组合滤波器如何对单位脉冲进行卷积（假设数组足够长以至于没有数据丢失），因此其中的值是一个滤波器的系数，它是其他滤波器的组合。

这里是示例代码：

#include <stdio.h>
#include <string.h>


#define NumberOf(a) (sizeof (a) / sizeof *(a))


/*  Convolve Signal with Filter.

    Signal must contain OutputLength + FilterLength - 1 elements.  Conversely,
    if there are N elements in Signal, OutputLength may be at most
    N+1-FilterLength.
*/
static void convolve(
    float *Signal,
    float *Filter, size_t FilterLength,
    float *Output, size_t OutputLength)
{
    for (size_t i = 0; i < OutputLength; ++i)
    {
        double sum = 0;
        for (size_t j = 0; j < FilterLength; ++j)
            sum += Signal[i+j] * Filter[FilterLength - 1 - j];
        Output[i] = sum;
    }
}


int main(void)
{
    //  Define a length for buffers that is long enough for this demonstration.
    #define LongEnough  128


    //  Define some sample filters.
    float Filter0[] = { 1, 2, -1 };
    float Filter1[] = { 1, 5, 7, 5, 1 };

    size_t Filter0Length = NumberOf(Filter0);
    size_t Filter1Length = NumberOf(Filter1);


    //  Define a unit impulse positioned so it captures all of the filters.
    size_t UnitImpulsePosition = Filter0Length - 1 + Filter1Length - 1;
    float UnitImpulse[LongEnough];
    memset(UnitImpulse, 0, sizeof UnitImpulse);
    UnitImpulse[UnitImpulsePosition] = 1;


    //  Calculate a filter that is Filter0 and Filter1 combined.
    float CombinedFilter[LongEnough];

    //  Set N to number of inputs that must be used.
    size_t N = UnitImpulsePosition + 1 + Filter0Length - 1 + Filter1Length - 1;

    //  Subtract to find number of outputs of first convolution, then convolve.
    N -= Filter0Length - 1;
    convolve(UnitImpulse,    Filter0, Filter0Length, CombinedFilter, N);

    //  Subtract to find number of outputs of second convolution, then convolve.
    N -= Filter1Length - 1;
    convolve(CombinedFilter, Filter1, Filter1Length, CombinedFilter, N);

    //  Remember size of resulting filter.
    size_t CombinedFilterLength = N;

    //  Display filter.
    for (size_t i = 0; i < CombinedFilterLength; ++i)
        printf("CombinedFilter[%zu] = %g.\n", i, CombinedFilter[i]);


    //  Define two identical signals.
    float Buffer0[LongEnough];
    float Buffer1[LongEnough];
    for (size_t i = 0; i < LongEnough; ++i)
    {
        Buffer0[i] = i;
        Buffer1[i] = i;
    }


    //  Convolve Buffer0 by using the two filters separately.

    //  Start with buffer length.
    N = LongEnough;

    //  Subtract to find number of outputs of first convolution, then convolve.
    N -= Filter0Length - 1;
    convolve(Buffer0, Filter0, Filter0Length, Buffer0, N);

    //  Subtract to find number of outputs of second convolution, then convolve.
    N -= Filter1Length - 1;
    convolve(Buffer0, Filter1, Filter1Length, Buffer0, N);

    //  Remember the length of the result.
    size_t ResultLength = N;


    //  Convolve Buffer1 with the combined filter.
    convolve(Buffer1, CombinedFilter, CombinedFilterLength, Buffer1, ResultLength);


    //  Show the contents of Buffer0 and Buffer1, and their differences.
    for (size_t i = 0; i < ResultLength; ++i)
    {
        printf("Buffer0[%zu] = %g.  Buffer1[%zu] = %g.  Difference = %g.\n",
            i, Buffer0[i], i, Buffer1[i], Buffer0[i] - Buffer1[i]);
    }

    return 0;
}

Answer 3

我发现这段代码 sn-p 对我不起作用（Visual Studio 2005）。

我最终发现卷积问题有一个很好的答案：

1d linear convolution in ANSI C code?

对于那些不知道的人 - 卷积与 FIR 滤波的操作完全相同 - “内核”是 FIR 滤波器脉冲响应，信号是输入信号。

我希望这可以帮助一些正在寻找 FIR 代码的可怜的 sap :-)