问下这段FIR滤波程序

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楼主 | 2016-4-19 10:17 | 只看该作者回帖奖励

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滤波程序, FIR, TI, ck, TE

#define       N                10       // number of FIR coefficients.
#define       BLKSIZE       20       // block size of samples filtered each time.
#define       M                (N + BLKSIZE - 1)

float workbuf[M] = {0};       // history + current input block, initialize to 0.

// float h[N]: array of fir filter coefficients.
// float x[BLKSIZE]: array of input samples block
// float y[BLKSIZE]: array of samples block after filtering
void fir(float h[N], float x[BLKSIZE], float y[BLKSIZE])
{

      int i, j;
      float *p;

      // copy current input block to work buffer
      // workbuf[N-1] = x[0], workbuf[N] = x[1],
      // workbuf[N+1] = x[2], ..., workbuf[M-1] = x[BLKSIZE-1];
      for (i=0; i<BLKSIZE; i++)
      {
            workbuf[i+N-1] = x;
      }

      // FIR filter loop
      for (i=0; i<BLKSIZE; i++)
      {
            p = &workbuf[i+N-1];// each y needs right starting x.
            for (j=0; j<N; j++)
            {
                     y += (h[j] * (*p--));
            }
      }

      // copy last N-1 samples of x[n] to workbuf
      // as history for next block
      p = &x[BLKSIZE-1];       // p points the last element;
      for (i=0; i<N-1; i++)
      {
            workbuf[N-2-i] = *p--;
      }

}

/////////////////////////////////////////////////
代码源自本论坛
没看明白，滤波的结果大了将近一倍的样子。

疑问是：如果一个周期的采样数BLKSIZE远远小于滤波器系数的个数N。
那后面
s the last element;
      for (i=0; i<N-1; i++)
            workbuf[N-2-i] = *p--;
这句话怎么理解。