pctorc.c

语音CELP压缩解压源代码（C语音)

/**************************************************************************
*
* NAME
*	pctorc 
*
* FUNCTION
*
*	Convert from lp-polynomial to reflection coefficients.
*
*	BEWARE: This code does not use memory efficiently.
*
* SYNOPSIS
*
*	subroutine pctorc(lpc, rc, n)
*
*   formal 
*                       data    I/O
*       name            type    type    function
*       -------------------------------------------------------------------
*       lpc(n+1)        float   i       Array of n+1 coefficients
*                                       a(0)+a(1)z**(-1) + a(2)Z**(-2) +
*                                       .... + a(n)z**(-n)
*       rc(n)           float   i/o       reflection coefficients (voiced-> +rc1)
*       n               int     i       Order of polynomial
*     
***************************************************************************
*       
* DESCRIPTION
*
*       This routine uses the Levinson recursion to compute reflection
*       coefficients from the LPC coefficients.  The first LPC
*	coefficient is assumed to be 1, and although it is passed
*	to the routine, it is not used in the calculations.
*       Note:  the dimension of the internal array t limits the value
*	of the maximum order.
*
*	CELP's LPC predictor coefficient convention is:
*              p+1         -(i-1)
*       A(z) = SUM   a   z          where a  = +1.0
*              i=1    i                    1
*
*	The sign convention used defines the first reflection coefficient
*	as the normalized first autocorrelation coefficient, which results
*	in positive values of rc(1) for voiced speech.
*
***************************************************************************
*
* CALLED BY
*
*	autohf	postfilter  specdist celp  intsynth
*
* CALLS
*
*
*
**************************************************************************/
#include "ccsub.h"
pctorc(lpc, rc, n)
int n;
float lpc[], rc[];
{
  float t[MAXNO+1], a[MAXNO+1];
  int i, j;
  
  for (i = 0; i <= n; i++)
    a[i] = lpc[i];
  for (i = n; i > 1; i--)
  {
    rc[i-1] = -a[i];
    for (j = 1; j < i; j++)
      t[i-j] = (a[i-j] + rc[i-1] * a[j]) / (1.0 - rc[i-1] * rc[i-1]);
    for (j = 1; j < i; j++)
      a[j] = t[j];
  }
  rc[0] = -a[1];
}