无法理解用于计算指数函数的CORDIC算法

时间:2015-11-13 17:19:11

标签: matlab math numerical-methods cordic elementary-functions

我读了一两篇关于CORDIC的论文,但不太明白。但是我从互联网上下载了一个使用这种算法计算指数函数的代码。它非常有用,可以帮助我在FPGA上触发一个函数的指数项。但现在我正在尝试编写报告,我无法解释CORDIC部分如何工作,我无法与一般CORDIC算法相关。请帮助我,并提前感谢你。

function fx = exp_cordic ( x, n )
 a_length = 25;

  a = [ ...
    1.648721270700128, ...
    1.284025416687742, ...
    1.133148453066826, ...
    1.064494458917859, ...
    1.031743407499103, ...
    1.015747708586686, ...
    1.007843097206488, ...
    1.003913889338348, ...
    1.001955033591003, ...
    1.000977039492417, ...
    1.000488400478694, ...
    1.000244170429748, ...
    1.000122077763384, ...
    1.000061037018933, ...
    1.000030518043791, ...
    1.0000152589054785, ...
    1.0000076294236351, ...
    1.0000038147045416, ...
    1.0000019073504518, ...
    1.0000009536747712, ...
    1.0000004768372719, ...
    1.0000002384186075, ...
    1.0000001192092967, ...
    1.0000000596046466, ...
    1.0000000298023228 ];
  e = 2.718281828459045;

  x_int = floor ( x );
%
%  Determine the weights.
%
  poweroftwo = 0.5;
  z = x - x_int;

  for i = 1 : n
    w(i) = 0.0;
    if ( poweroftwo < z )
      w(i) = 1.0;
      z = z - poweroftwo;
    end
    poweroftwo = poweroftwo / 2.0;
  end
%
%  Calculate products.
%
  fx = 1.0;

  for i = 1 : n

    if ( i <= a_length )
      ai = a(i);
    else
      ai = 1.0 + ( ai - 1.0 ) / 2.0;
    end

    if ( 0.0 < w(i) )
      fx = fx * ai;
    end

  end
%
%  Perform residual multiplication.
%
  fx = fx             ...
    * ( 1.0 + z       ...
    * ( 1.0 + z / 2.0 ...
    * ( 1.0 + z / 3.0 ...
    * ( 1.0 + z / 4.0 ))));
%
%  Account for factor EXP(X_INT).
%
  if ( x_int < 0 )

    for i = 1 : -x_int
      fx = fx / e;
    end

  else

    for i = 1 : x_int
      fx = fx * e;
    end

  end

  return
end

我做了一些修改并删除了一些代码,并试图让它更简单,它有效,我不知道我做了什么,为什么它仍然工作!!!!

a = [ ...
    1.648721270700128, ...
    1.284025416687742, ...
    1.133148453066826, ...
    1.064494458917859, ...
    1.031743407499103, ...
 ];


  e = 2.718281828459045;

  x_int = floor ( x );
  z = x - x_int;
  fx = 1.0;
  for i = 1 : n
    if ( 2^(-i) < z )
     z=z-2^(-i);
      fx = fx * a(i); 
    end
  end

  if ( x_int < 0 )
    for i = 1 : -x_int
      fx = fx / e;
    end
  else

    for i = 1 : x_int
      fx = fx * e;
    end
  end

  return
end

1 个答案:

答案 0 :(得分:1)

这使用了众所周知的事实

exp(x+y)=exp(x)*exp(y) and a^(x*y)=(a^x)^y.

输入数x首先被分解为整数和小数部分x = x_int + zx_int的指数可以通过任何整数幂算法轻松计算,所呈现的算法相当次优。

因子表是二进制表示中的小数部分

z = z[1]/2+z[2]/4+z[3]/8+…

其中z[i]01。然后第一个循环计算

exp(1/2)^z[1] * exp(1/4)^z[2] * exp(1/8)^z[3]*…

其中第二次取幂为

(z[i]==1) ? exp(1/2^i) : 1 

即,只有z[i]==1的因素实际存在于产品中。