计算具有sin/cos LUT的棕褐色

问题描述

#define PARTPERDEGREE 1
double mysinlut[PARTPERDEGREE * 90 + 1];
double mycoslut[PARTPERDEGREE * 90 + 1];
void MySinCosCreate()
{
    int i;
    double angle, angleinc;

    // Each degree also divided into 10 parts
    angleinc = (M_PI / 180) / PARTPERDEGREE;
    for (i = 0, angle = 0.0; i <= (PARTPERDEGREE * 90 + 1); ++i, angle += angleinc)
    {
        mysinlut[i] = sin(angle);
    }

    angleinc = (M_PI / 180) / PARTPERDEGREE;
    for (i = 0, angle = 0.0; i <= (PARTPERDEGREE * 90 + 1); ++i, angle += angleinc)
    {
        mycoslut[i] = cos(angle);
    }
}


double MySin(double rad)
{
    int ix;
    int sign = 1;
    double angleinc = (M_PI / 180) / PARTPERDEGREE;

    if(rad > (M_PI / 2))
        rad = M_PI / 2 - (rad - M_PI / 2);

    if(rad < -(M_PI / 2))
        rad = -M_PI / 2 - (rad + M_PI / 2);

    if(rad < 0)
    {
        sign = -1;
        rad *= -1;
    }

    ix = (rad * 180) / M_PI * PARTPERDEGREE;
    double h = rad - ix*angleinc;
    return sign*(mysinlut[ix] + h*mycoslut[ix]);
}

double MyCos(double rad)
{
    int ix;
    int sign = 1;
    double angleinc = (M_PI / 180) / PARTPERDEGREE;


    if(rad > M_PI / 2)
    {
        rad = M_PI / 2 - (rad - M_PI / 2);
        sign = -1;
    }
    else if(rad < -(M_PI / 2))
    {
        rad = M_PI / 2 + (rad + M_PI / 2);
        sign = -1;
    }
    else if(rad > -M_PI / 2 && rad < M_PI / 2)
    {
        rad = abs(rad);
        sign = 1;
    }

    ix = (rad * 180) / M_PI * PARTPERDEGREE;

    double h = rad - ix*angleinc;
    return sign*(mycoslut[ix] - h*mysinlut[ix]);
}

double MyTan(double rad)
{
    return MySin(rad) / MyCos(rad);
}

事实证明，使用除法计算tan甚至比原始的tan函数昂贵.

It turns out that calculating tan using division is even more expensive than original tan function.

有什么方法可以使用sin/cos查找表值来计算tan而不进行除法运算，因为除法在我的MCU上很昂贵.

Is there any way to calculate tan using sin/cos lookup table values without division operation, since division is expensive on my MCU.

拥有tan LUT并使用tan/sin或tan/cos提取结果更好吗?

Is it better to have tan LUT and extract result using tan/sin or tan/cos as it's done now for sin/cos?

推荐答案

尤其是在微控制器中，y/x除法通常可以通过log& exp表或使用迭代乘法，直到分母1 +-eps = 1 +-x ^(2 ^ n)为零.

In microcontrollers especially, the division y/x can often be optimized either by log & exp tables, or with iterative multiplications, until the denominator 1 +- eps = 1 +- x^(2^n) is zero.

y/X = y / (1-x)
    = (1+x)y / (1+x)(1-x)
    = (1+x)(1+x^2)y / (1-x^2)(1+x^2)
    = (1+x)(1+x^2)(1+x^4)y / (1-x^4)(1+x^4)

这篇关于计算具有sin/cos LUT的棕褐色的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！