请教下这两个公式的作用有啥用处?

83874634@qq.com

主要是func1函数

static double
    func1(double iSp0, double* iSp1)
{
    return (iSp0 * (iSp0 * (iSp0 * (iSp0 * (iSp0 * iSp1[5] + iSp1[4]) + iSp1[3]) + iSp1[2]) + iSp1[1]) + iSp1[0]);
}

static void
    func2()
{
    double t0, t1;

    t0 = func1(g_device.v10f0_adc4_float, &g_configs.v2b17_knb_adc4[0][0]);
    t1 = func1(g_device.v10f0_adc4_float, &g_configs.v2b17_knb_adc4[0][6]);
    g_device.v11f8_adcX_float_fliter = (g_device.v11f0_adcX_float - t0) / t1;


    t0 = func1(g_device.v10f0_adc4_float, &g_configs.v2b17_knb_adc4[1][0]);
    t1 = func1(g_device.v10f0_adc4_float, &g_configs.v2b17_knb_adc4[1][6]);
    g_device.v1238_adcY_float_fliter = (g_device.v1230_adcY_float - t0) / t1;

    t0 = func1(g_device.v10f0_adc4_float, &g_configs.v2b17_knb_adc4[2][0]);
    t1 = func1(g_device.v10f0_adc4_float, &g_configs.v2b17_knb_adc4[2][6]);
    g_device.v1278_adcZ_float_fliter = (g_device.v1270_adcZ_float - t0) / t1;
}

zouzhichao

粗略看了一下，好像是多项式拟合
第一个函数是多项式计算

83874634@qq.com

zouzhichao 发表于 2022-10-24 14:51
粗略看了一下，好像是多项式拟合
第一个函数是多项式计算
(引用自2楼)

感谢, 就是拟合曲线