代码之家 › 专栏 › 技术社区 › Ahmad Magdy

基于matlab的曲线拟合结果误差分析

curve-fitting interpolation matlab

Ahmad Magdy · 技术社区 · 7 年前

Linear model Poly6:
     f(x) = p1*x^6 + p2*x^5 + p3*x^4 + p4*x^3 + p5*x^2 + p6*x + p7
       where x is normalized by mean 100.7 and std 0.9139
Coefficients (with 95% confidence bounds):
       p1 =    -0.08382  (-0.4273, 0.2597)
       p2 =    -0.06449  (-0.4851, 0.3562)
       p3 =      0.3342  (-0.8434, 1.512)
       p4 =     0.09103  (-0.9764, 1.158)
       p5 =     -0.3258  (-1.459, 0.8071)
       p6 =       1.629  (0.9808, 2.278)
       p7 =       38.76  (38.49, 39.03)

Goodness of fit:
  SSE: 9.913
  R-square: 0.9322
  Adjusted R-square: 0.9239
  RMSE: 0.4498

temp=[36 36.1 36.2 36.3 36.4 36.5 36.6 36.7 36.8 36.9 37 37.1 37.2 37.3 37.4 37.5 37.6 37.7 37.8 37.9 38 38.1 38.2 38.3 38.4 38.5 38.6 38.7 38.8 38.9 39 39.1 39.2 39.3 39.4 39.5 39.6 39.7 39.8 39.9 40 40.1 40.2 40.3 40.4 40.5 40.6 40.7 40.8 40.9 41 41.1 41.2 41.3 41.4 41.5] ;
uncalibrated_temp=[99.132 99.185 99.052 99.162 99.203 99.142 99.650 99.720 99.610 99.561 99.764 99.961 99.942 99.863 99.825 99.941 100.127 100.156 100.462 100.323 100.381 100.392 100.527 100.582 100.549 100.362 100.488 100.656 100.792 100.953 100.891 101.095 101.161 101.182 101.161 101.224 101.537 101.491 101.392 101.539 101.565 101.749 101.704 101.764 101.707 101.910 101.968 101.805 101.807 101.791 101.771 102  101.892 101.731 101  101.581 ];
[][1]

它给了我一个图表:

但当我使用一般方程时,它给出了一条曲线,这条曲线与插值曲线有很大不同。

f_temp=uncalibrated_temp;
temp1 = -0.08382  *f_temp.^6  - 0.06449  *f_temp.^5 + 0.3342  *f_temp.^4 + 0.09103  *f_temp.^3 - 0.3258  *f_temp.^2 + 1.629  *f_temp + 38.76  
figure,plot (uncalibrated_temp,temp1)

它给出了右侧的曲线,其中左侧的曲线是由两个矩阵的真点生成的曲线

2 回复 | 直到 7 年前

Loamsiada 7 年前

我没有工具箱……但从纯数值的角度来看,将100量级的值放入6次多项式将是令人失望的:)

Matlab通过在进行拟合过程之前移动和重新缩放值来避免这个问题。信息

where x is normalized by mean 100.7 and std 0.9139

看起来x值已经移动了100.7,缩放了0.9139

f_temp=(uncalibrated_temp-100.7)/0.9139;
temp1 = -0.08382  *f_temp.^6  - 0.06449  *f_temp.^5 + 0.3342  *f_temp.^4 + 0.09103  *f_temp.^3 - 0.3258  *f_temp.^2 + 1.629  *f_temp + 38.76

f_temp=(uncalibrated_temp-100.7)*0.9139;

但我确信其中一个是正确的。如果你找到了解决方案,请告诉我。

祝你好运;)

Vahe Tshitoyan 7 年前

拟合结果中缺少一条重要的线

试试这个

f_temp=(uncalibrated_temp-100.7)./0.9139;

mean(uncalibrated_temp)

和

std(uncalibrated_temp)

您将看到模型给出的值(100.7..和0.91..)。要使数据达到0平均值和1标准差,请从数据中减去平均值,然后除以标准差。例如

f_temp=(uncalibrated_temp-mean(uncalibrated_temp))./std(uncalibrated_temp);