Matlab中复变量的曲线拟合

Question

我想解如下图所示的方程组，

矩阵系统

其中矩阵 A 的分量是复数，角度 (theta) 从具有 m 个分区的 0 to 2*pi 和 n = 9 开始。已知值z = x + iy.假设矩阵z的x和y为

z =

     0    1.0148
0.1736    0.9848
0.3420    0.9397
0.5047    0.8742
0.6748    0.8042
0.8419    0.7065
0.9919    0.5727
1.1049    0.4022
1.1757    0.2073
1.1999         0
1.1757   -0.2073
1.1049   -0.4022
0.9919   -0.5727
0.8419   -0.7065
0.6748   -0.8042
0.5047   -0.8742
0.3420   -0.9397
0.1736   -0.9848
     0   -1.0148

你如何迭代解决它们？请注意，所需常量的第一个分量的值必须等于 1。我正在使用 Matlab。

Answer 1

您可以为复杂值数据申请simple multilinear regression。

第 1 步。为线性回归准备好矩阵

你的线性系统

$\begin{bmatrix} z_1\ z_2\ z_3\ \vdots\ z_m \end{bmatrix} = \begin{bmatrix} \varphi_{1,1}&\varphi_{1,2}&\varphi_{1,3}&\cdots&\varphi_{1,n}\ \varphi_{2,1}&\varphi_{2,2}&\varphi_{2,3}&\cdots&\varphi_{2,n}\ \varphi_{3,1}&\varphi_{3,2}&\varphi_{3,3}&\cdots&\varphi_{3,n}\ \vdots&\vdots&\vdots&\ddots&\vdots\ \varphi_{m,1}&\varphi_{m,2}&\varphi_{m,3}&\cdots&\varphi_{m,n} \end{bmatrix} \begin{bmatrix} 1\ \alpha_2\ \alpha_3\ \vdots\ \alpha_n \end{bmatrix}$

不用矩阵写成

$\begin{cases} z_1 = \varphi_{1,1} + \varphi_{1,2}\alpha_2 + \varphi_{1,3}\alpha_3 + \cdots + \varphi_{1,n}\alpha_n\ z_2 = \varphi_{2,1} + \varphi_{2,2}\alpha_2 + \varphi_{2,3}\alpha_3 + \cdots + \varphi_{2,n}\alpha_n\ z_3 = \varphi_{3,1} + \varphi_{3,2}\alpha_2 + \varphi_{3,3}\alpha_3 + \cdots + \varphi_{3,n}\alpha_n\ \vdots\ z_m = \varphi_{m,1} + \varphi_{m,2}\alpha_2 + \varphi_{m,3}\alpha_3 + \cdots + \varphi_{m,n}\alpha_n\ \end{cases}$

重新排列的结果

$\begin{cases} z_1 - \varphi_{1,1} = \varphi_{1,2}\alpha_2 + \varphi_{1,3}\alpha_3 + \cdots + \varphi_{1,n}\alpha_n\ z_2 - \varphi_{2,1} = \varphi_{2,2}\alpha_2 + \varphi_{2,3}\alpha_3 + \cdots + \varphi_{2,n}\alpha_n\ z_3 - \varphi_{3,1} = \varphi_{3,2}\alpha_2 + \varphi_{3,3}\alpha_3 + \cdots + \varphi_{3,n}\alpha_n\ \vdots\ z_m - \varphi_{m,1} = \varphi_{m,2}\alpha_2 + \varphi_{m,3}\alpha_3 + \cdots + \varphi_{m,n}\alpha_n\ \end{cases}$

如果你用矩阵重写它你会得到

$\begin{bmatrix} z_1-\varphi_{1,1}\ z_2-\varphi_{2,1}\ z_3-\varphi_{3,1}\ \vdots\ z_m-\varphi_{m,1}\end{bmatrix} = \begin{bmatrix} \varphi_{1,2}&\varphi_{1,3}&\cdots&\varphi_{1,n}\ \varphi_{2,2}&\varphi_{2,3}&\cdots&\varphi_{2,n}\ \varphi_{3,2}&\varphi_{3,3}&\cdots&\varphi_{3,n}\ \vdots&\vdots&\ddots&\vdots\ \varphi_{m,2}&\varphi_{m,3}&\cdots&\varphi_{m,n} \end{bmatrix} \begin{bmatrix} \alpha_2\ \alpha_3\ \vdots\ \alpha_n \end{bmatrix}$

步骤 2. 应用多元线性回归

设上面的系统为

$\mathbf{Y} = \mathbf{R} \cdot \mathbf{\alpha}$

哪里

$\mathbf{Y} = \begin{bmatrix} z_1-\varphi_{1,1}\ z_2-\varphi_{2,1}\ z_3-\varphi_{3,1}\ \vdots\ z_m-\varphi_{m,1}\end{bmatrix} \qquad \mathbf{R} = \begin{bmatrix} \varphi_{1,2}&\varphi_{1,3}&\cdots&\varphi_{1,n}\ \varphi_{2,2}&\varphi_{2,3}&\cdots&\varphi_{2,n}\ \varphi_{3,2}&\varphi_{3,3}&\cdots&\varphi_{3,n}\ \vdots&\vdots&\vdots&\ddots&\vdots\ \varphi_{m,2}&\varphi_{m,3}&\cdots&\varphi_{m,n} \end{bmatrix} \qquad \mathbf{\alpha} = \begin{bmatrix} \alpha_2\ \alpha_3\ \vdots\ \alpha_n \end{bmatrix}$

现在您可以应用线性回归，当

时returns最适合α

$\hat{\mathbf{\alpha}} = \left (\mathbf{R}^* \mathbf{R} \right )^{-1} \mathbf{R}^* \mathbf{Y}$

其中

$\mathbf{R}^* = \bar{\mathbf{R}}^'$

是共轭转置。

在 MATLAB 中

Y = Z - A(:,1);             % Calculate Y subtracting the first col of A from Z
R = A(:,:); R(:,1) = [];    % Calculate R as an exact copy of A, just without first column
Rs = ctranspose(R);         % Calculate R-star (conjugate transpose of R)
alpha = (Rs*R)^(-1)*Rs*Y;   % Finally apply multiple linear regression
alpha = cat(1, 1, alpha);   % Add alpha1 back, whose value is 1

或者，如果您更喜欢 built-ins，请查看 regress 函数：

Y = Z - A(:,1);             % Calculate Y subtracting the first col of A from Z
R = A(:,:); R(:,1) = [];    % Calculate R as an exact copy of A, just without first column
alpha = regress(Y, R);      % Finally apply multiple linear regression
alpha = cat(1, 1, alpha);   % Add alpha1 back, whose value is 1

Matlab中复变量的曲线拟合

Curve fitting of complex variable in Matlab

matlab

curve-fitting

第 1 步。为线性回归准备好矩阵

步骤 2. 应用多元线性回归

在 MATLAB 中