Question

我想解决下图所示的以下方程组，

矩阵系统

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

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

你如何迭代地解决它们？请注意，所需常量的第一个组件的值必须等于1.我正在使用Matlab。

Answer 1

您可以为复杂值数据应用simple multilinear regression。

步骤1.准备矩阵以进行线性回归

您的线性系统

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

没有矩阵，写成

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

重新排列了yelds

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

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

让上面的系统

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

，其中

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

现在，您可以应用线性回归，在

时返回最适合α的回归

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

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，如果您更喜欢内置插件，请查看regress功能：

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

1 个答案:

步骤1.准备矩阵以进行线性回归

步骤2.应用多元线性回归

在MATLAB中