Using subs in a matrix of equations

2 次查看（过去 30 天）
Anthony Sirico
Anthony Sirico 2021-3-6
评论： Anthony Sirico 2021-3-6
If i have a matrix of eqns, how can i sub in certain letter values now. I am using a live script where the first section creates the matrix, the second section i want to sub in the values of the constants.
%section1
close all; clear; clc;
syms A C0 Dv ka kh kt x1 x2 x3 x4 v1 v2 v3 u1 u2
%each element of the linearizes matrix
f1 = 1/A * (ka*(u1+u2)-C0*sqrt(x1-x2));
f2 = 1/A * (C0*sqrt(x1-x2)-Dv*sqrt(x2)*v1);
f3 = 1/(A*x1) * ((v2-x3)*ka*u1+(v3-x3)*ka*u2);
f4 = 1/(A*x2) * (x3-x4)*C0*sqrt(x1-x2);
AJ = jacobian([f1;f2;f3;f4],[x1 x2 x3 x4])
BJ = jacobian([f1;f2;f3;f4],[u1 u2])
BvJ = jacobian([f1;f2;f3;f4],[v1 v2 v3])
CJ = [0 kh 0 0;
0 0 0 kt]
%section 2
A = 0.785;
Dv = 2.66;
C0 = 0.056;
ka = 0.004;
kh = 2;
kt = 0.1;

请先登录，再回答此问题。

采纳的回答

Walter Roberson
Walter Roberson 2021-3-6
Ran in:
%section1
close all; clear; clc;
syms A C0 Dv ka kh kt x1 x2 x3 x4 v1 v2 v3 u1 u2
%each element of the linearizes matrix
f1 = 1/A * (ka*(u1+u2)-C0*sqrt(x1-x2));
f2 = 1/A * (C0*sqrt(x1-x2)-Dv*sqrt(x2)*v1);
f3 = 1/(A*x1) * ((v2-x3)*ka*u1+(v3-x3)*ka*u2);
f4 = 1/(A*x2) * (x3-x4)*C0*sqrt(x1-x2);
AJ = jacobian([f1;f2;f3;f4],[x1 x2 x3 x4])
AJ = 
BJ = jacobian([f1;f2;f3;f4],[u1 u2])
BJ = 
BvJ = jacobian([f1;f2;f3;f4],[v1 v2 v3])
BvJ = 
CJ = [0 kh 0 0;
0 0 0 kt]
CJ = 
%section 2
A = 0.785;
Dv = 2.66;
C0 = 0.056;
ka = 0.004;
kh = 2;
kt = 0.1;
%section 3
subs(AJ)
ans = 
subs(BJ)
ans = 

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Calculus 的更多信息

标签

产品


版本

R2020b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by