How to store the results of a loop in matlab.

Question

Elyar Ghaffarian Dallali 2022-12-2

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1869607-how-to-store-the-results-of-a-loop-in-matlab

编辑： Matt J 2022-12-2

Actually, I need to solve consecutive eigen value problems which are time dependent; accordingly, I want to extract eigen value of each time step using a ''for loop'' in matlab; the point is, however, that results of each loop is needed to be a components of a vector associated with the specific time step; in other words, I want to write a loop for t=0:1:20 that in each loop, matlab calculate the associated eigen value and put it in the vector as a component of it. so at the end of the day we are supposed to have a vector of eigen values associated with vector of time. please find matlab code here, but when I run the code I encounter with the alarm which is showing ''Array indices must be positive integers or logical values'' and I cannot find a vector of eigenvalues in the workspace as a variable.

A = zeros(1,19);

for t = 0:1:19

m(t) = [1.0, 0.89*sin(0.17*t); 0, 120.0]

k(t) = [760, 0; -2.9*sin(0.19*t), 400]

c(t) = [0.44, 0; -0.96*sin(0.17*t), 140 ]

mk = -inv(m(t))*k(t)

mc = -inv(m(t))*c(t)

B(t) = [zeros(2), eye(2); mk(t), mc(t)]

C(t) = real(sqrt(eig(B(t))))

A(t) =C(1)

end

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Matt J 2022-12-2

编辑：Matt J 2022-12-2

Why are you interested only in the first eigenvalue A(t)=C(1)? How do you know the other eigenvalue is irrelevant, seeing as the order of the eigenvalues can be random?

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Answer 1

Matt J 2022-12-2

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1869607-how-to-store-the-results-of-a-loop-in-matlab#answer_1118597

编辑：Matt J 2022-12-2

在 MATLAB Online 中打开

[B,C,m,k,c]= deal(cell(1,20));
A=nan(1,20);
for t = 1:20
    
    z=t-1;    
    
    m{t}= [1.0, 0.89*sin(0.17*z); 0, 120.0];
    k{t} = [760, 0; -2.9*sin(0.19*z), 400] ;
    c{t} = [0.44, 0; -0.96*sin(0.17*z), 140 ];
    
    mk = -m{t}\k{t};
    mc = -m{t}\c{t};
    B{t} = [zeros(2), eye(2); mk, mc];
    C{t} = real(sqrt(eig(B{t})));
    A(t) =C{t}(1);
end
A
A = 1×20
    3.6978    3.6978    3.6978    3.6978    3.6978    3.6977    3.6977    3.6976    3.6976    3.6976    3.6976    3.6976    3.6977    3.6977    3.6977    3.6978    3.6978    3.6978    3.6978    3.6978