How to have matrix values as a function of another variable

I have a matrix P (defined below) that has some values (h and u) that are a function of a variable t. t have values ranging from 1 to 50. When I define h and n as a function of t and insert them into the matrix I get an error saying "Dimensions of matrices being concatenated are not consistent." How to I fix this issue?
t = 1:50;
h = 0.9*exp(-2*t);
u = 0.9*(1-exp(-2*t));
N = 4;
P = [0.1 h 0 0 0 ; ...
u 0.1 h 0 0; ...
0 u 0.1 h 0; ...
0 0 u 0.1 h; ...
0 0 0 u 0.1];

 採用された回答

Rahul K
Rahul K 2017 年 2 月 17 日
編集済み: Rahul K 2017 年 2 月 17 日

0 投票

h and u are vectors since t is also a vector, hence why you get the error when trying to put them into the matrix P.
Do you want a different P matrix for each value of t?

4 件のコメント

Andrew Poissant
Andrew Poissant 2017 年 2 月 17 日
Yes, I want a new P matrix for each value of t.
Rahul K
Rahul K 2017 年 2 月 17 日
編集済み: Rahul K 2017 年 2 月 17 日
If you want to write over the same P value,
t = 1:50;
O5 = ones(1,5);
O4 = ones(1,4);
h = 0.9*exp(-2*t);
u = 0.9*(1-exp(-2*t));
N = 4;
for i=1:50
P = diag(0.1*O5)+diag(h(i)*O4,1)+diag(u(i)*O4,-1);
end
Otherwise if you want to store everything in a 3D matrix,
P = zeros(5,5,50);
for i=1:50
P(:,:,i) = diag(0.1*O5)+diag(h(i)*O4,1)+diag(u(i)*O4,-1);
end
Andrew Poissant
Andrew Poissant 2017 年 2 月 17 日
The number of P matrices are what I am looking for (not looking for 3D matrix). However, all of the values in the matrices are the same, even though the values of h and u should change with the iteration number (aka time). Do you know why the values are remaining the same?
Andrew Poissant
Andrew Poissant 2017 年 2 月 17 日
nvm, the values just converge very soon in the iterations. thanks!

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