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Improving modal analysis code

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Pinco
Pinco 2012 年 8 月 6 日
Hi everyone!
My code resolves a dynamic problem with coupled equation, so I study this problem in the modal space.
I have the matrix of MODAL solution (each column is the solution for time t*, with t* = 0,1,2,.. T) but I want to know the NODAL SOLUTION, so I wrote this code to execute inverse modal transformation (PHI is the modal matrix while BI2 is a translation coordinates matrix, such as g = BI2*PHI*y, where g is in the nodal space while y in the modal):
% y is the modal solution matrix
% g is the nodal solution matrix
g = zeros(size(y));
for i=1:length(y)
g(:,i) = BI2*PHI*y(:,i);
end
In fact, the modal inverse relation is true column by column [y(:,i) -> g(:,i)].
How can I improve my code? This for loop is very slow when I have a lot of dof.. I think I have to rewrite it, but I don't know how.
Thanks in advance! Pinco

採用された回答

Matt Kindig
Matt Kindig 2012 年 8 月 6 日
Hi Pinco,
What are the dimensions of BI2, PHI, and y? You might be able to use matrix multiplication directly to eliminate the loop. You may just be able to write it as:
g = BI2*PHI*y;
  2 件のコメント
Pinco
Pinco 2012 年 8 月 8 日
Thanks a lot for your answer.
size(y) = (n,1)
size(BI2) = (n,n)
size(PHI) = (n,n)
size(g) = (n,t)
where n is the number of dof selected and t is the number of time-point used (i.e if T = 250s and I use a discretization dt = 1s, t=0:250).
Now I can test your code ;)
Pinco
Pinco 2012 年 8 月 18 日
It works very well! I'm sorry for this time in my answer. Thank you very much!!!

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