Improving modal analysis code

4 visualizaciones (últimos 30 días)
Pinco
Pinco el 6 de Ag. de 2012
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

Respuesta aceptada

Matt Kindig
Matt Kindig el 6 de Ag. de 2012
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 comentarios
Pinco
Pinco el 8 de Ag. de 2012
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 el 18 de Ag. de 2012
It works very well! I'm sorry for this time in my answer. Thank you very much!!!

Iniciar sesión para comentar.

Más respuestas (0)

Categorías

Más información sobre Loops and Conditional Statements en Help Center y File Exchange.

Community Treasure Hunt

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

Start Hunting!

Translated by