How to find state transition matrix with symbolic parameters efficiently
223 views (last 30 days)
Show older comments
xianjie Zhou
on 17 Feb 2016
Commented: khaled elmoshrefy
on 3 Jul 2020
Hi
I am going to find the state transition matrix of a 4x4 system. But matlab ran for a long time without giving a solution. Is there any other ways to find the solution?
Thanks
syms t
A=[-2.1 -0.98 -0.158 -2.05
2.22 -11.03 0 0
-27.64 0.19 -6.39 -82.6
0 0 1 0]
TM=expm(A*t)
0 Comments
Accepted Answer
Arnab Sen
on 23 Feb 2016
Hi xianjie,
I tried to execute the code you provided and found the same issue that it's running long time without giving the output. I investigated into it and found that the function 'expm' is defined as:
>> [V,D] = eig(X)
>> expm(X) = V*diag(exp(diag(D)))/V
Now I found that the right matrix division 'mrdivide'('/') is taking long time. I am not sure why it's taking so long. However, right matrix division '/' can be approximated as matrix inverse 'inv' followed by matrix multiplication. That is, A/B can be approximated by the operation A*inv(B). So, as a workaround, you may consider the following code snippet to achieve the same functionality:
>>syms t;
>>A=[-2.1 -0.98 -0.158 -2.05
2.22 -11.03 0 0
-27.64 0.19 -6.39 -82.6
0 0 1 0];
>>X=A*t;
>>[V,D]=eig(X);
>>TM=(V*diag(exp(diag(D))))* inv(V)
For more details, refer to the following link:
0 Comments
More Answers (1)
See Also
Categories
Find more on Linear Algebra in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!