How to find state transition matrix with symbolic parameters efficiently

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)

Answer 1

[Arnab Sen]

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:

http://www.mathworks.com/help/matlab/ref/expm.html

[xianjie Zhou]

Hi Arnab Sen, thanks for the answers. It still take a bit long but can achieve the solution in the end.

[khaled elmoshrefy]

hi, what is t in the exp ?

[ISLAM (伊兰沐）]

Thank you so much! I tried 2x2 matrix I am pasting my code here.

syms t;

A=[0 1; -4 -5];

%State Transition matrix

X=A*t;

[V,D]=eig(X);

STM=(V*diag(exp(diag(D))))* inv(V)

Output

[ (4*exp(-t))/3 - exp(-4*t)/3, exp(-t)/3 - exp(-4*t)/3]

[(4*exp(-4*t))/3 - (4*exp(-t))/3, (4*exp(-4*t))/3 - exp(-t)/3]

Its correct!