How can i calculate e^A*t
245 visualizzazioni (ultimi 30 giorni)
Mostra commenti meno recenti
How can i calculate e^A*t without using Markov Chain?
Where e=exp , A is a square matrix, and t is a variable
10 Commenti
Risposta accettata
Più risposte (5)
Kye Taylor
il 31 Mag 2012
Use the expm function for computing a matrix exponential
4 Commenti
KJ N
il 9 Nov 2017
exp() only does computes the exponential of A element-by-element, as shown above like this: >> a = [1 2 3 ; 2 5 2; 1 4 3]
a =
1 2 3
2 5 2
1 4 3
>> syms t
>> exp(a*t)
ans =
[ exp(t), exp(2*t), exp(3*t)]
[ exp(2*t), exp(5*t), exp(2*t)]
[ exp(t), exp(4*t), exp(3*t)]
If that's what you're going for, that's great, but not terribly difficult to compute by hand for even somewhat large n x n matrices with integer elements. However, the original poster said they wanted to avoid using the markov chain (a somewhat onerous process, especially when done by hand for large matrices, even with simple integer values as the elements), leading me to understand they were referring to the matrix exponential, not the element-by-element exponential, hence the correct answer in this case would be to use expm(). I had been looking for the same answer, and Kye Taylor was the only post saying use expm instead of exp, so I thought I would try to ensure those in the future looking for the same answer as myself would be helped by a clarification.
Walter Roberson
il 9 Nov 2017
We tried a number of times to get the original poster to clarify, but all we got was that they want the exp() solution and that they are looking for a "deeper reason" for something. The poster effectively defined the exp() solution as being the correct one.
Your analysis might well be what the poster really needed, but it is contrary to what little they defined as being correct for their needs.
Shenhai
il 20 Gen 2017
Modificato: Shenhai
il 20 Gen 2017
I guess it is not always possible to get the close form solution of exp(At)...
Sometimes I can get result with: exp(At) = iL(sI-A)^-1, where iL is the inverse Laplace transformation, like:
syms s t
A = [0 1;0 0];
expAt = ilaplace(inv(s*eye(size(A,1))-A),s,t);
This will give the result as: [1 t;0 1]
Any other ideas?
0 Commenti
Shahram Bekhrad
il 8 Giu 2012
As far as I'm aware you probably need it for finding the answer of a state space equation. I myself couldn't find any good function or command yet, so you might have to write a Script file (m-file) and find it. you can use about 3 or 4 way of calculating the said statement. These things are taught in courses like modern control theory. I used the following expression but still have some difficulties. exp(A.t)=I+At+ (At)^2/2! + (At)^3/3!+ (At)^4/4!+. . .
0 Commenti
ABCD
il 29 Set 2016
Dear Nick, do you mean this?
>> a = [1 2 3 ; 2 5 2; 1 4 3]
a =
1 2 3
2 5 2
1 4 3
>> syms t >> exp(a*t)
ans =
[ exp(t), exp(2*t), exp(3*t)] [ exp(2*t), exp(5*t), exp(2*t)] [ exp(t), exp(4*t), exp(3*t)]
1 Commento
ABCD
il 29 Set 2016
>> a = [1 2 3 ; 2 5 2; 1 4 3]
a =
1 2 3
2 5 2
1 4 3
>> syms t
>> exp(a*t)
ans =
[ exp(t), exp(2*t), exp(3*t)]
[ exp(2*t), exp(5*t), exp(2*t)]
[ exp(t), exp(4*t), exp(3*t)]
Vedere anche
Categorie
Scopri di più su Linear Algebra in Help Center e File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!