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I am given that for $A$ that is $n \times n$ matrix of full rank,

$$\int_{0}^{t}e^{A\sigma}d\sigma = (e^{At}-I)A^{-1}$$

Then I am using this to solve LTI system

$$\dot{x}=Ax+Bu$$

Here, $x(0) = x_{0}$ and u is a constant vector.

I went ahead and used the general solution for LTI system,

$$x(t)=e^{A(t-t_{0})}x_{0}+\int_{t_0}^{t}e^{A(t-t_{0})}B(\tau)u(\tau) \, d\tau$$

I have $B$ and $U$ constant and time from 0 to t so this reduces to

$$x(t)=e^{At}x_{0}+\int_{0}^{t}e^{A(t-t_{0})}Bu \, d\tau$$

I am kinda stuck here, what should I do with those constant matrix $B u$ to solve this system using $\int_{0}^{t}e^{A\sigma}d\sigma = (e^{At}-I)A^{-1}$ ?

I know I am not allowed to just pull out $Bu$ outside of the integral because I am dealing with matrices. Any ideas?

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1 Answer 1

Your formula for the LTI system is slightly incorrect.

$\int_{0}^{t}e^{A(t-\tau)}Bu \, d\tau = e^{At}(\int_{0}^{t}e^{-A \tau} d\tau) Bu$.

Now you can substitute your formula (with $-A$).

This gives: $e^{At}(\int_{0}^{t}e^{-A \tau} d\tau) Bu = e^{At} (I-e^{-At}) A^{-1} Bu = (e^{At}-I)A^{-1} Bu$.

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Ah!! Thanks a lot. I knew I was missing on something trivial! –  onion Oct 1 '12 at 5:25
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