# Laplace Transform of Vector Valued LTI system [closed]

I'm not really interested in proofs, but I have a LTI system that is vector valued and want to take the laplace transform of it and the inverse laplace transform of the transfer function. I'm wondering how to do this with vectors? A table of vector valued Laplace transforms would be nice too if you can find. I could not. Thanks.

Edit: In particular, I am wondering more about the z-transform. Mainly you have a matrix-valued transfer function $$H(z)$$. How mechanically do you perform the contour integral of $$H(z)$$ to get $$h(t)$$. In particular, how to translate the z-transform tables for vectors and (although I can look this up) how to perform the partial fractions approach for vector-valued functions in the numerator and denominator. One example illustrating what to do at every step starting with a vector-valued transfer function and ending up with $$h(t)$$ would probably be the most helpful.

## closed as unclear what you're asking by Mark Viola, user10354138, Cesareo, Lord Shark the Unknown, max_zornOct 21 '18 at 1:44

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• For $n=2$ and $L$ a LTI system, if you know $L[(\delta,0)]$ and $L[0,\delta]$ ($\delta$ the Dirac delta, or the Kronecker delta in the discrete case) or equivalently their Laplace transforms (Z-transform in the discrete case), then you know $L[ f]$ for every $f$. The (inverse) Laplace or Z-transform is entry-wise and works as usual, except that $Y(s) = H(s) X(s)$ becomes $Y(s) = H(s) \times X(s)$ – reuns Oct 22 '18 at 0:28

## 1 Answer

For $$f : \mathbb{R} \to \mathbb{C}^{n }$$ and $$h : \mathbb{R} \to \mathbb{C}^{n \times n}$$ some vector and matrix valued functions both in $$L^1$$ then $$h\ast f(t) = \int_{-\infty}^\infty h(\tau)\times f(t-\tau) d\tau$$ where $$h(\tau)\times f(t-\tau)$$ is the multiplication of a vector by a matrix, and all the LTI systems are of the form $$f \mapsto h \ast f$$ for some matrix valued distribution $$h$$.

For $$f,h$$ causal, their Laplace transform is as usual and they obey $$\mathcal{L}[h \ast f](s) =\mathcal{L}[h](s)\times\mathcal{L}[f](s)$$

Proof : the entries $$(h \ast f)_{k} = \sum_{l=1}^n f_l \ast h_{k,l}$$ are complex valued functions so their Laplace transform obey $$\mathcal{L}[(h \ast f)_{k}](s) = \sum_{l=1}^n\mathcal{L}[f_l](s)\mathcal{L}[h_{k,l}](s)=\sum_{l=1}^n\mathcal{L}[f]_l(s)\mathcal{L}[h]_{k,l}(s)$$. The columns of $$h$$ are obtained from plugging $$(\Delta_j)_j(t) =\delta(t)$$, $$(\Delta_j)_l(t) = 0$$ in the LTI system