94 Laplace transform of (A,B,C,D)

The Laplace transform of

$$\frac{dX}{dt}=AX+BU$$

$$Y=CX+DU$$

is

$$s\widehat{X}(s)-X(0)=A\widehat{X}(s)+B\widehat{U}(s)$$

$$\widehat{Y}(s)=C\widehat{X}(s)+D\widehat{U}(s).$$

Hence when $s$ is not an eigenvalue of $A$,

$$\widehat{X}(s)={(sI-A)}^{-1}B\widehat{U}(s)+{(sI-A)}^{-1}X(0)$$

$$\widehat{Y}(s)=C{(sI-A)}^{-1}B\widehat{U}(s)+D\widehat{U}(s)+C{(sI-A)}^{-1}X(0).$$

When $X(0)=0$ we get $\widehat{Y}(s)=T(s)\widehat{U}(s)$.