Lecture 2

Lecture 2- State Space Representation

symbols

$\dot{X}(t) = AX(t) + BU(t)$ $Y(t) = CX(t) + DU(t)$

$\dot{X}(t)$ derivative of state vector

$X(t)$ state vector | nx1

$Y(t)$ output vector | px1

$U(t)$ input/control vector | mx1

$A$ symstem matrix | nxn

$B$ input matrix | nxm

$C$ output matrix | pxn

$D$ feedforward matrix | pxm

set $X(t)|_{t=0} = 0$, do Laplace Transfrom

$sX(s) = AX(s) + BU(s)$ $Y(s) = CX(s) + DU(s)$

so,

$X(s) = (sI-A)^{-1} BU(s) = \frac{adj(sI-A)B}{det(sI-A)} U(s)$ $Y(s) = [\frac{Cadj(sI-A)B}{det(sI-A)} + D] U(s)$

thus, transfer function $G(s)$

$G(s) \equiv \frac{Y(s)}{U(s)} = \frac{Cadj(sI-A)B}{det(sI-A)} + D$

if $X(t)|_{t=0} = X(0)$

$sX(s)-X(0) = AX(s) + BU(s)\\ Y(s) = CX(s) + DU(s)$

then

$Y(s) = C(sI-A)^{-1}X(0) + C(sI-A)^{-1}BU(s) + DU(s)$

Here set $\Phi(t) \equiv L[(sI-A)^{-1}] = e^{At} = \sum_{k=0}^{\infty} \frac{t^k A^k}{k!}$

in other way

$(sI-A)^{-1} = \Bigg\{ \begin{aligned} -A^{-1}(I-sA^{-1})^{-1} = -A^{-1}[\sum^{\infty}_{k=0}s^kA^{-k}] && |s|< \lambda_{min}\\ s^{-1}(I-\frac{A}{s})^{-1} = \frac{1}{s}[\sum^{\infty}_{k=0}s^{-k}A^{k}] && |s|> \lambda_{max}\\ \end{aligned}$

Here $\frac{1}{s^{k+1}} = L^{-1}[\frac{t^k}{k!}]$

So we can get $(sI-A)^{-1}|_{s=0} = -A^{-1}, \lim_{s\to \infty} s(sI-A)^{-1}=I$