$\begin{pmatrix} x_1 \ x_2 \ \end{pmatrix}$ =$\begin{bmatrix}-3& 2& \ 1& 1&\\end{bmatrix}$$\begin{pmatrix} x_1 \ x_2 \ \end{pmatrix}$ $\begin{pmatrix} 0 \ 2 \ \end{pmatrix}$ u(t) µ(t)=[1 0]$\begin{pmatrix} x_1 \ x_2 \ \end{pmatrix}$ Here$ x_1$&$x_2$ are state variables, µ(t) is a force vector & µ(t) being the system response.

Mumbai University > Mechanical Engineering > Sem 5 > Mechanical measurements and control

Marks: 10M

Year: May 2015 ,Dec 2015 ,May 2016

The given state space system is

$\begin{pmatrix} \ x_1 \\ \ x_2 \\ \end{pmatrix} = \begin{bmatrix} \ -3 & 2 \\ \ 1 & 1 \\ \end{bmatrix} \begin{pmatrix} \ x_1 \\ \ x_2 \\ \end{pmatrix} \begin{bmatrix} \ 0 \\ \ 2\\ \end{bmatrix}$ u(t)

T.F=$\frac{(Y(S))}{(U(S))}$

=c[SI-A]$^-1$(B+D)

Here,

A=$\begin{bmatrix}-3& 2&\\ 1& 1&\\\end{bmatrix}$

B=$\begin{bmatrix} 0 \\ 2\\ \end{bmatrix}$;

C=[1 0] and D=0

[SI-A]=$\begin{bmatrix}S& 0&\\ 0& S&\\\end{bmatrix}$-$\begin{bmatrix}-3& 2&\\ 1& 1&\\\end{bmatrix}$

=$\begin{bmatrix}S+3& -2&\\ -1& S-1&\\\end{bmatrix}$

The inverse of [SI-A] is

$[SI-A]^(-1) $=$\frac{(adj[SI-A])}{(|SI-A|)}$

adj[SI-A] =$\begin{bmatrix}S-1& 2&\\ 1& S+3&\\\end{bmatrix}$

|SI-A|=(S-1)(S+3)-2

=$S^2+3S-S-3-2$

=$S^2+2S-5$