$\begin{pmatrix} x_1 \ x_2 \ x_3 \ \end{pmatrix}$= $\begin{bmatrix} 0& 1& 0& \ 0& 0& 1& \ -1& -3& -2& \ \end{bmatrix}$ $\begin{pmatrix}x_1 \ x_2 \ x_3 \ \end{pmatrix}$+ $\begin{pmatrix} 0 \ 0\ 1\ \end{pmatrix}$ u(t) u(t)=[ 1 0 0 ] $\begin{pmatrix} x_1 \ x_2 \ x_3 \ \end{pmatrix}$ Here $x_1$, $x_2$ and $x_3$ are state variables, μ(t) is a false vaector & μ(t) being the sytem response. Obtain transfer function of the system.

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

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Year: Dec 2014

Comparing the given equation with standard equation, we get

A=$\begin{bmatrix} 0& 1& 0& \\ 0& 0& 1& \\ -1& -3& -2& \\ \end{bmatrix}$

B=$\begin{bmatrix}0\\ 0\\ 1\\\end{bmatrix}$

C=$\begin{bmatrix} 1& 0& 0&\\\end{bmatrix}$

D=0

We know that

T.F=$\frac{(Y(S))}{(U(S))}$=C[SI-A]$^(-1)$ B+D

Now,

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