$\begin{pmatrix} x_1 \ x_2 \ x_3 \ \end{pmatrix}$= $\begin{bmatrix} 0& 1& 0& \ 0& 0& 1& \ -1& -3& -2& \ \end{bmatrix}$ $\begin{pmatrix}x_1 \ x_2 \ x_3 \ \end{pmatrix}$+ $\begin{pmatrix} 0 \ 0\ 1\ \end{pmatrix}$ 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

Marks:

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

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

Now,to find [SI-A]$^(-1)$ by adj method, we know

$P^(-1)$=$\frac{(Adj P)}{(|P|)}$

$[SI-A]^(-1) $=$\frac{1}{(S^2+2S-5)}$ $\begin{bmatrix} S-1 & 2 \\ 1 & S+3 & \\ \end{bmatrix}$

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

=[1 0] $\frac{1}{(S^2+2S-5)}\begin{bmatrix} \ s-1 & 2 \\ \ 1 & s+3 \\ \ \end{bmatrix} \begin{bmatrix} \ 0 \\ \ 2 \\ \end{bmatrix}$

=$\frac{\begin{bmatrix}S-1& 0&\\\end{bmatrix} \begin{bmatrix} 0 \\ 2 \\\end{bmatrix}}{(S^2+2S-5)}$

$\frac{(Y(S))}{(U(S))}$=$\frac{4}{(S^2+2S-5)}$

The cofactor matrix is, P(1,1)=S(S+2)+3=$S^2$+2S+3 P(1,2)=1 P(1,3)=+S P(2,1)=-S-2 P(2,2)=$S^2$+2S P(2,3)=3S-1 P(3,1)=1 P(3,2)=-S P(3,3)=$S^2$

Cofactors elements=$(-1)^(i+j) A_ij$

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

This is the matrix of cofactors

Adj(P)=$[Matrix of coefficient]^T$

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

Now

|P|=$S[S^2+2S+3]+1[-1]+0$

|P|=$S^3+2S^2+3S-1$

Substituting in equation, $[SI-A]^(-1)$=$\frac{\begin{bmatrix}S^2+2S+3& S+2& 1& \\ 1& S^2+2S& S& \\ S &-3S+1 &S^2& \\\end{bmatrix}}{(S^3+2S^2+3S-1)}$

$[SI-A]^(-1)$=$\frac{1}{(S^3+2S^2+3S-1)}$ $\begin{bmatrix}S^2+2S+3& S+2& 1& \\ 1& S^2+2S& S& \\ S&-3S+1& S^2&\\\end{bmatrix}$

Substituting above value in the equation of transfer function,

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

=[1 0 0] x $\frac{1}{(S^3+2S^2+3S-1)}$ $\begin{bmatrix}S^2+2S+3& S+2& 1& \\ 1& S^2+2S& S&\\ S&-3S+1&S^2&\\\end{bmatrix}$ $\begin{bmatrix}0 \\0 \\1 \\\end{bmatrix}$

=$\frac{1}{(S^3+2S^2+3S-1)}$ [S^2+2S+3 S+2 1] $\begin{bmatrix}0 \\0 \\1 \\\end{bmatrix}$

=$\frac{1}{(S^3+2S^2+3S-1)}$(1)

T.F=$\frac{1}{(S^3+2S^2+3S-1)}$