Question determine the transfer function of the system described by the state-space representation shown below 3. A partir de la representación en el espacio de estados, obtenga la función de transferencia del sistema. (30p) * = 135 = }}x+ y = [1 2]x

State space representation to transfer function \RightarrowState space form, and order is n\begin{array}{l}\dot{x}=A x+B U \\y=C x+D U\end{array}Here A - system metrix ( n \times n )B - imput matrix (n \times m)C - output matrix (PXh)D - Feed back or foruard matrix (P \times m)taking laplace transform of equation(i), (ii)\begin{aligned}s X(s)-X(0) & =A X(s)+B U(s) \\Y(s) & =C X(s)+D U(s)\end{aligned}taking zero initial condition x(0)=0\begin{array}{l}S X(s)=A X(s)+B U(s) \ldots \text { (iii) } \\y(s)=c X(s)+D U(s) \ldots \text { (iv) }\end{array}transfor function H(s)=\frac{Y(S)}{V(S)}=\frac{\text { output }}{\text { inpeut }}we noed to find y(s) and v(s), so need to romove Qr x(s) from (iv) o (i), (ii), (iii), (iv)from equation ( iii)\begin{array}{l}S \times(S)-A X(S)=B U(s) \quad \text { [Here I indentity } \\(S I-A) X(S)=B U(S)\end{array}muliplying both side by (S I-A)^{-1}X(s)=(S I-A)^{-1} B U(s)Let say Q(S)=(S I-A)^{-1}X(s)=\phi(s) B U(s)put in equation (iv)\begin{array}{l}y(s)=C \phi(s) B U(s)+D U(s) \\y(s)=[c \phi(s) B+D] U(s)\end{array}H(s)=\frac{Y(s)}{u(s)}=C \Phi(s) B+DHere \phi(S)=(S I-A)^{-1}, and I is inelentity matrieSolution \Rightarrowgiven state space form\begin{array}{l}\dot{x}=\left[\begin{array}{cc}-5 & -1 \\3 & -1\end{array}\right] x+\left[\begin{array}{l}2 \\5\end{array}\right] u \\y=\left[\begin{array}{ll}1 & 2\end{array}\right] x\end{array}Here\begin{array}{l}A=\left[\begin{array}{rr}-5 & -1 \\3 & -1\end{array}\right]_{2 \times 2}, B=\left[\begin{array}{l}2 \\5\end{array}\right]_{2 \times 1} \\C=[1,2]_{1 \times 2}, D=[0]\end{array}First find \phi(s)=(s I-A)^{-1}dimension of I is same as A, so 2 \times 2\begin{array}{l}I=\left[\begin{array}{ll}1 & 0 \\0 & 1\end{array}\right], \quad S I=\left[\begin{array}{ll}S & 0 \\0 & S\end{array}\right] \\S I-A=\left[\begin{array}{ll}5 & 0 \\0 & S\end{array}\right]-\left[\begin{array}{cc}-5 & -1 \\3 & -1\end{array}\right] \\S I-A=\left[\begin{array}{cc}S+5 & +1 \\-3 & S+1\end{array}\right]\end{array}determinant of S I-A or |S I-A|\begin{array}{l}\phi(s) \\\end{array}Nou premultuplying by matrix C and post multiplying by matrix B\begin{array}{l}\left.c \phi(s)=\frac{1}{[(s+1)(s+5)+3}\right] \cdot[1,2]\left[\begin{array}{cc}s+1 & -1 \\3 & s+5\end{array}\right] \\=\frac{1}{[(s+1)(s+5)+3]} \cdot[s+1+6 \quad-1+2 s+10] \\\end{array}\begin{array}{l}=\frac{1}{[(s+1)(s+5)+3]} \cdot[(s+7) 2+(2 s+9) 5] \\c \phi(s) B=\frac{2 s+14+10 s+45}{(s+1)(s+5)+3}=\frac{12 s+59}{(s+1)(s+5)+3} \\H(s)=\text { transferfunction }=C \phi(s) B+D \\H(s)=\frac{12 s+59}{(s+1)(s+5)+3}+0 \\\end{array}Ans - state space form \dot{x}=\left[\begin{array}{c}-5-1 \\ 3-1\end{array}\right] x+\left[\begin{array}{l}2 \\ 5\end{array}\right] xy=\left[\begin{array}{ll}1 & 2\end{array}\right] xtransfer function =\frac{12 S+59}{(s+1)(s+5)+3} ...