Question

ï = 2u – 48 - 8x (a) Use Laplace transform to solve for the transfer function (b) (show steps). (b) Bring into state space form. X = AX + BU Y = CX + DU Y = {0} and X = X;} and U = {4}} (c) Find the transfer function from state space form (show le steps). e there a transfer functons, and what is the X2)

Answer 1

Given the equation

x=24 - 48 - 8x

a)

Consider Laplace of x and u as,

Initial conditions as zero $x(0)=0,\dot{x}(0)=0,u(0)=0$

Take Laplace transformation of the equation

$L(\ddot{x})=L(2u-4\dot{x}-8x)$

SPX - Sx(0) - x(0) = 20 – 4(SX – x(0) - 8X S²X = 20 – 45X - 8X (52 +45 + 8)X = 20

The transfer function is :$\\ G(S)=\frac{U(S)}{X(S)}=\frac{S^2+4S+8}{2}$

b)Take

$u_1=u$

$\\ x_1=x, \ x_2=\dot{x} \\ \dot{x_1}=\dot{x} \Rightarrow \dot{x_1}=x_2\ \ \ \ \ Eq1 \\ \dot{x_2}=\ddot{x}$

Also from the given equation $\\ \ddot{x}=-8x-4\dot{x}+2u \Rightarrow \dot{x_2}=-8x_1-4x_2+2u_1\ \ \ Eq2$

also

Bringing Eq1 and Eq2 to state-space equation $\\ \dot{X}=AX+BU,\ \ \ \ Where \ X=\begin{Bmatrix} x_1\\x_2 \end{Bmatrix}, \ U=\begin{Bmatrix} u_1\\ 0 \end{Bmatrix}$

$\\ \begin{Bmatrix} \dot{x_1}\\ \dot{x_2} \end{Bmatrix}= \begin{Bmatrix} 0 &1 \\ -8 & -4 \end{Bmatrix}\begin{Bmatrix} x_1\\ x_2 \end{Bmatrix} +\begin{Bmatrix} 0 & 0 \\ 2 & 0 \end{Bmatrix}\begin{Bmatrix} u_1\\0 \end{Bmatrix} \\ or \dot{X}=AX+BU\ \ \ A= \begin{Bmatrix} 0 &1 \\ -8 & -4 \end{Bmatrix}\ \ \ B=\begin{Bmatrix} 0 & 0 \\ 2 & 0 \end{Bmatrix} \ \ \ \ Eq3$

Also state-space equation

$\\ Y=CX+DU\ \ \ Y=\begin{Bmatrix} x\\ 0 \end{Bmatrix} \\ It\ can\ be \ written \ as \\ using\ x_1=x \\ \begin{Bmatrix} x\\ 0 \end{Bmatrix}=\begin{Bmatrix} 1 &0 \\0 & 0 \end{Bmatrix}\begin{Bmatrix} x_1\\x_2 \end{Bmatrix} \ or \ Y=CX\ \ C=\begin{Bmatrix} 1 &0 \\ 0 &0 \end{Bmatrix} \ \ Eq4$

C)Take Laplace Eq3

$\\ SX(S)=AX(S)+BU(S) \\ (SI-A)X(S)=BU(S) \\ X(S)=(SI-A)^{-1}BU(S)\ \ \ \ Eq5 \\ Y(S)=CX(S)\ \ \ Eq6 \\Eq5\ in \ Eq6 \\ Y(S)=C(SI-A)^{-1}BU(S) \\ \begin{Bmatrix} X(S)\\0 \end{Bmatrix}=\begin{Bmatrix} 1 & 0\\0 & 0 \end{Bmatrix} (\begin{Bmatrix} S &-1 \\8 &S+4 \end{Bmatrix})^{-1}\begin{Bmatrix} 0 &0 \\2 &0 \end{Bmatrix}\begin{Bmatrix} U_1(S) \\ 0 \end{Bmatrix} \\$

$\\Y(S)=\begin{Bmatrix} 1 & 0\\0 & 0 \end{Bmatrix} \begin{Bmatrix} S+4 & 1 \\-8 &S \end{Bmatrix}\begin{Bmatrix} 0 &0 \\2 &0 \end{Bmatrix}U(S)/(S^2+4S+8) \\ Y(S)= \begin{Bmatrix} S+4 & 1 \\0 &0 \end{Bmatrix}\begin{Bmatrix} 0 &0 \\2 &0 \end{Bmatrix}U(S)/(S^2+4S+8)=\begin{Bmatrix} 2 \\0 \end{Bmatrix}U(S)/(S^2+4S+8) \\$

$\begin{Bmatrix} X(S)\\ 0 \end{Bmatrix}=\begin{Bmatrix} 2U_1(S) /(S^2+4S+8) \\0 \end{Bmatrix} \\$

The transfer function can be taken from the above equation as

$\\ G(S)=\frac{X(S)}{U_1(S)}=\frac{2}{S^2+4S+8}$