Question

For a system with the transfer function:

$H\left(s\right)=\frac{3(s-2)}{s^3+4s^2-2s+1}$

the matrix A in the state space form $\dot{x}=Ax+Bu$ is equal to

• A. $\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ -1& 2& -4\end{array}\right]$
• B. $\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ -1& 2& -4\end{array}\right]$
• C. $\left[\begin{array}{ccc}0& 1& 0\\ 3& -2& 1\\ 1& -2& 4\end{array}\right]$
• D. $\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 1\\ -1& 2& -4\end{array}\right]$

Answer 1

The correct option is B $\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ -1& 2& -4\end{array}\right]$

$H\left(s\right)=\frac{Y\left(s\right)}{U\left(s\right)}=\frac{3(s-2)}{s^3+4s^2-2s+1}$

$\frac{Y\left(s\right)}{X_1\left(s\right)}.\frac{X_1\left(s\right)}{U\left(s\right)}=3(s-2)\left(\frac{1}{S^3+4s^2-2s+1}\right)$

Let,

$\frac{X_1\left(s\right)}{U\left(s\right)}$ = $\frac{1}{s^3+4s^2-2s+1}$

$S^3{x}_1\left(s\right)+4s^2{x}_1\left(s\right)-2s{x}_1\left(s\right)+x_1\left(s\right)=u\left(s\right)$

Replacing ${}^{\prime }{s}^{\prime }$ by $\frac{d}{dt}$,

$\frac{d^3{x}_1}{d{t}^2}+4\frac{d^2{x}_1}{d{t}^2}-2\frac{d{x}_1}{dt}+x_1=u\left(t\right)......\left(i\right)$

Let,
$\frac{d{x}_1}{dt}$ = $x_2={\dot{x}}_1$

$\frac{d^2{x}_1}{dt}$ = ${\dot{x}}_2=x_3$

Replacing equation (i),

$\dot{x_3}+4x_3-2x_2+x_1=u\left(t\right)$

$\dot{x_3}=-x_1+2x_2-4x_3+u\left(t\right)$

$\dot{x_1}=x_2$

$\dot{x_2}=x_3$

$\dot{x_3}$ = -$x_1+2x_2-4x_3+u\left(t\right)$

$\left[\begin{array}{c}\dot{x_1}\\ \dot{x_2}\\ \dot{x_3}\end{array}\right]$ = $\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ -1& 2& -4\end{array}\right]$ $\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]$ + $\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]u\left(t\right)$

$So,A=\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ -1& 2& -4\end{array}\right]$