Question

A state space representation for the transfer function $\frac{y\left(s\right)}{u\left(s\right)}=\frac{s+6}{s^2+5s+6}$ is $\dot{x}=Ax=Bu,$ where
$A=\left[\begin{array}{cc}0& 1\\ -6& -5\end{array}\right],B=\left[\begin{array}{c}0\\ 1\end{array}\right]$, and the value of C is

• A. $\left[\begin{array}{c}1\\ 2\end{array}\right]$
• B. $$\left[\begin{array}{cc}6& 1\end{array}\right]$$
• C. $$\left[\begin{array}{cc}1& 2\\ -1& 0\end{array}\right]$$
• D. $$\left[\begin{array}{cc}1& -5\end{array}\right]$$

Answer 1

The correct option is B $$\left[\begin{array}{cc}6& 1\end{array}\right]$$

$\frac{Y\left(s\right)}{U\left(s\right)}=\frac{s+6}{s^2+5s+6}$
$or\frac{X_1\left(s\right)}{U\left(s\right)}\times \frac{Y\left(s\right)}{X_1\left(s\right)}=\frac{s+6}{s^2+5s+6}$
Hence,
$\frac{X_1\left(s\right)}{U\left(s\right)}=\frac{1}{s^2+5s+6}$

Let the state variable be defined as

$x_1\left(t\right)=y\left(t\right)$
${\dot{x}}_1=x_2$
$\ddot{x_1}=\dot{x_2}$

$Then\dot{x_1}+5\dot{x_1}+6x_1=u\left(t\right)$ ....(i)

$or\dot{x_2}=-5x_2-6x_1+u\left(t\right)$

$and\frac{Y\left(s\right)}{X_1\left(s\right)}=s+6$

$orY\left(t\right)=\dot{x_1}+6x_1=x_2+6x_2$ ....(ii)

Writing equation (i) and equation (ii) in matrix from, we have

$y=\left[61\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]$

and $\left[\begin{array}{c}\dot{x_1}\\ \dot{x_2}\end{array}\right]=\left[\begin{array}{cc}0& 1\\ -6& -5\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]+\left[\begin{array}{c}0\\ 1\end{array}\right]u\left(t\right)$

Hence, the value of C is = [6 1]