Question

Network is described by the state model as

${\stackrel{.}{x}}_1=x_2$

${\stackrel{.}{x}}_2=-9x_1-7x_2+u$

$y=x_1+2x_2$

The steady state value of response for step input is

• A. 0.20
• B. 0.19
• C. 0.11
• D. 0.09

Answer 1

The correct option is C 0.11

(c)

${\stackrel{.}{x}}_1=x_2$

${\stackrel{.}{x}}_2=-9x_1-7x_2+u$

$y=x_1+2x_2$

$\left(\begin{array}{c}\stackrel{.}{x_1}\\ \stackrel{.}{x_2}\end{array}\right]=\left(\begin{array}{cc}0& 1\\ -9& -7\end{array}\right]\left(\begin{array}{c}{x}_1\\ x_2\end{array}\right]+\left(\begin{array}{c}0\\ 1\end{array}\right]u$$y=\left[12\right]\left(\begin{array}{c}{x}_1\\ x_2\end{array}\right]$

Transfer function = $C{[sI-A]}^{-1}B$
$sI-A=\left(\begin{array}{cc}s& 0\\ 0& s\end{array}\right]-\left(\begin{array}{cc}0& 1\\ -9& -7\end{array}\right]=\left(\begin{array}{cc}s& -1\\ 9& s+7\end{array}\right]$

${[sI-A]}^{-1}=\frac{1}{s^2+7s+9}\left(\begin{array}{cc}s+7& 1\\ -9& s\end{array}\right]$

$T.F.=\frac{1}{s^2+7s+9}\left[12\right]\left(\begin{array}{cc}s+7& 1\\ -9& s\end{array}\right]\left(\begin{array}{c}0\\ 1\end{array}\right]$

$T.F.=\frac{2s+1}{s^2+7s+9}$

$U\left(s\right)=\frac{1}{s}$

$Y\left(s\right)=\frac{(2s+1)}{s(s^2+7s+9)}$

Steady state value Y(s)

${lim}_{s\to 0}sY\left(s\right)={lim}_{s\to 0}\frac{2s+1}{s^2+7s+9}=\frac{1}{9}=0.11$

Alternate solution:
In s domain,
$X_2=s{X}_1$

$-9X_1-7X_2+U=s{X}_2$

$Y=X_1+2X_2$,

$U=\frac{1}{s}$

Then , $Y=\frac{2s+1}{s^2+7s+9}$

Steady state value $y\left(\infty \right)=sY{|}_{s\to 0}=\frac{1}{9}=0.11$