Question

A second order LTI system is described by the following state equations
$\frac{d}{dt}{x}_1\left(t\right)-x_2\left(t\right)=0$
$\frac{d}{dt}{x}_2\left(t\right)+2x_1\left(t\right)+3x_2\left(t\right)=r\left(t\right)$
where $x_1\left(t\right)$ and $x_2\left(t\right)$ are the two state variables and $r\left(t\right)$ denotes the input. The output $c\left(t\right)=x_1\left(t\right)$. The system is

• A. undamped (oscillatory)
• B. underdamped
• C. critically damped
• D. overdamped

Answer 1

The correct option is D overdamped

Given state variable equations are as follows:
$\frac{d}{dt}{x}_1\left(t\right)-x_2\left(t\right)=0.....\left(i\right)$
$\frac{d}{dt}\left(x_2\right)+2x_2\left(t\right)+3x_2\left(t\right)=r\left(t\right)....\left(ii\right)$

Also given that, input = $r\left(t\right)$ and output = $x_1\left(t\right)$

By applying Laplace transform to equation (i), we get
$s{X}_1\left(s\right)=X_2\left(s\right)\left(iii\right)$

By applying Laplace transform to equation (ii), we get,

$s{X}_2\left(s\right)-2X_1\left(s\right)+3X_2\left(s\right)=R\left(s\right)$

$X_2\left(s\right)=\frac{R\left(s\right)-2X_1\left(s\right)}{s+3}...\left(iv\right)$

By substituting equation (iv) in equation (iii), we get,

$s{X}_1\left(s\right)=\frac{R\left(s\right)-2X_1\left(s\right)}{s+3}$

$(s^2+3s+2)X_1\left(s\right)=R\left(s\right)$

So, the transfer function of the given system is ,

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

${\omega }_n^2=2$

$2\xi {\omega }_n=3$

$\xi =\frac{3}{2{\omega }_n}=\frac{3}{2\sqrt{2}}=1.06$

$As\xi >1$ ,
The given system is overdamped.