Question

For the system governed by the set of equation:
$\frac{d{x}_1}{dt}=2x_1+x_2+u$
$\frac{d{x}_2}{dt}=-2x_1+u$
$y=3x_1$
The transfer function $\frac{Y\left(s\right)}{U\left(s\right)}$ is given by

• A. $\frac{3(s+1)}{(s^2-2s+2)}$
• B. $\frac{3(2s+1)}{(s^2-2s+1)}$
• C. $\frac{(s+1)}{(s^2-2s+1)}$
• D. $\frac{3(2s+1)}{(s^2-2s+2)}$

Answer 1

The correct option is A $\frac{3(s+1)}{(s^2-2s+2)}$

$\frac{d{x}_1}{dt}=2x_1+x_2+u$ .....(i)

$\frac{d{x}_2}{dt}=-2x_1+u$ ....(ii)

On applying Laplace transform for equation (i),

$s{X}_1=2X_1+X_2+U$ ...(iii)

On applying Laplace transform for equation (ii),

$s{X}_2=-2X_1+U$

$X_2=(-\frac{2X_1+U}{s})$

substitute $X_2$ in equation (iii)

$s{X}_1=2X_1+(-\frac{2X_1+U}{s})+U$

$s{X}_1-2X_1+\frac{2X_1}{s}=\frac{U}{s}+U$

$X_1(s-2+\frac{2}{s})=U(1+\frac{1}{s})$

$\frac{X_1}{U}=\frac{(1+1/s)}{(s-2+2/s)}$

$\Rightarrow \frac{Y}{U}=\frac{3X_1}{U}=\frac{3(1+1/s)}{(s-2+2/s)}$

$\Rightarrow \frac{Y}{U}=\frac{3(s+1)}{(s^2-2s+2)}$