Question

The following equation defines a separately excited dc motor in the form of a differential equation
$\frac{d^2\omega }{d{t}^2}+\frac{B}{J}\frac{d\omega }{dt}+\frac{K^2}{LJ}\omega =\frac{K}{LJ}{V}_a$
The above equation may be organized in the statespace form as follows
$\left[\begin{array}{c}\frac{d^2\omega }{d{t}^2}\\ \frac{d\omega }{dt}\end{array}\right]=P\left[\begin{array}{c}\frac{d\omega }{dt}\\ \omega \end{array}\right]+Q{V}_a$
Where the P matrix is given by

• A. $\left[\begin{array}{cc}-\frac{B}{J}& -\frac{K^2}{LJ}\\ 1& 0\end{array}\right]$
• B. $\left[\begin{array}{cc}-\frac{K^2}{LJ}& -\frac{B}{J}\\ 0& 1\end{array}\right]$
• C. $\left[\begin{array}{cc}0& 1\\ -\frac{K^2}{LJ}& -\frac{B}{J}\end{array}\right]$
• D. $\left[\begin{array}{cc}1& 0\\ -\frac{B}{J}& -\frac{K^2}{LJ}\end{array}\right]$

Answer 1

The correct option is A $\left[\begin{array}{cc}-\frac{B}{J}& -\frac{K^2}{LJ}\\ 1& 0\end{array}\right]$

Let,
$x_1=\frac{d\omega }{dt}and{x}_2=\omega$
${\dot{x}}_2=x_1$

$\frac{d^2\omega }{d{t}^2}+\frac{B}{J}\frac{d\omega }{dt}+\frac{K^2}{LJ}\omega =\frac{K}{LJ}{V}_a$

$\frac{d^2\omega }{d{t}^2}=-\frac{B}{J}\frac{d\omega }{dt}-\frac{K^2}{LJ}\omega +\frac{K}{LJ}{V}_a$

$\Rightarrow {\dot{x}}_1=-\frac{B}{J}{x}_1-\frac{K^2}{LJ}{x}_2+\frac{K}{LJ}{V}_a$

${\dot{x}}_2=x_1$

$\left[\begin{array}{c}{\dot{x}}_1\\ {\dot{x}}_2\end{array}\right]=P\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]+Q{V}_a$

where,
$P=\left[\begin{array}{cc}-\frac{B}{J}& -\frac{K^2}{LJ}\\ 1& 0\end{array}\right]andQ=\left[\begin{array}{c}\frac{k}{LJ}\\ 0\end{array}\right]$