Question

A system is decribed by the following state and output equations:

$\frac{d{x}_1\left(t\right)}{dt}=-3x_1\left(t\right)+x_2\left(t\right)+2u\left(t\right)$

$\frac{d{x}_1\left(t\right)}{dt}=-2x_2\left(t\right)+u\left(t\right)$

y(t) = $x_1$(t)

when u(t) is the input and y(t) is the output.

The state-trasition matrix of the above system is

• A. $\left[\begin{array}{cc}{e}^{-3t}& 0\\ e^{-2t}+e^{-3t}& e^{-2t}\end{array}\right]$
• B. $\left[\begin{array}{cc}{e}^{-3t}& e^{-2t}-e^{-3t}\\ 0& e^{-2t}\end{array}\right]$
• C. $\left[\begin{array}{cc}{e}^{-3t}& e^{-2t}+e^{-3t}\\ 0& e^{-2t}\end{array}\right]$
• D. $\left[\begin{array}{cc}{e}^{3t}& e^{-2t}-e^{-3t}\\ 0& e^{-2t}\end{array}\right]$

Answer 1

The correct option is B $\left[\begin{array}{cc}{e}^{-3t}& e^{-2t}-e^{-3t}\\ 0& e^{-2t}\end{array}\right]$

$[sI-A{]}^{-1}=\frac{\left[\begin{array}{cc}s+2& 1\\ 0& s+3\end{array}\right]}{(s+2)(s+3)}$

$\left[\begin{array}{cc}\frac{1}{s+3}& \frac{1}{(s+2)(s+3)}\\ 0& \frac{1}{s+2}\end{array}\right]$

$\left[\begin{array}{cc}\frac{1}{s+3}& \frac{1}{s+2}-\frac{1}{s+3}\\ 0& \frac{1}{s+2}\end{array}\right]$

State transition matrix

$=L^{-1}\left[\right(sI-A{)}^{-1}]$

$=L^{-1}\left[\begin{array}{cc}\frac{1}{s+3}& \frac{1}{s+2}-\frac{1}{s+3}\\ 0& \frac{1}{s+2}\end{array}\right]$

$=\left[\begin{array}{cc}{e}^{-3t}& e^{-2t}-e^{-3t}\\ 0& e^{-2t}\end{array}\right]$