Question

A state variable system $x\left(t\right)=\left[\begin{array}{cc}0& 1\\ 0& -3\end{array}\right]X\left(t\right)+\left[\begin{array}{c}1\\ 0\end{array}\right]U\left(t\right),$ with the initial condition $X\left(0\right)=[-13{]}^T$ and the unit step input $u\left(t\right)$ has

The state transition matrix

• A. $\left[\begin{array}{cc}1& \frac{1}{3}(1-e^{-3t})\\ 0& e^{-3t}\end{array}\right]$
• B. $\left[\begin{array}{cc}1& \frac{1}{3}(e^{-t}-e^{-3t})\\ 0& e^{-t}\end{array}\right]$
• C. $\left[\begin{array}{cc}1& \frac{1}{3}(e^{-t}-e^{-3t})\\ 0& e^{-3t}\end{array}\right]$
• D. $\left[\begin{array}{cc}1& (1-e^{-t})\\ 0& e^{-t}\end{array}\right]$

Answer 1

The correct option is A $\left[\begin{array}{cc}1& \frac{1}{3}(1-e^{-3t})\\ 0& e^{-3t}\end{array}\right]$

$\dot{x}\left(t\right)=\left[\begin{array}{cc}0& 1\\ 0& -3\end{array}\right]x\left(t\right)+\left[\begin{array}{c}1\\ 0\end{array}\right]V\left(t\right)$ ... (i)

$\dot{x}\left(t\right)=Ax\left(t\right)+Bu\left(t\right)$ ...... (ii)

Comparing equation (i) and (ii), we get

$A=Systemmatrix=\left[\begin{array}{cc}0& 1\\ 0& -3\end{array}\right]andB=\left[\begin{array}{c}1\\ 0\end{array}\right]$

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

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

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

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

State transition matrix

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

$\varphi \left(t\right)=\left[\begin{array}{cc}1& \frac{1}{3}(1-e^{-3t})\\ 0& e^{-3t}\end{array}\right]$