Question

For the system
$\dot{X}=\left[\begin{array}{cc}2& 0\\ 0& 4\end{array}\right]X+\left[\begin{array}{c}1\\ 1\end{array}\right]u;y=\left[\begin{array}{cc}4& 0\end{array}\right]X,$
with $u$ as unit impulse and with zero initial state, the output $y$ becomes

• A. $2e^{2t}$
• B. $4e^{2t}$
• C. $2e^{4t}$
• D. $4e^{4t}$

Answer 1

The correct option is B $4e^{2t}$

$x\left(t\right)=e^{At}X\left(0\right)+{\int }_0^t{e}^{A(t-\tau )}BU\left(\tau \right)d\tau$
$\because X\left(0\right)=0\left(given\right)$
$\therefore x\left(t\right)=0+{\int }_0^t\left[\begin{array}{cc}{e}^{2(t-\tau )}& 0\\ 0& e^{4(t-\tau )}\end{array}\right].\left[\begin{array}{c}1\\ 1\end{array}\right].\delta \left(t\right)dt$
$={\int }_0^t\left[\begin{array}{c}{e}^{2(t-\tau )}\\ e^{4(t-\tau )}\end{array}\right].\delta \left(t\right)dt=\left[\begin{array}{c}{e}^{2t}\\ e^{4t}\end{array}\right]$
$Outputy\left(t\right)=\left[40\right].x\left(t\right)$
$=\left[40\right].\left[\begin{array}{c}{e}^{2t}\\ e^{4t}\end{array}\right]$
Hence,
$y\left(t\right)=4e^{2t}$