The state equation of a second-order linear system is given by:ẋ(t) = Ax(t), x(0) = x0For ${x_0} = \left[ {\begin{array}{*{20}{c}} 1\ { - 1} \end{array}} \right],\;x\left( t \right) = \left[ {\begin{array}{*{20}{c}} {{e^{ - t}}}\ { - {e^{ - t}}} \end{array}} \right]$ and for${x_0}\left[ {\begin{array}{*{20}{c}} 0\ 1 \end{array}} \right],\;x\left( t \right) = \left[ {\begin{array}{*{20}{c}} {{e^{ - t}} - {e^{ - 2t}}}\ { - {e^{ - t}} + 2{e^{ - 2t}}} \end{array}} \right]$When ${x_0} = \left[ {\begin{array}{*{20}{c}} 3\ 5 \end{array}} \right]$, x(t) is

Question

The state equation of a second-order linear system is given by:ẋ(t) = Ax(t), x(0) = x0For ${x_0} = \left[ {\begin{array}{*{20}{c}} 1\ { - 1} \end{array}} \right],\;x\left( t \right) = \left[ {\begin{array}{*{20}{c}} {{e^{ - t}}}\ { - {e^{ - t}}} \end{array}} \right]$ and for${x_0}\left[ {\begin{array}{*{20}{c}} 0\ 1 \end{array}} \right],\;x\left( t \right) = \left[ {\begin{array}{*{20}{c}} {{e^{ - t}} - {e^{ - 2t}}}\ { - {e^{ - t}} + 2{e^{ - 2t}}} \end{array}} \right]$When ${x_0} = \left[ {\begin{array}{*{20}{c}} 3\ 5 \end{array}} \right]$, x(t) is

TuteeHUB · Accepted Answer

$\left[ {\begin{array}{*{20}{c}} {3{e^{ - t}} - 5{e^{ - 2t}}}\ { - 3{e^{ - t}} + 10{e^{ - 2t}}} \end{array}} \right]$

TuteeHUB · Answer

$\left[ {\begin{array}{*{20}{c}} { - 8{e^{ - t}} + 11{e^{ - 2t}}}\ {8{e^{ - t}} - 22{e^{ - 2t}}} \end{array}} \right]$

TuteeHUB · Answer

$\left[ {\begin{array}{*{20}{c}} {11{e^{ - t}} - 8{e^{ - 2t}}}\ { - 11{e^{ - t}} + 16{e^{ - 2t}}} \end{array}} \right]$

TuteeHUB · Answer

$\left[ {\begin{array}{*{20}{c}} {5{e^{ - t}} - 3{e^{ - 2t}}}\ { - 5{e^{ - t}} + 6{e^{ - 2t}}} \end{array}} \right]$

1.	The state equation of a second-order linear system is given by:ẋ(t) = Ax(t), x(0) = x0For \({x_0} = \left[ {\begin{array}{{20}{c}} 1\\ { - 1} \end{array}} \right],\;x\left( t \right) = \left[ {\begin{array}{{20}{c}} {{e^{ - t}}}\\ { - {e^{ - t}}} \end{array}} \right]\) and for\({x_0}\left[ {\begin{array}{{20}{c}} 0\\ 1 \end{array}} \right],\;x\left( t \right) = \left[ {\begin{array}{{20}{c}} {{e^{ - t}} - {e^{ - 2t}}}\\ { - {e^{ - t}} + 2{e^{ - 2t}}} \end{array}} \right]\)When \({x_0} = \left[ {\begin{array}{*{20}{c}} 3\\ 5 \end{array}} \right]\), x(t) is
A.	\(\left[ {\begin{array}{*{20}{c}} { - 8{e^{ - t}} + 11{e^{ - 2t}}}\\ {8{e^{ - t}} - 22{e^{ - 2t}}} \end{array}} \right]\)
B.	\(\left[ {\begin{array}{*{20}{c}} {11{e^{ - t}} - 8{e^{ - 2t}}}\\ { - 11{e^{ - t}} + 16{e^{ - 2t}}} \end{array}} \right]\)
C.	\(\left[ {\begin{array}{*{20}{c}} {3{e^{ - t}} - 5{e^{ - 2t}}}\\ { - 3{e^{ - t}} + 10{e^{ - 2t}}} \end{array}} \right]\)
D.	\(\left[ {\begin{array}{*{20}{c}} {5{e^{ - t}} - 3{e^{ - 2t}}}\\ { - 5{e^{ - t}} + 6{e^{ - 2t}}} \end{array}} \right]\)
Answer» C. \(\left[ {\begin{array}{*{20}{c}} {3{e^{ - t}} - 5{e^{ - 2t}}}\\ { - 3{e^{ - t}} + 10{e^{ - 2t}}} \end{array}} \right]\)

Discussion

No Comment Found

Related MCQs

Reply to Comment