The state equation of a second-order linear system is given by, \[\mathop x\limits^ \cdot \left( t \right) = Ax\left( t \right),\,x\left( 0 \right) = {x_0}\]
For \[{x_0} = \left[ \begin{array}{l}
1\\
- 1
\end{array} \right],\,x\left( t \right) = \left[ \begin{array}{l}
{e^{ - t}}\\
- {e^{ - t}}
\end{array} \right]\] and for \[{x_0} = \left[ \begin{array}{l}
0\\
1
\end{array} \right],\,x\left( t \right) = \left[ \begin{array}{l}
{e^{ - t}} - {e^{ - 2t}}\\
- {e^{ - t}} + {e^{ - 2t}}
\end{array} \right].\]
When \[{x_0} = \left[ \begin{array}{l}
3\\
5
\end{array} \right],\,x\left( t \right)\] is
A. \[\left[ \begin{array}{l} - 8{e^{ - t}} + 11{e^{ - 2t}}\\ 8{e^{ - t}} - 22{e^{ - 2t}} \end{array} \right]\]
B. \[\left[ \begin{array}{l} 11{e^{ - t}} - 8{e^{ - 2t}}\\ - 11{e^{ - t}} + 16{e^{ - 2t}} \end{array} \right]\]
C. \[\left[ \begin{array}{l} 3{e^{ - t}} - 5{e^{ - 2t}}\\ - 3{e^{ - t}} + 10{e^{ - 2t}} \end{array} \right]\]
D. \[\left[ \begin{array}{l} - 5{e^{ - t}} + 6{e^{ - 2t}}\\ - 5{e^{ - t}} + 6{e^{ - 2t}} \end{array} \right]\]
Answer: Option B
This Question Belongs to Electronics And Communications Engineering >> Control Systems
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