Question

The state variable representation of a system is given as

\(\begin{array}{l} {\rm{\dot x}} = \left[ {\begin{array}{*{20}{c}} 0&1\\ 0&{ - 1} \end{array}} \right]{\rm{x}};{\rm{x}}\left( 0 \right) = \left[ {\begin{array}{*{20}{c}} 1\\ 0 \end{array}} \right]\\ {\rm{y}} = \left[ {\begin{array}{*{20}{c}} 0&1 \end{array}} \right]{\rm{x}} \end{array}\)

The response y(t) is
1. sin t
2. 1 - e^t
3. 1 - cos t
4. 0

Answer 1

Correct Answer - Option 4 : 0

\({\rm{\dot X}} = {\rm{AX}}\), where \({\rm{A}} = \left[ {\begin{array}{*{20}{c}} 0&1\\ 0&{ - 1} \end{array}} \right]\) is state transition matrix.

Thus, we have

\(\begin{array}{l} {\rm{X}}\left( {\rm{s}} \right) = {\left( {{\rm{sI}} - {\rm{A}}} \right)^{ - 1}}{\rm{X}}\left( 0 \right)\\ \Rightarrow {\rm{X}}\left( {\rm{s}} \right) = {\left[ {\begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} {{\rm{s\;\;\;\;\;}}}&{ - 1} \end{array}}\\ {\begin{array}{*{20}{c}} 0&{{\rm{s}} + 1} \end{array}} \end{array}} \right]^{ - 1}}\left[ {\begin{array}{*{20}{c}} 1\\ 0 \end{array}} \right]\\ \Rightarrow {\rm{X}}\left( {\rm{s}} \right) = \frac{1}{{{\rm{s}}\left( {{\rm{s}} + 1} \right)}}\left[ {\begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} {{\rm{s}} + 1}&1 \end{array}}\\ {\begin{array}{*{20}{c}} 0&{{\rm{\;\;\;\;\;\;\;\;s\;}}} \end{array}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1\\ 0 \end{array}} \right]\\ \Rightarrow {\rm{X}}\left( {\rm{s}} \right) = \frac{1}{{{\rm{s}}\left( {{\rm{s}} + 1} \right)}}\left[ {\begin{array}{*{20}{c}} {s + 1}\\ 0 \end{array}} \right]\\ \Rightarrow {\rm{X}}\left( {\rm{s}} \right) = \left[ {\begin{array}{*{20}{c}} {\frac{1}{{\rm{s}}}}\\ 0 \end{array}} \right]\\ \Rightarrow {\rm{x}}\left( {\rm{t}} \right) = \left[ {\begin{array}{*{20}{c}} 1\\ 0 \end{array}} \right]{\rm{\;}} \end{array}\)

Now, 

\({\rm{y}}\left( {\rm{t}} \right) = \left[ {\begin{array}{*{20}{c}} 0&1 \end{array}} \right]{\rm{x}} = \left[ {\begin{array}{*{20}{c}} 0&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1\\ 0 \end{array}} \right] = 0\)