Correct Answer - Option 3 :
\(\left[ {\begin{array}{*{20}{c}} {{e^t}}&0\\ {t{e^t}}&e^t \end{array}} \right]\)
Concept:
State transition matrix:
It is defined as inverse Laplace transform of |sI - A|-1
⇒ L-1 |sI - A|-1 = eAt = ϕ(t)
General state equation:
\(\dot x\left( t \right) = A\;x\left( t \right) + BU\left( t \right)\)
Also, \(\dot x\left( t \right) = \frac{{dx}}{{dt}}\)
\(\left[ {\begin{array}{*{20}{c}} {{{\dot x}_1}\left( t \right)}\\ {{{\dot x}_2}\left( t \right)}\\ \vdots \\ \vdots \\ {{{\dot x}_n}\left( t \right)} \end{array}} \right] = \left[ A \right]\left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{x_2}}\\ \vdots \\ \vdots \\ {{x_n}} \end{array}} \right] + \left[ B \right]\left[ {\begin{array}{*{20}{c}} {{U_1}}\\ {{U_2}}\\ \vdots \\ {{U_n}} \end{array}} \right]\)
Where, x1, x2, x3 …. xn are state variables
A is state matrix
B is the input matrix
Calculation:
Given state equation
\(\left[ {\begin{array}{*{20}{c}} {{{\dot x}_1}}\\ {{{\dot x}_2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&0\\ 1&1 \end{array}} \right]\;\left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{x_2}} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 1\\ 1 \end{array}} \right]u\)
\(A = \left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right],\;B = \left[ {\begin{array}{*{20}{c}} 1\\ 1 \end{array}} \right]\)
State transition matrix.
\(\phi \left( t \right) = {L^{ - 1}}\;{\left( {sI - A} \right)^{ - 1}}\)
\(\left[ {sI - A} \right] = \left[ {\begin{array}{*{20}{c}} s&0\\ 0&s \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 1&0\\ 1&1 \end{array}} \right]\)
\(= \left[ {\begin{array}{*{20}{c}} {s - 1}&0\\ { - 1}&{s - 1} \end{array}} \right]\)
\({\left[ {sI - A} \right]^{ - 1}} = \frac{1}{{{{\left( {s - 1} \right)}^2}}}\left[ {\begin{array}{*{20}{c}} {s - 1}&0\\ 1&{s - 1} \end{array}} \right]\)
\(= \left[ {\begin{array}{*{20}{c}} {\frac{1}{{s - 1}}}&0\\ {\frac{1}{{{{\left( {s - 1} \right)}^2}}}}&{\frac{1}{{s - 1}}} \end{array}} \right]\)
\({L^{ - 1}}\;{\left[ {sI - A} \right]^{ - 1}} = \left[ {\begin{array}{*{20}{c}} {{e^t}}&0\\ {t{e^t}}&{{e^t}} \end{array}} \right]\)