What is the state-transition matrix Φ{t) of the following system?

\(\begin{bmatrix}\dot {x_1}\\\ \dot {x_2}\end{bmatrix}=\begin{bmatrix}0&1\\\ -2&-3\end{bmatrix}\begin{bmatrix} {x_1}\\\ {x_2}\end{bmatrix}\)

Concept:

State transition matrix:

It is defined as inverse Laplace transform of |sI - A|-1

⇒ L-1 |sI - A|-1 = eAt = ϕ(t)

Given as state model, ẋ = A x(t)

Comparing with standard equation ẋ = A x(t) + B u(t)

Calculation:

\(A = \left[ {\begin{array}{*{20}{c}} 0&1\\ -2&-3 \end{array}} \right]\)

\(\left| {sI - A} \right| = \left[ {\begin{array}{*{20}{c}} s&0\\ 0&s \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 0&1\\ -2&-3 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {s }&-1\\ { 2}&{s +3} \end{array}} \right]\)

\({\left| {sI - A} \right|^{ - 1}} = \frac{1}{{{s^2+3s+2}}}\;\left[ {\begin{array}{*{20}{c}} {s +3}&1\\ -2&{s} \end{array}} \right]\)

Taking inverse Laplace transform,

\(\Phi(t)=\begin{bmatrix}2e^{-t}-e^{-2t}&e^{-t}-e^{-2t}\\\ -2e^{-t}+2e^{-2t} &-e^{-t}+2e^{-2t}\end{bmatrix}\)

Additional Information

Properties:

State transition matrix, ϕ(t) = eAt

• ϕ(0) = eA0) = I, Identity matrix
• \({\phi ^{ - 1}}\left( t \right) = \phi \left( { - t} \right)\)
• ϕ(t1 + t2) = ϕ(t1) ϕ(t2)
• [ϕ(t)]n = ϕ(nt)
• ϕ(t2 – t1) ϕ (t2 – t0) = ϕ (t2 – t0) = ϕ (t1 – t0) ϕ (t2 – t1)      

Trick: The above question can also be solved with the help of trick.

\(\phi \left( 0 \right) = {e^{{A^{\left( 0 \right)}}}} = I\), the Identity matrix.

We can put t = 0, in all the options and can be found the answer.

After putting t = 0, if the matrix becomes an identity matrix, that will be the answer.