Control Systems Test 8

Number of state variables = order of the system

= number of independent energy storage elements

= number of poles of the system.

Given system is governed by

\(\frac{{{d^2}y}}{{d{t^2}}} + s\frac{{dy}}{{dt}} + 6y = u\left( t \right)\)

State transition matrix, ϕ(t) = e^At

From the properties of state transition matrix,

\(\phi \left( 0 \right) = {e^{A\left( 0 \right)}} = I\)

1. \(\begin{array}{*{20}{c}} {\phi \left( t \right) = \left[ {\begin{array}{*{20}{c}} {{e^{ - t}}}&0\\ 1&{{e^{ - 2t}}} \end{array}} \right]}\\ {\phi \left( o \right) = \left[ {\begin{array}{*{20}{c}} 1&0\\ 1&0 \end{array}} \right] \ne I} \end{array}\)

2. \(\begin{array}{*{20}{c}} {\phi \left( t \right) = \left[ {\begin{array}{*{20}{c}} {{e^t}}&t\\ 1&{2{e^{ - t}}} \end{array}} \right]}\\ {\phi \left( o \right) = \left[ {\begin{array}{*{20}{c}} 1&0\\ 1&2 \end{array}} \right] \ne I} \end{array}\)

3. \(\begin{array}{*{20}{c}} {\phi \left( t \right) = \left[ {\begin{array}{*{20}{c}} {{e^t} + {e^{ - t}}}&0\\ 0&{2{e^{ - t}}} \end{array}} \right]}\\ {\phi \left( o \right) = \left[ {\begin{array}{*{20}{c}} 2&0\\ 0&2 \end{array}} \right] \ne I} \end{array}\)

4. \(\begin{array}{*{20}{c}} {\phi \left( t \right) = \left[ {\begin{array}{*{20}{c}} {{e^t}}&0\\ 0&{{e^{ - 2t}}} \end{array}} \right]}\\ {\phi \left( o \right) = \left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right] = I} \end{array}\)

Hence option (d) can be state transition matrix.

The state transition matrix \(\phi \left( t \right) = {L^{ - 1}}\left[ {{{\left( {SI - A} \right)}^{ - 1}}} \right]\)

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

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

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

|SI - A| = s(s + 3) + 2

= s² + 3s + 2

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

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

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

\(= \left[ {\begin{array}{*{20}{c}}{2{e^{ - t}} - {e^{ - 2t}}}&{{e^{ - t}} - {e^{ - 2t}}}\\{2{e^{ - 2t}} - 2{e^{ - t}}}&{2{e^{ - 2t}} - {e^{ - t}}}\end{array}} \right]\)

From the given state space representation,

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

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

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

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

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

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

\(= \left[ {\begin{array}{*{20}{c}} {{e^{ - t}} + t{e^{ - t}}}&{t{e^{ - t}}}\\ { - t{e^{ - t}}}&{{e^{ - t}} - t{e^{ - t}}} \end{array}} \right]\)

\(x\left( t \right) = {e^{At}}\;x\left( 0 \right)\)