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The state space representation of a system is given by

\(\dot x = \left[ {\begin{array}{*{20}{c}} 0&1\\ 0&{ - 3} \end{array}} \right]x + \left[ {\begin{array}{*{20}{c}} 1\\ 0 \end{array}} \right]u,y = \left[ {\begin{array}{*{20}{c}} 1&0 \end{array}} \right]x\)

The transfer function \(\frac{{Y\left( s \right)}}{{U\left( s \right)}}\) of the system will be


1. \(\frac{1}{s}\)
2. \(\frac{1}{{s\left( {s + 3} \right)}}\)
3. \(\frac{1}{{s + 3}}\)
4. \(\frac{1}{{{s^2}}}\)

1 Answer

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Best answer
Correct Answer - Option 1 : \(\frac{1}{s}\)

\(\begin{array}{l} \dot x = \left[ {\begin{array}{*{20}{c}} 0&1\\ 0&{ - 3} \end{array}} \right]x + \left[ {\begin{array}{*{20}{c}} 1\\ 0 \end{array}} \right]u\\ y = \left[ {\begin{array}{*{20}{c}} 1&0 \end{array}} \right]x\\ A = \left[ {\begin{array}{*{20}{c}} 0&1\\ 0&{ - 3} \end{array}} \right]B = \left[ {\begin{array}{*{20}{c}} 1\\ 0 \end{array}} \right]C = \left[ {\begin{array}{*{20}{c}} 1&0 \end{array}} \right] \end{array}\)

\(\begin{array}{l} \frac{{y\left( s \right)}}{{u\left( s \right)}} = c{\left[ {sI - A} \right]^{ - 1}}B + D\\ \left[ {sI - A} \right] = \left[ {\begin{array}{*{20}{c}} s&{ - 1}\\ 0&{s + 3} \end{array}} \right]\\ {\left[ {sI - A} \right]^{ - 1}} = \frac{1}{{s\left( {s + 3} \right)}}\left[ {\begin{array}{*{20}{c}} {s + 3}&1\\ 0&s \end{array}} \right] \end{array}\)

\(\begin{array}{l} \frac{{y\left( s \right)}}{{u\left( s \right)}} = \left[ {\begin{array}{*{20}{c}} 1&0 \end{array}} \right]\frac{1}{{s\left( {s + 3} \right)}}\left[ {\begin{array}{*{20}{c}} {s + 3}&1\\ 0&s \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1\\ 0 \end{array}} \right]\\ = \frac{1}{{s\left( {s + 3} \right)}}\left[ {\begin{array}{*{20}{c}} {s + 3}&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1\\ 0 \end{array}} \right]\\ = \frac{1}{{s\left( {s + 3} \right)}}\left[ {s + 3} \right]\\ = \frac{1}{s} \end{array}\)

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