Question

State variable description of an LTI system is given by

\(\left[ {\begin{array}{*{20}{c}} {{{\dot x}_1}}\\ {{{\dot x}_2}}\\ {{{\dot x}_3}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0&{{a_1}}&0\\ 0&0&{{a_2}}\\ {{a_3}}&0&0 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{x_3}}\\ {{x_3}} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 0\\ 0\\ 1 \end{array}} \right]U\)

\(Y = \left[ {\begin{array}{*{20}{c}} 1&0&0 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{x_2}}\\ {{x_3}} \end{array}} \right]\)

Where Y is the output and u is input. System is controllable for

1. a₁ ≠ 0, a₂ = 0, a₃ ≠ 0
2. a₁ = 0, a₂ ≠ 0, a₃ = 0
3. a₁ = 0, a₂ ≠ 0, a₃ ≠ 0
4. a₁ ≠ 0, a₂ ≠ 0, a₃ = 0

Answer 1

Correct Answer - Option 4 : a₁ ≠ 0, a₂ ≠ 0, a₃ = 0

\({Q_c} = \left[ {\begin{array}{*{20}{c}} B&{AB}&{{A^2}B} \end{array}} \right]\)

\({Q_c} = \left[ {\begin{array}{*{20}{c}} 0&0&{{a_1}{a_2}}\\ 0&{{a_2}}&0\\ 1&0&0 \end{array}} \right]\)

System is controllable if |Q_C| ≠ 0

a₁a₂ (0 – a₂) ≠ 0

\(- {a_1} \cdot a_2^2 \ne 0\)

Hence condition for controllability is

a₁ ≠ 0, a₂ ≠ 0, a₃ = 0