Correct Answer - Option 1 : The system is uncontrollable and completely observable
Concept:
State space representation:
X(t) = A(t)x(t) + B(t)u(t)
y(t) = C(t)x(t) + D(t)u(t)
y(t) is output
u(t) is input
x(t) is a state vector
A is a system matrix
This representation is continuous time-variant.
Controllability:
A system is said to be controllable if it is possible to transfer the system state from any initial state x(t0) to any desired state x(t) in a specified finite time interval by a control vector u(t)
Kalman's test for controllability:
X = Ax + Bu
Qc = [B AB A2B....An-1 B]
Qc = controllability matrix
If IQcl = 0, system is not controllable
If |Qcl ≠ 0, the system is controllable
Observability:
A system is said to be observable if every state x(t0) can be completely identified by measurement of output y(t) over a finite time interval.
Kalman's test for observability:
Qo = [CT ATCT (AT)2CT .... (AT)n-1CT]
Qo = observability testing matrix
If |Qo| = 0, system is not observable.
If |Qo| ≠ 0, the system is observable.
Calculation:
Given that
\(A = \left[ {\begin{array}{*{20}{c}} { - 1}&0\\ 0&{ 2} \end{array}} \right], B = \left[ {\begin{array}{*{20}{c}} 0\\ 1 \end{array}} \right], C = \left[ {\begin{array}{*{20}{c}} 1&2 \end{array}} \right]\)
Controllability matrix, C = [B AB]
\(C = \left[ {\begin{array}{*{20}{c}} 0&0\\ 1&{ 2} \end{array}} \right]\)
\(\left| C \right| = \left| {\begin{array}{*{20}{c}} 0&0\\ 1&{ 2} \end{array}} \right| = 0\)
Hence, this is not controllable
Observability matrix, \(O = \left[ {\begin{array}{*{20}{c}} C\\ {CA} \end{array}} \right]\)
\(O = \left[ {\begin{array}{*{20}{c}} 1&2\\ { - 1}&{ 4} \end{array}} \right]\)
\(\left| O \right| = \left| {\begin{array}{*{20}{c}} 1&2\\ { - 1}&{ 4} \end{array}} \right| = 4 - \left( { - 2} \right) = 6 \ne 0\)
Hence, this system is observable.
