Question 12.9: Determine the observability of the system, described by A=......

(sI − A) = s\left [ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right ] -\left [ \begin{matrix} 0 & -1 \\ 2 & 3 \end{matrix} \right ] =\left [ \begin{matrix} s & 1 \\ -2 & s-3 \end{matrix} \right ]

The characteristic polynomial of the system, given by the determinant of this matrix, is

s^{2}-3s+2=(s-1)(s-2)

The roots of this equation are the two eigenvalues \{\lambda_{1}=1,\lambda_{2}=2\}. Therefore, the transformation matrix from the last example is

P={\left[\begin{array}{cc}\ \ {2\ 1}\\ {-1\ 1}\end{array}\right]}

The first and second columns are, respectively, the eigenvectors corresponding to eigenvalues 2 and 5.

P ^{−1}AP =\left [ \begin{matrix} \frac{1}{3} & -\frac{1}{3} \\ \frac{1}{3} & \frac{2}{3} \end{matrix} \right ] \left [ \begin{matrix} 3 & 2 \\ 1 & 4 \end{matrix} \right ] \left [ \begin{matrix} 2 & 1 \\ -1 & 1 \end{matrix} \right ] = Λ=\left [ \begin{matrix} 2 & 0 \\ 0 & 5 \end{matrix} \right ]

The rank of the matrix

\lbrack C^{T}\mid A^{T}C^{T}]=\left [ \begin{matrix} 1 & -2 \\ -1 & -4 \end{matrix} \right ]

is 2. Therefore, the system is observable.

C P=[2\ 3]

has no zero entries, and, therefore, the system is observable.