Question 10.22: Controllability and Observability Consider a dynamic system ...

a. The system matrices \mathbf{A, B, C}, and D are

\mathbf{A} = \begin{bmatrix} -4 & 0 \\ 0 & -1 \end{bmatrix},      \mathbf{B} = \begin{bmatrix} 2 \\ 0 \end{bmatrix},    \mathbf{C} = \begin{bmatrix} 1 & 3 \end{bmatrix},     D=0

The transfer function is

G(s) = \begin{bmatrix} 1  & 3 \end{bmatrix} \begin{bmatrix} s+4 & 0 \\ 0 & s+1 \end{bmatrix}^{-1}\begin{bmatrix} 2  \\ 0 \end{bmatrix} + 0 =  \frac{\begin{bmatrix} 1  & 3 \end{bmatrix}\begin{bmatrix} s+1 & 0 \\ 0 & s+4 \end{bmatrix}\begin{bmatrix} 2  \\ 0 \end{bmatrix}}{(s+1)(s+4)} = \frac{2}{s+4}

b. Similarly, we have

\mathbf{A} = \begin{bmatrix} -4 & 0 \\ 0 & -1 \end{bmatrix},      \mathbf{B} = \begin{bmatrix} 2 \\ 1 \end{bmatrix},    \mathbf{C} = \begin{bmatrix} 1 & 0 \end{bmatrix},     D=0

and

G(s) = \begin{bmatrix} 1  & 0 \end{bmatrix} \begin{bmatrix} s+4 & 0 \\ 0 & s+1 \end{bmatrix}^{-1}\begin{bmatrix} 2  \\ 1 \end{bmatrix} + 0 =  \frac{\begin{bmatrix} 1  & 0 \end{bmatrix}\begin{bmatrix} s+1 & 0 \\ 0 & s+4 \end{bmatrix}\begin{bmatrix} 2  \\ 1 \end{bmatrix}}{(s+1)(s+4)} = \frac{2}{s+4}

Note that the dynamic models in Parts (a) and (b) are second-order systems, with state-space forms that differ from the original first-order system. However, they both end up with the same transfer function as the given first-order system due to pole–zero   cancellation. As seen in Part (a), the second state cannot be affected by the input matrix \mathbf{B},

G(s) = \frac{\begin{bmatrix} s+1  & 3(s+4) \end{bmatrix}}{(s+1)(s+4)}\begin{bmatrix} 2  \\ 0 \end{bmatrix}

This implies that the second state is uncontrollable by the actuator defined by matrix \mathbf{B}. Similarly, in Part (b), the second state cannot be seen by the output matrix \mathbf{C},

G(s) = \begin{bmatrix} 1  & 0 \end{bmatrix} \frac{\begin{bmatrix} 2(s+1) \\  s+4 \end{bmatrix}}{(s+1)(s+4)}

This implies that the second state is unobservable by the sensor defined by matrix \mathbf{C}.