Solutions of Exercise 6.0

Solutions of Exercise 6.0#

Consider the following system:

(23)#\[\begin{split} \begin{dcases} \dot{x} = \begin{bmatrix} -2 & 4 \\ 0 & -3 \end{bmatrix} x + \begin{bmatrix} 0 \\ 2 \end{bmatrix} u \\ y = \begin{bmatrix} 0 & 3 \end{bmatrix} x \end{dcases} \end{split}\]

  1. Compute the poles of the state-space system (23), and check whether the system is asymptotially stable or not.

  2. Is the system controllable and observable?

  3. Compute the transfer function \(G(s)\) of system (23).

  4. Compute the poles of \(G(s)\), and discuss whether they match those of (23).

Inverting \(2 \times 2\) matrices

Consider a generic matrix \(A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\). Its inverse can be easily computed as

(24)#\[\begin{split} A^{-1} = \frac{1}{\det(A)} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \end{split}\]

Solution#

Question 1#

The poles of (23) are the eigenvalues of \(A\). To compute them, we evaluate the roots of the characteristic polynomial \(\varphi(\lambda) = \det (\lambda I - A)\).

\[\begin{split} \varphi(\lambda) = \det \begin{bmatrix} \lambda + 2 & -4 \\ 0 & \lambda + 3\end{bmatrix} = (\lambda + 2)(\lambda + 3) - (-4) \cdot 0 = (\lambda + 2)(\lambda + 3) \end{split}\]

By setting \(\varphi(\lambda) = 0\) we get the poles, \(\lambda=-2\) and \(\lambda=-3\). System (23) is asymptotically stable because all the poles have negative real part.

Question 2#

To determine whether the system is controllable, we need to check that the controllability matrix is full rank.

\[\begin{split} \mathcal{S} = \begin{bmatrix} B & AB \end{bmatrix} = \begin{bmatrix} 0 & 8 \\ 2 & - 6 \end{bmatrix} \end{split}\]

\(\mathcal{S}\) is full rank. In fact, it’s determinant is \(\det(S) = -16 \neq 0\).

When it comes to observability, we compute the observability matrix

\[\begin{split} \mathcal{O} = \begin{bmatrix} C \\ CA \end{bmatrix} = \begin{bmatrix} 0 & 3 \\ 0 & -9 \end{bmatrix} \end{split}\]

Note that the observability matrix is singular, because the first column is null. Also, \(\det(\mathcal{O}) = 0\). Therefore the system is not (fully) observable.

Question 3#

To compute \(G(s)\), we can apply the formula seen during Lecture 8:

\[ G(s) = C (s I - A)^{-1} B + D \]

Let’s compute it step by step with the help of (24).

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

\( C(sI-A)^{-1} = \frac{1}{(s+2)(s+3)} \begin{bmatrix} 0 & 3 \end{bmatrix} \left[\begin{array}{c|c} s+3 & 4 \\ 0 & s+2 \end{array}\right] = \frac{1}{(s+2)(s+3)} \begin{bmatrix} 0 & 3(s+2) \end{bmatrix} \)

\( G(s) = C(sI-A)^{-1} B = \frac{1}{(s+2)(s+3)} \begin{bmatrix} 0 & 3(s+2) \end{bmatrix} \begin{bmatrix} 0 \\ 2\end{bmatrix} = \frac{6 \cancel{(s+2)}}{\cancel{(s+2)}(s+3)} = \frac{6}{s+3} \)

Question 4#

The only pole of \(G(s)\) is \(s=-3\). The pole in \(s=-2\) has been cancelled while computing \(G(s)\), as it was matching a zero of the system (see the last step above).

In this case, we can say that the eigenvalue \(\lambda=-2\) is not observable (unobservable) because

  • It is an eigenvalue of \(A\) that doesn’t show up in \(G(s)\)

  • The system had one unobservable state (\(\textrm{rank}(O) = 2-1 = 1\)).

Warning

Pole-zero cancellations can happen when computing the transfer function representation of state-space systems. Always remember that

  • \(\textrm{poles}(G(s)) \subseteq \textrm{eig}(A)\)

  • Cancellations are especially problematic when the poles being cancelled are unstable!

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