Solutions of Exercise 6.1

Solutions of Exercise 6.1#

A state-space representation of \(G(s) = \frac{1}{s+1}\) is given by

(26)#\[\begin{split} \begin{dcases} \dot{x} = \begin{bmatrix} -1 & 0 \\ 0 & 2 \end{bmatrix} x + \begin{bmatrix} 1 \\ 1 \end{bmatrix} u \\ y = \begin{bmatrix} 1 & 0 \end{bmatrix} x \end{dcases} \end{split}\]

  1. Compute the poles of (26) and compare them with those of \(G(s)\).

  2. Is \(G(s)\) input-output stable? Is (26) asymptotically stable?

  3. Examine the controlalbility and observability of the system.

  4. Explain why (26) is not a good realization of \(G(s)\), and provide a better state-space realization.

Solution#

Question 1#

The poles of \(G(s)\) are the roots of the denominator, i.e., \(s = -1\). The poles of (26) are the eigenvalues of \(A\), and they can be computed by finding the roots of the characteristic polynomial

\[\begin{split} \varphi(\lambda) = \det \begin{bmatrix} \lambda + 1 & 0 \\ 0 & \lambda - 2 \end{bmatrix} = (\lambda + 1) (\lambda - 2) \end{split}\]

The roots of \(\varphi(\lambda) = 0\) are \(\lambda = -1\) and \(\lambda = 2\).

Question 2#

The system is input-output stable if all the poles of \(G(s)\) have negative real part. In this case, we just have one pole in \(s=-1\), which means that the system is actually input-output stable.

Asymptotic stability depends, instead, depends on the eigenvalues of \(A\) (i.e., on the poles of the state-space system). We have one positive eigenvalue (\(\lambda = 2\)), which means that the (26) is unstable.

How is this possible?#

To understand what’s happening, let’s assume \(x(0) = 0\) and \(u(t) = \textrm{step}(t)\). Computing the state trajectory via the Lagrange formula yields

\[\begin{split} x(t) = \cancel{x(0) e^{At}} + \int_{0}^{t} e^{A\tau} B u(t - \tau) d \tau = \int_{0}^{t} \begin{bmatrix} e^{-\tau} & 0 \\ 0 & e^{2 \tau} \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix} 1 d \tau \end{split}\]

Since \(e^{At}\) is diagonal, solving this integral is faily easy, and can be done independently for the two states:

  • \(x_1(t) = \int_{0}^{t} e^{-\tau} d \tau = [- e^{-\tau} ]_0^t = 1 - e^{-t}\)

  • \(x_2(t) = \int_{0}^{t} e^{2\tau} d \tau = [0.5 e^{2\tau} ]_0^t = 0.5 e^{2t} - 0.5\)

Note that \(\lim_{t \to \infty} x_{1}(t) = 1\), meaning that \(x_1(t)\) is converging to a constant, while \(\lim_{t \to \infty} x_2(t) = + \infty\), meaning that state \(x_2(t)\) is diverging. That’s why (26) is unstable.

If we compute, however, \(y(t)\), we get

\[ y(t) = 1 x_1(t) + 0 x_2(t) = 1 - e^{-t} \]

Because \(\lim_{t \to \infty} y(t) = 1\) is finite, we can say that the system is input-output stable.

Question 3#

Let’s check the controllability matrix:

\[\begin{split} \mathcal{S} = \begin{bmatrix} B & AB \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ 1 & 2 \end{bmatrix} \end{split}\]

The system is controllable, since \(S\) if full rank (\(\det(S) = 3 \neq 0\)).

Now we move on to the observability matrix:

\[\begin{split} \mathcal{O} = \begin{bmatrix} C \\ CA \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ -1 & 0 \end{bmatrix} \end{split}\]

The observabiltiy matrix is singular, as \(\det(\mathcal{O}) = 0\). Note that the second column is indeed null! This means that the system is not observable.

Question 4#

(26) is surely not a good realization of \(G(s)\) because it has an extra pole (\(\lambda = 2\)) that is unstable and not observable[1].

A much better realization would be any of the controllable canonical realizations of \(G(s)\), which would give the following first-order state-space model:

\[\begin{split} \begin{dcases} \dot{x} = - x + u \\ y = x \end{dcases} \end{split}\]
Contents