Solutions of Exercise 7.0

Solutions of Exercise 7.0#

Given the system

(35)#\[ Y(s) = \frac{s+1}{s^2+3s+5} U(s) \]

  1. Write the state-space observable canonical form

  2. Design a state observer yielding a state estimate \(\hat{x}\), such that the state estimation error \(\varepsilon\) converges to zero with eigenvalues \(-10\), \(-10\).

Solution#

Question 1#

First, let’s recall that the observable canonical form of a second-order system characterized by the transfer function

\[ G(s) = \frac{b_0 s^2 + b_1 s + b_2}{1 s^2 + a_1 s + a_2} \]

is[1]

\[\begin{split} \begin{dcases} \dot{x} = \begin{bmatrix} -a_1 & 1 \\ -a_2 & 0 \end{bmatrix} x + \begin{bmatrix} b_1 - a_1 b_0 \\ b_2 - a_2 b_0 \end{bmatrix} u \\ y = \begin{bmatrix} 1 & 0 \end{bmatrix} x + b_0 u \end{dcases} \end{split}\]

In our case, \(a_1 = 3\), \(a_2 = 5\), \(b_0 = 0\), \(b_1 = 1\), and \(b_2 = 1\). This means that the \(A, B, C, D\) matrices are

(36)#\[\begin{split} A = \begin{bmatrix} -3 & 1 \\ -5 & 0 \end{bmatrix} \quad B = \begin{bmatrix} 1 \\ 1 \end{bmatrix} \quad C = \begin{bmatrix} 1 & 0 \end{bmatrix} \quad D = 0 \end{split}\]

Note

System (36) is surely observable – the observable canonical form is, by definition, observable.

At the exam, do not forget to assess the observability of the system before solving the observer pole-placement problem!

Question 2#

The state observer structure used is

\[ \hat{x} = A x + B u + K (y - C \hat{x}) \]

The state estimation error \(\varepsilon\) evolves according to \(\dot{\varepsilon} = (A - KC) \varepsilon\), therefore the innovation \(K\) must be computed such that

\[ \textrm{eig}(A-KC) = \{ -10, -10 \} \]

This is done with the pole placement algorithm, where \(K\) is computed such that the characteristic polynomial \(\varphi(\lambda)\) has \(\lambda^\star = -10, -10\) has roots.

The characteristic polynomial is

\[\begin{split} \begin{aligned} \varphi(\lambda) &= \det (\lambda I - (A - KC)) = \det\left( \begin{bmatrix} \lambda & \\ & \lambda \end{bmatrix} - \begin{bmatrix} -3 & 1 \\ -5 & 0 \end{bmatrix} + \begin{bmatrix} \kappa_1 \\ \kappa_2 \end{bmatrix} \begin{bmatrix} 1 & 0\end{bmatrix}\right) \\ &= \det \begin{bmatrix} \lambda + 3 + \kappa_1 & - 1 \\ +5 + \kappa_2 & \lambda \end{bmatrix} = \lambda^2 + (\kappa_1 + 3) \lambda + \kappa_2 + 5 \end{aligned} \end{split}\]

Matching this polynomial to the desired characteristic polynomial \(\varphi^{\star} (\lambda) = (\lambda+10)^2 = \lambda^2 + 20\lambda + 100\) we get

\[\begin{split} \begin{dcases} \kappa_1 + 3 = 20\\ \kappa_2 + 5 = 100 \end{dcases} \qquad \rightarrow \qquad K = \begin{bmatrix} 17 \\ 95 \end{bmatrix} \end{split}\]
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