1.11 Lecture 9. Tuesday September 30 2014

1.11.1 Handout, controllability criterion

1.11.2 Lecture: Separation theorem, observer design

Oct. 2, no class. Things covered today not on exam. Exam covers up to HW3. Solution to HW3 will be send oct6. Test on Thursday Oct. 7.2014. Closed books, closed notes, open minds. Remember Mason rules and realization.

We studied controllers (state space feedback) and studied observers. What happens when we combine them? Recall\[ \tilde{x}^{\prime }=A\tilde{x}+Bu+L\left ( Cx-C\tilde{x}\right ) \] Where \(\tilde{x}\) is the full estimated state from the observer. The idea is to look a the error and use \(L\) to reduce the error.\[ \tilde{x}^{\prime }=\left ( A-LC\right ) \tilde{x}+LCx+Bu \] How good is this observer? Study error \(e=x-\tilde{x}\). In perfect world, \(e\rightarrow 0\) quickly with no overshoot. Find \[ e^{\prime }=x^{\prime }-\tilde{x}^{\prime }\] Do some algebra\[ e^{\prime }=\left ( A-LC\right ) e \] note on initial states: \(\tilde{x}\left ( 0\right ) \neq x\left ( 0\right ) \). We want to pick \(L\) which is \(n\times r\) dimensions so that \(A-LC\) has desired eigenvalues. We want \(L\) to be stabilizing. At bare minimum we want \(A-LC\) stable. What can we do to generate the eigenvalues of \(\left ( A-LC\right ) ?\).  If \(\left ( A,C\right ) \) is observable, then we can make \(eig\left ( A-LC\right ) \) any value we want. Consider SISO system where \(\left ( A,C\right ) \) is observable Then rank of the observability matrix \(\Theta \) is \(n.\) Then since \(\left ( A,C\right ) \) is observable, by duality, \(\left ( A^{T},C^{T}\right ) \) is controllable. We will use now the controllability results fro pole assignment that tells us we can select gain \(K\) s.t. \(\left ( A^{T}+C^{T}K\right ) \) with desired eigenvalues.

I wrote this below for a HW assignment, I copy it here. For example of pole assignment for the observer \(\left ( A-LC\right ) .\)

We need to determine \(L\) such that the eigenvalues of \(\left ( A-LC\right ) \) are \(\lambda _{1}=-1\) and \(\lambda _{2}=-2\). Before showing the design steps using the actual data given in the problem, the design steps are given below for the general case.

1.11.3 Design steps for finding \(L\)

1.

Input is \(A,C\) and set of desired eigenvalues \(\lambda _{i}\)

2.

Verify that \(\left ( A,C\right ) \) is observable. If so then let \(A_{o}=A^{T},B_{o}=C^{T}\), hence \(\left ( A_{o},B_{o}\right ) \) is controllable.

3.

Find controllability matrix \(\mathbb{C} \left ( A_{o},B_{o}\right ) \)

4.

Write down the controllability companion form for \(A_{o},B_{o}\). Let them be called \(\tilde{A}_{o},\tilde{B}_{o}\). To do this, we only need to find the characteristic polynomial for \(A_{o}\) and read the coefficients in reverse and change the signs. \(\tilde{B}_{o}\) will always have zeros other than the last row.

5.

Find controllability matrix \(\tilde{\mathbb{C}}\left ( \tilde{A}_{o},\tilde{B}_{o}\right ) \)

6.

Find \(T=\tilde{\mathbb{C}}\mathbb{C} ^{-1}\)

7.

Find the closed loop matrix \(\left [ \tilde{A}_{o}+\tilde{B}_{o}\tilde{K}\right ] \) where \(\tilde{K}=\left [ k_{0},k_{1},\cdots ,k_{n-1}\right ] \) is the gain matrix we looking to determine.

8.

Find the characteristic polynomial of \(\left [ \tilde{A}_{o}+\tilde{B}_{o}\tilde{K}\right ] \), it will be a function of \(k_{i}\)

9.

Set up the desired polynomial \(p\left ( \lambda \right ) =\left ( \lambda -\lambda _{o}\right ) \left ( \lambda -\lambda _{o}\right ) \cdots \left ( \lambda -\lambda _{n-1}\right ) \) where \(\lambda _{i}\) are the desired eigenvalues given.

10.

Compare coefficients of polynomial from step (9) with the polynomial of step (7) and solve for \(k_{i}\)

11.

Now we have found \(\tilde{K}=\left [ k_{0},k_{1},\cdots ,k_{n-1}\right ] \). Convert it to \(K\) using \(T\) as follows: \(K=\tilde{K}T\)

12.

Find \(L=-K^{T}\). This completes the design.

13.

The observer \(A\) matrix now becomes \(\left [ A-LC\right ] \)

Now we will start talking about combining controller/observer systems. The key result is separation theorem. This system

can have states we want to control, that can not be observed/measured. We need an observer

We now have \(2n\) equations, \(n\) for the observer and \(n\) for the original \(\sum \) system. They are cross coupled. Assuming \((A,B)\) and \((A,C)\) are controllable and observable, and without loss of generality, let \(v=0\) then we have\begin{align*} x' &= A x + B \overset{u}{\overbrace{k\tilde{x}}}\\ \tilde{x}^{\prime } &= A\tilde{x} + Bk\tilde{x}+LC\left ( x-\tilde{x}\right ) \end{align*}

So now we have \[\begin{pmatrix} x^{\prime }\\ \tilde{x}^{\prime }\end{pmatrix} =\overset{2n\times 2n\text{ called Augmented }A_{+}}{\overbrace{\begin{pmatrix} A & Bk\\ LC & A+Bk-LC \end{pmatrix} }}\begin{pmatrix} x\\ \tilde{x}\end{pmatrix} \] We have \(A+Bk\) stable and we know that \(A-LC\) is stable by design. But we do not know if \(A_{+}\) is stable. (the augmented \(A\) above). We know eigenvalues of \(A_{+}\) is the eigenvalues of \(\left ( TA_{+}T^{-1}\right ) \) if \(T\) is not singular. So let us make special \[ T=\begin{pmatrix} I & 0\\ I & -I \end{pmatrix} \] Hence \(T^{-1}=\begin{pmatrix} I & 0\\ -I & I \end{pmatrix} \), now calculate \begin{align*} \left ( TA_{+}T^{-1}\right ) & =\begin{pmatrix} I & 0\\ I & -I \end{pmatrix}\begin{pmatrix} A & Bk\\ LC & A+Bk-LC \end{pmatrix}\begin{pmatrix} I & 0\\ -I & I \end{pmatrix} \\ & =\begin{pmatrix} A & Bk\\ A-LC & LC-A \end{pmatrix}\begin{pmatrix} I & 0\\ -I & I \end{pmatrix} \\ & =\begin{pmatrix} A+Bk & -Bk\\ 0 & A-LC \end{pmatrix} \end{align*}

Since diagonal matrix, then the eigenvalues on the diagonal.  So the eigenvalues are the union of the eigenvalues of \(A+Bk\) and eigenvalues of \(A-LC\). So \(A_{+}\) is stable if \(A+Bk\) and \(A-LC\) are stable.

HW3 assigned.