Review: We talked about transformation while preserving transfer functions. Can we find transformation to a specific target? Role of \(\mathbb{C} \) (controllability matrix) and \(\mathbb{Q} \) (observability matrix). Reader: Suppose \(A\) is selected randomly and so is \(B\), example within normal distribution, find probability that rank \(\mathbb{C} \) is \(n.\) The probability is 1. So almost all \(\left ( A,B\right ) \) are controllable. But if some entries of \(A,B\) are hardwired to some specific values due to design, they the chance of getting uncontrollable \(\left ( A,B\right ) \) starts to increase. For example, the controllable canonical form of \(A\) has hardwired entries in \(A\). Even when we get close to be uncontrollable numerically we will get into more problems.
Now we talk about nice properties of companion forms. Let us use \(n=4\) for illustration. Pole assignment: Select \(k\) s.t. \(A+Bk\) has pre-specified eigenvalues. Here \(A\) has bad eigenvalues, but \(A+Bk\) will have good eigenvalues. Note, we are using controllable canonical form, also assume we have access to all states. This is simple\begin{align*} A_{closed} & =A+Bk=\begin{pmatrix} 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ -\alpha _{0} & -\alpha _{1} & -\alpha _{2} & -\alpha _{3}\end{pmatrix} +\begin{pmatrix} 0\\ 0\\ 0\\ 1 \end{pmatrix}\begin{pmatrix} k_{0} & k_{1} & k_{2} & k_{3}\end{pmatrix} \\ & =\begin{pmatrix} 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ -k_{0}-\alpha _{0} & k_{1}-\alpha _{1} & k_{2}-\alpha _{2} & k_{3}-\alpha _{3}\end{pmatrix} \end{align*}
Hence \begin{equation} \det \left ( \lambda I-A_{closed}\right ) =\lambda ^{4}+\left ( \alpha _{3}-k_{3}\right ) ^{3}\lambda ^{3}+\left ( \alpha _{2}-k_{2}\right ) ^{2}\lambda ^{2}+\left ( \alpha _{1}-k_{1}\right ) \lambda +\left ( \alpha _{0}-k_{0}\right ) \tag{1} \end{equation} Now pick target eigenvalues, say \(\lambda _{0},\lambda _{1},\lambda _{2},\lambda _{3}\) (if complex, use complex conjugates). The desired \begin{align} P\left ( \lambda \right ) & =\left ( \lambda -\lambda _{0}\right ) \left ( \lambda -\lambda _{1}\right ) \left ( \lambda -\lambda _{2}\right ) \left ( \lambda -\lambda _{3}\right ) \nonumber \\ & =\lambda ^{4}+\alpha _{3}^{\ast }\lambda ^{3}+\alpha _{2}^{\ast }\lambda ^{2}+\alpha _{1}^{\ast }\lambda +\alpha _{0}^{\ast } \tag{2} \end{align}
Equate (1) and (2)\[ \lambda ^{4}+\alpha _{3}^{\ast }\lambda ^{3}+\alpha _{2}^{\ast }\lambda ^{2}+\alpha _{1}^{\ast }\lambda +\alpha _{0}^{\ast }=\lambda ^{4}+\left ( \alpha _{3}-k_{3}\right ) ^{3}\lambda ^{3}+\left ( \alpha _{2}-k_{2}\right ) ^{2}\lambda ^{2}+\left ( \alpha _{1}-k_{1}\right ) \lambda +\left ( \alpha _{0}-k_{0}\right ) \] Equate like coefficients, we obtain\begin{align*} k_{0} & =\alpha _{0}-\alpha _{0}^{\ast }\\ k_{1} & =\alpha _{1}-\alpha _{1}^{\ast }\\ k_{2} & =\alpha _{2}-\alpha _{2}^{\ast }\\ k_{3} & =\alpha _{3}-\alpha _{3}^{\ast } \end{align*}
Example: See handout. Given system \(A\) is \(3\times 3\) and \(B\) is \(3\times 1\,\), \(\left ( A,B\right ) \) is not in controllable form. The open loop eigenvalues are \(-0.222,1.11\pm j1.8\). Then transform to controllable form, then design in the controllable form, then transform back to original system to set the \(k\) in the original system. To set controllable form, \(\det \left ( \lambda I-A\right ) \) and this gives the last row of \(A_{companion}\). This is all what we need.\begin{align*} A_{comp} & =\begin{pmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ -1 & -4 & 2 \end{pmatrix} \\ B_{comp} & =\begin{pmatrix} 0\\ 0\\ 1 \end{pmatrix} \end{align*}
Assuming original \(A\) is controllable, we find \(T=\mathbb{C} _{comp}\mathbb{C} _{original}^{-1}\)
Reader: Design \(k\) such that 3 eigenvalues are \(\lambda =-2\), so \(p^{\ast }=\left ( \lambda +2\right ) ^{3}\). compare to \(\det \left ( \lambda I-A_{comp}\right ) \). This gives \(k_{comp}\), now transform back to original \(A\).
closed loop \(\tilde{A}+\tilde{B}\tilde{k},\)but \(\tilde{A}=TAT^{-1},\tilde{B}=TB\), \(\tilde{k}=kT\)
We will now do observer design. State estimation and observer design.
We see \(y,u\) and need to estimate state \(x\). i.e. supposed we are given few states and we need from these to estimate all other states.
Use Lvenberger observer to build estimator.
Observer equations\[ \tilde{x}^{\prime }=A\tilde{x}+Bu+L\left ( y-\tilde{y}\right ) \] \(L\) is called the observer gain matrix, which we design.\[ \tilde{x}^{\prime }=\left ( A-LC\right ) \tilde{x}+Bu+LCx \] Reader: Adding \(Du\) to \(y\) do not add affect design.
How does this observer perform? The error is \(e=x-\tilde{x}\), so \begin{align*} e^{\prime } & =\left ( Ax+Bu\right ) -\left ( A\tilde{x}+Bu+L\left ( y-\tilde{y}\right ) \right ) \\ & =\left ( A-LC\right ) \left ( x-\tilde{x}\right ) \\ & =\left ( A-LC\right ) e \end{align*}
We want \(e\rightarrow 0\) fast. So we need to design \(L\) to make \(\left ( A-LC\right ) \) do so.
If \(\left ( A,C\right ) \) is observable pair, then eigenvalues for \(\left ( A-LC\right ) \) can put anywhere by choice of \(L\). If \(\left ( A,C\right ) \) is observable, then \(\left ( A^{T},C^{T}\right ) \) is controllable pair using duality.