Reader: Consider \(H\left ( s\right ) =\gamma +\frac{\beta _{3}s^{3}+\beta _{2}s^{2}+\beta s+\beta _{0}}{s^{4}+\alpha _{3}s^{3}+\alpha _{2}s^{2}+\alpha s+\alpha _{0}}\) and show that \(A=\begin{pmatrix} 0 & 0 & 0 & -\alpha _{0}\\ 1 & 0 & 0 & -\alpha _{1}\\ 0 & 1 & 0 & -\alpha _{2}\\ 0 & 0 & 1 & -\alpha _{3}\end{pmatrix} ,B=\begin{pmatrix} \beta _{0}\\ \beta _{1}\\ \beta _{2}\\ \beta _{3}\end{pmatrix} ,C=\begin{pmatrix} 0 & 0 & 0 & 1 \end{pmatrix} ,D=\left [ \gamma \right ] \) is a realization.
We established realization before, when we obtained the controllable form. One way is to use Mason rule. Another way is using syms and find \(C\left ( sI-A\right ) ^{-1}B+D.\)
This realization is called the observable canonical form. Some realizations are better than other for different things. For state estimation, observable form is better, for state control, the controllable form is better.
Reader: Generalized reader above to proper \(H\left ( s\right ) \), can use Mason.
Reader: Even more general. Suppose \(\sum \left ( A,B,C,D\right ) \) is a realization of SISO transfer function \(H\left ( s\right ) \), show that \(\sum _{\ast }\left ( A^{T},C^{T},B^{T},D^{T}\right ) \) is also realization. So \(H_{\ast }\left ( s\right ) =C_{\ast }\left ( sI-A_{\ast }\right ) ^{-1}B_{\ast }+D_{\ast }\) where \(C_{\ast }=B^{T},A_{\ast }=A^{T},B_{\ast }=C^{T}\). So \(H_{\ast }\left ( s\right ) =B^{T}\left ( SI-A^{T}\right ) ^{-1}C^{T}+D\). Since this is SISO, it is scalar, so its transpose do not change. Take the transpose of the above gives \begin{align*} H_{\ast }\left ( s\right ) & =C\left ( \left ( sI-A^{T}\right ) ^{-1}\right ) ^{T}B+D\\ & =C\left ( sI-A\right ) ^{-1}B+D \end{align*}
Reader: For MIMO get \(H_{\ast }\left ( s\right ) =H^{T}\left ( s\right ) \)
Transformation: \(\sum \rightarrow T\rightarrow \sum _{\ast }\) where \(\sum _{\ast }\) is equivalent. Design in \(\sum _{\ast }\), then when design is completed, transform back to \(\sum \).
Handout: Given \(H\left ( s\right ) _{\sum }=H\left ( s\right ) _{\sum _{\ast }}\) when can \(\sum \) be transformed to controllable or observable forms using \(T?\) Necessary conditions for existence of \(T\) is that \[\mathbb{C} _{\ast }=T\mathbb{C} \] Where \(\mathbb{C} \) is the controllability matrix for the original \(\sum \) and \(\mathbb{C} _{\ast }\) is controllability matrix for the \(\sum _{\ast }\). For SISO, if \(\mathbb{C} \) is invertible, then \(T=\mathbb{C} _{\ast }\mathbb{C} ^{-1}\). What is MIMO? system is good if rank \(\mathbb{C} \) is \(n.\) i.e. controllable system. When this is satisfied, we say \(\left ( A,B\right ) \) is controllable pair.
Reader: \(\left ( A,B\right ) \) is controllable pair implies \(\mathbb{C} \mathbb{C} ^{T}\) is invertible. Proof: Assume \(\rho \left ( \mathbb{C} \right ) =n\), show that \(\mathbb{C} \mathbb{C} ^{T}\) is invertible. Use proof by contradiction. Assume no inverse,hence this means there is non-zero vector \(\vec{x}\) s.t. \(\mathbb{C} \mathbb{C} ^{T}\vec{x}=0\), so \(x^{T}\mathbb{C} \mathbb{C} ^{T}x=0\) or \(y^{T}y=0,\) so \(y=0\) or \(\mathbb{C} ^{T}x=0\), but \(x\neq 0\) so contradiction.
There is also observability rank condition. Reader: Mimic the controllability analysis above for pair \(\left ( A,C\right ) \). Sketch steps: Develop \(\mathbb{Q}=\begin{pmatrix} C\\ CA\\ \vdots \\ CA^{n-1}\end{pmatrix} \). We need to relate \(\mathbb{Q} \) to \(\mathbb{Q} _{\ast }\) using \(T\). Since \(C_{\ast }=CT^{-1}\), then \(C_{\ast }A_{\ast }=\left ( CT^{-1}\right ) \left ( T^{-1}AT\right ) =CA\). So we get condition that \(\rho \left ( \mathbb{Q} \right ) =n\) is necessary condition for existence of \(T\).
If the controllability failed, try the observability.
Reader: Show the controllability canonical form always satisfies \(\rho \left ( \mathbb{C} \right ) =n\). i.e. if we can put the system in the controllable form, then it is controllable. To show, start with \(3\times 3~\) matrix and find its \(\mathbb{C} \) to show it is \(3\).
Reader: Can controllable form fail to be observable? (yes).
Reader: Consider \(H\left ( s\right ) =\frac{s+1}{s^{2}+2s+1}=\frac{1}{s+1}\) and study the controllability and observability rank.
More generally, if \(H\left ( s\right ) \) has no pole/zero cancellation, then it is minimal and \(\mathbb{C} \) and \(\mathbb{Q} \) both have rank \(n\) (full rank). So if there is no pole/zero cancellation, then it is both controllable and observable. Consider \(H\left ( s\right ) =\frac{\left ( s^{2}+s+1\right ) \left ( s^{3}+2s^{2}+4\right ) }{\left ( s^{2}+s+1\right ) ^{2}\left ( s+1\right ) ^{2}}\), try and see.