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1.8 Lecture 6. Thursday September 18 2014 (Realization theorem, MIMO, state feedback)

If the transfer function $H\left ( s\right )$ is proper, then is it realizable. (SISO for now). Reminder: Need to show this for the general case.

Proof: We must find ${\displaystyle \sum } \left ( A,B,C,D\right )$ such that $H_{\ast }\left ( s\right ) =H\left ( s\right )$ where $H_{\ast }\left ( s\right )$ is the transfer function obtained from ${\displaystyle \sum } \left ( A,B,C,D\right )$ and $H\left ( s\right )$ is the transfer function we are given. We propose $A=\begin{pmatrix} 0 & 1 & 0 & \cdots & 0\\ 0 & 0 & 1 & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & \cdots & 0 & 0 & 1\\ -\alpha _{0} & -\alpha _{1} & -\alpha _{2} & \cdots & -\alpha _{n-1}\end{pmatrix}$. Let $H\left ( s\right ) =\gamma +\frac{\beta _{n-1}s^{n-1}+\beta _{n-2}s^{n-2}+\cdots +\beta _{0}}{s^{n}+\alpha _{n-1}s^{n-1}+\cdots +\alpha _{0}}$ and propose $B=\begin{pmatrix} 0\\ 0\\ \vdots \\ 0\\ 1 \end{pmatrix}$ and propose $C=\begin{pmatrix} \beta _{0} & \beta _{1} & \cdots & \beta _{n-1}\end{pmatrix}$ and $D=\left [ \gamma \right ]$, now we need to show that $H_{\ast }\left ( s\right ) =H\left ( s\right )$ using Mason rule.

Reader: Use Mason rule to show that this realization works. Now what about MIMO? Assume we are given $H\left ( s\right ) =\begin{pmatrix} H_{11}\left ( s\right ) & H_{12}\left ( s\right ) \\ H_{21}\left ( s\right ) & H_{22}\left ( s\right ) \end{pmatrix}$. We can do each on its own, then need to ”patched” to show that the matrix of them all work. Example using $2\times 2$.

If each $H_{ij}\left ( s\right )$ is proper, let ${\displaystyle \sum \limits _{ij}} =A_{ij},B_{ij},C_{ij},D_{ij}$ be realization for $H_{ij}\left ( s\right )$. Note each $A_{ij}$ can be different size. Propose $A=\begin{pmatrix} A_{11} & 0 & 0 & 0\\ 0 & A_{12} & 0 & 0\\ 0 & 0 & A_{21} & 0\\ 0 & 0 & 0 & A_{22}\end{pmatrix}$ and $B=\begin{pmatrix} B_{11} & 0\\ B_{12} & 0\\ 0 & B_{21}\\ 0 & B_{22}\end{pmatrix}$ and $C=\begin{pmatrix} C_{11} & C_{12} & 0 & 0\\ 0 & 0 & C_{21} & C_{22}\end{pmatrix}$ and $D=\begin{pmatrix} D_{11} & D_{12}\\ D_{21} & D_{22}\end{pmatrix}$. Now we claim $\left ( A,B,C,D\right )$ is realization of $\begin{pmatrix} H_{11}\left ( s\right ) & H_{12}\left ( s\right ) \\ H_{21}\left ( s\right ) & H_{22}\left ( s\right ) \end{pmatrix}$. We need to calculate

$$\begin{align*} H_{\ast }\left ( s\right ) & =C\left ( sI-A\right ) ^{-1}B+D\\ & =\begin{pmatrix} C_{11} & C_{12} & 0 & 0\\ 0 & 0 & C_{21} & C_{22}\end{pmatrix} \left ( sI-\begin{pmatrix} A_{11} & 0 & 0 & 0\\ 0 & A_{12} & 0 & 0\\ 0 & 0 & A_{21} & 0\\ 0 & 0 & 0 & A_{22}\end{pmatrix} ^{-1}\right ) \begin{pmatrix} B_{11} & 0\\ B_{12} & 0\\ 0 & B_{21}\\ 0 & B_{22}\end{pmatrix} +\begin{pmatrix} D_{11} & D_{12}\\ D_{21} & D_{22}\end{pmatrix} \end{align*}$$

Reader: The above reduces to

$$H_{\ast }\left ( s\right ) =\begin{pmatrix} C_{11}\left ( sI-A_{11}\right ) ^{-1}B_{11}+D_{11} & C_{12}\left ( sI-A_{12}\right ) ^{-1}B_{12}+D_{12}\\ C_{21}\left ( sI-A_{21}\right ) ^{-1}B_{21}+D_{21} & C_{22}\left ( sI-A_{22}\right ) ^{-1}B_{22}+D_{22}\end{pmatrix}$$ What about other dimensions?

Reader: Propose realization with $H\left ( s\right )$ that is $3\times 2$. i.e. $\begin{pmatrix} H_{11}\left ( s\right ) & H_{12}\left ( s\right ) \\ H_{21}\left ( s\right ) & H_{22}\left ( s\right ) \\ H_{31}\left ( s\right ) & H_{32}\left ( s\right ) \end{pmatrix} \,.$ Try it. What should $A,B,C,D$ look like? Note: Even though $H_{ij}\left ( s\right )$ might each be minimal, when we obtain the realization, it might no longer be minimal. Some realization are ”nicer” than others for analysis and design.

Motivation example: $A=\begin{pmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ -\alpha _{0} & -\alpha _{1} & -\alpha _{2}\end{pmatrix} ,b=\begin{pmatrix} 0\\ 0\\ 1 \end{pmatrix} ,C=\begin{pmatrix} \beta _{0} & \beta _{1} & \beta _{2}\end{pmatrix}$. When we add feedback, we ask what is the effect of feedback? This system is nice to study feedback. We often add feedback to improve time performance. $u\left ( t\right ) =k_{1}x_{1}+k_{2}x_{2}+k_{3}x_{3}+v$ where $v$ is new input. We can pick $k_{i}$. The closed loop becomes

$$x^{\prime }=\begin{pmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ -\alpha _{0} & -\alpha _{1} & -\alpha _{2}\end{pmatrix} +\begin{pmatrix} 0\\ 0\\ 1 \end{pmatrix} \left ( k_{1}x_{1}+k_{2}x_{2}+k_{3}x_{3}+v\right )$$ Reader Determine the new $A$ matrix from the old. $$x^{\prime }=\begin{pmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ k_{1}-\alpha _{0} & k_{2}-\alpha _{1} & k_{3}-\alpha _{2}\end{pmatrix} +\begin{pmatrix} 0\\ 0\\ 1 \end{pmatrix} v$$ Notice: State feedback preservers the companion form of $A$ and $b$. Find closed form transfer function. $$H_{closed}=\frac{\beta _{0}+\beta _{1}s+\beta _{2}s^{2}}{s^{3}+\left ( \alpha _{2}-k_{3}\right ) s^{2}+\left ( \alpha _{1}-k_{2}\right ) s+\left ( \alpha _{0}-k_{1}\right ) }$$ Note: $k_{i}$ affects one coefficient each. So poles of closed loop can be arbitrarily assigned anywhere we want.