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If I have the following:

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How do I show the following: $P_{11} = G_{11} + G_{12}\hat Y\tilde MG_{21}$ is:

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I am stuck in this complicated system.

Or, the other simpler one: How do I show the following: $P_{12} = -G_{12}\hat M$ :

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It seems the answer is not derived from just series combination of both.

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  • $\begingroup$ It seems they are not obtained from just the series comb. of these systems. $\endgroup$ – sleeve chen Mar 27 '15 at 19:55
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Notice that G is a 2x2 system. The input/output group $ij$ enters via $B_i$ and goes out via $C_j$ hence $G_{ij}$ refers to $C_j(sI-A)^{-1}B_i$

Then,

$$ G_{11} = \begin{bmatrix}A&B_1\\C_1 &0\end{bmatrix}, G_{12} = \begin{bmatrix}A&B_2\\C_1 &D_{12}\end{bmatrix}, ... $$ and so on. Thus, now if you do the tedious multiplication exercise (which I have to omit) you'll get the answer.

Two state space systems admit multiplication $G_2G_1$ if output number of $G_1$ is equal to the input number of $G_1$. Then by substituting the $y$ equation of $G_2$ and rearranging terms give

$$ G_2G_1 = \begin{bmatrix}A_2&B_2C_1&B_2D_1\\0 &A_1&B_1\\C_2&D_2C_1&D_2D_1\end{bmatrix} $$

Thus we have, $$ \begin{align} P_{12} = G_{12}\hat M &=\begin{bmatrix}A&B_2\\C_1 &D_{12}\end{bmatrix} \begin{bmatrix}A+B_2K&B_2\\K &I\end{bmatrix}\\& = \begin{bmatrix}A&B_2K&B_2\\0&A+B_2K&B_2\\C_1&D_{12}K&D_{12}\end{bmatrix}\\ &= \begin{bmatrix}A_p&B_p\\C_p &D_{p}\end{bmatrix} \end{align} $$

The remaining trick is to see that the last expression has uncontrollable states and can be reduced. Let $$ T=\begin{bmatrix}I&0\\I&I \end{bmatrix} $$ Then if we do the state transformation, $\hat x=Tx$ $$ P_{12} = \begin{bmatrix}A_p&B_p\\C_p &D_{p}\end{bmatrix} = \begin{bmatrix}T^{-1}A_pT&T^{-1}B_p\\C_pT &D_{p}\end{bmatrix} = \begin{bmatrix}A+B_2K&B_2K&B_2\\0&A+B_2K&0\\C_1+D_{12}K&D_{12}K&D_{12}\end{bmatrix} $$ we see that the second subsystem is uncontrollable and can be removed and the minimal realization is as given in the problem.

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  • $\begingroup$ In fact, the key part I want to ask is the multiplication. Here is one I asked before: math.stackexchange.com/questions/893634/… $\endgroup$ – sleeve chen Apr 8 '15 at 6:07
  • $\begingroup$ However, if you use the way in that link, you won't get the answer in my article. So my question is, what on earth the multiplication is? how to do it in the state space form. $\endgroup$ – sleeve chen Apr 8 '15 at 6:08
  • $\begingroup$ In that link, the multiplication is the series of two system. The output of one system is the input of the next system. $\endgroup$ – sleeve chen Apr 8 '15 at 6:21
  • $\begingroup$ @sleevechen I've added the second example solution. $\endgroup$ – percusse Apr 8 '15 at 10:20
  • $\begingroup$ One question: How do you choose $T$ in that form? In the textbook, $T^{-1}$ is chosen as $q_1 ... q_{n_1}$, all of which are from the linear independent columns of controllability matrix, and $q_{n_1+1} ... q_n$ to make $T^{-1}$ nonsingular. So how to choose $T$ in your answer? $\endgroup$ – sleeve chen Apr 11 '15 at 7:45

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