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There is two types of state space models:

The classic one:

$$\dot{x} = Ax + Bu + \omega$$ $$y = Cx + Du + \upsilon $$

Where $A$ is the system matrix, $B$ is the insignal matrix, $C$ is the output matrix, $D$ is the direct matrix, $\omega$ is the disturbance vector and $\upsilon$ is the noise vector. Normaly those vectors often are multiplied with a diagonal matrix.

Then we have the...other state space model:

$$\dot{x} = Ax + Bu + \omega$$ $$ z = Mx + D_z u $$ $$y = Cx + D_y u + \upsilon $$

So...what are $$ z = Mx + D_z u $$

supposed to be? Let me guess! The $z$ variable is only for loop shaping controllers which look like this:

enter image description here

And this: $$y = Cx + D_y u + \upsilon $$

Is only for measurement and this:

$$ z = Mx + D_z u $$

Is only for optimizing showing which state are the state who going to the observable state?

For example:

If $C$ are:

$$ C = \begin{bmatrix} 1 & 0 &0 \\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix} $$

And we want to measure all states $x$ just by one state. Let's say $x_3$. That means that $M$ are going to be:

$$M = \begin{bmatrix} 0 & 0 & 1 \end{bmatrix}$$

Right? So the classical state space model's $C$ matrix is the new $M$ matrix?

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  • $\begingroup$ I assume this question is in the context of $H_\infty$? $\endgroup$ – Kwin van der Veen Aug 2 '17 at 23:18
  • $\begingroup$ Not actually. I want to separate them. H-inf for one question and state space models for the other question. $\endgroup$ – Daniel Mårtensson Aug 2 '17 at 23:44
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Performance variables, i.e. the variables you actually want to control, in contrast to $y$ which are the measured signals.

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  • $\begingroup$ Tack Johan. So let's say that I have a state vector $x \in \Re^4$ and $x_2$ is the velocity from $x_1$ and $x_1$ is the position. Also $x_3$ is a position and $x_4$ is the velocity of that position. My goal is to control all the positions: $x_1$ and $x_3$. That means $M$ will be $M = \begin{bmatrix} 1 & 0 & 0 &0 \\ 0 & 0& 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix}$ ? And the $D_z$ must be the same as $B$ ? I'm reading the book "Reglerteori av Torkel Glad och Lennart Ljung". $\endgroup$ – Daniel Mårtensson Aug 3 '17 at 20:21
  • $\begingroup$ Or if I want to control the whole vector $x$, why not set $M$ as an indentity matrix? $\endgroup$ – Daniel Mårtensson Aug 3 '17 at 21:08
  • $\begingroup$ If you want to do standard LQ minimizing $z^Tz = x^TQx + u^TRu$, you would have $M = \begin{pmatrix}Q^{1/2}\\0\end{pmatrix}$ and $D_z = \begin{pmatrix}0\\R^{1/2}\end{pmatrix}$ $\endgroup$ – Johan Löfberg Aug 4 '17 at 5:00
  • $\begingroup$ But this is for H infinity control. Have a look at this question math.stackexchange.com/questions/2380221/… I already know how to create LQG controllers :) $\endgroup$ – Daniel Mårtensson Aug 4 '17 at 11:49
  • $\begingroup$ What ever. $z$ are the variables you have performance requirements on (such as limited $H_{2}$ or $H_{\infty}$ norm from $w$) $\endgroup$ – Johan Löfberg Aug 4 '17 at 12:17

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