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Merge pull request #2 from CosimoRusso/fixes-25-05-2020
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Fix typos, add some additional information
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edomora97 authored May 25, 2020
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12 changes: 6 additions & 6 deletions lectures/2020-05-25.tex
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\chapter{Minimum variance control}
\newlecture{Sergio Savaresi}{19/05/2020}
\newlecture{Sergio Savaresi}{25/05/2020}

Design and analysis of feedback systems.
It's not system identification nor software sensing.
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Intuitively it's very difficult to control non-minimum phase systems.
You can take the wrong decision if you react immediately.

Also for human it's difficult, for example \emph{steer to roll} dynamics in a bicycle.
Also for human it's difficult, for example \emph{steer to roll} dynamics in a bicycle: if you want to steer left, you must first steer a little to the right and then turn left.

Design of controller for non-minimum phase is difficult and requires special design techniques (no MVC but generalized MVC).
\end{remark}
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\[
J = E\left[ (y(t) - y^0(t))^2 \right]
\]
It's the variance of the tracking error, because of the it's called Minimum Variance Control.
It's the variance of the tracking error, that's why it's called Minimum Variance Control.

Some additional (small) technical assumptions:
\begin{itemize}
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\[
S: y(t) = \frac{b_1+b_1z^{-1}}{1-az^{-1}}u(t-1) + \frac{1}{1-az^{-1}}e(t)
\]
Note that this is an $ARMAX(1,0,1+1)$.
Note that this is an $ARMAX(1,0,1+1)=ARX(1,2)$.
\[
k=1 \qquad B(z) = b_0+b_1z^{-1} \qquad A(z)=1-az^{-1} \qquad C(z) = 1
\]
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Now we can impose that $\hat{y}(t|t-1)=y^0(t)$
\[
b_0u(t) + b_1u(t-1) + ay(t) = y^0(t) \qquad u(t) = \left( y^0(t+1) - ay(t) \right)\frac{1}{b_0+b_1z^{-1}}
b_0u(t) + b_1u(t-1) + ay(t) = y^0(t+1) \qquad u(t) = \left( y^0(t+1) - ay(t) \right)\frac{1}{b_0+b_1z^{-1}}
\]
But we don't have $y^0(t+1)$, so we use $y^0(t)$.
\[
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\begin{tikzpicture}[node distance=2cm,auto,>=latex']
\node[sum] at (0,0) (sum) {};
\node[int] at (1.5,0) (b1) {$\frac{1}{b_0+b_1z^{-1}}$};
\node[int] at (5,0) (b2) {$z^{-1}\frac{b_0+b_1z^{-1}}{1-az^[-1]}$};
\node[int] at (5,0) (b2) {$z^{-1}\frac{b_0+b_1z^{-1}}{1-az^{-1}}$};
\node[int] at (3,-1.5) (b3) {$a$};
\node[int] at (7,1.5) (b4) {$\frac{1}{1-az^{-1}}$};
\node[sum] at (7,0) (sum2) {};
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