State-space system matrix Amat and input matrix Bmat (5 states and 1 input)

$\mathrm{Amat}≔\mathrm{Matrix}\left(\left[\left[1\,-2\,3\,-5\,12\right]\,\left[0\,4\,-1\,2\,-3\right]\,\left[2\,5\,7\,-4\,3\right]\,\left[-2\,4\,3\,8\,7\right]\,\left[3\,4\,1\,-4\,7\right]\right]\right)\:$

$\mathrm{Bmat}≔\mathrm{Matrix}\left(\left[\left[3\right]\,\left[9\right]\,\left[5\right]\,\left[4\right]\,\left[6\right]\right]\right)\:$

Desired poles

$\mathrm{p1}≔\left[-1+\mathrm{I}\,-0.1\,-2.44\,-1-\mathrm{I}\,-7.77\right]\:$

Obtain the state feedback gain Kc:

$\mathrm{Kc}≔\mathrm{StateFeedback}:-\mathrm{Ackermann}\left(\mathrm{Amat}\,\mathrm{Bmat}\,\mathrm{p1}\right)$

$\mathrm{Kc}≔\left[\begin{array}{ccccc}\mathrm{-3.367696219}& 3.289423763& 4.030559336& 8.644197464& \mathrm{-5.820218623}\end{array}\right]$

Verify the closed-loop system matrix has the desired poles

$\mathrm{LinearAlgebra}:-\mathrm{Eigenvalues}\left(\mathrm{Amat}-\mathrm{Bmat}·\mathrm{Kc}\right)$

$\left[\begin{array}{c}\mathrm{-7.77000009752235}+0.\mathrm{I}\\ \mathrm{-2.43999927357362}+0.\mathrm{I}\\ \mathrm{-1.00000013420824}+0.999999523527025\mathrm{I}\\ \mathrm{-1.00000013420824}-0.999999523527025\mathrm{I}\\ \mathrm{-0.100000368487564}+0.\mathrm{I}\end{array}\right]$

System object sys with symbolic entries (3 states, 1 input, 3 outputs)

$\mathrm{sys}≔\mathrm{StateSpace}\left(\mathrm{Matrix}\left(\left[\left[1\,y\,3\right]\,\left[x\,4\,-1\right]\,\left[2\,5\,7\right]\right]\right)\,\mathrm{Matrix}\left(\left[\left[3\right]\,\left[9\right]\,\left[5\right]\right]\right)\,\mathrm{Matrix}\left(\left[\left[1\,0\,0\right]\,\left[0\,1\,0\right]\,\left[0\,0\,1\right]\right]\right)\,\mathrm{Matrix}\left(3\,1\right)\right)\:$

$\mathrm{PrintSystem}\left(\mathrm{sys}\right)$

$\left[\begin{array}{l}\mathbf{State\; Space}\\ \mathrm{continuous}\\ \mathrm{3\; output(s);\; 1\; input(s);\; 3\; state(s)}\\ \mathrm{inputvariable}\=\left[\mathrm{u1}\left(t\right)\right]\\ \mathrm{outputvariable}\=\left[\mathrm{y1}\left(t\right)\,\mathrm{y2}\left(t\right)\,\mathrm{y3}\left(t\right)\right]\\ \mathrm{statevariable}\=\left[\mathrm{x1}\left(t\right)\,\mathrm{x2}\left(t\right)\,\mathrm{x3}\left(t\right)\right]\\ \mathrm{a}\=\left[\begin{array}{ccc}1& y& 3\\ x& 4& \mathrm{-1}\\ 2& 5& 7\end{array}\right]\\ \mathrm{b}\=\left[\begin{array}{c}3\\ 9\\ 5\end{array}\right]\\ \mathrm{c}\=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\\ \mathrm{d}\=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]\end{array}\right.$

Desired poles

$\mathrm{p2}≔\left[-4\,-5+3\mathrm{I}\,-5-3\mathrm{I}\right]$

$\mathrm{p2}≔\left[\mathrm{-4}\,\mathrm{-5}+3\mathrm{I}\,\mathrm{-5}-3\mathrm{I}\right]$

Obtain the state feedback gain Kc and the feed-forward gain Kr:

$\mathrm{Kc},\mathrm{Kr}≔\mathrm{StateFeedback}:-\mathrm{Ackermann}\left(\mathrm{sys}\,\mathrm{p2}\,'\mathrm{return\_Kr}'=\mathrm{true}\,'\mathrm{parameters}'=\left[x=0\,y=2\right]\right)$

$\mathrm{Kc},\mathrm{Kr}≔\left[\begin{array}{ccc}4.383962009& 0.2488707000& 2.121655535\end{array}\right],\left[\begin{array}{ccc}3.08169350000358& 0.648777578948122& \mathrm{-1.05426356579070}\end{array}\right]$

Verify the closed-loop system matrix has the desired poles

$\mathrm{sys}:-a≔\mathrm{eval}\left(\mathrm{sys}:-a\,\left[x=0\,y=2\right]\right)\:$

$\mathrm{eigen}≔\mathrm{LinearAlgebra}:-\mathrm{Eigenvalues}\left(\mathrm{sys}:-a-\mathrm{sys}:-b·\mathrm{Kc}\right)$

$\mathrm{eigen}≔\left[\begin{array}{c}\mathrm{-5.00000000205003}+2.99999999901668\mathrm{I}\\ \mathrm{-5.00000000205003}-2.99999999901668\mathrm{I}\\ \mathrm{-3.99999999789996}+0.\mathrm{I}\end{array}\right]$

$\mathrm{sys1}≔\mathrm{StateSpace}\left(\mathrm{Matrix}\left(\left[\left[4\,0\,-3\,5\right]\,\left[3\,0\,0\,3\right]\,\left[1\,0\,-1\,1\right]\,\left[0\,0\,0\,0\right]\right]\right)\,\mathrm{Matrix}\left(\left[\left[1\right]\,\left[3\right]\,\left[2\right]\,\left[0\right]\right]\right)\,\mathrm{Matrix}\left(\left[\left[1\,0\,3\,5\right]\,\left[-3\,0\,5\,7\right]\right]\right)\,\mathrm{Matrix}\left(\left[\left[1\right]\,\left[0\right]\right]\right)\right)\:$

$\mathrm{PrintSystem}\left(\mathrm{sys1}\right)$

$\left[\begin{array}{l}\mathbf{State\; Space}\\ \mathrm{continuous}\\ \mathrm{2\; output(s);\; 1\; input(s);\; 4\; state(s)}\\ \mathrm{inputvariable}\=\left[\mathrm{u1}\left(t\right)\right]\\ \mathrm{outputvariable}\=\left[\mathrm{y1}\left(t\right)\,\mathrm{y2}\left(t\right)\right]\\ \mathrm{statevariable}\=\left[\mathrm{x1}\left(t\right)\,\mathrm{x2}\left(t\right)\,\mathrm{x3}\left(t\right)\,\mathrm{x4}\left(t\right)\right]\\ \mathrm{a}\=\left[\begin{array}{cccc}4& 0& \mathrm{-3}& 5\\ 3& 0& 0& 3\\ 1& 0& \mathrm{-1}& 1\\ 0& 0& 0& 0\end{array}\right]\\ \mathrm{b}\=\left[\begin{array}{c}1\\ 3\\ 2\\ 0\end{array}\right]\\ \mathrm{c}\=\left[\begin{array}{cccc}1& 0& 3& 5\\ \mathrm{-3}& 0& 5& 7\end{array}\right]\\ \mathrm{d}\=\left[\begin{array}{c}1\\ 0\end{array}\right]\end{array}\right.$

Desired poles

$\mathrm{p3}≔\left[-a+\mathrm{I}\cdot 3\,-a-\mathrm{I}\cdot 3\,-b+\mathrm{I}\cdot 9\,-b-\mathrm{I}\cdot 9\right]$

$\mathrm{p3}≔\left[-a+3\mathrm{I}\,-a-3\mathrm{I}\,-b+9\mathrm{I}\,-b-9\mathrm{I}\right]$

Get the state-feedback gain

$\mathrm{Kc}≔\mathrm{StateFeedback}:-\mathrm{Ackermann}\left(\mathrm{sys1}\,\mathrm{p3}\right)$

Error, (in ControlDesign:-StateFeedback:-Ackermann) the given state-space realization is not controllable

Remove the uncontrollable states if possible. ReduceSystem will remove the structural uncontrollable states.

$\mathrm{rsys}≔\mathrm{ReduceSystem}\left(\mathrm{sys1}\,'\mathrm{reducedtype}'=\mathrm{controllable}\right)\:$

$\mathrm{PrintSystem}\left(\mathrm{rsys}\right)$

$\left[\begin{array}{l}\mathbf{State\; Space}\\ \mathrm{continuous}\\ \mathrm{2\; output(s);\; 1\; input(s);\; 3\; state(s)}\\ \mathrm{inputvariable}\=\left[\mathrm{u1}\left(t\right)\right]\\ \mathrm{outputvariable}\=\left[\mathrm{y1}\left(t\right)\,\mathrm{y2}\left(t\right)\right]\\ \mathrm{statevariable}\=\left[\mathrm{x1}\left(t\right)\,\mathrm{x2}\left(t\right)\,\mathrm{x3}\left(t\right)\right]\\ \mathrm{a}\=\left[\begin{array}{ccc}4& 0& \mathrm{-3}\\ 3& 0& 0\\ 1& 0& \mathrm{-1}\end{array}\right]\\ \mathrm{b}\=\left[\begin{array}{c}1\\ 3\\ 2\end{array}\right]\\ \mathrm{c}\=\left[\begin{array}{ccc}1& 0& 3\\ \mathrm{-3}& 0& 5\end{array}\right]\\ \mathrm{d}\=\left[\begin{array}{c}1\\ 0\end{array}\right]\end{array}\right.$

Verify the resulting subsystem is controllable

$\mathrm{Controllable}\left(\mathrm{rsys}\,'\mathrm{method}'=\mathrm{rank}\right)$

$\mathrm{true}$

Desired poles for the controllable subsystem

$\mathrm{p4}≔\left[-a\,-b+\mathrm{I}\cdot 9\,-b-\mathrm{I}\cdot 9\right]\:$

$\mathrm{Kc}≔\mathrm{StateFeedback}:-\mathrm{Ackermann}\left(\mathrm{rsys}\,\mathrm{p4}\right)$

$\mathrm{Kc}≔\left[\begin{array}{ccc}-\frac{1}{6}a{b}^2-\frac{4}{3}ab-\frac{2}{3}{b}^2-\frac{95}{6}a-\frac{14}{3}b-\frac{185}{3}& -\frac{1}{18}a{b}^2-\frac{9}{2}a& \frac{1}{6}a{b}^2+\frac{2}{3}ab+\frac{1}{3}{b}^2+\frac{91}{6}a+\frac{10}{3}b+\frac{97}{3}\end{array}\right]$