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Example 1: Minimum Frobenius Norm Static Output Feedback¶

The example system has been taken from [1, Example 1]. The eigenvalue placement with minimum Frobenium norm is discussed in [2, Example 1]. The poles should be placed at -1, -2, -3, i.e., we want to have distinct real eigenvalues.

Guney, S.; Atasoy, A. (2011). An approach to pole placement method with output feedback.

In UKACC Control Conference, URL: https://www.researchgate.net/publication/267424630_An_Approach_to_Pole_Placement_Method_with_Output_Feedback 2. Röbenack, K.; Gerbet, D.: Minimum Norm Partial Eigenvalue Placement for Static Output Feedback Control. International Conference on System Theory, Control and Computing (ICSTCC),
October 20-23, 2021, Iași, Romania

Polynomial ring

In [19]:

%display latex
R.<k11,k12,k21,k22,s,l0,l1,l2,l3,q> = PolynomialRing(QQ, order='lex')
R

Out[19]:

$\newcommand{\Bold}[1]{\mathbf{#1}}\Bold{Q}[k_{11}, k_{12}, k_{21}, k_{22}, s, l_{0}, l_{1}, l_{2}, l_{3}, q]$

State space system

In [20]:

A = matrix(R,[[-11.4, -3.5, 0],[4, 0, 0],[0, 1, 0]])
B = matrix(R,[[2,1],[0,-1],[0,0]])
C = matrix(R,[[1,0,1.425],[1,-1,0]])
(A,B,C)

Out[20]:

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\left(\begin{array}{rrr} -\frac{57}{5} & -\frac{7}{2} & 0 \\ 4 & 0 & 0 \\ 0 & 1 & 0 \end{array}\right), \left(\begin{array}{rr} 2 & 1 \\ 0 & -1 \\ 0 & 0 \end{array}\right), \left(\begin{array}{rrr} 1 & 0 & \frac{57}{40} \\ 1 & -1 & 0 \end{array}\right)\right)$

Symbolic feedback matrix

In [21]:

K = matrix(R,[[k11,k12],[k21,k22]])
K

Out[21]:

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rr} k_{11} & k_{12} \\ k_{21} & k_{22} \end{array}\right)$

Closed-loop characteristic polynomial

In [22]:

CP = det(s*matrix.identity(3)-(A-B*K*C))
CP

Out[22]:

$\newcommand{\Bold}[1]{\mathbf{#1}}-2 k_{12} k_{21} s + 2 k_{11} k_{22} s + 2 k_{11} s^{2} + 2 k_{12} s^{2} + k_{21} s^{2} + 2 k_{22} s^{2} + s^{3} - \frac{57}{20} k_{12} k_{21} + \frac{57}{20} k_{11} k_{22} - 8 k_{12} s + \frac{83}{40} k_{21} s + \frac{109}{10} k_{22} s + \frac{57}{5} s^{2} + \frac{57}{5} k_{11} - \frac{2109}{200} k_{21} + 14 s$

Remainders of polynomial division for an eigenvalue placement at -1, -2, -3

In [23]:

q1,r1 = CP.quo_rem(s+1)
q2,r2 = q1.quo_rem(s+2)
q3,r3 = q2.quo_rem(s+3)
r1,r2,r3

Out[23]:

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(-\frac{17}{20} k_{12} k_{21} + \frac{17}{20} k_{11} k_{22} + \frac{67}{5} k_{11} + 10 k_{12} - \frac{581}{50} k_{21} - \frac{89}{10} k_{22} - \frac{18}{5}, -2 k_{12} k_{21} + 2 k_{11} k_{22} - 6 k_{11} - 14 k_{12} - \frac{37}{40} k_{21} + \frac{49}{10} k_{22} - \frac{66}{5}, 2 k_{11} + 2 k_{12} + k_{21} + 2 k_{22} + \frac{27}{5}\right)$

Lagrangian function

In [24]:

L = q + l0*(k11^2+k12^2+k21^2+k22^2-q) + l1*r1 + l2*r2 + l3*r3
L

Out[24]:

$\newcommand{\Bold}[1]{\mathbf{#1}}k_{11}^{2} l_{0} + \frac{17}{20} k_{11} k_{22} l_{1} + 2 k_{11} k_{22} l_{2} + \frac{67}{5} k_{11} l_{1} - 6 k_{11} l_{2} + 2 k_{11} l_{3} + k_{12}^{2} l_{0} - \frac{17}{20} k_{12} k_{21} l_{1} - 2 k_{12} k_{21} l_{2} + 10 k_{12} l_{1} - 14 k_{12} l_{2} + 2 k_{12} l_{3} + k_{21}^{2} l_{0} - \frac{581}{50} k_{21} l_{1} - \frac{37}{40} k_{21} l_{2} + k_{21} l_{3} + k_{22}^{2} l_{0} - \frac{89}{10} k_{22} l_{1} + \frac{49}{10} k_{22} l_{2} + 2 k_{22} l_{3} - l_{0} q - \frac{18}{5} l_{1} - \frac{66}{5} l_{2} + \frac{27}{5} l_{3} + q$

Neccessary optimility condition and associated polynomial ideal

In [25]:

vars = [k11,k12,k21,k22,l0,l1,l2,l3,q]
PLIST = []
for ii in range(len(vars)):
    print(ii," : ",vars[ii]," : ",diff(L,vars[ii]))
    PLIST.append(diff(L,vars[ii]))
I = Ideal(PLIST)
I

0  :  k11  :  2*k11*l0 + 17/20*k22*l1 + 2*k22*l2 + 67/5*l1 - 6*l2 + 2*l3
1  :  k12  :  2*k12*l0 - 17/20*k21*l1 - 2*k21*l2 + 10*l1 - 14*l2 + 2*l3
2  :  k21  :  -17/20*k12*l1 - 2*k12*l2 + 2*k21*l0 - 581/50*l1 - 37/40*l2 + l3
3  :  k22  :  17/20*k11*l1 + 2*k11*l2 + 2*k22*l0 - 89/10*l1 + 49/10*l2 + 2*l3
4  :  l0  :  k11^2 + k12^2 + k21^2 + k22^2 - q
5  :  l1  :  17/20*k11*k22 + 67/5*k11 - 17/20*k12*k21 + 10*k12 - 581/50*k21 - 89/10*k22 - 18/5
6  :  l2  :  2*k11*k22 - 6*k11 - 2*k12*k21 - 14*k12 - 37/40*k21 + 49/10*k22 - 66/5
7  :  l3  :  2*k11 + 2*k12 + k21 + 2*k22 + 27/5
8  :  q  :  -l0 + 1

Out[25]:

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(2 k_{11} l_{0} + \frac{17}{20} k_{22} l_{1} + 2 k_{22} l_{2} + \frac{67}{5} l_{1} - 6 l_{2} + 2 l_{3}, 2 k_{12} l_{0} - \frac{17}{20} k_{21} l_{1} - 2 k_{21} l_{2} + 10 l_{1} - 14 l_{2} + 2 l_{3}, -\frac{17}{20} k_{12} l_{1} - 2 k_{12} l_{2} + 2 k_{21} l_{0} - \frac{581}{50} l_{1} - \frac{37}{40} l_{2} + l_{3}, \frac{17}{20} k_{11} l_{1} + 2 k_{11} l_{2} + 2 k_{22} l_{0} - \frac{89}{10} l_{1} + \frac{49}{10} l_{2} + 2 l_{3}, k_{11}^{2} + k_{12}^{2} + k_{21}^{2} + k_{22}^{2} - q, \frac{17}{20} k_{11} k_{22} + \frac{67}{5} k_{11} - \frac{17}{20} k_{12} k_{21} + 10 k_{12} - \frac{581}{50} k_{21} - \frac{89}{10} k_{22} - \frac{18}{5}, 2 k_{11} k_{22} - 6 k_{11} - 2 k_{12} k_{21} - 14 k_{12} - \frac{37}{40} k_{21} + \frac{49}{10} k_{22} - \frac{66}{5}, 2 k_{11} + 2 k_{12} + k_{21} + 2 k_{22} + \frac{27}{5}, -l_{0} + 1\right)\Bold{Q}[k_{11}, k_{12}, k_{21}, k_{22}, s, l_{0}, l_{1}, l_{2}, l_{3}, q]$

Elimination ideal

In [26]:

J0 = I.elimination_ideal([k11,k12,k21,k22,l0,l1,l2,l3])
J0

Out[26]:

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(82230276339214102137754000000000 q^{4} - 16573151331396038867639067132000000 q^{3} + 2096519019934073899613493346851250000 q^{2} - 290174420184422803500902782422838853400 q + 926701100980361302638540549730477878569\right)\Bold{Q}[k_{11}, k_{12}, k_{21}, k_{22}, s, l_{0}, l_{1}, l_{2}, l_{3}, q]$

Computation of possible solutions w.r.t. q

In [27]:

J0.change_ring(PolynomialRing(RR, 'q')).gens()[0].roots()

Out[27]:

$\newcommand{\Bold}[1]{\mathbf{#1}}\left[\left(3.26883937657574, 1\right), \left(170.916064276310, 1\right)\right]$

Selection of the minimum solution q

In [28]:

qmin = J0.change_ring(PolynomialRing(RR, 'q')).gens()[0].roots()[0][0]
qmin

Out[28]:

$\newcommand{\Bold}[1]{\mathbf{#1}}3.26883937657574$

Dimension of the polynomial ideal

In [29]:

I.dimension()

Out[29]:

$\newcommand{\Bold}[1]{\mathbf{#1}}1$

For the direct computation of the algebraic variety, we need a 0-dimensional ideal. To achieve this, we omit the variable s

In [30]:

I0 = I.change_ring(PolynomialRing(QQ,'k11,k12,k21,k22,l0,l1,l2,l3,q'))
sol = I0.variety(QQbar)
sol

Out[30]:

$\newcommand{\Bold}[1]{\mathbf{#1}}\left[\verb|{q:|\phantom{\verb!x!}\verb|3.268839376575746?,|\phantom{\verb!x!}\verb|l3:|\phantom{\verb!x!}\verb|0.4268284555997831?,|\phantom{\verb!x!}\verb|l2:|\phantom{\verb!x!}\verb|-0.3375305868756624?,|\phantom{\verb!x!}\verb|l1:|\phantom{\verb!x!}\verb|-0.2229755726084477?,|\phantom{\verb!x!}\verb|l0:|\phantom{\verb!x!}\verb|1,|\phantom{\verb!x!}\verb|k22:|\phantom{\verb!x!}\verb|-0.6992229440461313?,|\phantom{\verb!x!}\verb|k21:|\phantom{\verb!x!}\verb|-1.157344168706315?,|\phantom{\verb!x!}\verb|k12:|\phantom{\verb!x!}\verb|-1.174350365749620?,|\phantom{\verb!x!}\verb|k11:|\phantom{\verb!x!}\verb|-0.2477546058510917?}|, \verb|{q:|\phantom{\verb!x!}\verb|170.9160642763102?,|\phantom{\verb!x!}\verb|l3:|\phantom{\verb!x!}\verb|-25.89323591709195?,|\phantom{\verb!x!}\verb|l2:|\phantom{\verb!x!}\verb|7.090078463322560?,|\phantom{\verb!x!}\verb|l1:|\phantom{\verb!x!}\verb|0.3079126349941297?,|\phantom{\verb!x!}\verb|l0:|\phantom{\verb!x!}\verb|1,|\phantom{\verb!x!}\verb|k22:|\phantom{\verb!x!}\verb|6.174475682394953?,|\phantom{\verb!x!}\verb|k21:|\phantom{\verb!x!}\verb|-10.798376008389913?,|\phantom{\verb!x!}\verb|k12:|\phantom{\verb!x!}\verb|-3.990217664985441?,|\phantom{\verb!x!}\verb|k11:|\phantom{\verb!x!}\verb|0.5149299867854444?}|, \verb|{q:|\phantom{\verb!x!}\verb|13.68035394055291?|\phantom{\verb!x!}\verb|-|\phantom{\verb!x!}\verb|141.3649013768673?*I,|\phantom{\verb!x!}\verb|l3:|\phantom{\verb!x!}\verb|-38.02263838715278?|\phantom{\verb!x!}\verb|+|\phantom{\verb!x!}\verb|28.09491245948313?*I,|\phantom{\verb!x!}\verb|l2:|\phantom{\verb!x!}\verb|-16.37028291228165?|\phantom{\verb!x!}\verb|-|\phantom{\verb!x!}\verb|7.040331241381002?*I,|\phantom{\verb!x!}\verb|l1:|\phantom{\verb!x!}\verb|9.69553140531921?|\phantom{\verb!x!}\verb|-|\phantom{\verb!x!}\verb|6.198863621543925?*I,|\phantom{\verb!x!}\verb|l0:|\phantom{\verb!x!}\verb|1,|\phantom{\verb!x!}\verb|k22:|\phantom{\verb!x!}\verb|4.161920215448037?|\phantom{\verb!x!}\verb|-|\phantom{\verb!x!}\verb|2.131263818093075?*I,|\phantom{\verb!x!}\verb|k21:|\phantom{\verb!x!}\verb|-7.975570937866967?|\phantom{\verb!x!}\verb|+|\phantom{\verb!x!}\verb|2.989305095507171?*I,|\phantom{\verb!x!}\verb|k12:|\phantom{\verb!x!}\verb|1.572282513745333?|\phantom{\verb!x!}\verb|-|\phantom{\verb!x!}\verb|5.838511282216107?*I,|\phantom{\verb!x!}\verb|k11:|\phantom{\verb!x!}\verb|-4.446417260259886?|\phantom{\verb!x!}\verb|+|\phantom{\verb!x!}\verb|6.475122552555597?*I}|, \verb|{q:|\phantom{\verb!x!}\verb|13.68035394055291?|\phantom{\verb!x!}\verb|+|\phantom{\verb!x!}\verb|141.3649013768673?*I,|\phantom{\verb!x!}\verb|l3:|\phantom{\verb!x!}\verb|-38.02263838715278?|\phantom{\verb!x!}\verb|-|\phantom{\verb!x!}\verb|28.09491245948313?*I,|\phantom{\verb!x!}\verb|l2:|\phantom{\verb!x!}\verb|-16.37028291228165?|\phantom{\verb!x!}\verb|+|\phantom{\verb!x!}\verb|7.040331241381002?*I,|\phantom{\verb!x!}\verb|l1:|\phantom{\verb!x!}\verb|9.69553140531921?|\phantom{\verb!x!}\verb|+|\phantom{\verb!x!}\verb|6.198863621543925?*I,|\phantom{\verb!x!}\verb|l0:|\phantom{\verb!x!}\verb|1,|\phantom{\verb!x!}\verb|k22:|\phantom{\verb!x!}\verb|4.161920215448037?|\phantom{\verb!x!}\verb|+|\phantom{\verb!x!}\verb|2.131263818093075?*I,|\phantom{\verb!x!}\verb|k21:|\phantom{\verb!x!}\verb|-7.975570937866967?|\phantom{\verb!x!}\verb|-|\phantom{\verb!x!}\verb|2.989305095507171?*I,|\phantom{\verb!x!}\verb|k12:|\phantom{\verb!x!}\verb|1.572282513745333?|\phantom{\verb!x!}\verb|+|\phantom{\verb!x!}\verb|5.838511282216107?*I,|\phantom{\verb!x!}\verb|k11:|\phantom{\verb!x!}\verb|-4.446417260259886?|\phantom{\verb!x!}\verb|-|\phantom{\verb!x!}\verb|6.475122552555597?*I}|\right]$

Selection of the solution to be used

In [31]:

lx = sol[0]
lx

Out[31]:

$\newcommand{\Bold}[1]{\mathbf{#1}}\verb|{q:|\phantom{\verb!x!}\verb|3.268839376575746?,|\phantom{\verb!x!}\verb|l3:|\phantom{\verb!x!}\verb|0.4268284555997831?,|\phantom{\verb!x!}\verb|l2:|\phantom{\verb!x!}\verb|-0.3375305868756624?,|\phantom{\verb!x!}\verb|l1:|\phantom{\verb!x!}\verb|-0.2229755726084477?,|\phantom{\verb!x!}\verb|l0:|\phantom{\verb!x!}\verb|1,|\phantom{\verb!x!}\verb|k22:|\phantom{\verb!x!}\verb|-0.6992229440461313?,|\phantom{\verb!x!}\verb|k21:|\phantom{\verb!x!}\verb|-1.157344168706315?,|\phantom{\verb!x!}\verb|k12:|\phantom{\verb!x!}\verb|-1.174350365749620?,|\phantom{\verb!x!}\verb|k11:|\phantom{\verb!x!}\verb|-0.2477546058510917?}|$

Numerical feedback gain matrix

In [32]:

K0 = K.subs(k11=RR(lx['k11']),k12=RR(lx['k12']),
            k21=RR(lx['k21']),k22=RR(lx['k22']))
K0

Out[32]:

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rr} -0.247754605851092 & -1.17435036574962 \\ -1.15734416870631 & -0.699222944046131 \end{array}\right)$

Frobenius norm of the computed feedback matrix

In [33]:

K0.norm('frob')

Out[33]:

$\newcommand{\Bold}[1]{\mathbf{#1}}1.8079931904118847$

Verification of the closed-loop eigenvalues (over the rational field)

In [36]:

A0 = matrix(QQ,A-B*K0*C)
A0.eigenvalues()

Out[36]:

$\newcommand{\Bold}[1]{\mathbf{#1}}\left[-3.000000000000005?, -1.999999999999994?, -1.000000000000002?\right]$

Verification of the closed-loop eigenvalues (as floating point numbers)

In [39]:

A0 = matrix(RDF,A-B*K0*C)
A0.eigenvalues()

Out[39]:

$\newcommand{\Bold}[1]{\mathbf{#1}}\left[-2.999999999999992, -2.0000000000000075, -0.9999999999999986\right]$