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Example 2: Minimum Frobenius Norm Static Output Feedback¶

The example system has been taken from [1, Example 2]. The eigenvalue placement with minimum Frobenium norm is discussed in [2, Example 2]. The poles should be placed at -3, -3, -4, i.e., we want to have a double eigenvalue.

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 [29]:

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

Out[29]:

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

State space system

In [30]:

A = matrix(R,[[0, 1, 0],[19.62, 0, -8.86],[0, 0, -100]])
B = matrix(R,[[0, -1],[0, 1],[1, 0]])
C = matrix(R,[[1, 0, 2],[1, 1, 0]])
(A,B,C)

Out[30]:

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\left(\begin{array}{rrr} 0 & 1 & 0 \\ \frac{981}{50} & 0 & -\frac{443}{50} \\ 0 & 0 & -100 \end{array}\right), \left(\begin{array}{rr} 0 & -1 \\ 0 & 1 \\ 1 & 0 \end{array}\right), \left(\begin{array}{rrr} 1 & 0 & 2 \\ 1 & 1 & 0 \end{array}\right)\right)$

Symbolic feedback matrix

In [31]:

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

Out[31]:

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

Closed-loop characteristic polynomial

In [32]:

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

Out[32]:

$\newcommand{\Bold}[1]{\mathbf{#1}}2 k_{11} s^{2} - k_{21} s^{2} + s^{3} + \frac{461}{10} k_{12} k_{21} - \frac{461}{10} k_{11} k_{22} - \frac{443}{50} k_{12} s - 99 k_{21} s - \frac{931}{50} k_{22} s + 100 s^{2} - \frac{481}{10} k_{11} - \frac{443}{50} k_{12} + 100 k_{21} - 1862 k_{22} - \frac{981}{50} s - 1962$

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

In [33]:

q1,r1 = CP.quo_rem(s+3)
q2,r2 = q1.quo_rem(s+3)
q3,r3 = q2.quo_rem(s+4)
r1,r2,r3

Out[33]:

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\frac{461}{10} k_{12} k_{21} - \frac{461}{10} k_{11} k_{22} - \frac{301}{10} k_{11} + \frac{443}{25} k_{12} + 388 k_{21} - \frac{90307}{50} k_{22} - \frac{51507}{50}, -12 k_{11} - \frac{443}{50} k_{12} - 93 k_{21} - \frac{931}{50} k_{22} - \frac{29631}{50}, 2 k_{11} - k_{21} + 90\right)$

Lagrangian function

In [34]:

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

Out[34]:

$\newcommand{\Bold}[1]{\mathbf{#1}}k_{11}^{2} l_{0} - \frac{461}{10} k_{11} k_{22} l_{1} - \frac{301}{10} k_{11} l_{1} - 12 k_{11} l_{2} + 2 k_{11} l_{3} + k_{12}^{2} l_{0} + \frac{461}{10} k_{12} k_{21} l_{1} + \frac{443}{25} k_{12} l_{1} - \frac{443}{50} k_{12} l_{2} + k_{21}^{2} l_{0} + 388 k_{21} l_{1} - 93 k_{21} l_{2} - k_{21} l_{3} + k_{22}^{2} l_{0} - \frac{90307}{50} k_{22} l_{1} - \frac{931}{50} k_{22} l_{2} - l_{0} q - \frac{51507}{50} l_{1} - \frac{29631}{50} l_{2} + 90 l_{3} + q$

Neccessary optimility condition and associated polynomial ideal

In [35]:

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 - 461/10*k22*l1 - 301/10*l1 - 12*l2 + 2*l3
1  :  k12  :  2*k12*l0 + 461/10*k21*l1 + 443/25*l1 - 443/50*l2
2  :  k21  :  461/10*k12*l1 + 2*k21*l0 + 388*l1 - 93*l2 - l3
3  :  k22  :  -461/10*k11*l1 + 2*k22*l0 - 90307/50*l1 - 931/50*l2
4  :  l0  :  k11^2 + k12^2 + k21^2 + k22^2 - q
5  :  l1  :  -461/10*k11*k22 - 301/10*k11 + 461/10*k12*k21 + 443/25*k12 + 388*k21 - 90307/50*k22 - 51507/50
6  :  l2  :  -12*k11 - 443/50*k12 - 93*k21 - 931/50*k22 - 29631/50
7  :  l3  :  2*k11 - k21 + 90
8  :  q  :  -l0 + 1

Out[35]:

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(2 k_{11} l_{0} - \frac{461}{10} k_{22} l_{1} - \frac{301}{10} l_{1} - 12 l_{2} + 2 l_{3}, 2 k_{12} l_{0} + \frac{461}{10} k_{21} l_{1} + \frac{443}{25} l_{1} - \frac{443}{50} l_{2}, \frac{461}{10} k_{12} l_{1} + 2 k_{21} l_{0} + 388 l_{1} - 93 l_{2} - l_{3}, -\frac{461}{10} k_{11} l_{1} + 2 k_{22} l_{0} - \frac{90307}{50} l_{1} - \frac{931}{50} l_{2}, k_{11}^{2} + k_{12}^{2} + k_{21}^{2} + k_{22}^{2} - q, -\frac{461}{10} k_{11} k_{22} - \frac{301}{10} k_{11} + \frac{461}{10} k_{12} k_{21} + \frac{443}{25} k_{12} + 388 k_{21} - \frac{90307}{50} k_{22} - \frac{51507}{50}, -12 k_{11} - \frac{443}{50} k_{12} - 93 k_{21} - \frac{931}{50} k_{22} - \frac{29631}{50}, 2 k_{11} - k_{21} + 90, -l_{0} + 1\right)\Bold{Q}[k_{11}, k_{12}, k_{21}, k_{22}, s, l_{0}, l_{1}, l_{2}, l_{3}, q]$

Elimination ideal

In [36]:

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

Out[36]:

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(743984251427871486920144275832068491488281250 q^{4} - 6684322385321726860643482367467620059165725453125 q^{3} + 22013416240139431256173076827447995890610917423698875 q^{2} - 31706669025491173284411110344950631901748502702664953888 q + 16933085437845849430221649184469909408834771791804607019548\right)\Bold{Q}[k_{11}, k_{12}, k_{21}, k_{22}, s, l_{0}, l_{1}, l_{2}, l_{3}, q]$

Computation of possible solutions w.r.t. q

In [37]:

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

Out[37]:

$\newcommand{\Bold}[1]{\mathbf{#1}}\left[\left(2038.25815459246, 1\right), \left(3278.95026577783, 1\right)\right]$

Selection of the minimum solution q

In [38]:

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

Out[38]:

$\newcommand{\Bold}[1]{\mathbf{#1}}2038.25815459246$

Dimension of the polynomial ideal

In [39]:

I.dimension()

Out[39]:

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

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

In [40]:

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

Out[40]:

$\newcommand{\Bold}[1]{\mathbf{#1}}\left[\verb|{q:|\phantom{\verb!x!}\verb|2038.258154592421?,|\phantom{\verb!x!}\verb|l3:|\phantom{\verb!x!}\verb|41.95161902174205?,|\phantom{\verb!x!}\verb|l2:|\phantom{\verb!x!}\verb|-0.5492155735055070?,|\phantom{\verb!x!}\verb|l1:|\phantom{\verb!x!}\verb|-0.03108165258292408?,|\phantom{\verb!x!}\verb|l0:|\phantom{\verb!x!}\verb|1,|\phantom{\verb!x!}\verb|k22:|\phantom{\verb!x!}\verb|-0.8847562529184268?,|\phantom{\verb!x!}\verb|k21:|\phantom{\verb!x!}\verb|-0.1616471218549174?,|\phantom{\verb!x!}\verb|k12:|\phantom{\verb!x!}\verb|-2.273450734426870?,|\phantom{\verb!x!}\verb|k11:|\phantom{\verb!x!}\verb|-45.08082356092746?}|, \verb|{q:|\phantom{\verb!x!}\verb|3278.950265777833?,|\phantom{\verb!x!}\verb|l3:|\phantom{\verb!x!}\verb|283.4279275257919?,|\phantom{\verb!x!}\verb|l2:|\phantom{\verb!x!}\verb|-6.706680673787214?,|\phantom{\verb!x!}\verb|l1:|\phantom{\verb!x!}\verb|-0.3488968102848198?,|\phantom{\verb!x!}\verb|l0:|\phantom{\verb!x!}\verb|1,|\phantom{\verb!x!}\verb|k22:|\phantom{\verb!x!}\verb|-35.61327065206581?,|\phantom{\verb!x!}\verb|k21:|\phantom{\verb!x!}\verb|4.971119105352403?,|\phantom{\verb!x!}\verb|k12:|\phantom{\verb!x!}\verb|13.35872552049407?,|\phantom{\verb!x!}\verb|k11:|\phantom{\verb!x!}\verb|-42.51444044732380?}|, \verb|{q:|\phantom{\verb!x!}\verb|1833.643019842273?|\phantom{\verb!x!}\verb|-|\phantom{\verb!x!}\verb|207.9262264607811?*I,|\phantom{\verb!x!}\verb|l3:|\phantom{\verb!x!}\verb|22.92645275271472?|\phantom{\verb!x!}\verb|-|\phantom{\verb!x!}\verb|48.79668803123861?*I,|\phantom{\verb!x!}\verb|l2:|\phantom{\verb!x!}\verb|1.806582640038055?|\phantom{\verb!x!}\verb|-|\phantom{\verb!x!}\verb|0.4763577215930209?*I,|\phantom{\verb!x!}\verb|l1:|\phantom{\verb!x!}\verb|0.1972647661083527?|\phantom{\verb!x!}\verb|+|\phantom{\verb!x!}\verb|0.0661897043039494?*I,|\phantom{\verb!x!}\verb|l0:|\phantom{\verb!x!}\verb|1,|\phantom{\verb!x!}\verb|k22:|\phantom{\verb!x!}\verb|-10.044949010449721?|\phantom{\verb!x!}\verb|-|\phantom{\verb!x!}\verb|6.656486438006417?*I,|\phantom{\verb!x!}\verb|k21:|\phantom{\verb!x!}\verb|0.7271754583177050?|\phantom{\verb!x!}\verb|+|\phantom{\verb!x!}\verb|2.685324786258234?*I,|\phantom{\verb!x!}\verb|k12:|\phantom{\verb!x!}\verb|7.045889413219074?|\phantom{\verb!x!}\verb|-|\phantom{\verb!x!}\verb|16.01618243384714?*I,|\phantom{\verb!x!}\verb|k11:|\phantom{\verb!x!}\verb|-44.63641227084115?|\phantom{\verb!x!}\verb|+|\phantom{\verb!x!}\verb|1.342662393129117?*I}|, \verb|{q:|\phantom{\verb!x!}\verb|1833.643019842273?|\phantom{\verb!x!}\verb|+|\phantom{\verb!x!}\verb|207.9262264607811?*I,|\phantom{\verb!x!}\verb|l3:|\phantom{\verb!x!}\verb|22.92645275271472?|\phantom{\verb!x!}\verb|+|\phantom{\verb!x!}\verb|48.79668803123861?*I,|\phantom{\verb!x!}\verb|l2:|\phantom{\verb!x!}\verb|1.806582640038055?|\phantom{\verb!x!}\verb|+|\phantom{\verb!x!}\verb|0.4763577215930209?*I,|\phantom{\verb!x!}\verb|l1:|\phantom{\verb!x!}\verb|0.1972647661083527?|\phantom{\verb!x!}\verb|-|\phantom{\verb!x!}\verb|0.0661897043039494?*I,|\phantom{\verb!x!}\verb|l0:|\phantom{\verb!x!}\verb|1,|\phantom{\verb!x!}\verb|k22:|\phantom{\verb!x!}\verb|-10.044949010449721?|\phantom{\verb!x!}\verb|+|\phantom{\verb!x!}\verb|6.656486438006417?*I,|\phantom{\verb!x!}\verb|k21:|\phantom{\verb!x!}\verb|0.7271754583177050?|\phantom{\verb!x!}\verb|-|\phantom{\verb!x!}\verb|2.685324786258234?*I,|\phantom{\verb!x!}\verb|k12:|\phantom{\verb!x!}\verb|7.045889413219074?|\phantom{\verb!x!}\verb|+|\phantom{\verb!x!}\verb|16.01618243384714?*I,|\phantom{\verb!x!}\verb|k11:|\phantom{\verb!x!}\verb|-44.63641227084115?|\phantom{\verb!x!}\verb|-|\phantom{\verb!x!}\verb|1.342662393129117?*I}|\right]$

Selection of the solution to be used

In [41]:

lx = sol[1]
lx

Out[41]:

$\newcommand{\Bold}[1]{\mathbf{#1}}\verb|{q:|\phantom{\verb!x!}\verb|3278.950265777833?,|\phantom{\verb!x!}\verb|l3:|\phantom{\verb!x!}\verb|283.4279275257919?,|\phantom{\verb!x!}\verb|l2:|\phantom{\verb!x!}\verb|-6.706680673787214?,|\phantom{\verb!x!}\verb|l1:|\phantom{\verb!x!}\verb|-0.3488968102848198?,|\phantom{\verb!x!}\verb|l0:|\phantom{\verb!x!}\verb|1,|\phantom{\verb!x!}\verb|k22:|\phantom{\verb!x!}\verb|-35.61327065206581?,|\phantom{\verb!x!}\verb|k21:|\phantom{\verb!x!}\verb|4.971119105352403?,|\phantom{\verb!x!}\verb|k12:|\phantom{\verb!x!}\verb|13.35872552049407?,|\phantom{\verb!x!}\verb|k11:|\phantom{\verb!x!}\verb|-42.51444044732380?}|$

Numerical feedback gain matrix

In [42]:

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

Out[42]:

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rr} -42.5144404473238 & 13.3587255204941 \\ 4.97111910535240 & -35.6132706520658 \end{array}\right)$

Frobenius norm of the computed feedback matrix

In [43]:

K0.norm('frob')

Out[43]:

$\newcommand{\Bold}[1]{\mathbf{#1}}57.26211894243726$

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

In [44]:

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

Out[44]:

$\newcommand{\Bold}[1]{\mathbf{#1}}\left[-4.000000000010891?, -2.999999999994561? - 3.33366?e-6 \sqrt{-1}, -2.999999999994561? + 3.33366?e-6 \sqrt{-1}\right]$

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

In [45]:

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

Out[45]:

$\newcommand{\Bold}[1]{\mathbf{#1}}\left[-4.000000000015731, -2.9999999999921556 + 4.019993690152769 \times 10^{-06}i, -2.9999999999921556 - 4.019993690152769 \times 10^{-06}i\right]$

Verification of the closed-loop eigenvalues: Real parts of numerically computed eigenvlues

In [46]:

vector([x.real() for x in v])

Out[46]:

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(-4.000000000015731,\,-2.9999999999921556,\,-2.9999999999921556\right)$