nonlinear variable state space observer pole placement

Question

Bernd Pfeifer 2022-11-16

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评论： Paul 2022-11-27

Hi,

i want to place some observer poles in dependency of an variable. The variable v is speed and it is changing with the time.

Example:

option

syms v

A = [-1 2; v -1]

c = [0 1]

p = [-2; -3];

so,

L = place(A',c',p).'; %doesnt work

-----------

2.option

syms v lambda l1 l2

L =[l1;l2];

solve((det(lambda * [1 0; 0 1] - (A - L*c)) == (lambda+2)*(lambda+3)),[l1 l2]); %matlab doesn't eliminate lambda

l1 = ?

l2 = ?

do i need to calculate the det matrix by myself or can matlab help me in a faster way?

Thanks and Regards

Bernd

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Answer 1

Paul 2022-11-16

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在 MATLAB Online 中打开

Hi Bernd,

Matlab assumes by default that all sym variables are complex, and this assumption can sometimes lead to unexpected results.

Your code works if we explicitly assume that v and the estimator gains are real (my experience is to include all relevant assumptions)

syms v l1 l2 real
syms lambda
A = [-1 2; v -1];
c = [0  1];
p = [-2; -3];
L = [l1;l2];
sol = solve((det(lambda * [1 0; 0 1] - (A - L*c)) == (lambda+2)*(lambda+3)),[l1 l2])         %matlab doesn't eliminate lambda
Warning: Solutions are only valid under certain conditions. To include parameters and conditions in the solution, specify the 'ReturnConditions' value as 'true'.
sol = struct with fields:
    l1: (2*v + 2)/v
    l2: 3

If v == 0, then A,C are not an observable pair, so we should exclude v == 0 from the analysis

assumeAlso(v ~= 0);
sol = solve((det(lambda * [1 0; 0 1] - (A - L*c)) == (lambda+2)*(lambda+3)),[l1 l2])
sol = struct with fields:
    l1: (2*v + 2)/v
    l2: 3

Or more compactly without hardcoding anything

sol = solve(det(lambda*eye(2)- (A - L*c)) == poly2sym(poly(p),lambda),L)
sol = struct with fields:
    l1: (2*v + 2)/v
    l2: 3

Or, using Ackerman's formula

alpha(lambda) = poly2sym(poly(p),lambda)

alpha(lambda) =

O = [c ; c*A];

(A^2 + 5*A + 6*eye(2))*inv(O)*[0;1]

ans =

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Paul 2022-11-17

编辑：Paul 2022-11-18

It still has lambdas in the solution because we now have an undertermined system, i.e., four equations with eight unknowns. So you'll need to either specify four of the observer gains a priori or include four other equations (or a combination of both) to uniquely determine the gains. This sounds easier than it is.

For state feedback controller design, there are methods for the multivariable case for partial eigenstructure assigment. By partial here I mean that there is never enough degrees of freedom to fully specify the closed-loop eigenvectors and the closed-loop eigenvalues, but we can specify the eigenvalues and then get eigenvectors that are close, in some sense, to desired eigenvectors. I see no reason why a similar technique couldn't work for observer design. There may be other methods as for muti-output observer design, such as specifying the design parameters for a Kalman filter and then taking the steady-state gains, for example. Or the algorithm in the Reference [1] of place. Of course, implementing any of these techniques symbolically may be a chore.

Bernd Pfeifer 2022-11-27

One question left.

Is there a reason why you chose the left eigenvectors for the assignement and not the right eigenvectors?

Regards

Bernd

Paul 2022-11-27

Assuming you mean the right eigenvectors of A - LC, I didn't see how that could work, because we need L (or transpose(L) ) to multiply V, but with a right eigvector we get a term LCV.

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Answer 2

Sam Chak 2022-11-16

3
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https://ww2.mathworks.cn/matlabcentral/answers/1853003-nonlinear-variable-state-space-observer-pole-placement#answer_1101863

编辑：Sam Chak 2022-11-16

在 MATLAB Online 中打开

Hi @Bernd Pfeifer

Long time I didn't solve math puzzles like this one. Generally, you should be able find the analytical solution for L that satisfies the condition of eigenvalues of

that gives

and

.

The following is an attempt to solve the problem heuristically for

. You can attempt for

.

v = 1:10;

c = [0 1];

p = [-2; -3];

for j = 1:length(v)

A = [-1 2; v(j) -1];

L = place(A', c', p) % always return L = [m 3]

m(j) = L(:, 1);

end

L = 1×2

4.0000 3.0000

L = 1×2

3.0000 3.0000

L = 1×2

2.6667 3.0000

L = 1×2

2.5000 3.0000

L = 1×2

2.4000 3.0000

L = 1×2

2.3333 3.0000

L = 1×2

2.2857 3.0000

L = 1×2

2.2500 3.0000

L = 1×2

2.2222 3.0000

L = 1×2

2.2000 3.0000

plot(v, m, 'p'), xlabel('v'), ylabel('L_1')

The pattern can be intuitively recognived as:

vv = linspace(1, 10, 901);

L1 = (4 + 2*(vv - 1))./vv;

plot(vv, L1), grid on, xlabel('v'), ylabel('L_1')

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Bernd Pfeifer 2022-11-17

Hi Sam,

thanks for the answer, but i want to implement the observer in simulink so i need one equation.

Paul already helped a lot and i think we get the problem in his way solved :)

Regards

Bernd

Sam Chak 2022-11-18

编辑：Sam Chak 2022-11-18

@Bernd Pfeifer, don't mention it. @Paul is a helpful and knowledgeable person in control design.

For implementation in Simulink, both equations for

are exactly the same (for your original 2nd-order system):

My equation came from my recognition of the numerical pattern as something related to the arithmetic sequence (without employing the curve-fitting tool), while Paul's equation is the solution as a direct result from his analytical mind and the computational power of MATLAB.

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