Multi-Output Partial Nonlinear Observer Normal Form

Multi-Output Partial Nonlinear Observer Normal Form D. Boutat, with W. Saadi, G. Zheng and L. Sbita INSA-CVL PRISME

18 December 2015

D. Boutat INSA-CVL

54th IEEE CDC December 15-18, 2015, Osaka, Japan

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Problem statement Let us consider the following multi-inputs outputs dynamical system :

x˙ = f (x) + g(x)u y = h(x) x denotes the variable state, y = (y1 , ..., ym ) is the output and u represent the outputs. We assume that is partially observable. Find a set of geometrical conditions that guarantee a split into two dynamical subsystems : one of them is a nonlinear observer normal form and the other involves all state variables.

D. Boutat INSA-CVL


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Problem statement We will give necessary and sufficient conditions which guarantee the existence of a change of coordinates φ(x) = (ξ T , ζ T )T where ξ = (ξ1 , ·, ξm )T with ξi = (ξi,1 , · · · , ξi,ri )T for 1 ≤ i ≤ m and ζ = (ζ1 , · · · , ζn−r )T such that system can be transformed into the following two subsystems : ξ˙i = Ai ξi + βi (y) ζ˙ = η(ξ, ζ)

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y¯i = Ci ξi = ξi,ri = yi + ϕ(y1 , ..., yi−1 )

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where Ai and Ci are Brunovsky’s forms :  0  1  Ai =  .  ..

respectively the (ri × ri ) and (1 × ri ) well-known ··· ··· .. .

0 ··· D. Boutat INSA-CVL

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 0 0 0 0   .. ..  , Ci = . .  1 0

0, · · · , 0 1


.

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Problem statement

It is obvious that we can design a Luenberger-like observer for each subsystems of (1) as follows : ˙ ξî = Ai ξî + βi (y) + Ki (yi − Ci ξî ),

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where Ki is the gain which was chosen such that (Ai − Ki Ci ) is Hurwitz. Consequently , if we set ei = ξi − ξî , we obtain the following observation error dynamics : e˙ i = (Ai − Ki Ci ) ei ,

i = 1, ..., m

and it is clear that they are exponentially stable.

D. Boutat INSA-CVL


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Former works on the topic

This problem is solved for the case of single output in : Partial observer normal form for nonlinear system Automatica, Volume 64, February 2016, Pages 54-62 Ramdane Tami, Gang Zheng, Driss Boutat, Didier Aubry, Haoping Wang. For a particular form is solved in Jo, H. N., Seo, H. J. (2002). Observer design for non-linear systems that are not uniformly observable. International Journal of Control, 75(5), 369380.

D. Boutat INSA-CVL


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Observability Differential 1-Forms

We associate to x˙ = f (x) + g(x)u y = h(x) the observability 1-forms the following exact differential 1-formes : θj,k = dLk−1 f hj . And we define the co-distribution ∆ = span {θj,k ; j = 1, ..., m, kP = 1, ..., rj } k We assume that r = rank∆ = m i=1 ri ≤ n, and that dLf hj ∈ ∆ for any k.

D. Boutat INSA-CVL


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Some properties of the observability co-distribution ∆ and its kernel For system under study, we have

Lemma 1 1 2

3

the co-distribution ∆ and its kernel ∆> are involutive ; the co-distribution ∆ and the distribution ∆> are both invariant with respect to the vector field f , i.e. Lf ν ∈ ∆ for any 1-form ν ∈ ∆ and [f, H] ∈ ∆> for any vector field H ∈ ∆> ; there exist (n − r) vector fields, noted as {τr+1 , · · · , τn }, that span ∆> and commute, i.e. [τi , τj ] = 0 for all r + 1 ≤ i, j ≤ n.

D. Boutat INSA-CVL


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Observability frame

Now, we define a family of vector fields (τi,1 )1≤i≤m as one solution of the following algebraic equations : θi,ri (τi,1 ) = 1 for 1 ≤ i ≤ m

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θi,k (τi,1 ) = 0 for 1 ≤ k ≤ ri − 1

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θj,k (τi,1 ) = 0 for j < i and 1 ≤ k ≤ rj

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θj,k (τi,1 ) = 0 for j > i and1 ≤ k ≤ rj

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where θj,k = dLk−1 f hj .

D. Boutat INSA-CVL


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Observability frame For each family of vector fields (τi,1 )1≤i≤m , can yield r independent vector fields, then τ = (τi,j ; 1 ≤ i ≤ m, 1 ≤ j ≤ ri ) can be calculated by induction as follows : τi,k = −adf τi,k−1 for 1 ≤ i ≤ m and 2 ≤ k ≤ ri

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where adf τi,k−1 = [f, τi,k−1 denotes the Lie bracket of f with τi,k−1 . This family of vector fields, together with the family τk , k = r + 1 : n given by item 3) of Lemma, form a basis of the tangent bundle T X of X . Consider the following co-distributions defined for i = 1 : m by n o ri −1 ∆i = span dLk−1 h \dL h , for 1 ≤ i, j ≤ m, 1 ≤ k ≤ r (10) j i i f f where \ means to remove the term behind it.(see Xia and Gao, Krener and Respondek D. Boutat INSA-CVL


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Some properties of the observability frame

Lemma 2 Given a family of τi,j deduced by the above equations for 1 ≤ i ≤ m and 1 ≤ j ≤ ri , then for any H ∈ ∆> we have [τi,j , H] ∈ ∆> .

Corollary Given a family of τi,j given above for 1 ≤ i ≤ m and 1 ≤ j ≤ ri , then we have [τi,k + h1 , τj,s + h2 ] = [τi,k , τj,s ] modulo ∆> (11) for any Hi ∈ ∆> and Hj ∈ ∆> .

D. Boutat INSA-CVL


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Main result Theorem There exists a local change of coordinates (ξ T , ζ T )T = φ(x) which transforms system into the partially nonlinear observer normal form if and only if the following conditions are satisfied : 1

there exists a family vector fields τi,1 for 1 ≤ i ≤ m, such that : [τi,k , τj,s ] = 0 modulo ∆>

2

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for all 1 ≤ i ≤ m, 1 ≤ k ≤ ri we have [τi,k , τj ] = 0

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where the vector fields τj were deduced in the item 3) of Lemma 1 ; 3

for 1 ≤ i, j ≤ m and ri > rj + 1 we have r +k

dLfj D. Boutat INSA-CVL

hj (τi,1 ) = 0


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Example Let us consider the following nonlinear system :  x˙ 1 = −x23 + x31 x3 − 21 x33 + x52 − 5x54     x˙ 2 = x1 − 21 x23    x˙ 3 = −x3 + x31 − 21 x23 x˙ 4 = −2x4 − 21 x22      x˙ = −x5 − x21 + x24 − 21 x23   5 y1 = x2 and y2 = x4

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A simple calculation gives rank {dh1 , dLf h1 , dh2 } = 3, thus h1 has index r1 = 2 and the index of h2 is r2 = 1, and one has θ1,1 = dx2 , θ1,2 = dx1 − x3 dx3 and θ2,1 = dx4 One obtainsn∆ = span {θ1,1 , θ1,2 o , θ2,1 }, and ∂ ∂ ∂ ∆> = span x3 ∂x1 + ∂x3 , ∂x5 . Set τ4 = x3 ∂x∂ 1 and τ5 = D. Boutat INSA-CVL


∂ ∂x5

commute. 12 / 16

Example Now, let us take the following possible solutions of (5-8) : ∂ ∂ ∂ ∂ τ¯1,1 = + q1 (x) x3 + + q2 (x) . ∂x1 ∂x1 ∂x3 ∂x5 Now we consider the following vector field ∂ = τ¯1,1 modulo ∆> . τ1,1 = ∂x1 Using this vector field to compute ∂ ∂ ∂ ∂ 2 + 3x1 x3 + − 2x1 . τ¯1,2 = [τ1,1 , f ] = ∂x2 ∂x1 ∂x3 ∂x5 Then we take again τ1,2 =

∂ = τ¯1,2 modulo ∆> . ∂x2

Now, we take τ2,1 = D. Boutat INSA-CVL

∂ modulo ∆> . ∂x4


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Example

It is easy to see that τ1,1 , τ1,2 and τ2,1 commute twos. Now, we will compute Λ. To compute the coordinate change, we consider the following matrix ! Λ=

θ1,1 τ1,1 θ1,2 τ1,1 θ2,1 τ1,1

θ1,1 τ1,2 θ1,2 τ1,2 θ2,1 τ1,2

θ1,1 τ2,1 θ1,2 τ2,1 θ2,1 τ2,1

!

Λ=

0 1 0

which gives

D. Boutat INSA-CVL

1 0 0

0 0 1


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Example

Now, we have dξ1,1 = ω1,1 = dx1 − x3 dx3 , dξ1,2 = ω1,2 = dx2 and dξ2,1 = ω2,1 = dx4 . Therefore, by integration we obtain the change of coordinates you are looking for : ξ1,1 = x1 − 21 x23 , ξ1,2 = x2 and ζ1 = x3 and ζ2 = x5 . This yields to the following normal form : ξ˙1,1 = y15 − 5y25 ξ˙1,2 = ξ1,1 ξ˙2,1 = −2y2 − 21 y12 ζ˙1 = −ζ1 + (ξ1,1 + 21 ζ12 )3 − 21 ζ12 2 − 1 ζ2 ζ˙2 = −ζ2 − (ξ1,1 + 12 ζ12 )2 + ξ2,1 2 1 y1 = ξ1,2 and y2 = ξ2,1

D. Boutat INSA-CVL


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Thank you very much for your attention.

D. Boutat INSA-CVL


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