Detailed proof of the Lemma 1 used in the manuscript " Discontinuous model recovery anti-windup solution for image based visual servoing " submitted to Automatica

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Detailed proof of the Lemma 1 used in the manuscript ” Discontinuous model recovery anti-windup solution for image based visual servoing ” submitted to Automatica

Laurent Burlion, Luca Zaccarian, Henry de Plinval, Sophie Tarbouriech

To cite this version:

Laurent Burlion, Luca Zaccarian, Henry de Plinval, Sophie Tarbouriech. Detailed proof of the Lemma 1 used in the manuscript ” Discontinuous model recovery anti-windup solution for image based visual servoing ” submitted to Automatica. [Research Report] Rapport LAAS n° 17378, ONERA; LAAS;

University of Trento. 2017. �hal-01593178�

Detailed proof of the Lemma 1 used in the manuscript “Discontinuous model recovery anti-windup solution for image based visual servoing” [1]

September 25, 2017

As mentioned in [1], the proof of technical Lemma 1 arises from brute force calculations of all possible cases. The details were omitted due to page restrictions. The proof of this result is reported below:

Lemma 1 For any selection of (˜x, x) = (˜p,v, p, v)˜ ∈R⁴, it holds that

δp∈Υ(˜maxx,x)vδp≤k_p^gλm

λ0

2|˜x||v|+|˜x|²

+ 3¯up|˜x|. (1)

To better follow the proof of Lemma 1, it is important to keep in mind the definition of MI (defined in eq. (11),[1]) plus the following basic properties of the saturation function used in [1]: given ¯u_p >0,

• (P1): satu¯p is 1-Lipschitz

• (P2): ∀x,|sat_u¯p(x)| ≤u¯p

• (P3): ∀k_p^g >0, ζ >0, ∀p6= ˜p,

satu¯p(k^gpζp)−satu¯p(k^gpζp)˜

p−p˜ ≥0 (2)

• (P4): ∀χ >0,∀x, y then

xy ≥0 =⇒ysatu¯p(χx)≥0 (3)

xy ≤0 =⇒ysatu¯p(χx)≤0 (4)

Proof. Given any pair (˜x, x) = (˜p,˜v, p, v), consider set Υ(˜x, x) defined in eq. (44),[1]. In particular, consider any selection of ζ ∈ MI(pv), of ζaw ∈ MI((p−p)(v˜ −v)) and any˜ µ ∈ [λ0, λm]. (From the definition ofM_I and eq. (4),[1], it is clear that ζ, ζ_aw, µ >0).

Then denote:

ψ:=v satu¯p(k^g_pζµp)−satu¯p(k^g_pζawµ(p−p))˜

(5) wherek^gp >0.

The proof of the lemma amounts to showing that ψ≤k_p^gλm

λ0

2|˜x||v|+|˜x|²

+ 3¯up|˜x| (6)

This is done by way of a lengthy but simple study of five possible cases:

1. Suppose

(p−p)(v˜ −v)˜ <0 & pv ≥0 (7)

In this case,ζ_aw= _λ¹

m and (5) develops as follows:

ψ = v

satu¯p(k^g_pζµp)−sat¯up

k_p^g µ

λ_m(p−p)˜

(8)

≤ v

sat_u¯p

k_p^g µ λ_mp

−sat_u¯p

k_p^g µ

λ_mu(p−p)˜

+vsatu¯p(k_p^gζµp) (9)

≤ k_p^gλm

λ0

|˜x||v|+vsat¯up(k_p^gζµp) (10) where the penultimate inequality was obtained using (P4), i.e 0≤vsat_u¯p

k^gp µ λmp

. where the last inequality was obtained using (P1)and 0< _λ^µ

m ≤1< ^λ_λ^m

0. Moreover, it is readily seen that (p−p)(v˜ −v)˜ <0 and (P3)imply:

sat_u¯p k^g_pζµp

−sat_u¯p k^g_pζµp˜

(v−v)˜ ≤0 (11)

Thus,

0≤vsatu¯p k_p^gζµp

≤ satu¯p k_p^gζµ˜p

(v−v)˜ +sat¯up k^g_pζµp

˜

v (12)

Rewriting (7) as follows

0≤pv <p(v˜ −v) +˜ p˜v (13)

and combining (10),(12),(13) one successively gets (using(P1) and(P2)):

ψ ≤ k^g_pλ_m

λ₀ |˜x||v|+k_p^gλ_m

λ₀|˜p(v−˜v)|+ ¯u_p|˜v| (14)

≤ k^g_pλ_m

λ₀ |˜x||v|+k_p^gλ_m

λ₀|˜pv|+k^g_pλ_m

λ₀|˜p˜v|+ ¯u_p|˜v| (15)

≤ 2k_p^gλ_m λ0

|˜x||v|+k^g_pλ_m 2λ0

|˜x|²+ ¯u_p|˜x| (16) which means that (6) is satisfied in case (7).

2. Suppose

(p−p)(v˜ −v)˜ <0 & pv <0 (17) In this case,ζ = _λ¹

m and (5) develops as follows:

ψ ≤ v

sat_u¯p

k_p^g µ

λ_mp

−sat_u¯p

k_p^g µ

λ_m(p−p)˜

(18)

≤ k_p^g µ λm

|v||p| ≤˜ k_p^g|˜x||v| ≤k^g_pλ_m λ0

|˜x||v| (19)

Where the second inequality was obtained using property (P1). This means that (6) is satisfied in case (17).

3. Suppose

(p−p)(v˜ −v)˜ >0 & pv ≤0 (20)

In this case,ζ_aw= _λ¹

0 and (5) develops as follows

ψ = v

satu¯p(k^g_pζµp)−satu¯p

k_p^g µ

λ0

(p−p)˜

(21)

≤ 0−vsat_u¯p

k^g_p µ

λ₀(p−p)˜

(22)

≤ −(v−˜v)sat_u¯p

k^g_p µ

λ₀(p−p)˜

−˜vsat_u¯p

k_p^g µ λ0

(p−p)˜

(23)

≤ 0 + ¯u_p|˜v| ≤u¯_p|˜x| (24)

where the first inequality is obtained using (P4)

where the penultimate inequality is obtained using (P2) From the last inequality, (6) is thus satisfied in case (20).

4. Suppose

(p−p)(v˜ −v)˜ >0 & pv >0 (25) In this case,ζ =ζaw= _λ¹

0 and (5) develops as follows

ψ = v

sat_u¯p

k^g_p µ

λ₀p

−sat_u¯p

k_p^g µ

λ₀(p−p)˜

(26)

≤ k_p^gµ|vp|˜ λ0

≤k^g_pλ_m λ0

|˜x||v| (27)

which means that (6) is satisfied in case (25).

5. Suppose

(p−p)(v˜ −v) = 0˜ (28)

In this case, (5) develops as follows:

ψ = v(satu¯p(k^g_pζµp)−satu¯p(k^g_pζawµ(p−p)))˜ (29)

= vsat_u¯p(k_p^gζµp)−vsat˜ _u¯p(k^g_pζ_awµ(p−p))˜ (30)

≤ vsat_u¯p(k_p^gζµp) + ¯u_p|˜v| (31) Moreover, it is readily seen that (p−p)(v˜ −v) = 0 implies that˜ p= ˜p orv= ˜v, which implies:

(sat¯up(k_p^gζµp)−sat¯up(k_p^gζµ˜p))(v−v) = 0˜ (32) Combining (31) and (32), one successively gets:

ψ ≤ ˜vsatu¯p(k^g_pζµp) + satu¯p(k^g_pζµ˜p)(v−v) + ¯˜ up|˜v|

(33)

≤ u¯_p|˜x|+ sat_u¯p(k_p^gζµp)v˜ +|sat_u¯p(k_p^gζµp)˜˜v|+ ¯u_p|˜v|

(34)

≤ k_p^gλm

λ0

|˜x||v|+ 3¯up|˜x| (35)

where the last inequality is obtained using(P1)and(P2). which means that (6) is satisfied in case

References

[1] L. Burlion, L. Zaccarian, H. de Plinval and S. Tarbouriech”Discontinuous model recovery anti-windup solution for image based visual servoing”, submitted to Automatica, 2017.