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306                                    Autonomous Mobile Robots

                                Then the system (8.22) becomes


                                            ˙ ˜ η = Q( ˜γ)µ = Q( ˜γ)E −1 ( ˜γ)R( ˜ θ)  v d  =  (˜η, v d )  (8.25)
                                                            ¯
                                                                        0

                                                              ˙
                                   It is easy to check that the system ˜η =  (˜η, v d ) in (8.25) has an equilibrium
                                at ˜η = 0 and a linear approximation as follows [18]

                                                       ∂ (˜η, v d )
                                                  ˙
                                                  ˜ η =             ˜ η = A(v d )˜η       (8.26)
                                                          ∂ ˜η
                                                                 ˜ η=0
                                   If (8.26) is exponentially stable in the neighborhood of ˜η = 0, then (8.25)
                                is uniformly asymptotically stable. Furthermore, for a linear system like that
                                in (8.26), the condition to be exponentially stable is that all eigenvalues have
                                negative real part. With some calculations, it is straightforward to yield


                                                                    1
                                                                           
                                                         0
                                                                    a
                                                                           
                                               A(v d ) =        
             v d       (8.27)
                                                         1    1  1 + f   1  
                                                        −    −         +
                                                          lp   p    2l    a
                                with eigenvalues
                                                            
              √
                                                       v d    1 + f
                                                λ 1,2 =    −       a + l ±                (8.28)
                                                      2alp      2


                                                          1 + f
                                                        
            2
                                                      =        a + l  − 4alp
                                                            2
                                   With these eigenvalues, we have


                                                                  v 2 d
                                                           λ 1 λ 2 =
                                                                  alp




                                        λ 1 + λ 2   v d  1 + f          fv d  1 + f
                                               =−            a + l =−            a + fl
                                           2       2alp   2            2alp    2



                                 © 2006 by Taylor & Francis Group, LLC



                                FRANKL: “dk6033_c008” — 2006/3/31 — 16:43 — page 306 — #12
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