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Rapport de stage - Master 2 SAR ATIAM - Base des articles ...

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6.2.2 Recherche <strong>de</strong>s valeurs propresOn posePar définition,F(θ,Υ) = φ 1(θ 2 )−φ 1 (Υ)φ 2 (Υ)⎧⎪ ⎨⎪ ⎩σ l = √ Υσ r = √ Υ( √Υ )R r −R l cosh( √Υ ) ,R l sinh( √Υ )R l −R r cosh( √Υ ) .R r sinh(59)(60)On en déduit ⎧⎪ ⎨⎪ ⎩σ l =σ r =e θsinh(θ)−Φ(Υ) = F(θ,Υ)+φ 2 (Υ) φ 2 (Υ) ,e −θ(61)sinh(θ)−Φ(Υ) = F(θ,Υ)−φ 2 (Υ) φ 2 (Υ) .En injectant (61) dans (58), on obtient une nouvelle expression <strong>de</strong> la matrice Q(s), uniquementen fonction <strong>de</strong>s paramètres θ et Υ.(Q 1,1 = φ 1 Γ2 ) (+ F(θ,Υ)+ sinh(θ) )(φ 2 Γ2 )( φ 2 (Υ)Q 1,2 = −φ 2 Γ2 )(Q 2,1 = −2F(θ,Υ)φ 1 Γ2 ) ( (sinh(θ) ) )2+ −F(θ,Υ) 2 −Γ 2 (φ 2 Γ2 )φ 2 (Υ)(Q 2,2 = φ 1 Γ2 ) (+ F(θ,Υ)− sinh(θ)φ 2 (Υ)On vérifie bien que l’on a encoreLe polynôme caractéristique associé à cette matrice est)φ 2(Γ2 ) (62)<strong>de</strong>t(Q) = 1 (63)P(λ) = λ 2 −2b ′ λ+1 (64)avec b ′ = φ 1(Γ2 ) +F(θ,Υ)φ 2(Γ2 ) (65)et les valeurs propres <strong>de</strong> la matrice <strong>de</strong> transfert acoustique s’écrivent⎛ √ ⎞λ r = b ′ b⎝1−′2 −1⎠b ′ 2et⎛ √ ⎞λ l = b ′ b⎝1+′2 −1⎠b ′ 2.(66)Comme <strong>de</strong>tQ = 1, on sait que λ l = λ −1 r . On notera par la suite λ = λ r . Il existe un matrice <strong>de</strong>passage P telle que[ ] λ 0Q = P0 λ −1 P −1 . (67)23

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