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By Bernard Brogliato

This moment variation of Dissipative platforms research and regulate has been considerably reorganized to house new fabric and increase its pedagogical positive factors. It examines linear and nonlinear structures with examples of either in each one bankruptcy. additionally incorporated are a few infinite-dimensional and nonsmooth examples. all through, emphasis is put on using the dissipative homes of a approach for the layout of good suggestions keep watch over legislation.

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Extra resources for Dissipative Systems Analysis and Control: Theory and Applications (Communications and Control Engineering)

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We have a passive physical system if the force is the input and the velocity is the output, and then a PD controller from position corresponds to PI controller from velocity. For this reason we might have referred to the controllers in this section as PI controllers for velocity control. We consider a mass m with position x(·) and velocity v(·) = x(·). ˙ The dynamics is given by m¨ x(t) = u(t) where the force u is the input. The desired position is xd (·), while the desired velocity is vd (·) = x˙ d (·).

S(s + p1 )(s + p2 ) . . 53) where Re[pi ] > 0 and Re[zi ] > 0 which means that h(s) has one pole at the origin and the remaining poles in Re [s] < 0, while all the zeros are in Re [s] < 0. Then the system with transfer function h (s) is passive if and only if Re[h(jω)] ≥ 0 for all ω ∈ [−∞, +∞]. 4 Linear Systems Proof: The residual of the pole on the imaginary axis is z1 z2 . . Ress=0 h(s) = p1 p2 . . 54) Here the constants zi and pi are either real and positive, or they appear in complex conjugated pairs where the products zi zi∗ = |zi |2 and pi p∗i = |pi |2 are real and positive.

Assume that there is a δ so that Re[h(jω)] ≥ δ > 0 and a γ so that |h (jω)| ≤ γ for all ω ∈ [−∞, +∞]. 88) 4δ . Next assume that and the result follows with 0 < γ < min 1, (γ+1) 2 g(jω)|2 ≤ 1 − γ for all ω. 9. 85): the association of the new system with transfer function g(s) merely corresponds to writing t down uy = 14 (a + b)(a − b) = 14 (a2 − b2 ). Thus if 0 u(s)y(s)ds ≥ 0 one gets t 2 t 2 a (s)ds ≥ 0 b (s)ds: the L2 -norm of the new output b(t) is bounded by 0 the L2 -norm of the new intput a(t).

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