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Control Systems Design, SC42000 / EE4C04 Lecture 2 SC42000/EE4C04, Ton van den Boom, DCSC, TU Delft Subscribe to view the full document.

Lecture 2 Solving Differential Equations (4.1) Phase portraits (Sec. 4.2) Stability (Sec. 4.3) Lyapunov stability analysis (4.4) Parametric and nonlocal behavior (4.5) System properties (Sec. 5.1 + slides) Matrix exponential (Sec. 5.2) Lecture 2 SC42000/EE4C04, Ton van den Boom, DCSC, TU Delft 1 Solving Differential Equations (4.1) Consider a system with feedback: ˙ x = f ( x, u ) u = α ( x ) bracerightbigg ˙ x = f ( x, α ( x )) = F ( x ) Find solution for ˙ x = F ( x ) , x ( t 0 ) = x 0 , for all t 0 < t < t f Problems: Existence and uniqueness Lecture 2 SC42000/EE4C04, Ton van den Boom, DCSC, TU Delft 2 Subscribe to view the full document.

Example 1: ˙ x = x 2 , with x (0) = 1 solution: x ( t ) = 1 1 t Finite escape time. Example 2: ˙ x = x , with x (0) = 0 solution: x ( t ) = braceleftbigg 0 if 0 t a ( t a ) 2 if t > a Non-uniqueness. Existence and uniqueness can be guaranteed by Lipschitz continuity: c : bardbl F ( x ) F ( y ) bardbl < c bardbl x y bardbl Lecture 2 SC42000/EE4C04, Ton van den Boom, DCSC, TU Delft 3 Qualitative Analysis of ODE: phase portraits (4.2) Consider planar case, x R 2 Plot with the two states on the axis. By studying properties of vector field, build phase portrait Lecture 2 SC42000/EE4C04, Ton van den Boom, DCSC, TU Delft 4 Subscribe to view the full document.

Qualitative Analysis of ODE: phase portraits Examples: braceleftbigg ˙ x = y ˙ y = x braceleftbigg ˙ x = 0 . 1 x y ˙ y = x braceleftbigg ˙ x = 0 . 1 x y ˙ y = x + y Lecture 2 SC42000/EE4C04, Ton van den Boom, DCSC, TU Delft 5 Qualitative Analysis of ODE: phase portraits Multiple equilibria: Pendulum: ˙ x = bracketleftbigg x 2 sin x 1 c x 2 + u cos x 1 bracketrightbigg , x 1 = θ , x 2 = ˙ θ Lecture 2 SC42000/EE4C04, Ton van den Boom, DCSC, TU Delft 6 Subscribe to view the full document.

Qualitative Analysis of ODE: phase portraits Limit cycle: The solutions in the phase plane converge to a circular trajectory. In the time domain this corresponds to an oscillatory solution. Example: ˙ x = y + x (1 x 2 y 2 ) , ˙ y = x + y (1 x 2 y 2 ) Lecture 2 SC42000/EE4C04, Ton van den Boom, DCSC, TU Delft 7 Stability (4.3) Consider autonomous system: ˙ x = f ( x ) with initial state x 0 and trajectory x ( t, x 0 ) t . Many different “kinds.” Most relevant for this class BIBO (bounded-input-bounded-output stability) Lyapunov stability asymptotical Lyapunov stability local asymptotical Lyapunov stability Lecture 2 SC42000/EE4C04, Ton van den Boom, DCSC, TU Delft 8 Subscribe to view the full document.

Stability x e is an equilibrium point for ˙ x = f ( x ) if f ( x e ) = 0 (Note that for linear ODE the origin is an equilibrium point) An equilibrium point x e is Lyapunov stable if ǫ > 0 , δ > 0 : bardbl x 0 x e bardbl < δ → bardbl x ( t, x 0 ) x e bardbl < ǫ, t 0 An equilibrium point x e is asymptotically stable if 1. it is Lyapunov stable 2. x ( t, x 0 ) x e , as t → ∞ pay attention to: local vs global validity of the above notions (e.g., can quantify domains of attraction ) Lecture 2 SC42000/EE4C04, Ton van den Boom, DCSC, TU Delft 9 Lyapunov Stability Analysis (4.4) Let x R n . A function V ( x ) is called positive (semi)definite in R n if V ( x ) > 0 ( V ( x ) 0) for all x R n with Subscribe to view the full document. • Fall '19
• Stability theory, Lyapunov stability, T0, TU Delft

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