Nonlinear Control Lecture 7: Passivity Farzaneh Abdollahi Department of Electrical Engineering Amirkabir University of Technology Fall 2011 Farzaneh Abdollahi Nonlinear Control Lecture 7 1/26
Passivity Definition State Model L 2 and Lyapunov Stability Feedback and Passivity Theorems Feedback and L 2 Stability Feedback and A.S Farzaneh Abdollahi Nonlinear Control Lecture 7 2/26
Passivity Definition Consider a memoryless function where h : [0, ) R p R p u: input; y: output y = h(t, u) (1) Exp. Resistive element: u is voltage; y is current It is passive if the inflow of power is always nonneg. uy 0 for all (u, y) Geometrically it means the u y curves lie in first and third quadrant The simplest option is linear resistor (u = Ry) Farzaneh Abdollahi Nonlinear Control Lecture 7 3/26
If u and y are vectors, the power flow onto the network will be u T y = p u i y i = i=1 p u i h i (u) i=1 For time-varying system, as long as the passivity condition is satisfied for all time, it is called passive. Extreme case of passivity : u T y = 0, this system is lossless. Input strictly passivity: if a fcn. h satisfies u T y u T φ(u), and u T φ(u) > 0, u 0 since u T y = 0 only if u = 0 arzaneh Abdollahi Nonlinear Control Lecture 7 4/26
Input Feedforward Passive Let us define a new output: ỹ = y φ(u): u T ỹ = u T [y φ(u)] u T φ(u) u T φ(u) = 0 any fcs. satisfying u T y u T φ(u) can be transformed into a passive fcn. via input feedforward. This fcs is input feedforward passive Farzaneh Abdollahi Nonlinear Control Lecture 7 5/26
Output Feedback Passive Suppose u T y y T ρ(y). Let us define a new input: ũ = u ρ(y): ũ T y = [u ρ(y)] T y y T ρ(y) y T ρ(y) = 0 any fcs. satisfying u T y y T ρ(y) can be transformed into a passive fcn. via output feedback. This fcs is output feedback passive If y T ρ(y) > 0, y 0 the fcn. is called output strictly passive since u T y = 0 only if y = 0 Farzaneh Abdollahi Nonlinear Control Lecture 7 6/26
Passivity of Memoryless Fcn. The system y = h(t, u) is passive if u T y 0 lossless if u T y = 0 input-feedforward passive if u T y u T φ(u) for some fcn φ input strictly passive if u T y u T φ(u) and u T φ(u) > 0, u 0 output-feedback passive if u T y y T ρ(y) for some fcn ρ output strictly passive if u T y y T ρ(y) and y T ρ(y) > 0, y 0 arzaneh Abdollahi Nonlinear Control Lecture 7 7/26
State Model Consider a dynamical system with state model ẋ = f (x, u) (2) y = h(x, u) f : R n R p R n is local lip. h : R n R p R p is cont. f (0, 0) = 0 and h(0, 0) = 0 # inputs = # outputs arzaneh Abdollahi Nonlinear Control Lecture 7 8/26
Motivated Example: RLC Circuit Consider the RLC circuit with linear C and L and nonlinear R The nonlinear resistors are represented by: i 1 = h 1 (v 1 ); v 2 = h 2 (i 2 ); i 3 = h 3 (v 3 ) Input u: voltage; output y: current power flow into the network: uy Define x 1 : current through L; x 2 : voltage across C Farzaneh Abdollahi Nonlinear Control Lecture 7 9/26
state model Lẋ 1 = u h 2 (x 1 ) x 2 Cẋ 2 = x 1 h 3 (x 2 ) y = x 1 + h 1 (u) The system is passive if the absorbed energy by the network is greater than the stored energy in the network over the same period: t 0 u(s)y(s)ds V (x(t)) V (x(0)) (3) where V (x) = 1/2Lx1 2 + 1/2Cx 2 2 : stored energy Strict inequality of (3) yields difference between the absorbed energy and increased stored energy equals to dissipative energy in the resistors (3) hold for every t 0 for all t u(t)y(t) V (x(t), u(t)) i.e. the power flow must be greater than or equal to the rate of change of the stored energy arzaneh Abdollahi Nonlinear Control Lecture 7 10/26
Motivated Exp. Cont d Take the derivative of V along the system traj: V = uy uh 1 (u) x 1 h 2 (x 1 ) x 2 h 3 (x 2 ) uy = V + uh 1 (u) + x 1 h 2 (x 1 ) + x 2 h 3 (x 2 ) If h 1, h 2 and h 3 are passive uy V ; system is passive Otherwise Case 1: If h 1 = h 2 = h 3 = 0, uy = V no energy dissipation, system is lossless Case 2: If h 2, h 3 sector [0, ] ( passive fcns.) uy V + uh 1 (u) If uh 1(u) > 0 for all u 0 it is input strick passive (absorbed energy is greater than increased stored energy unless u(t) 0) If uh 1(u) < 0 for some u it can be made passive by an input ff Case 3: If h1 = 0 and h 3 : passive fcn ( sector [0, ]) uy V + yh 2 (y) If yh 2(y) > 0 for all y 0 it is output strick passive (absorbed energy is greater than increased stored energy unless y(t) 0) If yh 2(y) < 0 for some u it can be made passive by an output fb arzaneh Abdollahi Nonlinear Control Lecture 7 11/26
Motivated Exp. Cont d Case 4: If h 1 [0, ] and h 2, h 3 (0, ) uy V + x 1 h 2 (x 1 ) + x 2 h 3 (x 2 ) where x1 h 2 (x 1 ) + x 2 h 3 (x 2 ) is pos. def. It is state strict passive or simply strick passive (absorbed energy is greater than increased stored energy unless x(t) 0) arzaneh Abdollahi Nonlinear Control Lecture 7 12/26
Passivity Based on State Model The system (2) is passive if there exist a cont. diff. p.s.d fcn V (x) ( called storage fcn) s.t. u T y V = V f (x, u), x (x, u) Rn R n Moreover, it is Lossless if u T y = V Input-feedforward passive if u T y V + u T φ(u) for some fcn φ Input strictly passive if u T y V + u T φ(u) and u T φ(u) > 0, u 0 Output-feedback passive if u T y V + y T ρ(y) for some fcn ρ Output strictly passive if u T y V + y T ρ(y) and y T ρ(y) > 0, y 0 Strictly passive if u T y V + ψ(x) for some p.d. ψ In all cases, the inequality should hold for all (x, u) arzaneh Abdollahi Nonlinear Control Lecture 7 13/26
Example Consider a cascade connection of an integrator and a passive memoryless fcn. ẋ = u, y = h(x) h is passive x 0 h(σ)dσ 0, x Storage fcn: V (x) = x 0 h(σ)dσ V = h(x)ẋ = yu it is loss less Now replace the integrator with 1/(as + 1), a > 0 The state model is: aẋ = x + u, y = h(x) V = a x 0 h(σ)dσ V = h(x)( x + u) = yu xh(x) yu It is passive. When xh(x) > 0 it is strictly passive Farzaneh Abdollahi Nonlinear Control Lecture 7 14/26
L 2 and Lyapunov Stability Lemma: If the system (2) is output strictly passive with u T y V + δy T y for some δ > 0 then it is finite-gain L 2 stable with L 2 gain less than or equal to 1/δ Definition: The system (2) is zero-state observable if no solution of ẋ = f (x, 0) can stay identically in S = {x R n h(x, 0) = 0} other than the trivial solution x(t) 0 Example: For linear system ẋ = Ax, y = Cx Observability is equivalent to y(t) = Ce At x(0) 0 { x(0) = 0 x(t) 0 arzaneh Abdollahi Nonlinear Control Lecture 7 15/26
L 2 and Lyapunov Stability Lemma: If system (2), is passive with a p.d. storage fcn. V (x), then the origin of ẋ = f (x, 0) is stable. Lemma: For system (2), the origin of ẋ = f (x, 0) is a.s if the system is strictly passive or output strictly passive and zero-state observable Furthermore, if the storage fcn. is radially unbounded, then the origin will be g.a.s arzaneh Abdollahi Nonlinear Control Lecture 7 16/26
Example Consider SISO system a > 0, k > 0 ẋ 1 = x 2 ẋ 2 = ax1 3 kx 2 + u y = x 2 Consider p.d. radially unbounded V (x) = (1/4)ax1 4 + (1/2)x 2 2 V = ky 2 + yu By ρ(y) = ky, it is output strictly passive L 2 f.g.s. with gain less than or equal to 1/k When u = 0, y(t) 0 x 2 0 x 1 0 zero-state observable it is g.a.s arzaneh Abdollahi Nonlinear Control Lecture 7 17/26
Feedback and Passivity Theorems Consider a fb connection of H 1 and H 2 in time-invariant dynamical system represented by state model ẋ i = f i (x i, e i ) y i = h i (x i, e i ) (4) or (possibly time-varying) memoryless fcn y i = h i (t, e i ) (5) Objective: Analyze stability of fb connection, using passivity properties of fb components (H 1 and H 2 ) Farzaneh Abdollahi Nonlinear Control Lecture 7 18/26
Feedback and Passivity Theorems 1. If both components H 1 and H 2 are dynamical systems, the closed-loop state model is ẋ = f (x, u) y = h(x, u) (6) where x = [x1 x 2 ] T, u = [u 1 u 2 ] T, y = [y 1 y 2 ] T Assuming fi (0, 0) = 0 and h i (0, 0) = 0 f (0, 0) = 0 and h(0, 0) = 0 Assuming f i and h i are locally lip. f and h are locally lip. arzaneh Abdollahi Nonlinear Control Lecture 7 19/26
Feedback and Passivity Theorems 2. If H 1 is a dynamical system and H 2 is memoryless fcn., the closed loop state model is ẋ = f (t, x, u) y = h(t, x, u) (7) where x = x1, u = [u 1 u 2 ] T, y = [y 1 y 2 ] T Assume f (t, 0, 0) = 0 and h(t, 0, 0) = 0 Assume f is p.c. in t and locally lip. in (x, u), and h is p.c. in t and cont. in (x, u) 3. If both components are memoryless fcns It can be considered as a special case when x does not exit arzaneh Abdollahi Nonlinear Control Lecture 7 20/26
Feedback and L 2 Stability Theorem: The feedback connection of two passive system is passive Proof: Let V 1 (x 1 ) and V 2 (x 2 ) are storage fcns of H 1 and H 2 respectively. If either components are memoryless fcn, take V i = 0 Then e T i y i V i Considering fb. connection e1 T y 1 + e2 T y 2 = (u 1 y 2 ) T y 1 + (u 2 + y 1 ) T y 2 = u1 T y 1 + u2 T y 2 u T y = u1 T y 1 + u2 T y 2 V 1 + V 2 = V Lemma: The fb connection of two output strictly passive systems with e T i y i V i + δ i y T i y i + ɛ i e T i e i, i = 1, 2 is finite-gain L 2 stable with gain if ɛ 1 + δ 2 > 0, ɛ 2 + δ 1 > 0 arzaneh Abdollahi Nonlinear Control Lecture 7 21/26
Example { ẋ = f (x) + G(x)e1 Consider H 1 : y 1 = h(x) e i, y i R p and H 2 : y 2 = ke 2 where k > 0, Suppose there is a p.d. fcn V 1 (x) s.t. V 1 x f (x) 0, V 1 x G(x) = ht (x), x R n Both components are passive and e2 T y 2 = ke2 T e 2 = γke2 T e 2 + (1 γ) k y2 T y 2, 0 < γ < 1 ɛ 1 = δ 1 = 0, ɛ 2 = γk, δ 2 = (1 γ) k It is finite gain L 2 stable arzaneh Abdollahi Nonlinear Control Lecture 7 22/26
Feedback and A.S. Stability of origin is trivial if both components are passive. (Tell me why?. Let us focus on a.s. Theorem: Consider fb connection of two T.I. dynamical systems of the form (4). The origin of the closed-loop system (6) ( when u = 0) is a.s. if both components are strictly passive or both components are output strictly passive and zero-state observable or one component is strictly passive and the other one is output strictly passive and zero-state observable Furthermore, if storage fcn of each component is radially unbounded, the origin is g.a.s arzaneh Abdollahi Nonlinear Control Lecture 7 23/26
Example Consider fb connection with: ẋ 1 = x 2 ẋ 3 = x 4 H 1 : ẋ 2 = ax1 3 kx 2 + e 1, H 2 : ẋ 4 = bx 3 x4 3 + e 2 y 1 = x 2 y 2 = x 4 a, b, k > 0 Use V 1 = (a/4)x 4 1 + (1/2)x 2 2 V 1 = ky 2 1 + y 1e 1 H 1 is output strictly passive. when e 1 = 0, y 1 0 x 2 0 x 1 0 H 1 is zero-state observable Use V 2 = (b/2)x 2 3 + (1/2)x 2 4 V 2 = y 4 2 + y 2e 2 H 2 is output strictly passive. when e 2 = 0, y 2 0 x 4 0 x 3 0 H 2 is zero-state observable V 1, V 2 are radially unbounded the closed-loop system is g.a.s arzaneh Abdollahi Nonlinear Control Lecture 7 24/26
Example Reconsider the pervious system but change the output of H 1 to y 1 = x 2 + e 1 Hence V 1 = k(y 1 e 1 ) 2 e 2 1 + y 1e 1 H 1 is passive Not strictly passive or output strictly passive Consider Lyap fcn of closed-loop sys. V = V 1 + V 2 = 1 4 ax 4 1 + 1 2 x 2 2 + 1 2 bx 2 3 + 1 2 x 2 4 V = kx 2 2 x 4 4 x 2 4 0 Also, V = 0 x 2 = x 4 = 0 x 2 0 x 1 0 x 4 0 x 3 0 V is radially unbounded the closed-loop system is g.a.s arzaneh Abdollahi Nonlinear Control Lecture 7 25/26
Feedback and A.S. Theorem: Consider fb connection of a strictly passive, T.I. dynamical systems of the form (4) and a passive (possibly time-varying) memoryless fcn of the form (5). The origin of the closed-loop system (7) ( when u = 0) is u.a.s. Furthermore, if storage fcn of the dynamical system is radially unbounded, the origin is g.u.a.s arzaneh Abdollahi Nonlinear Control Lecture 7 26/26