Introduction to Nonlinear Control Lecture # 4 Passivity

p. 1/6 Introduction to Nonlinear Control Lecture # 4 Passivity

È p. 2/6 Memoryless Functions ¹ y È Ý Ù È È È È u (b) µ power inflow = uy Resistor is passive if uy 0

p. 3/6 y y y u u u (a) (b) (c) Passive Passive Not passive y = h(t, u), h [0, ] Vector case: [ ] y = h(t, u), h T = h 1, h 2,, h p power inflow = Σ p i=1 u iy i = u T y

p. 4/6 Definition: y = h(t, u) is passive if u T y 0 lossless if u T y = 0 input strictly passive if u T y u T ϕ(u) for some function ϕ where u T ϕ(u) > 0, u 0 output strictly passive if u T y y T ρ(y) for some function ρ where y T ρ(y) > 0, y 0

p. 5/6 Sector Nonlinearity: h belongs to the sector [α, β] (h [α, β]) if αu 2 uh(t, u) βu 2 y y= βu y y=βu y=αu u y=αu u (a) α > 0 (b) α < 0 Also, h (α, β], h [α, β), h (α, β)

p. 6/6 αu 2 uh(t, u) βu 2 [h(t, u) αu][h(t, u) βu] 0 Definition: A memoryless function h(t, u) is said to belong to the sector [0, ] if u T h(t, u) 0 [K 1, ] if u T [h(t, u) K 1 u] 0 [0, K 2 ] with K 2 = K T 2 > 0 if h T (t, u)[h(t, u) K 2 u] 0 [K 1, K 2 ] with K = K 2 K 1 = K T > 0 if [h(t, u) K 1 u] T [h(t, u) K 2 u] 0

p. 7/6 Example h(u) = [ h 1 (u 1 ) h 2 (u 2 ) K 1 = [ ] α 1 0 0 α 2, h i [α i, β i ], β i > α i i = 1, 2 ], K 2 = [ β 1 0 0 β 2 ] h [K 1, K 2 ] [ β 1 α 1 0 K = K 2 K 1 = 0 β 2 α 2 ]

p. 8/6 Example h(u) Lu γ u K 1 = L γi, K 2 = L + γi [h(u) K 1 u] T [h(u) K 2 u] = h(u) Lu 2 γ 2 u 2 0 K = K 2 K 1 = 2γI

p. 9/6 A function in the sector [K 1, K 2 ] can be transformed into a function in the sector [0, ] by input feedforward followed by output feedback + + K 1 y = h(t, u) + K 1 [K 1, K 2 ] Feedforward [0, K] K 1 [0, I] Feedback [0, ]

Ú ½ ½ È È È È È ¾ Ú ¾ ¾ ¾ µ ««««Ú ¾ Ä Ú Ú È È È È È p. 10/6 State Models Ä ¹ ¹ ¹ Ý Ù ½ ½ Ú ½ µ È È Ú µ Lẋ 1 = u h 2 (x 1 ) x 2 Cẋ 2 = x 1 h 3 (x 2 ) y = x 1 + h 1 (u)

p. 11/6 V (x) = 1 2 Lx2 1 + 1 2 Cx2 2 t 0 u(s)y(s) ds V (x(t)) V (x(0)) u(t)y(t) V (x(t), u(t)) V = Lx 1 ẋ 1 + Cx 2 ẋ 2 = x 1 [u h 2 (x 1 ) x 2 ] + x 2 [x 1 h 3 (x 2 )] = x 1 [u h 2 (x 1 )] x 2 h 3 (x 2 ) = [x 1 + h 1 (u)]u uh 1 (u) x 1 h 2 (x 1 ) x 2 h 3 (x 2 ) = uy uh 1 (u) x 1 h 2 (x 1 ) x 2 h 3 (x 2 )

p. 12/6 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 and the system is passive Case 1: If h 1 = h 2 = h 3 = 0, then uy = V ; no energy dissipation; the system is lossless Case 2: If h 1 (0, ] (uh 1 (u) > 0 for u 0), then uy V + uh 1 (u) The energy absorbed over [0, t] will be greater than the increase in the stored energy, unless the input u(t) is identically zero. This is a case of input strict passivity

p. 13/6 Case 3: If h 1 = 0 and h 2 (0, ], then y = x 1 and uy V + yh 2 (y) The energy absorbed over [0, t] will be greater than the increase in the stored energy, unless the output y is identically zero. This is a case of output strict passivity Case 4: If h 2 (0, ) and h 3 (0, ), then uy V + x 1 h 2 (x 1 ) + x 2 h 3 (x 2 ) x 1 h 2 (x 1 ) + x 2 h 3 (x 2 ) is a positive definite function of x. This is a case of state strict passivity because the energy absorbed over [0, t] will be greater than the increase in the stored energy, unless the state x is identically zero

p. 14/6 Definition: The system ẋ = f(x, u), y = h(x, u) is passive if there is a continuously differentiable positive semidefinite function V (x) (the storage function) such that u T y V = V x Moreover, it is said to be lossless if u T y = V f(x, u), (x, u) input strictly passive if u T y V + u T ϕ(u) for some function ϕ such that u T ϕ(u) > 0, u 0

p. 15/6 output strictly passive if u T y V + y T ρ(y) for some function ρ such that y T ρ(y) > 0, y 0 strictly passive if u T y V + ψ(x) for some positive definite function ψ Example ẋ = u, y = x V (x) = 1 2 x2 uy = V Lossless

p. 16/6 Example ẋ = u, y = x + h(u), h [0, ] V (x) = 1 2 x2 uy = V + uh(u) Passive Example h (0, ] uh(u) > 0 u 0 Input strictly passive ẋ = h(x) + u, y = x, h [0, ] V (x) = 1 2 x2 uy = V + yh(y) Passive h (0, ] Output strictly passive

p. 17/6 Example ẋ = u, y = h(x), h [0, ] V (x) = Example x 0 h(σ) dσ V = h(x)ẋ = yu Lossless aẋ = x + u, y = h(x), h [0, ] V (x) = a x 0 h(σ) dσ V = h(x)( x+u) = yu xh(x) yu = V + xh(x) Passive h (0, ] Strictly passive

p. 18/6 Positive Real Transfer Functions Definition: A p p proper rational transfer function matrix G(s) is positive real if poles of all elements of G(s) are in Re[s] 0 for all real ω for which jω is not a pole of any element of G(s), the matrix G(jω) + G T ( jω) is positive semidefinite any pure imaginary pole jω of any element of G(s) is a simple pole and the residue matrix lim s jω (s jω)g(s) is positive semidefinite Hermitian G(s) is called strictly positive real if G(s ε) is positive real for some ε > 0

p. 19/6 Scalar Case (p = 1): G(jω) + G T ( jω) = 2Re[G(jω)] Re[G(jω)] is an even function of ω. The second condition of the definition reduces to Re[G(jω)] 0, ω [0, ) which holds when the Nyquist plot of of G(jω) lies in the closed right-half complex plane This is true only if the relative degree of the transfer function is zero or one

p. 20/6 Lemma: A scalar transfer function G(s) is strictly positive real if and only if G(s) is Hurwitz Re[G(jω)] > 0, ω [0, ) G( ) > 0 or lim ω ω2 Re[G(jω)] > 0

p. 21/6 Example: G(s) = 1 s has a simple pole at s = 0 whose residue is 1 [ ] 1 Re[G(jω)] = Re = 0, ω 0 jω Hence, G is positive real. It is not strictly positive real since 1 (s ε) has a pole in Re[s] > 0 for any ε > 0

p. 22/6 Example: G(s) = 1, a > 0, is Hurwitz s + a Re[G(jω)] = a ω 2 > 0, ω [0, ) + a2 lim ω ω2 Re[G(jω)] = lim ω Example: G(s) = G is not PR ω 2 a ω 2 = a > 0 G is SPR + a2 1 s 2 + s + 1, Re[G(jω)] = 1 ω 2 (1 ω 2 ) 2 + ω 2

p. 23/6 Positive Real Lemma: Let G(s) = C(sI A) 1 B + D where (A, B) is controllable and (A, C) is observable. G(s) is positive real if and only if there exist matrices P = P T > 0, L, and W such that PA + A T P = L T L PB = C T L T W W T W = D + D T

p. 24/6 Kalman Yakubovich Popov Lemma: Let G(s) = C(sI A) 1 B + D where (A, B) is controllable and (A, C) is observable. G(s) is strictly positive real if and only if there exist matrices P = P T > 0, L, and W, and a positive constant ε such that PA + A T P = L T L εp PB = C T L T W W T W = D + D T

p. 25/6 Lemma: The linear time-invariant minimal realization ẋ = Ax + Bu y = Cx + Du with G(s) = C(sI A) 1 B + D is passive if G(s) is positive real strictly passive if G(s) is strictly positive real Proof: Apply the PR and KYP Lemmas, respectively, and use V (x) = 1 2 xt Px as the storage function

p. 26/6 u T y V (Ax + Bu) x = u T (Cx + Du) x T P(Ax + Bu) = u T Cx + 1 2 ut (D + D T )u 1 2 xt (PA + A T P)x x T PBu = u T (B T P + W T L)x + 1 2 ut W T Wu + 1 2 xt L T Lx + 1 2 εxt Px x T PBu = 1 2 (Lx + Wu)T (Lx + Wu) + 1 2 εxt Px 1 2 εxt Px In the case of the PR Lemma, ε = 0, and we conclude that the system is passive; in the case of the KYP Lemma, ε > 0, and we conclude that the system is strictly passive

p. 27/6 Connection with Lyapunov Stability Lemma: If the system ẋ = f(x, u), y = h(x, u) is passive with a positive definite storage function V (x), then the origin of ẋ = f(x, 0) is stable Proof: u T y V x f(x, u) V x f(x, 0) 0

p. 28/6 Lemma: If the system ẋ = f(x, u), y = h(x, u) is strictly passive, then the origin of ẋ = f(x, 0) is asymptotically stable. Furthermore, if the storage function is radially unbounded, the origin will be globally asymptotically stable Proof: The storage function V (x) is positive definite u T y V x f(x, u) + ψ(x) V x f(x, 0) ψ(x) Why is V (x) positive definite? Let φ(t; x) be the solution of ż = f(z, 0), z(0) = x

p. 29/6 V (φ(τ, x)) V (x) V ψ(x) τ 0 V (φ(τ, x)) 0 V (x) V ( x) = 0 τ 0 ψ(φ(t; x)) dt, τ [0, δ] τ 0 ψ(φ(t; x)) dt ψ(φ(t; x)) dt = 0, τ [0, δ] ψ(φ(t; x)) 0 φ(t; x) 0 x = 0

p. 30/6 Definition: The system ẋ = f(x, u), y = h(x, u) is zero-state observable if no solution of ẋ = f(x, 0) can stay identically in S = {h(x, 0) = 0}, other than the zero solution x(t) 0 Linear Systems ẋ = Ax, y = Cx Observability of (A, C) is equivalent to y(t) = Ce At x(0) 0 x(0) = 0 x(t) 0

p. 31/6 Lemma: If the system ẋ = f(x, u), y = h(x, u) is output strictly passive and zero-state observable, then the origin of ẋ = f(x, 0) is asymptotically stable. Furthermore, if the storage function is radially unbounded, the origin will be globally asymptotically stable Proof: The storage function V (x) is positive definite u T y V x f(x, u) + yt ρ(y) V x f(x, 0) yt ρ(y) V (x(t)) 0 y(t) 0 x(t) 0 Apply the invariance principle

p. 32/6 Example ẋ 1 = x 2, ẋ 2 = ax 3 1 kx 2 + u, y = x 2, a, k > 0 V (x) = 1 4 ax4 1 + 1 2 x2 2 V = ax 3 1 x 2 + x 2 ( ax 3 1 kx 2 + u) = ky 2 + yu The system is output strictly passive y(t) 0 x 2 (t) 0 ax 3 1 (t) 0 x 1(t) 0 The system is zero-state observable. V is radially unbounded. Hence, the origin of the unforced system is globally asymptotically stable

p. 33/6 Feedback Systems u 1 + e 1 H 1 y 1 y 2 H 2 e 2 + + u 2 ẋ i = f i (x i, e i ), y i = h i (x i, e i ) y i = h i (t, e i )

p. 34/6 Passivity Theorems Theorem 6.1: The feedback connection of two passive systems is passive Theorem 6.3: Consider the feedback connection of two dynamical systems. When u = 0, the origin of the closed-loop system is asymptotically stable if each feedback component is either strictly passive, or output strictly passive and zero-state observable Furthermore, if the storage function for each component is radially unbounded, the origin is globally asymptotically stable

p. 35/6 Theorem 6.4: Consider the feedback connection of a strictly passive dynamical system with a passive memoryless function. When u = 0, the origin of the closed-loop system is uniformly asymptotically stable. if the storage function for the dynamical system is radially unbounded, the origin will be globally uniformly asymptotically stable Prove using V = V 1 + V 2 as a Lyapunov function candidate Proof of Theorem 6.3: H 1 is SP; H 2 is OSP & ZSO e T 1 y 1 V 1 + ψ 1 (x 1 ), ψ 1 (x 1 ) > 0, x 1 0 e T 2 y 2 V 2 + y T 2 ρ 2(y 2 ), y T 2 ρ(y 2) > 0, y 2 0

p. 36/6 e T 1 y 1+e T 2 y 2 = (u 1 y 2 ) T y 1 +(u 2 +y 1 ) T y 2 = u T 1 y 1+u T 2 y 2 V (x) = V 1 (x 1 ) + V 2 (x 2 ) V u T y ψ 1 (x 1 ) y2 T ρ 2(y 2 ) u = 0 V ψ 1 (x 1 ) y2 T ρ 2(y 2 ) V = 0 x 1 = 0 and y 2 = 0 y 2 (t) 0 e 1 (t) 0 ( & x 1 (t) 0) y 1 (t) 0 y 1 (t) 0 e 2 (t) 0 By zero-state observability of H 2 : y 2 (t) 0 x 2 (t) 0 Apply the invariance principle

p. 37/6 Example ẋ 1 = x 2 ẋ 2 = ax 3 1 kx 2 + e 1 y 1 } = x 2 + e 1 {{ } H 1 a, b, k > 0 V 1 = 1 4 ax4 1 + 1 2 x2 2 ẋ 3 = x 4 ẋ 4 = bx 3 x 3 4 + e 2 y 2 } = x 4 {{ } H 2 V 1 = ax 3 1 x 2 ax 3 1 x 2 kx 2 2 + x 2e 1 = ky 2 1 + y 1e 1 With e 1 = 0, y 1 (t) 0 x 2 (t) 0 x 1 (t) 0 H 1 is output strictly passive and zero-state observable

p. 38/6 V 2 = 1 2 bx2 3 + 1 2 x2 4 V 2 = bx 3 x 4 bx 3 x 4 x 4 4 + x 4e 2 = y2 4 + y 2e 2 With e 2 = 0, y 2 (t) 0 x 4 (t) 0 x 3 (t) 0 H 2 is output strictly passive and zero-state observable V 1 and V 2 are radially unbounded The origin is globally asymptotically stable

p. 39/6 Loop Transformation Recall that a memoryless function in the sector [K 1, K 2 ] can be transformed into a function in the sector [0, ] by input feedforward followed by output feedback + K 1 + y = h(t, u) + K 1

p. 40/6 + H 1 H 2 H 1 is a dynamical system H 2 is a memoryless function in the sector [K 1, K 2 ]

p. 41/6 + + H 1 K 1 + H 2 K 1

p. 42/6 + H 1 K + K 1 + H 2 K 1 K 1

p. 43/6 H 1 + + H 1 K + + K 1 + H 2 K 1 + + K 1 H 2

p. 44/6 Example ẋ 1 = x 2 ẋ 2 = h(x 1 ) + bx 2 + e 1 y 1 } = x 2 {{ } H 1 y 2 = σ(e 2 ) }{{} H 2 σ [α, β], h [α 1, ], b > 0, α 1 > 0, k = β α > 0 ẋ 1 = x 2 ẋ 2 = h(x 1 ) ax 2 + ẽ 1 ỹ 2 = σ(ẽ 2 ) }{{} ỹ 1 = kx 2 + ẽ 1 H }{{} 2 H 1 σ [0, ], a = α b

p. 45/6 Assume a = α b > 0 and show that H 1 is strictly passive V 1 = k x1 0 h(s) ds + x T Px V 1 = k x1 0 h(s) ds + p 11 x 2 1 + 2p 12x 1 x 2 + p 22 x 2 2 V = kh(x 1 )x 2 + 2(p 11 x 1 + p 12 x 2 )x 2 2(p 12 x 1 + p 22 x 2 )[ h(x 1 ) ax 2 + ẽ 1 ] Take p 22 = k/2, p 11 = ap 12

p. 46/6 V = 2p 12 x 1 h(x 1 ) (ka 2p 12 )x 2 2 + kx 2 ẽ 1 + 2p 12 x 1 ẽ 1 = 2p 12 x 1 h(x 1 ) (ka 2p 12 )x 2 2 + (kx 2 + ẽ 1 )ẽ 1 ẽ 2 1 + 2p 12x 1 ẽ 1 ỹ 1 ẽ 1 = V + 2p 12 x 1 h(x 1 ) + (ka 2p 12 )x 2 2 Take 0 < p 12 < min + (ẽ 1 p 12 x 1 ) 2 p 2 12 x2 1 V + p 12 (2α 1 p 12 )x 2 1 + (ka 2p 12)x 2 2 { } ak 2, 2α 1 p 2 12 < 2p k 12 2 = p 11p 22 H 1 is strictly passive. By Theorem 6.4 the origin is globally asymptotically stable (when u = 0)

p. 47/6 Absolute Stability + r u y G(s) ψ( ) The system is absolutely stable if (when r = 0) the origin is globally asymptotically stable for all memoryless time-invariant nonlinearities in a given sector

p. 48/6 Circle Criterion Suppose G(s) = C(sI A) 1 B + D is SPR and ψ [0, ] ẋ = Ax + Bu y = Cx + Du u = ψ(y) By the KYP Lemma, P = P T > 0, L, W, ε > 0 PA + A T P = L T L εp PB = C T L T W W T W = D + D T V (x) = 1 2 xt Px

p. 49/6 V = 1 2 xt Pẋ + 1 2ẋT Px = 1 2 xt (PA + A T P)x + x T PBu = 1 2 xt L T Lx 1 2 εxt Px + x T (C T L T W)u = 1 2 xt L T Lx 1 2 εxt Px + (Cx + Du) T u u T Du x T L T Wu u T Du = 1 2 ut (D + D T )u = 1 2 ut W T Wu V = 1 2 εxt Px 1 2 (Lx + Wu)T (Lx + Wu) y T ψ(y) y T ψ(y) 0 V 1 2 εxt Px The origin is globally exponentially stable

p. 50/6 What if ψ [K 1, ]? + G(s) + + G(s) K 1 ψ( ) + ψ( ) K 1 ψ( ) ψ [0, ]; hence the origin is globally exponentially stable if G(s)[I + K 1 G(s)] 1 is SPR

p. 51/6 What if ψ [K 1, K 2 ]? + G(s) + + G(s) K K 1 + + ψ( ) + ψ( ) K 1 K 1 + + ψ( ) ψ [0, ]; hence the origin is globally exponentially stable if I + KG(s)[I + K 1 G(s)] 1 is SPR

p. 52/6 I +KG(s)[I +K 1 G(s)] 1 = [I +K 2 G(s)][I +K 1 G(s)] 1 Theorem (Circle Criterion): The system is absolutely stable if ψ [K 1, ] and G(s)[I + K 1 G(s)] 1 is SPR, or ψ [K 1, K 2 ] and [I + K 2 G(s)][I + K 1 G(s)] 1 is SPR Scalar Case: ψ [α, β], β > α The system is absolutely stable if Re 1 + βg(s) is Hurwitz and 1 + αg(s) [ ] 1 + βg(jω) > 0, ω [0, ] 1 + αg(jω)

p. 53/6 Case 1: α > 0 By the Nyquist criterion 1 + βg(s) 1 + αg(s) = 1 1 + αg(s) + βg(s) 1 + αg(s) is Hurwitz if the Nyquist plot of G(jω) does not intersect the point (1/α) + j0 and encircles it m times in the counterclockwise direction, where m is the number of poles of G(s) in the open right-half complex plane 1 + βg(jω) 1 + αg(jω) = 1 β + G(jω) 1 α + G(jω)

p. 54/6 Re [ 1 β + G(jω) ] 1 α + G(jω) > 0, ω [0, ] D(α,β) q θ 2 θ 1 1/α 1/β The system is absolutely stable if the Nyquist plot of G(jω) does not enter the disk D(α, β) and encircles it m times in the counterclockwise direction

p. 55/6 Case 2: α = 0 1 + βg(s) Re[1 + βg(jω)] > 0, ω [0, ] Re[G(jω)] > 1, ω [0, ] β The system is absolutely stable if G(s) is Hurwitz and the Nyquist plot of G(jω) lies to the right of the vertical line defined by Re[s] = 1/β

p. 56/6 Case 3: α < 0 < β [ ] 1 + βg(jω) [ 1 β + G(jω) ] Re 1 + αg(jω) > 0 Re 1 α + G(jω) < 0 The Nyquist plot of G(jω) must lie inside the disk D(α, β). The Nyquist plot cannot encircle the point (1/α) + j0. From the Nyquist criterion, G(s) must be Hurwitz The system is absolutely stable if G(s) is Hurwitz and the Nyquist plot of G(jω) lies in the interior of the disk D(α, β)

p. 57/6 Example G(s) = 4 (s + 1)( 1 2 s + 1)(1 3 s + 1) 6 4 Im G 2 0 2 Re G 4 5 0 5

p. 58/6 Apply Case 3 with center (0, 0) and radius = 4 Sector is ( 0.25, 0.25) Apply Case 3 with center (1.5, 0) and radius = 2.834 Sector is [ 0.227, 0.714] Apply Case 2 The Nyquist plot is to the right of Re[s] = 0.857 Sector is [0, 1.166] [0, 1.166] includes the saturation nonlinearity

p. 59/6 Example G(s) = 4 (s 1)( 1 2 s + 1)(1 3 s + 1) 0.4 Im G 0.2 0 0.2 Re G G is not Hurwitz Apply Case 1 0.4 4 2 0 Center = ( 3.2, 0), Radius = 0.168 [0.2969, 0.3298]

p. 60/6 Popov Criterion + G(s) ψ( ) ẋ = Ax + Bu y = Cx = ψ i (y i ), 1 i p u i ψ i [0, k i ], 1 i p, (0 < k i ) G(s) = C(sI A) 1 B Γ = diag(γ 1,..., γ p ), M = diag(1/k 1,, 1/k p )

p. 61/6 M H 1 + G(s) (I + sγ) + + ψ( ) (I + sγ) 1 + + H 2 M Show that H 1 and H 2 are passive

p. 62/6 M + (I + sγ)g(s) = M + (I + sγ)c(si A) 1 B = M + C(sI A) 1 B + ΓCs(sI A) 1 B = M + C(sI A) 1 B + ΓC(sI A + A)(sI A) 1 B = (C + ΓCA)(sI A) 1 B + M + ΓCB If M + (I + sγ)g(s) is SPR, then H 1 is strictly passive with the storage function V 1 = 1 2 xt Px, where P is given by the KYP equations PA + A T P = L T L εp PB = (C + ΓCA) T L T W W T W = 2M + ΓCB + B T C T Γ

p. 63/6 H 2 consists of p decoupled components: γ i ż i = z i + 1 k i ψ i (z i ) + ẽ 2i, ỹ 2i = ψ i (z i ) V 2i = γ i zi V 2i = γ i ψ i (z i )ż i = ψ i (z i ) 0 ψ i (σ) dσ = y 2i e 2i + 1 k i ψ i (z i ) [ψ i (z i ) k i z i ] [ z i + 1 ki ψ i (z i ) + ẽ 2i ] ψ i [0, k i ] ψ i (ψ i k i z i ) 0 V 2i y 2i e 2i H 2 is passive with the storage function V 2 = p i=1 γ zi i 0 ψ i(σ) dσ

p. 64/6 Use V = 1 2 xt Px + p i=1 γ i yi 0 ψ i (σ) dσ as a Lyapunov function candidate for the original feedback connection ẋ = Ax + Bu, y = Cx, u = ψ(y) V = 1 2 xt Pẋ + 1 2ẋT Px + ψ T (y)γẏ = 1 2 xt (PA + A T P)x + x T PBu + ψ T (y)γc(ax + Bu) = 1 2 xt L T Lx 1 2 εxt Px + x T (C T + A T C T Γ L T W)u + ψ T (y)γcax + ψ T (y)γcbu

p. 65/6 V = 1 2 εxt Px 1 2 (Lx + Wu)T (Lx + Wu) ψ(y) T [y Mψ(y)] 1 2 εxt Px ψ(y) T [y Mψ(y)] ψ i [0, k i ] ψ(y) T [y Mψ(y)] 0 V 1 2 εxt Px The origin is globally asymptotically stable Popov Criterion: The system is absolutely stable if, for 1 i p, ψ i [0, k i ] and there exists a constant γ i 0, with (1 + λ k γ i ) 0 for every eigenvalue λ k of A, such that M + (I + sγ)g(s) is strictly positive real

p. 66/6 Scalar case 1 + (1 + sγ)g(s) k is SPR if G(s) is Hurwitz and 1 k + Re[G(jω)] γωim[g(jω)] > 0, ω [0, ) If lim ω { 1 k we also need { 1 lim ω ω2 } + Re[G(jω)] γωim[g(jω)] k } + Re[G(jω)] γωim[g(jω)] = 0 > 0

p. 67/6 1 k + Re[G(jω)] γωim[g(jω)] > 0, ω [0, ) slope = 1/γ ωim[g(j ω)] 1/k Re[G(j ω)] Popov Plot

p. 68/6 Example ẋ 1 = x 2, ẋ 2 = x 2 h(y), y = x 1 ẋ 2 = αx 1 x 2 h(y) + αx 1, α > 0 G(s) = 1 s 2 + s + α, ψ(y) = h(y) αy h [α, β] ψ [0, k] (k = β α > 0) γ > 1 α ω2 + γω 2 (α ω 2 ) 2 > 0, ω [0, ) + ω2 and lim ω ω 2 (α ω 2 + γω 2 ) (α ω 2 ) 2 + ω 2 = γ 1 > 0

p. 69/6 The system is absolutely stable for ψ [0, ] (h [α, ]) ω Im G 0.2 slope=1 0 0.2 0.4 0.6 0.8 1 Re G 0.5 0 0.5 1 Compare with the circle criterion (γ = 0) 1 k + α ω 2 (α ω 2 ) 2 + ω 2 > 0, ω [0, ], for k < 1+2 α