Lecture 5 Input output stability or How to make a circle out of the point 1+0i, and different ways to stay away from it... k 2 yf(y) r G(s) y k 1 y y 1 k 1 1 k 2 f( ) G(iω)
Course Outline Lecture 1-3 Lecture 4-6 Lecture 7-8 Lecture 9-13 Lecture 14 Modelling and basic phenomena (linearization, phase plane, limit cycles) Analysis methods (Lyapunov, circle criterion, describing functions) Common nonlinearities (Saturation, friction, backlash, quantization) Design methods (Lyapunov methods, Backstepping, Optimal control) Summary
Today s Goal To understand signal norms system gain bounded input bounded output (BIBO) stability To be able to analyze stability using the Small Gain Theorem, the Circle Criterion, Passivity Material [Glad & Ljung]: Ch 1.5-1.6, 12.3 [Khalil]: Ch 5 7.1 lecture slides
History f(y) r G(s) y y f( ) For whatg(s) andf( ) is the closed-loop system stable? Lur e and Postnikov s problem (1944) Aizerman s conjecture (1949) (False!) Kalman s conjecture (1957) (False!) Solution by Popov (1960) (Led to the Circle Criterion)
Gain Idea: Generalize static gain to nonlinear dynamical systems u S y The gain γ ofsmeasures the largest amplification fromutoy HereScan be a constant, a matrix, a linear time-invariant system, a nonlinear system, etc Question: How should we measure the size ofuandy?
Norms A norm measures size. A norm is a function from a space Ω tor +, such that for all x,y Ω x 0 and x =0 x=0 x+y x + y αx = α x, for all α R Examples Euclidean norm: x = x1 2+ +x2 n Max norm: x =max{ x 1,..., x n }
Signal Norms A signalx(t) is a function fromr + tor d. A signal norm is a way to measure the size ofx(t). Examples 2-norm (energy norm): x 2 = 0 x(t) 2 dt sup-norm: x =sup t R + x(t) The space of signals with x 2 < is denotedl 2.
Parseval s Theorem Theorem Ifx,y L 2 have the Fourier transforms then X(iω)= 0 In particular 0 e iωt x(t)dt, Y(iω)= 0 e iωt y(t)dt, y T (t)x(t)dt= 1 Y (iω)x(iω)dω. 2π x 2 2 = 0 x(t) 2 dt= 1 X(iω) 2 dω. 2π x 2 < corresponds to bounded energy.
System Gain A systems is a map between two signal spaces:y=s(u). u S y The gain ofs is defined as y 2 S(u) 2 γ(s)=sup =sup u L 2 u 2 u L 2 u 2 Example The gain of a static relationy(t)= αu(t) is γ(α)=sup u L 2 αu 2 u 2 =sup u L 2 α u 2 u 2 = α
Example Gain of a Stable Linear System γ(g) 10 1 G(iω) γ ( G ) Gu 2 =sup = sup G(iω) u L 2 u 2 ω (0, ) 10 0 10 1 10 2 10 1 10 0 10 1 Proof: Assume G(iω) K for ω (0, ). Parseval s theorem gives y 2 2 = 1 Y(iω) 2 dω 2π = 1 G(iω) 2 U(iω) 2 dω K 2 u 2 2 2π This proves that γ(g) K. See [Khalil, Appendix C.10] for a proof of the equality.
2 minute exercise: Show that γ(s 1 S 2 ) γ(s 1 )γ(s 2 ). u S 2 S 1 y
Example Gain of a Static Nonlinearity u(t) f(x) K x, f(x )=Kx Kx y(t) f( ) x f(x) x y 2 2 = f 2( u(t) ) dt 0 u(t)=x,t (0, ) gives equality γ(f)=sup u L2 y 2 u 2 =K. 0 K 2 u 2 (t)dt=k 2 u 2 2
BIBO Stability u S y γ(s)=sup u L 2 y 2 u 2 Definition S is bounded-input bounded-output (BIBO) stable if γ(s) <. Example: Ifẋ=Ax is asymptotically stable then G(s)=C(sI A) 1 B+D is BIBO stable.
The Small Gain Theorem r 1 e 1 S 1 S 2 e 2 r 2 Theorem AssumeS 1 ands 2 are BIBO stable. If γ(s 1 )γ(s 2 )<1 then the closed-loop map from(r 1,r 2 ) to(e 1,e 2 ) is BIBO stable.
Proof of the Small Gain Theorem Existence of solution(e 1,e 2 ) for every(r 1,r 2 ) has to be verified separately. Then gives e 1 2 r 1 2 + γ(s 2 )[ r 2 2 + γ(s 1 ) e 1 2 ] e 1 2 r 1 2 + γ(s 2 ) r 2 2 1 γ(s 2 )γ(s 1 ) γ(s 2 )γ(s 1 )<1, r 1 2 <, r 2 2 < give e 1 2 <. Similarly we get so alsoe 2 is bounded. e 2 2 r 2 2 + γ(s 1 ) r 1 2 1 γ(s 1 )γ(s 2 )
Linear System with Static Nonlinear Feedback (1) r G(s) y Ky f(y) y f( ) G(s)= γ(g)=2and γ(f) K. 2 (s+1) 2 and 0 f(y) K y The small gain theorem gives thatk [0,1/2) implies BIBO stability.
The Nyquist Theorem G(s) Ω 2 1.5 1 0.5 0 0.5 G(Ω) 1 1.5 2 2 1.5 1 0.5 0 0.5 1 Theorem The closed loop system is stable iff the number of counter-clockwise encirclements of 1 by G(Ω) (note: ω increasing) equals the number of open loop unstable poles.
The Small Gain Theorem can be Conservative Letf(y)=Ky for the previous system. 1 0.5 0 0.5 1 G(iω) 1 0.5 0 0.5 1 1.5 2 The Nyquist Theorem proves stability when K [0, ). The Small Gain Theorem proves stability when K [0, 1/2).
The Circle Criterion Case 1:0<k 1 k 2 < k 2 yf(y) r y k G(s) 1 y y f( ) 1 k 1 1 k 2 G(iω) Theorem Consider a feedback loop with y = Gu and u= f(y)+r. AssumeG(s) is stable and that 0<k 1 f(y) y k 2. If the Nyquist curve of G(s) does not intersect or encircle the circle defined by the points 1/k 1 and 1/k 2, then the closed-loop system is BIBO stable fromrtoy.
Other cases G: stable system 0<k 1 <k 2 : Stay outside circle 0=k 1 <k 2 : Stay to the right of the line Res= 1/k 2 k 1 <0<k 2 : Stay inside the circle Other cases: MultiplyfandGwith 1. G: Unstable system To be able to guarantee stability,k 1 andk 2 must have same sign (otherwise unstable fork=0) 0<k 1 <k 2 : Encircle the circlep times counter-clockwise (if ω increasing) k 1 <k 2 <0: Encircle the circlep times counter-clockwise (if ω increasing)
Linear System with Static Nonlinear Feedback (2) Ky f(y) 1 y 0.5 0 0.5 1 K G(iω) 1 1 0.5 0 0.5 1 1.5 2 The circle is defined by 1/k 1 = and 1/k 2 = 1/K. min ReG(iω)= 1/4 so the Circle Criterion gives that ifk [0,4) the system is BIBO stable.
Proof of the Circle Criterion Letk=(k 1 +k 2 )/2 and f(y)=f(y) ky. Then f(y) y k 2 k 1 =:R 2 r 1 e 1 G(s) y 1 r 1 G k G y 1 y 2 f( ) e 2 r 2 f( ) r 2 r 1 =r 1 kr 2
Proof of the Circle Criterion (cont d) r 1 G(s) k f( ) r 2 R 1 G(iω) SGT gives stability for G(iω) R<1with G= R< 1 G(iω) = 1 G(iω) +k Transform this expression through z 1/z. G 1+kG.
Lyapunov revisited Original idea: Energy is decreasing ẋ=f(x), x(0)=x 0 V(x(T)) V(x(0)) 0 (+some other conditions on V) New idea: Increase in stored energy added energy ẋ=f(x,u), x(0)=x 0 y=h(x) V(x(T)) V(x(0)) T 0 ϕ(y,u) dt (1) }{{} external power
Motivation Will assume the external power has the form φ(y,u)=y T u. Only interested in BIBO behavior. Note that V 0withV(x(0))=0and (1) T 0 y T udt 0 Motivated by this we make the following definition
Passive System u S y Definition The systems is passive fromutoy if T 0 y T udt 0, for alluand allt>0 and strictly passive fromutoy if there ǫ>0such that T 0 y T udt ǫ( y 2 T + u 2 T ), for alluand allt>0
A Useful Notation Define the scalar product y,u T = T 0 y T (t)u(t)dt u S y Cauchy-Schwarz inequality: y,u T y T u T where y T = y,y T. Note that y = y 2.
2 minute exercise: u S 1 y u S 1 S 2 y S 2 S 1
Feedback of Passive Systems is Passive r 1 e 1 S 1 y 1 y 2 S 2 e 2 r 2 IfS 1 ands 2 are passive, then the closed-loop system from (r 1,r 2 ) to(y 1,y 2 ) is also passive. Proof: y,r T = y 1,r 1 T + y 2,r 2 T = y 1,r 1 y 2 T + y 2,r 2 +y 1 T = y 1,e 1 T + y 2,e 2 T 0 Hence, y,r T 0if y 1,e 1 T 0and y 2,e 2 T 0
Passivity of Linear Systems Theorem An asymptotically stable linear system G(s) is passive if and only if ReG(iω) 0, ω>0 It is strictly passive if and only if there exists ǫ>0such that ReG(iω) ǫ(1+ G(iω) 2 ), ω>0 Example G(s) = s+1 s+2 strictly passive, is passive and G(s) = 1 is passive but not s strictly passive. 0.6 0.4 0.2 0 0.2 0.4 G(iω) 0 0.2 0.4 0.6 0.8 1
A Strictly Passive System Has Finite Gain u S y IfS is strictly passive, then γ(s)<. Proof: Note that y 2 =lim T y T. ǫ( y 2 T + u 2 T ) y,u T y T u T y 2 u 2 Hence, ǫ y 2 T y 2 u 2, so lettingt gives y 2 1 ǫ u 2
The Passivity Theorem r 1 e 1 S 1 y 1 y 2 S 2 e 2 r 2 Theorem IfS 1 is strictly passive ands 2 is passive, then the closed-loop system is BIBO stable fromrtoy.
Proof of the Passivity Theorem S 1 strictly passive ands 2 passive give Therefore ǫ ( y 1 2 T + e 1 2 T) y1,e 1 T + y 2,e 2 T = y,r T or y 1 2 T + r 1 y 2,r 1 y 2 T 1 ǫ y,r T y 1 2 T + y 2 2 T 2 y 2,r 2 T + r 1 2 T 1 ǫ y,r T Finally y 2 T 2 y 2,r 2 T + 1 ǫ y,r T ( 2+ 1 ) y T r T ǫ LettingT gives y 2 C r 2 and the result follows
Passivity Theorem is a Small Phase Theorem r 1 e 1 S 1 y 1 y 2 e 2 r 2 S 2 φ 2 φ 1
Example Gain Adaptation Applications in channel estimation in telecommunication, noise cancelling etc. u θ Process G(s) y θ(t) G(s) y m Model Adaptation law: dθ dt = γu(t)[y m(t) y(t)], γ >0.
Gain Adaptation Closed-Loop System u θ G(s) y γ s θ θ(t) G(s) y m
Gain Adaptation is BIBO Stable (θ θ )u G(s) y m y θ θ γ s u S S is passive (Exercise 4.12), so the closed-loop system is BIBO stable if G(s) is strictly passive.
Simulation of Gain Adaptation LetG(s)= 1 s+1 +ǫ, γ =1,u=sint, θ(0)=0and γ =1 2 0 y,y m 2 0 5 10 15 20 1.5 1 θ 0.5 0 0 5 10 15 20
Storage Function Consider the nonlinear control system ẋ=f(x,u), y=h(x) A storage function is ac 1 functionv: R n R such that V(0)=0 and V(x) 0, V(x) u T y, x,u x =0 Remark: V(T) represents the stored energy in the system V(x(T)) }{{} stored energy att=t T>0 T y(t)u(t)dt 0 }{{} absorbed energy + V(x(0)), }{{} stored energy att=0
Storage Function and Passivity Lemma: If there exists a storage functionv for a system ẋ=f(x,u), y=h(x) withx(0)=0, then the system is passive. Proof: For allt>0, y,u T = T 0 y(t)u(t)dt V(x(T)) V(x(0))=V(x(T)) 0
Lyapunov vs. Passivity Storage function is a generalization of Lyapunov function Lyapunov idea: Energy is decreasing V 0 Passivity idea: Increase in stored energy Added energy V u T y
Example KYP Lemma Consider an asymptotically stable linear system ẋ=ax+bu, y=cx Assume there exists positive definite symmetric matrices P, Q such that A T P+PA= Q, and B T P=C ConsiderV=0.5x T Px. Then V=0.5(ẋ T Px+x T Pẋ)=0.5x T (A T P+PA)x+u T B T Px = 0.5x T Qx+u T y<u T y, x =0 (2) and hence the system is strictly passive. This fact is part of the Kalman-Yakubovich-Popov lemma.
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