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Module 7 p3

# Module 7 p3 - g replacements Today's Goal You should be...

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Unformatted text preview: g replacements Today's Goal You should be able to 2E1262 Nonlinear Control derive the gain of a system analyze stability using Small Gain Theorem Circle Criterion Passivity Lecture 5 Inputoutput stability Lecture 5 1 October 16, 2003 Lecture 5 2 October 16, 2003 History For what Gain Idea: Generalize static gain to nonlinear dynamical systems PSfrag replacements PSfrag replacements and is the closed-loop system stable? The gain of Here should tell what is the largest amplification from to Lure and Postnikov's problem (1944) can be a constant, a matrix, a linear time-invariant system, etc Aizerman's conjecture (1949) (False!) Question: How should we measure the size of and ? Kalman's conjecture (1957) (False!) Solution by Popov (1960) (Lead to the Circle Criterion) Lecture 5 3 October 16, 2003 Lecture 5 4 October 16, 2003 Norms Gain of a Matrix %4 % A norm Every matrix 2 3 2 97 8 65 measures size. has a singular value decomposition Definition: A norm is a function 0 0 A A 8 5 8 7 , such that for all where 0 # 6 5 @ @ ) %1 7 Examples: Euclidean norm: Max norm: and diag , , and . The "gain" of 2 is the largest singular value of 2 2 : # 2 DC @ @ EF B , for all Q PI H % G " ! \$ ! # where is the Euclidean norm. & 0 0 (' 0 # % ) 1 Lecture 5 5 October 16, 2003 Lecture 5 6 October 16, 2003 Parseval's Theorem dc b! denotes the space of signals with bounded energy: b! ! Signal Norms Theorem: If have the Fourier transforms V rs ` rs ` V R p qi p qi A signal is a function . A signal norm is a norm on the space of signals . Examples: YX 2-norm (energy norm): sup-norm: e fg t XY XY X fg X h W W h S then V u 8 t e 0 X Y V X X fg fg Yg V ! X v w ph h ! V W W U T EF V In particular, V V u ! ! ! X a PI ` G 0 e ! v w ph h V W X Y Yg X fg Lecture 5 7 October 16, 2003 Lecture 5 8 October 16, 2003 PSfrag replacements System Gain . A system replacements PSfrag is a map from The gain of ! ! Example: The gain of a static gain ! ! b! # # b! to : PSfrag replacements 2 minute exercise: Show that . ! # EF EF is defined as G G I I ! ! is X X ! ! EF EF G G ! I ! I Lecture 5 9 October 16, 2003 Lecture 5 10 October 16, 2003 Gain of a Static Nonlinearity Gain of a Stable Linear System 10 1 and Lemma: Lemma: A static nonlinearity has gain 8 such that . 0 Proof: where ! ! 8 8 ! X Y PSfrag replacements 10 ! PSfrag replacements -1 fg EF EF 10 G G s I ! V W I -2 X X 0 1 10 -1 10 10 10 8 g d fg fg and Proof: Assume for for some . Parseval's theorem gives 8 g V u ! Yg ! fg t V V ! ! X ! ! v w Y X , ph ! X V ! ! V u ! ! ! ! U U W W u 8 X X , , gives equality, so Yg fg fg 5 ! v w ph V EF G ! 8 E I g X Arbitrary close to equality by choosing close to X October 16, 2003 Lecture 5 12 . Lecture 5 11 October 16, 2003 Sfrag replacements g replacements BIBO Stable The Small Gain Theorem PSfrag replacements i # # dc # Definition: is bounded-input bounded-output (BIBO) stable if . Example: If ! ! i! ! EF G ! I ! u c ! # # p # i Theorem: Assume and are BIBO stable. If to then the closed-loop system (from ! i! # # , ) is BIBO stable. A is asymptotically stable then is BIBO stable. Lecture 5 13 October 16, 2003 Lecture 5 14 October 16, 2003 Example--Static Nonlinear Feedback "Proof" of the Small Gain Theorem i i ! ! # # # ! ! ! !# PSfrag replacements gives # ! ! ! ! i # ! u ! # u i dc dc dc c ! ! # # # , , ! give ! . ! Similarly we get v # # ! ! ! ! u i! ! u ! # v and . Small Gain Theorem gives BIBO stability for u v i! . so also is bounded. Note: Formal proof requires , see Khalil ! Lecture 5 15 October 16, 2003 Lecture 5 16 October 16, 2003 Sfrag replacements 1 To: Y(1) placements The Nyquist Theorem Small Gain Theorem can be Conservative From: U(1) d PSfrag replacements 0.5 u v Let in the previous example. Then the Nyquist Theorem proves stability for all , while the Small Gain Theorem only proves stability for . Theorem: If is stable and the Nyquist curve , does not encircle , then the closed-loop system is stable. 0 PSfrag replacements 1 -0.5 -1 -1 -0.5 0 0.5 1 PSfrag replacements 0.5 0 -0.5 d fg g , -1 -1 -0.5 0 0.5 1 1.5 2 u Lecture 5 17 October 16, 2003 Lecture 5 18 October 16, 2003 The Circle Criterion # # Example--Static Nonlinear Feedback (cont'd) PSfrag replacements S S PSfrag replacements Theorem: Assume If the Nyquist curve of defined by the points system is BIBO stable. PSfrag replacements 1 PSfrag replacements 0.5 0 fg -0.5 -1 -1 -0.5 0 is asymptotically stable and 0.5 1 1.5 2 # u u u d ! and . ! # The "circle" is defined by Since & u Re fg ! " ! u u does not encircle or intersect the circle and , then the closed-loop ! # the Circle Criterion gives that the system is BIBO stable if . Lecture 5 19 October 16, 2003 Lecture 5 20 October 16, 2003 PSfrag replacements replacements Sfrag replacements S S Proof of the Circle Criterion v # ! ! # # Let , PSfrag replacements , and : ! # v u PSfrag replacements fg fg # u c Small Gain Theorem gives stability if is stable. Hence, u , where # u ! u fg fg Lecture 5 21 October 16, 2003 Lecture 5 22 October 16, 2003 Scalar Product PSfrag replacements Scalar product for signals and p # 3 fg mapped ) YX X X 1 The curve through # # u fg h and the circle gives the result: W If and are interpreted as vectors, then E October 16, 2003 PSfrag replacements is the length of is the angle between and Cauchy-Schwarz Inequality: E . u X EX w and Example: because are orthogonal if , fg E Lecture 5 23 Lecture 5 24 October 16, 2003 Passive System PSfrag replacements PSfrag replacements Definition: Consider signals passive if PSfrag replacements . The system is ! for all # and strictly passive if there exists # p such that ! # 2 minute exercise: Assume and are passive. Are then the parallel connections and the series connections above passive? What about ? for all Warning: There exist many other definitions for strictly passive ! Lecture 5 25 October 16, 2003 Lecture 5 26 October 16, 2003 PSfrag replacements Feedback of Passive Systems is Passive i # # # Example--Passive Electrical Components # ! f X f YX f X f ! f X i! ! Y ! ! ! h W Y # and ! f Y X f X v XY XY h W Lemma: If to # # are passive then the closed-loop system from is also passive. ! Proof: ! f Yf Yf Y X v X f f YX YX h W i i! # # Hence, if and . ! Lecture 5 27 October 16, 2003 Lecture 5 28 October 16, 2003 PSfrag replacements ! Passivity of Linear Systems PSfrag replacements A Strictly Passive System Has Finite Gain is passive if g Theorem: An asymptotically stable linear system and only if Re It is strictly passive if and only if there exists EF fg Re Example: PSfrag replacements dc Lemma: If G is strictly passive then I . such that Proof: ! g fg 0.6 V ! V V V ! ! V 0.4 0.2 ! u ! ! Hence, ! fg 0 , so u but not strictly passive. -0.4 -0.2 is passive, u ! 0 0.2 0.4 0.6 0.8 1 ! Lecture 5 29 October 16, 2003 Lecture 5 30 October 16, 2003 Proof # strictly passive and ! # ! # i passive give The Passivity Theorem i i! # # ! i # # ! # Therefore u # # # ! # ! ! or u ! i! ! v ! ! ! ! # ! # ! Hence Theorem: If is strictly passive and closed-loop system is BIBO stable from ! # u u ! is passive, then the to . v v ! ! Let d and the result follows. Lecture 5 31 October 16, 2003 Lecture 5 32 October 16, 2003 g replacements The Passivity Theorem is a "Small Phase Theorem" PSfrag replacements i # ! # # # # ! i! ! PSfrag replacements ! v 2 minute exercise: Apply the Passivity Theorem and compare it with the Nyquist Theorem. What about conservativeness? [Compare the discussion on the Small Gain Theorem in Lecture 5.] E w ! ! passive ! v w E c # # # strictly passive Lecture 5 33 October 16, 2003 Lecture 5 34 October 16, 2003 PSfrag replacements Example--Gain Adaptation Applications in telecommunication channel estimation and in noise cancellation etc. PSfrag replacements Gain Adaptation--Closed-Loop System Process 8 8 X X Model Adaptation law: Y 0 X X X YX Lecture 5 35 October 16, 2003 Lecture 5 36 October 16, 2003 Gain PSfrag replacements u Adaptation is BIBO Stable u E Simulation of Gain Adaptation 8 X Let , , , and . u 2 8 , 0 -2 0 5 10 15 1.5 20 PSfrag replacements 0.5 1 is passive (see exercises). If is strictly passive, the closed-loop system is BIBO stable 0 0 5 10 15 20 Lecture 5 37 October 16, 2003 Lecture 5 38 October 16, 2003 Storage Function Consider the nonlinear control system Storage Function and Passivity 7 A storage function is a % Lemma: If there exists a storage function for a system # 7 function such that Remark: 7 7 and , with , then the system is passive. 7 , Proof: For all , 7 7 represents the stored energy in the system 7 7 7 XY X X h W ` stored energy at YX X X stored energy at 7 , W h W ` input energy Lecture 5 39 October 16, 2003 Lecture 5 40 October 16, 2003 Example--KYP Lemma Consider an asymptotically stable linear system Lyapunov vs. Passivity Storage function is a generalization of Lyapunov function Assume there exists positive definite matrices Lyapunov idea: "Energy is decreasing" such that Passivity idea: "Increase in stored energy 7 7 Consider Added energy" . If the linear system is strictly passive, then v 7 7 v c This fact is part of the Kalman-Yakubovich-Popov lemma Lecture 5 41 October 16, 2003 Lecture 5 42 October 16, 2003 Today's Goal You should be able to derive the gain of a system Next Lecture Analysis of periodic solutions using describing functions analyze stability using Small Gain Theorem Circle Criterion Passivity Lecture 5 43 October 16, 2003 Lecture 5 44 October 16, 2003 ...
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