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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 closedloop 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 timeinvariant 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 2norm (energy norm): supnorm: 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 boundedinput boundedoutput (BIBO) stable if . Example: If ! ! i! ! EF G ! I ! u c ! # # p # i Theorem: Assume and are BIBO stable. If to then the closedloop 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 ExampleStatic 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 closedloop 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
# # ExampleStatic 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 closedloop ! # 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 CauchySchwarz 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 # # # ExamplePassive 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 closedloop 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 closedloop 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 ExampleGain Adaptation Applications in telecommunication channel estimation and in noise cancellation etc. PSfrag replacements Gain AdaptationClosedLoop 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 closedloop 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 ExampleKYP 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 KalmanYakubovichPopov 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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This note was uploaded on 05/06/2010 for the course ECE 514 taught by Professor Chaoukit.abdallah during the Spring '09 term at University of New Brunswick.
 Spring '09
 ChaoukiT.Abdallah

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