Martin Corless - Project Jupyter · 2.1 Examples 2.1.1 Continuous-time A ﬁrst example Figure 2.1:...

1

AAE 666 Notes

Martin Corless

School of Aeronautics & AstronauticsPurdue University

West Lafayette, [email protected]

February 15, 2013

Contents

1 Introduction 9

2 State space representation of dynamical systems 112.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2.1.1 Continuous-time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.1.2 Discrete-time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

2.2 General representation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182.2.1 Continuous-time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182.2.2 Discrete-time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

2.3 Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212.3.1 Vector spaces and IRn . . . . . . . . . . . . . . . . . . . . . . . . . . 212.3.2 IR2 and pictures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232.3.3 Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

2.4 Vector representation of dynamical systems . . . . . . . . . . . . . . . . . . . 252.5 Solutions and equilibrium states . . . . . . . . . . . . . . . . . . . . . . . . . 27

2.5.1 Continuous-time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272.5.2 Discrete-time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

2.6 Numerical simulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302.6.1 MATLAB . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302.6.2 SIMULINK . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

3 First order systems 333.1 Continuous time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333.2 Solutions to scalar first order ODES . . . . . . . . . . . . . . . . . . . . . . . 353.3 Discrete time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

4 Second order systems 374.1 Linear systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 374.2 Linearization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 394.3 Isocline method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 424.4 Periodic solutions and limit cycles . . . . . . . . . . . . . . . . . . . . . . . . 42

5 Some general considerations 455.1 Existence of solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 455.2 Uniqueness of solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

3

4 CONTENTS

5.3 Indefinite extension of solutions . . . . . . . . . . . . . . . . . . . . . . . . . 46

6 Stability and boundedness 476.1 Boundedness of solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 476.2 Stability of equilibrium states . . . . . . . . . . . . . . . . . . . . . . . . . . 506.3 Asymptotic stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

6.3.1 Global asymptotic stability . . . . . . . . . . . . . . . . . . . . . . . 526.3.2 Asymptotic stability . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

6.4 Exponential stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 546.5 LTI systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 566.6 Linearization and stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

7 Stability and boundedness: discrete time 617.1 Boundedness of solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 617.2 Stability of equilibrium states . . . . . . . . . . . . . . . . . . . . . . . . . . 627.3 Asymptotic stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63


7.4 Exponential stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 647.5 LTI systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 667.6 Linearization and stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

8 Basic Lyapunov theory 698.1 Lyapunov conditions for stability . . . . . . . . . . . . . . . . . . . . . . . . 71

8.1.1 Locally positive definite functions . . . . . . . . . . . . . . . . . . . . 718.1.2 A stability result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

8.2 Asymptotic stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 768.3 Boundedness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

8.3.1 Radially unbounded functions . . . . . . . . . . . . . . . . . . . . . . 788.3.2 A boundedness result . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

8.4 Global asymptotic stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . 818.4.1 Positive definite functions . . . . . . . . . . . . . . . . . . . . . . . . 818.4.2 A result on global asymptotic stability . . . . . . . . . . . . . . . . . 82

8.5 Exponential stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 858.5.1 Global exponential stability . . . . . . . . . . . . . . . . . . . . . . . 858.5.2 Proof of theorem 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 868.5.3 Exponential stability . . . . . . . . . . . . . . . . . . . . . . . . . . . 878.5.4 A special class of GES systems . . . . . . . . . . . . . . . . . . . . . 888.5.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

9 Basic Lyapunov theory: discrete time 919.1 Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 929.2 Asymptotic stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 939.3 Boundedness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 949.4 Global asymptotic stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

CONTENTS 5

9.5 Exponential stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

10 Lyapunov theory for linear time-invariant systems 9910.1 Positive and negative (semi)definite matrices . . . . . . . . . . . . . . . . . . 99

10.1.1 Definite matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9910.1.2 Semi-definite matrices . . . . . . . . . . . . . . . . . . . . . . . . . . 101

10.2 Lyapunov theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10310.2.1 Asymptotic stability results . . . . . . . . . . . . . . . . . . . . . . . 10310.2.2 MATLAB. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10710.2.3 Stability results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108

10.3 Mechanical systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11010.4 Rates of convergence for linear systems . . . . . . . . . . . . . . . . . . . . . 11310.5 Linearization and exponential stability . . . . . . . . . . . . . . . . . . . . . 114

11 Quadratic stability 11711.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11711.2 Polytopic nonlinear systems . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

11.2.1 LMI Control Toolbox . . . . . . . . . . . . . . . . . . . . . . . . . . . 11911.2.2 Generalization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12211.2.3 Robust stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123

11.3 Another special class of nonlinear systems . . . . . . . . . . . . . . . . . . . 12411.3.1 The circle criterion: a frequency domain condition . . . . . . . . . . . 12811.3.2 Sector bounded nonlinearities . . . . . . . . . . . . . . . . . . . . . . 12911.3.3 Generalization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131

11.4 Yet another special class . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13411.4.1 Generalization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13511.4.2 Strictly positive real transfer functions and a frequency domain condition137

12 Invariance results 14312.1 Invariant sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14312.2 Limit sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14512.3 LaSalle’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14812.4 Asymptotic stability and LaSalle . . . . . . . . . . . . . . . . . . . . . . . . 151


12.5 LTI systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15412.6 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155

12.6.1 A simple stabilizing adaptive controller . . . . . . . . . . . . . . . . . 15512.6.2 PI controllers for nonlinear systems . . . . . . . . . . . . . . . . . . . 156

13 Stability of nonautonomous systems 15913.1 Stability and boundedness . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160

13.1.1 Boundedness of solutions . . . . . . . . . . . . . . . . . . . . . . . . . 16013.1.2 Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16013.1.3 Asymptotic stability . . . . . . . . . . . . . . . . . . . . . . . . . . . 162

6 CONTENTS

13.1.4 Global asymptotic stability . . . . . . . . . . . . . . . . . . . . . . . 16213.1.5 Exponential stability . . . . . . . . . . . . . . . . . . . . . . . . . . . 163

14 Lyapunov theory for nonautonomous systems 16514.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16514.2 Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167

14.2.1 Locally positive definite functions . . . . . . . . . . . . . . . . . . . . 16714.2.2 A stability theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167

14.3 Uniform asymptotic stability . . . . . . . . . . . . . . . . . . . . . . . . . . . 16814.4 Global uniform asymptotic stability . . . . . . . . . . . . . . . . . . . . . . . 169

14.4.1 Positive definite functions . . . . . . . . . . . . . . . . . . . . . . . . 16914.4.2 A result on global uniform asymptotic stability . . . . . . . . . . . . 16914.4.3 Proof of theorem 28 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170

14.5 Exponential stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17214.5.1 Exponential stability . . . . . . . . . . . . . . . . . . . . . . . . . . . 17214.5.2 Global uniform exponential stability . . . . . . . . . . . . . . . . . . 172

14.6 Quadratic stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17214.7 Boundedness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17314.8 Attractive and invariant regions . . . . . . . . . . . . . . . . . . . . . . . . . 17514.9 Systems with disturbance inputs . . . . . . . . . . . . . . . . . . . . . . . . . 176

14.9.1 Linear systems with bounded disturbance inputs . . . . . . . . . . . . 17814.10Regions of attraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180

14.10.1A special class of nonlinear systems . . . . . . . . . . . . . . . . . . . 18114.11Linear time-varying systems . . . . . . . . . . . . . . . . . . . . . . . . . . . 183

14.11.1Lyapunov functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18314.12Linearization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185

15 The describing function method 18715.1 Periodic functions and Fourier series . . . . . . . . . . . . . . . . . . . . . . 18815.2 Describing functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18915.3 The describing function method . . . . . . . . . . . . . . . . . . . . . . . . . 193

15.3.1 Rational transfer functions . . . . . . . . . . . . . . . . . . . . . . . . 19715.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198

16 Aerospace/Mechanical Systems 20316.1 Aerospace/Mechanical systems . . . . . . . . . . . . . . . . . . . . . . . . . . 20316.2 Equations of motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20416.3 Some fundamental properties . . . . . . . . . . . . . . . . . . . . . . . . . . 207

16.3.1 An energy result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20716.3.2 Potential energy and total energy . . . . . . . . . . . . . . . . . . . . 20816.3.3 A skew symmetry property . . . . . . . . . . . . . . . . . . . . . . . . 210

16.4 Linearization of aerospace/mechanical systems . . . . . . . . . . . . . . . . . 21216.5 A class of GUAS aerospace/mechanical systems . . . . . . . . . . . . . . . . 21316.6 Control of aerospace/mechanical systems . . . . . . . . . . . . . . . . . . . . 215

16.6.1 Computed torque method . . . . . . . . . . . . . . . . . . . . . . . . 215

CONTENTS 7

16.6.2 Linearization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215

16.6.3 Globally stabilizing PD controllers . . . . . . . . . . . . . . . . . . . 216

17 Quadratic stabilizability 219

17.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219

17.2 Some special classes of nonlinear systems . . . . . . . . . . . . . . . . . . . . 220

17.2.1 A special class of nonlinear systems . . . . . . . . . . . . . . . . . . . 221

17.2.2 Generalization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226

17.3 Another special class of nonlinear systems . . . . . . . . . . . . . . . . . . . 228

17.3.1 Generalization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 230

17.4 Yet another special class . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232

17.4.1 Generalization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233

18 Singularly perturbed systems 235

18.1 Introduction and examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235

18.1.1 High gain output feedback . . . . . . . . . . . . . . . . . . . . . . . . 235

18.1.2 Single-link manipulator with flexible joint . . . . . . . . . . . . . . . 236

18.2 Singularly perturbed systems . . . . . . . . . . . . . . . . . . . . . . . . . . 238

18.3 Linear systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241

18.4 Exact slow and fast subsystems . . . . . . . . . . . . . . . . . . . . . . . . . 242

18.4.1 Approximations of the slow and fast systems . . . . . . . . . . . . . . 244

18.4.2 A special class of systems . . . . . . . . . . . . . . . . . . . . . . . . 245

19 Input output systems 251

19.1 Input-output systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251

19.1.1 Causality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253

19.2 Signal norms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253

19.3 Input-output stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255

19.4 General convolution systems . . . . . . . . . . . . . . . . . . . . . . . . . . . 256

19.4.1 Systems with L1 impulse responses . . . . . . . . . . . . . . . . . . . 256

19.4.2 Systems with impulses in impulse response . . . . . . . . . . . . . . . 261

19.4.3 L2 gain and H∞ norm . . . . . . . . . . . . . . . . . . . . . . . . . . 262

19.4.4 Finite dimensional linear systems . . . . . . . . . . . . . . . . . . . . 263

19.4.5 Linear differential delay systems . . . . . . . . . . . . . . . . . . . . . 267

19.5 A class of Lp stable nonlinear systems . . . . . . . . . . . . . . . . . . . . . . 269

19.6 Small gain theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 270

19.6.1 Stability of feedback systems . . . . . . . . . . . . . . . . . . . . . . . 270

19.6.2 Truncations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 270

19.6.3 Small gain theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271

19.7 Application of the small gain theorem: circle criterion . . . . . . . . . . . . . 272

19.7.1 A more general circle criterion . . . . . . . . . . . . . . . . . . . . . . 274

8 CONTENTS

20 Nonlinear H∞ 27720.1 Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277

20.1.1 The HJ Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27720.2 Control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28020.3 Series solution to control problem . . . . . . . . . . . . . . . . . . . . . . . . 281

20.3.1 Linearized problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28120.3.2 Nonlinear problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282

21 Performance 28521.1 Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28521.2 Polytopic systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 286

21.2.1 Polytopic models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28621.2.2 Performance analysis of polytopic systems . . . . . . . . . . . . . . . 289

21.3 Control for performance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29021.3.1 Linear time-invariant systems . . . . . . . . . . . . . . . . . . . . . . 29021.3.2 Control of polytopic systems . . . . . . . . . . . . . . . . . . . . . . . 29121.3.3 Multiple performance outputs . . . . . . . . . . . . . . . . . . . . . . 292

22 Appendix 29322.1 The Euclidean norm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293

Chapter 1

Introduction

HI!(How is it going?)

This is a course on the analysis and control of nonlinear systems.

Why study nonlinear systems?

• Most physical systems are nonlinear.• Nonlinear systems exhibit many types of behavior which do not occur in linear systems:several equilibrium conditions; isolated periodic solutions; chaos.• Some systems are not linearizable.• Linearization yields local behavior only.

Text: Khalil, K. H., Nonlinear Systems, 2nd Edition, Prentice Hall, 1996.

ReferencesSlotine J.-J. E., and Li, W., Applied Nonlinear Control, Prentice Hall, 1991.Vidyasagar, M., Nonlinear Systems Analysis, 2nd Edition, Prentice Hall, 1993.Boyd, S., El Ghaoui, L., Feron, E., and Balakrishnan, V., Linear Matrix Inequalities inSystem and Control Theory, SIAM, 1994.

9

10 CHAPTER 1. INTRODUCTION

Chapter 2

State space representation ofdynamical systems

2.1 Examples

2.1.1 Continuous-time

A first example

Figure 2.1: The first nonlinear system

Consider the motion of a small block of mass m moving along a straight horizontal line.Suppose there is Coulomb (or dry) friction between the block and surface with coefficent offriction µ > 0. Letting t ∈ IR represent time and v(t) ∈ IR represent the velocity of theblock at time t, the motion of the block can be described by the following first order ordinarydifferential equation (ODE):

mv = −µmg sgm (v)

where the signum function, sgm, is defined by

sgm (v) :=

−1 if v < 00 if v = 01 if v > 0

11

12 CHAPTER 2. STATE SPACE REPRESENTATION OF DYNAMICAL SYSTEMS

Figure 2.2: signum function

and g is the gravitational acceleration constant of the planet on which the block resides.Introducing x := v results in

x = −α sgm (x)

and α := µg.

Planar pendulum

Figure 2.3: Planar pendulum

Jθ +Wl sin θ = 0

Introducing the state variables,

x1 := θ and x2 := θ ,

results in the state space description:

x1 = x2

x2 = −a sin x1

where a :=Wl/J > 0.

Attitude dynamics of a rigid body

The equations describing the rotational dynamics of a rigid body are given by Euler’s equa-tions of motion. If we choose the axes of a body-fixed reference frame along the principal

2.1. EXAMPLES 13

Figure 2.4: Attitude dynamics

axes of inertia of the rigid body with origin at the center of mass, Euler’s equations of motiontake the simplified form

I1ω1 = (I2 − I3)ω2ω3

I2ω2 = (I3 − I1)ω3ω1

I3ω3 = (I1 − I2)ω1ω2

where ω1, ω2, ω3 denote the components of the body angular velocity vector with respectto the body principal axes, and the positive scalars I1, I2, I3 are the principal moments ofinertia of the body with respect to its mass center.

Body in central force motion

Figure 2.5: Central force motion

r − rω2 + g(r) = 0rω + 2rω = 0

For the simplest situation in orbit mechanics ( a “satellite” orbiting YFHB)

g(r) = µ/r2 µ = GM

where G is the universal constant of gravitation and M is the mass of YFHB.


Ballistics in drag

Figure 2.6: Ballistics in drag

x = v cos γ

y = v sin γ

mv = −mg sin γ −D(v)

mvφ = −mg cos γ

2.1. EXAMPLES 15

Double pendulum on cart

Figure 2.7: Double pendulum on cart

The motion of the double pendulum on a cart can be described by

(m0 +m1 +m2)y −m1l1 cos θ1 θ1 −m2l2 cos θ2 θ2 + m1l1 sin θ1θ21 +m2l2 sin θ2θ

22 = u

−m1l1 cos θ1 y +m1l21θ1 + m1l1g sin θ1 = 0

−m2l2 cos θ2 y +m2l22θ2 + m2l2g sin θ2 = 0

Figure 2.8: Double pendulum on cart


Two-link robotic manipulator

u1

q1

q2

u2

lc1

lc2

Payload

g^

Figure 2.9: A simplified model of a two link manipulator

The coordinates q1 and q2 denote the angular location of the first and second links relativeto the local vertical, respectively. The second link includes a payload located at its end. Themasses of the first and the second links are m1 and m2, respectively. The moments of inertiaof the first and the second links about their centers of mass are I1 and I2, respectively. Thelocations of the center of mass of links one and two are determined by lc1 and lc2, respectively;l1 is the length of link 1. The equations of motion for the two arms are described by:

[m1lc21 +m2l

21 + I1]q1 +[m2l1lc2 cos(q1 − q2)]q2 +m2l1lc2 sin(q1 − q2)q

22 − [m1lc1 +m2l1]g sin(q1) = u1

[m2l1lc2 cos(q1 − q2)]q1 +[m2lc22 + I2]q2 −m2l1lc2 sin(q1 − q2)q

21 −m2glc2 sin(q2) = u2

2.1. EXAMPLES 17

Traffic flow

Two roads connecting two points. Let x1 and x2 be the traffic flow rate (number of cars perhour) along the two routes.

x1 = −σ(c1(x)− c2(x))x1 + σ(c2(x)− c1(x))x2

x2 = σ(c1(x)− c2(x))x1 − σ(c2(x)− c1(x))x2

where

σ(y) =

{

y if y ≥ 00 if y < 0

ci(x) is the travel time along route i.

c1(x) = x21 + 1 c2(x) = 2x2

2.1.2 Discrete-timeExample 1 (Population dynamics)

x(k+1) = ax(k)

Example 2 (Computer control)

Example 3 (Iterative numerical algorithm) Recall Newton’s algorithm for solving theequation g(x) = 0:

x(k+1) = x(k)− g(x(k))

g′(x(k))

Example 4 (Consensus algorithm)


2.2 General representation


All of the preceding systems can be described by a bunch of first order ordinary differentialequations of the form

x1 = f1(x1, x2, . . . , xn)x2 = f2(x1, x2, . . . , xn)

...xn = fn(x1, x2, . . . , xn)

where the scalars xi(t), i = 1, 2, . . . , n, are called the state variables and the real scalar t iscalled the time variable.

Higher order ODE descriptions

Single equation. (Recall the planar pendulum.) Consider a system described by a singlenth- order differential equation of the form

F (q, q, . . . , q(n)) = 0

where q(t) is a scalar and q(n) := dnqdtn

. To obtain a state space description of this system, weneed to assume that we can solve for q(n) as a function of q, q, . . . , q(n−1). So suppose thatthe above equation is equivalent to

q(n) = a(q, q, . . . , q(n−1)) .

Now let

x1 := q

x2 := q...

xn := q(n−1)

to obtain the following state space description:

x1 = x2

x2 = x3...

xn−1 = xn

xn = a(x1, x2, . . . , xn)

2.2. GENERAL REPRESENTATION 19

Multiple equations. (Recall the two link manipulator.) Consider a system described bym scalar differential equations in N variables:

F1(q1, q1, . . . , q(n1)1 , q2, q2, . . . , q

(N2)2 , . . . , qN , qN , . . . , q

(nN )N ) = 0

F2(q1, q1, . . . , q(n1)1 , q2, q2, . . . , q

(n2)2 , . . . , qN , qN , . . . , q

(nN )N ) = 0

...

FN(q1, q1, . . . , q(n1)1 , q2, q2, . . . , q

(n2)2 , . . . , qN , qN , . . . , q

(nN )N ) = 0

where t, q1(t), q2(t), . . . , qN(t) are real scalars. Note that q(ni)i is the highest order derivative

of qi which appears in the above equations.First solve for the highest order derivatives, q

(n1)1 , q

(n2)2 , . . . , q

(nN )N , to obtain:

q(n1)1 = a1( q1, q1, . . . q

(n1−1)1 , q2, q2, . . . , q

(n2−1)2 , . . . , qN , qN , . . . , q

(nN−1)N )

q(n2)2 = a2( q1, q1, . . . q

(n1−1)1 , q2, q2, . . . , q

(n2−1)2 , . . . , qN , qN , . . . , q

(nN−1)N )

...

q(nN )N = aN ( q1, q1, . . . q

(n1−1)1 , q2, q2, . . . , q

(n2−1)2 , . . . , qN , qN , . . . , q

(nN−1)N )

Now let

x1 := q1 x2 := q1 . . . xn1 := q(n1−1)1

xn1+1 := q2, xn1+2 := q2 . . . xn1+n2 := q(n2−1)2

...

xn1+...+nm−1+1 := qN xn1+...+nm−1+2 := qN . . . xn := q(nN−1)N

wheren := n1 + n2 + . . .+ nN

to obtain

x1 = x2

x2 = x3...

xn1−1 = xn1

xn1 = a1(x1, x2, . . . , xn)

xn1+1 = xn1+2

...

xn1+n2 = a2(x1, x2, . . . , xn)...

xn = an(x1, x2, . . . , xn)

Example 5

q1 + q2 + 2q1 = 0

−q1 + q1 + q2 + 4q2 = 0


2.2.2 Discrete-time

A general state space description of a discrete-time system consists of a bunch of first orderdifference equations of the form

x1(k+1) = f1(x1(k), x2(k), . . . , xn(k))x2(k+1) = f2(x1(k), x2(k), . . . , xn(k))

...xn(k+1) = fn(x1(k), x2(k), . . . , xn(k))

where the scalars xi(k), i = 1, 2, . . . , n, are called the state variables and the integer k iscalled the time variable.

Higher order DE descriptions

Story is similar to continuous-time.

2.3. VECTORS 21

2.3 Vectors

2.3.1 Vector spaces and IRn

A scalar is a real or a complex number. The symbols IR and C represent the set of real andcomplex numbers, respectively.

In this section all the definitions and results are given for real scalars. However, they alsohold for complex scalars; to get the results for complex scalars, simply replace ‘real’ with‘complex’ and IR with C.

Consider any positive integer n. A real n-vector x is an ordered n-tuple of real numbers,x1, x2, . . . , xn. This is is usually written as follows:

x =

x1x2...xn

or x =(

x1, x2, · · · , xn)

The real numbers x1, x2, . . . , xn are called the scalar components of x; xi is called the i-thcomponent. The symbol IRn represents the set of ordered n-tuples of real numbers.

Addition. The addition of any two elements x, y of IRn yields an element of IRn and isdefined by:

x+ y =

x1x2...xn

+

y1y2...yn

:=

x1 + y1x2 + y2

...xn + yn

Zero element of IRn:

0 :=

00...0

Note that we are using the same symbol, 0, for a zero scalar and a zero vector.

The negative of x:

−x :=

−x1−x2...

−xn


Properties of addition

(a) (Commutative). For each pair x, y in IRn,

x+ y = y + x

(b) (Associative). For each x, y, z in IRn,

(x+ y) + z = x+ (y + z)

(c) There is an element 0 in IRn such that for every x in IRn,

x+ 0 = x

(d) For each x in IRn, there is an element −x in IRn such that

x+ (−x) = 0

Scalar multiplication. The multiplication of an element x of IRn by a real scalar α yieldsan element of IRn and is defined by:

αx = α

x1x2...xn

:=

αx1αx2...

αxn

Properties of scalar multiplication

(a) For each scalar α and pair x, y in IRn

α(x+ y) = αx+ αy

(b) For each pair of scalars α, β and x in IRn,

(α + β)x = αx+ βx

(c) For each pair of scalars α, β, and x in IRn,

α(βx) = (αβ)x

(d) For each x in IRn,

1x = x

2.3. VECTORS 23

Vector space. Consider any set V equipped with an addition operation and a scalarmultiplication operation. Suppose the addition operation assigns to each pair of elementsx, y in V a unique element x + y in V and it satisfies the above four properties of addition(with IRn replaced by V). Suppose the scalar multiplication operation assigns to each scalarα and element x in V a unique element αx in V and it satisfies the above four propertiesof scalar multiplication (with IRn replaced by V). Then this set (along with its additionand scalar multiplication) is called a vector space. Thus IRn equipped with its definitionsof addition and scalar multiplication is a specific example of a vector space. We shall meetother examples of vectors spaces later.

An element x of a vector space is called a vector.

A vector space with real (complex) scalars is called a real (complex) vector space.

Subtraction in a vector space is defined by:

x− y := x+ (−y)

Hence in IRn,

x− y =

x1x2...xn

−

y1y2...yn

=

x1 − y1x2 − y2

...xn − yn

2.3.2 IR2 and pictures

An element of IR2 can be represented in a plane by a point or a directed line segment.


2.3.3 Derivatives

Suppose x(·) is a function of a real variable t ∈ IR where x(t) is in IRn. Then

x :=dx

dt:=

dx1

dt

dx2

dt

...

dxn

dt

=

x1

x2

...

xn

2.4. VECTOR REPRESENTATION OF DYNAMICAL SYSTEMS 25

2.4 Vector representation of dynamical systems

Recall the general description of a dynamical system given in Section 2.2.1. For a systemwith n state variables, x1, · · · , xn, we define the state (vector) x to be the n-vector given by

x :=

x1x2...xn

.

We also introduce the n-vector valued function f defined by

f(x) :=

f1(x1, x2, . . . , xn)f2(x1, x2, . . . , xn)

...fn(x1, x2, . . . , xn)

.

Continuous time The general representation of a continuous-time dynamical system cannow be compactly described by the following first order vector differential equation:

x = f(x) (2.1)

where x(t) is an n-vector and the scalar t is the time variable. A system described bythe above equation is called autonomous or time-invariant because the right-hand side of theequation does not depend explicitly on time t. For the first part of the course, we will onlyconcern ourselves with these systems.

However, one can have a system containing time-varying parameters or “disturbanceinputs” which are time-varying. In this case the system might be described by

x = f(t, x)

that is, the right-hand side of the differential depends explicitly on time. Such a system iscalled non-autonomous or time-varying. We will look at them later.

Discrete time The general representation of a discrete-time dynamical system can nowbe compactly described by the following first order vector differential equation:

x(k+1) = f(x(k)) (2.2)

where x(k) is an n-vector and the scalar k is the time variable. A system described bythe above equation is called autonomous or time-invariant because the right-hand side of theequation does not depend explicitly on time k. For the first part of the course, we will onlyconcern ourselves with these systems.

However, one can have a system containing time-varying parameters or “disturbanceinputs” which are time-varying. In this case the system might be described by

x(k+1) = f(k, x(k))


that is, the right-hand side of the differential depends explicitly on time. Such a system iscalled non-autonomous or time-varying. We will look at them later.

2.5. SOLUTIONS AND EQUILIBRIUM STATES 27

2.5 Solutions and equilibrium states


A solution of (2.1) is any continuous function x(·) which is defined on some time interval andwhich satisfies

x(t) = f(x(t))

for all t in the time interval.An equilibrium solution is the simplest type of solution; it is constant for all time, that is,

it satisfiesx(t) ≡ xe

for some fixed state vector xe. The state xe is called an equilibrium state. Since an equilibriumsolution must satisfy the above differential equation, all equilibrium states are given by:

f(xe) = 0

or, in scalar terms,f1(x

e1, x

e2, . . . , x

en) = 0

f2(xe1, x

e2, . . . , x

en) = 0

...fn(x

e1, x

e2, . . . , x

en) = 0

Sometimes an equilibrium state is referred to as an equilibrium point, a stationary point, or,a singular point.

Example 6 (Planar pendulum) Here, all equilibrium states are given by

xe =

(

mπ0

)

where m is an arbitrary integer. Physically, there are only two distinct equilibrium states

xe =

(

00

)

and xe =

(

π0

)

An isolated equilibrium state is an equilibrium state with the property that there is aneighborhood of that state which contains no other equilibrium states.

Figure 2.10: Isolated equilibrium state


Example 7 The systemx = x− x3

has three equilibrium states: 0,−1, 1. Clearly, these are isolated equilibrium states.

Example 8 Every state of the system

x = 0

is an equilibrium state. Hence none of these equilibrium states are isolated.

Example 9 Consider the system described by

x =

{

sin(1/x) if x 6= 00 if x = 0

The origin is an equilibrium state, but, it is not isolated. All other equilibrium states areisolated.

Higher order ODEs

Consider a system described by an ordinary differential equation of the form

F (y, y, . . . , y(n)) = 0

where y(t) is a scalar. An equilibrium solution is a solution y(·) which is constant for all time,that is,

y(t) ≡ ye

for some constant ye. Hence,F (ye, 0, . . . , 0) = 0 (2.3)

For the state space description of this system introduced earlier, all equilibrium statesare given by

xe =

ye

0...0

where ye solves (2.3).

2.5. SOLUTIONS AND EQUILIBRIUM STATES 29

Multiple higher order ODEs Equilibrium solutions

yi(t) ≡ yei , i = 1, 2, . . . , m

HenceF1(y

e1, 0, . . . , y

e2, 0, . . . , . . . , y

eN , . . . , 0 ) = 0

F2(ye1, 0, . . . , y

e2, 0, . . . , . . . , y

eN , . . . , 0 ) = 0

...FN (y

e1, 0, . . . , y

e2, 0, . . . , . . . , y

eN , . . . , 0 ) = 0

(2.4)

For the state space description of this system introduced earlier, all equilibrium statesare given by

xe =

ye10...ye20......yeN...0

where ye1, ye2, . . . , y

eN solve (2.4).

Example 10 Central force motion in inverse square gravitational field

r − rω2 + µ/r2 = 0rω + 2rω = 0

Equilibrium solutionsr(t) ≡ re , ω(t) ≡ ωe

Hence,r, r, ω = 0

This yields−re(ωe)2 + µ/(re)2 = 0

0 = 0

Thus there are infinite number of equilibrium solutions given by:

ωe = ±√

µ/(re)3

where re is arbitrary. Note that, for this state space description, an equilibrium state corre-sponds to a circular orbit.

2.5.2 Discrete-time


2.6 Numerical simulation

2.6.1 MATLAB

>> help ode45

ODE45 Solve non-stiff differential equations, medium order method.

[T,Y] = ODE45(’F’,TSPAN,Y0) with TSPAN = [T0 TFINAL] integrates the

system of differential equations y’ = F(t,y) from time T0 to TFINAL with

initial conditions Y0. ’F’ is a string containing the name of an ODE

file. Function F(T,Y) must return a column vector. Each row in

solution array Y corresponds to a time returned in column vector T. To

obtain solutions at specific times T0, T1, ..., TFINAL (all increasing

or all decreasing), use TSPAN = [T0 T1 ... TFINAL].

[T,Y] = ODE45(’F’,TSPAN,Y0,OPTIONS) solves as above with default

integration parameters replaced by values in OPTIONS, an argument

created with the ODESET function. See ODESET for details. Commonly

used options are scalar relative error tolerance ’RelTol’ (1e-3 by

default) and vector of absolute error tolerances ’AbsTol’ (all

components 1e-6 by default).

[T,Y] = ODE45(’F’,TSPAN,Y0,OPTIONS,P1,P2,...) passes the additional

parameters P1,P2,... to the ODE file as F(T,Y,FLAG,P1,P2,...) (see

ODEFILE). Use OPTIONS = [] as a place holder if no options are set.

It is possible to specify TSPAN, Y0 and OPTIONS in the ODE file (see

ODEFILE). If TSPAN or Y0 is empty, then ODE45 calls the ODE file

[TSPAN,Y0,OPTIONS] = F([],[],’init’) to obtain any values not supplied

in the ODE45 argument list. Empty arguments at the end of the call list

may be omitted, e.g. ODE45(’F’).

As an example, the commands

options = odeset(’RelTol’,1e-4,’AbsTol’,[1e-4 1e-4 1e-5]);

ode45(’rigidode’,[0 12],[0 1 1],options);

solve the system y’ = rigidode(t,y) with relative error tolerance 1e-4

and absolute tolerances of 1e-4 for the first two components and 1e-5

for the third. When called with no output arguments, as in this

example, ODE45 calls the default output function ODEPLOT to plot the

solution as it is computed.

[T,Y,TE,YE,IE] = ODE45(’F’,TSPAN,Y0,OPTIONS) with the Events property in

OPTIONS set to ’on’, solves as above while also locating zero crossings

2.6. NUMERICAL SIMULATION 31

of an event function defined in the ODE file. The ODE file must be

coded so that F(T,Y,’events’) returns appropriate information. See

ODEFILE for details. Output TE is a column vector of times at which

events occur, rows of YE are the corresponding solutions, and indices in

vector IE specify which event occurred.

See also ODEFILE and

other ODE solvers: ODE23, ODE113, ODE15S, ODE23S, ODE23T, ODE23TB

options handling: ODESET, ODEGET

output functions: ODEPLOT, ODEPHAS2, ODEPHAS3, ODEPRINT

odefile examples: ORBITODE, ORBT2ODE, RIGIDODE, VDPODE

>>

2.6.2 SIMULINK


Exercises

Exercise 1 By appropriate definition of state variables, obtain a first order state spacedescription of the following systems where y1 and y2 are real scalars.

(i)

2y1 + y2 + sin y1 = 0

y1 + 2y2 + sin y2 = 0

(ii)

y1 + y2 + y31 = 0

y1 + y2 + y32 = 0

Exercise 2 By appropriate definition of state variables, obtain a first order state spacedescription of the following system where q1 and q2 are real scalars.

q1(k+2) + q1(k) + 2q2(k+1) = 0

q1(k+2) + q1(k+1) + q2(k) = 0

Exercise 3 Consider the Lorenz system described by

x1 = σ(x2 − x1)

x2 = rx1 − x2 − x1x3

x3 = −bx3 + x1x2

with σ = 10, b = 83, and r = 28. Simulate this system with initial states

010

and

01 + eps

0

where eps is the floating point relative accuracy in MATLAB. Comment on your results forthe integration interval [0 60].

Chapter 3

First order systems

The simplest type of nonlinear systems is one whose state can be described by a single scalar.We refer to such a system as a first order system.

3.1 Continuous time

A first order continuous-time system is described by

x = f(x) (3.1)

where x(t) is a scalar. As an example, recall the first nonlinear system.

State portrait. As illustrated in Figure 3.1, one may readily determine the qualitativebehavior of a first order system from the graph of f .

Figure 3.1: Qualitative behavior of a first order system

Example 11 Consider the first order system described by

x = x− x3

This system has three equilibrium states: −1, 0 and 1.

33

34 CHAPTER 3. FIRST ORDER SYSTEMS

Figure 3.2: State portrait for x = x− x3

Linear systems. A linear time-invariant first order system is described by

x = ax (3.2)

where a is a constant real scalar. All solutions of this system are of the form

x(t) = ceat

where c is an a constant real scalar. Thus the qualitative behavior of (3.2) is completelydetermined by the sign of a.

Linearization. Suppose xe is an equilibrium state for system (3.1). Then

f(xe) = 0 .

Suppose that the function f is differentable at xe with derivative f ′(xe). When x is “close”to xe, it follows from the definition of the derivative that

f(x) ≈ f(xe) + f ′(xe)(x− xe) = f ′(xe)(x− xe) .

If we introduce the “perturbed state”,

δx = x− xe,

thenδx = f(x) ≈ f ′(xe)δx .

So, the linearization of system (3.1) about an equilibrium state xe is defined to be thefollowing system:

δx = aδx where a = f ′(xe) . (3.3)

One can demonstrate the following result.

If f ′(xe) 6= 0, then the local behavior of the nonlinear system (3.1) about xe is qualitativelythe same as that of the linearized system about the origin.

Example 12

x = x− x3

Example 13

x = ax2

3.2. SOLUTIONS TO SCALAR FIRST ORDER ODES 35

3.2 Solutions to scalar first order ODES

Consider

x = f(x)

with initial condition x(0) = x0. Suppose that f(x0) 6= 0; then f(x(t)) 6= 0 for some interval[0, t1) and over this interval we have

1

f(x(t))

dx

dt= 1 .

Let

g(x) :=

∫ x

x0

1

f(η)dη

Thend

dt(g(x(t))) =

d

dx(g(x))

dx

dt=

1

f(x)

dx

dt= 1 ,

that is,d

dt(g(x(t))) = 1 .

Integrating over the interval [0, t] and using the initial condition x(0) = x0 yields g(x(t)) = t,that is,

∫ x(t)

x0

1

f(η)dη = t . (3.4)

One now solves the above equation for x(t).

Example 14 (Finite escape time) This simple example illustrates the concept of a finiteescape time.

x = x2

Here,∫ x

x0

1

η2dη = −1

x+

1

x0.

Hence, all nonzero solutions satisfy

− 1

x(t)+

1

x0= t .

Then, provided x0t ≤ 1, the above equation can be uniquely solved for x(t) to obtain

x(t) =x0

1− x0t. (3.5)

Suppose x0 > 0. Then x(t) goes to infinity as t approaches 1/x0; the solution blows up in afinite time. This cannot happen in a linear system.

36 CHAPTER 3. FIRST ORDER SYSTEMS

3.3 Discrete time

A first order continuous-time system is described by

x(k+1) = f(x(k))

Example 15

x(k+1) = λx(k)(1− x(k))

. Consider λ > 3.56994

Exercises

Exercise 4 Draw the state portrait of the first nonlinear system.

Exercise 5 Draw the state portrait for

x = x4 − x2 .

Exercise 6 Obtain an explicit expression for all solutions of

x = −x3 .

Chapter 4

Second order systems

In this section, we consider systems whose state can be described by two real scalar variablesx1, x2. We will refer to such systems as second order systems. They are described by

x1 = f1(x1, x2)x2 = f2(x1, x2)

(4.1)

• State plane, state portrait.

• Vector field f

Quite often we consider second order systems described by

x1 = x2x2 = f2(x1, x2)

In this case, the state plane and the state portrait are sometimes referred to as the phaseplane and phase portrait, respectively.

4.1 Linear systems

A linear time-invariant second order system is described by

x1 = a11x1 + a12x2x2 = a21x1 + a22x2

where each aij is a constant real scalar. The qualitative behavior of this system is determinedby the eigenvalues λ1, λ2 of the matrix

A =

[

a11 a12a21 a22

]

.

37

38 CHAPTER 4. SECOND ORDER SYSTEMS

stable node real λ1, λ2 < 0unstable node real 0 < λ1, λ2saddle real λ1 < 0 < λ2stable focus complex ℜ(λ1) < 0unstable focus complex ℜ(λ1) > 0center complex ℜ(λ1) = 0

See Khalil.

4.2. LINEARIZATION 39

4.2 Linearization

Suppose xe = (xe1, xe2) is an equilibrium state of system (4.1). Then the linearization of (4.1)

about xe is given by:

δx1 = a11δx1 + a12δx2 (4.2a)

δx2 = a21δx1 + a22δx2 (4.2b)

where

aij =∂fi∂xj

(xe1, xe2)

for i, j = 1, 2.

The behavior of a second order nonlinear system near an equilibrium state is qualitatively thesame as the behavior of the corresponding linearized system about 0, provided the eigenvaluesof the linearized system have non-zero real part.

Example 16 Van der Pol oscillator. Consider

w + φ(w) + w = 0

where φ is a nonlinear function. If one lets y = w and differentiates the above equation, oneobtains

y + φ′(y)y + y = 0 .

Considering φ(y) = µ(y3/3− y) yields

y − µ(1− y2)y + y = 0

Introducing state variables x1 = y and x2 = y, this system can be described by

x1 = x2

x2 = −x1 + µ(1− x21)x2

This system has a single equilibrium state at the origin. Linearization about the originresults in

δx1 = δx2

δx2 = −δx1 + µδx2 .

The eigenvalues λ1, λ2 of this linearized system are given by

λ1 =µ

2−

√

1− µ2

4and λ2 =

µ

2+

√

1− µ2

4.

If we consider 0 < µ < 2, then λ1 and λ2 are complex numbers with positive real parts.Hence, the origin is an unstable focus of the linearized system. Thus, the origin is an unstablefocus of the the original nonlinear system. Figure 4.1 contains some state trajectories of thissystem for µ = 0.3.

Note that, although the system has a single equilibrium state and this is unstable, allsolutions of the system are bounded. This behavior does not occur in linear systems. If theorigin of a linear system is unstable, then the system will have some unbounded solutions.


−3 −2 −1 0 1 2 3 4−3

−2

−1

0

1

2

3

y

ydot

Phase portrait of Van der Pol oscillator

Figure 4.1: Van der Pol oscillator

Example 17 Duffing’s equation Consider

y + φ(y) = 0

where φ is a nonlinear function. With φ(y) = −y + y3, we have

y − y + y3 = 0

The above differential equation has three equilibrium solutions:

ye = 0,−1, 1 .

Linearization about the zero solution yields

δy − δy = 0 .

A state space description of this linearized system is given by

δx1 = δx2

δx2 = δx1

The eigenvalues λ1, λ2 and corresponding eigenvectors v1, v2 associated with this linear sys-tem are given by

λ1 = 1 , v1 =

[

11

]

, λ2 = −1 , v1 =

[

1−1

]

.

So the origin of the linearized system is a saddle; see Figure 4.2. Hence the origin is a saddlefor the original nonlinear system.

Linearization about either of the two non-zero equilibrium solutions results in

δy + 2δy = 0 .

The eigenvalues associated with any state space realization of this system are

λ1 = √2 λ1 = −

√2 .


Figure 4.2: Duffing saddle

−3 −2 −1 0 1 2 3−3

−2

−1

0

1

2

3

y

ydot

Phase portrait of ddot y = y − y3

−15 −10 −5 0 5 10 15−15

−10

−5

0

5

10

15

y

ydot

Phase portrait of ddot y = y − y^3

Figure 4.3: Phase portraits of Duffing system


4.3 Isocline method

See Khalil.

4.4 Periodic solutions and limit cycles

A solution x(·) of x = f(x) is a periodic solution if it is non-constant and there is a T > 0such that

x(t + T ) = x(t) for all t

The smallest T for which the above identity holds is called the period of the solution. Periodicsolutions are sometimes called limit cycles.

Note that a limit cycle corresponds to a closed trajectory in the state plane. If a linearsystem has a periodic solution with period T , it will have an infinite number of periodicsolutions with that period. However, an nonlinear system can have just a single periodicsolution; recall the Van der Pol oscillator. Also, a nonlinear system may have an infinitenumber of periodic solutions, but, the period may be different for different amplitudes; thisoccurs for Duffing’s equation.

Figure 4.4: limit cycle

In the analysis of second order systems, it is sometimes useful to introduce polar coordi-nates r, φ which are implicity defined by

x1 = r cosφ and x2 = r sinφ .

Note that

r = (x21 + x22)12

and, when (x1, x2) 6= (0, 0),

φ = tan−1(x2/x1)

The scalar φ is not defined when (x1, x2) = (0, 0).

Figure 4.5: Polar coordinates

4.4. PERIODIC SOLUTIONS AND LIMIT CYCLES 43

One may show that

r = (x1x1 + x2x2)/r (4.3a)

φ = (x1x2 − x1x2)/r2 (4.3b)

Example 18 The simple harmonic oscillator.

x1 = x2

x2 = −x1

Here

r = 0

φ = −1

Figure 4.6: Simple harmonic oscillator

Example 19 An isolated periodic solution

x1 = x2 + x1(1− x21 − x22)

x2 = −x1 + x2(1− x21 − x22)

Here

r = r(1− r)2

φ = −1

Exercises

Exercise 7 Determine the nature of each equilibrium state of the damped duffing system

y + 0.1y − y + y3 = 0

Numerically obtain the phase portrait.


Exercise 8 Determine the nature (if possible) of each equilibrium state of the simple pen-dulum.

y + sin y = 0

Numerically obtain the phase portrait.

Exercise 9 Numerically obtain a state portrait of the following system:

x1 = x22x2 = −x1x2

Based on the state portrait, predict the stability properties of each equilibrium state.

Chapter 5

Some general considerations

Consider the differential equationx = f(x) (5.1)

where x is an n-vector. By a solution of (5.1) we mean any continuous function x(·) : [0, t1) →IRn, with t1 > 0, which satisfies (5.1). Consider an initial condition

x(0) = x0 , (5.2)

where x0 is some specified initial state. The corresponding initial value problem associatedwith (5.1) is that of finding a solution to (5.1) which satisfies the initial condition (5.2). Ifthe system is linear then, for every initial state x0, there is a solution to the correspondinginitial value problem, this solution is unique and is defined for all t ≥ 0, that is, t1 = ∞.One cannot make the same statement for nonlinear systems in general.

5.1 Existence of solutions

Example 20 Example of nonexistence

x = −s(x) , x(0) = 0

where

s(x) =

{

1 if x ≥ 0−1 if x < 0

Fact. Continuity of f guarantees existence.

The above function is not continuous at 0.

5.2 Uniqueness of solutions

Example 21 Example of nonuniqueness: I’m falling

x =√

2g|x| , x(0) = 0

45

46 CHAPTER 5. SOME GENERAL CONSIDERATIONS

The above initial value problem has an infinite number of solutions. They are given by

x(t) =

{

0 if 0 ≤ t ≤ Tg2(t− T )2 if T ≤ t

where T ≥ 0 is arbitrary.

Fact.Differentiability of f guarantees uniqueness.

The f above is not differentiable at 0.

5.3 Indefinite extension of solutions

Example 22 Finite escape time

x = 1 + x2 , x(0) = 0

Fact. If a solution cannot be extended indefinitely, that is over [0,∞), then it must havea finite escape time Te, that is, Te is finite and

limt→Te

x(t) = ∞ .

Hence, if a solution is bounded on bounded time intervals then, it cannot have a finiteescape time and, hence can be extended over [0,∞).

The following condition guarantees that solutions can be extended indefinitely. Thereare constants α and β such that

‖f(x)‖ ≤ α‖x‖+ β

for all x. Here one can show that when α 6= 0

‖x(t)‖ ≤ eα(t−t0)‖x(t0)‖+β

α(eαt − 1)

When α = 0,‖x(t)‖ ≤ ‖x(t0)‖+ β(t− t0)

Chapter 6

Stability and boundedness

Consider a general nonlinear system described by

x = f(x) (6.1)

where x(t) is a real n-vector and t is a real scalar. By a solution of (6.1) we mean anycontinuous function x(·) : [0, t1) → IRn with t1 > 0, which satisfies x(t) = f(x(t)) for0 ≤ t < t1.

6.1 Boundedness of solutions

DEFN. (Boundedness) A solution x(·) is bounded if there exists β ≥ 0 such that

||x(t)|| ≤ β for all t ≥ 0

A solution is unbounded if it is not bounded.

It should be clear from the above definitions that a solution x(·) is bounded if and onlyif each component xi(·) of x(·) is bounded. Also, a solution is unbounded if and only ifat least one of its components is unbounded. So, x(t) = [ e−t e−2t ]T is bounded whilex(t) = [ e−t et ]T is unbounded.

Example 23 All solutions of

x = 0

are bounded.

Example 24 All solutions of

x = x− x3

are bounded.

47

48 CHAPTER 6. STABILITY AND BOUNDEDNESS

Example 25 All solutions (except the zero solution) of

x = x

are unbounded.

Example 26 Considerx = x2 , x(0) = x0

If x0 > 0, the corresponding solution has a finite escape time and is unbounded. If x0 < 0,the corresponding solution is bounded.

Example 27 Undamped oscillator.

x1 = x2

x2 = −x1

All solutions are bounded.

Boundedness and linear time-invariant systems. Consider a general LTI (linear time-invariant) system

x = Ax (6.2)

Recall that every solution of this system has the form

x(t) =

l∑

i=1

ni−1∑

j=0

tjeλitvij

where λ1, . . . , λl are the eigenvalues of A, ni is the index of λi, and the constant vectors vij

depend on initial state.We say that an eigenvalue λ of A is non-defective if its index is one; this means that the

algebraic multiplicity and the geometric multiplicity of λ are the same. Otherwise we say λis defective.

Hence we conclude that all solutions of (6.2) are bounded if and only if for each eigenvalueλi of A:

(b1) ℜ(λi) ≤ 0 and

(b2) if ℜ(λi) = 0 then λi is non-defective.

If there is an eigenvalue λi of A such that either

(u1) ℜ(λi) > 0 or

(u2) ℜ(λi) = 0 and λi is defective

then, the system has some unbounded solutions.

6.1. BOUNDEDNESS OF SOLUTIONS 49

Example 28 Unattached mass

x1 = x2

x2 = 0

Here

A =

(

0 10 0

)

has a single eigenvalue 0. This eigenvalue has algebraic multiplicity 2 but geometric multi-plicity 1; hence some of the solutions of the system are unbounded. One example is

x(t) =

(

t1

)

Example 29

x1 = 0

x2 = 0

Here

A =

(

0 00 0

)

has a single eigenvalue 0. This eigenvalue has algebraic multiplicity 2 and geometric mul-tiplicity 2. Hence all the solutions of the system are bounded. Actually every state is anequilibrium state and every solution is constant.

Example 30 (Resonance) Consider a simple linear oscillator subject to a sinusoidal inputof amplitude W :

q + q = W sin(ωt+φ)

Resonance occurs when ω = 1. To see this, let

x1 := q , x2 := q , x3 := W sin(ωt+φ) , x4 := ωW cos(ωt+φ)

to yieldx = Ax

where

A =

0 1 0 0−1 0 1 00 0 0 10 0 −ω2 0

If ω = 1 then, A has eigenvalues and −. These eigenvalues have algebraic multiplicity twobut geometric multiplicity one; hence the system has unbounded solutions.


6.2 Stability of equilibrium states

Suppose xe is an equilibrium state of the system x = f(x). Then, whenever x(0) = xe, wehave (assuming uniqueness of solutions) x(t) = xe for all t ≥ 0. Roughly speaking, we saythat xe is a stable equilibrium state for the system if the following holds. If the initial stateof the system is close to xe, then the resulting solution is close to xe. The formal definitionis as follows.

DEFN. (Stability) An equilibrium state xe is stable if for each ǫ > 0 there exists δ > 0 suchthat whenever ||x(0)−xe|| < δ one has ||x(t)−xe|| < ǫ for all t ≥ 0.

Figure 6.1: Stability

An equilibrium state xe is said to be unstable if it is not stable.

Example 31

x = 0

Every equilibrium state is stable. (Choose ǫ = δ.)

Example 32

x = x

The origin is unstable.

Example 33

x = x− x3

The origin is unstable; the remaining equilibrium states 1 and −1 are stable.

Example 34 Undamped oscillatorThe origin is stable.

6.2. STABILITY OF EQUILIBRIUM STATES 51

Example 35 Simple pendulum

x1 = x2

x2 = − sin x1

(0, 0) is stable; (π, 0) is unstable.

Example 36 Van der Pol oscillator

x1 = x2

x2 = (1− x21)x2 − x1

Figure 6.2: Van der Pol oscillator

The origin is unstable. However, all solutions are bounded.

Stability of linear time-invariant systems. It can be shown that every equilibriumstate of a LTI system (6.2) is stable if and only if all eigenvalues λi of A satisfy conditions(b1) and (b2) above. Hence every equilibrium state of a LTI system is stable if and only ifall solutions are bounded.It can also be shown that every equilibrium state is unstable if and only if there is aneigenvalue λi of A which satisfies condition (u1) or (u2) above. Hence every equilibriumstate of a LTI system is unstable if and only if there are unbounded solutions.


6.3 Asymptotic stability

6.3.1 Global asymptotic stability

DEFN. (Global asymptotic stability) An equilibrium state xe is globally asymptotically stable(GAS) if

(a) It is stable

(b) Every solution x(·) converges to xe with increasing time, that is,

limt→∞

x(t) = xe

If xe is a globally asymptotically stable equilibrium state, then there are no other equi-librium states and all solutions are bounded. In this case we say that the system x = f(x)is globally asymptotically stable.

Example 37 The systemx = −x

is GAS.

Example 38 The systemx = −x3

is GAS.

6.3.2 Asymptotic stability

In asymptotic stability, we do not require that all solutions converge to the equilibrium state;we only require that all solutions which originate in some neighborhood of the equilibriumstate converge to the equilibrium state.

DEFN. (Asymptotic stability) An equilibrium state xe is asymptotically stable (AS) if

(a) It is stable.

(b) There exists R > 0 such that whenever ||x(0)−xe|| < R one has

limt→∞

x(t) = xe . (6.3)

The region of attraction of an equilibrium state xe which is AS is the set of initial stateswhich result in (13.3), that is it is the set of initial states which are attracted to xe. Thus, theregion of attraction of a globally asymptotically equilibrium state is the whole state space.

6.3. ASYMPTOTIC STABILITY 53

Figure 6.3: Asymptotic stability

Example 39

x = −xExample 40

x = x− x3

The equilibrium states −1 and 1 are AS with regions of attraction (−∞, 0) and (0,∞),respectively.

Example 41 Damped simple pendulum

x1 = x2

x2 = − sin x1 − x2

The zero state is AS but not GAS.

Example 42 Reverse Van der Pol oscillator

x1 = −x2x2 = −(1− x21)x2 + x1

The zero state is AS but not GAS. Also, the system has unbounded solutions.

LTI systems. For LTI systems, it should be clear from the general form of the solutionthat the zero state is AS if and only if all the eigenvalues λi of A have negative real parts,that is,

ℜ(λi) < 0

Also AS is equivalent to GAS.

Example 43 The system

x1 = −x1 + x2

x2 = −x2is GAS.


x1 = −x1 + x2

x2 = −x1 − x2

is GAS.


Figure 6.4: Reverse Van der Pol oscillator

6.4 Exponential stability

We now present the “strongest” form of stability considered in this section.

DEFN. (Global exponential stability) An equilibrium state xe is globally exponentially stable(GES) with rate of convergence α > 0 if there exists β > 0 such that every solution satisfies

||x(t)−xe|| ≤ β||x(0)−xe|| exp(−αt) for all t ≥ 0

Example 45

x = −x

GES with rate α = 1.

Note that global exponential stability implies global asymptotic stability, but, in general,the converse is not true. This is illustrated in the next example. For linear time-invariantsystems, GAS and GES are equivalent.

Example 46

x = −x3

Solutions satisfy

x(t) =x0

√

1 + 2x20twhere x0 = x(0) .

GAS but not GES

Example 47 (General Example) Consider a general system

x = f(x)

6.4. EXPONENTIAL STABILITY 55

and suppose that there exists α > 0 such that

x′f(x) ≤ −α‖x‖2

for all x. Then this system is GES about zero with rate of convergence α.

As a specific example, consider

x = −2x− 7x3

As another specific example, consider

x1 = −x1 + cos(x1)x2

x2 = −2x2 − cos(x1)x1

DEFN. (Exponential stability) An equilibrium state xe is exponentially stable (ES) with rate ofconvergence α > 0 if there exists R > 0 and β > 0 such that whenever ||x(0)−xe|| < R onehas

||x(t)−xe|| ≤ β||x(0)−xe|| exp(−αt) for all t ≥ 0

Note that exponential stability implies asymptotic stability, but, in general, the converse isnot true.

Example 48

x = −x3

GAS but not even ES

Example 49

x = − x

1 + x2

GAS, ES, but not GES


Figure 6.5: Exponential stability

6.5 LTI systems

Consider a LTI systemx = Ax (6.4)

Recall that every solution of this system has the form

x(t) =l∑

i=1

ni−1∑

j=0

tjeλitvij

where λ1, . . . , λl are the eigenvalues of A, the integer ni is the index of λi, and the constantvectors vij depend on initial state. From this it follows that the stability properties of thissystem are completely determined by the location of its eigenvalues; this is summarized inthe table below.

The following table summarizes the relationhip between the stability properties of a LTIsystem and the eigenproperties of its A-matrix. In the table, unless otherwise stated, aproperty involving λ must hold for all eigenvalues λ of A.

Stability properties Eigenproperties

Global exponential stability ℜ(λ) < 0and boundedness

Stability and ℜ(λ) ≤ 0boundedness If ℜ(λ) = 0 then λ is non-defective.

Instability and some There is an eigenvalue λ with ℜ(λ) > 0 orunbounded solutions ℜ(λ) = 0 and λ is defective.


x1 = −x1 + x2

x2 = −x2

6.5. LTI SYSTEMS 57

is GES.


x1 = −x1 + x2

x2 = −x1 − x2

is GES.


x1 = 0

x2 = 0

is stable about every equilibrium point.


x1 = x2

x2 = 0

is unstable about every equilibrium point.

Figure 6.6: The big picture. The concepts in each dashed box are equivalent for linearsystems.


6.6 Linearization and stability

Consider a nonlinear time-invariant system described by

x = f(x)

where x(t) is an n-vector at each time t. Suppose xe is an equilibrium state for this system,that is, f(xe) = 0, and consider the linearization of this system about xe:

δx = Aδx where A =∂f

∂x(xe) .

The following results can be demonstrated using nonlinear Lyapunov stability theory.

Stability. If all the eigenvalues of the A matrix of the linearized system have negative realparts, then the nonlinear system is exponentially stable about xe.

Instability. If at least one eigenvalue of the A matrix of the linearized system has a positivereal part, then the nonlinear system is unstable about xe.

Undetermined. Suppose all the eigenvalues of the A matrix of the linearized system havenon-positive real parts and at least one eigenvalue of A has zero real part. Then, based onthe linearized system alone, one cannot predict the stability properties of the nonlinear systemabout xe.

Note that the first statement above is equivalent to the following statement. If thelinearized system is exponentially stable, then the nonlinear system is exponentially stableabout xe.

Example 54 (Damped simple pendulum.) Physically, the system

x1 = x2

x2 = − sin x1 − x2

has two distinct equilibrium states: (0, 0) and (π, 0). The A matrix for the linearization ofthis system about (0, 0) is

A =

(

0 1−1 −1

)

Since all the eigenvalues of this matrix have negative real parts, the nonlinear system isexponentially stable about (0, 0). The A matrix corresponding to linearization about (π, 0)is

A =

(

0 11 −1

)

Since this matrix has an eigenvalue with a positive real part, the nonlinear system is unstableabout (π, 0)

The following example illustrates the fact that if the eigenvalues of the Amatrix have non-positive real parts and there is at least one eigenvalue with zero real part, then, one cannotmake any conclusions on the stability of the nonlinear system based on the linearization.

6.6. LINEARIZATION AND STABILITY 59

Example 55 Consider the scalar nonlinear system:

x = ax3

This origin is GAS if a < 0, unstable if a > 0 and stable if a = 0. However, the linearizationof this system about zero, given by

δx = 0

is independent of a and is stable.

The following example illustrates that instability of the linearized system does not implyinstability of the nonlinear system.

Example 56 Using nonlinear techniques, one can show that the following system is GASabout the origin.

x1 = x2

x2 = −x31 − x32

However, the linearization of this system, given by

δx1 = δx2

δx2 = 0 ,

is unstable about the origin.


Exercises

Exercise 10 For each of the following systems, determine (from the state portrait) thestability properties of each equilibrium state. For AS equilibrium states, give the region ofattraction.

(a)

x = −x − x3

(b)

x = −x+ x3

(c)

x = x− 2x2 + x3

Exercise 11 If possible, use linearization to determine the stability properties of each ofthe following systems about the zero equilibrium state.

(i)

x1 = (1 + x21)x2

x2 = −x31

(ii)

x1 = sin x2

x2 = (cosx1)x3

x3 = ex1x2

Exercise 12 If possible, use linearization to determine the stability properties of each equi-librium state of the Lorenz system.

Chapter 7

Stability and boundedness: discretetime

7.1 Boundedness of solutions

Consider a general discrete-time nonlinear system described by

x(k+1) = f(x(k)) (7.1)

where x(k) is an n-vector and k is an integer. By a solution of (7.1) we mean a sequencex(·) = ( x(0), x(1), x(2), . . . ) which satisfies (7.1) for all k ≥ 0.

DEFN. (Boundedness of solutions) A solution x(·) is bounded if there exists β ≥ 0 such that

||x(k)|| ≤ β for all k ≥ 0 .


It should be clear from the above definitions that a solution x(·) is bounded if and onlyif each component xi(·) of x(·) is bounded. Also, a solution is unbounded if and only if atleast one of its components is unbounded. So, x(k) = [ (0.5)k (−0.5)k ]T is bounded whilex(k) = [ (0.5)k 2k ]T is unbounded.

Example 57

x(k+1) = 0


Example 58

x(k+1) = x(k)


Example 59

x(k+1) = −2x(k)

All solutions (except the zero solution) are unbounded.

61

62 CHAPTER 7. STABILITY AND BOUNDEDNESS: DISCRETE TIME

Linear time invariant (LTI) systems. All solutions of the LTI system

x(k+1) = Ax(k) (7.2)

are bounded if and only if for each eigenvalue λ of A:

(b1) |λ| ≤ 1 and

(b2) if |λ| = 1 then λ is non-defective.

If there is an eigenvalue λ of A such that either

(u1) |λ| > 1 or

(u2) |λ| = 1 and λ is defective.

then the system has some unbounded solutions.

Example 60 Discrete unattached mass. Here

A =

(

1 10 1

)

has a single eigenvalue 1. This eigenvalue has algebraic multiplicity 2 but geometric mul-tiplicity 1; hence this eigenvalue is defective. So, some of the solutions of the systemx(k+1) = Ax(k) are unbounded. One example is

x(k) =

(

k1

)

7.2 Stability of equilibrium states

Suppose xe is an equilibrium state of the system x(k+1) = f(x(k)). Then, wheneverx(0) = xe, we have x(k) = xe for all k ≥ 0. Roughly speaking, we say that xe is a stableequilibrium state for the system if the following holds. If the initial state of the system isclose to xe, then the resulting solution is close to xe. The formal definition is as follows.

DEFN. (Stability) An equilibrium state xe is stable if for each ǫ > 0 there exists δ > 0 suchthat whenever ||x(0)−xe|| < δ, one has ||x(k)−xe|| < ǫ for all k ≥ 0.xe is unstable if it is not stable.

Example 61

x(k+1) = −x(k) .

The origin is stable. (Consider δ = ǫ.)


Example 62

x(k+1) = −2x(k) .

The origin is unstable.

Example 63

x(k+1) = −x(k)3 .

The single equilibrium at the origin is stable, but, the system has unbounded solutions.

LTI systems. Every equilibrium state of a LTI system (7.2) is stable if and only if alleigenvalues λ of A satisfy conditions (b1) and (b2) above. Hence every equilibrium state ofa LTI system is stable if and only if all solutions are bounded.

Every equilibrium state is unstable if and only if there is an eigenvalue λ of A whichsatisfies condition (u1) or (u2) above. Hence every equilibrium state of a LTI system isunstable if and only if there are unbounded solutions.



DEFN. (Global asymptotic stability (GAS)) An equilibrium state xe is globally asymptoticallystable (GAS) if

(a) It is stable

(b) Every solution x(·) converges to xe, that is,

limk→∞

x(k) = xe .

If xe is a globally asymptotically stable equilibrium state, then there are no other equi-librium states. In this case we say the system (7.1) is globally asymptotically stable.

Example 64

x(k+1) =1

2x(k)

GAS

Example 65

x(k+1) =x(k)

2 + x(k)2



DEFN. (Asymptotic stability (AS)) An equilibrium state xe is asymptotically stable (AS) if

(a) It is stable

(b) There exists R > 0 such that whenever ||x(0)− xe|| < R one has

limk→∞

x(k) = xe (7.3)

The region of attraction of an equilibrium state xe which is AS is the set of initial stateswhich result in (7.3), that is it is the set of initial states which are attracted to xe. Thus, theregion of attraction of a globally asymptotically equilibrium state is the whole state space.

Example 66

x(k+1) = x(k)3

Origin is AS with region of attraction [−1 1].

LTI systems. For LTI systems, it should be clear from the general form of the solutionthat the zero state is AS if and only if all the eigenvalues λi of A have magnitude less thanone, that is,

|λi| < 1 .

Also AS is equivalent to GAS.


We now present the “strongest” form of stability considered in this section.

DEFN. (Global exponential stability) An equilibrium state xe is globally exponentially stable(GES) with there exists 0 ≤ λ < 1 and β > 0 such that every solution satisfies

||x(k)−xe|| ≤ βλk||x(0)−xe|| for all k ≥ 0

Example 67

x(k+1) =1

2x(k)

GES with γ = 12.


Note that global exponential stability implies global asymptotic stability, but, in general,the converse is not true. This is illustrated in the next example. For linear time-invariantsystems, GAS and GES are equivalent.

Lemma 1 Consider a system described by x(k+1) = f(x(k)) and suppose that for somescalar λ ≥ 0,

‖f(x)‖ ≤ λ‖x‖

for all x. Then, every solution x(·) of the system satisfies

‖x(k)‖ ≤ λk‖x(0)‖

for all k ≥ 0. In particular, if λ < 1 then, the system is globally exponentially stable.

Example 68

x(k+1) =x(k)

√

1 + 2x(k)2

Solutions satisfy

x(k) =x0

√

1 + 2kx20where x0 = x(0) .

GAS but not GES

DEFN. (Exponential stability) An equilibrium state xe is exponentially stable (ES) if thereexists R > 0, 0 ≤ λ < 1 and β > 0 such that whenever ||x(0)−xe|| < R one has

||x(k)−xe|| ≤ βλk||x(0)−xe|| for all k ≥ 1

Note that exponential stability implies asymptotic stability, but, in general, the converseis not true.

Example 69

x(k+1) =x(k)

√

1 + 2x(k)2

Solutions satisfy

x(k) =x0

√

1 + 2kx20where x0 = x(0) .

GAS but not even ES


7.5 LTI systems

The following table summarizes the relationhip between the stability properties of a LTIsystem and the eigenproperties of its A-matrix. In the table, unless otherwise stated, aproperty involving λ must hold for all eigenvalues λ of A.

Stability properties eigenproperties

Asymptotic stability |λ| < 1and boundedness

Stability and |λ| ≤ 1boundedness If |λ| = 1 then λ is non-defective

Instability and some There is an eigenvalue of A withunbounded solutions |λ| > 1 or

|λ| = 1 and λ is defective

7.6. LINEARIZATION AND STABILITY 67

7.6 Linearization and stability

Consider a nonlinear time-invariant system described by

x(k+1) = f(x(k))

where x(k) is an n-vector at each time t. Suppose xe is an equilibrium state for this system,that is, f(xe) = xe, and consider the linearization of this system about xe:

δx(k+1) = Aδx(k) where A =∂f

∂x(xe) .

The following results can be demonstrated using nonlinear Lyapunov stability theory.

Stability. If all the eigenvalues of the A matrix of the linearized system have magnitude lessthan one, then the nonlinear system is exponentially stable about xe.

Instability. If at least one eigenvalue of the A matrix of the linearized system has magnitudegreater than one, then the nonlinear system is unstable about xe.

Undetermined. Suppose all the eigenvalues of the A matrix of the linearized system havemagnitude less than or equal to one and at least one eigenvalue of A has magnitude one.Then, based on the linearized system alone, one cannot predict the stability properties of thenonlinear system about xe.

Note that the first statement above is equivalent to the following statement. If thelinearized system is exponentially stable, then the nonlinear system is exponentially stableabout xe.

Example 70 (Newton’s method) Recall that Newton’s method for a scalar function canbe described by

x(k+1) = x(k)− g(x(k))

g′(x(k))

Here

f(x) = x− g(x)

g′(x)

So,

f ′(x) = 1− 1 +g(x)g′′(x)

g′(x)2=g(x)g′′(x)

g′(x)2.

At an equilibrium state xe, we have g(xe) = 0; hence f ′(xe) = 0 and the linearization aboutany equilibrium state is given by

δx(k+1) = 0 .

Thus, every equilibrium state is exponentially stable.

Chapter 8

Basic Lyapunov theory

Suppose we are interested in the stability properties of the system

x = f(x) (8.1)

where x(t) is a real n-vector at time t. If the system is linear, we can determine its stabilityproperties from the properties of the eigenvalues of the system matrix. What do we do fora nonlinear system? We could linearize about each equilibrium state and determine thestability properties of the resulting linearizations. Under certain conditions this will tell ussomething about the local stability properties of the nonlinear system about its equilibriumstates. However there are situations where linearization cannot be used to deduce even thelocal stability properties of the nonlinear system. Also, linearization tells us nothing aboutthe global stability properties of the nonlinear system.

In general, we cannot explicitly obtain solutions for nonlinear systems. However, Lya-punov theory allows to say something about the stability properties of a system withoutknowing the form or structure of its solutions. Lyapunov theory is based on Lyapunov func-tions which are scalar-valued functions of the system state.

Suppose V is a scalar-valued function of the state, that is V : IRn → IR. If V iscontinuously differentiable then, at any time t, the derivative of V along a solution x(·) ofsystem (8.1) is given by

dV

dt(x(t)) = DV (x(t))x(t)

= DV (x(t))f(x(t))

where DV (x) is the derivative of V at x and is given by

DV (x) =(

∂V∂x1

(x) ∂V∂x2

(x) . . . ∂V∂xn

(x))

Note that

DV (x)f(x) =∂V

∂x1(x)f1(x) +

∂V

∂x2(x)f2(x) + . . .+

∂V

∂xn(x)fn(x)

In what follows, if a condition involves DV , then it is implicitly assumed that V is continu-ously differentiable. Sometimes DV is denoted by

∂V

∂xor ∇V (x)T .

69

70 CHAPTER 8. BASIC LYAPUNOV THEORY

Also, when the system under consideration is fixed, DV f is sometimes denoted by

V

This a slight but convenient abuse of notation.In the following sections, we use Lyapunov functions to present sufficient conditions for

the stability and boundedness concepts previously introduced. To guarantee a specific typeof stability using a Lyapunov function, one has to obtain a Lyapunov function V which hascertain properties and whose “time-derivative” V also has certain properties.

8.1. LYAPUNOV CONDITIONS FOR STABILITY 71

8.1 Lyapunov conditions for stability

8.1.1 Locally positive definite functions

DEFN. (Locally positive definite function) A function V is locally positive definite (lpd) abouta point xe if

V (xe) = 0

and there is a scalar R > 0 such that

V (x) > 0 whenever x 6= xe and ||x− xe|| < R

Basically, a function is lpd about a point xe if it is zero at xe has a strict local minimumat xe.

Figure 8.1: A locally positive definite function

Example 71 (Scalar x) The following functions are lpd about zero.

V (x) = x2

V (x) = 1− e−x2

V (x) = 1− cosx

V (x) = x2 − x4

Example 72

V (x) = ||x||2

= x21 + x22 + . . .+ x2n

Lpd about the origin.

Example 73

V (x) = ||x||1= |x1|+ |x2|+ . . .+ |xn|

Lpd about the origin.


Quadratic forms. Suppose P is a real n × n symmetric matrix and is positive definite.Consider the quadratic form defined by

V (x) = xTPx =n∑

i=1

n∑

j=1

Pijxixj .

Clearly V (0) = 0. Recalling the definition of a positive definite matrix, it follows thatV (x) = xTPx > 0 for all nonzero x. Hence V is locally positive definite about the origin.

• The second derivative of V at x is the square symmetric matrix given by:

D2V (x) :=

∂2V∂2x1

(x) ∂2V∂x1∂x2

(x) . . . ∂2V∂x1∂xn

(x)

∂2V∂x2∂x1

(x) ∂2V∂2x2

(x) . . . ∂2V∂x2∂xn

(x)

......

...

∂2V∂xn∂x1

(x) ∂2V∂xn∂x2

(x) . . . ∂2V∂2xn

(x)

that is,

D2V (x)ij =∂2V

∂xi∂xj(x)

Sometimes D2V (x) is written as ∂2V∂x2 (x) and is referred to as the Hessian of V .

The following lemma is sometimes useful in demonstrating that a function is lpd.

Lemma 2 Suppose V is twice continuously differentiable and

V (xe) = 0

DV (xe) = 0

D2V (xe) > 0

Then V is a locally positive definite about xe.

Example 74 Consider V (x) = x2 − x4 where x is a scalar. Since V (0) = DV (0) = 0 andD2V (0) = 2 > 0, V is lpd about zero.

Example 75 Consider

V (x) = 1− cosx1 +1

2x22

ClearlyV (0) = 0

SinceDV (x) =

(

sin x1 x2)


we haveDV (0) = 0

Also,

D2V (x) =

(

cosx1 00 1

)

Hence,

D2V (0) =

(

1 00 1

)

> 0

Since V satisfies the hypotheses of the previous lemma with xe = 0, it is lpd about zero.

8.1.2 A stability result

If the equilibrium state of a nonlinear system is stable but not asymptotically stable, thenone cannot deduce the stability properties of the equilibrium state of the nonlinear systemfrom the linearization of the nonlinear system about that equilibrium state. The followingLyapunov result is useful in demonstrating stability of an equilibrium state for a nonlinearsystem.

Theorem 1 (Stability) Suppose there exists a function V and a scalar R > 0 such that V islocally positive definite about xe and

DV (x)f(x) ≤ 0 for ||x− xe|| < R

Then xe is a stable equilibrium state.

If V satisfies the hypotheses of the above theorem, then V is said to be a Lyapunovfunction which guarantees the stability of xe.

Example 76

x = 0

ConsiderV (x) = x2

as a candidate Lyapunov function. Then V is lpd about 0 and

DV (x)f(x) = 0

Hence (it follows from Theorem 1 that) the origin is stable.

Example 77 (Undamped linear oscillator.)

x1 = x2

x2 = −kx1 k > 0


Consider the total energy,

V (x) =1

2kx21 +

1

2x22

as a candidate Lyapunov function. Then V is lpd about the origin and

DV (x)f(x) = 0

Hence the origin is stable.

Example 78 (Simple pendulum.)

x1 = x2

x2 = − sin x1

Consider the total energy,

V (x) = 1− cosx1 +1

2x22

as a candidate Lyapunov function. Then, as we have allready shown, V is lpd about the origin;also

DV (x)f(x) = 0


Example 79 (Stability of origin for attitude dynamics system.) Recall

x1 =(I2 − I3)

I1x2x3

x2 =(I3 − I1)

I2x3x1

x3 =(I1 − I2)

I3x1x2

whereI1, I2, I3 > 0

Consider the kinetic energy

V (x) =1

2

(

I1x21 + I2x

22 + I3x

23

)

as a candidate Lyapunov function. Then V is lpd about the origin and

DV (x)f(x) = 0


Example 80 (Undamped Duffing system)

x1 = x2

x2 = x1 − x31


As candidate Lyapunov function for the equilibrium state xe = [ 1 0 ]T consider the totalenergy

V (x) =1

4x41 −

1

2x21 +

1

2x22 +

1

4.

Since

DV (x) =(

x31 − x1 x2)

and D2V (x) =

(

3x21 − 1 00 1

)

we have V (xe) = 0, DV (xe) = 0 and D2V (xe) > 0, and it follows that V is lpd about xe.One can readily verify that DV (x)f(x) = 0. Hence, xe is stable.

Exercises

Exercise 13 Determine whether or not the following functions are lpd. (a)

V (x) = x21 − x41 + x22

(b)V (x) = x1 + x22

(c)V (x) = 2x21 − x31 + x1x2 + x22

Exercise 14 (Simple pendulum with Coulomb damping.) By appropriate choice of Lya-punov function, show that the origin is a stable equilibrium state for

x1 = x2

x2 = − sin x1 − c sgm (x2)

where c > 0.

Exercise 15 By appropriate choice of Lyapunov function, show that the origin is a stableequilibrium state for

x1 = x2

x2 = −x31

Note that the linearization of this system about the origin is unstable.

Exercise 16 By appropriate choice of Lyapunov function, show that the origin is a stableequilibrium state for

x1 = x2

x2 = −x1 + x31



The following result presents conditions which guarantee that an equilibrium state is asymp-totically stable.

Theorem 2 (Asymptotic stability) Suppose there exists a function V and a scalar R > 0such that V is locally positive definite about xe and

DV (x)f(x) < 0 for x 6= xe and ||x− xe|| < R

Then xe is an asymptotically stable equilibrium state for x = f(x).

Example 81

x = −x3

ConsiderV (x) = x2

Then V is lpd about zero and

DV (x)f(x) = −2x4 < 0 for x 6= 0

Hence the origin is AS.

Example 82

x = −x+ x3

ConsiderV (x) = x2


DV (x)f(x) = −2x2(1− x2) < 0 for |x| < 1, x 6= 0

Hence the origin is AS. Although the origin is AS, there are solutions which go unboundedin a finite time.

Example 83

x = − sin x

ConsiderV (x) = x2


DV (x)f(x) = −2x sin(x) < 0 for |x| < π, x 6= 0



Example 84 (Simple pendulum with viscous damping.) Intuitively, we expect the origin tobe an asymptotically stable equilibrium state for the damped simple pendulum:

x1 = x2

x2 = − sin x1 − cx2

where c > 0 is the damping coefficient. If we consider the total mechanical energy

V (x) = 1− cosx1 +1

2x22

as a candidate Lyapunov function, we obtain

DV (x)f(x) = −cx22 .

Since DV (x)f(x) ≤ 0 for all x, we have stability of the origin. Since DV (x)f(x) = 0whenever x2 = 0, it follows that DV (x)f(x) = 0 for points arbitrarily close to the origin;hence V does not satisfy the requirements of the above theorem for asymptotic stability.

Suppose we modify V to

V (x) =1

2λc2x21 + λcx1x2 +

1

2x22 + 1− cosx1

where λ is any scalar with 0 < λ < 1. Letting

P =1

2

(

λc2 λcλc 1

)

note that P > 0 and

V (x) = xTPx+ 1− cosx1

≥ xTPx

Hence V is lpd about zero and we obtain

DV (x)f(x) = −λcx1 sin x1 − (1− λ)cx22 < 0 for ||x|| < π, x 6= 0

to satisfy the requirements of above theorem; hence the origin is AS.

Exercises

Exercise 17 By appropriate choice of Lyapunov function, show that the origin is an asymp-totically stable equilibrium state for

x1 = x2

x2 = −x51 − x2

Exercise 18 By appropriate choice of Lyapunov function, show that the origin is a asymp-totically stable equilibrium state for

x1 = x2

x2 = −x1 + x31 − x2


8.3 Boundedness

8.3.1 Radially unbounded functions

DEFN. A scalar valued function V of the state x is said to be radially unbounded if

limx→∞

V (x) = ∞

Example 85

V (x) = x2 yes

V (x) = 1− e−x2no

V (x) = x2 − x yesV (x) = x4 − x2 yesV (x) = x sin x noV (x) = x2(1− cosx) no

Example 86 Suppose P is a real n × n matrix and is positive definite symmetric andconsider the quadratic form defined by

V (x) = xTPx

Since P is real symmetric,

xTPx ≥ λmin(P )||x||2

for all x, where λmin(P ) is the minimum eigenvalue of P . Since P is positive definite,λmin(P ) > 0. From this it should be clear that V is radially unbounded.

The following lemma can be useful for guaranteeing radial unboundedness.

Lemma 3 Suppose V is twice continuously differentiable, and there is a positive definitesymmetric matrix P and a scalar R ≥ 0 such that

D2V (x) ≥ P for ||x|| ≥ R

Then, V is radially unbounded.

8.3.2 A boundedness result

Theorem 3 Suppose there exists a radially unbounded function V and a scalar R ≥ 0 suchthat

DV (x)f(x) ≤ 0 for ||x|| ≥ R

Then all solutions of x = f(x) are bounded.

8.3. BOUNDEDNESS 79

Note that, in the above theorem, V does not have to be positive away from the origin; itonly has to be radially unbounded.

Example 87 Recall

x = x− x3 .

Consider

V (x) = x2

Since V is radially unbounded and

DV (x)f(x) = −2x2(x2 − 1)

the hypotheses of the above theorem are satisfied with R = 1; hence all solutions are bounded.Note that the origin is unstable.

Example 88 Duffing’s equation

x1 = x2

x2 = x1 − x31

Consider

V (x) =1

2x22 −

1

2x21 +

1

4x41

It should be clear that V is radially unbounded; also

DV (x)f(x) = 0 ≤ 0 for all x

So, the hypotheses of the above theorem are satisfied with any R; hence all solutions arebounded.

Example 89

x = −x− x3 + w

where |w| ≤ β.

Example 90

x = Ax+Bw

where A is Hurwitz and |w| ≤ β.


Exercises

Exercise 19 Determine whether or not the the following function is radially unbounded.

V (x) = x1 − x31 + x41 − x22 + x42

Exercise 20 (Forced Duffing’s equation with damping.) Show that all solutions of thesystem

x1 = x2

x2 = x1 − x31 − cx2 + 1 c > 0

are bounded.Hint: Consider

V (x) =1

2λc2x21 + λcx1x2 +

1

2x22 −

1

2x21 +

1

4x41

where 0 < λ < 1. Letting

P =1

2

(

λc2 λcλc 1

)

note that P > 0 and

V (x) = xTPx− 1

2x21 +

1

4x41

≥ xTPx− 1

4

Exercise 21 Recall the Lorenz system

x1 = σ(x2 − x1)

x2 = rx1 − x2 − x1x3

x3 = −bx3 + x1x2

with b > 0. Prove that all solutions of this system are bounded. (Hint: Consider V (x) =rx21 + σx22 + σ(x3 − 2r)2.)

8.4. GLOBAL ASYMPTOTIC STABILITY 81

8.4 Global asymptotic stability

8.4.1 Positive definite functions

DEFN. (Positive definite function.) A function V is positive definite (pd) if

V (0) = 0

V (x) > 0 for all x 6= 0

limx→∞

V (x) = ∞

In the above definition, note the requirement that V be radially unbounded.

Example 91 Scalar x

V (x) = x2 pd

V (x) = 1− e−x2lpd but not pd

V (x) = 1− cosx lpd but not pd

Example 92

V (x) = x41 + x22

Example 93

V (x) = ||x||2= x21 + x22 + . . .+ x2n

Example 94 Suppose P is a real n × n matrix and is positive definite symmetric andconsider the quadratic form defined by

V (x) = xTPx

V is a positive definite function.

The following lemma can be useful in demonstrating that a function has the first twoproperties needed for a function to be pd.

Lemma 4 Suppose V is twice continuously differentiable and

V (0) = 0

DV (0) = 0

D2V (x) > 0 for all x

Then V (x) > 0 for all x.


If V satisfies the hypotheses of the above lemma, it is not guaranteed to be radially un-bounded, hence it is not guaranteed to be positive definite. Lemma 3 can be useful forguaranteeing radial unboundedness. We also have the following lemma.

Lemma 5 Suppose V is twice continuously differentiable,

V (0) = 0

DV (0) = 0

and there is a positive definite symmetric matrix P such that

D2V (x) ≥ P for all x

Then

V (x) ≥ 1

2xTPx

for all x.

8.4.2 A result on global asymptotic stability

Theorem 4 (Global asymptotic stability) Suppose there exists a positive definite function Vsuch that

DV (x)f(x) < 0 for all x 6= 0

Then the origin is a globally asymptotically stable equilibrium state for x = f(x).

Example 95

x = −x3

V (x) = x2

DV (x)f(x) = −2x4

< 0 for all x 6= 0

We have GAS. Note that linearization of this system about the origin cannot be used todeduce the asymptotic stability of this system.

Example 96 The first nonlinear system

x = −sgm (x) .

This system is not linearizable about its unique equilibrium state at the origin. Considering

V (x) = x2

we obtain

DV (x)f(x) = −2|x|< 0 for all x 6= 0 .

Hence, we have GAS.


Example 97 Consider

x1 = −x1 + β sin(x1 − x2)

x1 = −2x2 − β sin(x1 − x2)

We will show that this system is GAS provided β is small enough. Considering the positivedefinite function

V (x) = x21 + x22 .

as a candidate Lyapunov function, we have

V = 2x1x1 + 2x2x2

Example 98 (Linear globally stabilizing controller for inverted pendulum.)

x1 = x2

x2 = sin x1 + u

Consideru = −k1x1 − k2x2 with k1 > 1, k2 > 0

Closed loop system:

x1 = x2

x2 = −k1x1 + sin x1 − k2x2

Consider

V (x) =1

2λk22x

21 + λk2x1x2 +

1

2x22 +

1

2k1x

21 + cosx1 − 1

where 0 < λ < 1. Then V is pd (apply lemma 5) and

DV (x)f(x) = λk2(−k1x21 + x1 sin x1)− (1− λ)k2x22

Since| sin x1| ≤ |x1| for all x1


it follows that

x1 sin x1 ≤ x21 for all x1

hence

DV (x)f(x) ≤ −λk2(k1 − 1)x21 − (1− λ)k2x22 < 0 for all x 6= 0

The closed loop system is GAS.

Example 99 (Stabilization of origin for attitude dynamics system.)

x1 =(I2 − I3)

I1x2x3 +

u1I1

x2 =(I3 − I1)

I2x3x1 +

u2I2

x3 =(I1 − I2)

I3x1x2 +

u3I3

where

I1, I2, I3 > 0

Consider any linear controller of the form

ui = −kxi i = 1, 2, 3, k > 0

Closed loop system:

x1 =

(

I2 − I3I1

)

x2x3 − kx1I1

x2 =

(

I3 − I1I2

)

x3x1 − kx2I2

x3 =

(

I1 − I2I3

)

x1x2 − kx3I3

Consider the kinetic energy

V (x) =1

2

(

I1x21 + I2x

22 + I3x

23

)

as a candidate Lyapunov function. Then V is pd and

DV (x)f(x) = −k(x21 + x22 + x23)

< 0 for all x 6= 0

Hence the origin is GAS for the closed loop system.


Exercises

Exercise 22 Determine whether or not the following function is positive definite.

V (x) = x41 − x21x2 + x22

Exercise 23 Consider any scalar system described by

x = −g(x)

where g has the following properties:

g(x) > 0 for x > 0

g(x) < 0 for x < 0

Show that this system is GAS.

Exercise 24 Recall the Lorenz system

x1 = σ(x2 − x1)

x2 = rx1 − x2 − x1x3

x3 = −bx3 + x1x2

Prove that ifb > 0 and 0 ≤ r < 1 ,

then this system is GAS about the origin. (Hint: Consider V (x) = x21 + σx22 + σx23.)

Exercise 25 (Stabilization of the Duffing system.) Consider the Duffing system with ascalar control input u(t):

x1 = x2

x2 = x1 − x31 + u

Obtain a linear controller of the form

u = −k1x1 − k2x2

which results in a closed loop system which is GAS about the origin. Numerically simulatethe open loop system (u = 0) and the closed loop system for several initial conditions.


8.5.1 Global exponential stability

Theorem 5 (Global exponential stability) Suppose there exists a function V and positivescalars α, β1, β2 such that for all x,

β1||x−xe||2 ≤ V (x) ≤ β2||x−xe||2


and

DV (x)f(x) ≤ −2αV (x) .

Then, for the system x = f(x), the state xe is a globally exponentially stable equilibriumstate with rate of convergence α. In particular, all solutions x(·) of the system satisfy

‖x(t)−xe‖ ≤√

β2/β1 ‖x(0)−xe‖ e−αt for all t ≥ 0 .

Proof. See below.

Example 100

x = −x

Considering

V (x) = x2

we have

DV (x)f(x) = −2x2

= −2V (x)

Hence, we have GES with rate of convergence 1.

Example 101

x = −x− x3

Considering

V (x) = x2

we have

DV (x)f(x) = −2x2 − 2x4

≤ −2V (x)

Hence, we have GES with rate of convergence 1.

8.5.2 Proof of theorem 6

Consider any solution x(·) and let v(t) = V (x(t)). Then

v ≤ −2αv .


8.5.3 Exponential stability

Theorem 6 (Exponential stability) Suppose there exists a function V and positive scalars R,α, β1, β2 such that, whenever ‖x−xe‖ ≤ R, one has

β1||x−xe||2 ≤ V (x) ≤ β2||x−xe||2

andDV (x)f(x) ≤ −2αV (x)

Then, for system x = f(x), the state xe is an exponentially stable equilibrium state with rateof convergence α.

Exercise 26 Consider the scalar system

x = −x+ x3

As a candidate Lyapunov function for exponential stability, consider V (x) = x2. Clearly,the condition on V is satisfied with β1 = β2 = 1. Noting that

V = −2x2 + 2x4 = −2(1− x2)x2 ,

and considering R = 1/2, we obtain that whenever |x| ≤ R, we have V ≤ −2αV whereα = 3/4. Hence, we have ES.


8.5.4 A special class of GES systems

Consider a system described byx = f(x) (8.2)

and suppose that there exist two positive definite symmetric matrices P and Q such that

xTPf(x) ≤ −xTQx .

We will show that the origin is GES with rate

α := λmin(P−1Q)

where λmin(P−1Q) > 0 is the smallest eigenvalue of P−1Q.

As a candidate Lyapunov function, consider

V (x) = xTPx .

Thenλmin(P )||x||2 ≤ V (x) ≤ λmax(P )||x||2

that isβ1||x||2 ≤ V (x) ≤ β2||x||2

withβ1 = λmin(P ) > 0 and β2 = λmax(P ) > 0 .

Now note that

DV (x)f(x) = 2xTPf(x)

≤ −2xTQx

For any two positive-definite matrices P and Q, one can show that all the eigenvalues ofP−1Q are real positive and

xTQx ≥ λmin(P−1Q) xTPx

where λmin(P−1Q) > 0 is the smallest eigenvalue of P−1Q. Thus,

DV (x)f(x) ≤ −2λmin(P−1Q)xTPx

= −2αV (x)

Hence GES with rate α.

Example 102 Recall example 99.Closed loop system:

x1 =

(

I2 − I3I1

)

x2x3 − kx1I1

x2 =

(

I3 − I1I2

)

x3x1 − kx2I2

x3 =

(

I1 − I2I3

)

x1x2 − kx3I3


Considering

P =

I1 0 00 I2 00 0 I3

we havexTPf(x) = −kxTx

that is, Q = kI. Hence, GES with rate

α = λmin(P−1Q)

= λmin(kP−1)

= kλmin(P−1)

=k

λmax(P )

=k

max{I1, I2, I3}


8.5.5 Summary

The following table summarizes the results of this chapter for stability about the origin.

V V ⇒ stability properties

lpd ≤ 0 for ||x|| ≤ R ⇒ S

lpd < 0 for ||x|| ≤ R, x 6= 0 ⇒ AS

ru ≤ 0 for ||x|| ≥ R ⇒ B

pd < 0 for x 6= 0 ⇒ GAS

β1||x||2 ≤ V (x) ≤ β2||x||2 ≤ −2αV (x) ⇒ GES

β1||x||2 ≤ V (x) ≤ β2||x||2 for ||x|| ≤ R ≤ −2αV (x) for ||x|| ≤ R ⇒ ES

Figure 8.2: Lyapunov table

Chapter 9

Basic Lyapunov theory: discrete time

Suppose we are interested in the stability properties of the system,

x(k + 1) = f(x(k)) (9.1)

where x(k) ∈ IRn and k ∈ IR. If the system is linear, we can determine its stability propertiesfrom the properties of the eigenvalues of the system matrix. What do we for a nonlinearsystem? We could linearize about each equilibrium state and determine the stability prop-erties of the resulting linearizations. Under certain conditions (see later) this will tell ussomething about the local stability properties of the nonlinear system about its equilibriumstates. However there are situations where linearization cannot be used to deduce even thelocal stability properties of the nonlinear system. Also, linearization tells us nothing aboutthe global stability properties of the nonlinear system.

In general, we cannot explicitly obtain solutions for nonlinear systems. Lyapunov theoryallows to say something about the stability properties of a system without knowing the formor structure of the solutions.

• In this chapter, V is a scalar-valued function of the state, i.e. V : IRn → IR. At any timek, the one step change in V along a solution x(·) of system (9.1) is given by

V (x(k + 1))− V (x(k)) = ∆V (x(k))

where∆V (x) := V (f(x))− V (x)

91

92 CHAPTER 9. BASIC LYAPUNOV THEORY: DISCRETE TIME

9.1 Stability

Theorem 7 (Stability) Suppose there exists a locally positive definite function V and ascalar R > 0 such that

∆V (x) ≤ 0 for ||x|| < R

Then the origin is a stable equilibrium state.

• If V satisfies the hypotheses of the above theorem, then V is said to be a Lyapunov functionwhich guarantees the stability of origin.

Example 103

x(k + 1) = x(k)

ConsiderV (x) = x2

as a candidate Lyapunov function. Then V is a lpdf and

∆V (x) = 0

Hence (it follows from theorem 7 that) the origin is stable.



Theorem 8 (Asymptotic stability) Suppose there exists a locally positive definite functionV and a scalar R > 0 such that

∆V (x) < 0 for x 6= 0 and ||x|| < R

Then the origin is an asymptotically stable equilibrium state.

Example 104

x(k + 1) =1

2x(k)

ConsiderV (x) = x2

Then V is a lpdf and

∆V (x) = (1

2x)2 − x2

= −3

4x2

< 0 for x 6= 0


Example 105

x(k + 1) =1

2x(k) + x(k)2

ConsiderV (x) = x2

Then V is a lpdf and

∆V (x) = (1

2x+ x2)2 − x2

= −x2(32+ x)(

1

2− x)

< 0 for |x| < 1

2, x 6= 0



9.3 Boundedness

Theorem 9 Suppose there exists a radially unbounded function V and a scalar R ≥ 0 suchthat

∆V (x) ≤ 0 for ||x|| ≥ R

Then all solutions of (9.1) are bounded.

Note that, in the above theorem, V does not have to positive away from the origin; it onlyhas to be radially unbounded.

Example 106

x(k + 1) =2x(k)

1 + x(k)2

ConsiderV (x) = x2

Since V is radially unbounded and

∆V (x) =

(

2x

1 + x2

)2

− x2

= −x2(x2 + 3)(x2 − 1)

(x2 + 1)2

≤ 0 for |x| ≥ 1

the hypotheses of the above theorem are satisfied with R = 1; hence all solutions are bounded.Note that the origin is unstable.


9.4 Global asymptotic stability

Theorem 10 (Global asymptotic stability) Suppose there exists a positive definite functionV such that

∆V (x) < 0 for all x 6= 0

Then the origin is a globally asymptotically stable equilibrium state.

Example 107

x(k + 1) =1

2x(k)

Example 108

x(k + 1) =x(k)

1 + x(k)2

ConsiderV (x) = x2

Then

∆V (x) =

(

x

1 + x2

)2

− x2

= − 2x4 + x6

(1 + x2)2

< 0 for all x 6= 0

Hence this system is GAS about zero.

Exercise 27 Consider any scalar system described by

x(k + 1) = g(x(k))

where g has the following properties:

|g(x)| < |x| for x 6= 0

Show that this system is GAS.



Theorem 11 (Global exponential stability.) Suppose there exists a function V and scalarsα, β1, β2 such that for all x,

β1||x||2 ≤ V (x) ≤ β2||x||2 β1, β2 > 0

and

V (f(x)) ≤ α2V (x) 0 ≤ α < 1

Then, every solution satisfies

||x(k)|| ≤√

β2β1αk||x(0)|| for k ≥ 0

Hence, the origin is a globally exponentially stable equilibrium state with rate of convergenceα.

Proof.

Example 109

x(k + 1) = −1

2x(k)


ConsideringV (x) = x2

we have

V (f(x)) =1

4x2

= (1

2)2V (x)

Hence, we have GES with rate of convergence α = 12.

Example 110

x(k + 1) =1

2sin(x(k))

ConsideringV (x) = x2

we have

V (f(x)) = (1

2sin x)2

= (1

2)2| sin x|2

≤ (1

2)2|x|2

= (1

2)2V (x)

Hence, we have GES with rate of convergence α = 12

Chapter 10

Lyapunov theory for lineartime-invariant systems

The main result of this section is contained in Theorem 12.

10.1 Positive and negative (semi)definite matrices

10.1.1 Definite matrices

For any square matrix P we can define an associated quadratic form:

x′Px =n∑

i=1

n∑

j=1

pijx′ixj

If P is hermitian (P ′ = P ) then the scalar x′Px is real for all x ∈ Cn.

DEFN. A hermitian matrix P is positive definite (pd) if

x′Px > 0

for all nonzero x. We denote this by P > 0. The matrix P is negative definite (nd) if −P ispositive definite; we denote this by P < 0.

Example 111 For

P =

(

1 −1−1 2

)

we have (note the completion of squares trick)

x′Px = x′1x1 − x′1x2 − x′2x1 + 2x′2x2

= (x1 − x2)′(x1 − x2) + x′2x2

= |x1 − x2|2 + |x2|2

Clearly, x′Px ≥ 0 for all x. If x′Px = 0, then x1 − x2 = 0 and x2 = 0; hence x = 0. So,P > 0.

99

100CHAPTER 10. LYAPUNOV THEORY FOR LINEAR TIME-INVARIANT SYSTEMS

Fact 1 The following statements are equivalent for any hermitian matrix P .

(a) P is positive definite.

(b) All the eigenvalues of P are positive.

(c) All the leading principal minors of P are positive.

10.1. POSITIVE AND NEGATIVE (SEMI)DEFINITE MATRICES 101

Example 112 Consider

P =

(

1 −1−1 2

)

Since p11 = 1 > 0 and det(P ) = 1 > 0, we must have P > 0.

>> eig(P)

ans =

0.3820

2.6180

Note the positive eigenvalues.

10.1.2 Semi-definite matrices

DEFN. A hermitian matrix P is positive semi-definite (psd) if

x′Px ≥ 0

for all non-zero x. We denote this by P ≥ 0P is negative semi-definite (nsd) if −P is positive semi-definite; we denote this by P ≤ 0

Fact 2 The following statements are equivalent for any hermitian matrix P .

(a) P is positive semi-definite.

(b) All the eigenvalues of P are non-negative.

(c) All the leading minors of P are positive.

Example 113 This example illustrates that non-negativity of the leading principal minorsof P is not sufficient for P ≥ 0.

P =

(

0 00 −1

)

We have p11 = 0 and det(P ) = 0. However,

x′Px = −|x2|2

hence, P is not psd. Actually, P is nsd.

Fact 3 Consider any m× n complex matrix M and let P =M ′M . Then

(a) P is hermitian and P ≥ 0

(b) P > 0 iff rank M = n.


Example 114

P =

(

1 11 1

)

Since

P =

(

11

)

(

1 1)

andrank

(

1 1)

= 1

P ≥ 0 but P is not pd.

Exercise 28 (Optional)Suppose P is hermitian and T is invertible. Show that P > 0 iff T ′PT > 0.

Exercise 29 (Optional)Suppose P and Q are two hermitian matrices with P > 0. Show that P + λQ > 0 for

all real λ sufficiently small; i.e., there exists λ > 0 such that whenever |λ| < λ, one hasP + λQ > 0.

10.2. LYAPUNOV THEORY 103

10.2 Lyapunov theory

10.2.1 Asymptotic stability results

The scalar systemx = ax

is asymptotically stable if and only if a+a′ < 0. Consider now a general linear time-invariantsystem described by

x = Ax (10.1)

The generalization of a + a′ < 0 to this system is A + A′ < 0. We will see shortly that thiscondition is sufficient for asymptotic stability; however, as the following example illustrates,it is not necessary.

Example 115 The matrix

A =

(

−1 30 −1

)

is asymptotically stable. However, the matrix

A+ A′ =

(

−2 33 −2

)

is not negative definite.

Since stability is invariant under a similarity transformation T , i.e., the stability of A andT−1AT are equivalent for any nonsingular T , we should consider the more general condition

T−1AT + T ′A′T−′ < 0

Introducing the hermitian matrix P := T−′T−1, and pre- and post-multiplying the aboveinequality by T−′ and T−1, respectively, yields

P > 0 (10.2a)

PA+ A′P < 0 (10.2b)

We now show that the existence of a hermitian matrix P satisfying these conditions guar-antees asymptotic stability.

Lemma 6 Suppose there is a hermitian matrix P satisfying (10.2). Then system (10.1) isasymptotically stable

Proof. Suppose there exists a hermititian matrix P which satisfies inequalities (10.2).Consider any eigenvalue λ of A. Let v 6= 0 be an eigenvector corresponding to λ, i.e.,

Av = λv

Thenv′PAv = λv′Pv ; v′A′Pv = λv′Pv


Pre- and post-multiplying inequality (10.2b) by v′ and v, respectively, yields

λv′Pv + λv′Pv < 0

i.e.,

2Re(λ)v′Pv < 0

Since P > 0, we must have v′Pv > 0; hence Re(λ) < 0. Since the above holds for everyeigenvalue of A, system (10.1) is asymptotically stable.

Remark 1 A hermitian matrix P which satisfies inequalities (10.2) will be referred to as aLyapunov matrix for (10.1) or A.

Remark 2 Conditions (10.2) are referred to as linear matrix inequalities (LMI’s). In recentyears, efficient numerical algorithms have been developed to solve LMI’s or determine thata solution does not exist.

Lyapunov functions. To obtain an alternative interpretation of inequalities (10.2), con-sider the quadratic function V of the state defined by

V (x) := x′Px

Since P > 0, this function has a strict global minimum at zero. Using inequality (10.2b), itfollows that along any non-zero solution x(·) of (10.1),

dV (x(t))

dt= x′Px+ x′P x

= x′ (A′P + PA)x

< 0

i.e., V (x(·)) is strictly decreasing along any non-zero solution. Intuitively, one then expectsevery solution to asymptotically approach the origin. The function V is called a Lyapunovfunction for the system. The concept of a Lyapunov function readily generalizes to nonlinearsystems and has proved a very useful tool in the analysis of the stability properties ofnonlinear systems.

Remark 3 Defining the hermitian matrix S := P−1, inequalities (10.2) become

S > 0 (10.3a)

AS + SA′ < 0 (10.3b)

Hence the above lemma can be stated with S replacing P and the preceding inequalitiesreplacing (10.2).

So far we have shown that if a LTI system has a Lyapunov matrix, then it is AS. Is theconverse true? That is, does every AS LTI system have a Lyapunov matrix? And if this is


true how does one find a Lyapunov matrix? To answer this question note that satisfactionof inequality (10.2b) is equivalent to

PA+ A′P +Q = 0 (10.4)

where Q is a hermitian positive definite matrix. This linear matrix equation is known as theLyapunov equation. So one approach to looking for Lyapunov matrices could be to choose apd hermitian Q and determine whether the Lyapunov equation has a pd hermitian solutionfor P .

We first show that if the system x = Ax is asymptotically stable and the Lyapunovequation (10.4) has a solution then, the solution is unique. Suppose P1 and P2 are twosolutions to (10.4). Then,

(P2 − P1)A + A′(P2 − P1) = 0 .

Hence,eA

′t(P2 − P1)AeAt + eA

′tA′(P2 − P1)eAt = 0

that is,d(

eA′t(P2 − P1)e

At)

dt= 0

This implies that, for all t,

eA′t(P2 − P1)e

At = eA′0(P2 − P1)e

A0 = P2 − P1 .

Since x = Ax is asymptotically stable, limt−→∞ eAt = 0. Hence

P2 − P1 = limt−→∞

eA′t(P2 − P1)e

At = 0 .

From this it follows that P2 = P1.The following lemma tells us that every AS LTI system has a Lyapunov matrix and a

Lyapunov matrix can be obtained by solving the Lyapunov equation with any pd hermitianQ.

Lemma 7 Suppose system (10.1) is asymptotically stable. Then for every matrix Q, theLyapunov equation (10.4) has a unique solution for P . If Q is positive definite hermitianthen this solution is positive definite hermitian.

Proof. Suppose system (10.1) is asymptotically stable and consider any matrix Q. Let

P :=

∫ ∞

0

eA′tQeAtdt

This integral exists because each element of eAt is exponentially decreasing.To show that P satisfies the Lyapunov equation (10.4), use the following properties of

eAt,

eAtA =deAt

dtA′eA

′t =deA

′t

dt


to obtain

PA+ A′P =

∫ ∞

0

(

eA′tQeAtA+ A′eA

′tQeAt)

dt

=

∫ ∞

0

(

eA′tQ

deAt

dt+deA

′t

dtQeAt

)

dt

=

∫ ∞

0

d(

eA′tQeAt

)

dtdt

= limt→∞

∫ t

0

deA′tQeAt

dtdt

= limt→∞

eA′tQeAt − Q

= −Q

We have already demonstrated uniqueness of solutions to (10.4),Suppose Q is pd hermitian. Then it should be clear that P is pd hermitian. .

Using the above two lemmas, we can now state the main result of this section.

Theorem 12 The following statements are equivalent.

(a) The system x = Ax is asymptotically stable.

(b) There exist positive definite hermitian matrices P and Q satisfying the Lyapunov equa-tion (10.4).

(c) For every positive definite hermitian matrix Q, the Lyapunov equation (10.4) has aunique solution for P and this solution is hermitian positive definite.

Proof. The first lemma yields (b) =⇒ (a). The second lemma says that (a) =⇒ (c).Hence, (b) =⇒ (c).

To see that (c) =⇒ (b), pick any positive definite hermitian Q. So, (b) is equivalent to(c). Also, (c) =⇒ (a); hence (a) and (c) are equivalent.

.

Example 116

x1 = x2

x2 = −x1 + cx2

Here

A =

(

0 1−1 c

)


Choosing

Q =

(

1 00 1

)

and letting

P =

(

p11 p12p12 p22

)

(note we have taken p21 = p12 because we are looking for symmetric solutions) the Lyapunovequation results in

−2p12 + 1 = 0

p11 + cp12 − p22 = 0

2p12 + 2cp22 + 1 = 0

We consider three cases:

Case i) c = 0. No solution; hence no GAS.

Case ii) c = −1. A single solution; this solution is pd; hence GAS.

Case iii) c = 1. A single solution; this solution is not pd; hence no GAS.

Remark 4 One can state the above theorem replacing P with S and replacing (10.4) with

AS + SA′ +Q = 0 (10.5)

10.2.2 MATLAB.

>> help lyap

LYAP Lyapunov equation.

X = LYAP(A,C) solves the special form of the Lyapunov matrix

equation:

A*X + X*A’ = -C

X = LYAP(A,B,C) solves the general form of the Lyapunov matrix

equation:

A*X + X*B = -C

See also DLYAP.

Note. In MATLAB, A’ is the complex conjugate transpose of A (i.e. A′). Hence, to solve(10.4) one must use P= lyap(A’, Q).


>> q=eye(2);

>> a=[0 1; -1 -1];

>> p=lyap(a’, q)

p =

1.5000 0.5000

0.5000 1.0000

>> p*a + a’*p

ans =

-1.0000 -0.0000

0 -1.0000

>> det(p)

ans =

1.2500

>> eig(p)

ans =

1.8090

0.6910

Exercise 30 Consider the Lyapunov equation

PA+ A′P + 2αP +Q = 0

where A is a square matrix, Q is some positive definite hermitian matrix and α is somepositive scalar. Show that this equation has a unique solution for P and this solution ispositive definite hermitian if and only if for every eigenvalue λ of A,

Re(λ) < −α

10.2.3 Stability results

For stability, the following lemma is the analog of lemma 10.2.

Lemma 8 Suppose there is a hermitian matrix P satisfying

P > 0 (10.6a)

PA+ A′P ≤ 0 (10.6b)

Then system (10.1) is stable.

We do not have the exact analog of lemma 7. However the following can be shown.


Lemma 9 Suppose system (10.1) is stable. Then there exist a positive definite hermitianmatrix P and a positive semi-definite hermitian matrix Q satisfying the Lyapunov equation(10.4).

The above lemma does not state that for every positive semi-definite hermitian Q theLyapunov equation has a solution for P . Also, when the Lyapunov equation has a solution,it is not unique. This is illustrated in the following example.

Example 117

A =

(

0 1−1 0

)

With

P =

(

1 00 1

)

we obtainPA+ A′P = 0

In this example the Lyapunov equation has a pd solution with Q = 0; this solution isnot unique; any matrix of the form

P =

(

p 00 p

)

(where p is arbitrary) is also a solution.If we consider the psd matrix

Q =

(

1 00 0

)

the Lyapunov equation has no solution.

Combining the above two lemmas results in the following theorem.


(a) The system x = Ax is stable.

(b) There exist a positive definite hermitian matrix P and a positive semi-definite matrixQ satisfying the Lyapunov equation (10.4).


10.3 Mechanical systems

Example 118 A simple structure

Recall

m1q1 + (c1 + c2)q1 − c2q2 + (k1 + k2)q1 − k2q2 = 0m2q2 − c2q1 + c2q2 − k2q1 + k2q2 = 0

Letting

q =

(

q1q2

)

this system can be described by the following second order vector differential equation:

Mq + Cq +Kq = 0

where the symmetric matrices M,C,K are given by

M =

(

m1 00 m2

)

C =

(

c1 + c2 −c2−c2 c2

)

K =

(

k1 + k2 −k2−k2 k2

)

Since m1, m2 > 0, we have M > 0.

If k1, k2 > 0 then K > 0 (why?) and if c1, c2 > 0 then C > 0.

Note also that

kinetic energy = 12m1q

21 +

12m2q

22 = 1

2q′Mq

potential energy = 12k1q

21 +

12k2(q2 − q1)

2 = 12q′Kq

10.3. MECHANICAL SYSTEMS 111

Consider now a general mechanical system described by

Mq + Cq +Kq = 0

where q(t) is a real N -vector of generalized coordinates which describes the configuraton ofthe system. The real matrix M is the inertia matrix and satisfies

M ′ =M > 0

We call K and C the ‘stiffness’ matrix and ‘damping’ matrix respectively. The kinetic energyof the system is given by

1

2q′Mq

With x = (q′, q′)′, this system has a state space description of the form x = Ax with

A =

(

0 I−M−1K −M−1C

)

A first Lyapunov matrix. Suppose

K ′ = K > 0 and C ′ = C ≥ 0 .

The potential energy of the system is given by

1

2q′Kq

Consider the following candidate Lyapunov matrix

P =1

2

(

K 00 M

)

Then P ′ = P > 0 andPA+ A′P +Q = 0

where

Q =

(

0 00 C

)

It should be clear thatPA+ A′P ≤ 0

iffC ≥ 0

Hence,

K ′ = K > 0 and C ′ = C ≥ 0 imply stability.

Note that

V (x) = x′Px =1

2q′Mq +

1

2q′Kq = total energy

V (x) = −q′Cq


A second Lyapunov matrix.

P =1

2

(

K + λC λMλM M

)

=1

2

(

K 00 M

)

+λ

2

(

C MM 0

)

For sufficiently small λ, the matrix P is positive definite.Now,

PA+ A′P +Q = 0

where

Q =

(

λK 00 C − λM

)

For sufficiently small λ > 0, the matrix C − λM is pd and, hence, Q is pd. So,

K ′ = K > 0 and C ′ = C > 0 imply asymptotic stability.

10.4. RATES OF CONVERGENCE FOR LINEAR SYSTEMS 113

10.4 Rates of convergence for linear systems

Theorem 14 Consider an asymptotically stable linear system described by x = Ax and letα be any real number satisfying

0 < α < α

whereα := −max{ℜ(λ) : λ is an eigenvalue of A } .

Then x = Ax is GES with rate α.

Proof: Consider any α satisfying 0 < α < α. As a consequence of the definition of α, allthe eigenvalues of the matrix A+αI have negative parts. Hence, the Lyapunov equation

P (A+ αI) + (A+αI)′P + I = 0 .

has a unique solution for P and P = P ′ > 0. As a candidate Lyapunov for the systemx = Ax, consider V (x) = x′Px. Then,

V = x′P x+ x′Px

= x′PAx+ (Ax)′Px

= x′(PA+ A′P )x .

From the above Lyapunov matrix equation, we obtain that

PA+ A′P = −2αP − I ;

hence,

V = −2αx′Px− x′x

≤ −2αV (x) .

Hence the system is globally exponentially stable with rate of convergence α.


10.5 Linearization and exponential stability

Consider a nonlinear systemx = f(x) (10.7)

and suppose that xe is an equilibrium state for this system. We assume that f is differentialat xe and let

Df(xe) =

∂f1∂x1

(xe)∂f1∂x2

(xe) · · · ∂f1∂xn

(xe)

∂f2∂x1

(xe)∂f2∂x2

(xe) · · · ∂f2∂xn

(xe)

......

...

∂fn∂x1

(xe)∂fn∂x2

(xe) · · · ∂fn∂xn

(xe)

where x = (x1, x2, · · · , xn). Then the linearization of x = f(x) about xe is the linear systemdefined by

δx = Aδx where A = Df(xe) . (10.8)

Theorem 15 Suppose xe is an equilibrium state of a nonlinear system of the form x = f(x)and the corresponding linearization is exponentially stable. Then the nonlinear system isexponentially stable about xe.

Proof. Suppose the linearization is exponentially stable and let

α := −max{ℜ(λ) : λ is an eigenvalue of A } .

Since all the eigenvalues have negative real parts, we have α > 0. Consider now any αsatisfying 0 < α < α. As a consequence of the definition of α, all the eigenvalues of thematrix A+αI have negative parts. Hence, the Lyapunov equation

P (A+ αI) + (A+αI)′P + I = 0 .

has a unique solution for P and P = P ′ > 0. As a candidate Lyapunov function for thenonlinear system, consider

V (x) = (x−xe)′P (x−xe) .Recall that

f(x) = f(xe) +Df(xe)(x−xe) + o(x−xe)where the “remainder term” has the following property:

limx→xe, x 6=xe

o(x−xe)‖x−xe‖ = 0 .

Hencef(x) = A(x−xe) + o(x−xe)

10.5. LINEARIZATION AND EXPONENTIAL STABILITY 115

and

V = 2x′P x

= 2x′Pf(x)

= 2x′PAx+ 2x′Po(x)

≤ x′(PA+ A′P )x+ 2‖x‖‖P‖‖o(x)‖ .

From the above Lyapunov matrix equation, we obtain that

PA+ A′P = −2αP − I ;

hence,

V = −2αx′Px− x′x+ 2‖x‖‖P‖‖o(x)‖= −2αV (x)− ‖x‖2 + 2‖x‖‖P‖‖o(x)‖ .

As a consequence of the properties of o, there is a scalar R > 0 such that

2‖P‖‖o(x)‖ ≤ ‖x‖ when ‖x‖ ≤ R .

Hence, whenever ‖x‖ ≤ R we obtain that

V ≤ −2αV (x) .

Hence the nonlinear system is exponentially stable about xe with rate of convergence α.

Chapter 11

Quadratic stability

11.1 Introduction

[3] In this chapter, we introduce numerical techniques which are useful in finding Lyapunovfunctions for certain classes of globally exponentially stable nonlinear systems. We restrictconsideration to quadratic Lyapunov functions. For specific classes of nonlinear systems, wereduce the search for a Lyapunov function to that of solving LMIs (Linear matrix inequalities).

The results in this chapter are also useful in proving stability of switching linear systems.To establish the results in this chapter, recall that a system

x = f(x) (11.1)

is globally exponentially stable about the origin if there exist positive definite symmetricmatrices P and Q which satisfy

xTPf(x) ≤ −xTQxfor all x. When this is the case we call P a Lyapunov matrix for system (11.1).

11.2 Polytopic nonlinear systems

Here we consider nonlinear systems described by

x = A(x)x (11.2)

where the n-vector x is the state. The state dependent matrix A(x) has the followingstructure

A(x) = A0 + ψ(x)∆A (11.3)

where A0 and ∆A are constant n× n matrices and ψ is a scalar valued function of the statex which is bounded above and below, that is

a ≤ ψ(x) ≤ b (11.4)

for some constants a and b. Examples of functions satisfying the above conditions are givenby ψ(x) = g(c(x)) where c(x) is a scalar and g(y) is given by sin y, cos y, e−y2 or sgm (y).The signum function is useful for modeling switching systems.

117

118 CHAPTER 11. QUADRATIC STABILITY

Example 119 Inverted pendulum under linear feedback. Consider an inverted pendulumunder linear control described by

x1 = x2 (11.5a)

x2 = −2x1 − x2 + γ sin x1 γ > 0 (11.5b)

This system can be described by (11.2)-(11.4) with

A0 =

(

0 1−2 −1

)

, ∆A =

(

0 01 0

)

and

ψ(x) =

{

γ sin x1/x1 if x1 6= 0γ if x1 = 0

Since | sinx1| ≤ |x1|, we have−γ ≤ ψ(x) ≤ γ;

hence a = −γ and b = γ.

The following theorem provides a sufficient condition for the global exponential stability ofsystem (11.2)-(11.4). These conditions are stated in terms of linear matrix inequalities andthe two matrices corresponding to the extreme values of ψ(x), namely

A1 := A0 + a∆A and A2 := A0 + b∆A

Theorem 16 Suppose there exists a positive-definite symmetric matrix P which satisfies thefollowing linear matrix inequalities:

PA1 + AT1 P < 0

PA2 + AT2 P < 0

(11.6)

Then system (11.2)-(11.4) is globally exponentially stable (GES) about the origin with Lya-punov matrix P .

Proof. As a candidate Lyapunov function for GES, consider V (x) = xTPx. Then

V = 2xTP x

= 2xTPA0x+ 2ψ(x)xTP∆Ax .

For each fixed x, the above expression for V is a linear affine function of the scalar ψ(x).Hence an upper bound for V occurs when ψ(x) = a or ψ(x) = b which results in

V ≤ 2xTPA1x = xT (PA1 + AT1 P )x

or

V ≤ 2xTPA2x = xT (PA2 + AT2 P )x

11.2. POLYTOPIC NONLINEAR SYSTEMS 119

respectively. As a consequence of the matrix inequalities (11.6), there exist positive scalarsα1 and α2 such that

PA1 + AT1 P ≤ −2α1P

PA2 + AT2 P ≤ −2α2P

Letting α = min{α1, α2}, we now obtain that

V ≤ −2αxTPx = −2αV .

This guarantees the system is GES with rate α.

Exercise 31 Prove the following result: Suppose there exists a positive-definite symmetricmatrix P and a positive scalar α which satisfy

PA1 + AT1 P + 2αP ≤ 0 (11.7a)

PA2 + AT2 P + 2αP ≤ 0 (11.7b)

where A1 := A0 + a∆A and A2 := A0 + b∆A. Then system (11.2)-(11.4) is globally expo-nentially stable about the origin with rate of convergence α.

11.2.1 LMI Control Toolbox

Recall the pendulum of Example 119. We will use the Matlab LMI Control Toolbox to see ifwe can show exponential stability of this system. First note that the existence of a positivedefinite symmetric P satisfying inequalities (11.6) is equivalent to the existence of anothersymmetric matrix P satisfying

PA1 + AT1 P < 0

PA2 + AT2 P < 0

P > I

For γ = 1, we determine the feasibility of these LMIs using the following Matlab program.

% Quadratic stability of inverted pendulum under linear PD control.

%

gamma=1;

A0 = [0 1; -2 -1];

DelA = [0 0; 1 0];

A1 = A0 - gamma*DelA;

A2 = A0 + gamma*DelA;

%

%

setlmis([])

%

p=lmivar(1, [2,1]);

%


lmi1=newlmi;

lmiterm([lmi1,1,1,p],1,A1,’s’)

%

lmi2=newlmi;

lmiterm([lmi2,1,1,p],1,A2,’s’)

%

Plmi= newlmi;

lmiterm([-Plmi,1,1,p],1,1)

lmiterm([Plmi,1,1,0],1)

%

lmis = getlmis;

%

[tfeas, xfeas] = feasp(lmis)

%

P = dec2mat(lmis,xfeas,p)

Running this program yields the following output

Solver for LMI feasibility problems L(x) < R(x)

This solver minimizes t subject to L(x) < R(x) + t*I

The best value of t should be negative for feasibility

Iteration : Best value of t so far

1 0.202923

2 -0.567788

Result: best value of t: -0.567788

f-radius saturation: 0.000% of R = 1.00e+09

tfeas = -0.5678

xfeas = 5.3942

1.0907

2.8129

P = 5.3942 1.0907

1.0907 2.8129

Hence the above LMIs are feasible and the pendulum is exponentially stable with Lyapunovmatrix

P =

(

5.3942 1.09071.0907 2.8129

)


By iterating on γ, the largest value for which LMIs (11.6) were feasible was found tobe γ ≈ 1.3229. However, from previous considerations, we have shown that this system isexponentially stable for γ < 2. Why the difference?

Exercise 32 What is the supremal value of γ > 0 for which Theorem 16 guarantees thatthe following system is guaranteed to be stable about the origin?

x1 = −2x1 + x2 + γe−x21x2

x2 = −x1 − 3x2 − γe−x21x1

Exercise 33 Consider the pendulum system of Example 119 with γ = 1. Obtain the largestrate of exponential convergence that can be obtained using the results of Exercise 31 andthe LMI toolbox.


11.2.2 Generalization

One can readily generalize the results of this section to systems described by

x = A(x)x (11.8)

where the state dependent matrix A(x) has the following structure

A(x) = A0 + ψ1(x)∆A1 + · · ·+ ψl(x)∆Al , (11.9)

where A0,∆A1, · · · ,∆Al are constant n× n matrices and each ψi is a scalar valued functionof the state x which satisfies

ai ≤ ψi(x) ≤ bi (11.10)

for some known constants ai and bi.

To obtain a sufficient condition for exponential stability, introduce the following set of 2l

“extreme matrices”:

A = {A0 + ψ1∆A1 + · · ·+ ψl∆Al : ψi = ai or bi for i = 1, · · · , l}

For example, with l = 2, A consists of the following four matrices:

A0 + a1∆A1 + a2∆A2

A0 + a1∆A1 + b2∆A2

A0 + b1∆A1 + a2∆A2

A0 + b1∆A1 + b2∆A2 (11.11)

We have now the following result.

Theorem 17 Suppose there exists a positive-definite symmetric matrix P which satisfies thefollowing linear matrix inequalities:

PA+ ATP < 0 for all A in A (11.12)

Then system (11.8)-(11.10) is globally exponentially stable about the origin with Lyapunovmatrix P .

Exercise 34 Consider the double inverted pendulum described by

θ1 + 2θ1 − θ2 + 2kθ1 − kθ2 − sin θ1 = 0

θ2 − θ1 + θ2 − kθ1 + kθ2 − sin θ2 = 0

Using the results of Theorem 17, obtain a value of the spring constant k which guaranteesthat this system is globally exponentially stable about the zero solution.


11.2.3 Robust stability

Consider an uncertain system described by

x = f(x, δ) (11.13)

where δ is some uncertain parameter vector. suppose that the only information we have on δis its set of possible values, that is,

δ ∈ ∆

where the set ∆ is known. So we do not know what δ is, we only know a set of possiblevalues of δ. We say that the above system is robustly stable if it is stable for all allowablevalues of δ, that is, for all δ ∈ ∆.

We say that the uncertain system is quadratically stable if there are positive definitesymmetric matrices P and Q such

xTPf(x, δ) ≤ −xTQx (11.14)

for x and δ ∈ ∆. Clearly, quadratic stability guarantees robust exponentially stability aboutzero.

One of the useful features of the above concept, is that one can sometimes guarantee that(11.14) holds for all δ by just checking that is holds for a finite number of δ. For, exampleconsider the linear uncertain system described by

x = A(δ)x .

Suppose that the uncertain matrix A(δ) has the following structure

A(δ) = A0 + δ1∆A1 + · · ·+ δ∆Al , (11.15)

where A0,∆A1, · · · ,∆Al are constant n× n matrices and each δi is a scalar which satisfies

ai ≤ δi ≤ bi (11.16)

Introduce the following set of 2l “extreme matrices”:

A = {A0 + ψ1∆A1 + · · ·+ ψl∆Al : ψi = ai or bi for i = 1, · · · , l}

Suppose there exists a positive-definite symmetric matrix P which satisfies the followinglinear matrix inequalities:

PA+ ATP < 0 for all A in A (11.17)

Then, for all allowable δ, the uncertain system is globally exponentially stable about theorigin with Lyapunov matrix P .


11.3 Another special class of nonlinear systems

Here we consider nonlinear systems which have the following structure:

x = Ax+Bφ(Cx) (11.18)

where the n-vector x(t) is the state; B and C are constant matrices with dimensions n×mand p× n, respectively; and φ is a function which satisfies

||φ(z)|| ≤ γ||z|| (11.19)

for some γ ≥ 0.Note that if we introduce a fictitious output z = Cx and a fictitious input w = φ(Cx),

the nonlinear system (11.18) can be described as a feedback combination of a linear timeinvariant system

x = Ax+Bwz = Cx

and a memoryless nonlinearityw = φ(z)

This is illustrated in Figure 11.1.

Figure 11.1: A feedback system

Example 120 Inverted pendulum under linear feedback. Recall the inverted pendulum ofthe previous example described by (11.5). This system can be described by the above generaldescription with

A =

(

0 1−2 −1

)

B =

(

01

)

C =(

1 0)

φ(z) = γ sin z

Also,||φ(z)|| = |γ sin z| ≤ γ|z| = γ||z||

The following theorem provides a sufficient condition for the global exponential stabilityof system (11.18).

Theorem 18 Suppose there exists a positive-definite symmetric matrix P which satisfies thefollowing quadratic matrix inequality (QMI):

PA+ ATP + γ2PBBTP + CTC < 0 (11.20)


11.3. ANOTHER SPECIAL CLASS OF NONLINEAR SYSTEMS 125

Proof. Along any solution of system (11.18)-(11.19) we have

2xTP x = 2xTPAx+ 2xTPBφ = xT (PA+ ATP )x+ 2xTPBφ . (11.21)

Note that for any two real numbers a and b,

2ab = a2 + b2 − (a− b)2 ;

hence2ab ≤ a2 + b2 .

Considering the last term in (11.21), we now obtain

2xTPBφ ≤ 2||BTPx||||φ||≤ 2||BTPx||γ||Cx||≤ γ2||BTPx||2 + ||Cx||2= γ2xTPBBTPx+ xTCTCx . (11.22)

Combining (11.21) and (11.22), it now follows that

2xTP x ≤ −2xTQx

where −2Q = PA + ATP + γ2PBBTP + CTC. By assumption, the matrix Q is positivedefinite. Hence system (11.18)-(11.19) is globally exponentially stable about the origin withLyapunov matrix P .

A linear matrix inequality (LMI). Using a Schur complement result (see Appendix atthe end of this chapter), one can show that satisfaction of quadratic matrix inequality (11.20)is equivalent to satisfaction of the following matrix inequality:

(

PA+ ATP + CTC PBBTP −γ−2I

)

< 0 (11.23)

Note that this inequality is linear in P and γ−2. Suppose one wishes to compute the supremalvalue γ of γ which guarantees satisfaction of the hypotheses of Theorem 18 and hence stabilityof system (11.32). This can be achieved by solving the following optimization problem:

minimize β subject to

(

PA+ ATP + CTC PBBTP −βI

)

< 0

0 < P

0 < β

and then letting γ = β− 12 where β is the infimal value of β. The following program uses the

LMI toolbox to compute γ for the inverted pendulum example.


%Quadratic stability radius of inverted pendulum

%

A = [0 1; -2 -1];

B=[0;1];

C=[1 0];

%

setlmis([])

%

pvar=lmivar(1,[2,1]); %P

betavar=lmivar(1,[1,1]); %beta

%

lmi1=newlmi;

lmiterm([lmi1 1 1 pvar],1,A,’s’); %PA+A’P

lmiterm([lmi1 1 1 0],C’*C); %C’C

lmiterm([lmi1 1 2 pvar],1,B); %PB

lmiterm([lmi1 2 2 betavar],-1,1); %-beta I

%

lmi2=newlmi;

lmiterm([-lmi2 1 1 pvar],1,1) %P

%

lmi3=newlmi;

lmiterm([-lmi3 1 1 betavar],1,1); %beta

%

lmis=getlmis;

%

c=mat2dec(lmis,0,1); %specify weighting

options=[1e-5 0 0 0 0];

[copt,xopt]=mincx(lmis,c,options) %optimize

%

gamma=1/sqrt(copt)

Running this program yields the following output:

Solver for linear objective minimization under LMI constraints

Iterations : Best objective value so far

1 0.862772

2 0.807723

3 0.675663

*** new lower bound: -0.031634

4 0.645705

*** new lower bound: 0.148578

5 0.626163



6 0.598055


7 0.580987


8 0.572209


9 0.571637


10 0.571450


11 0.571438


12 0.571429


13 0.571429


14 0.571429


Result: feasible solution of required accuracy

best objective value: 0.571429

guaranteed absolute accuracy: 2.86e-06

f-radius saturation: 0.000% of R = 1.00e+09

copt = 0.5714

xopt = 1.1429

0.2857

0.5714

0.5714

gamma = 1.3229

Thus γ = 1.3229 which is the same as that achieved previously.

A Riccati equation. It should be clear that P satisfies the above QMI if and only if itsatisfies the following Riccati equation for some positive definite symmetric matrix Q:

PA+ ATP + γ2PBBTP + CTC +Q = 0 (11.24)

Using properties of QMI (11.20) (see Ran and Vreugdenhil, 1988), one can demonstrate thefollowing.

Lemma 10 There exists a positive-definite symmetric matrix P which satisfies the quadraticmatrix inequality (11.20) if and only if for any positive-definite symmetric matrix Q there


is an ǫ > 0 such that for all ǫ ∈ (0, ǫ] the following Riccati equation has a positive-definitesymmetric solution for P

PA+ ATP + γ2PBBTP + CTC + ǫQ = 0 (11.25)

Using this corollary, the search for a Lyapunov matrix is reduced to a one parameter search.

Exercise 35 Prove the following result: Suppose there exist a positive-definite symmetricmatrix P and a positive scalar α which satisfy

(

PA+ ATP + CTC + 2αP PBBTP −γ−2I

)

≤ 0 . (11.26)

Then system (11.18)-(11.19) is globally exponentially stable about the origin with rate ofconvergence α.

11.3.1 The circle criterion: a frequency domain condition

In many situations, one is only interested in whether a given nonlinear system is stable ornot; one may not actually care what the Lyapunov matrix is. To this end, the followingfrequency domain characterization of quadratic stability is useful. Consider system (11.18)and define the transfer function G by

G(s) = C(sI −A)−1B (11.27)

and let‖G‖∞ := sup

ω∈IR‖G(jω)‖ (11.28)

Then we have the following result from the control literature.

Lemma 11 (KYP1) [6, 4] There exists a positive-definite symmetric matrix P which sat-isfies the linear matrix inequality (11.23) if and only if

(i) A is asymptotically stable and

(ii)

γ‖G‖∞ < 1 (11.29)

The last lemma and Theorem 18 tell us that γ , the supremal value of γ for stability ofsystem (11.18)-(11.19) via the method of this section, is given by

γ = 1/||G||∞Example 121 Consider Example 120 again. Here the matrix A is asymptotically stableand

G(s) = 1/(s2 + s+ 2)

One may readily compute that‖G‖∞ = 2/

√7 < 1

Hence, this nonlinear system is exponentially stable. Also, γ = 1/||G||∞ =√7/2 ≈ 1.3229.

This is the same as before.


-0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

Real Axis

Imag

Axi

s

Figure 11.2: A nice figure

11.3.2 Sector bounded nonlinearities

Suppose φ is a scalar valued function of a scalar variable (m = p = 1). Then the bound(11.19) is equivalent to

−γz2 ≤ zφ(z) ≤ γz2

That is, φ is a function whose graph lies in the sector bordered by the lines passing throughthe origin and having slopes γ and −γ; see Figure 11.3. Consider now a function φ whose

Figure 11.3: A norm bounded nonlinearity

graph lies in the sector bordered by the lines passing through the origin and having slopes aand b, with a < b; see Figure 11.4. Analytically, this can be expressed as

Figure 11.4: A general sector bounded nonlinearity

az2 ≤ zφ(z) ≤ bz2 (11.30)


Example 122 The saturation function which is defined below is illustrated in Figure 11.5.

φ(z) =

z if |z| ≤ 1−1 if z < −11 if z > 1

Since 0 ≤ zφ(z) ≤ z2, the saturation function is sector bounded with a = 0 and b = 1.

Figure 11.5: Saturation

The following result is sometimes useful in showing that a particular nonlinear functionis sector bounded.

Fact 4 Suppose φ is a differentiable scalar valued function function of a scalar variable,φ(0) = 0 and for all z

a ≤ φ′(z) ≤ b

for some scalars a and b. Then φ satisfies (11.30) for all z.

We can treat sector bounded nonlinearities using the results of the previous section byintroducing some transformations. Specifically, if we introduce

φ(z) := φ(z)− kz where k := (a + b)/2 ,

then,

−γz2 ≤ zφ(z) ≤ γz2 with γ := (b− a)/2

and system (11.32) is described by

x = Ax+Bφ(Cx) (11.31)

with A = A + kBC. Since |φ(z)| ≤ γ|z|, we can apply the results of the previous section tosystem (11.31) to obtain stability results for the original nonlinear system, which containsthe nonlinearity satisfying (11.30).



Here we consider systems whose nonlinearity is characterized by several nonlinear functions,φi : IR

pi → IRmi , i = 1, 2, . . . , l :

x = Ax+B1φ(C1x) + . . .+Blφl(Clx) (11.32)

where A0 and Bi, Ci, i = 1, 2, · · · , l are constant matrices of appropriate dimensions. Eachφi satisfies

||φi(zi)|| ≤ γ||zi|| (11.33)

for some γ ≥ 0.If we introduce fictitious outputs zi = Cix and a fictitious inputs wi = φi(Cix), the above

nonlinear system can be described as a feedback combination of a linear time invariant system

x = Ax+B1w1 + . . .+Blwl

z1 = C1x...

zl = Clx

and multiple memoryless nonlinearities:

wi = φi(zi) i = 1, 2, · · · , l


Figure 11.6: A multi-loop feedback system

We will use the results of the previous section to obtain results for the systems of thissection. Letting m = m1 + · · ·+ml and p = p1 + · · ·+ pl, suppose we introduce a functionφ : Rp → Rm defined by

φ(z) =

φ1(z1)φ2(z2)

...φl(zl)

where z =

z1z2...zl

with zi ∈ IRpi (11.34)


Then system (11.32) can also be described by (11.18) with

B :=(

B1 B2 . . . Bl

)

, C :=

C1

C2...Cl

Also,

||φ(z)||2 =l∑

i=1

||φi(zi)||2 ≤l∑

i=1

γ2||zi||2 = γ2||z||2

that is ||φ(z)|| ≤ γ||z||. We can now apply the results of the previous section to obtainsufficient conditions for global exponential stability. However, these results will be overlyconservative since they do not take into account the special structure of φ; the function φ isnot just any function satisfying the bound ‖φ(z)‖ ≤ γ ‖z‖; it also has the structure indicatedin (11.34). To take this structure into account, consider any l positive scalars λ1, λ2, . . . , λland introduce the nonlinear function

φ(z) =

λ1φ1(λ−11 z1)

λ2φ2(λ−12 z2)...

λlφl(λ−1l zl)

where z =

z1z2...zl

with zi ∈ IRpi

Then system (11.32) can also be described by (11.18), specifically, it can be described by

x = Ax+ Bφ(Cx) (11.35)

with

B :=(

λ−11 B1 λ−1

2 B2 . . . λ−1l Bl

)

, C :=

λ1C1

λ2C2...

λlCl

Noting that ||λiφi(λ−1i zi)|| = λi||φ(λ−1

i zi)|| ≤ λiγ||λ−1i zi|| = γ||zi||, we obtain

||φ(z)||2 =l∑

i=1

||λiφ(λ−1i zi)||2 ≤ γ2

l∑

i=1

||zi||2 = γ2||z||2

that is, ||φ(z)|| ≤ γ||z||.To obtain sufficient conditions for stability of (11.32), we simply apply the results of the

previous section to (11.35). Using Theorem 18 one can readily obtain the following result:

Theorem 19 Suppose there exists a positive-definite symmetric matrix P and l positivescalars µ1, µ2, . . . , µl which satisfy the following matrix inequality:

PA+ ATP +

l∑

i=1

(

µ−1i γ2PBiBi

TP + µiCTi Ci

)

< 0 (11.36)

Then system (11.32) is globally exponentially stable about the origin with Lyapunov matrixP .


Proof. If we let λi =√µi for i = 1, 2, · · · , l, then (11.36) can be written as

PA+ ATP + γ2PBBTP + CT C < 0

Since ||φ(z)|| ≤ γ||z||, it follows from representation (11.35) and Theorem 18 that system(11.32) is globally exponentially stable about the origin with Lyapunov matrix P .

An LMI. Note that, using a Schur complement result, inequality (11.36) is equivalent tothe following inequality which is linear in P and the scaling parameters µ1, · · · , µl:

PA+ ATP +∑l

i=1 µiC′iCi γPB1 γPB2 · · · γPBl

γBT1 P −µ1I 0 · · · 0

γBT2 P 0 −µ2I · · · 0...

.... . .

...γBT

l P 0 0 · · · −µlI

< 0

It should be clear that one may also obtain a sufficient condition involving a Riccatiequation with scaling parameters µi using Corollary 10 and a H∞ sufficient condition usingLemma 11.

Exercise 36 Recall the double inverted pendulum of Example 34. Using the results of thissection, obtain a value of the spring constant k which guarantees that this system is globallyexponentially stable about the zero solution.


11.4 Yet another special class

Here we consider systems described by

x = Ax−Bφ(Cx) (11.37)

wherez′φ(z) ≥ 0 (11.38)

for all z. Examples of φ include φ(z) = z, z3, z5, sat(z), sgm (z).

Figure 11.7: A first and third quadrant nonlinearity

Note that such a system can also be regarded as a feedback combination of the LTIsystem

x = Ax+Bw

z = Cx

and a memoryless nonlinear system

w = −φ(z) .

Theorem 20 Suppose there exists a positive-definite symmetric matrix P which satisfies

PA+ A′P < 0B′P = C

(11.39)


Proof.

2x′P x = x′(PA+ A′P )x− 2x′PBφ(Cx)

= x′(PA+ A′P )x− 2(Cx)′φ(Cx)

≤ x′(PA+ A′P )x .

Hence2x′P x ≤ −2xTQx

11.4. YET ANOTHER SPECIAL CLASS 135

where 2Q = −PA− ATP . Since Q is positive definite, we have global exponential stabilityabout zero.

Now note that P > 0 satisfies (11.39) if and only if the following optimization problemhas a minimum of zero:

Minimize β subject to

PA+ A′P < 0

P > 0

(

βI B′P − CPB − C ′ βI

)

≥ 0

Example 123 Considerx1 = −3x1 +x2x2 = x1 −3x2 −x32

Here

A =

(

−3 11 −3

)

, B =

(

01

)

, C =(

0 1)

and φ(z) = z3. Clearly zφ(z) ≥ 0 for all z. Also, conditions (11.39) are assured with P = I.Hence, the system of this example is globally exponentially stable. Performing the aboveoptimization with the LMI toolbox yields a minimum of β to equal to “zero” and

P =

(

22.8241 00 1

)

This P also satisfies conditions (11.39).

Exercise 37 Prove the following result: Suppose there exists a positive-definite symmetricmatrix P and a positive scalar α which satisfy which satisfies

PA+ A′P + 2αP ≤ 0 (11.40a)

B′P = C (11.40b)

Then system (11.37)-(11.38) is globally exponentially stable about the origin with rate αand with Lyapunov matrix P .



x = Ax+Bwz = Cx+Dww = −φ(z)

(11.41)

wherez′φ(z) ≥ 0 (11.42)


for all z. In order for the above system to be well-defined, one must have a unique solutionto the equation

z +Dφ(z) = Cx

for each x.

Theorem 21 Consider a system described by (11.41) and satisfying (11.42). Suppose thereexist a positive-definite symmetric matrix P and a positive scalar α which satisfy

(

PA+ A′P + 2αP PB − C ′

B′P − C −(D +D′)

)

≤ 0 (11.43)

Then system (11.41) is globally exponentially stable about the origin with rate α and Lyapunovmatrix P .

Proof.

2x′P x = x′(PA+ A′P )x+ 2x′PBw =

(

xw

)′(PA+ A′P PB

B′P 0

)(

xw

)

We also have

2z′φ(z) = 2(Cx+Dw)′w =

(

xw

)′(0 C ′

C D +D′

)(

xw

)

It now follows from (11.43) that

x′(PA+ A′P + 2αP )x+ 2x′PBw − 2(Cx+Dw)′w ≤ 0 ,

that is,

2xTP x+ 2αxTPx+ z′φ(z) ≤ 0 .

Hence

2x′P x ≤ −2αx′Px− z′φ(z)

≤ −2αx′Px .

• Note that (11.43) is linear in P .

If D +D′ > 0, then (11.43) is equivalent to

PA+ A′P + 2αP + (PB − C ′)(D +D′)−1(B′P − C) ≤ 0


11.4.2 Strictly positive real transfer functions and a frequency

domain condition

We consider here square transfer functions G given by

G(s) = C(sI − A)−1B +D (11.44)

where A,B,C,D are real matrices of dimensions n× n, n ×m,m× n and m ×m. We willsay the G is stable if all its poles have negative real parts. Clearly, if A is Hurwitz (all itseigenvalues have negative real parts) then G is stable.

Definition 1 (SPR) A square transfer function G is strictly positive real (SPR) if thereexists a real scalar ǫ > 0 such that G has no poles in a region containing the set {s ∈ C :ℜ(s) ≥ −ǫ} and

G(ω−ǫ) + G(ω−ǫ)′ ≥ 0 for all ω ∈ IR . (11.45)

We say that G is regular if det[G(ω)+G(ω)′] is not identically zero for all ω ∈ IR.

Requirement (11.45) is not very satisfactory; it involves an apriori unknown parameterǫ. How do we get rid of ǫ?

Lemma 12 [5] A transfer function G is SPR and regular if and only if the following con-ditions hold.

(a) [Stability] G is stable.

(b) [Dissipativity]G(ω) + G(ω)′ > 0 for all ω ∈ IR (11.46)

(c) [Asymptotic side condition]

lim|ω|→∞

ω2ρ det[G(ω) + G(ω)′] 6= 0 (11.47)

where ρ is the nullity of D +D′.

In either case, the above limit is positive.

Note that D = G(∞) := lim|ω|→∞ G(ω) and the nullity of D + D′ is the dimension ofthe null space of D +D′.

Remark 5 Suppose G(s) = C(sI − A)−1B +D is a scalar valued transfer function. Thenρ = 0 if D 6= 0 and ρ = 1 is D = 0. Hence, the above side condition reduces to

D 6= 0 or lim|ω|→∞

ω2[G(ω) + G(ω)′] 6= 0

When D = 0,lim

|ω|→∞ω2[G(ω) + G(ω)′] = −CAB


Example 124 Consider the scalar transfer function

g(s) =s+ 1

s2 + bs + 1.

We claim that this transfer function is SPR if and only if b > 1. This is illustrated in Figures11.8 and 11.9 for b = 2 and b = 0.5 respectively.

−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−0.5

−0.4

−0.3

−0.2

−0.1

0

0.1

0.2

0.3

0.4

0.5

Nyquist Diagram

Real Axis

Imag

inar

y A

xis

Figure 11.8: Nyquist plot of s+1s2+2s+1

−1 −0.5 0 0.5 1 1.5 2 2.5 3−2.5

−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

2.5

Nyquist Diagram

Real Axis

Imag

inar

y A

xis

Figure 11.9: Nyquist plot of s+1s2+0.5s+1

To prove the above claim, we first note that g is stable if and only if b > 0. Now notethat

g(ω) =1 + ω

1− ω2 + bω

Hence

g(ω) + g(ω)′ =2[1 + (b− 1)ω2]

(1− ω2)2 + (bω)2

It should be clear from the above expression that g(ω) + g(ω)′ > 0 for all finite ω if andonly if b ≥ 1. Also,

limω→∞

ω2[g(ω) + g(ω)′] = b− 1 .

Thus, in order for the above limit to be positive, it is necessary and sufficient that b > 1.


The next lemma tells us why SPR transfer functions are important in stability analysisof the systems under consideration in the last section.

Lemma 13 (KYPSPR) Suppose (A,B) is controllable and (C,A) is observable. Thenthere exists a matrix P = P ′ > 0 and scalar α > 0 such that LMI (11.43) holds if and onlyif the transfer function G(s) = D + C(sI − A)−1B is strictly positive real.

Corollary 1 Consider a system described by (11.41) and satisfying (11.42) and suppose that(A,B) is controllable, (C,A) is observable and the transfer function C(sI − A)−1B +D isSPR. Then this system is globally exponentially stable about the origin.

Remark 6 Sometimes the KYPSPR Lemma is stated as follows: Suppose (A,B) is control-lable and (C,A) is observable. Then G(s) = D+C(sI−A)−1B is strictly positive real if andonly if there exists a matrix P = P ′ > 0, a positive scalar ǫ and matrices W and L such that

PA+ A′P = −L′L− ǫP (11.48)

PB = C ′ − L′W (11.49)

W ′W = D +D′ (11.50)

To see that this result is equivalent to Lemma 13 above, let α = ǫ/2 and rewrite theabove equations as

PA+ A′P + 2αP = −L′L

PB − C ′ = −L′W

−(D +D′) = −W ′W ,

that is,

M :=

(

PA+ A′P + 2αP PB − C ′

B′P − C −(D +D′)

)

= −(

L′L L′WW ′L W ′W

)

= −(

LW

)′(LW

)

The existence of matrices L and W satisfying

M = −(

LW

)′(LW

)

is equivalent to M ≤ 0.

Exercise 38 Consider the transfer function

g(s) =βs+ 1

s2 + s+ 2

Using Lemma 12, determine the range of β for which this transfer function is SPR. Verifyyour results with the KYPSPR lemma.


Appendix: A Schur complement result

The following result is useful in manipulating matrix inequalities. Suppose

Q =

(

Q11 Q12

Q21 Q22

)

is a hermitian 2× 2 block matrix. Then

Q > 0

if and only if

Q22 > 0 and Q11 −Q12Q−122 Q21 > 0

To see this, note that Q′22 = Q22, Q

′21 = Q12 and

(

Q11 Q12

Q21 Q22

)

=

(

I 0Q−1

22 Q21 I

)′(Q11 −Q12Q

−122 Q21 0

0 Q22

)(

I 0Q−1

22 Q21 I

)

(11.51)

Note that Q11 and Q22 do not have to have the same dimensions.

Clearly, we can readily obtain the following result:

Q < 0

if and only ifQ22 < 0 and Q11 −Q12Q

−122 Q21 < 0

The following result also follows from (11.51). Suppose Q22 > 0. Then

Q ≥ 0

if and only ifQ11 −Q12Q

−122 Q21 ≥ 0

Bibliography

[1] Boyd, S. and El Ghaoui, L. and Feron, E. and Balakrishnan, V., Linear Matrix Inequal-ities in System and Control Theory, SIAM, Philadelphia, 1994.

[2] Gahinet, P., Nemirovski, A., Laub, A.J., and Chilali, M., LMI Contol Toolbox UsersGuide, The MathWorks Inc., Natick, Massachussets, 1995.

[3] Corless, M., “Robust Stability Analysis and Controller Design With Quadratic Lya-punov Functions,” in Variable Structure and Lyapunov Control, A. Zinober, ed.,Springer-Verlag, 1993.

[4] Acıkmese, A.B., and Corless, M., “Stability Analysis with Quadratic Lyapunov Func-tions: Some Necessary and Sufficient Multiplier Conditions,“ Systems and Control Let-ters, Vol. 57, No. 1, pp. 78-94, January 2008.

[5] Corless, M. and Shorten, R., “A correct characterization of strict positive realness”,submitted for publication.

[6] A. Rantzer, “On the Kalman-Yakubovich-Popov lemma,” Systems & Control Letters,vol. 28, pp. 7-10, 1996.

[7] Shorten, R., Corless, M., Wulff, K., Klinge, S. and R. Middleton, R., “Quadratic Sta-bility and Singular SISO Switching Systems,” submitted for publication.

[8] Ran, A.C.M., R. Vreugdenhil, R. 1988, Existence and comparison theorems for algebraicRiccati equations for continuous- and discrete-time systems, Linear Alg. Appl. Vol. 99,pp. 63–83, 1988

141

142 BIBLIOGRAPHY

Chapter 12

Invariance results

Recall the simple damped oscillator described by

x1 = x2

x2 = −(k/m)x1 − (c/m)x2

with m, c, k positive. This system is globally asymptotically stable about the origin. If weconsider the total mechanical energy, V (x) = 1

2kx21 +

12mx22, of this system as a candidate

Lyapunov function, then along any solution we obtain

V (x) = −cx22 ≤ 0 .

Using our Lyapunov results so far, this will only guarantee stability; it will not guaranteeasymptotic stability, because, regardless of the value of x1, we have V (x) = 0 wheneverx2 = 0. However, one can readily show that there are no non-zero solutions for which x2(t)is identically zero. Is this sufficient to guarantee asymptotic stability? We shall shortly seethat it is sufficient; hence we can use the energy of this system to prove asymptotic stability.

Consider a system described byx = f(x) (12.1)

By a solution of system (12.1) we mean a continuous function x(·) : [0,∞) → IRn whichidentically satisfies x(t) = f(x(t)).

12.1 Invariant sets

A subset M of the state space IRn is an invariant set for system (12.1) if it has the propertythat every solution which originates in M remains in M for all future time, that is, for everysolution x(·),

x(0) ∈ M implies x(t) ∈ M for all t ≥ 0

The simplest example of an invariant set is a set consisting of a single equilibrium point. Asanother example of an invariant set, consider any linear system x = Ax and let v be any realeigenvector of the matrix A. Then the line consisting of all vectors of the form cv where c isany real number is an invariant set for the linear system.

143

144 CHAPTER 12. INVARIANCE RESULTS

Figure 12.1: Invariant sets

Lemma 14 SupposeDV (x)f(x) ≤ 0 when V (x) < c

ThenM = {x : V (x) < c}

is an invariant set.

Figure 12.2: Invariant set lemma

Example 125

x1 = −x1 + x31

Considering V (x) = x2 and c = 1, one can readily show that the interval (−1, 1) is aninvariant set.

Largest invariant set. It should be clear that the union and intersection of two invariantsets is also invariant. Hence given any subset S of IRn, we can talk about the largest invariantset M contained in S, that is M is an invariant set and contains every other invariant setwhich is contained in S. As an example, consider the Duffing system

x1 = x2

x2 = x1 − x31

and let S correspond the the set of all points of the form (x1, 0) where x1 is arbitrary. If asolution starts at one of the equilibrium states

(−1, 0) (0, 0) (1, 0)

12.2. LIMIT SETS 145

it remains there. Hence, the set

{(−1, 0), (0, 0), (1, 0)}

is an invariant set for the system under consideration. We claim that this is the largestinvariant set in S. To see this, let M be the largest invariant set in S and consider anysolution (x1(·), x2(·)) which lies in M. Then this solution lies in S from which it follows thatx2(t) is zero for all t. This implies that x2(t) is zero for all t. From the system description,we must have x1(t)− x31(t) = 0. Hence x1(t) must be −1, 0 , or 1.

Figure 12.3: Largest invariant set

12.2 Limit sets

A state x∗ is called a positive limit point of a solution x(·) if there exists a sequence {tk}∞k=0

of times such that:

limk→∞

tk = ∞limk→∞

x(tk) = x∗

The set of all positive limit points of a solution is called the positive limit set of the solution.

The solution x(t) = e−t has only zero as a positive limit point, whereas x(t) = cos t hasthe interval [−1, 1] as its positive limit set. What about x(t) = e−t cos t? The positive limitset of x(t) = et is empty. Note that the positive limit set of a periodic solution consists ofall the points of the solution.

Suppose a solution converges to a single state x∗; then the positive limit set of this solutionis the set consisting of that single state, that is, {x∗}. For another example, consider theVan der Pol oscillator. Except for the zero solution, the positive limit set of every solutionis the set consisting of all the states in the limit cycle.

Exercise 39 What are the positive limit sets of the following solutions?

(a) x(t) = sin(t2)

(b) x(t) = et sin(t)


• Distance of a point x from a set M:

d(x,M) := inf{||x− y|| : y ∈ M}

Figure 12.4: Distance of a point from a set

Convergence to a set: A solution x(·) converges to a set M if

limt→∞

d(x(t),M) = 0

Figure 12.5: Convergence to a set

Lemma 15 (Positive limit set of a bounded solution.) The positive limit set L+ of abounded solution to (12.1) is nonempty and has the following properties.

1. It is closed and bounded.

2. It is an invariant set.

3. The solution converges to L+.

Proof. See Khalil.

12.2. LIMIT SETS 147

Example 126 (Van der Pol oscillator)


12.3 LaSalle’s Theorem

In the following results, V is a scalar valued function of the state, that is, V : IRn → IR, andis continuously differentiable.

Lemma 16 (A convergence result for bounded solutions.) Suppose x(·) is a boundedsolution of (12.1) and there exists a function V such that

DV (x(t))f(x(t)) ≤ 0

for all t ≥ 0. Then x(·) converges to the largest invariant set M contained in the set

S := {x ∈ IRn | DV (x)f(x) = 0}

Proof. Since x(·) is a bounded solution of (12.1), it follows from the previous lemma thatit has a nonempty positive limit set L+. Also, L+ is an invariant set for (12.1) and x(·)converges to this set.

We claim that V is constant on L+. To see this, we first note that

d

dt(V (x(t)) = DV (x(t))f(x(t) ≤ 0 ,

hence, V (x(t)) does not increase with t. Since x(·) is bounded and V is continuous, it followsthat V (x(·)) is bounded below. It now follows that there is a constant c such that

limt→∞

V (x(t)) = c .

Consider now any member x∗ of L+. Since L+ is the positive limit set of x(·), there is asequence {tk}∞k=0 such that

limk→∞

tk = ∞ and limk→∞

x(tk) = x∗ .

Since V is continuous,V (x∗) = lim

k→∞V (x(tk)) = c .

We have now shown that V (x∗) = c for all x∗ in L+. Thus V is constant on L+.Since L+ is an invariant set for (12.1) and V is constant on L+, it follows thatDV (x∗)f(x∗)

is zero for all x∗ in L+. Hence L+ is contained in S and, since L+ is an invariant set, it mustbe contained in the largest invariant set M in S.

Finally, since x(·) converges to L+ and L+ is contained in M, the solution x(·) alsoconverges to M.

The above result tells us that the solution converges to largest invariant set which iscontained in the set in which V is zero. Thus, the solution also converges to the set in whichV is zero.

Remark 7 Note that the above result does not require that V have any special propertiessuch as positive definiteness or radial unboundedness.

12.3. LASALLE’S THEOREM 149

Example 127 Consider the damped simple pendulum described by

x1 = x2

x2 = − sin x1 − x2 .

Suppose V is the total mechanical energy of this system, that is,

V (x) = 1− cosx1 +1

2x22 .

Then V (x) = −x22 ≤ 0. Since V (x) = 0 is equivalent to x2 = 0, it follows that every boundedsolution of this system converges to the largest invariant set M in

S = {(x1, 0) : x1 is an arbitrary real number } .

Consider any solution x(·) contained inM. Since this solution is in S, we must have x2(t) = 0for all t. Hence x2(t) is zero for all t. It now follows from the system description that

sin x1(t) = 0 for all t .

Thus, x1(t) must belong to the set E of equilibrium states, that is,

E = {(nπ, 0) : n is an integer } .

Thus M is contained in E . Since E is invariant, E = M. Thus all bounded solutions convergeto E .

Theorem 22 (LaSalle’s Theorem.) Suppose there is a radially unbounded function Vsuch that

DV (x)f(x) ≤ 0

for all x. Then all solutions of (12.1) are bounded and converge to the largest invariant setM contained in the set

S := {x ∈ IRn | DV (x)f(x) = 0}

Proof. We have already seen that the hypothesis of the theorem guarantee that allsolutions are bounded. The result now follows from Lemma 16.

Example 128 The damped mass.

x1 = x2

x2 = −x2

Consider

V (x) =1

2x21 + x1x2 + x22


Figure 12.6: The damped mass

Since V is a positive definite quadratic form, it is radially unbounded. Also,

V (x) = −x22 ≤ 0

Hence all solutions approach the set

M = S = {[

x1 0]T

: x1 ∈ IR}

that is, the x1 axis.

Figure 12.7: Phase portrait of the damped mass

Example 129 (Damped Duffing oscillator.)

x1 = x2

x2 = x1 − x31 − cx2

with c > 0.

Exercise 40 Using LaSalle’s Theorem, show that all solutions of the system

x1 = x22x2 = −x1x2

must approach the x1 axis.

12.4. ASYMPTOTIC STABILITY AND LASALLE 151

Exercise 41 Consider the scalar nonlinear mechanical system

q + c(q) + k(q) = 0

If the term −c(q) is due to damping forces it is reasonable to assume that c(0) = 0 and

c(q)q > 0 for all q 6= 0

Suppose the term −k(q) is due to conservative forces and define the potential energy by

P (q) =

∫ q

0

k(η) dη

Show that if limq→∞ P (q) = ∞, then all motions of this system must approach one of itsequilibrium positions.

12.4 Asymptotic stability and LaSalle


Using LaSalle’s Theorem, we can readily obtain the following result on global asymptoticstability. This result does not require V to be negative for all nonzero states.

Theorem 23 (Global asymptotic stability) Suppose there is a positive definite functionV with

DV (x)f(x) ≤ 0

for all x and the only solution for which

DV (x(t))f(x(t)) ≡ 0

is the zero solution. Then the origin is a globally asymptotically stable equilibrium state.

Proof.

Example 130 A damped nonlinear oscillator

x1 = x2

x2 = −x31 − cx2

with c > 0.

Example 131 A nonlinearly damped oscillator

x1 = x2

x2 = −x1 − cx32

with c > 0.


Exercise 42 Consider a nonlinear mechanical system described by

mq + cq + k(q) = 0

where q is scalar, m, c > 0 and k is a continuous function which satisfies

k(0) = 0

k(q)q > 0 for all q 6= 0

limq→∞

∫ q

0

k(η) dη = ∞

(a) Obtain a state space description of this system.

(b) Prove that the state space model is GUAS about the state corresponding to the equi-librium position q = 0.

(i) Use a La Salle type result.

(ii) Do not use a La Salle type result.

Figure 12.8: Single link manipulator

Exercise 43 Consider the inverted pendulum (or one link manipulator) as illustrated whereu is a control torque. This system can be described by

q − a sin q = bu

with a = mgl/I, b = 1/I, where m is the mass of B, I is the moment of inertia of B aboutits axis of rotation through O, l is the distance between O and the mass center of B, and gis the gravitational acceleration constant of YFHB. We wish to stabilize this system aboutthe position corresponding to q = 0 by a linear feedback controller of the form

u = −kpq − kdq

Using the results of the last problem, obtain the least restrictive conditions on the con-troller gains kp, kd which assure that the closed loop system is GUAS about the state corre-sponding to q(t) ≡ 0. Illustrate your results with numerical simulations.

12.4. ASYMPTOTIC STABILITY AND LASALLE 153


Theorem 24 (Asymptotic stability.) Suppose there is a locally positive definite functionV and a scalar R > 0 such that

DV (x)f(x) ≤ 0 for ||x|| ≤ R

and the only solution for which

DV (x(t))f(x(t)) ≡ 0 and ||x(t)|| ≤ R

is the zero solution. Then the origin is an asymptotically stable equilibrium state.

Example 132 Damped simple pendulum.

x1 = x2

x2 = −mglI

sin x1 −c

Ix2

with c > 0.


12.5 LTI systems

x = Ax


(a) The system x = Ax is asymptotically stable.

(b) There exist a positive definite hermitian matrix P and a matrix C with (C, A) observ-able which satisfy the Lyapunov equation

PA+ A∗P + C∗C = 0 (12.2)

(c) For every matrix C with (C, A) observable, the Lyapunov equation (12.2) has a uniquesolution for P and this solution is hermitian positive-definite.

12.6. APPLICATIONS 155

12.6 Applications

12.6.1 A simple stabilizing adaptive controller

Consider a simple scalar system described by

x = ax+ bu (12.3)

where the state x and the control input u are scalars. The scalars a and b are unknown butconstant. The only information we have on these parameters is that

b > 0 .

We wish to obtain a controller which guarantees that, regardless of the values of a and b, allsolutions of the resulting closed loop system satisfy

limt→∞

x(t) = 0 .

To achieve the above goal, we propose the following adaptive controller

u = −kx (12.4a)˙k = γx2 (12.4b)

where γ is some positive scalar.We now show that the above controller results in the desired goal and in addition, the

adaptive parameter k is bounded. Let k any scalar for which

a := a− bk < 0 .

Letting δk = k − k, the closed loop system resulting from the proposed adaptive controllercan be described by

x = ax− bδkx (12.5a)

δk = γx2 (12.5b)

As a candidate Lyapunov function, consider,

V (x, δk) = x2 + (b/γ)(δk)2

Along any solution of the closed loop system (12.5), we have

V = 2xx+ 2(b/γ)δkδk

= 2ax2 − 2x2bδk + 2bδkx2

= 2ax2 ≤ 0 .

Since b and γ are positive, V is radially unbounded; also V ≤ 0. It now follows that x and δkare bounded. Also, all solutions approach the largest invariant set in the set correspondingto x = 0. This implies that x(t) approaches zero as t goes infinity.


Exercise 44 (a) Consider an uncertain system described by

x = −3x+ δ sin x+ u (12.6)

where t ∈ IR is “time”, x(t) ∈ IR is the state, and u(t) ∈ IR is the control input. Theuncertainty in the system is due to the unknown constant real scalar parameter δ. We wishto design a controller generating u(t) which assures that, for all initial conditions and for allδ ∈ IR, the closed-loop system satisfies

limt→∞

x(t) = 0 (12.7)

If δ were known, one could letu = −δ sin x

to obtain a GUAS closed-loop system and, hence, the desired behavior. Since δ is unknown,we propose the following parameter adaptive controller

u = −δ sin x (12.8a)˙δ = γx sin x (12.8b)

where γ > 0 is a constant.

(i) Prove that the closed-loop system (12.6),(12.8) has property (12.7).

(ii) Numerically simulate the open loop system (system (12.6) with u = 0) and the closedloop system ((12.6),(12.8)) for different initial conditions, and for different values of δand γ.

(b) Consider now an uncertain described by

x = f(x) + δg(x) + u (12.9)

where t, x, u, δ are as before and f and g are continuous functions with f satisfying

f(x) > 0 if x < 0f(x) < 0 if x > 0

Design a controller generating u which assures that, for all initial conditions and for all δ,the resulting closed-loop system has property (12.7). Prove that your design works.

12.6.2 PI controllers for nonlinear systems


x = f(x) + g(x)(u+ w) (12.10)

where the n-vector x(t) is the state, the m-vector u(t) is the control input while the constantm-vector w is an unknown constant disturbance input.

12.6. APPLICATIONS 157

We assume that there is a positive definite Lyapunov function V such that

DV (x)f(x) < 0 for all x 6= 0 .

This implies that the nominal uncontrolled undisturbed system

x = f(x)

is GAS about zero.We wish to design a controller for u which assures that, for every constant disturbance

input, one haslimt→∞

x(t) = 0 .

To this end, we propose the following PI-type controllers:

u(t) = −KP y(t)−KI

∫ t

0

y(τ) dτ where y = g(x)′DV (x)′ (12.11)

and the gain matrices KP and KI are arbitrary positive definite symmetric matrices. Weclaim that these controllers achieve the desired behavior.

To prove our claim, note that the closed loop system can be described by

x = f(x)− g(x)KPy + g(x)(w − xc)

xc = KIy

y = g(x)′DV (x)′

and consider the candidate Lyapunov function

W (x, xc) = V (x) +1

2(xc−w)′K−1

I (xc−w) .

Exercise 45 Consider the system of exercise 43 subject to an unknown constant disturbancetorque w, that is,

q − a sin q = b(u+ w) .

Design a controller which assures that, for any initial conditions, the angle q approachessome desired constant angle qe as time goes to infinity. Consider a controller structure ofthe form:

u = −kp(q − qe)− kdq − kI

∫

(q − qe) .

Prove your design works and illustrate your results with numerical simulations.

Chapter 13

Stability of nonautonomous systems

Here, we are interested in the stability properties of systems described by

x = f(t, x) (13.1)

where the state x(t) is an n-vector at each time t. By a solution of the above system we meanany continuous function x(·) : [t0, t1) → IRn (with t0 < t1) which satisfies x(t) = f(t, x(t))for t0 ≤ t < t1. We refer to t0 as the initial time associated with the solution.

We will mainly focus on stability about the origin. However, suppose one is interested instability about some nonzero solution x(·). Introducing the new state e(t) = x(t)− x(t), itsevolution is described by

e = f(t, e) (13.2)

with f(t, e) = f(t, x(t) + e)− ˙x(t). Since e(t) = 0 corresponds to x(t) = x(t), one can studythe stability of the original system (13.1) about x(·) by studying the stability of the errorsystem (13.2) about the origin.

Scalar linear systems. All solutions of

x = a(t)x

satisfy

x(t) = e∫ tt0

a(τ) dτx(t0) .

Considerx = −e−tx

Here a(t) = −e−t. Since a(t) < 0 for all t, one might expect that all solutions converge tozero. This does not happen. Since,

x(t) = e(e−t−e−t0)

we havelimt→∞

x(t) = e−e−t0x(t0) .

Hence, whenever x(t0) 6= 0, the solution does not converge to zero.

159

160 CHAPTER 13. STABILITY OF NONAUTONOMOUS SYSTEMS

13.1 Stability and boundedness

13.1.1 Boundedness of solutions

DEFN. Boundedness. A solution x(·) is bounded if there exists β ≥ 0 such that

||x(t)|| ≤ β for all t ≥ t0


DEFN. Global uniform boundedness (GUB). The solutions of (13.1) are globally uniformlybounded if for each initial state x0 there exists a scalar β(x0) such that for all t0

x(t0) = x0 =⇒ ||x(t)|| ≤ β(x0) for all t ≥ t0

Note that, in the above definition, the bound β(x0) is independent of the initial time t0.

13.1.2 Stability

Suppose the origin is is an equilibrium state of system (13.1), that is,

f(t, 0) = 0

for all t.

DEFN. Stability (S). The origin is stable if for each ǫ > 0 and each t0 there exists δ > 0 suchthat

||x(t0)|| < δ =⇒ ||x(t)|| < ǫ for all t ≥ t0

Figure 13.1: Stability

The origin is unstable if it is not stable.

13.1. STABILITY AND BOUNDEDNESS 161

Note that, in the above definition, the scalar δ may depend on the initial time t0. When δcan be chosen independent of t0, we say that the stability is uniform.

DEFN. Uniform stability (US). The origin is uniformly stable if for each ǫ > 0 there existsδ > 0 such that for all t0,

||x(t0)|| < δ =⇒ ||x(t)|| < ǫ for all t ≥ t0

Example 133 Stable but not uniformly stable. Consider the scalar linear time-varying sys-tem:

x = (6t sin t− 2t)x

We will analytically show that this system is stable about the origin. We will also show thatnot only is it not uniformly stable about zero, the solutions are not uniformly bounded. Ifone simulates this system with initial condition x(t0) = x0 for different initial times t0 whilekeeping x0 the same one will obtain a sequence of solutions whose peaks go to infinity as t0goes to infinity. This is illustrated in Figure 13.2.

Figure 13.2: Growing peaks

All solutions x(·) satisfy

x(t) = x(t0)e[α(t)−α(t0)] where α(t) = 6 sin t− 6t cos t− t2

Clearly, for each initial time t0, there is a bound β(t0) such that

e[α(t)−α(t0)] ≤ β(t0) for all t ≥ t0

Hence every solution x(·) satisfies |x(t)| ≤ β(t0)|x(t0)| for t ≥ t0 and we can demonstratestability of the origin by choosing δ = ǫ/β(t0) for any ǫ > 0.

To see that this system is not uniformly stable consider t0 = 2nπ for any nonnegativeinteger n. Then α(t0) = −12nπ − 4n2π2 and for t = (2n+ 1)π we have α(t) = 6(2n+ 1)π −(2n+ 1)2π2. Hence

α(t)− α(t0) = (6− π)(4n+ 1)π

So for any solution x(·) and any nonnegative integer n, we have

x(2nπ + π) = x(t0)e(6−π)(4n+1)π

So, regardless of how small a nonzero bound one places on x(t0), one cannot place a boundon x(t) which is independent of t0. Hence the solutions of this system are not uniformlybounded. Also, this system in not uniformly stable about the origin.



DEFN. Asymptotic stability (AS). The origin is asymptotically stable if

(a) It is stable.

(b) For each t0, there exists R(t0) > 0 such that

||x(t0)|| < R(t0) =⇒ limt→∞

x(t) = 0 (13.3)

Figure 13.3: Asymptotic stability

Note that, in the above definition, R may depend on t0.

DEFN. Uniform asymptotic stability (UAS). The origin is uniformly asymptotically stable if

(a) It is uniformly stable.

(b) There exists R > 0 such that the following holds for each initial state x0 with ||x0|| < R.

i) There exists β(x0) such that for all t0,

x(t0) = x0 =⇒ ||x(t)|| ≤ β(x0) for all t ≥ t0

ii) For each ǫ > 0, there exists T (ǫ, x0) ≥ 0 such that for all t0,

x(t0) = x0 =⇒ ||x(t)|| < ǫ for all t > t0 + T (ǫ, x0)


DEFN. Global asymptotic stability (GAS). The origin is globally asymptotically stable if

(a) It is stable.

(b) Every solution satisfies

limt→∞

x(t) = 0

13.1. STABILITY AND BOUNDEDNESS 163

DEFN. Global uniform asymptotic stability (GUAS). The origin is globally uniformly asymptot-ically stable if

(a) It is uniformly stable.

(b) The solutions are globally uniformly bounded.

(c) For each initial state x0 and each each ǫ > 0, there exists T (ǫ, x0) ≥ 0 such that for allt0,

x(t0) = x0 =⇒ ||x(t)|| < ǫ for all t > t0 + T (ǫ, x0)


DEFN. Uniform exponential stability (UES). The origin is uniformly exponentially stable withrate of convergence α > 0 if there exists R > 0 and β > 0 such that whenever ||x(t0)|| < Rone has

||x(t)|| ≤ β||x(t0)||e−α(t−t0) for all t ≥ t0

Note that exponential stability implies asymptotic stability, but, in general, the converse isnot true.

DEFN. Global uniform exponential stability (GUES). The origin is globally exponentially stablewith rate of convergence α > 0 if there exists β > 0 such that every solution satisfies

||x(t)|| ≤ β||x(t0)||e−α(t−t0) for all t ≥ t0

Chapter 14

Lyapunov theory for nonautonomoussystems

14.1 Introduction

Here, we are interested in the stability properties of the system,

x = f(t, x) (14.1)

where x(t) ∈ IRn and t ∈ IR. We will focus on stability about the origin. However, supposeone is interested in stability about some nonzero solution x(·). Introducing the new statee(t) = x(t)− x(t), its evolution is described by

e = f(t, e) (14.2)

with f(t, e) = f(t, x(t) + e)− ˙x(t). Since e(t) = 0 corresponds to x(t) = x(t), one can studythe stability of the original system (14.1) about x(·) by studying the stability of the errorsystem (14.2) about the origin.

In this chapter, V is a scalar-valued function of the time and state, that is, V : IR×IRn →IR. Suppose V is continuously differentiable. Then, at any time t, the derivative of V alonga solution x(·) of system (14.1) is given by

dV (t, x(t))

dt=

∂V

∂t(t, x(t)) +

∂V

∂x(t, x(t))x(t)

=∂V

∂t(t, x(t)) +

∂V

∂x(t, x(t))f(t, x(t))

where ∂V∂t

is the partial derivative of V with respect to t and ∂V∂x

is the partial derivative ofV with respect to x and is given by

∂V

∂x=

[

∂V

∂x1

∂V

∂x2. . .

∂V

∂xn

]

Note that∂V

∂t+∂V

∂xf =

∂V

∂t+∂V

∂x1f1 +

∂V

∂x2f2 + . . .+

∂V

∂xnfn

165

166 CHAPTER 14. LYAPUNOV THEORY FOR NONAUTONOMOUS SYSTEMS

In what follows, if a condition involves ∂V∂t

or ∂V∂x, then it is implicitly assumed that V is

continuously differentiable.

14.2. STABILITY 167

14.2 Stability

14.2.1 Locally positive definite functions

A function V is said to be locally positive definite (lpd) if there is a locally positive definitefunction V1 : IRn → IR and a scalar R > 0 such that

V1(x) ≤ V (t, x) for all t and for ||x|| < R

A function V is said to be locally decresent (ld) if there is a locally positive definite functionV2 : IR

n → IR and a scalar R > 0 such that

V (t, x) ≤ V2(x) for all t and for ||x|| < R

Example 134

(a)

V (t, x) = (2 + cos t)x2

lpd and ld

(b)

V (t, x) = e−t2x2

ld but not lpd

(c)

V (t, x) = (1 + t2)x2

lpd but not ld

14.2.2 A stability theorem

Theorem 26 (Uniform stability) Suppose there exists a function V with the followingproperties.

(a) V is locally positive definite and locally decresent.

(b) There is a scalar R > 0 such that

∂V

∂t(t, x) +

∂V

∂xf(t, x) ≤ 0 for all t and for ||x|| < R

Then the origin is a uniformly stable equilibrium state for x = f(t, x).

If V satisfies the hypotheses of the above theorem, then V is said to be a Lyapunovfunction which guarantees the stability of origin for system (14.1).

Example 135

x = a(t)x

where a(t) ≤ 0 for all t.


14.3 Uniform asymptotic stability

Theorem 27 (Uniform asymptotic stability) Suppose there exists a function V withthe following properties.

(a) V is locally positive definite and locally decresent.

(b) There is a continuous function W and a scalar R > 0 such that for all t and all nonzerox with ||x|| < R,

∂V

∂t(t, x) +

∂V

∂x(t, x)f(t, x) ≤ −W (x) < 0

Then the origin is an uniformly asymptotically stable equilibrium state for system (14.1).

Example 136 To see that V < 0 is not sufficient, consider

x = −e−tx

With V (x) = x2, we have V = −2e−tx2 < 0 for all nonzero x. However, with x(0) = x0 6= 0,we have

x(t) = e(e−t−1)x0

Hence limt→∞ x(t) = e−1x0 6= 0 and this system is not asymptotically stable about zero.

Example 137

x = (−2 + sin t)x+ x2

14.4. GLOBAL UNIFORM ASYMPTOTIC STABILITY 169

14.4 Global uniform asymptotic stability

14.4.1 Positive definite functions

A function V is said to be positive definite (pd) if there is a positive definite function V1 :IRn → IR such that

V1(x) ≤ V (t, x) for all t and x

A function V is said to be decresent (d) if there is a positive definite function V2 such that

V (t, x) ≤ V2(x) for all t and x

Example 138 Recall Example 134.

14.4.2 A result on global uniform asymptotic stability

Theorem 28 (Global uniform asymptotic stability) Suppose there exists a function Vwith the following properties.

(a) V is positive definite and decresent.

(b) There is a continuous function W such that for all t and all nonzero x,

∂V

∂t(t, x) +

∂V

∂xf(t, x) ≤ −W (x) < 0

Then the origin is a globally uniformly asymptotically stable equilibrium state for x = f(t, x).

Example 139

x1 = −x1 −1

4(sin t)x2

x2 = x1 − x2

As a candidate Lyapunov function, consider

V (t, x) = x21 + (1 +1

4sin t)x22

Since

x21 +3

4x22 ≤ V (t, x) ≤ x21 +

5

4x22

it follows that V is positive definite and decresent. Along any solution of the above system,we have

V =1

4cos t x22 + 2x1(−x1 −

1

4sin t x2) + (2 +

1

2sin t)x2(x1 − x2)

= −2x21 − (2 +1

2sin t− 1

4cos t)x22 + 2x1x2

≤ −W (x)

where

W (x) = 2x21 − 2x1x2 +5

4x22

Since W (x) is positive for all nonzero x, it follows from the previous theorem that the abovesystem is globally uniformly stable about the origin.


14.4.3 Proof of theorem 28

In what follows x(·) represents any solution and t0 is its initial time.

Uniform stability. Consider any ǫ > 0. We need to show that there exists δ > 0 suchthat whenever ||x(t0)|| < δ, one has ||x(t)|| < ǫ for all t ≥ t0.

Since V is positive definite, there exists a positive definite function V1 such that

V1(x) ≤ V (t, x)

for all t and x. Since V1 is positive definite, there exists c > 0 such that

V1(x) < c =⇒ ||x|| < ǫ

Since V is decresent, there exists a positive definite function V2 such that

V (t, x) ≤ V2(x)

for all t and x. Since V2 is continuous and V (0) = 0, there exists δ > 0 such that

||x|| < δ =⇒ V2(x) < c

Along any solution x(·), we have

dV (t, x(t))

dt≤ 0

for all t ≥ t0; hence

V1(x(t)) ≤ V (t, x(t)) ≤ V (t0, x(t0)) ≤ V2(x(t0)) for t ≥ t0.

that is,V1(x(t)) ≤ V2(x(t0)) t ≥ t0

It now follows that if ||x(t0)|| < δ, then for all t ≥ t0:

V1(x(t)) ≤ V2(x(t0)) < c

that is, V1(x(t)) < c; hence, ||x(t)|| < ǫ.

Global uniform boundedness. Consider any initial state x0. We need to show thatthere is a bound β such that whenever x(t0) = x0, one has ||x(t)|| ≤ β for all t ≥ t0.

Since V1 is a radially unbounded function, there exists β > 0 such that

V1(x) ≤ V2(x0) =⇒ ||x|| ≤ β

We have already seen that along any solution x(·) we have

V1(x(t)) ≤ V2(x(t0)) t ≥ t0

It now follows that if x(t0) = x0, then for all t ≥ t0:

V1(x(t)) ≤ V2(x0) ;

hence ||x(t)|| ≤ β.

14.4. GLOBAL UNIFORM ASYMPTOTIC STABILITY 171

Global uniform convergence. Consider any initial state x0 and any ǫ > 0. We needto show that there exists T ≥ 0 such that whenever x(t0) = x0, one has ||x(t)|| < ǫ for allt ≥ t0 + T .

Consider any solution with x(t0) = x0. From uniform stability, we know that there existsδ > 0 such that for any t1 ≥ t0,

||x(t1)|| < δ =⇒ ||x(t)|| < ǫ for t ≥ t1

We have shown that there exists β ≥ 0 such that

||x(t)|| ≤ β for t ≥ t0

We also havedV (t, x(t))

dt≤ −W (x(t))

for all t ≥ t0. Letγ := min{W (x) : δ ≤ ||x|| ≤ β}

Since W is continuous and W (x) > 0 for x 6= 0, the above minimum exists and γ > 0. Let

T := V2(x0)/γ

We now show by contradiction, there exists t1 with t0 ≤ t1 ≤ t2, t2 := t0 + T , such that||x(t1)|| < δ; from this it will follow that ||x(t)|| < ǫ for all t ≥ t1 and hence for t ≥ t0 + T .

So, suppose, on the contrary, that

||x(t)|| ≥ δ for t0 ≤ t ≤ t2

Then V1(x(t2)) > 0, and for t0 ≤ t ≤ t2,

δ ≤ ||x(t)|| ≤ β ;

hencedV (t, x(t))

dt≤ −γ

So,

V1(x(t2)) ≤ V (x(t2), t2)

= V (t0, x(t0)) +

∫ t2

t0

dV (t, x(t))

dtdt

≤ V2(x0)− γT

≤ 0

that is,V1(x(t2)) ≤ 0

This contradicts V1(x(t2)) > 0. Hence, there must exist t1 with t0 ≤ t1 ≤ t0 + T such that||x(t1)|| < δ.




Theorem 29 (Uniform exponential stability) Suppose there exists a function V andpositive scalars α, β1, β2, and R such that for all t and all x with ‖x‖ < R,

β1||x||2 ≤ V (t, x) ≤ β2||x||2

and

∂V

∂t(t, x) +

∂V

∂xf(t, x) ≤ −2αV (t, x)

Then, for system (14.1), the origin is a uniformly exponentially stable equilibrium state withrate of convergence α.

14.5.2 Global uniform exponential stability

Theorem 30 (Global uniform exponential stability) Suppose there exists a functionV and positive scalars α, β1, β2 such that for all t, x,

β1||x||2 ≤ V (t, x) ≤ β2||x||2

and

∂V

∂t(t, x) +

∂V

∂xf(t, x) ≤ −2αV (t, x)

Then, for system (14.1), the origin is a globally exponentially uniformly stable equilibriumstate with rate of convergence α.

Proof.

Example 140

x = −(2 + cos t)x− x3

14.6 Quadratic stability

From the above it should be clear that all the results in the section on quadratic stabilityalso hold for time-varying systems.

14.7. BOUNDEDNESS 173

14.7 Boundedness

A scalar valued function V of time and state is said to be uniformly radially unbounded ifthere are radially unbounded functions V1 and V2 of state only such that

V1(x) ≤ V (t, x) ≤ V2(x)

for all t and x.

Example 141

V (t, x) = (2 + sin t)x2 yes

V (t, x) = (1 + e−t2)x2 − x yes

V (t, x) = e−t2x2 noV (t, x) = (1 + t2)x2 no

Theorem 31 Suppose there exists a uniformly radially unbounded function V and a scalarR ≥ 0 such that that for all t and x with ||x|| ≥ R,

∂V

∂t(t, x) +

∂V

∂xf(t, x) ≤ 0

Then the solutions of x = f(t, x) are globally uniformly bounded.

Note that, in the above theorem, V does not have to be positive away from the origin.

Example 142 Consider the disturbed nonlinear system,

x = −x3 + w(t)

where w is a bounded disturbance input. We will show that the solutions of this system areGUB.

Let β be a bound on the magnitude of w, that is, |w(t)| ≤ β for all t. Consider V (x) = x2.Since V is (uniformly) radially unbounded and

V = −2x4 + 2xw(t)

≤ −2|x|4 + 2β|x|

the hypotheses of the above theorem are satisfied with R = β13 ; hence GUB.

Exercise 46 (Forced Duffing’s equation with damping.) Show that all solutions of the sys-tem

x1 = x2

x2 = x1 − x31 − cx2 + w(t) c > 0

|w(t)| ≤ β


are bounded.Hint: Consider

V (x) =1

2λc2x21 + λcx1x2 +

1

2x22 −

1

2x21 +

1

4x41

where 0 < λ < 1. Letting

P =1

2

[

λc2 λcλc 1

]

note that P > 0 and

V (x) = xTPx− 1

2x21 +

1

4x41

≥ xTPx− 1

4

14.8. ATTRACTIVE AND INVARIANT REGIONS 175

14.8 Attractive and invariant regions

Suppose E is a nonempty subset of the state space IRn. Recall that E is invariant for thesystem

x = f(t, x) (14.3)

if E has the property that, whenever a solution starts in E , it remains therein thereafter,that is, if x(t0) is in E , then x(t) is in E for all t ≥ t0.

We say that the set E is attractive for the above system if every solution x(·) of (14.3)converges to E , that is,

limt→∞

d(x(t), E) = 0 .

Theorem 32 Suppose there is a continuously differentiable function V , a continuous func-tion W and a scalar c such that the following hold.

1) V is radially unbounded.

2) Whenever V (x) > c, we have

DV (x)f(t, x) ≤ −W (x) < 0 (14.4)

for all t.

Then, the solutions of the system x = f(t, x) are globally uniformly bounded and the set

{x ∈ IRn : V (x) ≤ c}

is an invariant and attractive set.

Figure 14.1: An invariant and attractive set

Example 143

x = −x3 + cos t


14.9 Systems with disturbance inputs

Here we consider disturbed systems described by

x = F (t, x, w) (14.5)

where the m-vector w(t) is the disturbance input at time t. As a measure of the size of adisturbance w(·) we shall consider its peak norm:

‖w(·)‖∞ = supt≥0

‖w(t)‖ .

Figure 14.2: The peak norm of a signal

We wish to consider the following type of problem: Given the peak norm of the dis-turbance, obtain an ultimate bound on the norm of the state or some performance outputdefined by

z = H(t, x, w) . (14.6)

We first have the following result.

Theorem 33 Consider the disturbed system (14.5) and suppose there exists a continuouslydifferentiable function V , a scalar µ1 > 0 and a continuous function U such that the followinghold.

1) V is radially unbounded.

2) For all t,

DV (x)F (t, x, w) ≤ −U(x, w) < 0 whenever V (x) > µ1‖w‖2 (14.7)

Then, for every bounded disturbance w(·), the state x(·) is bounded and the set

{

x ∈ IRn : V (x) ≤ µ1‖w(·)‖2∞}

is invariance and attractive for system (14.5).

Example 144

x = −x− x3 + w

14.9. SYSTEMS WITH DISTURBANCE INPUTS 177

Considering V (x) = x2, we have

DV (x)x = −x2 − x4 + xw

≤ −|x|2 − |x|4 + |x||w|= −|x|

(

|x|+ |x|3 − |w|)

= −U(x, w)

where U(x, w) = |x| (|x|+ |x|3 − |w|). Clearly U(x, w) > 0 when |x| > |w|, that is V (x) >|w|2. Hence, taking µ1 = 1, it follows from the above theorem that the following interval isinvariant and attractive for this system:

[−ρ ρ] where ρ = ‖w(·)‖∞ .

Proof of Theorem 33 Consider the disturbed system (14.5) subject to a specific boundeddisturbance w(·). This disturbed system can be described by x = f(t, x) with f(t, x) =F (t, x, w(t)). Let

ρ = ‖w(·)‖∞ = supt≥0

‖w(t)‖ .

We shall apply Theorem 32 with c = µ1ρ2.

Consider any x which satisfies V (x) > c. Then, whenever ‖w‖ ≤ ρ, we have V (x) >µ1‖w‖2. If we define

W (x) = min {U(x, w) : ‖w‖ ≤ ρ}then, it follows from condition (14.7) that W (x) > 0 whenever V (x) > c.

Since ‖w(t)‖ ≤ ρ for all t ≥ 0, it follows that whenever V (x) > c, we have

DV (x)f(t, x) = DV (x)F (t, x, w(t))

≤ −U(x, w(t))≤ −W (x) < 0

It now follows from Theorem (32) that the state x(·) is bounded and the set

{

x ∈ IRn : V (x) ≤ µ1‖w(·)‖2∞}

is invariance and attractive.


Corollary 2 Consider the disturbed system (14.5) equipped with a performance output spec-ified in (14.6). Suppose the hypotheses of Theorem 33 are satisfied and there exists a scalarµ2 such that for all t, x and w,

‖H(t, x, w)‖2 ≤ V (x) + µ2‖w‖2 (14.8)

Then for every disturbance bounded disturbance w(·), the output z(·) is bounded and satisfies

lim supt→∞

‖z(t)‖ ≤ γ‖w(·)‖∞

where

γ =√µ1 + µ2 . (14.9)

If in addition, V (0) = 0 then

‖z(·)‖∞ ≤ γ‖w(·)‖∞

when x(0) = 0.

14.9.1 Linear systems with bounded disturbance inputs

Consider a disturbed linear system:

x = Ax+Bw

z = Cx+Dw

where all the eigenvalues of A have negative real part and w is a bounded input.

Suppose there exists a positive real scalar α such that

(

PA+ ATP + αP PBBTP −αµ1I

)

< 0 (14.10a)

(

CTC − P CTDDTC DTD − µ2I

)

≤ 0 (14.10b)

Then

lim supt→∞ ‖z(t)‖ ≤ γ‖w(·)‖∞

where

γ =√µ1 + µ2 . (14.11)

Also, when x(0) = 0, we have

‖z(·)‖∞ ≤ γ‖w(·)‖∞

Proof.

14.9. SYSTEMS WITH DISTURBANCE INPUTS 179

For each fixed α we have LMIs in P , µ1 and µ2. One can minimize µ1 + µ2 for each αand do a line search over α.

The above results can be generalized to certain classes of nonlinear systems like polytopicsystems. Above results can also be used for control design.


14.10 Regions of attraction

Suppose that the origin is an asymptotically stable equilibrium state of the system

x = f(t, x) (14.12)

where the state x(t) is an n-vector. We say that a non-empty subset A of the state spaceIRn is a region of attraction for the origin if every solution which originates in A converges tothe origin, that is,

x(t0) ∈ A implies limt→∞

x(t) = 0 .

Figure 14.3: Region of attraction

The following result can be useful in computing regions of attraction.

Theorem 34 Consider system (14.12) and suppose there is a continuously differentiablefunction V , a continuous function W and a scalar c with the following properties.

1) V is positive definite.

2) Whenever x 6= 0 and V (x) < c, we have

DV (x)f(t, x) ≤ −W (x) < 0

for all t.

Then, the origin is a uniformly asymptotically stable equilibrium state with

A = {x ∈ IRn : V (x) < c}

as a region of attraction. Also, A is invariant.

Figure 14.4:

Example 145

x = −x(1 − x2)

14.10. REGIONS OF ATTRACTION 181

Example 146

x1 = −x1(1− x21 − x22) + x2

x2 = −x1 − x2(1− x21 − x22)

Considering V (x) = x21 + x22, we have

V = −2V (1− V )

14.10.1 A special class of nonlinear systems

We consider here nonlinear systems described by

x = A(t, x)x (14.13)

where

A(t, x) = A0 + ψ(t, x)∆A and − a(||Cx||) ≤ ψ(t, x) ≤ b(||Cx||) (14.14)

while a, b are non-decreasing functions and C is a matrix. We have previously consideredthese systems where a and b are constants and obtained sufficient conditions for globaluniform exponential stability using quadratic Lyapunov functions.


x1 = x2

x2 = −x1 − (1− x21)x2

This system can be expressed as(

x1x2

)

=

[(

0 1−1 −1

)

+ x21

(

0 00 1

)](

x1x2

)

(14.15)

HereC =

[

1 0]

, a(µ) = 0 , b(µ) = µ2 .

Theorem 35 Suppose there is a positive definite symmetric P and a positive scalar µ suchthat

PA0 + AT0 P − a(µ)(P∆A+∆ATP ) < 0 (14.16a)

PA0 + AT0 P + b(µ)(P∆A+∆ATP ) < 0 (14.16b)

CTC − P ≤ 0 (14.16c)

Then the origin is a uniformly asymptotically stable equilibrium state with

A = {x ∈ IRn : xTPx < µ2}as a region of attraction. Also A is invariant. If, in addition,

P − βI ≤ 0 (14.17)

for positive number β, then the ball

{x ∈ IRn : ||x|| < β− 12µ}

is a region of attraction.


Proof. First note that the inequality, CTC − P ≤ 0, implies that

‖Cx‖2 ≤ xTPx

for all x. So, consider any x for which xTPx < µ2. Then

‖Cx‖ < µ .

Using the nondecreasing property of b, we now obtain that

ψ(t, x) ≤ b(‖Cx‖) ≤ b(µ) .

In a similar fashion, we also obtain that

ψ(t, x) ≥ −a(‖Cx‖) ≥ −a(µ) .

Hence, whenever xTPx < µ2, we have

−a(µ) ≤ ψ(t, x) ≤ b(µ) .

Considering V (x) = xTPx as a candidate Lyapunov function, one may use inequalities(14.16a)-(14.16b) to show that whenever xTPx < µ2 and x 6= 0, one has

DV (x)x ≤ −W (x) < 0 .

Hence the setA = {x ∈ IRn : xTPx < µ2}

is invariant and a region of attraction for the origin.Now note that the inequality, P − βI ≤ 0, implies that

xTPx ≤ β−1‖x‖2

for all x. So whenever, ‖x‖ < β− 12µ, we must have xTPx < µ2, that is x is in A. Hence the

set of states of norm less than β− 12µ is a region of attraction for the origin.

14.11. LINEAR TIME-VARYING SYSTEMS 183

14.11 Linear time-varying systems


x = A(t)x (14.18)

All solutions satisfy

x(t) = Φ(t, τ)x(τ)

where Φ is the state transition matrix.

Computation of the state transition matrix. Let X be the solution of

X(t) = A(t)X(t) (14.19a)

X(0) = I (14.19b)

where I is the n× n identity matrix. Then

Φ(t, τ) = X(t)X(τ)−1 (14.20)

Example 148 (Markus and Yamabe, 1960.) This is an example of an unstable secondorder system whose time-varying A matrix has constant eigenvalues with negative real parts.Consider

A(t) =

[

−1 + a cos2 t 1− a sin t cos t−1− a sin t cos t −1 + a sin2 t

]

with 1 < a < 2. Here,

Φ(t, 0) =

[

e(a−1)t cos t e−t sin t−e(a−1)t sin t e−t cos t

]

Since a > 1, the system corresponding to this A(t) matrix has unbounded solutions. However,for all t, the characteristic polynomial of A(t) is given by

p(s) = s2 + (2− a)s+ 2

Since 2− a > 0, the eigenvalues of A(t) have negative real parts.

14.11.1 Lyapunov functions

As a candidate Lyapunov function for a linear time-varying system consider a time-varyingquadratic form V given by:

V (t, x) = xTP (t)x

where P (·) is continuously differentiable and P (t) is symmetric for all t.To guarantee that V is positive definite and decrescent, we need to assume that there

exists positive real scalars c1 and c2 such that for all t,

0 < c1 ≤ λmin(P (t)) and λmax(P (t)) ≤ c2 (14.21)


This guarantees thatc1‖x||2 ≤ V (t, x) ≤ c2‖x||2 .

Considering the rate of change of V along a solution, we have

V (t, x) =∂V

∂t(t, x) +

∂V

∂x(t, x)x

= x′P (t)x+ 2x′P (t)A(t)x

= −x′Q(t)xwhere Q(t) is given by

P + PA+ A′P +Q = 0 (14.22)

To satisfy the requirements on V (t, x) for GUAS and GUES we need

0 < c3 ≤ λmin(Q(t)) (14.23)

for all t. This guarantees thatV (t, x) ≤ −c3‖x||2 .

HenceV (t, x) ≤ −2αV (t, x) where α = c3/2c2 .

So, if there are matrix functions P (·) and Q(·) which satisfy (14.21)-(14.23) , we can concludethat the LTV system (14.18) is GUES. The following theorem provides the converse result.

Theorem 36 Suppose A(·) is continuous and bounded and the corresponding linear time-varying system (14.18) is globally uniformly asymptotically stable. Then, for all t, the matrix

P (t) =

∫ ∞

t

Φ(τ, t)TΦ(τ, t) dτ (14.24)

is well defined, where Φ is the state transition matrix associated with (14.18), and

P + PA+ A′P + I = 0 . (14.25)

Also, there exists positive real scalars c1 and c2 such that for all t,

c1 ≤ λmin(P (t)) and λmax(P (t)) ≤ c2

Proof. See Khalil or Vidyasagar.A consequence of the above theorem is that GUES and GUAS are equivalent for systems

where A(·) is continuous and bounded.

Example 149 The following system has a time varying quadratic Lyapunov function butdoes not have a time-invariant one.

x = (−1 + 2 sin t)x

HereΦ(τ, t) = e

∫ τt(−1+2 sin s) ds = e(t−τ+2 cos t−2 cos τ) = e(t−τ)e2(cos t−cos τ)

Hence,

P (t) =

∫ ∞

t

e2(t−τ)e4(cos t−cos τ) dτ =

∫ ∞

0

e−2τe4(cos t−cos (t+τ )) dτ .


14.12 Linearization

x = f(t, x)

Suppose that x(·) is a specific solution.Let

δx = x− x

Linearization about x(·)δx = A(t)δx (14.26)

where

A(t) =∂f

∂x(t, x(t)) .

Chapter 15

The describing function method

Here we consider the problem of determining whether or not a nonlinear system has a limitcycle (periodic solution) and approximately finding this solution.

We consider systems which can be described by the negative feedback combination ofa SISO LTI (linear time invariant) system and a SISO memoryless nonlinear system; seeFigure 15.2. If the LTI system is represented by its transfer function G and the nonlinear

Figure 15.1: Systems under consideration

system is represented by the function φ, then the system can be described by

y(s) = G(s)u(s) (15.1a)

u(t) = −φ(y(t)) (15.1b)

where u and y are the input and output, respectively for the LTI system and their respectiveLaplace transforms are u and y.

As a general example, consider any system described by

x = Ax− Bφ(Cx) (15.2)

where x(t) is an n-vector. In this example

G(s) = C(sI −A)−1B

A solution y(·) is periodic with period T > 0 if

y(t+ T ) = y(t) for all t .

We are looking for periodic solutions to (15.1).

187

188 CHAPTER 15. THE DESCRIBING FUNCTION METHOD

15.1 Periodic functions and Fourier series

Consider any piecewise continuous signal s : IR → IR which is periodic with period T . Thens has a Fourier series expansion, that is, s can be expressed as

s(t) = a0 +∞∑

k=1

ak cos(kωt) +∞∑

k=1

bk sin(kωt)

where ω = 2π/T is called the natural frequency of s. The Fourier coefficients a0, a1, . . . andb1, b2, . . . are uniquely given by

a0 =1

T

∫ T2

−T2

s(t) dt

ak =2

T

∫ T2

−T2

s(t) cos (kωt) dt and bk =2

T

∫ T2

−T2

s(t) sin (kωt) dt for k = 1, 2, . . .

If s is odd, that is, s(−t) = −s(t) then,

ak = 0 for k = 0, 1, 2, . . .

and

bk =4

T

∫ T2

0

s(t) sin (kωt) dt for k = 1, 2, . . .

Example 150 Sometimes it is easier to compute Fourier coefficients without using the aboveintegrals. Consider

s(t) = sin3 t .

Clearly, this is an odd periodic signal with period T = 2π. Since

sin t =ejt − e−jt

2j

we have

sin3(t) =

(

ejt − e−jt

2j

)3

=e3jt − 3ejt + 3e−jt − e−3jt

−8j

=3(ejt − e−jt)

8j− e3jt − e−3jt

8j

=3

4sin t− 1

4sin 3t .

So,

b1 =3

4, b3 = −1

4and all other Fourier coefficients are zero.

Exercise 47 Obtain the Fourier series for s(t) = sin5 t

15.2. DESCRIBING FUNCTIONS 189

15.2 Describing functions

The describing function for a static nonlinearity is an approximate description of its frequencyresponse. Suppose φ : IR → IR is a piecewise continuous function and consider the signal

s(t) = φ(a sinωt)

for a, ω > 0. This signal is the output of the nonlinear system defined by φ and subject toinput a sinωt.

The signal s is piecewise continuous and periodic with period T = 2π/ω. Hence it has aFourier series expansion and its Fourier coefficients are given by

a0(a) =1

T

∫ T2

−T2

φ(a sinωt) dt =1

2π

∫ π

−π

φ(a sin θ) dθ (θ = ωt = 2πt/T )

and for k = 1, 2, . . .,

ak(a) =1

π

∫ π

−π

φ(a sin θ) cos kθ dθ and bk(a) =1

π

∫ π

−π

φ(a sin θ) sin kθ dθ

Real describing functions. Suppose that the function φ : IR → IR is piecewise continuousand odd, that is,

φ(−y) = −φ(y)If φ is odd, then s is odd; hence

ak(a) = 0 for k = 0, 1, . . .

and

bk(a) =2

π

∫ π

0

φ(a sin θ) sin kθ dθ .

If we approximate s by neglecting its higher harmonics, that is,

s(t) ≈ b1(a) sin(ωt)

thens(t) ≈ N(a)a sinωt

where N is called the describing function for φ and is given by

N(a) = b1(a)/a =2

πa

∫ π

0

φ(a sin θ) sin θ dθ

Basically, N(a) is an approximate system gain for the nonlinear system when subject to asinusoidal input of amplitude a. Notice that, unlike a dynamic system, this gain is indepen-dent of the frequency of the input signal. For a linear system, the gain is independent ofamplitude.


Example 151 Cubic nonlinearity. Consider

φ(y) = y3

Then, recalling the last example, we see that

φ(a sin θ) = a3 sin3 θ =3a3

4sin θ − a3

4sin 3θ

Hence b1(a) = 3a3/4 and

N(a) =b1(a)

a=

3a2

4.

Example 152 Signum function. Recall the signum function given by

φ(y) =

−1 if y < 00 if y = 01 if y > 0

Its describing function is given by

N(a) =2

πa

∫ π

0

φ(a sin θ) sin θ dθ =2

πa

∫ π

0

sin θ dθ =4

πa.

Example 153 Consider the following approximation to the signum switching function.

Figure 15.2: Approximation to signum function

φ(y) =

−1 if y < −e0 if −e ≤ y ≤ e1 if y > e

(15.3)

where e is some positive real number. This is an odd function. Clearly, N(a) = 0 if a ≤ e.If a > e, let θe be the unique number between 0 and π/2 satisfying θe = arcsin(e/a) that is,a sin(θe) = e. Then,

∫ π

0

φ(a sin θ) sin θ dθ =

∫ π−θe

θe

sin θ dθ = − cos(π − θe) + cos θe = 2 cos θe

15.2. DESCRIBING FUNCTIONS 191

Since sin θe = e/a and 0 ≤ θe < π/2, we have cos θe =√

1− (e/a)2. Thus

b1(a) =4√a2 − e2

πa

Hence,

N(a) =

0 if a ≤ e

4√a2 − e2

πa2if a > e

(15.4)

One can readily show that the maximum value of N(a) is 2/(πe) and this occurs at a =√2e.

Example 154 Consider a sector bounded nonlinearity satisfying

αy2 ≤ yφ(y) ≤ βy2

for some constants α and β. We will show that

α ≤ N(a) ≤ β

Complex describing functions. If the function φ is not odd then, its describing functionmay be complex valued; see [1] for an example. Assuming a0(a) = 0, in this case, we let

N(a) = b1(a)/a+ a1(a)/a (15.5)

where

a1(a) =1

π

∫ π

−π

φ(a sin θ) cos θ dθ and b1(a) =1

π

∫ π

−π

φ(a sin θ) sin θ dθ

Note thatN(a) = |N(a)|ejφ(a)

where

|N(a)| =√

a1(a)2 + b1(a)2

a

and φ(a) is the unique angle in [0, 2π) given by

cos(φ(a)) =b1(a)

√

a1(a)2 + b1(a)2sin(φ(a)) =

a1(a)√

a1(a)2 + b1(a)2


Here, the approximate output of the nonlinearity due to input a sin(ωt) is given by

s(t) ≈ a1(a) cos(ωt) + b1(a) sin(ωt)

= |N(a)| a [sin(φ(a)) cos(ωt) + cos(φ(a)) sin(ωt)]

= ≈ |N(a)| a sin(ωt+ φ(a))

that is,s(t) ≈ |N(a)|a sin(ωt+ φ(a)) (15.6)

where φ(a) is the argument of N(a). Notice the phase shift in s due to φ(a).

Exercise 48 Compute the describing function for φ(y) = y5.

Exercise 49 Consider the saturation function sat described by

sat(y) =

−1 if y < −1y if −1 ≤ y ≤ 11 if y > 1

Show that its describing function is given by

N(a) =

1 if 0 ≤ a ≤ 1

2π

[

sin−1( 1a) +

√a2−1a2

]

if 1 < a

15.3. THE DESCRIBING FUNCTION METHOD 193

15.3 The describing function method

Recall the nonlinear system described in (15.1) Suppose we are looking for a periodic solutionfor y which can be approximately described by

y(t) ≈ a sin(ωt)

where a > 0.

Real describing functions. Suppose φ is odd,

u(t) ≈ −φ(a sin(ωt)) = −∞∑

k=1

bk(a) sin(kωt)

where

bk(a) =2

π

∫ π

0

φ(a sin θ) sin(kθ) dθ

So,

u(t) ≈ −b1(a) sin(ωt) = −N(a)a sin(ωt)

where N is the describing function for φ. Since u(t) ≈ −N(a)a sin(ωt) and y(t) ≈ a sin(ωt)and y = Gu, it follows that

G(ω) ≈ −1/N(a) .

So, if for some pair a, ω > 0, the condition

1 + G(ω)N(a) = 0 (15.7)

is satisfied, it is likely that the nonlinear system will have a periodic solution with

y(t) ≈ a sinωt

When N(a) is real the above condition can be expressed

ℑ(G(ω)) = 0 (15.8a)

1 + ℜ(G(ω))N(a) = 0 (15.8b)

The first condition in (15.8) simply states that the imaginary part of G(ω) is zero, that is,G(ω) is real. This condition is independent of a and can be solved for values of ω. Thevalues of a are then determined by the second condition in (15.8).

Example 155

y + y3 = 0


With φ(y) := y3, this system is described by

y = u

u = −φ(y)

Hence

G(s) =1

s2

and

N(a) =3a2

4

Here condition (15.7) is

1− 3a2

4ω2= 0

Hence, we predict that this system has periodic solutions of all amplitudes a and withapproximate periods

T =2π

ω=

2π√3a/2

=4π√3a

Example 156 Consider the nonlinear system,

x1 = x2

x2 = −4x1 − 5x2 − sgm (x1 − x2)

This can be represented as the LTI system

x1 = x2

x2 = −4x1 − 5x2 + u

y = x1 − x2

subject to negative feedback from the memoryless nonlinearity u = −sgm (y). Hence

G(s) =−s+ 1

s2 + 5s+ 4

and N(a) = 4/πa. The condition 1 + G(ω)N(a) = 0 results in

1 +−ω + 1

−ω2 + 4 + 5ωN(a) = 0

that is−ω2 + 4 + 5ω + (−ω + 1)N(a) = 0

or

4− ω2 +N(a) = 0

5ω − ωN(a) = 0


0 2 4 6 8 10 12 14 16 18 20−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

Figure 15.3: Limit cycle for system in example 156

Solving yields ω = 3 and N(a) = 5. This results in

a =4

5π= 0.2546 , T =

2π

ω= 2.0944

The y-response of this system subject to initial conditions x1(0) = 0.5 and x2(0) = 0 isillustrated in Figure 15.3

Example 157 Consider the system of the previous example with the signum function re-placed with an approximation as described in Example 153, that is,

x1 = x2

x2 = −4x1 − 5x2 − φ(x1 − x2)

where φ is given by (15.3) for some e > 0. Proceeding as in the previous example, thedescribing function method still gives the following conditions for a limit cycle

ω = 3 and N(a) = 5 .

Recalling the expression for N(a) in 15.4, we must have a > e and

4√a2 − e2

πa2= 5

Solving this for a2 yields two solutions

a2 =8± 4

√

4− (5πe)2

(5π)2

For these solutions to be real, we must have 4− (5πe)2 ≥ 0 that is

e ≤ 2

5π= 0.1273 .


Complex describing functions. We still have the criterion (15.7) for a limit cycle, thatis,

1 + G(ω)N(a) = 0

However, we cannot usually simplify things as in (15.8).

Example 158 (Van der Pol oscillator) Here we demonstrate a generalization of the method.

y + µ(y2 − 1)y + y = 0

We can describe this system by

y − µy + y = µu

u = −y2y

Here

G(s) =µ

s2 − µs+ 1

If y(t) = a sinωt then

y2y = a3ω sin2 ωt cosωt ≈ a3ω

4cosωt ≈ a3ω

4sin(ωt+ π/2)

So here

N(a, ω) =a2ωeπ/2

4=a2ω

4.

Solving 1 + G(ω)N(a, ω) = 0 yields

ω = 1 a = 2 .


15.3.1 Rational transfer functions

When the describing function of the nonlinearity is real, the approximate frequencies ω atwhich a limit cycle occurs are a subset of those frequencies for which G(ω) is real, thatis, the imaginary part of G(ω) is zero. Here we consider strictly proper rational transferfunctions G and show how one can easily determine the frequencies ω for which G(ω) isreal. This involves a state space representation (A,B,C, 0) of G.

Suppose G(s) = C(sI −A)−1B is a scalar transfer function with real A,B,C and A hasno imaginary eigenvalues. Noting that

(ωI −A)−1 = (−ωI − A)(−ωI − A)−1(ωI − A)−1

= (−ωI − A)(ω2I + A2)−1

= −A(ω2I + A2)−1 − ω(ω2I + A2)−1 ,

we obtain thatG(ω) = −CA(ω2I + A2)−1B − ωC(ω2I + A2)−1B . (15.9)

The above expression splits G(ω) into its real and imaginary part, in particular the imagi-nary part of G(ω) is given by

ℑ(

G(ω))

= −ωC(ω2I + A2)−1B (15.10)

To determine the nonzero values (if any) of ω for which the imaginary part of G(ω) is zero,we note that

det

(

−ω2I − A2 BC 0

)

= det[

−ω2I − A2]

det[

C(ω2I + A2)−1B]

= det[

−ω2I − A2]

C(ω2I + A2)−1B .

Introducing the matrix pencil given by

P (λ) =

(

−A2 − λI BC 0

)

(15.11)

we have shown that, for ω 6= 0,

detP (ω2) = − det [− ω2I −A2]ℑ(

G(ω))

/ω

Since it is assumed that A does not have any purely imaginary eigenvalues, A2 does not haveany negative real eigenvalues. This implies that det(−ω2I−A2) is non-zero for all ω. Hence,for nonzero ω,

ℑ(

G(ω))

= 0 if and only if detP (ω2) = 0 (15.12)

Note that

P (λ) =

(

−A2 BC 0

)

− λ

(

I 00 0

)


The values of λ for which detP (λ) = 0 are the finite generalized eigenvalues of the matrixpencil P . You can use the Matlab command eig to compute generalized eigenvalues. Thuswe have the following conclusion:

G(ω) is real if and only if ω =√λ and λ is a positive, real, finite, generalized eigenvalue

of the matrix pencil P

Example 159 For example 156 we have

A =

(

0 1−4 −5

)

B =

(

01

)

C =(

1 −1)

Hence

P (λ) =

(

−A2 − λI BC 0

)

=

4− λ 5 0−20 −21− λ 11 −1 0

and detP (λ) = −λ + 9. Hence 9 is the only finite generalized eigenvalue of P and it ispositive real. Thus, there is a single nonzero value of ω for which G(ω) is real and this isgiven by ω = 3.

15.4 Exercises

Exercise 50 Obtain the describing function of the nonlinear function

φ(y) = y5

Exercise 51 Determine whether or not the following Duffing system has a periodic solution.Determine the approximate amplitude and period of all periodic solutions.

y − y + y3 = 0 .

Exercise 52 Determine whether or not the following system has a periodic solution. De-termine the approximate amplitude and period of all periodic solutions.

y + µ(y3/3− y) + y = 0 .

where µ is a positive real number.

Exercise 53 Use the describing function method to predict period solutions to

y + sgm (y) = 0

Illustrate your results with numerical simulations.

Exercise 54 Use the describing function method to predict period solutions to

x(t) = −x(t)− 2 sgm (x(t− h))

Illustrate your results with numerical simulations.

15.4. EXERCISES 199

Exercise 55 Consider the double integrator

q = u

subject to a saturating PID controller

u = sat(u) where u = −kP q − kI

∫

q − kDq

(a) For kP = 1 and kD = 2 determine the largest of kI ≥ 0 for which the closed loopsystem is stable about q(t) ≡ 0.

(b) For kP = 1 and kD = 2, use the describing function method to determine the smallestvalue kI ≥ 0 for which the closed loop system has a periodic solution.

Bibliography

[1] Boniolo, I., Bolzern, P., Colaneri,P., Corless, M., Shorten, R. “Limit Cycles in SwitchingCapacitor Systems: A Lure Approach”

201

202 BIBLIOGRAPHY

Chapter 16

Aerospace/Mechanical Systems

16.1 Aerospace/Mechanical systems

Let the real scalar t represent time. At each instant of time, the configuration of theaerospace/mechanical systems under consideration can be described by a real N -vector q(t).We call this the vector of generalized coordinates; each component is usually an angle, alength, or a displacement. It is assumed that there are no constraints, either holonomic ornon-holonomic, on q. So, N is the number of degrees-of-freedom of the system. We let thereal scalar T represent the kinetic energy of the system. It is usually given by

T =1

2qTM(q)q =

N∑

j=1

N∑

k=1

Mjk(q)qj qk

where the symmetric N × N matrix M(q) is called the system mass matrix. Usually, itsatisfies

M(q) > 0 for all q

The real N -vector Q is the sum of the generalized forces acting on the system. It includesconservative and non-conservative forces. It can depend on t, q, q.

Example 160 Two cart system.

Here

T =1

2m1q

21 +

1

2m2q

22

So,

M =

(

m1 00 m2

)

203

204 CHAPTER 16. AEROSPACE/MECHANICAL SYSTEMS

Example 161 Single link manipulator (inverted pendulum).

q = θ

T =1

2Jθ2

So, M = J . Also, Q = Wl sin θ.

Example 162 Particle undergoing central force motion.

Figure 16.1: Central force motion

Consider a particle P of mass m undergoing a central force motion. The force center isO which is fixed in inertial reference frame i. Here,

q = (r θ)T

and the kinetic energy is given by

T =1

2m(r2 + r2θ2)

Hence

M(q) =

(

m 00 mr2

)

Also, the generalized forces are given by

Q =

(

F0

)

16.2 Equations of motion

La Grange’s Equation

d

dt

(

∂T

∂qi

)

− ∂T

∂qi= Qi for i = 1, 2, · · · , N

In vector form:d

dt

(

∂T

∂q

′)

− ∂T

∂q

′= Q

16.2. EQUATIONS OF MOTION 205

Example 163 Single link manipulator (inverted pendulum). Application of La Grange’sequation yields:

Jθ −Wl sin θ = 0 (16.1)

Example 164 Particle undergoing central force motion. Application of La Grange’s equa-tion yields

mr − mrθ2 −F = 0

mr2θ + 2mrrθ = 0

The following result makes a statement about the structure of the differential equationsobtained from the application of La Grange’s equation.

Proposition 1 Application of La Grange’s equation yields

M(q)q + c(q, q)−Q = 0 (16.2)

where the i-th component of c is a quadratic function of q and is given by

ci(q, q) =

N∑

j=1

N∑

k=1

(

∂Mij

∂qk− 1

2

∂Mjk

∂qi

)

qj qk

Proof. Since

T =1

2qTM(q)q

it follows that∂T

∂q

′=M(q)q

Hence

d

dt

(

∂T

∂q

′)

=Mq + Mq

and application of La Grange’s equation yields

Mq + c−Q = 0

where

c = Mq − ∂T

∂q

′. (16.3)

Since

T =1

2

N∑

j=1

N∑

k=1

Mjk(q)qj qk ,

it follows that∂T

∂qi=

1

2

N∑

j=1

N∑

k=1

∂Mjk

∂qiqj qk (16.4)


Noting that

(M q)i =N∑

j=1

Mij qj =N∑

j=1

(

N∑

k=1

∂Mij

∂qkqk

)

qj =N∑

j=1

N∑

k=1

∂Mij

∂qkqj qk (16.5)

we have

ci = (M q)i −∂T

∂qi=

N∑

j=1

N∑

k=1

(

∂Mij

∂qk− 1

2

∂Mjk

∂qi

)

qj qk

16.3. SOME FUNDAMENTAL PROPERTIES 207

16.3 Some fundamental properties

In this section, we present some fundamental properties of aerospace/mechanical systemswhich are useful in the analysis and control design for these systems.

16.3.1 An energy result

First we have the following result.

Proposition 2

q′(

c− 1

2M q

)

= 0 (16.6)

Proof. It follows from (16.3) that

q′(

c− 1

2M q

)

= q′(

1

2M q − ∂T

∂q

′)

=1

2q′Mq − ∂T

∂qq (16.7)

Using (16.4),

∂T

∂qq =

N∑

i=1

∂T

∂qiqi =

N∑

i=1

(

1

2

N∑

j=1

N∑

k=1

∂Mjk

∂qiqj qk

)

qi =1

2

N∑

i=1

N∑

j=1

N∑

k=1

∂Mjk

∂qiqiqj qk . (16.8)

Using (16.5),

1

2q′M q =

1

2

N∑

i=1

qi(M q)i =1

2

N∑

i=1

(

N∑

j=1

N∑

k=1

∂Mij

∂qkqj qk

)

qi =1

2

N∑

i=1

N∑

j=1

N∑

k=1

∂Mij

∂qkqiqj qk

By interchangining the indices i and k and using the symmetry of M , we obtain that

1

2q′M q =

1

2

N∑

i=1

N∑

j=1

N∑

k=1

∂Mjk

∂qiqiqj qk . (16.9)

Substituting (16.8) and (16.9) into (16.7) yields

q′(

c− 1

2Mq

)

= 0


Now we have the main result of this section.

Theorem 37 The time rate of change of the kinetic energy equals the power of the general-ized forces, that is,

dT

dt= q′Q (16.10)

Proof. Since the kinetic energy is given by

T =1

2q′Mq

application of (16.2) and (16.6) yields:

dT

dt= q′Mq +

1

2q′M q

= q′(−c +Q) +1

2q′M q = q′Q− q′

(

c− 1

2Mq

)

= q′Q .

16.3.2 Potential energy and total energy

A generalized force is said to be conservative if it satisfies the following two properties:

(a) It depends only on q.

(b) It can be expressed as the negative of the gradient of some scalar valued function ofq. This function is called the potential energy associated with the force.


Q =Wl sin θ and q = θ

Here Q = −∂U∂q

′where

U = Wl(cos θ − 1) .


Q =

(

F0

)

q =

(

rθ

)

F = −µmr2

Here Q = −∂U∂q

′where

U = −µmr.


Suppose one splits the generalized force vector Q into a conservative piece −K and a‘non-conservative’ piece −D, thus

Q = −K(q)−D(q, q) (16.11)

where −K is the sum of the conservative forces and −D is the sum of the ‘non-conservative’forces. Let U be the total potential energy associated with the conservative forces, then

K =∂U

∂q

′. (16.12)

Introducing the LagrangianL = T − U (16.13)

Lagrange’s equation can now be written as

d

dt

(

∂L

∂q

′)

− ∂L

∂q

′+D = 0 .

In particular, if all the forces are conservative, we have

d

dt

(

∂L

∂q

′)

− ∂L

∂q

′= 0 .

Introducing the total mechanical energy

E = T + U (16.14)

we obtain thatdE

dt= −q′D (16.15)

that is, the time rate of change of the total energy equals the power of the non-conservativeforces. To see this:

dE

dt=

dT

dt+dU

dt

= q′Q+∂U

∂qq

= −q′∂U′

∂q− q′D(q, q) +

∂U

∂qq

= −q′D(q, q) .

In particular, if all the forces are conservative, we have

dE

dt= 0 (16.16)

thus E is constant, that is, energy is conserved.


16.3.3 A skew symmetry property

Proposition 3 For each q, q ∈ IRN , there is a skew-symmetric matrix S(q, q) such that

c =1

2

(

M − S)

q (16.17)

Proof. Using (16.3),

c =1

2Mq +

1

2M q − ∂T

∂q

′

From (16.5),

(M q)i =

N∑

j=1

N∑

k=1

∂Mij

∂qkqj qk =

N∑

j=1

N∑

k=1

∂Mik

∂qjqj qk

Thus, using (16.4),

(

1

2M q − ∂T

∂q

′)

i

=1

2(M q)i −

∂T

∂qi

′=

1

2

N∑

j=1

N∑

k=1

(

∂Mik

∂qj− ∂Mjk

∂qi

)

qj qk

= −N∑

j=1

1

2Sij qj

where

Sij :=

N∑

k=1

(

∂Mjk

∂qi− ∂Mik

∂qj

)

qk (16.18)

Hence Sji = −Sij ,

1

2M q − ∂T ′

∂q= −1

2Sq

and

c =1

2

(

M − S)

q

with S skew-symmetric.

Corollary 3 For each q, q ∈ IRN , there is a matrix C(q, q) such that

c = Cq (16.19)

and M − 2C is skew-symmetric.

Proof. Let

C =1

2(M − S) (16.20)

where S is skew-symmetric as given by previous result. Then c = Cq and

M − 2C = S


Remark 8 Substituting (16.19) into (16.2) yields

M(q)q + C(q, q)q −Q = 0, (16.21)

and using (16.20) to (16.18) we obtain

Cij =1

2

N∑

k=1

(

∂Mij

∂qk+∂Mik

∂qj− ∂Mjk

∂qi

)

qk . (16.22)


16.4 Linearization of aerospace/mechanical systems

Here we consider aerospace/mechanical systems described by (16.2), that is,

M(q)q + c(q, q)−Q(q, q) = 0 (16.23)

The equilibrium configurations q(t) ≡ qe of the above system are given by

Q(qe, 0) = 0

Linearization about q(t) ≡ qe results in

Mδq + Dδq + Kδq = 0 (16.24)

where

M =M(qe), D = −∂Q∂q

(qe, 0), K = −∂Q∂q

(qe, 0)

Note that the c term does not contribute anything to the linearization; the contribution ofM(q)q is simply M(qe)δq.

We have now the following stability result.

Theorem 38 Consider the linearization (16.24) of system (16.23) about q(t) ≡ qe andsuppose that M and K are symmetric and positive definite. Suppose also that for all q,

δq′Dδq ≥ 0

andδq(t)′Dδq(t) ≡ 0 ⇒ δq(t) ≡ 0

then the nonlinear system (16.23) is asymptotically stable about the solution q(t) ≡ qe.

Proof. As a candidate Lyapunov function for the linearized system (16.24), consider

V (δq, δq) =1

2δq′Mδq +

1

2δq′Kδq

Example 167m1q1 + dq1 +(k1 + k2)q1 −k2q2 = 0m2q2 − k2q1 +k2q2 = 0

This system is asymptotically stable.

Example 168

mq1 + dq1 − dq2 +(k1 + k2)q1 −k2q2 = 0mq2 − dq1 + dq2 − k2q1 +(k1 + k2)q2 = 0

This system is not asymptotically stable.

16.5. A CLASS OF GUAS AEROSPACE/MECHANICAL SYSTEMS 213

Figure 16.2: An example

Figure 16.3: Another example

16.5 A class of GUAS aerospace/mechanical systems

Here we consider aerospace/mechanical systems described by

M(q)q + c(q, q) +D(q, q) +K(q) = 0 (16.25)

where q(t) is a real N -vector and

M(q)′ =M(q) > 0

for all q. So, here Q = −D(q, q)−K(q)

Assumption 1 There is an N -vector qe such that:

(i)

K(qe) = 0 and K(q) 6= 0 for q 6= qe

(ii)

D(q, 0) = 0 for all q

This first assumption guarantees that q(t) ≡ qe is a unique equilibrium solution of the abovesystem. The next assumption requires that the generalized force term −K(q) be due topotential forces and that the total potential energy associated with these forces is a positivedefinite function.


Assumption 2 There is a function U , which is positive definite about qe, such that

K(q) =∂U

∂q

′(q)

for all q.

The next assumption requires that the generalized force term −D(q, q) be due to dampingforces and that these forces dissipate energy along every solution except constant solutions.

Assumption 3 For all q, q,q′D(q, q) ≥ 0

If q(·) is any solution for which

q(t)′D(q(t), q(t)) ≡ 0

thenq(t) ≡ 0

Theorem 39 Suppose the mechanical system (16.25) satisfies assumptions 1-3. Then thesystem is GUAS about the equilibrium solution q(t) ≡ qe.

Proof Consider the total mechanical energy (kinetic + potential)

V =1

2q′M(q)q + U(q)

as a candidate Lyapunov function and use a LaSalle type result.

Example 169

I1q1 + dq1 +(k1 + k2)q1 −k2q2 = 0I2q2 − k2q1 +k2q2 −Wl sin(q2) = 0

where all parameters are positive and

Wl <k1k2k1 + k2

=1

1k1

+ 1k2

Exercise 56 Consider the sprung pendulum subject to aerodynamic drag described by

Iθ + d|θ|θ +Wl sin θ + kθ = 0

where k > Wl and d > 0.

(i) Linearize this system about the zero solution. What are the stability properties of thelinearized system? Based on this linearization, what can you say about the stability ofthe nonlinear system about the zero solution?

(ii) Use Theorem 39 to show that the nonlinear system is GUAS about the zero solution.

16.6. CONTROL OF AEROSPACE/MECHANICAL SYSTEMS 215

16.6 Control of aerospace/mechanical systems

Here we consider aerospace/mechanical systems with control inputs described by

M(q)q + c(q, q) + F (q, q) = u (16.26)

where q(t) and the control input u(t) are real N -vectors. So, Q = −F + u. The salientfeature of the systems considerd here is that the number of scalar control inputs is the sameas the number of degrees of freedom of the system.

16.6.1 Computed torque method

This is a special case of the more general technique of feedback linearization. Suppose welet

u = c(q, q) + F (q, q) +M(q)v (16.27)

where v is to be regarded as a new control input. Then, M(q)q =M(q)v and since M(q) isnonsingular, we obtain

q = v (16.28)

This can also be expressed as a bunch of N decoupled double integrators:

qi = vi for i = 1, 2, · · · , NNow use YFLTI (your favorite linear time invariant) control design method to design con-trollers for (16.28).

Disadvantages: Since this method requires exact cancellation of F and exact knowledge ofM , it may not be robust with respect to uncertainty in M and F .

16.6.2 Linearization

Consider any desired controlled equilibrium configuration

q(t) ≡ qe

for the systemM(q)q + c(q, q) + F (q, q) = u

This can be achieved with the constant input

u(t) ≡ ue := F (qe, 0)

If we linearize the above system about (q(t), u(t)) ≡ (qe, ue), we obtain

Mδq + Dδq + Kδq = δu (16.29)

where

M =M(qe), D =∂F

∂q(qe, 0), K =

∂F

∂q(qe, 0)

Now use YFLTI control design method to design controllers for δu and let

u = ue + δu (16.30)


16.6.3 Globally stabilizing PD controllers

Here we consider mechanical systems described by

M(q)q + c(q, q) +D(q, q) +K(q) = u (16.31)

So, F (q, q) = D(q, q) +K(q).

Assumption 4 There is an N -vector qe uch that

K(qe) = 0 and D(q, 0) = 0 for all q

This first assumption guarantees that q(t) ≡ qe is an equilibrium solution of the uncontrolled(u = 0) system. It does not rule out the possibility of other equilibrium solutions.

Assumption 5 There is a function U and a scalar β such that for all q

K(q) =∂U

∂q

′(q)

and∂K

∂q(q) ≥ −βI

Assumption 6 For all q and q,q′D(q, q) ≥ 0

PD controllers

u = −Kp(q − qe)−Kdq (16.32a)

with

K ′p = Kp > βI , Kd +K ′

d > 0 (16.32b)

Theorem 40 Consider a system described by (16.31) which satisies assumptions 4-6 and issubject to PD controller (16.32). Then the resulting closed system is GUAS about q(t) ≡ qe.

Proof. We show that the closed loop system

M(q)q + c(q, q) +D(q, q) +Kdq +K(q) +Kp(q − qe) = 0 (16.33)

satisfies the hypotheses of Theorem 39.

Example 170 Single link manipulator with drag.

Iθ + dθ|θ| −Wl sin θ = u

with d ≥ 0. Here q = θ,

K(θ) = −Wl sin θ and D(θ, θ) = dθ|θ|

16.6. CONTROL OF AEROSPACE/MECHANICAL SYSTEMS 217

Hence∂K

∂q(q) = −Wl cos q ≥ −Wl

So β =Wl and a globally stabilizing controller is given by

u = −kpq − kdq

kp > Wl , kd > 0

0 2 4 6 8 10 12 14 16 18 20-0.5

0

0.5

1

time (sec)

link a

ngle

(rad)

q1(0) =1, q1dot(0) = 0

0 2 4 6 8 10 12 14 16 18 20-0.2

0

0.2

0.4

0.6

time (sec)

link a

ngle

(rad)

q1(0) =0, q1dot(0) = 1

Figure 16.4: Response of closed-loop, rigid, single-link manipulator


Exercise 57 Consider the two link manipulator with two control input torques, u1 and u2:

[m1lc21 +m2l

21 + I1]q1 +[m2l1lc2 cos(q1 − q2)]q2 +m2l1lc2 sin(q1 − q2)q

22 − [m1lc1 +m2l1]g sin(q1) = u1

[m2l1lc2 cos(q1 − q2)]q1 +[m2lc22 + I2]q2 −m2l1lc2 sin(q1 − q2)q

21 −m2glc2 sin(q2) = u2

Using each of the 3 control design methods of this section, obtain controllers which asymptot-ically stabilize this system about the upward vertical configuration. Design your controllersusing the following data:

m1 l1 lc1 I1 m2 l2 lc2 I2kg m m kg ·m2 kg m m kg ·m2

10 1 0.5 10/12 10 1 0.5 5/12

Table 16.1: Parameters of the two-link manipulator

Simulate the closed loop system with each of your three controller and for the followingvalues of m2: 5 kg and 10 kg.

Chapter 17

Quadratic stabilizability

17.1 Preliminaries

Here we consider the problem of designing stabilizing state feedback controllers for specificclasses of nonlinear systems. Controller design is based on quadratic Lyapunov functions.

First recall that a systemx = f(t, x) (17.1)

is quadratically stable if there exist positive definite symmetric matrices P and Q which satisfy

2x′Pf(t, x) ≤ −2x′Qx (17.2)

for all t and x. When this is the case, the system is globally uniformly exponentially stable(GUES) about the origin and the scalar α = λmin(P

−1Q) is a rate of exponential convergence.Also, we call P a Lyapunov matrix for (17.1).

Consider now a system with control input u where u(t) is an m-vector:

x = F (t, x, u) . (17.3)

Suppose u is given by a static linear state feedback controller

u = Kx ; (17.4)

then the resulting closed loop system is described by

x = F (t, x,Kx) . (17.5)

We refer to K as a state feedback gain matrix.We say that system (17.3) is quadratically stabilizable via linear state feedback if there exists

a matrix K such that the closed loop system (17.5) is quadratically stable, that is, exist amatrix K and positive definite symmetric matrices P and Q which satisfy

2x′PF (t, x,Kx) ≤ −2x′Qx (17.6)

for all t and x.

219

220 CHAPTER 17. QUADRATIC STABILIZABILITY

If we let x = Px, S = P−1 and L = KP−1 and Q = P−1QP−1, the above requirementcan we expressed as:

2x′F (t, Sx, Lx) ≤ −2x′Qx (17.7)

for all t and x. Then

P = S−1 K = LS−1 . (17.8)

17.2 Some special classes of nonlinear systems

Here we consider nonlinear systems with control inputs described by

x = A(t, x)x+B(t, x)u (17.9)

where the real n-vector x(t) is the state and the real m-vector u(t) is the control input. Thefollowing observation is sometimes useful in reducing control design problems to the solutionof linear matrix inequalities (LMIs).

• Suppose there exist a matrix L and positive definite symmetric matrices S and Q whichsatisfy

A(t, x)S +B(t, x)L+ SA(t, x)′ + L′B(t, x)′ ≤ −Q (17.10)

for all t and x. Then the linear state feedback controller

u = LS−1x (17.11)

renders system (17.9) GUES about the origin with Lyapunov matrix P = S−1.

To see this, suppose (17.10) holds and let P = S−1 and K = LS−1. Pre- and post-multiply inequality (17.10) by S−1 to obtain

P [A(t, x) +B(t, x)K] + [A(t, x) +B(t, x)K]′P ≤ −PQP (17.12)

Now notice that the closed loop system resulting from controller (17.11) is described by

x = [A(t, x) +B(t, x)K]x (17.13)

that is, it is described by (17.1) with f(t, x) = [A(t, x) +B(t, x)K]x. Thus, using inequality(17.12), the closed loop system satisfies

2x′Pf(t, x) = 2x′P [A(t, x) +B(t, x)K]x

= x′P [A(t, x) +B(t, x)K] + [A(t, x) +B(t, x)K]′Px

≤ −x′PQPx .

Thus, the closed loop system satisfies inequality (17.6) with PQP replacing Q. Therefore,the closed loop system (17.13) is GUES with Lyapunov matrix P = S−1.

17.2. SOME SPECIAL CLASSES OF NONLINEAR SYSTEMS 221

17.2.1 A special class of nonlinear systems

We first consider systems described by (17.9) where the time/state dependent matricesA(t, x) and B(t, x) have the following structure:

A(t, x) = A0 + ψ(t, x)∆A , B(t, x) = B0 + ψ(t, x)∆B (17.14)

The matrices A0, B0 and ∆A,∆B are constant and ψ is a scalar valued function of time tthe state x which is bounded above and below, that is,

a ≤ ψ(t, x) ≤ b (17.15)

for some constants a and b. Examples of such functions are given by sin x, cosx, e−x2and

sin t.

Example 171 (Single-link manipulator) A single-link manipulator (or inverted pendu-lum) subject to a control torque u is described by

Jθ −Wl sin θ = u

with J, Wl > 0. Letting x1 = θ and x2 = θ this system has the following state spacedescription:

x1 = x2 (17.16a)

Jx2 = Wl sin x1 + u (17.16b)

This description is given by (17.9) and (17.14) with

A0 =

(

0 10 0

)

, B0 =

(

01/J

)

, ∆A =

(

0 01 0

)

, ∆B = 0

and

ψ(t, x) =

{

(Wl/J) sinx1/x1 if x1 6= 0(Wl/J) if x1 = 0

Since | sinx1| ≤ |x1|, we have

−Wl/J ≤ ψ(t, x) ≤ Wl/J ;

hence ψ satisfies inequalities (17.15) with a = −Wl/J and b = Wl/J .

The following theorem yields controllers for the global exponential stabilization of systemsdescribed by (17.9) and (17.14) and satisfying (17.15).

Theorem 41 Consider a system described by (17.9) and (17.14) and satisfying (17.15).Suppose there exist a matrix L and a positive-definite symmetric matrix S which satisfy thefollowing linear matrix inequalities:

A1S +B1L+ SA′1 + L′B1 < 0 (17.17a)

A2S +B2L+ SA′2 + L′B2 < 0 (17.17b)


with

A1 := A0 + a∆A , B1 := B0 + a∆B (17.18a)

A2 := A0 + b∆A , B2 := B0 + b∆B . (17.18b)

Then the controller given by (17.11) renders system (17.9),(17.14) globally uniformly expo-nentially stable about the origin with Lyapunov matrix S−1.

Proof. Suppose S and L satisfy inequalities (17.17) and let Q be any positive definitematrix which satisfies

A1S +B1L+ SA′1 + L′B1 ≤ −Q (17.19a)

A2S +B2L+ SA′2 + L′B2 ≤ −Q . (17.19b)

For each t and x,

N(t, x) := A(t, x)S +B(t, x)L+ SA(t, x)′ + L′B(t, x)′

= N0 + ψ(t, x)∆N

where

N0 = A0S +B0L+ SA′0 + L′B′

0 and ∆N = ∆AS +∆BL+ S∆A′ + L′∆B′

Since N(t, x) = N0 + ψ(t, x)∆N and a ≤ ψ(t, x) ≤ b, it follows that

N(t, x) ≤ −Q

for all t and x ifN0 + a∆N ≤ −Q and N0 + b∆N ≤ −Q

These last two inequalities are precisely inequalities (17.19).

Exercise 58 Prove the following result: Consider a system described by (17.9),(17.14) andsatisfying (17.15). Suppose there exists a positive-definite symmetric matrix S, a matrix Land a positive scalar α which satisfy

A1S +B1L+ SA′1 + L′B1 + 2αS ≤ 0 (17.20a)

A2S +B2L+ SA′2 + L′B2 + 2αS ≤ 0 (17.20b)

where A1, B1, A2, B2 are given by (17.18). Then controller (17.11) renders system (17.9),(17.14), GUES about the origin with rate of convergence α and Lyapunov matrix S = P−1.

Note that the existence of a positive definite symmetric S and a matrix L satisfyinginequalities (17.17) or (17.20) is equivalent to the existence of another symmetric matrix Sand another matrix L satisfying S ≥ I and (17.17) or (17.20), respectively.

Suppose B1 = B2 and (S, L) is a solution to (17.17) or (17.20), then (S, L−γB′1) is also

a solution for all γ ≥ 0. Since the feedback gain matrix corresponding to the latter pair is


K = L0S−1−γB′

1S−1, this results in large gain matrices for large γ. So how do we restrict

the size of K? One approach is to minimize β subject to

KSK ′ ≤ βI (17.21)

If we also require I ≤ S, we obtain that KK ′ ≤ KSK ′ ≤ βI. Thus, β is an upper bound on||K||2. If we minimize β subject to inequalities (17.17), the resulting controller may yieldan unacceptably small rate of convergence. To avoid this, we can choose a specific desiredrate of convergence α > 0 and minimize β subject to (17.20). Finally, note that inequality(17.21) can be written as LS−1L′ ≤ βI, or,

(

βI LL′ S

)

≥ 0

To summarize, we can globally stabilize system (17.9),(17.14),(17.15) with rate of conver-gence α by solving the following optimization problem:


A1S +B1L+ SA′1 + L′B1 + 2αS ≤ 0

A2S +B2L+ SA′2 + L′B2 + 2αS ≤ 0

I ≤ S(

βI L′

L S

)

≥ 0

(17.22)

and letting u = LS−1x.

Example 172 Recall the single-link manipulator of Example 171. For J = Wl = 1, thefollowing Matlab program uses the results of this section and the LMI toolbox to obtainstabilizing controllers.

% Quadratic stabilization of single-link manipulator

%

clear all

% Data and specs

alpha = 0.1; %The specified rate of convergence

J = 1;

Wl = 1;

%

% Description of manipulator

A0 = [0 1; 0 0];

DelA = [0 0; 1 0];

B0 = [0; 1/J];

DelB = 0;

a= -Wl/J

b= Wl/J


A1 = A0 + a*DelA;

A2 = A0 + b*DelA;

B1 = B0 + a*DelB;

B2 = B0 + b*DelB;

%

% Form the system of LMIs

setlmis([])

%

S=lmivar(1, [2,1]);

L=lmivar(2, [1,2]);

beta=lmivar(1, [1,1]);

%

lmi1=newlmi;

lmiterm([lmi1,1,1,S], A1,1,’s’) %A1*S + S*A1’

lmiterm([lmi1,1,1,L], B1,1,’s’) %B1*L + L’*B1’

lmiterm([lmi1,1,1,S], 2*alpha, 1) %2*alpha*S

%

lmi2=newlmi;

lmiterm([lmi2,1,1,S], A2,1,’s’) %A2*S + S*A2’

lmiterm([lmi2,1,1,L], B2,1,’s’) %B2*L + L’*B2’

lmiterm([lmi2,1,1,S], 2*alpha, 1) %2*alpha*S

%

Slmi= newlmi;

lmiterm([-Slmi,1,1,S],1,1)

lmiterm([Slmi,1,1,0],1)

%

lmi4=newlmi;

lmiterm([-lmi4,1,1,beta],1,1) %beta*I

lmiterm([-lmi4,1,2,L],1,1) %L

lmiterm([-lmi4,2,2,S],1,1) %S

lmis = getlmis;

%

c=mat2dec(lmis,0,0,1); %specify weighting

options=[1e-5 0 0 0 0]; %Minimize

[copt,xopt]=mincx(lmis,c,options)

%

S = dec2mat(lmis,xopt,S)

L = dec2mat(lmis,xopt,L)

%

K =L*inv(S) %Controller gain matrix

Running this program yields:

K =(

−1.3036 −1.2479)

for α = 0.1

K =(

−3.6578 −2.8144)

for α = 1


K =(

−152.9481 −20.1673)

for α = 10

For initial conditionsθ(0) = π θ(0) = 0 ,

Figure 17.1 illustrates the closed loop behavior of θ corresponding to α = 1 and α = 10.

0 1 2 3 4 5 6−0.5

0

0.5

1

1.5

2

2.5

3

3.5

t

theta

Figure 17.1: Closed loop single-link manipulator

Exercise 59 (PID regulation of single-link manipulator) Consider a single-link ma-nipulator subject to a unknown constant disturbance torque w:

Jθ −Wl sin θ = w + u .

Suppose we wish to stabilize this system about some desired constant motion θ(t) ≡ θd.The corresponding steady state value of u is unknown and is given by ud = −Wl sin θd −w.Consider now the following PID controller:

u = −kp(θ − θd)− kI

∫ t

0

(θ − θd) dτ − kdθ (17.23)

The resulting closed loop system is described by

Jθ + kdθ + kp(θ − θd)−Wl sin θ − w + kIη = 0

η = θ − θd

The equilibrium solutions θ(t) ≡ θd, η(t) ≡ ηd of this system satisfy

θ = θd

ud + kIη = 0


So, when ud 6= 0, one needs kI 6= 0 and the equilibrium ηd value of η is given by ηd = −ud/kI .Introduce states

x1 = θ − θd

x2 = θ

x3 = η − ηd

to obtain the following state space description:

x1 = x2 (17.24a)

Jx2 = −kpx1 − kdx2 − kIx3 +Wl[sin(θd + x1)− sin(θd)] (17.24b)

x3 = x1 (17.24c)

If one now obtains a matrixK =

(

−kp −kd −kI)

so that system (17.24) is GUES about the origin, then the PID controller (17.23) will guar-antee that the angle θ of the manipulator will exponentially approach the desired value θdand all other variables will be bounded. Note that a stabilizing K will have kI 6= 0.

Using Theorem 41 and the LMI toolbox, obtain a stabilizing K for J = Wl = 1. Simulatethe resulting closed loop system for several initial conditions and values of θd and w.


One can readily generalize the results of this section to systems described by (17.9) wherethe time/state dependent matrices A(t, x) and B(t, x) have the following structure

A(t, x) = A0 + ψ1(t, x)∆A1 + · · ·+ ψl(t, x)∆Al (17.25a)

B(t, x) = B0 + ψ1(t, x)∆B1 + · · ·+ ψl(t, x)∆Bl (17.25b)

Here, A0,∆A1, · · · ,∆Al are constant n×n matrices and B0,∆B1, · · · ,∆Bl are constant n×mmatrices. Also, each ψi is a scalar valued function of the time t and state x which satisfies

ai ≤ ψi(t, x) ≤ bi (17.26)

for some known constants ai and bi.

Example 173 (Robust control of single-link manipulator) Recall from Example 171that the motion of a single-link manipulator (or inverted pendulum) subject to a controltorque u can be described by

x1 = x2 (17.27a)

x2 = (Wl/J) sin x1 + (1/J)u (17.27b)

with J, Wl > 0. Suppose that we do not know the parameters J and Wl exactly and onlyhave knowledge on their upper and lower bounds, that is,

0 < J ≤ J ≤ J and Wl ≤Wl ≤ Wl


where the bounds J, J, Wl, Wl are known. Introducing

ψ1(t, x) =

{

(Wl/J) sin x1/x1 if x1 6= 0(Wl/J) if x1 = 0

and ψ2(t, x) = 1/J ,

the above state space description is given by (17.9) and (17.25) with l = 2 and

A0 =

(

0 10 0

)

, B0 =

(

00

)

, ∆A1 =

(

0 01 0

)

, ∆B1 = 0 , ∆A2 = 0 , ∆B2 =

(

01

)

.

Since | sinx1| ≤ |x1|, we have

−Wl/J ≤ ψ1(t, x) ≤Wl/J

1/J ≤ ψ2(t, x) ≤ 1/J ;

hence ψ1 and ψ2 satisfy inequalities (17.26) with a1 = −Wl/J, b1 = Wl/J and a2 =1/J, b2 = 1/J .

To obtain a sufficient condition for a system given by (17.9), (17.25) and (17.26), weintroduce the following set of 2l “extreme matrix pairs”:

AB = { (A0 + δ1∆A1 + · · ·+ δl∆Al, B0 + δ1∆B1 + · · ·+ δl∆Bl) : δi = ai or bi for i = 1, · · · , l}

For example, with l = 2, AB consists of the following four matrix pairs:

(A0 + a1∆A1 + a2∆A2, B0 + a1∆B1 + a2∆B2)

(A0 + a1∆A1 + b2∆A2, B0 + a1∆B1 + b2∆B2)

(A0 + b1∆A1 + a2∆A2, B0 + b1∆B1 + a2∆B2)

(A0 + b1∆A1 + b2∆A2, B0 + b1∆B1 + b2∆B2)

We have now the following result.

Theorem 42 Consider a system described by (17.9),(17.25) and satisfying (17.26). Supposethere exist a matrix L and a positive-definite symmetric matrix S which satisfy

AS +BL+ SA′ + L′B′ < 0 (17.28)

for every pair (A,B) in AB. Then the controller given by (17.11) renders system (17.9),(17.25)globally uniformly exponentially stable about the origin with Lyapunov matrix S−1.

Recalling the discussion in the previous section, we can globally stabilize system (17.9),(17.14),(17.26)with rate of convergence α by solving the following optimization problem:


AS +BL+ SA′ + L′B′ + 2αS ≤ 0 for all (A,B) in ABS ≥ I

(

βI L′

L S

)

≥ 0

(17.29)



Exercise 60 Consider the double inverted pendulum described by

θ1 + kθ1 − kθ2 − sin θ1 = u

θ2 − kθ1 + kθ2 − sin θ2 = 0

with k = 1. Using the above results, obtain a state feedback controller which renders thissystem globally exponentially stable about the zero solution.

17.3 Another special class of nonlinear systems

Here we consider nonlinear systems which have the following structure:

x = Ax+B1φ(t, Cx+Du) +B2u (17.30)

where the real n-vector x(t) is the state and the real m2-vector u(t) is the control input. Thematrices B1, B2 and C are constant with dimensions n×m1, n×m2, and p×n, respectively.The function φ : IR× IRp → IRm1 satisfies

||φ(t, z)|| ≤ γ||z|| (17.31)

Note that if we introduce a fictitious output z = Cx + Du and a fictitious input w =φ(t, Cx+Du), the nonlinear system (17.30) can be described as a feedback combination ofa linear time invariant system

x = Ax+B1w +B2uz = Cx+Du

and a memoryless nonlinearityw = φ(t, z)


Figure 17.2: A feedback system

Example 174 Single link manipulator. Recall the single-link of Example 171 described by(17.16). This system can be described by the above general description where

A =

(

0 10 0

)

B1 =

(

0Wl/J

)

B2 =

(

01/J

)

C =(

1 0)

D = 0

and φ(t, z) = sin z. Also,||φ(t, z)|| = | sin z| ≤ |z| = ||z||


The following theorem yields controllers for for the global exponential stabilization ofsystem (17.30).

Theorem 43 Consider a system described by (17.30) and satisfying (17.31). Suppose thereexist a matrix L, a positive-definite symmetric matrix S and a positive scalar µ which satisfythe following matrix inequality :

AS +B2L+ SA′ + L′B′2 + µγ2B1B

′1 + µ−1(CS +DL)′(CS +DL) < 0 (17.32)

Then the controller given by (17.11) renders system (17.30) globally uniformly exponentiallystable about the origin with Lyapunov matrix S−1.

Proof. Consider any K and u = Kx. Then

x = (A+B2K)x+B1φ(t, (C +DK)x)

Recalling quadratic stability results, this system is GUES with Lyapunov matrix P if thereexists µ > 0 such that

P (A+B2K) + (A+B2K)′P + µγ2PB1B′1P + µ−1(C +DK)′(C +DK) < 0 = 0

Letting S = P−1 and L = KS, this is equivalent to


′1 + µ−1(CS +DL)′(CS +DL) < 0

Exercise 61 Consider a system described by (17.30) and satisfying (17.31). Suppose thereexist a matrix L, a positive-definite symmetric matrix S and a positive scalar µ which satisfy


′1 + 2αS + µ−1(CS +DL)′(CS +DL) ≤ 0 (17.33)

Then the controller given by (17.11) renders system (17.30) globally uniformly exponentiallystable about the origin with rate of convergence α and Lyapunov matrix S−1

Using a Schur complement result, one can show that satisfaction of quadratic matrixinequality (17.32) is equivalent to satisfaction of the following matrix inequality :

(


′1 (CS +DL)′

CS +DL −µI

)

< 0 (17.34)

Note that this inequality is linear in S, L, and µ. Recalling the discussion in Section 17.2,we can globally stabilize system (17.30) with rate of convergence α by solving the followingoptimization problem:


(


′1 + 2αS (CS +DL)′

CS +DL −µI

)

≤ 0

S ≥ I

(

βI L′

L S

)

≥ 0

(17.35)




Here we consider systems whose nonlinearity is characterized by several nonlinear functions,φi : IR

pi → IRmi , i = 1, 2, . . . , l :

x = Ax+B11φ1(t, C1x+D1u) + . . .+B1lφl(t, Clx+D1u) +B2u (17.36)

where A, B2 and B1i, Ci, Di (i = 1, 2, · · · , l) are constant matrices of appropriate dimensions.Each φi satisfies

||φi(t, zi)|| ≤ γ||zi|| (17.37)

If we introduce fictitious outputs zi = Cix+Diu and fictitious inputs wi = φi(t, Cix+Diu),the above nonlinear system can be described as a feedback combination of a linear timeinvariant system

x = Ax+B11w1 + . . .+B1lwl +B2uz1 = C1x+D1u

...zl = Clx+Dlu

and multiple memoryless nonlinearities:

wi = φi(t, zi) for i = 1, 2, · · · , l


Figure 17.3: A multi-loop feedback system

One can readily obtain the following result:

Theorem 44 Consider a system described by (17.36) and satisfying (17.37). Suppose thereexist a matrix L, a positive-definite symmetric matrix S and l positive scalars µ1, µ2, . . . , µl

which satisfy the following matrix inequality :

AS +B2L+ SA′ + L′B′2 +

l∑

i=1

(

µiγ2B1iB1i

′ + µ−1i (CiS +DiL)

′(CiS +DiL))

< 0 (17.38)

Then the controller given by (17.11) renders system (17.36) globally uniformly exponentiallystable about the origin with Lyapunov matrix S−1.


Proof. Similar to proof of last theorem.Note that, using a Schur complement result, inequality (17.38) is equivalent to the fol-

lowing inequality which is linear in L, P and the scaling parameters µ1, · · · , µl:

AS +B2L+ SA′ + L′B′2 +

∑li=1 µiγ

2B1iB′1i (C1S +D1L)

′ (C2S +D2L)′ · · · (ClS +DlL)

′

C1S +D1L −µ1I 0 · · · 0C2S +D2L 0 −µ2I · · · 0

......

.... . .

...ClS +DlL 0 0 · · · −µlI

< 0

Recalling previous discussions, we can globally stabilize system (17.36) with rate of con-vergence α by solving the following optimization problem:


AS + B2L + SA′ + L′B′

2+

∑

l

i=1µiγ

2B1iB′

1i+ 2αS (C1S + D1L)′ (C2S + D2L)′ · · · (ClS + DlL)′

C1S + D1L −µ1I 0 · · · 0C2S + D2L 0 −µ2I · · · 0

.

.

.

.

.

.

.

.

....

.

.

.ClS + DlL 0 0 · · · −µlI

≤ 0

S ≥ I

(

βI L′

L S

)

≥ 0

(17.39)and letting u = LS−1x.


17.4 Yet another special class

x = Ax+B1φ(t, Cx+Du) +B2u (17.40)

wherez′φ(t, z) ≥ 0 (17.41)

for all z.

Figure 17.4: A first and third quadrant nonlinearity

Theorem 45 Consider a system described by (17.40) and satisfying (17.41). Suppose thereexist a matrix L, a positive-definite symmetric matrix S and a positive scalar µ which satisfy

AS +B2L+ SA′ + L′B′2 < 0 (17.42a)

µB′1 = CS +DL (17.42b)

Then controller (17.11) renders system (17.40) globally exponentially stable about the originwith Lyapunov matrix S−1.

Proof. Apply corresponding analysis result to closed loop system.

• Note that (17.42) has a solution with S, µ > 0 if and only if the following optimizationproblem has a minimum of zero:


AS +B2L+ SA′ + L′B′2 < 0

S ≥ I

µ > 0

(

βI µB′1 − CS −DL

µB1 − SC ′ − L′D′ βI

)

≥ 0


Exercise 62 Prove the following result: Consider a system described by (17.40) and satis-fying (17.41). Suppose there exist a matrix L, a positive-definite symmetric matrix S, andpositive scalars µ, α which satisfy

AS +B2L+ SA′ + L′B′2 + 2αS ≤ 0 (17.43a)

µB′1 = CS +DL (17.43b)

Then controller (17.11) renders system (17.40) globally uniformly exponentially stable aboutthe origin with rate α and with Lyapunov matrix S−1.



x = Ax+B1w +B2uz = Cx+D1w +D2uw = −φ(t, z)

(17.44)

wherez′φ(t, z) ≥ 0 (17.45)

for all z.

Theorem 46 Consider a system described by (17.44) and satisfying (17.45). Suppose thereexist a matrix L, a positive-definite symmetric matrix S and positive scalars α and µ whichsatisfy

(

AS + SA′ +B2L+ L′B′2 + 2αS (CS +D2L)

′ − µB1

CS +D2L− µB′1 −µ(D1 +D′

1)

)

≤ 0 (17.46)

Then controller (17.11) renders system (17.40) globally uniformly exponentially stable aboutthe origin with rate α and with Lyapunov matrix S−1.

Proof.

• Note that (17.46) is linear in L, S, µ.If D +D′ > 0, then (17.46) is equivalent to

AS+SA′ +B2L+L′B′2+2αS+µ−1(µB1−SC ′ −L′D′

2)(D1+D′1)

−1(µB′1−CS−D2L) ≤ 0

Chapter 18

Singularly perturbed systems

18.1 Introduction and examples

Roughly speaking, a singularly perturbed system is a system which depends on a scalarparameter in such a fashion that for a particular value of the parameter (usually taken tobe zero) the system order reduces.

18.1.1 High gain output feedback

Consider a scalar-input scalar-output system described by

x1 = x2

x2 = u

y = 2x1 + x2

This system has the following transfer function

G(s) =s+ 2

s2

Suppose we subject this system to high-gain static output feedback, that is, we let

u = −ky

where k is large. Letting µ = k−1, the closed loop system is described by

x1 = x2 (18.1)

µx2 = −2x1 − x2 (18.2)

We are interested in the behavior of this system for µ > 0 small.The eigenvalues of the closed loop system matrix

[

0 1−2µ−1 −µ−1

]

235

236 CHAPTER 18. SINGULARLY PERTURBED SYSTEMS

are λs(µ) and µ−1λf(µ) where

λs(µ) =−4

1 +√1− 8µ

, λf (µ) =1 +

√1− 8µ

−2

and

limµ→0

λs(µ) = −2 , limµ→0

λf (µ) = −1

Thus, for small µ > 0, system responses are characterized by a “slow” mode eλs(µ)t and a

“fast” mode eλf (µ)t/µ.

For µ = 0, the differential equation (18.2) reduces to the algebraic equation

−2x1 − x2 = 0

Substitution for x2 into (18.1) yields

x1 = −2x1

Note that the dynamics of this system are described by the limiting slow mode.

18.1.2 Single-link manipulator with flexible joint

Consider the single-link manipulator illustrated in Figure 18.1. There is a flexible joint orshaft between the link and the rotor of the actuating motor; this motor exerts a controltorque u. The equations of motion for this system are:

u J

m,I

l

k

q1

q2

Figure 18.1: Single link manipulator with flexible joint

18.1. INTRODUCTION AND EXAMPLES 237

Iq1 = Wl sin q1 − c(q1 − q2)− k(q1 − q2)Jq2 = u + c(q1 − q2) + k(q1 − q2)

where W is the weight of the link and k > 0 and c > 0 represent the stiffness and dampingcoefficients, respectively, of the shaft. We want to design a feedback controller for u whichguarantees that the resulting closed loop system is globally asymptotically stable aboutq1(t) ≡ q2(t) ≡ 0. Suppose for the purpose of controller design, we simplify the problem andmodel the shaft as rigid. This yields the following rigid model:

(I + J)q1 = Wl sin q1 + u

A controller which globally exponentially stabilizes this model about q1(t) ≡ 0 is given by

u = p(q1, q1) = −kpq1 − kdq1

withkp > Wl, kd > 0

Will this controller also stabilize the flexible model provided, the flexible elements aresufficiently stiff? To answer this question we let

k = µ−2k0 , c = µ−1c0

where k0, c0 > 0 and consider system behavior as µ→ 0. Introducing state variables

x1 = q1 , x2 = q1 , y1 = µ−2(q2 − q1) , y2 = µ−1(q2 − q1)

the closed loop flexible model is described by

x1 = x2x2 = I−1Wl sin x1 + I−1(k0y1 + c0y2)µy1 = y2µy2 = −I−1Wl sin x1 + J−1p(x) − I−1(k0y1 + c0y2)

where I−1 := I−1 + J−1.Suppose we let µ = 0 in this description. Then the last two equations are no longer

differential equations; they are equivalent to

y2 = 0

y1 = k−10 I(−I−1Wl sin x1 + J−1p(x))

Substituting these two equations into the first two equations of the above state space de-scription yields

x1 = x2

x2 = (I + J)−1(Wl sin x1 + p(x))

Note that this is a state space description of the closed loop rigid model.


18.2 Singularly perturbed systems

The state space description of the above two examples are specific examples of singularlyperturbed systems. In general a singularly perturbed system is described by

x = f(x, y, µ) (18.3a)

µy = g(x, y, µ) (18.3b)

where x(t) ∈ IRn and y(t) ∈ IRm describe the state of the system at time t ∈ IR and µ > 0is the singular perturbation parameter. We suppose 0 ≤ µ < µ.

Reduced order system. Letting µ = 0 results in

x = f(x, y, 0)

0 = g(x, y, 0)

If we assume that there is a continuously differentiable function h0 such that

g(x, h0(x), 0) ≡ 0 ,

then y = h0(x). This yields the reduced order system:

x = f(x, h0(x), 0)

A standard question is whether the behavior of the original full order system (18.3) can bepredicted by looking at the behavior of the reduced order system. To answer this questionone must also look at another system called the boundary layer system.

Boundary layer system. First introduce “fast time” described by

τ = t/µ

and let

ξ(τ) := x(µτ) = x(t)

η(τ) := y(µτ)− h0(x(µτ)) = y(t)− h0(x(t))

to obtain a regularly perturbed system:

ξ′ = µf(ξ, h0(ξ) + η, µ)

η′ = g(ξ, h0(ξ) + η, µ)− µ∂h0∂x

(ξ)f(ξ, h0(ξ) + η, µ)

Setting µ = 0 results in ξ(τ) ≡ 0; hence ξ(τ) = ξ(0) = x(0) =: x0 and

η′ = g(x0, h0(x0) + η, 0)

18.2. SINGULARLY PERTURBED SYSTEMS 239

This is a dynamical system parameterized by x0. It has a unique equilibrium state η = 0.We will refer to it as the boundary layer system.

A standard result is the following: Suppose the boundary layer system is exponentiallystable and this stability is uniform with respect to x0; then on any compact time interval notincluding 0, the behavior of x for the full order system approaches that of x for the reducedorder system as µ→ 0.

Another result is the following: Suppose the reduced order system is exponentially stableabout 0 and the boundary layer is exponentially stable for all x0. Then, under certainregularity conditions, the full order system is exponentially stable for µ sufficiently small.

Recalling the first example of the last section, the boundary layer system is given by theexponentially stable system

η′ = −ηSince the reduced order system is exponentially stable, the full order system is exponentiallystable for µ > 0 small, that is, for k > 0 large.

The boundary layer system for the second example is given by

η′1 = η2

η′2 = −I−1(k0η1 + c0η2)

Clearly this system is exponentially stable. Since the reduced order system is exponentiallystable, the closed loop flexible system is exponentially stable for µ > 0 sufficiently small,that is, for k and c sufficiently large. This is illustrated by numerical simulations in Figure18.2.


0 2 4 6 8 10 12 14 16 18 20-1

0

1

time (sec)

link

angl

e (ra

d)

rigid

0 2 4 6 8 10 12 14 16 18 20-1

0

1

time (sec)

link

angl

e (ra

d)

flexible: k=900, c=30

0 2 4 6 8 10 12 14 16 18 20-1

0

1

time (sec)

link

angl

e (ra

d)

flexible: k=9, c=3

0 2 4 6 8 10 12 14 16 18 20-10

0

10

time (sec)

link

angl

e (ra

d)

flexible: k=1, c=1

Figure 18.2: Comparison of rigid case and flexible case with large and small stiffnesses

18.3. LINEAR SYSTEMS 241

18.3 Linear systems

A linear time invariant singularly perturbed system is described by

x = A(µ)x+B(µ)yµy = C(µ)x+D(µ)y

(18.4)

where A(·), B(·), C(·), and D(·) are differentiable matrix valued functions defined on someinterval [0, µ). For µ > 0, this is a LTI system with “A-matrix”

A(µ) =

[

A(µ) B(µ)µ−1C(µ) µ−1D(µ)

]

. (18.5)

We assume that

D(0) is invertible

Letting µ = 0 results in the reduced order system:

x = Ax

where

A := A(0)−B(0)D(0)−1C(0)

The boundary layer system is described by:

η′ = D(0)η

Let

λr1, λr2, . . . , λ

rn

and

λbl1 , λbl2 , . . . , λ

blm

be the eigenvalues of A and D(0), respectively, where an eigenvalue is included p times if itsalgebraic multiplicity is p. Then we have the following result.

Theorem 47 If D(0) is invertible, then for each µ > 0 sufficiently small, A(µ) has neigenvalues

λs1(µ), λs2(µ), . . . , λsn(µ)

with

limµ→0

λsi (µ) = λri , i = 1, 2, . . . , n

and m eigenvalues

λf1(µ)/µ, λf2(µ)/µ, . . . , λfm(µ)/µ

with

limµ→0

λfi (µ) = λbli , i = 1, 2, . . . , m


Recalling the high gain feedback example we have

λr1 = −2 , λbl1 = −1

Hence the closed loop system has two eigenvalues λs1(µ) and λf1(µ)/µ with the followingproperties

limµ→0

λs1(µ) = −2 , limµ→0

λf1(µ) = −1

The above theorem has the following corollary: If all the eigenvalues of A and D(0) havenegative real parts then, for µ sufficiently small, all the eigenvalues of A(µ) have negative realparts. If either A or D(0) has an eigenvalue with a positive real part, then for µ sufficientlysmall, A(µ) has an eigenvalue with positive real part.

18.4 Exact slow and fast subsystems

Suppose a singularly perturbed system has a boundary layer system which is just marginallystable, that is, just stable but not exponentially stable. Then we cannot apply our previousresults. Further analysis is necessary.

The slow system. Consider a singularly perturbed system described by (18.3). A subsetM of IRn× IRm is an invariant set for (18.3) if every solution originating in M remains there.A parameterized setMµ, is a slow manifold for (18.3), if for each µ ∈ [0, µ),Mµ is an invariantset for (18.3) and there is a continuous function h : IRn × [0, µ) → IRm such that

Mµ = {(x, y) ∈ IRn × IRm : y = h(x, µ)} . (18.6)

When system (18.3) has a slow manifold described by (18.6), then the behavior of the systemon this manifold is governed by

x = f(x, h(x, µ), µ) (18.7a)

y = h(x, µ) . (18.7b)

We refer to system (18.7a) as the slow system associated with singularly perturbed system(18.3). Thus the slow system describes the motion of (18.3) restricted to its slow invariantmanifold.

The slow manifold condition. Now note that if Mµ, given by (18.6), is an invariantmanifold for (18.3), then h must satisfy the following slow manifold condition:

g(x, h(x, µ), µ)− µ∂h

∂x(x, µ)f(x, h(x, µ), µ) = 0 (18.8)

For each µ > 0, this condition is a partial differential equation for the function h(·, µ); forµ = 0, it reduces to a condition we have seen before:

g(x, h(x, 0), 0) = 0 .

18.4. EXACT SLOW AND FAST SUBSYSTEMS 243

If we introduce the boundary layer state z(t) given by

z := y − h(x, µ) , (18.9)

then the original system (18.3) is equivalently described by

x = f(x, z, µ) (18.10a)

µz = g(x, z, µ) (18.10b)

with

f(x, z, µ) := f(x, h(x, µ) + z, µ) (18.11a)

g(x, z, µ) := g(x, h(x, µ) + z, µ)− µ∂h

∂x(x, µ)f(x, h(x, µ) + z, µ) (18.11b)

Note that Mµ is an invariant manifold for system (18.3) if and only if the manifold

{(x, z) ∈ IRn × IRm : z = 0} (18.12)

is invariant for (18.10). Also, the slow manifold condition is equivalent to

g(x, 0, µ) = 0 . (18.13)

From (18.10) and (18.13), it should be clear that if the slow manifold condition holds then,for any x(0) = x0 and z(0) = 0, the resulting system behavior satisfies (assuming uniquenessof solutions), z(t) ≡ 0. Hence the manifold described by (18.12) is an invariant manifold. So,the slow manifold condition is necessary and sufficient for the existence of a slow invariantmanifold.

The fast system. Introducing the fast time variable

τ := t/µ

and letting

η(τ) := x(t) = x(µτ)

ξ(τ) := z(t) = z(µτ)

system (18.10) is described by the following regularly perturbed system:

η′ = µf(η, ξ, µ)

ξ′ = g(η, ξ, µ)

We define

ξ = g(η, ξ, µ)

to be the fast system.


18.4.1 Approximations of the slow and fast systems

Supposef(x, h(x, µ), µ) = f0(x) + µf1(x) + o(µ; x)

wherelimµ→0

o(µ; x)/µ = 0 .

Thenx = f0(x)

andx = f0(x) + µf1(x)

are called the zero and first order slow systems, respectively. Note that the zero order slowsystem is the reduced order system.

To compute these approximations, we suppose that

h(x, µ) = h0(x) + µh1(x) + o(µ; x)

Then,

f0(x) = f(x, h0(x), 0)

f1(x) =∂f

∂y(x, h0(x), 0)h1(x) +

∂f

∂µ(x, h0(x), 0)

The following expressions for h0 and h1 can be obtained from the slow manifold condition

g(x, h0(x), 0) = 0 (18.14a)

∂g

∂y(x, h0(x), 0)h1(x)−

∂h0∂x

(x)f0(x) +∂g

∂µ(x, h0(x), 0) = 0 (18.14b)

To approximate the fast system, suppose that

g(η, ξ, µ) = g0(η, ξ) + µg1(η, ξ) + o(µ; η, ξ)

Thenξ′ = g0(η, ξ)

andξ′ = g0(η, ξ) + µg1(η, ξ) (18.15)

are called the zero and first order fast systems, respectively. Note that ξ = 0 is an equilibriumstate for both. Using (18.11b) one can obtain

g0(η, ξ) = g(η, h0(η) + ξ, 0)

g1(η, ξ) =∂g

∂y(η, h0(η) + ξ, 0)h1(η)−

∂h0∂x

(η)f(η, h0(η) + ξ, 0) +∂g

∂µ(η, h0(η) + ξ, 0)

Note that the zero order fast system is the boundary layer system.A reasonable conjecture is the following: if the reduced order system is exponentially

stable and the first order fast system is exponentially stable for µ > 0 sufficiently small, thenthe full order system is exponentially stable for small µ > 0.


18.4.2 A special class of systems

Here we consider a special class of singularly perturbed systems described by

x = f(x, µ) +B(µ)yµy = g(x, µ) +D(µ)y

(18.16)

with D(0) invertible.Utilizing (18.14a) the function h0 is given by

g(x, 0) +D(0)h0(x) = 0 ;

henceh0(x) = −D(0)−1g(x, 0)

andf0(x) = f(x, 0)− B(0)D(0)−1g(x, 0) .

Utilizing (18.14b) the function h1 is given by

D(0)h1(x)−∂h0∂x

(x)f0(x) +∂g

∂µ(x, 0) +

∂D

∂µ(0)h0(x) = 0

The function g0 is given byg0(η, ξ) = D(0)ξ

and the function g1 is given by

g1(η, ξ) = D(0)h1(η)−∂h0∂x

(η)[f0(η) +B(0)ξ] +∂g

∂µ(x, 0) +

∂D

∂µ(0)[h0(η) + ξ]

= D(0)h1(η)−∂h0∂x

(η)f0(η) +∂g

∂µ(x, 0) +

∂D

∂µ(0)h0(η) + [−∂h0

∂x(η)B(0) +

∂D

∂µ(0)]ξ

= D1(η)ξ

where

D1(η) = D(0)−1 ∂g

∂x(η, 0)B(0) +

∂D

∂µ(0)

So the first order fast system is described by

ξ′ = [D(0) + µD1(η)]ξ

Example 175 Consider now the closed loop flexible one link manipulator with

k = µ−2k0 , c = c0

This is described by

x1 = x2x2 = I−1Wl sin x1 + I−1(k0y1 + µc0y2)µy1 = y2µy2 = −I−1Wl sin x1 + J−1p(x) − I−1(k0y1 + µc0y2)


where I−1 = I−1 + J−1. This is in the form of (18.16) with

B(µ) =

[

0 0I−1k0 I−1µc0

]

, D(µ) =

[

0 1

−I−1k0 −I−1µc0

]

and

g(x, µ) =

[

0−I−1Wl sin x1 + J−1p(x)

]

Hence

D1 =

[

(I + J)−1kd 0

0 −I−1c0

]

and

D(0) + µD1 =

[

µ(I + J)−1kd 1

−I−1k0 −µI−1c0

]

A necessary and sufficient condition for this sytem to be exponentially for small µ > 0 is

kd < (I + J)I−1c0

Figures 18.3 and 18.4 illustrate numerical simulations with the above condition unsatisfiedand satisfied, respectively.


0 2 4 6 8 10 12 14 16 18 20-1

0

1

time (sec)

link

angl

e (ra

d)

rigid

0 2 4 6 8 10 12 14 16 18 20-20

0

20

time (sec)

link

angl

e (ra

d)

flexible: k=900, c=0.1

0 2 4 6 8 10 12 14 16 18 20-50

0

50

time (sec)

link

angl

e (ra

d)


0 2 4 6 8 10 12 14 16 18 20-5

0

5x 10

4

time (sec)

link

angl

e (ra

d)


Figure 18.3: Flexible case with low damping


0 2 4 6 8 10 12 14 16 18 20-1

0

1

time (sec)

link

angl

e (ra

d)

rigid

0 2 4 6 8 10 12 14 16 18 20-2

0

2

time (sec)

link

angl

e (ra

d)


0 2 4 6 8 10 12 14 16 18 20-1

0

1

time (sec)

link

angl

e (ra

d)


0 2 4 6 8 10 12 14 16 18 20-500

0

500

time (sec)

link

angl

e (ra

d)


Figure 18.4: Flexible case with “medium” damping


Suppose we colocate the rate sensor with the actuator. Then

u = p(q1, q2)

= −kpq1 − kdq2

= −kpx1 − kdx2 − µkdy2

= p(x)− µkdy2

andx1 = x2x2 = I−1Wl sin x1 + I−1(k0y1 + µc0y2)µy1 = y2µy2 = −I−1Wl sin x1 + J−1p(x) − I−1k0y1 − µ(I−1c0 + J−1kd)y2

Now

D(µ) =

[

0 1

−I−1k0 −µ(I−1c0 + J−1kd)

]

This yields

D(0) + µD1 =

[

µ(I + J)−1kd 1

−I−1k0 −µ(I−1c0 + J−1kd)

]

This is exponentially stable for µ > 0 small.Figure 18.5 illustrates numerical simulations with c0 = 0.


0 2 4 6 8 10 12 14 16 18 20-1

0

1

time (sec)

link a

ngle

(rad)

rigid

0 2 4 6 8 10 12 14 16 18 20-1

0

1

time (sec)

link a

ngle

(rad)

flexible: k=900, c=0

0 2 4 6 8 10 12 14 16 18 20-1

0

1

time (sec)

link a

ngle

(rad)

flexible: k=10, c=0

0 2 4 6 8 10 12 14 16 18 20-5000

0

5000

time (sec)

link a

ngle

(rad)

flexible: k=1, c=0

Figure 18.5: Rigid and flexible case with colocated rate feedback

Chapter 19

Input output systems

19.1 Input-output systems

To describe an input-output system, we need several ingredients. First we have the time setT . Usually

T = [0,∞)

for continuous-time systems orT = { 0, 1, 2, . . . , }

for discrete-time systems. We have the set U of input signals u : T → U where U is the setof input values. Common examples are U = IRm and U is the set of piecewise continuousfunctions u : [0,∞) → IRm. We also have the set Y of output signals y : T → Y where Uis the set of output values. Common examples are Y = IRo and Y is the set of piecewisecontinuous functions y : [0,∞) → IRo.

By saying that a function s : [0,∞) → IRn is piecewise continuous we will mean that on everyfinite interval [0, T ], the function is continuous except (possibly) at isolated points.

Example 176 Two examples of piecewise continuous functions.

(i) Square wave function.

s(t) =

1 if N ≤ t < N + 12

−1 if N + 12≤ t ≤ N + 1

N = 0, 1, 2, . . .

(ii)

s(t) =

{

(1− t)−1 if 0 ≤ t < 10 if 1 ≤ t

An input-output system is a map

G : U → Y

251

252 CHAPTER 19. INPUT OUTPUT SYSTEMS

Figure 19.1: Input-output system

that is, for each input signal u ∈ U the system produces an output signal y = G(u) withy ∈ Y .

Example 177 Consider a linear time-invariant state space system described by

x = Ax+Bu

y = Cx+Du

x(0) = x0

where x0 ∈ IRn is fixed and t ∈ [0,∞), u(t) ∈ IRm, y(t) ∈ IRo. For each piecewise contin-uous input signal u, this system produces a unique piecewise continuous output signal y,specifically

y(t) = CeAtx0 +

∫ t

0

CeA(t−τ)Bu(τ) dτ +Du(t) for t ≥ 0 .

Example 178 Consider a discrete-time linear time-invariant system described by

x(k + 1) = Ax(k) +Bu(k)

y(k) = Cx(k) +Du(k)

x(0) = x0

where k ∈ { 0, 1, 2, . . . , }, u(k) ∈ IRm, y(k) ∈ IRo, and x0 ∈ IRn is fixed.

y(k) = CAkx0 +k−1∑

j=0

CA(k−1−j)Bu(j) +Du(k)

Example 179 Consider a linear time-invariant system described by a delay-differential equa-tion:

x(t) = A0x(t) + A1x(t−h) +Bu(t)

y = Cx(t) +Du(t)

x(t) = x0(t) for 0 ≤ t ≤ h

where the continuous function x0(·) is fixed and t ∈ [0,∞), u(t) ∈ IRm, y(t) ∈ IRo. For eachpiecewise continuous input signal u, this system produces a unique piecewise continuousoutput signal y,

19.2. SIGNAL NORMS 253

19.1.1 Causality

Basically, an input-output system is causal if the current value of the output depends onlyon the previous values of the input. Mathematically, we express this as follows. A system Gis causal, if

u1(t) = u2(t) for t ≤ T

impliesG(u1)(t) = G(u2)(t) for t ≤ T .

The systems in the above examples are causal. As examples of noncausal systems, con-sider

y(t) = u(t+ 1)

y(t) =

∫ ∞

t

u(τ) dτ

19.2 Signal norms

To talk about the size of signals, we introduce signal norms. Consider any real scalar p ≥ 1.We say that a piecewise continuous signal s : [0,∞) → IRn is an Lp signal if

∫ ∞

0

||s(t)||p dt <∞

where ||s(t)|| is the Euclidean norm of the n-vector s(t). If S is any linear space of Lp signals,then the scalar valued function || · ||p defined by

||s||p :=(∫ ∞

0

||s(t)||p dt)

1p

is a norm on this space. We call this the p-norm of s. Common choices of p are one and two.For p = 2 we have

||s||2 =(∫ ∞

0

||s(t)||2 dt)

12

.

This is sometimes called the rms (root mean square) value of the signal. For p = 1,

||s||1 =∫ ∞

0

‖s(t)‖ dt .

We say that s is an L∞ signal if

ess supt≥0

||s(t)|| <∞ .

By ess sup we mean that supremum is taken over the set of times t where s is continuous. IfS is a any linear space of L∞ signals, then the scalar valued function || · ||∞ defined by

||s||∞ := ess supt≥0

||s(t)||


is a norm on this space. We call this the ∞-norm of s. Also ||s(t)|| ≤ ||s||∞ a.e (almosteverywhere) that is, everywhere except(possibly) where s is not continuous.

Example 180 Lp or not Lp

(i) Lp for all p.s(t) = e−αt with α > 0 .

Since |e−αt| ≤ 1 for all t ≥ 0, this signal is an L∞ signal; also ‖s‖∞ = 1. For 1 ≤ p <∞,we have

∫∞0

|e−αt|p dt = 1/αp. Hence, this is an Lp signal with ‖s‖p = (1/αp)1/p.

(ii) Not Lp for all p.s(t) = et

(iii) L∞ but not Lp for any other p.s(t) ≡ 1

(iv) Not L∞ but Lp for any other p

s(t) =

{

(1− t)−12 for 0 ≤ t < 1

0 for 1 ≤ t

(v) Not L1 but Lp for any p > 1s(t) = (1 + t)−1

For 1 ≤ p < ∞, we say that a piecewise continuous signal s is an extended Lp signal oran Lp

e signal if∫ T

0

||s(t)||p dt <∞

for all T ≥ 0. A piecewise continuous signal s is an extended L∞ signal or an L∞e signal if

ess sup0≤t≤T

||s(t)|| <∞

for all T ≥ 0. Clearly, if s is Lp, then it is Lpe.

Example 181 Lpe or not Lp

e

(i)

s(t) = et

Lpe for all p.

(iv)

s(t) =

{

(1− t)−12 for 0 ≤ t < 1

0 for 1 ≤ t

Not L∞e but Lp for any other p

Note that if a signal is L∞e then it is Lp

e for all p.

19.3. INPUT-OUTPUT STABILITY 255

19.3 Input-output stability

Suppose U is a set of Lpe signals u : [0,∞) → IRm and Y is a set of Lp

e signals y : [0,∞) → IRo.Consider an input-output system

G : U → Y .

DEFN. The system G is Lp stable if it has the following properties.

(i) Whenever u is an Lp signal, the output y = G(u) is an Lp signal.

(ii) There are scalars β and γ such that, for every Lp signal u ∈ U , we have

||G(u)||p ≤ β + γ||u||p (19.1)

The Lp gain of a system is the infimum of all γ such that there is a β which guarantees (19.1)for all Lp signals u ∈ U .

Example 182 Consider the memoryless nonlinear SISO system defined by

y(t) = sin(u(t)) + 1

Since

|y(t)| = 1 + | sin(u(t))| ≤ 1 + |u(t)|

it follows that for any 1 ≤ p ≤ ∞ and any Lp signal u we have

||y||p ≤ 1 + ||u||p

Hence this system is Lp stable with gain 1 for 1 ≤ p ≤ ∞.

Example 183 Consider the simple delay system with time delay h ≥ 0:

y(t) =

{

0 for 0 ≤ t ≤ hu(t− h) for h ≤ t

One can readily show that for any 1 ≤ p ≤ ∞ and any Lp signal u we have

||y||p = ||u||p

Hence this system is Lp stable for 1 ≤ p ≤ ∞.

Example 184 (Simple integrator is not io stable.)

x = u

y = x

x(0) = 0


For 1 ≤ p <∞ consider

u(t) =

{

1 for 0 ≤ t ≤ 10 for 1 ≤ t

Then

y(t) =

{

t for 0 ≤ t ≤ 11 for 1 ≤ t

Since u is an Lp signal and y is not, this system is not Lp stable.For p = ∞ consider

u(t) ≡ 1

Theny(t) = t

Since u is an L∞ signal and y is not, this system is not L∞ stable.

19.4 General convolution systems

Suppose U is a set of Lpe signals u : [0,∞) → IRm and Y is a set of Lp

e signals y : [0,∞) → IRo.Consider the input-output system defined by the convolution integral

y(t) =

∫ t

0

H(t−τ)u(τ) dτ (19.2)

where H : [0,∞) → IRo×m is a piecewise continuous matrix valued function of time. H iscalled the impulse response of the system.

As a specific example, consider a finite-dimensional linear system described by

x = Ax+Bu

y = Cx

x(0) = 0

Then y(t) =∫ t

0CeA(t−τ)Bu(τ) dτ . Hence this is a convolution system with

H(t) = CeAtB .

19.4.1 Systems with L1 impulse responses

Consider a convolution system with impulse response H . We say that H is L1 if∫ ∞

0

||H(t)|| dt <∞

where||H(t)|| = σmax[H(t)]

and we define

||H||1 :=∫ ∞

0

||H(t)|| dt .

19.4. GENERAL CONVOLUTION SYSTEMS 257

Example 185 Consider the SISO system

x = −αx+ u

y = x

x(0) = 0

with α > 0. Here

y(t) =

∫ t

0

H(t−τ)u(τ) dτ

with

H(t) = e−αt

For any T > 0,∫ T

0

|H(t)| dt = 1

α(1− e−αT )

Hence H is L1 and

||H||1 =1

α

Remark 9 Suppose there are constants α > 0 and β ≥ 0 such that

‖H(t)‖ ≤ βe−αt (19.3)

for all t ≥ 0. Then H is L1.


Remark 10 Consider

x = Ax+Bu

y = Cx

where A is Hurwitz. Then H(t) = CeAtB is L1. To see this, first choose any α > 0 satisfying

α < α := −max {ℜ(λ) : λ is an eigenvalue of A} . (19.4)

Then there exists β > 0 such that ‖H(t)‖ ≤ βe−αt.

Theorem 48 Consider a convolution system (19.2) where the impulse response H is L1.For every p ∈ [1,∞], this system is Lp stable and for every Lp signal u ∈ U ,

||y||p ≤ ||H||1||u||p

Remark 11 The above result is a nice general result. However, except for p = ∞, it usuallyyields a conservative estimate for the Lp gain of a convolution system. Below we present anexact characterization of the L2 norm of a convolution system.

Remark 12 The main problem in applying the above result is in calculating ‖H‖1.

Before proving Theorem 48, we need the following result.

Holders inequality. Suppose f, g : [0,∞) → IR are piecewise continuous and considerany positive real scalars p and q which satisfy

1

p+

1

q= 1

Then, for all T > 0,

∫ T

0

|f(t)g(t)| dt ≤(∫ T

0

|f(t)|p dt)

1p(∫ T

0

|g(t)|q dt)

1q

.


Proof of Theorem 48. First note that, for any input signal u and any time t, we have

||y(t)|| =

∥

∥

∥

∥

∫ t

0

H(t−τ)u(τ) dt∥

∥

∥

∥

≤∫ t

0

||H(t−τ)u(τ)|| dt

≤∫ t

0

||H(t−τ)|| ||u(τ)|| dt .

Case : p = ∞. Suppose u is L∞, then ||u(τ)|| ≤ ||u||∞ a.e. and

||y(t)|| ≤∫ t

0

||H(t−τ)|| ||u||∞ dt

≤∫ t

0

||H(t−τ)|| dτ ||u||∞

=

∫ t

0

||H(τ)|| dτ ||u||∞

≤∫ ∞

0

||H(τ)|| dτ ||u||∞= ||H||1||u||∞ .

Hence y is L∞ and

||y||∞ ≤ ||H||1||u||∞ .

Case : p <∞. Suppose u is Lp. Choose q such that

1

p+

1

q= 1 .

Then, using Holders inequality, we obtain

||y(t)|| ≤∫ t

0

||H(t−τ)||1/q||H(t−τ)||1/p||u(τ)|| dt

≤(∫ t

0

||H(t−τ)||dτ)1/q (∫ t

0

||H(t−τ)||||u(τ)||p dt)1/p

≤ ||H||1/q1

(∫ t

0

||H(t−τ)||||u(τ)||p dt)1/p

.

Hence, for any T > 0, we have

∫ T

0

||y(t)||p dt ≤ ||H||p/q1

∫ T

0

∫ t

0

||H(t−τ)||||u(τ)||p dτdt .


By changing the order of integration, we can write

∫ T

0

∫ t

0

||H(t−τ)||||u(τ)||p dτdt =

∫ T

0

∫ T

τ

||H(t−τ)||||u(τ)||p dtdτ

=

∫ T

0

∫ T

τ

||H(t−τ)|| dt ||u(τ)||p dτ

≤∫ T

0

||H||1||u(τ)||p dτ

= ||H||1∫ T

0

||u(τ)||p dτ .

Since 1 + p/q = p, we now obtain

∫ T

0

||y(t)||p ≤ ||H||p1∫ T

0

||u(τ)||p dτ .

If u is Lp then∫∞0

||u(τ)||p is finite and

∫ T

0

||y(t)||p dτ ≤ ||H||p1∫ ∞

0

||u(τ)||p dτ ;

hence y is Lp and||y||p ≤ ||H||1||u||p .

Remark 13 Here we provide a way of obtaining an upper bound on ‖H‖1 for single outputsystems using a Lyapunov equation. Consider a scalar output system described by

x = Ax+Bu

y = Cx

where A is Hurwitz, that is, all eigenvalues of A have negative real parts. Consider any αsatisfying (19.4) and let S be the solution to

AS + SA′ + 2αS +BB′ = 0 . (19.5)

Then

‖H‖1 ≤(

1

2αCSC ′

)12

(19.6)

To see this first note that the above Lyapunov equation can be written as

(A+ αI)S + S(A+ αI)′ +BB′ = 0

Since A+ αI is Hurwitz, S is uniquely given by

S =

∫ ∞

0

e(A+αI)tBB′e(A+αI)′t dt =

∫ ∞

0

e2αteAtBB′eAt dt


Hence

CSC ′ =

∫ ∞

0

e2αtCeAtBB′eAtC dt =

∫ ∞

0

e2αtH(t)H(t)′ dt =

∫ ∞

0

e2αt‖H(t)‖2 dt .

We now use Holders inequality to obtain:

∫ ∞

0

‖H(t)‖ dt =

∫ ∞

0

e−αteαt‖H(t)‖dt

≤(∫ ∞

0

(e−αt)2dt

)12(∫ ∞

0

(eαt‖H(t)‖)2dt)

12

=

(∫ ∞

0

e−2αtdt

)12(∫ ∞

0

e2αt‖H(t)‖2dt)

12

=

(

1

2αCSC ′

)12

.

This implies that

‖H‖1 ≤(

1

2αCSC ′

)12

.

19.4.2 Systems with impulses in impulse response

H(t) = H0(t) + δ(t−τ1)H1 + δ(t−τ2)H2 + · · · (19.7)

where H0 is piecewise continuous, the matrices H1, H2, . . . are constant matrices, δ is theunit impulse function and τ1, τ2, . . . ≥ 0 are constants.

Example 186

x = Ax+ Bu

y = Cx+Du

HereH(t) = CeAtB + δ(t)D .

Note that when H is given by (19.7),

y(t) =

∫ t

0

H0(t−τ)u(τ) dτ +H1u(t−τ1) +H2u(t−τ2) + · · ·

Here we generalize the L1 norm to

‖H‖A = ‖H0‖1 + ‖H1‖+ ‖H2‖+ · · ·

Also, whenever ‖H‖A is bounded, we have

‖y‖p ≤ ‖H‖A‖u‖p . (19.8)


19.4.3 L2 gain and H∞ norm

Consider a convolution system described by (19.2) and suppose the impulse response H ispiecewise continuous. The Laplace Transform of H is a matrix valued function of a complexvariable s and is given by

H(s) =

∫ ∞

0

H(t)e−stdt .

for all complex s where the integral is defined. If u and y are the Laplace transforms of uand y, respectively, then the convolution system can be described by

y(s) = H(s)u(s) (19.9)

and H is called the transfer function of the system.Suppose H has no poles in the closed right half of the complex plane. If H is L1, we have

||H(s)|| ≤ ||H||1

for all s with ℜ(s) ≥ 0.

H∞ norm of a transfer function. Consider a transfer function H : C → Cm×o. SupposeH is analytic in the open right half complex plane. Then, the H∞-norm of H is defined by

||H||∞ = supω∈IR

||H(ω)||

where||H(ω)|| = σmax[H(ω)] .

Note that if H is L1 then||H||∞ ≤ ||H||1

Theorem 49 Consider a convolution system (19.2) where the impulse response H is piece-wise continuous and let H be the Laplace transform of H. Then, this system is L2 stable ifand only if H is defined for all s with ℜ(s) > 0 and ||H||∞ is finite. In this case, wheneveru is L2,

||y||2 ≤ ||H||∞||u||2 (19.10)

Actually, one can show that ||H||∞ is the L2 gain of the system in the sense that

‖H‖∞ = supu∈L2, u 6=0

‖y‖2‖u‖2

Example 187 Consider the SISO system,

x = −αx+ u

y = x

x(0) = 0


with α > 0. Here

H(s) =1

s+ α

Since H(jω) = 1/(ω + α), it follows that |H(ω)| = 1/(ω2 + α2)12 . Hence

‖H‖∞ =1

α.

Thus, the L2 gain of this system is 1/α.

19.4.4 Finite dimensional linear systems

x = Ax+Bu (19.11a)

y = Cx+Du (19.11b)

x(0) = x0 (19.11c)

Here,

y(t) = CeAtx0 +

∫ t

0

CeA(t−τ)Bu(τ) dτ +Du(t) .

The transfer function of this system is given by

H = C(sI −A)−1B +D

Theorem 50 Suppose (A,B) is stabilizable and (C,A) is detectable. Then the followingstatements are equivalent.

(i) All the eigenvalues of the matrix A have negative real parts.

(ii) All the poles of the transfer function H have negative real parts.

(iii) The input-output system (19.11) is Lp stable for some p ∈ [1,∞].

(iv) The input-output system (19.11) is Lp stable for all p ∈ [1,∞].

LMI characterization of L2 gain

Here we provide an upper bound on the L2 gain of a finite-dimensional linear time-invariantsystem described by (19.11).

Suppose there is a symmetric matrix P ≥ 0 and a scalar γ ≥ 0 such that(

PA+ A′P + C ′C PB + C ′DB′P +D′C −γ2I +D′D

)

≤ 0 . (19.12)

Then,∫ T

0

‖y(t)‖2 dt ≤ x′0Px0 + γ2∫ T

0

‖u(t)‖2 dt

for all T ≥ 0. Hence, whenever u is L2 then so is y and

‖y‖2 ≤ β + γ‖u‖2 where β =√

x′0Px0


Remark 14 Inequality (19.12) is an LMI in the variables P and γ2. The scalar γ is anupper bound on the L2 gain of the system.

Proof. Consider any solution x(·) of (19.11) and let v(t) = x(t)′Px(t). Then v0 := v(0) =x′0Px0 and

v = 2x′P x = 2x′PAx+ 2x′PBu .

It now follows from the matrix inequality (19.12) that

x′(PA+ A′P + CC ′)x+ x′(PB + C ′D)u+ u′(B′P +D′C)x+ u′(−γ2I +DD′)u ≤ 0 ,

that is,2x′PAx+ 2x′PBu+ ‖y‖2 − γ2‖u‖2 ≤ 0 .

Recalling the definition of v and the expression for v, we now obtain that

v(t) + ‖y‖2 − γ2‖u‖2 ≤ 0

for all t ≥ 0. Hence, for any T ≥ 0,

v(t)− v(0) +

∫ T

0

‖y(t)‖2 dt− γ2∫ T

0

‖u(t)‖ dt ≤ 0 .

Since v(t) ≥ 0 and v(0) = x′0Px0, we obtain that

∫ T

0

‖y(t)‖2 dt ≤ x′0Px0 + γ2∫ T

0

‖u(t)‖ dt


LMI characterization of L∞ gain

Here we provide an upper bound on the L∞ gain of a finite-dimensional linear time-invariantsystem described by (19.11).

Suppose there is a symmetric matrix P ≥ 0 and scalars α > 0, µ1, µ2 ≥ 0 such that(

PA+ A′P + 2αP PBB′P −2αµ1I

)

≤ 0 (19.13)

and(

−P + C ′C C ′DD′C −µ2I +D′D

)

≤ 0 (19.14)

and letγ = (µ1 + µ2)

12 .

Then, for all t ≥ 0,

‖y(t)‖ ≤ βe−αt + γ‖u‖∞ (19.15)

where β = (x′0Px0)12 . Hence, whenever u is L∞ then so is y and

‖y‖∞ ≤ β + γ‖u‖∞ (19.16)

Remark 15 For a fixed α, inequalities (19.13) and (19.14) are LMIs in the variables P , µ1

and µ2. The scalar γ is an upper bound on the L∞ gain of the system.

Proof. Consider any solution x(·) of (19.11) and let v(t) = x(t)′Px(t). Then v0 := v(0) =x′0Px0 and

v = 2x′P x = 2x′PAx+ 2x′PBu .

It now follows from the matrix inequality (19.13) that

x′(PA+ A′P + 2αP )x+ x′PBu+ u′B′Px− 2αµ1u′u ≤ 0 ,

that is,2x′PAx+ 2x′PBu ≤ −2αx′Px+ 2αµ1‖u‖2 .

Recalling the definition of v and the expression for v, we now obtain that

v(t) ≤ −2αv(t) + 2αµ1‖u(t)‖2

for all t ≥ 0. Suppose now u(·) is L∞. Then ‖u(t)‖ ≤ ‖u‖∞ for all t ≥ 0 and

v(t) ≤ −2αv(t) + 2αµ1‖u‖2∞ .

for all t ≥ 0. It now follows that

v(t) ≤ v(0)e−2αt +

∫ t

0

e−2α(t−τ)2αµ1‖u‖2∞ dτ

= v0e−2αt + µ1‖u‖2∞(1− e−2αt)

≤ v0e−2αt + µ1‖u‖2∞ .


Thus,v(t) ≤ v0e

−2αt + µ1‖u‖2∞ . (19.17)

Recalling the matrix inequality (19.14) we obtain that

x′(−P + C ′C)x+ x′CDu+ u′D′Cx+ u′(−µ2I +D′D)u ≤ 0 ,

that is,‖y‖2 = x′CCx+ 2x′CDu+ u′DDu ≤ x′Px+ µ2‖u‖2 .

Hence, for all t ≥ 0,

‖y(t)‖2 ≤ v(t) + µ2‖u(t)‖2 ≤ v0 e−2αt + µ1‖u‖2∞ + µ2‖u‖2∞ ,

that is,‖y(t)‖2 ≤ v0 e

−2αt + γ2‖u‖2∞ .

Taking the square root of both sides on this inequality yields

‖y(t)‖ ≤ β e−αt + γ‖u‖∞

where β = v120 = (x′0Px0)

12 . This is the desired result.


19.4.5 Linear differential delay systems

Consider a system with a delay described by

x(t) = A0x(t) + A1x(t−h) +Bu(t) (19.18a)

y(t) = Cx(t) (19.18b)

x(t) = 0 − h ≤ t ≤ 0 (19.18c)

Suppose Φ is the unique solution to

Φ(t) = A0Φ(t) + A1Φ(t−h)Φ(0) = I

Φ(t) = 0 for − h ≤ t < 0 .

Then,

y(t) =

∫ t

0

H(t−τ)u(τ) dτ where H(t) = CΦ(t)B . (19.19)

Proof: Consider

x(t) =

∫ t

0

Φ(t−τ)Bu(τ) dτ

for t ≥ −h. We will show that this is the unique solution to

x(t) = A0x(t) + A1x(t−h) +Bu(t) (19.20)

x(t) = 0 − h ≤ t ≤ 0 (19.21)

Clearly x(t) = 0 for −h ≤ t ≤ 0 . Now note that for t ≥ 0,

x(t−h) =∫ t−h

0

Φ(t−h−τ)Bu(τ) dτ =

∫ t

0

Φ(t−h−τ)Bu(τ) dτ

Now differentiate x to obtain

x(t) = Φ(0)Bu(t) +

∫ t

0

Φ(t−τ)Bu(τ) dτ

= Bu(t) +

∫ t

0

(A0Φ(t−τ) + A1Φ(t−τ−h))Bu(τ) dτ

= Bu(t) + A0

∫ t

0

Φ(t−τ)Bu(τ) dτ + A1

∫ t

0

Φ(t−τ−h)Bu(τ) dτ

= A0x(t) + A1x(t−h) +Bu(t) ,

that is, x satisfies (19.20). Thus

y(t) = Cx(t) = C

∫ t

0

Φ(t−τ)Bu(τ) dτ =

∫ t

0

H(t−τ)Bu(τ) dτ

where H(t) = CΦ(t)B.


Example 188 Consider a system with delay h described by

x(t) = −ax(t) − bx(t−h) + u

y(t) = x(t)

x(t) = 0 for − h ≤ t ≤ 0

with|b| < a , h ≥ 0

The transfer function of this system is given by

H(s) =1

d(s)where d(s) = s+ a + be−hs

We first show that H is analytic for ℜ(s) ≥ 0. Suppose λ = α+ ω is a pole of H ; then λ isa zero of d, that is,

α + ω + a+ be−h(α+ω) = 0

Hence,α = −a− be−αh cos(ωh) ≤ −a + |b|e−αh ≤ −a + |b| < 0 .

We now obtain an upper bound on the H∞ norm of H.

19.5. A CLASS OF LP STABLE NONLINEAR SYSTEMS 269

19.5 A class of Lp stable nonlinear systems

Consider an input-output system described by

x = F (x, u)

y = H(x, u)

x(0) = x0

We make the following assumptions.There is a continuously differentiable function V and a positive scalars β1, β2, β3 and α

such that

β1‖x||2 ≤ V (x) ≤ β2||x||2

DV (x)F (x, 0) ≤ −2αV (x)

||DV (x)|| ≤ β3||x||

There are non-negative scalars β4, β5, and β6 such that

||F (x, u)− F (x, 0)|| ≤ β4||u||

||H(x, u)|| ≤ β5||x||+ β6||u|| .

Claim: The above system is Lp stable for all p ∈ [1,∞].

Example 189

x = −x− x3 + u

y = sin x

Proof of claim:


19.6 Small gain theorem

19.6.1 Stability of feedback systems

Figure 19.2: A Feedback system

Figure 19.3: Feedback system for stability definition

19.6.2 Truncations

Suppose s : T → IRn is a signal and consider any time T . The corresponding truncation of sis defined by

sT (t) :=

{

s(t) for 0 ≤ t < T0 for T ≤ t

Note that a system G is causal if and only if for every input u and every time T , we have

G(u)T = G(uT )T .

19.6. SMALL GAIN THEOREM 271

For any Lp signal s, we have ‖sT‖p ≤ ‖s‖p. A signal s is Lpe if and only if sT is Lp for all

T . Also, an Lpe signal s is an Lp signal if there exists a constant β such that ‖sT‖p ≤ β for

all T .

19.6.3 Small gain theorem

Suppose U1 and U2 are vector spaces of Lpe signals with the property that if they contain

a signal, they also contain every truncation of the signal. Suppose G1 : U1 → U2 andG2 : U2 → U1 are causal Lp-stable systems which satisfy

||G1(u1)||p ≤ β1 + γ1||u1||p||G2(u2)||p ≤ β2 + γ2||u2||p (19.22)

Consider the feedback system described by

y1 = G1(u1)y2 = G2(u2)u1 = r1 + y2u2 = r2 + y1

(19.23)

We assume that for each pair (r1, r2) of signals with r1 ∈ U1 and r2 ∈ U2, the aboverelations uniquely define a pair (y1, y2) of outputs with y1 ∈ U2 and y2 ∈ U1.

Theorem 51 (Small gain theorem) Consider a feedback system satisfying the above con-ditions and suppose

γ1γ2 < 1

Then, whenever r1 and r2 are Lp signals, the signals u1 u2, y1, y2 are Lp and

||u1||p ≤ (1− γ1γ2)−1(β2 + γ2β1 + ||r1||p + γ2||r2||p)

||u2||p ≤ (1− γ1γ2)−1(β1 + γ1β2 + ||r2||p + γ1||r1||p)

Hence,

||y1||p ≤ β1 + γ1(1− γ1γ2)−1(β2 + γ2β1 + ||r1||p + γ2||r2||p)

||y2||p ≤ β2 + γ2(1− γ1γ2)−1(β1 + γ1β2 + ||r2||p + γ1||r1||p)

and the feedback system is Lp stable.

Proof. Suppose r1 and r2 are Lp signals. Since r1, r2 and y1, y2 are Lpe signals, it follows

that u1 and u2 are Lpe signals. Consider any time T > 0. Since all of the signals r1, r2, u1,

u2, y1, y2 are Lpe, it follows that their respective truncations r1T , r2T , u1T , u2T , y1T , y2T are

Lp signals. Since G1 is causal,

y1T = G1(u1)T = G1(u1T )T

Hence,‖y1T‖p = ‖G1(u1T )T‖p ≤ ‖G1(u1T )‖p ≤ β1 + γ1‖u1T‖p


Since u2T = r2T + y1T , it now follows that

‖u2T‖p ≤ ‖r2T‖p + ‖y1T‖p≤ β1 + ‖r2T‖p + γ1‖u1T‖p

In a similar fashion we can show that

‖u1T‖p ≤ β2 + ‖r1T‖p + γ2‖u2T‖p

Hence,‖u1T‖p ≤ β2 + ‖r1T ‖p + γ2 (β1 + ‖r2T‖p + γ1‖u1T‖p)

Rearranging yields

‖u1T‖p ≤ (1− γ1γ2)−1 (β2 + γ2β1 + ‖r1T‖p + γ2‖r2T‖p)

≤ (1− γ1γ2)−1 (β2 + γ2β1 + ‖r1‖p + γ2‖r2‖p)

Since the above inequality holds for all T > 0, it follows that u1 is an Lp signal and

‖u1‖p ≤ (1− γ1γ2)−1 (β2 + γ2β1 + ‖r1‖p + γ2‖r2‖p)

In a similar fashion we can show that

‖u2‖p ≤ (1− γ1γ2)−1 (β1 + γ1β2 + ‖r2‖p + γ1‖r1‖p)

19.7 Application of the small gain theorem: circle cri-

terion

Figure 19.4: A Feedback system

Consider a nonlinear system which is a feedback combination of a convolution system

y(t) =

∫ t

0

H(t−τ)u(τ) dτ

and a memoryless nonlinear system

u(t) = φ(t, y(t)) .

19.7. APPLICATION OF THE SMALL GAIN THEOREM: CIRCLE CRITERION 273

Suppose that the impulse response H of the convolution system is L1 and, for some γ ≥ 0,the nonlinearity satisfies

‖φ(t, y)‖ ≤ γ‖y‖ (19.24)

for all t and y.Since the impulse response H of the convolution system is L1, the L2 gain of the con-

volution system is ‖H‖∞ where H is the system transfer function. We now claim that thenonlinear system is L2 stable with “gain” γ. To see this, consider u(t) = φ(t, y(t)) and notethat, for any T > 0,

∫ T

0

‖u(t)‖2 dt =∫ T

0

‖φ(t, y(t))‖2 dt ≤∫ T

0

(γ‖y(t)‖)2 dt = γ2∫ T

0

‖y(t)‖2 dt .

If y is L2 then∫ T

0

‖u(t)‖2 dt ≤ γ2∫ ∞

0

‖y(t)‖2 dt

for all T ≥ 0; hence∫ ∞

0

‖u(t)‖2 dt ≤ γ2∫ ∞

0

‖y(t)‖2 dt

This implies that u is L2 and ‖u‖2 ≤ γ‖y‖2. Hence the nonlinear system is L2 stable with“gain” γ. From the small gain theorem we obtain that the feedback system is L2 stable if

γ‖H‖∞ < 1 (19.25)

For a SISO system, the above condition is equivalent to the requirement that the Nyquistplot of the system transfer function lie within the circle of radius 1/γ centered at the origin.As a consequence, the above condition is sometimes referred to as the circle criterion.

Example 190 Considerx = −ax(t) + b sin(x(t−h))

where a > |b|. Letting y(t) = x(t−h) this system can be described by

x(t) = −ax(t) + u(t)

y(t) = x(t−h)u(t) = b sin(y(t))

HereH(s) = e−sh/(s+ a) and φ(t, y) = b sin(y) .

Thus,|H(ω)|2 = |eω|2/(ω2 + a2) = 1/(ω2 + a2) ;

so, ‖H‖∞ = 1/a. Also,|b sin(y)| ≤ |b||y| ;

hence the condition (19.24) is satisfied with γ = |b|. Since a > |b|, we obtain that

γ‖H‖∞ = |b|/a < 1.


Figure 19.5: Another feedback system

19.7.1 A more general circle criterion

Consider a nonlinear system which is a feedback combination of a SISO convolution system

y(t) =

∫ t

0

H(t−τ)u(τ) dτ

and a SISO memoryless nonlinear system

u(t) = −φ(t, y(t)) .

Suppose that the impulse response H of the convolution system is L1 and, for some scalarsa and b, the nonlinearity satisfies the sector condition:

ay2 ≤ yφ(t, y) ≤ by2 (19.26)

for all t and y.We will convert this system into one considered in the previous section. To this end,

introduce a new input

w = u+ ky where k =a+ b

2.

Thenw = φ(t, y) (19.27)

where φ(t, y) := −φ(t, y) + ky and, it follows from (19.26) that

−γy2 ≤ yφ(t, y) ≤ γy2 where γ =b− a

2

Thus,|φ(t, y)| ≤ γ|y| (19.28)

Since y = Hu and w = u+ ky, we obtain that

y = H(w − ky)

Hencey = Gw (19.29)

whereG = (1 + kH)−1H .

19.7. APPLICATION OF THE SMALL GAIN THEOREM: CIRCLE CRITERION 275

Thus the systems under consideration here can be described by (19.29) and (19.27) whereφ satisfies (19.28). Using the circle criterion of the last section, this system is L2 stableprovided all the poles of G have negative real parts and

γ‖G‖∞ < 1 (19.30)

This last condition can be expressed as

γ2G(ω)′G(ω) < 1 .

for all ω ∈ IR ∪ {∞}, that is,

γ2[

[1 + kH(ω)]−1H(ω)]′ [

[1 + kH(ω)]−1H(ω)]

< 1

for all ω ∈ IR ∪ {∞}, Recalling the definitions of γ and k, the above inequality can berewritten as

−abH(ω)′H(ω)− k[H(ω) + H(ω)′] < 1 (19.31)

We have three cases to consider:

Case 1: ab > 0 In this case, inequality (19.31) can be rewritten as

[H(ω) + c]′[H(ω) + c] > R2

where

c =1

2

(

1

a+

1

b

)

, R =1

2

∣

∣

∣

∣

1

a− 1

b

∣

∣

∣

∣

This is equivalent to|H(ω) + c| > R .

that is, the Nyquist plot of H must lie outside the circle of radius R centered at −c. Notethat this is the circle which intersects the real axis at −1/a and −1/b.

Figure 19.6: Circle criterion for 0 < a < b

Case 2: ab < 0: In this case (19.31) can be rewritten as

[H(ω) + c]′[H(ω) + c] < R2

This is equivalent to|H(ω) + c| < R .

that is, the Nyquist plot of H must lie inside the circle of radius R centered at −c. Again,this is the circle which intersects the real axis at −1/a and −1/b.


Figure 19.7: Circle criterion for a < 0 < b

Case 3: ab = 0: Suppose a = 0 and b > 0. In this case k = b/2 and (19.31) can berewritten as

−(b/2)(H(ω) + H(ω)′) < 1

that is,

ℜ(

H(ω))

> −1

b(19.32)

This means that the Nyquist plot of H must lie to the right of the vertical line whichintersects the real axis at −1/b.

Figure 19.8: “Circle” criterion for a = 0 and 0 < b

References

1. Desoer, C.A. and Vidyasagar, M., “Feedback Systems: Input-Output Properties,” Aca-demic Press, 1975.

Chapter 20

Nonlinear H∞

20.1 Analysis

Consider a system,

x = F (x) +G(x)w

z = H(x)

with initial conditionx(0) = x0

where w : [0,∞) → IRm is a disturbance input and z is a performance output with z(t) ∈ IRp.

Desired performance. We wish that the system has the following property for someγ > 0. For every initial state x0 there is a real number β0 such that for every disturbanceinput w one has

∫ ∞

0

||z(t)||2 dt ≤ γ2∫ ∞

0

||w||2 dt+ β0

20.1.1 The HJ Inequality

Theorem 52 Suppose there is a function V which for all x, satisfies

DV (x)F (x) +1

4γ2DV (x)G(x)G(x)TDV T (x) +H(x)TH(x) ≤ 0 (20.1)

V (x) ≥ 0

for some γ > 0. Then for every initial state x0 and for every disturbance input w one has∫ ∞

0

||z(t)||2 dt ≤ γ2∫ ∞

0

||w||2 dt+ V (x0) (20.2)

277

278 CHAPTER 20. NONLINEAR H∞

Proof. Consider any initial state x0 and any disturbance input w. Along any resultingsolution x(·), we have

dV (x(t))

dt= L(x(t), w(t))

where

L(x, w) = DV (x)F (x) +DV (x)G(x)w

= DV (x)F (x) +1

4γ2||G(x)TDV (x)T ||2 + γ2||w||2 − ||γw − 1

2γG(x)TDV (x)T ||2

≤ DV (x)F (x) +1

4γ2||G(x)TDV (x)T ||2 + γ2||w||2

Using inequality (20.1) we have:

DV (x)F (x) +1

4γ2||G(x)TDV (x)T ||2 ≤ −||H(x)||2

HenceL(x, w) ≤ −||H(x)||2 + γ2||w||2

i.e.,dV (x(t))

dt≤ −||z(t)||2 + γ2||w(t)||2

Integrating from 0 to T yields:

V (x(T ))− V (x0) ≤ −∫ T

0

||z(t)||2dt+ γ2∫ T

0

||w(t)||2 dt (20.3)

Rearranging terms and recalling that V (x(T )) ≥ 0, we obtain

∫ T

0

||z(t)||2dt ≤ γ2∫ T

0

||w(t)||2 dt+ V (x0)− V (x(T ))

≤ γ2∫ T

0

||w(t)||2 dt+ V (x0)

If w is not L2, then∫∞0

||w(t)||2 dt = ∞ and (20.2) trivially holds. If w is L2, the integral∫∞0

||w(t)||2 dt is finite, hence∫∞0

||z(t)||2dt is finite and satisfies (20.2).

Hamilton Jacobi equation

DV (x)F (x) +1

4γ2DV (x)G(x)G(x)TDV T (x) +H(x)TH(x) = 0 (20.4)

The “worst case” w is given in feedback form as

w = w∗(x) =1

2γ2G(x)TDV (x)T

20.1. ANALYSIS 279

Quadratic V . SupposeV (x) = xTPx

where P is a symmetric real matrix. Then DV (x) = 2xTP and the HJ inequality is

2xTPF (x) + γ−2xTPG(x)G(x)TPx+H(x)TH(x) ≤ 0

LTI system. Suppose

x = Ax+Bw

z = Cx

Then the HJ is

DV (x)Ax+1

4γ2DV (x)BBTDV (x) + xTCTCx ≤ 0

If V (x) = xTPx, this results in

2xTPAx+ γ−2xTPBBTPx+ xTCTCx ≤ 0

This is satisfied for all x iff P satisfies

PA+ ATP + γ−2PBBTP + CTC ≤ 0


20.2 Control

x = F (x) +G1(x)w +G2(x)u

z =

[

H(x)u

]

with initial conditionx(0) = x0

Desired performance. We wish to choose a state feedback controller

u = k(x)

such that the resulting closed loop system

x = F (x) +G2(x)k(x) +G1(x)w

z =

[

H(x)k(x)

]

has the following property for some γ > 0. For every initial state x0 there is a real numberβ0 such that for every disturbance input w one has

∫ ∞

0

||z(t)||2 dt ≤ γ2∫ ∞

0

||w||2 dt+ β0

• We can guarantee the desired performance if there is a function V ≥ 0 satisfying

DV (F +G2k) +1

4γ2DVG1G

T1DV

T + CTC + kTk ≤ 0

Suppose V ≥ 0 satisfies The Hamilton-Jacobi Issacs (HJI) inequality

DV F +1

4DV ( γ−2G1G

T1 −G2G

T2 )DV T + CTC ≤ 0 (20.5)

Then letting

k = −12GT

2DVT

the desired inequality holds.

20.3. SERIES SOLUTION TO CONTROL PROBLEM 281

20.3 Series solution to control problem

We wish to solve the HJI equation,

DV F + 14DV ( γ−2G1G

T1 −G2G

T2 )DV T + CTC = 0 (20.6)

for V ≥ 0. Letting

G(x) := [G1(x) G2(x)] Q(x) := H(x)TH(x), R :=

[

−γ2 00 1

]

the HJI equation (20.6) can be rewritten as

DV F − 1

4DVGR−1BTDV T +Q = 0 (20.7)

Also letting

v∗ :=

[

u∗w∗

]

w∗ =1

2γ2GT

1DVT

we have

v∗ = −1

2R−1GTDV T (20.8)

20.3.1 Linearized problem

x = Ax+Bv (20.9a)

z =

[

Cxu

]

(20.9b)

withA = DF (0) , B = G(0), C = DH(0)

Here the HJI equation (20.6) becomes

DV (x)Ax− 1

4DV (x)BR−1BTDV (x) + xTCTCx = 0

Considering a quadratic formV (x) = xTPx

with P symmetric, as a candidate solution to the above HJE we obtain

xT [PA+ ATP − PBR−1BTP + CTC]x = 0

This is satisified for all x iff the matrix P solves the following Algebraic Riccati Equation(ARE)

PA+ ATP − PBR−1BTP + CTC = 0 (20.10)

From standard linear H∞ theory, a necessary and sufficient condition for the linear system tohave L2-gain less than γ is that the above Algebraic Riccati Equation (ARE) has a solutionP with A−BR−1BP Hurwitz. When (C,A) is observable, this solution P is positive definite.

In this case v∗ is given byv∗(x) = −R−1BTPx (20.11)


20.3.2 Nonlinear problem

First note that the HJI equation (20.7) can be written as

DV F − vT∗ Rv∗ +Q = 0 (20.12a)

v∗ +1

2R−1GTDV T = 0 (20.12b)

Suppose

F = F [1] + F [2] + F [3] + · · ·G = G[0] +G[1] +G[2] + · · ·H = H [1] +H [2] +H [3] + · · ·

where F [k], G[k] and H [k] are homogeneous functions of order k. Note that

F [1](x) = Ax, G[0](x) = B, H [1](x) = Cx

We consider a series expansion for V of the form

V = V [2] + V [3] + V [4] + · · · (20.13)

where V [k] is a homogeneous function of order k. Substituting (20.13) in equation (20.12b)one obtains that

v∗ = v[1]∗ + v[2]∗ + v[3]∗ + · · · (20.14)

where v[k]∗ is the homogeneous function of order k given by

v[k]∗ = −1

2R−1

k−1∑

j=0

G[j]TDV [k+1−j]T (20.15)

Substitution into (20.12a) and equating terms of order m ≥ 2 to zero yields

m−2∑

k=0

DV [m−k]F [k+1] −m−1∑

k=1

v[m−k]T∗ Rv[k]∗ + Q[m] = 0 (20.16)

For m = 2 equation (20.16) simplifies to

DV [2]F [1] − v[1]∗ Rv[1]∗ + Q[2] = 0

Since F [1](x) = Ax,

v[1]∗ = −1

2R−1G[0]TDV [2]T

= −1

2R−1BTDV [2]T

andQ[2](x) = xTCTCx

20.3. SERIES SOLUTION TO CONTROL PROBLEM 283

we obtain

DV [2](x)Ax− 1

4DV [2] T (x)BR−1BTDV [2](x) + xTCTCx = 0

which is the HJI equation for the linearized problem. Hence

V [2](x) = xTPx

where P T = P > 0 solves the ARE with A∗ := A− BR−1BTP Hurwitz. Hence

v[1]∗ (x) = −R−1BTPx

Consider now any m ≥ 3. Using

v[m−1]∗ = −1

2R−1

m−2∑

k=0

G[k]TDV [m−k]T

the sum of the terms in (20.16) depending on V [m] is given by:

DV [m]F [1] − v[m−1]T∗ Rv[1]∗ − v[1]T∗ Rv[m−1]

∗ = DV [m]F [1] − 2v[m−1]T∗ Rv[1]∗

= DV [m]F [1] +DV [m]G[0]v[1]∗ +m−2∑

k=1

DV [m−k]G[k]v[1]∗

= DV [m]f [1] +m−2∑

k=1

DV [m−k]G[k]v[1]∗

wheref = F +Gv[1]∗ (20.17)

andf [1] = A∗x (20.18)

The sum of the terms on the lefthandside of (20.16) can now be written as:

DV [m]f [1] +m−2∑

k=1

DV [m−k]F [k+1] +m−2∑

k=1

DV [m−k]G[k]v[1]∗ −m−2∑

k=2

v[m−k]T∗ Rv[k]∗ +Q[m]

= DV [m]f [1] +m−2∑

k=1

DV [m−k]f [k+1] −m−2∑

k=2

v[m−k]T∗ Rv[k]∗ +Q[m]

Hence, usingf [k+1] = F [k+1] +G[k]v∗[1]

(20.16) can be written as

DV [m]f [1] = −m−2∑

k=1

DV [m−k]f [k+1] +

m−2∑

k=2

v[m−k]T∗ Rv[k]∗ − Q[m]


For example, with Q[m] = 0,

DV [3]f [1] = −DV [2]f [2]

= −DV [2](F [2] +G[1]v[1]∗ )

hence

v[2]∗ = −1

2R−1(BTDV [3] +G[1]TDV [2])

For m = 4:DV [4]f [1] = −DV [3]f [2] −DV [2]f [3] + v[2]T∗ Rv[2]∗

and

v[3]∗ = −1

2R−1(BTDV [4] +G[1] TDV [3] +G[2]TDV [2])

IfF [2], G[1] = 0

ThenV [3], v[2]∗ = 0

andDV [4]f [1] = −DV [2]f [3]

v[3]∗ = −1

2R−1(BTDV [4] +G[2]TDV [2])

Example 191

x = x− x3 + w + u

z = u

HJI equation

Chapter 21

Performance

21.1 Analysis

Consider a system with output described by

x = f(t, x) (21.1a)

z = h(t, x) (21.1b)

where z is a performance output.

Lemma 17 Suppose that there exists a differentiable function V such that V (x) ≥ 0 and

DV (x)f(t, x) + h(t, x)′h(t, x) ≤ 0 (21.2)

for all t and x. Then every solution of system (21.1) satisfies∫ ∞

t0

‖z(t)‖2 dt ≤ V (x0) (21.3)

where x0 = x(t0).

285

286 CHAPTER 21. PERFORMANCE


x = −αxz = x

where α > 0. Let V (x) = x2/2α.

Example 193

x = −x3z = x2

Let V (x) = x2/2.

Example 194

x = −2x+ sin x

z = x

Let V (x) = x2/2.

Linear time-invariant systems. Consider a LTI system described by

x = Ax (21.4)

z = Cx (21.5)

where all the eigenvalues of A have negative real part. The Lyapunov equation

PA+ A′P + C ′C = 0 (21.6)

has a unique solution for P ; moreover P is symmetric and positive semi-definite. Post- andpre-multiplying this equation by x and its transpose yields

2x′PAx+ (Cx)′Cx = 0 ;

thus, (21.2) holds with V (x) = x′Px. Actually, in this case one can show that∫ ∞

t0

‖z(t)‖2 dt = V (x0) .

21.2 Polytopic systems

21.2.1 Polytopic models

Here V is a vector space; so V could be IRn or the set of real n×m matrices or some moregeneral space. Suppose v1, . . . , vl is a finite number of elements of V. A vector v is a convexcombination of v1, . . . , vl if it can be expressed as

v = λ1v1 + · · ·+ λlvN

where the real scalars λ1, . . . , λN satisfy λj ≥ 0 for j = 1, . . . , N and

λ1 + · · ·+ λl = 1 .

The set of all convex combinations of v1, . . . , vN is called a polytope. The vectors v1, . . . , vNare called the vertices of the polytope.

21.2. POLYTOPIC SYSTEMS 287

Polytopes and LMIs. Suppose that

L(v) ≤ 0

is an LMI in v. Then this LMI holds for all v in a polytope if and only if it holds for thevertices v1, . . . , vN of the polytope, that is,

L(vj) ≤ 0 for j = 1, . . . , N . (21.7)

To see this, suppose that (21.7) holds. Since L(v) is affine in v, we can express it as

L(v) = L0 + L1(v)

where L1(v) is linear in v. Suppose v is in the polytope; then v =∑∞

j=1 λjvj where λj ≥ 0and

∑∞j=1 λj = 1. Hence

L(v) = L0 + L1(Σ∞j=1λjvj) = Σ∞

j=1λjL0 + Σ∞j=1λjL1(vj) = Σ∞

j=1L(vj) ≤ 0 .

Obtaining polytopic models. Consider a parameter δ which satisfies

δ ≤ δ ≤ δ .

Then, δ can be expressed asδ = δ + µ(δ − δ)

where µ = (δ − δ)/(δ − δ). Clearly 0 ≤ µ ≤ 1. If we let λ1 = 1 − µ and λ2 = µ then,λ1, λ2 ≥ 0, λ1 + λ2 = 1 and

δ = λ1δ + λ2δ ,

that is δ can be written as a convex combination of its two extreme values δ and δ. This isa special case of the general result contained in the next remark.

Remark 16 This remark is useful in obtaining polytopic models. Suppose G is a matrixvalued function of some variable ξ and for all ξ,

G(ξ) = G(δ(ξ))

where δ is some parameter vector (which can depend on ξ). Suppose that G depends in amulti-affine fashion on the components of the l-vector δ and each component of δ is bounded,that is,

δk ≤ δk(ξ) ≤ δk for k = 1, . . . , l

where δ1, . . . , δl and δ1, . . . , δl are known constants. Then, for all ξ, the vector G(ξ) can beexpressed as a convex combination of the N = 2l matrices G1, . . . , GN corresponding to theextreme values of the components of δ; these vertices Gj are given by

G(δ) where δk = δk or δk for k = 1, . . . , l . (21.8)

In applying to show satisfaction of Condition 1 below, let ξ = (t, x), G(ξ) = (A(t, x) C(t, x)).Then (Aj Cj) = Gj for j = 1, . . . , N .



G(ξ) =(

cos ξ1 sin ξ2 + cos ξ1 1 + cos ξ1 + sin ξ2)

For all ξ, the matrix G(ξ) can be expressed as a convex combination of the following fourmatrices:

(

0 −1)

,(

−2 1)

,(

0 1)

,(

2 3)

.

Remark 17 The following remark is also useful for obtaining polytopic models. Consideran n-vector-valued function g of two variables t and ξ which is differential with respect toits second variable and satisfies the following two conditions:

(a)

g(t, 0) ≡ 0 .

(b) There are matrices, G1, . . . , GN such that, for all t and ξ, the derivative matrix

∂g

∂ξ(t, ξ)

can be expressed as a convex combination of G1, . . . , GN .

Then, for each t and ξ,

g(t, ξ) = Gξ

where G is some convex combination of G1, . . . , GN .

In applying this to show satisfaction of Condition 1, let ξ = x, g(t, ξ) = (f(t, x) h(t, x)).Then

∂g

∂ξ(t, ξ) =

[

∂f∂x(t, x) ∂h

∂x(t, x)

]

and (Aj Cj) = Gj for j = 1, . . . , N .


g(t, ξ) =

(

sin ξ1ξ2

1+|ξ2|

)

.

Here g(t, 0) = 0 and

∂g

∂ξ=

(

cos ξ1 00 1

(1+|ξ2|)2

)

.

Hence, g(t, ξ) = Gξ where G can be expressed as a convex combination of the following fourmatrices:

(

−1 00 0

)

,

(

−1 00 1

)

,

(

1 00 0

)

,

(

1 00 1

)

.

21.2. POLYTOPIC SYSTEMS 289

21.2.2 Performance analysis of polytopic systems

In this section, we consider a specific type of uncertain/nonlinear system described by (21.1).Specifically, we consider systems which satisfies the following condition.

Condition 1 There are matrices, Aj , Cj, j = 1, . . . , N so that for each t and x,

f(t, x) = Ax (21.9)

h(t, x) = Cx (21.10)

where the matrix (A C) (which can depend on t, x) is a convex combination of (A1 C1) , . . . , (AN CN).

So, when Condition 1 is satisfied, system 21.1 can be described by

x = A(t, x)x (21.11a)

z = C(t, x)x (21.11b)

where (A(t, x) C(t, x)) is contained in the polytope whose vertices are (A1 C1), . . . , (AN CN).Hence, we sometimes refer to a system satisfying Condition 1 as a polytopic uncertain/nonlinearsystem.

Suppose thatPAj + A′

jP + C ′jCj ≤ 0 for j = 1, · · · , N . (21.12)

Using a Schur complement result, the above inequalities can be expressed as

(

PAj + A′jP C ′

j

Cj −I

)

≤ 0 for j = 1, · · · , N . (21.13)

Since the above inequalities are affine in (Aj Cj), and (A(t, x) C(t, x)) is a convex combinationof (A1 C1), . . . , (AN CN) it now follows that

(

PA(t, x) + A(t, x)′P C(t, x)′

C(t, x) −I

)

≤ 0

for all t and x. Reusing the Schur complement now results in

PA(t, x) + A(t, x)′P + C(t, x)′C(t, x) ≤ 0

for all t and x. Post- and pre-multiplying this equation by x and its transpose yields

2x′PA(t, x)x+ (C(t, x)x)′C(t, x)x ≤ 0

that is,2x′Pf(t, x) + h(t, x)′h(t, x) ≤ 0

for all t and x; thus, (21.2) holds with V (x) = x′Px. Hence every solution of the systemsatisfies

∫ ∞

t0

‖z(t)‖2 dt ≤ x′0Px0 (21.14)


where x0 = x(t0).To obtain a performance estimate for a fixed initial state one could minimize β subject

to LMIs (21.13) andx′0Px0 − β ≤ 0 . (21.15)

Note that, if we let S = P−1 then, (21.13) can be expressed as

(

AjS + A′jS SC ′

j

CjS −I

)

≤ 0 for j = 1, · · · , N (21.16)

Also, using Schur complements, inequality (21.15) can be expressed as

(

−S x0x′0 −β

)

≤ 0 for j = 1, · · · , N (21.17)

Suppose x0 lies in a polytope with vertices x1, . . . , xM . Then we minimize β subject to(21.16) and

(

−S xkx′k −β

)

≤ 0 for k = 1, · · · ,M (21.18)

21.3 Control for performance

21.3.1 Linear time-invariant systems

Consider

x = Ax+ Bu (21.19a)

z = Cx+Du (21.19b)

In this case∫ ∞

0

‖z(t)‖2 dt =∫ ∞

0

x(t)′C ′Cx(t) + 2x(t)′C ′Du(t) + u(t)′D′Du(t) dt .

When C ′D = 0, this reduces to the usual performance encountered in LQR control, that is,

∫ ∞

0

‖z(t)‖2 dt =∫ ∞

0

x(t)′Qx(t) + u(t)′Ru(t) dt ;

hereQ = C ′C and R = D′D .

Supposeu = Kx (21.20)

The resulting closed loop system is described by

x = (A+BK)x (21.21a)

z = (C +DK)x (21.21b)

21.3. CONTROL FOR PERFORMANCE 291

and inequality (21.2) holds for all x if and only if

P (A+BK) + (A+BK)′P + (C +DK)′(C +DK) ≤ 0 ,

that is,

PA+ A′P + C ′C + (PB + C ′D)K +K ′(B′P +D′C) +K ′D′DK ≤ 0 .

Assuming that D′D is invertible, this can be expressed as

PA + A′P + C ′C − (PB + C ′D)(D′D)−1(B′P +D′C) + Q ≤ 0 (21.22)

where P = P and

Q = [K + (D′D)−1(B′P +D′C)]′D′D[K + (D′D)−1(B′P +D′C)] ≥ 0 .

Suppose P is a stabilizing solution to the following LQR Riccati equation:

PA+ A′P + C ′C − (PB + C ′D)(D′D)−1(B′P +D′C) = 0 . (21.23)

Then it can be shown that P ≤ P where P is any matrix satisfying inequality (21.22) witha stabilizing K. In this case

K = −(D′D)−1(B′P +D′C) . (21.24)

This is the LQR control gain matrix. Note that P is a stabilizing solution ifA−B(D′D)−1(B′P+D′C) is asymptotically stable.

21.3.2 Control of polytopic systems

In this section, we consider a specific type of uncertain/nonlinear system described by

x = F (t, x, u) (21.25a)

z = H(t, x, u) (21.25b)

Specifically, we consider systems which satisfies the following condition.

Condition 2 There are matrices, Aj , Bj, Cj, Dj, j = 1, . . . , N so that for each t, x and u,

F (t, x) = Ax+Bu (21.26)

H(t, x) = Cx+Du (21.27)

where the matrix (A B C D) (which can depend on t, x, u) is a convex combination of(A1 B1 C1 D1) , . . . , (AN BN CN DN ).

So, when Condition 3 is satisfied, a system under consideration can be described by

x = A(t, x)x+ B(t, x)u (21.28a)

z = C(t, x)x+D(t, x)u (21.28b)

where (A(t, x) B(t, x) C(t, x) D(t, x)) is contained in the polytope whose vertices are(A1 B1 C1 D1), . . . , (AN BN CN DN). Hence, we sometimes refer to a system satisfyingCondition 3 as a polytopic uncertain/nonlinear system.


Linear state feedback controllers. We consider

u = Kx (21.29)

This results in the following closed loop system

x = [A(t, x) +B(t, x)K]x (21.30a)

z = [C(t, x) +D(t, x)K]x (21.30b)

Note that this system satisfies Condition 1 with vertex matrices (A1+B1K, C1+D1K), . . . , (AN+BNK). Hence

∫ ∞

t0

‖z(t)‖2 dt ≤ x′0Px0

where S = P−1,

(

AjS +BjL+ SA′j + L′B′

j SC ′j + L′D′

j

CjS +DjL −I

)

≤ 0 for j = 1, . . . , N (21.31)

andK = LS−1 (21.32)

21.3.3 Multiple performance outputs

x = F (t, x, u) (21.33a)

zi = Hi(t, x, u) for i = 1, . . . , p (21.33b)

Condition 3 There are matrices, Aj, Bj , C1j, D1j . . . , Cpj, Dpj, j = 1, . . . , N so that for eacht, x and u,

F (t, x) = Ax+Bu (21.34)

Hi(t, x) = Cix+Diu for i = 1, . . . , p (21.35)

where the matrix (A B C1 D1 , . . . , Cp Dp) (which can depend on t, x, u) is a convex combi-nation of (A1 B1 Ci1 Di1 , . . . , Cp1 Dp1) , (AN BN CiN DiN , . . . , CpN DpN) .

∫ ∞

t0

‖zi(t)‖2 dt ≤ x′0Px0 for i = 1, . . . , p

where S = P−1,

(

AjS +BjL+ SA′j + L′B′

j SC ′ij + L′D′

ij

CijS +DijL −I

)

≤ 0 for i = 1, . . . , p and j = 1, . . . , N(21.36)

andK = LS−1 (21.37)

Chapter 22

Appendix

22.1 The Euclidean norm

What is the size or magnitude of a vector? Meet norm.

• Consider any complex n vector x.

The Euclidean norm or 2-norm of x is the nonegative real number given by

||x|| := (|x1|2 + . . .+ |xn|2)12

Note that

||x|| = (x1x1 + . . .+ xnxn)12

= (x∗x)12

If x is real, these expressions become

||x|| = (x21 + . . .+ x2n)12

= (xTx)12

>> norm([3; 4])

ans = 5

>> norm([1; j])

ans = 1.4142

Note that in the last example xTx = 0, but x∗x = 2.

Properties of || · ||. The Euclidean norm has the following properties.

293

294 CHAPTER 22. APPENDIX

(i) For every vector x,||x|| ≥ 0

and||x|| = 0 iff x = 0

(ii) (Triangle Inequality.) For every pair of fectors x, y,

||x+ y|| ≤ ||x||+ ||y||

(iii) For every vector x and every scalar λ.

||λx|| = |λ|||x||

• Any real valued function on a vector space which has the above three properties is definedto be a norm.

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