Dynamic Systems and Feedback

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ME 132, Dynamic Systems and Feedback Class Notes Andrew Packard, Roberto Horowitz, Kameshwar Poolla, Francesco Borrelli

Fall 2018 Instructor: Andy Packard

Department of Mechanical Engineering University of California Berkeley CA, 94720-1740

copyright 1995-2018 Packard, Horowitz, Poolla, Borrelli

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Contents 1 Introduction

1

1.1

Structure of a closed-loop control system . . . . . . . . . . . . . . . . . . . .

4

1.2

Example: Temperature Control in Shower . . . . . . . . . . . . . . . . . . .

5

1.3

Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

8

2 Block Diagrams

13

2.1

Common blocks, continuous-time . . . . . . . . . . . . . . . . . . . . . . . .

13

2.2

Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

14

2.3

Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

16

2.4

Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

17

3 Mathematical Modeling and Simulation

24

3.1

Systems of 1st order, coupled differential equations . . . . . . . . . . . . . .

24

3.2

Remarks about Integration Options in simulink . . . . . . . . . . . . . . . .

26

3.3

Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

26

4 State Variables

30

4.1

Definition of State . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

30

4.2

State-variables: from first order evolution equations . . . . . . . . . . . . . .

31

4.3

State-variables: from a block diagram . . . . . . . . . . . . . . . . . . . . . .

32

4.4

Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

32

5 First Order, Linear, Time-Invariant (LTI) ODE

34

5.1

The Big Picture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

34

5.2

Solution of a First Order LTI ODE . . . . . . . . . . . . . . . . . . . . . . .

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5.2.1

Free response . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

36

5.2.2

Forced response, constant inputs . . . . . . . . . . . . . . . . . . . .

37

5.2.3

Forced response, bounded inputs . . . . . . . . . . . . . . . . . . . .

38

5.2.4

Stable system, Forced response, input approaching 0 . . . . . . . . .

38

5.2.5

Linearity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

40

5.2.6

Forced response, input approaching a constant limit . . . . . . . . . .

40

Forced response, Sinusoidal inputs . . . . . . . . . . . . . . . . . . . . . . . .

41

5.3.1

Forced response, input approaching a Sinusoid . . . . . . . . . . . . .

43

5.4

First-order delay-differential equation: Stability . . . . . . . . . . . . . . . .

44

5.5

Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

45

5.6

Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

46

5.3

6 Feedback systems

53

6.1

First-order plant, Proportional control . . . . . . . . . . . . . . . . . . . . .

53

6.2

Proportional Plant, first-order controller . . . . . . . . . . . . . . . . . . . .

54

6.3

Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

55

7 Two forms of high-order Linear ODEs, with forcing 7.1

Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

8 Jacobian Linearizations, equilibrium points

68 69 73

8.1

Jacobians and the Taylor Theorem . . . . . . . . . . . . . . . . . . . . . . .

73

8.2

Equilibrium Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

74

8.3

Deviation Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

75

8.4

Tank Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

76

8.5

Output Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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8.6

Calculus for systems not in standard form . . . . . . . . . . . . . . . . . . .

81

8.7

Another common non-standard form . . . . . . . . . . . . . . . . . . . . . .

82

8.8

Linearizing about general solution . . . . . . . . . . . . . . . . . . . . . . . .

83

8.9

Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

85

8.10 Additional Related Problems . . . . . . . . . . . . . . . . . . . . . . . . . . .

87

9 Linear Algebra Review

105

9.1

Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

9.2

Determinants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

9.3

Inverses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108

9.4

Solving Linear equations: Gaussian Elimination . . . . . . . . . . . . . . . . 109

9.5

Matrix functions of Time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110

10 Linear Systems and Time-Invariance

112

10.1 Linearity of solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112 10.2 Time-Invariance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 11 Matrix Exponential

114

11.1 Diagonal A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116 11.2 Block Diagonal A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116 11.3 Effect of Similarity Transformations . . . . . . . . . . . . . . . . . . . . . . . 116 11.4 Solution To State Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 117 11.5 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118 11.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120 12 Eigenvalues, eigenvectors, stability

123

12.1 Diagonalization: Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . 123

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12.2 Eigenvalues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123 12.3 Diagonalization Procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126 12.4 eAt as t → ∞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126 12.5 Complex Eigenvalues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127 12.6 Alternate parametrization with complex eigenvalues . . . . . . . . . . . . . . 130 12.6.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133 12.7 Step response . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136 12.8 Quick estimate of unit-step-response of 2nd order system . . . . . . . . . . . 137 12.9 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138 13 Frequency Response for Linear Systems: State-Space representations

147

13.1 Theory for Stable System: Complex Input Signal . . . . . . . . . . . . . . . 147 13.2 MIMO Systems: Response due to real sinusoidal inputs . . . . . . . . . . . . 148 13.3 Experimental Determination . . . . . . . . . . . . . . . . . . . . . . . . . . . 149 13.4 Steady-State response

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149

14 Important special cases for designing closed-loop systems

150

14.1 Roots of 2nd-order monic polynomial . . . . . . . . . . . . . . . . . . . . . . 150 14.2 Setting the coefficients to attain certain roots . . . . . . . . . . . . . . . . . 151 14.3 1st-order plant, 1st-order controller . . . . . . . . . . . . . . . . . . . . . . . 151 14.4 2nd-order plant, constant-gain controller with derivative feedback . . . . . . 156 14.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157 15 Step response

161

15.1 Quick estimate of unit-step-response of 2nd order system . . . . . . . . . . . 162 16 Stabilization by State-Feedback

164

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16.1 Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164 17 State-Feedback with Integral Control

165

17.1 Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165 17.2 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168 17.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173 18 Linear-quadratic Optimal Control

177

18.1 Learning more . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178 19 Single, high-order, linear ODES (SLODE)

179

19.1 Linear, Time-Invariant Differential Equations . . . . . . . . . . . . . . . . . 179 19.2 Importance of Linearity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180 19.3 Solving Homogeneous Equation . . . . . . . . . . . . . . . . . . . . . . . . . 180 19.3.1 Interpretation of complex roots to ODEs with real-coefficients . . . . 182 19.4 General Solution Technique . . . . . . . . . . . . . . . . . . . . . . . . . . . 184 19.5 Behavior of Homogeneous Solutions as t → ∞ . . . . . . . . . . . . . . . . . 185 19.6 Response of stable system to constant input (Steady-State Gain) . . . . . 186 19.7 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187 19.8 Stability Conditions for 2nd order differential equation . . . . . . . . . . . . 188 19.9 Important 2nd order example . . . . . . . . . . . . . . . . . . . . . . . . . . 190 19.10Summary for SLODEs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197 19.10.1 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197 19.10.2 General Solution Technique . . . . . . . . . . . . . . . . . . . . . . . 198 19.10.3 Behavior of Homogeneous Solutions as t → ∞ . . . . . . . . . . . . . 198 19.10.4 Stability of Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . 198

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19.10.5 2nd order differential equation . . . . . . . . . . . . . . . . . . . . . . 199 19.10.6 Solutions of 2nd order differential equation . . . . . . . . . . . . . . . 199 19.11Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200 20 Frequency Responses of Linear Systems

207

20.1 Complex and Real Particular Solutions . . . . . . . . . . . . . . . . . . . . . 208 20.2 Response due to real sinusoidal inputs . . . . . . . . . . . . . . . . . . . . . 209 20.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209 21 Derivatives appearing on the inputs: Effect on the forced response

212

21.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213 21.2 Other Particular Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213 21.3 Limits approaching steps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214 21.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217 22 Distributions

218

22.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 218 22.2 Procedure to get step response . . . . . . . . . . . . . . . . . . . . . . . . . . 221 22.3 Summary: Solution of SLODEs with Derivatives on the inputs . . . . . . . . 223 22.3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223 22.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227 23 Transfer functions

228

23.1 Linear Differential Operators (LDOs) . . . . . . . . . . . . . . . . . . . . . . 228 23.2 Algebra of Linear differential operations . . . . . . . . . . . . . . . . . . . . 230 23.3 Feedback Connection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231 23.4 More General Feedback Connection . . . . . . . . . . . . . . . . . . . . . . . 232

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23.5 Cascade (or Series) Connection . . . . . . . . . . . . . . . . . . . . . . . . . 234 23.6 Parallel Connection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 236 23.7 General Connection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 238 23.8 Systems with multiple inputs . . . . . . . . . . . . . . . . . . . . . . . . . . 238 23.9 Poles and Zeros of Transfer Functions . . . . . . . . . . . . . . . . . . . . . . 238 23.10Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239 24 Arithmetic of Feedback Loops

246

24.1 Tradeoffs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 250 24.2 Signal-to-Noise ratio . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 250 24.3 What’s missing? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251 24.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252 25 Robustness Margins

261

25.1 Gain Margin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263 25.2 Time-Delay Margin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 266 25.3 Percentage Variation Margin . . . . . . . . . . . . . . . . . . . . . . . . . . . 267 25.3.1 The Small-Gain Theorem . . . . . . . . . . . . . . . . . . . . . . . . 269 25.3.2 Necessary and Sufficient Version . . . . . . . . . . . . . . . . . . . . . 271 25.3.3 Application to Percentage Variation Margin . . . . . . . . . . . . . . 274 25.4 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 275 25.5 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 276 25.5.1 Generic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 276 25.5.2 Missile . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277 25.5.3 Application to percentage variation margin . . . . . . . . . . . . . . . 281 25.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285

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26.1 Sylvester’s determinant identity . . . . . . . . . . . . . . . . . . . . . . . . . 299 26.2 Setup . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 299 26.3 Gain margin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 299 26.4 Time delay margin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302 26.5 Appendix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304 27 Connection between Frequency Responses and Transfer functions

305

27.1 Interconnections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 305 28 Decomposing Systems into Simple Parts

307

28.1 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307 29 Unfiled problems

310

30 Recent exams

317

30.1 Fall 2017 Midterm 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 317 30.2 Fall 2017 Midterm 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 328 30.3 Fall 2017 Final . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333 30.4 Fall 2015 Midterm 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 350 30.5 Fall 2015 Midterm 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 357 30.6 Fall 2015 Final . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 362 30.7 Spring 2014, Midterm 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 381 30.8 Spring 2014, Midterm 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 388 30.9 Spring 2014 Final Exam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 392 31 Older exams

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31.1 Spring 2012, Midterm 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 414 31.2 Spring 2012, Midterm 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 418 31.3 Spring 2012, Final Exam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 423 31.4 Spring 2009, Midterm 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 437 31.5 Spring 2009, Midterm 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 445 31.6 Spring 2009, Final Exam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 445 31.7 Spring 2005 Midterm 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 445 31.8 Spring 2005 Midterm 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 451 31.9 Spring 2004 Midterm 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 462 31.10Fall 2003 Midterm 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 468 31.11Fall 2003 Midterm 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 475 31.12Fall 2003 Final . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 480

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Introduction

In this course we will learn how to analyze, simulate and design automatic control strategies (called control systems) for various engineering systems. The system whose behavior is to be controlled is called the plant. This term has its origins in chemical engineering where the control of chemical plants or factories is of concern. On occasion, we will also use the terms plant, process and system interchangeably. As a simple example which we will study soon, consider an automobile as the plant, where the speed of the vehicle is to be controlled, using a control system called a cruise control system. The plant is subjected to external influences, called inputs which through a cause/effect relationship, influence the plant’s behavior. The plant’s behavior is quantified by the value of several internal quantities, often observable, called plant outputs. Initially, we divide these external influences into two groups: those that we, the owner/operator of the system, can manipulate, called them control inputs; and those that some other externality (nature, another operator, normally thought of as antagonistic to us) manipulates, called disturbance inputs. By control, we mean the manipulation of the control inputs in a way to make the plant outputs respond in a desirable manner. The strategy and/or rule by which the control input is adjusted is known as the control law, or control strategy, and the physical manner in which this is implemented (computer with a real-time operating system; analog circuitry, human intervention, etc.) is called the control system. The most basic objective of a control system are: • The automatic regulation (referred to as tracking) of certain variables in the controlled plant to desired values (or trajectories), in the presence of typical, but unforseen, disturbances An important distinguishing characteristic of a strategy is whether the controlling strategy is open-loop or closed-loop. This course is mostly (completely) about closed-loop control systems, which are often called feedback control systems. • Open-loop control systems: In an open-loop strategy, the values of the control input (as a function of time) are decided ahead-of-time, and then this input is applied to the system. For example “bake the cookies for 8 minutes, at 350◦ .” The design of the open-loop controller is based on inversion and/or calibration. While open-loop

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systems are simple, they generally rely totally on calibration, and cannot effectively deal with exogenous disturbances. Moreover, they cannot effectively deal with changes in the plant’s behavior, due to various effects, such as aging components. They require re-calibration. Essentially, they cannot deal with uncertainty. Another disadvantage of open-loop control systems is that they cannot stabilize an unstable system, such as balancing a rocket in the early stages of liftoff (in control terminology, this is referred to as “an inverted pendulum”). • Closed-loop control systems: In order to make the plant’s output/behavior more robust to uncertainty and disturbances, we design control systems which continuously sense (measure) the output of the plant (note that this is an added complexity that is not present in an open-loop system), and adjust the control input, using rules, which are based on how the current (and past) values of the plant output deviate from its desired value. These feedback rules (or strategy) are usually based on a model (ie., a mathematical description) of how the plant behaves. If the plant behaves slightly differently than the model predicts, it is often the case that the feedback helps compensate for these differences. However, if the plant actually behaves significantly different than the model, then the feedback strategy might be unsuitable, and may cause instability. This is a drawback of feedback systems. In reality, most control systems are a combination of open and closed-loop strategies. In the trivial cookie example above, the instructions look predominantly open-loop, though something must stop the baking after 8 minutes, and temperature in the oven should be maintained at (or near) 350◦ . Of course, the instructions for cooking would even be more “closed-loop” in practice, for example “bake the cookies at 350◦ for 8 minutes, or until golden-brown.” Here, the “until golden-brown” indicates that you, the baker, must act as a feedback system, continuously monitoring the color of the dough, and remove from heat when the color reaches a prescribed “value.” In any case, ME 132 focuses most attention to the issues that arise in closed-loop systems, and the benefits and drawbacks of systems that deliberately use feedback to alter the behavior/characteristics of the process being controlled. Some examples of systems which benefit from well-designed control systems are • Airplanes, helicopters, rockets, missiles: flight control systems including autopilot, pilot augmentation • Cruise control for automobiles. Lateral/steering control systems for future automated highway systems • Position and speed control of mechanical systems:

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1. AC and/or DC motors, for machines, including Disk Drives/CD, robotic manipulators, assembly lines. 2. Elevators 3. Magnetic bearings, MAGLEV vehicles, etc. • Pointing control (telescopes) • Chemical and Manufacturing Process Control: temperature; pressure; flow rate; concentration of a chemical; moisture content; thickness. Of course, these are just examples of systems that we have built, and that are usually thought of as “physical” systems. There are other systems we have built which are not “physical” in the same way, but still use (and benefit from) control, and additional examples that occur naturally (living things). Some examples are • The internet, whereby routing of packets through communication links are controlled using a conjestion control algorithm • Air traffic control system, where the real-time trajectories of aircraft are controlled by a large network of computers and human operators. Of course, if you “look inside”, you see that the actual trajectories of the aircraft are controlled by pilots, and autopilots, receiving instructions from the Air Traffic Control System. And if you “look inside” again, you see that the actual trajectories of the aircraft are affected by forces/moments from the engine(s), and the deflections of movable surfaces (ailerons, rudders, elevators, etc.) on the airframe, which are “receiving” instructions from the pilot and/or autopilot. • An economic system of a society, which may be controlled by the federal reserve (setting the prime interest rate) and by regulators, who set up “rules” by which all agents in the system must abide. The general goal of the rules is to promote growth and wealthbuilding. • All the numerous regulatory systems within your body, both at organ level and cellular level, and in between. A key realization is the fact that most of the systems (ie, the plant) that we will attempt to model and control are dynamic. We will later develop a formal definition of a dynamic system. However, for the moment it suffices to say that dynamic systems “have memory”, i.e. the current values of all variables of the system are generally functions of previous inputs, as well as the current input to the system. For example, the velocity of a mass particle at time t depends on the forces applied to the particle for all times before t. In the general case, this means that the current control actions have impact both at the current time (ie., when they are applied) and in the future as well, so that actions taken now have later consequences.

ME 132, Fall 2018, UC Berkeley, A. Packard

1.1

4

Structure of a closed-loop control system

The general structure of a closed-loop system, including the plant and control law (and other components) is shown in Figure 1. A sensor is a device that measures a physical quantity like pressure, acceleration, humidity, or chemical concentration. Very often, in modern engineering systems, sensors produce an electrical signal whose voltage is proportional to the physical quantity being measured. This is very convenient, because these signals can be readily processed with electronics, or can be stored on a computer for analysis or for real-time processing. An actuator is a device that has the capacity to affect the behavior of the plant. In many common examples in aerospace/mechanical systems, an electrical signal is applied to the actuator, which results in some mechanical motion such as the opening of a valve, or the motion of a motor, which in turn induces changes in the plant dynamics. Sometimes (for example, electrical heating coils in a furnace) the applied voltage directly affects the plant behavior without mechanical motion being involved. The controlled variables are the physical quantities we are interested in controlling and/or regulating. The reference or command is an electrical signal that represents what we would like the regulated variable to behave like. Disturbances are phenomena that affect the behavior of the plant being controlled. Disturbances are often induced by the environment, and often cannot be predicted in advance or measured directly. The controller is a device that processes the measured signals from the sensors and the reference signals and generates the actuated signals which in turn, affects the behavior of the plant. Controllers are essentially “strategies” that prescribe how to process sensed signals and reference signals in order to generate the actuator inputs. Finally, noises are present at various points in the overall system. We will have some amount of measurement noise (which captures the inaccuracies of sensor readings), actuator noise (due for example to the power electronics that drives the actuators), and even noise affecting the controller itself (due to quantization errors in a digital implementation of the control algorithm). Note that the sensor is a physical device in its own right, and also subject to external disturbances from the environment. This cause its output, the sensor reading, to generally be different from the actual value of the physical veriable the sensor is “sensing.” While this difference is usually referred to as noise, it is really just an additional disturbance that acts on the overall plant. Throughout these notes, we will attempt to consistently use the following symbols (note -

ME 132, Fall 2018, UC Berkeley, A. Packard

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nevertheless, you need to be flexible and open-minded to ever-changing notation): P u d r

plant input disturbance reference

K y n

controller output noise

Arrows in our block diagrams always indicate cause/effect relationships, and not necessarily the flow of material (fluid, electrons, etc.). Power supplies and material supplies may not be shown, so that normal conservations laws do not necessarily hold for the block diagram. Based on our discussion above, we can draw the block diagram of Figure 1 that reveals the structure of many control systems. Again, the essential idea is that the controller processes measurements together with the reference signal to produce the actuator input u(t). In this way, the plant dynamics are continually adjusted so as to meet the objective of having the plant outputs y(t) track the reference signal r(t). Disturbances

Measurement noise

Actuator noise

Actuators

Plant

Sensors

Controller Controller noise

Commands

Figure 1: Basic structure of a control system.

1.2

Example: Temperature Control in Shower

A simple, slightly unrealistic example of some important issues in control systems is the problem of temperature control in a shower. As Professor Poolla tells it, “Every morning I wake up and have a shower. I live in North Berkeley, where the housing is somewhat run-down, but I suspect the situation is the same everywhere. My shower is very basic. It has hot and cold water taps that are not calibrated. So I can’t exactly preset the shower temperature that I desire, and then just step in. Instead, I am forced to use feedback control. I stick my hand in the shower to measure the temperature. In my brain, I have an idea of

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6

what shower temperature I would like. I then adjust the hot and cold water taps based on the discrepancy between what I measure and what I want. In fact, it is possible to set the shower temperature to within 0.5◦ F this way using this feedback control. Moreover, using feedback, I (being the sensor and the compensatory strategy) can compensate for all sorts of changes: environmental changes, toilets flushing, etc. This is the power of feedback: it allows us to, with accurate sensors, make a precision device out of a crude one that works well even in changing environments.” Let’s analyze this situation in more detail. The components which make up the plant in the shower are • Hot water supply (constant temperature, TH ) • Cold water supply (constant temperature, TC ) • Adjustable valve that mixes the two; use θ to denote the angle of the valve, with θ = 0 meaning equal amounts of hot and cold water mixing. In the units chosen, assume that −1 ≤ θ ≤ 1 always holds. • 1 meter (or so) of piping from valve to shower head If we assume perfect mixing, then the temperature of the water just past the valve is C C + TH −T θ(t) Tv (t) := TH +T 2 2 = c1 + c2 θ(t)

The temperature of the water hitting your skin is the same (roughly) as at the valve, but there is a time-delay based on the fact that the fluid has to traverse the piping, hence T (t) = Tv (t − ∆) = c1 + c2 θ(t − ∆) where ∆ is the time delay, about 1 second. Let’s assume that the valve position only gets adjusted at regular increments, every ∆ seconds. Similarly, lets assume that we are only interested in the temperature at those instants as well. Hence, we can use a discrete notion of time, indexed by a subscript k, so that for any signal, v(t), write vk := v(t)|t=k∆ In this notation, the model for the Temperature/Valve relationship is Tk = c1 + c2 θk−1

(1.1)

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Now, taking a shower, you have (in mind) a desired temperature, Tdes , which may even be a function of time Tdes,k . How can the valve be adjusted so that the shower temperature approaches this? Open-loop control: pre-solve for what the valve position should be, giving θk =

Tdes,k − c1 c2

(1.2)

and use this – basically calibrate the valve position for desired temperature. This gives Tk = Tdes,(k−1) which seems good, as you achieve the desired temperature one ”time-step” after specifying it. However, if c1 and/or c2 change (hot or cold water supply temperature changes, or valve gets a bit clogged) there is no way for the calibration to change. If the plant behavior changes to Tk = c˜1 + c˜2 θk−1 (1.3) but the control behavior remains as (1.2), the overall behavior is Tk+1 = c˜1 +

c˜2 (Tdes,k − c1 ) c2

which isn’t so good. Any percentage variation in c2 is translated into a similar percentage error in the achieved temperature. How do you actually control the temperature when you take a shower: Again, the behavior of the shower system is: Tk+1 = c1 + c2 θk Closed-loop Strategy: If at time k, there is a deviation in desired/actual temperature of Tdes,k − Tk , then since the temperature changes c2 units for every unit change in θ, the valve angle should be increased by an amount c12 (Tdes,k − Tk ). That might be too aggressive, trying to completely correct the discrepancy in one step, so choose a number λ, 0 < λ < 1, and try λ θk = θk−1 + (Tdes,k − Tk ) (1.4) c2 (of course, θ is limited to lie between −1 and 1, so the strategy should be written in a more complicated manner to account for that - for simplicity we ignore this issue here, and return to it later in the course). Substituting for θk gives 1 1 λ (Tk+1 − c1 ) = (Tk − c1 ) + (Tdes,k − Tk ) c2 c2 c2 which simplifies down to Tk+1 = (1 − λ) Tk + λTdes,k

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Starting from some initial temperature T0 , we have

If Tdes,n

T1 = (1 − λ)T0 + λTdes,0 T2 = (1 − λ)T1 + λTdes,1 = (1 − λ)2 T0 + (1 − λ)λTdes,0 + λTdes,1 .. . . = .. P n Tk = (1 − λ)k T0 + k−1 n=0 (1 − λ) λTdes,k−1−n is a constant, T¯, then the summation simplifies to   Tk = (1 − λ)k T0 + 1 − (1 − λ)k T¯   = T¯ + (1 − λ)k T0 − T¯

which shows that, in fact, as long as 0 < λ < 2, then the temperature converges (convergence rate determined by λ) to the desired temperature. Assuming your strategy remains fixed, how do unknown variations in TH and TC affect the performance of the system? Shower model changes to (1.3), giving   ˜ Tk + λT ˜ des,k Tk+1 = 1 − λ ˜ := c˜2 λ. Hence, the deviation in c1 has no effect on the closed-loop system, and the where λ c2 deviation in c2 only causes a similar percentage variation in the effective value of λ. As long ˜ < 2, the overall behavior of the system is acceptable. This is good, and shows as 0 < λ that small unknown variations in the plant are essentially completely compensated for by the feedback system. On the other hand, large, unexpected deviations in the behavior of the plant can cause problems for a feedback system. Suppose that you maintain the strategy in equation (1.4), but there is a longer time-delay than you realize? Specifically, suppose that there is extra piping, so that the time delay is not just ∆, but m∆. Then, the shower model is Tk+m−1 = c1 + c2 θk−1 and the strategy (from equation 1.4) is θk = θk−1 +

λ c2

(1.5)

(Tdes,k − Tk ). Combining, gives

Tk+m = Tk+m−1 + λ (Tdes,k − Tk ) This has some very undesirable behavior, which is explored in problem 5 at the end of the section.

1.3

Problems

1. For any β ∈ C and integer N , consider the summation α :=

N X k=0

βk

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If β = 1, show that α = N + 1. If β 6= 1, show that α=

1 − β N +1 1−β

If |β| < 1, show that ∞ X

βk =

k=0

1 1−β

2. Consider the equations relating variables r, e, y, n, u and d. Assume P and C are given numbers. e = r − (y + n) u = Ce y = P (u + d) So, this represents 3 linear equations in 6 unknowns. Solve these equations, expressing e, u and y as linear functions of r, d and n. The linear relationships will involve the numbers P and C. 3. For a function F of a many variables (say two, for this problem, labeled x and y), the “sensitivity of F to x” is defined as “the ratio of the percentage change in F due to a percentage change in x.” Denote this by SxF . (a) Suppose x changes by δ, to x + δ. The percentage change in x is then % change in x =

δ (x + δ) − x = x x

Likewise, the subsequent percentage change in F is % change in F =

F (x + δ, y) − F (x, y) F (x, y)

Show that for infinitesimal changes in x, the sensitivity is SxF = (b) Let F (x, y) =

xy . 1+xy

x ∂F F (x, y) ∂x

What is SxF .

xy (c) If x = 5 and y = 6, then 1+xy ≈ 0.968. If x changes by 10%, using the quantity F Sx derived in part (10b), approximately what percentage change will the quantity xy undergo? 1+xy

(d) Let F (x, y) =

1 . xy

What is SxF .

(e) Let F (x, y) = xy. What is SxF .

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4. Consider the difference equation pk+1 = αpk + βuk with the following parameter values, initial condition and terminal condition:   R α= 1+ , β = −1, uk = M for all k, p0 = L, p360 = 0 12

(1.6)

(1.7)

where R, M and L are constants. (a) In order for the terminal condition to be satisfied (p360 = 0), the quantities R, M and L must be related. Find that relation. Express M as a function of R and L, M = f (R, L). (b) Is M a linear function of L (with R fixed)? If so, express the relation as M = g(R)L, where g is a function you can calculate. (c) Note that the function g is not a linear function of R. Calculate dg dR R=0.065 (d) Plot g(R) and a linear approximation, defined below dg gl (R) := g(0.065) + [R − 0.065] dR R=0.065 for R is the range 0.01 to 0.2. Is the linear approximation relatively accurate in the range 0.055 to 0.075? (e) On a 30 year home loan of $400,000, what is the monthly payment, assuming an annual interest rate of 3.75%. Hint: The amount owed on a fixed-interestrate mortgage from month-to-month is represented by the difference equation in equation (1.6). The interest is compounded monthly. The parameters in (1.7) all have appropriate interpretations. 5. Consider the shower example. Suppose that there is extra delay in the shower’s response, but that your strategy is not modified to take this into account. We derived that the equation governing the closed-loop system is Tk+m = Tk+m−1 + λ (Tdes,k − Tk ) where the time-delay from the water passing through the mixing value to the water touching your skin is m∆. Using calculators, spreadsheets, computers (and/or graphs) or analytic formula you can derive, determine the values of λ for which the system is stable for the following cases: (a) m = 2, (b) m = 3, (c) m = 5. Remark

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1: Remember, for m = 1, the allowable range for λ is 0 < λ < 2. Hint: For a first attempt, assume that the water in the piping at k = 0 is all cold, so that T0 , T1 , . . . , Tm−1 = TC , and that Tdes,k = 12 (TH + TC ). Compute, via the formula, Tk for k = 0, 1, . . . , 100 (say), and plot the result. 6. In this class, we will deal with differential equations having real coefficients, and real initial conditions, and hence, real solutions. Nevertheless, it will be useful to use complex numbers in certain calculations, simplifying notation, √ and allowing us to write only 1 equation when there are actually two. Let j denote −1. Recall that if γ is a p 2 2 complex number, then |γ| = γR + γI , where γR := Real(γ) and γI := Imag(γ), and γ = γR + jγI and γR and γI are real numbers. If γ 6= 0, then the angle of γ, denoted ∠γ, satisfies cos ∠γ =

γR , |γ|

sin ∠γ =

γI |γ|

and is uniquely determinable from γ (only to within an additive factors of 2π). (a) Draw a 2-d picture (horizontal axis for Real part, vertical axis for Imaginary part) (b) Suppose A and B are complex numbers. Using the numerical definitions above, carefully derive that |AB| = |A| |B| ,

∠ (AB) = ∠A + ∠B

7. (a) Given a real numbers θ1 and θ2 , using basic trigonometry and show that [cos θ1 + j sin θ1 ] [cos θ2 + j sin θ2 ] = cos(θ1 + θ2 ) + j sin(θ1 + θ2 ) (b) How is this related to the identity ejθ = cos θ + j sin θ 8. Given a complex number G, and a real number θ, show that (here, j :=
Re Gejθ = |G| cos (θ + ∠G) 9. Given a real number ω, and real numbers A and B, show that
1/2 A sin ωt + B cos ωt = A2 + B 2 sin (ωt + φ) for all t, where φ is an angle that satisfies cos φ =

A (A2 +

B 2 )1/2

,

sin φ =

B (A2 + B 2 )1/2

√

−1)

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Note: you can only determine φ to within an additive factor of 2π. How are these conditions different from saying just tan φ =

B A

10. Draw the block diagram for temperature control in a refrigerator. What disturbances are present in this problem? 11. Take a look at the four journals • IEEE Control Systems Magazine • IEEE Transactions on Control System Technology • ASME Journal of Dynamic Systems Measurement and Control • AIAA Journal on Guidance, Navigation and Control All are in the Engineering Library. More importantly, they are available online, and if you are working from a UC Berkeley computer, you can access the articles for free (see the UCB library webpage for instructions to configure your web browser at home with the correct proxy so that you can access the articles from home as well, as needed). (a) Find 3 articles that have titles which interest you. Make a list of the title, first author, journal/vol/date/page information. (b) Look at the articles informally. Based on that, pick one article, and attempt to read it more carefully, skipping over the mathematics that we have not covered (which may be alot/most of the paper). Focus on understanding the problem being formulated, and try to connect it to what we have discussed. (c) Regarding the paper’s Introduction section, describe the aspects that interest you. Highlight or mark these sentences. (d) In the body of the paper, mark/highlight figures, paragraphs, or parts of paragraphs that make sense. Look specifically for graphs of signal responses, and/or block diagrams. (e) Write a 1 paragraph summary of the paper. Turn in the paper with marks/highlights, as well as the title information of the other papers, and your short summary.

ME 132, Fall 2018, UC Berkeley, A. Packard

2

13

Block Diagrams

In this section, we introduce some block-diagram notation that is used throughout the course, and common to control system grammar.

2.1

Common blocks, continuous-time

The names, appearance and mathematical meaning of a handful of blocks that we will use are shown below. Each block maps an input signal (or multiple input signals) into an output signal via a prescribed, well-defined mathematical relationship. Name

Diagram u-

γ

(example) w Sum w Difference u-

7.2 + -e +6 z + -e −6 z

R

R

Static Nonlinearity u(example)

y(t) = γu(t), ∀t

γ = 7.2

y(t) = γu(t), ∀t

y -

y -

y(t) = w(t) + z(t), ∀t y -

y(t) = w(t) − z(t), ∀t y -

Rt

u(τ )dτ, ∀t ≥ t0

y0 , t0 given

y(t) = y0 +

y0 , t0 given

y(t0 ) = y0 ; y(t) ˙ = u(t) ∀t ≥ t0

t0

y -

Integrator u-

γ∈R

y

Integrator u-

Continuous

-

Gain u-

Info

y

Ψ

-

sin

-

Ψ:R→R

y(t) = Ψ(u(t)), ∀t

Ψ(·) = sin(·)

y(t) = sin(u(t)), ∀t

y

ME 132, Fall 2018, UC Berkeley, A. Packard

u - delay, T Delay

2.2

14

y -

T ≥0

y(t) = u(t − T ), ∀t

Example

Consider a toy model of a stick/rocket balancing problem, as shown below. d B B B

u

) B Bθ B B -B

Imagine the stick is supported at it’s base, by a force approximately equal to it’s weight. This force is not shown. A sideways force can act at the base as well, this is denoted by u, and a sideways force can act at the top, denoted by d. This is similar to the dynamic instabilities of a rocket, just after launch, when the velocity is quite slow, and the only dominant forces/moments are from gravity and the engine thrust: 1. The large thrust of the rocket engines essentially cancels the gravitational force, and the rocket is effectively “balanced” in a vertical position; 2. If the rocket rotates away from vertical (for example, a positive θ), then the moment/torque about the bottom end causes θ to increase; 3. The vertical force of the rocket engines can be “steered” from side-to-side by powerful motors which move the rocket nozzles a small amount, generating a horizontal force (represented by u) which induces a torque, and causes θ to change; 4. Winds (and slight imbalances in the rocket structure itself) act as disturbance torques (represented by d) which must be compensated for; So, without a control system to use u to balance the rocket, it would tip over. For the purposes of this example, the differential equation “governing” the angle-of-orientation is taken to be ˙ = θ(t) + u(t) − d(t). θ(t)

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where θ is the angle-of-orientation, u is the horizontal control force applied at the base, and d is the horizontal disturbance force applied at the tip. Remark: This is not the correct equation, as Newton’s laws would correctly involve the 2nd derivative, θ¨ and terms with cos θ, sin θ, θ˙2 and so on. However this simple model does yield an interesting ˙ negative θ contributes to a negative unstable system (positive θ contributes to a positive θ; ˙ which can be controlled with proportional control. As an exercise, try balancing a stick θ) or ruler in your hand (or better yet, on the tip of your finger). Here, using a simple Matlab code, we will see the effect of a simple proportional feedback control strategy u(t) = 5 [θdes (t) − θ(t)] . This will result in a stable system, with a steerable rocket trajectory (the actual rocket inclination angle θ(t) will generally track the reference inclination angle θdes (t)). Interestingly, although the strategy for u is very simple, the actual signal u(t) as a function of t is somewhat complex, for instance, when the conditions are: θ(0) = 0, θdes (t) = 0 for 0 ≤ t ≤ 2, θdes (t) = 1 for 2 < t, and d(t) = 0 for 0 ≤ t ≤ 6, d(t) = −0.6 for 6 < t. The Matlab files, and associated plots are shown at the end of this section, after Section 2.4. Depending on your point-of-view, the resulting u(t), and it’s affect on θ might seem almost magical to have arisen from such a simple proportional control strategy. This is a great illustration of the power of feedback. Nevertheless, let’s return to the main point of this section, block diagrams. How can this composite system be represented in block diagram form? Use an “integrator” to transform θ˙ into θ, independent of the governing equation. θ˙ -

R

θ -

Then, “create” θ˙ in terms of θ, u and d, as the governing equation dictates. This requires summing junctions, and simple connections, resulting in d u

˙ g - g θ- ? 6

R

θ -

We discussed a proportional control strategy, namely u(t) = 5 [θdes (t) − θ(t)] which looks like

ME 132, Fall 2018, UC Berkeley, A. Packard

θdes + -e −6 θ

-

5

16

u -

Putting these together yields the closed-loop system. See problem 1 in Section 2.4 for an extension to a proportional-integral control strategy.

2.3

Summary

It is important to remember that while the governing equations are almost always written as differential equations, the detailed block diagrams almost always are drawn with integrators (and not differentiators). This is because of the mathematical equivalence shown in the Integrator entry in the table in section 2.1. Using integrators to represent the relationship, the figure conveys how the derivative of some variable, say x˙ is a consequence of the values of other variables. Then, the values of x evolve simply through the “running integration” of this quantity.

ME 132, Fall 2018, UC Berkeley, A. Packard

2.4

17

Problems

1. This question extends the example we discussed in class. Recall that the process was governed by the differential equation ˙ = θ(t) + u(t) − d(t). θ(t) The proportional control strategy u(t) = 5 [θdes (t) − θ(t)] did a good job, but there was room for improvement. Consider the following strategy u(t) p(t) a(t) ˙ e(t)

= = = =

p(t) + a(t) Ke(t) Le(t) (with a(0) = 0) θdes (t) − θmeas (t)

where K and L are constants. Note that the control action is made up of two terms p and a. The term p is proportional to the error, while term a’s rate-of-change is proportional to the error. (a) Convince yourself (and me) that a block diagram for this strategy is as below (all missing signs on summing junctions are + signs). Note: There is one minor issue you need to consider - exchanging the order of differentiation with multiplication by a constant...

-

θdes

-f −6

-

R

K

-

L

? -f

u -

θmeas (b) Create a Simulink model of the closed-loop system (ie., process and controller, hooked up) using this new strategy. The step functions for θdes and d should be namely 0 for t ≤ 1 θdes (t) = 1 for 1 < t ≤ 7 , 1.4 for t > 7

0 for t ≤ 6 d(t) = −0.4 for 6 < t ≤ 11 0 for t > 11

Make the measurement perfect, so that θmeas = θ. (c) Simulate the closed-loop system for K = 5, L = 9. The initial condition for θ should be θ(0) = 0. On three separate axis (using subplot, stacked vertically, all with identical time-axis so they can be “lined” up for clarity) plot θdes and d versus t; θ versus t; and u versus t.

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(d) Comment on the “performance” of this control strategy with regards to the goal of “make θ follow θdes , even in the presence of nonzero d.” What aspect of the system response/behavior is insensitive to d? What signals are sensitive to d, even in the steady-state? (e) Suppose the process has up to 30% “variablity” due to unknown effects. By that, suppose that the process ODE is ˙ = γθ(t) + βu(t) − d(t). θ(t) where γ and β are unknown numbers, known to satisfy 0.7 ≤ γ ≤ 1.3 and 0.7 ≤ β ≤ 1.3. Using for loops, and rand (this generates random numbers uniformly distributed between 0 and 1, hence 0.7 + 0.6*rand generates a random number uniformly distributed between 0.7 and 1.3. Simulate the system 50 times (using different random numbers for both γ and β, and plot the results on a single 3axis (using subplot) graph (as in part 1c above). What aspect of the closed-loop system’s response/behavior is sensitive to the process variability? What aspects are insensitive to the process variability? (f) Return to the original process model. Simulate the closed-loop system for K = 5, and five values of L, {1, 3.16, 10, 31.6, 100}. On two separate axis (using subplot and hold on), plot θ versus t and u versus t, with 5 plots (the different values of L) on each axis. (g) Discuss how the value of the controller parameter L appears to affect the performance. (h) Return to the case K = 5, L = 9. Now, use the “transport delay” block (found in the Continuous Library in Simulink) so that θmeas is a delayed (in time) version of θ. Simulate the system for 3 different values of time-delay, namely T = {0.001, 0.01, 0.1}. On one figure, superimpose all plots of θ versus t for the three cases. (i) Comment on the effect of time-delay in the measurement in terms of affecting the regulation (ie., θ behaving like θdes . (j) Return to the case of no measurement delay, and K = 5, L = 9. Now, use the “quantizer” block (found in the Nonlinear Library in Simulink) so that θmeas is the output of the quantizer block (with θ as the input). This captures the effect of measuring θ with an angle encoder. Simulate the system for 3 different levels of quantization, namely {0.001, 0.005, 0.025}. On one figure, make 3 subplots (one for each quantization level), and on each axis, graph both θ and θmeas versus t. On a separate figure, make 3 subplots (one for each quantization level), graphing u versus t. (k) Comment on the effect of measurement quantization in terms of limiting the accuracy of regulation (ie., θ behaving like θdes , and on the “jumpiness” of the control action u).

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NOTE: All of the computer work (parts 1c, 1e, 1f, 1h and 1j) should be automated in a single, modestly documented script file. Turn in a printout of your Simulink diagrams (3 of them), and a printout of the script file. Also include nicely formatted figure printouts, and any derivations/comments that are requested in the problem statement.

ME 132, Fall 2018, UC Berkeley, A. Packard

3 3.1

24

Mathematical Modeling and Simulation Systems of 1st order, coupled differential equations

Consider an input/output system, with m inputs (denoted by d), q outputs (denoted by e), and governed by a set of n, 1st order, coupled differential equations, of the form     f1 (t, x1 (t), x2 (t), . . . , xn (t), d1 (t), d2 (t), . . . , dm (t)) x˙ 1 (t)  x˙ 2 (t)   f2 (t, x1 (t), x2 (t), . . . , xn (t), d1 (t), d2 (t), . . . , dm (t))    .   ..   .   .   .       f (t, x (t), x (t), . . . , x (t), d (t), d (t), . . . , d (t)) x ˙ (t)   n   n 1 2 n 1 2 m = (3.1)      e1 (t)   h1 (t, x1 (t), x2 (t), . . . , xn (t), d1 (t), d2 (t), . . . , dm (t))       e2 (t)   h2 (t, x1 (t), x2 (t), . . . , xn (t), d1 (t), d2 (t), . . . , dm (t))    .   ..   ..   . eq (t) hq (t, x1 (t), x2 (t), . . . , xn (t), d1 (t), d2 (t), . . . , dm (t)) where the functions fi , hi are given functions of t as well as the n variables x1 , x2 , . . . , xn , the m variables d1 , d2 , . . . , dm . For shorthand, we write (3.1) as x(t) ˙ = f (t, x(t), d(t)) e(t) = h (t, x(t), d(t)) Given an initial condition vector x0 , and for the solutions  x1 (t)  x2 (t)  x(t) =  ..  . xn (t)

(3.2)

a forcing function d(t) for t ≥ t0 , we wish to solve    e1 (t)   e2 (t)     e(t) =  ..  ,   .  eq (t)

on the interval [t0 , tF ], given the initial condition x (t0 ) = x0 . and the input forcing function d(·). ode45 solves for this using numerical integration techniques, such as 4th and 5th order RungeKutta formulae. You should have learned about this in E7, and used ode45 extensively. You can learn more about numerical integration by taking Math 128. We will not discuss this important topic in detail in this class - please review your E7 material.

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Remark: The GSI will give 2 interactive discussion sections on how to use Simulink, a graphical-based tool to easily build and numerically solve ODE models by interconnecting individual components, each of which are governed by ODE models. Simulink is part of Matlab, and is available on the computers in Etcheverry. The Student Version of Matlab also comes with Simulink (and the Control System Toolbox). The Matlab license from the UC Berkeley software licensing arrangement has both Simulink (and the Control System Toolbox included. However, to quickly recap (in an elementary manner) how ODE solvers work, consider the Euler method of solution. If the functions f and d are “reasonably” well behaved in x and t, then the solution, x(·) exists, is a continuous function of t, and in fact, is differentiable at all points. Hence, it is reasonable that a Taylor series for x at a given time t will be predictive of the values of x(τ ) for values of τ close to t. If we do a Taylor’s expansion on a function x, and ignore the higher order terms, we get an approximation formula x(t + δ) ≈ x(t) + δ x(t) ˙ = x(t) + δf (t, x(t), d(t)) Roughly, the “smaller” δ is, the closer the left-hand-side is to the actual value of x(t + δ). Euler’s method propogates a solution to (3.1) by using this approximation repeatedly for a fixed δ, called “the stepsize.” Hence, Euler’s method gives that for any integer k >= 0, the solution to (3.1) approximately satisfies x((k + 1)δ) = x(kδ) + δ f (kδ, x(kδ), d(kδ)) | {z } | {z } | {z } n×1

n×1

n×1

Writing out the first 4 time steps (ie., t = 0, δ, 2δ, 3δ, 4δ) gives x(δ) x(2δ) x(3δ) x(4δ)

= = = =

x(0) + δf (0, x(0), d(0)) x(δ) + δf (δ, x(δ), d(δ)) x(2δ) + δf (2δ, x(2δ), d(2δ)) x(3δ) + δf (3δx(3δ), d(3δ))

(3.3)

and so on. So, as long as you have a subroutine that can evaluate f (t, x, d), given any values of t, x and d, you can quickly propogate an approximate solution simply by calling the subroutine once for every timestep. Computing the output, e(t) simply involves evaluating the function h(t, x(t), d(t)) at the solution points. In the Runge-Kutta method, a more sophisticated approximation is made, which results in more computations (4 function evaluations of f for every time step), but much greater

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accuracy. In effect, more terms of the Taylor series are used, involving matrices of partial derivatives, and even their derivatives, df d2 f d3 f , , dx dx2 dx3 but without actually requiring explicit knowledge of these derivatives of the function f .

3.2

Remarks about Integration Options in simulink

The Simulation → SimulationParameters → Solver page is used to set additional optional properties, including integration step-size options, and may need to be used to obtain smooth plots. Additional options are Term RelTol

AbsTol

MaxStep

3.3

Meaning Relative error tolerance, default 1e-3, probably leave it alone, though if instructions below don’t work, try making it a bit smaller Absolute error tolerance, default 1e-6, probably leave it alone, though if instructions below don’t work, try making it a bit smaller maximum step size. I believe the default is (StopTimeStartTime)/50. In general, make it smaller than (StopTime-StartTime)/50 if your plots are jagged.

Problems

1. A sealed-box loudspeaker is shown below.

ME 132, Fall 2018, UC Berkeley, A. Packard

Top Plate

27

Voice Coil Former

Magnet Back Plate Vent

Signal From Accelerometer

From Amplifier Dust Cap Spider Pole Piece Voice Coil

Woofer Cone

Basket Subwoofer Electrical Leads

Surround Sealed Enclosure

Ignore the wire marked “accelerometer mounted on cone.” We will develop a model for accoustic radiation from a sealed-box loudspeaker. The equations are • Speaker Cone: force balance, m¨ z (t) = Fvc (t) + Fk (t) + Fd (t) + Fb + Fenv (t) • Voice-coil motor (a DC motor): ˙ − RI(t) − Bl (t)z(t) Vin (t) − LI(t) ˙ = 0 Fvc (t) = Bl (t)I(t) • Magnetic flux/Length Factor: Bl (t) =

BL0 1 + BL1 z(t)4

• Suspension/Surround: Fk (t) = −K0 z(t) − K1 z 2 (t) − K2 z 3 (t) Fd (t) = −RS z(t) ˙ • Sealed Enclosure: 

Az(t) Fb (t) = P0 A 1 + V0

−γ

• Environment (Baffled Half-Space) x(t) ˙ = Ae x(t) + Be z(t) ˙ Fenv (t) = −AP0 − Ce x(t) − De z(t) ˙

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where x(t) is 6 × 1, and  −474.4 4880 0 0 0 0  −4880 −9376 0 0 0 0   0 0 −8125 7472 0 0 Ae :=   0 0 −7472 −5.717 0 0   0 0 0 0 −3515 11124 0 0 0 0 −11124 −2596     Be :=    

Ce :=



−203.4 −594.2 601.6 15.51 213.6 140.4

       

       

203.4 −594.2 −601.6 15.51 213.6 −140.4



De = 46.62 This is an approximate model of the impedance “seen” by the face of the loudspeaker as it radiates into an “infinite half-space.” The values for the various constants are Symbol Value A 0.1134 meters2 V0 0.17 meters3 P0 1.0133 × 105 Pa m 0.117 kg L 7 × 10−4 H R 3Ω BL0 30.7 Tesla · meters BL1 107 meters−4 K0 5380 N/meter K1 0 K3 2 × 108 N/meter3 RS 12.8 N · sec/meter (a) Build a Simulink model for the system. Use the Subsystem grouping capability to manage the complexity of the diagram. Each of the equations above should represent a different subsystem. Make sure you think through the question “what is the input(s) and what is the output(s)?” for each subsystem.

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(b) Write a function that has two input arguments (denoted V¯ and Ω) and two output arguments, zmax,pos and zmax,neg . The functional relationship is defined as follows: Suppose Vin (t) = V¯ sin Ω · 2πt, where t is in seconds, and V¯ in volts. In other words, the input is a sin-wave at a frequency of Ω Hertz. Simulate the loudspeaker behavior for about 25/Ω seconds. The displacement z of the cone will become nearly sinusoidal by the end of the simulation. Let zmax,pos and zmax,neg be the maximum positive and negative values of the displacement in the last full cycle (i.e., the last 1/Ω seconds of the simulation, which we will approximately decree as the “steady-state response to the input”). (c) Using bisection, determine the value (approximately) of V¯ so that the steadystate maximum (positive) excursion of z is 0.007meters if the frequency of the excitation is Ω = 25. What is the steady-state minimum (negative) excursion of z?

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State Variables

See the appendix for additional examples on mathematical modeling of systems.

4.1

Definition of State

For any system (mechanical, electrical, electromechanical, economic, biological, acoustic, thermodynamic, etc.) a collection of variables q1 , q2 , . . . , qn are called state variables, if the knowledge of • the values of these variables at time t0 , and • the external inputs acting on the system for all time t ≥ t0 , and • all equations describing relationships between the variables qi and the external inputs is enough to determine the value of the variables q1 , q2 , . . . , qn for all t ≥ t0 . In other words, past history (before t0 ) of the system’s evolution is not important in determining its evolution beyond t0 – all of the relevant past information is embedded in the variables value’s at t0 . Example: The system is a point mass, mass m. The point mass is acted on by an external force f (t). The position of the mass is measured relative to an inertial frame, with coordinate w, velocity v, as shown below in Fig. 2.

v(t) w(t)

: velocity : position

f(t)

Figure 2: Ideal force acting on a point mass

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Claim #1: The collection {w} is not a suitable choice for state variables. Why? Note that for t ≥ t0 , we have  Z τ Z t 1 v(t0 ) + f (η)dη dτ w(t) = w(t0 ) + t0 t0 m Hence, in order to determine w(t) for all t ≥ t0 , it is not sufficient to know w(t0 ) and the entire function f (t) for all t ≥ t0 . You also need to know the value of v(t0 ). Claim #2: The collection {v} is a legitimate choice for state variables. Why? Note that for t ≥ t0 , we have Z t 1 f (τ )dτ v(t) = v(t0 ) + t0 m Hence, in order to determine v(t) for all t ≥ t0 , it is sufficient to know v(t0 ) and the entire function f (t) for all t ≥ t0 . Claim #3: The collection {w, v} is a legitimate choice for state variables. Why? Note that for t ≥ t0 , we have Rt w(t) = w(t0 ) + t0 v(η)dη Rt v(t) = v(t0 ) + t0 m1 f (τ )dτ Hence, in order to determine v(t) for all t ≥ t0 , it is sufficient to know v(t0 ) and the entire function f (t) for all t ≥ t0 . In general, it is not too hard to pick a set of state-variables for a system. The next few sections explains some rule-of-thumb procedures for making such a choice.

4.2

State-variables: from first order evolution equations

Suppose for a system we choose some variables (x1 , x2 , . . . , xn ) as a possible choice of state variables. Let d1 , d2 , . . . , df denote all of the external influences (ie., forcing functions) acting on the system. Suppose we can derive the relationship between the x and d variables in the form x˙ 1 = f1 (t, x1 (t), x2 (t), . . . , xn (t), d1 (t), . . . , df (t)) x˙ 2 = f2 (t, x1 (t), x2 (t), . . . , xn (t), d1 (t), . . . , df (t)) (4.1) .. .. . . x˙ n = fn (t, x1 (t), x2 (t), . . . , xn (t), d1 (t), . . . , df (t)) Then, the set {x1 , x2 , . . . , xn } is a suitable choice of state variables. Why? Ordinary differential equation (ODE) theory tells us that given • an initial condition x(t0 ) := x0 , and • the forcing function d(t) for t ≥ t0 ,

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there is a unique solution x(t) which satisfies the initial condition at t = t0 , and satisfies the differential equations for t ≥ t0 . Hence, the set {x1 , x2 , . . . , xn } constitutes a state-variable description of the system. The equations in (4.1) are called the state equations for the system.

4.3

State-variables: from a block diagram

Given a block diagram, consisting of an interconnection of • integrators, • gains, and • static-nonlinear functions, driven by external inputs d1 , d2 , . . . , df , a suitable choice for the states is • the output of each and every integrator Why? Note that • if the outputs of all of the integrators are labled x1 , x2 , . . . , xn , then the inputs to the integrators are actually x˙ 1 , x˙ 2 , . . . , x˙ n . • The interconnection of all of the base components (integrators, gains, static nonlinear functions) implies that each x˙ k (t) will be a function of the values of x1 (t), x2 (t), . . . , xn (t) along with d1 (t), d2 (t), . . . , df (t). • This puts the equations in the form of (4.1). We have already determined that that form implies that the variables are state variables.

4.4

Problems

1. Shown below is a block diagram of a DC motor connected to an load inertia via a flexible shaft. The flexible shaft is modeled as a rigid shaft (inertia J1 ) inside the motor, a massless torsional spring (torsional spring constant Ks ) which connects to the load inertia J2 . θ is the angular position of the shaft inside the motor, and ψ is the angular position of the load inertia.

ME 132, Fall 2018, UC Berkeley, A. Packard

- 1

J2

33

-

R

-

R

ψ(t)

Ks 

- y(t)

− ? e + 6

- −1 J1

u(t)

-

β

R + -? e+ +6

-

R θ(t)

−γ  Choose state variables (use rule given in class for block diagrams that do not contain differentiators). Find matrices A, B and C such that the variables x(t), y(t) and u(t) are related by x(t) ˙ = Ax(t) + Bu(t) y(t) = Cx(t) Hint: There are 4 state variables. 2. Using op-amps, resistors, and capacitors, design a circuit to implement a PI controller, with KP = 4, KI = 10. Use small capacitors (these are more readily available) and large resistors (to keep currents low).

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First Order, Linear, Time-Invariant (LTI) ODE

5.1

The Big Picture

We shall study mathematical models described by linear time-invariant input-output differential equations. There are two general forms of such a model. One is a single, high-order differential equation to describe the relationship between the input u and output y, of the form y [n] (t) + a1 y [n−1] (t) + · · · + an−1 y(t) ˙ + an y(t) = b0 u[m] (t) + b1 u[m−1] (t) + · · · + bm−1 u(t) ˙ + bm u(t) (5.1) dk y(t) [k] th where y denotes the k derivative of the signal y(t): dtk . These notes refer to equation (5.1) as a HLODE (High-order, Linear Ordinary Differential Equation). An alternate form involves many first-order equations, and many inputs. The general case of this situation has n dependent variables, x1 , x2 , . . . , xn , and m inputs, d1 , d2 , . . . , dm . The differential equations governing the evolution of the xi variables is x˙ 1 (t) = a11 x1 (t) + a12 x2 (t) + · · · + a1n xn (t) + b11 d1 (t) + b12 d2 (t) . . . + b1m dm (t) x˙ 2 (t) = a21 x1 (t) + a22 x2 (t) + · · · + a2n xn (t) + b21 d1 (t) + b22 d2 (t) . . . + b2m dm (t) .. . . = .. x˙ n (t) = an1 x1 (t) + an2 x2 (t) + · · · + ann xn (t) + bn1 d1 (t) + bn2 d2 (t) . . . + bnm dm (t) We will learn how to solve these differential equations, and more importantly, we will discover how to make broad qualitative statements about these solutions. Much of our intuition about control system design and its limitations will be drawn from our understanding of the behavior of these types of equations. A great many models of physical processes that we may be interested in controlling are not linear, as above. Nevertheless, as we shall see much later, the study of linear systems is a vital tool in learning how to control even nonlinear systems. Essentially, feedback control algorithms make small adjustments to the inputs based on measured outputs. For small deviations of the input about some nominal input trajectory, the output of a nonlinear system looks like a small deviation around some nominal output. The effects of the small input deviations on the output is well approximated by a linear (possibly time-varying) system. It is therefore essential to undertake a study of linear systems. In this section, we review the solutions of linear, first-order differential equations with constant coefficients and time-dependent forcing functions. The concepts of • stability

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• time-constant • sinusoidal steady-state • frequency response functions are introduced. A significant portion of the remainder of the course generalizes these to higher order, linear ODEs, with emphasis on applying these concepts to the analysis and design of feedback systems.

5.2

Solution of a First Order LTI ODE

Consider the following system x(t) ˙ = a x(t) + b u(t)

(5.2)

where u is the input, x is the dependent variable, and a and b are constant coefficients (ie., numbers). For example, the equation 6x(t) ˙ + 3x(t) = u(t) can be manipulated into 1 1 x(t) ˙ = − 2 x(t) + 6 u(t). Given the initial condition x(0) = x0 and an arbitrary input function u(t) defined for t ∈ [0, ∞), a solution of Eq. (5.2) must satisfy Z t at xs (t) = e x0 + ea(t−τ ) b u(τ ) dτ . (5.3) | {z } 0 {z } free resp. | forced resp. You should derive this with the integrating factor method (problem 1). Also note that the derivation makes it evident that given the initial condition and forcing function, there is at most one solution to the ODE. In other words, if solutions to the ODE exist, they are unique, once the initial condition and forcing function are specified. We can also easily just check that (5.3) is indeed the solution of (5.2) by verifying two facts: the function xs satisfies the differential equation for all t ≥ 0; and xs satisfies the given initial condition at t = 0. In fact, the theory of differential equations tells us that there is one and only one function that satisfies both (existence and uniqueness of solutions). For this ODE, we can prove this directly, for completeness sake. Above, we showed that if x satisfies (5.2), then it must be of the form in (5.3). Next we show that the function in (5.3) does indeed satisfy the differential equation (and initial condition). First check the value of xs (t) at t = 0: Z 0 a0 xs (0) = e x0 + ea(t−τ ) b u(τ ) dτ = x0 . 0

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Taking the time derivative of (5.3) we obtain   Z t d −aτ at at e b u(τ ) dτ e x˙ s (t) = a e x0 + dt 0 Z t at at = a e x0 + a e e−aτ b u(τ ) dτ +eat e−at b u(t) | {z0 } axs (t)

= a xs (t) + b u(t) . as desired. 5.2.1

Free response

Fig. 3 shows the normalized free response (i.e. u(t) = 0) of the solution Eq. (5.3) of ODE (5.2) when a < 0. Since a < 0, the free response decays to 0 as t → ∞, regardless of the 1 0.9 0.8

Normalized State [x/xo]

0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

0.5

1

1.5

2 2.5 3 Normalized Time [t/|a|]

3.5

4

4.5

5

Figure 3: Normalized Free response of first order system (a < 0) initial condition. Because of this (and a few more properties that will be derived in upcoming sections), if a < 0, the system in eq. (5.2) is called stable (or sometimes, to be more precise, asymptotically stable). Notice that the slope at time t = 0 is x(0) ˙ = ax0 and T = 1/|a| is the time that x(t) would cross 0 if the initial slope is continued, as shown in the figure. The time 1 T := |a|

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is called the time constant of a first order asymptotically stable system (a < 0). T is expressed in the units of time, and is an indication of how fast the system responds. The larger |a|, the smaller T and the faster the response of the system. Notice that xf ree (T ) x0 xf ree (2T ) x0 xf ree (3T ) x0 xf ree (4T ) x0

1 ≈ .37 = 37% e 1 = 2 ≈ .13 = 13% e 1 = 3 ≈ .05 = 5% e 1 = 4 ≈ .018 ≈ 2% e =

This calculation is often summarized informally as ”in a relative sense, the free-response of a stable, first-order system decays to zero in approximately 3 time-constants.” Of course, in that usage, the notion of “decays to zero” is quite vague, and actually means that 95% of the initial condition has decayed away after 3 time-constants. Obviously, a different notion of “decays to zero” yields a different rule-of-thumb for the time to decay. If a > 0 the free response of ODE (5.2) is unstable, i.e. limt→∞ |x(t)| = ∞. When a = 0, x(t) = x0 for all t ≥ 0, and we can say that this system is limitedly stable or limitedly unstable. 5.2.2

Forced response, constant inputs

We first consider the system response to a step input. In this case, the input u(t) is given by  0 if t < 0 u(t) = um µ(t) = um if t ≥ 0 where um is a constant and x(0) = 0. The solution (5.3) yields
b 1 − eat um . x(t) = −a b If a < 0, the steady state output xss is xss = −a um . Note that for any t x(t + −1 ) − xss = 1 |x(t) − xss | ≈ 0.37 |x(t) − xss | e a

time-units, the solution moves 63% closer to its So, when the input is constant, every −1 a final, limiting value. Exercise 5 gives another interesting, and useful interpretation of the time-constant.

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5.2.3

38

Forced response, bounded inputs

Rather than consider constant inputs, we can also consider inputs that are bounded by a constant, and prove, under the assumption of stability, that the response remains bounded as well (and the bound is a linear function of the input bound). Specifically, if a < 0, and if u(t) is uniformly (in time) bounded by a positive number M , then the resulting solution . To derive this, suppose that |u(τ )| ≤ M for all τ ≥ 0. x(t) will be uniformly bounded by bM |a| Then for any t ≥ 0, we have Z t a(t−τ ) |x(t)| = e b u(τ ) dτ 0 Z t a(t−τ ) e ≤ b u(τ ) dτ Z0 t ≤ ea(t−τ ) b M dτ 0


bM ≤ 1 − eat −a bM . ≤ −a Thus, if a < 0, x(0) = 0 and u(t) ≤ M , the output is bounded by x(t) ≤ |bM/a|. This is called a bounded-input, bounded-output (BIBO) system. If the initial condition is non-zero, the output x(t) will still be bounded since the magnitude of the free response monotonically converges to zero, and the response x(t) is simply the sum of the free and forced responses. Note: Assuming b 6= 0, the system is not bounded-input/bounded-output when a ≥ 0. In that context, from now on, we will refer to the a = 0 case (termed limitedly stable or limitedly unstable before) as unstable. See problem 6 in Section 5.6.

5.2.4

Stable system, Forced response, input approaching 0

Assume the system is stable, so a < 0. Now, suppose the input signal u is bounded and approaches 0 as t → ∞. It seems natural that the response, x(t) should also approach 0 as t → ∞, and deriving this fact is the purpose of this section. While the derivation is interesting, the most important is the result: For a stable system, specifically (5.2), with a < 0, if the input u satisfies limt→∞ u(t) = 0, then the solution x satisfies limt→∞ x(t) = 0. Onto the derivation: First, recall what limt→∞ z(t) = 0 means: for any  > 0, there is a T > 0 such that for all t > T , |z(t)| < .

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Next, note that for any t > 0 and a 6= 0, Z t
1 ea(t−τ ) dτ = − 1 − eat a 0 If a < 0, then for all t ≥ 0, Z Z t Z t aτ a(t−τ ) e dτ ≤ e dτ = 0

0

0

∞

1 1 eaτ dτ = − = a |a|

Assume x0 is given, and consider such an input. Since u is bounded, there is a positive constant B such that |u(t)| ≤ B for all t ≥ 0. Also, for every  > 0, there is • a T,1 > 0 such that |u(t)| < t

• a T,2 > 0 such that ea 2 < • a T,3 > 0 such that eat
T , the response x(t) satisfies Rt x(t) = eat x0 + 0 ea(t−τ ) bu(τ )dτ Rt Rt = eat x0 + 02 ea(t−τ ) bu(τ )dτ + t ea(t−τ ) bu(τ )dτ 2

We can use the information to bound each term individually, namely 1. Since t ≥ T,3 at e x0 ≤

  |x0 | = 3|x0 | 3

2. Since |u(t)| ≤ B for all t, R t Rt 2 a(t−τ ) bu(τ )dτ ≤ 02 ea(t−τ ) |bu(τ )| dτ 0 e Rt ) ≤ B|b| 02 ea(t−τ dτ   t 1 a 2t = B|b| −a e 1 − ea 2 t

1 a2 ≤ B|b| −a e t

Since t ≥ T,2 , ea 2 ≤

|a| , 3B|b|

which implies Z t 2  ea(t−τ ) bu(τ )dτ < 0 3

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3. Finally, since t ≥ T,1 , the input |u(τ )| < 3 |a| for all τ ≥ 2t . This means |b| R R t a(t−τ ) t a(t−τ ) bu(τ )dτ < |b| 3 |a| dτ te t e |b| 2 2 R t ea(t−τ ) dτ ≤ |b| 3 |a| |b| 0 1 ≤ |b| 3 |a| |b| |a| = 3 Combining implies that for any t > T , |x(t)| < . Since  was an arbitrary positive number, this complete the proof that limt→∞ x(t) = 0. 5.2.5

Linearity

Why is the differential equation in (5.2) called a linear differential equation? Suppose x1 is the solution to the differential equation with the initial condition x0,1 and forcing function u1 , and x2 is the solution to the differential equation with the initial condition x0,2 and forcing function u2 . In other words, the function x1 satisfies x1 (0) = x1,0 and for all t > 0, x˙ 1 (t) = ax1 (t) + bu1 (t). Likewise, the function x2 satisfies x2 (0) = x2,0 and for all t > 0, x˙ 2 (t) = ax2 (t) + bu2 (t). Now take two constants α and β. What is the solution to the differential equation with initial condition x(0) = αx0,1 + βx0,2 , and forcing function u(t) = αu1 (t) + βu2 (t)? It is easy (just plug into the differential equation, or the integral form of the solution) to see that the solution is x(t) = αx1 (t) + βx2 (t) This is often called the superposition property. In this class, we will more typically use the term linear, indicating that the solution of the differential equation is a linear function of the (initial condition, forcing function) pair.

5.2.6

Forced response, input approaching a constant limit

If the system is stable (a < 0), and the input u(t) has a limit, limt→∞ u(t) = u¯, then by combining the results of Sections 5.2.4, 5.2.2 and 5.2.5, it is easy to conclude that x(t) approaches a limit as well, namely b lim x(t) = − u¯ t→∞ a

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An alternate manner to say this is that b x(t) = − u¯ + q(t) a where limt→∞ q(t) = 0.

5.3

Forced response, Sinusoidal inputs

Consider the linear dynamical system x(t) ˙ = Ax(t) + Bu(t) y(t) = Cx(t) + Du(t)

(5.4)

We assume that A, B, C and D are scalars (1 × 1). If the system is stable, (ie., A < 0) it might be “intuitively” clear that if u is a sinusoid, then y will approach a steady-state behavior that is sinusoidal, at the same frequency, but with different amplitude and shifted in-time. In this section, we make this idea precise. Take ω ≥ 0 as the input frequency, and (although not physically relevant) let u¯ be a fixed complex number and take the input function u(t) to be u(t) = u¯ejωt for t ≥ 0. Note that this is a complex-valued function of t. Then, the response is Rt x(t) = eAt x0 + 0 eA(t−τ ) Bu(τ )dτ Rt = eAt x0 + eAt 0 e−Aτ B u¯ejωτ dτ Rt = eAt x0 + eAt 0 e(jω−A)τ dτ B u¯ Now, since A < 0, regardless of ω, (jω − A) 6= 0, and we can solve the integral as   B u¯ B At x(t) = e x0 − + ejωt jω − A jω − A

(5.5)

(5.6)

Hence, the output y(t) = Cx(t) + Du(t) would satisfy     B u¯ CB At y(t) = Ce x0 − + D+ u¯ejωt jω − A jω − A In the limit as t → ∞, the first term decays to 0 exponentially, leaving the steady-state response   CB yss (t) = D + u¯ejωt jω − A

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Hence, we have verified our initial claim – if the input is a complex sinusoid, then the steadystate output is a complex sinusoid at the same exact frequency, but “amplified” by a complex i h CB gain of D + jω−A . The function G(ω) G(ω) := D +

CB jω − A

(5.7)

is called the frequency response function of the system in (5.4). Hence, for stable, first-order systems, we have proven u(t) := u¯ejωt ⇒ yss (t) = G(ω)¯ uejωt G can be calculated rather easily using a computer, simply by evaluating the expression in (5.7) at a large number of frequency points ω ∈ R. The dependence on ω is often graphed using two plots, namely • log10 |G(ω)| versus log10 (ω), and • ∠G(ω) versus log10 (ω). This plotting arrangement is called a Bode plot, named after one of the modern-day giants of control system theory, Hendrik Bode. What is the meaning of a complex solution to the differential equation (5.4)? Suppose that functions u, x and y are complex, and solve the ODE. Denote the real part of the function u as uR , and the imaginary part as uI (similar for x and y). Then, for example, xR and xI are real-valued functions, and for all t x(t) = xR (t) + jxI (t). Differentiating gives dxR dxI dx = +j dt dt dt Hence, if x, u and y satisfy the ODE, we have (dropping the (t) argument for clarity) dxR dt

I + j dx = A (xR + jxI ) + B (uR + juI ) dt yR + jyI = C (xR + jxI ) + D (uR + juI )

But the real and imaginary parts must be equal individually, so exploiting the fact that the coefficients A, B, C and D are real numbers, we get dxR dt

yR and

dxI dt

yI

= AxR + BuR = CxR + DuR = AxI + BuI = CxI + DuI

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Hence, if (u, x, y) are functions which satisfy the ODE, then both (uR , xR , yR ) and (uI , xI , yI ) also satisfy the ODE. Finally, we need to do some trig/complex number calculations. Suppose that H ∈ C is not equal to zero. Recall that ∠H is the real number (unique to within an additive factor of 2π) which has the properties cos ∠H =

ImH ReH , sin ∠H = |H| |H|

Then, Re Hejθ




= Re [(HR + jHI ) (cos θ + j sin θ)] = HR cos h θ − HI sin θ i HI R = |H| H cos θ − sin θ |H| |H| = |H| [cos ∠H cos θ − sin ∠H sin θ]
= |H| cos (θ + ∠H) jθ Im He = Im [(HR + jHI ) (cos θ + j sin θ)] = HR sin i h θ + HI cos θ HI R sin θ + |H| cos θ = |H| H |H| = |H| [cos ∠H sin θ + sin ∠H cos θ] = |H| sin (θ + ∠H)

Now consider the differential equation/frequency response case. Let G(ω) denote the frequency response function. If the input u(t) = cos ωt = Re (ejωt ), then the steady-state output y will satisfy y(t) = |G(ω)| cos (ωt + ∠G(ω)) A similar calculation holds for sin, and these are summarized below. Input 1 cos ωt sin ωt

5.3.1

Steady-State Output G(0) = D − CB A |G(ω)| cos (ωt + ∠G(ω)) |G(ω)| sin (ωt + ∠G(ω))

Forced response, input approaching a Sinusoid

If the system in (5.4) is stable (A < 0), combine the results of Sections 5.2.4, 5.3 and 5.2.5, to obtain the following result: Suppose A < 0, ω ≥ 0, and u¯ is a constant. If the input u is of the form u(t) = u¯ejωt + z(t)

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and limt→∞ z(t) = 0, then the response y(t) is of the form   CB u¯ejωt + q(t) y(t) = D + jω − A where limt→∞ q(t) = 0. Informally, we conclude eventually sinusoidal inputs lead to eventually sinusoidal outputs, and say that the system has the steady-state, sinusoidal gain (SStG) property. Note that the relationship between the sinuosoidal terms is the frequency-response function.

5.4

First-order delay-differential equation: Stability

In actual feedback systems, measurements from sensors are used to make decisions on what corrective action needs to be taken. Often in analysis, we will assume that the time from when the measurement occurs to when the corresponding action takes place is negligible (ie., zero), since this is often performed with modern, high-speed electronics. However, in reality, there is a time-delay, so that describing the system’s behavior involves relationships among variables at different times. For instance, a simple first-order delay-differential equation is x(t) ˙ = a1 x(t) + a2 x(t − T )

(5.8)

where T ≥ 0 is a fixed number. We assume that for T = 0, the system is stable, so a1 +a2 < 0. Since we are studying the effect of delay, we also assume that a2 6= 0. When T = 0, the homogeneous solutions are of the form x(t) = e(a1 +a2 )t x(0), which decay exponentially to zero. As the constant number T increases from 0, the homogeneous solutions change, becoming complicated expressions that are challenging to derive. It is a fact that there is a critical value of T , called Tc such that • for all T satisfying 0 ≤ T < Tc , the homogeneous solutions of (5.8) decay to zero as t→∞ • for some ω ≥ 0 (denoted ωc ) the function ejωt is a homogeneous solution when T = Tc . Hence, we can determine Tc (and ωc ) by checking the conditions for ejωt to be a homogeneous solution of (5.8). Plugging in give jωejωt = a1 ejωt + a2 ejω(t−T ) for all t. Since ejωt 6= 0, divide, leaving jω = a1 + a2 e−jωT .

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Since this equality relates complex numbers (which have a real and imaginary part), it can be thought of as 2 equations −jωTin 2 unknowns (ω and T ). We know that regardless of ω and = 1, so it must be that T , it always holds that e |jω − a1 | = |a2 | which implies ωc = a1 + a2 e−jωTc .

5.5

p a22 − a21 . Then Tc is the smallest positive number such that jωc =

Summary

In this section, we studied the free and forced response of linear, first-order differential equations. Several concepts and properties were established, including • linearity of solution to initial condition and forcing; • stability; • time-constant; • response to step inputs; • response to sinusoidal inputs; • response to inputs which go to 0; • effect of additional terms from delays. These concepts will be investigated for higher-order differential equations in later sections. Many of the principles learned here carry over to those as well. For that reason, it is important that you develop a mastery of the behavior of forced, first order systems. In upcoming sections, we study simple feedback systems that can be analyzed using only 1st-order differential equations, using all of the facts about 1st-order systems that have been derived.

ME 132, Fall 2018, UC Berkeley, A. Packard

5.6

46

Problems

1. Use the integrating factor method to derive the solution given in equation (5.3) to the differential equation (5.2). 2. Suppose f is a piecewise continuous function. Assume A < B. Explain (with pictures, or equations, etc) why Z B Z B ≤ f (x)dx |f (x)| dx A

A

This simple idea is used repeatedly when bounding the output response in terms of bounds on the input forcing function. 3. Work out the integral in the last line of equation (5.5), deriving equation (5.6). 4. A stable first-order system (input u, output y) has differential equation model x(t) ˙ = ax(t)   + bu(t) b u(t) = a x(t) − −a where a < 0 and b are some fixed numbers. (a) Let τ denote the time-constant, and γ denote the steady-state gain from u → x. Solve for τ and γ in terms of a and b. Also, invert these solutions, expressing a and b in terms of the time-constant and steady-state gain. (b) Suppose τ > 0. Consider a first-order differential equation of the form τ x(t) ˙ = −x(t) + γu(t). Is this system stable? What is the time-constant? What is the steady-state gain from u → x? Note that this is a useful manner to write a firstorder equation, since the time-constant and steady-state gain appear in a simple manner. (c) Suppose τ = 1, γ = 2. Given the initial condition x(0) = 4, and the input signal u u(t) = 1 for 0 ≤ t < 5 u(t) = 2 for 5 ≤ t < 10 u(t) = 6 for 10 ≤ t < 10.1 u(t) = 3 for 10.1 ≤ t < ∞. sketch a reasonably accurate graph of x(t) for t ranging from 0 to 20. The sketch should be based on your understanding of a first-order system’s response (using its time-constant and steady-state gain), not by doing any particular integration.

ME 132, Fall 2018, UC Berkeley, A. Packard

(d) Now suppose τ = 0.001, γ = 2. input signal u u(t) = 1 u(t) = 2 u(t) = 6 u(t) = 3

47

Given the initial condition x(0) = 4, and the for for for for

0 ≤ t < 0.005 0.005 ≤ t < 0.01 0.01 ≤ t < 0.0101 0.0101 ≤ t < ∞.

sketch a reasonably accurate graph of x(t) for t ranging from 0 to 0.02. The sketch should be based on your understanding of a first-order system’s response, not by doing any particular integration. In what manner is this the “same” as the response in part 4c? 5. Consider the first-order linear system x(t) ˙ = A [x(t) − u(t)] 1 , and the with A < 0. Note that the system is stable, the time constant is τc = −A steady-state gain from u to x is 1. Suppose the input (for t ≥ 0) is a ramp input, that is u(t) = βt

(where β is a known constant) starting with an initial condition x(0) = x0 . Show that the solution for t ≥ 0 is x(t) = β(t − τc ) + (x0 + βτc ) eAt {z } | | {z } shifted ramp

decaying exponential

Hint: you can do this in 2 different manners - carry out the convolution integral, or verify that the proposed solution satisfies the initial condition (at t = 0) and satisfies the differential equation for all t > 0. Both are useful exercises, but you only need to do one for the assignment. Note that if we ignore the decaying exponential part of the solution, then the “steady-state” solution is also a ramp (with same slope β, since the steady-state gain of the system is 1), but it is delayed from the input by τc time-units. This gives us another interpretation of the time-constant of a first-order linear system (i.e., “ramp-input leads to ramp-output, delayed by τc ”). Make an accurate sketch of u(t) and x(t) (versus t) on the same graph, for x0 = 0, β = 3 and A = −0.5. 6. In the notes and lecture, we established that if a < 0, then the system x(t) ˙ = ax(t) + bu(t) is bounded-input/bounded-output (BIBO) stable. In this problem, we show that the linear system x(t) ˙ = ax(t) + bu(t) is not BIBO stable if a ≥ 0. Suppose b 6= 0. (a) (a = 0 case): Show that there is an input u such that |u(t)| ≤ 1 for all t ≥ 0, but the response x(t) satisfying x(t) ˙ = 0 · x(t) + bu(t), with initial condition x(0) = 0

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is not bounded as a function of t. Hint: Try the constant input u(t) = 1 for all t ≥ 0. What is the response x? Is there a finite number which bounds |x(t)| uniformly over all t? (b) Take a > 0. Show that there is an input u such that |u(t)| ≤ 1 for all t ≥ 0, but the response x(t) satisfying x(t) ˙ = ax(t) + bu(t), with initial condition x(0) = 0, grows exponentially (without bound) with t. 7. Consider the system x(t) ˙ = ax(t) + bu(t) y(t) = cx(t) + du(t) Suppose a < 0, so the system is stable. (a) Starting from initial condition x(0) = 0, what is the response for t ≥ 0 due to the unit-step input 0 for t ≤ 0 u(t) = 1 for t > 0 Hint: Since x(0) = 0 and u(0) = 0, it is clear from the definition of y that y(0) = 0. For t > 0, x converges exponentially to its limit, and y differs from x only by scaling (c) and the addition of du(t), which for t > 0 is just d. (b) Compute, and sketch the response for a = −1; b = 1; c = −1; d = 1 (c) Compute and sketch the response for a = −1; b = 1; c = 2; d = −1 (d) Explain/justify the following terminology: • the “steady-state-gain” from u → y is d − • the “instantaneous-gain” from u → y is d

cb a

8. (a) Suppose ω > 0 and ψ > 0. Let y1 (t) := sin ωt, and y2 (t) = sin(ωt − ψ). Explain ψ what is meant by the statement that “the signal y2 lags the signal y1 by 2π of a period.” (b) Let ψ := π3 . On 3 separate graphs, plot 4 periods of sine-signals listed below. i. sin 0.1t and sin(0.1t − ψ) ii. sin t and sin(t − ψ) iii. sin 10t and sin(10t − ψ) (c) Explain how the graphs in part 8b confirm the claim in part 8a. 9. For the first-order linear system (constant coefficients a, b, and c) x(t) ˙ = ax(t) + bu(t) y(t) = cx(t)

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(a) Using the convolution integral for the forced response, find the output y(t) for t ≥ 0 starting from the initial condition x(0) = 0, subject to input • u(t) = 1 for t ≥ 0 • u(t) = sin(ωt) for t ≥ 0 (you will probably need to do two steps of integrationby-parts). (b) For this first-order system, the frequency-response function G(ω) is G(ω) =

cb jω − a

Make a plot of log10 |G(ω)| versus log10 ω for 0.001 ≤ ω ≤ 1000 for two sets of values: system S1 with parameters (a = −1, b = 1, c = 1) and system S2 with parameters (a = −10, b = 10, c = 1). Put both magnitude plots on the same axis. Also make a plot of ∠G(ω) versus log10 ω for 0.001 ≤ ω ≤ 1000 for both systems. You can use the Matlab command angle which returns (in radians) the angle of a nonzero complex number. Put both angle plots on the same axis. (c) What is the time-constant and steady-state gain (from u → y) of each system? How is the steady-state gain related to G(0)? (d) For each of the following cases, compute and plot y(t) versus t for the : i. ii. iii. iv. v. vi. vii. viii.

S1 S2 S1 S2 S1 S2 S1 S2

with with with with with with with with

x(0) = 0, u(t) = 1 for t ≥ 0 x(0) = 0, u(t) = 1 for t ≥ 0 x(0) = 0, u(t) = sin(0.1 ∗ t) for t ≥ 0 x(0) = 0, u(t) = sin(0.1 ∗ t) for t ≥ 0 x(0) = 0, u(t) = sin(t) for t ≥ 0 x(0) = 0, u(t) = sin(t) for t ≥ 0 x(0) = 0, u(t) = sin(10 ∗ t) for t ≥ 0 x(0) = 0, u(t) = sin(10 ∗ t) for t ≥ 0

Put cases (i), (ii) on the same graph, cases (iii), (iv) on the same graph, cases (v), (vi) on the same graph and cases (vii), (viii) on the same graph. Also, on each graph, also plot u(t). In each case, pick the overall duration so that the limiting behavior is clear, but not so large that the graph is cluttered. Be sure and get the steady-state magnitude and phasing of the response y (relative to u) correct.

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(e) Compare the steady-state sinusoidal responses of the response you computed and plotted in 9d with the frequency-response functions that are plotted in part 9b. Illustrate out how the frequency-response function gives, as a function of frequency, the steady-state response of the system to a sin-wave input of any frequency. Mark the relevant points of the frequency-response curves. 10. With regards to your answers in problem 9, (a) Comment on the effect parameters a and b have on the step responses in cases (a)-(b). (b) Comment on the amplification (or attenuation) of sinusodal inputs, and its relation to the frequency ω. (c) Based on the speed of the response in (a)-(b) (note the degree to which y “follows” u, even though u has an abrupt change), are the sinusoidal responses in (c)-(h) consistent? 11. Consider a first-order system, where for all t, x(t) ˙ = ax(t) + bu(t) y(t) = cx(t)

(5.9)

under the action of delayed feedback u(t) = −Ky(t − T ) where T ≥ 0 is a fixed number, representing a delay in the feedback path. (a) Eliminate u and y from the equations to obtain a delay-differential equation for x of the form x(t) ˙ = A1 x(t) + A2 x(t − T ) The parameters A1 and A2 will be functions of a, b, c and K. (b) Assume T = 0 (ie., no delay). Under what condition is the closed-loop system stable? (c) Following the derivation in section 5.4 (and the slides), derive the value of the smallest delay that causes instability for the five cases i. ii. iii. iv. v.

a = 0, b = 1, c = 1, K = 5 a = −1, b = 1, c = 1, K = 4 a = 1, b = 1, c = 1, K = 6 a = −3, b = 1, c = 1, K = 2 a = −3, b = 1, c = 1, K = −2

Also determine the frequency at which instability will occur.

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(d) Confirm your findings using Simulink, implementing an interconnection of the first-order system in (5.9), a Transport Delay block, and a Gain block for the feedback. Show relevant plots. 12. Consider a system with input u, and output y governed by the differential equation y(t) ˙ + a1 y(t) = b0 u(t) ˙ + b1 u(t)

(5.10)

This is different than what we have covered so far, because the derivative of the input shows up in the right-hand-side (the overall function forcing y, from the ODE point of view). Note that setting b0 = 0 gives an ODE more similar to what we considered earlier in the class. (a) Let q(t) := y(t) − b0 u(t). By substitution, find the differential equation governing the relationship between u and q. This should look familar. (b) Assume that the system is at rest (ie., y ≡ 0, u ≡ 0, and hence q ≡ 0 too), and at some time, say t = 0, the input u changes from 0 to u¯ (eg., a “step”-function input), specifically 0 for t ≤ 0 u(t) = u¯ for t > 0 Solve for q, using the differential equation found in part 12a, using initial condition q(0) = 0. (c) Since y = q + u, show that the step-response of (5.10), starting from y(0) = 0 is   b1 b1 y(t) = u¯ + b0 u¯ − u¯ e−a1 t for t > 0 a1 a1 This can be written equivalently as
b 1 − a1 b 0 u¯ 1 − e−a1 t + b0 u¯ a1 (d) Take a1 = 1. Draw the response for b0 = 2 and five different values of b1 , namely b1 = 0, 0.2, 1, 2, 4. (e) Take a1 = 1. Draw the response for b0 = −1 and five different values of b1 , namely b1 = 0, 0.2, 1, 2, 4. (f) Take a1 = 1. Draw the response for b0 = 0 and five different values of b1 , namely b1 = 0, 0.2, 1, 2, 4. (g) Suppose that a1 > 0 and b0 = 1. Draw the step response for two cases: b1 = 0.9a1 and b1 = 1.1a1 . Comment on the step response for 0 < a1 ≈ b1 . What happens if a1 < 0 (even if b1 ≈ a1 , but not exactly equal)?

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13. (a) So, consider the cascade connection of two, first-order, stable, systems x˙ 1 (t) y1 (t) x˙ 2 (t) y(t)

= = = =

A1 x1 (t) + B1 u(t) C1 x1 (t) + D1 u(t) A2 x2 (t) + B2 y1 (t) C2 x2 (t) + D2 y1 (t)

By stable, we mean both A1 < 0 and A2 < 0. The cascade connection is shown pictorially below. u -

y1

S1

-

S2

y -

Suppose that the frequency response of System 1 is M1 (ω), φ1 (ω) (or just the complex G1 (ω)), and the frequency response of System 2 is M2 (ω), φ2 (ω) (ie., the complex G2 (ω)). Now, suppose that ω is a fixed real number, and u(t) = sin ωt. Show that the steady-state behavior of y(t) is simply yω,ss (t) = [M1 (ω)M2 (ω)] sin (ωt + φ1 (ω) + φ2 (ω)) (b) Let G denote the complex function representing the frequency response (forcingfrequency-dependent amplitude magnification A and phase shift φ, combined into a complex number) of the cascaded system. How is G related to G1 and G2 ? Hint: Remember that for complex numbers G and H, |GH| = |G| |H| ,

∠ (GH) = ∠G + ∠H

14. Re-read “Leibnitz’s” rule in your calculus book, and consider the time-varying differential equation x(t) ˙ = a(t)x(t) + b(t)u(t) with x(0) = xo . Show, by substitution, or integrating factor, that the solution to this is Z t R Rt t a(ξ)dξ x(t) = e 0 xo + e τ a(ξ)dξ b(τ )u(τ )dτ 0

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6

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Feedback systems

6.1

First-order plant, Proportional control

Plant dynamics: x(t) ˙ = ax(t) + b1 d(t) + b2 u(t) y(t) = cx(t) + d1 d(t) Sensor noise model: ym (t) = y(t) + n(t) Control Law (strategy) u(t) = K1 r(t) − K2 ym (t) All of a, b1 , . . . , d1 , K1 , K2 are constants. Closed-loop dynamics are obtained by eliminating u and ym , obtaining ODE for how x is affected by x and (r, d, n). With that acomplished, the “outputs” of the closed-loop system are usually considered to by y (the plant output) and u (the control action). Simple substitution gives x(t) ˙ = (a − b2 K2 c)x(t) + (b2 K11 )r(t) + (b1 − b2 K2 d1 )d(t) + (−b2 K2 )n(t) y(t) = cx(t) + 0r(t) + d1 d(t) + 0n(t) u(t) = (−K2 c)x(t) + K1 r(t) + (−K2 d1 )d(t) + (−K2 )n(t) Closed-loop stability is the condition a − b2 K2 c < 0. All steady-state gains from (r, d, n) to (y, u) can easily be computed, as can the individual frequency-response functions from each input (r, d, n) to each output (y, u). Design parameters (from the control engineer’s view) are K1 and K2 . Typical goals are: • the closed-loop system must be stable (giving a − b2 K2 c < 0 as a stability constraint on the choice of K2 ); • achieve a desired closed-loop time constant (the closed-loop time constant is τclosed−loop = 1 , setting a quantitative constraint on the choice of K2 ); b2 K2 c−a • achieve a desired (small, usually) closed-loop steady-state gain from d → y (the steadyd1 a−cb1 state gain from d → y is a−b , again setting a quantitative constraint on the choice 2 K2 c of K2 ); • the steady-state gain from r → y should be 1 (this gives on K1 , once K2 has been chosen).

b2 K1 c b2 K2 c−a

= 1 as a constraint

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6.2

54

Proportional Plant, first-order controller

Plant model: y(t) = αu(t) + βd(t) Sensor model ym (t) = y(t) + n(t) Controller x(t) ˙ = ax(t) + b1 r(t) + b2 ym (t) u(t) = cx(t) + d1 r(t) Perform the same elimination, obtaining the closed-loop dynamics as x(t) ˙ = (a + b2 αc)x(t) + (b1 + b2 αd1 )r(t) + (b2 β)d(t) + (b2 )n(t) y(t) = (αc)x(t) + (αd1 )r(t) + (β)d(t) + (0)n(t) u(t) = (c)x(t) + (d1 )r(t) + (0)d(t) + (0)n(t) The stability condition is a + b2 αc < 0. The steady-state gain from d → y is just β+

β(a + b2 αc) − αcb2 β βa −αcb2 β = = a + b2 αc a + b2 αc a + b2 αc

A common goal is to make the steady-state gain from d → y identically equal to 0 perfect steady-state disturbance rejection! Assuming β 6= 0 (ie., for the plant by itself, the disturbance has an effect on the output), the only solution is to set a := 0. With this choice, the stability condition is b2 αc < 0. Also with a = 0, the steady-state gain from r → y is −b1 . In order to make this equal to 1, it must be that b1 = −b2 . Compute all 6 closed-loop b2 frequency-responses - note that with regard to b2 and c, only the product b2 c appears. So, only their product is important, and therefore without loss of generality, set b2 = −1. The number c is still a design parameter, and the stability condition is αc > 0. The feedback law is now x(t) ˙ = r(t) − ym (t),

u(t) = cx(t) + d1 r(t)

where c and d1 are the design parameters. These are often relabled as KI and KF , giving the feedback law as x(t) ˙ = r(t) − ym (t),

u(t) = KI x(t) + KF r(t)

The stability condition is αKI > 0. This is called integral control. Super important, and we will generalize it to more complex situations as the course progresses. See slides.

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Problems

1. Design Problem (stable plant/unstable plant): Consider a first-order system P , with inputs (d, u) and output y, governed by x(t) ˙ = ax(t) + b1 d(t) + b2 u(t) y(t) = cx(t) A proportional control, u(t) = K1 r(t) − K2 ym (t) is used, and the measurement is assumed to the be the actual value, plus measurement noise, ym (t) = y(t) + n(t) As usual, y is the process output, and is the variable we want to regulate, r is the reference signal (the desired value of y), d is a process disturbance, u is the control variable, n is the measurement noise (so that y + n is the measurement of y), K1 and K2 are gains to be chosen. For simplicity, we will choose some nice numbers for the values, specifically b1 = b2 = c = 1. There will be two cases studied: stable plant, with a = −1, and unstable plant, with a = 1. You will design the feedback gains as described below, and look at closed-loop properties. The goal of the problem is to start to see that unstable plants are intrinsically harder to control than stable plants. This problem is an illustration of this fact (but not a proof). (a) Keeping a, K1 and K2 as variables, substitute for u, write the differential equation for x in the form x(t) ˙ = Ax(t) + B1 r(t) + B2 d(t) + B3 n(t) Also, express the output y and control input u as functions of x and the external inputs (r, d, n) as y(t) = C1 x(t) + D11 r(t) + D12 d(t) + D13 n(t) u(t) = C2 x(t) + D21 r(t) + D22 d(t) + D23 n(t) Together, these are the “closed-loop governing equations.” Note that all of the symbols (A, B1 , . . . , D23 ) will be functions of a and the controller gains, K1 and K2 . Below, we will “design” K1 and K2 two different ways, and assess the performance of the overall system. (b) Under what conditions is the closed-loop system is stable? Under those conditions, i. What is the time-constant of the closed-loop system? ii. What is the steady-state gain from r to y (assuming d ≡ 0 and n ≡ 0)? iii. What is the steady-state gain from d to y (assuming r ≡ 0 and n ≡ 0)?

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(c) First we will consider the stable plant case, so a = −1. If we simply look at the plant (no controller), u and d are independent inputs, and the steady-state 1 which in this particular instance happens to be 1. Find gain from d to y is cb −a the value of K2 so that: the closed-loop system is stable; and the magnitude of the closed-loop steady-state gain from d to y is 15 of the magnitude of the openloop steady-state gain from d to y. That will be our design of K2 , based on the requirement of closed-loop stability and this disturbance rejection specification. (d) With K2 chosen, choose K1 so that the closed-loop steady-state gain from r to y is equal to 1 (recall, the goal is that y should “track” r, as r represents the desired value of y). (e) Temporarily, assume the feedback is delayed, so u(t) = K1 r(t) − K2 ym (t − T ) for some T ≥ 0. Using the numerical values, determine the smallest T > 0 such that the closed-loop system is unstable. You have to rewrite the equations, accounting for the delay in the feedback path. Since stability is property of the system independent of the external inputs, you can ignore (r, d, n) and obtain an equation of the form x(t) ˙ = A1 x(t) + A2 (x(t − T ). Then look back to problem 11 for reference. (f) Next, we will assess the design in terms of the closed-loop effect that the external inputs (r, d, n) have on two main variables of interest (y, u). For a change-of-pace, we will look at the frequency-response functions, not time-domain responses. For notational purposes, let Hv→q denote the frequency-response function from a input signal v to an output signal q (for example, from r to y). We will be making plots of |Hv→q (ω)| vs ω and ∠Hv→q (ω) vs ω You will need to mimic code in the lab exercises, using the commands abs and angle. As mentioned, the plots will be arranged in a 2 × 3 array, organized as |Hr→y | ∠Hr→y |Hr→u |

|Hd→y |

|Hn→y |

|Hd→u |

|Hn→u |

These are often referred to as the “‘gang of six.” The plot shows all the important cause/effects, in the context of sinusoidal steady-state response, within the closedloop system, namely how (references, disturbances, measurement noise) affect the (regulated variable, control variable). Note that because one of the entries actually has both the magnitude and angle plotted, there will be 7 axes. If you forget how to do this, try the following commands and see what happens in Matlab. a11T = subplot(4,3,1); a11B = subplot(4,3,4); a12 = subplot(2,3,2);

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a13 a21 a22 a23

= = = =

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subplot(2,3,3); subplot(2,3,4); subplot(2,3,5); subplot(2,3,6);

Plot the frequency-response functions. Use (for example) 100 logarithmically spaced points for ω, varying from 0.01 to 100. Make the linetype (solid, black) (g) Next we move to the unstable plant case, a = 1. In order make a fair comparison, we need to make some closed-loop property identical to the previous case’s closed-loop property, and then compare other closed-loop properties. For the unstable plant design, choose K1 and K2 so that i. the closed-loop system is stable ii. the closed-loop time-constant is the same as the closed-loop time constant in the stable plant case, iii. the closed-loop steady-state gain from r to y is 1. With those choices, plot the closed-loop frequency response functions on the existing plots, using (dashed/red) linetypes for comparison. (h) Note that several curves are the same as the stable plant case. However in all the other cases (d → u, n → y, n → u) the unstable plant case has higher gains. This means that in order to get the same r → y tracking and closed-loop timeconstant, the system with the unstable plant uses “more” control input u, and is more sensitive to noise at all frequencies. (i) Again, temporarily, assume the feedback is delayed, so u(t) = K1 r(t)−K2 ym (t−T ) for some T ≥ 0. Using the numerical values, determine the smallest T > 0 such that the closed-loop system is unstable. How does this compare to the earlier calculation of the time-delay margin in the open-loop stable case? (j) Look at the paper entitled “Respect the Unstable” by Gunter Stein. It is in the IEEE Control Systems Magazine, August 2003, pp. 12-25. You will need to ignore most of the math at this point, but there are some good paragraphs that reiterate what I am saying here, and good lessons to be learned from the accidents he describes. Please write a short paragraph (a few sentences) about one of the accidents he describes. 2. Open-loop versus Closed-loop control: Consider a first-order system P , with inputs (d, u) and output y, governed by x(t) ˙ = ax(t) + b1 d(t) + b2 u(t) y(t) = cx(t) (a) Assume P is stable (ie., a < 0). For P itself, what is the steady-state gain from u to y (assuming d ≡ 0)? Call this gain G. What is the steady-state gain from d to y (assuming u ≡ 0)? Call this gain H.

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(b) P is controlled by a “proportional” controller of the form u(t) = K1 r(t) + K2 [r(t) − (y(t) + n(t))] Here, r is the reference signal (the desired value of y), n is the measurement noise (so that y + n is the measurement of y), K1 and K2 are gains to be chosen. By substituting for u, write the differential equation for x in the form x(t) ˙ = Ax(t) + B1 r(t) + B2 d(t) + B3 n(t) Also, express the output y and control input u as functions of x and the external inputs (r, d, n) as y(t) = C1 x(t) + D11 r(t) + D12 d(t) + D13 n(t) u(t) = C2 x(t) + D21 r(t) + D22 d(t) + D23 n(t) All of the symbols (A, B1 , . . . , D23 ) will be functions of the lower-case given symbols and the controller gains. Below, we will “design” K1 and K2 two different ways, and assess the performance of the overall system. (c) Under what conditions is the closed-loop system is stable? What is the steadystate gain from r to y (assuming d ≡ 0 and n ≡ 0)? What is the steady-state gain from d to y (assuming r ≡ 0 and n ≡ 0)? (d) Design #1: In this part, we design a feedback control system that actually had no feedback (K2 = 0). The control system is called “open-loop” or ‘feed-forward”, and will be based on the steady-state gain G (from u → y) of the plant. The open-loop controller is simple - simply invert the gain of the plant, and use that for K1 . Hence, we pick K1 := G1 , and K2 := 0. Call this Design #1. Note that we are now considering a feedback control system that actually i. For Design #1, compute the steady-state gains from all external inputs (r, d, n) to the two “outputs” (y, u). ii. Comment on the steady-state gain from r → y. iii. (See problem 24 for the definition of “sensitivity”). What is the sensitivity of the steady-state gain from r → y to the parameter b2 ? What about the sensitivity to a? Here you should treat K1 as a fixed number. iv. Comment on the relationship between the steady-state gain from d → y without any control (ie., H computed above) and the steady-state gain from d → y in Design #1, as computed in part 2(d)i. v. Comment on the steady-state gain from d to u in Design #1. Based on d’s eventual effect on u, is the answer in part 2(d)iv surprising? vi. Comment on the steady-state gain from n to both y and u in Design #1. Remember that Design #1 actually does not use feedback...

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vii. What it the time-constant of the system with Design #1. viii. In this part we have considered a control system that actually had no feedback (K2 = 0). Consequently, this is called open-loop control, or feedforward control (since the control input is just a function of the reference signal r, “fed-forward to the process”), or control-by-calibration since the reciprical of the value of of G is used in the control law. Write a short, concise (4 bullet points) quantitative summary of the effect of this strategy. Include a comparison of the process time-constant, and the resulting time-constant with the controller in place, as well as the tracking capabilities (how y follows r), the sensitivity of the tracking capabilities to parameter changes, and the disturbance rejection properties. (e) Now design a true feedback control system. This is Design #2. Pick K2 so that the closed-loop steady-state gain from d → y is at least 5 times less than the uncontrolled steady-state gain from d → y (which we called H). Constrain your choice of K2 so that the closed-loop system is stable. Since we are working fairly general, for simplicity, you may assume a < 0 and b1 > 0, b2 > 0 and c > 0. i. With K2 chosen, pick K1 so that the closed-loop steady-state gain from r → y is 1. ii. With K1 and K2 both chosen as above, what is the sensitivity of the steadystate gain from r → y to the parameter b2 ? iii. What is the time-constant of the closed-loop system? iv. What is the steady-state gain from d → u? How does this compare to the previous case (feedforward)? v. With K2 6= 0, does the noise n now affect y? (f) Let’s use specific numbers: a = −1, b1 = 1, b2 = 1, c = 1. Summarize all computations above in a table – one table for the feedforward case (Design #1), and one table for the true feedback case (Design #2). Include in the table all steady-state gains, time constant, and sensitivity of r → y to b2 . (g) Plot the frequency responses from all external inputs to both outputs. Do this in a 2 × 3 matrix of plots that I delineate in class. Use Matlab, and the subplot command. Use a frequency range of 0.01 ≤ ω ≤ 100. There should be two lines on each graph. (h) Mark your graphs to indicate how Design #2 accomplishes tracking, disturbance rejection, and lower time-constant, but has increased sensitivity to noise. (i) Keeping K1 and K2 fixed, change b2 from 1 to 0.8. Redraw the frequency responses, now including all 4 lines. Indicate on the graph the evidence that Design #2 accomplishes good r → y tracking that is more insensitive to process parameter changes than Design #1 .

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3. At this point, we can analyze (stability, steady-state gain, sinusoidal steady-state gains (FRFs), time-constant, etc.) of first-order, linear dynamical systems. In a previous problem, we analyzed a 1st-order process model, and a proportional-control strategy. In this problem, we try a different situation, where the process is simply proportional, but the controller is a 1st-order, linear dynamical system. Specifically, suppose the process model is nondynamic (“static”) simply y(t) = αu(t) + βd(t) where α and β are constants. Depending on the situation (various, below) α may be considered known to the control designer, or unknown. In the case it is unknown, we will still assume that the sign (ie., ±) is known. The control strategy is dynamic x(t) ˙ = ax(t) + b1 r(t) + b2 ym (t) u(t) = cx(t) + d1 r(t) where ym (t) = y(t) + n(t) and the various “gains” (a, b1 , . . . , d1 ) constitute the design choices in the control strategy. Be careful, notation-wise, since (for example) d1 is a constant parameter, and d(t) is a signal (the disturbance). There are alot of letters/parameters/signals to keep track of. (a) Eliminate u and ym from the equations to obtain a differential equation for x of the form x(t) ˙ = Ax(t) + B1 r(t) + B2 d(t) + B3 n(t) which governs the closed-loop behavior of x. Note that A, B1 , B2 , B3 are functions of the parameters a, b1 , . . . in the control strategy, as well as the process parameters α and β. (b) What relations on (a, b1 , . . . , d1 , α, β) are equivalent to closed-loop system stability? (c) As usual, we are interested in the effect (with feedback in place) of (r, d, n) on (y, u), the regulated variable, and the control variable, respectively. Find the coefficients (in terms of (a, b1 , . . . , d1 , α, β)) so that y(t) = C1 x(t) + D11 r(t) + D12 d(t) + D13 n(t) u(t) = C2 x(t) + D21 r(t) + D22 d(t) + D23 n(t) (d) Suppose that Tc > 0 is a desired closed-loop time constant. Show that the following design objectives can be met with one design, assuming that the value of α is known to the designer. • closed-loop is stable • closed-loop time constant is Tc • steady-state gain from d → y is 0

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• steady-state gain from r → y is 1 A few things to look out for: the conditions above do not uniquely determine all of the parameters: indeed, only the product b2 c can be determined; and any arbitrary value for d1 is acceptable (although it’s particular value does affect other properties, like r → u, for instance). (e) Assuming the choices above have been satisfied, what is the steady-state gain from d → u? Given that the steady-state gain from d → y is 0, does this make sense, in retrospect? (f) Show that 1 , d1 = arbitrary αTc is one acceptable choice. Note that to achieve the desired time-constant, the value of α must be known to the control designer. Write the controller equations with all these simplifications. a = 0, b1 = 1, b2 = −1; c =

(g) Assume that α is not known, but it is known that 0 < αL ≤ α ≤ αU , where αL and αU are known bounds (and both are positive, as indicated). Suppose that Tc > 0 is a desired closed-loop time constant. Show that the following design objectives can be met with one design. • • • •

closed-loop is stable actual closed-loop time constant is guaranteed ≤ Tc ; steady-state gain from d → y is 0 steady-state gain from r → y is 1

(h) Again assume that α is not known, but it is known that 0 < αL ≤ α ≤ αU , where αL and αU are known bounds (and both are positive, say). Suppose that Tc > 0 is a desired closed-loop time constant. Show that the following design objectives can be met with one design. • • • •

closed-loop is stable actual closed-loop time constant is guaranteed ≥ Tc ; steady-state gain from d → y is 0 steady-state gain from r → y is 1

4. Let the plant (process to be controlled) be governed by the static (no differential equations) model y(t) = αu(t) + βd(t) as in the lectures. Suppose nominal values of α and β are known. Consider an integral controller, of the form x(t) ˙ = r(t) − ym (t) u(t) = KI x(t) + D1 r(t)

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where KI and D1 are two design parameters to be chosen, and ym = y + n where n is an additive sensor noise. (a) Derive the closed-loop differential equation governing x, with inputs r, d, n. Under what conditions is the closed-loop stable? What is the time constant of the system. (b) Suppose the nominal values of the plant parameters are α ¯ = 2.1, β¯ = 0.9. Design KI such that the closed-loop system is stable, and the nominal closed-loop time constant is 0.25 time units. (c) Simulate (using ode45) the system subject to the following conditions • Use KI as designed in part 4b, and set D1 = 0. • Initial condition, x(0) = 0. • Reference input is a series of steps: r(t) = 0 for 0 ≤ t < 1; r(t) = 2 for 1 ≤ t < 2; r(t) = 3 for 2 ≤ t < 6; r(t) = 0 for 6 ≤ t < 10. • Disturbance input is a series of steps: d(t) = 0 for 0 ≤ t < 3; d(t) = 1 for 3 ≤ t < 4; d(t) = 2 for 4 ≤ t < 5; d(t) = 3 for 5 ≤ t < 6. d(t) = 4 for 6 ≤ t < 7. d(t) = 0 for 7 ≤ t < 10. • Noise n(t) = 0 all t. Plot y versus t and u versus t is separate, vertically stacked, axes. We refer to these as the nominal, closed-loop responses. (d) Assume that the true value of α (unknown to the control designer) is different from α ¯ (= 2.1). Write an expression for actual the closed-loop time-constant. Your answer should depend on α, α ¯ and the desired time constant (in this case, 0.25). (e) Repeat the simulation above, with all parameters the same, except that the α, in the plant itself, should take on some off-nominal values, in order to study the robustness of the closed-loop system to variations in the process (plant) behavior. Keep the value of KI fixed from the design step in part 4b. Do 5 simulations, for α taking on values from 1.5 to 2.5. Plot these with dashed-lines, and include the nominal closed-loop responses (single, thick solid line) for comparison. (f) For the all 5 of the systems (from the collection of perturbed plant models) make a “gang-of-six” frequency-response function plot, using linear (not log) scales, arranged as |Hr→y | ∠Hr→y |Hr→u |

|Hd→y |

|Hn→y |

|Hd→u |

|Hn→u |

Choose your frequency range appropriately, probably [0 20] should be adequate, so that the plots look good.

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(g) One informal goal of this class is to learn to make connections between the frequency-domain analysis (eg., the frequency-response function plots) and the time-domain analysis (ODE simulations with specific, often non-sinusoidal, inputs). Make a list of 5 connections between the frequency-response plots (which precisely give sinusoidal, steady-state response information) and the time-response plots (which were non-steady state, non-sinusoidal input responses). (h) Finally, repeat the nominal simulation from part 4c, with D1 = 0.25 and separately D1 = −0.25 (just two simulations). Again, plot these with dashed-lines, and include the nominal closed-loop responses (single, thick solid line) for comparison. i. How does the value of D1 affect the response (instantaneous, steady-state, time-constant) of y and u due to reference input r? ii. How does the value of D1 affect the response (instantaneous, steady-state, time-constant) of y and u due to disturbance input d? 5. Suppose an input/output relationship is given by y(t) = Φ(u(t)) + d(t)

(6.1)

where Φ is a monotonically increasing, differentiable function. This will be used to model a plant, with control input u, disturbance input d, and output (to be regulated) y. Regarding Φ, specifically, assume there exist positive α and β such that α ≤ Φ0 (v) ≤ β for all v ∈ R. One such Φ is given below in part 5e. Note that the model in (6.1) generalizes the linear plant model in problem 4, to include nonlinear dependence of y on u. We’ll ignore sensor noise in this problem. An integral controller is used, of the form x(t) ˙ = r(t) − y(t) u(t) = KI x(t) As usual, we want to understand how e(t) := r(t) − y(t) behaves, even in the presence of nonzero disturbances d(t). (a) Show that e(t) = r(t) − Φ(KI x(t)) − d(t) for all t. (b) If r(t) = r¯, a constant, and d(t) = d¯ a constant, show that e(t) ˙ = −Φ0 (KI x(t))KI (¯ r − y(t)) which simplifies to e(t) ˙ = −Φ0 (KI x(t))KI e(t).

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(c) By chain rule, it is always true that d(e2 ) = 2e(t)e(t). ˙ dt Substitute in expression for e˙ to show that d(e2 ) = −2KI Φ0 (KI x(t))e2 (t) dt (d) Assume KI > 0. Define z(t) := e2 (t). Note that z(t) ≥ 0 for all t, and show that z(t) ˙ ≤ −2KI αz(t) for all t. Hence z evolves similarly to the stable, 1st order system w(t) ˙ = −2KI αw(t), but may approach 0 even “faster” at times. Hence, we conclude that z approaches 0 at least as fast as w would, and hence can be thought of as having a maximum time-constant of 2K1I α . (e) Take, for example, Φ to be Φ(v) :=

2v + 0.1v 3 for v ≥ 0 . 2v − 0.2v 2 for v < 0

Plot this Φ function on the domain −5 ≤ v ≤ 5. (f) Use ode45 or Simulink to simulate the closed-loop system, with KI = 0.25. Initialize the system at x(0) = 0, and consider the reference and disturbance inputs as defined below: r(t) = 3 for 0 ≤ t < 10; r(t) = 6 for 10 ≤ t < 20; r(t) = 6 + 0.25(t − 20) for 20 ≤ t < 40; r(t) = 11 − 0.4(t − 40) for 40 ≤ t ≤ 60, and π t) for 0 ≤ t < 60. One on graph, plot y and r versus t. On another d(t) = sin( 15 axes, plot the control input u versus t. (g) Consider an exercise machine that can be programmed to control a person’s heart rate by measuring the heart-rate, and adjusting the power input (from the person) that must be delivered (by the person) to continually operate the machine (eg, an elliptical trainer or “stairmaster”, where the resistance setting dictates the rate at which a person must take steps). Assume people using the machine are in decent enough shape to be exercising (as all machines have such warnings printed on the front panel). Explain, in a point-by-point list, how the problem you have just solved above can be related to a simple integral-control strategy for this type of workout machine. Make a short list of the issues that would come up in a preliminary design discussion about such a product. 6. A feedback system is shown below. All unmarked summing junctions are “plus” (+).

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d r

+ - f e- K −6

u- f ? v-

y

P

? f

n

Figure 4: Closed-loop system (a) The plant P , is governed by the ODE y(t) ˙ = y(t) + v(t). Note that the plant is unstable. The controller is a simple proportional control, so u(t) = Ke(t), where K is a constant-gain. Determine the range of values of proportional gain K for which the closed-loop system is stable. (b) Temporarily, suppose K = 4. Confirm that the closed-loop system is stable. What is the time-constant of the closed-loop system? (c) The control must be implemented with a sampled-data system (sampler, discrete control logic, zero-order hold) running at a fixed sample-rate, with sample time TS . The proportional feedback uk = Kek is implemented, as shown below. d r

+ - f e - Sample TS −6

ek -

K

uk - z.o.h. u- ? fvTS

y

P

? f

n

Figure 5: Closed-loop, sampled-data system The plant ODE is as before, y(t) ˙ = y(t) + v(t). Determine a relationship between TS and K (sample-time and proportional gain) such that the closed-loop system is stable. (d) Return the situation where K = 4. Recall the rule-of-thumb described in class 1 of the closed-loop time constant. that the sample time TS should be about 10 Using this sample time, determine the allowable range of K, and show that the choice K = 4 is safely in that range. (e) Simulate the overall system (Lab on Wednesday/Thursday will describe exactly how to do this, and it will only take a few minutes to do this) and confirm that the behavior with the sampled-data implementation is approximately the same as the ideal continuous-time implementation. 7. Suppose two systems are interconnected, with individual equations given as S1 : y(t) ˙ = [u(t) − y(t)] S2 : u(t) = 2 [y(t) − r(t)]

(6.2)

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(a) Consider first S1 (input u, output y): Show that for any initial condition y0 , if u(t) ≡ u¯ (a constant), then y(t) approaches a constant y¯, that only depends on the value of u¯. What is the steady-state gain of S1 ? (b) Next consider S2 (input (r, y), output u): Show that if r(t) ≡ r¯ and y(t) ≡ y¯ (constants), then u(t) approaches a constant u¯, that only depends on the values (¯ r, y¯). (c) Now, assume that the closed-loop system also has the steady-state behavior – that is, if r(t) ≡ r¯, then both u(t) and y(t) will approach limiting values, u¯ and y¯, only dependent on r¯. Draw a block-diagram showing how the limiting values are related, and solve for u¯ and y¯ in terms of r¯. (d) Now check your answer in part 4c. Suppose y(0) = 0, and r(t) = 1 =: r¯ for all t ≥ 0. Eliminate u from the equations 24.5, and determine y(t) for all t. Make a simple graph. Does the result agree with your answer in part 4c? Lesson: since the assumption we made in part 4c was actually not valid, the analysis in part 4c is incorrect. That is why, for a closed-loop steady-state analysis to be based on the separate component’s steady-state properties, we must know from other means that the closed-loop system also has steady-state behavior. 8. Suppose two systems are interconnected, with individual equations given as S1 : y(t) ˙ = [u(t) + y(t)] S2 : u(t) = 2 [r(t) − y(t)]

(6.3)

(a) Consider first S1 (input u, output y): If u(t) ≡ u¯ (a constant), then does y(t) approach a constant y¯, dependent only on the value of u¯? (b) Next consider S2 (input (r, y), output u): If r(t) ≡ r¯ and y(t) ≡ y¯ (constants), then does u(t) approach a constant u¯, dependent only on the values r¯, y¯? (c) Suppose y(0) = y0 is given, and r(t) =: r¯ for all t ≥ 0. Eliminate u from the equations 24.6, and determine y(t) for all t. Also, plugging back in, determine u(t) for all t. Show that y and u both have limiting values that only depend on the value r¯, and determine the simple relationship between r¯ and (¯ y , u¯). Lesson: Even though S1 does not have steady-state behavior on its own, in feedback with S2 , the overall closed-loop system does. 9. Consider the equations relating variables r, e, y, n, u and d. Assume P and C are given numbers. e = r − (y + n) u = Ce y = P (u + d)

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So, this represents 3 linear equations in 6 unknowns. Solve these equations, expressing e, u and y as linear functions of r, d and n. The linear relationships will involve the numbers P and C. 10. For a function F of a many variables (say two, for this problem, labeled x and y), the “sensitivity of F to x” is defined as “the ratio of the percentage change in F due to a percentage change in x.” Denote this by SxF . (a) Suppose x changes by δ, to x + δ. The percentage change in x is then % change in x =

δ (x + δ) − x = x x

Likewise, the subsequent percentage change in F is % change in F =

F (x + δ, y) − F (x, y) F (x, y)

Show that for infinitesimal changes in x, the sensitivity is SxF = (b) Let F (x, y) =

xy . 1+xy

x ∂F F (x, y) ∂x

What is SxF .

xy (c) If x = 5 and y = 6, then 1+xy ≈ 0.968. If x changes by 10%, using the quantity F Sx derived in part (10b), approximately what percentage change will the quantity xy undergo? 1+xy

(d) Let F (x, y) =

1 . xy

What is SxF .

(e) Let F (x, y) = xy. What is SxF .

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Two forms of high-order Linear ODEs, with forcing

The governing equations for the car, with integral controller can be expressed in two distinct, but equivalent, forms. By writing down the governing equations for each component individually, and then eliminating the interconnection variable u gives v(t) ˙ = m1 [−αv(t) + EKI z(t) + Gw(t)] z(t) ˙ = vdes (t) − v(t) This is two 1st order (only first derivative occurs), coupled (z affects v˙ and v affects z) ˙ linear differential equations, with (in this case) two external inputs, vdes and w. The general case of this situation has n dependent variables, x1 , x2 , . . . , xn , and m inputs, d1 , d2 , . . . , dm . The differential equations governing the evolution of the xi variables is x˙ 1 (t) = a11 x1 (t) + a12 x2 (t) + · · · + a1n xn (t) + b11 d1 (t) + b12 d2 (t) . . . + b1m dm (t) x˙ 2 (t) = a21 x1 (t) + a22 x2 (t) + · · · + a2n xn (t) + b21 d1 (t) + b22 d2 (t) . . . + b2m dm (t) .. . . = .. x˙ n (t) = an1 x1 (t) + an2 x2 (t) + · · · + ann xn (t) + bn1 d1 (t) + bn2 d2 (t) . . . + bnm dm (t) In matrix notation, the equations can be written very concisely, x(t) ˙ = Ax(t) + Bd(t) where x, A, B, and d are vectors and matrices      d1 (t) a11 a12 x1 (t)  d2 (t)   a21 a22  x2 (t)       x(t) :=  ..  , d(t) :=  ..  , A =  .. ..  .   .  .  . xn (t) dm (t) an1 an2

  · · · a1n b11   · · · a2n   b21 , B =  .. . . . ..   . .  · · · ann bn1

· · · b1m · · · b2m .. ... . · · · bnm

    

This is called a linear, state-space representation. For the cruise-control equations, it appears as        −α EK   G I v(t) vdes (t) v(t) ˙ 0 m m m = + z(t) w(t) z(t) ˙ 1 0 −1 0 | {z } | {z } | {z } | {z } | {z } x(t) ˙

A

x(t)

B

d(t)

We began the class by defining dynamical systems in this form, without the restriction of linearity. So, this state-space form is a general, and useful system representation. But, on the other hand, there was an alternative manner to write the system equations. In this other approach, we eliminated z from the original equations, to yield v¨(t) +

α EKI EKI G v(t) ˙ + v(t) = vdes (t) − w(t) ˙ m m m m

(7.1)

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So, now the closed-loop system behavior is equivalently described with one 2nd order (2 derivatives) differential equation, with two external inputs, vdes and w, although for w, only w˙ explicitly appears. This is mathematically equivalent to the state-space form above, but of a different form. First some notation, if y denotes a function (of a variable t, say), then y [k] or y (k) denotes the k’th derivative of the function y, y [k] =

dk y . dtk

Then, for a system with (say) two inputs (u, w) and one output (y), the n’th order differential equation relating the inputs to outputs is y [n] (t) + a1 y [n−1] (t) + · · · + an−1 y [1] (t) + an y(t) = b0 u[n] (t) + b1 u[n−1] (t) + · · · + bn−1 u[1] (t) + bn u(t) +c0 w[n] (t) + c1 w[n−1] (t) + · · · + cn−1 w[1] (t) + cn w(t) We will refer to this as the SLODE (single, linear ODE) representation. Both representations are useful, and in certain situations, advantageous over each other. We will study each in detail, starting with the state-space representation. In section 19, we cover much of the mathematics to analyze the behavior of systems governed by the SLODE representation. In section ??, we revisit the cruise control with integral control, and analyze the dynamic properties we had observed. We also propose a more complex control architecture which has more desirable properties.

7.1

Problems

1. Suppose that (A1 , B1 , C1 , D1 ) are the state-space matrices of a linear system with n1 states, m1 inputs and q1 outputs. Hence A1 ∈ Rn1 ×n1 , B1 ∈ Rn1 ×m1 , C1 ∈ Rq1 ×n1 , D1 ∈ Rq1 ×m1 . Let x1 denote the n1 × 1 state vector (so here, the subscript “1” does not mean the first element, rather it is itself a vector, perhaps indexed as   (x1 )1  (x1 )2    x1 =   ..   . (x1 )n1 Similarly, let u1 denote the m1 × 1 input vector and y1 denote the q1 × 1 input vector. Likewise, suppose that (A2 , B2 , C2 , D2 ) are the state-space matrices of a linear system

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with n2 states, m2 inputs and q2 outputs. Hence A2 ∈ Rn2 ×n2 , B2 ∈ Rn2 ×m2 , C2 ∈ Rq2 ×n2 , D2 ∈ Rq2 ×m2 . Let x2 denote the n2 × 1 state vector. Also let u2 denote the m2 × 1 input vector and y2 denote the q2 × 1 input vector. (a) Assume q2 = m1 , in other words, the number of outputs of system 2 is equal to the number of inputs to system 1. Hence the systems can be cascaded as shown, u2 -

S2

-

y1

S1

-

with input u2 and output y1 . Define x as   x1 (t) x(t) := x2 (t) which is an (n1 + n2 ) × 1 vector. Find matrices (defined in terms of vertical and horizontal concatenations of various products of the individual state-space matrices) such that the equations x(t) ˙ = Ax(t) + Bu2 (t) y1 (t) = Cx(t) + Du2 (t) (b) In Matlab, the * operation (”multiplication”) is the cascade operation for systems (tf, ss, and so on), and the above operation would be written as S1 ∗S2 . The actual name of the * operator is mtimes. Execute >> open +ltipack\@ssdata\mtimes.m within Matlab, and find the lines of code which implement the operation above. (c) Assume q1 = q2 and m1 = m2 , in other words, the number of outputs of system 1 is equal to the number of outputs of system 2, and the number of inputs of the two systems are equal as well. Hence the systems can be connected in parallel as shown, with input u and output y. -

S1

y1 ? d++ 6

u -

Define x as

S2

 x(t) :=

y2 x1 (t) x2 (t)



y

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which is an (n1 + n2 ) × 1 vector. Find matrices (defined in terms of vertical and horizontal concatenations of various products of the individual state-space matrices) such that the equations x(t) ˙ = Ax(t) + Bu(t) y(t) = Cx(t) + Du(t) (d) In Matlab, the + operation (”addition”) is the parallel interconnection operation for systems (tf, ss, and so on), and the above operation would be written as S1 +S2 . The actual name of the + operator is plus. Execute >> open +ltipack\@ssdata\plus.m within Matlab, and find the lines of code which implement the operation above. (e) Assume that q1 = m2 and m1 = q2 , in other words, the number of outputs of one system is equal to the number of outputs of the other system. Hence the systems can be connected in feedback as shown, with input u and output y. Assume further that D1 = 0q1 ×m1 . u

+d −6

y

S1

-

S2  Define x as

 x(t) :=

x1 (t) x2 (t)



which is an (n1 + n2 ) × 1 vector. Find matrices (defined in terms of vertical and horizontal concatenations of various products of the individual state-space matrices) such that the equations x(t) ˙ = Ax(t) + Bu(t) y(t) = Cx(t) + Du(t) (f) The code for feedback implements this, but it is buried deeper within other subroutines. Let’s just confirm that this is what it does. >> s1 = ss(-10,3.3,1.1,0,’StateName’,{’xT’}); >> s2 = ss(-5,2.2,5.5,0,’StateName’,{’xB’}); >> feedback(s1,s2) Verify that the state-ordering is as we defined, and that the entries are correct.

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Be sure to check that all of your matrix manipulations have the correct dimensions, and that the concatenations have compatible dimensions (horizontal concatenations must have the same number of rows, vertical concatenation must have the same number of columns). 2. This problem may be out-of-place. It has some transfer function questions At one point, NASA was developing booster rockets and a crew exploration vehicle to replace the Space Shuttle. One main component of the Orion Crew Exploration Vehicle (CEV) is a conical crew module. This module has several thrusters to control the vehicle attitude on re-entry to the earth’s orbit. A linear model for the short-period mode of the CEV pitch dynamics during re-entry is given by:     2 1 0 x(t) ˙ = x(t) + u(t) −36 2 1     y(t) = 0 1 x(t) + 0 u   α(t) where x(t) := , which are the angle-of-attack and the pitch-rate, respectively, q(t) and u(t) is the pitch torque generated by the thrusters. The output y is the pitch rate, so y = x2 = q. (a) Using the formula we derived in class, G(s) = D + C(sI − A)−1 B, derive the transfer function from u to y of this state-space model. (b) Enter the matrices into Matlab, form a state-space object using the ss constructor. Type >> disp(G) >> size(G) >> class(G) (c) (d)

(e) (f)

and paste the results into your assignment. Use the command tf as a converter, and convert the model respresentation to a transfer function object. Confirm that your answer in part 4a is correct. If x is eliminated (through clever substitutions - which you know, and are key to the state-space-to-transfer function conversion), what is the differential equation relating u to y. With the transfer function obtained in part 4a, use the procedure derived in class to obtain a state-space model for the system. Note that the state-space model you obtain is not the same as the state-space model we started with. Enter both systems into Matlab, and confirm with step that the input/output behaviors of the two systems are indeed identical. Hence we have seen that a system can have different state-space models (that yield the exact same input/output behavior).

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73

Jacobian Linearizations, equilibrium points

In modeling systems, we see that nearly all systems are nonlinear, in that the differential equations governing the evolution of the system’s variables are nonlinear. However, most of the theory we have developed has centered on linear systems. So, a question arises: “In what limited sense can a nonlinear system be viewed as a linear system?” In this section we develop what is called a “Jacobian linearization of a nonlinear system,” about a specific operating point, called an equilibrium point. The section begins with a short review of derivatives, especially for functions of many variables.

8.1

Jacobians and the Taylor Theorem

First consider a scalar values function φ(x1 , x2 , · · · , xn ) of n variables. The Jacobian of φ is the 1 × n row vector i h ∂φ ∂φ · · · ∇φ(x1 , · · · , xn ) = ∂x ∂xn 1 For example for the function φ(x1 , x2 , x3 ) of three variables, φ(x1 , x2 , x3 ) = x1 x22 − 1/x1 + sin(x3 ) the Jacobian is ∇φ =

h

∂φ ∂x1

=



x22 + 1/x21 2x1 x2 cos(x3 )

∂φ ∂x2

∂φ ∂x3

i 

The Jacobian is a function of the variables x1 , x2 , x3 . We can evaluate the Jacobian at a point ξ and get actual numbers. In this example, we could evaluate the Jacobian at (x1 , x2 , x3 ) = (1, 2, 0) to get   ∇φ(x)|x=(1,2,0) = 5 4 1 Now consider a vector-valued function φ : Rn → Rm . What this actually means is that we have m individual functions in n variables. These functions are collectively represented by φ, and the variables are collectively represented by x. Let us write this function as   φ1 (x1 , · · · , xn )   .. φ(x) = φ(x1 , x2 , · · · , xn ) =   . φm (x1 , · · · , xn )

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The Jacobian of φ is the m × n matrix of functions  ∂φ1 ··· ∂x1  .. ∇φ =  . ··· ∂φm ∂x1

···

∂φ1 ∂xn

.. .

∂φm ∂xn

  

Fix a vector ξ ∈ Rn . We can evaluate the Jacobian at ξ to Our notation for this is  ∂φ1 ∂φ1 · · · ∂x ∂x1 n  .. ∇φ(x)|x=ξ =  ... ··· . ∂φm ∂x1

···

∂φm ∂xn

get an m × n matrix of numbers.    @ξ

For example, consider the function  φ(x1 , x2 , x3 ) =

φ1 (x1 , x2 , x3 ) φ2 (x1 , x2 , x3 )



 =

x1 sin(x2 ) x22 cos(x3 )



The Jacobian of φ(x) evaluated at (1, π/2, 0) is     1 1 0 sin(x2 ) x1 cos(x2 ) 0 = ∇φ|x=(1,1,0) = 0 π 0 0 2x2 cos(x3 ) x22 sin(x3 ) x=(1,π/2,0) Analogous to the classical Taylor series expansion of a scalar function of one variable, we can write the Taylor series expansion of φ around ξ as φ(x) = φ(ξ) + ∇φ|x=(ξ) (x − ξ) + higher order terms The first two terms above are the linear approximation of φ around ζ. We can therefore approximate φ as φ(x) ≈ φ(ξ) + ∇φ(ξ) (x − ξ) and this approximation will be good in some (possibly small) neighborhood of ξ.

8.2

Equilibrium Points

Consider a nonlinear differential equation x(t) ˙ = f (x(t), u(t))

(8.1)

where f is a function mapping Rn × Rm → Rn . A point x¯ ∈ Rn is called an equilibrium point if there is a specific u¯ ∈ Rm (called the equilibrium input) such that f (¯ x, u¯) = 0n

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Suppose x¯ is an equilibrium point (with equilibrium input u¯). Consider starting the system (8.1) from initial condition x(t0 ) = x¯, and applying the input u(t) ≡ u¯ for all t ≥ t0 . The resulting solution x(t) satisfies x(t) = x¯ for all t ≥ t0 . That is why it is called an equilibrium point.

8.3

Deviation Variables

Suppose (¯ x, u¯) is an equilibrium point and input. We know that if we start the system at x(t0 ) = x¯, and apply the constant input u(t) ≡ u¯, then the state of the system will remain fixed at x(t) = x¯ for all t. What happens if we start a little bit away from x¯, and we apply a slightly different input from u¯? Define deviation variables to measure the difference. δx (t) := x(t) − x¯ δu (t) := u(t) − u¯ In this way, we are simply relabling where we call 0. Now, the variables x(t) and u(t) are related by the differential equation x(t) ˙ = f (x(t), u(t)) Substituting in, using the constant and deviation variables, we get δ˙x (t) = f (¯ x + δx (t), u¯ + δu (t)) This is exact. Now however, let’s do a Taylor expansion of the right hand side, and neglect all higher (higher than 1st) order terms ∂f ∂f δ (t) + δ˙x (t) ≈ f (¯ x, u¯) + x x=¯x δu (t) x=¯ x ∂x u=¯ ∂u u u=¯ u But f (¯ x, u¯) = 0, leaving ˙δx (t) ≈ ∂f δ (t) x=¯ x x ∂x u=¯ u

+

∂f δ (t) x=¯ x u ∂u u=¯ u

This differential equation approximately governs (we are neglecting 2nd order and higher terms) the deviation variables δx (t) and δu (t), as long as they remain small. It is a linear, time-invariant, differential equation, since the derivatives of δx are linear combinations of the δx variables and the deviation inputs, δu . The matrices ∂f ∂f n×n n×m A := ∈R , B := (8.2) x=¯x ∈ R x=¯ x ∂x u=¯ ∂u u u=¯ u

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are constant matrices. With the matrices A and B as defined in (8.2), the linear system δ˙x (t) = Aδx (t)

+

Bδu (t)

is called the Jacobian Linearization of the original nonlinear system (8.1), about the equilibrium point (¯ x, u¯). For “small” values of δx and δu , the linear equation approximately governs the exact relationship between the deviation variables δu and δx . For “small” δu (ie., while u(t) remains close to u¯), and while δx remains “small” (ie., while x(t) remains close to x¯), the variables δx and δu are related by the differential equation δ˙x (t) = Aδx (t)

+

Bδu (t)

In some of the rigid body problems we considered earlier, we treated problems by making a small-angle approximation, taking θ and its derivatives θ˙ and θ¨ very small, so that certain terms were ignored (θ˙2 , θ¨ sin θ) and other terms simplified (sin θ ≈ θ, cos θ ≈ 1). In the context of this discussion, the linear models we obtained were, in fact, the Jacobian linearizations around the equilibrium point θ = 0, θ˙ = 0. If we design a controller that effectively controls the deviations δx , then we have designed a controller that works well when the system is operating near the equilibrium point (¯ x, u¯). We will cover this idea in greater detail later. This is a common, and somewhat effective way to deal with nonlinear systems in a linear manner.

8.4

Tank Example

qH

qC

?

?

6

T h

? ?

Consider a mixing tank, with constant supply temperatures TC and TH . Let the inputs be the two flow rates qC (t) and qH (t). The equations

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for the tank are ˙ h(t) = T˙T (t) =

1 AT



qC (t) + qH (t) − cD Ao

1 h(t)AT

 p 2gh(t)

(qC (t) [TC − TT (t)] + qH (t) [TH − TT (t)])

Let the state vector x and input vector u be defined as     h(t) qC (t) x(t) := , u(t) := TT (t) qH (t)
√ f1 (x, u) = A1T u1 + u2 − cD Ao 2gx1 f2 (x, u) = x11AT (u1 [TC − x2 ] + u2 [TH − x2 ]) ¯ > 0 and any tank temperature T¯T satisfying Intuitively, any height h TC ≤ T¯T ≤ TH should be a possible equilibrium point (after specifying the correct values of the equilibrium ¯ and T¯T chosen, the equation f (¯ inputs). In fact, with h x, u¯) = 0 can be written as      √ 1 1 u¯1 cD Ao 2g¯ x1 = TC − x¯2 TH − x¯2 u¯2 0 The 2 × 2 matrix is invertible if and only if TC 6= TH . Hence, as long as TC 6= TH , there is a unique equilibrium input for any choice of x¯. It is given by      √ 1 TH − x¯2 −1 cD Ao 2g¯ x1 u¯1 = 0 u¯2 TH − TC x¯2 − TC 1 This is simply √ cD Ao 2g¯ x1 (TH − x¯2 ) u¯1 = TH − TC

√ cD Ao 2g¯ x1 (¯ x2 − TC ) u¯2 = TH − TC

,

Since the ui represent flow rates into the tank, physical considerations restrict them to be nonegative real numbers. This implies that x¯1 ≥ 0 and TC ≤ T¯T ≤ TH . Looking at the differential equation for TT , we see that its rate of change is inversely related to h. Hence, the differential equation model is valid while h(t) > 0, so we further restrict x¯1 > 0. Under those restrictions, the state x¯ is indeed an equilibrium point, and there is a unique equilibrium input given by the equations above. Next we compute the necessary partial derivatives. "

∂f1 ∂x1 ∂f2 ∂x1

∂f1 ∂x2 ∂f2 ∂x2

#

" =

D Ao √ − Agc T 2gx1

2 [TH −x2 ] − u1 [TC −xx2 ]+u 2A T 1

0 −(u1 +u2 ) x1 A T

#

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"

∂f1 ∂u1 ∂f2 ∂u1

∂f1 ∂u2 ∂f2 ∂u2

#

78

 =

1 AT TC −x2 x1 A T

1 AT TH −x2 x1 AT



The linearization requires that the matrices of partial derivatives be evaluated at the equilibrium points. Let’s pick some realistic numbers, and see how things vary with different equilibrium points. Suppose that TC = 10◦ , TH = 90◦ , AT = 3m2 , Ao = 0.05m, cD = 0.7. Try ¯ = 1m and h ¯ = 3m, and for T¯T , try T¯T = 25◦ and T¯T = 75◦ . That gives 4 combinations. h Plugging into the formulae give the 4 cases
¯ T¯T = (1m, 25◦ ). The equilibrium inputs are 1. h, u¯1 = q¯C = 0.126

,

The linearized matrices are   −0.0258 0 A= 0 −0.517

u¯2 = q¯H = 0.029

 ,

B=

0.333 0.333 −5.00 21.67




¯ T¯T = (1m, 75◦ ). The equilibrium inputs are 2. h, u¯1 = q¯C = 0.029

,

The linearized matrices are   −0.0258 0 A= 0 −0.0517

u¯2 = q¯H = 0.126

 ,

B=

0.333 0.333 −21.67 5.00




¯ T¯T = (3m, 25◦ ). The equilibrium inputs are 3. h, u¯1 = q¯C = 0.218

,

The linearized matrices are   −0.0149 0 A= 0 −0.0298

u¯2 = q¯H = 0.0503

 ,

B=

0.333 0.333 −1.667 7.22




¯ T¯T = (3m, 75◦ ). The equilibrium inputs are 4. h, u¯1 = q¯C = 0.0503 The linearized matrices are   −0.0149 0 A= 0 −0.0298

,

u¯2 = q¯H = 0.2181

 ,

B=

0.333 0.333 −7.22 1.667



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We can try a simple simulation, both in the exact nonlinear equation, and the linearization, and compare answers. We will simulate the system x(t) ˙ = f (x(t), u(t)) subject to the following conditions  x(0) = and





0.022 for 0 ≤ t ≤ 25 0.043 for 25 < t ≤ 100



0.14 for 0 ≤ t ≤ 60 0.105 for 60 < t ≤ 100

u1 (t) = u2 (t) =

1.10 81.5

This is close to equilibrium condition #2. So, in the linearization, we will use linearization #2, and the following conditions     1 0.10 δx (0) = x(0) − = 75 6.5 and



−0.007 for 0 ≤ t ≤ 25 0.014 for 25 < t ≤ 100



0.014 for 0 ≤ t ≤ 60 −0.021 for 60 < t ≤ 100

δu1 (t) := u1 (t) − u¯1 = δu2 (t) := u2 (t) − u¯2 =

To compare the simulations, we must first plot x(t) from the nonlinear simulation. This is the “true” answer. For the linearization, we know that δx approximately governs the deviations from x¯. Hence, for that simulation we should plot x¯ + δx (t). These are shown below for both h and TT .

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Water Height: Actual (solid) and Linearization (dashed) 1.3 1.25

Meters

1.2 1.15 1.1 1.05 1 0

10

20

30

40

50

60

70

80

90

100

90

100

Water Temp: Actual (solid) and Linearization (dashed) 82 80 78

Degrees

76 74 72 70 68 66 0

8.5

10

20

30

40

50 Time

60

70

80

Output Equations

Add simple section for a nonlinear output equation y(t) = h(x(t), u(t))

(8.3)

where h is a function mapping n + m real variables into q variables. Associated with the equilibrium pair (¯ x, u¯), define the equilibrium output y¯ := h(¯ x, u¯). Let δy (t) measure the

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difference between actual output, and equilibrium output, so δy (t) := y(t) − y¯ What is the approximate relation between δx , δu and δy , while they remain small? Do a Taylor series approximation of equation (8.3) to obtain ∂h ∂h δx (t) + δu (t) δy (t) ≈ ∂x x=¯x ∂u x=¯x u=¯ u

u=¯ u

The matrices

∂h ∈ Rq×n C := x=¯ x ∂x u=¯u are constant matrices.

8.6

∂h D := ∈ Rq×m x=¯ x ∂u u=¯u

,

(8.4)

Calculus for systems not in standard form

Often times, the system equations will not naturally be in the standard form, for example, they may be of the form M (x(t))x(t) ˙ = q(x(t), u(t)) where M is a function mapping n variables into n × n invertible matrices, and q maps n + m variables into n. Clearly, the equations can be rewritten as x(t) ˙ = M −1 (x(t))q(x(t), u(t)) This naturally suggests defining a function f (x, u) := M −1 (x)q(x, u), and then proceeding with the linearization process, working with f . But, this creates alot of unnecessary calculations, all of which cancel out in the end. Use the notation



∂f1 ∂xj ∂f2 ∂xj

 ∂f  :=  ∂xj  ··· ∂fn ∂xj

    

Then by chain rule, we have ∂f ∂M −1 ∂q = −M −1 (x) M (x)q(x, u) + M −1 (x) ∂xj ∂xj ∂xj Since M (x) is invertible for all x, it is clear that f (¯ x, u¯) = 0n×1

⇔

q(¯ x, u¯) = 0n×1

Therefore, evaluating the partial of f at the equilibrium point leads to a very simple expression ∂f ∂q −1 = M (¯ x) x=¯ x x=¯ x ∂xj u=¯ ∂xj u=¯ u u

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82

Another common non-standard form

It is also common that the equations can be in the form   x(t) ˙ M (x(t)) = q(x(t), u(t)) v(t) where v are other dependent-variables (ie., “outputs”) of the problem. For example, in applying Newton’s laws to a system of interconnected rigid links, the internal forces at each link are unknowns that must be introduced into the equations of motion, regardless of whether they are being sought after (in the case, Lagrange’s formulation may be simpler, as the internal forces do not explicitly play a role in that methodology). Assume M (x) is invertible for all x, and treat v as an output. The state/output equations are     x(t) ˙ f (x(t), u(t)) −1 = M (x(t))q(x(t), u(t)) =: v(t) h(x(t), u(t)) An equilibrium point (¯ x, u¯, v¯) satisfies  M (¯ x)

0 v¯

 = q(¯ x, u¯)

The linearization about the equilibrium point is obtained by taking partial derivatives of f and h with respect to components of x and u, and evaluating the expressions at (¯ x, u¯, v¯). Since the horizontal stacking of f above h has a nice representation (ie., M −1 h), it is easiest to write expressions for     A B , C D Specifically, for any matrix W , let W[i] represent the i’th column of W . Then " #      ∂f  ∂f B A i k , = ∂u , = ∂x ∂h ∂h D C [i] ∂xi ∂uk [k] x=¯ x,u=¯ u

x=¯ x,u=¯ u

Direct calculation gives     ∂M −1 ∂q A −1 = M (x) − M (x)q(x, u) + C [i] ∂xi ∂xi x=¯x,u=¯u   0 Clearly, = M −1 (¯ x)q(¯ x, u¯), so this simplifies to v¯       ∂M 0 ∂q A −1 = M (¯ x) − + C [i] ∂xi v¯ ∂xi x=¯x,u=¯u Similarly 

B D

 =M [k]

−1

∂q (x) ∂uk x=¯x,u=¯u

ME 132, Fall 2018, UC Berkeley, A. Packard

8.8

83

Linearizing about general solution

In section 8.3, we discussed the linear differential equation which governs small deviations away from an equilibrium point. This resulted in a linear, time-invariant differential equation. Often times, more complicated situations arise. Consider the task of controlling a rocket trajectory from Earth to the moon. By writing out the equations of motion, from Physics, we obtain state equations of the form x(t) ˙ = f (t, x(t), u(t), d(t))

(8.5)

where u is the control input (thrusts), d are external disturbances. Through much computer simulation, a preplanned input schedule is developed, which, under ideal circumstances (ie., d(t) ≡ 0), would get the rocket from Earth to the moon. Denote ¯ (which we assume is 0). this preplanned input by u¯(t), and the ideal disturbance by d(t) This results in an ideal trajectory x¯(t), which solves the differential equation,
¯ x¯˙ (t) = f t, x¯(t), u¯(t), d(t) Now, small nonzero disturbances are expected, which lead to small deviations in x, which must be corrected by small variations in the pre-planned input u. Hence, engineers need to have a model for how a slightly different input u¯(t) + δu (t) and slightly different disturbance ¯ + δd (t) will cause a different trajectory. Write x(t) in terms of a deviation from x¯(t), d(t) defining δx (t) := x(t) − x¯(t), giving x(t) = x¯(t) + δx (t) Now x, u and d must satisfy the differential equation, which gives x(t) ˙ = f (t, x(t), u(t), d(t))
¯ + δd (t) ˙x¯(t) + δ˙x (t) = f t, x¯(t) + δx (t), u¯(t) + δu (t), d(t) ∂f ≈ f (t, x¯(t), u¯(t), 0) + ∂f x(t) δx (t) + ∂u x(t)=¯ x(t) δu (t) + ∂x x(t)=¯ u(t)=¯ u(t) ¯ d(t)=d(t)

u(t)=¯ u(t) ¯ d(t)=d(t)

∂f x(t) δd (t) ∂d x(t)=¯ u(t)=¯ u(t) ¯ d(t)=d(t)



But the functions x¯ and u¯ satisfy the governing differential equation, so x¯˙ (t) = f (¯ x(t), u¯(t), 0, t), leaving the (approximate) governing equation for δx ∂f ∂f ∂f x(t)=¯x(t) δx (t) + x(t)=¯x(t) δu (t) + x(t)=¯x(t) δd (t) δ˙x (t) = ∂x u(t)=¯u(t) ∂u u(t)=¯u(t) ∂d u(t)=¯ u(t) ¯ d(t)=d(t)

¯ d(t)=d(t)

Define time-varying matrices A, B1 and B2 by ∂f ∂f A(t) := B1 (t) := x(t)=¯ x(t) x(t)=¯ x(t) ∂x u(t)=¯ ∂u u(t)=¯ u(t) u(t) ¯ d(t)=d(t)

¯ d(t)=d(t)

¯ d(t)=d(t)

∂f B2 (t) := x(t)=¯ x(t) ∂d u(t)=¯ u(t) ¯ d(t)=d(t)

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The deviation variables are approximately governed by δ˙x (t) = A(t)δx (t) + B1 (t)δu (t) + B2 (t)δd (t) = A(t)δx (t) +



B1 (t) B2 (t)





δu (t) δd (t)




This is called the linearization of system (8.5) about the trajectory x¯, u¯, d¯ . This type of linearization is a generalization of linearizations about equilibrium points. Linearizing about an equilibrium point yields a LTI system, while linearizing about a trajectory yields an LTV system. In general, these types of calculations are carried out numerically, ¯ • simulink to get solution x¯(t) given particular inputs u¯(t) and d(t) • Numerical evaluation of

∂f , ∂x

etc, evaluated at the solution points

• Storing the time-varying matrices A(t), B1 (t), B2 (t) for later use In a simple case it is possible to analytically determine the time-varying matrices.

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8.9

85

Problems

1. This purpose of this problem is to remind you how first-order Taylor series (ie., linear approximation) are used to approximate a function of many variables. The temperature in a particular 3-dimensional solid is a function of position, and is known to be T (x, y, z) = 42 + (x − 2)2 + 3 (y − 4)2 − 5 (z − 6)2 + 2yz (a) Find the first order approximation (linearization) of the temperature near the location (¯ x = 4, y¯ = 6, z¯ = 0). Use δx , δy and δz as your deviation variables. (b) What is the maximum error between the actual temperature and the first order approximation formula for |δx | ≤ 0.3, |δy | ≤ 0.2, |δz | ≤ 0.1? Solve this numerically, by simply sampling a dense grid over the 3-dimesional cube, and determining the maximum error. (c) More generally, suppose that x¯ ∈ R, y¯ ∈ R, z¯ ∈ R. Find the first order approximation of the temperature near the location (¯ x, y¯, z¯). 2. The pitching-axis of a tail-fin controlled missile is governed by the nonlinear state equations α(t) ˙ = K1 M fn (α(t), M ) cos α(t) + q(t) q(t) ˙ = K2 M 2 [fm (α(t), M ) + Eu(t)] Here, the states are x1 := α, the angle-of-attack, and x2 := q, the angular velocity of the pitching axis. The input variable, u, is the deflection of the fin which is mounted at the tail of the missile. K1 , K2 , and E are physical constants, with E > 0. M is the speed (Mach) of the missile, and fn and fm are known, differentiable functions (from wind-tunnel data) of α and M . Assume that M is a constant, and M > 0. (a) Show that for any specific value of α ¯ , with |¯ α| < π2 , there is a unique pair (¯ q , u¯) such that   α ¯ , u¯ q¯ is an equilibrium point of the system (this represents a turn at a constant rate). Your answer should clearly show how q¯ and u¯ are functions of α ¯ , and will most likely involve the functions fn and fm . (b) Calculate the Jacobian Linearization of the missile system about the equilibrium point. Your answer will mostly be symbolic, and depend on partial derivatives of the functions fn and fm . Be sure to indicate where the various terms are evaluated. 3. A magnetic levitation system is modeled by the nonlinear differential equations x˙ 1 (t) = x2 (t)

ME 132, Fall 2018, UC Berkeley, A. Packard

x˙ 2 (t) = g −

86

γx23 (t) [2x1 (t) + β]2

2x1 (t) + β 2x2 (t)x3 (t) [u(t) − Rx3 (t)] + α 2x1 (t) + β Here, α, β, γ, R and g are positive constants. The input u(t) is a voltage applied to the electromagnet. The state variables are x1 = position of suspended mass, x2 = velocity of suspended mass, x3 = current in armature. x˙ 3 (t) =

(a) Take x¯1 to be a positive number. Find numbers x¯2 , x¯3 and u¯ such that (¯ x, u¯) is an equilibrium point. NOTE: Your answers for x¯2 , x¯3 and u¯ will, in general, be functions of x¯1 , and involve the given constants. (b) Define deviation variables η and v by η(t) := x(t) − x¯ v(t) := u(t) − u¯ Find the linear differential equations η(t) ˙ = Aη(t) + Bv(t) which approximately governs the relationship between v and η, while they remain “small” 4. In almost every situation, the sensor used to obtain the feedback measurement is a dynamic system itself. A rate-gyro is an inertial instrument used to measure the angular velocity of a rigid body relative to an inertial frame. You can look up schematic pictures of rate-gyros on the web. There is an outerframe casing which is attached to the rigid-body in question. This casing/outer-frame supports another internal frame, with bearings. The internal frame is free to rotate relative to casing frame. Let θ denote that angular rotation. The outer frame is rigidly attached to the rigid body. θ is the angle that the gimbal makes with the frame, and is easily measured (eg., using an angle encoder). The angular rate that the rigid-body (and hence the outer frame) is rotated about a vertical axis ˙ ˙ is denoted ψ(t). By measuring θ(t), we can obtain information about ψ(t). The equation governing the relationship between θ and ψ˙ is ¨ + cθ(t) ˙ + kθ(t) = co Iz ψ(t) ˙ cos θ(t) − I ψ˙ 2 (t) cos θ(t) sin θ(t) I θ(t) ˙ Obviously, since they are related, measuring θ gives information about ψ. ˙ and input u := ψ. ˙ Rewrite the governing (a) Define state variables x1 := θ, x2 := θ, equations in first-order form, x˙ 1 (t) = f1 (x1 (t), x2 (t), u(t)) x˙ 2 (t) = f2 (x1 (t), x2 (t), u(t))

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(b) Show that (¯ x1 = 0, x¯2 = 0, u¯ = 0) is an equilibrium point. (c) Compute the Jacobian-linearization of the system about the equilibrium point. (d) For the linearized system, what is the steady-state value of δx1 due to a steadystate input δu = α.

8.10

Additional Related Problems

These are related to linearization, but may require other concepts that are developed in later chapters. 1. (Model taken from “Introduction to Dynamic Systems Analysis,” T.D. Burton, 1994, pg. 212, problem 23) Let f1 (x1 , x2 ) := x1 − x31 + x1 x2 , and f2 (x1 , x2 ) := −x2 + 1.5x1 x2 . Consider the 2-state system x(t) ˙ = f (x(t)). (a) Note that there are no inputs. Find all equilibrium points of this system. Hint: In this problem, there are 4 equilibrium points. (b) Derive the Jacobian linearization which describes the solutions near each equilibrium point. You will have 4 different linearizations, each of the form η(t) ˙ = Aη(t) with different A matrices dependent on which equilibrium point the linearization has been computed. (c) Using eigenvalues, determine the stability of each of 4 Jacobian linearizations. Note: (for part 10f below) • If the linearization is stable, it means that while the deviation of x(t) from x¯ remains small, the variables x(t) − x¯ approximately evolve by a linear differential equation whose homogenous solutions all decay to zero. So, we would expect that initial conditions near the equilibium point would converge to the equilibrium point. • Conversely, if the linearization is unstable, it means that while the deviation of x(t) from x¯ remains small, the variables x(t) − x¯ approximately evolve by a linear differential equation that has some homogenous solutions which grow. So, we would expect that some initial conditions near the equilibium point would initially diverge away from the equilibrium point. (d) Using Simulink, or ode45, simulate the system starting from 50 random initial conditions satisfying −3 ≤ x1 (0) ≤ 3, and −3 ≤ x2 (0) ≤ 3. Plot the resulting solutions in x2 versus x1 plane (each solution will be a curve - parametrized by time t). This is called a phase-plane graph, and we already looked at such plots,

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for linear systems, a few weeks ago. Make the axis limits −4 ≤ xi ≤ 4. On each curve, hand-draw in arrow(s) indicating which direction is increasing time. Mark the 4 equilibrium points on your graph. See the “Critical Remarks” section below before starting this calculation. (e) At each equilibrium point, draw the two eigenvectors of the linearization, and notate (next to each eigenvector) what the associated eigenvalue is. If the eigenvalues/eigenvectors are complex, draw the real-part of the eigenvector, and the imaginary part (recall that in linear systems, real-valued solutions oscillate between the real-part and the imaginary part of the eigenvectors). (f) For each equilibrium point, describe (2 sentences) the behavior of the solution curves near the equilibrium points in relation to the behavior of the linearized solutions, and how the curves relate to the stability computation, of the linearization, in part 10c. Critical (!) remark: Since some initial conditions will lead to diverging solutions, it is important to automatically stop the simulation when either |x1 (t)| ≥ 8 (say) or |x2 (t)| ≥ 8. If using Simulink, use a Stop Simulation block (from Sinks), with its input coming from the Relational Operator (from Math). If using ode45, you can use events, which are programmed to automatically stop the numerical simulation when values of the solution, x(t) reach/exceed certain values. 2. A hoop (of radius R) is mounted vertically, and rotates at a constant angular velocity Ω. A bead of mass m slides along the hoop, and θ is the angle that locates the bead location. θ = 0 corresponds to the bead at the bottom of the hoop, while θ = π corresponds to the top of the hoop, as shown below.

The nonlinear, 2nd order equation (from Newton’s law) governing the bead’s motion is mRθ¨ + mg sin θ + αθ˙ − mΩ2 R sin θ cos θ = 0 All of the parameters m, R, g, α are positive.

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˙ (a) Let x1 (t) := θ(t) and x2 (t) := θ(t). Write the 2nd order nonlinear differential equation in the state-space form x˙ 1 (t) = f1 (x1 (t), x2 (t)) x˙ 2 (t) = f2 (x1 (t), x2 (t)) (b) Show that x¯1 = 0, x¯2 = 0 is an equilibrium point of the system. (c) Find the linearized system η(t) ˙ = Aη(t) which governs small deviations away from the equilibrium point (0, 0). (d) Under what conditions (on m, R, Ω, g) is the linearized system stable? (e) Show that x¯1 = π, x¯2 = 0 is an equilibrium point of the system. (f) Find the linearized system η(t) ˙ = Aη(t) which governs small deviations away from the equilibrium point (π, 0). (g) Under what conditions is the linearized system stable? (h) It would seem that if the hoop is indeed rotating (with angular velocity Ω) then there would other equilibrium point (with 0 < θ < π/2). Do such equilibrium points exist in the system? Be very careful, and please explain your answer. (i) Find the linearized system η(t) ˙ = Aη(t) which governs small deviations away from this equilibrium point. (j) Under what conditions is the linearized system stable? 3. Car Engine Model (the reference for this problem is Cho and Hedrick, “Automotive Powertrain Modelling for Control,” ASME Journal of Dynamic Systems, Measurement and Control, vol. 111, No. 4, December 1989): In this problem we consider a 1-state model for an automotive engine, with the transmission engaged in 4th gear. The engine state is ma , the mass of air (in kilograms) in the intake manifold. The state of the drivetrain is the angular velocity, ωe , of the engine. The input is the throttle angle, α, (in radians). The equations for the engine is m ˙ a (t) = c1 T (α(t)) − c2 ωe (t)ma (t) Te = c3 ma (t) where we treat ωe and α as inputs, and Te as an output. The drivetrain is modeled ω˙ e (t) =

1 [Te (t) − Tf (t) − Td (t) − Tr (t)] Je

The meanings of the terms are as follows:

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• T (α) is a throttle flow characteristic depending on throttle angle,  0.00032 for α < 0  1 − cos (1.14α − 0.0252) for 0 ≤ α ≤ 1.4 T (α) =  1 for α > 1.4 • Te is the torque from engine on driveshaft, c3 = 47500 Nm/kg. • Tf is engine friction torque (Nm), Tf = 0.106ωe + 15.1 • Td is torque due to wind drag (Nm), Td = c4 ωe2 , with c4 = 0.0026 Nms2 • Tr is torque due to rolling resistance at wheels (Nm), Tr = 21.5. • Je is the effective moment of inertia of the engine, transmission, wheels, car, Je = 36.4kgm2 • c1 = 0.6kg/s, c2 = 0.095 • In 4th gear, the speed of car, v (m/s), is related to ωe as v = 0.129ωe . (a) Combine these equations into state variable form. x(t) ˙ = f (x(t), u(t)) y(t) = h(x(t)) where x1 = ma , x2 = ωe , u = α and y = v. (b) For the purpose of calculating the Jacobian linearization, explicitly write out the function f (x, u) (without any t dependence). Note that f maps 3 real numbers (x1 , x2 , u) into 2 real numbers. There is no explicit time dependence in this particular f . (c) Use Matlab to plot the throttle flow characteristic T (α) for 0 ≤ α ≤ 1.4 (d) Let v¯ > 0 denote an equilibium car speed. Find expressions for the corresponding equilibrium values m ¯ a, ω ¯ e and α ¯ . The expressions should all be functions of v¯. (In upcoming problems, for clarity, you will express various answers in terms of these expressions; there is no specific need to ”by-hand” subsitute the expression into your answers.) (e) Based on the throttle flow characteristic, T , what is the maximum equilibrium speed of the car? (f) Compute the equilibrium values of m ¯ a, ω ¯ e and α ¯ so that the car travels at a constant speed of 22 m/s (let v¯ denote the equilibrium speed). Repeat calculation for an equilibrium speed of v¯ = 32 m/s. Make sure your formula are clear enough that you repeat this at any value of v¯.

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(g) Consider deviations from these equilibrium values, α(t) ωe (t) ma (t) y(t)

= = = =

α ¯ + δα (t) ω ¯ e + δωe (t) m ¯ a + δma (t) y¯ + δy (t)

Find the (linear) differential equations that approximately govern these deviation variables (Jacobian Linearzation discussed in class). Your answer should consist of 4 matrices which, in general, depend on v¯. Use the notation A(¯ v ), B(¯ v ), C(¯ v ), D(¯ v) for the 4 matrices. Call this system Jv¯, to denote its dependence on the equilibrium speed. (h) Compute the numerical values of the Jacobian Linearization at two specific cases, namely v¯ = 22 m/s and v¯ = 32 m/s. (i) Return attention to the full nonlinear model of the car/engine. Using Simulink, starting from the initial condition ωe (0) = ω ¯ e , ma (0) = m ¯ a , apply constant input of α(t) = α ¯+β for 6 values of β, β = ±0.01, ±0.04, ±0.1 and obtain the response for v and ma . Do this for two cases, v¯ = 22 and v¯ = 32. (j) Compute (using a State-Space block in Simulink, or ode45) the response of the Jacobian linearization starting from initial condition δωe (0) = 0 and δma (0) = 0, with the input δα (t) = β for for 6 values of β, β = ±0.01, ±0.04, ±0.1. Add the responses δωe (t) and δma (t) to the equilibrium values ω ¯e, m ¯ a and plot these sums, which are approximations of the actual response. Compare the linearized response with the actual response from part 3i Comment on the difference between the results of the “true” nonlinear simulation and the results from the linearized analysis. Note: do this for both the equilibrium points – at 22 m/s and at 32 m/s. (k) Repeat this nonlinear/linear comparison, using a sinusoidal input of the form α(t) = α ¯ + β sin(0.5t) for 3 values of β, β = 0.01, 0.04, 0.1 and obtain the response for v and ma . Do an analogous simulation for the linearized system, and appropriately compare the responses. Comment on the difference between the results of the “true” nonlinear simulation and the results from the linearized analysis. Again, do this for both cases, 22 m/s and 32 m/s..

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4. This problem extends the results of 3. In this problem, will approximate the 2-state linearized model by a lower-order (1-state) linear system. In problem 3, you have established that the linear system Jv¯ is governed by state-space matrices of the form       a11 (¯ v ) a12 (¯ v) b1 (¯ v) A(¯ v) = , B(¯ v) = , C(¯ v ) = 0 0.129 , D(¯ v) = 0 a21 (¯ v ) a22 (¯ v) 0 (a) Write the transfer function (from δ u to δ y ) of the linearized car model at v¯ m/s. Denote this transfer function as Gv¯(s). For clarity, express the answer in terms of the entries of the state-space matrices. (b) Find the quantities (γ, τ1 , τ2 ) (all of which depend on v¯) such that Gv¯ is of the form γ 1 . Gv¯(s) = (τ1 s + 1) (τ2 s + 1) For concreteness, choose the convention that τ1 ≤ τ2 . Hint: In terms of a statespace model, the steady-state gain is given by D − CA−1 B, and the poles of the corresponding transfer function are the eigenvalues of the A matrix. In our proposed expression for G, clearly γ is the steady-state gain, and the poles are at −1 and −1 respectively. τ1 τ2 (c) (d) Plot γ, τ1 and τ2 as functions of v¯ for 18 ≤ v¯ ≤ 36. (e) Hence, at each specific equilibrium speed v¯, Gv¯ is the cascade of two first-order systems: • A system with steady-state gain 1, and time constant τ1 • A system with steady-state gain γ, and time constant τ2 Since τ1 uLimits(2) && KI*eInput>0 a1 = 0; % make zdot=0 to avoid windup elseif % Some analogous condition goes here a1 = 0; % make zdot=0 to avoid windup else a1 = eInput; % standard zdot in PI control end case 3 ... end Resimulate, and verify that the antiwindup controller reduces the undershoot at t = 50. (e) The controller gains, and various limit-values in the antiwindup scheme are all based on the nominal values of the parameters in the car model, and the equilibrium speed at which the linearization was calculated. Keeping the parameters within the controller fixed, do a family of 50 simulations, randomly varying the car/engine parameters (c1 , c2 , ...Je ,.. etc) by ±20%. Use the desired-speed profile as in the previous simulation. Plot three variables: the velocity response, the control input α, and the controller integrator signal z. Note the robustness of the feedback system to variations in the car/engine behavior. Also notice that the speed (v) is relatively insensitive to the parameter variations, but that the actual control input α is quite dependent on the variations. This is the whole point of feedback control – make the regulated variable insensitive to variations, through the use of feedback strategies which “automatically” cause the control variable to accommodate these variations. 6. A common actuator in industrial and construction equipment is a hydraulic device, called a hydraulic servo-valve. A diagram is below.

ME 132, Fall 2018, UC Berkeley, A. Packard

p0

95

pS

p0

-h

q4

q1

q2

A K




A


















-y

q3 A 
AAAU
AA A A A A A A A A A A A A A A A A



p1

Fe

p2

Two resevoirs, one at high pressure, ps , called the supply, and one at low pressure, p0 , called the drain. The small piston can be moved relatively easily, its position h(t) is considered an input. As is moves, its connects the supply/drain to opposite sides of the power piston, which causes significant pressure differences across the power piston, and causes it to move. Constant pS p0 c1 c0 Ap K ρ L cD mp

Value 1.4 × 107 N/m2 1 × 105 N/m2 0.02m 4 × 10−7 m2 0.008m2 6.9 × 108 N/m2 800kg/m3 1m 0.7 2.5kg

The volume in each cylinder is simply V1 (y) = Ap (L + y) V2 (y) = Ap (L − y)

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The mass flows are q1 (t) q2 (t) q3 (t) q4 (t)

= = = =

p cD A1 (h(t))pρ (pS − p1 (t)) cD A2 (h(t))pρ (pS − p2 (t)) cD A3 (h(t))pρ (p2 (t) − p0 ) cD A4 (h(t)) ρ (p1 (t) − p0 )

The valve areas are all given by the same underlying function A(h), 4.0 × 10−7 for h ≤ −0.0002 −7 A(h) = 4.0 × 10 + 0.02(u + 0.0002) for −0.0002 < h < 0.005 1.044 × 10−4 for 0.005 < h By symmetry, we have A1 (h) = A(−h);, A2 (h) = A(h);, A3 (h) = A(−h);, A4 (h) = A(h). Mass continuity (with compressibility) gives   1 ˙ q1 (t) − q4 (t) = ρ p˙1 (t)V1 (y(t)) + Ap y(t) K and



 1 q2 (t) − q3 (t) = ρ p˙2 (t)V2 (y(t)) − Ap y(t) ˙ K

Finally, Newton’s law on the power piston gives mp y¨(t) = Ap (p1 (t) − p2 (t)) − Fe (t) (a) Why can the small piston be moved easily? (b) Define states x1 := p1 , x2 := p2 , x3 := y, x4 := y. ˙ Define inputs u1 := h, u2 := Fe . Define outputs y1 := y, y2 := y, ˙ y3 := y¨. Write state equations of the form x(t) ˙ = f (x(t), u(t)) y = g(x(t), u(t)) (c) Suppose that the power piston is attached to a mass M , which itself is also acted on by an external force w. This imposes the condition that Fe (t) = M y¨(t) + w(t), where w is the external force acting on the mass. Substitute this in. The “inputs” to your system will now be h and w. Although it’s bad notation, call this pair u, with u1 = h, u2 = w. You will still have 4 state equations. (d) Develop a Simulink model of the hydraulic servovalve, attached to a mass, M . Have y and y˙ as outputs (used for feedback later), and h and w as inputs.

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(e) Let y¯ be any number satisfying −L < y¯ < L. With the mass attached, compute the equilibrium values p¯1 and p¯2 such that   p¯1    p¯2  0   x¯ :=  , u¯ :=  y¯ 0 0 is an equilibrium point. Hint: The value of M should not affect your answers for p¯1 and p¯2 (f) Derive (by hand) the linearization of the system, at the equilibrium point. with u and w as the inputs. The matrices will be parametrized by all of the area, fluid parameters, as well as M and y¯. Make y the output. (g) Ignoring the disturbance force w, compute the transfer function of the linearized model from h to y. Denote this by Gy¯,M (s), since it depends on y¯ and M . Let M vary from 10kg to 10000kg, and let y¯ vary from −0.8 to 0.8. Plot the Bode plot of the frequency-response function from h → y for at least 35 combinations of M and y¯, namely 7 log-spaced values of M and 5 lin-spaced values of y¯. Place all on one plot. (h) There are several simulations here. Consider two different cases with M = 30kg and separately M = 10000kg. Do short duration (Final Time is small, say 0.15) response with step inputs with h := 0.0001, w = 0, and h := 0.0004, w = 0, and h := 0.001, w = 0, and h := 0.004, w = 0, and h := 0.01, w = 0. The power piston should accelerate to the left. Plot the power-piston y(t), normalized (ie., divided by) by the constant value of h. (You should have 10 plots on one sheet). Note, had the system been linear, (which it is not), after normalizing by the size of the step, the outputs would be identical, which they are not. Comment on the differences as the the forcing function step size gets larger (more deviation from equilibrium input of h = 0). Hints: Be careful with maximum step size (in Parameters window), using small sizes (start with 0.0001) as well as the simulation time. Make sure that y(t) does not exceed L. Note that a dominant aspect of the behavior is that y is the integral (with a negative scaling factor) of h. (i) On one figure, using many different values for M and y¯, plot Gy¯,M (jω) − G0,10000 (jω) G0,10000 (jω) versus ω, for ω in the range 0.01 ≤ ω ≤ 1000. This is a plot of percentage-variation of the linearized model away from the linearized model G0,10000 . (j) On one figure, using many different values for M and y¯, plot Gy¯,M (jω) − Gnom (jω) Gnom (jω)

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for ω in the range 0.01 ≤ ω ≤ 1000, using Gnom :=

−325 s

This is a plot of percentage-variation from Gnom . (k) Using Gnom , design a proportional controller h(t) = KP [r(t) − y(t)] so that the nominal closed-loop time constant is 0.25. On the same plot as in part 6j, plot the percentage variation margin. Do all of the variations lie underneath the percentage variation margin curve? (l) With the nominal closed-loop time constant set to 0.25, we know that Lnom has magnitude less than 1 for frequencies beyond about 4 rad/sec. We can reduce the gain of L even more beyond that without changing the dominant overall response of the closed system, but we will improve the percentage variation margin. Keeping KP fixed from before, insert a first-order filter in the controller, so that (in transfer function form) KP [R − Y ] H= τf s + 1 1 Try using τf = 10 0.25. Again on the same plot as in part 6j, plot the new percentage variation margin. Has the margin improved?

(m) Using this controller, along with the true nonlinear model, simulate the closedloop system with 3 different reference signals, given by the (time/value) pairs listed below. Do this for three different values for M , namely 30, 2000, and 10000 kg. Time Case 1 Case 2 Case 3 0 0 0 0 1 0.05 0.25 0.75 3 0.05 0.25 0.75 5 -0.025 -0.125 -0.375 10 -0.025 -0.125 -0.375 12 0 0 0 14 0 0 0 Between points, the reference signal changes linearly, connecting the points with straight lines. Plot the resulting trajectory y, along with r. Comment on the performance of the closed-loop system. 7. Recall the power piston modeling problem from Section ??. A diagram is below.

ME 132, Fall 2018, UC Berkeley, A. Packard

p0

99

pS

p0

-h

q4 A K




A


















q1

q2

-y

q3 A 
AAAU
AA A A A A A A A A A A A A A A A A



p1

qL

Fe

p2 -

(a) Define states x1 := p1 , x2 := p2 , x3 := y, x4 := y. ˙ Define inputs u1 := h, u2 := Fe . Define outputs y1 := y, y2 := y, ˙ y3 := y¨. Write state equations of the form x(t) ˙ = f (x(t), u(t)) y = g(x(t), u(t)) (b) Suppose y¯ is a constant, with −L < y¯ < L. With a mass M attached, compute the equilibrium values p¯1 and p¯2 such that   p¯1    p¯2  0  x¯ :=  u¯ :=  y¯  , 0 0 is an equilibrium point. Hint: The values of M and y¯ will not affect your answers for p¯1 and p¯2 (c) Derive (by hand) the linearization of the system, at the equilibrium point. with u and w as the inputs. The matrices will be parametrized by all of the area, fluid parameters, as well as M and y¯. Make y the outputs, (d) Ignoring the disturbance force w, compute the transfer function of the linearized model from h to y. Denote this by Gy¯,M (s), since it depends on y¯ and M . Let M vary from 10kg to 10000kg, and let y¯ vary from −0.8 to 0.8. Plot the Bode plot for at least 35 combinations of M and y¯, namely 7 log-spaced values of M and 5 lin-spaced values of y¯. Place all on one plot.

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(e) On one figure, using many different values for M and y¯, plot Gy¯,M (jω) − G0,10000 (jω) G0,10000 (jω) for ω in the range 0.01 ≤ ω ≤ 1000. This is a plot of percentage-variation from G0,10000 . (f) On one figure, using many different values for M and y¯, plot Gy¯,M (jω) − Gnom (jω) Gnom (jω) for ω in the range 0.01 ≤ ω ≤ 1000, using Gnom :=

−325 s

This is a plot of percentage-variation from Gnom . (g) Using the linearization as the model, (h) Using the simple model Gnom , and the percentage-variation margin, design a proportional control to accomplish good positioning control of the mass. The closed-loop time constant should be on the order of 0.25-0.5 seconds. (i) Using the controller on the true nonlinear model, simulate the closed-loop response to the following inputs (T = 1). r(t) 6

0.05

T −0.025

J J J J J 5T J 3T J J

7T

J J J J J 5T J 3T J J

7T

8T -

t

r(t) 6

0.25

T −0.125

8T -

t

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r(t) 6

0.75 J J J J J 5T J 3T J J

T −0.375

7T

8T -

t

(j) Simulate the system subjected to a constant disturbance force, w(t) = w, ¯ staring from all 0 initial conditions. Determine the maximum value for w ¯ so that h(t) ≤ L ¯ and submit plots of h and y 0.005 and |y(t)| ≤ 2 for all t. Give the value for w, as they transition to their steady-state values. 8. In 7, we derived a complex model for a hydraulic servo-valve, taking compressibility of the hydraulic fluid into account. In this problem, we derive a model for incompressible fluid, which is much simpler, and gives us a good, simplified, “starting point” model for the servovalve’s behavior. The picture is as before, but with ρ constant (though p1 and p2 are still functions of time – otherwise the piston would not move...). (a) Assume that the forces from the environment are such that we always have p0 < p1 (t) < ps and p0 < p2 (t) < ps . Later, we will check what limits this imposes on the that flow into side 1 is p p environment forces. Suppose that u > 0. Recall ch(t) ρ (ps − p1 (t)), and flow out of side 2 is ch(t) ρ (p2 (t) − p0 ), where is c is some constant reflecting the discharge coefficient, and the relationship between orifice area and spool displacement (h). Show that it must be that ps + p0 = p1 (t) + p2 (t) (b) Let ∆(t) := p1 (t) − p2 (t). Show that p1 (t) =

ps + p0 + ∆(t) , 2

p2 (t) =

ps + p0 − ∆(t) 2

(c) Assuming that the mass of the piston is very small, show that Ap ∆(t) + FE (t) = 0 (d) Show that p0 < p1 (t) < ps for all t if and only if |FE (t)| ≤ Ap (ps − p0 ) Show that this is also necessary and sufficient for p0 < p2 (t) < ps to hold for all t.

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(e) Show that the velocity of the piston y˙ is given by s   ps − p0 ∆(t) ρAp y(t) ˙ = ch(t) ρ − 2 2 (f) Manipulate this into the form s   c ps − p0 FE (t) h(t) ρ + y(t) ˙ = ρAp 2 2Ap

(8.6)

Hence, the position of the piston y(t) is simply the integral of a nonlinear function of the two “inputs,” h and FE . (g) Go through the appropriate steps for the case h < 0 to derive the relationship in that case. ¯ = 0, F¯ ) is an equi(h) Suppose y¯ and F¯ are constants. Show that the “triple” (¯ y, h librium point of the system described in equation (8.6). (i) What is the linear differential equation governing the behavior near the equilibrium point? 9. A satellite is in a planar orbit around the earth, with coordinates as shown below.

The equations of motion are m [¨ r(t) − r(t)ω(t)] = − r2K(t) m [ω(t) ˙ + 2r(t)ω(t)] ˙ = u(t) where u is a force input provided by thrusters mounted in the tangential direction. Define states as x1 := r, x2 := r, ˙ x3 := ω. (a) Define states for the system (with u as the input), and write down state equations.

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(b) For a given value x¯3 = ω ¯ > 0, determine values for x¯1 , x¯2 , and u¯ to give an equilibrium point. (c) What is the physical interpretation of this particular equilibrium point? (d) Calculate the linearization about the equilibrium point. (e) Is the linearized system stable? (f) Propose, derive, and fully explain a feedback control system which stabilizes the linearized system. Document all steps, and mention any sensors (measurements) your control system will need to make. 10. The equations of motion for a wheel/beam system, assuming rolling without slipping, representing a simplified Segway device are of the form   x(t) ˙ M (x(t)) = h(x(t), u(t)) v(t) where



 Ωw x :=  Ωb  , θb



 Fr s :=  Fθ  , Ff

u=T

and     M (x) :=    

Iw 0 0 Ip 0 0 mp R sin x3 0 mp R cos x3 −mp L −mw R 0

 0 0 0 R 0 0 L 0   1 0 0 0  , 0 1 0 0   0 0 1 0  0 sin x3 cos x3 1

    h(x, u) :=    

u u x2 mp (g cos x3 − Lx22 ) −mp g sin x3 0

Here Ωw is the angular velocity of wheel; Ωb and θb are anglular velocity and angle (relative to vertical) of the rigid beam (represents the rider). The parameters are: R, radius of wheel; L, length of beam; mp mass of beam; mw mass of wheel; Iw massmoment of inertia of wheel; Ip mass-moment of inertia of beam. The input T , denoted by u, represents the internal torque applied between the wheel/beam assembly, by an electric motor. Fr and Fθ are the internal forces acting at the bearing of the wheel/beam assembly, and Ff is the force at the road/tire interface (required for the no-slip behavoir). The model was derived using basic dynamics (MechE 104). (a) Let v¯ be a positive constant, representing a forward velocity of the center of the ¯ w, Ω ¯ b , θ¯b ) and T¯ that are an wheel. Show that for any v¯, there exist unique (Ω equilibrium point of the system, and correspond to a constant forward velocity of v¯.

       

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(b) Implement the equilibrium point calulation in Matlab. Write an m-file called SegwayLinearize.m, with function declaration line

(c) Find the linearization of the system about the equilibrium point (which is parametrized by v¯). For this, follow the procedure in section 8.7. Do all partial-differentiation analytically (ie., by hand), but implement the overall procedure numerically, with Matlab.

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9

105

Linear Algebra Review

9.1

Notation

1. R denotes the set of real numbers, C denotes the set of complex numbers. If we use F, it means that within a definition/theorem/proof every occurance of F can be interpreted either as R or C. 2. Rn will denote the set of all n × 1 vectors with real entries. Cn will denote the set of all n × 1 vectors with complex entries. If v ∈ Fn , then vi denotes the i’th entry of v. 3. Rn×m denotes the set of all real, n × m matrices. Here, the first integer denotes the number of rows, and the second integer denotes the number of columns. Similarly, Cn×m denotes the set of complex n × m matrices. If A ∈ Fn×m , then Aij (usually) denotes the (i, j)’th entry of v (i’th row, j’th column). 4. Addition of vectors: if v ∈ Fn and u ∈ Fn , then the sum u + v ∈ Fn is defined by (v + u)i := vi + ui 5. Addition of matrices: if A ∈ Fn×m and B ∈ Fn×m , then the sum A + B ∈ Fn×m is defined by (A + B)ij := Aij + Bij 6. Scalar multiplication: if α ∈ F and v ∈ Fn , then the product αv ∈ Fn is defined by (αv)i := αvi 7. Scalar multiplication: if α ∈ F and A ∈ Fn×m , then the product αA ∈ Fn×m is defined by (αA)ij := αAij 8. Matrix-Vector multiplication: if A ∈ Fn×m and v ∈ Fm , then the product Av ∈ Fn is defined by m X (Av)i := Aij vj j=1

9. Matrix-Matrix Multiplication: if A ∈ Fn×m and B ∈ Fm×p , then the product AB ∈ Fn×p is defined by m X (AB)ij := Aik Bkj k=1

10. Suppose x, y, z ∈ Fn and α, β ∈ F, then

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• Addition is commutative and associative x + y = y + x,

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

• The zero vector 0n is the only vector such that x + 0n = x for all x ∈ Fn • For each x ∈ Rn the vector −x is the unique vector such that x + (−x) = 0 • Scalar multiplication is associative and distributive with respect to scalar addition α(βx) = (αβ)x,

(α + β)x = αx + βx

• Scalar multiplication is distributive with respect to vector addition α(x + y) = αx + αy 11. Matrix addition is commutative, so for every matrix A and B of the same dimensions, A+B =B+A 12. For every A, B ∈ Fn×m and every x, y ∈ Fm , A(x + y) = Ax + Ay,

(A + B)x = Ax + Bx

Hence vector addition distributes across matrix-vector multiplication, and matrix addition distributes across matrix-vector multiplication. 13. Matrix multiplication is associative, so for every A ∈ Fn×p , B ∈ Fp×q , A ∈ Fq×m A(BC) = (AB)C 14. For every A ∈ Fn×m , α ∈ F, and x ∈ Fm , A(αx) = α(Ax) = (αA)x 15. Block Partitioned Matrices: Suppose that n1 + n2 = n, m1 + m2 = m, p1 + p2 = p. Let A ∈ Fn×m and B ∈ Fm×p be partitioned as     A11 A12 B11 B12 A= B= A21 A22 B21 B22 with each Aij ∈ Fni ×mj and Bjk ∈ Fmj ×pk . Then   A11 B11 + A12 B21 A11 B12 + A12 B22 AB = A21 B11 + A22 B21 A21 B12 + A22 B22 Why is this easy to remember?

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9.2

107

Determinants

1. Definition: If A ∈ F1×1 , A = a11 , then det (A) := a11 . 2. Definition: If A ∈ Fn×n , then the (n − 1) × (n − 1) matrix obtained by eliminating the i’th row and j’th column is denoted A(i;j) . 3. Definition: If A ∈ Fn×n , then det (A) is defined in terms of determinants of (n − 1) × (n − 1) matrices, n X   det (A) := a1j (−1)1+j det A(1;j) j=1

4. Theorem: Given A ∈ Fn×n , and integers p, q with 1 ≤ p, q ≤ n, det (A) =

n X

p+j

apj (−1)

n   X   det A(p;j) = aiq (−1)i+q det A(i;q)

j=1

i=1

5. Given A ∈ Fn×n . The following are true: (a) if all elements of any row (or column) are zero, then det (A) = 0 ˜ (b) if all elements   of any row (or column) are multiplied by a constant c to yield A, then det A˜ = c det (A) (c) if c is a scalar, then det (cA) = cn det (A) (d) if A has two equal columns (or rows), then det (A) = 0 (e) If one column (or row) is a multiple of a different column, then det (A) = 0   ˜ (f) Interchanging two columns (or rows) of A to yield A gives det A˜ = − det (A).
(g) det (A) = det AT 6. X, Y ∈ Cn×n , then det (XY ) = det (X) det (Y ) 7. If X ∈ Cn×n , Z ∈ Cm×m Y ∈ Cn×m , then   X Y det = det (X) det (Z) 0 Z 8. Suppose X ∈ Cn×m and Y ∈ Cm×n . Then det (In + XY ) = det (Im + Y X) Note: This holds for any dimensions n and m.

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9.3

108

Inverses

1. Definition: A ∈ Fn×m . If there is a matrix G ∈ Fm×n such that GA = Im , then A is left-invertible, and G is a left inverse of A. 2. Definition: A ∈ Fn×m . If there is a matrix H ∈ Fm×n such that AH = In , then A is right-invertible, and H is a right inverse of A. 3. Let A ∈ Fn×m and b ∈ Fn be given. Consider the linear equation Ax = b. (a) If A has a left inverse, then there is at most one solution x to the equation Ax = b (b) If A has a right inverse, then there is a solution x to the equation Ax = b (though it may not be unique) 4. Theorem: If A has a left and right inverse, then they are the same, and are unique (and denoted A−1 ). Furthermore, it must be that A is square. 5. Definition: If A is square, and has a left and right inverse, then A is called nonsingular and/or invertible; if A does not have an inverse, then A is called singular. 6. If A and B are square nonsingular matrices of the same dimension, then • A−1 is invertible, and (A−1 )

−1

=A

• (AB) is invertible, and (AB)−1 = B −1 A−1 7. For any X ∈ Cn×n , there is a matrix adj(X) ∈ Cn×n such that adj(X)X = Xadj(X) = det(X)In . The (i, j) element of adj(X) is   (−1)i+j det A(j;i) 8. For any X ∈ Cn×n , X is nonsingular (invertible) if and only if det (X) 6= 0. 9. If X ∈ Cn×n and Z ∈ Cm×m are invertible, then for every Y ∈ Cn×m the matrix   X Y 0 Z is invertible, and the inverse is  −1  −1  X Y X −X −1 Y Z −1 = 0 Z 0 Z −1

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10. Suppose X ∈ Cn×m and Y ∈ Cm×n . Then (In + XY ) is invertible if and only if (Im + Y X) is invertible. Moreover, if either (and hence both) is invertible, (In + XY )−1 = In − X (Im + Y X)−1 Y (Im + Y X)−1 = Im − Y (In + XY )−1 X Also (In + XY )−1 X = X (Im + Y X)−1

9.4

Solving Linear equations: Gaussian Elimination

Consider the linear equations 2x1 x1 3x1 x1

− 3x2 − x2 + 2x2 + x2

+ 2x3 + x3 + 2x3 − 3x3

+ 5x4 + 2x4 + x4 − x4

= = = =

3 1 0 0

Gaussian Elimination reduces this system of 4 equations in 4 unknowns to a system of 3 equations in 3 unknowns by using 1 of the equations to eliminate one of the unknowns from the other 3 equations. Then the set of 3 equations/unknowns is reduced to a set of 2 equations/unknowns, and finally reduced to 1 equation/unknown, which is solved, and the recursively backsubstituted into the larger problems, getting all of the solution. One possible path to the solution is as follows: Multiply the 1st equation by 1/2, and subtract appropriate multiples from the other equations to get x1 − 1.5x2 + x3 + 2.5x4 = 1.5 − 0.5x2 + − 0.5x4 = −0.5 6.5x2 − x3 − 6.5x4 = −4.5 2.5x2 − 4x3 − 3.5x4 = −1.5 Multiply the 2nd equation by −2, and subtract tions to get x1 + x3 x2 −x3 − 4x3

appropriate multiples from the other equa+ x4 = 0 − x4 = −1 = 2 − x4 = 1

Multiply the 3rd equation by −1, and subtract appropriate multiples from the other equa-

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tions to get x1 x2 x3

+ x4 = 2 − x4 = −1 = −2 − x4 = −7

Multiply the 4th equation by −1, and subtract appropriate multiples from the other equations to get x1 = −5 x2 = 6 x3 = −2 x4 = 7 What we really were doing here was operating on the matrix [A b] ∈ R4×5 ,   2 −3 2 5 3  1 −1 1 2 1     3 2 2 1 0  1 1 −3 −1 0 with elementary row operations (EROs), to reduce it to row-echeleon form. Recall that elementary row operations are 1. multiply a row by a scalar 2. interchange 2 rows 3. replace a row by (itself + scalar multiple of another row)

9.5

Matrix functions of Time

Suppose that A is a matrix function of time, so that for each t ∈ R, A(t) ∈ Rn×m . We write this as A : R → Rn×m . If each entry Aij of A is differentiable, then the derivative of A is simply the matrix of derivatives of the individual entries,   d d d A (t) A (t) · · · A (t) 11 12 1m dt dt dt  d A21 (t) d A22 (t) · · · d A2m (t)    dt dt dt ˙ :=  A(t)  .. .. .. . .   . . . . d d d A (t) dt An2 (t) · · · dt Anm (t) dt n1 Hence, the notation A˙ ij is an ok notation, since you can interpret it to mean either the (i, j) ˙ or the derivative of the (i, j) entry of A (since they are the same entry of the matrix A, thing).

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Similarly, assume that B : R → Rm×q . Hence, we can define the product C := AB, simply by taking the matrix product at each time, so that C : R → Rn×q . with C(t) = A(t)B(t) for each t ∈ R. It is easy to verify that the product rule holds, namely, ˙ ˙ ˙ C(t) = A(t)B(t) + A(t)B(t) Indeed, the (i, j) entry of C is Cij =

m X

Aik Bkj

k=1

These are all scalar functions, so if we differentiate, we get C˙ ij =

m  X

A˙ ik Bkj + Aik B˙ kj



k=1

˙ Note that the first term is just the (i, j) entry of the matrix product AB, and the second ˙ Hence for all 1 ≤ i ≤ n and 1 ≤ j ≤ q, term is just the (i, j) entry of the matrix product AB. h i h i ˙ ˙ ˙ Cij = AB + AB ij

which is longhand for writing ˙ + AB˙ C˙ = AB

ij

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10 10.1

112

Linear Systems and Time-Invariance Linearity of solution

Consider a vector differential equation x(t) ˙ = A(t)x(t) + B(t)d(t) x(t0 ) = x0

(10.1)

where for each t, A(t) ∈ Rn×n , B(t) ∈ Rn×m , and for each time t, x(t) ∈ Rn and d(t) ∈ Rm . Claim: The
solution function x(·), on any interval [t0 , t1 ] is a linear function of the pair x0 , d(t)[t0 t1 ] . The precise meaning of this statement is as follows: Pick any constants α, β ∈ R, • if x1 is the solution to (10.1) starting from initial condition x1,0 (at t0 ) and forced with input d1 , and • if x2 is the solution to (10.1) starting from initial condition x2,0 (at t0 ) and forced with input d2 , then αx1 + βx2 is the solution to (10.1) starting from initial condition αx1,0 + βx2,0 (at t0 ) and forced with input αd1 (t) + βd2 (t). This can be easily checked, note that for every time t, we have x˙ 1 (t) = A(t)x1 (t) + B(t)d1 (t) x˙ 2 (t) = A(t)x2 (t) + B(t)d2 (t) Multiply the first equation by the constant α and the second equation by β, and add them together, giving d dt

[αx1 (t) + βx2 (t)] = αx˙ 1 (t) + β x˙ 2 (t) = α [Ax1 (t) + Bd1 (t)] + β [Ax2 (t) + Bd2 (t)] = A [αx1 (t) + βx2 (t)] + B [αd1 (t) + βd2 (t)]

which shows that the linear combination αx1 +βx2 does indeed solve the differential equation. The initial condition is easily checked. Finally, the existence and uniqueness theorem for differential equations tells us that this (αx1 + βx2 ) is the only solution which satisfies both the differential equation and the initial conditions. Linearity of the solution in the pair (initial condition, forcing function) is often called The Principal of Superposition and is an extremely useful property of linear systems.

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Note that in this setup, the coefficients of the differential equation (ie., the entries of the matrix A) are allowed to be functions of time. Linearity of solution to the pair (initial conditions, forcing) still followed. Time-invariance is a separate issue from linearity, and is discussed next.

10.2

Time-Invariance

A separate issue, unrelated to linearity, is time-invariance. A system described by x(t) ˙ = f (x(t), d(t), t) is called time-invariant if (roughly) the behavior of the system does not depend explicitly on the absolute time. In other words, shifting the time axis does not affect solutions. Precisely, suppose that x1 is a solution for the system, starting from x10 at t = t0 subject to the forcing d1 (t), defined for t ≥ t0 . Now, let x˜ be the solution to the equations starting from x0 at t = t0 + ∆, subject to the ˜ := d(t − ∆), defined for t ≥ t0 + ∆. Suppose that for all choices of t0 , ∆, x0 and forcing d(t) d(·), the two responses are related by x˜(t) = x(t − ∆) for all t ≥ t0 + ∆. Then the system described by x(t) ˙ = f (x(t), d(t), t) is called time-invariant. In practice, the easiest manner to recognize time-invariance is that the right-hand side of the state equations (the first-order differential equations governing the process) do not explicitly depend on time. For instance, the system x˙ 1 (t) = 2 ∗ x2 (t) − sin [x1 (t)x2 (t)d2 (t)] x˙ 2 (t) = − |x2 (t)| − x2 (t)d1 (t) is nonlinear, yet time-invariant.

ME 132, Fall 2018, UC Berkeley, A. Packard

11

114

Matrix Exponential

Recall that for the scalar differential equation x(t) ˙ = ax(t) + bu(t) x(t0 ) = x0 the solution for t ≥ t0 is given by the formula Z t a(t−t0 ) ea(t−τ ) bu(τ )dτ x(t) = e + t0

What makes this work is the special structure of the exponential function, namely that d at e = aeat . dt Now consider a vector differential equation x(t) ˙ = Ax(t) + Bu(t) x(t0 ) = x0

(11.1)

where A ∈ Rn×n , B ∈ Rn×m are constant matrices, and for each time t, x(t) ∈ Rn and u(t) ∈ Rm . The solution can be derived by proceeding analogously to the scalar case. For a matrix A ∈ Cn×n define a matrix function of time, eAt ∈ Cn×n as At

e

:=

∞ k X t

k! k=0

Ak

= I + tA +

t2 2 A 2!

+

t3 3 A 3!

+ ···

This is exactly the same as the definition in the scalar case. Now, for any T > 0, every element of this matrix power series converges absolutely and uniformly on the interval [0, T ]. Hence, it can be differentiated term-by-term to correctly get the derivative. Therefore d At e dt

:=

∞ X k tk−1 k A k! k=0 2

= A + tA2 + + t2! A3 + · · ·  2 = A I + tA + t2! A2 + · · · = AeAt This is the most important property of the function eAt . Also, in deriving this result, A could have been pulled out of the summation on either side. Summarizing these important identities, d At eA0 = In , e = AeAt = eAt A dt

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So, the matrix exponential has properties similar to the scalar exponential function. However, there are two important facts to watch out for: • WARNING: Let aij denote the (i, j)th entry of A. The (i, j)th entry of eAt IS NOT EQUAL TO eaij t . This is most convincingly seen with a nontrivial example. Consider " # 1 1 A= 0 1 A few calculations show that " # 1 2 A2 = 0 1

" A3 =

1 3 0 1

#

" ···

Ak =

1 k 0 1

#

The definition for eAt is eAt = I + tA +

t2 2 t3 3 A + A + ··· 2! 3!

Plugging in, and evaluating on a term-by-term basis gives # " t3 t2 t3 t2 + + · · · 0 + t + 2 + 3 + · · · 1 + t + 2! 3! 2! 3! eAt = 2 3 0 + 0 + 0 + 0 + ··· 1 + t + t2! + t3! + · · · The (1, 1) and (2, 2) entries are easily seen to be the power series for et , and the (2, 1) entry is clearly 0. After a few manipulations, the (1, 2) entry is tet . Hence, in this case " # et tet At e = 0 et which is very different than the element-by-element exponentiation of At, " # " # ea11 t ea12 t et et = ea21 t ea22 t 1 et • WARNING: In general, e(A1 +A2 )t 6= eA1 t eA2 t unless t = 0 (trivial) or A1 A2 = A2 A1 . However, the identity eA(t1 +t2 ) = eAt1 eAt2 is always true. Hence eAt e−At = e−At eAt = I for all matrices A and all t, and therefore for all A and t, eAt is invertible.

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11.1

116

Diagonal A

If A ∈ Cn×n is diagonal, then eAt is to see that for any k,  β1 0 · · ·   0 β2 · · · A= .. . .  .. . .  . 0

easy to compute. Specifically, if A is diagonal, it is easy

0 0 .. .

· · · βn

0





    

  A =   k

β1k 0 0 β2k .. .. . . 0 0

··· ··· .. .

0 0 .. .

· · · βnk

     

In the power series definition for eAt , any off-diagonal terms are identically zero, and the i’th diagonal term is simply 

11.2

eAt

 ii

= 1 + tβi +

t2 2 t3 3 βi + βi + · · · = eβi t 2! 3!

Block Diagonal A

If A1 ∈ Cn1 ×n1 and A2 ∈ Cn2 ×n2 , define " A :=

A1 0 0 A2

#

Question: How is eAt related to eA1 t and eA2 t ? Very simply – note that for any k ≥ 0, # " #k " k A 0 A 0 1 1 Ak = = 0 A2 0 Ak2 Hence, the power series definition for eAt gives " # A1 t 0 e eAt = 0 eA2 t

11.3

Effect of Similarity Transformations

For any invertible T ∈ Cn×n , eT

−1 AT t

Equivalently, T eT

= T −1 eAt T

−1 AT t

T −1 = eAt

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This can easily be shown from the power series definition. eT

−1 AT t


t2 −1
2 t3 −1
3 = I + t T −1 AT + T AT + T AT + · · · 2! 3!

It is easy to verify that for every integer k T −1 AT


k

= T −1 Ak T

Hence, we have (with I written as T −1 IT ) eT

−1 AT t

= T −1 IT + tT −1 AT +

t2 −1 2 t3 T A T + T −1 A3 T + · · · 2! 3!

Pull T −1 out on the left side, and T on the right to give   t2 2 t3 3 T −1 AT t −1 e =T I + tA + A + A + · · · T 2! 3! which is simply eT

−1 AT t

= T −1 eAt T

as desired.

11.4

Solution To State Equations

The matrix exponential is extremely important in the solution of the vector differential equation x(t) ˙ = Ax(t) + Bu(t) (11.2) starting from the initial condition x(t0 ) = x0 . Now, consider the original equation in (11.2). We can derive the solution to the forced equation using the “integrating factor” method, proceeding in the same manner as in the scalar case, with extra care for the matrix-vector operations. Suppose a function x satisfies (11.2). Multiply both sides by e−At to give e−At x(t) ˙ = e−At Ax(t) + e−At Bu(t) Move one term to the left, leaving, e−At Bu(t) = e−At x(t) ˙ − e−At Ax(t) = e−At x(t) ˙ − Ae−At x(t)   = dtd e−At x(t)

(11.3)

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Since these two functions are equal at every time, we can integrate them over the interval [t0 t1 ]. Note that the right-hand side of (11.3) is an exact derivative, giving Z t1 Z t1  d  −At −At e x(t) dt e Bu(t)dt = t0 dt t0 t −At = e x(t) t10 = e−At1 x(t1 ) − e−At0 x(t0 ) Note that x(t0 ) = x0 . Also, multiply both sides by eAt1 , to yield Z t1 At1 e e−At Bu(t)dt = x(t1 ) − eA(t1 −t0 ) x0 t0

This is rearranged into x(t1 ) = e

A(t1 −t0 )

Z

t1

x0 +

eA(t1 −t) Bu(t)dt

t0

Finally, switch variable names, letting τ be the variable of integration, and letting t be the right end point (as opposed to t1 ). In these new letters, the expression for the solution of the (11.2) for t ≥ t0 , subject to initial condition x(t0 ) = x0 is Z t A(t−t0 ) x(t) = e x0 + eA(t−τ ) Bu(τ )dτ t0

consisting of a free and forced response.

11.5

Examples

Given β ∈ R, define " A := Calculate

"

0 β −β 0

#

A2 =

−β 2 0 0 −β 2

and hence for any k, " # k 2k (−1) β 0 A2k = , 0 (−1)k β 2k

" A2k+1 =

#

k

0 k+1 2k+1

(−1)

β

(−1) β 0

2k+1

#

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Therefore, we can write out the first few terms of the power series for eAt as # " 1 2 2 1 4 4 1 3 3 1 5 5 1 − β t + β t − · · · βt − β t + β t − · · · 2 4! 3! 5! eAt = −βt + 3!1 β 3 t3 − 5!1 β 5 t5 + · · · 1 − 12 β 2 t2 + 4!1 β 4 t4 − · · · which is recognized as " At

e

=

cos βt sin βt − sin βt cos βt

#

Similarly, suppose α, β ∈ R, and " A :=

α β −β α

#

Then, A can be decomposed as A = A1 + A2 , where " # " # α 0 0 β A1 := , A2 := 0 α −β 0 Note: in this special case, A1 A2 = A2 A1 , hence e(A1 +A2 )t = eA1 t eA2 t Since A1 is diagonal, we know eA1 t , and eA2 t follows from our previous example. Hence " # αt αt e cos βt e sin βt eAt = −eαt sin βt eαt cos βt This is an important case to remember. Finally, suppose λ ∈ F, and " # λ 1 A= 0 λ A few calculations show that

" Ak =

λk kλk−1 0 λk

#

Hence, the power series gives " eAt =

eλt teλt 0 eλt

#

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11.6

120

Problems

1. Consider the differential equation x(t) ˙ = Ax(t), from initial condition x(0) = x0 . Here A ∈ Rn×n and x0 ∈ Rn . We know that the solution involves the matrix exponential, and is x(t) = eAt x0 . (a) For each 1 ≤ k ≤ n, let ek denote the unit vector whose k’th element is equal to 1, and all other elements are 0, so       0 0 1  0   1   0        e1 =  ..  , e2 =  ..  , ··· en =  ..   .   .   .  0 0 1 Show that for each k, the k’th column of eAt is the solution of the differential equation with initial condition x(0) = ek . (b) Consider A of the form 

0 0 0 .. .

1 0 0 .. .

0 1 0 .. .

0 0 1 .. .

··· ··· ··· .. .

0 0 0 .. .

0 0 0 .. .

     A=   0 0 0 0 ··· 1 0   0 0 0 0 ··· 0 1 0 0 0 0 ··· 0 0

          

(11.4)

Write an expression x˙ k in terms of x1 , x2 , . . . , xn (c) Using the A matrix in (11.4), what is the solution of the differential equation x(t) ˙ = Ax(t) subject to the initial condition x(0) = ek . (d) For the A matrix in (11.4), what is eAt ? 2. Suppose β ∈ C and A ∈ Cn×n . We want to derive the expression for e(βIn +A)t , in terms of β and eAt . (a) Consider the differential equation x(t) ˙ = Ax(t), with initial condition x(0) = x0 . βt Define a function z(t) := e x(t). i. What is z(0)? ii. Show that z satisfies z(t) ˙ = (βIn + A)z(t) for all t ≥ 0. (b) What is e(βIn +A)t , in terms of β and eAt ? Justify your answer.

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(c) What is eAt for 

 −2 1 0 1  A =  0 −2 0 0 −2 3. Take α ∈ R, and define

 A=

0 α −α 0



(a) What is A2 ? (b) What is A3 ? (c) What is A4 ? (d) What is A5 ? (e) For any nonnegative integer k, what is A2k ? (f) For any nonnegative integer k, what is A2k+1 ? (g) What is the power-series of the (1, 1) and (2, 2) elements of eAt ? (h) What is the power-series of the (1, 2) element of eAt ? (i) What is eAt ? (j) Write the expression for eAt for the A matrices listed       −2 1 2 3 −4 −5 A1 := , A2 := , A3 := , −1 −2 −3 2 5 −4

 A4 :=

−6 −1 1 −6

(k) Plot the solution (both components, x1 (t) and x2 (t) versus t) for x(t) ˙ = A1 x(t), with the initial condition   1 x(0) = 0 4. Suppose A ∈ Rn×n , and AP = P D, where P ∈ Cn×n is invertible, and D ∈ Cn×n is diagonal. Let pk ∈ Cn×1 denote the k’th column of P , and dk denote the k’th diagonal element of D, so   d1 0 · · · 0  0 d2 · · · 0      P = p1 p2 · · · pn , D =  .. .. . . .   . . . ..  0 0 · · · dn (a) For the differential equation x(t) ˙ = Ax(t), subject to the initial condition x(0) = pk , what is the solution?



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(b) Suppose {αk }nk=1 are constants. Find the Pnsolution of the differential equation subject to the initial condition x(0) = k=1 αk pk . Note here that the initial condition is a linear combination of the columns of P . 5. Let



 −7 2 −2 0 −3  A =  −5 10 −4 1

(a) Verify that 

 1 2 0 P =  2 5 1 , −1 0 1



 −1 0 0 0  D =  0 −2 0 0 −3

satisfy AP = P D, and that P is invertible. In fact, verify that   −5 2 −2 1  P −1 =  3 −1 −5 2 −1 (b) What is the solution of the differential equation x(t) ˙ = Ax(t) subject to the initial condition   3 x0 =  7  −1 Hint: What linear combination of the columns of P give this particular x0 ? (c) Using the expression derived above, plot all three components of x versus time t. (d) Use ode45 and a one-line anonymous function to numerically solve the ODE from this initial condition. Plot the response (x1 (t), x2 (t) and x3 (t) versus t), and compare to the answer derived analytically in part 5b.

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123

Eigenvalues, eigenvectors, stability Diagonalization: Motivation For diagonal matrices Λ ∈ Fn×n ,   0 e λ1 t 0 · · · 0  0 e λ2 t · · · 0 0    eΛt =  .. ..  ⇒ .. .. ..  . . .  . . λn t · · · λn 0 0 ··· e

Recall two facts from Section 11:  λ1 0 · · ·  0 λ2 · · ·  Λ =  .. .. . .  . . . 0

0

    

and: If A ∈ Fn×n , and T ∈ Fn×n is invertible, and A˜ := T −1 AT , then ˜

eAt = T eAt T −1 Clearly, for a general A ∈ Fn×n , we need to study the invertible transformations T ∈ Fn×n such that T −1 AT is a diagonal matrix. Suppose that T is invertible, and Λ is a diagonal matrix, and T −1 AT = Λ. Moving the T −1 to the other side of the equality gives AT = T Λ. Let ti denote the i’th column of the matrix T . SInce T is assumed to be invertible, none of the columns of T can be identically zero, hence ti 6= Θn . Also, let λi denote the (i, i)’th entry of Λ. The i’th column of the matrix equation AT = T Λ is just Ati = ti λi = λi ti This observation leads to the next section.

12.2

Eigenvalues

Definition: Given a matrix A ∈ Fn×n . A complex number λ is an eigenvalue of A if there is a nonzero vector v ∈ Cn such that Av = vλ = λv The nonzero vector v is called an eigenvector of A associated with the eigenvalue λ. Remark: Consider the differential equation x(t) ˙ = Ax(t), with initial condition x(0) = v. λt Then x(t) = ve is the solution (check that it satisfies initial condition and differential equation). So, an eigenvector is “direction” in the state-space such that if you start in the direction of the eigenvector, you stay in the direction of the eigenvector.

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Fact 1: Note that if λ is an eigenvalue of A, then there is a vector v ∈ Cn , v 6= Θn such that Av = vλ = (λI) v Hence (λI − A) v = Θn . Since v 6= Θn , it must be that det (λI − A) = 0. Definition: For an n × n matrix A, define a polynomial, pA (·), called the characteristic polynomial of A by pA (s) := det (λI − A) Here, the symbol s is simply the indeterminate variable of the polynomial. Example: Take 

 2 3 −1 A =  −1 −1 −1  0 2 0 Straightforward manipulation gives pA (s) = s3 − s2 + 3s − 2. Hence, we have shown that the eigenvalues of A are necessarily roots of the equation pA (s) = 0. Fact 2: For a general n × n matrix A, we will write pA (s) = sn + a1 sn−1 + · · · + an−1 s + an where the a1 , a2 , . . . , an are complicated products and sums involving the entries of A. Since the characteristic polynomial of an n × n matrix is a n’th order polynomial, the equation pA (s) = 0 has at most n distinct roots (some roots could be repeated). Therefore, a n × n matrix A has at most n distinct eigenvalues. Fact 3: Conversely (to Fact 1), suppose that λ ∈ C is a root of the polynomial equation pA (s)|s=λ = 0 Question: Is λ an eigenvalue of A? Answer: Yes. Since pA (λ) = 0, it means that det (λI − A) = 0 Hence, the matrix λI − A is singular (not invertible). Therefore, by the matrix facts, the equation (λI − A) v = Θn has a nonzero solution vector v (which you can find by Gaussian elimination). This means that λv = Av

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for a nonzero vector v, which means that λ is an eigenvalue of A, and v is an associated eigenvector. Important Summary: We summarize these facts as: • A is a n × n matrix • The characteristic polynomial of A is pA (s) := det (sI − A) = sn + a1 sn−1 + · · · + an−1 s + an • A complex number λ is an eigenvalue of A if and only if λ is a root of the “characteristic equation” pA (λ) = 0. Next, we have a useful fact from linear algebra: Suppose A is a given n × n matrix, and (λ1 , v1 ) , (λ2 , v2 ) , . . . , (λn , vn ) are eigenvalue/eigenvector pairs. So, for each i, vi 6= Θn and Avi = vi λi . Fact: If all of the {λi }ni=1 are distinct, then the set of vectors {v1 , v2 , . . . , vn } are a linearly independent set. In other words, the matrix V := [v1 v2 · · · vn ] ∈ Cn×n is invertible. Proof: We’ll prove this for 3 × 3 matrices – check your linear algebra book for the generalization, which is basically the same proof. Suppose that there are scalars α1 , α2 , α3 , such that 3 X

αi vi = Θ3

i=1

This means that Θ3 = = = =

(A − λ3 I) Θ3 P (A − λ3 I) 3i=1 αi vi α1 (A − λ3 I) v1 + α2 (A − λ3 I) v2 + α3 (A − λ3 I) v3 α1 (λ1 − λ3 ) v1 + α2 (λ2 − λ3 ) v2 + Θ3

(12.1)

Now multiply by (A − λ2 I), giving Θ3 = (A − λ2 I) Θ3 = (A − λ2 I) [α1 (λ1 − λ3 )v1 + α2 (λ2 − λ3 )v2 ] = α1 (λ1 − λ3 )(λ1 − λ2 )v1 Since λ1 6= λ3 , λ1 6= λ2 , v1 6= Θ3 , it must be that α1 = 0. Using equation (12.1), and the fact that λ2 6= λ3 , v2 6= Θ3 we get that α2 = 0. Finally, α1 v1 + α2 v2 + α3 v3 = Θ3 (by assumption), and v3 6= Θ3 , so it must be that α3 = 0. ]

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12.3

126

Diagonalization Procedure

In this section, we summarize all of the previous ideas into a step-by-step diagonalization procedure for n × n matrices. 1. Calculate the characteristic polynomial of A, pA (s) := det (sI − A). 2. Find the n roots of the equation pA (s) = 0, and call the roots λ1 , λ2 , . . . , λn . 3. For each i, find a nonzero vector ti ∈ Cn such that (A − λi I) ti = Θn 4. Form the matrix T := [t1 t2 · · · tn ] ∈ Cn×n (note that if all of the {λi }ni=1 are distinct from one another, then T is guaranteed to be invertible). 5. Note that AT = T Λ, where    Λ= 

λ1 0 0 λ2 .. .. . . 0 0

··· ··· .. .

0 0 .. .

· · · λn

6. If T is invertible, then T −1 AT = Λ. Hence  e λ1 t 0  0 eλ2 t  eAt = T eΛt T −1 = T  .. ..  . . 0 0

    

··· ··· .. .

0 0 .. .

· · · e λn t

   −1 T 

We will talk about the case of nondistinct eigenvalues later.

12.4

eAt as t → ∞

For the remainder of the section, assume A has distinct eigenvalues. However, please note that the results we obtain are true for the case of repeated eigenvalues as well.

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• if all of the eigenvalues (which may be complex) of A satisfy

then eλi t

Re (λi ) < 0   → 0 as t → ∞, so all entries of eAt decay to zero

• If there is one (or more) eigenvalues of A with Re (λi ) ≥ 0 then eλi t →

bounded 6= 0 as → ∞ ∞

Hence, some of the entries of eAt either do not decay to zero, or, in fact, diverge to ∞. So, the eigenvalues are an indicator (the key indicator) of stability of the differential equation x(t) ˙ = Ax(t) • if all of the eigenvalues of A have negative real parts, then from any initial condition x0 , the solution x(t) = eAt x0 decays to Θn as t → ∞ (all coordinates of x(t) decay to 0 as t → ∞). In this case, A is said to be a Hurwitz matrix. • if any of the eigenvalues of A have nonnegative real parts, then from some initial conditions x0 , the solution to x(t) ˙ = Ax(t) does not decay to zero. In the case of a repeated eigenvalue, say λ is repeated r times, then like repeated roots of the characteristic polynomial, these repeated eigenvalues generally produce terms like tk eλt into eAt , where k in a integer, satisfying 0 ≤ k ≤ r − 1. As we have seen earlier, if Re(λ) < 0, this type of term decays to zero as t → ∞, regardless of k, whereas if Re(λ) ≥ 0, then this type of term does not decay to zero. This is consistent with the notion of stability covered above.

12.5

Complex Eigenvalues

In many systems, the eigenvalues are complex, rather than real. This seems unusual, since the system itself (ie., the physical meaning of the state and input variables, the coefficients

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in the state equations, etc.) is very much Real. The procedure outlined in section 12.3 for matrix diagonalization is applicable to both real and complex eigenvalues, and if A is real, all intermediate complex terms will cancel, resulting in eAt being purely real, as expected. However, in the case of complex eigenvalues it may be more advantageous to use a different similarity transformation, which does not lead to a diagonalized matrix. Instead, it leads to a real block-diagonal matrix, whose structure is easily interpreted. Let us consider a second order system, with complex eigenvalues x(t) ˙ = Ax(t)

(12.2)

pA (λ) = det(λI2 − A) = (λ − σ)2 + ω 2 .

(12.3)

where A ∈ Rn×n , x ∈ R2 and

The two eigenvalues of A are λ1 = σ + jω and λ2 = σ − jω, while the two eigenvectors of A are given by (λ1 I2 − A)v1 = Θ2

(λ2 I2 − A)v2 = Θ2 .

(12.4)

Notice that, since λ2 is the complex conjugate of λ1 , v2 is the complex conjugate vector of v1 , i.e. if v1 = vr + jvi , then v2 = vr − jvi . This last fact can be verified as follows. Assume that λ1 = σ + jω, v1 = vr + jvi and insert these expressions in the first of Eqs. (12.4). [(σ + jω)I2 − A)] [vr + jvi ] = Θ2 , Separating into its real and imaginary parts we obtain [σvr − ωvi ] + j [σvi + ωvr ] = Avr + jAvi [σvr − ωvi ] = Avr [σvi + ωvr ] = Avi .

(12.5)

Notice that Eqs. (12.5) hold if we replace ω by −ω and vi by −vi . Thus, if λ1 = σ + jω and v1 = vr + jvi are respectively an eigenvalue and eigenvector of A, then λ2 = σ − jω and v2 = vr − jvi are also respectively an eigenvalue and eigenvector. Eqs. (12.5) can be rewritten in matrix form as follows       σ ω A vr vi = vr vi . −ω σ Thus, we can define the similarity transformation matrix   T = vr vi ∈ R2×2

(12.6)

(12.7)

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and the matrix  Jc =

σ ω −ω σ

 (12.8)

such that A = T Jc T −1 ,

eAt = T eJc t T −1 .

(12.9)

The matrix exponential eJc t is easy to calculate. Notice that   σ ω Jc = = σI2 + S2 −ω σ where  I2 =

1 0 0 1



 and S2 =

0 ω −ω 0



is skew-symmetric, i.e. S2T = −S2 . Thus,   cos(ωt) sin(ωt) S2 t e = . − sin(ωt) cos(ωt) This last result can be verified by differentiating with respect to time both sides of the equation: d S2 t e = S2 eS2 t dt and d dt



cos(ωt) sin(ωt) − sin(ωt) cos(ωt)





 −ω sin(ωt) ω cos(ωt) = −ω cos(ωt) −ω sin(ωt)     0 ω cos(ωt) sin(ωt) = = S2 eS2 t . −ω 0 − sin(ωt) cos(ωt)

Since σI2 S2 = S2 σI2 , then e

Jc t

=e

σt



cos(ωt) sin(ωt) − sin(ωt) cos(ωt)

 .

(12.10)

ME 132, Fall 2018, UC Berkeley, A. Packard

12.6

130

Alternate parametrization with complex eigenvalues

Suppose σ ± jω is a complex-conjugate pair of eigenvalues with σ < 0 and ω > 0. These are the roots of the polynomial p(λ) := λ2 − 2σλ + (σ 2 + ω 2 ) = (λ − σ)2 + ω 2 Define two new constants (ξ, ωn ), in terms of σ and ω, as ωn :=

√ σ2 + ω2,

ξ :=

−σ ωn

Since σ < 0 and ω > 0 it is clear that ωn > 0,

0> help eig, and the online documentation). Repeat problem 2 using eig, and explain any differences that you get. Hint: if a matrix has distinct eigenvalues, in what sense are the eigenvectors not unique? 5. Consider the differential equation x(t) ˙ = Ax(t), where x(t) is (3 × 1) and A is from problem (2) above. Suppose that the initial condition is   1 x(0) =  −1  0 Write the initial condition as a linear combination of the eigenvectors, and find the solution x(t) to the differential equation, written as a time-dependent, linear combination of the eigenvectors. 6. Suppose that we have a 2nd order system (x(t) ∈ R2 ) governed by the differential equation   0 −2 x(t) ˙ = x(t) 2 −5 Let A denote the 2 × 2 matrix above. (a) Find the eigenvalues an eigenvectors of A. In this problem, the eigenvectors can be chosen to have all integer entries (recall that eigenvectors can be scaled) (b) On the grid below (x1 /x2 space), draw the eigenvectors.

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140 x2 6

6

6 6 6 6

6 6

6 6 6 - x1 -

(c) A plot of x2 (t) vs. x1 (t) is called a phase-plane plot. The variable t is not explicitly plotted on an axis, rather it is the parameter of the tick marks along the plot. On the grid above, using hand calculations, draw the solution to the equation x(t) ˙ = Ax(t) for the initial conditions         3 −2 −3 4 x(0) = , x(0) = , x(0) = , x(0) = 3 2 −3 2 HINT: Remember that if vi are eigenvectors, and λi are eigenvalues of A, then P2 the solution to x(t) ˙ = Ax(t) from the initial condition x(0) = i=1 αi vi is simply x(t) =

n X

αi eλi t vi

i=1

(d) Use Matlab to create a similar picture with many (say 20) different initial conditions spread out in the x1 , x2 plane 7. Suppose A is a real, n × n matrix, and λ is an eigenvalue of A, and λ is not real, but λ is complex. Suppose v ∈ Cn is an eigenvector of A associated with this eigenvalue λ. Use the notation√ λ = λR + jλI and v = vr + jvI for the real and imaginary parts of λ, and v ( j means −1). (a) By equating the real and imaginary parts of the equation Av = λv, find two equations that relate the various real and imaginary parts of λ and v. ¯ (complex conjugate of λ) is also an eigenvalue of A. What is the (b) Show that λ associated eigenvector?

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(c) Consider the differential equation x˙ = Ax for the A described above, with the eigenvalue λ and eigenvector v. Show that the function x(t) = eλR t [cos (λI t) vR − sin (λI t) vI ] satisfies the differential equation. What is x(0) in this case? (d) Fill in this sentence: If A has complex eigenvalues, then if x(t) starts on the part of the eigenvector, the solution x(t) oscillates between the and parts of the eigenvector, with frequency associated with the part of the eigenvalue. During the motion, the solution also increases/decreases exponentially, based on the part of the eigenvalue. (e) Consider the matrix A  A= It is possible to show that " A

√1 2 j √12

√1 2 −j √12

#

" =

√1 2 j √12

1 2 −2 1

√1 2 −j √12



#

1 + 2j 0 0 1 − 2j



2 Sketch the trajectory of the solution  x(t) in R to the differential equation x˙ = Ax 1 for the initial condition x(0) = . 0

(f) Find eAt for the A given above. NOTE: eAt is real whenever A is real. See the notes for tricks in making this easy. 8.

a) Suppose β1 and β2 are real numbers. Recall that the roots {λ1 , λ2 } of the polynomial equation λ2 + β1 λ + β2 = 0 satisfy Re(λ1 ) < 0 and Re(λ2 ) < 0 if and only if β1 > 0 and β2 > 0. Give three (3) separate examples (specific values of β1 and β2 ) of polynomials whose roots do not satisfy this condition, specifically such that the two roots satisfy i. Both are real, and both are positive ii. Both are real; one is positive, and one is negative iii. Both are complex, with positive real-part. b) What polynomial is associated with the stability of the differential equation x˙ 1 (t) = a11 x1 (t) + a12 x2 (t) x˙ 2 (t) = a21 x1 (t) + a22 x2 (t) subject to initial condition x(0) = x0 ?

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c) Combine parts (a) and (b) to show that a 2-state system is stable if and only if a11 a22 − a12 a21 > 0,

a11 + a22 < 0

c) In a general linear system, expressed as x(t) ˙ = Ax(t) each state evolves with rate-of-change determined by a linear combination of the values of all the states. For example x˙ 1 (t) = A11 x1 (t) + A12 x2 (t) + · · · + A1n xn (t). An abstract notion of “adding damping” to the dynamics of x1 is to replace A11 by A11 − D1 , where D1 ≥ 0. Hence the dynamics become x˙ 1 (t) = (A11 − D1 )x1 (t) + A12 x2 (t) + · · · + A1n xn (t). If we do this for all the states, then the new dynamic equation becomes x(t) ˙ = (A − D)x(t) where D is a diagonal matrix, with non-negative  D1 0 · · ·  0 D2 · · ·  D =  .. .. . .  . . . 0

0

entries on the diagonal  0 0   ..  .  · · · Dn

Show (by constructing an example) that this form of “adding damping” can (somewhat surprisingly) change a system from being stable to being unstable. Hint: Can this be done for system with only 1 state? If not, try a 2-state system. 9. In this problem, we look at the “damping-ratio, natural-frequency” parametrization of complex roots, as opposed to the real/imaginary parametrization. The “dampingratio, natural-frequency” description is a very common manner in which the location of complex eigenvalues is described. For this problem, suppose that 0 < ξ < 1, and ω > 0. (a) What are the roots of the equation s2 + 2ξωn s + ωn2 = 0

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(b) Let λ be the complex number λ := −ξωn + jωn

p 1 − ξ2

(note that this is one of the roots you computed above). Show that |λ| = ωn , regardless of 0 < ξ < 1. (c) The complex number λ is plotted in the complex plane, as shown below. λ×

Im 6

A Aψ A A

C Re -

Express sin ψ in terms of ξ and ωn . (d) Run the commands A = randn(5,5); damp(eig(A)) several times, and copy/paste the output into the assignment. Explain the displayed output and its connection to the results you derived here. 10. (Model taken from “Introduction to Dynamic Systems Analysis,” T.D. Burton, 1994, pg. 212, problem 23) Let f1 (x1 , x2 ) := x1 − x31 + x1 x2 , and f2 (x1 , x2 ) := −x2 + 1.5x1 x2 . Consider the 2-state system x(t) ˙ = f (x(t)). (a) Note that there are no inputs. Find all equilibrium points of this system. Hint: In this problem, there are 4 equilibrium points. (b) Derive the Jacobian linearization which describes the solutions near each equilibrium point. You will have 4 different linearizations, each of the form η(t) ˙ = Aη(t) with different A matrices dependent on which equilibrium point the linearization has been computed. (c) Using eigenvalues, determine the stability of each of 4 Jacobian linearizations. Note: (for part 10f below) • If the linearization is stable, it means that while the deviation of x(t) from x¯ remains small, the variables x(t) − x¯ approximately evolve by a linear differential equation whose homogenous solutions all decay to zero. So, we would expect that initial conditions near the equilibium point would converge to the equilibrium point.

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• Conversely, if the linearization is unstable, it means that while the deviation of x(t) from x¯ remains small, the variables x(t) − x¯ approximately evolve by a linear differential equation that has some homogenous solutions which grow. So, we would expect that some initial conditions near the equilibium point would initially diverge away from the equilibrium point. (d) Using Simulink, or ode45, simulate the system starting from 50 random initial conditions satisfying −3 ≤ x1 (0) ≤ 3, and −3 ≤ x2 (0) ≤ 3. Plot the resulting solutions in x2 versus x1 plane (each solution will be a curve - parametrized by time t). This is called a phase-plane graph, and we already looked at such plots, for linear systems, a few weeks ago. Make the axis limits −4 ≤ xi ≤ 4. On each curve, hand-draw in arrow(s) indicating which direction is increasing time. Mark the 4 equilibrium points on your graph. See the “Critical Remarks” section below before starting this calculation. (e) At each equilibrium point, draw the two eigenvectors of the linearization, and notate (next to each eigenvector) what the associated eigenvalue is. If the eigenvalues/eigenvectors are complex, draw the real-part of the eigenvector, and the imaginary part (recall that in linear systems, real-valued solutions oscillate between the real-part and the imaginary part of the eigenvectors). (f) For each equilibrium point, describe (2 sentences) the behavior of the solution curves near the equilibrium points in relation to the behavior of the linearized solutions, and how the curves relate to the stability computation, of the linearization, in part 10c. Critical (!) remark: Since some initial conditions will lead to diverging solutions, it is important to automatically stop the simulation when either |x1 (t)| ≥ 8 (say) or |x2 (t)| ≥ 8. If using Simulink, use a Stop Simulation block (from Sinks), with its input coming from the Relational Operator (from Math). If using ode45, you can use events, which are programmed to automatically stop the numerical simulation when values of the solution, x(t) reach/exceed certain values. 11. Hand-computation of unit-step-response of stable systems of the form x(t) ˙ = Ax(t) + Bu(t), y(t) = Cx(t) from x(0) = 0. Procedure: (a) Calculate eigenvalues of A (easy if A ∈ R2 ). Verify the system is stable. (b) If eigenvalues are both real, compute the time-constant associated with each one. Note the slow (longest time constant) one, as that will tend to dominate the time for the response to converge to the final value. If the eigenvalues are repeated, then there may be terms of the form teλt in eAt . This term, despite the linear t portion, still decays to zero. (c) If the eigenvalues are complex, express them in α ± jβ form and in (ξ, ωn ) form. Recall that the time-constant associated with the exponential decay is

ME 132, Fall 2018, UC Berkeley, A. Packard

−1 α

=

145

1 . ξωn αt

Also remember that the “number of visible periods” in terms of the p form e cos βt = e−ξωn t cos(ωn 1 − ξ 2 t) is approximately p 1 − ξ2 β # visible periods ≈ = 2ξ −2α (d) Since x(0) = 0 and there is no D-term, it is clear that y(0) = 0. What is y(0)? ˙ Differentiating y(t) = Cx(t) gives y(t) ˙ = C x(t) ˙ = CAx(t) + CBu(t). For a unitstep response from x(0) = 0, we have x(0) = 0, u(0) = 1 giving y(0) ˙ = CB. This final result is easy to calculate and easy-to-remember. (e) Compute the final value of y, namely limt→∞ y(t). Since u(t) = 1 for all t, the convergent value of y is simply the steady-state gain. We know the steady-state gain is −CA−1 B, hence lim y(t) = −CA−1 B t→∞

This is an easy calculation to make for A ∈ R2 . In summary, the step-response, from x(0) = 0 is a function of the form y(t) = c0 + c1 eλ1 t + c2 eλ2 t

or

y(t) = c0 + c1 eλ1 t + c2 teλ1 t

or, if the eiegnvalues are complex, y(t) = c0 + c1 eαt cos βt + c2 eαt sin βt or (expressed with (ξ, ωn ) parametrization) p p y(t) = c0 + c1 e−ξωn t cos(ωn 1 − ξ 2 t)t + c2 e−ξωn t sin(ωn 1 − ξ 2 t)t Note that this type of function has limited complexity. In the above (simple) calculations, you have computed • the approximate time for the exponential terms to decay • the frequency of the (if present) sinusoidal terms and a relative measure of the oscillation period and the decay time • the final value of y, and • the initial rate-of-change of y. This list of quantitative properties should allow you to make a decent sketch (not numerically accurate at intermediate times, but qualitatively accurate, and quantitatively representative) of the step-response. For each of the systems below, carry out, by hand, the steps above, and make a sketch of the step-response. Then enter the system in Matlab (using the Control System Toolbox ss object), and use the step command to get an accurate plot.

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(a) S1  A=

−19 15 −20 15



 ,

0 −2

B=

 ,

C=



−1 2



(b) S2  A=



33 18 −65 −35

 ,

B=

−1 −1





0 2

,

C=



1 −5





2 0



−1 −1



(c) S3  A=

395 45 −3529 −402

 ,

B=

 ,

C=

(d) S4  A=

0 3 −1 −3



 ,

B=

1 −4

 ,

C=



(e) S5  A=

11 −3 64 −17



 ,

B=

0 3

 ,

C=



6 0



ME 132, Fall 2018, UC Berkeley, A. Packard

13

13.1

147

Frequency Response for Linear Systems: State-Space representations Theory for Stable System: Complex Input Signal

Consider the linear dynamical system x(t) ˙ = Ax(t) + Bu(t) y(t) = Cx(t) + Du(t)

(13.1)

We assume that there are n states, m inputs, and q outputs (so A ∈ Rn×n , B ∈ Rn×m , C ∈ Rp×n , D ∈ Rp×m ). If the system is stable, (ie., all of A’s eigenvalues have negative real parts) it is “intuitively” clear that if u is a sinusoid, then y will approach a steady-state behavior that is sinusoidal, at the same frequency, but with different amplitude and phase. In this section, we make this idea precise. Take ω ≥ 0 as the input frequency, and let u¯ ∈ Cm be a fixed complex vector (as before, we’ll derive results very easily with complex-valued inputs, and then reinterpret the results for real-valued sine-wave inputs afterward). Take the input function u(t) to be u(t) = u¯ejωt for t ≥ 0. Then, the response is Rt x(t) = eAt x0 + 0 eA(t−τ ) Bu(τ )dτ Rt = eAt x0 + eAt 0 e−Aτ B u¯ejωτ dτ Rt = eAt x0 + eAt 0 e(jωI−A)τ dτ B u¯ Now, since A is stable, all of the eigenvalues have negative real parts. This means that all of the eigenvalues of (jωI − A) have positive real parts. Hence (jωI − A) is invertible, and we can write the integral as Rt x(t) = eAt x0 + eAt 0 e(jωI−A)τ dτ B u¯ t B u¯ = eAt + eAt (jωI − A)−1 e(jωI−A)τ 0  −1  (jωI−A)t At At = e x0 + e (jωI − A) e − I B u¯  (jωI−A)t  −1 At At −I (jωI − A) B u¯ = e x0 + e e At At jωt At = e x0 + e e e  − I (jωI − A)−1 B u¯ = eAt x0 + ejωt − eAt (jωI − A)−1 B u¯  = eAt x0 − (jωI − A)−1 B u¯ + (jωI − A)−1 B u¯ejωt

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Hence, the output y(t) is   uejωt y(t) = CeAt x0 − (jωI − A)−1 B u¯ + C (jωI − A)−1 B u¯ejωt + D¯ In the limit as t → ∞, the first term decays to 0 exponentially, leaving the steady-state response   yss (t) = D + C (jωI − A)−1 B u¯ejωt Hence, we have verified our initial claim – if the input is a complex sinusoid, then the steadystate output is a complex sinusoid at the same exact frequency, but amplified by a complex gain of D + C (jωI − A)−1 B. The function G(ω) G(ω) := D + C (jωI − A)−1 B

(13.2)

is called the frequency response of the linear system in (13.1). Hence, for stable systems, we have proven u(t) := u¯ejωt ⇒ yss (t) = G(ω)¯ uejωt More precisely, this is the frequency response from u to y, so we might also write Gyu (ω) to indicate what is the input (u) and what is the output (y). Gyu can be calculated rather easily using a computer, simply by evaluating the matrix expression in (13.2) at a large number of frequency points ω ∈ R.

13.2

MIMO Systems: Response due to real sinusoidal inputs

In the case where the system has multiple inputs and outputs, it is a bit more complicated to write out the response due to sinusoidal inputs at a fixed frequency, since the different inputs may all have different magnitudes and phases. As before, suppose that there are m inputs, and q outputs (so B ∈ Rn×m , C ∈ Rp×n , D ∈ Rp×m ). Take a ∈ Rm , b ∈ Rm , and ω ≥ 0. Consider the input u(t) = a cos ωt + b sin ωt Note that this is the real part(Re) of a complex input, namely   u(t) = Re (a − jb)ejωt Hence, the steady state output must be the real part of a function, specifically,   y(t) = Re G(ω)(a − jb)ejωt So, in summary: To determine the steady-state response due to an input u(t) = a cos ωt + b sin ωt,

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let c ∈ Rq and d ∈ Rq be real vectors so that c − jd = G(ω)(a − jb) Then, the steady-state output is yss (t) = c cos ωt + d sin ωt

13.3

Experimental Determination

Since the frequency response has the interpretation as a representation of the frequencydependent amplitude gain and phase shift between a sinusoidal input and steady-state output, it is easy to obtain experimentally for a given physical system. This can be done by performing several different forced-response experiments, with the forcing being a sinusoid, each experiment using a different frequency. We will try this procedure in lab (both simulated on the computer, and with the EV3 systems)

13.4

Steady-State response

Suppose the input is a constant, u(t) = a, where a ∈ Rm . This is in the form of the sinusoid as derived above with ω = 0. Hence, the steady-state response is
yss (t) = G(0)a = D − CA−1 B a For this reason, D − CA−1 B is called the steady-state gain of a stable system.

ME 132, Fall 2018, UC Berkeley, A. Packard

14

150

Important special cases for designing closed-loop systems

We now have the tools to study high-order closed-loop systems, but in this section, we look at two situations which lead to 2nd-order closed-loop systems, and derive some very useful design rules. Although the systems considered appear simple, this is an extremely important section for students to learn in an introductory course on feedback systems. Understanding these examples, and the purpose of the control architecture is critical to developing the proper intuition of feedback systems.

14.1

Roots of 2nd-order monic polynomial

The roots of λ2 + c1 λ + c2 = 0 are given by the quadratic formula, and are p −c1 ± c21 − 4c2 λ1,2 = 2 Theorem: Both roots have negative real-parts if and only if c1 > 0 and c2 > 0. Proof: If c1 > 0 and c2 > 0, then c21 − 4c2 is either • nonegative, but less than c21 , or • negative. In the first case, the square-root term is real, but less than c1 . Hence, regardless of ±, the numerator is a negative, real-number. Hence both roots are indeed negative real numbers. In the second case, the square-root is imaginary, and hence the real-part of both roots is − c21 which is negative. Conversely, first suppose c1 ≤ 0. There are two cases: • The quantity −4c2 ≥ 0. In this case, the square root is nonnegative, and the + of ± produces a real-root, whose value is nonnegative. • The quantity −4c2 < 0. Hence the square-root is imaginary, and both roots have real-part equal to − c21 which is nonnegative. The alternative is that c2 ≤ 0 (and take c1 > 0, since the case c1 ≤ 0 has already been addressed). The square-root is equal to c1 , and hence the + in ± yields a root equal to 0. This completes the proof.

ME 132, Fall 2018, UC Berkeley, A. Packard

14.2

151

Setting the coefficients to attain certain roots

Suppose λ1 and λ2 are the real-valued, desired roots of a 2nd-order polynomial. Then, the polynomial, in monic form, must be (λ − λ1 )(λ − λ2 ) = λ2 + (−λ1 − λ2 )λ + λ1 λ2 Therefore, defining c1 := −λ1 − λ2 and c2 := λ1 λ2 , the polynomial λ2 + c1 λ + c2 has roots at (λ1 , λ2 ). Likewise, if γ is a complex-valued number (with non-zero imaginary part), then the monic polynomial with roots at λ1 = γ and λ2 = γ¯ is (λ − γ)(λ − γ¯ ) = λ2 − 2Re(γ)λ + |γ|2 Therefore (again) defining c1 := −2Re(γ) and c2 := |γ|2 , the polynomial λ2 + c1 λ + c2 has roots at (γ, γ¯ ). The message here is: if you have enough degrees-of-freedom to select the coefficients of a polynomial, then you can make the polynomial have any desired root-pattern. Use the roots to determine the coefficients (as functions of the roots, not the other way around), and then use the degrees-of-freedom to set the coefficients as needed. In control design, this is called “eigenvalue placement.”

14.3

1st-order plant, 1st-order controller

1st-order plant, 1st-order controller. 2nd-order closed-loop system. Closed-loop dynamics, stability conditions, frequency-response functions, design-degrees of freedom. Plant dynamics x(t) ˙ = Ax(t) + B1 d(t) + B2 u(t) y(t) = Cx(t) + D1 d(t) Controller dynamics q(t) ˙ = F q(t) + G1 r(t) + G2 ym (t) u(t) = Hq(t) + J1 r(t) + J2 ym (t) Measurement noise ym (t) = y(t) + n(t) Assumption: u has an effect on x, so B2 6= 0. We also assume measuring y gives some information about x, so C 6= 0.

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Careful, but elementary, substitution gives closed-loop dynamics as 









x(t) ˙ A + B2 J2 C B2 H B2 J1 B1 + B2 J2 D1 B2 J2   q(t)    G2 C F G1 G2 D1 G2   ˙ =   y(t)   C 0 0 D1 0   u(t) J2 C H J1 J2 D1 J2

x(t) q(t) r(t) d(t) n(t)

     

The closed-loop characteristic polynomial is    A + B2 J2 C B2 H p(λ) = det λI2 − G2 C F = (λ − A − B2 J2 C)(λ − F ) − B2 HG2 C = λ2 + (−A − B2 J2 C − F )λ + (A + B2 J2 C)F − B2 HG2 C Since this is a 2nd-order equation, the exact conditions for both roots to have negative real parts are simply (−A − B2 J2 C − F ) > 0,

(A + B2 J2 C)F − B2 HG2 C > 0 {z } | :=∆

Clearly, since B2 C 6= 0, regardless of the plant parameters, and for any choice of F , the choice of J2 sets the value of the first coefficient, while the subsequent choice of the product HG2 sets the value of the 2nd coefficient. Conclusion: complete freedom in “designing” the closed-loop characteristic polynomial, regardless of the plant data, and regardless of the choice of controller parameter F . Assuming stability, the steady-state gain from d → y is       A + B2 J2 C B2 H −1 B1 + B2 J2 D1 SSGd→y = D1 − C 0 G2 C F G2 D1 This is just SSGd→y = D1 −



 1 C 0 ∆



F −B2 H −G2 C A + B2 J2 C



B1 + B2 J2 D1 G2 D1



Making the substitutions gives SSGd→y =

F (D1 A − CB1 ) D1 (A − B2 J2 C)F − CF (B1 + B2 J2 D1 ) = ∆ ∆

Note that if F = 0, then the steady-state gain from d to y is exactly 0. This is robust, in that with this choice, even if the plant parameters (A, B1 , B2 , C, D1 ) change a bit, the steady-state gain from d to y remains exactly 0. Moreover, if

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F 6=, then the steady-state gain is generally not 0 (unless D1 A − CB1 = 0, and then this condition is not robust to changes in the plant parameters). So, the only choice for robustly achieving SSGd→y = 0 is F = 0. Assuming F = 0 is the choice, the steady-state gain from r to y is (after simplification) −

G1 G2

Hence G2 := −G1 is a design rule, which renders the steady-state gain r to y equal to 1, also “robustly”, in that deviations in the plant parameters do not affect this steady-state gain! At this point, the controller equations are q(t) ˙ = G1 (r(t) − ym (t)) u(t) = Hq(t) + J1 r(t) + J2 ym (t) A block-diagram is shown below. r -

J1

R +g ?G1 - dt - H −6

? -g +6

u -

ym -

J2

Note that since G1 is a constant, for any signal e(t), Z t Z t H G1 e(τ )dτ = HG1 e(τ )dτ 0

0

hence, only the product HG1 is important. Therefore without loss in generality, we take G1 = 1. Traditionally, H is notated as KI , and J2 is denoted −KP . Finally, J1 is noted as KF . In terms of these three constants, the controller appears as

ME 132, Fall 2018, UC Berkeley, A. Packard

r -

+g ? −6

-

R

dt

154

KF

-

KI

? -g −6

u -

ym -

KP

The closed-loop characteristic equation is λ2 + (−A + B2 KP C)λ + B2 KI C It is still clear that by choice of KP and KI , the closed-loop characteristic equation can be designed arbitrarily. KF does not affect the closed-loop characteristic equation, but it does play a part in how r affects u (and subsequently y). Problem-Statement Summary: • Given a 1st-order plant; and • a specified, desired closed-loop characteristic polynomial, λ2 + f1 λ + f2 , with roots in the open-left-half plane (for stability) • Design a feedback control system such that 1. The closed-loop system is stable; more specifically, achieve the given closed-loop characteristic polynomial, λ2 + f1 λ + f2 ; 2. The steady-state gain from d → y is 0; 3. The steady-state gain from r → y is 1. Solution: • In order to achieve the two steady-state gain objectives, the PI-architecture above must be used, as just derived. • In order to achieve the correct closed-loop polynomial, pick KI and KP such that −A + B2 KP C = f1 , This is always possible, since B2 C 6= 0.

B2 KI C = f2

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• Implement control strategy as shown. • There are no constraints on KF . Adjust it to achieve a desired transient response of y (and u) due to reference comands r. The effect that sensor noise n has on the plant output y and the control signal u was not explicitly considered. The choice of desired closed-loop characteristic equation affects the subsequent effect that n has on y and u. Tradeoffs must be considered when accounting for n.

ME 132, Fall 2018, UC Berkeley, A. Packard

14.4

156

2nd-order plant, constant-gain controller with derivative feedback

The plant dynamics are x˙ 1 (t) = x2 (t) x˙ 2 (t) = A1 x1 (t) + A2 x2 (t) + B1 d(t) + B2 u(t) y(t) = x1 (t) (again, assume B2 6= 0). The regulated variable is y = x1 . In this example though, both y and y˙ are measured, and available for use in the feedback law. The presence of y˙ in the feedback law is the reason this is referred to as derivative feedback. u(t) = K0 r(t) + K1 y(t) + K2 y(t) ˙ = K0 r(t) + K1 x1 (t) + K2 x2 (t) Note that for this initial analysis, we will not focus on sensor noise. The closed-loop equations are         x˙ 1 (t) 0 1 x1 (t) 0 0 r(t) = + x˙ 2 (t) A1 + B2 K1 A2 + B2 K2 x2 (t) B2 K0 B1 d(t) The closed-loop characteristic polynomial is    0 1 det λI2 − = λ(λ − (A2 + B2 K2 )) − (A1 + B2 K1 ) A1 + B2 K1 A2 + B2 K2 = λ2 + (−A2 − B2 K2 )λ + (−A1 − B2 K1 ) Clear that by proper choice of K1 , K2 , the closed-loop characteristic polynomial can be assigned arbitrarily. Write the polynomial in the (ξ, ωn ) parametrization, λ2 + 2ξωn λ + ωn2 , and equate coefficients, giving 2ξωn = −A2 − B2 K2 ,

ωn2 = −A1 − B2 K1

Design equations K1 = −

A1 + ωn2 , B2

K2 = −

A2 + 2ξωn B2

Note that in this case, • Position-feedback, which refers to the K1 y(t) term in u(t), sets ωn , the speed-ofresponse, and, with that chosen... • Velocity feedback, refering to the K2 y(t) ˙ term in u(t), provides the damping, ξ.

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In the homework, you will consider the following issues: • The steady-state gain from d → y can also easily be derived. It will also be a function of K1 and K2 . Obvously, these 3 important closed-loop quantities 1. closed-loop ωn 2. closed-loop damping ratio, ξ, and 3. closed-loop steady-state gain from d → y cannot be set independently using the 2 feedback controller parameters K1 and K2 . Accounting for the effect of noise, n, only increases the tradeoffs that must be considered. • By contrast, the steady-state gain from r → y is a function of K0 , as well as the system parameters, K1 and K2 . Assuming system parameters are known, the steady-state gain from r → y can be adjusted to any desired value, once choices for K1 and K2 have been made.

14.5

Problems

1. Consider the Proportional-Integral control strategy derived in lecture q(t) ˙ = r(t) − ym (t) u(t) = KI q(t) + KF r(t) − KP ym (t) as applied to a first-order plant with two separate disturbances, d1 and d2 , specifically x(t) ˙ = Ax(t) + B1 d1 (t) + B2 u(t) y(t) = Cx(t) + D2 d2 (t) with ym (t) = y(t) + n(t). (a) Find the elements of the 4 × 6 matrix such that the equations below represent the closed-loop system   x(t)      q(t)  x(t) ˙    q(t)     r(t)  ˙       y(t)  =    d1 (t)     d2 (t)  u(t) n(t) (b) What is the closed-loop A matrix (from your array above)

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(c) What is the closed-loop characteristic polynomial? (d) Suppose the desired closed-loop eigenvalues are described in terms of the (ξ, ωn ) parametrization. What are the design equations for gains KI and KP such that the eigenvalues of the closed-loop system are at this location. Your answer should be KI and KP as functions of ξ, ωn , A, B1 , . . . , D2 . (e) Ignoring u as an output, and n as an input, write the closed-loop equations in the form     x(t)     x(t) ˙    q(t)    r(t)   q(t) = ˙     d1 (t)  y(t) d2 (t) This is simply drawn from your answer in part (a). Call the matrices Aclp ∈ R2×2 , Bclp ∈ R2×3 , Cclp ∈ R1×2 , Dclp ∈ R1×3 . (f) Using the matrices from part (e), form Dclp − Cclp A−1 clp Bclp which is the steady-state gain matrix from   r  d1  → y d2 Are these steady-state gains as expected? (g) What is the instantaneous-gain (ie., FRF at ω = ∞) from r → u? 2. Please read the problem though before starting. Pay special attention to the question asked at the end in part (l). Consider the general PI control architecture shown below r

-

+? −

f 6

R

KF

-

KI

? -f −6

u -

ym -

KP

Suppose the model of the process being controlled is x(t) ˙ = u(t) + d(t), y(t) = x(t) where y is the process output, and d is a disturbance that enters additively to the control input u. For simplicity, assume there is no measurement noise, so ym = y.

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(a) Under what conditions on KP and KI and KF is the closed-loop system stable? (b) What is the closed-loop frequency-response function from r to y? (c) What is the steady-state gain from r to y? (d) What is the closed-loop frequency-response function from d to y. (e) What is the steady-state gain from d to y? (f) Which frequency-response functions are unaffected by the value of KF . (g) Choose the values of KP and KI so that the closed-loop roots of the characteristic are described by (ξ = 0.707, ωn = 1). (h) Show that regardless of the value of KF , the steady-state gain from r to y is 1. (i) Consider the case when r is a unit-step, and all initial conditions are 0. Use Matlab (ode45 or Simulink or step) to compute and plot the response of the control input u and the process output y for different values of KF , namely KF = KP , 0.5KP , 0.25KP , 0 and even −0.25KP (this is probably not a good idea, but it still works... explain why). Quantitatively describe how KF affects the response from r to both u and y. (j) In what sense is the KF = KP case easier to implement than the general case shown in the figure? (k) Consider 3 different designs: • the values of KP and KI chosen so that the closed-loop roots of the characteristic are described by (ξ = 0.707, ωn = 1) • the values of KP and KI chosen so that the closed-loop roots of the characteristic are described by (ξ = 0.707, ωn = 2) • the values of KP and KI chosen so that the closed-loop roots of the characteristic are described by (ξ = 0.707, ωn = 4) all with KF = 0.25KP . i. One one plot (with two axes), use Matlab to plot the FRF of r to y, Gr→y (ω) for all three designs. The top axes should be |Gr→y (ω)| versus ω, and the bottom axes should be ∠Gr→y (ω) versus ω. In both plots, use log-scale for ω, and in the magnitude plot, also use log-scale for |G|. Read the help on loglog and semilogx if you have forgotten how to make these plots. Make sure the linetypes you use for the three designs are different, and include a legend (using legend) on the plots. Also label the axis using xlabel and ylabel, and include a title (with title). ii. One another plot, use Matlab to plot the FRF of d to y, Gd→y (ω), specifically |Gd→y (ω)| versus ω. Use log-scale for ω and log-scale for |G|. Make sure the linetypes you use for the three designs are different, and should coincide with the linetypes used in the r → y plots. Also include a legend and title, and label the axis.

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iii. One another plot, use Matlab to plot the response y(t) versus t due to a unitstep reference input r. Take d(t) = 0 ∀t. The initial condition in the plant and controller should be 0. Make sure the linetypes you use for the three designs are different, and should coincide with the linetypes used in previous plots. Include a legend and title, and label the axis. Choose a final time of 8 , which is 8 time-constants of the slowest system. 0.707·1 iv. One another plot, use Matlab to plot the response y(t) versus t due to a unitstep disturbance input d. Take r(t) = 0 ∀t. The initial condition in the plant and controller should be 0. Make sure the linetypes you use for the three designs are different, and should coincide with the linetypes used in previous plots. Include a legend and title, and label the axis. Choose a final time of 8 , which is 8 time-constants of the slowest system. 0.707·1 v. Compare the time and frequency response plots. Make a short (4 items) list of comments about the general “consistency” between the information presented across the various plots. (l) Comment only: Suppose you are asked to repeat the entire problem for a process with mathematical model x(t) ˙ = x(t) + u(t) + d(t), y(t) = x(t) and for a process with mathematical model x(t) ˙ = −x(t) + u(t) + d(t), y(t) = x(t). Have you set up your Matlab script so that completely repeating this problem for this new process would be quite easy (eg., 5-10 minutes of hand-calculations followed by changing 1 or 2 lines in a the script)? If your answer in no, consider rearranging your hand-calculations and script file so this is the case. If your answer is yes, consider carrying out these calculations, for your own practice.

ME 132, Fall 2018, UC Berkeley, A. Packard

15

161

Step response

Suppose the input is a constant, u(t) = u¯, where u¯ ∈ Rm . What is the response starting from an initial condition x(0) = x0 ? Rt x(t) = eAt x0 + 0 eA(t−τ ) Bu(τ )dτ Rt = eAt x0 + eAt 0e−Aτ dτ B u¯  t = eAt x0 + −eAt A−1 e−Aτ 0 B u¯  −At  −1 e − I = eAt x0 + −eAt A   −1 B u¯ − I = eAt x0 + −eAt e−At  −1 A B u¯ At At = e x0 + e − I A B u¯ Hence, the output y(t) = Cx(t) + Du(t) equals y(t) = CeAt x0 − CA−1 B u¯ + CeAt A−1 B u¯ + D¯ u Expressed differently y(t) − ((D − CA−1 B)¯ u) = CeAt (x0 − (−A−1 B u¯)) Assuming stability, eAt decays to 0 as t → ∞. In that case, this expression again (as in the 1st-order case) shows that the difference between y(t) and the limiting value of y decays as the product of • C, • eAt , which is decaying to 0, • difference between x0 and final value of x It is important to note that even in the case that both u and y are scalar, the function CeAt (x0 − (−A−1 B u¯)) has • initial value equal to y(0) − yfinal ; • converges to a final value equal to 0; but... • its magnitude does not necessarily monotonically decrease to zero, since eAt is a matrix, with different linear combinations of exponentials, and these can grow in magnitude before they eventually decay. So, the responses can look more interesting/complex than the responses we observed in first-order systems. The eigenvalues of A tell some of the story, as do the eigenvectors, in an indirect manner. For exact value of response at a specific t, one needs to compute the response (numerically or analytically).

ME 132, Fall 2018, UC Berkeley, A. Packard

15.1

162

Quick estimate of unit-step-response of 2nd order system

Governing equation x(t) ˙ = Ax(t) + Bu(t); y(t) = Cx(t) + Du(t), x(0) = 02 1. Determine characteristic polynomial, pA (λ) := det(λI2 − A) = λ2 + a1 λ + a2 2. System is stable if and only if a1 > 0 and a2 > 0. 3. Are eigenvalues real or complex? Eigenvalues are real if and only is a21 − 4a2 ≥ 0. If eigenvalues are real, compute them. If the eigenvalues are complex, try the (ξ, ωn ) parametrization (as opposed to real/imag). Solve for (ξ, ωn ) as 2ξωn = a1 ,

ωn2 = a2

4. Compute steady-state gain, −CA−1 B 5. Compute the value of y(0), ˙ which is just CB. 6. The solution must transition from y(0) = 0, with the starting slope equal to the computed value of y(0) ˙ to the final value −CA−1 B, with terms involving eλ1 t , eλ2 t

eigenvalue description

or e−ξωn t cos(ωn

p

1 − ξ 2 t), e−ξωn t sin(ωn

p 1 − ξ 2 t)

In the eigenvalue representation, the slowest (least negative) eigenvalue should (roughly) dominate the response, and its time-constant will determine the total elapsed time to “convergence” to the final value. In the (ξ, ωn ) description, p the time constant of the 1 exponential envelope if ξωn , the frequency of oscillation is ωn 1 − ξ 2 , and the ratio p 1 − ξ2 2ξ

time to decay ≈ period of oscillation Put another way, • Settling time (Time to decay)=

3 ξωn

• The period of oscillation = √2π

.

ωn

1−ξ 2

≈

1 for ξ < 0.4 2ξ

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• The number of oscillations that can be observed (during the settling time) is p 3 1 − ξ2 Time to decay N := = Period of oscillation 2πξ • Everywhere ωn appears together in a term ωn t. Hence, ωn simply “scales” the response yH (t) in t. The larger value of ωn , the faster the response. The timeconstant of the exponential decay is ξω1n

ME 132, Fall 2018, UC Berkeley, A. Packard

16

164

Stabilization by State-Feedback

This material generalizes section 14.4

16.1

Theory

Consider the linear dynamical system x(t) ˙ = Ax(t) + Bu(t) As usual, let x(t) ∈ Rn , and input u(t) ∈ Rm . Suppose that the states x(t) are available for measurement, so that a control law of u(t) = Kx(t) is possible. Dimensions dictate that K ∈ Rm×n . How can the values that make up the gain matrix K be chosen to ensure closed-loop stability? An obvious approach is to 1. Pick n desired closed-loop eigenvalues, λ1 , λ2 , . . . , λn 2. Calculate the coefficients of the desired closed-loop characteristic polynomial, pdes (s) := (s − λ1 ) (s − λ2 ) · · · (s − λn ) = sn + c1 sn1 + · · · + cn Here the ci are complicated functions of the numbers λ1 , λ2 , . . . , λn . 3. Explicitly calculate the closed-loop characteristic polynomial symbolically in the entries of K, pA+BK (s) = sn + f1 (K)sn−1 + f2 (K)sn−2 + · · · + fn−1 (K)s1 + fn (K) 4. Choose K so that for each 1 ≤ i ≤ n, the equation fi (K) = ci

(16.1)

is satisfied. Suppose that u(t) ∈ R is a single input (m = 1). Then the gain matrix K ∈ R1×n . In this case, we can actually show that the coefficients of the closed-loop characteristic equation are affine (linear plus constant) functions of the entries of the K matrix. This means that solving the n equations in (16.1) will be relatively “easy,” involving a matrix inversion problem. pA+BK (s) := = = = = = =

det [sI − (A + BK)] det [sI − (A + BK)] det [(sI − A)− BK]  det (sI − A) I − (sI − A)−1 BK  det (sI − A) det I − (sI − A)−1 BK  det (sI − A) 1 − K (sI − A)−1 B det (sI − A) − Kadj (sI − A) B

ME 132, Fall 2018, UC Berkeley, A. Packard

17

165

State-Feedback with Integral Control

In this section, we generalize the results from section 14.3

17.1

Theory

Consider the linear dynamical system x(t) ˙ = Ax(t) + Bu(t) + Ed(t) y(t) = Cx(t)

d

-

E

u

-

B

-? b x˙6

R

dt x- C

-

y

A 

As usual, let x(t) ∈ Rn , disturbance d(t) ∈ Rnd , control input u(t) ∈ Rnu , and output y(t) ∈ Rny . We make the assumption that nu ≥ ny . This is important for the results we will state (and partially prove). Suppose that the states x(t) are available for measurement, and we want to control y, that is, we want each component of y, yi (t), to track the i’th component, ri (t), of a reference signal, r(t), with no steady-state effect from constant, but unknown, disturbances. In many applications, the reference signal r and disturbance signal d will consist of step-like functions, taking constant values over intervals. Hence, we want zero steady-state tracking error ei (t) := ri (t) − yi (t) for step inputs ri (t) = µ(t), and step disturbance inputs dj . Experience tells us that the controller must, among other things, integrate each component of the error e(t). Hence, we are led to the control structure shown in the figure below,

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166 Plant

d Controller R b er -- − 6

-

KI

-

E

u B

-b 6

- b? 6

R

dt A 

-

C

-y

x

KSF  

The controller is everything in the dashed box. The dimensions are KSF ∈ Rnu ×n , KI ∈ Rnu ×ny . We know that if the gains KSF and KI are chosen so that the closed-loop system is stable, then for step inputs r(t) and step disturbances d, the steady-state error e will be zero. The reasoning is as before – if the system is stable, and is subjected to step inputs, then all signals in the loop approach constant values. This includes the outputs and inputs of the integrators. If the input of an integrator approaches a constant value, then the output would be approaching a ramp, with slope equal to the limiting input value. However, the output approaches a constant value, so the slope of the ramp must actually be 0. Let ξ(t) be the outputs of the integrators in the controller. The state equations of the closed-loop system are          x(t) ˙ A + BKSF BKI x(t) 0 E = + r(t) + d(t) ˙ −C 0 ξ(t) Iny 0 ξ(t) We need to pick the gains KSF and KI so that the closed-loop system is stable. By regouping the closed-loop state equations we can put the problem into an “extended state feedback problem.”               x(t) x(t) ˙ x(t) B  0 E A 0 K K + + r(t) + d(t) = ˙ ξ(t) 0 | SF{z I } ξ(t) I 0 −C 0 ξ(t) | {z } | {z } Ke Ae

Be

Hence, the closed-loop system “Aclp ” matrix looks like Aclp := Ae + Be Ke where • Ae and Be are completely known • Ke is completely free to choose

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Hence, the stabilization problem is simply a state-feedback stabilization problem for the extended plant. Given A, B and C, one can form Ae and Be , and follow the derivation in section 16 to obtain the gain matrix Ke . Once calculated, Ke is partitioned into the appropriate state feedback, KSF , and integral gain, KI , and the controller is implemented as shown below b er -- − 6

R

-

KI

-b 6

-u

x y

KSF  

Note that the feedback structure, which contains the integrated error loop, ensures that there will be 0 steady-state errors to unit step reference inputs. It also ensures that there will be 0 steady-state error due to step disturbances to the plant. Let’s prove this... Theorem: Suppose KSF and KI are chosen so that   A + BKSF BKI −C 0 has all eigenvalues in open-left-half plane (ie., negative real parts). Hence it is an invertible matrix. For a general, stable, linear system (A, B, C,), the expression for steady-state gain is D − CA−1 B. Applying that here gives       A + BKSF BKI −1 0 SteadyStateGainr→y = − C 0 =I −C 0 I and SteadyStateGaind→y = −



C 0





A + BKSF BKI −C 0

Proof: Denote 

− C 0





A + BKSF BKI −C 0

−1 =:



−1 

X Y

E 0

 =0



In other words, X and Y are the unique matrices which satisfy       A + BKSF BKI − C 0 = X Y −C 0 Simple trial and error shows that X = 0, Y = I is a solution, and hence must be the solution. Plugging those in completes the proof.

ME 132, Fall 2018, UC Berkeley, A. Packard

17.2

168

Example

Consider a tank level control system, for illustrative purposes only. The physical setup is shown in the figure below. u

d ?

? 6 h

2

6 h3

6 h

1

R1

R12

R23

R3

For simplicity, everything is modeled as linear, giving the state equations     1 1 1 1   ˙       − + 0 h1 (t) 0 1  A1 R1 R12  h1 (t) A1 R12   1 1 1 1 1  h˙ 2 (t)  =      − A2 R12 + R23   h2 (t) + 1 u(t)+ 0  d(t) A2 R12 A2 R23     h3 (t) 0 0 h˙ 3 (t) 1 1 1 1 − + 0 A3 R23 A3 R23 R3 and



y(t) =



 h1 (t) 0 1 0  h2 (t)  h3 (t) 

Hence, in terms of extended matrices, we have    1 1 1 1 − + 0  A1 R1 R12 A1 R12   1 1 − A12 R112 + R123  A2 R12 Ae =  A2 R23  1 0 − A13 R123 +  A3 R23 0 −1 0

0



  0  ,   1 0  R3 0



 0  1   Be =   0  0

In order to keep the notation down to a minimum, define constants a, . . . , g so that   a b 0 0  c d e 0   Ae =   0 f g 0  0 −1 0 0 It is easy to verify (expanding along top row for instance) that det (λI4 − Ae ) = λ4 + (−a − d − g)λ3 + (dg − ef + ad + ag − bc)λ2 + (aef − adg + bcg)λ

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and (computing only the 2nd column of adj (λI − Ae ))  bλ2 − bgλ  λ3 + (−a − g)λ2 + agλ adj (λI − Ae ) Be =   f λ2 − af λ −λ2 + (a + g)λ − ag

   

Hence, if we let ci be the coefficients of the closed-loop characteristic equation, we have λ4 + c1 λ3 + c2 λ2 + c3 λ + c4 = det (λI − Ae ) − Ke adj (λI − Ae ) Be In matrix  c1  c2   c3 c4

form, we get  

  −a − d − g 0 1 0 0   dg − ef + ad + ag − bc   b −a − g f −1 = −     −bg aef − adg + bcg ag −af a + g 0 0 0 0 −ag



 Ke1   Ke2      Ke3  Ke4

If we pick the desired closed-loop eigenvalues, then we can solve for the desired closed-loop characteristic equation coefficients ci , and by simple matrix inversion, we can then solve for the gains that make up the controller matrix Ke . In this example, the tank parameters are A1 A2 A3 R1 R12 R23 R3

4 8 5 1 2 0.4 1

I choose the closed-loop eigenvalues to be at −0.45 ± j0.218, −0.4, −0.5 resulting in a desired closed-loop characteristic equation λ4 + 1.8λ3 + 1.26λ2 + 0.405λ + 0.05 Solving for the gain matrix Ke yields   Ke = −0.052 −0.35 −0.37 0.19

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which partitions into KSF =



−0.052 −0.35 −0.37



KI = 0.19

A simulation is shown below. The initial condition is hi (0) = 0 for all tanks. The reference input is r(t) = 5 for 0 ≤ t ≤ 80 r(t) = 8 for 80 < t and the disturbance flow is d(t) = 0 d(t) = 2 d(t) = −2 d(t) = 0 d(t) = 4

for for for for for

0 ≤ t ≤ 30 30 < t ≤ 50 50 < t ≤ 70 70 < t ≤ 90 90 < t

This works great, in part because the control u enters tank 2 directly, which is the tank whose height we are trying to control, while the disturbance enters tank 1. So, the control has a more direct effect. Next, we simulate a different configuration, using the same control gains. In this simulation, the disturbance is modified: • Instead of entering the 1st tank, the disturbance enters the 2nd tank. • The magnitude of the disturbance flow is reduced by a factor of 5. Simulations are shown below:

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171

Reference, Height 10

Meters

8 6 4 2 0 0

20

40

60

20

40

60

80

100

120

140

160

80 100 Time, seconds

120

140

160

Control and Dist Flow

4 3 2 1 0 −1 −2 0

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Reference, Height 10

Meters

8 6 4 2 0 0

20

40

60

20

40

60

80

100

120

140

160

80 100 Time, seconds

120

140

160

Control and Dist Flow

1.5 1 0.5 0 −0.5 0

ME 132, Fall 2018, UC Berkeley, A. Packard

17.3

173

Problems

1. Consider a system of the form x˙ 1 (t) = x2 (t) x˙ 2 (t) = A21 x1 (t) + A22 x2 (t) + B2 u(t) + E2 d(t) y(t) = x1 (t)

(17.1)

We want to design a controller that uses measurements of y and y˙ (note y˙ = x2 ), as well as a reference input r, such that • the closed-loop system eigenvalues are at desired, specified location • the steady-state gain from r → y equals 1, robustly (ie., even with modest variations in A21 , A22 , B2 , E2 ) • the steady-state gain from d → y equals 0, robustly (ie., even with modest variations in A21 , A22 , B2 , E2 ) The controller architecture is a particular special case of Section 17, namely feedback of the states (x1 and x2 ), as well as the integral of r − y, η(t) ˙ = r(t) − ym (t) u(t) = KI η(t) + K0 r(t) + K1 ym (t) + K2 y˙ m (t)

(17.2)

where ym (t) = x1 (t) + n1 (t) and y˙ m (t) = x2 (t) + n2 (t), and n1 and n2 are noises due to the individual sensors measuring x1 and x2 . This is called a “PI controller with rate-feedback”, since the feedback consists of a proportional and integral feedback of y, as well as feedback of y. ˙ With this architecture, the control designer’s job is to decide on appropriate values for the gains, KI , K0 , K1 , K2 . (a) Find state-space model (4 inputs, 3 outputs, 3 states) of the closed-loop system combining the plant model (17.1) and controller dynamics (17.2). The inputs should be (r, d, n1 , n2 ) and outputs (y, y, ˙ u). (b) Find closed-loop characteristic equation in terms of the plant parameters and controller gains. (c) Show that by choice of {KI , K1 , K2 }, the 3rd-order closed-loop characteristic polynomial can be made equal to any 3rd-order polynomial. Assume that B2 6= 0. (d) Follow the linear algebra derivation in Section 14.1 to conclude that if {KI , K1 , K2 } are chosen so that the closed-loop system is stable, then • steady-state gain from r → y is 1 (insensitive to any changes in parameters, as long as closed-loop remains stable!) • steady-state gain from d → y is 0 (insensitive to any changes in parameters, as long as closed-loop remains stable!)

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(e) There are 3 eigenvalues, with one of two possibilities • a complex-conjugate pair (real and imaginary part, or (ωn , ξ) description), and one real eigenvalue • 3 real eigenvalues In either case, we have 3 parameters to describe the locations. For this purpose, introduce 3 parameters: (ωn , ξ, β), and take the characterstic equation as pclosed−loop (λ) = (λ2 + 2ξωn λ + ωn2 )(λ + βωn ) so that the roots are at λ1,2 = −ξωn ± jωw

p

1 − ξ 2 , λ3 = −βωn

Again, ωn sets the “speed-of-response” of the complex-pair, ξ sets the damping, and β sets the speed-of-response of the real eigenvalue, relative to the complexconjugate pair. Because of the ease in interpreting ωn , we can pick a good candidate value for ωn , as well as a sensible candidate value for ξ, and then try varying β from (say) 0.33 to 3 and see the effect. Multiply out pclosed−loop , and develop “design equations” for {KI , K1 , K2 } in terms of plant parameters and desired eigenvalue locations. (f) Assume A21 = A22 = 0 and B2 = E2 = 1. Let the desired eigenvalue locations be described by ωn = 1, ξ = 0.707, β = {0.33, 1, 3} (3 different choices for β). Find KI , K1 , K2 for each of the 3 designs. (g) For each design, temporarily set r = 0, and focus on disturbance rejection. Note that K0 is not important (it only influences the effect r has on y and u), and can be (for example) set to 0. Use ss objects, and step to plot the step-response of y and u due to a unit-step disturbance d. (h) Next, set d = 0, and focus on the effect r has on y (and u). Again do 3 simulations, with step-inputs for r, for the different values of β, all with K0 := −K1 . This means that the proportional feedback is of the form K0 (r − ym ), in other words, also proportional to error. As usual, plot y and u. (i) Repeat the reference step-responses with K0 := 0, K0 = −0.5K1 and K0 = −1.5K1 . So, this is now 9 plots (3 different β and 3 different K0 ). Comment on the effect of the immediate aggressiveness of the action of u in driving y towards r = 1. Again, plot y and u. (j) Implement the example in Simulink, and verify that the results are identical. 2. This problem builds on the development in problem 1, using a plant model closer to the EV3 motor, and just doing the implementation in Simulink. Connection to EV3

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motor control: The Lego EV3 motor, using x1 := θ and x2 := Ω, we have indeed x˙ 1 = x2 and 1 x˙ 2 = (−αΩ + T + d) J where T is the torque acting on shaft from EV3 electronics (as we can command it, so u) and d represents any other (external) torque (from the “environment”, eg., an additional inertia attached to the shaft, an external disturbance torque, like your fingers , B2 = J1 . grabbing the motor shaft, and trying to stop it...). so that A21 = 0, A22 = −α J In the initial lab, you identified the time-constant τ and steady-state gain γ, (from T to Ω) which relate to these parameters as τ=

J , α

γ=

1 α

α=

1 , γ

J=

τ γ

which is equivalently

The units we used were degrees/sec (for Ω) and LegoTorque for T . Hence the units of J are not SI, but rather units for J =

LegoTorque · s2 degree

As mentioned, later we will do an experiment to estimate the correct conversion from LegoTorque to N · m, and then we can compute J in SI units. For consistency across everyone’s assignments, take J = 0.01 and α = 0.125. (a) Suppose an additional inertia is attached (rigidly, let’s say for now) to the motor (common in a mechanical motion control application). Let JL denote the additional inertia. The equation for Ω becomes x˙ 2 =

1 (−αΩ + T + d) J + JL

which changes both the steady-state gain and time-constant, in a known manner. Task: Build a Simulink Model of the EV3 motor, with additional inertia, using these equations. It should constructed in such a manner that it is easy to modify the value of JL in the workspace, and have this propagate into the model. Once constructed, group as a subsystem, and (if desired, not required), mask the subsystem block so that the user would enter J, JL and α to characterize the system. (b) Add the controller architecture to the Simulink model. The controller has one state, 3 inputs, (r, ym , y˙ m ) and one output, u. The controller equations are

ME 132, Fall 2018, UC Berkeley, A. Packard

η(t) ˙ = r(t) − ym (t) u(t) = KI η(t) + K0 r(t) + K1 ym (t) + K2 y˙ m (t)

176

(17.3)

It is defined by 4 parameters, (KI , K1 , K2 , K0 ). Again, group this as a subsystem, and (if desired), mask it so the user cleanly enters the 4 parameters in a nice dialog box. Important Remark: Even though the EV3 only has a θ measurement, and we will be inferring Ω from numerical differentiation of θ, start in Simulink as though you have independent measurements of θ and Ω in order to implement the feedback law. (c) Assume JL = 0.01. Design the controller gains so that the closed-loop eigenvalues are at   2 3 ωn = 8, ξ = 0.9, β = , 1, 3 2 (d) Simulate the system, from zero initial conditions, with a 150◦ reference step input (step occurs at t = 1), and a step-disturbance torque of 70 (in LegoTorque units) which occurs at t = 4. The simulation time should be 6 seconds. Plot both y and u. In all cases, define K0 = −0.25K1 . 3. We will add several layers of “non-idealized” behavior to the Simulink model, and see the effect this has on the overall control system performance. (a) Create a new Simulink model (by copying, and then modifying as described below): • Quantized measurement of y, with quantization level equal to 1 • Only sampled (in time) measurements of the quantized y is available. Use a Zero-Order-Hold block to model the sampling, with a sample-time of 0.05. • No direct measurement of y. ˙ Instead, estimate y˙ as in the lab, using a Discrete State-Space Block, implementing a 1st-order, backwards finite difference estimate of y, ˙ from the quantized, sampled measurement of y. • Quantize (also at quantization level equal to 1) the output of controller (LegoTorque command), and use Sample-Hold block too, with sample-time of 0.05. (b) Use Simulink to simulate this situation, with the same inputs (reference and disturbance) as previous, using the gains obtained in the 3 cases. In all cases, define K0 = −0.25K1 .

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177

Linear-quadratic Optimal Control

Picking appropriate locations of the desired eigenvalues can be challenging. We only know vaguely how they affect the response. The eigenvector matrix has a big influence on the behavior of T eΛt T −1 . In this section, we introduce an approach whereby the designer picks/selects/dictates a different goal, and mathematics then determines what the statefeedback gain should be to optimize the goal. In many cases, this is more natural. More advanced theory further tries to make the designer’s specifications even more natural, and again use mathematics to turn those specifications into a control strategy. The system dynamics are as usual, x(t) ˙ = Ax(t) + Bu(t),

x(0) = x0

where x(t) ∈ Rn and u(t) ∈ Rm . Goal of control is to drive x(t) → 0n , but the desired manner in which it gets to zero depends on the system and objectives. Suppose Q = QT ∈ Rn×n is a positive-semidefinite matrix, and any two non-zero values of x, say x[1] ∈ Rn and x[2] are considered “equally far away from 0” if xT[1] Qx[1] = xT[2] Qx[2] . Additionally, in getting x → 0, we desire to not use too “much” control action, as this is “costly” in some sense. Suppose R = RT ∈ Rm×m is a positive-definite matrix, and two non-zero values of u, say u[1] and u[2] are considered “equally far away from 0” if uT[1] Ru[1] = uT[2] Ru[2] . Finally, futher assume that a nonzero value of x and a nonzero value of u are considered “equally far away from 0” if xT Qx = uT Ru. With this level of “penalization” for nonzero x and u, one criterion that measures how far away from 0 a particular (x(·), u(·)) trajectory is Z ∞ J(x0 , u(·)) := xT (t)Qx(t) + uT (t)Ru(t)dt 0

Note that the cost criterion depends on the initial condition and the entire control trajectory u (since the resulting x()˙ is a consequence of x0 and u). It is a quadratic function of the resulting state trajectory x(·) and the input u(·). If one adopts J(x0 , u(·)) as the quantity to make as small as possible, by choice of u, the optimal control problem is: For a given initial condition x0 , the optimization problem is Z ∞ ∗ J (x0 ) := min xT (t)Qx(t) + uT (t)Ru(t)dt, subject to x(t) ˙ = Ax(t) + Bu(t) u(·)

0

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Clearly, the optimal control-input signal, u(t)|t∈[0,∞) , depends on the initial condition x0 . A particular input may be the optimal input starting from a particular initial condition. Starting from a different initial condition almost surely requires a different input. Fact: Given the system model A and B, and the cost-function matrices Q and R, there exists a matrix P = P T ∈ Rn×n and a matrix F ∈ Rm×n such that • for all initial conditions x0 , the optimal cost J(x0 ) is simply xT0 P x0 ; • for all initial conditions, the optimal input trajectory u(t)|t∈[0,∞) can be calculated via a feedback law, namely u(t) = −F x(t). This form, expressed as a state-feedback, is referred to as the optimal linear-quadratic regulator. Moreover P and F are easily calculated using matrix operations. We don’t have enough time to go into the specifics, but will be able to use the ideas in homework and lab. The command lqr in Matlab’s Control System toolbox makes the calculation as [F,P] = lqr(A,B,Q,R).

18.1

Learning more

In the next section, we consider an easier problem, still called linear quadratic control. The two simplifying assumptions are discrete-time (the differential equation is replaced with a difference equation, so no calculus) and we only consider a fixed, finite-time interval. The finite-time interval is easier than the infinite-time horizon considered in this section, but of course, in the finite-time derivation, we only determine what the control is over that period, hence in an actual implementation, the problem must be “over” at the end of the period.

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179

Single, high-order, linear ODES (SLODE)

Throughout this section, if y denotes a function (of time, say), then y [k] or y (k) denotes the k’th derivative of the function y, dk y y [k] = k dt In the case of k = 1 and k = 2, it will be convenient to also use y˙ and y¨.

19.1

Linear, Time-Invariant Differential Equations

Often, in this class, we will analyze a closed-loop feedback control system, and end up with an equation of the form y (n) (t) + a1 y (n−1) (t) + · · · + an y(t) = v(t)

(19.1)

where y is some variable of the plant that is of interest to us, and v is a forcing function, usually either a reference signal (ydes (t)) or a disturbance (ie., inclination of hill), or a combination of such signals. One job of the control designer is to analyze the resulting equation, and determine if the behavior of the closed-loop system is acceptable. The differential equation in (19.1) is called a forced, linear, time-invariant differential equation. For now, associate the fact that the {ai }ni=1 are constants with the term time-invariant, and the fact that the left-hand side (which contains all y terms) is a linear combination of y and its derivatives with the term linear. The right-hand side function, v, is called the forcing function. For a specific problem, it will be a given, known function of time. Sometimes, we are given an initial time t0 and initial conditions for differential equation, that is, real numbers (n−1) y0 , y˙ 0 , y¨0 , . . . , y0 (19.2) and we are looking for a solution, namely a function y that satisfies both the differential equation (19.1) and the initial condition constraints y(t0 ) = y0 ,

y(t ˙ 0 ) = y˙0 ,

y¨(t0 ) = y¨0 ,

(n−1)

. . . , y (n−1) (t0 ) = y0

.

(19.3)

For essentially all differential equations (even those that are not linear, and not timeinvariant), there is a theorem which says that solutions always exist, and are unique: Theorem (Existence and Uniqueness of Solutions): Given a forcing function v, defined for t ≥ t0 , and initial conditions of the form (19.3). Then, there exists a unique function y which satisfies the initial conditions and the differential equation (19.1).

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19.2

180

Importance of Linearity

Suppose that yP is a function which satisfies (19.1) so that (n)

(n−1)

yP (t) + a1 yP

(t) + · · · + an yP (t) = v(t)

(19.4)

and yH is a function which, for all t satisfies (n)

(n−1)

yH (t) + a1 yH

(t) + · · · + an yH (t) = 0

(19.5)

yH is called a homogeneous solution of (19.1), and the differential equation in (19.5) is called the homogeneous equation. The function yP is called a particular solution to (19.1), since it satisfies the equation with the forcing function in place. The derivative of yP + yH is the sum of the derivatives, so we add the equations satisfied by yP and yH to get [yP + yH ](n) (t) + a1 [yP + yH ](n−1) (t) + · · · an [yP + yH ] (t) = v(t) This implies that the function yP + yH also satisfies (19.1). Hence, adding a particular solution to a homogeneous solution results in a new particular solution. Conversely, suppose that yP1 and yP2 are two functions, both of which solve (19.1). Consider the function yd := yP1 − yP2 . Easy manipulation shows that this function yd satisfies the homogeneous equation. It is a trivial relationship that yP1 = yP2 + (yP1 − yP2 ) = yP2 + yd We have shown that any two particular solutions differ by a homogeneous solution. Hence all particular solutions to (19.1) can be generated by taking one specific particular solution, and adding to it every homogeneous solution. In order to get the correct initial conditions, we simply need to add the “right” homogeneous solution. Remark: The main points of this section rely only on linearity, but not time-invariance.

19.3

Solving Homogeneous Equation

Let’s try to solve (19.5). Take a fixed complex number r ∈ C, and suppose that the function yH , defined as yH (t) = ert

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(k)

is a solution to (19.5). Substituting in, using the fact that for any integer k > 0, yH (t) = rk ert , gives
rn + a1 rn−1 + · · · + an ert = 0 for all t. Clearly, ert 6= 0, always, so it can be divided out, leaving rn + a1 rn−1 + · · · + an = 0

(19.6)

Thus, if ert is a solution to the homogeneous equation, it must be that the scalar r satisfies (19.6). Conversely, suppose that r is a complex number which satisfies (19.6), then simple substitution reveals that ert does satisfy the homogeneous equation. if r is a repeated  rtMoreover, rt root, say l times, then substitution shows that the functions e , te , . . . , tl−1 ert all satisfy the homogeneous differential equation. This leads to the following nomenclature: Let r1 , r2 , . . . , rn be the roots of the polynomial equation λn + a1 λn−1 + · · · + an = 0 This polynomial is called the characteristic polynomial associated with (19.1). Fact 1 (requires proof ): If the {ri }ni=1 are all distinct from one another , then yH satisfies (19.5) if and only if there exist complex constants c1 , c2 , . . . , cn such that yH (t) =

n X

ci eri t

(19.7)

i=1

Fact 2 (requires proof ): If the {ri }ni=1 are not distinct from one another, then group the roots {r1 , r2 , · · · , rn } as p1 , p1 , . . . , p1 , p2 , p2 , . . . , p2 , · · · , pf , pf , . . . , pf | {z } | {z } | {z } l1

l2

(19.8)

lf

Hence, p1 is a root with multiplicity l1 , p2 is a root with multiplicity l2 and so on. Then yH satisfies (19.5) if and only if there exist complex constants cij (i = 1, . . . f , j = 0, . . . li − 1) such that f li −1 X X yH (t) = cij epi t tj (19.9) i=1 j=0

So, Fact 2 includes Fact 1 as a special case. Both indicate that by solving for the roots of the characteristic equation, it is easy to pick n linearly independent functions which form a basis for the set of all homogeneous solutions to the differential equation. Here, we explicitly have

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used the time-invariance (ie., the coefficients of the ODE are constants) to generate a basis (the exponential functions) for all homogeneous solutions. However, the fact that the space of homogenous solutions is n-dimensional only relies on linearity, and not time-invariance. Basic idea: Suppose there are m solutions to the homogeneous differential equation, labeled y1,H , y2,H , . . . , ym,H , with m > n. Then, look at the n × m matrix of these solutions initial conditions,   (0) (0) (0) y1,H y2,H · · · ym,H  (1)  (1) (1)  y1,H y2,H · · · ym,H   M :=  .. .. .. ..   . . . .   (n−1) (n−1) (n−1) y1,H y2,H · · · ym,H Since m > n, this must have linearly dependent columns, so there is a nonzero m × 1 vector α such that M α = 0n×1 . Define yz := α1 y1,H + α2 y2,H + · · · + αm ym,H . Since this is a sum of homogeneous solutions, it itself is a homogeneous solution. Moreover, yz satisfies yz(0) (t0 ) = 0, yz(1) (t0 ) = 0, · · · , yz(n−1) (t0 ) = 0 Note that the function yI (t) ≡ 0 for all t also satisfies the same initial conditions as yz , and it satisfies the homogeneous differential equation as well. By uniqueness of solutions, it must be that yz (t) = yI (t) for all t. Hence, yz is actually the identically zero function. Hence, we have shown that the set of homogeneous solutions is finite dimensional, and of dimension at most n. Moreover, by the simple substitutions above, we already know how to construct n linearly independent solutions to the homogeneous differential equation, so those must form a basis for all homogeneous solutions. The reason that terms like tk ert appear for differential equations whose characteristic equation has repeated roots is partially explored in Problem 6.

19.3.1

Interpretation of complex roots to ODEs with real-coefficients

In all cases we consider, the differential equation has real coefficients (the a1 , a2 , . . . , an ). This means that the characteristic equation has real coefficients, but it does not mean that the roots are necessarily real numbers (eg., even a real quadratic equation may have complex roots). However, since the coefficients are real, then complex roots will come in complexconjugate pairs, so if λ := α + jβ is a root, and α and β are real numbers, and β 6= 0, then it will also be true that α − jβ is a root of the characteristic equation as well. Given the roots, we know that all homogeneous solutions are given by equation 19.9, where the constants vary over all choices of complex numbers. We will almost exclusively be interested in solutions starting from real initial conditions, with real forcing functions, and

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hence will only need to consider real homogeneous solutions. This raises the question of how to interpret the complex functions eλt which arise when λ (a root) is complex. Suppose λ1 ∈ C is a complex root. Since the characteristic equation has real coefficients, it follows that that complex conjugate of λ1 , λ2 = λ¯1 is also a root. Let α and β be real numbers such that λ1 = α + jβ, λ2 = α − jβ, β 6= 0 The terms of the homogeneous solution associated with these roots are c1 e λ 1 t + c2 e λ 2 t where c1 and c2 are any complex numbers. We want to study under what conditions (on c1 and c2 ) will this function be purely real. Plugging in gives c1 eαt [cos βt + j sin βt] + c2 eαt [cos βt − j sin βt] Regrouping gives (eαt [(c1 + c2 ) cos βt + j(c1 − c2 ) sin βt] Since cos and sin are linearly independent functions, the only conditions for this to be a purely real function is that c1 + c2 be real, and j(c1 − c2 ) be real. This requires that c1 and c2 be complex-conjugates of one another. In other words, for the function c1 eαt [cos βt + j sin βt] + c2 eαt [cos βt − j sin βt] is real-valued, for all t if and only if c1 = c¯2 . Let c2 be any complex number, and let c1 := c¯2 . Then c1 + c2 = 2A, where A is the real part of c2 . Also, j(c1 − c2 ) = 2B, where B is the imaginary part of c2 . Hence, the purely real homogeneous solutions are of the form (eαt [2A cos βt + 2B sin βt] where A and B are free real constants. Since they are free, it serves no purpose to write 2A and 2B, and we can simply write that all real homogeneous solutions arising from the two roots λ1 = α + jβ and λ2 = α − jβ are (eαt [A cos βt + B sin βt] where A and B are any real numbers. With this is mind, we can rewrite a specialized version of Fact 2, for the case when the coefficients of the ODE are real, and we are interested in writing down a description of just all real homogeneous solutions.

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Fact 3: Assume that all coefficients a1 , a2 , . . . , an are real numbers. Group the n roots of the characteristic equation as p1 , p1 , . . . , p1 , · · · , pd , pd , . . . , pd , {z } | {z } | l1

ld

α1 + jβ1 , α1 + jβ1 , . . . , α1 + jβ1 , α1 − jβ1 , α1 − jβ1 , . . . , α1 − jβ1 , | {z } | {z } h1

h1

.. .

(19.10)

αc + jβc , αc + jβc , . . . , αc + jβc , αc − jβc , α1 − jβc , . . . , αc − jβc {z } | {z } | hc

hc

where each p, α and β are real, and each β 6= 0. Then yH is a real-valued function which satisfies (19.5) if and only if there exist real constants cij (i = 1, . . . d, j = 0, . . . li − 1), Aks , Bks (k = 1, . . . c, s = 0, . . . hk − 1)such that yH (t) =

li −1 d X X

pi t j

cij e t +

c hX k −1 X

i=1 j=0

19.4

eαk t ts [Aks cos βk t + Bks sin βk t]

(19.11)

k=1 s=0

General Solution Technique

The general technique (conceptually, at least) for finding particular solutions of (19.1) which also satisfy the initial conditions is a combination of all of the above ideas. It is very important conceptually, and somewhat important in actual use. 1. Find any particular solution yP 2. Choose constants cij so that the function yP (t) +

f li −1 X X

cij tj epi t

i=1 j=0

satisfies the initial conditions. The resulting function is the solution. Note that in step 2, there are n equations, and n unknowns.

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19.5

185

Behavior of Homogeneous Solutions as t → ∞

In this section, we study the behavior of the class of homogeneous solutions as t → ∞. If we can show that all homogeneous solutions decay to 0 as t → ∞, then it must be that for a give forcing function, all particular solutions, regardless of the initial conditions, approach each other. This will be useful in many contexts, to quickly understand how a system behaves. Suppose r is a complex number, r ∈ C, and we decompose it into its real and imaginary parts. Let α := Real(r) and β := Imag(r). Hence, both α and β are real numbers, and √ r = α + jβ. We will always use j := −1. The exponential function ert can be expressed as ert = eαt [cos βt + j sin βt] Note that the real part of r, namely α, determines the qualitative behavior of ert as t → ∞. Specifically, • if α < 0, then limt→∞ ert = 0 • if α = 0, then ert does not decay to 0, and does not “explode,” but rather oscillates, with |ert | = 1 for all t • if α > 0, then limt→∞ |ert | = ∞ When all of the roots of the chracteristic polynomial are distinct, all homogeneous solutions are of the form ert where r is a root of the characteristic polynomial. Therefore, we see that the roots of the characteristic polynomial determine the qualitative nature of the homogeneous solutions. If the characteristic polynomial has a root rp of multiplicity lp ≥ 1, then according to Eq. (19.9), the homogeneous solution modes associated the root rp are erp t , t erp t , · · · , t(lp −1) erp t . Notice that for any real constants m < ∞, and α < 0, lim tm eαt = 0 .

t→∞

Therefore, the real part of the repeated root rp , namely αp , also determines the qualitative behavior of the homogeneous solution modes associated with this repeated root as t → ∞. Specifically, • if αp < 0, then limt→∞ tm erp t = 0 for m = 0, · · · lp − 1.

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• if αp = 0, then |erp t | = 1 for all t, but • if αp ≥ 0, then limt→∞ |tm ert | = ∞ for m = 1, · · · lp − 1 where lp ≥ 1 is the multiplicity of the root rp . We summarize all of these results as follows: • If all of the roots, {ri }ni=1 , of the characteristic polynomial satisfy Real(ri ) < 0 then every homogeneous solution decays to 0 as t → .∞ • If any of the roots, {ri }ni=1 , of the characteristic polynomial satisfy Real(ri ) ≥ 0 then there are homogeneous solutions that do not decay to 0 as t → ∞ .

19.6

Response of stable system to constant input (Steady-State Gain)

Suppose the system (input u, output y) is governed by the SLODE y [n] (t) + a1 y [n−1] (t) + · · · + an−1 y [1] (t) + an y(t) = b0 u[n] (t) + b1 u[n−1] (t) + · · · + bn−1 u[1] (t) + bn u(t) Suppose initial conditions for y are given, and that the input u is specified to be a constant, u(t) ≡ u¯ for all t ≥ t0 . What is the limiting behavior of y? If the system is stable, then this is easy to compute. First notice that the constant function yP (t) ≡ abnn u¯ is a particular solution, although it does not satisfy any of the initial conditions. The actual solution y differs from this particular solution by some homogeneous solution, yH . Hence for all t, y(t) = yP (t) + yH (t) bn = u¯ + yH (t) an Now, take limits, since we know (by the stability assumption) that limt→∞ yH (t) = 0, giving lim y(t) =

t→∞

Hence, the steady-state gain of the system is

bn u¯ an

bn . an

ME 132, Fall 2018, UC Berkeley, A. Packard

19.7

187

Example

Consider the differential equation y¨(t) + 4y(t) ˙ + y(t) = 1

(19.12)

subject to the initial conditions y(0) = y0 , y(0) ˙ = y˙ 0 . The characteristic equation is λ2 + 4λ + 1 = 0, which has roots at √ λ = −2 ± 3 ≈ {−3.7, −0.3} Hence, all homogeneous solutions are of the form yH (t) = c1 e−3.7t + c2 e−0.3t Terms of the form e−3.7t take about 0.8 time units to decay, while terms of the form e−0.3t take about 10 time units to decay. In general then (though not always - it depends on the initial conditions) homogeneous solutions will typically take about 10 time units to decay. A particular solution to (19.12) is simply yP (t) = 1 for all t ≥ 0. Note that this choice of yP does not satisfy the initial conditions, but it does satisfy the differential equation. As we have learned, all solutions to (19.12) are any particular solution plus all homogeneous solutions. Therefore, the general solution is √

y(t) = 1 + c1 e(−2−

3)t

√

+ c2 e(−2+

3)t

which has as its derivative y(t) ˙ = (−2 −

√ √ √ √ 3)c1 e(−2− 3)t + (−2 + 3)c2 e(−2+ 3)t

Evaluating these at t = 0, and equating to the given initial conditions yields y(0) = 1 + c1 + c2 = y0 √
√
y(0) ˙ = −2 − 3 c1 + −2 + 3 c2 = y˙ 0 In matrix form, we have 

1√ 1√ −2 − 3 −2 + 3



c1 c2







d −b −c a

=

y0 − 1 y˙ 0

This is easy to invert. recall 

a b c d

−1

1 = ad − bc





ME 132, Fall 2018, UC Berkeley, A. Packard

Hence 

1√ 1√ −2 − 3 −2 + 3

188

−1

1 = √ 2 3

so that 

c1 c2



 =

1√ 1√ −2 − 3 −2 + 3



√  −2 +√ 3 −1 2+ 3 1

−1 

y0 − 1 y˙ 0



For the case at hand, let’s take y0 := 0, y˙ 0 := 0, hence √     1 2 − √3 c1 = √ c2 2 3 −2 − 3 A plot of y(t) is shown in the figure below. 1 0.9 0.8

Response, y(t)

0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

5

10

15

Time, t

Note that indeed, as we had guessed, the homogeneous solution (which connects the initial values at t = 0 to the final behavior at t → ∞ takes about 10 time units to decay.

19.8

Stability Conditions for 2nd order differential equation

Given real numbers a0 , a1 and a2 , with a0 6= 0, we wish to determine if the roots of the equation a0 λ 2 + a1 λ + a2 = 0

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have negative real parts. This question is important in determining the qualitative nature (exponential decay versus exponential growth) of the homogeneous solutions to a0 y¨(t) + a1 y(t) ˙ + a2 y(t) = 0 Since a0 6= 0 (so that we actually do have a quadratic, rather than linear equation) divide out by it, giving a2 a1 ˙ + y(t) = 0 y¨(t) + y(t) a0 a0 a1 a2 Call b1 := a0 , b2 := a0 . The characteristic equation is λ2 + b1 λ + b2 = 0. The roots are p p −b1 + b21 − 4b2 −b1 − b21 − 4b2 λ1 = , λ2 = 2 2 Consider 4 cases: 1. b1 > 0, b2 > 0. In this case, the term b21 − 4b2 < b21 , hence either p • b21 − 4b2 is imaginary p p • b21 − 4b2 is real, but b21 − 4b2 < b1 . In either situation, both Re(λ1 ) < 0 and Re(λ2 ) < 0. 2. b1 ≤ 0: Again, the square-root is either real or imaginary. If it is imaginary, then √ −b1 + b21 −4b2 −b1 Re(λi ) = 2 ≥ 0. If the square-root is real, then then Re(λ1 ) = ≥ 0. In 2 either case, at least one of the roots has a nonnegative real part. 3. bp 2 ≤ 0: In this case, the square root is real, and hence both roots are real. However, b21 − 4b2 ≥ |b1 |, hence λ1 ≥ 0. so one of the roots has a non-negative real part. This enumerates all possibilities. We collect these ideas into a theorem. Theorem: For a second order polynomial equation λ2 +b1 λ+b2 = 0, the roots have negative real parts if and only if b1 > 0, b2 > 0. If the leading coefficient is not 1, then we have Theorem (RH2): For a second order polynomial equation b0 λ2 + b1 λ + b2 = 0, the roots have negative real parts if and only if all of the bi are nonzero, and have the same sign (positive or negative). Higher-order polynomials have more complicated conditions. Theorem (RH3): For a third order polynomial equation λ3 + b1 λ2 + b2 λ + b3 = 0, all of the roots have negative real parts if and only if b1 > 0, b3 > 0 and b1 b2 > b3 .

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Note that the condition b1 b2 > b3 , coupled with the first two conditions, certainly implies that b2 > 0. However, the simpler conditions {b1 > 0, b2 > 0, b3 > 0} are only necessary, not sufficient, as an example illustrates: the roots of λ3 + λ2 + λ + 3 are {−1.575, 0.2874 ± j1.35}. So, use the exact conditions in theorem RH3 to determine the stability of a 3rd order system by simple examination of its characteristic equation coefficients. Theorem (RH4): For a fourth order polynomial equation λ4 +b1 λ3 +b2 λ2 +b3 λ+b4 = 0, all of the roots have negative real parts if and only if b1 > 0, b4 > 0, b1 b2 > b3 and (b1 b2 − b3 )b3 > b21 b4 . These theorems are specific examples of the Routh-Hurwitz theorem which is often taught in undergraduate control courses. The Routh-Hurwitz theorem, proven around 1880, independently by Routh and Hurwitz, is interested in the determining if all roots of the polynomial equation λn + b1 λn−1 + · · · + bn−1 λ + bn = 0 have negative real-parts. The theorem states that all roots have negative real parts if and only if n simple inequalities, involving polynomial expressions of the coefficients (the bk ) are satisfied. Note that the theorem does not find the roots - it merely proves, unequivocally, that the roots all lie in the open, left-half plane if and only if some easy-to-evaluate expressions, involving the coefficients, are negative. There is a pattern by which one can “simply” derive what the polynomial inequalities are. The three theorems above (RH2, RH3, and RH4) give the inequalities for n = 2, n = 3 and n = 4, respectively. As you notice, the conditions become more complex with increasing order. For systems beyond 4th order, it is not clear how useful (beyond the theoretical elegance) the formula are, especially in the context of modern laptop computational power, where high-quality root-finding programs are easily accessible (eg., Matlab, Mathematica, etc). We encourage the student to memorize the 2nd, 3rd and 4th order conditions and through their use, determine if knowing higher-order versions would be useful. The interested student can find more information about Routh-Hurwitz on the web.

19.9

Important 2nd order example

It is useful to study a general second order differential equation, and interpret it in a different manner. Start with y¨(t) + a1 y(t) ˙ + a2 y(t) = bv(t)

(19.13)

with y(0) = 0, y(0) ˙ = 0 and v(t) = 1 for all t ≥ 0. Assume that the homogeneous solutions are exponentially decaying, which is equivalent to a1 , a2 > 0. Rewrite using new variables

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(instead of a1 and a2 ) ξ, ωn as y¨(t) + 2ξωn y(t) ˙ + ωn2 y(t) = bv(t) where both ξ, ωn > 0. In order to match up terms here, we must have 2ξωn = a1 ,

ωn2 = a2

which can be inverted to give a1 ξ= √ , 2 a2

ωn =

√

a2

Note that since a1 , a2 > 0, we also have ξ, ωn > 0. With these new variables, the homogeneous equation is y¨(t) + 2ξωn y(t) ˙ + ωn2 y(t) = 0 which has a characteristic polynomial λ2 + 2ξωn λ + ωn2 = 0 The roots are at

p λ = −ξωn ± ωn pξ 2 − 1 = −ξωn ± jωn 1 − ξ 2

There are three cases to consider: • ξ > 1: Roots are distinct, and both roots are real, and negative. The general homogeneous solution is most easily written in the form     √ √ −ξωn −ωn ξ 2 −1 t −ξωn +ωn ξ 2 −1 t + c2 e yH (t) = c1 e • ξ = 1, which results in repeated real roots, at λ = −ξωn , so that the general form of homogeneous solutions are yH (t) = c1 e−ξωn t + c2 te−ξωn t • 0 < ξ < 1: Roots are distinct, and complex (complex-conjugate pair), with negative real part, and the general homogeneous solution is easily expressable as h i h i √ √ −ξωn +jωn 1−ξ 2 t −ξωn −jωn 1−ξ 2 t yH (t) = c1 e + c2 e This can be rewritten as h i √ √ 2 2 yH (t) = e−ξωn t c1 ejωn 1−ξ t + c2 e−jωn 1−ξ t

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Recall ejβ = cos β + j sin β. Hence p p   c1 cos ωnp1 − ξ 2 t + jc1 sin ωn p1 − ξ 2 t −ξωn t yH (t) = e +c2 cos ωn 1 − ξ 2 t − jc2 sin ωn 1 − ξ 2 t which simplifies to −ξωn t

yH (t) = e

h

i p p 2 2 (c1 + c2 ) cos ωn 1 − ξ t + j(c1 − c2 ) sin ωn 1 − ξ t

For a general problem, use the initial conditions to determine c1 and c2 . Usually, the initial conditions, y(0), y(0), ˙ are real numbers, the differential equation coefficients (ap 1 , a2 or ξ, ωn ) p 2 are real, hence the solution must be real. Since cos ωn 1 − ξ t and sin ωn 1 − ξ 2 t are linearly independent functions, it will always then work out that c1 + c2 = purely real c1 − c2 = purely imaginary In other words, Im(c1 ) = −Im(c2 ), and Re(c1 ) = Re(c2 ), which means that c1 is the complex conjugate of c2 . Under this condition, call c := c1 . The homogeneous solution is i h p p −ξωn t 2 2 yH (t) = e 2Re(c) cos ωn 1 − ξ t − 2Im(c) sin ωn 1 − ξ t Use A := 2Re(c), and B := −2Im(c), and the final form of the real homogeneous solution is h i p p yH (t) = e−ξωn t A cos ωn 1 − ξ 2 t + B sin ωn 1 − ξ 2 t A and B are two, free, real parameters for the real homogeneous solutions when 0 < ξ < 1. p Of course, it is true that −ξωn is the real part of the roots, and ωn 1 − ξ 2 is the imaginary part. The solution is made up of two terms: 1. An exponentially decaying envelope, e−ξωn t . Note that this decays to zero in approximately ξω3n p p 2. sinosoidal oscillation terms, sin ωn 1 − ξ 2 t and cos ωn 1 − ξ 2 t. The period of oscillation is √2π 2 . ωn

1−ξ

Comparing these times, we expect “alot of oscillations before the homogeneous solution decays” if 2π 3 p 0, b3 > 0 and b1 b2 > b3 . • Theorem (RH4): For a fourth order polynomial equation λ4 +b1 λ3 +b2 λ2 +b3 λ+b4 = 0, all of the roots have negative real parts if and only if b1 > 0, b4 > 0, b1 b2 > b3 and (b1 b2 − b3 )b3 > b21 b4 .

ME 132, Fall 2018, UC Berkeley, A. Packard

19.10.5

199

2nd order differential equation

• General form is y¨(t) + a1 y(t) ˙ + a2 y(t) = bv(t) • For stable systems (a1 > 0, a2 > 0), rewrite using new variables ξ, ωn as y¨(t) + 2ξωn y(t) ˙ + ωn2 y(t) = bv(t) where both ξ, ωn > 0. • We have 2ξωn = a1 ,

ωn2 = a2

which can be inverted to give a1 ξ= √ , 2 a2

ωn =

√

a2

Note that since a1 , a2 > 0, we also have ξ, ωn > 0. • The characteristic polynomial λ2 + 2ξωn λ + ωn2 = 0 • The roots are at

19.10.6

p λ = −ξωn ± ωn pξ 2 − 1 = −ξωn ± jωn 1 − ξ 2

Solutions of 2nd order differential equation

• ξ > 1: Roots are distinct, and both roots are real, and negative.     √ √ −ξωn +ωn ξ 2 −1 t −ξωn −ωn ξ 2 −1 t yH (t) = c1 e + c2 e Settling time (Time to decay)=

3√ |−ξωn +ωn ξ 2 −1|

• ξ = 1: Roots are repeated real at λ = −ξωn , yH (t) = c1 e−ξωn t + c2 te−ξωn t Settling time (Time to decay)=

3 ξωn

(19.18)

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• 0 < ξ < 1: Roots are distinct, and complex (complex-conjugate pair), with negative real part i h p p yH (t) = e−ξωn t A cos ωn 1 − ξ 2 t + B sin ωn 1 − ξ 2 t or equivalently yH (t) = e−ξωn t

hp

(A2 + B 2 ) sin(ωn

i p 1 − ξ 2 t + φ)

where φ = atan(A/B) – Settling time (Time to decay)=

3 ξωn

– The period of oscillation = √2π

.

ωn

1−ξ 2

– The number of oscillations that can be observed (during the settling time) is p 3 1 − ξ2 Time to decay N := = Period of oscillation 2πξ – Everywhere ωn appears together in a term ωn t. Hence, ωn simply “scales” the response yH (t) in t. The larger value of ωn , the faster the response.

19.11

Problems

1. Consider the differential equation y¨(t) + 3y(t) ˙ + 2y(t) = 0

(19.19)

Let yH,1 (t) := 2e−t − e−2t , and yH,2 (t) := e−t − e−2t . (a) Compute yH,1 (0) and y˙ H,1 (0). Plot yH,1 (t) versus t. Show that the function yH,1 satisfies the differential equation in (19.19). (b) Compute yH,2 (0) and y˙ H,2 (0). Plot yH,2 (t) versus t. Show that the function yH,2 satisfies the differential equation in (19.19). (c) For any two constants α and β, define yc (t) := αyH,1 (t) + βyH,2 (t) i. Does yc satisfy the ODE? ii. What is yc (0)? Your answer will involve α and β. iii. What is y˙ c (0)? Your answer will involve α and β. (d) What is the solution to the ODE with initial conditions y(0) = 6, y(0) ˙ = −4?

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2. Consider the example equation from Section 19.7, y¨(t) + 4y(t) ˙ + y(t) = 1, with initial conditions y(0) = y0 , y(0) ˙ = v0 . Consider all possible combinations of initial conditions from the lists below: y0 = {−2, −1, 0, 1, 2} and v0 = {−2, −1, 0, 1, 2} (Now, do part (a) first!) (a) Note that without explicitly solving the differential equation, one can easily compute 4 things: limiting value of y (as t → ∞), time constant of the “slowest” homogeneous solution, initial value (given) and initial slope (given). With these numbers, carefully sketch on graph paper how you think all 25 solutions will look. (b) In the notes, the solution is derived for general initial conditions. Use Matlab (or similar) to plot these expressions. Compare to your simplistic approximations in part 2a. 3. The response (with all appropriate initial conditions set to 0) of the systems listed below is shown. Match the ODE with the solution graph. Explain your reasoning. (a) y(t) ˙ + y(t) = 1 (b) y(t) ˙ + 5y(t) = 5 (c) y¨(t) + 2y(t) ˙ + y(t) = 1 (d) y¨(t) − 2y(t) ˙ − y(t) = −1 (e) y¨(t) − 2y(t) ˙ + 9y(t) = 9 (f) y¨(t) + 0.4y(t) ˙ + y(t) = 1 (g) y¨(t) + 0.12y(t) ˙ + 0.09y(t) = 0.09 (h) y¨(t) + 11y(t) ˙ + 10y(t) = 10 (i) y¨(t) + 0.3y(t) ˙ + 0.09y(t) = 0.09 (j) y¨(t) + 3y(t) ˙ + 9y(t) = 9 (k) y¨(t) + 4.2y(t) ˙ + 9y(t) = 9 (l) y¨(t) + 0.2y(t) ˙ + y(t) = 1

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2

1

1

0.5

0

0

20

40

60

80

100

0

1

2

0.5

1

0

0

2

4

6

1

0

0

2

4

6

0

10

20

30

5 0

0.5 −5 0

0

0.5

1

1.5

2

−10

0

0.5

1

1.5

2

2.5

1.5 1

1 0.5 0

0

20

40

60

0

1.5

0

1

−0.5

0.5

−1

0

0

1

2

3

1

−1.5

0

0

10

0.2

20

0.4

30

0.6

0.8

40

1

1.5 1

0.5 0.5 0

0

2

4

6

0

0

1

4. Consider the homogeneous differential equation y [3] (t) + 9¨ y (t) + 24y(t) ˙ + 20y(t) = 0 (a) What is the characteristic polynomial of the ODE? (b) What are the roots of the characteristic polynomial.

2

3

4

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(c) Write the general form of a homogeneous solution. Explain what are the free parameters. (d) Show, by direct substitution, that yH (t) = te−2t is a solution. (e) Show, by direct substitution, that yH (t) = t2 e−2t is not a solution. (f) Find the solution which satisfies initial conditions y(0) = 3, y(0) ˙ = 1, y [2] (0) = 0. (g) Find the solution which satisfies initial conditions y(0) = 3, y(0) ˙ = 1, y [3] (0) = 0. 5. Suppose a < 0, and consider the function teat for t ≥ 0. (a) For what value of t does the maximum occur? (b) At what value(s) of t is the function equal to 0.05 of its maximum value. For comparison, recall that for the function eat , the function is equal to 0.05 of the 3 maximum value at about −a . 6. (a) Suppose λ ∈ C and k ≥ 0 is an integer. Show that x(t) := solution to the differential equation

1 k+1 −λt t e k+1

is a

x(t) ˙ + λx(t) = tk e−λt (b) Suppose y and z satisfy the differential equations z(t) ˙ + λz(t) = y(t),

y(t) ˙ + λy(t) = 0

Eliminate y, and find a 2nd-order differential equation governing z (c) Suppose q satisfies the differential equation q¨(t) + 2λq(t) ˙ + λ2 q(t) = 0 Define r(t) := q(t) ˙ + λq(t). What differential equation does r satisfy? Hint: What is r(t) ˙ + λr(t) 7. In section 19.9, we considered differential equations of the form y¨(t) + a1 y(t) ˙ + a2 y(t) = bv(t). √ If a1 > 0 and a2 > 0, and a1 < 2 a2 , then we chose to write the solution in terms of the (ωn , ξ) parameters, which are derived from a1 and a2 . If the forcing function v is a constant, v(t) ≡ v¯, we derived that all particular solutions are of the form i h p p b v¯ + e−ξωn t A cos ωn 1 − ξ 2 t + B sin ωn 1 − ξ 2 t a2 where A and B are free parameters. Suppose the initial conditions are y(0) = y0 and y(0) ˙ = y˙0 . Find the correct values for A and B so that the initial conditions are satisfied. Your answer should be in terms of the given initial conditions, and system parameters (ωn , ξ, b).

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8. In this problem, we look at the “damping-ratio, natural-frequency” parametrization of complex roots, as opposed to the real/imaginary parametrization. The “dampingratio, natural-frequency” description is a very common manner in which the location of complex eigenvalues is described. For this problem, suppose that 0 < ξ < 1, and ω > 0. (a) What are the roots of the equation s2 + 2ξωn s + ωn2 = 0 (b) Let λ be the complex number λ := −ξωn + jωn

p 1 − ξ2

(note that this is one of the roots you computed above). Show that |λ| = ωn , regardless of 0 < ξ < 1. (c) The complex number λ is plotted in the complex plane, as shown below. λ ×A

Aψ A A

Im 6

C Re -

Express sin ψ in terms of ξ and ωn . (d) Run the commands A = randn(5,5); damp(eig(A)) several times, and copy/paste the output into the assignment. Explain the displayed output and its connection to the results you derived here. 9. The cascade of two systems is shown below. The relationship between the inputs and outputs are given. Differentiate and eliminate the intermediate variable v, obtaining a differential equation relationship between u and y. u -

S1

v-

y S2

S2 : y(t) ˙ + c1 y(t) = d1 v(t)

-

S1 : v¨(t) + a1 v(t) ˙ + a2 v(t) = b1 u(t) ˙ + b2 u(t) Repeat the calculation for the cascade in the reverse order, as shown below.

ME 132, Fall 2018, UC Berkeley, A. Packard

u -

S2

v-

y S1

205

S2 : v(t) ˙ + c1 v(t) = d1 u(t)

-

S1 : y¨(t) + a1 y(t) ˙ + a2 y(t) = b1 v(t) ˙ + b2 v(t) 10. Compute (by analytic hand calculation) and plot the solutions to the differential equations below. Before you explicitly solve each differential equation, make a table listing • each root of the characteristic equation • the damping ratio ξ, and natural frequency ωn for each pair (if there is one) of complex roots. • the final value of y, ie., limt→∞ y(t). for each case. For the plots, put both cases in part (a) on one plot, and put both cases for part (b) on another plot. √
√
3 (a) i. dtd 3 y(t) + 1 + 10 2 y¨(t) + 100 + 10 2 y(t) ˙ + 100y(t) = 100u(t), subject to the initial conditions y¨(0) = y(0) ˙ = y(0) = 0, and u(t) = 1 for all t > 0. Hint: One of the roots to the characteristic equation is −1. Given that you can easily solve for the other two. ii. y(t) ˙ + y(t) = u(t) subject to the initial conditions y(0) = 0, and u(t) = 1 for all t > 0. (b)

d3 y(t) dt3

+ 10.6¨ y (t) + 7y(t) ˙ + 10y(t) = 10u(t), subject to the initial conditions y¨(0) = y(0) ˙ = y(0) = 0, and u(t) = 1 for all t > 0. Hint: One of the roots to the characteristic equation is −10. ii. y¨(t) + 0.6y(t) ˙ + y(t) = u(t), subject to the initial conditions y(0) ˙ = y(0) = 0, and u(t) = 1 for all t > 0. i.

11. We have studied the behavior of the first-order differential equation x(t) ˙ = − τ1 x(t) + τ1 u(t) v(t) = x(t) which has a “time-constant” of τ , and a steady-state gain (to step inputs) of 1. Hence, if τ is “small,” the output v of system follows u quite closely. For “slowly-varying” inputs u, the behavior is approximately v(t) ≈ u(t). (a) With that in mind, decompose the differential equation in (10)(a)(i) into the cascade of i. a “fast” 2nd order system, with steady-state gain equal to 1 ii. “slow” first order system whose steady-state gain is 1. Is one of these decomposed systems similar to the system in (10)(a)(ii)? Are the two plots in (10)(a) consistent with your decomposition?

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(b) Do a similar decomposition for (10)(b)(i), and again think/comment about the response of the 3rd order system in (10)(b)(i) and the 2nd order system’s response in (10)(b)(ii). 12. (a) Suppose α is a fixed real number. If p(s) is a n’th order polynomial, with roots λ1 , λ2 , . . . , λn , what are the roots of the n’th order polynomial g(s) defined by g(s) := p(s − α) (b) Do all of the roots of s3 + 7s2 + 22s + 16 have real parts less than 0? (c) Do all of the roots of s3 + 7s2 + 22s + 16 real parts less than −1?

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207

Frequency Responses of Linear Systems

In this section, we consider the steady-state response of a linear system due to a sinusoidal input. The linear system is the standard one, y [n] (t) + a1 y [n−1] (t) + · · · + an−1 y [1] (t) + an y(t) = b0 u[n] (t) + b1 u[n−1] (t) + · · · + bn−1 u[1] (t) + bn u(t)

(20.1)

with y the dependent variable (output), and u the independent variable (input). Assume that the system is stable, so that the roots of the characteristic equation are in the open left-half of the complex plane. This guarantees that all homogeneous solutions decay exponentially to zero as t → ∞. Suppose that the forcing function u(t) is chosen as a complex exponential, namely ω is a fixed real number, and u(t) = ejωt . Note that the derivatives are particularly easy to compute, namely u[k] (t) = (jω)k ejωt It is easy to show that for some complex number H, one particular solution is of the form yP (t) = Hejωt How? Simply plug it in to the ODE, leaving H [(jω)n + a1 (jω)n−1 + · · · + an−1 (jω) + an ] ejωt = [b0 (jω)n + b1 (jω)n−1 + · · · + bn−1 (jω) + bn ] ejωt For all t, the quantity ejωt is never zero, so we can divide out leaving H [(jω)n + a1 (jω)n−1 + · · · + an−1 (jω) + an ] = [b0 (jω)n + b1 (jω)n−1 + · · · + bn−1 (jω) + bn ] Now, since the system is stable, the roots of the polynomial λn + a1 λn−1 + · · · + an−1 λ + an = 0 all have negative real part. Hence, λ = jω, which has 0 real part, is not a root. Therefore, we can explicitly solve for H as H=

b0 (jω)n + b1 (jω)n−1 + · · · + bn−1 (jω) + bn (jω)n + a1 (jω)n−1 + · · · + an−1 (jω) + an

(20.2)

Moreover, since actual solution differs from this particular solution by some homogeneous solution, we must have y(t) = yP (t) + yH (t)

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In the limit, the homogeneous solution decays, regardless of the initial conditions, and we have lim y(t) − yP (t) = 0 t→∞

Since yP is periodic, and y tends towards it asymptotically (the homogeneous solutions are decaying), we call this specific particular solution the steady-state behavior of y, and denote it yss . The explanation we have given was valid at an arbitrary value of the forcing frequency, ω. The expression for H in (20.2) is still valid. Hence, we often write H(ω) to indicate the dependence of H on the forcing frequency. H(ω) :=

b0 (jω)n + b1 (jω)n−1 + · · · + bn−1 (jω) + bn (jω)n + a1 (jω)n−1 + · · · + an−1 (jω) + an

(20.3)

This function is called the “frequency response” of the linear system in (20.1). Sometimes it is referred to as the “frequency response from u to y,” written as Hu→y (ω). For stable systems, we have proven for fixed value u¯ and fixed ω u(t) := u¯ejωt ⇒ yss (t) = H(ω)¯ uejωt

20.1

Complex and Real Particular Solutions

What is the meaning of a complex solution to the differential equation (20.1)? Suppose that functions u and y are complex, and solve the ODE. Denote the real part of the function u as uR , and the imaginary part as uI (similar for y). Then uR and uI are real-valued functions, and for all t u(t) = uR (t) + juI (t). Differentiating this k times gives [k]

[k]

u[k] (t) = uR (t) + juI (t) Hence, if y and u satisfy the ODE, we have h i h i [n] [n] [n−1] [n−1] yR (t) + jyI (t) + a1 yR (t) + jyI (t) + · · · + an [yR (t) + jyI (t)] = h i h i [n] [n] [n−1] [n−1] = b0 uR (t) + juI (t) + b1 uR (t) + juI (t) + · · · + bn [uR (t) + juI (t)] But the real and imaginary parts must be equal individually, so exploiting the fact that the coeffcients ai and bj are real numbers, we get [n]

[n−1]

[1]

[n]

[n−1]

[1]

yR (t) + a1 yR (t) + · · · + an−1 yR (t) + an yR (t) [n] [n−1] [1] = b0 uR (t) + b1 uR (t) + · · · + bn−1 uR (t) + bn uR (t) and

yI (t) + a1 yI (t) + · · · + an−1 yI (t) + an yI (t) [n] [n−1] [1] = b0 uI (t) + b1 uI (t) + · · · + bn−1 uI (t) + bn uI (t) Hence, if (u, y) are functions which satisfy the ODE, then both (uR , yR ) and (uI , yI ) also satisfy the ODE.

ME 132, Fall 2018, UC Berkeley, A. Packard

20.2

209

Response due to real sinusoidal inputs

Suppose that H ∈ C is not equal to zero. Recall that ∠H is the real number (unique to within an additive factor of 2π) which has the properties cos ∠H =

ImH ReH , sin ∠H = |H| |H|

Then, Re Hejθ




= Re [(HR + jHI ) (cos θ + j sin θ)] = HR cos h θ − HI sin θ i HR HI = |H| |H| cos θ − |H| sin θ = |H| [cos ∠H cos θ − sin ∠H sin θ]
= |H| cos (θ + ∠H) Im Hejθ = Im [(HR + jHI ) (cos θ + j sin θ)] = HR sin i h θ + HI cos θ HI R sin θ + |H| cos θ = |H| H |H| = |H| [cos ∠H sin θ + sin ∠H cos θ] = |H| sin (θ + ∠H)

Now consider the differential equation/frequency response case. Let H(ω) denote the frequency response function. If the input u(t) = cos ωt = Re (ejωt ), then the steady-state output y will satisfy y(t) = |H(ω)| cos (ωt + ∠H(ω)) A similar calculation holds for sin, and these are summarized below. Input 1 cos ωt sin ωt

20.3

Steady-State Output H(0) = abnn |H(ω)| cos (ωt + ∠H(ω)) |H(ω)| sin (ωt + ∠H(ω))

Problems

1. Write a Matlab function which has three input arguments, A, B and Ω. A and B are row vectors, of the form of the form A = [a0 a1 a2 · · · an ] ,

B = [b0 b1 b2 · · · bn ]

where a0 6= 0. These represent the input/output system a0 y [n] (t) + a1 y [n−1] (t) + · · · + an−1 y [1] (t) + an y(t) = b0 u[n] (t) + b1 u[n−1] (t) + · · · + bn−1 u[1] (t) + bn u(t)

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with y the dependent variable (output), and u the independent variable (input). Ω is a 1-by-N vector of real frequencies. The function should return one argument, H, which will be the same dimension as Ω, but in general will be complex. The value of H(i) should be the frequency-response function of the system above at frequency ω = Ω(i). 2. Using the Matlab function from above, draw the Bode plot of the frequency response function for the 3rd order system in problem 10(a)(i) in Section 19.11. On the same graph, plot the frequency response function for the 1st order system in problem 10(a)(ii). Comment on the similarities and differences (eg., in what frequency ranges are they appreciably different?; in the frequency ranges where they are different, what is the magnitude of the response function as compared to the largest value of the response magnitude? 3. Using the Matlab function from above, draw the Bode plot of the frequency response function for the 3rd order system in problem 10(b)(i) in Section 19.11. On the same graph, plot the frequency response function for the 2nd order system in problem 10(b)(ii). Comment on the similarities and differences (eg., in what frequency ranges are they appreciably different?; in the frequency ranges where they are different, what is the magnitude of the response function as compared to the largest value of the response magnitude? 4. Suppose the ODE for a system is y¨(t) + 2ξωn y(t) ˙ + ωn2 y(t) = ωn2 u(t) where u is the input, and y is the output. Assume that ξ > 0 and ωn > 0. (a) Derive the frequency response function of the system. Let ω denote frequency, and denote the frequency response function as H(ω). (b) What are |H(ω)| and |H(ω)|2 . Is |H(ω)| (or |H(ω)|2 ) ever equal to 0? (c) Work with |H(ω)|2 . By dividing numerator and denominator by ωn4 , derive |H(ω)|2 =  1−



ω ωn

1  2 2

+ 4ξ 2



ω ωn

2

What is |H(ω)| approximately equal to for for ω ωn , |H(ω)| ≈ ωωn . As a function of log ω, what is log |H(ω)| approximately equal to for for ω >> ωn ? Specifically, show that for ω >> ωn , log |H(ω)| ≈ 2 log ωn − 2 log ω (e) What is ∠H(ω) for ω > ωn ? (g) What are |H(ω)| and ∠H(ω) for ω = ωn ? d |F (ω)| = 0 at exactly (h) For a general function F (ω) that is never 0, show that dω d the same values of ω as when |F (ω)|2 = 0. dω d (i) Find the frequencies (real numbers) ω where |H(ω)| = 0. dω (j) Show that if ξ < √12 , then the maximum of |H(ω)| occurs at a non-zero frequency. p Denote the critical frequency by ωcrit , and show namely ωcrit = ωn 1 − 2ξ 2 . (k) Assuming that ξ
0 must be y (t) = αt + β + c1 eλ1 t + c2 eλ2 t (21.4) where c1 and c2 are uniquely chosen to satisfy the initial conditions y (0) = Y0 , y˙  (0) = Y˙ 0 . Differentiating gives that y˙  (t) = α + c1 λ1 eλ1 t + c2 λ2 eλ2 t (21.5) Satisfying the initial conditions at t = 0 gives conditions that the constants c1 and c2 must satisfy      Y0 − β 1 1 c1 = ˙ λ1 λ2 c2 Y0 − α which can be solved as 

c1 c2



1 = λ2 − λ1



λ2 −1 −λ1 1



Y0 − β Y˙ 0 − α



In terms of c1 and c2 , the value of y and y˙  at t =  are (using equations (21.4 and 21.5))    λ     y () e 1 eλ2  c1 α + β = + y˙  () λ1 eλ1  λ2 eλ2  c2 α Substituting, gives  λ        1 Y0 − β y () e 1 eλ2  λ2 −1 α + β = + −λ1 1 α y˙  () Y˙ 0 − α λ2 − λ1 λ1 eλ1  λ2 eλ2 

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Rearranging gives        1 Y0 − β λ2 eλ1  − λ1 eλ2 
eλ2  − eλ1  y () α + β = + y˙  () α Y˙ 0 − α λ2 − λ1 λ2 λ1 eλ1  − eλ2  λ2 eλ2  − λ1 eλ1  For notational purposes, let 1 M := λ2 − λ1



λ2 eλ1  − λ1 eλ2 
eλ2  − eλ1  λ1  λ2  λ2 λ1 e − e λ2 eλ2  − λ1 eλ1 



Then, we have 

y () y˙  ()





Y0 − β Y˙ 0 − α

= M



 +

α + β α



This is further manipulated into          Y0 1  β y () = M ˙ + −M + y˙  () 0 1 α Y0 But recall, α and β depend on , so that should be substituted,    "     1 Y0 1  y () −M + = M ˙ + 0 1 y˙  () Y0 

b1 a2 −a1 b2 a22 b2 a2

#

Note that lim M = I2 →0

and using L’Hospital’s rule,        1 1  0 0 0 0 −M + = = lim 0 1 λ2 λ1 −λ2 − λ1 a2 a1 →0  Hence, 

y(0+ ) y(0 ˙ +)



 := lim

→0,>0

y () y˙  ()



 =

Y0 ˙ Y 0 + b1



 =

y(0− ) y(0 ˙ −)



 +

0 b1



In summary, suppose the input waveform is such that the right-hand side (ie., forcing function) of the differential equation has singularities at certain discrete instants in time. Between those instants, all terms of the differential equation are well behaved, and we solve it in the classical manner. However, at the instants of the singularity, some specific high order derivative of y experiences a discontinuity in its value, and the discontinuity can be determined

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from the differential equation. Essentially, we get “new” initial conditions for the differential equation just after the singularity in terms of the solution of the ODE just before the singularity and the ODE itself. Key take-home points: 1. We have seen that the values of y and y˙ at a time just before the input singularity are related to (but not equal to) the values of y and y˙ just after the singularity. 2. Another point to notice: this calculation shows that the term b1 u(t), ˙ which appears on the right-hand side of the ODE in equation (21.3) plays a role in the solution. So, both the left and right sides of an input/output ODE have qualitative and quantitative effects of the solution. We will formalize this as we proceed... In the next section we learn some tricks that allow us to redo this type of calculation for general systems (not just second order). Distributions are mathematical constructions that, in a rigorous manner, allow us to repeat the rigorous calculation we just did in this section in a less tedious fashion.

21.4

Problems

1. Consider the differential equation y¨(t) + 4y(t) ˙ + 3y(t) = b1 u(t) ˙ + 3u(t), subject to the forcing function u(t) := µ(t), and initial conditions y(0− ) = 0, y(0 ˙ − ) = 0. (a) Determine: • the conditions of y (and y) ˙ at t = 0+ , • the roots of the homogeneous equation (and their values of ξ and ωn if complex) • the final value, limt→∞ y(t). Some of these will be dependent on the parameter b1 . (b) Based on your answers, sketch a guess of the solution for 8 cases: b1 = −6, −3, −0.3, −0.03, 0.03, 0.3, 3, 6. Sketch them on the same axes. (c) Determine the exact solution (by hand). Do as much of this symbolically as you can (in terms of b1 ) so that you only need to the majority of the work once, and then plug in the values of b1 several times. Use the plot command in MatLab (or other) to get a clean plot of the solution y(t) from 0+ to some suitable final time. Plot them on the same graph.

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218

Distributions

In this section we learn some tricks that allow us to redo the calculation in the previous section for general systems (not just second order). Distributions are mathematical constructions that, in a rigorous manner, allow us to repeat the rigorous calculation we just did in section 21 in a less tedious fashion.

22.1

Introduction

Recall our goal: determine the response of y [n] (t) + a1 y [n−1] (t) + a2 y [n−2] (t) + · · · + an y(t) = b0 u[n] (t) + b1 u[n−1] (t) + b2 u[n−2] (t) + · · · + bn u(t)

(22.1) [0]

subject to u(t) = µ(t) (a unit step at t = 0) and initial conditions y(0− ) = y0 , y [1] (0− ) = [1] [2] [n−1] y0 , y [2] (0− ) = y0 , . . . , y [n−1] (0− ) = y0 . Here 0− refers to the time just before the unitstep input is applied. So, the system is placed in initial conditions, and released, the release time being denoted 0− . At that instant, the input’s value is 0, so u(0− ) = 0. An infinitesimal time later, at 0+ , the input value is changed to +1. The input is actually at unit-step at t = 0, and the initial conditions are known just before the step-input is applied. The difficulty is that the right-hand side of (22.1) is not classically defined, since u is not differentiable. However, we may obtain the solution by considering a sequence of problems, with smooth approximations to the unit step, and obtain the solution as a limiting process. Such a procedure goes as follows: 1. Define a n-times differentiable family of functions µ that have the property  0 for t < 0 µ (t) = 1 for t >  and which converge to the unit step µ as  → 0. 2. Compute solution y (t) to the differential equation y [n] (t) + a1 y [n−1] (t) + a2 y [n−2] (t) + · · · + an y(t) = b0 u[n] (t) + b1 u[n−1] (t) + b2 u[n−2] (t) + · · · + bn u(t) [0]

subject to the forcing function u(t) = µ (t) and initial conditions y(0) = y0 , y [1] (0) = [1] [2] [n−1] y0 , y [2] (0) = y0 , . . . , y [n−1] (0) = y0 .

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[n−1]

3. Look at the values of y (), y˙  (), y¨ () . . . , y (). Take the limit as  → 0, and get a relationship between the values of y, y, ˙ y¨, . . . , y [n−1] at t = 0− and t = 0+ . 4. For t > 0, the right hand side of the ODE in (22.1) is well defined (in fact, it’s a constant, bn ), and the solution can be determined from finding any particular solution, and combining with the family of homogeneous solutions to correctly match the conditions at t = 0+ . This procedure is tedious, though the following can be proven: 1. It gives the correct answer for the solution of the ODE subject to the unit step input [k]

2. The final answer for the limits y () as  → 0 are the same regardless of the form of µ , as long as it satisfies the properties given. Moreover, you can accomplish this in an easier fashion by using generalized functions, called distributions. The most common distribution that is not a normal function is the Dirac-δ function. To gain intuition about this, consider an -approximation to the unit-step function of the form   0 for t < 0 1 t for 0 < t <  µ (t) =   1 for t ≥ 1 The derivative of this is

  0 for t < 0 d 1 for 0 < t <  µ (t) =   dt 0 for t ≥ 1

Call this function δ . Note that for all values of  > 0, Z ∞ δ (t)dt = 1 −∞

and that for t < 0 and t > , δ (t) = 0. Moreover, for any continuous function f Z ∞ lim f (t)δ (t)dt = f (0) →0

−∞

and for any t 6= 0, lim→0 δ (t) = 0. Hence, in the limit we can imagine a “function” δ whose value at nonzero t is 0, whose value at t = 0 is undefined, but whose integral is finite, namely 1.

ME 132, Fall 2018, UC Berkeley, A. Packard

Now, we can start another level smoother. Let δ  0 for    t for 2 δ (t) := 2−t for    2 0 for

220

be defined as t 0, Z

∞

δ (t)dt = 1 −∞

and that for t < 0 and t > , δ (t) = 0. • Also, for any continuous function f Z ∞ lim f (t)δ (t)dt = f (0) →0

−∞

and for any t 6= 0, lim→0 δ (t) = 0. • The function δ(t) = lim→0 δ is the Dirac-δ function. • The “function” δ is equal to 0 at all times with the exception of t = 0. At t = 0 it is undefined, but the integral is finite, namely 1. • We proceed similarly and define the first derivative of the function δ(t), the second derivative and so on.. δ [0] (t) := δ(t) :=

dµ , dt

δ [k] (t) =

dδ [k−1] dt

The procedure • Consider our general system differential equation y [n] (t) + a1 y [n−1] (t) + a2 y [n−2] (t) + · · · + an y(t) = b0 u[n] (t) + b1 u[n−1] (t) + b2 u[n−2] (t) + · · · + bn u(t) and suppose that u(t) = µ(t).

(22.5)

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• By using distribution properties we can argue that y [n] is of the form y [n] = e1 µ[n] + e2 µ[n−1] + · · · + en µ[1] + en+1 µ + fn (t) where fn (t) is a continuous function, and the constants e1 , e2 , . . . , en+1 need to be determined. • Integrating, we get y [n] = e1 δ [n−1] + e2 δ [n−2] + · · · + en δ + en+1 µ + fn (t) y = e1 δ [n−2] + · · · + en−1 δ + en µ + fn−1 (t) .. . [n−1]

y [1] = y [0] =

e1 δ

+ e2 µ + e1 µ

+ f2 (t) + f0 (t)

where each of the fi (t) are continuous functions. • Since fi (t) are continuous functions, they are continuous at 0. f0 (0− ) = f0 (0+ ) = y(0− ) f1 (0− ) = f1 (0+ ) = y(0 ˙ −) f2 (0− ) = f2 (0+ ) = y¨(0− ) .. .. .. . . . − + f[n−1] (0 ) = f[n−1] (0 ) = y [n−1] (0− ) Therefore the ei give the discontinuity in each derivative of y at t = 0+ : y(0+ ) = y(0− ) + e1 y(0 ˙ + ) = y(0 ˙ − ) + e2 [2] + y (0 ) = y [2] (0− ) + e3 .. .. .. . . . [k] + y (0 ) = y [k] (0− ) + ek+1 .. .. .. . . . y [n−1] (0+ ) = y [n−1] (0− ) + en • Plugging into the ODE, and equating the different δ, δ [1] , . . .,δ [n−1] functions gives n equations in n unknowns, expressed in matrix form below      1 0 0 ··· 0 e1 b0  a1     1 0 ··· 0     e 2   b1   a2     a1 1 · · · 0   e 3  =  b2  (22.6)    .. .. .. . . ..   ..   ..   . . .  .   .  . . an−1 an−2 · · · a1 1 en bn−1

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The matrix is always invertible, and the system of equations can be solved, yielding the e vector. • It is easy to see that it can be solved recursively, from e1 through en as e 1 = b0 e2 = b1 − a1 e1 e3 = b2 − a1 e2 − a2 e1 .. .. . . ek = bk−1 − a1 ek−1 − a2 ek−2 − · · · − ak−1 e1 • Compute the new initial conditions at t = 0+ : y(0+ ) = y(0− ) + e1 y(0 ˙ + ) = y(0 ˙ − ) + e2 y [2] (0+ ) = y [2] (0− ) + e3 .. .. .. . . . y [k] (0+ ) = y [k] (0− ) + ek+1 .. .. .. . . . y [n−1] (0+ ) = y [n−1] (0− ) + en • Given these “new” initial conditions at t = 0+ , we can proceed with the solution of y [n] (t) + a1 y [n−1] (t) + a2 y [n−2] (t) + · · · + an y(t) = bn

(22.7)

with the initial conditions at 0+ as shown in previous chapters. Note that since we are interested in t ≥ 0+ , the right-hand side (remember u is a unit-step input at t = 0), is simply bn . • Note that in Matlab one can compactly solve the e quantities by using the command “A\b” where   1 0 0 ··· 0  a1 1 0 ··· 0     0  a1 1 ··· 0  A =  a2  , B = b0 b1 b2 · · · bn−1 bn  .. .. .. . . ..   . . .  . . an−1 an−2 · · · a1 1

ME 132, Fall 2018, UC Berkeley, A. Packard

22.4

227

Problems

1. Consider a system with input u, and output y governed by the differential equation y(t) ˙ + a1 y(t) = u(t) ˙ + b1 u(t)

(22.8)

Find the unit-step response with initial conditions y(0− ) = 0. Compare to answer (should be the same) as that obtained in problem 12 in section 5.6. 2. Consider y [3] (t) + 3.8¨ y (t) + 6.8y(t) ˙ + 4y(t) = b2 u(t) ˙ + 3u(t) subject to the forcing function u(t) := µ(t), and initial conditions y(0− ) = 0, y(0 ˙ −) = 0, y¨(0− ) = 0. Follow the same instructions as in problem 1 in Section 21, handling the 7 cases b2 = −6, −3, −0.3, −0.03, 0.03, 0.3, 3, 6. Since this is a higher order problem, you will need to also determine y¨(0+ ). Hint: One of the roots of the characteristic equation is −1. Also, if you proceed symbolically, you end up with the coefficients of the homogeneous components being of the form c = M −1 v(b2 ) where M is a 3 × 3 matrix made up of the three roots of the characteristic polynomial, and v is a 3 × 1 vector that depends on b2 . On paper, leave it as that (don’t bother computing the inverse). Then, for each of the 6 cases, plug in a particular value for b2 , and let MatLab compute the coefficients automatically. Set up your plotting script file to accept a 3 × 1 vector of homogeneous coefficients. In your solutions, include any useful M-files that you write. 3. Consider the three differential equations y [4] (t) + 5.8y [3] (t) + 14.4¨ y (t) + 17.6y(t) ˙ + 8y(t) = u(t) [4] [3] y (t) + 5.8y (t) + 14.4¨ y (t) + 17.6y(t) ˙ + 8y(t) = 2¨ u(t) + 2u(t) ˙ + u(t) [4] [3] y (t) + 5.8y (t) + 14.4¨ y (t) + 17.6y(t) ˙ + 8y(t) = 2¨ u(t) − 2u(t) ˙ + u(t) Suppose that each is subject to the forcing function u(t) := µ(t), and initial conditions y(0− ) = 0, y(0 ˙ − ) = 0, y¨(0− ) = 0, y [3] (0− ) = 0. Compute the roots (hint: one at −1, one at −2), get final value of y(t), compute “new” conditions of y (and derivatives) at 0+ , and sketch solutions. Then, derive the exact expression for the solutions, and plot using MatLab.

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228

Transfer functions

Associated with the linear system (input u, output y) governed by the ODE y [n] (t) + a1 y [n−1] (t) + · · · + an−1 y [1] (t) + an y(t) = b0 u[n] (t) + b1 u[n−1] (t) + · · · + bn−1 u[1] (t) + bn u(t)

(23.1)

we write “in transfer function form” Y =

b0 sn + b1 sn−1 + · · · + bn−1 s + bn U sn + a1 sn−1 + · · · + an−1 s + an

(23.2)

The expression in (23.2) is interpreted to be equivalent to the ODE in (23.1), just a different way of writing the coefficients. The notation in (23.2) is suggestive of multiplication, and we will see that such an interpretation is indeed useful. The function G(s) :=

b0 sn + b1 sn−1 + · · · + bn−1 s + bn sn + a1 sn−1 + · · · + an−1 s + an

is called the transfer function from u to y, and is sometimes denoted Gu→y (s) to indicate this. At this point, the expression in equation (23.2), Y = Gu→y (s)U is nothing more than a new notation for the differential equation in (23.1). The differential equation itself has a well-defined meaning, and we understand what each term/symbol (derivative, multiplication, sum) represents, and the meaning of the equality sign, =. By contrast, in the transfer function expression, (23.2), there is no specific meaning to the individual terms, or the equality symbol. The expression, as a whole, simply means the differential equation to which it is associated. Nevertheless, in this section, we will see that, in fact, we can assign proper equality, and make algebraic substitutions and manipulations of transfer function expressions, which will aid our manipulation of linear differential equations. But all of that requires proof, and that is the purpose of this section.

23.1

Linear Differential Operators (LDOs)

Note that in the expression (23.2), the symbol s plays the role of dtd , and higher powers of s k mean higher order derivatives, ie., sk means dtd k . If z is a function of time, let the notation   dn dn−1 d dn z dn−1 z dz b0 n + b1 n−1 + · · · + bn−1 + bn (z) := b0 n + b1 n−1 + · · · + bn−1 + bn z dt dt dt dt dt dt

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We will call this type of operation a linear differential operation, or LDO. For the purposes of this section, we will denote these by capital letters, say h n i n−1 L := dtd n + a1 dtd n−1 + · · · + an−1 dtd + an i h n dn−1 d d R := b0 dtn + b1 dtn−1 + · · · + bn−1 dt + bn Using this shorthand notation, we can write the original ODE in (23.5) as L(y) = R(u) With each LDO, we naturally associate a polynomial. Specifically, if  n  d dn−1 d L := + a1 n−1 + · · · + an−1 + an dtn dt dt then pL (s) is defined as pL (s) := sn + a1 sn−1 + · · · + an−1 s + an Similarly, with each polynomial, we associate an LDO – if q(s) := sm + b1 sm−1 + · · · + bm−1 s + bm then Lq is defined as dm dm−1 d + b + · · · + bm−1 + bm Lq := 1 m m−1 dt dt dt 



Therefore, if a linear system is governed by an ODE of the form L(y) = R(u), then the transfer function description is simply Y =

pR (s) U pL (s)

Similarly, if the transfer function description of a system is V = then the ODE description is Ld (v) = Ln (w).

n(s) W d(s)

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23.2

230

Algebra of Linear differential operations

Note that two successive linear differential operations can be done in either order. For example let   2 d d L1 := +5 +6 dt2 dt and

 d3 d2 d L2 := −2 2 +3 −4 dt3 dt dt 

Then, on a differentiable signal z, simple calculations gives h 2 i h 3 i i d d2 d L1 (L2 (z)) = dtd 2 + 5 dtd + 6 − 2 + 3 − 4 (z) 3 dt2 dt h 2 i dt
= dtd 2 + 5 dtd + 6 z [3] − 2¨ z + 3z˙ − 4z = z [5] − 2z [4] + 3z [3] − 4z [2] 5z [4] − 10z [3] + 15z [2] − 20z [1] 6z [3] − 12z [2] + 18z [1] − 24z [5] [4] [3] = z + 3z − z − z [2] − 2z [1] − 24z which is the same as L2 (L1 (z)) =

h

d3 dt3

=

h

d3 dt3 [5]

i h 2 i i 2 d d − 2 dtd 2 + 3 dtd − 4 + 5 + 6 (z) 2 dt i dt
2 − 2 dtd 2 + 3 dtd − 4 z [2] + 5z˙ + 6z

+ 5z [4] + 6z [3] −2z [4] − 10z [3] − 12z [2] z [3] + 15z [2] + 18z [1] −4z [2] − 20z [1] − 24z [5] [4] [3] [2] = z + 3z − z − z − 2z [1] − 24z

= z

This equality is easily associated with the fact that multiplication of polynomials is a commutative operation, specifically (s2 + 5s + 6) (s3 − 2s2 + 3s − 4) = (s3 − 2s2 + 3s − 4) (s2 + 5s + 6) = s5 + 3s4 − s3 − s2 − 2s + 24 We will use the notation [L1 ◦ L2 ] to denote this composition of LDOs. The linear differential operator L1 ◦ L2 is defined as operating on an arbitrary signal z by [L1 ◦ L2 ] (z) := L1 (L2 (z)) Similarly, if L1 and L2 are LDOs, then the sum L1 + L2 is an LDO defined by its operation on a signal z as [L1 + L2 ] (z) := L1 (z) + L2 (z).

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It is clear that the following manipulations are always true for every differentiable signal z, L (z1 + z2 ) = L (z1 ) + L (z2 ) and [L1 ◦ L2 ] (z) = [L2 ◦ L1 ] (z) In terms of LDOs and their associated polynomials, we have the relationships p[L1 +L2 ] (s) = pL1 (s) + pL2 (s) p[L1 ◦L2 ] (s) = pL1 (s)pL2 (s) In the next several subsections, we derive the LDO representation of an interconnection from the LDO representation of the subsystems.

23.3

Feedback Connection

The most important interconnection we know of is the basic feedback loop. It is also the easiest interconnection for which we derive the differential equation governing the interconnection from the differential equation governing the components. Consider the simple unity-feedback system shown below in Figure 8 + d u−6

r

S

-y

Figure 8: Unity-Feedback interconnection Assume that system S is described by the LDO L(y) = N (u). The feedback interconnection yields u(t) = r(t) − y(t). Eliminate u by substitution, yielding an LDO relationship between r and y L(y) = N (r − y) = N (r) − N (y) This is rearranged to the closed-loop LDO (L + N )(y) = N (r). That’s a pretty simple derivation. Based on the ODE description of the closed-loop, we can immediately write the closed-loop transfer function, Y

pN (s) R p[L+N ] (s) pN (s) = R. pL (s) + pN (s) =

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Additional manipulation leads to further interpretation. Let G(s) denote the transfer func(s) tion of S, so G = ppNL (s) . Starting with the transfer function expression we just derived, we can formally manipulate to an easily recognizable expression. Specifically Y

=

=

=

pN (s) R pL (s) + pN (s)

1

pN (s) pL (s) R (s) + ppNL (s)

G(s) R 1 + G(s)

This can be interpreted rather easily. Based on the original system interconnection, redraw, replacing signals with their capital letter equivalents, and replacing the system S with its transfer function G. This is shown below. r

+ d u−6

-y

S

R

+ d U−6

G

-Y

The diagram on the right is interpreted as a diagram of the equations U = R − Y , and Y = GU . Note that manipulating these as though they are arithmetic expressions gives Y = G(R − Y ) after substituting for U (1 + G)Y = GR moving GY over to left − hand − side G R solving for Y. Y = 1+G This is is precisely what we want!

23.4

More General Feedback Connection

Consider the more general single-loop feedback system shown below. For diversity, we’ll use a positive-feedback convention here. r

+d +6

S1

y -

q

v

S2 

Assume that system S1 is described by the LDO L1 (y) = N1 (u), and S2 is governed by L2 (v) = N2 (q). Note that by definition then, the transfer function descriptions of S1 and S2

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are Y =

pN1 U, pL1 |{z}

233

pN 2 Q pL2 |{z}

V =

:=G1

:=G2

The interconnection equations are u = r+v, and q = y. In order to eliminate v (for example), first substitute for u and q, leaving L1 (y) = N1 (r + v) L2 (v) = N2 (y)

governing equation for S1 governing equation for S2

Now apply L2 to first equation, and N1 to second equation, yielding (after using the ◦ notation for composing linear differential operators), L2 ◦ L1 (y) = L2 ◦ N1 (r) + L2 ◦ N1 (v) N1 ◦ L2 (v) = N1 ◦ N2 (y) The expressions involving v cancel, leaving [L2 ◦ L1 − N1 ◦ N2 ] (y) = L2 ◦ N1 (r) In transfer function terms, this is Y =

pL2 ◦N1 pL2 ◦L1 −N1 ◦N2

R

which is easily rewritten as

pL2 pN1 R pL2 pL1 − pN1 pN2 But, again, this can be formally manipulated into a recognizable expression involving the the individual transfer functions of S1 and S2 . Divide top and bottom of the transfer function by pL1 pL2 , to give Y =

Y =

1

=

pL2 pN1 pL2 pL1 p p − pNL1 pNL 2 2 1

R

G1 R 1 − G2 G1

Again, this can be interpreted rather easily. Based on the original system interconnection, redraw, replacing signals with their capital letter equivalents, and replacing the systems Si with their transfer functions Gi . This is shown below. r

+d +6

S1

y -

q

v

S2 

R

+d +6

G1

Y Q

V G2 

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The diagram on the right is interpreted as a diagram of the equations U = R − G2 Y , and Y = G1 U . Note that manipulating these as though they are arithmetic expressions gives Y = G1 (R + G2 Y ) after substituting for U (1 − G1 G2 )Y = G1 R moving G1 G2 Y over to left − hand − side G1 R solving for Y. Y = 1−G 2 G1 This is is precisely what we want!

23.5

Cascade (or Series) Connection

Suppose that we have two linear systems, as shown below,

u -

S1

y -

S2

v-

with S1 governed by y [n] (t) + a1 y [n−1] (t) + · · · + an y(t) = b0 u[n] (t) + b1 u[n−1] (t) + · · · + bn u(t) and S2 governed by v [m] (t) + c1 v [m−1] (t) + · · · + cm v(t) = d0 y [m] (t) + d1 y [m−1] (t) + · · · + dm y(t) Let G1 (s) denote the transfer function of S1 , and G2 (s) denote the transfer function of S2 . Define the differential operations  n  d dn−1 d L1 := + a1 n−1 + · · · + an−1 + an dtn dt dt   dn dn−1 d R1 := b0 n + b1 n−1 + · · · + bn−1 + bn dt dt dt and  m  d dm−1 d L2 := + c1 m−1 + · · · + cm−1 + cm dtm dt dt   dm dm−1 d R2 := d0 m + d1 m−1 + · · · + dm−1 + dm dt dt dt Hence, the governing equation for system S1 is L1 (y) = R1 (u), while the governing equation for system S2 is L2 (v) = R2 (y). Moreover, in terms of transfer functions, we have G1 (s) =

pR1 (s) , pL1 (s)

G2 (s) =

pR2 (s) pL2 (s)

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Now, apply the differential operation R2 to the first system, leaving R2 (L1 (y)) = R2 (R1 (u)) Apply the differential operation L1 to system 2, leaving L1 (L2 (v)) = L1 (R2 (y)) But, in the last section, we saw that two linear differential operations can be applied in any order, hence L1 (R2 (y)) = R2 (L1 (y)). This means that the governing differential equation for the cascaded system is L1 (L2 (v)) = R2 (R1 (u)) which can be rearranged into L2 (L1 (v)) = R2 (R1 (u)) or, in different notation [L2 ◦ L1 ] (v) = [R2 ◦ R1 ] (u) In transfer function form, this means V

=

p[R2 ◦R1 ] (s) U p[L2 ◦L1 ] (s)

=

pR2 (s)pR1 (s) U pL2 (s)pL1 (s)

= G2 (s)G1 (s)U Again, this has a nice interpretation. Redraw the interconnection, replacing the signals with the capital letter equivalents, and the systems by their transfer functions.

u -

S1

y -

S2

v-

U -

G1

Y-

G2

V-

The diagram on the right depicts the equations Y = G1 U , and V = G2 Y . Treating these as arithmetic equalities allows substitution for Y , which yields V = G2 G1 U , as desired. Example: Suppose S1 is governed by y¨(t) + 3y(t) ˙ + y(t) = 3u(t) ˙ − u(t)

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and S2 is governed by v¨(t) − 6v(t) ˙ + 2v(t) = y(t) ˙ + 4y(t) Then for S1 we have   2 d d +3 +1 , L1 = dt2 dt while for S2 we have  2  d d L2 = −6 +2 , dt2 dt



 d R1 = 3 − 1 , dt

G1 (s) =

 d R2 = +4 , dt

3s − 1 s2 + 3s + 1



G2 (s) =

s2

s+4 − 6s + 2

The product of the transfer functions is easily calculated as G(s) := G2 (s)G1 (s) =

3s2 + 11s − 4 s4 − 3s3 − 15s2 + 2

so that the differential equation governing u and v is v [4] (t) − 3v [3] (t) − 15v [2] (t) + 2v(t) = 3u[2] (t) + 11u[1] (t) − 4u(t) which can also be verified again, by direct manipulation of the ODEs.

23.6

Parallel Connection

Suppose that we have two linear systems, as shown below,

-

S1

u -

S2

y1 +d? + 6

y

y2

System S1 is governed by [n]

[n−1]

y1 (t) + a1 y1

(t) + · · · + an y1 (t) = b0 u[n] (t) + b1 u[n−1] (t) + · · · + bn u(t)

and denoted as L1 (y1 ) = R1 (u). Likewise, system S2 is governed by [m]

[m−1]

y2 (t) + c1 y2

(t) + · · · + cm y2 (t) = d0 u[m] (t) + d1 u[m−1] (t) + · · · + dm u(t)

and denoted L2 (y2 ) = R2 (u).

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Apply the differential operation L2 to the governing equation for S1 , yielding L2 (L1 (y1 )) = L2 (R1 (u))

(23.3)

Similarly, apply the differential operation L1 to the governing equation for S2 , yielding and L1 (L2 (y2 )) = L1 (R2 (u)) But the linear differential operations can be carried out is either order, hence we also have L2 (L1 (y2 )) = L1 (L2 (y2 ))

(23.4)

Add the expressions in (23.3) and (23.4), to get L2 (L1 (y)) = = = = =

L2 (L1 (y1 + y2 )) L2 (L1 (y1 )) + L2 (L1 (y2 )) L2 (R1 (u)) + L1 (R2 (u)) [L2 ◦ R1 ] (u) + [L1 ◦ R2 ] (u) [L2 ◦ R1 + L1 ◦ R2 ] (u)

In transfer function form this is Y

=

p[L2 ◦R1 +L1 ◦R2 ] (s) U p[L2 ◦L1 ] (s)

=

p[L2 ◦R1 ] (s) + p[L1 ◦R2 ] (s) U pL2 (s)pL1 (s)

pL2 (s)pR1 (s) + pL1 (s)pR2 (s) U pL2 (s)pL1 (s)   pR1 (s) pR2 (s) + U = pL1 (s) pL2 (s) =

= [G1 (s) + G2 (s)] U So, the transfer function of the parallel connection is the sum of the individual transfer functions. This is extremely important! The transfer function of an interconnection of systems is simply the algebraic gain of the closed-loop systems, treating individual subsystems as complex gains, with their “gain” taking on the value of the transfer function.

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23.7

238

General Connection

The following steps are used for a general interconnection of systems, each governed by a linear differential equation relating their inputs and outputs. • Redraw the block diagram of the interconnection. Change signals (lower-case) to upper case, and replace each system with its transfer function. • Write down the equations, in transfer function form, that are implied by the diagram. • Manipulate the equations as though they are arithmetic expressions. Addition and multiplication commute, and the distributive laws hold.

23.8

Systems with multiple inputs

Associated with the multi-input, single-output linear ODE L(y) = R1 (u) + R2 (w) + R3 (v) we write Y =

pR (s) pR (s) pR1 (s) U+ 2 W+ 3 V pL (s) pL (s) pL (s)

(23.5)

(23.6)

This may be manipulated algebraically.

23.9

Poles and Zeros of Transfer Functions

Consider to Linear Differential operators, L and R, and the input/output system described by L(y) = R(u). Associated with L and R are polynomials, pL (s) and pR (s), and the transfer (s) function of the system is G := ppRL (s) . As we already know, the roots of pL (s) = 0 give us complete information regarding the form of homogeneous solutions to the differential equation L(y) = 0. These roots, which are roots of the denominator of the transfer function G are called the poles of the transfer function. The zeros of the transfer function are defined as the roots of the numerator, in other words, roots of pR (s). Obviously, these roots yield complete information regarding homogeneous solutions of the differential equation R(u) = 0. In sections 21 and 22, we saw the importance of the right-hand-side of the differential equation on the forced response. Later we will learn how to interpret this in terms of the number and location of the transfer function zeros.

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23.10

239

Problems

1. Consider the feedback interconnection shown below. The signals r and w represent a reference and disturbance input. r

+e 6 u −6 v

y

S1

-

? e+  w S2  q +



v

The system S1 is described by the LDO L1 (y) = N1 (u), and S2 is governed by L2 (v) = N2 (q). The interconnection equations are u = r − v and q = y + w. (a) Find the differential equation relating (r and w) to y. The ODE will involve sums/differences/compositions of the Linear Differential Operators L1 , N1 , L2 , N2 . (b) Find the differential equation relating (r and w) to v. The ODE will involve sums/differences/compositions of the Linear Differential Operators L1 , N1 , L2 , N2 . (c) Find the differential equation relating (r and w) to u. The ODE will involve sums/differences/compositions of the Linear Differential Operators L1 , N1 , L2 , N2 . What is the characteristic equation in each case? Note, in fact, that the characteristic equation is the same for all cases, whether the equations are written as y as influenced by (r, w), or v as influenced by (r, w), or u as influenced by (r, w). 2. Find the transfer function from u to y for the systems governed by the differential equations (a) y(t) ˙ =

1 τ

[u(t) − y(t)]

(b) y(t) ˙ + a1 y(t) = b0 u(t) ˙ + b1 u(t) (c) y(t) ˙ = u(t) (explain connection to Simulink icon for integrator...) (d) y¨(t) + 2ξωn y(t) ˙ + ωn2 y(t) = ωn2 u(t) 3. Consider the interconnection below. The transfer functions of systems S1 and S2 are G1 (s) =

3 , s+6

G2 (s) =

s+2 s+1

Determine the differential equation governing the relationship between u and y. u -

S1

-

S2

y -

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4. (a) F, G, K and H are transfer functions of systems. A block diagram of an interconnection is shown below. The input r and output y are labeled with their corresponding capital letters. Find the transfer function from r to y. R F

+d +6

G

-

H

Y -

K  (b) For a single-loop feedback system, a rule for determining the closed-loop transfer function from an specific input to a specific output is forward path gain 1 − loop gain Explain how part 4a above is a “proof” of this fact. 5. (a) F, G, K and H are transfer functions of systems. A block diagram of an interconnection is shown below. The input r and output y are labeled with their corresponding capital letters. Find the transfer function from r to y. R F

+d −6

G

-

H

Y -

K  (b) For a single-loop feedback system, with negative feedback, a rule for determining the closed-loop transfer function from an specific input to a specific output is forward path gain 1 + Loop gain Explain how part 5a above is a “proof” of this fact. 6. G and K are transfer functions of systems. A block diagram of an interconnection is shown below. The input r and output y are labeled with their corresponding capital letters. Find the transfer function from r to y, and express it in terms of NG , DG , NK , DK , the numerators and denominators of the transfer functions G and K. Be sure to “clear” all fractions, cancelling common terms. +d R G −6 K 

Y -

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Remark: Why the cancellation? Recall the derivation of the transfer function for a feedback loop in section 23.3. Simple manipulations reveal the LDO relationship, N (r) = (L+N )(y), which is immediately translated into a transfer function relationship pN (s) R pN (s) + pL (s)

Y =

The final step was “artificial”, dividing both numerator and denominator by additional common factor, pL , so as to get the desired expression Y =

1

pN (s) pL (s) R (s) + ppNL (s)

=

G R 1+G

Reversing this last step simply requires clearing the fraction, and canceling identical common factors. 7. A feedback connection of 4 systems is shown below. Let the capital letters also denote the transfer functions of each of the systems. F  +? +d

r -d E −6

-

H

y -

G  (a) Break this apart as shown below. d +? +d

r -d E −6

x -

G  What is the transfer function from r and d to x? Call your answers G1 and G2 . (b) Now, draw the overall system as r

-

G1

-

G2 F 

+- d H +6

-y

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In terms of G1 , G2 and F and H, what is the transfer function from r to y? Substitute for G1 and G2 , and get the transfer function from r to y in terms of the original subsystems. (c) In terms of numerators and denominators of the individual transfer functions NG (s) , for example), what is the characteristic equation of the closed(G(s) = D G (s) loop system? 8. (a) Suppose that the transfer function of a controller, relating reference signal r and measurement y to control signal u is U = C(s) [R − Y ] Suppose that the plant has transfer function relating control signal u and disturbance d to output y as Y = G3 (s) [G1 (s)U + G2 (s)D] Draw a simple diagram, and determine the closed-loop transfer functions relating r to y and d to y. (b) Carry out the calculations for C(s) = KP +

KI , s

G1 (s) =

E , τs + 1

G2 (s) = G,

G3 (s) =

1 ms + α

Directly from this closed-loop transfer function calculation, determine the differential equation for the closed-loop system, relating r and d to y. (c) Given the transfer functions for the plant and controller in (8b), i. Determine the differential equation for the controller, which relates r and y to u. ii. Determine the differential equation for the plant, which relates d and u to y. iii. Combining these differential equations, eliminate u and determine the closedloop differential equation relating r and d to y. 9. Find the transfer function from e to u for the PI controller equations z(t) ˙ = e(t) u(t) = KP e(t) + KI z(t) 10. Suppose that the transfer function of a controller, relating reference signal r and measurement ym to control signal u is U = C(s) [R − YM ]

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Suppose that the plant has transfer function relating control signal u and disturbance d to output y as Y = [G1 (s)U + G2 (s)D] Suppose the measurement ym is related to the actual y with additional noise (n), and a filter (with transfer function F ) YM = F (s) [Y + N ] (a) Draw a block diagram (b) In one calculation, determine the 3 closed-loop transfer functions relating inputs r, d and n to the output y. (c) In one calculation, determine the 3 closed-loop transfer functions relating inputs r, d and n to the control signal u. 11. A first order system has a transfer function G(s) =

γ τs + 1

(a) What is the differential equation relating the input and output? (b) Under what conditions is the system stable? (c) If the system is stable, what is the time-constant of the system? 12. Assume G1 , G2 and H are transfer functions of linear systems. (a) Compute the transfer function from R to Y in the figure below. R+ j - G 1 −6 H -

Y -



+ ? j

− 6

G2

(b) Suppose that the transfer functions are given in terms of numerator and denominator pairs, so N1 N2 NH G1 = , G2 = , H= D1 D2 DH where all of the N and D are polynomials. Assume each denominator is of higher order than its associated numerator. Carefully express the transfer function from R to Y in terms of the individual numerators and denominators.

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(c) What is the characteristic equation of the closed-loop system? Be sure that its order is the sum of the individual orders of the the 3 subsystems. 13. Consider a plant P with transfer function P (s) =

s s2 − 1

The plant is a challenging to control. It is a simple model of an inverted pendulum, where the only control action u, represents the angular speed of a reaction wheel, mounted at the free-end (top) of the pendulum. By “torquing” the reaction wheel (relative to the pendulum), equal/opposite torque act on the pendulum, which if properly done, can stabilize it in the “up” position. (a) Suppose the input to P is labeled u and the output y. What is the differential equation governing the relationship between u and y. (b) Is the plant stable? (c) Consider a controller, C, with transfer function C(s) =

as + b s+c

For the feedback loop consisting of P and C (with the usual negative (−) feedback convention), what is the closed-loop characteristic equation? (d) Design: Find a, b and c (ie., parameters of the control law) so that the roots of the closed-loop characteristic equation are −4, −2 ± j2 (e) The controller stabilizes P , in that the feedback loop consisting of P and C is stable. However, as we saw earlier, P is not stable on its own. Is C stable? (f) Assume we relax the choice of the desired closed-loop roots, and only restrict them to all have negative real-parts (since we want the closed-loop system to be stable). Are there any choice of such desired roots so that the controller (which yields those closed-loop roots) is itself a stable system? 14. Read about the command tf in Matlab. Use the HTML help (available from the menubar, under Help, as well as the command-line (via >> help tf). 15. Read about the command step in Matlab. Use the HTML help (available from the menubar, under Help, as well as the command-line (via >> help step). 16. Execute the following commands

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>> >> >> >>

245

sys1 = tf(1,[1 3.8 6.8 4]) sys2 = tf([3 1],[1 3.8 6.8 4]) sys3 = tf([-3 1],[1 3.8 6.8 4]) step(sys1,sys2,sys3)

Relate these to problem 2 in Section 22.4. 17. Execute the following commands >> >> >> >>

sys1 = tf(1,[1 5.8 14.4 17.6 8]) sys2 = tf([2 2 1],[1 5.8 14.4 17.6 8]) sys3 = tf([2 -2 1],[1 5.8 14.4 17.6 8]) step(sys1,sys2,sys3)

Relate these to problem 3 in Section 22.4. 18. Execute the commands >> >> >> >> >> >> >>

sys1.den class(sys1.den) sys1.den{1} class(sys1.den{1}) size(sys1.den{1}) roots(sys1.den{1}) pole(sys1)

Explain what is being referenced and displayed. Recall that the “poles” of a transfer function are the roots of the associated characteristic polynomial.

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24

246

Arithmetic of Feedback Loops

Many important guiding principles of feedback control systems can be derived from the arithmetic relations, along with their sensitivities, that are implied by the figure below.

Process to be controlled d

Controller r

- g −6

e

-

C

u

-

H

-

G

- ? g

-y

?

S yf

F



ym

? g

n

Sensor Filter The analysis in this section is oversimplified, and at a detail-oriented level, not realistic. Nevertheless, the results we derive will reappear throughout the course (in more precise forms) as we introduce additional realism and complexity into the analysis. In this diagram, • lines represent variables, and • rectangular block represent operations that act on variables to produce a transformed variable. Here, r, d and n are independent variables. The variables e, u, y, ym , yf are dependent, being generated (caused) by specific values of (r, d, n). The blocks with upper-case letters represent multiplication operations, namely that the input variable is transformed into the output variable via multiplication by the number represented by the upper case letter in the block. For instance, the block labeled “Filter” indicates that the variables ym and yf are related by yf = F ym . Each circle represents a summing junction, where variables are added (or subtracted) to yield an output variable. Addition is always implied, with subtraction explicitly denoted by a negative sign (−) next to the variable’s path. For example, the variables r, e and yf are related by e = r − yf .

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Each summing junction and/or block can be represented by an equation which relates the inputs and outputs. Writing these (combining some steps) gives e = r − yf u = Ce y = Gu + Hd ym = Sy + n yf = F ym

generate the regulation error control strategy process behavior sensor behavior filtering the measurement

(24.1)

Each block (or group of blocks) has a different interpretation. Process: The process-to-be-controlled has two types of input variables: a freely manipulated control input variable u, and an unknown disturbance input variable d. These inputs affect the output variable y. As users of the process, we would like to regulate y to a desired value. The relationship between (u, d) and y is y = Gu + Hd. G is usually considered known, but with some modest potential error. Since d is unknown, and G is slightly uncertain, the relationship between u and y is not perfectly known. Controller: The controller automatically determines the control input u, based on the difference between the desired value of y, which is the reference input r, and the filtered measurement of the actual value of y. Sensor: Measured variable is the noisy output of another system, called the sensor. Like the process, the sensor is also subjected to external disturbances. Because these corrupt the output of the sensor, which is supposed to represent the process output variable, the disturbance to the sensor is often called noise. Filter: Electrical/Mechanical/Computational element to separate (as best as possible) the noise n from the signal ym . The goal (unattainable) of feedback (the S, F and C) is: for all reasonable (r, d, n), make y ≈ r, independent of d and n, and this behavior should be resilent to modest/small changes in G (once C is fixed). Note that there is a cycle in the cause/effect relationships - specifically, starting at yf we have r, yf cause e e causes u u, d cause y y, n cause ym ym causes yf This is called a feedback loop, and can be beneficial and/or detrimental. For instance,

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• Note that d (and u) also affects y, and through the the feedback loop, ultimately affects u, which in turn again affects y. So, although u explicitly only depends on e, through the feedback loop the control action, u, may in actuality compensate for disturbances d. • However, through the feedback loop, y is affected by the imperfection to which it is measured, n. Eliminating the intermediate variables (such as e, yf and ym ) yields the explicit dependence of y on r, d, n. This is called the closed-loop relationship. y =

H −GCF GC r + d + n |1 + GCF |1 + GCF |1 + GCF {z S} {z S} {z S} (r→y)CL

(d→y)CL

(24.2)

(n→y)CL

Note that y is a linear function of the independent variables (r, d, n), but a nonlinear function of the various component behaviors (the G, H, F , C, etc). Each term which multiplies one of the external variables is called a closed-loop gain, and the notation for a closed-loop gain is given. Now, in more mathematical terms, the goals are: 1. Make the magnitude of (d → y)CL significantly smaller than the uncontrolled effect that d has on y, which is H. 2. Make the magnitude of (n → y)CL “small,” relative to S1 . 3. Make (r → y)CL gain approximately equal to 1 4. Generally, behavior should be insensitive to G. Implications • Goal 1 implies H 1 + GCF S 1.

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• Goal 2 implies that any noise injected at the sensor output should be significantly attenuated at the process output y (with proper accounting for unit changes by S). This requires GCF 1 1 + GCF S 1.

(24.3)

On the other hand, if Goal 2 is satisfied, then |GCF S| is small (relative to 1), so GC ≈ GC 1 + GCF S Therefore, the requirement of Goals 2 and 3 are |F S| 13 . Note that if |GCF S| >> 1 or |GCF S| > 1), the quotient made about equal to the Noise-to-Signal. The lesson is that if the disturbance is very large relative to the noise, then feedback should be used, and the effect of the disturbance can be reduced, but ratio of noise-to-disturbance limits the overall reduction possible.

24.3

What’s missing?

The most important thing missing from the analysis above is that the relationships are not in fact constant multiplications, and the special nature of the types of signals. Nevertheless, many, if not all, of the ideas presented here will be applicable even in the more general setting.

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24.4

252

Problems

1. (a) For complex numbers A and B, it is a fact that |A + B| ≤ |A| + |B|. This is called the triangle inequality. Draw a picture (in the complex plane) to illustrate this in a few examples. (b) Using the triangle inequality, show that for any complex numbers A and B, |A ± B| ≥ |A| − |B| What is the simple reason that |A ± B| ≥ |B| − |A| is also true? 2. Suppose L is a complex number, and 0 <  < 1. (a) Show that 1 1 + L ≤ 

1 −1 

⇒

|L| ≥

⇒

1 1 + L ≤ 

(b) Show that 1 |L| ≥ + 1  (c) If  is very small compared to 1, then 1 1 1 −1≈ +1≈    Combine the first two parts of this problem into the following “quasi”-theorem: If L is a complex number, and 0 <  0 (since complex roots come in conjugate pairs, if they cross the imaginary axis, one will cross positively, and one negatively - we can focus on one or the other, and for no specific reason, we look at the positive crossing). Hence, in assessing the gain margin, we are looking for values γ and β, with γ close to 1, and β ≥ 0 such that λ = jβ is a root of pγ . Plugging in gives (jβ)n + (a1 + γb1 )(jβ)n−1 + (a2 + γb2 )(jβ)n−2 + · · · + (an−1 + γbn−1 )(jβ) + (an + γbn ) = 0 Rearranging gives [(jβ)n + a1 (jβ)n−1 + a2 (jβ)n−2 + · · · + an−1 (jβ) + an ] . + γ [b1 (jβ)n−1 + b2 (jβ)n−2 + · · · + bn−1 (jβ) + bn ] = 0

(25.2)

This is two equations (since it relates complex quantities) in two real unknowns (β and γ). We need to find all solutions, and then isolate the solutions whose γ coordinate is closest to 1. First, we can look for solutions to (25.2) by looking for imaginary-axis roots (ie., real β) to pγ=0 (jβ) = 0, namely (jβ)n + a1 (jβ)n−1 + a2 (jβ)n−2 + · · · + an−1 (jβ) + an = 0. If there are roots, then coupled with γ = 0, we have found some solutions to equation (25.2). There likely are others, with γ 6= 0, which are described below. We can also look for solutions to (25.2) with (jβ)n + a1 (jβ)n−1 + a2 (jβ)n−2 + · · · + an−1 (jβ) + an 6= 0, so that after dividing, we manipulate equation (25.2) into −1 = γ

b1 (jβ)n−1 + b2 (jβ)n−2 + · · · + bn−1 (jβ) + bn (jβ)n + a1 (jβ)n−1 + a2 (jβ)n−2 + · · · + an−1 (jβ) + an

But, this is recognized as ˆ −1 = γ L(s)

s=jβ

ˆ This can be solved graphically, finding values of β for which L(s)

is a real number (and s=jβ

then taking a negative reciprocal to find the associated γ value). Summarizing, a simple method to determine all solutions to equation (25.2) is 1. Let βi denote all (if any) of the real solutions to (jβ)n + a1 (jβ)n−1 + a2 (jβ)n−2 + · · · + an−1 (jβ) + an = 0. If there are any, then the pairs (βi , 0) are solutions to equation 25.2.

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2. Next, plot (in the complex plane, or on separate magnitude/phase plots) the value of ˆ L(jβ) as β ranges from 0 to ∞. By assumption, this plot will not pass through the −1 point in the complex plane (can you explain why this is the case?). ˆ 3. Mark all of the locations, {β1 , β2 , . . . , βk }, where L(jβ) is a real number. The frequencies βk are called the phase-crossover frequencies. Sometimes, the terminology ˆ phase-crossover frequencies only refers to frequencies where where L(jβ) is a negative real number. 4. At each phase crossover point, βk , determine the associated value of γ by computing γk :=

−1 ˆ L(s)

s=jβk

5. Make a list of all of the γ values obtained in the calculations above (step 1 and 4). For all solutions with γk < 1, find the one closest to 1. Label this γ. Similarly, for all solutions with γk > 1, find the one closest to 1. Label this γ¯ . In the event that there are no solutions less than 1, then γ = −∞. Similarly, if there are no solutions greater than 1, then γ¯ = ∞. Graphically, we can determine the quantity γ¯ easily. Find the smallest number > 1 such that if the plot of L(jβ) were scaled by this number, it would intersect the −1 point. That ˆ is γ¯ . This is easily done by computing the closest intersection of the curve L(jβ) (as β varies) with the real axis, to the right of −1, but less than 0. If the intersection occurs at ˆ by 1 will cause the intersection. By choosing the location −α, then scaling the plot of L α the closest intersection, α is “close” to 1, and hence γ¯ := α1 is the desired value. Note: It ˆ is possible that there are no intersections of the curve L(jβ) of the real axis to the right of −1 but less than 0. In that case, the closed-loop system is stable for all values of γ in the interval [1, ∞), hence we define γ¯ := ∞. ˆ If the plot of L(jβ) as β varies intersects the negative real axis to the left of −1, then γ > 0, and it is easy to determine. Simply find the largest positive number < 1 such that if the plot ˆ of L(jβ) were scaled by γ, it would intersect the −1 point. This is easily done by computing ˆ the closest intersection of the curve L(jβ) (as β varies) with the real axis, to the left of −1. ˆ by 1 will cause the If the intersection occurs at the location −α, then scaling the plot of L α intersection. By choosing the closest intersection, α is “close” to 1, and hence γ := α1 is the desired value. ˆ In the above computations, each intersection of the curve L(jβ) (as β varies) with the negative real axis is associated with two numbers: 1. The location (in the complex plane) of the intersection, of which the negative reciprocal determines the gain γ that would cause instability.

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2. The value of β at which the intersection takes place, which determines the crossing point (on the imaginary axis) where the closed-loop root migrates from the left-half plane into the right-half plane, and hence determines the frequency of oscillations just at the onset of instability due to the gain change.

25.2

Time-Delay Margin

The homogeneous equation, with γ = 1, but T > 0 of equation (25.1) is y [n] (t) + a1 y [n−1] (t) + · · · + an−1 y [1] (t) + an y(t) + b1 y [n−1] (t − T ) + · · · + bn−1 y [1] (t − T ) + bn y(t − T ) = 0

(25.3)

Recall that we assume that for T = 0, the system is stable, and hence all homogeneous solutions decay exponentially to 0 as t → ∞. It is a fact (advanced) that as T increases the system may eventually become unstable, and if it does, then at the critical value of T for which stability is violated, sinusoidal solutions to the homogeneous equation exist. In other words, the system “goes unstable” by first exhibiting purely sinusoidal homogeneous solutions. The frequency of this sinusoidal solution is not known a-priori, and needs to be determined, along with the critical value of the time-delay. The delay terms may cause a non-decaying homogeneous solution to exist. We pose the question: what is the minimum T and corresponding frequency ω such that a sinusoidal solution to (25.3) of the form yH (t) = ejωt exists? If such a solution exists, then by plugging it into (25.3), and canceling the common ejωt term, which is never 0, we get a necessary algebraic condition n−1 (jω)n + a1 (jω) + · · · + an−1 jω + an   −jωT e b1 (jω)n−1 + · · · + bn−1 jω + bn = 0

Rearranging gives −1 = e−jωT

b1 (jω)n−1 + b2 (jω)n−2 + · · · + bn−1 (jω) + bn (jω)n + a1 (jω)n−1 + a2 (jω)n−2 + · · · + an−1 (jω) + an

But, this is recognized as −jωT

−1 = e

ˆ L(s)

s=jω

This is two equations (since it relates complex quantities) in two real unknowns (T and ω). We want to solve for solutions with T close to 0 (recall that 0 was the nominal value of T , and we assumed that the closed-loop system was stable for that). Since ωT is real, it follows −jωT ˆ that e = 1. This gives conditions on the possible values of ω, namely L(jω) = 1. Once these values of ω are known, the corresponding values of T can be determined, and the smallest such T becomes the time-delay margin.

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In order to understand the graphical solution techniques described, recall that for any complex number Ψ, and any real number θ, we have ∠ejθ Ψ = θ + ∠Ψ Hence, at any value of ω, we have ˆ ˆ ∠e−jωT L(jω) = −ωT + ∠L(jω) Also, ∠(−1) = ±π. Hence, a simple method to determine solutions is: ˆ 1. Plot (in the complex plane, or on separate magnitude/phase plots) the quantity L(jω) as ω varies from 0 to ∞. ˆ 2. Identify all of the frequencies ωi > 0 where L(jω i ) = 1. These are called the gaincrossover frequencies. ˆ 3. At each gain-crossover frequency ωi , (which necessarily has L(jωi ) = 1), determine ˆ the value of T > 0 such that e−jωi T L(jω i ) = −1. Recall that multiplication of any −jωi T complex number by e simply rotates the complex number ωi T radians in the clockwise (negative) direction. Hence, each T can be determined by calculating the ˆ angle from L(jω) to −1, measured in a clockwise direction, and dividing this angle by ω. On a phase plot, determine the net angle change in the negative direction (downward) to get to the closest odd-multiple of π (or 180◦ ). 4. Repeat this calculation for each gain-crossover frequency, and choose the smallest of all of the T ’s obtained.

25.3

Percentage Variation Margin

Suppose that P and C are the transfer functions of a plant and controller, respectively. Assume that the closed-loop system, shown below, is stable. u2

u1

e1

- j -

− 6

y1 C

? - j

y2

e2 -

P

-

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Hence all of the closed-loop transfer functions (from ui to ej and from ui to yk ) have poles in the open-left-half plane. Suppose that due to uncertainty, the actual plant transfer function is Pactual = P (1 + ∆(s)) where ∆ is an unknown, stable system, representing the uncertainty in P . In block-diagram form, Pactual looks like

-

∆ -? e -

P

-

With this model, it follows that Pactual − P P so in some sense, ∆ represents a relative uncertainty. ∆=

In this case, the closed-loop system is actually of the form

-e −6

C

-? e

∆ -? e -

P

-

−P C Note that ∆ is effectively in positive feedback with the transfer function 1+P . In other C words, the perturbed closed-loop system can be thought of as the feedback interconnection −P C PC of two stable systems, namely ∆ and 1+P . Since the nominal closed-loop is stable, 1+P is C C stable, and hence the addition of the stable transfer function ∆ can only lead to instability if the loop introduced by that ∆ introduces is unstable.

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∆

-

−P C 1+P C

? 6 -e −6

e -?

-? e

C

P

-

Therefore, stability of the perturbed closed-loop can be assessed by studying the stability of -e 6

∆

-

−P C 1+P C





? e

where only limited information about ∆ is available. For this type of analysis, we

25.3.1

The Small-Gain Theorem

This section presents an important theorem from system theory, called the Small Gain Theorem. We introduce it in its simplest form, and then derive a test for establishing stability in the presence of uncertain components. ˆ Small Gain Theorem: Suppose G is a stable system. Let G(s) be its transfer function. If ˆ max G(jω) (25.4) γ, ω∈R

then there is a stable system H such that 1 ˆ max H(jω) < ω∈R γ such that the feedback connection of G and H is unstable. Theorem NS2: Suppose G is a given, stable system. Assume G(jω) 6= 0 for all ω ∈ R. For any frequency ω ¯ ∈ R with ω ¯ > 0, there exists a stable system Hω¯ such that 1. for all ω ∈ R, ω 6= ω ¯ 1 |Hω¯ (jω)| < G(jω) 2. at ω ¯, 1 |Hω¯ (j ω ¯ )| = G(j ω ¯) 3. The feedback connection of G and H is unstable.

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There are two independent facts about transfer functions that are used to prove these theorems Fact 1: Given a positive ω ¯ > 0, and a complex number δ, with Imag (δ) 6= 0, there is a β > 0 such that by proper choice of sign s − β =δ ± |δ| s + β s=j ω¯ Remark: In other words, given any positive frequency ω ¯ > 0, and any complex number δ, there is a stable, first-order system, ∆(s) that has a flat frequency-response magnitude (across all frequency) and satisfies ∆(j ω ¯ ) = δ. s−β as s s+β takes on values of jω, with ω ranging from 0 to +∞, traverses a circular arc in the complex plane, from −1 (at ω = 0) to 1 (as ω → +∞) clockwise, centered at 0, and in the top-half of s−β as s takes on values of jω, with ω ranging the complex plane. Similarly, the plot of − s+β from 0 to +∞, traverses from 1 (at ω = 0) to −1 (as ω → +∞) clockwise, in the bottom-half plane. Therefore, the choice of + or − sign depends on the imaginary part of δ. Here is a constructive procedure for the case Imag(δ) > 0 (verify that this is indeed correct). The proof is as follows (details provided in class): For a positive β, the plot of

1. Write δ as δ = |δ| ejφ for a real number φ. q φ 2. Pick β := ω ¯ 1−cos 1+cos φ If Imag(δ) < 0, use above procedure with −δ replacing δ. Then, add the (−) sign to the transfer function. Fact 2: For any ω ¯ > 0 and any 0 < ξ < 1, define the transfer function I(s) :=

s2

2ξ ω ¯s + 2ξ ω ¯s + ω ¯2

Then 1. for all ω ∈ R, ω 6= ω ¯ 1 |I(jω)| < G(jω) 2. at ω ¯, I(j ω ¯) = 1

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3. I is stable. The proof of this is just simple verification of the claims. Now, returning Theorem NS1. Suppose G is a given, stable system, γ > 0, and ˆ max G(jω) > γ, ω∈R

ˆ Pick ω ¯ > 0 so that G(j ω ¯ ) > γ. Next, construct H using the “∆” construction above, with δ :=

−1 . ˆ ω G(j ¯)

It follows that 1 1 ˆ < max H(jω) = ω∈R γ ˆ ω ¯ ) G(j

and



 ˆ H(s) ˆ 1 + G(s)

= 0, implying that s=j ω ¯

1 ˆH ˆ 1+G

has a pole at s = j ω ¯.

Alternatively, for Theorem NS2, since G(jω) is never 0, it follows that you can write G(s) =

nL (s)nR (s) d(s)

where the roots of nL (s) = 0 are in the open left-half plane, and the roots of nR (s) = 0 are in the open right-half plane. Define Ω(s) :=

d(s) nL (s)nR (−s)

There are two relevant facts about the polynomial nR (−s), which are: 1. since the roots of nR (s) = 0 are in the open right-half plane, it follows that the roots of nR (−s) = 0 are in the open left-half plane; 2. since the coefficients of nR (s) are real, it follows that nR (−jω) is the complex conjugate of nR (jω) and consequently the magnitudes are the same. Therefore Ω is stable, and for all ω |Ω(jω)| =

1 |G(jω)|

Now pick β > 0, and the proper choice of ± so that s−β 1 ± Ω(s) = s+β G(j ω ¯) s=j ω ¯

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Finally with any 0 < ξ < 1, define H as H(s) := (±)

s2

s−β 2ξ ω ¯s Ω(s) 2 + 2ξ ω ¯s + ω ¯ s+β

It is easy to verify that H has all of the properties claimed in Theorem NS2. 25.3.3

Application to Percentage Variation Margin

Recall that the stability of the perturbed closed-loop can be assessed by studying the stability of -e 6

∆ −P C 1+P C



-



? e

We can now apply the various small-gain theorems, with playing the role of H.

−P C 1+P C

playing the role of G, and ∆

The first small gain theorem implies that if ∆ is stable, and satisfies 1 + P C(jω) 1 |∆(jω)| < −P C(jω) = P C(jω) 1+P C(jω)

for all ω, then stability is assured. By contrast, Theorem NS2 implies that for any ω ¯ , there exists a stable ∆ such that 1. for all ω ∈ R, ω 6= ω ¯ 1 + P C(jω) |∆(jω)| < P C(jω) 2. at ω ¯, 1 + P C(jω) |∆(j ω ¯ )| = P C(jω) 3. The feedback connection of

−P C 1+P C

and ∆ is unstable, with a closed-loop pole at s = j ω ¯.

Hence, the function 1 + P C(jω) MP V (ω) := P C(jω) is a robustness margin, representing how much relative uncertainty, frequency-by-frequency, that can tolerated in the plant without loss of stability in the closed-loop system.

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25.4

275

Summary

We have introduced 3 margins. Each measure, in different senses, how close the loop gain, L, gets to the (-1) point. The gain margin measures the distance along the real axis; the timedelay margin measures the distance along a circular arc (the unit circle) and normalizes by frequency; while the Percentage-Variation margin quantifies the distance in a more absolute manner. All 3 margins are easy to compute, knowing the loop-gain, and should be computed and compared when understanding tradeoffs between multiple competing control system designs.

ME 132, Fall 2018, UC Berkeley, A. Packard

25.5

Examples

25.5.1

Generic

276

Consider a system with ˆ L(s) =

4 (0.5s + 1)3

ˆ ˆ ˆ The plots of L(jω) as ω varies are shown below. We will do calculations , ∠L(jω), and L(jω) in class. Loop Gain

1

Nyquist Plot

10

1

0.5

0 0

10

−0.5

−1

−1.5

−1

10

−2

−2.5

−2

10

0

10

1

10

Loop Phase 0

−50

−100

−150

−200

−250

0

10

1

10

−3 −1

−0.5

0

0.5

1

1.5

2

2.5

3

3.5

4

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25.5.2

277

Missile

A missile is controlled by deflecting its fins. The transfer function of a the Yaw axis of a tail-fin controlled missile is Pˆ (s) =

−0.5(s2 − 2500) (s − 3)(s2 + 50s + 1000)

A PI controller, with transfer function 10(s + 3) ˆ C(s) = s is used. This results in a stable closed-loop system, with closed-loop roots at −8.1 ± j14.84, −18.8, −6.97 ˆ are shown below, along with some time responses. Plots of the L

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2

Log Magnitude

10

1

10

0

10

0

1

10

10

2

10

Frequency (radians/sec) 220

Phase (degrees)

200 180 160 140 120 100 80

0

1

10

10 Frequency (radians/sec)

2

10

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Nyquist Plot 2

1.5

1

0.5

0

−0.5

−1

−1.5

−2 −5

−4

−3

−2

−1

0

1

Open−Loop Missile Step Response 0.5

0.4

G’s

0.3

0.2

0.1

0

−0.1 0

0.05

0.1

0.15 Time: seconds

0.2

0.25

0.3

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Missile Acceleration Response 2

Yaw Acceleration: G’s

1.5

1

0.5

0

−0.5

0

0.1

0.2

0.3

0.4

0.5 0.6 Time: Seconds

0.7

0.8

0.9

1

0.7

0.8

0.9

1

Fin Deflection 12

10

8

6

Degrees

4

2

0

−2

−4

−6

−8

0

0.1

0.2

0.3

0.4

0.5 0.6 Time: Seconds

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25.5.3

281

Application to percentage variation margin

Take the missile example from Section 17, P (s) =

−0.5(s2 − 2500) , (s − 3)(s2 + 50s + 1000)

C(s) =

10(s + 3) s

We will also try Pperturbed (s) = P (s) Note this is the same as

1 τs + 1

  −τ s 1 = P (s) 1 + P (s) τs + 1 τs + 1

Hence, the percentage variation in going from P to Pperturbed is Pperturbed − P −τ s = P τs + 1 Let’s plot the magnitude of this versus the percentage-variation margin for different values of τ .

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3

10

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Problems

1. Suppose C(s) = 4s and P (s) = 2 in a standard Controller/Plant feedback architecture, with negative feedback. (a) Is the closed-loop system stable? (b) What is the closed-loop transfer function from r to y. (c) What is the steady-state gain from r → y? (d) What is the time-constant of the closed-loop system. (e) What is the time-delay margin? Denote it by Td . At what frequency will the self-sustaining (ie., unstable) oscillations occur? (f) Verify your answers with Simulink, using time delays of 0, 9 T , 9.9 T . 10 d 10 d

1 T , 3T , 5T , 7T , 10 d 10 d 10 d 10 d

2. Suppose C(s) = Ks and P (s) = β in a standard Controller/Plant feedback architecture, with negative feedback. Assume both β and K are positive. (a) Is the closed-loop system stable? (b) What is the time-constant of the closed-loop system. (c) What is the time-delay margin? (d) What is the ratio of the time-constant of the closed-loop system to the time-delay margin? Note that this ratio is not a function of β and/or K. 3. The transfer functions of several controller/process (C/P ) pairs are listed below. Let L(s) := P (s)C(s) denote the loop transfer function. For each pair, consider the closedloop system r(t)

e(t) + - i −6

u(t) C

-

P

-

y(t)

(a) determine the closed-loop characteristic polynomial; (b) compute the roots of the closed-loop characteristic polynomial, and verify that the closed-loop system is stable; (c) for the gain-margin problem, find all of the phase-crossover frequencies, and the value of L at those frequencies; (d) determine the gain-margin of the closed-loop system; (e) for the time-delay-margin problem, find all of the gain-crossover frequencies, and the value of hL at those frequencies;

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(f) determine the time-delay-margin of the closed-loop system; The systems are:

(b) C(s) =

2.4s+1 , s 0.4s+1 , s

(c) C(s) =

10(s+3) , s

(a) C(s) =

P (s) = P (s) =

1 s−1 1 s+1

P (s) =

−0.5(s2 −2500) (s−3)(s2 +50s+1000)

4. In problem 3 above, the first two cases, (a) and (b), have identical closed-loop characteristic polynomials, and hence identical closed-loop roots. Nevertheless, they have different stability margins. In one case, the plant P is unstable, and in the other case it is stable. Check which case has better stability margins in each of the different measures. Make a conclusion (at least in the case) that all other things being equal, it is “harder” to reliably control an unstable plant than it is a stable one. 5. Let L(s) be the transfer function of a system L. Consider the diagram below. z6 d

−6

q

-? d

p 6 -

L

Let N (s) denote the transfer function from q to z. Let S(s) denote the transfer function from q to p. (a) Derive N (s) and S(s) in terms of L(s). (b) Derive L(s) in terms of N (s). (c) Derive L(s) in terms of S(s). Application: As we have learned, the gain-margin and time-delay margin can be computed from a frequency-response plot of the open-loop system L (which is often P C). In this problem, we see that that through simple algebra/arithmetic, the value of L can be inferred from certain closed-loop frequency responses, namely N or S. In many industries, a working, stable closed-loop system is available for experimentation, so that the frequency-response of N and/or S can be determined experimentally. With that data in hand, the value of L(jω) can be solved for, and then the margin analysis carried out. In this manner, the margins are detemined from closed-loop frequencyresponse experiments, rather than from differential equation models of P and C. 6. In the diagram below,

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d

−6

- sKP +KI s

- d 1−6

287

1 s

-

1 s

-

KD 2 3

derive S, N and L at each marked location. In each case, verify (after the derivation) that N 1−S L=− = 1+N S 7. Find Leff for determining time-delay and/or gain margins at the locations marked by 1, 2 and 3. d

−6

-

C1

-d 1 −6

A

-

C2 

G1

-

G2

-

2 3

8. Consider the diagram below, which depicts a position control system, using PI control and rate-feedback (as we did with the motor). Additionally, there is a 1st-order model of the actuator, which produces the forces based on the commands from the controller blocks. Here, m = 2 and τ = 0.0312. If we choose KD = 16, KP = 52.48 and KI = 51.2, it is possible (you do not need to) to verify that the closed-loop system is W stable. R

- e - KP s+KI s −6

1 - e Aτ s+1 −6

-? e -

KD 

D

1 ms

B C

E-

1 s

Y -

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In order to assist you in the question below, Bode plots of certain transfer functions listed below are given in the following pages (not all may be useful...). H1 (s) = H3 (s) =

KP s + KI 4 mτ s + ms3 + KD s2

H2 (s) =

KD s2 mτ s4 + ms3 + KP s + KI

KD s2 + KP s + KI KD s2 + KP s + KI H (s) = 4 mτ s4 + ms3 + KD s2 + KP s + KI mτ s4 + ms3 KP s + KI KD H (s) = H5 (s) = 6 mτ s2 + ms mτ s4 + ms3

Using the graphs (estimate values as best as you can...), answer the following margin questions. Explain any work you do, and make relevant marks on the Bode plots that you use in your calculations. (a) What is the gain margin at location A? (Hint – first determine what is the appropriate L for margin calculations at A, match with the H’s, and do calculation from supplied graphs). (b) What is the time-delay margin at location A? (c) What is the gain margin at location B? (d) What is the time-delay margin at location B? (e) What is the gain margin at location C? (f) What is the time-delay margin at location C? (g) What is the gain margin at location D? (h) What is the time-delay margin at location D? (i) What is the gain margin at location E? (j) What is the time-delay margin at location E?

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Bode Plot of H1

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9. Consider the multiloop interconnection of 4 systems shown below. F  r -d E −6

+? +d

-d 6 ? ?

p

G 

y

H

-

z

q

(a) What is the transfer function from q to z? Denote the transfer function as Gq→z (b) What is the transfer function from q to p? Denote the transfer function as Gq→p (c) Verify that 1 + Gq→p = Gq→z . 10. A closed-loop feedback system consisting of plant P and controller C is shown below. r(t)

e(t) + - i −6

u(t) C

-

P

-

y(t)

It is known that the nominal closed-loop system is stable. In the presence of gainvariations in P and time-delay in the feedback path, the closed-loop system changes to r(t)

+ - i −6

e(t)

u(t) -

C

f (t) = y(t − T )

-

γ

-

P

-

y(t)

delay, T 

In this particular system, there is both an upper and lower gain margin - that is, for no time-delay, if the gain γ is decreased from 1, the closed-loop system becomes unstable at some (still positive) value of γ; and, if the gain γ is increased from 1, the closed-loop system becomes unstable at some value of γ > 1. Let γl and γu denote these two values, so 0 < γl < 1 < γu . For each fixed value of γ satisfying γl < γ < γu the closed-loop system is stable. For each such fixed γ, compute the minimum time-delay that would cause instability. Specifically, do this for several (say 8-10) γ values satisfying γl < γ < γu , and plot below.

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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 2.1 2.2 2.3 2.4 2.5 GAMMA

ˆ The data on the next two pages are the magnitude and phase of the product Pˆ (jω)C(jω). They are given in both linear and log spacing, depending on which is easier for you to read. Use these graphs to compute the time-delay margin at many fixed values of γ satisfying γl < γ < γu . 2

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3.75 3.5 3.25 3 1

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11. An unstable plant, P , with differential equation relating its input u and output y, y¨(t) − 16y(t) ˙ + 15y(t) = u(t) ˙ + 5u(t) is given. (a) Calculate the range of the parameter KP for which the closed-loop system is stable. + e er KP −6

u-

y

P

-

(b) If KP = 30, what is the steady state error, ess , due to a unit-step reference input?

(c) A integral controller will reduce the steady-state error to 0, assuming that the closed-loop system is stable. Using any method you like, show that the closedloop system shown below, is unstable for all values of KI . + e e- R r −6

-

KI

u-

P

y -

(d) Find a PI controller, that results in a stable closed-loop system.

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(e) Consider the system with just Proportional-control, along with a time-delay, T , in the feedback path. We wish to determine the maximum allowable delay before instability occurs. Find the equation (quadratic in ω 2 ) that ω must satisfy for there to exist homogeneous solutions yH (t) = ejωt for some ω. The equation should involve ω, KP , and the parameters in the plant, but not the time delay, T . + e er KP −6 

u-

P

y -

delay, T 

(f) For two cases of proportional control: KP = 20 and KP = 30; determine in each case the time delay T that will just cause instability and the frequency of the oscillations as instability is reached.

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299

Gain/Time-delay margins: Alternative derivation

26.1

Sylvester’s determinant identity

Suppose there are two matrices, X ∈ Cn×m and Y ∈ Cm×n . Then: det(In − XY ) = det(Im − Y X) The proof is in section 26.5.

26.2

Setup

For both problems, we start with four assumptions: A1 : L is a linear system (in state space or transfer function form) A2 : L is not necessarily stable A3 : The feedback connection pictured below is stable A4 : L is a one-input, one-output system. u −

y L

The derivations presented below are based on the state-space form of L, denoted as: x(t) ˙ = Ax(t) + bu(t) y(t) = cx(t) Since there is one-input and one-output (A4), it follows that A ∈ Rn×n , b ∈ Rn×1 and c ∈ R1×n .

26.3

Gain margin

For the connection pictured above: u(t) = −y(t) x(t) ˙ = (A − bc)x(t)

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Since the feedback connection is stable (A3), the eigenvalues of A − bc all have negative real parts. The gain margin is the range of γ, containing γ = 1 for which the system pictured below remains stable, where γ ≥ 0. u

γ

−

y L

This means that the new feedback connection equations are: x(t) ˙ = Ax(t) + γbu(t) y(t) = cx(t) u(t) = −y(t) So we know that: 1. x(t) ˙ = (A − γbc)x(t) (from equations above) 2. By A3, the eigenvalues of A − bc all have negative real parts (from 1.1) 3. The eigenvalues are a continuous function of γ (since the matrix is a continuous function of γ and the eigenvalues of a matrix are a continuous function of the entries of the matrix) So, for γ = 1, all of the eigenvalues have negative real-parts (they are in the open left-half plane). As γ varies (both less than 1, and greater than 1), the only way an eigenvalue can become unstable for some value of γ 6= 1, it must cross the imaginary axis at an intermediate value of γ (ie., closer to 1). Therefore, to compute the gain margin, we find all real values of γ and ω ≥ 0 where A − γbc has an eigenvalue on the imaginary axis (ie., at jω). Let (γ1 , ω1 ), (γ2 , ω2 ), · · · (γN , ωN ) be all these pairs. From this list, select • the smallest value of γ that is greater than 1, and call this the upper-gain margin, γU . If there is no γ > 1 in the list, then γU := ∞. • the largest value of γ that is less than 1, and call this the lower-gain margin, γL . If there is no γ ∈ [0, 1) in the list, then γL := 0.

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To find all of the (γk , ωk ) pairs, make a further assumption (denoted A5, which not necessary, but simplifies the derivation) that A has no imaginary-axis eigenvalues (but A can be unstable, as in A4). Then we find all (γ, ω) pairs where γ ∈ R, γ ≥ 0, ω ∈ R and det(jωIn − (A − γbc)) = 0 ⇔ det((jωIn − A) + γbc) = 0 ⇔ det [(jωIn − A)(In + (jωIn − A)−1 (γbc))] = 0 ⇔ det(jωIn − A) · det(In + (jωIn − A)−1 (γbc)) = 0 We already have assumed (A5) that det(jωIn − A) 6= 0, so the pairs (γ, ω) must satisfy det(In + γ(jωIn − A)−1 bc) = 0 Define X := (jωIn − A)−1 b and Y := c, then use Sylvester’s determinant identity to rewrite this equation, reducing its dimension to 1 × 1 (so it no longer involves a determinant). det(I1 + γc(jωI1 − A)−1 b) = 0 ⇔ 1 + γc(jω − A)−1 b = 0 ⇔ 1 + γL(jω) = 0 Since the unknown value of γ satisfies γ ∈ R, γ > 0, it must be that the value of ω is such that L(jω) ∈ R and L(jω) < 0. With such a frequency identified, the associated value of γ −1 . is clearly γ := L(jω) Summary of the steps: To find all positive, real values of γ such that A−γbc has an eigenvalue on the imaginary axis 1. Find all ω ≥ 0 such that L(jω) is real and negative. Recall, a complex number β is real and negative if and only if ∠β = ±π, ±3π, ±5π, · · · radians. On a Bode phase plot of L, identify all frequencies where ∠L(jω) = ±π, ±3π, ±5π, · · · . • These are called the phase-crossover frequencies (since the phase is “crossing” a particular value - namely an odd multiple of π). • In Matlab, with allmargin, these frequencies are listed in the GMFrequency field. Denote these as {ω1 , ω2 , · · · , ωN }. 2. Associated with each phase-crossover frequency, define γk := all γk are real-valued, and positive. This gives several pairs,

−1 . L(jωk )

By construction,

{ω1 , γ1 }, {ω2 , γ2 }, . . . , {ωN , γN } to consider. 3. The lower gain-margin, γL is the largest γk smaller than 1. If there are no γk ∈ [0, 1), then γL := 0.

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4. The upper gain-margin, γU is the smallest γk larger than 1. If there are no γk > 1, then γU := ∞. 5. The system is stable for the range γ ∈ (γL , γS ) An example carrying out the steps is shown in the script file gainMarginExample.m, which is posted at bCourses, under the Matlab Skills Exercises. The script file also illustrates the use of allmargin.

26.4

Time delay margin

Make the same assumptions, A1, A2, A3, A4 as before. The time-delay margin is smallest T > 0 for which the system pictured below becomes unstable. u −

y L

time delay, T

Here, the feedback connection equations are: x(t) ˙ = Ax(t) + bu(t) y(t) = cx(t) u(t) = −y(t − T ) Goal: find the smallest T > 0 where x(t) ˙ = Ax(t) − bcx(t − T ) is unstable. By unstable, we mean from some initial conditions, the solution does not decay to zero. As mentioned in class, the delay-differential equations are difficult to analyze. We will invoke a fact, similar to the fact I invoked when we derived time-delay margin for a single-state system back in week #2. Fact: if there is a T > 0 for which x(t) ˙ = Ax(t) − bcx(t − T ) is unstable, then at the minimum T for which it is unstable, there is a non-decaying solution that is exactly sinusoidal. With this fact, our search for existence of these solutions is much simpler. We are looking for the smallest T > 0, ω ≥ 0, and x¯n×1 6= 0n×1 where x(t) = x¯ejωt is a solution. Plugging this form of x into x(t) ˙ = Ax(t) − bcx(t − T ) gives jω¯ xejωt = A¯ xejωt − bc¯ xejω(t−T ) ∀t ⇔ jω¯ xejωt = A¯ xejωt − bc¯ xejωt e−jωT ∀t −jωT ⇔ jω¯ x = A¯ x − bc¯ xe ⇔ (jωIn − A + bce−jωT )¯ x=0

(1.3)

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Note that we used the fact that for any ω, and any t, the scalar ejωt 6= 0, and it was divided out. For fixed values of ω and T , the expression (jωIn − A + bce−jωT )¯ x = 0n is true for some x¯ 6= 0 if and only if det(jωIn − A + bcejωT ) = 0 Invoke A5 again (A has no imaginary axis eigenvalues), and simplify as   det(jωIn − A + bcejωT ) = 0 ⇔ det (jωIn − A)(In + (jωIn − A)−1 bce−jωT ) = 0 ⇔ det(jωIn − A) · det(In + (jωIn − A)−1 bce−jωT ) = 0 ⇔ det(In + (jωIn − A)−1 bce−jωT ) = 0 ⇔ det(I1 + c(jωI1 − A)−1 be−jωT ) = 0 ⇔ 1 + c(jω − A)−1 be−jωT ) = 0 ⇔ e−jωT L(jω) = −1 Even though we don’t know the value of T or ω, both e−jωT and −1 both have a magnitude of 1. Therefore, we are interested in values of ω for which |L(jω)| = 1. So, suppose |L(jω)| = 1. So L(jω) is a complex number on the perimeter of the unit circle in C (radius of 1, centered at 0). The factor e−jωT , when multiplying L(jω) rotates the complex number L(jω) by ωT radians, clockwise. We are interested in a rotation that makes e−jωT L(jω) = −1. Important: Hence we define θ, satisfying 0 < θ < 2π, as the angle of a bf clockwise rotation that takes L(jω) to −1. With θ defined, the corresponding T is simply T := ωθ . Summary of the steps: To find all T > 0, and associated ω ≥ 0 such that e−jωT L(jω) = −1 1. Find all ω ≥ 0 such that |L(jω)| = 1. On a Bode magnitude plot of L, identify all frequencies where |L(jω)| = 1. • These are called the gain-crossover frequencies (since the gain is “crossing” a particular value - namely 1). • In Matlab, with allmargin, these frequencies are listed in the DMFrequency field. Denote these as {ω1 , ω2 , · · · , ωN }. 2. Associated with each gain-crossover frequency, define θk , satisfying 0 < θk < 2π, as the angle of a clockwise rotation that takes L(jωk ) exactly to −1. With θk defined, the corresponding Tk is simply Tk := ωθkk . 3. The time-delay margin, T¯ is the smallest such T . If there are no gain-crossover frequencies, then the time-delay margin is ∞

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An example carrying out the steps is shown in the script file TimeDelayMarginExample.m, which is posted at bCourses, under the Matlab Skills Exercises. The script file also illustrates the use of allmargin.

26.5

Appendix

To prove Sylvester’s determinant identity, we define the block matrix of size (n+m)×(n+m) as:   In X Z := Y Im Then we can show, using the determinants of block matrices, that:      In −X In X In − XY 0n×m = Y Im 0m×n Im Y Im       In −X In X In − XY 0n×m det det = det 0m×n Im Y Im Y Im   
I X det(In + X(Im )−1 0m×n ) det(Im ) det n = det((In − XY ) − 0n×m (Im )−1 Y ) det(Im ) Y Im   In X det = det(In − XY ) (1.1) Y Im Similarly, we can show that: 

    0n×m In X In X = Im Y Im 0m×n Im − Y X       In 0n×m In X In X det det = det Y Im 0m×n Im − Y X −Y Im   
I X det(In − 0n×m (Im )−1 (−Y )) det(Im ) det n = det(In − X(Im − Y X)−1 0m×n ) det(Im − Y X) Y Im   In X det = det(Im − Y X) (1.2) Y Im In −Y

Since det(Z) is equal to (1.1) as well as (1.2), it must be that det(In − XY ) = det(Im − Y X)

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305

Connection between Frequency Responses and Transfer functions

Consider the standard linear system y [n] (t) + a1 y [n−1] (t) + · · · + an−1 y [1] (t) + an y(t) = b0 u[n] (t) + b1 u[n−1] (t) + · · · + bn−1 u[1] (t) + bn u(t)

(27.1)

with y the dependent variable (output), and u the independent variable (input). Assuming that the system is stable, we know that the steady-state response to a sinusoidal input is also a sinusoid, with magnitude and phase determined by the system-dependent frequency response function

H(ω) :=

b0 (jω)n + b1 (jω)n−1 + · · · + bn−1 (jω) + bn (jω)n + a1 (jω)n−1 + · · · + an−1 (jω) + an

(27.2)

For stable systems, we have proven for fixed value u¯ and fixed ω ∈ R u(t) := u¯ejωt ⇒ yss (t) = H(ω)¯ uejωt Recall that the transfer function from u to y is the rational function G(s) given by G(s) :=

b0 sn + b1 sn−1 + · · · + bn−1 s + bn sn + a1 sn−1 + · · · + an−1 s + an

Note that H(ω) = G(s)|s=jω . So, we can immediately write down the frequency response function once we have derived the transfer function. Sometimes, we do not use different letters to distinguish the transfer function and frequency response, typically writing G(s) to denote the transfer function and G(jω) to denote the frequency response function.

27.1

Interconnections

Frequency Responses are a useful concept when working with interconnections of linear systems. Since the frequency response function, H(ω) turned out to be the transfer function G(s) evaluated at s = jω, frequency response functions of interconnections follow the same rules as transfer functions of interconnections. This is extremely important, so we reiterate it: The frequency response of a stable interconnection of systems (which are individually possibly unstable)

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is simply the algebraic gain of the closed-loop systems, treating individual subsystems as complex gains, with their “gain” taking on the value of the frequency response function. This is true, even if some of the subsystems are not themselves stable. The frequency response of the parallel connection, shown below -

S1

y1 ? d++ 6

u -

S2

y

y2

is simply H(ω) = G1 (jω) + G2 (jω), where G1 (s) and G2 (s) are the transfer functions of the dynamic systems S1 and S2 respectively. For the cascade of two stable systems, u -

S1

v-

y

S2

-

the frequency response is H(ω) = G2 (jω)G1 (jω). The other important interconnection we know of is the basic feedback loop. Consider the general single-loop feedback system shown below. r

+ d u−6

S1

y

R + - dU- G1 −6

-

S2 

Y -

G2 

The diagram on the right is interpreted as the equations U = R − G2 Y , and Y = G1 U . As derived earlier, manipulating these as though they are arithmetic expressions gives the correct closed-loop transfer function description Y =

G1 (s) R 1 + G2 (s)G1 (s)

Consequently, the closed-loop frequency-response function is just Hr→y (ω) =

G1 (jω) 1 + G2 (jω)G1 (jω)

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307

Decomposing Systems into Simple Parts

See slides, specifically SystemApproximations.pdf.

28.1

Problems

1. In class, we tallked about the possibility of neglecting “fast” first-order systems whose steady-state gain is equal to 1. In this problem, we will investigate to what extent this is possible and in what cases can it cause a problem. Consider a second order system, with transfer function P (s) =

s2

500 5 100 5 1 500 = = = + 99s − 100 (s + 100)(s − 1) (s − 1) (s + 100) (s − 1) (0.01s + 1)

(a) Note that P is a cascade of an unstable system with pole at 1 (representing a time-to-double of about 0.7 time units, since e1·0.7 ≈ 2), and a stable system with time-constant equal to 0.01 and steady-state gain equal to 1. Approximate P (using time-scale separation) by a first-order system, denoting the approximation by Pa , P (s) =

5 100 5 1 5 = ≈ =: Pa (s) (s − 1) (s + 100) (s − 1) (0.01s + 1) s−1

On one graph, plot a step-response of P and Pa on the time-interval [0, 1.4]. Note how similar the responses are on this time-scale. (b) Using the approximation, Pa , design a PI-controller, with transfer function s · KP + KI s such that the closed-loop system consisting of Pa and C1 in the standard feedback configuration, has poles at −5 ± j5 (time-constant about 0.2 and period of oscillation about 1.3 times units). Note that since the approximation is 1st order, and the controller is 1st order, the closed-loop system consisting of Pa and C1 is indeed 2nd order. C1 (s) =

(c) With the controller C1 designed, determine the closed-loop characteristic equation for the closed-loop system consisting of P and C1 . Is this closed-loop system stable? Where are the closed-loop poles? (There should be 3 of them.) (d) Using step and feedback compute the step response of the closed-loop systems for two responses: r to y, and d to y. For each case, do this for the two closedloop systems - one with (C1 , Pa ) and one with (C1 , P ), labeled “approximation” and “true”, respectively, and plot them on the same graphs. Use a final time of TF = 1.4.

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(e) Finally, with the usual inputs (r, d, n) and output (y, u), make a 2 × 3 array of Bode plots, with two lines/axes (one for (C, Pa ) and one for (C1 , P )). For r → y, plot both magnitude and phase; for all others, just plot magnitude. Look back in the early chapters where I called this the “gang-of-six” to see how it should be plotted. (f) Based on time-domain, and frequency-domain plots, would you say that Pa is a good approximation of P for the purposes of this particular control design for C1 , where closed-loop poles are placed at −5 ± j5? (g) Repeat this entire control design/analysis process for a different control design, namely again using the approximation, Pa , and designing a PI-controller, with transfer function s · KP + KI C2 (s) = s such that the closed-loop system consisting of Pa and C2 in the standard feedback configuration, has poles at −50 ± j50 (time-constant about 0.02 and period of oscillation about 0.13 times units). Make the same time-domain and frequencydomain comparisons. In time-domain, use TF = 0.14. (h) Based on time-domain, and frequency-domain plots, would you say that Pa is a good approximation of P for the purposes of this particular control design for C2 , where closed-loop poles are placed at −50 ± j50? (i) Finally, plot a step-response of P and Pa on the time-interval [0, 0.14]. On this short time-scale, are there some more obvious differences? Compare this to the similarities you saw in part 1a, which was on a longer time-scale. Note that on this short time-scale, we can see that the response of P looks similar (but not exactly) to a delayed (by 0.01 time-units) version of Pa . (j) Based on the observation (approximate time-delay interpretation) on part 1i, and the oscillatory responses obtained in part 1g, compute the time-delay margin of the closed-loop system consisting of (C2 , Pa ). Note that the value of the margin is actually similar to “approximate delay” that P seems to have relative to Pa . This partly explains why Pa is not a good approximation to P for the purposes of the C2 design. (k) By contrast, compute the time-delay margin of the closed-loop system consisting of (C1 , Pa ). Note that the delay margin is much longer in this case, and hence the small “delay” between Pa and P is not as significant for the C1 design. 2. (a) Factor the transfer function below into a cascade of lower order systems. G(s) =

0.11s4

+

2.025s3

2s + 1 − 3.26s2 + 4.935s + 2.7

(b) Determine (this is a judgement call, but here the answer should be “yes”) if there is a good, lower-order (2nd-order, in fact) approximation, GA .

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(c) Find the approximation, and use Matlab to plot the step-response of the original system and the approximation. As both are unstable, limit the step-response time to 3 time-units. (d) Using GA , design a 2nd-order controller with integral-action, C(s) =

as2 + bs + c s(s + d)

such that the closed-loop poles are at {−32.5, −1.4 ± j2.8, −0.23} (e) Consider the generic closed-loop system described by equations e = r − ym , ym = y + n, u = C(e), and y = P (d + u), where C represents the controller, and P represents the plant (here P = G or P = GA ). Confirm (using feedback and pole) that the closed-loop poles using the controller along with the approximate plant are as desired. (f) Using this controller with original plant, G, what are the actual closed-loop poles? (g) On the same axes, plot step-responses from r to y for both closed-loop systems. (h) On the same axes, plot step-responses from d to y for both closed-loop systems. (i) On the same axes, plot Bode-magnitude responses from n to u for both closed-loop systems. (j) For the particular design, comment on the suitability of the approximation.

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310

Unfiled problems

1. This problem is motivated by thinking about a physical tug-of-war game, but played in a virtual environment, by individuals in different locations, where the players are “linked” by a feedback system, which gives each one infomation about the other. The goal is, with (say) motors, to simulate the effect of player 1 on player 2, and visa-versa. Ultimately, we can build a small scale version of this with the EV3 system, using two motors and one EV3 brick. Two identical systems, G1 and G2 , are operated in separate locations, but “controlled” by one central controller, K, as shown below. All unmarked summing junctions are “+”. F1- e − 6 u

+ ? -e

F2

-

G1

1 2



-

G2

y1-

K 

+ ? e 6 − -

y2

The external inputs F1 and F2 represent inputs by operators to the individual systems at each location. The goal of the “cooperative control” is: • synchronization: the difference, ydiff := y1 − y2 should remain small relative to independent forcings F1 and F2 , and 2 • preserving open-loop dynamics: the average, yavg := y1 +y should be approxi2 F1 +F2 mately equal to G 2 , written concisely as yavg ≈ GFavg , where G denotes the (common) input/output behavior of G1 and G2 .

Since G1 and G2 are separate, but identical, assume their models are of the form G1 :

x˙ 1 (t) = Ax1 (t) + Bu1 (t) y1 (t) = Cx1 (t)

G2 :

x˙ 2 (t) = Ax2 (t) + Bu2 (t) y2 (t) = Cx2 (t)

and

where for each t, xi (t) ∈ Rn , ui (t) ∈ Rnu and yi (t) ∈ Rny . Note that each system, G1 and G2 , has n state variables.

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(a) Write an n-state model which describes how yavg behaves, as driven by Favg := F1 +F2 . Note that I am asking for only n states, which seems like it might be 2 impossible, since the overall device clearly has 2n states, plus any additional states that there are in the controller. But, we’re only interested in how Favg affects yavg , so perhaps some simplification is possible... It is useful to look at a simpler picture, merely showing the control architecture (what is measured, and how the control signal is routed to each system). Hint: Let x := 12 (x1 + x2 ), and then add the state equations for the two systems... F1- e − 6  -? e+

F2

-

y1-

G1

+ y ? e − 6

u



-

G2

-

y2

(b) Take G(s) = 1s (ie., an integrator) and K = sKPs+KI . For the closed-loop system, what is the transfer function from Favg to yavg ? (c) Let the state-equations for the controller be      ¯ η η˙ A¯ B = ¯ ¯ C D y u Let m denote the state-dimension of the controller. Write an (n + m)-state model which describes how ydiff behaves. Again, you need to be very careful in defining a “minimal” state-variable for this model. Hint: Define ξ := x1 − x2 , and see what transpires by defining the overall state to be   ξ x := η (d) Suppose G(s) = 1s (ie., an integrator). Let K = sKPs+KI . For the closed-loop system, what is the transfer function from Fdiff := F1 − F2 to ydiff ? 1 (e) Now suppose G(s) = s(Js+b) . Let K = sKPs+KI . For the closed-loop system, what is the transfer function from Fdiff := F1 − F2 to ydiff ? Also, what is the transfer function from Favg to yavg ?

2. In lab, we discovered a simple architecture that would lead to some form of approximate coordination and synchronization of two identical systems, being acted upon by separate external forcing functions. In this problem, we extend the idea to 3 systems, and carry out the same analysis. For simplicity, this problem should be solved

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completely in transfer function notation, simply applying arithmetic rules. We could verify everything in terms of state-space models (as we did in the lab), but there is not time for that here. Again, just manipulate with addition, subtraction, multiplication, and division, as appropriate for transfer functions. The 3 identical systems are each forced with a control input, Ui and an external input Fi . Each of the systems has a transfer function G, hence Yi = G(Ui + Fi ),

i = 1, 2, 3

The architecture for control generalizes the 2-system case. Let K be the transfer function of a linear system, and use U1 = U2 = U3 =

1 K(Y2 3 1 K(Y3 3 1 K(Y1 3

− Y1 ) + 31 K(Y3 − Y1 ) = 13 K(−2Y1 + Y2 + Y3 ) − Y2 ) + 31 K(Y1 − Y2 ) = 13 K(−2Y2 + Y3 + Y1 ) − Y3 ) + 31 K(Y2 − Y3 ) = 13 K(−2Y3 + Y1 + Y2 )

(a) Find Y1 + Y2 + Y3 in terms of F1 , F2 , F3 . (b) Based on part (a), fill in the sentence: The equals the transfer function times the inputs.

of the systems outputs of the external

(c) By careful (but simple) manipulation, find a closed-loop expression for Y2 − Y1 , in terms of K, G and F1 , F2 , F3 . (s) and G(s) = (d) If K(s) = ndKK (s) of the closed-loop system?

nG (s) , dG (s)

what do you think is the characteristic equation

(e) By symmetry, what are the closed-loop expressions for Y3 − Y2 and Y3 − Y1 ? 3. Consider a plant P with transfer function P (s) =

s2

s −1

The plant is a challenging to control. It is a simple model of an inverted pendulum, where the only control action u, represents the angular speed of a reaction wheel, mounted at the free-end (top) of the pendulum. By “torquing” the reaction wheel (relative to the pendulum), equal/opposite torque act on the pendulum, which if properly done, can stabilize it in the “up” position. (a) Suppose the input to P is labeled u and the output y. What is the differential equation governing the relationship between u and y. (b) Is the plant stable?

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(c) Consider a controller, C, with transfer function C(s) =

as + b s+c

For the feedback loop consisting of P and C (with the usual negative (−) feedback convention), what is the closed-loop characteristic equation? (d) Design: Find a, b and c (ie., parameters of the control law) so that the roots of the closed-loop characteristic equation are −4, −2 ± j2 (e) The controller stabilizes P , in that the feedback loop consisting of P and C is stable. However, as we saw earlier, P is not stable on its own. Is C stable? (f) Assume we relax the choice of the desired closed-loop roots, and only restrict them to all have negative real-parts (since we want the closed-loop system to be stable). Are there any choice of such desired roots so that the controller (which yields those closed-loop roots) is itself a stable system? 4. This problem may be out-of-place. It has some transfer function questions At one point, NASA was developing booster rockets and a crew exploration vehicle to replace the Space Shuttle. One main component of the Orion Crew Exploration Vehicle (CEV) is a conical crew module. This module has several thrusters to control the vehicle attitude on re-entry to the earth’s orbit. A linear model for the short-period mode of the CEV pitch dynamics during re-entry is given by:     2 1 0 x(t) ˙ = x(t) + u(t) −36 2 1 y(t) =



0 1



x(t) +



 0 u



 α(t) where x(t) := , which are the angle-of-attack and the pitch-rate, respectively, q(t) and u(t) is the pitch torque generated by the thrusters. The output y is the pitch rate, so y = x2 = q. (a) Using the formula we derived in class, G(s) = D + C(sI − A)−1 B, derive the transfer function from u to y of this state-space model. (b) Enter the matrices into Matlab, form a state-space object using the ss constructor. Type >> disp(G) >> size(G) >> class(G)

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and paste the results into your assignment. (c) Use the command tf as a converter, and convert the model respresentation to a transfer function object. Confirm that your answer in part 4a is correct. (d) If x is eliminated (through clever substitutions - which you know, and are key to the state-space-to-transfer function conversion), what is the differential equation relating u to y. (e) With the transfer function obtained in part 4a, use the procedure derived in class to obtain a state-space model for the system. (f) Note that the state-space model you obtain is not the same as the state-space model we started with. Enter both systems into Matlab, and confirm with step that the input/output behaviors of the two systems are indeed identical. Hence we have seen that a system can have different state-space models (that yield the exact same input/output behavior). 5. Suppose that (A1 , B1 , C1 , D1 ) are the state-space matrices of a linear system with n1 states, m1 inputs and q1 outputs. Hence A1 ∈ Rn1 ×n1 , B1 ∈ Rn1 ×m1 , C1 ∈ Rq1 ×n1 , D1 ∈ Rq1 ×m1 . Let x1 denote the n1 × 1 state vector (so here, the subscript “1” does not mean the first element, rather it is itself a vector, perhaps indexed as   (x1 )1  (x1 )2    x1 =   ..   . (x1 )n1 Similarly, let u1 denote the m1 × 1 input vector and y1 denote the q1 × 1 input vector. Likewise, suppose that (A2 , B2 , C2 , D2 ) are the state-space matrices of a linear system with n2 states, m2 inputs and q2 outputs. Hence A2 ∈ Rn2 ×n2 , B2 ∈ Rn2 ×m2 , C2 ∈ Rq2 ×n2 , D2 ∈ Rq2 ×m2 . Let x2 denote the n2 × 1 state vector. Also let u2 denote the m2 × 1 input vector and y2 denote the q2 × 1 input vector. (a) Assume q2 = m1 , in other words, the number of outputs of system 2 is equal to the number of inputs to system 1. Hence the systems can be cascaded as shown, u2 -

S2

-

S1

with input u2 and output y1 . Define x as   x1 (t) x(t) := x2 (t)

y1 -

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which is an (n1 + n2 ) × 1 vector. Find matrices (defined in terms of vertical and horizontal concatenations of various products of the individual state-space matrices) such that the equations x(t) ˙ = Ax(t) + Bu2 (t) y1 (t) = Cx(t) + Du2 (t) (b) In Matlab, the * operation (”multiplication”) is the cascade operation for systems (tf, ss, and so on), and the above operation would be written as S1 ∗S2 . The actual name of the * operator is mtimes. Execute >> open +ltipack\@ssdata\mtimes.m within Matlab, and find the lines of code which implement the operation above. (c) Assume q1 = q2 and m1 = m2 , in other words, the number of outputs of system 1 is equal to the number of outputs of system 2, and the number of inputs of the two systems are equal as well. Hence the systems can be connected in parallel as shown, with input u and output y. -

S1

y1 ? d++ 6

u -

Define x as

S2 

x(t) :=

y

y2 x1 (t) x2 (t)



which is an (n1 + n2 ) × 1 vector. Find matrices (defined in terms of vertical and horizontal concatenations of various products of the individual state-space matrices) such that the equations x(t) ˙ = Ax(t) + Bu(t) y(t) = Cx(t) + Du(t) (d) In Matlab, the + operation (”addition”) is the parallel interconnection operation for systems (tf, ss, and so on), and the above operation would be written as S1 +S2 . The actual name of the + operator is plus. Execute >> open +ltipack\@ssdata\plus.m within Matlab, and find the lines of code which implement the operation above. (e) Assume that q1 = m2 and m1 = q2 , in other words, the number of outputs of one system is equal to the number of outputs of the other system. Hence the systems can be connected in feedback as shown, with input u and output y. Assume further that D1 = 0q1 ×m1 .

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u

+d −6

316

y

S1

-

S2  Define x as

 x(t) :=

x1 (t) x2 (t)



which is an (n1 + n2 ) × 1 vector. Find matrices (defined in terms of vertical and horizontal concatenations of various products of the individual state-space matrices) such that the equations x(t) ˙ = Ax(t) + Bu(t) y(t) = Cx(t) + Du(t) (f) The code for feedback implements this, but it is buried deeper within other subroutines. Let’s just confirm that this is what it does. >> s1 = ss(-10,3.3,1.1,0,’StateName’,{’xT’}); >> s2 = ss(-5,2.2,5.5,0,’StateName’,{’xB’}); >> feedback(s1,s2) Verify that the state-ordering is as we defined, and that the entries are correct. Be sure to check that all of your matrix manipulations have the correct dimensions, and that the concatenations have compatible dimensions (horizontal concatenations must have the same number of rows, vertical concatenation must have the same number of columns). 6. Consider the interconnection below. The transfer functions of systems S1 and S2 are G1 (s) =

3 , s+6

G2 (s) =

s+2 s+1

Determine the differential equation governing the relationship between u and y. u -

S1

-

S2

y -

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317

Recent exams Fall 2017 Midterm 1

#1

#2

#3

#4

#5

#6

25

10

12

25

20

8

NAME

1. A closed-loop feedback system is shown below. Signals are labeled and equations for each component (controller, plant, sensor) are given.

Keep the plant parameter a as a general value, but assume b1 = b2 = c = 1, for simplicity. K1 and K2 are gains (constants). Keep these as variables. Specific values will be designed in part (b) of the problem. (a) For the closed-loop system above, fill in the 3 × 4 matrix which relates (x, r, d, n) to (x, ˙ y, u). Your expressions should involve variables (K1 , K2 , a).





  x(t) ˙  y(t)  =  u(t)



x(t)   r(t)       d(t)  n(t) 

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(b) Now take a = −1 so that the plant, by itself, is stable, and hence has a timeconstant of τP = 1. Suppose one goal of feedback control is to achieve closedloop stability, and make the closed-loop system respond faster, so that the closed-loop time constant is less than τP . Specifically, work in ratios, expressing this design requirement that the closed-loop time-constant, τCL , should be a fraction γ of the plant time constant, namely Design Requirement #1 :

τCL = γ · τP

where 0 < γ < 1 is a given design target. The other design requirement is Design Requirement #2 :

SSGr→y = 1

in words, the steady-state gain from r → y should equal 1. Task: As a function of γ, find expressions for K1 and K2 which simultaneously achieve the two Design Requirements.

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(c) For the closed-loop system, what is the instantaneous gain from r → u, as a function of the design parameter γ? Is this gain increasing or decreasing as γ decreases? Explain this relationship intuitively (ie., “if we require the system to respond more quickly, with perfect steady-state behavior from r → y, the instantaneous effect that r must have on u....”).

(d) For the closed-loop system, what is the steady-state gain from r → u, as a function of the design parameter γ? How is this affected as γ decreases? Explain this relationship intuitively (ie., “if we require the system to respond more quickly, with perfect steady-state behavior from r → y, the steady-state effect that r must have on u....”).

(e) For the closed-loop system, what is the steady-state gain from d → y, as a function of the design parameter γ? How is this affected as γ decreases?

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2. Basic System Properties: The equations    x(t) ˙ −2 1  y1 (t)   1 0     y2 (t)  =  2 4 y3 (t) 1 −1

320

governing a 3-input, 3-output system are   −1 3 x(t)   0 −1    u1 (t)    −3 0 u2 (t)  −2 1 u3 (t)

(a) Is the system stable?

(b) What is the time-constant of the system?

(c) What is the steady-state gain from u3 to y2 ?

(d) What is the instantaneous-gain from u2 to y2 ?

(e) What is the frequency-response function G(ω) from u1 to y3 ?

(f) Suppose x(0) = 10 and lim u1 (t) = 2,

t→∞

What is value of lim y3 (t) t→∞

lim u2 (t) = 1,

t→∞

lim u3 (t) = 1.

t→∞

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3. (a) Define the complex number G =

321

19.5 . j5+12

i. Find the value of |G|

ii. Find the value of ∠G

(b) Sketch the final output (in the axes) of the Matlab code. Carefully focus on the period, amplitude and time-alignment of the input signal (dashed) and response signal (solid). w = 5; TF = 10*2*pi/w; x0 = -2; uH = @(z) sin(w*z); fH = @(t,x) -12*x + 19.5*uH(t); [tSol, xSol] = ode45(fH,[0 TF],x0); plot(tSol, uH(tSol),’--’, tSol, xSol); % input=dashed; solution=solid xlim((2*pi/w)*[7 8]); % reset horz limits to exactly cover 1 period

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4. A closed-loop feedback system is shown below. Signals are labeled. Note that the one marked summing junction has a − sign, as typical of our negative feedback convention. Treat all unmarked summing junctions as +.

Fill in the 3 × 4 matrix which relates (x, r, d, n) to (x, ˙ y, u) as shown below





  ˙ x(t)  y(t)  =  u(t) (a) What is the time-constant of the closed-loop system?

(b) What is the steady-state gain from r → y?

(c) What is the steady-state gain from d → y?

(d) What is the instantaneous gain from r → y?

(e) What is the instantaneous gain from d → y?



x(t)   r(t)       d(t)  n(t) 

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(f) The two axes below show a specific reference input r (solid) and disturbance input d (dashed). These are the same in both axes. Assume n(t) ≡ 0 for all t. The closed-loop system starts from x(0) = −0.2, and is forced by this reference and disturbance input. Make careful sketches of y(t) and u(t) in the top and bottom axes, respectively (note that they are individually marked with task “Sketch Output y” and “Sketch Input u”).

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5. Consider the delay-differential equation x(t) ˙ = A1 x(t) + A2 x(t − T ) where A1 , A2 and T are real-valued constants. T ≥ 0 is called the “delay.” Depending on the values, there are 3 cases: • The system is unstable for T = 0; or • The system is stable for all T ≥ 0; or • The system is stable for T = 0, but unstable for some positive value of T . In this case, we are interested in the smallest T > 0 for which instability occurs, and the frequency of the nondecaying oscillation that occurs at this critical value of delay. Fill in the table below. In each row, please mark/check one of the first three columns (from the three cases above). If you check the 3rd column, then include numerical values in the 4th and 5th columns associated with the instability. Show work below. unstable stable for stable at T = frequency at smallest T at at T = 0 all T ≥ 0 0, but unstable at which instability which instability some finite T > 0 occurs occurs A1 = −1, A2 = −3 A1 = −4, A2 = −2 A1 = 1, A2 = −3 A1 = −2, A2 = 3

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6. Three first-order systems, Sys1, Sys2, Sys3, all stable, have the familar form x˙ 1 (t) = a1 x1 (t) + b1 u1 (t) y1 (t) = c1 x1 (t) + d1 u1 (t) {z } | Sys1

x˙ 2 (t) = a2 x2 (t) + b2 u2 (t) y2 (t) = c2 x2 (t) + d2 u2 (t) | {z } Sys2

x˙ 3 (t) = a3 x3 (t) + b3 u3 (t) y3 (t) = c3 x3 (t) + d3 u3 (t) | {z } Sys3

The Magnitude plot of the associated frequency-response functions G1 (ω), G2 (ω) and G3 (ω) are shown below (note, G1 is the frequency-response function of Sys1, etc).

The step-responses of the systems, labeled SR:A, SR:B and SR:C are shown below.

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Match each step-response with the corresponding Frequency-response magnitude plot (eg., is SR:A the step response of Sys1, Sys2 or Sys3?)

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30.2

328

Fall 2017 Midterm 2

1. For each matrix below, write the expression for eAt (a)  A1 =

1 3 −3 1



−2 0 0 4



,

eA1 t =

,

eA2 t =

(b)  A2 =

(c)  A3 =

−3 1 0 −3



eA3 t =

,

(d)  A4 =

−2 + j5 0 0 −2 − j5

 ,

eA4 t =

(e)  A5 =

0 −4 4 0

 ,

eA5 t =

2. Consider the quadratic polynomial p(s), which depends on two real-valued parameters β1 and β2 , p(s) = s2 − s + 1 + β1 (7s + 4) + β2 (5s + 3) Find the values of β1 and β2 so that the roots of p(s) are at {−2 + j1, −2 − j1}. Hint: What quadratic polynomial has roots at {−2 + j1, −2 − j1}

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3. When we studied studied how Simulink worked, we saw that in order to simulate an interconnection of dynamical systems, the code simply needed to “call” each individual system, often in a specific order, in order to determine the entire state-derivative, and then do this repeatedly to compute an approximate, numerical solution to the ODEs. This strategy of keeping all of the systems separate provides generality that is especially useful when simulating interconnections of systems that are nonlinear. For interconnections of linear systems (governed by state equations), we can often explicitly determine the state equations of the interconnection, since all of the necessary substitutions are simple (because of linearity). That is the task in this problem. Suppose the plant P is described by P:

x(t) ˙ = Ax(t) + Ed(t) + Bu(t) y(t) = Cx(t)

where A, E, B, C are matrices with dimensions A ∈ Rn×n ,

E ∈ Rn×v ,

B ∈ Rn×m ,

C ∈ Rq×n

u(t) ∈ Rm ,

y(t) ∈ Rq

and the signals are of dimensions x(t) ∈ Rn ,

d(t) ∈ Rv ,

Suppose the controller is also described by a linear system model, namely C:

z(t) ˙ = F z(t) + Gr(t) + Hym (t) u(t) = Jz(t) + Kr(t) + Lym (t)

where F, G, H, J, K, L are matrices of dimension F ∈ Rw×w ,

G ∈ Rw×f ,

H ∈ Rw×q ,

J ∈ Rm×w ,

K ∈ Rm×f ,

L ∈ Rm×q ,

and the signals are of dimensions z(t) ∈ Rw ,

r(t) ∈ Rf

The simple model for sensor-noise is ym (t) = y(t) + η(t) where η(t) ∈ Rq . The (familiar) block diagram is: d r C

-

y

P

-

-

ym

? u

+ ? g+

η

The task in this problem is to find the state-equation model for the closed-loop system,

ME 132, Fall 2018, UC Berkeley, A. Packard

with



 r inputs =  d  , η

330

 states =

x z



 ,

outputs =

y u



Task/Question: Fill in the “’block” 4 × 5 matrix the correctly describes the closedloop system.

 





  x(t)           z(t)            r(t)            d(t)         η(t)



 x(t)   ˙             z(t) ˙       =       y(t)             u(t)

Make sure that your matrix products are in the correct order (nothing is scalar here, so you should not be sloppy about order). If you have time, convince yourself that all the dimensions make sense! That will also help you find errors... 4. Suppose A ∈ R3×3 and AV = V Λ, where   −1 + j4 −1 − j4 0 2 + j1 1 , V =  2 − j1 6 6 −1



 −2 + j2 0 0 0 −2 − j2 0  Λ= 0 0 −3

Find matrices W ∈ R3×3 and Γ ∈ R3×3 (note - these are real-valued matrices, in contrast to V and Λ, which are complex) such that AW = W Γ where W is invertible, and Γ is “block-diagonal”. Note: You do not have to prove that W is invertible, but whatever you write down should be invertible. 5. A one-state plant, P is governed by equations x(t) ˙ = Ax(t) + B1 d(t) + B2 u(t),

y(t) = Cx(t)

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where x is the state of the plant, d is an external disturbance, and u is the control variable. The plant output is y. The constants A, B1 , B2 , C are referred to as the plant parameters, and are assumed known, with B2 6= 0 and C 6= 0. A feedback control system is proposed, which uses the reference input r and measures y (no measurement noise for this problem, to keep the notation to a minimum) to produce u. The goal of control is Goal1: closed-loop should be stable Goal2: the eigenvalues of the closed-loop system can be assigned to desired values by appropriate choices of the parameters within the controller’s equations. Goal3: steady-state gain from r → y should equal 1 Goal4: steady-state gain from d → y should equal 0 Goal5: the objective in Goal3 should be robust to “modest” changes in the plant parameters. Obviously, if Goal3 is unachievable, then Goal5 is also unachievable. Goal6: the objective in Goal4 should be robust to “modest” changes in the plant parameters. Obviously, if Goal4 is unachievable, then Goal6 is also unachievable. (a) Consider a proportional controller of the form u(t) = KP (r(t) − y(t)) Which goals are achievable (by proper choice of KP ), and which goals are unachievable (regardless of the choice)? Hint: if you are unsure about acheiving Goal1 and/or Goal2 for any of these problems, consider the plans x(t) ˙ = x(t) + u(t), y(t) = x(t), which is a simple unstable plant on which you can gain insight.

(b) Consider a proportional controller of the form u(t) = K1 r(t) + K2 y(t) Which goals are achievable (by proper choice of K1 and K2 ), and which goals are unachievable (regardless of the choice)?

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(c) Consider an integral controller of the form q(t) ˙ = r(t) − y(t),

u(t) = KI q(t)

Which goals are achievable (by proper choice of KI ), and which goals are unachievable (regardless of the choice)? (d) Consider a proportional/integral controller of the form q(t) ˙ = r(t) − y(t),

u(t) = KI q(t) + KP (r(t) − y(t))

Which goals are achievable (by proper choice of KI and KP ), and which goals are unachievable (regardless of the choice)?

6. A one-state plant, P is governed by equations x(t) ˙ = −4x(t) + d(t) + 3u(t),

y(t) = 2x(t)

where x is the state of the plant, d is an external disturbance, and u is the control variable. The plant output is y. A reference input r is available to the controller. (a) Design a PI controller of the form q(t) ˙ = r(t) − y(t),

u(t) = KI q(t) + KP (r(t) − y(t))

such that the closed-loop eigenvalues are given by (ξ = 0.9, ωn = 10).

(b) In the closed-loop system, what is the steady-state gain from r → y? (c) In the closed-loop system, what is the steady-state gain from d → y? (d) In the closed-loop system, what is the steady-state gain from d → u?

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30.3

333

Fall 2017 Final

#1

#2

#3

#4

#5

#6

#7

#8

20

15

20

15

15

18

15

20

#9

# 10

# 11

# 12

# 13

# 14

# 15

# 16

18

12

12

15

12

20

18

15

NAME

Facts: 1. 3rd order stability test: All roots of the third-order polynomial λ3 + a1 λ2 + a2 λ + a3 have negative real-parts if and only if a1 > 0, a3 > 0 and a1 a2 > a3 . 2. 4ht order stability test: All roots of the fourth-order polynomial λ4 + b1 λ3 + b2 λ2 + b3 λ+b4 have negative real-parts if and only if b1 > 0, b4 > 0, b1 b2 > b3 and (b1 b2 −b3 )b3 > b21 b4 . 3. If u(t) = u¯ (a constant) for all t ≥ 0, and A ∈ Rn×n is invertible, then the response of x(t) ˙ = Ax(t) + Bu(t) from initial condition x(0) = x0 is x(t) = eAt x0 + (eAt − I)A−1 B u¯. 4. The characteristic polynomial of the 1st order (vector) differential equation x(t) ˙ = Ax(t) is det(λIn − A), where n is the dimension of x. 5. The block diagram below is referred to as the “standard (P, C) feedback loop”, d r

+ - j −6

C

u- ? j -

P

-

? j 

y

n

6. The block diagram below is referred to as the “standard (P, C, F ) feedback loop”, d r

+ - j −6

C

u- ? j -

F



P

-

? j 

y

n

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1. Consider the following plant/controller transfer function pairs, for the standard (P, C) feedback configuration (see front page) 1 P1 = 2 , s P3 =

s + 13 , C1 = s+3

1 , (s − 1)(s + 20)

C3 =

P2 = 16(4s + 1) , s

1 , s(s + 10) P4 =

C2 = 1 , s2

20s + 8 s

C4 =

KP s + KI s

(a) What is the closed-loop characteristic polynomial for the (P1 , C1 ) pair (b) Is the (P1 , C1 ) closed-loop system stable? (c) What is the closed-loop characteristic polynomial for the (P2 , C2 ) pair (d) Is the (P2 , C2 ) closed-loop system stable? (e) What is the closed-loop characteristic polynomial for the (P3 , C3 ) pair (f) Is the (P3 , C3 ) closed-loop system stable? (g) What is the closed-loop characteristic polynomial for the (P4 , C4 ) pair (h) Are there any values of KP , KI such that closed-loop (P4 , C4 ) system is stable? (i) True/False: All 2nd order plants can be stabilized by a PI controller (j) True/False: Some unstable 2nd order plants can be stabilized by a PI controller 2. A stable linear system has a frequency-response function, denoted H(ω). (a) It is known that H(2) = 3 − 1j. What precise/concrete statement can be made about a particular response of the system?

(b) The state-space model for the system is the usual x(t) ˙ = Ax(t) + Bu(t),

y(t) = Cx(t) + Du(t)

where u and y are the input, and output, respectively. How are A, B, C, D and H(ω) related? (c) The transfer function of the system is denoted G(s) = function related to the frequency-response function?

n(s) . d(s)

How is the transfer

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3. The equations for a satellite orbiting a stationary mass are x˙ 1 (t) = x2 (t),

x˙ 2 (t) = x1 (t)x23 (t) −

β x21 (t)

,

x˙ 3 (t) =

1 (u(t) − 2x2 (t)x3 (t)) , x1 (t)

where x1 is the radius, x2 is the rate-of-change of radius, and x3 is the orbital angular velocity. The control u is a force applied in the tangential direction. The gravitational constant and mass of the stationary mass are collected into the single positive constant β. (a) Show that for any constant x¯1 > 0, there exist constants x¯2 , x¯3 and u¯ such that   x¯1  x¯2  , u¯ x¯3 is an equilibrium point of the system. (Note - you can choose the equilibrium point that has x¯3 > 0). (b) Find the Jacobian linearization of the system about the equilibrium point. 4. Suppose a linear system x(t) ˙ = Ax(t) + Bu(t) has state-space data     0 1 0 0    a 0 b , B = 0 , A= 0 −c 0 d where a, b, c, d are positive constants. (a) What is the characteristic polynomial of A? (b) Given only the information that a, b, c, d are all positive, can you conclude anything about the stability (ie., all eigenvalues have negative real-parts) of the system? (c) Define an output y(t) = Cx(t) with C=



0 0 1

What is the transfer function from u to y? 5. Take the plant P , described by its transfer function, s2 − 4 P (s) = s(s2 + 1) which is quite challenging to control.



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(a) Let u and y denote the input and output, respectively. What is the differential equation governing the relationship between u and y (b) Consider the standard (P, C) closed-loop configuration (see front page for definition). Using a proportional controller, C(s) = KP , a constant-gain, can the closed-loop system be made stable, by proper choice of KP ? (c) In terms of transfer functions, what is the simplest controller form, C(s) you can propose, such that if the controller coefficients are chosen properly, will render the closed-loop system (using standard (P, C) closed-loop configuration) stable? 6. Consider the standard (P, C) configuration. Let the transfer functions of P and C be denoted nP (s) nC (s) P (s) = , C(s) = dP (s) dC (s) Match the defined transfer functions Mi or Ni with the closed-loop transfer functions (eg., Gr→y means closed-loop transfer function from r to y)

M1 :=

dP dC nP dC dP nC nP nC , M2 := , M3 := , M4 := dP dC + nP nC dP dC + nP nC dP dC + nP nC dP dC + nP nC N1 = −M1 ,

(a) Gr→y = (b) Gr→u = (c) Gd→y = (d) Gd→u = (e) Gn→y = (f) Gn→u =

N2 = −M2

N3 = −M3

N4 = −M4

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7. Consider the standard (P, C, F ) configuration. Let the transfer functions of P and C be denoted nC (s) nF (s) nP (s) , C(s) = , F (s) = P (s) = dP (s) dC (s) dF (s) We are interested in the closed-loop transfer functions, expressed similarly to the previous problem (ie., simple fractions involving the individual n and d polynomials) (a) What is the transfer function from r → u (b) What is the transfer function from d → y (c) What is the transfer function from r → y (d) What is the closed-loop characteristic polynomial of the system 8. The governing equations for a DC motor are V (t) − RI(t) − Kω(t) = 0,

J ω(t) ˙ = KI(t) − αω(t) + T (t)

• V (t) is the voltage across the motor winding, at the terminals; I(t) is the current flowing through the motor windings; ω(t) is the angular velocity of the shaft; and T (t) is the sum of all torques applied externally to the motor shaft (interactions with other linkages/inertias, disturbances, etc) • J, K, α, R are positive constants, which are properties of the motor itself, namely the shaft/gear inertia, the motor-constant, the viscous friction coefficient of the bearings, and the electrical resistance in the windings. Consider two behaviors of the motor shorted: the terminals are connected together so that V (t) = 0 for all t open: the terminals are not connected to anything or to each other, so I(t) = 0 for all t (a) Treating T as the single input, and (ω, I) as the two outputs, write state equations for the shorted system (hint: there is 1 state, and in this case, 1 input, and 2 outputs) (b) For the shorted system, what is the time-constant? (c) For the shorted system, what is the transfer-function from T to ω? (d) For the shorted system, what is the steady-state gain from T to ω?

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(e) Treating T as the single input, and (ω, V ) as the 2 outputs, write state equations for the open system (f) For the open system, what is the transfer-function from T to ω? (g) For the open system, what is the steady-state gain from T to ω? (h) For the open system, what is the time-constant? (i) If you, with your hands/fingers, apply a torque to the two separate systems, which will be easier to turn? Why?

9. Consider the plant/controller pair (for the standard (P, C) configuration described on the front page), described in transfer-function form P (s) =

1 , s−α

C=

√

5 · α (≈ 2.24α)

where α > 0 is a constant. Note that C is a constant gain. Remark: For the questions below, if you have trouble, first work out the answers for the case α = 1, then go back and see how to generalize to an arbitrary, fixed, positive α. (a) Is the plant P a stable system? Is the closed-loop system stable? (b) Since the closed-loop system is a first-order system, what is the time-constant of the closed-loop system. Your answer will be in terms of α. (c) Define L(s) := P (s)C. Using mathematical manipulations, find the frequency ωc ≥ 0 such that |L(jωc )| = 1. Your answer will be in terms of α. (d) What is ∠L(jωc ), in radians? Hint: Your answer will not depend on α. (e) Make a simple sketch, in the complex-plane C, showing the unit-circle, and mark the value of L(jωc ). (f) What is the time-delay margin (in terms of α) for the closed-loop system?

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10. The Bode plots (Magnitude and Phase) for 4 simple transfer functions, G1 (s) =

1 , −s + 1

G2 (s) = −s + 1,

G3 (s) =

1 , s+1

are shown below. Match each system to its corresponding plot

G4 (s) = s + 1

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341

11. The Bode plots (Magnitude and Phase) for 4 simple transfer functions, G1 (s) =

20 , s + 20

G2 (s) =

0.2 , s + 0.2

G3 (s) =

0.05 , s + 0.05

are shown below. Match each system to its corresponding plot

G4 (s) =

5 s+5

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12. Make a straight-line Bode plot (magnitude and phase) for the transfer function G(s) = 9

(−s + 100)(s + 1) (s + 3)(s + 10)(s + 30)

on the graph paper below. Be sure to normalize the terms properly (ie., write (s + 30) s as ( 30 + 1), and account for all the accumulated scaling factors at the end with a single gain adjustment). Important: The axes-limits on this graph paper are appropriate for the final result. Use the graph-paper on the next page (which has wider limits, and easier to get started with) to work out your solution, then transfer it to this page.

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13. State whether each statement is True or False. No reasons need be given... (a) For all A1 , A2 ∈ Rn×n , and all t ≥ 0, e(A1 +A2 )t = eA1 t eA2 t (b) For all A ∈ Rn×n , and all t ∈ R, e(−A)t = eA(−t) (c) For all A ∈ Rn×n , and all t ≥ 0, eAt e−At = In (d) For all A ∈ Rn×n , and all t ∈ R, the matrix eAt is invertible. (e) For all A ∈ Rn×n , and all t1 , t2 ∈ R, eA(t1 +t2 ) = eAt2 eAt1 (f) For all A ∈ Rn×n , and all t ∈ R, AeAt = eAt A (g) For all A1 , A2 ∈ Rn×n , and all t ≥ 0, A1 eA2 t = eA1 t A2 (h) If A ∈ Rn×n is invertible, then for all t ∈ R, A−1 eAt = eAt A−1 (i) If A ∈ Rn×n , then for all t ∈ R, and all ω ∈ R, (jωI − A)eAt = eAt (jωI − A) (j) If A ∈ Rn×n has no imaginary eigenvalues, then for all t ∈ R, and all ω ∈ R, (jωI − A)−1 eAt = eAt (jωI − A)−1

(k) Consider the two expressions E1 := (I + tA +

t2 t2 2 A )(I + tB + B 2 ), 2 2

E2 := I + t(A + B) +

t2 (A + B)2 2

where A, B ∈ Rn×n . Expand both expressions, and obtain the coefficient-matrix associated with the t2 term. Are they equal, in general? (l) Which disequality (circle your answer) comes from the reasoning in the previous problem e(A−B)t 6= eAt e−Bt , BeAt 6= eBt A, eABt 6= eAt eBt

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14. A 2nd-order system, which represents the dynamical equations for a feedback controller, has state equations        z˙1 (t) 1 2 z1 (t) 5 = + (r(t) − ymeas (t)) z˙2 (t) 3 4 z2 (t) 6 and output equation u(t) =



7 8





z1 (t) z2 (t)

 + 9(r(t) − ymeas (t))

The numbers are not necessarily realistic, but that is not the focus of the problem. Using the same ideas as in the lab, this control “strategy” is to implemented at a sample-rate of 10 milliseconds (0.01 seconds). Accessing the reference r and the measurement y is accomplished with 2 external functions that we are free to call as needed: • The reference signal is computed by an externally defined function named getReference. That function has one input argument (time, as an integer, in units of milliseconds), and it returns a float. • At any time, the process output, y can be measured and its value returned with the function makeMeasurement. This function has no input arguments, and returns a float. The control action, u(t) can be “sent” to the actuator with the command setControlValue. This function has one input argument, the value of the control action (as a float). The behavior of the function is that the control signal immediately gets set to the specified value, and is held constant (at this value), until the function is called again. On the next page, fill in the blank lines of the code to complete this program.

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int T1Val, ExpLength = 30000, SampleRate = _____ ; float rVal, yVal, eVal, uVal; float z1, z2, z1Dot, z2Dot; z1 = 0.0; z2 = 0.0; float SampleTime; SampleTime = 0.001*SampleRate; clearTimer(T1); clearTimer(T2); // Begin control at t=0 setControlValue(9*(getReference(0)-makeMeasurement)); while (time1[T1] < ExpLength) { T1Val = time1[T1]; if (time1[T2]>= ______________ ) { clearTimer(_______); ______ = makeMeasurement; rVal = ________________________; _______ = rVal - yVal; uVal = _______________________________________; setControlValue(_______________); z1Dot = ___________________________________________________________; z2Dot = ___________________________________________________________; z1 = ___________

+

________________________________;

z2 = ___________

+

________________________________;

} } 15. Let σ, β ∈ R, with σ 2 + β 2 > 0 (in other words, at least one of them is nonzero).

ME 132, Fall 2018, UC Berkeley, A. Packard

Consider

349

 A :=

σ β −β σ



(a) What are the eigenvalues of A

(b) Is A invertible?

(c) What is the inverse of A

(d) A is a special matrix we studied in class. What is eAt

(e) What is (sI2 − A)−1

(f) What is A−1 eAt − In




16. Consider the linear system x(t) ˙ = Ax(t) + Bu(t), y(t) = Cx(t) for       −4 3 −2 A := , B := , C= 1 0 −3 −4 3 (a) What is the transfer function from u to y

(b) What is the exact expression for the response y(t) due to a unit-step input, u(t) = 1 for all t, starting from initial condition x(0) = 0. See problem 15, and Facts on front page if needed.

ME 132, Fall 2018, UC Berkeley, A. Packard

30.4

350

Fall 2015 Midterm 1

1. Consider the delay-differential equation x(t) ˙ = A1 x(t) + A2 x(t − T ) where A1 , A2 and T are real-valued constants. T ≥ 0 is called the “delay.” Depending on the values, there are 3 cases: • The system is unstable for T = 0; or • The system is stable for all T ≥ 0; or • The system is stable for T = 0, but unstable for some positive value of T . In this case, we are interested in the smallest T > 0 for which instability occurs, and the frequency of the nondecaying oscillation that occurs at this critical value of delay. Fill in the table below. In each row, please mark/check one of the first three columns (from the three cases above). If you check the 3rd column, then include numerical values in the 4th and 5th columns associated with the instability. Show work below. unstable stable for stable at T = frequency at smallest T at at T = 0 all T ≥ 0 0, but unstable at which instability which instability some finite T > 0 occurs occurs A1 = −2, A2 = 3 A1 = −4, A2 = 1 A1 = −2, A2 = −1 A1 = −1, A2 = −2 A1 = 1, A2 = −4 A1 = 0, A2 = −1 2. A stable first-order system has • a time-constant of 1; • a steady-state gain of 1.5; and • an instantaneous gain of −0.5.

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(a) Sketch the approximate response of the system (starting from 0-initial condition) to the input shown. Remember that the steady-state gain is not equal to 1.

(b) Find values a, b, c, d such that system x(t) ˙ = ax(t) + bu(t) y(t) = cx(t) + du(t) has the specified properties (ie., time-constant, steady-state gain, and instantaneous gain) as given above. Hint: Correct answer is not unique - different combinations of (a, b, c, d) all combine to achieve these 3 specified properties. 3. (a) Define the complex number g = i. ii. iii. iv.

Find Find Find Find

the the the the

value value value value

of of of of

5 . j4+3

Re(g) Im(g) |g| ∠g

(b) Sketch the final output (in the axes) of the Matlab code uH = @(z) sin(4*z); fH = @(t,x) -3*x + 5*uH(t); [tSol, xSol] = ode45(fH,[0 100],-6); plot(tSol, uH(tSol),’--’, tSol, xSol); % input=dashed; solution=solid xlim((2*pi/4)*[20 21]); % resets horizontal limits

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4. A process, with input u, disturbance d and output y is governed by x(t) ˙ = 2x(t) + d(t) + 3u(t),

y(t) = x(t)

(a) Is the process stable?

(b) Suppose x(0) = 1, and u(t) = d(t) ≡ 0 for all t ≥ 0. What is the solution x(t) for t ≥ 0.

(c) Consider a proportional-control strategy, u(t) = K1 r(t) + K2 [r(t) − y(t)]. Determine the closed-loop differential equation relating the variables (x, r, d).

(d) For what values of K1 and K2 is the closed-loop system stable?

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(e) As a function of K2 , what is the steady-state gain from d → y in the closed-loop system?

(f) As a function of K1 and K2 , what is the steady-state gain from r → y in the closed-loop system?

(g) Choose K1 and K2 so that the steady-state gain from r → y equals 1, and the steady-state gain from d → y equals 0.1.

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(h) With those gains chosen, sketch (try to be accurate) the two responses y(t) and u(t) for the following situation:   0 for 0 ≤ t ≤ 1 0 for 0 ≤ t ≤ 2 x(0) = 0, r(t) = , d(t) = 1 for 1 < t 1 for 2 < t

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5. Consider our standard feedback interconnection, consisting of a 1st-order linear controller and proportional plant described by controller :

x(t) ˙ = Ax(t) + B1 r(t) + B2 ym (t) u(t) = Cx(t) + D1 r(t)

plant : y(t) = αu(t) + βd(t) The measurement equation is ym (t) = y(t) + n(t). (a) Under what conditions (on the parameters A, B1 , . . . , D1 , α, β) is the closed-loop system stable? Carefully justify your answer.

(b) Is it possible that the controller, as a system by itself, is unstable, but the closedloop system is stable. If your answer is “No”, please give a careful explanation. If your answer is “Yes”, please give a concrete example.

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(c) Suppose α = β = 1. Design the controller parameters so that • the closed-loop is stable, with specified time-constant, τdesired = 0.2. • The steady-state gain from r → y is 1, even if α, β change by modest amounts (but do not change sign) after the controller has been designed and implemented? • The steady-state gain from d → y is 0, even if α, β change by modest amounts (but do not change sign) after the controller has been designed and implemented? Show your work, and clearly mark your answers.

(d) In the design above, the two steady-state gains (r → y and d → y) are completely insensitive to modest changes in α and β. What are some important closed-loop properties that do change if α and β vary?

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Fall 2015 Midterm 2

1. Remark: This is not a hard problem. Do not be scared off by the long description. A state-estimator, E, is a dynamical system that attempts to estimate the internal state x of a system G, by observing only the inputs, u, and outputs, y, of G. The estimator does not have access to the initial condition, x(0) of the process. The system G is assumed to be governed by a known, linear, state-space model, namely x(t) ˙ = Ax(t) + Bu(t) y(t) = Cx(t) Since A, B, C are known, and the estimator has access to u and y, the strategy is that the estimator should be a mathematical copy of the process G, with state xˆ. The goal is to get xˆ to converge to x, asymptotically in time (regardless of u). Hence, it makes sense that the estimator’s input should be the same input as the process (u), and the state, xˆ, inside the estimator, should be adjusted in some manner proportional to the difference between the process output y, and the estimator’s prediction of y, in a way that makes xˆ(t) − x(t) → 0 as t → ∞, regardless of u and x(0). A diagram of such a system is shown below. Note that the estimator E has two inputs, “receiving” both u and y, and produces one output, namely producing the estimate, xest = xˆ of x. u

-

B

- f x˙ 6

R

y

xC

A  G − ? f + 6

L  -

B

?x ˆ˙ -f 6

R

A 

xˆ C

yˆ

est

x E

R All boxed quantities, A, B, C, L, F are matrices. The boxes labeled are integrators (note the signal definitions for x and xˆ and their time derivatives). All summing junction sign conventions, unless otherwise marked, are positive.

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(a) Define  z(t) :=

x(t) xˆ(t)



Fill in the matrices below, to complete the state-space model which governs the entire (process and estimator) system       z(t) ˙ = 

     z(t) +   

   u(t) 

and "

#

est

x (t) =

z(t)

(b) Define the estimation error, e as e(t) := x(t)− xˆ(t). Show that e satisfies a simple, linear differential equation, whose right-hand side only involves, A, C, L and e.

(c) What is one important property that the matrix L must have in order for xˆ(t) − x(t) → 0 as t → ∞, regardless of u, x(0) and xˆ(0)?

(d) What are some issues that are not addressed here, that might impact (and distinguish among) several various choices for L?

2. Suppose 0 < ξ < 1 and ωn > 0. Consider the system ˙ y¨(t) + 2ξωn y(t) ˙ + ωn2 y(t) = ωn u(t) subject to the initial conditions y(0− ) = 0, y(0 ˙ − ) = 0, and the unit-step forcing function, namely u(t) = 0 for t = 0, and u(t) = 1 for t > 0. Show that the response is p  1 −ξωn t 2 p e sin 1 − ξ ωn t y(t) = 1 − ξ2 Hint: Recall that the set of all real-valued homogeneous solutions ˙ + p  p of y¨(t) + 2ξω  n y(t) 2 −ξωn t −ξω t ωn y(t) = 0 is yH (t) = Ae cos 1 − ξ 2 ωn t + Be n sin 1 − ξ 2 ωn t where A and B are any real numbers.

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3. Consider the 2-state system governed by the equation x(t) ˙ = Ax(t). Shown below are the phase-plane plots (x1 (t) vs. x2 (t)) for 4 different cases. Match the plots with the A matrices, and correctly draw in arrows indicating the evolution in time.

 A1 =

−2 0 3 1



 ,

A2 =

1 3 −3 1



 ,

A3 =

5

−3 2 −1 0



 ,

A4 =

−1 3 −3 −1



2

4

4

x2

x2

2 0

0 −2

−5 −5

0 x1

−4 −4

5

0

−5 −5

0 x1

5

x2

x2

5

−2

0 x1

5

0

−5 −5

0 x1

4. (a) What is the general form of the solution to the differential equation x¨(t) + 6x(t) ˙ + 5x(t) = 0

5

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(b) What is the general form of the solution to the differential equation x¨(t) + 6x(t) ˙ + 5x(t) = −10

Your expressions both should have two free constants. 5. A first-order process, with state x, input u, disturbance d and output y is governed by x(t) ˙ = x(t) + u(t) + d(t),

y(t) = x(t)

(a) Is the process stable?

(b) Suppose x(0) = −3, and u(t) = d(t) ≡ 0 for all t ≥ 0. What is the solution y(t) for t ≥ 0.

(c) A PI (Proportional plus Integral) controller is proposed u(t) = KP [r(t) − y(t)] + KI z(t) z(t) ˙ = r(t) − y(t) Define the state-vector q as  q(t) :=

x(t) z(t)



Write the state-space model of the closed-loop system, with state q, inputs (d, r) and outputs (y, u).

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(d) What is the closed-loop characteristic polynomial?

(e) For what values of KP and KI is the closed-loop system stable?

(f) The closed-loop system is 2nd order. What are the appropriate values of KP and KI so that the closed-loop system eigenvalues are described by ξ = 0.8, ωn = 0.5?

(g) What are the appropriate values of KP and KI so that the closed-loop system eigenvalues are described by ξ = 0.8, ωn = 1.0?

(h) What are the appropriate values of KP and KI so that the closed-loop system eigenvalues are described by ξ = 0.8, ωn = 2.0?

6. A 2nd-order, unstable process, with control input u, disturbance input d, and output y, is governed by the equation y¨(t) + y(t) ˙ − y(t) = u(t) + d(t) A PI (Proportional plus Integral) controller is proposed, both to stabilize the system, and provide good disturbance rejection, u(t) = KP [r(t) − y(t)] + KI z(t) z(t) ˙ = r(t) − y(t) Here r is a reference input. (a) Using just the controller equations, express u(t) ˙ in terms of r, r, ˙ y and y. ˙

(b) By differentiating the process equation, and substituting, derive the closed-loop differential equation relating r and d (and possibly their derivatives) to the output variable y (and its derivatives). The variable u should not appear in these equations.

(c) Using the 3rd-order test for stability, determine the conditions on KP and KI such that the closed-loop system is stable.

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Fall 2015 Final

1. For each question, state whether the claim is True or False. Give a concise justification. (a) True/False: First-order (1-state) linear systems can have an oscillatory freeresponse

(b) True/False:The steady-state response of a stable linear system, due to a sinusoidal input, depends on the initial condition

(c) True/False:The 2-degree-of-freedom PI controller (with gains K1 , K2 , KI ) z(t) ˙ = r(t) − y(t) u(t) = K1 r(t) − K2 y(t) + KI z(t) can stabilize some unstable linear plants that the 1-degree-of-freedom PI controller (with gains KP , KI ), z(t) ˙ = r(t) − y(t) u(t) = KP (r(t) − y(t)) + KI z(t) cannot stabilize.

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(d) Fact: Plants (call these “class L”) whose input signals (u) are limited between fixed upper and lower bounds (for example, it must be that −1 ≤ u(t) ≤ 1) are more challenging to control than plants whose input signals can take on any values (call these plant, with no limit on u, “class U”). i. True/False: AntiWindup logic fixes this deficiency, and make the two types of plants (class L and class U) equally easy to control, regarding reference tracking, disturbance rejection and noise insensitivity.

ii. True/False: AntiWindup logic can be especially useful in dealing with plants from class L, when being controlled with a proportional controller.

iii. True/False: AntiWindup logic can be especially useful in dealing with plants from class L, when being controlled with an integral controller.

iv. True/False: AntiWindup logic can be especially useful in dealing with plants from class L, when being controlled with an proportional+integral (PI) controller.

v. True/False: For AntiWindup logic to be effective, the lower limit on u must be equal and opposite to the upper limit on u

vi. True/False: Implementing AntiWindup logic in a PI controller is very complicated and computationally expensive, making it difficult and challenging to implement in many cases.

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(e) True/False: For a nonlinear system x(t) ˙ = f (x(t), u(t)), with equilibrium point (¯ x, u¯), the Jacobian linearization w(t) ˙ = Aw(t) + Bv(t) describes the approximate behavior of w(t) := x(t) − x¯ and v(t) := u(t) − u¯, while w and v remain small.

(f) True/False: For a nonlinear system x(t) ˙ = f (x(t), u(t)), with equilibrium point (¯ x, u¯), the Jacobian linearization w(t) ˙ = Aw(t) + Bv(t) describes the approximate behavior of w(t) := x(t) − x¯ and v(t) := u(t) − u¯, while x and u remain small.

(g) True/False: Since many control systems are used to regulate a system near an equilibrium point, the combination of Jacobian linearization and control techniques which apply to linear systems are a powerful combination to design control systems for nonlinear plants that are regulated near an equilibrium point.

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(h) True/False: The plant P (s) =

(s + 3)(s − 10) (s − 1)(s + 2)(s2 − 0.4s + 1)

can be stabilized with a controller of the form b0 s 3 + b1 s 2 + b2 s + b3 C(s) = 3 s + a1 s 2 + a2 s + a3 by proper choice of the {bi , ai } coefficients. (i) True/False: The plant P (s) =

(s + 3)(s − 10) (s − 1)(s + 2)(s2 − 0.4s + 1)

cannot be stabilized with a controller of the form C(s) =

b0 s 4 + b1 s 3 + b2 s 3 + b 3 s + b4 s(s3 + a1 s2 + a2 s + a3 )

by any choice of the {bi , ai } coefficients, because of the extra “s” term in the denominator. (j) True/False: Any 3rd-order plant of the form P (s) =

b1 s 2 + b 2 s + b3 s 3 + a1 s 2 + a2 s + a3

can be stabilized with a controller of the form e0 s + e1 C(s) = s + f1 by proper choice of the {e0 , e1 , f1 } coefficients, but the roots of the closed-loop characteristic equation (which is 4th order) cannot be arbitrarily assigned since there are only 3 free parameters, (e0 , e1 , f1 ), to work with. 2. Consider the system described by state equations x˙ 1 (t) = x1 (t) + 3x2 (t) + u(t) x˙ 2 (t) = −x1 (t) − 2x2 (t) + u(t) y(t) = x2 (t) with state x, input u and output y. (a) Is the system stable? (b) Eliminate x from the state equations below to obtain the differential equation relating input u and output y (c) What is the peak-magnitude of the steady-state response of the system due to the input u(t) = sin t?

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3. Find A, B C and D matrices, of appropriate dimension, such that the relationship between u and y in the state-space model x(t) ˙ = Ax(t) + Bu(t) y(t) = Cx(t) + Du(t) is y [3] (t) + 4y [2] (t) − y [1] (t) + 7y(t) = −2u[1] (t) + 7u(t) 4. A plant P is described by the differential equation y(t) ˙ = 2y(t) + u(t) + d(t), where u represents a control-input, and d is a disturbance. (a) Using a proportional controller, u(t) = KP (r(t) − y(t)), what is the condition on KP for closed-loop stability?

(b) Assuming KP is chosen so that the closed-loop system is stable, what are the following quantities i. closed-loop time constant

ii. closed-loop steady-state gain from r → y

iii. closed-loop steady-state gain from d → y

(c) Using a 2 degree-of-freedom proportional controller, u(t) = K1 r(t) − K2 y(t), what are the conditions on K1 and K2 for closed-loop stability?

(d) Assuming (K1 , K2 ) are chosen so that the closed-loop system is stable, what are the following quantities i. closed-loop time constant

ii. closed-loop steady-state gain from r → y

iii. closed-loop steady-state gain from d → y

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(e) Using a integral controller, u(t) = KI z(t); z(t) ˙ = r(t) − y(t), what is the condition on KI for closed-loop stability?

(f) Using a PI controller, u(t) = KI z(t) + KP (r(t) − y(t)); z(t) ˙ = r(t) − y(t), what are the conditions on (KI , KP ) for closed-loop stability

(g) Assuming (KI , KP ) are chosen so that the closed-loop system is stable, what are the following quantities i. closed-loop damping ratio

ii. closed-loop steady-state gain from r → y

iii. closed-loop steady-state gain from d → y

5. In this problem, we see how an adjustment to the right-hand-side of an ODE model can approximately model the effect of a small time-delay. (a) Suppose b is a fixed real number. Solve (for c1 and c2 ) the system of simultaneous linear equations below −1 + c1 + c2 = 0,

−c1 + c2 = b

(b) Calculate the unit-step response (input u, output y) for the differential equation y¨(t) − y(t) = bu(t) ˙ + u(t) Initial conditions are y(0−1 ) = 0, y(0 ˙ − ) = 0. Your answer should involve b and t.

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(c) Take b = 0. Plot (accurately, use calculator to compute y(t) at t = 0.1, 0.2, . . . , 1.0, 1.5 - this may take a few minutes to do correctly) the step response on the graph paper provided. While plotting this, also plot the response delayed (in time) by 0.1, which would be the step-response of the system y¨(t) − y(t) = u(t − 0.1)

(d) On the same plot (again, more calculator computations), put hollow dots (o) for the response of y¨(t) − y(t) = −0.1u(t) ˙ + u(t) Note that the inclusion of the −0.1u(t) ˙ term almost “perfectly” mimics the effect of the delay

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(e) What is the transfer function, from u to y of the system described by y¨(t) − y(t) = −0.1u(t) ˙ + u(t) Call this Papp . (f) Design a 1st-order controller C so that the closed-loop system with Papp and C has poles at {−1, −3, −20}. 6. A nonlinear first-order system is governed by x(t) ˙ = sin x(t) cos x(t) + tan u(t) where u is the input (restricted to − π2 < u(t) < π2 ). (a) Show that (¯ x = 0, u¯ = 0) is an equilibrium point of the system (b) What is the Jacobian linearization of the system about the equilibrium point (¯ x = 0, u¯ = 0)? (c) Find an equilibrium point whose Jacobian linearization is stable. 7. The governing equations for a DC motor, with shaft moment-of-inertia J are V (t) − I(t)R − Kω(t) = 0,

J ω(t) ˙ = KI(t) − αω(t) + T (t)

where V is the voltage across the terminals, I is the current flowing in the windings, ω is the angular velocity, the term −αω represents a bearing friction, and T is any additional (external) torque on the shaft. A schematic is shown below

In class, some students pointed out that by connecting two motors together, rotating one shaft (with your hand, say) causes the other shaft to rotate (without any external voltage applied). If we connect the electrical windings of two motors together, then the voltage across each set of terminals is the same (V1 = V2 and the currents are opposite (I1 = −I2 ). A schematic is below

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Make the following assumptions • The physical properties of the two motors are identical (so R1 = R2 =: R, K1 = K2 =: K and α1 = α2 =: α) • No external torque acts on shaft #2, so T2 = 0 (only torques from the current (the KI term) and the frictional torque (−αω2 ) act). • Ignore the equation J1 ω˙ 1 (t) = K1 I1 (t) − α1 ω1 (t) + T (t). Instead, we will treat ω1 as an “input” to the overall system

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Answer the questions below for this two-motor system. (a) Treating ω1 as the input, and ω2 as the output, eliminate I from these equations and get a differential equation for ω2 , with ω1 appearing as a forcing function. (b) What is the transfer function from ω1 to ω2 ? (c) All physical constants (R, J, K, α) are positive. Show that the steady-state gain from ω1 to ω2 is less than 1. (d) If I rotate shaft #1 with my finger at a constant angular velocity, what can you say about the resulting angular velocity of shaft # 2. 8. Consider the 2-state linear system        x˙ 1 (t) 1 3 x1 (t) 0 = + u(t) x˙ 2 (t) −1 −2 x2 (t) 1 Three LQR (linear quadratic regulator) designs were performed, with cost functions Z ∞ J1 = 20x21 (t) + 20x22 (t) + u2 (t)dt, 0 ∞

Z

x21 (t) + x22 (t) + u2 (t)dt,

J2 = 0

Z J3 =

∞

0.05x21 (t) + 0.05x22 (t) + u2 (t)dt

0

resulting in optimal feedback laws u(t) = F1 x(t),

u(t) = F2 x(t),

u(t) = F3 x(t)

Below are simulations ˙ = (A+BFi )x(t), from the initial   of the closed-loop systems, x(t) 1 condition x0 = . Also shown is the optimal input, u(t) versus t. 1 The state-response simulations are labeled StateSim X, StateSim Y and StateSim Z (one corresponds to F1 , one to F2 and one to F3 ). The input simulations are labeled InputSim D, InputSim E and InputSim F (again, one corresponds to F1 , one to F2 and one to F3 ). The feedback gains (listed after the plots) are labeled Gain H, Gain L and Gain N (one of these equals F1 , one equals F2 and one equals F3 ). (a) Match the state responses (X, Y, Z) to the cost function (J1 , J2 , J3 ) (b) Match the input responses (D, E, F) to the cost function (J1 , J2 , J3 ) (c) Match the feedback gain (H, L, N) to the cost function (J1 , J2 , J3 ) Briefly explain your answer.

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GainN =



 0.1498 0.2245 ,

GainH =

372



 1.2974 1.5755 ,

GainL =



6.3530 5.8815



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9. In lab, we discovered a simple architecture that would lead to some form of approximate coordination and synchronization of two identical systems, being acted upon by separate external forcing functions. In this problem, we extend the idea to 3 systems, and carry out the same analysis. For simplicity, this problem should be solved completely in transfer function notation, simply applying arithmetic rules. We could verify everything in terms of state-space models (as we did in the lab), but there is not time for that here. Again, just manipulate with addition, subtraction, multiplication, and division, as appropriate for transfer functions. The 3 identical systems are each forced with a control input, Ui and an external input Fi . Each of the systems has a transfer function G, hence Yi = G(Ui + Fi ),

i = 1, 2, 3

The architecture for control generalizes the 2-system case. Let K be the transfer function of a linear system, and use U1 = U2 = U3 =

1 K(Y2 3 1 K(Y3 3 1 K(Y1 3

− Y1 ) + 31 K(Y3 − Y1 ) = 13 K(−2Y1 + Y2 + Y3 ) − Y2 ) + 31 K(Y1 − Y2 ) − Y3 ) + 31 K(Y2 − Y3 )

(a) Find Y1 + Y2 + Y3 in terms of F1 , F2 , F3 .

(b) Based on part (a), fill in the sentence: The equals the transfer function times the inputs.

of the systems outputs of the external

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(c) By careful (but simple) manipulation, find a closed-loop expression for Y2 − Y1 , in terms of K, G and F1 , F2 , F3 . (s) and G(s) = (d) If K(s) = ndKK (s) of the closed-loop system?

nG (s) , dG (s)

what do you think is the characteristic equation

(e) By symmetry, what are the closed-loop expressions for Y3 − Y2 and Y3 − Y1 ? 10. Consider the diagram below. Here, m = 0.1 and τ = 0.004. If we choose KD = 5.1, KP = 120.4, KI = 708.2 it is possible (you do not need to) to verify that the closed-loop system is stable. W R

- e - KP s+KI s −6

1 - e Aτ s+1 −6

-? e -

KD 

D

1 ms

E-

1 s

Y -

B C

In order to assist you in the question below, Bode plots of certain transfer functions listed below are given in the following pages (not all may be useful...). L1 (s) =

KD s2 + KP s + KI mτ s4 + ms3 + KD s2 + KP s + KI KD mτ s2 + ms KP s + K I L5 (s) = mτ s4 + ms3

L3 (s) =

L2 (s) =

KD s2 + KP s + KI mτ s4 + ms3

KD s2 mτ s4 + ms3 + KP s + KI KP s + KI L6 (s) = 4 mτ s + ms3 + KD s2

L4 (s) =

Using the graphs (estimate values as best as you can...), answer the following margin questions. Explain any work you do, and make relevant marks on the Bode plots that you use in your calculations. (a) What is the gain margin at location A? (Hint – first determine what is the appropriate L for margin calculations at A, match with the L’s, and do calculation from supplied graphs). Write your answers and calculations on the page containing the relevant Bode plot, and write “LOCATION A” on that pager as well. (b) What is the time-delay margin at location A?

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(c) What is the gain margin at location B? Write your answers and calculations on the page containing the relevant Bode plot, and write “LOCATION B” on that page as well. (d) What is the time-delay margin at location B? (e) What is the gain margin at location C? Write your answers and calculations on the page containing the relevant Bode plot, and write “LOCATION C” on that pager as well. (f) What is the time-delay margin at location C? (g) What is the gain margin at location D? Write your answers and calculations on the page containing the relevant Bode plot, and write “LOCATION D” on that pager as well. (h) What is the time-delay margin at location D? (i) What is the gain margin at location E? Write your answers and calculations on the page containing the relevant Bode plot, and write “LOCATION E” on that pager as well. (j) What is the time-delay margin at location E?

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Spring 2014, Midterm 1

#1

#2

#3

#4

#5

8

14

10

8

10

1. Consider the following Matlab code: wH = @(t) t>4;

NAME

381

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f = @(x,u) -x+2*u; f45 = @(t,xt) f(xt,wH(t)); [tSol,xSol] = ode45(f45,[0 6],1); plot(tSol, xSol, ’--’, tSol, wH(tSol)); In the axes below, sketch the result produced by the plot command above.

382

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2. A plant is described by a simple proportional model relating the control input (u) and disturbance input (d) to the output, namely y(t) = αu(t) + βd(t). An integral controller is implemented, of the form x(t) ˙ = r(t) − y(t),

u(t) = KI x(t)

(a) In terms of α, β and KI , what are the conditions such that closed-loop system is stable?

(b) Assuming the the closed-loop is stable, in terms of α, β and KI , what is the steady-state gain from r to y?

(c) Assuming the the closed-loop is stable, in terms of α, β and KI , what is the steady-state gain from d to y?

(d) Assuming the the closed-loop is stable, in terms of α, β and KI , what is the steady-state gain from d to u?

(e) Explain how the answer in part 2d is consistent with the answer in part 2c.

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(f) Suppose α = 2 and β = 1 are the parameters of the plant. Design KI so that the closed-loop time-constant is 41 . Assuming an initial condition of x(0) = 0, a reference input r and disturbance d, shown below are applied. On the two graphs provided, accurately sketch the response (u and y).

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3. What is the smallest value of T > 0 such that for some real-valued ω, the function z(t) = sin(ωt) satisfies the differential equation z(t) ˙ = −4z(t − T )

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4. Consider our standard feedback interconnection, consisting of a 1st-order linear controller and proportional plant described by controller :

x(t) ˙ = Ax(t) + B1 r(t) + B2 ym (t) u(t) = Cx(t) + D1 r(t)

plant : y(t) = αu(t) + βd(t) The measurement equation is ym (t) = y(t) + n(t). (a) Under what conditions (on the parameters A, B1 , . . . , D1 , α, β) is the closed-loop system stable? Carefully justify your answer.

(b) Is it possible that the controller, as a system by itself, is unstable, but the closedloop system is stable. If your answer is “No”, please give a careful explanation. If your answer is “Yes”, please give a concrete example.

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5. A stable linear system of the form z(t) ˙ = Az(t) + Bu(t) is forced with a sine-wave input, u(t) = sin 3t. (a) What is the frequency of the resulting sinusoidal steady-state response z(t)?

(b) The resulting sinusoidal steady-state response z(t) has a magnitude of “lags” the input u by 18 of a period. Determine A and B.

firstOrderSystemID

√

2, and

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388

Spring 2014, Midterm 2

1. Consider the following Matlab code: P = ss(1,2,3,0); P.u = ’uP’; P.y = ’yP’; C = ss(0,1,2,1); C.u = ’e’; C.y = ’ucmd’; S1 = sumblk(’e=r-ym’); S2 = sumblk(’uP = ucmd + d’); S3 = sumblk(’ym = yP + n’); H = connect(P,C,S1,S2,S3,{’r’,’d’,’n’},{’yP’,’ucmd’}); H.a H.d (without worrying about the format) Write down, separately, the two matrices that will be displayed in the Command Window, due to the last two commands. 2. An unstable plant is described by a 1st-order linear differential equation x(t) ˙ = x(t) + d(t) + u(t) y(t) = x(t) The variables are: x is the state of the plant, d is the disturbance input, u is the control input, and y is the plant output. There is a measurement noise, n, so that the measured output ym is defined by ym (t) = y(t) + n(t). Design a 1st-order linear control system with the following properties: • There are two inputs to the controller: a reference input r, and the measured plant output, ym ; The controller has one state, you can call it z, or whatever your favorite letter is for a controller state; There is one output of the controller, u, which becomes an input to the plant (ie., the “control input”) • The 2nd-order closed-loop system is stable. In fact, expanding on closed-loop stability, the eigenvalues of the closed-loop “A” matrix should be at −2 and −3. • The steady-state gain from r to y should be 1. The steady-state gain from d to y should be 0. The instantaneous-gain from r to u should be 0. 3. The matrix A is

 A=

(a) What are the eigenvalues of A?

−3 2 −1 0



ME 132, Fall 2018, UC Berkeley, A. Packard

(b) Find a 2 × 2 diagonal matrix Λ such that AV = V Λ, where V is given by   1 2 V = 1 1

(c) Confirm that the inverse matrix of V is   −1 2 −1 V = 1 −1

(d) Find eAt .

389

ME4.132, 2018, UC Berkeley, A.plot Packard TheFall step-response and Bode of the system x(t) ˙ = −x(t) + u(t), y(t) = x(t)390 is shown below. It is labeled TimeResponse, #1. Step-responses of 5 other systems are shown as well (labeled #2 through #6).

Frequency-response functions of the same 6 systems are shown on the next page. The Frequency-response for System #1 is marked. The others are not marked. Match (on the next page) each system’s Step-response with its associated FrequencyResponse. Within each frequency-response axes boundaries, give a short (5-10 words) justification for your choice.

Horizontal axis is frequency, in “rads/time-unit”, consistent with previous page. ME 132, Fall 2018, UC Berkeley, A. Packard 391

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392

Spring 2014 Final Exam

1. A plant and controller are described by their transfer functions, as P (s) =

1 , s2 − 2s + 1

C(s) =

167.9s2 + 245.6s + 85 0.0439s3 + 2.56s2 + 35.6s + 1

It is verified (separately) that the standard, single-loop closed-loop system is indeed stable. A Bode plot of the product L := P C is shown below.

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(a) Determine the phase-crossover frequencies, and the gain-margin of the system. (b) Determine the gain-crossover frequencies, and the time-delay margin of the system 2. Suppose 0 < ξ < 1 and ωn > 0. Consider the system y¨(t) + 2ξωn y(t) ˙ + ωn2 y(t) = ωn u(t) ˙ subject to the initial conditions y(0− ) = 0, y(0 ˙ − ) = 0, and the unit-step forcing function, namely u(t) = 0 for t = 0, and u(t) = 1 for t > 0. Show that the response is p  1 y(t) = p e−ξωn t sin 1 − ξ 2 ωn t 1 − ξ2 Hint: Recall that the set of all real-valued homogeneous solutions ˙ +  p of y¨(t) + 2ξω  n y(t) p −ξω t 2 −ξωn t cos ωn y(t) = 0 is yH (t) = Ae 1 − ξ 2 ωn t + Be n sin 1 − ξ 2 ωn t where A and B are any real numbers. 3. A block diagram is shown below. Each system is represented by its transfer function. Signals (such as r, y, e or v) are represented by lower-case signals, and associated capital letters are used in the transfer function descriptions. Assume the parameter m > 0.

ME 132, Fall 2018, UC Berkeley, A. Packard - KI s

R

-d − 6E

KP

394

V - d? −6

-

1 ms

-

1 s

Y -

KD 

(a) Under what conditions (on KP , KI , KD , m) is the closed-loop system stable?

(b) Assuming closed-loop stability, and all initial conditions equal to 0, consider the unit-step response (ie., r(t) = 0 for t ≤ 0; r(t) = 1 for t > 0). What is limt→∞ y(t)?

(c) Under the input described in 3b, determine the final limiting value of v, namely limt→∞ v(t)

(d) Explain why for the response in part 3b, there is always “overshoot”, namely at some times, the value of y is larger than its final value, so maxt>0 y(t) > 1. 4. A popular recipe from the 1940’s for designing PID controllers (PI control, with innerloop rate-feedback) is the Ziegler-Nichols method. It is based on simple experiments with the actual process, not requiring ODE models of the process. Nevertheless, we can analyze the method on specific process transfer functions. The method is as follows: Step 1: Connect plant, P , in negative feedback with a proportional-gain controller

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Step 2: Slowly increase gain of proportional controller. At some value of gain, the closed-loop system will become unstable, and start freely oscillating. Denote the value of this critical proportional gain as Kc and the period of the oscillations as Tc . c , KD = Step 3: For the actual closed-loop system, use KP = 0.6Kc , KI = 1.2 K Tc 3 KT. 40 c c

(a) Suppose the plant has transfer function 1 s(τ s + 1)

P (s) =

where τ is a fixed, positive number. What difficulties arise in attempting to use the Zeigler-Nichols design method? (b) Suppose the plant has transfer function P (s) =

−τ1 s + 1 s(τ2 s + 1)

where τ1 and τ2 are some fixed, positive numbers. Imagine that you carry out the Step 1 & 2 of the procedure directly on the plant. What will the parameters Kc and Tc be equal to? Your answers should be exclusively in terms of τ1 and τ2 .

5. Block diagrams for two systems are shown below. Two of the blocks are just gains, (KP and KD ) and the other blocks are described by their transfer functions. The constant β is positive, β > 0. The system on the left is stable if and only if KP > 0 and KD > β (no need to check this – it is correct). What are the conditions on KP , KD and τ , such that the system on the right is stable? Hint: Note that τ is the time-constant of the filter in the approximate differentiation used to obtain y˙ app from y. The stability requirements will impose some relationship between it’s cutoff frequency τ1 and the severity (eg., speed) of the unstable dynamics of the process, namely β. R

-c −6

KP

-c - 1 − 6 s−β

KD 

V- 1 s

Y -

R

-c −6

KP

-c - 1 − 6 s−β

V- 1 s

s  KD  τ s+1 Vapp

Y -

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6. Consider the 2-state system governed by the equation x(t) ˙ = Ax(t). Shown below are the phase-plane plots (x1 (t) vs. x2 (t)) for 4 different cases. Match the plots with the A matrices, and correctly draw in arrows indicating the evolution in time.  A1 =

−2 0 3 1



 ,

A2 =

1 3 −3 1



 ,

A3 =

5

−3 2 −1 0



 ,

A4 =

−1 3 −3 −1

4

x2

x2

2 0

0 −2

−5 −5

0 x1

−4 −4

5

0

−5 −5

0 x1

2

4

5

x2

x2

5

−2

0 x1

5

0

−5 −5

0 x1

7. (a) What is the general form of the solution to the differential equation x¨(t) + 6x(t) ˙ + 5x(t) = 0

(b) What is the general form of the solution to the differential equation x¨(t) + 6x(t) ˙ + 5x(t) = −10

Your expressions both should have two free constants.

5



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8. 12 different input(u)/output(y) systems are given below. The unit-step response, starting from zero initial conditions at t = 0− , are shown. Match the system with the step response. (a) y¨(t) + 8.4y(t) ˙ + 36y(t) = −36u(t) (b) y¨(t) + 1.4y(t) ˙ + y(t) = −5u(t) ˙ − u(t) (c) y¨(t) + 0.4y(t) ˙ + y(t) = −4u(t) ˙ (d) y¨(t) + 8.4y(t) ˙ + 36y(t) = −12u(t) ˙ − 36u(t) (e) y¨(t) + 1.4y(t) ˙ + y(t) = −4u(t) ˙ + u(t) (f) y¨(t) + 2y(t) ˙ + 25y(t) = 6u(t) ˙ + 25u(t) (g) y¨(t) + 1.4y(t) ˙ + y(t) = u(t) (h) y¨(t) + 2y(t) ˙ + 25y(t) = 6u(t) ˙ (i) y¨(t) + 8.4y(t) ˙ + 36y(t) = −12u(t) ˙ + 36u(t) (j) y¨(t) + 0.4y(t) ˙ + y(t) = 5u(t) ˙ + u(t) (k) y¨(t) + 0.4y(t) ˙ + y(t) = 4u(t) ˙ − u(t) (l) y¨(t) + 2y(t) ˙ + 25y(t) = −8u(t) ˙ + 25u(t) 4

4

4

2

2

2

0

0

0

−2

−2 0

2

4

−2 0

2

4

0

4

4

4

2

2

2

0

0

0

−2

−2

−2

0

5

10

15

0

1

2

4

4

4

2

2

2

0

0

0

−2

−2

−2

0

1

2

0

2

4

4

4

4

2

2

2

0

0

0

−2

−2 0

5

10

5

10

15

0

5

10

0

5

10

−2 0

1

2

0

5

10

15

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9. A hoop (of radius R) is mounted vertically, and rotates at a constant angular velocity Ω. A bead of mass m slides along the hoop, and θ is the angle that locates the bead location. θ = 0 corresponds to the bead at the bottom of the hoop, while θ = π corresponds to the top of the hoop, as shown below.

The nonlinear, 2nd order equation (from Newton’s law) governing the bead’s motion is mRθ¨ + mg sin θ + αθ˙ − mΩ2 R sin θ cos θ = 0 All of the parameters m, R, g, α are positive. ˙ (a) Let x1 (t) := θ(t) and x2 (t) := θ(t). Write the 2nd order nonlinear differential equation in the state-space form x˙ 1 (t) = f1 (x1 (t), x2 (t)) x˙ 2 (t) = f2 (x1 (t), x2 (t)) (b) Show that x¯1 = 0, x¯2 = 0 is an equilibrium point of the system. (c) Find the linearized system η(t) ˙ = Aη(t) which governs small deviations away from the equilibrium point (0, 0). (d) Under what conditions (on m, R, Ω, g) is the linearized system stable? (e) Show that x¯1 = π, x¯2 = 0 is an equilibrium point of the system. (f) Find the linearized system η(t) ˙ = Aη(t) which governs small deviations away from the equilibrium point (π, 0). (g) Under what conditions is the linearized system stable? (h) It would seem that if the hoop is indeed rotating (with angular velocity Ω) then there would other equilibrium point (with 0 < θ < π/2). Do such equilibrium points exist in the system? Be very careful, and please explain your answer.

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(i) Find the linearized system η(t) ˙ = Aη(t) which governs small deviations away from this equilibrium point. (j) Under what conditions is the linearized system stable? 10. A process, with input u, disturbance d and output y is governed by y(t) ˙ = 2y(t) + 3u(t) + d(t) (a) Is the process stable?

(b) Suppose y(0) = 1, and u(t) = d(t) ≡ 0 for all t ≥ 0. What is the solution y(t) for t ≥ 0.

(c) Consider a proportional-control strategy, u(t) = K1 r(t) + K2 [r(t) − y(t)]. Determine the closed-loop differential equation relating the variables (y, r, d).

(d) For what values of K1 and K2 is the closed-loop system stable?

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(e) As a function of K2 , what is the steady-state gain from d → y in the closed-loop system?

(f) As a function of K1 and K2 , what is the steady-state gain from r → y in the closed-loop system?

(g) Choose K1 and K2 so that the steady-state gain from r → y equals 1, and the steady-state gain from d → y equals 0.1.

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(h) With those gains chosen, sketch (try to be accurate) the two responses y(t) and u(t) for the following situation:   0 for 0 ≤ t ≤ 1 0 for 0 ≤ t ≤ 2 y(0) = 0, r(t) = , d(t) = 1 for 1 < t 1 for 2 < t 1.4 1.2

y(t) Response

1 0.8 0.6 0.4 0.2 0 0

0.5

1

1.5

2

2.5

3

2

2.5

3

Time, t

4

u(t) Response

3 2 1 0 −1 0

0.5

1

1.5 Time, t

11. A process, with input u, disturbance d and output y is governed by y(t) ˙ = y(t) + u(t) + d(t)

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(a) Is the process stable?

(b) Suppose y(0) = −3, and u(t) = d(t) ≡ 0 for all t ≥ 0. What is the solution y(t) for t ≥ 0.

(c) A PI (Proportional plus Integral) controller is proposed u(t) = KP [r(t) − y(t)] + KI z(t) z(t) ˙ = r(t) − y(t) Eliminate z and u, and determine the closed-loop differential equation relating the variables (y, r, d).

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(d) For what values of KP and KI is the closed-loop system stable?

(e) The closed-loop system is 2nd order. What are the appropriate values of KP and KI so that the closed-loop system characteristic polynomial has roots described by ξ = 0.8, ωn = 0.5?

(f) What are the appropriate values of KP and KI so that the closed-loop system characteristic polynomial has roots described by ξ = 0.8, ωn = 1.0?

(g) What are the appropriate values of KP and KI so that the closed-loop system characteristic polynomial has roots described by ξ = 0.8, ωn = 2.0?

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(h) For each of the 3 cases above (part e, f and g), accurately sketch the response of y due to a unit-step disturbance d, assuming r is identically zero, and assuming all initial conditions are zero. 1.2

Response, (for part e)

1 0.8 0.6 0.4 0.2 0 −0.2 0

0.5

1

1.5 Time

2

2.5

3

0.5

1

1.5 Time

2

2.5

3

1.2

Response, (for part f)

1 0.8 0.6 0.4 0.2 0 −0.2 0

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1.2

Response, (for part g)

1 0.8 0.6 0.4 0.2 0 −0.2 0

0.5

1

1.5 Time

2

2.5

3

12. Recall that if systems are connected in parallel (same input, and outputs add together) then the transfer function of the parallel connection is the sum of the transfer functions. Consider the three different, complicated transfer functions G1 (s) =

0.05s4 + 0.394s3 + 7.868s2 + 14.43s + 64 0.04s5 + 1.184s4 + 7.379s3 + 73.19s2 + 95.36s + 64

0.05s4 + 2.536s3 + 64.36s2 + 87.87s + 64 G2 (s) = 0.04s5 + 1.184s4 + 7.379s3 + 73.19s2 + 95.36s + 64 0.9s4 + 4.27s3 + 65.97s2 + 88.42s + 64 G3 (s) = 0.04s5 + 1.184s4 + 7.379s3 + 73.19s2 + 95.36s + 64 The step responses and frequency-response magnitude of all three systems are shown. Although it may not be physically motivated, mathematically, each Gi can be decomposed additively as as (you do not need to verify this) G1 (s) = 0.9

1 64 1 + 0.05 + 0.05 s2 + 1.4s + 1 s2 + 3.2s + 64 0.04s + 1

1 64 1 + 0.9 2 + 0.05 + 1.4s + 1 s + 3.2s + 64 0.04s + 1 1 64 1 G3 (s) = 0.05 2 + 0.05 2 + 0.9 s + 1.4s + 1 s + 3.2s + 64 0.04s + 1 Based on this information, match up each Gi to its step response and frequency response magnitude. Document your reasoning. G2 (s) = 0.05

s2

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Step Response of G1, G2, G3

Output Response

1.5

1

0.5

0 0

1

1

2

3 Time

4

5

6

Frequency Response of G1, G2, G3

10

0

Magnitude

10

−1

10

−2

10

0

1

10

10

2

10

Frequency

13. An unstable process, with control input u, disturbance input d, and output y, is governed by the equation y¨(t) + y(t) ˙ − y(t) = u(t) + d(t) A PI (Proportional plus Integral) controller is proposed, both to stabilize the system, and provide good disturbance rejection, u(t) = KP [r(t) − y(t)] + KI z(t) z(t) ˙ = r(t) − y(t) Here r is a reference input. (a) Using just the controller equations, express u(t) ˙ in terms of r, r, ˙ y and y. ˙

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(b) By differentiating the process equation, and substituting, derive the closed-loop differential equation relating r and d (and possibly their derivatives) to the output variable y (and its derivatives). The variable u should not appear in these equations. (c) Using the 3rd order test for stability, determine the conditions on KP and KI such that the closed-loop system is stable. 14. A process is governed by 1 [u(t) + d(t)] m Here, u is the control input, d is the disturbance input, and y is the output. m is a positive constant. (one interpretation – the signals u and d are forces acting on a mass m, whose position is y). y¨(t) =

We assume that there are two sensors: one sensor to measure y, and one sensor to measure y. ˙ The goal of control is to make the process output y follow a reference input r, even in the presence of nonzero disturbances d, and slight unknown variations in m. In order to achieve this, we propose the control law u(t) = KP (r(t) − y(t)) − KD y(t) ˙ Note that the control law, as written, uses both sensor measurements. It is a P-control with inner-loop rate feedback. (a) Fill in transfer functions (some blocks may just be gains) in the block diagram below, so that it represents the overall system. D R

-e −6

-

-e −6

-? e -

-

Y -



(b) Use any method you like to obtain the closed-loop differential equation governing the relationship between y and the inputs r and d.

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(c) Under what condition (on the controller gains KP and KD and the system parameter m) is the closed-loop system stable?

(d) Assume that the controller gains are chosen so that the closed-loop system stable. If r(t) ≡ r¯ and d(t) ≡ d¯ for all t ≥ 0, what are the steady-state values of y and u, lim y(t)

t→∞

lim u(t)

t→∞

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(e) Assume that m = 1. Choose the controller gains so that the roots of the closedloop characteristic polynomial are given by ξ = 0.7, ωn = 4.

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(f) Using the gains calculated above, neatly sketch your best “guess” (you can calculate/compute/derive as much as you like) of the responses y(t) and u(t) versus t, when the system starts with all initial conditions equal to 0, for the case when r is a unit step and d is identically 0. You should get the time-scale correct (mark time-axis) as well as the qualitative aspects of the plots (recall interpretation a mass is being accelerated from rest and then brought back to rest in a new position...).

15. A block diagram is shown below. Each system is represented by its transfer function.

ME 132, Fall 2018, UC Berkeley, A. Packard

411 D

R

- d - KP s+KI s −6

-d - 1 τ s+1 −6

- d? -

1 s

-

1 s

Y -

KD 

(a) In terms of KP , KI , KD , τ , what is the transfer function from R to Y (hint: the denominator should be 4th order).

(b) In terms of KP , KI , KD , τ , what is the transfer function from D to Y (hint: same as above).

(c) In terms of KP , KI , KD , τ , what is the characteristic polynomial of the closed-loop system?

(d) In terms of KP , KI , KD , τ , what is the differential equation relating r and d to y?

16. In computing gain and time-delay margins, we solve equations of the form −1 = γL(jω)

− 1 = e−jωT L(jω)

using an appropriate L, depending on the system under consideration. For the system below, what is the appropriate L in order to compute gain and timedelay margins at the point marked by ×.

r

-c −6

KP

-c - 1 − 6 s−β

KD  ×

y˙- 1 s

y -

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17. A 1st order process y(t) ˙ = u(t) + d(t) is controlled by a proportional control u(t) = KP [r(t) − ym (t)] where ym (t) = y(t) + n(t). The interpretation of signals is: u is control input; y is process output; d is external disturbance on process; r is a reference input, representing a desired value of y; n is measurement noise. (a) Eliminate u from the equations, and get the closed-loop differential equation relating (r, d, n) to y.

(b) Under what conditions on KP is the closed-loop system stable?

(c) How is the time-constant of the closed-loop system related to KP ?

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(d) Shown below are the closed-loop frequency responses from (r, d, n) → y, as KP increases from 0.1 to 10. Indicate on each graph with an arrow “cutting” across the plots, the direction of increasing KP . 0

1

10

0

10

10

−1

10

−2

10

0

−1

10

10

−3

10

−2

10

0

10

2

10

0 −1

−20

−2

10

10

−40 −60 −80 −100 −2 10

−2

0

10 Frequency

2

−3

10

10

−2

10

0

10 Frequency

2

10

10

−2

10

0

10 Frequency

2

10

(e) Shown below are the closed-loop frequency responses from (r, d, n) → u, as KP increases from 0.1 to 10. Indicate on each graph with an arrow “cutting” across the plots, the direction of increasing KP . 1

0

10

0

10

−1

10

0

10

−1

10

−2

10

−1

10

−2

10

1

10

10

−3

−2

10

0

10 Frequency

2

10

10

−2

−2

10

0

10 Frequency

2

10

10

−2

10

0

10 Frequency

2

10

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31 31.1

414

Older exams Spring 2012, Midterm 1

#1

#2

#3

#4

TOTAL

20

14

6

10

50

NAME

1. A plant, with control input u, disturbance input d, internal state x, and output y is governed by the equations x(t) ˙ = Ax(t) + B1 d(t) + B2 u(t) y(t) = Cx(t) where A, B1 , B2 and C are constant numbers. A proportional controller, designed to regulate the output y, even in the presence of disturbances is of the form u(t) = −Kym (t) where ym (t) = y(t) + n(t), and n represents additive measurement noise. (a) Express the equations for the closed-loop interconnection (plant/controller/sensor) in the form x(t) ˙ = ax(t) + b1 d(t) + b2 n(t) y(t) = c1 x(t) + d11 d(t) + d21 n(t) u(t) = c2 x(t) + d21 d(t) + d22 n(t) where the entries a, b1 , b2 , . . . , d22 are constants (and generally all are functions of A, B1 , B2 , C, K).

(b) Under what condition is the closed-loop system stable? Your condition should be expressed in terms of (A, B1 , B2 , C, K).

(c) Assuming closed-loop stability, what is the steady-state gain from d to y. Your answer should be expressed in terms of (A, B1 , B2 , C, K).

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(d) Assuming closed-loop stability, what is the time-constant of the closed-loop system? Your answer should be expressed in terms of (A, B1 , B2 , C, K).

(e) Suppose the values are A = B1 = B2 = C = 1 and K = 3. Starting from initial condition x(0) = 0, a disturbance d (with n = 0) acts on the system. The timedependence for d(t) is shown below. On the same axis, make accurate sketches of y(t) and u(t). Mark your curves, so I know which is which. 2 Disturbance, d(t) 1.5

Disturbance and Responses (Y,U)

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(f) Separately, suppose the sensor noise n becomes nonzero (but now with d = 0). Specifically, take n(t) = sin 10t for all t. Both y and u will converge to sine-waves in this case. Approximately determine the steady-state magnitude of the resultant sinusoidal responses of y and of u due to this sinusoidal measurement noise.

2. Albert is building a control system to control an unstable plant. The feedback information (from the measurement to the controller) has a time-delay of T time units. Unfortunately, the control action (from the controller to the plant) also has a delay of T time units. The equations are given

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• The plant is described by differential equations. For all t, the relationship between the signals are (as usual) x(t) ˙ = Ax(t) + B1 d(t) + B2 u(t) y(t) = Cx(t) where A, B1 , B2 and C are constant numbers. • The measurement signal, available at the controller, is denoted ym and for all t is given by ym (t) = y(t − T ) + n(t − T ) where T ≥ 0 is the fixed time-delay in the measurement system, and n represents additive measurement noise. • The controller derives a requested control action, ur based on the reference input r and measurement signal ym . For all t, it is ur (t) = K1 r(t) − K2 ym (t) where K1 and K2 are constant numbers. • Finally, the actual control signal is a delayed version of the requested control action, namely u(t) = ur (t − T ) for all t (a) Find the delay-differential equation governing the relationship between x, ˙ x, r, d and n. Because of the time-delay, the value of x(t) ˙ can be expressed as combinations of the values of x, r, d and n at t, t − T and t − 2T (although not all combinations will appear).

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(b) Suppose A = 3, B1 = B2 = C = 1, K1 = 2, K2 = 5. Show that the closed-loop system is stable for T = 0.

(c) What is the smallest positive T such that the closed-loop system is unstable?

(d) At the critical value of T for which instability occurs (determined by you in part c), what is the frequency of oscillation of the sinusoidal homogeneous solutions?

3. The 5 lines of code below are executed in the Matlab command window f = @(tau,z) -z+1; v0 = 5; [T,V] = ode45(f,[0 1],v0); N = length(V); disp(V(N)) % equivalently disp(X(end)) Recall that the disp command is a simple formatted display command, and it will print out a number (for example) in a nice form. What is the approximate value that will be printed by the line disp(V(N))? 4. A first-order system x(t) ˙ = Ax(t) + Bu(t) is forced with a sin-wave input (u(t) = sin t) as shown below (the experimenter cannot set the initial condition, and the initial condition is effectively random each time the experiment is run). The resulting value of x is also shown. Approximately determine the values of A and B (you can assume that the experimenter knows that B is positive, on physical grounds).

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2 Input, u(t) Output x(t) 1.5

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Spring 2012, Midterm 2

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1. (a) System Q, with input v and output y is governed by the differential equation y¨(t) + 4y(t) ˙ − y(t) = v(t) ˙ + 3v(t)

Q:

For the feedback interconnection below, what is the differential equation governing the relationship between inputs (r, d) and output w.

r

+ d v−6

Q

d y +? d -w +

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(b) System G, with input z and output w is governed by the relationship w(t) ˙ + 3w(t) = z(t). System H, with input u and output y is governed by the relationship y¨(t) + y(t) ˙ + 2y(t) = 2u(t) ˙ + 5u(t). If the systems are cascaded in series, via the interconnection equation u(t) = w(t) (ie., output of G becomes input to H), what is the differential equation relating z to y? z-

G

w = uH

y -

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(c) Three systems (with inputs denoted u1 , u2 and u, respectively; and corresponding outputs y1 , y2 and y) are given in differential equation form: S1 : S2 : SP :

y˙1 (t) + 5y1 (t) = u1 (t) y˙2 (t) + 2y2 (t) = 2u2 (t) y¨(t) + 7y(t) ˙ + 10y(t) = u(t) ˙ + 8u(t)

Find the simplest interconnection (allow any combinations of cascade, parallel, feedback, signs on summing junctions, etc) of systems S1 and S2 which gives the behavior defined by SP . Clearly draw a block-diagram of the interconnection you propose, and justify your answer with equations. 2. A feedback system using a PI controller is proposed to control an unstable plant, P . The plant transfer function is 1 P (s) = s−1 The controller transfer function is C(s) =

KP s + KI s

The feedback interconnection is shown. d r

+ - k −6

C

u- ? k -

P

-

? k 

y

n

(a) What is the closed-loop characteristic polynomial (your answer should involve the two controller gains, KP and KI ).

(b) Under what conditions (on KP and KI ) is the closed-loop system stable?

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(c) Assume the controller is implemented, satisfying the stability conditions. Fill in the table with the correct closed-loop values. Here Gn→u (s) refers to the closedloop transfer function from n to u. Property

Value

Steady-state gain, d → y Steady-state gain, r → y Steady-state gain, d → u limω→∞ |Gn→u (jω)| (d) Design the controller gains, KP and KI so that the closed-loop poles (i.e., the roots of the closed-loop characteristic polynomial) are at −β ± jβ, where β > 0 is a new design parameter. Your answer should be formulae for KP and KI , in terms of β. Hint/help: in order to check your answer, it is true that if β = 1, then the correct values are KP = 3, KI = 2. 3. A modification to problem 2 is taken here. The setup is the same, with KP s + K I 1 , C(s) = P (s) = s−1 s The measurement if filtered by a first-order system, with steady-state gain of 1, and time constant τ . The transfer function of the filter is 1 F (s) = τs + 1 or equivalently, F (s) =

1 τ

s+

1 τ

The feedback interconnection is shown. d r

+

- k -

−6

C

u- ? k -

F



P

-

? k 

y

n

(a) What is the closed-loop characteristic polynomial (your answer should involve the two controller gains, KP and KI , and the filter time-constant, τ ).

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(b) Assume that KP and KI are designed as in Problem 2, with β = 1, so KP = 3, KI = 2. What is the allowable range of values of τ which result in closed-loop stability? (c) Assume the closed-loop system is implemented, with appropriate values for KP , KI and τ such that closed-loop stability holds. Fill in the table with the correct closed-loop values. Property

Value

Steady-state gain, d → y Steady-state gain, r → y Steady-state gain, d → u limω→∞ |Gn→u (jω)|

4. 12 different input(u)/output(y) systems are given below. The unit-step response, starting from zero initial conditions at t = 0− , are shown. Match the system with the step response (write equation letter, a, b, ..., inside the top-right-corner of the appropriate graph). (a) y¨(t) + 2y(t) ˙ + 25y(t) = 6u(t) ˙ (b) y¨(t) + 0.4y(t) ˙ + y(t) = 5u(t) ˙ + u(t) (c) y¨(t) + 2y(t) ˙ + 25y(t) = −8u(t) ˙ + 25u(t) (d) y¨(t) + 8.4y(t) ˙ + 36y(t) = −36u(t) (e) y¨(t) + 1.4y(t) ˙ + y(t) = −5u(t) ˙ − u(t) (f) y¨(t) + 0.4y(t) ˙ + y(t) = −4u(t) ˙ (g) y¨(t) + 8.4y(t) ˙ + 36y(t) = −12u(t) ˙ − 36u(t) (h) y¨(t) + 1.4y(t) ˙ + y(t) = −4u(t) ˙ + u(t) (i) y¨(t) + 2y(t) ˙ + 25y(t) = 6u(t) ˙ + 25u(t) (j) y¨(t) + 1.4y(t) ˙ + y(t) = u(t)

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(k) y¨(t) + 0.4y(t) ˙ + y(t) = 4u(t) ˙ − u(t) (l) y¨(t) + 8.4y(t) ˙ + 36y(t) = −12u(t) ˙ + 36u(t) 4

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Spring 2012, Final Exam

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1. Consider the PI controller architecture shown below.

NAME

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ME 132, Fall 2018, UC Berkeley, A. Packard

r

-

+? −

f 6

R

424

Kf

-

KI

? -f −6

u -

y -

KP

For simplicity, assumed the plant is governed by y(t) ˙ = u(t) + d(t), and that there is no measurement noise. (a) What is the differential equation relating (r, d) to y?

(b) What are the conditions (on Kf , KP and KI ) such that the closed-loop system is stable?

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(c) Assume closed-loop stability. What is the steady-state gain from d to y? (d) Take KP := 1.4 and KI = 1. Sketch, on a single graph (provided) the response of y, assuming r is a unit-step, and d ≡ 0. Assume all initial conditions, just before the step in r, are 0. You should sketch 3 responses, for the following values of Kf : Kf = −3; Kf = 0; Kf = 3. Hint: The vertical axis limits are almost exactly set so as to just include the 3 responses. This should help you graph the responses reasonably accurately without much work. 1.5

Response

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Time

2. Variables x1 , x2 , u and y are related by the equations x˙ 1 (t) = −6x1 (t) + x2 (t) x˙ 2 (t) = x1 (t) + u(t) y(t) = x1 (t) + x2 (t) (a) Using state-space theory we learned, eliminate x1 and x2 , and obtain a differential equation relating u and y. (b) If u is the independent input, and y is the output, is the system stable? 3. The pitching-axis of a tail-fin controlled missile is governed by the nonlinear state equations α(t) ˙ = K1 M fn (α(t), M ) cos α(t) + q(t) q(t) ˙ = K2 M 2 [fm (α(t), M ) + Eu(t)] Here, the states are x1 := α, the angle-of-attack, and x2 := q, the angular velocity of the pitching axis. The input variable, u, is the deflection of the fin which is mounted at the tail of the missile. K1 , K2 , and E are physical constants, with E > 0. M is the speed (Mach) of the missile, and is also a constant, and M > 0. Finally, the functions fn and fm are known, differentiable functions (from wind-tunnel data) of α and M .

8

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(a) Show that for any specific value of α ¯ , with |¯ α| < π2 , there is a pair (¯ q , u¯) such that   α ¯ , u¯ q¯ is an equilibrium point of the system (this represents a turn at a constant rate). Your answer should clearly show how q¯ and u¯ are functions of α ¯ , and will involve the functions fn and fm . Again - the functions fn and fm are assumed known, so your answer can depend on these.

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(b) Let v represent deviations from the equilibrium state, and r represent deviations from the equilibrium input, so       α(t) α ¯ v1 (t) = + q(t) q¯ v2 (t) and u(t) = u¯ + r(t) Find the (linear) differential equations which approximately govern the relationship between r and v, while these variables remain small (referred to as the Jacobian Linearization of the missile system about the equilibrium point). Your answer is fairly symbolic, and may depend on partial derivatives of the functions fn and fm . Be sure to indicate where the various terms are evaluated. 4. In this problem, we consider the standard feedback loop as shown below. d r

+

- k -

−6

C

u- ? k -

P

-

? k 

y

n

The plant P is given in transfer function form, P (s) =

−s + 10 . s2

(a) A simple proportional control strategy is proposed, C(s) = K, where K is a realnumber representing the proportional gain. Show that for all possible values of K, the closed-loop system will be unstable. Hint: form the closed-loop characteristic polynomial, which will depend on K, and work from there).

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(b) Next, a PI strategy (proportional plus integral) is proposed, as as + b C(s) = s (here, a denotes the proportional gain, and b denotes the integral gain). Show that for all possible values of a and b, the closed-loop system will be unstable.

(c) Finally, a general 1st order controller, of the form as + b C(s) = s+c is proposed. Find values for (a, b, c) such that the three roots of the closed-loop characteristic equation are at −1 and −1 ± j. (Time-saving hint: the roots of λ2 + 2λ + 2 are −1 ± j). 5. Two closed-loop systems are shown below, as described by the transfer functions given in each block. d d ? ? + + u u 3 1 - g- g- 1 - y1 - g- g- 1 - y2 r r s s −6 −6 ? g

? g

n

Closed-Loop System #1

Closed-Loop System #2

(a) A disturbance, d, consisting of successive step functions is shown below. 1.5

Disturbance, d(t)

1

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0

−0.5 0

1

2

3 Time, t

4

5

6

n

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The resulting output y (with r = 0, n = 0) is shown below. Two separate graphs are shown. One is the response of closed-loop system #1, and the other is the response of closed-loop system #2. Clearly mark which y response corresponds to which system. Justify your answer below (answers will receive no credit without adequate justification). 1

Response to Disturbance

Response to Disturbance

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ME 132, Fall 2018, UC Berkeley, A. Packard

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(b) A noise signal, n, consisting of high-frequency, random noise is shown below. 0.5 0.4 0.3

Noise, n(t)

0.2 0.1 0 −0.1 −0.2 −0.3 −0.4 −0.5 0

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The resulting output y (with r = 0, d = 0) is shown. Two separate graphs are shown. One is the response of closed-loop system #1, and the other is the response of closed-loop system #2. Clearly mark which y response corresponds to which system. Justify your answer below (answers will receive no credit without adequate justification). 0.2

0.2

0.15

0.15 0.1

Response to Noise

Response to Noise

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(c) For the response to n, since the n signal is very high-frequency, approximately how are the two responses related?

6. Consider the transfer function H(s) =

βs + 1 s+β

6

ME 132, Fall 2018, UC Berkeley, A. Packard

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(a) Sketch a Magnitude plot of H, for β = 5 on the graph below 1

Magnitude

10

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Frequency

1

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ME 132, Fall 2018, UC Berkeley, A. Packard

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(b) Sketch a Phase plot of H, for β = 5 on the graph paper provided. 120 100

Angle (degrees)

80 60 40 20 0 −20 −40 −2

10

−1

10

0

10

Frequency

1

10

2

10

ME 132, Fall 2018, UC Berkeley, A. Packard

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(c) Consider a feedback system, as below d r

+

- k -

−6

C

u- ? k -

P

-

? k 

where

1 s2 Under what conditions (on β) is the closed-loop system stable? C(s) = H(s),

P (s) =

y

n

ME 132, Fall 2018, UC Berkeley, A. Packard

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(d) Continue to employ H as the controller C, with β = 5. What is the value of the Frequency-Response function, from N -to-U , at very high-frequency? (e) Continue to employ H as the controller C, with β = 5. What is the value of the Frequency-Response function, from D-to-Y , at very low-frequency? 7. The standard closed-loop system, as below, is used. d r

+

- k -

−6

u- ? k -

C

P

-

? k 

Two specific controller/plant pairs are used: (a) Controller/Plant pair # 1 C1 (s) =

3.8s + 4 , s

P1 (s) =

1 s−1

The Bode plot for P1 C1 is shown below 3

10

2

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280 260 240

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220 200 180 160 140 120 100 80 −2 10

−1

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10

y

n

ME 132, Fall 2018, UC Berkeley, A. Packard

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(b) Controller/Plant pair # 2 C2 (s) =

1.8s + 4 , s

P2 (s) =

1 s+1

The Bode plot for P2 C2 is shown below 3

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2

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−90

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−95

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You will answer 4 questions for each system

1

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ME 132, Fall 2018, UC Berkeley, A. Packard

For system # 1, determine (a) the closed-loop characteristic polynomial

(b) the closed-loop transfer function from D to Y

(c) the Gain-Margin (Bode plots of L := P C are provided)

(d) the Time-Delay-Margin (Bode plots of L := P C are provided)

436

ME 132, Fall 2018, UC Berkeley, A. Packard

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For system # 2, determine (a) the closed-loop characteristic polynomial

(b) the closed-loop transfer function from D to Y

(c) the Gain-Margin (Bode plots of L := P C are provided)

(d) the Time-Delay-Margin (Bode plots of L := P C are provided)

31.4

Spring 2009, Midterm 1

1. The solutions to the differential equations listed below are shown on the next page. Match the ODE with the solution graph. Make a table with pertinent information that justifies your answers. In each case, all appropriate initial conditions are 0. By that, I mean that if the differential equation is first-order, then the initial condition is y(0) = 0. If the differential equation is second order, then the initial conditions are y(0) ˙ = 0, y(0) = 0. (a) y(t) ˙ + y(t) = 1 (b) y(t) ˙ + 5y(t) = 5 (c) y(t) ˙ − y(t) = 1 (d) y(t) ˙ + 10y(t) = 10 (e) y¨(t) − 2y(t) ˙ − y(t) = −1 (f) y¨(t) − 2y(t) ˙ + 9y(t) = 9 (g) y¨(t) + 0.4y(t) ˙ + y(t) = 1 (h) y¨(t) + 0.12y(t) ˙ + 0.09y(t) = 0.09 (i) y¨(t) + 6y(t) ˙ + 5y(t) = 5 (j) y¨(t) + 0.3y(t) ˙ + 0.09y(t) = 0.09 (k) y¨(t) + 3y(t) ˙ + 9y(t) = 9 (l) y¨(t) + 1.8y(t) ˙ + 9y(t) = 9

ME 132, Fall 2018, UC Berkeley, A. Packard

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2. An unstable process, with control input u, disturbance input d and output y is governed by y(t) ˙ = y(t) + 2u(t) + d(t) (a) Consider a proportional-control strategy, u(t) = K1 r(t) − K2 [y(t) + n(t)]. Here,

ME 132, Fall 2018, UC Berkeley, A. Packard

439

r is a reference-command input, and n is measurement noise. Combine with the process model to eliminate u, and determine the closed-loop differential equation relating the variables (y, r, d, n).

(b) For what values of K1 and K2 is the closed-loop system stable?

(c) Assuming closed-loop stability, what is the time-constant of the closed-loop system (in terms of K1 and K2 )?

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(d) As a function of K2 , what is the steady-state gain from d → y in the closed-loop system?

(e) As a function of K1 and K2 , what is the steady-state gain from r → y in the closed-loop system?

(f) Choose K1 and K2 so that the steady-state gain from r → y equals 1, and the steady-state gain from d → y equals 0.2.

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(g) With those gains chosen, sketch the two responses y(t) and u(t) for the following situation: measurement noise n(t) = 0 for all t, and    0 for 0 ≤ t ≤ 1.5 0 for 0 ≤ t ≤ 0.5 1 for 1.5 < t ≤ 2.5 y(0) = 0, r(t) = , d(t) = 1 for 0.5 < t  0 for 2.5 < t 1.4 1.2

y(t) Response

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3 2 1 0 −1 0

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3. Your answers in part 3a-3d both should have two free constants. I would like there to √ be no −1 in the answers, just constants, exponentials, cos and sin.

ME 132, Fall 2018, UC Berkeley, A. Packard

442

(a) What is the general form of all real solutions to the differential equation x¨(t) + 5x(t) ˙ + 6x(t) = 0

(b) What is the general form of all real solutions to the differential equation x¨(t) + 5x(t) ˙ + 6x(t) = 12

(c) What is the general form of all real solutions to the differential equation y¨(t) + 2y(t) ˙ + 17y(t) = 0

(d) What is the general form of all real solutions to the differential equation y¨(t) + 2y(t) ˙ + 17y(t) = −51

(e) The solutions in part 3c are made up of an exponentially decaying envelope, superimposed on a sinusoid. What (approximately) is the ratio Time − to − Decay Period − of − Oscillation

4. An unstable process, with control input u, disturbance input d and output y is governed by y(t) ˙ = y(t) + u(t) + d(t) (a) A PI (Proportional plus Integral) controller is proposed u(t) = KP [r(t) − y(t) − n(t)] + KI z(t) z(t) ˙ = r(t) − y(t) − n(t) Here z is the variable representing the integrator, r is a command-reference input, and n is measurement noise. Combine with the governing equation for the process, and eliminate z and u, determining the closed-loop differential equation relating the variables (y, r, d, n). Note that derivatives of the external inputs (r, d, n) may appear.

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(b) For what values of KP and KI is the closed-loop system stable?

(c) The closed-loop system is 2nd order. What are the appropriate values of KP and KI so that the closed-loop system characteristic polynomial has roots described by ξ = 0.8, ωn = 0.5?

(d) What are the appropriate values of KP and KI so that the closed-loop system characteristic polynomial has roots described by ξ = 0.8, ωn = 1.0?

(e) What are the appropriate values of KP and KI so that the closed-loop system characteristic polynomial has roots described by ξ = 0.8, ωn = 2.0?

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5. A stable, 1st order system has a frequency-response function G(ω). The magnitude, |G(ω)| and angle ∠G(ω) are shown below, as functions of ω. The units of frequency are radians/second. 1

10

0 −0.2 −0.4

0

Magnitude

Angle, Radians

10

−1

−0.6 −0.8 −1

10

−1.2 −1.4 −2

10

−1

0

10

10

1

2

10 Frequency, rad/sec

10

−1.6 −1 10

0

10

1

10 Frequency, rad/sec

A sin-wave input is applied to the system, and after some time, the system’s response reaches a steady-state sinsoidal response. The forcing function is shown below (it has been applied consistently since t = 0). On this same axis, sketch the steady-state response of the system. Note: You need to compute some numbers reading some things from the graphs. I said you did not need a calculator, so these calculations can be approximate.

Forcing and Response

1

0.5

0

−0.5

−1

8

8.5

9 Time (sec)

9.5

2

10

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445

Spring 2009, Midterm 2

Cannot find this one...

31.6

Spring 2009, Final Exam

Cannot find this one...

31.7

Spring 2005 Midterm 1

1. Your answers √ in part 1a and 1c both should have two free constants. I would like there to be no −1 in the answers, just exponentials, cos and sin. (a) What is the general form of real (as opposed to complex) solutions to the differential equation y¨(t) + 2y(t) ˙ + 17y(t) = 0

(b) The solutions in part 1a are made up of an exponentially decaying envelope, superimposed on a sinusoid. What (approximately) is the ratio Period − of − Oscillation Time − to − Decay Explain.

(c) What is the general form of real solutions to the differential equation y¨(t) + 2y(t) ˙ + 17y(t) = −51

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2. A process has a very simple model, y(t) = Hu(t) + w(t). The control input is u, the disturbance input is w and the output is y. Here, H is simply a gain, ie., the behavior of the system is “static” - it is not governed by a differential equation. The goal of control is to make the process output y follow a reference input r, even in the presence of nonzero disturbances w, and slight unknown variations in H. In order to achieve this, we use an integral controller z(t) ˙ = r(t) − y(t) u(t) = KI z(t) Here, r is the reference input. (a) Combine the the process model, and the controller equations to (eliminating z and u) get a relationship between the process output y, the two “forcing” functions r and w, and any of their derivatives.

(b) Under what conditions (on H and KI ) is the closed-loop system stable?

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(c) Assume that KI is chosen so that the closed-loop system is stable. Does a 20% change in H (ie., H changing to 0.8H or 1.2H) affect stability of the closed-loop system?

(d) Assume that KI is chosen so that the closed-loop system is stable. If r(t) ≡ r¯ and w(t) ≡ w¯ for all t ≥ 0 (¯ r and w¯ are some fixed constant values), what are the steady-state values (in terms of r¯, w, ¯ KI , H) of y and u, defined as lim y(t)

t→∞

lim u(t)

t→∞

(e) How does a 20% change in H affect (approximately) the steady-state values of y and u derived in part (2d) above?

(f) Assume that KI is chosen so that the closed-loop system is stable. What is the time-constant of the closed-loop system? Approximately how does a 20% change in H (ie., H changing to 0.8H or 1.2H) affect (approximately) the time constant?

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3. A closed-loop system is shown below. Here H and K are positive constants, namely H = 4, and K = 1.

(a) By eliminating y and u from the equations, find the constants A, B1 , B2 and B3 such that the internal variable z is related to the external forcing functions (r, w, n) in the form z(t) ˙ = Az(t) + B1 r(t) + B2 w(t) + B3 n(t)

(b) Express the variable y as a combination of z, r, w, n, namely find the constants C1 , D11 , D12 and D13 such that y(t) = C1 z(t) + D11 r(t) + D12 w(t) + D13 n(t)

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(c) Express the variable u as a combination of z, r, w, n, namely find the constants C2 , D21 , D22 and D23 such that u(t) = C2 z(t) + D21 r(t) + D22 w(t) + D23 n(t)

(d) Shown on the next page is the frequency-response matrix (2 × 3) from the three external inputs (r, w, n) to the two “outputs-of-interest” (y, u). In 5 of the cases, only the magnitude is shown. In the case of input/output pair (r, y), both magnitude and phase are shown. In each axes, 3 or 4 lines are shown, though only one is correct. In each case, mark the correct frequency response curve.

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1

1

10

10

0

10

0

0

10

10

−2

10

−2

10

0

10

2

10

−1

50 0 −50 −100 −150

10

−2

10

−2

−2

10 −2

10

0

10

0

10

2

10

1

0

0

10

−1

10

−1

10

−1

10

−2

10

−2

0

2

10

10

0

10

10

0

10

1

10

−2

−2

10

2

1

10

10

10

10

10

−1

10

2

10

10

−2

−2

10

0

10

2

10

10

−2

10

0

10

2

10

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451

Spring 2005 Midterm 2

1. Consider the closed loop system shown below.

The transfer function for the controller is C(s) =

s + 25 3 s

(31.1)

3 s+5

(31.2)

The transfer function for the process is P (s) =

(a) Assume all initial conditions are zero. Suppose that the reference input, r, is a unit step function, and the disturbance input, d is identically zero. Shown below are several possible responses for the output variable y. Only one of them is correct. Clearly mark the correct one. Show work/reasoning.

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Step Response

Step Response 1.5

Amplitude

Amplitude

1.5

1

0.5

0

0

2

4 Time (sec)

1

0.5

0

6

0

Step Response

Amplitude

Amplitude

1.5

1

0.5

0

0.5 1 Time (sec)

1

0.5

0

1.5

0

Step Response

0.1

0.2 Time (sec)

0.3

Step Response

2

1.5

1.5

Amplitude

Amplitude

1.5

Step Response

1.5

0

0.5 1 Time (sec)

1

1

0.5

0.5 0

0

2

4 Time (sec)

6

0

0

0.1

0.2 Time (sec)

0.3

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(b) Assume all initial conditions are zero. Suppose that the disturbance input, d, is a unit step function, and the reference input, r, is identically zero. Shown below are several possible responses for the output variable y. Only one of them is correct. Clearly mark the correct one. Show work/reasoning. Step Response

0.3

0.3

0.2

0.2

Amplitude

Amplitude

Step Response

0.1 0 −0.1

0.1 0

0

0.1

0.2 Time (sec)

−0.1

0.3

0

0.3

0.2

0.2

0.15

0.1 0 −0.1

0.05

0

0.5 1 Time (sec)

0

1.5

0

0.5 1 Time (sec)

1.5

Step Response

0.6

1.5

0.4

1

Amplitude

Amplitude

6

0.1

Step Response

0.2 0 −0.2

4 Time (sec)

Step Response

Amplitude

Amplitude

Step Response

2

0.5 0

0

0.5 1 Time (sec)

1.5

−0.5

0

0.5 1 Time (sec)

2. Consider the diagram below, which depicts a position control system, using PI control

1.5

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and rate-feedback (as we did with the motor). Additionally, there is a 1st-order model of the actuator, which produces the forces based on the commands from the controller blocks. Here, m = 2 and τ = 0.0312. If we choose KD = 16, KP = 52.48 and KI = 51.2, it is possible (you do not need to) to verify that the closed-loop system is stable. W R

- e - KP s+KI s −6

1 - e Aτ s+1 −6

-? e -

D

1 ms

E-

1 s

Y -

B

KD 

C

In order to assist you in the question below, Bode plots of certain transfer functions listed below are given in the following pages (not all may be useful...). H1 (s) = H3 (s) =

KP s + KI 4 mτ s + ms3 + KD s2

H2 (s) =

KD s2 mτ s4 + ms3 + KP s + KI

KD s2 + KP s + KI KD s2 + KP s + KI H (s) = 4 mτ s4 + ms3 + KD s2 + KP s + KI mτ s4 + ms3 KD KP s + KI H5 (s) = H6 (s) = 2 mτ s + ms mτ s4 + ms3

Using the graphs (estimate values as best as you can...), answer the following margin questions. Explain any work you do, and make relevant marks on the Bode plots that you use in your calculations. (a) What is the gain margin at location A? (Hint – first determine what is the appropriate L for margin calculations at A, match with the H’s, and do calculation from supplied graphs). (b) What is the time-delay margin at location A? (c) What is the gain margin at location B? (d) What is the time-delay margin at location B? (e) What is the gain margin at location C? (f) What is the time-delay margin at location C? (g) What is the gain margin at location D?

ME 132, Fall 2018, UC Berkeley, A. Packard

(h) What is the time-delay margin at location D? (i) What is the gain margin at location E? (j) What is the time-delay margin at location E?

455

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Bode Plot of H1

2

10

1

Magnitude

10

0

10

−1

10

−2

10

−2

10

−1

10

0

10

1

10

2

10

−120 −130 −140 −150 Angle (degrees)

−160 −170 −180 −190 −200 −210 −220 −230 −240 −250 −260 −2 10

−1

10

0

10 Frequency, Radians/Second

1

10

2

10

ME 132, Fall 2018, UC Berkeley, A. Packard

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Bode Plot of H2

2

10

1

Magnitude

10

0

10

−1

10

−2

Angle (degrees)

10

−2

−1

10

10

250 240 230 220 210 200 190 180 170 160 150 140 130 120 110 100 −2 10

10

−1

0

10

0

10 Frequency, Radians/Second

1

10

1

10

2

10

2

10

ME 132, Fall 2018, UC Berkeley, A. Packard

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Bode Plot of H3

2

10

1

Magnitude

10

0

10

−1

10

−2

10

−2

Angle (degrees)

10

20 10 0 −10 −20 −30 −40 −50 −60 −70 −80 −90 −100 −110 −120 −130 −140 −150 −160 −170 −180 −2 10

−1

10

−1

10

0

10

0

10 Frequency, Radians/Second

1

10

1

10

2

10

2

10

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Bode Plot of H4

2

10

1

Magnitude

10

0

10

−1

10

−2

Angle (degrees)

10

−2

−1

10

10

240 230 220 210 200 190 180 170 160 150 140 130 120 110 100 90 80 −2 10

10

−1

0

10

0

10 Frequency, Radians/Second

1

10

1

10

2

10

2

10

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Bode Plot of H5

2

10

1

Magnitude

10

0

10

−1

10

−2

10

−2

10

−1

10

0

10

1

10

2

10

−90 −100

Angle (degrees)

−110 −120 −130 −140 −150 −160 −170 −2 10

−1

10

0

10 Frequency, Radians/Second

1

10

2

10

ME 132, Fall 2018, UC Berkeley, A. Packard

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Bode Plot of H6

2

10

1

Magnitude

10

0

10

−1

10

−2

10

−2

10

−1

10

0

10

1

10

2

10

170 160

Angle (degrees)

150 140 130 120 110 100 90 −2 10

−1

10

0

10 Frequency, Radians/Second

1

10

2

10

ME 132, Fall 2018, UC Berkeley, A. Packard

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462

Spring 2004 Midterm 1

#1

#2

#3

#4

#5

10

20

10

15

25

TOTAL

80

NOTE: Any unmarked summing junctions are positively signed (+). Shown below is a graph of 1 p

−ξx

1 − ξ2

e

sin

p

1−

ξ2x



versus x, for 7 values of ξ evenly spaced between 0.3 and 0.9.

0.6

0.4

0.2

0

−0.2

−0.4 0

2

4

6

8

10

1. (a) What is the general form of real (as opposed to complex) solutions to the differential equation x¨(t) + 4x(t) ˙ + 13x(t) = 0

ME 132, Fall 2018, UC Berkeley, A. Packard

463

(b) What is the general form of real solutions to the differential equation x¨(t) + 4x(t) ˙ + 13x(t) = −26

√ Your expressions both should have two free constants. I would like there to be no −1 in the answers, just exponentials, cos and sin. HINT: Although the roots are complex, and the (ξ, ωn ) parametrization certaintly may be used, it will be “less messy” to compute the roots as complex numbers, and examine their real/imaginary parts. 2. Shown below are two systems. The system on the left is the nominal system, while the system on the right represents a deviation from the nominal (the insertion of the dashed box) and is called the perturbed system. - KP

- KP

R r -d - R du KI - ? −6

y -

r -d - R du - d KI - ? −6

−6

1 β

-

R

-

R

y -

Based on the values of KP an KI , and some analysis, you should have a general idea of how the nominal system behaves (eg., the effect of r on u and y). Consider 3 different possibilities (listed below) regarding the relationship between the nominal and perturbed systems: (a) The perturbed system behaves pretty much the same as the nominal system. (b) The perturbed system behaves quite differently from the nominal system, but is still stable. (c) The perturbed system is unstable. For each row in the table below, which description from above applies? Write a, b, or c in each box. Show work below. KP 2.8 1.4 14 70

KI 4 1 100 2500

β 0.02 1 0.2 0.02

Your Answer

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3. Neatly/accurately sketch the solution (for t ≥ 0) to the differential equation x(t) ˙ = −2x(t) + u(t) subject to the initial condition x(0) = 2, and forcing function u(t) = 0 for 0 ≤ t ≤ 2.5 u(t) = −2 for 2.5 < t ≤ 4.5 u(t) = 8 for 4.5 < t ≤ 6

6

5

4

3

2

1

0

−1

0

1

2

3

4

5

6

4. Suppose 0 < ξ < 1 and ωn > 0. Consider the system y¨(t) + 2ξωn y(t) ˙ + ωn2 y(t) = ωn u(t) ˙ subject to the initial conditions y(0− ) = 0, y(0 ˙ − ) = 0, and the unit-step forcing function, namely u(t) = 0 for t = 0, and u(t) = 1 for t > 0. Show that the response is p  1 −ξωn t 2 y(t) = p e sin 1 − ξ ωn t 1 − ξ2 Hint: Recall that the set of all real-valued homogeneous solutions ˙ +  p of y¨(t) + 2ξω  n y(t) p 2 −ξωn t −ξω t 1 − ξ 2 ωn t + Be n sin 1 − ξ 2 ωn t where A ωn y(t) = 0 is yH (t) = Ae cos and B are any real numbers.

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5. A process, with input u, disturbance d and output y is governed by y(t) ˙ = y(t) + u(t) + d(t) (a) Is the process stable?

(b) Suppose y(0) = −3, and u(t) = d(t) ≡ 0 for all t ≥ 0. What is the solution y(t) for t ≥ 0.

(c) A PI (Proportional plus Integral) controller is proposed u(t) = KP [r(t) − y(t)] + KI z(t) z(t) ˙ = r(t) − y(t) Eliminate z and u, and determine the closed-loop differential equation relating the variables (y, r, d).

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(d) For what values of KP and KI is the closed-loop system stable?

(e) The closed-loop system is 2nd order. What are the appropriate values of KP and KI so that the closed-loop system characteristic polynomial has roots described by ξ = 0.8, ωn = 0.5?

(f) What are the appropriate values of KP and KI so that the closed-loop system characteristic polynomial has roots described by ξ = 0.8, ωn = 1.0?

(g) What are the appropriate values of KP and KI so that the closed-loop system characteristic polynomial has roots described by ξ = 0.8, ωn = 2.0?

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(h) For each of the 3 cases above (part e, f and g), accurately sketch the response of y due to a unit-step disturbance d, assuming r is identically zero, and assuming all initial conditions are zero. 1.2

Response, (for part e)

1 0.8 0.6 0.4 0.2 0 −0.2 0

0.5

1

1.5 Time

2

2.5

3

0.5

1

1.5 Time

2

2.5

3

1.2

Response, (for part f)

1 0.8 0.6 0.4 0.2 0 −0.2 0

ME 132, Fall 2018, UC Berkeley, A. Packard

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1.2

Response, (for part g)

1 0.8 0.6 0.4 0.2 0 −0.2 0

31.10

0.5

1

1.5 Time

2

2.5

3

Fall 2003 Midterm 1

1. Neatly/accurately sketch the solution (for t ≥ 0) to the differential equation x(t) ˙ = −3x(t) + 6u(t) subject to the initial condition x(0) = −1, and forcing function u(t) = 0 u(t) = 3 u(t) = 2 u(t) = 1

for for for for

0≤t≤2 2 0 (here, u¯ is just some constant value). Compute the response x(t), for t ≥ 0.

(e) With x(t) computed above, compute the output y(t), and sketch below.

5 4.5 4 3.5

Response

3 2.5 2 1.5 1 0.5 0 0

0.5

1

1.5

2 2.5 3 Time (seconds)

3.5

4

4.5

5

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(f) Suppose that the initial condition is x(0) = 0, let τ = 0.2. Apply a ramp input (with slope 3) u(t) = 3t for t ≥ 0. Compute the response y(t), and plot. If you cannot derive the expression for y, guess what it should look like, and plot it below. 5 4.5 4 3.5

Response

3 2.5 2 1.5 1 0.5 0 0

0.5

1

1.5

2 2.5 3 Time (seconds)

3.5

4

4.5

5

2. Block diagrams for two systems are shown below. Two of the blocks are just gains, (KP and KD ) and the other blocks are described by their transfer functions. The constant β is positive, β > 0. The system on the left is stable if and only if KP > 0 and KD > β (no need to check this – it is correct). What are the conditions on KP , KD and τ , such that the system on the right is stable? Hint: Note that τ is the time-constant of the filter in the approximate differentiation used to obtain y˙ app from y. The stability requirements will impose some relationship between it’s cutoff frequency τ1 and the severity (eg., speed) of the unstable dynamics of the process, namely β. R

-c −6

KP

V- 1

-c - 1 − 6 s−β

s

Y -

R

-c −6

KP

-c - 1 − 6 s−β

V- 1

s  KD  τ s+1 Vapp

KD 

3. In computing gain and time-delay margins, we solve equations of the form −1 = γL(jω)

s

− 1 = e−jωT L(jω)

using an appropriate L, depending on the system under consideration. For the system below, what is the appropriate L in order to compute gain and timedelay margins at the point marked by ×.

Y -

ME 132, Fall 2018, UC Berkeley, A. Packard

r

-c −6

KP

483

y˙- 1

-c - 1 − 6 s−β

s

y -

KD ×

4. (a) For the system below, are the gain and time-delay margins at the point marked by 1 the same as the gain and time-delay margins at the point marked by 2? Justify your answer. (b) For the system below, are the gain and time-delay margins at the point marked by 2 the same as the gain and time-delay margins at the point marked by 3? Justify your answer.

r

-c −6

KP

y˙- 1

- c 3- 1 − 6 s−β

2

KD 

s

y -

1

5. A hoop (of radius R) is mounted vertically, and rotates at a constant angular velocity Ω. A bead of mass m slides along the hoop, and θ is the angle that locates the bead location. θ = 0 corresponds to the bead at the bottom of the hoop, while θ = π corresponds to the top of the hoop, as shown below.

ME 132, Fall 2018, UC Berkeley, A. Packard

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The nonlinear, 2nd order equation (from Newton’s law) governing the bead’s motion is mRθ¨ + mg sin θ + αθ˙ − mΩ2 R sin θ cos θ = 0 All of the parameters m, R, g, α are positive. ˙ (a) Let x1 (t) := θ(t) and x2 (t) := θ(t). Write the 2nd order nonlinear differential equation in the state-space form x˙ 1 (t) = f1 (x1 (t), x2 (t)) x˙ 2 (t) = f2 (x1 (t), x2 (t)) (b) Show that x¯1 = 0, x¯2 = 0 is an equilibrium point of the system. (c) Find the linearized system η(t) ˙ = Aη(t) which governs small deviations away from the equilibrium point (0, 0). (d) Under what conditions (on m, R, Ω, g) is the linearized system stable? (e) Show that x¯1 = π, x¯2 = 0 is an equilibrium point of the system. (f) Find the linearized system η(t) ˙ = Aη(t) which governs small deviations away from the equilibrium point (π, 0). (g) Under what conditions is the linearized system stable? (h) It would seem that if the hoop is indeed rotating (with angular velocity Ω) then there would other equilibrium point (with 0 < θ < π/2). Do such equilibrium points exist in the system? Be very careful, and please explain your answer. (i) Find the linearized system η(t) ˙ = Aη(t) which governs small deviations away from this equilibrium point. (j) Under what conditions is the linearized system stable? 6. A closed-loop feedback system consisting of plant P and controller C is shown below. r(t)

e(t) + - h - C −6

u(t) -

P

-

y(t)

In this problem, it is known that the nominal closed-loop system is stable. The plots ˆ below are the magnitude and phase of the product Pˆ (jω)C(jω), given both in linear and log scales, depending on which is easier for you to read. Use these graphs to compute the time-delay margin and the gain margin. Clearly indicate the gaincrossover and phase-crossover frequencies which you determine in these calculations.

ME 132, Fall 2018, UC Berkeley, A. Packard

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2

4

10

3.75 3.5 3.25 3 1

10

2.75

Magnitude

Magnitude

2.5 2.25 2 1.75 0

10

1.5 1.25 1 0.75 0.5

−1

10

−1

0

10

1

10

2

10

3

10

4

5

6

7

8

Frequency, RAD/SEC

220

210

210

207.5

9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Frequency, RAD/SEC

205 200

202.5 190

200 197.5

170

Phase (DEGREES)

Phase (DEGREES)

180

160 150 140 130

195 192.5 190 187.5 185 182.5 180

120

177.5 110

175 100

172.5 −1

0

10

1

10

170

2

10

10

3

4

5

6

7

8

Frequency, RAD/SEC

9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Frequency, RAD/SEC

7. A closed-loop feedback system consisting of plant P and controller C is shown below. r(t)

+ - i −6

e(t)

u(t) -

C

-

P

-

y(t)

It is known that the nominal closed-loop system is stable. In the presence of gainvariations in P and time-delay in the feedback path, the closed-loop system changes to r(t)

e(t) + - i −6

u(t) C

f (t) = y(t − T )

-

γ

-

P

delay, T 

-

y(t)

ME 132, Fall 2018, UC Berkeley, A. Packard

486

In this particular system, there is both an upper and lower gain margin - that is, for no time-delay, if the gain γ is decreased from 1, the closed-loop system becomes unstable at some (still positive) value of γ; and, if the gain γ is increased from 1, the closed-loop system becomes unstable at some value of γ > 1. Let γl and γu denote these two values, so 0 < γl < 1 < γu .

Min Time−Delay

For each fixed value of γ satisfying γl < γ < γu the closed-loop system is stable. For each such fixed γ, compute the minimum time-delay that would cause instability. Specifically, do this for several (say 8-10) γ values satisfying γl < γ < γu , and plot below.

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 2.1 2.2 2.3 2.4 2.5 GAMMA

ˆ The data on the next two pages are the magnitude and phase of the product Pˆ (jω)C(jω). They are given in both linear and log spacing, depending on which is easier for you to read. Use these graphs to compute the time-delay margin at many fixed values of γ satisfying γl < γ < γu . 2

4

10

3.75 3.5 3.25 3 1

10

2.75

Magnitude

Magnitude

2.5 2.25 2 1.75 0

10

1.5 1.25 1 0.75 0.5

−1

10

−1

10

0

1

10

10 Frequency, RAD/SEC

2

10

3

4

5

6

7

8

9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Frequency, RAD/SEC

ME 132, Fall 2018, UC Berkeley, A. Packard

487

220

210

210

207.5 205

200

202.5 190

200 197.5

170

Phase (DEGREES)

Phase (DEGREES)

180

160 150 140 130

195 192.5 190 187.5 185 182.5 180

120

177.5 110

175 100

172.5 −1

10

0

1

10

10

170

2

10

3

4

5

6

Frequency, RAD/SEC

7

8

9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Frequency, RAD/SEC

8. A popular recipe from the 1940’s for designing PID controllers (PI control, with innerloop rate-feedback) is the Ziegler-Nichols method. It is based on simple experiments with the actual process, not requiring ODE models of the process. Nevertheless, we can analyze the method on specific process transfer functions. The method is as follows: Step 1: Connect plant, P , in negative feedback with a proportional-gain controller Step 2: Slowly increase gain of proportional controller. At some value of gain, the closed-loop system will become unstable, and start freely oscillating. Denote the value of this critical proportional gain as Kc and the period of the oscillations as Tc . c , KD = Step 3: For the actual closed-loop system, use KP = 0.6Kc , KI = 1.2 K Tc 3 KT. 40 c c

(a) Suppose the plant has transfer function 1 s(τ s + 1)

P (s) =

where τ is a fixed, positive number. What difficulties arise in attempting to use the Zeigler-Nichols design method? (b) Suppose the plant has transfer function P (s) =

−τ1 s + 1 s(τ2 s + 1)

where τ1 and τ2 are some fixed, positive numbers. Imagine that you carry out the Step 1 & 2 of the procedure directly on the plant. What will the parameters Kc and Tc be equal to? Your answers should be exclusively in terms of τ1 and τ2 .