2E1262 2006

2E1262 Nonlinear ControlAutomatic Control

• Disposition

5 credits, lp 2

28h lectures, 28h exercises, 3 home-works

• Instructors

Bo Wahlberg, professor,bo@ee.kth.se

Krister Jacobsson, teaching assistant,krister.jacobsson@ee.kth.se

STEX (entrance floor in S3 building), administration,stex@s3.kth.se

Lecture 1 1

2E1262 2006

Course Goal

To provide participants with a solid theoretical foundation of nonlinearcontrol systems combined with a good engineering understanding

You should after the course be able to

• understand common nonlinear control phenomena

• apply the most powerful nonlinear analysis methods

• use some practical nonlinear design methods

Lecture 1 2

2E1262 2006

2E1262 Nonlinear Control

Lecture 1

• Practical information

• Course outline

• Nonlinear control phenomena

• Nonlinear differential equations

Lecture 1 3

2E1262 2006

Today’s Goal

You should be able to

• Recognize some phenomena in nonlinear systems

• Transform differential equation to first-order form

• Mathematically describe saturation, dead-zone, relay, back-lash

• Derive equilibrium points

Lecture 1 4

2E1262 2006

Course Information

• All info and handouts are available at

http://www.s3.kth.se/control/kurser/2E1262

• Get an E computer account for the computer exercise (see STEX)

• Homeworks are mandatory and have to be handed in on time

Lecture 1 5

2E1262 2006

Material

• Textbook: Khalil, Nonlinear Systems, Prentice Hall, 3rd ed.,2002. Optional but highly recommended.

• Lecture notes: Copies of transparencies

• Exercises: Class room and home exercises

• Homeworks: 3 computer exercises to hand in

• Software: Matlab (provided by KTH Library)

Alternative textbooks (decreasing mathematical brilliance):Sastry, Nonlinear Systems: Analysis, Stability and Control; Vidyasagar,Nonlinear Systems Analysis; Slotine & Li, Applied Nonlinear Control; Glad &Ljung, Reglerteori, flervariabla och olinjara metoder.Only references to Khalil will be given.

Two course compendia sold by STEX.

Lecture 1 6

2E1262 2006

Course Outline

• Introduction: nonlinear models, computer simulation (L1-L2)

• Feedback analysis: linearization, stability theory, describingfunction (L3-L6)

• Control design: compensation, high-gain design, Lyapunovmethods (L7-L10)

• Alternatives: gain scheduling, optimal control, neural networks,fuzzy control (L11-L13)

• Summary (L14)

Lecture 1 7

2E1262 2006

Linear Systems

Definition: Let M be a signal space. The system S : M → M islinear if for all u, v ∈ M and α ∈ R

S(αu) = αS(u) scaling

S(u + v) = S(u) + S(v) superposition

Example: Linear time-invariant systems

x(t) = Ax(t) + Bu(t), y(t) = Cx(t), x(0) = 0

y(t) = g(t) � u(t) =

∫ t

0

g(τ)u(t − τ)dτ

Y (s) = G(s)U(s)

Notice the importance to have zero initial conditions

Lecture 1 8

2E1262 2006

Linear Systems Have Nice Properties

Local stability=global stability Stability if all eigenvalues of A (orpoles of G(s)) are in the left half-plane

Superposition Enough to know a step (or impulse) response

Frequency analysis possible Sinusoidal inputs give sinusoidaloutputs: Y (iω) = G(iω)U(iω)

Lecture 1 9

2E1262 2006

Linear Models are not Enough

Example:Positioning of a ball on a beam

φ x

Nonlinear model: mx(t) = mg sin φ(t), Linear model: x(t) = gφ(t)

Lecture 1 10

2E1262 2006

Can the ball move 0.1 meter in 0.1 seconds from steady state?

Linear model (step response with φ = φ0) gives

x(t) ≈ 10t2

2φ0 ≈ 0.05φ0

so that

φ0 ≈ 0.1

0.05= 2 rad = 114◦

Unrealistic answer. Clearly outside linear region!

Linear model valid only if sin φ ≈ φ

Must consider nonlinear model. Possibly also include othernonlinearities such as centripetal force, saturation, friction etc.

Lecture 1 11

2E1262 2006

2 minute exercise: Find a simple system x = f(x, u) that is stablefor a small input step but unstable for large input steps.

Lecture 1 12

2E1262 2006

Stability Can Depend on Reference SignalExample: Control system with valve characteristic f(u) = u2

Σ 1s

1(s+1)2

Motor Valve Process

−1

r y

Simulink block diagram:

1

(s+1)(s+1)y

s

1

−1

u2

Lecture 1 13

2E1262 2006

STEP RESPONSES

0 5 10 15 20 25 300

0.2

0.4

Time tO

utpu

t y

0 5 10 15 20 25 300

2

4

Time t

Out

put y

0 5 10 15 20 25 300

5

10

Time t

Out

put y

r = 0.2

r = 1.68

r = 1.72

Stability depends on amplitude of the reference signal!

(The linearized gain of the valve increases with increasing amplitude)

Lecture 1 14

2E1262 2006

Stable Periodic Solutions

Example: Position control of motor with back-lash

y

Sum

5

P−controller

1

5s +s2

Motor

0

Constant

Backlash

−1

Motor: G(s) = 1s(1+5s)

Controller: K = 5

Lecture 1 15

2E1262 2006

Back-lash induces an oscillation

Frequency and amplitude independent of initial conditions:

0 10 20 30 40 50−0.5

0

0.5

Time t

Out

put y

0 10 20 30 40 50−0.5

0

0.5

Time t

Out

put y

0 10 20 30 40 50−0.5

0

0.5

Time t

Out

put y

How predict and avoid oscillations?

Lecture 1 16

2E1262 2006

Automatic Tuning of PID ControllersRelay induces a desired oscillation whose frequency and amplitudeare used to choose PID parameters

Σ Process

PID

Relay

A

T

u y

− 1

0 5 10

−1

0

1

Time

uy

Lecture 1 17

2E1262 2006

Harmonic Distortion

Example: Sinusoidal response of saturation

a sin t y(t) =∑∞

k=1 Ak sin(kt)Saturation

100

10−2

100

Frequency (Hz)

Ampli

tude

y

0 1 2 3 4 5−2

−1

0

1

2

Time t

100

10−2

100

Frequency (Hz)

Ampli

tude

y

0 1 2 3 4 5−2

−1

0

1

2

Time t

a = 1

a = 2a = 2a = 2a = 2

Lecture 1 18

2E1262 2006

Example: Electrical power distribution

Nonlinearities such as rectifiers, switched electronics, andtransformers give rise to harmonic distortion

Total Harmonic Distortion =

∑∞k=2 Energy in tone k

Energy in tone 1

Example: Electrical amplifiers

Effective amplifiers work in nonlinear region

Introduces spectrum leakage, which is a problem in cellular systems

Trade-off between effectivity and linearity

Lecture 1 19

2E1262 2006

Subharmonics

Example: Duffing’s equation y + y + y − y3 = a sin(ωt)

0 5 10 15 20 25 30−0.5

0

0.5

0 5 10 15 20 25 30−1

−0.5

0

0.5

1

Time t

Time t

ya

sin

ωt

Lecture 1 20

2E1262 2006

Nonlinear Differential Equations

Definition: A solution to

x(t) = f(x(t)), x(0) = x0 (1)

over an interval [0, T ] is a C1 function x : [0, T ] → Rn such that (1)

is fulfilled.

• When does there exists a solution?

• When is the solution unique?

Example: x = Ax, x(0) = x0, gives x(t) = exp(At)x0

Lecture 1 21

2E1262 2006

Existence Problems

Example: The differential equation x = x2, x(0) = x0

has solution x(t) =x0

1 − x0t, 0 ≤ t <

1

x0

Solution not defined for tf =1

x0

Solution interval depends on initial condition!

Recall the trick: x = x2 ⇒ dx

x2= dt

Integrate ⇒ −1

x(t)− −1

x(0)= t ⇒ x(t) =

x0

1 − x0t

Lecture 1 22

2E1262 2006

Finite Escape Time

Simulation for various initial conditions x0

0 1 2 3 4 50

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

Time t

x(t)

Finite escape time of dx/dt = x2

Lecture 1 23

2E1262 2006

Uniqueness Problems

Example: x =√

x, x(0) = 0, has many solutions:

x(t) =

{(t − C)2/4 t > C

0 t ≤ C

0 1 2 3 4 5−1

−0.5

0

0.5

1

1.5

2

Time t

x(t)

Lecture 1 24

2E1262 2006

Physical Interpretation

Consider the reverse example, i.e., the water tank lab process with

x = −√x, x(T ) = 0

where x is the water level. It is then impossible to know at what timet < T the level was x(t) = x0 > 0.

Hint: Reverse time s = T − t ⇒ ds = −dt and thus

dx

ds= −dx

dt

Lecture 1 25

2E1262 2006

Lipschitz Continuity

Definition: f : Rn → R

n is Lipschitz continuous if there existsL, r > 0 such that for allx, y ∈ Br(x0) = {z ∈ R

n : ‖z − x0‖ < r},

‖f(x) − f(y)‖ ≤ L‖x − y‖

Euclidean norm is given by

‖x‖2 = x21 + · · · + x2

n

Slope L

Lecture 1 26

2E1262 2006

Local Existence and UniquenessTheorem:If f is Lipschitz continuous, then there exists δ > 0 such that

x(t) = f(x(t)), x(0) = x0

has a unique solution in Br(x0) over [0, δ].

Proof: See Khalil, Appendix C.1. Based on the contraction mappingtheorem

Remarks

• δ = δ(r, L)

• f being C0 is not sufficient (cf., tank example)

• f being C1 implies Lipschitz continuity(L = maxx∈Br(x0) f ′(x))

Lecture 1 27

2E1262 2006

State-Space Models

State x, input u, output y

General: f(x, u, y, x, u, y, . . .) = 0

Explicit: x = f(x, u), y = h(x)

Affine in u: x = f(x) + g(x)u, y = h(x)

Linear: x = Ax + Bu, y = Cx

Lecture 1 28

2E1262 2006

Transformation to Autonomous System

A nonautonomous system

x = f(x, t)

is always possible to transform to an autonomous system byintroducing xn+1 = t:

x = f(x, xn+1)

xn+1 = 1

Lecture 1 29

2E1262 2006

Transformation to First-Order System

Given a differential equation in y with highest derivative dnydtn

,

express the equation in x =(y dy

dt. . . dn−1y

dtn−1

)T

Example: Pendulum

MR2θ + kθ + MgR sin θ = 0

x =(θ θ

)Tgives

x1 = x2

x2 = − k

MR2x2 − g

Rsin x1

Lecture 1 30

2E1262 2006

Equilibria

Definition: A point (x∗, u∗, y∗) is an equilibrium, if a solution startingin (x∗, u∗, y∗) stays there forever.

Corresponds to putting all derivatives to zero:

General: f(x∗, u∗, y∗, 0, 0, . . .) = 0

Explicit: 0 = f(x∗, u∗), y∗ = h(x∗)Affine in u: 0 = f(x∗) + g(x∗)u∗, y∗ = h(x∗)Linear: 0 = Ax∗ + Bu∗, y∗ = Cx∗

Often the equilibrium is defined only through the state x∗

Lecture 1 31

2E1262 2006

Multiple Equilibria

Example: Pendulum

MR2θ + kθ + MgR sin θ = 0

θ = θ = 0 gives sin θ = 0 and thus θ∗ = kπ

Alternatively in first-order form:

x1 = x2

x2 = − k

MR2x2 − g

Rsin x1

x1 = x2 = 0 gives x∗2 = 0 and sin(x∗

1) = 0

Lecture 1 32

2E1262 2006

Some Static and Dynamic Nonlinearities

Sign

Saturation

Relay

eu

MathFunction

Look−UpTable

Dead Zone

Coulomb &Viscous Friction

Backlash

|u|

Abs

Lecture 1 33

2E1262 2006

2 minute exercise: Construct a rate limiter (i.e., a system that limitthe rate of change of a signal) using one of the previous models

?−1s

Hint: Try a saturation!

Lecture 1 34

2E1262 2006

Next Lecture

• Simulation in Matlab

• Linearization

Lecture 1 35