Chapter 3: Analisys of closed-loop systemsalamo/Archivos/Chapter_3.pdf · 2009. 9. 23. · Teoría...

1

Chapter 3: Analisys of closed-loop systems

Control Automático

3º Curso. Ing. IndustrialEscuela Técnica Superior de Ingenieros

Universidad de Sevilla

Course 2008-09

2

Control of SISO systems

Control around an operation point (u0,y0)

Process

+u0

u(t)

-

y(t)Δu(t)

Controllerr(t) e(t)

(Reference is not a deviation variable )

Automatic control

Controllere(t)Δu(t)

Controller

How to chose the value of the controller parameters (Kp and Ti) in a way such that the closed-loop system has an appropriate performance?

PI

3


Heuristic designTuning based on experiments

Real systemModel

Design based on tablesTuning based on a set of experiments and a table that determines thevalue of each parameter

Zieger-Nichols(Chapter 6)

Mathematical designThe tuning is based on a mathematical analisys of the closed-loop systemand provides guaranteed properties

Transient responseSteady state responseRobustness

Analitic design techniquesRoot–locus design techniquesLoop-shaping design techniques 4


Process

+u0

u(t)

-

y(t)Δu(t)

Controllerr(t)e(t)

Closed-loop system

It is hard to analize the dynamics of the closed-loop system if the process or thecontroller are nonlinear systems

The design techniques studied in this course are based on the linearized modelof the system around a given operation point (u0,y0)

5

Incremental variables model

The linearized model can be compared with the incremental variables modelfor a given operating point which is defined as follows:

System+ -

Incremental variables model (u0,y0)

Remark: The modeldepends on the operatingpoint

Assumption: The initial stateis the operation point

6

Incremental variables model

Linearized model around (u0,y0)

Linearized model

-

Δu(t)Controller

e(t)

Analisys of the closed-loop system. In this case, both the controller and the system are LTI systems.

Assumption: Zero initial conditions.

Teoría de sistemas

Assumption: The properties of this (simplified) system are similar to the ones of the real closed-loop system if the trajectories are close to the operating point- Speed of the transient- Tracking of time-varying references- Disturbance rejection

7

Teoría de sistemasTEMA 1. Introducción y fundamentos.

Sistemas dinámicos. Conceptos básicos. Ecuaciones y evolución temporal. Linealidad en los sistemas dinámicos.

TEMA 2. Representación de sistemas.Clasificación de los sistemas. Clasificación de comportamientos. Señales de prueba. Descripción externa e interna. Ecuaciones diferenciales y en diferencias. Simulación.

TEMA 3. Sistemas dinámicos lineales en tiempo continuo. Transformación de Laplace. Descripción externa de los sistemas dinámicos. Función de transferencia. Respuesta impulsional. Descripción interna de los sistemas dinámicos.

TEMA 4. Modelado y simulación. Modelado de sistemas. Modelado de sistemas mecánicos. Modelado de sistemas hidráulicos. Modelado de sistemas eléctricos. Modelado de sistemas térmicos. Linealización de modelos no lineales. Modelos lineales. Álgebra de bloques. Simulación.

TEMA 5. Respuesta temporal de sistemas lineales. Sistemas dinámicos lineales de primer orden. Ejemplos. Sistemas dinámicos lineales de segundo orden. Respuesta ante escalón. Sistemas de orden n.

TEMA 6. Respuesta frecuencial de sistemas lineales.Función de transferencia en el dominio de la frecuencia. Transformación de Fourier. Representación gráfica de la función de transferencia. Diagramas más comunes. Diagrama de Bode.

TEMA 7. Estabilidad.Estabilidad de sistemas lineales. Criterios relativos a la descripción externa de los sistemas dinámicos. Criterio de Routh-Hurwitz. Criterio de Nyquist. Criterios relativos a la descripción interna.

Analysis of linear time invariant (LTI) systems 8

Closed-loop transfer function

Laplace transform (assuming zero initial conditions)

Remark: From now on, the variables “y” and “u” denote incremental values; that is, the deviation of the input and the output from the operating point

Linear time invariant systems (LTI)

Properties used: Linearity, transform of the time derivative

Transfer function

9

Proportional term

Controllere(t)u(t)

The value of the input u(t) is proportional to the error

Transfer function

C(s)E(s)U(s)

Time domain

Frequency domain

Desing parameter: Kp

10

Integral term

Controllere(t)u(t)

The value of the input u(t) is proportional to the error and its integral

Transfer function

C(s)E(s)U(s)

Temporal domain

Frequency domain

Desing parameter: Kp, Ti

Lag compensation net

Controller with propertiessimilar to the PI

11

Derivative term

Controllere(t)u(t)

Transfer function

C(s)E(s)U(s)

Time domain

Frequency domainDesign parameters: Kp, Td

Lead compensator net

The value of the input u(t) is proportional to the error and its time derivative

Controller with propertiessimilar to the PD

12

PID controller

Controllere(t) u(t)

Widely used in industry

The value of the input u(t) is proportional to the error, its time derivative and its integral

13

PID controller

Controllere(t)u(t)

Transfer function

C(s)E(s)U(s)

Time domain

Frequency domainDesign parameter: Kp, Td, Ti

Lead-lag compensator

The value of the input u(t) is proportional to the error, its time derivative and itsintegral

Controller with properties similar to the PID

14

Index

Closed Loop Transfer Function

Tuning a Controller

Stability analysis of system

Steady-state response of a closed-loop systems

Transient response of a stable system.

Closed-loop poles and zeros vs. Controller parameters

15


Block algebra (Ogata 3.3, Tema 3, Teoría de sistemas)

+

-S1(s)

S2(s)

S3(s) = S1(s)-S2(s) S1(s) S2(s) = S1(s)

S3(s) = S1(s)

Signal sum Signal bifurcation

G(s)S1(s) S2(s)=G(s)S1(s)

LTI system

16


G(s)

-Y(s)C(s)

E(s) U(s)R(s)

R(s) Y(s)Gbc(s) Gbc(s) models how the ouput of the closed-loopsystem reacts to changes in the references

The properties of a controller are defined based onthe response of the closed-loop system

Closed-loop system

17

Other transfer functions

C(s) G(s)

H(s)

R(s)

Ym(s)

U(s)

Y(s)

Sensor dynamics

C(s) G(s)

H(s)

R(s)

Ym(s)

U(s)

Y(s)

Gd(s)

D(s)Disturbances

-

+

+

++

-

18

Index


Tuning a Controller





19

Controller tuning

Obtain a set of controller parameter (C(s)) in order to guarantee that the closed-loop systems satisfies a given set of conditions (specifications)

SpecificationsStabilityRaise time (step reference change)Steady state error

Specifications on thelinearized model are

relevant to thebehaviour of the real closed-loop system

Mathematical designTuning based on the anlisys of the closed-loop dynamics (which depend onthe controller parameters)

Gbc(s) is not defined if parametersof C(s) are not fixed

20

Example

Closed-loop system

Closed-loop response? Depends on the value of Kp

Three poles that depend on KpStatic gain depens on Kp

Reference signal: Step change - Simulations performed in Simulink/Matlab

21

Example

Kp=0.1

Kp=1

Kp=10

Kp=1522

Example

0 10 20 30 40 50 600

0.5

1Kp = 0.1, Td = 0, 1/Ti = 0

y(t)

0 10 20 30 40 50 600

0.05

0.1

u(t)

0 10 20 30 40 50 600

0.5

1

e(t)

0 10 20 30 40 50 600

20

40

∫ 0t e(τ

)dτ

Respuesta del sistema en BC

0 10 20 30 40 50 600

0.5

1

1.5Kp = 1, Td = 0, 1/Ti = 0

y(t)

0 10 20 30 40 50 60−0.5

0

0.5

1

u(t)

0 10 20 30 40 50 60−0.5

0

0.5

1

e(t)

0 10 20 30 40 50 600

5

10

∫ 0t e(τ

)dτ


0 10 20 30 40 50 600

1

2Kp = 10, Td = 0, 1/Ti = 0

y(t)

0 10 20 30 40 50 60−10

0

10

u(t)

0 10 20 30 40 50 60−1

0

1

e(t)

0 10 20 30 40 50 60−0.5

0

0.5

1

∫ 0t e(τ

)dτ


0 10 20 30 40 50 60−10

0

10

20Kp = 15, Td = 0, 1/Ti = 0

y(t)

0 10 20 30 40 50 60−200

0

200

u(t)

0 10 20 30 40 50 60−20

−10

0

10

e(t)

0 10 20 30 40 50 60−5

0

5

10

∫ 0t e(τ

)dτ


23

Example

0 10 20 30 40 50 600

0.5

1Kp = 0.1, Td = 0, 1/Ti = 0

y(t)

0 10 20 30 40 50 600

0.05

0.1

u(t)

0 10 20 30 40 50 600

0.5

1

e(t)

0 10 20 30 40 50 600

20

40

∫ 0t e(τ

)dτ

Respuesta del sistema en BC 24

Example

0 10 20 30 40 50 600

0.5

1

1.5Kp = 1, Td = 0, 1/Ti = 0

y(t)

0 10 20 30 40 50 60−0.5

0

0.5

1

u(t)

0 10 20 30 40 50 60−0.5

0

0.5

1

e(t)

0 10 20 30 40 50 600

5

10∫ 0t e

(τ)d

τ


25

Example

0 10 20 30 40 50 600

1

2Kp = 10, Td = 0, 1/Ti = 0

y(t)

0 10 20 30 40 50 60−10

0

10

u(t)

0 10 20 30 40 50 60−1

0

1

e(t)

0 10 20 30 40 50 60−0.5

0

0.5

1

∫ 0t e(τ

)dτ


Example

0 10 20 30 40 50 60−10

0

10

20Kp = 15, Td = 0, 1/Ti = 0

y(t)

0 10 20 30 40 50 60−200

0

200

u(t)

0 10 20 30 40 50 60−20

−10

0

10

e(t)

0 10 20 30 40 50 60−5

0

5

10

∫ 0t e(τ

)dτ


27

Types of behaviour

Unit step responseClasification of the output signal Δy(t) depending on the input signal

Unit Step (the most widely used).Ramp.Sinusoidal.

Provides information about the dynamic proterties of the system

Model expressed in error variables (u0,y0)

Assumption: Initial conditions in the operatingpoint.

Unit Step: Behaviours:• Overdamped• Underdamped• Unstable• Oscilatory

28

Types of behaviour

Overdamped

0 5 10 150

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Step Response

Time (sec)

Am

plitu

de

Delay LGain KRaise Time ts

Lts

KRaise Time: Time to reach 63% ofthe steady state value.

Delay: Time of reaction for theoutput with respect to a change in the input.

Gain: Quotient of the output andthe input values.

29

Types of behaviour

Underdamped

K

tetpts

Delay LGain KRaise Time tsPeak Time tpSettling Time teOvershoot Mp

Raise Time: Time to reachthe steady state value for thefirst time.Peak Time: Time to reach themaximum value.Settling Time: Time toconfine the output within a band of 5% arounf the steadystate value.Overshoot: PercentageIncrement of the peak valuewith respect to the steadystate value..

Mp

30

Types of behaviour

Unstable

31

Index


Tuning a Controller





32

Stability (Chapter 7. Stability)

Stability Criterion:Gbc(s) is stable if and only if all poles are located on the left-halfcomplex plane.

Closed loop poles are the roots of (depend on C(s))

A closed-loop system might become unstable if the controller is not properlydesigned.

The controller design must guarantee closed loop stability

Example: Kp=15

Poles: -5.65, 0.0500 + 1.8272i, 0.0500 - 1.8272i

33

Stability

Analitic Procedure (Try & Error)• Evalute closed loop poles for every combination of the controller parameters(Kp, Td, Ti) using the model of the system.

Routh-Hurwitz stability criterion

• A tool to evaluate if a polinomial has roots on the right-half complex plane. • The method prevents form computing the whole set of roots of a higher

order polinomial• It can be used to evaluate stability conditions

Nyquist Stability Criterion (to be studied in chapter 5)

34

Routh-Hurwitz Stability Criterion

Allows to determine if there exists a root in the right-half complex plane

Important: Note the notation

1 – If there exists a negative parameter, then the polinomial has at least one root in the right-half plane.

2 – Build the Routh-Hurwitz table. If there exists a negative component on the firstcolumn, then the polinomial has at least one root in the right-half plane.

There are special rules to deal withdegenerate cases (See Chapter 7)

35

Example

Closed loop system

The poles are the solution of the following equation (depends on Kp)

Gains range

36

Example

Closed loop system

The poles are the solution of the following equation (depend on Kp y Ti)

Not vey useful for multipleparameters

37

Index


Tuning a Controller

Stability analysis of systems

Steady-state response of a closed-loop system



38

Steady State Response

Analysis of system response as time tends to infinity (We assume the closed loop system is stable)

Steady state error

Final Value Theorem (property of Laplace transform)

Important: Depends on R(s)Different references define different steady-state error parameters.

39

Error for Step input

0 1 2 3 4 5 6 7 8 9 100

0.2

0.4

0.6

0.8

1

Error for a constant steady-state input

Position error constant

All stable systems have bounded steady-state errorsFor the error to be null (The system reaches the reference)

40

Error for Ramp input

Steady state error for ramp input

Velocity Error

Bounded Velocity Error ⇔ Null position Error(C(s)G(s) has at least one integrator)

For the velocity error to be null (the system reached the reference)

0 1 2 3 4 5 6 70

1

2

3

4

5

6

7

41

Error for parabolic input

Steady state error for parabolic input

Acceleration error constant

Bounded acceleration error ⇔ Null position error ⇔ Null velocity error (C(s)G(s) has at least two integrators)

For the parabolic error to be null (The system reaches the reference)

0 0.5 1 1.5 2 2.5 30

0.5

1

1.5

2

2.5

3

3.5

4

4.5

42

Error Table

Type of a system = Number of integrators

Parabolic

0Ramp

00Step

210Error

Type

43

Example

Type I System

Proportional Controller

Proportional Controller affects the Bode Gain of the system, but can not changeits Type.

Improves (quantitatively) steady statebehaviour.

Depends on Kp.

44

Example

0 10 20 30 40 50 600

0.5

1

1.5Kp = 1, Td = 0, 1/Ti = 0

y(t)

0 10 20 30 40 50 60−0.5

0

0.5

1

u(t)

0 10 20 30 40 50 60−0.5

0

0.5

1

e(t)

0 10 20 30 40 50 600

5

10

∫ 0t e(τ

)dτ


Position error. Constant reference (Step)

45

Example

Velocity error. Increasing reference (ramp)

0 10 20 30 40 50 600

20

40

60Kp = 1, Td = 0, 1/Ti = 0

y(t)

0 10 20 30 40 50 600

5

10

u(t)

0 10 20 30 40 50 600

5

10

e(t)

0 10 20 30 40 50 600

100

200

300

∫ 0t e(τ

)dτ


Example

Type I system

PI Controller

P controllers affect the Bode gain ofthe system and increases the systemtype

Improves (qualitatively) steady statebehaviour

Depends on Kp and Ti

(Lag controllers allow to increase Bode gain)

47

Example

Position error. Constant reference (Step)

0 10 20 30 40 50 600

0.5

1

1.5Kp = 1, Td = 0, 1/Ti = 0.1

y(t)

0 10 20 30 40 50 60−1

0

1

2

u(t)

0 10 20 30 40 50 60−0.5

0

0.5

1

e(t)

0 10 20 30 40 50 600

1

2

3

∫ 0t e(τ

)dτ


Example

Velocity error. Increasing reference (ramp)

0 10 20 30 40 50 600

20

40

60Kp = 1, Td = 0, 1/Ti = 0.1

y(t)

0 10 20 30 40 50 600

5

10

u(t)

0 10 20 30 40 50 600

1

2

3

e(t)

0 10 20 30 40 50 600

50

100

∫ 0t e(τ

)dτ


The integral term isintroduced to improvesteady state response.

(It might unstabilize thesystem.Ex. Try simulation withKp=1, Ti=1)

49

Index


Tuning a Controller

Stability analysis of systems

Steady-state response of a closed-loop system



50

Transient Response

Response to unit step inputClassification of output signal Δy(t) depending on the input signal.

Unit Step (the most widely used).Ramp.Sinusoidal.

Provides information about the dynamical properties of the system

Y(s)G(s)

-C(s)

E(s) U(s)R(s)

Response in y(t) when a reference r(t) is applied

Reference signal: Unit Step signal. Shows the speed of response of the system

(In general the reference signal will be different than the unit step)

51

Transient Response

TEMA 5. Respuesta temporal de sistemas lineales. Sistemas dinámicos lineales de primer orden. Ejemplos. Sistemas dinámicos lineales de segundo orden. Respuesta ante escalón. Sistemas de orden n.

We are interested in the output y(t) as r(t) varies in time (Closed-loop behavior)

The trnasient response of a LTI system depends on the closed-loop transferfunction (Gbc(s))

One option: Try & ErrorGiven a system, simulate or aplly inverde Laplace transformIt is difficult to characterize propoerties as raise time or overshootIdentify the effect of parameters in the response.

52

First Order Systems

sK

yuKydtdy

τ

τ

+==

==+

1U(s)Y(s)G(s)

0)0( ,

units) in time (measured Constant Time :

output) &input toaccording (units uyGain Static :K

τ∞

∞

ΔΔ

0 1 2 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 29 300

1

2

3

4

5

tiempo

2=Δu

0 1 2 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 29 300

0.51

1.5

22.5

33.5

44.5

5

5.56

6.57

7.58

8.59

9.510

tiempo

y

6=Δy

τ

78.363.0 =Δ⋅ y

53

Second order Systems

rad/s) (frequency Natural :nal)(adimensiot Coefficien Damping :

U)Y/dim (dimGain Static :K

2

n

222

2

1212

2

ωδ

ωωωδ uKydtdy

dtyd

ubyadtdya

dtyd

nnn =++

=++

22

2

2U(s)Y(s)G(s)

nn

n

ssK

ωωδω

++==

54


1:Poles 2 −±− δωωδ nn

Im

Re ⎪⎩

⎪⎨

⎧

<=>

:1:1 :1

δδδ Overdamped

Critically damped.

Underdamped

Im

Re

55


..=OS(∞yttt

21 δωαπ−

−=

n

st

21 δωπ−

=n

pt

21100.(%). δ

πδ

−−

⋅= eOS

δωnet

3=

Underdamped system

00

Tiemp

y(t)

)()()(

..∞

∞−=

yyty

OS p

)(∞y

etptst56

)(lim gain static theisK where

)]1cos()1([)(

)2()(

)('1)(

0

22

11

22

11

1

sGK

tctsenbeeaKty

ssps

csk

ssY

s

kkkk

r

k

tt

j

tpj

kkk

r

kj

t

j

i

m

i

kkj

→

=

−

=

−

==

=

=

⋅−+⋅−++=

++∏+∏

+∏⋅=

∑∑ δωδω

ωωδ

ωδ

Higher order Systems

nnnn

mmm

mmm

m

m

m

nnn

n

n

n

asasasbsbsbsG

tubdt

tdubdt

tudbdt

tudbyadt

tdyadt

tydadt

tyd

+++++++

=

++++=++++

−−

−

−−

−

−−

−

11

1

110

11

1

1011

1

1

......)(

)()(...)()()(...)()(

57

• In practice, some poles have more influence in the response than others. Thesepoles are called dominant poles

• The dominant poles are those yielding the slowest reponse

• The response speed is given by the exponent of the exponential terms (the real part of the pole). Remember:

Dominant Poles

Dominant dynamics: poles with the slowest response

In practice, the dominat poles are determined from their relativedistance to the imaginary axis.

Re

Imp1

p’1

p2

p’2

d2

d1

Re

Im

p1

p2

p’2

d2

d1

p1 is dominant if d2/d1>5

The static gain mustremain the same

58

Dominant Poles

12

)17)(16)(1(544

)17)(16)(1(544)(

+=

+≈

+++=

ssssssG

-1 is el dominant. The remaining poles are neglected

Re

Im

-1-16

-17

Tiempo(s)0 1 2 3 4 5 6

0

0.5

1

1.5

2

2.5y(t)

59

Effect of zeros in the output

0 0.5 1 1.50

1

2

3

4

5

6 Step Response

Time (sec)

Am

plitu

de

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.5

1

1.5

2

2.5Step Response

Time (sec)

Ampl

itude

Zeros have an influence in the response

60

Effect of zeros in the output

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

1

2

3

4

5

6

Step Response

Time (sec)

Ampl

itude

y(t)dy(t)/dtyc(t)

Qualitatively:

Effect of adding a zero

61

Non-minimum phase zeros

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2Step Response

Time (sec)

Ampl

itude

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2Step Response

Time (sec)

Ampl

itude

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.5

1

1.5

2

2.5Step Response

Time (sec)

Ampl

itude

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

1

2

3

4

5

6Step Response

Time (sec)

Ampl

itude

-20 -15 -10 -5 0 5-1

0

1

x xo o o o

62

Non-minimum phase zeros

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2-2.5

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2Step Response

Time (sec)

Ampl

itude

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2-1

-0.5

0

0.5

1

1.5

2Step Response

Time (sec)

Ampl

itude

-20 -15 -10 -5 0 5-1

0

1

x x o o

63

Dynamics cancellation

-7 -6 -5 -4 -3 -2 -1 0-1

0

1

x xo

The closer the zero is to the pole, the less it influences system responseAffects the dominant dynamics (in transient regime)Settling time is not significantly affected.

0 1 2 3 4 5 60

0.2

0.4

0.6

0.8

1

Step Response

Time (sec)

Ampl

itude

64

Design Hypothesis

Quick review of concepts• Time response of first order systems• Time response of second order systems• Time response of higher order systems• Effect of zeros

It is difficult to obtain explicit results in general

Design Hypothesis• Explicit expressions for the effect of controller parameters on the

transient response for step inputs are required.• A pair of conjugate complex poles dominate the closed loop

response• Zeros are difficult to deal with, in general. Not considered.

Two design tools:• Root Locus design• Frequency domain design

65

Index


Tuning a Controller





66

Poles and Zeros of Gbc(s)


Zeros of the closed-loop system The same zeros of the open-loop plant plus those of thecontroller

Poles of the closed-loop system Depend on the design parameter

• In some cases it is possible to explicit expressions for 2nd order systems withP and PD controllers

• Not possible in generalApproximate methodologies

67

ExampleP Controller

The poles depend on Kp

Complex plane representation

Root Locus

68

Example

−6 −5 −4 −3 −2 −1 0 1−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

Kp=0.1

Kp=1

Kp=10Kp=15

69

Illustratuve example: Magnetic levitation system

Description ValueBall material SteelBall diameter 25 mmCoil diameter 80 mmWinding turns 2850Resistence 22 ΩInductance 277 mH a 1 kHz

442 mH a 120 kHz70

Illustratuve example: Magnetic levitation system

Nonlinear model of the system

2

2

XIkmgXm −=&&

m : Ball massg : Gravity constantX : Distance between ball and coil

(magnitudes to be controlled)I : Coil current (control action)K : constant coefficient

X Fm

Fg

System LinealizationWe assume the operating point X0 with control action I0 and consider error variables

XXXIIIΔ+=

Δ+=

0

0

71

The incremental variables depend on theselected operating point.

System Linearization

72

System Linearization

Chapter 3: Analisys of closed-loop systemsalamo/Archivos/Chapter_3.pdf · 2009. 9. 23. · Teoría...

Documents

Transcript of Chapter 3: Analisys of closed-loop systemsalamo/Archivos/Chapter_3.pdf · 2009. 9. 23. · Teoría...