fundamentals of control systemshthoang/cstd/fundctrlsys_chapter6.pdf · fundamentals of control...

Lecture NotesLecture Notes

Fundamentals of Control SystemsFundamentals of Control Systems

Instructor: Assoc. Prof. Dr. Huynh Thai HoangDepartment of Automatic Control

Faculty of Electrical & Electronics EngineeringHo Chi Minh City University of Technology

Email: [email protected]@yahoo.com

Homepage: www4.hcmut.edu.vn/~hthoang/

6 December 2013 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/ 1

Chapter 6Chapter 6

DESIGN OF CONTINUOUS DESIGN OF CONTINUOUS CONTROL SYSTEMSCONTROL SYSTEMS

6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 2

Introduction

ContentContent

Introduction Effect of controllers on system performance Control systems design using the root locus method Control systems design using the root locus method Control systems design in the frequency domain Design of PID controllersg Control systems design in state-space Design of state estimators


IntroductionIntroduction


Introduction to design processIntroduction to design process

Design is a process of adding/configuring hardware as well assoftware in a system so that the new system satisfies thed i d ifi tidesired specifications.


The controller is connected in series with the plant

Series compensatorSeries compensator

The controller is connected in series with the plant.

R(s)G(s)+

Y(s)GC(s)

Controllers: phase lead, phase lag, lead-lag compensator, P,PD PI PIDPD, PI, PID,…

Design method: root locus frequency response


Design method: root locus, frequency response

State feedback controlState feedback control

All the states of the system are fed back to calculate the control All the states of the system are fed back to calculate the controlrule.

+r(t) u(t)

C y(t))()()( tutt BAxx

x(t)

K

State feedback controller: )()()( ttrtu Kx

Design method: pole placement LQR

nkkk 21K


Design method: pole placement, LQR,…

Effects of controller on systemEffects of controller on systemEffects of controller on system Effects of controller on system performanceperformance


The addition of a pole (in the left-half s-plane) to the open-

Effects of the addition of polesEffects of the addition of poles The addition of a pole (in the left half s plane) to the open

loop transfer function has the effect of pushing the root locusto the right, tending to lower the system’s relative stability andt l d th ttli f thto slow down the settling of the response.

Im s Im s Im s

Re s Re s Re s


The addition of a zero (in the left-half s-plane) to the open-

Effects of the addition of zerosEffects of the addition of zeros

The addition of a zero (in the left half s plane) to the openloop transfer function has the effect of pulling the root locus tothe left, tending to make the system more stable and to speed

th ttli f thup the settling of the response.

Im s Im s Im s

Re s Re s Re s


Effects of lead compensatorsEffects of lead compensators

Transfer function:

)1( 1

1)(

TsTsKsG CC

Transfer function:

Frequency response:

TjTjKjG CC

11)(

Tj1

1sin 1

Characteristics of the Bode plots:

1

sinmax

T1

max

lg10lg20)( max CKL

The lead compensators improve


the transient response (POT, ts,..)

Lead compensator implementationLead compensator implementation Lead compensator transfer function:

)1( 1

1 11

)()(

22111142 CRCR

TsTsK

sCRsCR

RRRR

sEsU

C

Lead compensator transfer function:

11)( 2231 TssCRRRsE

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


Effects of lag compensatorsEffects of lag compensators

Transfer function:

)1( 1

1)(

TsTsKsG CC

Transfer function:

Frequency response:

TjTjKjG CC

11)(

Tj1

1sin 1

Characteristics of the Bode plots:

1

sinmin

T1

min

lg10lg20)( min CKL

The lag compensators


greduce the steady-state error.

Lag compensator implementationLag compensator implementation Lag compensator transfer function:

)1( 1

1 11

)()(

221122

11

31

42 CRCRTsTsK

sCRsCR

RRRR

sEsU

C

Lag compensator transfer function:

11)( 2231 TssCRRRsE

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


Effects of leadEffects of lead--lag compensatorslag compensators11

sTsT )1,1(

11

11)( 21

2

22

1

11

sTsT

sTsTKsG CC Transfer function:

Bode diagram

The lead lag compensators improve transient response and


The lead-lag compensators improve transient response andreduces the steady-state error.

Effects of proportional controller (P)Effects of proportional controller (P)

G )(T f f ti

Increasing proportional gain leads to decreasing steady-stateh h b l bl d h POT

PC KsG )( Transfer function:

error, however, the system become less stable, and the POTincreases.

(t) Ex: response of a

proportional control t h

y(t)

system whose plant has the transfer functiontransfer function below:

10)(G


)3)(2()(

sssG

Effects of proportional derivative controller (PD)Effects of proportional derivative controller (PD)

Bode diagram Transfer function: Bode diagram Transfer function:)1()( sTKsKKsG DPDPC

The PD controller is a special case of phase lead

t thcompensator, the maximum phase lead is max=900 at the frequency max q ymax=+.

The PD controller speed upthe response of the system,however it also makes thesystem more sensitive to


system more sensitive tohigh frequency noise.

Effects of proportional derivative controller (PD)Effects of proportional derivative controller (PD)

Note: The larger the derivative constant the faster the Note: The larger the derivative constant, the faster theresponse of the system.

y(t)y( )

unompensated


PD controller implementationPD controller implementation PD controller transfer function:

sKKsCRRRRR

sEsU

DP )1()()(

1142

PD controller transfer function:

RRsE )( 31

E(s)( )U(s)


Effects of proportional integral controller (PI)Effects of proportional integral controller (PI) Transfer function: Bode diagram Transfer function:

)11()(sT

Ks

KKsGI

PI

PC

Bode diagram

The PI controller is a special case of phase lag

t th i icompensator, the minimum phase lag is min= 900 at the frequency min=+.q y min

PI controllers eliminate stead state error to stepsteady state error to step input, however it can increase POT and settling


time.

Effects of proportional integral controller (PI)Effects of proportional integral controller (PI) Note: The larger the integral constant the larger the POT Note: The larger the integral constant, the larger the POT

of response of the system.y(t)y( )

uncompensated


PI controller implementationPI controller implementation PI controller transfer function:

1)()( 2242

sKK

sCRsCR

RRRR

sEsU I

P

PI controller transfer function:

)( 2231 ssCRRRsE

E(s)U( )U(s)


Effects of proportional integral controller (PID)Effects of proportional integral controller (PID) Transfer function:

Bode diagram Transfer function:

sKs

KKsG DI

PC )(

1 )11()( sTsT

KsG DI

PC

TKG 111)(

sT

sTKsG D

IPC 2

1

11)(

Effects of PID controllers:

speed up response of the system

Eliminate steady-state


error to step input.

Comparison of PI, PD and PID controllersComparison of PI, PD and PID controllers

(t)y(t)

U t dUncompensated


Control systems design Control systems design y gy gusing the root locus methodusing the root locus method


Procedure for designing lead compensator using Procedure for designing lead compensator using the root locusthe root locus

)/1( Ts )1( )/1()/1()(

TsTsKsG CCLead compensator:

Step 1: Determine the dominant poles from desired*s

(POT)Overshoot

2* 1 j

Step 1: Determine the dominant poles from desired transient response specification:

2,1s

Step 2: Determine the deficiency angle so that the dominant

ts timeSettling

( )

n 2

2,1 1 nn js

Step 2: Determine the deficiency angle so that the dominant poles lie on the root locus of the compensated system: *

2,1s

mn

zsps **0* )arg()arg(180

i

ii

i zsps1

11

1 )arg()arg(180

where pi and zi are poles & zeros of G(s) before compensation.


*1

0* to from angle180 spi *1to from angle szi

Procedure for designing lead compensator using the root locusProcedure for designing lead compensator using the root locus

Step 3: Determine the pole & zero of the lead compensator Step 3: Determine the pole & zero of the lead compensatorDraw 2 arbitrarily rays starting from the dominant pole such that the angle between the two rays equal to *. The

*1s

intersection between the two rays and the real axis are the positions of the pole and the zero of the lead compensator. Two methods often used for drawing the rays:Two methods often used for drawing the rays: Bisector method Pole elimination methodPole elimination method

Step 4: Calculate the gain KC using the formula:

1)()( *1

ssC sGsG


Example of designing a lead compensator using RLExample of designing a lead compensator using RL

( )R(s)+

Y(s)GC(s)

)5(50ss

Objective: design the compensator GC(s) so that the j g p C( )response of the compensated system satisfies: POT<20%; ts< 0,5sec (2% criterion).

Solution: Because the design objective is to improve the transient

)1()/1()(

TsKsG

response, we need to design a lead compensator:


)1( )/1(

)(

Ts

KsG CC

Example of designing a lead compensator using RL (cont’)Example of designing a lead compensator using RL (cont’)

Step 1 Determine the dominant poles Step 1: Determine the dominant poles:

2.01

exp2

POT 6,12.0ln

1

2

45,0

1 2

1 2

707,0Chose

5,04

nt

qñ

5,04 n 4,11 n

15nChose

The dominant poles are:22*

2,1 707,011515707,01 jjs nn

*

p


5,105,10*2,1 js


Step 2: Determine the deficiency angle: Step 2: Determine the deficiency angle:

)]5()510510arg[(]0)510510arg[(1800* jjMethod 1:

)]5()5,105,10arg[(]0)5,105,10arg[(180 jj

5,55,10arctan

5,105,10arctan1800

5,55,10)6,117135(1800

0* 6,72Im s

s*j10,5s

Method 2:)(1800*

Re s12O

)(180 210

)6,117135(180 000


5 O10,50* 6,72


Step 3: Determine the pole and the zero of the compensator Step 3: Determine the pole and the zero of the compensator(bisector method)

s* PxIm s

j10,5

1O

AB C Re s

510,5 5

12,28ˆ22

ˆsin

*

*

xPO

OPOB 0,822

ˆsin

*

xPO

OPOC,

22ˆ

sin*

xPO

0,8

22ˆ

sin*

xPO

OPOC

8s


288)(

ssKsG CC


Step 4: Determine the gain of the compensator: Step 4: Determine the gain of the compensator:

1)()( * ssC sGsG

1)5510510)(510510(

50.2851051085,105,10

jjjjKC )55,105,10)(5,105,10(285,105,10 jjj

18511154120

5079,10

CK

85,111541,20 C

7,6 CK

876)( sG

Conclusion: The transfer function of the lead compensator is:


287,6)(

ssGC

Root locus of the systemRoot locus of the system

Root locus of the Root locus of the


uncompensated system compensated system

Transient response of the systemTransient response of the system

y(t)

uncompensatedcompensated

Transient response of the system



Procedure for designing lag compensator using the root locusProcedure for designing lag compensator using the root locus

)/1( Ts )1( )/1()/1()(

TsTsKsG CCLag compensator:

Step 1: Determine to meet the steady-state error requirement: Step 1: Determine to meet the steady-state error requirement:

*P

KK

*V

KK

*a

KK

or orPK VK aK

Step 2: Chose the zero of the lag compensator: )Re(1 *21s Step 2: Chose the zero of the lag compensator: )( 2,1T

11 Step 3: Calculate the pole of the compensator:

TT . Step 3: Calculate the pole of the compensator:

Step 4: Calculate K satisfying the condition: 1)()( * sGsG


Step 4: Calculate KC satisfying the condition: 1)()( *2,1

ssC sGsG

Example of designing a lag compensator using RLExample of designing a lag compensator using RL

( )R(s)+

Y(s)GC(s)

)4)(3(10

sss

Objective: design the compensator GC(s) so that the j g p C( )compensated system satisfies the following performances: steady state error to ramp input is 0,02 and transient response of the compensated system is nearly unchangedresponse of the compensated system is nearly unchanged.

Solution:

)1()/1()(

TsKsG

The compensator to be design is a lag compensator:


)1( )/1(

)(

Ts

KsG CC

Example of designing a lag compensator using RL (cont’)Example of designing a lag compensator using RL (cont’)

Step 1: Determine

The velocity constant of uncompensated system :

83.0)4)(3(

10lim)(lim00

sss

sssGKssV

5011*

* VK

The desired velocity constant:

5002,0*

xlV e

K

83.0 VKThen:

50* VK

Then:

0170


017,0


Step 2: Chose the zero of the lag compensator Step 2: Chose the zero of the lag compensatorThe pole of the uncompensated system:

10 1 js0)(1 sG 0)4)(3(

101

sss

51

3

2,1

sjs

Th d i t l f th t d t j1 The dominant poles of the uncompensated system: js 12,1

1Re11 s

TChose: 1,01

T

T T

Step 3: Calculate the pole of the compensator:11 1

)1,0)(017,0(11

TT 0017,01

T

1,0)(

sKsG


0017,0)(

sKsG CC


Step 4: Determine the gain of the compensator

1)()( * ssC sGsGss

1)4)(3(

10.0017,0

1,0 1

jC ssss

sK))((7, 1 js

1)41)(31)(1(

10.)001701(

)1,01(

jjjjjKC )41)(31)(1()0017,01( jjjj

10042,1 CK

1,0)(

ssG


0017,0)(

ssGC

Root locus of the systemRoot locus of the system

Root locus of the Root locus of the


uncompensated system compensated system

Transient response of the systemTransient response of the system

( )y(t)

uncompensatedt dcompensated



Procedure for designing lead lag compensator using the RLProcedure for designing lead lag compensator using the RL

The compensator to be designed

)()()( 21 sGsGsG CCC

The compensator to be designed

phaselead

phaselag

Step 1: Design the lead compensator GC1(s) to satisfy thetransient response performances.

Step 2: Let G1(s)= G (s). GC1(s)D i h l G ( ) i i i h G ( )Design the lag compensator GC2(s) in series with G1(s) tosatisfy the steady-state performances (and not to degrade thetransient response obtained after phase lead compensating)


transient response obtained after phase lead compensating)

Example of designing a lead lag compensator using RLExample of designing a lead lag compensator using RL

( )R(s)+

Y(s)GC(s)

)5.0(4ss

Objective: design the compensator GC(s) so that thej g p C( )compensated system has the dominant poles with = 0.5,n =5 (rad/sec) and the velocity constant KV =80.

Solution The compensator to be designed is a lead lag compensatorp g g p

because the design objective is to improve the transientresponse and to reduce the steady-state error.


)()()( 21 sGsGsG CCC

Example of designing a lead lag compensator using RL (cont’)Example of designing a lead lag compensator using RL (cont’)

Step 1: Design the lead compensator GC1(s)

The dominant poles:22*

2,1 5,01555,01 jjs nn

33452* js 33,45,22,1 js

The deficiency angle:0* )(180 210*

)115120(180 000 0* 0* 55



Chose the zero of the lead compensator so that it eliminatesChose the zero of the lead compensator so that it eliminatesthe pole at –0.5 of G(s) (pole elimination method)

501

5,01

T

5,0OA

5.460sin55sin76.4

sin

ˆsin0

0

PAB

BPAPAAB

5,0OA

AB51

1 ABOA

T–1/T1–1/T1

1

55,0)( 11

sKsG CC


5s


C l l KCalculate KC1: 1)()( *1 ssC sGsG

145,0sK 1)5,0(

.5,

33,45,21

jsC sss

K

256K 25,61 CK

5,025,6)(1

ssGC

The lead-compensated open-loop system:

525,6)(1 s

sGC

)5(25)()()( 11

ss

sGsGsG C

The lead-compensated open-loop system:


)5( ss


Step 2: Design the lag compensator GC2(s)1T

s

2

222 1)(

Ts

TKsG CC

2T Determine :

525lim)(lim 1 sssGKV 5)5(

lim)(lim010 sssssGK

ssV

80* VKV

161

805

* V

KK


1680VK


D i h f h l Determine the zero of the lag compensator:

5,2)33,45,2Re()Re(1 * jsT 2T

16,01

Chose: ,

2T

)16,0.(1611.1

Calculate the pole of the lag compensator:

),(1622 TT

01.01


2T


Calculate KC2 using the gain condition: 1)()( *12 C sGsGCalculate KC2 using the gain condition: 1)()( *12 ssC sGsG

1)()( ** 12 ssssC sGsG

101,033,45,216,033,45,2

2

jjKC

The transfer function of the lag compensator:

01.12 CK

)010()16,0(01,1)(2

sssGC

The transfer function of the lag compensator:

)16,0)(5,0(31,6)()()( 21

sssGsGsG CCCFinal result:

)01,0( s


)01,0)(5(31,6)()()( 21 ss

sGsGsG CCC

Control system Control system design in design in yy ggfrequency domainfrequency domain


Procedure for designing lead compensators in frequency domainProcedure for designing lead compensators in frequency domain

1Ts )1( 11)(

TsTsKsG CCThe lead compensator:

Step 1: Determine KC to meet the steady-state error requirement: Step 1: Determine KC to meet the steady state error requirement:

PPC KKK /* VVC KKK /* aaC KKK /*or or

Step 2: Let G (s)=K G(s) Plot the Bode diagram of G (s) Step 2: Let G1(s)=KCG(s). Plot the Bode diagram of G1(s) Step 3: Determine the gain crossover frequency of G1(s):

0)(1 CL 1)(1 CjG or0)(1 CL 1)(1 CjG or

Step 4: Determine the phase margin of G1(s) (phase margin of uncompensated system): )(180 1 CM of uncompensated system): )(180 1 CM

Step 5: Determine the necessary phase lead angle to be added to the system: MM *


MMmax

is the desired phase margin,*M 00 205

Procedure for designing lead compensators in frequency domainProcedure for designing lead compensators in frequency domain

sin1

St 7 D t i th i f ( f

Step 6: Calculate :max

max

sin1sin1

Step 7: Determine the new gain crossover frequency (ofthe compensated open-loop system) using the conditions:

lg10)(1 CL /1)( jGor lg10)(1 CL /1)(1 CjGor

Step 8: Calculate the time constant T: T 1pC

Step 9: Check if the compensated system satisfies the gainmargin? If not repeat the design procedure from step 5margin? If not, repeat the design procedure from step 5.

Note: It is possible to determine C (step 3), M (step 4) and’C (step 7) by using Bode diagram instead of using analytic


C (step 7) by using Bode diagram instead of using analyticcalculation.

Design lead compensator in frequency domain Design lead compensator in frequency domain -- ExampleExample

( )R(s)+

Y(s)GC(s))2(

4ss

Objective: Design the compensator GC(s) so that thej g p C( )compensated system satisfies the performances:

;20* VK ;500* M dBGM 10*

Solution:

The transfer function of the lead compensator to be designed:p g

TsTsKsG CC

11)( )1(


Design lead compensator in frequency domain Design lead compensator in frequency domain –– Example (cont’)Example (cont’)

Step 1: Determine K Step 1: Determine KC

The velocity constant of the uncompensated system:

24li)(li GK 2)2(

lim)(lim00

ss

sssGKssV

The desired velocity constant: 20* VK

220*

V

VC K

KK 10CK

y V

V

Step 2: Denote)2(

4.10)()(1

sssGKsG C

)15,0(20)(1

ss

sG


Draw the Bode diagram of G1(s)


-20dB/dec26

-40dB/dec

c=62

160

M


-160


Step 3: The gain crossover frequency of G (s) Step 3: The gain crossover frequency of G1(s)According to the Bode diagram: 6C (rad/sec)

Step 4: The phase margin of G1(s)

According to the Bode diagram:00

1 160)( C0

1 20)(180 CM

Step 5: The necessary phase lead angle to be added: MM * (chose =7)

000max 72050

0

MMmax (chose 7)


0max 37


Step 6: Calculate Step 6: Calculate

0

0max

37sin137sin1

sin1sin1

4

Step 7: Determine the new gain crossover frequency using Bode plot

max 37sin1sin1

dBL 64lg10lg10)( p dBL C 64lg10lg10)(1

The abscissa of the intersection between Bode magnitudediagram and the horizontal line with ordinate of 6dB is thed ag a a d e o o a e o d a e o 6d s enew gain crossover frequency. According to the plot (in slide54), we have:

9C (rad/sec)9C (rad/sec) Step 8: Calculate T

11T


)4)(9(

C

T 056,0T 224,0T


-20dB/dec

+20dB/dec

-20dB/dec-40dB/dec

-20dB/dec

-40dB/dec

-6-40dB/dec

c=6 ’c=9 1/T=181/T=4.5

160

MM *


-160


Step 9: Check the gain margin of the compensated system

According to the compensated Bode diagram, GM* = +, thenthe compensated system fulfills the design requirementsthe compensated system fulfills the design requirements.

Conclusion: The designed lead compensator is:

sssGC 056,01

224,0110)(


Design lead compensator in frequency Design lead compensator in frequency domain domain –– Example 2Example 2

( )R(s)+

Y(s)GC(s) G(s)

16)(02.0eG

s

Objective: Design the compensator G (s) so that the

)2510)(2()( 2

ssssG

Objective: Design the compensator GC(s) so that thecompensated system has: andsteady-state error to unit step input ;05.0* sse

;500* M dBGM 10* steady state error to unit step input ;05.0sse

Solution:


Procedure for designing lag compensators in frequency domainProcedure for designing lag compensators in frequency domain

1Ts )1( 11)(

TsTsKsG CCThe lag compensator:

Step 1: Determine KC to meet the steady-state error requirement: Step 1: Determine KC to meet the steady state error requirement:

PPC KKK /* VVC KKK /* aaC KKK /*or or

Step 2: Let G (s)=K G(s) Plot the Bode diagram of G (s) Step 2: Let G1(s)=KCG(s). Plot the Bode diagram of G1(s) Step 3: Determine the new gain crossover frequency

satisfying the following condition:C

satisfying the following condition:

is the desired phase margin*M 00 205 *0

1 180)( MC

is the desired phase margin,M 205 Step 4: Calculate using the condition:

l20)( L 1)( jGor


lg20)(1 CL

)(1 CjGor

Procedure for designing lag compensators in frequency domainProcedure for designing lag compensators in frequency domain

St 5 Ch th f th l t th t Step 5 : Chose the zero of the lag compensator so that:

CT

1 TT

Step 6: Calculate the time constant T:

TT 11

T

Step 7: Check if the compensated system satisfies the gainmargin? If not, repeat the design procedure from step 3.

Note: It is possible to determine , (step 3), (step 4) by using Bode diagram instead of using analytic

C)(1 C )(1 CL


calculation.

Design lag compensator in frequency domain Design lag compensator in frequency domain –– ExampleExample

( )R(s)+

Y(s)GC(s))15.0)(1(

1 sss

Objective: design the lag compensator GC(s) so that thatj g g p C( )compensated system satisfies the following performances:

;5* VK ;400* M dBGM 10*

Solution The transfer function of the lag compensator to be designed: The transfer function of the lag compensator to be designed:

TsTsKsG CC

11)( )1(


Ts1

Design lag compensator in frequency domain Design lag compensator in frequency domain –– Example (cont’)Example (cont’)

Step 1: Determine KC Step 1: Determine KC

The velocity constant of the uncompensated system:

11lim)(lim GK 1)15.0)(1(

lim)(lim00

sss

sssGKssV

The desired velocity constant: 5* VK

5*

V

VC K

KK

y V

V

Step 2: Denote )()(1 sGKsG C5

)15.0)(1(

5)(1

ssssG


Draw the Bode diagram of G1(s)


-20dB/dec

60dB/dec

-40dB/dec14

-60dB/dec

21



Step 3: Determine the new gain crossover frequency:

*01 180)( MC

01 135)( C

0001 540180)( C

According to the Bode diagram: 5.0C (rad/sec)

St 4 C l l t i th diti

1 135)( C

Step 4: Calculate using the condition: lg20)(1 CL

( )According the Bode diagram: (dB)18)(1 CL lg2018 9,0lg 9,010


126,0


Step 5: Chose the zero of the lag compensator: Step 5: Chose the zero of the lag compensator:

5.01 CT

Chose 05.01

T 20T

159T

Step 6: Calculate the time constant T

00630050126011 159T0063,005,0126,0

TT

Step 7: It can be verified in the Bode diagram that the Step 7: It can be verified in the Bode diagram that thecompensated system satisfies the gain margin requirement.

)120(5)(

ssGConclusion


)1159(5)(

ssGCConclusion


-20dB/dec

60dB/dec

-40dB/dec14L1(’c)

L’(’)

GM*

-60dB/dec

21’c=0.50.0067 0.05 ’

( )

-135


Design lag Design lag compensator in frequency compensator in frequency domain domain –– Example 2Example 2

( )R(s)+

Y(s)GC(s) G(s)

16)(02.0eG

s

Objective: Design the compensator G (s) so that the

)2510)(2()( 2

ssssG

Objective: Design the compensator GC(s) so that thecompensated system has: andsteady-state error to unit step input ;05.0* sse

;500* M dBGM 10* steady state error to unit step input ;05.0sse

Solution:


Comparison of phase lead and phase lag compensatorComparison of phase lead and phase lag compensator


(Dorf and Bishop (2008), Modern control system –p.729)

D i f PIDD i f PID t llt llDesign of PID Design of PID controllers controllers


Zeigler Zeigler Nichols method 1Nichols method 1

Determine the PID parameters based on the step response of Determine the PID parameters based on the step response ofthe open-loop system.

Plantu(t) y(t)

y(t)

K

t


T1 T2

Zeigler Zeigler Nichols method 1 (cont’) Nichols method 1 (cont’)

R( )R(s)+

Y(s)PlantPID

1PID controller:

sT

sTKsG D

IPC

11)(

Controller KP TI TD

P T /(T K) 0P T2/(T1K) 0

PI 0.9T2/(T1K) 0.3T1 0


PID 1.2T2/(T1K) 2T1 0.5T1

Zeigler Zeigler Nichols method 1 Nichols method 1 –– Example Example

Problem: Design a PID y(t) Problem: Design a PID controller to control a furnace providing the open-loop h t i ti f th f

y(t)

characteristic of the furnace obtained from a experiment beside.

150

t (min)( )

8 24150K

sec480min81 T

024014402121 2 TKP

ssGPID 240110240)(

sec1440min242 T

024.0150480

2.12.11 KT

KP

sec96048022 1 TTI

ss

sGPID 240960

1024.0)(


sec2404805.05.0 1 TTD

Zeigler Zeigler Nichols method 2Nichols method 2 Determine the PID parameters based on the response of the Determine the PID parameters based on the response of the

closed-loop system at the stability boundary.

Plant+ KKcr

y(t)

Tcr


t

Zeigler Zeigler Nichols method 2 (cont’)Nichols method 2 (cont’)

R(s)+

Y(s)PlantPID

1PID controller:

sT

sTKsG D

IPC

11)(

Controller KP TI TD

P 0 5K 0P 0.5Kcr 0

PI 0.45Kcr 0.83Tcr 0


PID 0.6Kcr 0.5Tcr 0.125Tcr

Zeigler Zeigler Nichols method 2 Nichols method 2 –– ExampleExample

Problem: Design a PID controller to control the angle position Problem: Design a PID controller to control the angle positionof a DC motor, providing that by experiment the critical gain ofthe system is 20 and the critical cycle is T= 1 sec.

According to the given data: Solution:

20crKsec1crT

g g

Applying Zeigler – Nichols method 2:

12206.06.0 crP KK

s

ssGPID 5.0

125.01112)(

crP

sec5.015.05.0 crI TTsec1250112501250 TT


sec125.01125.0125.0 crD TT

Analytical method for designing PID controller Analytical method for designing PID controller

St 1 E t bli h ti ( ) ti th l ti hi Step 1: Establish equation(s) representing the relationship between the controller to be designed and the desired performances.

Step 2: Solve the equation(s) obtained in step 1 for the t ( ) f th t llparameter(s) of the controller.


Analytical method for designing PID controller Analytical method for designing PID controller

Example: Design PID controller so that the control system Example: Design PID controller so that the control system satisfies the following requirements: Closed-loop complex poles with =0.5 and n=8. Velocity constant KV = 100.

R(s) Y(s)100( )+

( )GPID(s)

10010100

2 ss

Solution: The transfer function of the PID controller to be Solution: The transfer function of the PID controller to be designed

sKKKsG DI

PC )(


sKs

KsG DPC )(

Analytical method for designing PID controller (cont’)Analytical method for designing PID controller (cont’)

Velocity constant of the controlled system:

10010100lim)()(lim 200 ss

sKs

KKssGssGK DI

PsCsV 1001000 sssss

IV KK

According to the design requirement: KV = 100

100IK

The characteristic equation of the controlled system:

100 K 010010

1001 2

sssK

sKK D

IP

0100)100100()10010( 23 KKK (1)


0100)100100()10010( 23 IPD KsKsKs (1)

Analytical method for designing PID controller (cont’)Analytical method for designing PID controller (cont’)

The desired characteristic equation: The desired characteristic equation:

0)2)(( 22 nnssas

0)648)(( 2 0)648)(( 2 ssas

064)648()8( 23 asasas (2)

Balancing the coefficients of the equations (1) and (2), we have:

aK 810010 25156

aKaK

aK

P

D

64100648100100

810010

54114,1225.156

P

KKa

aKI 64100 54,1DK

Conclusion: G 5411006412)(


Conclusion: ss

sGC 54,164,12)(

Manual tuning of PID controllersManual tuning of PID controllers

Effect of increasing a parameter of PID controller independently on closed-loop performance:

Para-meter Rise time POT Settling

time

Steady-stateerror

Stabilityerror

KP Decrease Increase Smallchange Decrease Degrade

KI Decrease Increase Increase Eliminate Degrade

KDMinor

change Decrease Decrease No effect Improve ifK smallD change KD small

6 December 2013 82© H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/

A d f l t i f PID t ll

Manual tuning of PID controllers (cont.)Manual tuning of PID controllers (cont.)

A procedure for manual tuning of PID controllers:

1. Set KI and KD to 0, gradually increase KP to the critical i K (i th i k th l d l tgain Kcr (i.e. the gain makes the closed-loop system

oscilate)

2. Set KP Kcr /2

3. Gradually increase KI until the steady-state error is eliminated in a sufficient time for the process (Note that too much KI will cause instability).

4. Increase KD if needed to reduce POT and settling time (Note that too much KD will cause excessive response and overshoot)

6 December 2013 83© H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/

Control systems design in stateControl systems design in state--spacespaceusing pole placement methodusing pole placement method


ControllabilityControllability

)()()( tutt BAxxC id t

)()()()()(

ttytutt

CxBAxx

Consider a system:

The system is complete state controllable if there exists anunconstrained control law u(t) that can drive the system froman initial state x(t0) to a arbitrarily final state x(tf) in a finitetime interval t0 t tf . Qualitatively, the system is statetime interval t0 t tf . Qualitatively, the system is statecontrollable if each state variable can be influenced by theinput.

y(t)


Signal flow graph of an incomplete state controllable system

Controllability conditionControllability condition

)()()( ttt BAxx

)()()()()(

ttytutt

CxBAxx

System:

Controllability matrix

][ 12 BABAABB nC ][ BABAABB C

The necessary and sufficient condition for the controllability is:

nrank )(C

N t th t “ t ll bl ” i t d f “ l t Note: we use the term “controllable” instead of “complete state controllable” for short.


Controllability Controllability –– ExampleExample

)()()( tutt BAxx

)()()()()(

ttytutt

CxBAxx

Consider a system

where: 10 5where:

3210

A

25

B 31C

E l h ll bili f h

Solution: Controllability matrix:

Evaluate the controllability of the system.

ABBC Because:

162

25

C

Because:

84)det( C 2)( Crank


The system is controllable

State feedback controlState feedback controlr(t) u(t) y(t)x(t)

+r(t) u(t)

C y(t))()()( tutt BAxx

x(t)

K

Consider a system described by the state equations:

)()()()()(

ttytutt

CxBAxx

Consider a system described by the state equations:

The state feedback controller: )()()( ttrtu Kx )()( tty Cx

)()(][)( trtt BxBKAx

The state equations of the closed-loop system:


)() ty(t Cx

Pole placement methodPole placement method

If the system is controllable then it is possible to determineIf the system is controllable, then it is possible to determinethe feedback gain K so that the closed-loop system has thepoles at any location.

Step 1: Write the characteristic equation of the closed-loopsystem 0]det[ BKAIs (1)0]det[ BKAIs ( )

Step 2: Write the desired characteristic equation:n

0)(1

n

iips

)1( nip th d i d l

(2)

Step 3: Balance the coefficients of the equations (1) and (2),

),1( , nipi are the desired poles


we can find the state feedback gain K.

Pole placement method Pole placement method –– ExampleExample

Problem: Given a system described by the state state Problem: Given a system described by the state-state equation:

)()()( tutt BAxx

)()( tty Cx

010 0 100C

100010

A

30

B 374 1

Determine the state feedback controller )()()( ttrtu Kx Determine the state feedback controller so that the closed-loop system has complex poles with

and the third pole at 20. 10;6,0 n

)()()( ttrtu Kx


n

Pole placement method Pole placement method –– Example (cont’)Example (cont’)

Solution

The characteristic equation of the closed-loop system:0]det[ BKAIs

Solution

0]det[ BKAIs

030

100010

010001

d t

kkk 013

374100

100010det 321

kkks

The desired characteristic equation:

(1)0)12104()211037()33( 313212

323 kkskkkskks


0)2)(20( 22 nnsss (2)23


(2)0200034032 23 sss

Pole placement method Pole placement method –– Example (cont’)Example (cont’)

B l th ffi i t f th ti (1) d (2) h Balance the coefficients of the equations (1) and (2), we have:

3233 32 kk

200012104340211037

21

321

kkkkk

Solve the above set of equations, we have:

578220k

48217839,3

578,220

2

1

kkk

482,173k

482178393578220K Conclusion:


482,17839,3578,220K Conclusion:

D i f t t ti tD i f t t ti tDesign of state estimatorsDesign of state estimators


The concept of state estimationThe concept of state estimation

To be able to implement state feedback control system it is To be able to implement state feedback control system, it isrequired to measure all the states of the system.

However in some applications we can only measure the However, in some applications, we can only measure theoutput, but cannot measure the states of the system.

The problem is to estimate the states of the system from thep youtput measurement.

State estimator (or state observer)


ObservabilityObservability

)()()( tutt BAxx

)()()()()(

ttytutt

CxBAxx

Consider a system:

The system is complete state observable if given the control y p glaw u(t) and the output signal y(t) in a finite time interval t0 t tf , it is possible to determine the initial states x(t0). Qualitatively the system is state observable if all state variableQualitatively, the system is state observable if all state variable x(t) influences the output y(t).

y(t)


Signal flow graph of an incomplete state observable system

Observability conditionObservability condition

)()()( tutt BAxxS t

)()()()()(

ttytutt

CxBAxx System

It is necessary to estimate the state from mathematical )(ˆ tx

Observability matrix: C

ymodel of the system and the input-output data.

)(

Observability matrix:

2CACAC

O

1nCA

CA

O

CA

The necessary and sufficient condition for the observability is:


nrank )(O

Observability Observability –– ExampleExample

)()()( tutt BAxx

)()()()()(

ttytutt

CxBAxx

Consider the system

10 1where:

32

10A

21

B 31C

Solution: Observability matrix:

Evaluate the observability of the system.

CAC

O

8

36

1O

Because 10)det( O 2)( Orank


The system is observable

State estimatorState estimator

u(t) y(t)x(t)r(t) u(t) y(t))()()( tutt BAxx x(t)

C

+

r(t)

CB

L)(ˆ tx

+++ C

)(ˆ tyB

A

+

K

)(ˆ)(ˆ))(ˆ)(()()(ˆ)(ˆ

ttytytytutt

xCLBxAx State estimator:


yT

nlll ][ 21 Lwhere:

Design of state estimatorsDesign of state estimators Requirements: Requirements:

The state estimator must be stable, estimation errorshould approach to zero.

Dynamic response of the state estimator should be fast Dynamic response of the state estimator should be fastenough in comparison with the dynamic response of thecontrol loop.

All the roots of the equation locates i th h lf l ft l

It is required to chose L satisfying:0)det( LCAsI

in the half-left s-plane. The roots of the equation are further

from the imaginary axis than the roots of the equation0)det( LCAsI

0)det( BKAsI Depending on the design of L, we have different state estimator:

Luenberger state observer


Luenberger state observer Kalman filter

Procedure for designing the Luenberger state observerProcedure for designing the Luenberger state observer

St 1 W it th h t i ti ti f th t t b Step 1: Write the characteristic equation of the state observer

0]det[ LCAIs (1)

Step 1: Write the desired characteristic equation:

0)( n

ips (2))(1i

ip

),1( , nipi are the desired poles of the state estimator

Step 3: Balance the coefficients of the characteristicequations (1) and (2), we can find the gain L.


Design of state estimators Design of state estimators –– ExampleExample

Problem: Given a system described by the state equation: Problem: Given a system described by the state equation:

)()()()()( tutt

CBAxx

)()( tty Cx

010 0 001C

100010

A

30

B 374 1

Assuming that the states of the system cannot be directly Assuming that the states of the system cannot be directlymeasured. Design the Luenberger state estimator so that thepoles of the state estimator lying at 20, 20 and 50.


p y g ,

Design of state estimators Design of state estimators –– Example (cont’)Example (cont’)

Solution The characteristic equation of the Luenberger state estimator:

0]det[ LCAIs

Solution

0]det[ LCAIs

0001100010

010001

d t1

ll

0001374

100100010det

3

2

lls


(1)0)457()73()3( 321212

13 lllsllsls


0)50()20( 2 ss23


(2)020000240090 23 sss

Design of state estimators Design of state estimators –– Example (cont’)Example (cont’)

B l i th ffi i t f th (1) d (2) l d t Balancing the coefficients of the equ. (1) and (2) leads to:

9031l

20000437240073

321

21

lllll

Solve the above set of equations, we have:

87l

12991213287

2

1

lll

129913l

T12991213287L Conclusion


12991213287L Conclusion

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