Kontrol Teorisi

7/26/2019 Kontrol Teorisi

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Otomatik Kontrol I

P(oransal)I(integral)D(trevsel) kontrol

Dr. Vasfi Emre mrl


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By Vasfi Emre mrl, Ph.D., 2005 2

PID kontrol matematii

Doru akm motoru zerinde uygulama


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PID kontrol

Kullanm kolayl dolaysyla endstride ounluklakullanlmaktadr.

Oransal

ntegral Trevsel PIDPID

Oransal ntegral TrevselOransal ntegral Trevsel


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PID blok diagramPIDdenetleyicie(t) u(t)

1

TdsKp

1/(Tis)

e(t) u(t)


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Oransal kontrol

Hatann sabit deeri iin sabit denetleyici k retilir. Hatanndevam etme durumunda kontrol k deimez.

Hata miktarna ve Kp katsaysna bal olarak oransalkontrol, denetleyici kn retir

Sistemin statik doruluunu ve dinamik cevabn artrr

Hatann ve oransal kontrol katsaysnn dorudanfonksiyonudur

up = Kp . (hata miktar)


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ntegral kontrol birikimli/kmlatif

Hatann sfrdan farkl olma durumunda kontrol miktarn artrr

Ki kontrol katsaysna ve hata miktarna gre denetleyicik ayarlanr.

Dinamik cevapdan feragat ederek statik doruluk

miktarn artrr Hata birikiminin ve integral kontrolc katsaysnnfonksiyonudur.


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Trevsel kontrol

Hatann sabit olma durumunda, trevsel kontrol bir kretmez. Fakat kontrol hatann deime hzna bal olarakretilir.

Hatann deime hzna ve Kd trevsel kontrolkatsaysna bal olarak, denetleyici k ayarlanr

Dinamik cevab artrr veya gelitirir Hatann deime hznn ve trevsel kontrol katsaysnndorudan fonksiyonudur


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Oransal ntegral Trevsel

up = Kp . (hata) uI =Ki (hata).dt uD =Kd .(de/dt)

PID kontrol


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PID denetleyicinin uyarlamas daha mevcuttur


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k

b

x

M

F

rnek

Bu sistemin dinamik denklemi


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Bu sistemin dinamik denklemi


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Let,

M=1kg

B= 10 N.s/m

K=20 N/m

F(s)=1


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Ak evrim sistem davran yle olacaktr,


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Sadece oransal kontrol uygulayacak olursak

X(s)F(s) KP

s2 + 10s + (20+KP)

Oransal kontrol katsays KP=300


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Oransal-Trevsel kontrol uygulayacak olursak

PD kontrol ile beraber kapal evrim sistemin transfer fonksiyonu

KP=300, KD =10


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PI kontrol uygulayacak olursak

PI kontrol ile beraber sistemin transfer fonksiyonu

KP=30, KI =70

PID k t l


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PID kontrol

PID kontrol ile beraber kapal evrim sistemin transfer fonksiyonu

Kp=350, Ki=300, Kd=50


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Grafiklerin yorumu Oransal kontrol ykselme

zamann ve kararl halhatasn azaltm, stam

artrm ve yerlemezamann azaltmtr

Trevsel kontrol stam veyerleme zamann drr.

Fakat ykselme zamanna vekararl hal hatasna az etkisi

vardr.

ntegral kontrol kararl halhatasn azaltacaktr. Yalnz

oransal kontrol katsaysdrlr ki integral kontroldeoransal kontrol gibi stam

artrc etkiye sahiptir.

PID kontrol uygulanmasylastamsz, hzl ykselme

zamanl ve kararl hal hatasolmayan bir cevap eldeedilmitir.


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Gerekli olmad takdirde oransal, integral ve trevselkontroln nn ayn anda uygulanmasna gerek yoktur.

Kontrolcnn mmkn olduunca basit tutulmasnda

da fayda vardr.

Sonu


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)(

)()()(

)()(

)()(

)()(

)()(

)()(

)()(1

)()(

)()(

)(1

)()(

)(

)()()()(

)()()()(

)()()()(

)(

)(

2

)(

)(

2

2

)(

sTKKBRsJRBLsJL

sLRsV

KKBRsJRBLsJL

Ks

sTK

sLRs

K

KKBRsJRBLsJLsV

sTK

sLRsK

K

BsJsLRsV

sKsTK

sK

BsJsLRsV

sTK

sK

BsJsI

sKsIsLRsV

sTsBsIKssJ

sKsIsLsIRsV

TBiKJ

vdt

di

LiRv

L

fon ksiyonutransferarasrasntorkuykvehzmotor

sT

s

tvaaaa

aaex

fon ksiyonutransferarasrasnvoltajtahrikvehzmotor

sV

s

tvaaaa

t

Lt

aa

t

tvaaaa

ex

L

t

aav

t

aaex

vL

tt

aaex

Ltt

a

vaaaex

Lat

vaaaaex

Lat

tK

bemf

a

aaaex

Lex

v

PM DC Motor Modelleme

Elektrikselksmndiferansiyel denklemi

Mekanikksmndiferansiyel denklemi

Yukardaki denklemlerin sfr balang artyla

Laplacednmlerinin alnmas

C k i bl k


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PM DC motorun ak evrim blokdiagram

Yukardaki denklemler bu blok diagram ileifade edilebilir. Bu blok diagram zersenizayn diferansiyel denklemlerle karlarsnz


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PM DC Motor Kontrol Problem

DC motorda neyi kontrol ederiz?1. Hz kontrol2. Sistem veriminin artrlmas3. Bozucu etkilerin etkisinin azaltlmas


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PM DC Motor Kontrol Problem

Kapal evrim transfer fonksiyonu


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DC motor kontrol


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DC motorun kapal evrim kontrol

Ak evrim ve kapal evrim sistem bozucu etkiye nasl cevap verir?

LT = 0 , OL

LT 0 , OL

LT = 0 , CL

L

T 0 , CL V


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Oransal kontroln etkisi _Think of proportional controller whith proportionality constant of K

Vex = K.r in steady_state, rAKyss .. if ryA

K ss 1

If 0LT

_ Lss TBArKy ... ifA

K1

Lss TBry .

IfL

T = 0 , CL

_ Vex = K.(r-y)

11.

1

11

.

21

21

ss

AK

ss

AK

sR

sY

sTBsRKAsssY L...11 21

ryAKrKA

KAy ssss

1

.1

.

V_

)(11

)(11

.

2121

sTss

BsRss

AKsY L

Lss TAK

Br

AK

AKy

.1.1

.


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PM DC Motor Control - Result

OL CL

TL= 0 0LT TL= 0 0LT

ryss

if disturbance is

zero andmodel of

the systemcorrect,

selecting control

constant

Lss TBry .

if disturbance is not

zero, we will

observe the

amplified effect of

disturbance of the

output

rKA

KAyss

.1

.

if no disturbance

increasing gain K

willresult inreducing

steady_state error,

ryss

Lss TAK

Br

AK

AKy

.1.1

.

in case of disturbance,

increasing K will decrease

the effect of disturbance


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PM DC Motor Control - ResultFor the system, increasing gain K decreases steady_state error. However, is there a limit

increasing gain K ?

Physical system limitations, stability

Check stability by locking at the pole locations of the system for increasing gain K


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PM DC Motor Control Problem

For given example (DC motor speed control), proportional control does not result in 0

sse Apply integral control

t

I

dteT

Kptu

0

.)( sT

Kp

sE

sU

I

)(

)(

IT = integtal time (reset time)

t

I

p

dtyrT

K

Vex0

feedback control

Lt

I

pTBdtyr

T

KAyyy .)(

0

2121

L

Iy

TBrT

KpAy

T

KpAyyy .

.)( 2121

in steady_state ifL

T =0 ryss

PM DC Motor Control Problem


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PM DC Motor Control Problem -

Stability


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Example

If we have mechanic system with transfer function 1

1)(

ssT

Kt G(s)+-

(s)T

PIDr(s)

Kt+-

(s)T

r(s)

sT

sTsTTK

i

idip

12

1

1

s


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Example cont.

For a DC motor we studied transfer function (relation) betweenexcitation voltage and output angular velocity (s)/Vex(s).However we know that armature current is directly (almost)related to generated torque.

Kt: torque constant (Nm/A).

If we think of only bearing damping and carried load, a conveyortransfer function is simply J: motor inertia

B:bearing damping

T:excitation torque

This is a linear system, yet conveyor load is subject to change, sodisturbance changes.

BJss

T

1)(


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ExampleOpen Loop Response

0 1 2 3 4 5 6 7 8 9 100

0.005

0.01

0.015

0.02

0.025

time (sec)

Angularvelocity(r

ad/sec)

Unit Step Response of the Open Loop DC motor System (Electrical Dynamicsare neglected)


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ExamplePerformance?

ess = ? =%Mp

tr = 1,8/nts = 4,6/ n.tp = / dIn standard 2nd order form

If TD is selected to be 1;

Char. eq. of the system

sTss rs

1...lim0

itpiD

tp

itpiD

iitp

iiD

tpiD

tp

TKKTT

KKs

TKKTT

TTKKs

sTsTTKKTT

KK

sT

...

.

...

..

1.....

.

)(2

2

)1..(

.

/1.1.

.

)(2

2

tpi

tp

i

tp

tp

KKT

KKss

TssKK

KK

sT

)1..(

.,12,0

)1..(

. 22

tpi

tp

nn

tpi

tp

KKT

KK

KKT

KKss


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ExampleLittle Talk

If KpKt is very large compared to 1, then, numerator gain will be almost 1.Also, if Ti is close to 1 then, numerator has one zero at s=0.

Q: If ts is able to be set?

A: No, since ts is already set to 9,2 sec. Because of TD =1 and so 2n=1

)1.(

.2

1

tpi

tp

KKT

KK


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Examplecont.

In what condition ess will be minimum?

In this case

For r(s) = 1/s unit stepess = 0

For r(s) = 1/s2 unit ramp

ess = Type I

If Ti is very small Kp is large enough, ess 0

sTsse rsss 1...lim0

)1.(

.

1.

1

1.

1

)](1[2

2

KtKpTi

KtKpss

sKtKp

sKtKp

sT

KtKp

Ti

.


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Examplecont.

Performance specs as, %Mp = %10, tr 1 sectr= 1,8/n 1 sec, n 1,8

.

)1.(718.0

.

)1.(24.3

.

)1.(718.0

.

)1.(718.0.

59.0

)1.(

.2

159.0

3.5

3.2

)1,0(ln

1,0ln

)1.0(ln)1.0(ln,1/1.0ln,1.0

)1.(24.3

.),1.(24.3.,8.1

)1.(

.

222

2222221/2

tp

tp

i

tp

tp

i

tp

tp

itpitp

tpi

tp

tp

tp

itpitp

tpi

tp

n

KK

KKT

KK

KKT

KK

KK

TKKTKK

KKT

KK

e

KK

KKTKKTKK

KKT

KK

or


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ExampleClosed Loop Results

0 5 10 150

0.2

0.4

0.6

0.8

1

1.2

1.4

time(sec)

angularvelocity(rad/sec)

Closed Loop PID control of a DC motor

(Electrical Dynamics are Neglected)

Kp=10, Td=1

Ti=0.0293 black

Ti=0.0351 brown

Ti=0.0410 light blue

Ti=0.0468 red

Ti=0.0527 green

Ti=0.0585 blue


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Example%Mp = %1, ts 6 secis required. Find

constants K and b. a=1


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Example

What is the system steady state error for unit step input and system type

882.068.1

)67.1)(1(67.1)(

)(

)(2

ss

sssT

sR

sY

882.068.1

)67.167.2(67.11

1.lim)(1)(.lim

2

2

00ss

ss

sssTsRse ssss

16.2882.068.1

9.178.267.0lim

2

2

0

ss

sss type 0


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Example

if = 0,1 is archievable for various K in the system, find it b = 1.67 a = 1

various ways to solve1_ from roots of char. eq.

2_ from standart nd2 order dif. eq.

2_ 02 22 nnss assuming K>0

01

67.1

1

67.22

K

Ks

K

Ks 067.167.2

1

2

mmssK

Km

mn 67.12 mn 67.22

03.167.12

67.2 m

m

m

K

K

196.01.0

K

K

1096.0 2

K = 0.01

Dependence on settling time on PID parameters

K and bDependencies of Percent overshoot on PID

parameters K and b


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0.1

1.1

1.7 0.10

0.90

1.700

20

40

60

80

100

settlin

gtime(sec)

b

K

80.00-100.00

60.00-80.00

40.00-60.00

20.00-40.00

0.00-20.00

0.

10

0.

50

0.

90

1.

30

1.

61

1.

65

1.

69

2.

00

0.10

1.300

5

10

15

20

25

30

35

40

PercentOvershoot

b

K

35.00-40.00

30.00-35.00

25.00-30.00

20.00-25.00

15.00-20.00

10.00-15.00

5.00-10.00

0.00-5.00

0.

10

0.

50

0.

90

1.

30

1.

61

1.6

5

1.

69

2.

00

0.10

1.200.00

10.00

20.00

30.00

40.00

50.00

60.00

70.00

peaktime

K

b

Dependencies of peak time on PID parameters K

and b

60.00-70.00

50.00-60.00

40.00-50.00

30.00-40.00

20.00-30.00

10.00-20.00

0.00-10.00

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