kontrol teorisi
TRANSCRIPT
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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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By Vasfi Emre mrl, Ph.D., 2005 3
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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By Vasfi Emre mrl, Ph.D., 2005 21
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