Nghiên Cứu Thiết Kế Bộ Điều Khiển PID Mờ

7/26/2019 Nghin Cu Thit K B iu Khin PID M

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S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn

1


I HC THI NGUYNTRNG I HC K THUT CNG NGHIP

----------------------------------

LUN VN THC SK THUTNGNH: TNGHO

NGHIN CU THIT K B IU

KHIN PID M

NGUYN VN THIN

THI NGUYN - 2010


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2

I HC THI NGUYN

TRNG HKT CNG NGHIPCNG HO X HI CH NGHA VIT NAM

clp- Tdo - Hnhphc

THUYTMINH

LUN VN THC SK THUT

TI:

NGHIN CU THITKBIUKHINPID M .

Ngnh: T NG HO.

Hc vin: NGUYN VN THIN

Ngi hng dn Khoa hc:TS. NGUYN VN V

NGI HNG DN KHOA HC

TS. Nguyn Vn V

HC VIN

Nguyn Vn Thin

BAN GIM HIU KHOA T SAU I HC


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LI CAM OAN

Ti xin cam oan lun vn ny l cng trnh do ti

tng hp v nghin cu. Trong ln vn c s dng mt s

ti liu tham kho nh nu trong phn ti liu tham kho .

Tc gi lun vn

NguynVn Thin


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4

LI NI U

Ngy nay vi s pht trin ca khoa hc k thut vic ng dng lthuyt iu khin hin i vo thc t ang ngy cng pht trin mnh m

trong c l thuyt iu khin m. Trong cng nghip hin nay n 90%

cc b iu khin trong thc t l da trn lut iu khin PID, b iu

khin PID pht huy tt hiu qu ca n l th vic xc nh v hiu chnh cc

tham s ca n l rt quan trng tuy nhin vic hiu chnh cc tham s ca b

iu khin PID cn th ng. V vy vic nghin cu ng dng l thuyt m

xc nh v hiu chnh tham s cho b iu khin PID cho ph hp vi cc

trng thi lm vic l cn thit v hin nay ang c nghin cu v pht trin

mnh m .

Vi ti Nghin cu thit k b iu khin PID mc chia lm

3 chng nh sau:

Chng I :Tng quan v b iu khin PID

Chng II :B iu khin m

Chng III :Thit k b iu khin PID m

Lnh vc nghin cu ng dng l thuyt m xc nh v hiu chnh

tham s cho b iu khin PID l mt lnh vc kh phc tp mt khc do trnh

v thi gian c hn nn bn than lun vn ca em khng trnh khi nhng

thiu st. Em rt mong c s ng gp kin ca cc thy, c bn than

lun vn ca em c hon thin hn to tin cho nhng bc nghin cu

tip theo.

Em xin gi li cm n chn thnh n thy Ts. Nguyn Vn V tn

tnh gip cho em hon thnh lun vn ng thi hn . Em xin chn thnh

cm n cc thy c ca khoa in, trng i hc Thi Nguyn trang b

cho em nhng kin thc cn thit hon thnh bn lun vn ny cng nh

qu trnh cng tc sau ny.


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5

MC LC

Li cam oan

Li ni u

Danh mc cc ch vit tt, cc k hiu

Danh mc cc bng

Danh mc cc hnh v, th

M U......................................................................................................... 14

1. L do chn ti ................................................................................... 144

2. ngha khoa hc v thc tin............................................................... 144

2.1. ngha khoa hc............................................................................. 144

2.2. ngha thc tin............................................................................. 144

Chng 1.TNG QUAN V B IU KHIN PID.................................... 15

1.1. CU TRC CHUNG CA H IU KHIN................................... 15

1.2.CC CH TIU NH GI CHT LNGH IU KHIN.......... 15

1.2.1. Ch tiu cht lng tnh.................................................................. 151.2.2. Ch tiu cht lng ng................................................................ 16

1.2.2.1. Lng qu iu chnh.............................................................. 16

1.2.2.2. Thi gian qu ...................................................................... 17

1.2.2.3. S ln dao ng........................................................................ 17

1.3. CC LUT IU KHIN ....................................................................................17

1.3.1. Quy lut iu chnh t l (P).......................................................... 171.3.2. Quy lut iu chnh tch phn (I)................................................... 18

1.3.3. Quy lut iu chnh t l vi phn (PD).......................................... 19

1.3.4. Quy lut iu chnh t l tch phn (PI)......................................... 20

1.3.5. Quy lut iu chnh t l vi tch phn (PID).................................. 22

1.4. CC PHNG PHP XC NH THAM S PID.......................... 24

1.4.1. Phng php Ziegler - Nichols ...................................................... 26


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1.4.2. Phng php Chien HronesReswick....................................... 29

1.4.3. Phng php tng T ca Kuhn...................................................... 31

1.4.4. Phng php ti u........................................................................ 32

1.4.4.1. Phng php ti u ln...................................................... 32

1.4.4.2. Phng php ti u i xng.................................................. 39

1.4.5. Xc nh tham s PID da trn qu trnh ti u trn my tnh....... 44

1.5. KT LUN CHNG 1..................................................................... 45

Chng 2. B IU KHIN M.................................................................. 47

2.1. LCH S PHT TRIN CA LOGIC M........................................ 472.2. MT S KHI NIM C BN V LOGIC M.............................. 47

2.2.1. nh ngha tp m .......................................................................... 47

2.2.2. Cc hm lin thuc thng c s dng..................................... 49

2.2.3. Bin ngn ng v gi tr ca bin ngn ng.................................. 49

2.3. B IU KHIN M......................................................................... 50

2.3.1. Khu m ha.................................................................................. 512.3.2. Khu thc hin lut hp thnh ....................................................... 52

2.3.3. Khu gii m.................................................................................. 55

2.4. B IU KHIN M TNH............................................................... 59

2.4.1. Khi nim....................................................................................... 59

2.4.2. Thut ton tng hp mt b iu khin m tnh............................ 59

2.4.3. Tng hp b iu khin m tuyn tnh tng on......................... 602.5. B IU KHIN M NG............................................................. 61

2.6. B IU KHIN M LAI PID.......................................................... 64

2.6.1. Gii thiu chung............................................................................. 64

2.6.2. B iu khin m lai kinh in...................................................... 65

2.6.3. B iu khin m lai cascade......................................................... 65

2.6.4. B iu khin m chnh nh tham s b iu khin PID............. 66


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2.6.5. B iu khin m t chnh cu trc............................................... 66

2.7. KT LUN CHNG 2..................................................................... 67

Chng 3.THIT KB IU KHIN PID M....................................... 68

3.1. T VN ...................................................................................... 68

3.2. THIT K B IU KHIN M CHNH NH THAM S PID......... 70

3.2.1. Cu trc b iu khin................................................................... 70

3.2.2. Thit k b iu khin.................................................................... 70

3.2.3. Kt qu m phng.......................................................................... 77

3.3. NG DNG PID M IU KHIN H TRUYN NGT-D ....... 783.3.1. Cc yu cu i vi h truyn ng T-D ....................................... 78

3.3.2.Tng hp mch vng iu chnh dng in RI............................... 80

3.3.3.Tng hp mch vng iu chnh tc .......................................... 82

3.3.3.1. iu chnh tc dng b iu chnh tc t l.................. 82

3.3.3.2. iu chnh tc dng b iu chnh tc tch phn t l PI.. 85

3.3.4. Bi ton ng dng c th................................................................ 863.3.4.1. Tnh ton tham s mch vng dng in................................. 88

3.3.4.2. Tnh ton tham s b iu khin tc PI.............................. 89

3.3.5. Thit k h iu khin m lai......................................................... 90

3.3.5.1. Xc nh cc bin vo ra.......................................................... 91

3.3.5.2. Xc nh gi tr cho cc bin vo v ra.................................... 92

3.3.6. M phng nh gi cht lng ...................................................... 993.3.6.1. Xy dng s m phng....................................................... 99

3.3.6.2. Kt qum phng.................................................................. 100

3.4. KT LUN CHNG 3 ................................................................... 106

TI LIU THAM KHO............................................................................. 109

TM TT ..................................................................................................... 110


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DANH MC CC CH VIT TT, CC K HIU

STT K hiu Din gii

1 TT i tng iu khin

2 TBK Thit b iu khin

3 TBL - CTH Thit b o lng v chuyn i tn hiu

4 exl Sai s xc lp

5 max Lng qu iu chnh

6 tqd Thi gian qu

7 n S ln dao ng

8 K H s khuch i

9 TI Hng s thi gian tch phn

10 Td Hng s thi gian vi phn

11 L Hng s thi gian tr

12 T Hng s thi gian qun tnh

13 h qu iu chnh

14 e(t) Tn hiu u vo

15 u(t) Tn hiu u ra

16 T-D H truyn ng my pht ng c

17 ng c mt chiu

18 B B bin i xoay chiu - mt chiu c iu khin

19 RI B iu chnh dng in

20 R B iu chnh tc

21 Si Xenx dng in

22 F Mch lc tn hiu

23 Tf Hng s thi gian ca mch lc

24 Tvo Hng s thi gian s chuyn mch chnh lu


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25 Tk Hng s thi gian mch iu khin chnhlu

26 Tu Hng s thi gian mch phn ng

27 Ti Hng s thi gian xenx dng in

28 Ru in tr mch phn ng

29 Mc Mmen ti

30 T Hng s thi gian mch lc

31 L in cm mch phn ng

32 Icp Dng in cho php ln nht

33 KFi T thngnh mc

34 J Mmen qun tnh

35 CL Chnh lu

36 KCL H s chnh lu

37 Urcm Bin my pht xung rng ca

38 Kbd T s bin i dng

39 FT My pht tc

40 E Sc in ng ca ng c in mt chiu


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DANH MC CC BNG

STT K hiu Din gii

1 Bng 3.1 Lut iu khin cho h s Kp

2 Bng 3.2 Lut iu khin cho h s Kd

3 Bng 3.3 Lut iu khin cho h s

4 Bng 3.4 Hm lin thuc ca bin u vo

5 Bng 3.5 Hm lin thuc ca bin u ra

6 Bng 3.6 Lut iu khincho HsKP

7 Bng 3.7 Lut iu khin cho HsKI


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DANH MC CC HNH V, TH

STT K hiu Din gii tn hnh v1 Hnh1.1 Cu trc h thng iu khin

2 Hnh1.2 Th hin c tnh ca sai s xc lp

3 Hnh1.3 Th hin c tnh ca lng qu iu chnh

4 Hnh1.4 Th hin c tnh ca thi gian qu

5 Hnh1.5 Th hin c tnh ca s ln dao ng

6 Hnh1.6 Cc c tnh ca quy lut iu chnh t l vi phn

7 Hnh1.7 Cc c tnh ca quy lut iu chnh t l tch phn8 Hnh1.8 Cc c tnh ca quy lut iu chnh t l tch phn

9 Hnh1.9 iu khin vi b iu khin PID

10 Hnh1.10 Nhim v ca b iu khin PID

11 Hnh1.11 Xc nh tham s cho m hnh xp x

12 Hnh1.12 Xc nh hng s khuch i ti hn

13 Hnh1.13 Hm qu i tng thch hpcho phng php Chien -

Hrones - Reswick

14 Hnh1.14 Quan h gia din tch v tng cc hng s thi gian

15 Hnh1.15 Di tn s m c bin hm t tnh bng 1, cng rng

cng tt

16 Hnh1.16 iu khin khu qun tnh bc nht

17 Hnh1.17 Minh ho t tng thit k b iu khin PID ti u i xng

18 Hnh2.1 M ho bin Tc

19 Hnh2.2 S khi ca b iu khin m

20 Hnh2.3 Hm lin thuc ca lut hp thnh

21 Hnh2.4 Gii m bng phng php cc i

22 Hnh2.5 Gii m theo nguyn l trung bnh

23 Hnh2.6 Gii m theo nguyn l cn tri

24 Hnh2.7 Gii m theo nguyn l cn phi


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25 Hnh2.8 Gii m theo phng php im trng tm

26 Hnh2.9 c tnh vo ra cho trc

27 Hnh2.10 Hm lin thuc ca cc bin ngn ng vo ra

28 Hnh2.11 H iu khin m theo lut PI

29 Hnh2.12 H iu khin m theo lut PD

30 Hnh2.13 H iu khin m theo lut PID

31 Hnh2.14 M hnh b iu khin m lai kinh in

32 Hnh2.15 Cu trc h m lai Cascade

33 Hnh3.1 H iu khin vi b iu khin PID m34 Hnh3.2 Cu trc b iu khin

35 Hnh3.3 Cu trc b iu khin PID m

36 Hnh3.4 Hm lin thuc ca e(t) v de(t)/dt

37 Hnh3.5 Hm lin thuc ca bin Kp, Kd

38 Hnh3.6 Hm lin thuc ca bin

39 Hnh3.7 c tnh qu thng gp ca h iu khin dng PID

40 Hnh3.8 Giao din m phng m41 Hnh3.9 Hm lin thuc ca tn hiu e(t) v de/dt

42 Hnh3.10 Hm lin thuc ca bin Kp, Kd

43 Hnh3.11 Hm lin thuc ca bin

44 Hnh3.12 c tnh iu chnh PID ti u vi i tng bc hai

45 Hnh3.13 c tnh iu chnh PID m (Kp= 20.1; Kd = 20.1; Ki=8.6)

so vi c tnh PID ti u

46 Hnh3.14 S khi ca h truyn ng T-D

47 Hnh3.15 Cu trc mch vng iuchnh dng in.

48 Hnh3.16 S khi ca mch vng dng in.

49 Hnh3.17 S khi h iu chnh tc

50 Hnh3.18 S khi ca h iu chnh tc

51 Hnh3.19 Qu trnh dng in v tc khi c nhiu ti


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52 Hnh3.20 S cu trc h truyn ng T- D mt chiu

53 Hnh3.21 S cu trc mch vng iu chnh dng in.

54 Hnh3.22 Cu trc mch vng iu chnh tc .

55 Hnh3.23 Cu trc bn trong b chnh nh m

56 Hnh3.24 M hnh cu trc h iu khin chnh nh m tham s b

iu khin PI

57 Hnh3.25 Cu trc b chnh nh m

58 Hnh3.26 Xc nh tp m cho bin vo ERROR

59 Hnh3.27 Xc nh tp m cho bin vo dw/dt60 Hnh3.28 Xc nh tp m cho bin ra HsKP

61 Hnh3.29 Xc nh tp m cho bin ra HsKI

62 Hnh3.30 c tnh qu thng gp ca h iu khin dng PID

63 Hnh3.31 Cc lut hp thnh.

64 Hnh3.32 Cu trc ca h iu khin m lai PI

65 Hnh3.33 Cu trc ca khu m

66 Hnh3.34 Cu trc ca b iu khin PI

67 Hnh3.35 Cu trc ca i tng

68 Hnh3.36 c tnh ca b iu khin PI khi mmen ti hng s

69 Hnh3.37 c tnh ca b iu khin PI-m khi mmen ti hng s

70 Hnh3.38 c tnh ca b iu khin PI-m so vi b iu khin PI khi

mmen ti hng s

71 Hnh3.39 c tnh ca b iu khin PI khi mmen ti thay i72 Hnh3.40 c tnh ca b iu khin PI- m khi mmen ti thay i

73 Hnh3.41 c tnh ca cc b iu khin khi mmen ti thay i

74 Hnh3.42 c tnh ca b iu khin PI khi tc t thay i

75 Hnh3.43 c tnh ca b iu khin PI-m khi tc t thay i

76 Hnh3.44 c tnh ca cc b iu khin khi tc t thay i


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Chng 1

TNG QUAN V B IU KHIN PID

1.1. CU TRC CHUNG CA H IU KHIN

Cu trc chung ca h thng iu khin t ng nh Hnh1.1.

Trong :

TT : i tng iu khin.

TBK : Thit b iu khin.

TBL - CTH : Thit b o lng v chuyn i tn hiu.

U(t) : L tn hiu vo ca h thng - cn gi l tn hiu t hay lng

ch o xc nh im lm vic ca h thng.

y(t) : Tn hiu u ra ca h thng. y chnh l i lng c iu chnh.

x(t) : L tn hiu iu khin tc ng ln i tng.

e(t) : L sai lch iu khin.

Z(t) : L tn hiu phn hi.

Thit b iu khin l thnh phnquan trng nht duy tr ch lm

vic cho c h thng iu khin.

1.2. CC CH TIU NH GI CHT LNG H IU KHIN

1.2.1. Ch tiu cht lng tnh

Ch tiu cht lng tnh c nh gi bngsai s xc lp (sai lch tnh):

l sai lch ca lng ra so vi yu cu khi qu trnh iu khin kt thc.

TBK TK

TBLCTH

U(t) y(t)

Z(t)

e(t) x(t)

Hnh1.1: Cu trc h thng iu khin


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Sai s xc lp :L sai s ca h thng khi thi gian tin n v cng

exl= )(lim tet

exl= )(lim ssEs

(1.1)

1.2.2. Ch tiu cht lng ng

Cht lng ng ca h thng c nh gi qua 3 ch tiu c bn :

-Lng qu iu chnh.

-Thi gian qu .

-S ln dao ng.

1.2.2.1. Lng qu iu chnh

Lng qu iu chnh: L lng sai lch ca p ng ca h thng so

vi gi tr xc lp ca n.

Lng qu iu chnh max ( Percent of OvershootPOT ) c tnh

bng cng thc :

max=xl

xlma

c

cc x x100% (1.2)

G(s)

H(s)

R(s) E(s) C(s)

r t

cht t

e t

exl

exl t0

Hnh1.2: Th hin c tnh ca sai s xc lp

t0

cxl

c(t)cmax

cxl

max

Hnh1.3: Th hin c tnh ca lng qu iu chnh


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1.2.2.2. Thi gian qu

Thi gian qu ( tqd) : L thi gian k t khi c tc ng vo h thng

(khi ng h thng) cho n khi sai lch ca qu trnh iu khin nm trong

gii hn cho php % . % thng chn l 2% (0.02) hoc 5% (0.05)

1.2.2.3. S ln dao ng

n l s ln dao ng ca y(t) xung quanh gi tr yxl

Gi tr n cng nh cng tt. Gi tr n do yu cu thit k t ra, thng n 3

1.3. CC LUT IU KHIN

1.3.1. Quy lut iu chnh t l (P)

Trong quy lut iu chnh t l tc ng iu chnh c xc nh theo

cng thc:

U = K.e (1.3)

Trong , K l tham s iu chnh gi l h s khuch i. Hm truyn

t ca b iu chnh t l c dng:

y(t)

Hnh1.4: Th hin c tnh ca thi gian qu

yxl

0tqd

t

Hnh1.5: Th hin c tnh ca s ln dao ng

0

yxl

y(t)n

t


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W(p) = K (1.4)

-Hm truyn tn s ca n l : W(j) = K.

-c tnh pha tn s : () = 0

T cc c tnh trn ta thy quy lut t l phn ng nh nhau i vi tn

hiu mi tn s. Gc lch pha gia tn hiu ra v tn hiu vo bng khng.

V vy, tn hiu iu khin s xut hin ngay khi c tn hiu sai lch. Gi tr

v tc thay i ca tn hiu iu khin U t l vi gi tr v tc thay i

ca tn hiu vo

u im c bn ca quy lut t l l tc tc ng nhanh. H thng iuchnh s dng quy lut t l c tnh n nh cao, thi gian iu chnh ngn.

Nhc im c bn ca quy lut t l l khng c kh nng trit tiu sai

lch tnh.

1.3.2. Quy lut iu chnh tch phn (I)

Quy lutiu chnh tch phn c m t bi phng trnh vi phn :

U =IT1 edthoc dt

du = K.e (1.5)

Trong , TI=K

1 l hng s thi gian tch phn

-Hm truyn t c dng: W(p) =pTI.

1

-Hm truyn tn s: W(j) =Tj

1 = -jT

1 =T

1 e-j 2

-c tnh bin tn s: A() =T

1

-c tnh pha tn s: () = -2

R rng quy lut tch phn phn ng km vi tn hiu c tn s cao.

Trong c di tn s tn hiu ra chm pha so vi tn hiu vo mt gc bng2

,

nh vy quy lut tch phn phn ng chm.


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u im c bn ca quy lut iu chnh tch phn l c kh nng trit

tiu sai lch d v quy lut iu chnh (I) ch ngng tc ng khi sai lch e = 0

Nhc im c bn ca quy lut tch phn l tc tc ng chm nn

h thng iu chnh t ng s dng quy lut tch phn s km n nh. Thi

gian iu khin ko di. Trong thc t, quy lut iu chnh tch phn ch s

dng cho cc i tng c tr v hng s thi gian nh.

1.3.3. Quy lut iu chnh t l vi phn (PD)

L quy lut iu chnh cm t bi phng trnh vi phn:

U = K1.e + K2dt

de = Km

dt

deTe d (1.6)

Trong , Km= K1l h s khuch i

Td=1

2

K

Kl hng s thi gian vi phn

Cc tham s hiu chnh ca quy lut PD l Kmv Td

-Hm qu : h(t) = Km[ 1(t) + Td.(t)]

-Hm truyn t ca quy lut PD c dng : W(p) = Km(1+Td.p)

-Hm truyn tn s : W(j) = Km(1+jTd.) = A().ej()

Vi A() = 2)(1 dT v () = arctgTd nh vy 0


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Quy lut PD c hai tham s hiu chnh l Kmv Td. Nu Td= 0 th quy

lut PD tr thnh quy lut t l, nu Km= 0 th quy lut PD tr thnh quy lut

vi phn.

Trong ton di tn s, tn hiu ra lun lun vt trc tn hiu vo nn

quy lut PD tc ng nhanh hn quy lut t l nhng qu trnh iu chnh vn

khng c kh nng trit tiu sai lch d ging nh quy lut t l. Phn t vi

phn tng tc tc ng nhng ng thi cng rt nhy cm vi nhiu tn

s cao. V vy, trong cng nghip, quy lut t l vi phn ch s dng khi quy

trnh cng ngh cho php c sai lch d v i hi tc tc ng rt nhanh.1.3.4. Quy lut iu chnh t l tch phn (PI)

Quy lut PI l s kt hp ca hai quy lut P v I c m t bng

phng trnh vi phn sau :

U = K1.e + K2edt = Km

edtT

eI

1 (1.7)

Trong , Km= K1l h s khuch i ca PI.

TI=2

1

K

K l hng s thi gian tch phn.

Thi gian tch phn l khong thi gian cn thit cho tc ng tch

phn bng tc ng t l, v vy n cn c gi l thi gian gp i. Hm

truyn t v hm truyn tn s ca quy lut t l tch phn c dng:

- Hm qu ca quy lut PI:

h(t) = Km

dttTt I

)(11)(1 = Km

t

TI11

-Hm truyn t: W(p) = Km

pTI

11

-Hm truyn tn s: W(j) = Km

ITj

11

-c tnh bin tn s: A() = 1)( 2

I

I

m TT

K


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Nu ta chn c tham s Km, TIthch hp th quy lut iu chnh PI c th

p dng cho phn ln cc i tng trong cngnghip.

Nhc im ca quy lut tch phn l tc tc ng nh hn quy lut

t l. V vy, nu i tng yu cu tc tc ng nhanh do nhiu thay i

lin tc th quy lut tch phn khng p ng c yu cu.

1.3.5. Quy lut iu chnh t l vi tch phn (PID)

Quy lut iu chnh t l vi tch phn c m t bi phng trnh:

U = K1.e + K2edt +K3dt

de = Km

dt

deTedt

T

e DI

1 (1.8)

Trong , Km= K1l h s khuch i ca PI.

TI=2

1

K

Kl hng s thi gian tch phn

TD=1

3

K

Kl hng s thi gian vi phn

-Hm qu : h(t) = Km

)(1

1 tTtT DI

-Hm truyn t: W(p) = Km

pT

pT D

I

11

-Hm truyn tn s: W(j) = Km

)

1(1

I

DT

Tj

-c tnh bin tn s: A() = 222 )1()(

IDI

I

m TTTT

K

-c tnh pha tn s: () = arctg

I

DT

T1

Nh vy, 2/ 0 < () < 2/


23/110


23

Cc c tnh ca quy lut iu chnh PID c m t trn Hnh1.8.

Ta nhn thy di tn s thp c tnh ca quy lut PID gn ging vi

quy lut PI, di tn s cao PID gn ging vi quy lut PD, ti 0=DITT

1

PID mang c tnh ca P.

Quy lut PID c ba tham s hiu chnh Km, TI, TD. Xt nh hng ca

ba tham s ta thy:

- Khi TD= 0 v TI= quy lut PID tr thnh quy lut P

- Khi TD= 0 quy lut PID tr thnh quy lut PI

- Khi TI= quy lut PID tr thnh quy lut PD

u im ca quy lut PID l tc tc ng nhanh v c kh nng trit

tiu sai lch tnh. V tc tc ng, quy lut PID cn c th nhanh hn c

quy lut t l. iu ph thuc vo thng s TI, TD.

Nu ta chn c tham s ti u th quy lut PID s p ng c mi

yu cu v iu chnh cht lng ca cc quy trnh cng ngh. Tuy nhin,

vic chn c b ba thng s ti u l rt kh khn. Do trong cng

nghip, quy lut PID thng ch c s dng khi i tng iu chnh c

Hnh1.8: Cc c tnh ca quy lut iu chnh t l tch phn

BT

A()

Km

PT

2/

R()

TBP

= 0

Km

h(t)

tKm

()

I()

0=DITT

1

2/

TI

*

*


24/110


24

nhiu thay i lin tc v quy trnh cng ngh i hi chnh xc cao m

quy lut PI khng p ng c.

1.4. CC PHNG PHP XC NH THAM S PID

Tn gi PID l ch vit tt ca ba thnh phn c bn c trong b iu

khin Hnh1.9a gm khu khuch i (P), khu tch phn (I), v khu vi phn

(D). Ngui ta vn thng ni rng PID l mt tp th hon ho bao gm ba

tnh cch khc nhau:

- Phc tng v thc hin chnh xc nhim v c giao (t l)

- Lm vic v c tch lu kinh nghim thc hin tt nhim v (tch phn).- Lun c sang kin v phn ng nhanh nhy vi s thay i tnh

hung trong qu trnh thc hin nhim v (vi phn).

B iu khin PID c s dng kh rng ri iu khin i tng

SISO theo nguyn l hi tip Hnh1.9b. L do b PID c s dng rng ri

l tnh n gin ca n c v cu trc ln nguyn l lm vic. B PID cnhim v a sai lch e(t) ca h thng v 0 sao cho qu trnh qu tha

mn cc yu cu c bn v cht lng:

- Nu sai lch e(t) cng ln th thng qua thnh phn up(t), tn hiu iu

chnh u(t) cng ln (vai tr ca khuch i kp).

- Nu sai lch e(t) cha bng 0 th thng qua thnh phn uI(t), PID vn

cn tn ti tn hiu iu chnh (vai tr ca tch phn TI).

Hnh1.9:iu khin vi b iu khin PID

a) b)

kpsTI

1

TDs

e

up

uI

uD

u i tngiu khin

e_

yuPID


25/110


26/110


26

R(s) = kp( 1 +sTI

1 ) (1.12)

1.4.1. Phng php Ziegler - NicholsZiegler v Nichols a ra hai phng php thc nghim xc nh

tham s b iu khin PID. Trong khi phng php th nht s dng dng m

hnh xp x qun tnh bc nht c tr ca i tng iu khin:

S(s) =Ts

ke Ls

1 ( 1.13)

th phng php th hai ni tri hn ch hon ton khng cn n m hnh

ton hc ca i tng. Tuy nhin, n c hn ch l ch p dng c cho

mt lp cc i tng nht nh.

Phng php Ziegler Nichols thnht:

Phng php thc nghim ny c nhim v xc nh cc tham s kp, TI,

TDcho b iu khin PID trn c s xp x hm truyn t S(s) ca i tng

thnh dng (1.13), h kn nhanh chng tr v ch xc lp v qu iu

chnh h khng vt qu mt gii hn cho php, khong 40% so vi h=

tlim h(t), tc l c

h

h 0,4.

Ba tham s L (hng s thi gian tr), k (h s khuch i) v T (hng s

thi gian qun tnh) ca m hnh xp x (1.13) c th c xc nh gn ng

t th hm qu h(t) ca i tng. Nu i tng c hm qu dng

nh Hnh1.11a th t th hm h(t) ta c ra c ngay:

Hnh1.10:Nhim v ca b iu khin PID

PID S(s)e

_yu

a)

40%

h(t)

t

1

b)


27/110


27

-L l khon thi gian u ra h(t) cha c phn ng ngay vi kch thch

1(t) ti u vo.

-k l gi tr gii hn h =t

lim h(t)

-Gi A l im kt thc thi gian tr, tc l im trn trc honh c

honh bng L. Khi T l khonh thi gian cn thit sau L tip tuyn

ca h(t) ti A t gi tr k.

Trng hp hm qu h(t) khng c dng l tng nh Hnh1.11a,

song c dng gn ging l hnh ch S ca khu qun tnh bc hai hoc bc n

nh Hnh1.11b m t, th ba tham s k, L, T ca m hnh (1.13) c xc

nh xp x nh sau:

-k l gi tr gii hn h =t

lim h(t).

-K ng tip tuyn ca h(t) ti im un ca n. Khi L s l

honh giao im ca tip tuyn vi trc honh v T l khong thi gian

cn thit ng tip tuyn i c t gi tr 0 n gi tr k.

Nh vy ta c th thy, iu kin p dng c phng php xp x

m hnh bc nht c tr ca i tng l i tng phi n nh, khng c

giao ng v t nht hm qu ca n phi c dng ch S.

Sau khi c cc tham s cho m hnh xp x (1.13) ca i tng

Ziegler Nichols ngh s dng cc tham s kp, TI, TD cho b iu

khin nh sau:

Hnh1.11: Xc nh tham s cho m hnh xp x

h(t)

L T

k

t

a)

h(t)

k

t

L T b)


28/110


28

- Nu ch s dng b iu khin khuch i R(s) = kp, th chn kp=kL

T

- Nu s dng b PI vi R(s)=kp( 1+ sTI1 ) th chn kp= kL

T9,0 v TI= L310

-Nu s dng PID c R(s) = kp(1+sTI

1 +TDs ) th chn kp=kL

T2,1 , TI= 2L, TD=2

L

Phng php Ziegler Nichols th hai.

Phng php thc nghim th hai ny c c im l khng s dng

m hnh ton hc ca i tng, ngay c m hnh xp x gn ng (1.13)

Phng php Ziegler Nichols th hai c ni dung nh sau:-Thay b iu khin PID trong h kn Hnh1.12a bng b khuch i. Sau

tng h s khuch i ti gi tr ti hn kth h kn bin gii n nh, tc l

h(t) c dng dao ng iu ho Hnh1.12b xc nh chu k Tthca dao ng...

-Xc nh tham s cho b iu khin P, PI hay PID nh sau:

+ Nu s dng R(s) = kpth chn kp= thk2

1

+ Nu s dng R(s) =kp(1 +sTI

1 ) th chn kp=0,45kthv TI= 0,85Tth

+ Nu s dng PID th chn kp= 0,6kth, TI= 0,5Tth, TD= 0,12Tth

Phng php thc nghim th hai c mt nhc im l ch p dng

c cho nhng i tng c c ch bin gii n nh khi hiu chnh

hng s khuch i trong h kn.

kth

i tng

iu khin e

_

y

Hnh1.12:Xc nh hng s khuch i ti hn

Tth

h(t)

t

2

1,51

0,5

1 2 3 5 7 9a) b)


29/110


29

1.4.2. Phngphp ChienHronesReswick

V mt nguyn l phng php Chien Hrones Reswick gn ging

vi phng php th nht ca ZieglerNichols. Song n khng s dng m

hnh tham s (1.13) gn ng dng qun tnh bc nht c tr cho i tng m

thay vo l trc tip dng hm qu h(t) ca n.

Phng php Chien HronesReswick cng phi c gi thit rng i

tng l n nh, hm qu h(t) khng giao ng v c dnghnh ch S

Hnh1.13 tc l lun c o hm khng m:dt

tdh )( = g(t) 0 .

Tuy nhin, phng php ny thch ng vi nhng i tng bc cao

nh qun tnh bc n:

S(s) = nsT

k

1

V c hm qu h(t) tho mn:a

b > 0

Trong a l honh giao im tip tuyn ca h(t) ti im un Uvi trc thi gian Hnh1.13 v b l khong thi gian cn thit tip tuyn

i c t 0 ti gi tr xc lp k =t

lim h(t).

T dng hm qu h(t) i tng vi hai tham s a, b tho mn,

Chien Hrones Reswick a bn cch bn cch xc nh tham s b

iu khin cho bn yu cu cht lng nh sau:

-Yu cu ti u theo nhiu (gim nh hng nhiu) v h kn khng c

qu iu chnh.

Hnh1.13:Hm qu cho phng php ChienHronesReswick

h(t)

k

t

a b

U a

b>3


30/110


30

+ B iu khin P : Chn kp=ak

b

10

3

+ B iu khin PI : Chn kp= akb

106 v TI= 4a

+ B iu khin PID : Chn kp=ak

b

20

19 , TI=5

12a v TD=50

21a

- Yu cu ti u theo nhiu (gim nh hng nhiu) v h kn c qu

iu chnh h khng vt qu 20% so vi h =t

lim h(t).


b

10

7

+ B iu khin PI : Chn kp=ak

b

10

7 v TI=10

23a


b

20

6 , TI= 2a v TD=50

21a

-Yu cu ti u theo tn hiu t trc (gim sai lch bm) v h kn

khng c qu iu chnh h.

+ B iu khin P : Chn kp= akb

103


b

20

7 v TI=5

6b


b

5

3 , TI= b v TD=2

a

- Yu cu ti u theo tn hiu t trc (gim sai lch bm) v h kn

c qu iu chnh h khng vt qu 20% so vi h =

t

lim h(t):


b

10

7


b

5

6 v TI= b


b

20

19 , TI=20

27b v TD=100

47a


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31

1.4.3. Phng php tng T ca Kuhn

Cho i tng c hm truyn t

S(s) = k)1).......(1)(1(

)1).......(1)(1(

21

21

sTsTsT

sTsTsT

nmmm

mttt

e-sT, ( m< n ) (1.14 )

Gi thit rng hm qu h(t) ca n c dng hnh ch S nh m t

Hnh1.14, vy th (1.14) phi tho mn nh l.

Giao im ca ng qu o bin pha A(j) ca a thc Hurwitz A(s)

vi trc thc phi nm xen k gia nhng giao im ca n vi trc o.

Gi tr ti hai giao im k nhau ca A(j) vi trc thc ca a thcHurwitz A(s) phi tri du nhau.

Gi tr ti hai giao im k nhau A(j) vi trc o ca a thc Hurwitz

A(s) phi tri du nhau.

Tc l cc hng s t s tiT phi c gi thit l nh hn hng s thi

gian tng ng vi n mu s mjT . Ni cch khc nu nh c s sp xp:

tm

tt TTT ...21 vm

nmm TTT ...21

th cng phi c.mt

TT 11 ,mt

TT 22 , ,m

m

t

m TT

y cc ch ci t v m trong tiT , m

jT . khng c ngha lu tha m ch l

k hiu ni rng n thuc v a thc t s hay mu s trong hm truyn t S(s).

Gi A l din tch bao bi ng cong h(t) v k =t

lim h(t) vy th ta s c.

h(t

k

A

Hnh1.14:Quan h gia din tch v tng cc hng s thi

t


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33

L h thng lun c c p ng y(t) ging nh tn hiu lch c

a ra u vo (t) ti mi im tn s hoc t ra thi gian qu y(t)

bm c vo (t) cng ngn cng tt. Ni cch khc, b iu khin l tng

R(s) cn phi mang n cho h thng kh nng:

)( jG = 1 vi mi (1.18)

Nhng trong thc t, v nhiu l do m yu cu R(s) tho mn (1.18) khc p ng. Chng hn nh v h thng thc lun cha trong n bn cht

qun tnh, tnh cng li lch tc ng t ngoi vo. Song tnh xu ca

h thng li c gim bt mt cch t nhin ch lm vic c tn s ln,

nn ngi ta thng tho mn vi b iu khin R(s) khi n mang li c

cho h thng tnh cht (1.18) trong mt di tn s rng ln cn thuc 0.

B iu khin R(s) tho mn:)( jG 1 (1.19)

trong di tn s tn s c rng ln c gi l b iu khin ti u ln.

Hnh1.15 l v d minh ho cho nguyn tc iu khin ti u ln. B iu

khin R(s) cn phi c chn sao cho min tn s ca biu Bole hm

truyn h kn G(s) tho mn:

L() = 20lg )( jG 0

Hnh1.15: Di tn s m c bin hm t tnh bng1, cng rng cng tt

R(s) S(s)

e

_

yu

a)

Cng rng cng ttL()

0

-20

- 40

0,1

10c b)


34/110


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37

dng c kt lun 1 ta chn b iu khin PI thay v b iu khin I nh

lm vi i tng bc nht:

R(s) = kp(1+sTI

1 ) =sT

sTk

I

Ip )1( =sT

sT

R

I )1( , TR =p

I

k

T ( 1.25)

Gh(s) = R(s)S(s) =)1)(1(

)1(

21

1

sTsTsT

sTk

R

( 1.26)

nhm thc hin vic b hng s thi gian T1ca biu thc (1.24) theo ngha

TI= T1

vi cch chn tham s TIny, hm truyn t h h (1.26) tr thnh.Gh(s) =

)1( 2sTsT

k

R

V n hon ton ging (1.22) tc l ta li c c TRtheo kt lun 1:

TR=p

I

k

T= 2kT2 kp =

22kT

TI =2

1

2kT

T

Vy:

Kt lun 4:

Nu i tng iu khin l khu qun tnh bc hai (1.24), th b iu khin

PI (1.25) vi cc tham s TI= T1, kp=2

1

2kT

Tth s l b iu khin ti u ln.

Nu i tng khng phi l khu qun tnh bc hai m li c hm

truyn t S(s) dng (1.23) vi cc hng s thi gian T2, T3,.Tnrt nh so

vi T1th do n c th xp x bng:S(s) =

)1)(1( 1 TssT

k

trong T =

n

i

iT2

nh phng php tng cc hng s thi gian nh ta cn c:

Kt lun 5:

Nui tng iu khin(1.23) c mt hng s thi gian T1ln vt

tri v cc hng s thi gian cn li T2, T3,.Tnrt nh, th b iu khin PI


38/110


38

(1.25) c cc tham s TI= T1, kp=

n

i

iTk

T

2

1

2

s l b iu khin ti u ln.

iu khin i tng qun tnh bc ba.

i tng l khu qun tnh bc ba:

S(s) =)1)(1)(1( 321 sTsTsT

k

(1.27)

Ta s s dng b iu khin PID

R(s) = kp(1+sTI

1 +TDs) =sT

sTsT

R

BA )1)(1( TR=p

I

k

T (1.28)

vi : TA+TB= TIv TATB= TITD

Khi hm truyn t h h s tr v dng (1.22) nu ta chn.

TA = T1, TB= T2 TI= T1+ T2, TD=21

21

TT

TT

Suy ra :

TR=p

I

kT = 2kT3 kp=

32kTTI =

3

21

2kTTT

Vy ta c kt lun tip theo.

Kt lun 6:

Nu i tng iu khin l khu qun tnh bc ba(1.27) th b iu

khin PID (1.28) vi cc tham s TI= T1+ T2, TD=21

21

TT

TT

, kp=

3

21

2kT

TT s l

b iu khin ti u ln.

Trong trng hp i tng li c dng hm truyn t (1.23) nhng

cc hng s thi gian T3, T4, ... Tnrt nh so vi hai hng s cn li T1, T2th

khi s dng phng php tng cc hng s thi gian nh, xp x n v

dng qun tnh bc ba:

S(s) =

)1)(1)(1( 21 TssTsT

k

trong T =

n

i

iT

3


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41

S(s) =)1( 1sTs

k

(1.31)

th vi b iu khin PI:

R(s) = kp(1+sTI

1 ) (1.32)

h h s c hm truyn t ging nh (1.30) l:

Gh(s) = R(s)S(s) =)1(

)1(

1

2sTsT

sTkk

I

Ip

(1.33)

R rng l trong vng I, hm Gh(s) theo (1.33) tho mn (1.29).

vng II, biu bin Bole ca Gh(s) c nghing -20db/dec xung quanh

im tn s ct cth phi c:

I=IT

1 < 1=1

1

T TI> T1 (1.34)

v

)( Ih jG > )( ch jG = 1 > )( 1jGh (1.35)

T m hnh (1.33) ca h h ta c gc pha

h() = arcGh(j) = arctan(TI) - arctan(T1)-

Nhm nng cao d tr n nh cho h kn, cc tham s b iu

khin cn phi c chn sao cho ti tn sct cgc pha h(c) l ln nht

iu ny dn n:

d

d ch )( = 0 2

)(1 Ic

I

T

T

-2

1

1

)(1 T

T

c

= 0

c=1

1

TTI

lg(c) =2

)lg()lg( 1 I (1.36)

Kt qu (1.36) ny ni rng trong biu Bole, im tn s ct ccn

phi nm gia hai tn s gy Iv 1. cng l l do ti sao phng php

c tn l i xng.


42/110


42

Gi khong cch gia Iv 1o trong h trc ta biu Bole l a,

ta c:

lga = lg1-lgI= lg1T

TI a =1T

TI (1.37)

Nh vy, r rng s c (1.34) nu c a>1

Thay ccho trong (1.36) vo (1.35) ta s c vi (1.33) v (1.37):

)( ch jG = 1 2

1

2

2

)(1

)(1

CCI

CIp

TT

Tkk

= 1

kp= akT11 (1.38)

Ni cch khc nu c a>1 v ( 1.38) th cng c (1.35)

Khong cch a gia Iv 1cn l mt i lng c trng cho qu

iu chnh h ca h kn nu h c dao ng. C th l a cng ln, qu

iu chnh h cng nh. iu ny ta thy c nh sau:

Trong vng II, hm truyn t h h Gh(s) c thay th gn ng bng:

Gh(s) )1(

1

1sTsTC vi TC=

C

1

Khi h kn s c hm truyn t.

G(s) =)(1

)(

sG

sG

h

h

2

11

1

sTTsT CC =

2)(21

1

TsDTs

vi T = 1TTC v 2D =1T

TC

lg2D = (lgTClgT1) =2

lga (v tnh cht i xng ca C)

D =2

a < 1 nu 4>a>1

Vy trong vng II, hm qu h kn c dng dao ng tt dn khi

4>a>1. qu iu chnh ca hm qu h kn s l.

h = exp

21 D

D a =)(ln

)(ln4

22

2

h

h

(1.39)


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47

Chng 2

B IU KHIN M

2.1. LCH S PHT TRIN CA LOGIC M

Lch s ca iu khin m bt u t nm 1965, khi gio s Lofti A

Zadeh trng i hc California - M a ra khi nim v l thuyt tp m

(Fuzzy set theory). T tr i cc nghin cu l thuyt v ng dng tp m

pht trin mt cch mnh m. Vi nhng thi im ng ch sau:

-

Nm 1972, cc gio s Terano v Asai thit lp ra c s nghincu h thng iu khin m Nht.

-Nm 1974, Mamdani nghin cu iu khin m cho l hi.

-Nm 1980, hng Smidth Co. nghin cu iu khin m cho l xi mng.

-Nm 1983, hng Fuji Electric nghin cu ng dng m cho nh my

s l nc.

-Nm 1984, Hip hi h thng m quc t (IFSA) c thnh lp.

-Nm 1989, phng th nghim quc t nghin cu ng dng k thut

m u tin c thnh lp.

Cho n nay, h thng iu khin m c cc nh khoa hc, cc k s

v sinh vin trong mi lnh vc khoa hc k thut c bit quan tm v ng

dng trong sn xut v i sng, c rt nhiu ti liu nghin cu l thuyt

v cc kt qa ng dng logic m trong iu khin h thng. Tuy nhin logic

m vn ang ha hn pht trin mnh m.

2.2. MT S KHI NIM C BN V LOGIC M

2.2.1. nh ngha tp m

Logic m bt u vi khi nim tp m.

Khi nim v tp hp c hnh thnh trn nn tng logic v c

nh ngha nh mt s xp t chung cc vt, cc i tng c cng chung

mt tnh cht, c gi l phn t ca tp hp . ngha logic ca khi


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48

nim tp hp c xc nh ch mt vt hoc mt i tng bt k ch c

th c hai kh nng hoc l phn t ca tp ang xt hoc khng.

Xt tp hp A trn. nh x A{0,1} nh ngha trn tp A nh sau:

A(x) = 0 nu x A v

A(x) = 1 nu x A (2.1)

c gi l hm lin thuc ca tp hp A. Mt tp X lun c X(x)=1,

vi mi x c gi l khng gian nn (tp nn).

Mt tp hp A c dng A = {xX x} tha mn mt s tnh cht no

th c gi l c tp nn X, hay c nh ngha trn tp nn X.

Nh vy trong l thuyt kinh in, hm lin thuc hon ton tng

ng vi nh ngha mt tp hp. T nh ngha v mt tp hp A bt k ta

c th xc nh c hm lin thuc A(x) cho tp hp v ngc li t

hm lin thuc A(x) ca tp hp A cng hon ton suy ra c nh ngha

cho tp hp A.

Tuy nhin, cch biu din hm lin thuc nh vy khng ph hp vi

nhng tp hp c m t m nh tp B gm cc s thc nh hn nhiu so

vi 6: B = {x R x


49/110


49

Nh vy, khc vi tp hp kinh in A, t nh ngha kinh in ca

tp m B hoc C khng suy ra c hm lin thuc B(x) hoc C(x) ca

chng. Do , ta c nh ngha v tp m nh sau:

Tp m F xc nh trn tp kinh in X l mt tp m mi phn t ca

n l mt cp cc gi tr (x, F(x) trong x X v F l nh x.

F:X[0,1].

nh x Fc gi l hm lin thuc ca tp m F. Tp kinh in X

c gi l tp nn (hay v tr) ca tp m F.

2.2.2. Cc hm lin thuc thng c s dng

Hm lin thuc c xy dng da trn cc ng thng : Dng ny c

u im l n gin. Chng gm hai dng chnh l: tam gic v hnh thang.

Hm lin thuc c xy dng da trn ng cong phn b Gauss:

kiu th nht l ng cong Gauss dng n gin v kiu th hai l s kt

hp hai ng cong Gauss khc nhau hai pha. C hai ng cong ny u

c u im l trn v khng gy mi im nn chng l phng php ph

bin xc nh tp m.

Ngoi ra, hm lin thuc cn c th c mt s dng t ph bin (ch

c s dng trong mt s ng dng nht nh). l cc dng sigma v

dng ng cong Z, Pi v S.

2.2.3. Bin ngn ng v gi tr ca bin ngn ng

Mt bin c th gn bi cc t trong ngn ng t nhin lm gi tr can gi l bin ngn ng.

Mt bin ngn ng thng bao gm 4 thng s: X, T, U, M vi :

+ X : Tn ca bin ngn ng.

+ T : Tp ca cc gi tr ngn ng.

+ U : Khng gian nn m trn bin ngn ng X nhn cc gi tr r.

+ M : Ch ra s phn b ca T trn U


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V d:bin ngn ng Tc xe c tp cc gi tr ngn ng l rt

chm, chm, trung bnh, nhanh, rt nhanh, khng gian nn ca bin l tp cc

s thc dng. Vy bin tc xe c 2 min gi tr khc nhau:

- Min cc gi tr ngn ng N: [rt chm, chm, trung bnh, nhanh, rt nhanh]

-Min cc gi tr vt l V = {x R (x0 )}

Mi gi tr ngn ng (mi phn t ca N) c tp nn l min gi tr vt

l V. T mt gi tr vt l ca bin ngn ng ta c c mt vc t gm cc

ph thuc ca x: X T = [ rt chm chm trung binh nhanh rt nhanh ]

nh x trn c gi l qu trnh Fuzzy ho gi tr r x.V d : ng vi tc 50 km/h ta c.

Vc t (50) =

0

0

5,0

5,0

0

2.3. B IU KHIN M

S khi ca b iu khin m trn Hnh2.2. bao gm 4 khi:

- Khi m ha (fuzzifiers).

- Khi hp thnh.

- Khi lut m.

- Khi gii m (defuzzifiers).

Khi m ha(fuzzifiers)

Khi hpthnh

Gii m

Khi lut m

u vox

u ray

Hnh2.2: S khi ca b iu khin m.

1

50

Tc

RC CTBNH RN

Hnh2.1:M ho bin Tc


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Ta thy m ho n tr cho php tnh ton v sau rt n gin nhng

khng kh c nhiu u vo, m ho Gaus hoc m ho hnh tam gic

khng nhng cho php tnh ton v sau tng i n gin m cn ng thi

c th kh nhiu u vo.

2.3.2. Khu thc hin lut hp thnh

Khu thc hin lut hp thnh gm 2 khi l khi lut m v khi

hp thnh.

Khi lut m (suy lun m) bao gm tp cc lut Nu Th da vo

cc lut m c s c ngi thit k vit ra cho thch hp vi tng bin vgi tr ca cc bin ngn ng theo quan h m Vo/Ra.

Khi hp thnh dng bin i cc gi tr m ho ca bin ngn ng

u vo thnh cc gi tr m ca bin ngn ng u ra theo cc lut hp thnh

no .

Khu thc hin lut hp thnh, c tn gi l thit b hp thnh, x l

vector v cho gi tr m B ca tp bin u ra.Cho hai bin ngn ng v . Nu bin nhn gi tr (m) A vi hm

lin thuc A(x) v nhn gi tr (m) B vi hm lin thuc B(y) th biu thc:

= A c gi l mnh iu kin v = B c gi l mnh kt lun.

Nu k hiu mnh = A l p v mnh = B l q th mnh hp thnh:

p q (t p suy ra q) (2.3)

hon ton tng ng vi lut iu khin:

Nu = A th = B (2.4)

Mnh hp thnh trn l mt v d n gin v b iu khin m. N

cho php t mt gi tr u vo xohay c th l t ph thuc A(xo) i vi

tp m A ca gi tr u vo xoxc nh c h s tha mn mnh kt

lun q ca gi tr u ra y. H s tha mn mnh kt lun ny c gi l

gi tr ca mnh hp thnh khi u vo bng A v gi tr ca mnh hp


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thnh (2.3) l mt gi tr m. Biu din gi tr m l tp hp C th mnh

hp thnh m (2.4) chnh l mt nh x:

A(xo) C(y)

Ta c cng thc xc nh hm lin thuc cho mnh hp thnh

B=AB.

B'(y) = min{A, B(y)}, c gi l quy tc hp thnh MIN

B'(y) = A.B(y), c gi l quy tc hp thnh PROD

y l hai quy tc hp thnh thng c dng trong l thuyt iu

khin m m t mnh hp thnh A B.

Hm lin thuc AB(y) ca mnh hp thnh A B s c k hiu l

R. Ta c lut hp thnh l tn chung gi m hnh R biu din mt hay nhiu hm

lin thuc cho mt hay nhiu mnh hp thnh, ni cch khc lut hp thnh

c hiu l mt tp hp ca nhiu mnh hp thnh. Mt lut hp thnh ch c

mt mnh hp thnh c gi l lut hp thnh n. Ngc li nu n c

nhiu hn mt mnh hp thnh ta s gin l lut hp thnh kp. Phn ln cc

h m trong thc t u c m hnh l lut hp thnh kp. Ngoi ra R cn c mt

s tn gi khc ph thuc vo cch kt hp cc mnh hp thnh (max hay sum)

v quy tc s dng trong tng mnh hp thnh (MIN hay PROD):

- Lut hp thnh max-PROD, nu cc hm lin thuc thnh phn c

xc nh theo quy tc hp thnh PROD v php hp gia cc mnh hp

thnh c ly theo lut max.

- Lut hp thnh max-MIN, nu cc hm lin thuc thnh phn c

xc nh theo quy tc hp thnh MIN v php hp gia cc mnh hp

thnh c ly theo lut max.

- Lut hp thnh sum-MIN, nu cc hm lin thuc thnh phn c

xc nh theo quy tc hp thnh MIN v php hp c ly theo cng thc

Lukasiewicz.


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- Lut hp thnh sum-PROD, nu cc hm lin thuc thnh phn c

cc nh theo quy tc hp thnh PROD v php hp c ly theo cng thc

Lukasiewicz.

Tng qut, ta xt thut ton xy dng lut hp thnh c nhiu mnh

hp thnh. Xt lut hp thnh gm p mnh hp thnh:

R1: Nu = A1Th = B1hoc

R2: Nu = A2Th = B2hoc

. . .

RP: Nu = AP, Th = BP

Trong cc gi tr m A1, A2,..., APc cng tp nn X v B1, B2,...,

BPc cng tp nn Y.

x0

A=>B(y)

x

A(x)

y

B(x)

A=>B(y)

x0x

A(x)

y

B(y)

Hnh2.3:Hm lin thuc ca lut hp thnh : (a) Hm lin thuc A(x)v B(y).(b) AB(y) xc nh theo quy tc min.(c) AB(y) xc nh theo

quy tc PROD.

x

A(x)

y

B(x)


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B1 B2

y

B

H

yHnh2.7: Gii m theo nguyn l cn phi

B1 B2

y

B

H

y

Hnh2.6: Gii m theo nguyn l cn tri

B1 B2

y

y

y

B

H

y

Hnh2.5: Gii m theo nguyn l trung bnh

Trong Hnh2.4 th G l khong [y1, y2] ca min gi tr ca tp m u

ra B2ca lut iu khin R2.

Ba cch tnh l: Nguyn l cn tri, cn phi v trung bnh. K hiu

y1, y2l im cn tri v cn phi ca G.

- Nguyn l trung bnh:Theo nguyn l trung bnh, gi tr r y s l:

y =2

21 yy (2.8)

-Nguyn l cn tri:Gi tr r y c ly bng cn tri y1ca G.

- Nguyn l cn phi:Gi tr r y c ly bng cn phi y2ca G.


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Gii m theo phng php trng tm

Phng php trng tm s cho ra kt qu y l honh im trng

tm ca min c bao bi trc honh v ng B(y). Cng thc xc nh

y theo phng php im trng tm nh sau:

y =

S

'

S

'

)(

)(

dyy

dyyy

B

B

(2.9)

Trong S l min xc nh ca tp m B. Cng thc ny cho php

xc nh gi tr y vi s tham gia ca tt c cc tp m u ra mt cch bnhng v chnh xc, tuy nhin li khng n tha mn ca lut iu

khin quyt nh v thi gian tnh ton lu.

Phng php trng tm c u im l c tnh n nh hng ca tt c

cc lut iu khin n gi tr u ra, tuy vy cng c nhc im l khi gp

cc dng hm lin thuc hp thnh c dng i xng th kt qu sai nhiu. V

gi tr tnh c li ng vo ch hm lin thuc c gi tr thp nht, thm ch

bng 0, iu ny hon ton sai v suy ngha v thc t. trnh iu ny, khi

nh ngha cc hm lin thuc cho tng gi tr m ca mt bin ngn ng nn

ch sao cho lut hp thnh u ra trnh c dng ny, c th bng cch

kim tra s b qua m phng.

B1 B2

y

B

yS

Hnh2.8: Gii m theo phng php imtrng tm


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Gii m theo phng php trung bnh tm.

Nu gi thit mi tp m Bk(y) c xp x bng mt cp gi tr (yk ,

Hk) duy nht (singleton) trong Hkl cao ca Bk(y) v ykl mt im

mu trong min gi tr ca Bk(y) c Bk(y)=Hk.

th: y =

q

k

k

q

k

kk

H

Hy

1

1 (2.10)

y l cng thc tnh xp x y theo phng php cao. Nhiu trng

hp s dng u ra dng singleton rt c hiu qu trong qu trnh gii m v

n gin c cng vic tnh ton cn thit. Cng thc ny p dng c cho

mi lut hp thnh nh max-MIN, max-PROD, sum-MIN v sum-PROD.

2.4. B IU KHIN M TNH

2.4.1. Khi nim

B iu khin tnh l b iu khin c quan h vo/ra y(x), vi x l u

vo y l u ra, theo dng mt phng trnh i s (tuyn tnh hoc phituyn). B iu khin m tnh khng xt ti cc yu t ng ca i tng

(vn tc, gia tc, ). Cc b iu khin tnh in hnh l b iu khin

khuch i P, b iu khin Rle hai v tr, ba v tr,

2.4.2. Thut ton tng hp mt b iu khin m tnh

Cc bc tng hp mt b iu khin m tnh v c bn ging cc

bc chung tng hp b iu khin m nh trnh by trn. hiu k

hn ta xt vd c th sau:

V d:

Hy thit k b iu khin m tnh SISO c hm truyn t y = f(x)

trong khong x = [1,2] tng ng vi y trong khong y = [1, 2].

Bc 1: nh ngha cc tp m vo ra.

nh ngha N tp m u vo: A1, A2, ANtrn khong [1,2] ca x

c hm lin thuc Ai(x) ( i = 1,2,N) dng hnh tam gic cn.


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nh ngha N tp m u ra : B1,B2,BNtrn khong [1, 2] ca y c

hm lin thuc Bi(x) ( j = 1,2,N) dng hnh tam gic cn.

Bc 2: Xy dng lut iu khin

Vi N hm lin thuc u vo ta s xy dng c N lut iu khin

theo cu trc:

Ri: nu = Aith = Bi

Bc 3: Chn thit b hp thnh

Gi thit chn nguyn tc trin khai SUM PROD cho mnh hp

thnh, v cng thc Lukasiewicz cho php hp th tp m u raB khi uvo l mt gi tr r x0s l:

B(y) = MIN

N

i

AiBi xy1

0 )()(,1 (2.11)

V Bi(y) l mt hm Kronecker Bi(y)Ai(x0) = Ai(x0), khi :

B(y) = MIN

N

i

AiBi xy1

0 )()(,1 (2.12)

Bc 4: Chn phng php gii m

Chn phngphp cao gii m, ta c:

y(x0) =

N

i

i

N

i

ii

H

Hy

1

1 =)(

)(

1

0

1

0

N

iiA

N

iiAi

x

xy

(2.13)

Quan h truyn t ca b iu khin m c dng:

y(x0) =)(

)(

1

1

N

i

iA

N

i

iAi

x

xy

(2.14)

2.4.3. Tng hp b iu khin m tuyn tnh tng on

Trong k thut nhiu khi ta phi thit k b iu khin m vi c tnh

vo ra cho trc tuyn tnh tng on. Chng hn, cn thit k b iu

khin m c c tnh vo ra nh Hnh2.9.


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Thut ton tng hp b iu khin ny ging nh thut ton tng hp

b iu khin m vi hm truyn t y(x) bt k. Tuy nhin, cc on c

tnh thng v ni vi nhau mt cch lin tc ti cc nt th cn tun th mt s

nguyn tc sau:

+ Mi gi tr r u vo phi lm tch cc 2 lut iu khin.

+ Cc hm lin thuc u vo c dng hm tam gic c nh l mt

im nt k, c min xc nh l khong [xk-1, xk+1] nh Hnh2.10a.

+ Cc hm lin thuc u ra c dng hm singleton ti cc im

nt ykHnh2.10b.

+ Ci c lut hp thnh Max Min vi lut iu khin tng qut:

Rk: Nu = Akth = Bk

+ Gii m bng phng php cao.

2.5. B IU KHIN M NG

B iu khin m ng l b iu khin m m u vo c xt ti cc

trng thi ng ca i tng nh vn tc, gia tc, o hm ca gia tc, V

d i vi h iu khin theo sai lch th u vo ca b iu khin m ngoi

A1 A2 A3 A4 A5 B1,B2 B3,B4 B5

x y

Hnh 2.10:Hm lin thuc ca cc bin ngn ng vo ra

a) b)

y5

y3,y4

x3 x4 x5x

y1,y2

x2x1

Hnh2.9:c tnh vo ra cho trc

y


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tn hiu sai lch e theo thi gian cn c cc o hm ca sai lch gip cho b

iu khin phn ng kp thi vi cc bin ng t xut ca i tng.

Cc b iu khin m ng hay c dung hin nay l b iu khin

m theo lut t l tch phn (PI), t l vi phn (PD) v t l vi tch phn (PID).

Mt b iu khin m theo lut I c th thit k t mt b m theo lut

P (b iu khin m tuyn tnh) bng cch mc ni tip mt khu tch phn

vo trc hoc sau khi m . Do tnh phi tuyn ca h m, nn vic mc

khu tch phn trc hay sau h m hon ton khc nhau Hnh2.11a,b.

Khi mc thm mt khu vi phn u vo ca mt b iu khin m theo

lut t l s c c mt b iu khin m theo lut t l vi phn PD Hnh2.12.

Cc thnh phn ca b iu khin ny cng ging nh b iu khin theo

lut PD thng thng bao gm sai lch gia tn hiu cho v tn hiu ra ca

h thng e v o hm ca sai lch e. Thnh phn vi phn gip cho h thng

phn ng chnh xc hn vi nhng bin i ln ca sai lch theo thi gian.

B iu khinm

i tng-

E

a)

B iu

khin m i tng-b)

Hnh2.11:H iu khin m theo lut PI

dt

d

B iu khinm

i tng-

Ey

x

Hnh2.12:H iu khin m theo lut PD


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65

-Kh nng 2: B iu khin kinh in PID c s dng mch vng

iu khin trong; mch vng iu khin ngoi s dng b iu khin m

chnh nh tham s cho b iu khin PID. B iu khin m lai xy dng

theo phng php ny c gi l b iu khin m lai chnh nh m tham

s b iu khin PID.

2.6.2. B iu khin m lai kinh in

Khi thit k b iu khin m lai kinh in, trc ht ta c th thit k b

iu khin m mch vng trong m cha cn quan tm n iu kin n nh

ca h thng. Sau , khi thit k b iu khin PID mch vng ngoi ta micn xt n vn n nh. Nh vy b iu khin PID mch vng ngoi s

thc hin chc nng gim st n nh ca h thng, cn b iu khin m

mch vng trong s m bo cht lng iu chnh cho h thng. Chc nng

gim st ca b iu khin PID mch vng ngoi c l gii nh sau: nu b

iu khin m mch vng trong hot ng tt tc l m bo cht lng iu

chnh cho h thng th b iu khin PID mch vng ngoi khng tham giavo cng vic iu chnh. Khi b iu khin m mch vng trong hot ng

khng tt, c khuynh hng gy mt n nh cho h thng th b iu khin PID

mch vng ngoi s can thip nhm a h thng v trng thi n nh.

B iukhin PID

i tngiu khin

x u yB iukhin m

-

e

Thit b olng

Hnh2.14:M hnh b iu khin m lai kinh in

2.6.3. B iu khin m lai cascade

Mt h m lai khc c biu din nh Hnh2.15, phn b tn hiu

iu chnh u c ly ra t b iu khin m.

Trong trng hp h thng ccu trc nh trn th vic chn cc i


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lng u vo ca h m ph thuc vo tng ng dng c th. Tt nhin cc

i lng thng c s dng lm tn hiu vo ca h m l tn hiu ch o

x, sai lc e, tn hiu ra y cng vi cc o hm hoc cc tch phn ca cc i

lng ny. V nguyn tc c th s dng cc i lng khc ca i tng

cng nh s dng cc nhiu xc dnh c.

B iukhin m

i tngB iu khinkinh in

+

u

u

-x

Hnh2.15: Cu trc h m lai Cascade

2.6.4. B iu khin m chnhnh tham s b iu khin PID

Cs ca phng php ny l da vo vic phn tch sai lch e(t) v o

hm ca sai lch, cc tham s KP, TI, TDca b iu khin PID s c t ng

chnh nh theo phng php chnh nh m ca Zhao, Tomizuka v Isaka .

2.6.5. B iu khin m t chnh cu trc

B iu khin m t chnh nh cc lut iu khin c gi l b iu

khin m t chnh cu trc. B chnh nh c thit k m bo u ra l gi tr

hiu chnh ca tn hiu iu khin u(t) (tn hiu ra ca b iu khin). thay

i lut iu khin trc tin l phi xc nh c quan h gia gi tr chiu chnh u ra ca b iu khin vi gi tr bin i u vo. Do vy cn

c m hnh th ca i tng, m hnh ny dng tnh ton gi tr u vo

tng ng vi mt gi tr u ra cn t c ca b iu khin. Da trn tn

hiu ra mong mun v tn hiu vo tng ng ca b iu khin c th xc nh

v hiu chnh cc nguyn tc iu khin, cc nguyn tc ny m bo cht lng

iu khin ca h thng. i vi nhng i tng bc cao c thi gian tr ln c


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th c thi gian chnh nh chm, cn i vi cc h thng bc thp c thi gian

tr nh yu cu thi gian chnh nh nhanh. Vic chnh nh ch c ngha khi

qu trnh chnh nh kt thc trc khi h thng kt thc qu trnh qu .

2.7. KT LUN CHNG 2

Ni dungchng 2, tm hiunhngvnchung v l thuyt m tng

qut nh logic m, tp m, bin ngn ng, lut hp thnh v biukhin

m.Lunvn tp trung nghin cu cu trc ca b iu khin m v phn

tch lm r chc nng cc khi trong b iu khin m,phn tch cc b iu

khin m tnh, m ng phm vi ng dng v u nhcim ca biukhin. Trn csphn tch u nhcimcabiukhinmcng nh b

iu khin m lai PID, xuthngthitkbiukhinlai PID ml c

s cho vic thit k tnh ton chng 3.

Biukhinmhinang cnghin cungdngrngri nhcc

u imcan. Hiukhinmmbokhng cnkhilngtnh ton

lnv phctpnhcc biukhinkhc v c thtnghpbiukhinmvihm truyntphi tuynbtk. Biukhinmc nhiuu im,

cbitkhi iukhini tngm ta cha bitnhiuvi tng, thiu

thng tin, thng tin khng tin cy.

Tuy nhin trong thc t ng dng ca k thut iu khin m cho thy

rng khng phi c thay th mt b iu khin kinh in bng mt b iu

khin m th s c mt h thng tt hn. Trong nhiu trng hp, h thngc c tnh ng hc tt v bn vng cn phi thit k thit b iu khin lai

gia b iu khin m v b iu khin kinh in. T dn n khi nim

"h m lai" v lnh vc thit k, ng dng b iu khin m lai nng cao

cht lng iu khin ca h thng. H iu khin m lai s pht huy ht cc

u im ca b iu khin m v b iu khin r.


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Chng 3

THIT K B IU KHIN PID M

3.1. T VN

Do cu trc n gin v bn vng nn cc b iu khin PID c

dng ph bin trong cc h iu khin cng nghip v p ng c cc yu

cu t ra. Hm truyn tca b iu khin PID l:

sK

s

KKsG d

ip )(

(3.1)Trong Kp, Ki, Kd, l cc h s t l, tch phn, o hm.

Nu vit theo hm thi gian th tn hiu ra ca b iu khin PID l:

dt

tdeTde

TteKtu D

t

IP 0

1

(3.2)

Trong :

-e(t) l tn hiu u vo

- u(t) l tn hiu u ra

- KP l h s khuch i

- TIl hng s tch phn

- TDl hng s vi phn.

Cht lng ca h thng ph thuc vo cc tham s KP, TI, TDca b

iu khin PID. Nhng v cc h s ca b iu khin PID ch c tnh ton

cho mt ch lm vic c th ca h thng vi cc tham s ca i tng l

xc nh c. V vy trong qu trnh lm vic, nu tham s ca h thng

thay i th, lng ra ca h thng cng thay i, ngha l b iu khin PID

khng cn m bo cht lng ra ca h nh mong mun c na.

Cc h cn iu khin trong thc t ch yu l cc h phi tuyn, c

cha cc tham s khng bit trc, cc tham s ca h thng bin thin theo


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70

rt ph bin trong cc h thng truyn ng in cht lng cao, di cng

sut ng c mt chiu t vi W n hng MW. y l loi ng c c th

p ng yu cu mmen khi ng ln, o chiu nhanh, ch tiu iu chnh

tc cao : iu chnh trn, di iu chnh rng, sai lch tnh nh H

T-D cng l h phi tuyn v trong qu trnh lm vic mch t b bo ho, in

tr , in cm, m men qun tnh ca h l thay i.

3.2. THIT K B IU KHIN M CHNH NH THAM S PID

3.2.1. Cu trc b iu khin

Cu trc b iu khin mchnhnhtham sPID p dngcho itngl khu dao ngbc hai nhsau:

B iu khin bn trong dung PID truyn thng, cn bn ngoi dng b

iu khin m t ng chnh nh tham s ca b PID.

3.2.2. Thit k b iu khin

Gi thit h s t l cho php thay i trong khong [ Kpmin, Kpmax], h

s o hm thay i trong khong [ Kdmin, Kdmax].

tin li trong vic tnh ton ta bin i chng v n v tng i.

Kp =minmax

min

pp

pp

KK

KK

, Kd =

minmax

min

dd

dd

KK

KK

Hng s thi gian tch phn: TI= Td.

Tng t ta c : KI= .dp

T

K

= .

2

d

p

K

K

Hnh3.2: Cu trc b iu khin

17.02.0

25.12

ss

PID

B iukhin m


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71

Nh vy nhim v c th ca ta l thit k b iu khin m chnh

nh t ng ba tham s : Kp, Kd, .

Gi thit tn hiu vo ca b iu khin m l e(t), e(t) = de/dt. Th cu

trc b iu khin m nh Hnh3.3.

Suy lun m c thc hin theo cc lut sau :

Nu e(t) l A* v e(t) l B* th Kp l C*, Kd l D*, l E*

Trong : A*, B*, C*, D*, E* l cc tp m v * = 1, 2, 3, M.

Gi thit min xc nh ca e(t) l [e1, e2] v e(t) l [e1, e2]

V mi loi dng 7 tp m nh hnh v. ( y M = 7, cc tp m S3,

S2, S1, Z0, B1, B2, B3).

Kp, Kddng hai tp m : Nh v ln (S, B).

Hm

Hm

Hm

PID

Kp

Kd

KI

e(t)

e(t)

Hnh3.3: Cu trc b iu khin PID m

S1S2S3 B1 B2 B3Z0

e2e2 e2e2

Hnh3.4:Hm lin thuc ca e(t) v de(t)/dt


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73

Ta i xc nh cc lut iu khin tng ng : Khi bt u khi ng,

khong thi gian al, lc ny cn tn hiu iu khin ln tn hiu ra tng

nhanh, suy ra lc ny Kpln, Kdnh v Kiln ( nh) ta c lut:

Nue(t) ln v e(t) l Zero ThKpln, Kdnh v nh

Xung quanh khong thi gian b1 ta mun tn hiu iu khin nh

khng qu iu chnh, ngha l Kpnh, Kd ln v Ki nh vy ta c lut:

Nue(t) l Zero v e(t) l m ln ThKp nh, Kd ln, ln

Cc tc ng iu khin xung quanh khong thi gian c1v d1tng t

nh xung quanh a1v b1.Dng lut hp thnh MAX PROD, m ho n tr, gii m theo trung

bnh tm lc ta c cc h s Kp, Kd, c tnh ton lc iu khin.

Kp(t) =

n

i

BA

n

i

BAp

tete

tetey

1

1

))('())((

))('())((

, Kd(t) =

n

i

BA

n

i

BAd

tete

tetey

1

1

))('())((

))('())((

(t) =

n

i

BA

n

i

BA

tete

tetey

1

1

))('())((

))('())((

Bng 3.1. Lut iu khin cho h s Kp

Kpde/dt

S3 S2 S1 Z0 B1 B2 B3

E(t)

S3 B B B B B B BS2 S B B B B B S

S1 S S B B B S S

Z0 S S S B S S S

B1 S S B B B S S

B2 S B B B B B S

B3 B B B B B B B


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74

Bng3.2. Lut iu khin cho h s Kd

Kd

de/dt

S3 S2 S1 Z0 B1 B2 B3

e(t)

S3 S S S S S S S

S2 B B S S S B B

S1 B B B S B B B

Z0 B B B B B B B

B1 B B B S B B B

B2 B B S S S B BB3 S S S S S S S

Bng3.3. Lut iu khin cho h s

de/dt

S3 S2 S1 Z0 B1 B2 B3

e(t)

S3 S S S S S S S

S2 MS MS S S S MS MS

S1 M MS MS S MS MS M

Z0 B M MS MS MS M B

B1 M MS MS S MS MS M

B2 MS MS S S S MS MS

B3 S S S S S S S


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75

Da trn ngn ng Matlab ta i m phng b iu khin m vi dao

din nh sau:

Hnh3.8: Giao din m phng m

Hnh3.9:Hm lin thuc ca tn hiu e(t) v de/dt


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77

3.2.3. Kt qu m phng

Hnh3.12:c tnh iu chnh PID ti u vi i tng bc hai

(Kp= 130;Kd= 10; Ki= 435.6)

Hnh3.13:c tnh iu chnh PID m (Kp= 20.1; Kd = 20.1;

Ki= 8.6) so vi c tnh PID ti u

c tnh PID c tnh PIDm


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78

Nhn xt :T b iu khin m xy dng khi p dng iu khin

i tng hm bc hai ta xc nh b tham s PID m (Kp = 20.1; Kd=20.1;

Ki = 8.6) so snh vi b tham s PID ti u (Kp = 132, Kd = 10, Ki = 435.6).

Ta thy so snh c tnh iu chnh m v c tnh iu chnh theo PID tiu

th b iu khin PID m khng c qu iu chnh thi gian n nh nhanh

hn, sai s tnh nh.

3.3. NG DNG PID M IU KHIN H TRUYNNGT-D

3.3.1. Cc yu cu i vi h truyn ng T-D

S nguyn l h truyn ng T- D rt gn trn Hnh 3.14.

UR Ri FX B

Si

S

HCD

Ui

Ui

-

U

Uk Mc

I

U

-

Hnh3.14: S khi ca h truyn ng T-

: ng c mt chiu.

B : B bin i xoay chiu - mt chiu c iu khin.

RI: B iu chnh dng in.

R: B iu chnh tc .

Lng vo ca h l in p iu khin Udk, lng ra l tc n ca

ng c. Cht lng ca h c nh gi bng cc ch tiu cht lng ng

v ch tiu cht lng tnh.

-Ch tiu cht lng ng l:

+ Thi gian qu tqd

+ Lng qu iu chnh max

+ S ln dao ng n


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79

-Ch tiu cht lng tnh l:

+ Sai lc tnh

+ Phm vi iu chnh

+ trn khi iu chnh

Tu theo yu cu cng ngh m ngi ta t ra cc ch tiu ng v ch

tiu tnh c th khc nhau v h c thit k vi cc s cu trc khc

nhau : c th l h h, h kn vi mthoc nhiu mch vng phn hi. Thng

thng h truyn ng T- D c thit k vi 2 mch vng kn : mch vng

iu chnh dng in v mch vng iu chnh tc . Mch vng iu chnhdng in c coi l mt khu trong h thng iu chnh tc v xcnh

mmen ko ca ng c.

Mt phng n n gin nht iu chnh dng in c cu trc nh

Hnh3.15a dng b iu chnh tc c dng b khuych i v mch phn

hi dng in phi tuyn P. Khi tn hiu dng in cha khu phi tuyn

ra khi vng km nhy th b iu chnh lm vic nh b iu chnh tc m khng c s tham gia ca mch phn hi dng in. Khi dng in ln,

khu P s lm vic vng tuyn tnh ca c tnh v pht huy tc dng hn

ch dng ca b iu chnh R.

Phng n thhai c m t trn Hnh 3.15b. C hai mch vng vi

hai b iu chnh ring bit R1, R2, trong R2l b iu chnh dng in vi

gi tr t I. Cu trc kiu ny cho php iu chnh c lp tng mch vng.Phng n iu chnh dng in c s dng rng ri nht trong

truyn ng in t ng nh trn Hnh3.15c, trong R1 l b iu chnh

dng in, Rl b iu chnh tc . Mi mch vng c b iu chnh ring

c tng hp t i tng ring v theo cc tiu chun ring.


80/110


80

R S01 S02

MC

I-

-

p

a)

R1 S01 S02

MC

I-

b)

R2-

I

MC

R S01 S02I

-

c)

RI-

I

Hnh 3.15: Cu trcmch vng iu chnh dng in

3.3.2.Tng hp mch vng iu chnh dng in RI

Trong cc h thng truyn ng t ng cng nh cc h thng chp

hnh th mch vng iu chnh dng in l mch vng c bn. Chc nng c

bn ca mch vng dng in trong cc h thng truyn ng mt chiu v

xoay chiu l trc tip (hoc gin tip) xc nh mmen ko ca ng c,ngoi ra cn c chc nng bo v, iu chnh gia tc v.v

S khi ca mch vng iu chnh dng in nh trn Hnh3.16,

trong F l mch lc tn hiu, Ril b iu chnh dng in, B l b bin

i xoay chu - mt chiu, Sil xenx dng in.

Xenx dng in c th thc hin bng cc bin dng mch xoay

chiu hoc bng in tr sun hoc cc mch do cch ly trong mch mt chiu.


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82

c th ly a = 2.

pTR

KK

pTpR

s

u

icl

ui

2.

1)(

(3.6)

Cui cng hm truyn ca mch vng dng in l:

22221

11

112

11

)(

)(

pTpTKpTpTKpU

pI

ssissii

(3.7)

3.3.3.Tng hp mch vng iu chnh tc

H thng iu chnh tc l h thng m i lng c iu chnh ltc gc ca ng c in, cc h ny rt thng gp trong thc t k thut.

H thng iu chnh tc c hnh thnh t h thng iu chnh dng in.

Cc h thng ny c th l o chiu hoc khng o chiu. Do cc yu cu

cng ngh m h cn t v sai cp mt hoc v sai cp hai.

Tu theo yu cu ca cng ngh m cc b iu chnh tc Rc th

c tng hp theo hai tn hiu iu khin hoc theo nhiu ti M c. Trong

trng hp chung h thng phi c c tnh iu chnh tt c t pha tn hiu

iu khin ln t pha tn hiu nhiu lon.

S khi chc nng c trnh by trn Hnh3.17.

UR Ri FX B

Si

S

HCD

Ui

Ui

-

U

Uk Mc

I

U

-

Hnh3.17: S khi h iu chnh tc

3.3.3.1. iu chnh tc dng b iu chnh tc t l

phn trn ta tng hp c mch dng in, trong phn ny s s

dng biu thc kt qu trong b qua nh hng ca s.. ca ng c:


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85

3.3.3.2. iu chnh tc dng b iu chnh tc tch phn t l PI

Trong nhiu thit b cng ngh thng c yu cu h thng iu chnh

v sai cp cao, khi ny c th s dng phng php ti u i xng tng

hp cc b iu chnh. Vi mch vng iu chnh tc hm truyn ca b

iu chnh c dng:

pKT

pTpR

o

o1

)( (3.17)

V hm truyn mch h s l:

)1'2(1..1)(

pTpTKKKR

pKTpTpF

sci

u

o

oo

(3.18)

T (3.18) c th tm c hm truyn mch kn F(p), ng nht F(P)

vi hm chun ti u i xng ta tm c tham s ca b iu chnh.

Nu chn Ts = Ts th:

To = 8Ts

sci

u

s

s

ci

u TTKK

KRTT

TKKKRK '4.

.'8)'2(8.

.

2

).'8

11(

'4

1.

.)(

pTTKR

TKKpR

ssu

ci

(3.19)

Thy rng thnh phn t l ca b iu chnh (3.19) ng bng h s

khuych i ca b khuych i (3.12).

Khi tng hp h thng theo phng php ti u i xng thng phi

dng thm khu to tn hiu t trnh qu iu chnh. Khu to tn hiu t

ny thng c hm truyn t ca khu lc thng thp bc nht, c hng s

thi gian lc tu thuc vo gia tc cho php ca h thng. Tt nhin khu to

tn hiu t ny phi t bn ngoi mch vng iu chnh tc .

Hm truyn mch kn ca h thng:

1]1)'21('4['8

'81

)(

)()(

pTpTpT

pT

pU

pUpF

sss

s

(3.20)


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86

Cn c vo cc biu thc nu trn ta c th tnh c hm truyn

vi tn hiu nhiu lon l dng in ti:

1]1)'21('4['8

'81

)(

)()(

pTpTpT

pT

pI

pIpF

sss

s

ci

(3.21)

v cng tnh c sai s tc tng ng khi nhiu ti c dng hng s:

1]1)'21('4['8

)'21('8

.

'4

..

)]()([)(

pTpTpT

pTpT

TK

IRT

RpTK

pIpIp

sss

ss

c

cus

uc

c

(3.22)Kt qu l, mch vng iu chnh tc l v sai cp hai i vi tn

hiu iu khin (3.20) v l v sai cp mt i vi tn hiu nhiu (3.22). Nh

vy khi n nh th sai lch tc s bng khng.

3.3.4. Bi ton ng dng c th

c v d c th tc gi chn h T- D c s nguyn l nh

Hnh3.20 v cc phn t c tham s nh sau :

Ui

UW

Rd

Phtxung

12

ng b

Uk

+

-

FT

Lc

Kbd

CL

Hnh3.20: S cu trc h truyn ng T-D mt chiu

ng c : 2,2kW220V-12A- 1500vng/pht.

2,1uR ; mHLu 31


87/110


87

36cp

I A

50:1bd

K

2

1

( )( )

( ) 1 .iI

i

i

KU pF p

I p T p

3

0 0,166 10vT

9rcmU v

30,2 10k

T

1002

50

di

bd

RK

K

. 0, 0005i d dT R C

2,34 2,43 22057,2

9cl

rc

UK

U

Dng in cho php ln nht:

Hng s thi gian mch phn ng:

T thng nh mc:

33

10.83,252,1

10.31

u

uu

R

LT

.

m

uuu

m

ui

W

IRU

W

EKF

.

= 1,3089.

Mmen qun tnh tnh ton k c roto ca ng c:2.016,0 mkgJ .

- Chnh lu CL: Chnh lu thc hin nhim v bin i dng in

xoay chiu thnh dng in mt chiu. Chnh lu CL s dng chnh lu cu 3

pha c iu khin dng van tiristo. Khi pht xung iu khin cc tiristo s

dng h thng pht xungng b nhiu knh, trong vic ng b c

thc hin nh vic ng b ho in p ta vi li. in p ta c dng hnh

rng ca qut ngc c to ra t my pht in p ta.

+ H s KCL: V chnh lu l chnh lu cu 3 pha m = 3.

+ Hng s thi gian mch chnh lu :

+ My pht xung rng ca c bin :

+ Hng s thi gian mch iu khin : .

-o lng dng in: S dng 3 bin dng lp t ba pha. in p

s cp bin dng qua mch chnh lu cu it ba pha, mch lc RC lc thnhphn xoay chiu sau chnh lu.

+ Thng s mch lc RC: Rd= 100 ; Cd= 0,000005.

+ T s bin i dng: .

+ Hm truyn c cu o dng in:

Trong : H s t l:

Hng s thi gian b lc:


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88

- o lng tc : S dng my pht tc mt chiu FT. m bo

yu cu l in p mt chiu c cha t thnh phn xoay chiu tn s cao v t

l vi tc ng c, khng b tr nhiu v gi tr v du so vi bin i i

lng o, ta s dng my pht tc mt chiu c t thng khng i trong ton

vng iu chnh tc . V vy phi hn ch tn tht mch t bng vic s

dng vt liu t c t tr hp v s dng l thp k thut in mng (hn ch

tn tht dng in xoy). loi b sng iu ho tn s cao s dng b lc

lp u ra my pht tc.

+ Mypht tc FT: vUphtvngn mm 24;/3000 .+ Hm truyn ca my pht tc khi c b lc s l:

pT

K

p

pUpFf

.1)(

)()(

Trong : 30002455,955,9

m

m

n

UK

= 0,0764

+ Hng s thi gian nh:0005,0T

.3.3.4.1. Tnh ton tham s mch vng dng in

- H thng lm vic ch dng in lin tc.

- Coi sc in ng E khng nh hng n qu trnh iu chnh ca

mch vng dng in.

- B iu chnh dng in RI thit k theo tiu chun ti u mun nh

trnh by trn.

Tnh ton thng s ca khu chnh lu:

vokcl TTT 2 = 0,0002 x 0,000166 = 3,32 e-8

vokcl TTT 1 = 0,0002 + 0,000166 = 0,000366

B iu chnh dng in RI thit k theo tiu chun ti u mun.

paT

R

KK

pTpR

siu

icl

ui

.

1)(

= pT

pT

ri

u1


89/110


89

vokisi TTTT = 0,0005+ 0,0002 + 0,000166 = 0,000866

2,1

22,57000866,022

u

iclsiri R

KKT

T = 0,16512

M hnh cu trc mch vng dng in vi b iu chnh dng in RI

thit k theo tiu chun ti u mun nh Hnh3.21.

Hnh3.21: S cu trc mch vng iu chnh dng in.

3.3.4.2. Tnh ton tham s b iu khin tc PI

- B iu chnh tc Rdng b iu chnh tc tch phn t l PI

thit k theo tiu chun ti u i xng nh trnh by mc 3.3.3.2.

pT

K

pU

pI

si

i

i 21

1

.)(

)(

= p001732,015,0

B iu chnh tc R thit k theo tiu chun ti u i xng.

pTTKR

TKFKpR

swswu

cii

8

11

4

1)(

pTTKR

TKFK

TKR

TKFKpR

swswu

cii

swu

cii

8

1

4

1

4

1)(

2

2 wsisw

TTT

= 0,001116

2i

uc

KF

RJT

= 0,011207


90/110


90

swu

ciin

TKR

TKFKKP

4

1

= 71,685

swswu

ciin

TTKR

TKFKKI

8

1

4

1

= 8029,3

hn ch dng in trong qu trnh qu ta dng khu bo ho v

khu ny c t u ra b iu chnh tc . M hnh cu trc mch vng

tc vi b iu chnh tc RI thit k theo tiu chun ti u mun nh

Hnh3.22.

Hnh 3.22: Cu trc mch vng iu chnh tc .

3.3.5. Thitkhiukhinmlai

Nh tng hp h truyn ng T-Db diu khin PID vng trong

c thit k cho h truyn ng l khu PI, nh vy b iu khin m

vng ngoi c nhim v l phi t ng chnh nh c 2 tham s KP , KI

ca b PI. C s thit k b iu khin m l da vo vic phn tch sai

lch e(t) v o hm ca tn hiu ra. Cc tham s KP, KIca b iu khin PI

s c t ng chnh nh theo phng php chnh nh m. Cu trc bn

trong b iu khin m c dng:

B chnh nh m 1

B chnh nh m 2

HsKP

HsKIe

dw/dt

Hnh3.23: Cutrc bn trong b chnh nh m


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92

Hnh3.25: Cu trc b chnh nh m

3.3.5.2. Xc nh gi tr cho cc bin vo v ra

- Xc nh min gi tr vt l cho cc bin vo v ra:

+ Sai lch ERROR c chn trong min gi tr [-15,+15].

+ Tc bin i dw/dt c chn trong min gi tr [-5000,+5000]

+ u ra Hs KPc min gi tr nm trong khong [1,10].+ u ra Hs KIc min gi tr nm trong khong [1,10].

- Xc nh s lng tp m (cc gi tr ngn ng) cn thit cho cc

bin: Nguyn lchung l s lng cc gi tr ngn ng cho mi bin nn nm

trong khong t 3 n 10 gi tr. Nu s lng t hn 3 th qu th v t c

ngha v khng thc hin c vic ly vi phn. Nu ln hn 10 th qu mn

con ngi kh c kh nng cm nhn qu chi ly, bao qut ht cc trng hpxy ra v nh hng n b nh, tc tnh ton. V vy, chn s lng tp

m cho mi bin u vo l 7 v mi bin u ra l 4, c th nh sau:

+ ERROR {NB, NM, NS, ZE, PS, PM, PB}.

+ dw/dt {NB, NM, NS, ZE, PS, PM, PB}.

+ Hs KP {S, MS, M, B}.

+ Hs KI {S, MS, M, B}.


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93

Trong k hiu:

+ NBm nhiu ; NM m va ; NS m t ; ZEZero.

+ PBDng nhiu ; PM Dng va ; PS Dng t.

+ SNh ; MS Nh va ; M Va ; B Ln.

Xc nh hm lin thuc: Trong k thut iu khin thng u tin

chn hm lin thuc kiu hnh tam gic hoc hnh thang. Cc loi ny c biu

thc n gin, tnh ton d dng, tuy nhin cc hm lin thuc ny ch gm

cc on thng nn khng mm mi cc im gy.

Bng 3.4. Hm lin thuc ca bin u vo

Bin

ngn ng

Hm lin

thuc ca

bin ERROR

Thng s

ca bin

ERROR

Hm lin

thuc ca

bin dw/dt

Thng s ca bin dw/dt

NB Trimf -20, -15, -10 Trapmf -1e+009, -5167,-4833, -3501

NM Trimf -15 , -10 , -5 Trimf -5000 , -3334 , -1666

NS Trimf -10 , -5 , 0 Trimf -3334 , -1666 , 0ZE Trimf -5 , 0 , 5 Trimf -1666 , 0 , 1666

PS Trimf 0 , 5 , 10 Trimf 0 , 1666 , 3334

PM Trimf 5 , 10 , 15 Trimf 1666 , 3334 , 5000

PB Trimf 10 , 15 , 20 Trapmf 3510, 4833,5167, 1e+009

Bng 3.5. Hm lin thuc ca bin u ra

Bin

ngn ng

Hm lin thuc

ca bin HsKP

Thng s ca

bin HsKP

Hm lin thuc

ca bin HsKI

Thng s ca

bin HsKI

S Trimf -1.999 , -1 , 4 Trimf -1.999 , -1 , 4

MS Trimf 1 , 4 , 7 Trimf 1 , 4 , 7

M Trimf 4 , 7 , 10 Trimf 4 , 7 , 10

B Trimf 7 , 10 , 13.01 Trimf 7 , 10 , 13.01


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94

Hnh3.26:Xc nh tp m cho bin vo ERROR

Hnh3.27:Xc nh tp m cho bin vo dw/dt


95/110


95

Hnh3.28:Xc nh tp m cho bin ra HsKP

Hnh3.29:Xc nh tp m cho bin ra HsKI


96/110


96

Xy dng cc lut iu khin.

Da vo bn cht vt l, cc s liu vo ra c c, kinh nghim, v

da vo c tnh qu thng gp ca h thng iu khin dng PID nh

Hnh3.30, ta xc nh cc lut iu khin tng ng. Chng hn:

a1

a2b1b

2

c1d1

tn hiu ra

t

Hnh3.30:c tnh qu thng gp ca h iu khin dng PID

Khi bt u khi ng, khong thi gian al, lc ny cn tn hiu iu

khin ln tn hiu ra tng nhanh, suy ra lc ny Kpln v KIln , ta c lut:

Nue(t) ln v dw/dt l Zero ThKpln KIln.

Xung quanh khong thi gian b1 ta mun tn hiu iu khin nh

khng qu iu chnh, ngha l Kpnh, KInh , vy ta c lut:

Nue(t) l Zero v dw/dt ln ThKpnh, KInh.

Cc tc ng iu khin xung quanh khong thi gian c1v d1tng t

nh xung quanh a1v b1. Vi suy lun tng t, mi mt bin ra ta c t hp

ca 7 x 7 = 49 lut nh cc bng 3.6 v bng 3.7.


97/110


97

Bng 3.6. Lut iu khin cho HsKP

HsKPdw/dt

NB NM NS ZE PS PM PB

ERROR

NB B B B B B B B

NM M M B B B M M

NS S S MS M MS S S

ZE S S S MS S S S

PS S S MS M MS S S

PM M M B B B M M

PB B B B B B B B

Bng3.7. Lut iu khin cho HsKI

HsKIdw/dt

NB NM NS ZE PS PM PB

ERROR

NB B B B B B B B

NM M M B B B M M

NS S MS M M M MS S

ZE S S MS MS MS S SPS S MS M M M MS S

PM M M B B B M M

PB B B B B B B B


98/110


98

Hnh3.31: Cc lut hp thnh


99/110


99

Dng lut hp thnh Max-Prod, gii m theo phng php trng tm,

Khi cc h s HsKp, HsKI c tnh ton lc iu khin l:

49

1

49

1

/.)(

/.)(

)(

l BA

l BA

lp

p

dtdwte

dtdwtey

tHsK

ll

ll

(3.23)

49

1

49

1

/.)(

/.)(

)(

l BA

l BA

lI

I

dtdwte

dtdwtey

tHsK

ll

ll

(3.24)

Trong l

Il

p yy ,

l tm ca cc tp m tng ng.

3.3.6. M phng nh gi cht lng

3.3.6.1. Xy dng s m phng

M hnh m phng c xy dng trn phn mm Matlab Simulink.

Hnh3.32: Cu trc ca h iu khin m lai PI


100/110


100

2

Hs KI

1

Hs KP

sailech e

dw /dt1

u

HsKPKI

m

2

dw/dt

1

e

Hnh3.33: Cu trc ca khu m

Hnh3.34: cu trc ca b iu khin PI

Hnh3.35: Cu trc ca i tng

3.3.6.2. Kt qu m phng

Tc t n = 1500 vng/pht.

B iu khin PI y c tng hp theo phng php mun i xng.


101/110


101

Kt qu m phng v so snh vi phng php

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