nghiên cứu thiết kế bộ Điều khiển pid mờ
TRANSCRIPT
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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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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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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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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/
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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
*
*
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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
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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)
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-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)
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- 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)
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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
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+ 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 iu khin P : Chn kp=ak
b
10
7
+ B iu khin PI : Chn kp=ak
b
10
7 v TI=10
23a
+ B iu khin PID : Chn kp=ak
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 iu khin PI : Chn kp=ak
b
20
7 v TI=5
6b
+ B iu khin PID : Chn kp=ak
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 iu khin P : Chn kp=ak
b
10
7
+ B iu khin PI : Chn kp=ak
b
5
6 v TI= b
+ B iu khin PID : Chn kp=ak
b
20
19 , TI=20
27b v TD=100
47a
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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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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)
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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
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(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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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.
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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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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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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
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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
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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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-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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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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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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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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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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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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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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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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-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.
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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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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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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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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
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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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- 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
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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
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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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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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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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Hnh3.26:Xc nh tp m cho bin vo ERROR
Hnh3.27:Xc nh tp m cho bin vo dw/dt
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Hnh3.28:Xc nh tp m cho bin ra HsKP
Hnh3.29:Xc nh tp m cho bin ra HsKI
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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.
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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
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Hnh3.31: Cc lut hp thnh
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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
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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.
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Kt qu m phng v so snh vi phng php