Ứng dụng mạng nơ ron Điều khiển quá trình thay Đổi mực chất lỏng
DESCRIPTION
PID FuzzyTRANSCRIPT
-
i
B GIO DC V O TO
I HC NNG
PHM TH DIU HIN
NG DNG MNG N RON IU KHIN
QU TRNH THAY I MC CHT LNG
Chuyn ngnh : T ng ha
M s: 60.52.60
TM TT LUN VN THC S K THUT
Nng - Nm 2012
-
ii
Cng trnh c hon thnh ti
I HC NNG
Ngi hng dn khoa hc: TS. V NH TIN
Phn bin 1: TS. NGUYN B
Phn bin 2: PGS.TS. ON QUANG VINH
Lun vn c bo v ti Hi ng chm lun vn tt nghip Thc
s k thut hp ti i hc Nng vo ngy 5 thng 01 nm
2012.
* C th tm hiu lun vn ti:
- Trung tm Thng tin - Hc liu, i hc Nng
- Trung tm Hc liu, i hc Nng
-
1
M U
1. Tinh cp thit cua ti:
Hin nay, trong cng nghip ha lc du , cng nghip ha cht ,
cng nghip x l nc , sn xut giy , sn xut in nng ,Vn
iu khin mc , lu lng dong chay cn ap ng vi chinh xac
cao phuc vu cho qua trinh san xut at hiu qua tt hn.
Vi b iu khin m v b iu khin s dng mng Nron c
thm mt hng phat trin mi trong lnh vc nghin cu thit k
iu khin h thng, c rt nhiu ng dng trong lnh vc iu khin
trong cng nghip hi n nay. B iu khin m v b iu khin s
dng mng Nron v nguyn tc u l nhng b iu khin tnh phi
tuyn. Chng c th c thit k vi cht lng h thng cho trc
theo mt chnh xc tuy v lm vic theo nguyn l t duy ca
con ngi. Tnh nng ca mng Nron c quyt nh bi chng
loi Nron s dng v cu trc mng ghp ni cc Nron vi
nhau. N hon ton c lp vi i tng iu khin. Thm ch
nhng ngi thit k nu c kin thc thit k v hiu bit v i
tng th iu cng khng gip ch g cho vic la chn Nron v
xy dng cu trc mng. Ngc li, i vi ngi thit k b iu
khin m th nhng kin thc hiu bit v i tng li rt cn thit.
Ngay khi mi c thit k, mng Nron cha c tri thc. Tri
thc ca n c hnh thnh qua cc giai on theo cc mu hc .
Mu hc cng tt, cng a dng v cang nhiu trng hp th tri thc
ban u se cng gn vi thc t . Song nu iu l cha th tri
thc ca mng vn c th c b sung, v hon thin thm trong
-
2
qu trnh lm vic vi i tng. Vi b iu khin m th hon ton
ngc li. Khi c thit k xong, b iu khin m c ngay mt c
ch lm vic nht nh v c ch ny s khng thay i v c gi
c nh trong sut thi k lm vic. Ni cch khc mng Nron c
kh nng hc cn b iu khin m th khng.
han ch nhng nhc im ma cac b iu khin ring le
trn cha ap ng c va k tha nhng u im cua mang Nron
v Logic m , kt hp chung lai tao ra mt cng cu manh nhm giai
quyt cac bai toan phi tuyn phc tap.
Vn t ra nh th , hng nghin cu xy dng tai cua
tc gi y l nghin cu ng dung h M N ron iu khin
mc cht long cho h ba bn nc. Vi hng nghin cu o, tn
ti c chn:
ng dung mng nron iu khin qu trnh thay i mc
cht long
2. Muc tiu nghin cu
- ng dng mng nron nhm tao ra mt cng cu manh giai
quyt bai toan iu khin phi tuyn trong iu khin qua trinh.
- C th xy dng cu trc b iu khin vi kt hp gia m
v mng nron iu khin mc cht long cho h ba bn nc.
- S dng phn mm MATLAB lam cng cu m phong kt qua
nghin cu.
3. i tng v pham vi nghin cu
i tng nghin cu
Kt hp gia m v mng nron xy dng thut toan iu
khin cho i tng phi tuyn trong iu khin qua trinh.
Phm vi nghin cu
-
3
ng dung tri tu nhn tao vi s lai ghep h m va mang nron
iu khin mc cht long cho h ba bn nc , nghin cu s kt
hp gia h m va mang nron tao nn thut toan iu khin qua
trnh lm hng nghin cu chnh.
Nghin cu xy dng b iu kh in cho i tng la h ba bn
nc va cu th la iu khin gi c n inh mc cht long
trong ba bn.
4. Phng phap nghin cu
- Nghin cu tng quan mang nron va h m nron . Trong o,
nghin cu kt hp h m v mn g nron phuc vu cho nghin cu
chnh ca ti.
- Xc nh v gii quyt vn nghin cu chnh ca ti:
+ Xem xet va a ra cac dang m hinh h i tng . Tm hiu
i tng h ba bn nc vi mt cu hinh cu th v m hnh ton
hc ca h i tng c tnh cht phi tuyn nhiu u vo , nhiu u
ra.
+ Nghin cu thut toan iu khin dung h m nron , ng
dng to ra cng c mnh gii quyt cc bi ton phi tuyn trong
iu khin qua trinh.
+ Xy dng cu truc b iu khin vi h m nron cho i
tng h ba bn nc a chon cu th.
+ ng dng phn mm Matlab m phong kt qu thit k ,
chng minh tinh ung n thut toan iu khin.
- Nhn xet kt qua nghin cu.
5. Y nghia khoa hc v thc tin cua ti
- Y nghia khoa hoc: Nghin cu tri tu nhn tao tao ra cng
c iu khin mnh trong iu khin qu trnh.
-
4
- Y nghia thc tin : Kt hp h m v mng nron iu
khin mc cht long cho h ba bn nc.
6. B cuc ti
Ngoi phn m u, kt lun v ti liu tham kho. Lun vn
gm c cc chng nh sau:
Chng 1: Khi qut h m.
Chng 2: Khi qut mng nron.
Chng 3: M hinh ton h i tng.
Chng 4: Thit k b iu khin m v mng nron iu
khin mc cht long cho h ba bn nc.
Chng 5: M phong kt qua thit k.
CHNG 1. KHI QUT H M
1.1. GII THIU CHUNG V H M.
1.2. LOGIC M-TP M
1.3. CC HM THUC THNG GP
- Hm thuc kiu tam gic
- Hm thuc kiu hnh thang.
- Hm thuc kiu hnh chung c xc nh bi 3 tham s {a,
b, c}: 1
( ; , , )2
1
x a b cA b
x c
a
trong b thng l s dng
- Hm thuc Gaus (Hnh 1.3b) c xc nh bi 2 tham s
{ , c}: 2
2
( )
( ; , , )
x c
x a cA
e
-
5
1.4. BIN M VA BIN NGN NG.
1.4.1. Bin m
1.4.2. Bin ngn ng
1.5. SUY LUN M V LUT HP THNH.
1.5.1. Suy lun m.
1.5.2. Mnh hp thanh.
1.5.3. Lut hp thanh MAX-MIN, MAX-PROD
a) Lut hp thnh mt iu kin R: A B.
* Lut hp thnh MAX-MIN
* Lut hp thnh MAX-PROD.
* Thut ton xy dng R.
b) Lut hp thnh ca mnh nhiu iu kin.
1.5.4. Lut cua nhiu mnh hp thnh.
a) Lut chung ca hai mnh hp thnh.
b) Thut ton xy dng lut chung ca nhiu mnh hp
thnh.
1.6. GIAI M (R HA)
C hai phng php gii m chnh l: phng php cc i v
phng php im trng tm
1.6.1. Phng php cc ai
1.6.2. Phng php im trng tm
1.7. KT LUN CHNG 1
-
6
CHNG 2. KHI QUT MNG NRON
2.1. GII THIU
2.2. T BO NRON NHN TO
2.3. CC LOI MNG NRON NHN TO THNG GP
V PHNG PHAP HUN LUYN MANG
2.3.1. Mang nron truyn thng mt lp
2.3.2. Mang nron truyn thng nhiu lp
2.3.3. Mang nron hi quy mt lp
2.3.4. Mang nron hi quy nhiu lp
2.3.5. Cc phng phap hun luyn mang nron nhn tao.
a) Hoc co gim st
b) Hoc cng c
c) Hoc khng co gim st
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7
2.4. H THNG TICH HP H M VI MNG NRON.
B iu khin m-nron vi cac lut m duy nht (mng nron
Singleton) (Hnh 2.17) c dng nh sau:
Lut hoc th i la Ri c dng:
NU x1 l 1iA VA x2 l 2
iA VA... VA xn l inA THI y l wi
Trong o: xj l cc bin u vo (j=1,2,3,...,n), y la bin u ra ,
( )ij jA x l bin ngn ng m ca bin u vo x i vi ham lin thuc
( )i jjA
x ; Kt qua cua lut hoc th i (i=1,2,...,h).
- Lp 1: L lp gm c n tn hiu x u vo ,
1 2[ ... ]T
nx x x x .
- Lp 2: L lp m ha , gm co cac nut thc hin gia tri ham
lin thuc. Mi nut co ngo ra la:
1
( )n
ii jj jA
x
-
8
- Lp 3: Lp thc hin lut m . Mi nut co ngo ra la gia tri
vecto c s m:
1
1 1
( )
( )
( )
n
i jjji
nh
i jji j
A
A
x
x
x
- Lp 4: L lp gii m. Nt i din ngo ra ca mng l y:
1 1
1 1
( )
( ) ( )
( )
nhi
i jji j T
nh
i jji j
A
A
x
y x x
x
Trong o : ( )i jjA
x l gi tr hm lin thuc ca bin m x j ;
[ , ,..., ]h h hT T
l vecto trng s l in kt gia lp 3 v lp
ngo ra.
2.5. KT LUN CHNG 2
CHNG 3. M HINH TOAN H I TNG
3.1. GII THIU CHUNG
i iu khin trong ti c xc nh theo hnh 3.1, y l
h c 3 ngo vo, 3 ngo ra. Ta xc nh c 3 ngo vo u1(t), u2(t), u3(t)
iu khin lu lng ngo vo 3 bn qin1, qin2, qin3 v 3 tn hiu ngo ra
l mc nc ca 3 bn h1(t), h2(t), h3(t).
Hnh 3.1. M hnh h ba bn
nc
-
9
3.2. XC NH M HNH TON HC CHO H BA BN
NC
Gi: A1, A2, A3 ln lt l tit din ngang bn cha 1, 2 v 3.
h1(t), h2(t), h3(t) l chiu cao mc nc trong bn cha 1, 2
v 3.
Th V1 = A1h1(t), V2 = A2h2(t) v V3 = A3h3(t) ln lt l th
tch cht long ca bn 1, 2 v 3.
a1, a2, a3, a12, a13, a23 ln lt l din tch ca val A,B,C, AB, BC
v AC, vi iu kin cc van x ny l mt hng s cho trc khng
i.
k1, k2, k3 ln lt l h s t l vi cng sut ca my bm 1, 2
v 3.
qi1, qi2, qi3, qo1, qo2, qo3 ln lt l lu lng dng chy vo v ra
ca bn 1, 2 v 3.
qo12, qo23, qo13 ln lt l lu lng dng chy t bn nc 1
sang bn nc 2, t bn nc 2 sang bn nc 3 v t bn nc 1
sang bn nc 3.
CdA, CdB, CdC ln lt l h s x ca van A ra ngoi bn 1, van
B ra ngoi bn 2 v van C ra ngoi bn 3.
CdAB, CdBC, CdBC ln lt l h s x van lin kt gia bn 1 v
bn 2, bn 2 v bn 3, bn 1 v bn 3.
.))()(2))()(sgn(
)()(2))()(sgn()(2)((1
)(
))()(2))()(sgn(
)()(2))()(sgn()(2)((1
)(
))()(2))()(sgn(
)()(2))()(sgn()(2)((1
)(
131313
2323233333
3
3
121212
3232232222
2
2
313113
2121121111
1
1
ththgththCa
ththgththCatghCatukA
th
ththgththCa
ththgththCatghCatukA
th
ththgththCa
ththgththCatghCatukA
th
dAC
dBCdC
dAB
dBCdB
dAC
dABdA
-
10
3.3. CHN CC THNG S CHO M HNH
Chn din tch ngang bn cha 1, 2 v 3 l: A1 = A2 = A3 = 150
cm2.
Chn chiu cao thc ca bn 1, 2 v 3 l H1 = H2 = H3 = 80 cm.
Chn tit din ca van x bn 1, bn 2, bn 3, gia bn 1 v
bn 2, gia bn 2 v bn 3, gia bn 1 v bn 3 l: a1 = a2 = a3 = a12
= a13 = a23 = 2,5 cm2.
Chn h s t l vi cng sut ca my bm 1, bm 2 v bm 3
l: k1 = k2 = k3 = 160.
Chn h s x CdA = CdB = CdC = CdAB = CdBC = CdAC = 0,6.
3.4. XY DNG M HNH I TNG TRN MATLAB-
SIMULINK
3.5. KT LUN CHNG 3
Hnh 3.2. M hinh toan hoc h ba bn
nc
-
11
sTsT
ksR DI
p
11)(
CHNG 4 - THIT K B IU KHIN M NRON
IU KHIN MC CHT LONG CHO H BA
BN NC
4.1. XY DNG B IU KHIN PID
4.1.1. C s l thuyt b iu khin PID
B iu khin PID c m t bng m hnh vo-ra:
T m hnh vo ra trn, ta c c hm truyn t ca b iu
khin PID:
4.1.2. Xy dng b iu khin PID cho qu trnh cht lng
Tc gi m t bng mt khu qun tnh bc nht c hm s
truyn:
Ta ly: k = 6.52, T = 200 (s), =30(s)
S dng phng php Ziegler-Nichols th nht, s dng b
iu khin PI, ta c:
4.2. C S THIT K B IU KHIN M NRON VI
LUT IU KHIN THICH NGHI.
Xt m hnh ton hc l h i tng phi tuyn MIMO. Phng
trnh ng hc c dng:
])(
)(1
)([)(0
dt
tdeTde
Ttektu D
t
I
p
seTs
ksW
1)(
ss es
eTs
ksW 30
2001
52.6
1)(
0092.0100
92.010030
3
10
3
10
92.030*52.6
200*9.0
I
p
II
p
T
KKT
k
TK
-
12
11 1 11
1
1
( ) ( )
( ) ( )
.
.
.
jj
p
p p j ppj
pm
j
pm
j
x x u d
x x u d
y f g
y f g
(4.1)
Trong o:
fk v gkj (vi k=1 p) l cc hm phi tuyn.
1 2, ,...,T
p
pu u u u R l vecto tn hiu iu khin ngo vo h
i tng.
1 2, ,...,T
p
py y y y R l vecto tn hiu ngo ra ca h i
tng.
1 2, ,...,T
p
pd d d d R l vecto tn hiu nhiu t ngoi tc ng
vo.
Vecto trang thai ( 1 1) ( 1)1 1 ,1, ,..., ,..., ,...,
. . Tm mp n
p ppx y y y y y y R
Trong bai toan nay , yu cu thit k b iu khin co tin hiu
ngo ra y s bm theo tn hiu t 1 2, ,...,
Tp
r r r rpy y y y R .
T (4.1) c th biu din phng trnh trng thi h i tng
c rut gon nh sau:
[ ( ) ( ) ]0
A x B F x G x u d
Ty C x
x (4.2)
, ,A B C ln lt la ma trn cheo cua cac ma trn 0 0 0, ,k k kA B C , vi
0 01 02 0
1 2
1 2
2
1 2
1 2
( )1
[ , ,..., ]
[ , ,..., ]
[ , ,..., ]
( ) [ , ( ),..., ( )]
( ) [ ( ), ( ),..., ( )]
( ) [ ( ), ( ),..., ( )]
nxn
p
nxp
p
nxp
p
T p
p
T pxp
p
T p
k k k pk
f x
A diag A A A R
B diag B B B R
C diag C C C R
F x f x f x R
G x G x G x G x R
G x g x g x g x R
-
13
Ta co inh nghia:
Sai s bam: ;r re Y x e Y x
Trong o e v x l c lng ca e v x
( 1 1) ( 1)
1 1 ,1, ,..., ,..., ,...,. . T
m mp n
r r r rp rpr rpY y y y y y y R
( ) ( ) ( 2) ( )
1 2, ,...,T
m m m mp p
r r r rpy y y y R
( 1 1) ( 1)
1 1 1 11 12 1 1 2, ,..., ,..., , ,..., , ,..., ,..., , ,...,. . T T
m mp n
p p p n p p pne e e e e e e e e e e e e R
1 2 11 12 1, ,..., , ,...,T T
p
p pe e e e E E E R
Nu ham f k(x) v g kj(x) a bit chc chn va khng co nhiu
ngoi d th theo tiu chun Lyapunov lut iu khin la:
1* ( ) ( ) ( )( )
m T
r c m T
r c
F x y K eu G x F x y K e
G x
c lng vecto sai s trang thai e :
0 0 1 1
1
( ) ( ).
T
c
T
e A B K e K E
C
Trong o0 01 02 0, ,...,
nxp
pK diag K K K R l vecto khuch i
b quan sat, v 0 0 1 0 2 0, ,...,
kT
k k k kn
mK K K K R c chon sao cho
0
T
k k k ckA A B K thoa Hurwitz.
Sai s cua b quan sat c xac inh:
1 1 1
E E E
e e e
Lut iu khin c ra vi tin hiu iu khin u la tng tin
hiu xp xi theo lut iu khin (4.3) v tn hiu thnh phn kh
nhiu ngoai va sai s cua m hinh: f
u u v
-
14
Trong o : 1 2, ,...,f f fpfpu u u u R l tn hiu dung h m -
nron Singleton xp xi lut iu khin ly tng
1 2, ,...,p
pv v v Rv l thnh phn bu sai s ca m hnh v kh
nhiu ngoai.
Mng nron Singleton vi cu trc ca h m dung xp x
lut iu khin ly tng .
B xp xi lut iu khin ly tng m-nron Singleton
S dung lut suy din max -prod, m hoa singleton va giai m
theo phng phap trung binh trong tm.
Vi ngo vao cua mang ( e e e ) l c lng sai s ca e , do o hinh 4.3 di y la mang nron Singleton co vecto ngo vao la
1 2 [ , ,..., ]ne e e e
Nn ngo ra cua h m-nron dung xp xi lut iu khin luc
ny l yk=ufk (k=1 p):
-
15
1 1
1 1
( )
( )
( )
nhi
i jkkji j T
fk k knh
i jkji j
A
A
u
Trong o:
1 21 2 1 2( / ) , ,..., ( ), ( ),..., ( )
TTT T T
pf f f f fp pu u u u u
1 2, ,..., p
pR
i
k l im vch m ti ( ) 1
i
i kk
B
1 2, , ...,( )
Th h
k k k k R l vecto c s m , trong o
ik c
inh nghia
1
1 1
( )
( )
( )
n
i jkjji
k nh
i jkji j
A
A
Lut cp nht c chon:
1 1
1 1
( ) ( ( ) 0)
Pr( ( )) ( ) 0
k k
k
k k k k k k k
k
k k k k k k
E e neu m hay m va E e
E e neu m va E e
k c cp nht bi lut cp nht (4.15) v 0
k l thng s
thch nghi thit k.
Khi:
1
1 1 2
( )
2 Pr( ( )) ( )k k
T
k k k
k k k k k k k k k k
k
E em va m th E e E e
Trong o 1( ) ( )[ ( )]k k ke L s e
-
16
Thnh phn bu sai s ca m hnh v kh nhiu ngoi c x c
inh:
1 1
1 1
1 1
0
0
/
k k k
k k k k
k k k
k
k
k k
neu E va E
v neu E va E
E neu E
(4.17)
Lut iu khi n thch nghi trc tip c xac inh :
( / )k fk k
u u e v (4.18)
4.3. THIT K B IU KHIN CHO H BA BN NC
t
.)()()(
)()()(
)()()(
3333
2222
1111
thhthth
thhthth
thhthth (4.19)
.)()(2)sgn(
2)sgn(2
)()(2)sgn(
)(2)sgn(2
)()(2)sgn(
)()((2)sgn(2
3
3
3313113
232323
3
333
2
1
2
2
211212
2
323223
2
22
2
1
1
1
1
313113
1
212112
1
111
uA
kththghhCa
hhghhCaA
ghCah
uA
k
A
ththghhCa
A
hhghhCa
A
ghCah
uA
k
A
ththghhCa
A
ththghhCa
A
ghCah
dAC
dBC
dC
dC
dBCdB
dAC
dABdA
(4.20)
Phng trinh trang thai cua h i tng nh sau:
333
222
111
3
2
1
3
2
1
)()(
)()(
)()(
100
010
001
000
000
000
uhghf
uhghf
uhghf
h
h
h
h
h
h
-
17
3
2
1
3
2
1
000
000
000
h
h
h
y
y
y
Cc bc thit k b iu khin nh sau:
Bc 1: Chn h s khuch i hi tip v h s khuch i b
quan sat trang thai nh sau:
1500
0150
0015
cK v
5.16.518
1440
125.220
0K
Bc 2: Xy dng b quan sat trang thai theo (4.4) xc nh
sai s c lng (t).
Bc 3: Chn cc thng s ca thnh phn bu sai s ca m
hnh v kh nhiu ngoi theo (4.17) c chon:
11 2 1 2 1 2
10,005 ; 0,02 ; 1,215 ; ( )
12
bo loc c chon L ss
Bc 4: Xy dng cac lut m cho sai s c lng (t) v sau
o tinh vecto c s m theo (4.14).
-
18
2,1.22
1
1
2
301
1
1
1)()()()(
111
312
11
d
c
bAAAA
2,1.22
2
2
2
21
1
1
1)()()()(
222
322
21
d
c
bAAAA
2,1.22
3
3
21
1
1
1)()()()(
333
332
31
d
c
bAAAA
2,1.22
4
4
2
21
1
1
1)()()()(
444
342
41
d
c
bAAAA
Bc 5: Thc hin lut iu khin (4.18) v lut thh nghi
(4.15).
4.3. KT LUN CHNG 4
CHNG 5 - M PHONG KT QUA THIT K
5.1. XY DNG B IU KHIN DUNG PID
5.1.1. Xy dng b iu khin trn Matlab-Simulink
Hnh 5.1. B iu khin dung PID
5.1.2. Thc hin iu khin h ba bn nc trn Matlab -
Simulink
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Hnh 5.2. iu khin h ba bn nc dung PID
5.2. XY DNG B IU KHIN DNG M NRON
5.2.1. Xy dng b iu khin trn Matlab-Simulink
Hnh 5.8. Khi thc hin lut thich nghi
5.2.1. Thc hin iu khin h ba bn nc trn Matlat -
Simulink
Hnh 5.13. M hinh iu khin h ba bn nc
5.3. KT QUA M PHONG
5.3.1. Kt qua m phng vi b iu khin PID
a. Kt qua m phong bn 1 (Hnh 5.14)
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Hnh 5.14. Kt qua m phong bn 1 dung b iu khin PID
b. Kt qua m phong bn 2 (Hnh 5.15)
Hnh 5.15. Kt qua m phong bn 2 dung b iu khin PID
c. Kt qua m phong bn 3 (Hnh 5.16)
Hnh 5.16. Kt qua m phong bn 3 dung b iu khin PID
5.3.2. Kt qua m phong vi b iu khin dung h m
nron
a. Kt qua m phong bn nc 1 (Hnh 5.17)
Hnh 5.17. Kt qu m phong bn 1 dung b iu khin m nron
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b. Kt qua m phong bn nc 2 (Hnh 5.20)
Hnh 5.20 Kt qua m phong bn 2 dung b iu khin m nron
c. Kt qua m phong bn nc 3 (Hnh 5.23)
Hnh 5.23. Kt qua m phong bn 3 dung b iu khin M Nron
5.4. KT LUN CHNG 5
Tn hiu t l sng vung , thi im u thi h thng
dao ng, cha xac lp nhanh c , nhng sau o n inh tim cn
vi tin hiu t, c th:
- T c t nh sai lch e 1 vi mc nc t h 1d=50cm (Hnh
5.18), tc gi ly 4 khong thi im xem xt s thay i mc
nc t trc thi thy lng qua iu chinh % v sai lch % e1 ln
nht giam dn tim cn v 0 (Bng 5.1). n khoang thi im th 4
th mc nc h 1 ngo ra ca i tng n nh vi sai lch % ln
nht la e1%=0.5486% (Bng 5.1).
- Cng tng t c tnh sai lch e1 vi mc nc t h1d=25cm
(Hnh 5.19), tc gi ly 4 khong thi im xem xet s thay i
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mc nc t trc thi thy lng qua iu chinh % v sai lch % e1
ln nht giam dn tim cn v 0 (Bng 5.2). n khoang thi im
th 4 th mc nc h1 ngo ra ca i tng n inh vi sai lch %
ln nht la e1%=0.333% (Bng 5.2).
- T c tinh sai lch e 2 vi mc nc t h 2d=40cm (Hnh
5.21), tc gi ly 4 khong thi im xem xt s thay i mc
nc t trc thi thy lng qua iu chi nh % v sai lch % e2 ln
nht giam dn tim cn v 0 (Bng 5.3). n khoang thi im th 4
th mc nc h 2 ngo ra ca i tng n nh vi sai lch % ln
nht la e2%=0.1225% (Bng 5.3).
- Cng tng t c tnh sai lch e2 vi mc nc t h2d=20cm
(Hnh 5.22), tc gi ly 4 khong thi im xem xt s thay i
mc nc t trc thi thy lng qua iu chinh % v sai lch % e2
ln nht giam dn tim cn v 0 (Bng 5.4). n khong thi im
th 4 th mc nc h2 ngo ra ca i tng n nh vi sai lch %
ln nht la e2%=0.4479% (Bng 5.4).
- T c tinh sai lch e 3 vi mc nc t h 3d=30cm (Hnh
5.24), tc gi ly 4 khong thi im x em xet s thay i mc
nc t trc thi thy lng qua iu chinh % v sai lch % e3 ln
nht giam dn tim cn v 0 (Bng 5.5). n khoang thi im th 4
th mc nc h 2 ngo ra ca i tng n nh vi sai lc h % ln
nht la e3%=0.1455% (Bng 5.5).
- Cng tng t c tnh sai lch e3 vi mc nc t h3d=15cm
(Hnh 5.25), tc gi ly 4 khong thi im xem xt s thay i
mc nc t trc thi thy lng qua iu chinh % v sai lch % e3
ln nht giam dn tim cn v 0 (Bng 5.6). n khoang thi im
th 4 th mc nc h3 ngo ra ca i tng n nh vi sai lch %
ln nht la e3%=0.1585% (Bng 5.6).
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Nh vy vi s phn tich sai lch theo cac khoang thi gian nh
trn thi tin hiu ngo ra la mc nc h 1, h2, h3 bm theo tn hiu t
vi sai lch rt nho, thi gian xac lp nhanh, kt qua ngo ra bam theo
tn hiu t tt.
Kt qua iu khin dung b iu khin m nron (Hnh
5.17), (Hnh 5.20), (Hnh 5.23), cho ra kt qua bam theo tin hiu t
tt hn so vi vic dung b iu khin PID (Hnh 5.14), (Hnh 5.15),
(Hnh 5.16).
Vy vic s dung h m nron iu khin mc cht long cho h
ba bn nc tao ra b iu khin ap ng tt vi s thay i tin hiu
t cua m hinh i tng.
KT LUN VA HNG PHAT TRIN TAI
* Kt lun
- Nghin cu i tng phi tuyn nhiu u vao , nhiu u ra
trong iu khin qua trinh vi vic xy dng c m hinh toan hoc
i tng h ba bn nc.
- Vic kt hp h m va mang nron tao ra c b iu
khin a khc phuc c nhng nhc im vn co cua cac b iu
khin ring le.
- S dung c tri tu nhn tao phuc vu trong iu khin la vic
to ra h lai vi s kt hp ca iu khin m v mng nron . Gii
quyt c bai toan iu khin qua trinh vi i tng h ba bn
nc, l h phi tuyn phc tp c 3 ngo vo, 3 ngo ra (MIMO).
- Kim tra c tinh ung n cua thut toan iu khin qua
vic m phong kt qua nghin cu trn Matlab -Simulink, cho ra kt
qu iu khin tt.
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* Hng phat trin cua tai
- Lun vn vi tai c tac gia nghin cu mi dng li
mc tim hiu xy dng b iu khin vi mang nron Singtelon
c cu trc h m dung xp x lut iu khin l tng , v a
kim tra thut t on iu khin trn Matlab -Simulink. Do o d inh
pht trin tip tc ti:
+ Tin n xy m hinh thi nghim thc cho h ba bn nc ,
ng thi cung tao ra cac m hinh thi nghim khac iu khin cho
cc h phi tuyn MIMO nh iu khin nhit , p sut, lu lng,
... hoc cac h tay may.
+ Tm kim cc thut ton iu khin khc vi hy vng to ra
cng cu iu khin manh me hn nh cac h m nron CANFIS
(Coactive Adaptive Neural Fuzzy Inference Systems ), cc b iu
khin dung mang iu khin hoc thich nghi m FALCON (Fuzzy
Adaptive Learning Control Network ) c kh nng hc thng s v
hc cu trc, ...