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Chng 2 iu khin thch nghi
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Chng 2
IU KHIN THCH NGHI
2.1 Khi nim 2.1.1 nh ngha
Thch nghi l qu trnh thay i thng s v cu trc hay tc ng iu khin trn
c s lng thng tin c c trong qu trnh lm vic vi mc ch t c mt trng thi
nht nh, thng l ti u khi thiu lng thng tin ban u cng nh khi iu kin
lm vic thay i hay : iu khin thch nghi l tng hp cc k thut nhm t ng
chnh nh cc b iu chnh trong mch iu khin nhm thc hin hay duy tr mt mc
nht nh cht lng ca h khi thng s ca qu trnh c iu khin khng bit trc
hay thay i theo thi gian.
H thng c m t trong hnh di y gm 2 vng: - Vng hi tip thng thng -
Vng hi tip iu khin thch nghi
Kt lun
1. iu khin thch nghi lin quan n: - S khc nhau trong cc qu trnh
ng hc - S khc nhau trong cc nhiu 2. Cc h thng thch nghi l phi
tuyn
2.1.2 Nhn dng h thng
Lm th no c c m hnh?
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Chng 2 iu khin thch nghi
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- Vt l (hp trng) - Kinh nghim (hp en) - Kt hp ( hp xm)
K hoch ho thc nghim Chn la cu trc m hnh
- Cc hm chuyn i - p ng xung - Cc m hnh trng thi
Tham s thch nghi - Thng k - Cc vn nghch o
S hp l
2.1.3 c lng tham s thch nghi thi gian thc
1. Gii thiu 2. Bnh phng cc tiu v hi qui 3. H thng ng 4. Cc iu
kin thc nghim 5. Cc v d 6. Cc kt lun
2.1.4 Phn loi
C th phn loi cc h thch nghi theo cc tiu chun sau : 1. H thch
nghi m hnh tham chiu ( MRAS ) 2. B t chnh nh ( STR ) 3. Lch trnh li
4. H t hc 5. H t t chc
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Chng 2 iu khin thch nghi
Trang 94
2.1.5 ng dng
T chnh nh Lch trnh li Thch nghi lin tc
Hnh 2.1 S cc ng dng
Qu trnh ng hc
Bin i Hng s
S dng b iu khin vi cc thng s bin i
S dng b bin i vi cc thng s hng
S bin thin khng bit trc
S bin thin bit trc
S dng b iu khin thch nghi
S dng lch trnh li
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Chng 2 iu khin thch nghi
Trang 95
2.2 H thch nghi m hnh tham chiu MRAS (Model Reference Adaptive
Systems)
2.2.1 S chc nng H thng thch nghi s dng m hnh chun l mt trong
nhng phng php chnh ca iu khin thch nghi. Nguyn l c bn c trnh by hnh
2.2
Hnh 2.2 S khi ca mt h thng thch nghi m hnh tham chiu
M hnh chun s cho p ng ng ra mong mun i vi tn hiu t (yu cu). H
thng c mt vng hi tip thng thng bao gm i tng v b iu khin. Sai s e l
sai lch gia ng ra ca h thng v ca m hnh chun e = y - ym. B iu khin c
thng s thay i da vo sai s ny. H thng c hai vng hi tip:hi tip trong
l vng hi tip thng thng v vng hi tip bn ngoi hiu chnh tham s cho vng
hi tip bn trong. Vng hi tip bn trong c gi s l nhanh hn vng hi tip
bn ngoi. Hnh 2.2 l m hnh MRAS u tin c ngh bi Whitaker vo nm 1958 vi
hai tng mi c a ra: Trc ht s thc hin ca h thng c xc nh bi mt m hnh,
th hai l sai s ca b iu khin c chnh bi sai s gia m hnh chun v h
thng. M hnh chun s dng
u
y uc
M hnh
C cu hiu chnh
B iu khin i tng
Tham s iu khin
ym
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Chng 2 iu khin thch nghi
Trang 96
trong h thch nghi bt ngun t h lin tc sau c m rng sang h ri rc c
nhiu ngu nhin. Chng ny tp trung vo tng c bn. vn c trnh by mt cch r
rng, ta ch tp trung vo cu hnh trong hnh 2.2 c gi l h MRAS song song
. y l mt trong nhiu cch c th xy dng m hnh chun. Chng ny cp chnh n h
lin tc theo phng php trc tip c ngha l tham s c cp nht mt cch trc
tip.
2.2.2 Lut MIT (Massachusetts Institude Technology) ( MIT =
Massachusetts Institute Technology : Vin cng ngh Massachusetts)
Hnh 2.3 M hnh sai s H thng thch nghi m hnh tham chiu u tin c a
ra gii quyt vn : cc c im ca mt m hnh tham chiu yu cu ng ra l qu
trnh l tng cn c p ng i vi tn hiu iu khin nh th no. th minh ha trong
hnh 2.2. Trong trng hp ny, m hnh tham chiu mang tnh song song hn l
ni tip, ging nh cho SOAS (Self Oscillating Adaptive Systems). B iu
khin c th c xem nh bao gm hai vng: mt vng pha trong gi l vng hi tip
thng thng c qu trnh v b iu khin. Cc thng s ca b iu khin c chnh nh
bi vng ngoi sao cho sai s e gia ng ra y v ng ra m hnh ym l nh nht.
V vy vng ngoi cn c gi l vng chnh nh. Vn l xc nh c cu chnh nh cho h
thng n nh, ngha l sai s bng zero. iu ny khng th thc hin c. C cu
chnh nh vi thng s sau c gi l lut MIT, c s dng cho h MRAS u tin:
=
ee
dtd
s
pipipipi pipipipi
Khu tch phn
u
yuC e
e
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Chng 2 iu khin thch nghi
Trang 97
Trong phng trnh ny e l sai s ca m hnh e = y ym. Cc thnh phn ca
vector e/ l o hm nhy ca sai s i vi cc thng s chnh nh .Thng s xc nh
tc thch nghi. Lut MIT c th c gii thch nh sau. Gi s rng cc thng s
thay i chm hn nhiu so vi cc bin khc ca h thng. bnh phng sai s l b
nht, cn thay i cc thng s theo hng gradient m ca bnh phng sai s e2.
Gi s mun thay i thng s ca b iu khin sao cho sai s gia ng ra ca i
tng v ca m hnh chun tin ti zero. t e l sai s v l thng s hiu chnh.
Ch tiu cht lng :
J( ) = 21
e2 (2.1)
lm cho J() MIN th cn phi thay i cc thng s theo hng m ca gradient
J, c ngha l :
=
=
e
eJ
t (2.2)
Gi s rng cc thng s cn thay i thay i chm hn nhiu so vi cc
bin khc ca h thng. V vy o hm
e c tnh vi gi thit l
hng s. Biu thc o hm
e gi l hm nhy ca h thng. Lut iu
chnh theo phng trnh (2.2) vi
e l nhy th c lin h ging nh
lut MIT. Cch chn hm tn tht theo phng trnh (2.1) c th l tu . Nu
chn J( ) = e (2.3) Khi lut hiu chnh s l :
)(esignedtd
= (2.4)
Hoc
)(esignesigndtd
=
y gi l gii thut du - du. H ri rc s dng gii thut ny c ng dng
trong vin thng ni i hi tnh ton nhanh v thc hin n gin.
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Chng 2 iu khin thch nghi
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Phng trnh (2.2) cn c p dng trong trng hp c nhiu thng s hiu chnh,
khi tr thnh mt vector v
e
l gradient ca sai s i
vi cc thng s tng ng. ng dng ca lut MIT c biu din bng hai v d sau
: V d 2.1 - Hiu chnh li nui tin Xt vn hiu chnh li nui tin vi m hnh
v i tng u c hm truyn l G(S). Sai s l:
e = y ym = G(p) uc G(p) uc vi uc l tn hiu t, ym l ng ra m hnh, y
l ng ra i tng, l thng s hiu chnh, v p = d/dt l ton t vi phn. nhy
khi y bng :
e
= G(p)uc = ym /
Lut MIT c cho :
dtd
= - yme/
Nu du ca c bit, khi y a ra = / S thay i ca tham s t l vi tch sai
s e v ng ra ca m hnh ym. V d trn khng dng vic xp x : Khi lut MIT c
p dng vo nhng vn phc tp hn th cn phi c xp x tnh c nhy.
V d 2.2 MRAS cho h bc nht Xt h thng c m t bi phng trnh:
buaydtdy
+= (2.5)
vi u l bin iu khin, y l ng ra c o lng. Gi s mong mun c c h vng
kn c m t bi:
dtdym
= - amym + bmuc
M hnh km theo hon ho c th t c vi b iu khin : u(t) = 0t uc(t) 0s
y(t) (2.6) vi tham s t0 = bm / b ; s0 = (am a)/b
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Chng 2 iu khin thch nghi
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Ch hi tip s l dng nu am < a, ngha l m hnh mong mun th chm hn
qu trnh. p dng lut MIT , s dng sai s e = y ym , vi y l ng ra h kn.
Theo phng trnh (2.5) v (2.6) th:
y = 0
0
bsapbt
++uc
vi p l ton t vi phn. nhy c th tnh c bng cch ly o hm ring phn
theo tham s ca b iu khin s0 v t0 :
0t
e
=
0bsapb++
uc
0s
e
= - 20
02
)( bsaptb++
uc = -0bsap
b++
y
Cc cng thc ny khng th dng v thng s i tng a v b cha bit. V vy cn
phi lm xp x c c lut hiu chnh tham s thc t. thc hin iu ny, u tin cn
quan st vi gi tr ti u ca tham s b iu khin, ta c :
p + a + bs0 = p + am Hn na cn ch l b c th c bao gm trong h s tc
thch nghi . Bi v n xut hin trong tch b, iu ny i hi du ca b phi c
bit. Sau khi xp x, lut cp nht cc tham s iu khin c c l:
(2.7)
V d trn ch cch s dng lut MIT to c lut hiu chnh thng s. Bi tp v
nh (dng lm bi tp trong phn Cu hi n tp v bi tp cui chng): M phng bng
Matlab h MRAS trong v d 2.2 (V d 4.2 TLTK[1]) vi a = 1, b = 0.5, am
= 2 v bm = 2. Tn hiu vo l sng vung vi bin bng 1 v = 2.
eyapdt
ds
euapdt
dt
m
c
m
+=
+=
1
1
0
0
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Chng 2 iu khin thch nghi
Trang 100
Vi tnh cht sau cn ch : 1. Khng cn thit i hi mt m hnh km theo hon
ho. Cc th tc c th c p dng cho h phi tuyn. Phng php ny cng c th c
dng iu khin cho h bit trc mt phn.
2. Cu trc nh hnh 2.3 c mt php nhn gia e v
e.
Ly tch phn phng trnh (2.7) s cho ra cc tham s v c truyn n b iu
khin s dng php nhn th hai. 3. S xp x l cn thit c c lut iu khin hiu
chnh tham s thc t.
Lut MIT c th thc hin tt nu li thch nghi l nh. ln tu thuc vo bin
ca tn hiu chun v li ca i tng. V vy khng th c mt gii hn c nh m bo an
ton do lut MIT c th cho mt h vng kn khng an ton. Lut hiu chnh b
sung c th c dng bng l thuyt n nh. Nhng lut ny tng t lut MIT nhng cc
hm nhy th ng nhin l khc. ny c trnh by nhiu hn trong mc 2.2.4
2.2.3 Ni dung, phng php thit k MRAS C ba phng php c bn phn tch v
thit k h MRAS :
Phng php tip cn Gradient Hm Lyapunov L thuyt b ng Phng php
gradient c dng bi Whitaker u tin cho h MRAS. Phng php ny da vo gi s
tham s ca b hiu chnh thay i chm hn cc bin khc ca h thng. Gi s ny
tha nhn c s n nh gi cn thit cho vic tnh ton nhy v cho c cu hiu chnh
thch nghi. Phng php tip cn gradient khng cho kt qu cn thit cho h
thng kn n nh. B quan st c a ra p dng l thuyt n nh Lyapunov v l
thuyt b ng c dng b sung cho c cu thch nghi. i vi h thng c tham s iu
chnh c nh trong hnh 2.2, phng php thch nghi s dng m hnh chun cho mt
cch hiu chnh tham s tng qut c c hm truyn h thng vng kn gn vi m hnh.
y gi l vn m hnh km theo. Mt cu hi t ra l chng ta lm cho sai
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Chng 2 iu khin thch nghi
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lch nh nh th no, iu ny ph thuc bi m hnh, h thng v tn hiu t. Nu c
th lm cho sai s bng 0 i vi mi tn hiu yu cu th gi l m hnh km theo
hon ho.
M hnh km theo Vn m hnh km theo c th c gii quyt bng thit k phn s
cc (miu t ngn gn v thit k phn cc c cho trong ph lc A (TLTK[1])). M
hnh km theo l cch n gin thit lp hay gii mt vn iu khin tu ng. M hnh
s dng c th l tuyn tnh hay phi tuyn. Cc tham s trong h thng c hiu
chnh c c y cng gn vi ym cng tt i vi mt tp cc tn hiu vo. Phng php
thch nghi l mt cng c thit k h MRAS, vn ny c trnh by trong mc 2.2.4.
Mc d m hnh km theo hon ho ch c th t c trong iu kin l tng nhng phn
tch trng hp ny s cho hiu bit su sc vo vn thit k. Xt h 1 u vo,1 u ra
c th l lin tc hay ri rc c phng trnh:
y(t) = )(tuAB
(2.8)
vi u l tn hiu iu khin, y l ng ra. K hiu A, B l nhng a thc theo
bin S hay Z. Gi s bc ca A bc ca B ngha l h thng l hp thc (i vi h
lin tc) v nhn qu i vi h ri rc. Gi s h s bc cao nht ca A l 1.Tm b iu
khin sao cho quan h gia tn hiu t uc v tn hiu ra mong mun ym c cho
bi :
)(tuAB
y cm
m
m = (2.9)
vi Am, Bm cng l nhng a thc theo bin S hoc Z. Lut iu khin tng qut
c cho bi :
(2.10) vi R, S, T l cc a thc. Lut iu khin ny c xem nh va c thnh
phn hi tip m vi hm truyn S/R v thnh phn nui tin vi hm truyn T/R.
Xem hnh 2.4
SyTuRu c =
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Chng 2 iu khin thch nghi
Trang 102
Hnh 2.4 H vng kn vi b iu khin tuyn tnh tng qut
Kh u 2 phng trnh (2.8) v (2.10) c phng trnh sau cho h thng vng
kn : cBTuyBSAR =+ )( (2.11) t c p ng vng kn mong mun, th AR + BS
phi chia ht cho Am, cc zero ca i tng, khi cho B = 0, s l zero ca h
kn nu khng b kh bi cc vng kn. Bi v cc im zero khng n nh khng th b
kh nn c th phn tch thnh B = B+B-, trong B+ cha nhng thnh phn c th
kh i, B- l thnh phn cn li. Theo phng trnh (2.11) AR + BS l a thc c
trng ca h thng c phn tch thnh ba thnh phn : kh zero ca i tng:B+ ;
cc mong mun ca m hnh c cho bi Am; cc cc ca b quan st A0. V th : AR
+ BS = B+A0Am (2.12) gi l phng trnh Diophantine ( hay l phng trnh
nhn dng Benzout). V B+ c th kh nn :
(2.13) Chia phng trnh (2.12) cho B+ s c: A .R1 + B -.S = A0Am
(2.14) V yu cu l phi ging p ng mong mun nn t s (2.11) phi chia ht
cho Bm, nu khng th s khng c li gii cho bi ton thit k. V vy : Bm = B
-.Bm (2.15) T = A0Bm iu kin m bo tn ti li gii l :
bc( A0) 2 bc(A) - bc( Am) - bc(B+) - 1
Cu y
B iu khin Qu trnh u
SyTuRu C = AB
1RBR+
=
-
Chng 2 iu khin thch nghi
Trang 103
bc( Am) - bc (Bm) bc( A) - bc(B) Nhng iu kin ny c cho trong ph
lc A (TLTK[1]). Gi s tt c cc zero u b kh, khi c th vit (2.14) li nh
sau :
A0Am = AR1 + b0S Nhn 2 v cho y v dng thm phng trnh (2.8) ta c
:
A0.Am.y = BR1u + b0Sy = b0(Ru + Sy) (2.16) Cc thng s v tri bit,
v phi cha bit. a thc T c c trc tip t phng trnh (2.15). Cc tham s m
hnh ca phng trnh (2.16) by gi c th c dng c lng cc tham s cha bit ca
b iu khin (chng 3 TLTK[1]). iu ny dn n h MRAS trc tip. Li gii tng
qut c trnh by trong chng 4 TLTK[1].
H tuyn tnh tng qut H SISO c m t bi phng trnh sau:
Ay = Bu Vi c tnh h thng mong mun t c l:
Amym = Bmuc B iu khin: Ru = Tuc - Sy (*) H vng kn c m t:
CuBSARBTy+
=
Thay y vo (*) ta tnh c:
CuBSARAT
u+
=
Sai s l: e = y - ym By gi cn phi xc nh cc o hm ring ca sai s i
vi tng tham s hiu chnh tm lut chnh nh thng s cc hm nhy. t ri , si ,
ti l cc h s ca a thc R, S, T. Cc hm nhy c cho bi:
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Chng 2 iu khin thch nghi
Trang 104
m
CmC A
uBu
BSARBT
e +
=
=+
=
C
ik
i
uBSAR
BTApr
e2)( uBSAR
Bp ik
+
i = 1,. . , k
li ,,0 =
C
im
i
uBSAR
Bpt
e
+=
i = 0,,m
Trong k = bc(R), l = bc(S), m = bc(T). V phi cc phng trnh trn cn
cha A, B l cc thng s cha bit nn khng tnh c cc hm nhy. Mt cch xp x c
c lut cp nht c thc t l:
AR + BS A0AmB+
Suy ra cc hm nhy:
uAApB
r
e
m
ik
i 0
Tng t cho si v ti Tuy nhin v phi vn cn B- l cha bit. Nu tt c cc
zero u c kh, khi ta c B- = b0. Nu du ca b0 bit c th c th thc hin c
lut cp nht thng s. Thnh phn b0 c th c bao gm trong c . Nn c th suy
ra lut cp nht hiu chnh cc thng s nh sau:
uAA
pe
dtdr
m
iki
0
= i = 1,, k = bc(R )
yAA
pe
dtds
m
ili
0
= li ,...,0= = bc(S)
Cm
imi u
AAp
edtdt
0
= mi ,...,0= = bc(T)
yBSAR
Bpu
BSARBTBp
s
eil
C
il
i +=
+=
2)(
-
Chng 2 iu khin thch nghi
Trang 105
Nhn xt:
- Cn phi xy dng 3 trng thi ca b lc mAA0
1cho lut hiu chnh trn.
- S thay i cc tham s ny t l vi tch sai s e v tn hiu b lc
mAA0
1
- c c lut iu chnh cc tham s trn cn phi gi s cc zero phi n nh v
du ca b0 phi c bit. - C th trnh c gi s ny bng cch s dng cc thut ton
phc tp hn nh c lng trng thi
Tiu chun cc tiu ho - Lut MIT c th c s dng cho cc hm tn tht khc.
- Lut hiu chnh cc thams s c th t c bng cch tnh gradient
hm tn tht i vi cc tham s v s thay i cc tham s phi ngc du vi
gradient.
- Phng php ny cn bit cc tham s ca m hnh i tng tnh ton nhy. Tuy
nhin iu ny l khng c thc v do c th s dng phng php xp x hay bng cc b
c lng thng s.
Sai s v s hi t tham s H thng thch nghi s dng m hnh chun da vo
tng l lm cho sai s e = y ym tin ti zero. iu ny khng c ngha l cc
tham s iu khin tin ti gi tr ng ca n (v d nh trng hp tn hiu =
0).
V d 2.3 Hi t sai s Gi s h thng c s nh hnh 2.5: Ng ra: y = u Lut
iu khin: u = uc M hnh: ym = 0uc Sai s: e = y ym = uc - 0uc = ( -
0)uc Lut hiu chnh tham s theo phng php gradient:
-
Chng 2 iu khin thch nghi
Trang 106
)( 02
=
= cue
edtd
Li gii cho phng trnh vi phn trn l:
tIet += ])0([)( 00 (*) Trong : duI
t
ct )(02
=
(0) l gi tr ban u ca . V v vy sai s e tr thnh:
e(t) = uc(t) tIe ])0([ 0 Do It >0 nn khi t th e(t) 0 ngay c
khi tn hiu iu khin uc(t) 0.
Hnh 2.5 M hnh hi t sai s Gi tr gii hn ca ph thuc vo tnh cht ca
uc() (hi t hoc phn k) ( do (t) tnh theo biu thc (*) ). V d trn cho
bit c sai s e 0 tuy nhin tham s khng tin n gi tr ng ca n. y l tnh
cht ca h thng thch nghi s dng m hnh chun. iu kin chnh xc hi t tham
s l tn hiu kch thch phi lun tn ti.
0G(s)
-
s
pipipipi
pipipipi
G(s)
M hnh
y i tng
u
e
uc
+
-
ym
-
Chng 2 iu khin thch nghi
Trang 107
n nh ca vng iu khin thch nghi
v d trn bin thin tham s t l vi bnh phng tn hiu iu khin uc. iu ny
hp l trong mt s trng hp l khi tn hiu iu khin uc cng ln th cng d pht
hin gi tr b sai ca . Tuy nhin thay i ca tham s iu chnh ph thuc vo
bin ca tn hiu iu khin c th dn n khng n nh. V d sau y cho lut iu
khin khng ph thuc vo uc: V d 2.4 Gi s h thng c m hnh hnh 2.6:
Hnh 2.6 H thng thch nghi m hnh tham chiu cho vic chnh nh li nui
tin
Vn l iu chnh 0. Gi s hm truyn c cho bi:
212
1)(asas
sG++
=
Sai s e = G(p)( - 0 ) uc
M hnh Gm
0 G
G pipipipi
pipipipi
-
s
cu
my
e
y +
-
C cu hiu chnh
-
Chng 2 iu khin thch nghi
Trang 108
Trong p biu th cho php ly o hm. V vy:
e
= G(p)uc = 0my
iu chnh tham s theo lut MIT:
m
m yey
ee
edtd
==
= 0 vi 0 =
H thng iu khin thch nghi v vy biu din c bi cc phng trnh vi phn
sau:
cm
mm uyadt
dya
dtyd 0
212
2
=++ (I)
cuyadtdy
adt
yd =++ 2122
(II)
mmm yyyyedtd )( == (III)
Phng trnh (I) c th gii c nu cho sn hm uc , xem nh bin ym bit
trc
o hm (II) ta c:
dtdu
tudtd
dtdy
adt
yda
dtyd c
c )(222
13
3
+=++
Thay (III) vo ta c:
dtdu
tutytytuty
dtdu
tuyyydtdy
adt
yda
dtyd
c
cmcm
c
cmm
)()()()()(
)()(
2
2
2
1
3
++=
+=++
Suy ra:
)()()()()()( 2222
13
3
tytudt
duttytytu
dtdy
adt
yda
dtyd
mc
c
mc +=+++
y l phng trnh vi phn tuyn tnh bin thin theo thi gian. hiu c h
thng, ta thc hin cch th nh sau: - u tin gi s cu l hng s
0cu
-
Chng 2 iu khin thch nghi
Trang 109
- Ng ra m hnh khi s c gi tr cn bng l 0my . Gi s c cu hiu chnh
thch nghi c ni vo khi t n im cn bng (trng thi cn bng). Khi phng
trnh (II) trn s c cc h s hng v c li gii trng thi cn bng l: 2
000 /)( auyty cm ==
n nh nu 21aa > 20
2
00 )( cmc ua
yu =
Lut hiu chnh b sung Nhng hiu bit c c t vic tnh ton trong v d 2.3
ch ra rng cn phi b sung cho lut MIT. Lut MIT l phng php gradient c
bn. gim c c bng lut MIT c quyt nh bi tham s , s ny l do ngi dng
chn. C th t c phng php gradient b sung m t l hiu chnh khng ph thuc
vo bin ca tn hiu (t) yu cu. Mt kh nng l lm chun ho v thay th lut
MIT bi:
+
=
ee
ee
dtd
T
Tham s > 0 c a vo trnh trng hp chia cho 0. C th nhn thy rng t
l hiu chnh tham s ph thuc vo bin ca tn hiu yu cu mt lng nh bi v do
nhiu o lng.
2.2.4 Thit k MRAS dng l thuyt n nh ca Lyapunov Vi lut hiu chnh
tham s c c t phng php Gradient c trnh by trong mc 2.2.3 ly gn ng l
c c lut hiu chnh tham s da vo kinh nghim c v hp l ri chng ta th ch
ra rng sai s m hnh s tin n 0. Mt kh nng khc c c vng ngoi ca h thng
thch nghi s dng m hnh chun l tm ra lut hiu chnh m m bo sai s tin v
0. Nhng nghin cu cho lut hiu chnh nh vy c thc hin trong mt khong
thi gian di. tng c bn thit k lut hiu chnh da vo l thuyt n nh c trnh
by trong mc ny v c th hin theo lch s pht trin.
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Chng 2 iu khin thch nghi
Trang 110
tp trung vo vn chnh trnh nhng chi tit khng cn thit, t hiu chnh
li nui tin ca h thng c bit trc c dng trong mc ny. H thng dng y ging
nh hnh 2.6 nhng c cu thch nghi th khc. Vn l tm lut hi tip bo m sai
s e = y ym trong hnh 2.6 tin n 0, cn bit rng vn iu khin h thng vi c
tnh ng hc bit trc v h s li cha bit th khng qu kh. Vn ring bit c chn
trnh by tng hn l trnh by mt vn thc t. Mt khi tng c bn c pht trin, s
m rng n nhng cu hnh tng qut th tng i d hiu hn, chi tit c trnh by
trong TLTK[1].
Phng php th hai ca Lyapunov Minh ha bng th phng php Lyapunov Hnh
2.7 (a), (b) v (c) biu din cc trng thi cn bng v nhng ng cong tiu
biu tng ng i vi h thng n nh, n nh tim cn v khng n nh. Trong hnh 2.7
(a), (b) hoc (c), vng S() gii hn cho trng thi ban u x0, v vng S()
tng ng vi gii hn cho qi o xut pht ti x0. Ch rng nhng nh ngha c cp
trc y khng ch ra chnh xc vng ca iu kin cho php ban u. V vy cc nh
ngha p dng cho vng ln cn ca trng thi cn bng (l trng thi ti mi o hm
u trit tiu), tr khi S() tng ng vi trng thi ban u ca i tng. Ch l
trong hnh 2.7 (c), ng cong ri vng S() v dn n trng thi cn bng khng n
nh. Tuy nhin, chng ta khng th ni rng ng cong s i n v tn bi v n c th
n gn mt vng trn gii hn pha ngoi vng S(). (Nu mt h thng tuyn tnh bt
bin theo thi gian l khng n nh, cc ng cong bt u gn vi trng thi cn
bng khng n nh i n v cc. Nhng trong trng hp ca h thng phi tuyn, iu
ny tht s khng cn thit). S hiu bit v cc nh ngha ni trn l yu cu ti
thiu hiu vic phn tch n nh ca cc h thng tuyn tnh v phi tuyn c mt
trong phn ny. Ch rng nhng nh ngha ny khng ch hn ch cc khi nim v s n
nh ca mt trng thi cn bng. Thc ra, nhng cch nh ngha khc cng c s
dng.Chng hn, trong cc l thuyt iu khin thng thng hoc kinh in, ch c
cc h thng n nh tim cn mi c gi l h thng n nh, cn cc h thng khc n nh
theo Lyapunov, nhng khng n nh tim cn, c gi l khng n nh.
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Chng 2 iu khin thch nghi
Trang 111
(a) (b) (c) Hnh 2.7 (a) Trng thi cn bng n nh
(b)Trng thi cn bng tim cn (c)Trng thi cn bng khng n nh V d 2.5
Xt h thng c m t bi phng trnh trng thi sau:
1x = x2 - x1( 21x + 22x )
2x = - x1 - x2( 21x + 22x ) Trng thi cn bng (o hm = 0) ti gc ta
(x1 = 0, x2 = 0). Nu chng ta nh ngha mt hm v hng V(x) nh sau:
V(x) = 21x + 22x l hm xc nh dng, sao cho o hm theo thi gian hm
V(x) theo mt ng cong bt k
V (x) = 2 1x 1x + 2 2x 2x = -2( 21x + 22x )2
l hm xc nh m. iu ny cho thy rng V(x) tng lin tc theo ng cong bt
k; v vy V(x) l hm Lyapunov. Hm V(x) tr thnh v hn vi lch v hn t trng
thi cn bng, trng thi cn bng gc ca h thng l n nh tim cn trong vng
rng. Ch rng nu chng ta V(x) nhn gi tr hng s 0, C1, C2,. . . (0 <
C1 < C2
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Chng 2 iu khin thch nghi
Trang 112
Khi vng trn V(x) = Ck nm hon ton trong vng trn V(x) = Ck+1, mt
ng cong i din i qua vng bin gii ca cc ng vin V t ngoi vo trong. T
y, biu din hnh hc ca hm Lyapunov c th c pht biu nh sau: V(x) l thc
o khong cch ca bin trng thi x t gc to ca trng thi trung gian. Nu
khong cch gia gc v bin trng thi tc thi x(t) tng lin tc khi t tng
{V[x(t)] < 0 } th x(t) 0. Qu o (1) trn hnh 2.8 l chuyn ng n nh
tim cn v gc ta , song khng tho tiu chun n nh th 2 ca Lyapunov: hm
)(xV khng phi l hm xc nh m vi mi bin trng thi x. Tiu chun n nh th 2
ca Lyapunov l iu kin , khng phi l iu kin cn nh gi tnh n nh ca nghim
phng trnh vi phn phi tuyn. Nu tho tiu chun th h n nh. Nu khng tho,
vn kt lun v tnh n nh cn b ng, ph thuc vo: 1.Chn hm V(x) 2.Chn bin
trng thi x
Hnh 2.8 Cc vng trn hng s V v hai qu o n nh
V=C1
V=C2
V=C3
V tng
x1
x2 (1) (2)
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Chng 2 iu khin thch nghi
Trang 113
V d thit k MRAS dng Lyapunov Gi s tt c cc bin trng thi ca h thng
u o lng c, nh l v n nh ca Lyapunov c th dng thit k lut iu khin thch
nghi m bo s n nh cho h thng vng kn, v d sau trnh by tng ny.
V d 2.6 H MRAS bc nht da vo l thuyt n nh.
Xt bi ton nh trong v d 2.2. Khi tham s ca i tng c bit lut iu
khin theo phng trnh 2.6 cho kt qu mong mun. Mt h thch nghi s dng m
hnh chun m c th tm ra cc h s t0 v s0 khi tham s a, b khng c bit c
th t c nh sau :
Sai s : e = y - ym
Ly o hm v s dng phng trnh 2.5, 2.14 v m hnh mong mun kh o hm y v
ym , ta c :
dtde
= -ame + (am a b 0s )y + (b 0t - bm)uc
Ch rng sai s e s tin n 0 nu cc tham s ny bng vi gi tr mong mun.
By gi ta cn c gng xy dng mt c cu hiu chnh tham s sao cho cc thng s
t0 v s0 tin n gi tr mong mun. S dng cho mc ch ny, hm Lyapunov c dng
:
V(e, 0t , 0s ) = 21 [e2 +
b1 (b 0s + a - am)2 + b
1 (b 0t bm)2]
Hm ny s bng 0 khi e = 0 v cc tham s b iu khin bng vi gi tr ti u.
o hm ca V l :
dtdV
= edtde
+ 1 (b 0s + a am) dt
ds0 +
1 ( b 0t - bm) dt
dt0
= -ame2 +
1 ( 0bs + a am )( dt
ds0 - ye ) +
1 ( 0bt bm )( dt
dt0 + uce )
Nu cc tham s c cp nhp bi:
dtdt0
= -uce (2.17)
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Chng 2 iu khin thch nghi
Trang 114
dtds0
= ye
ta c
2eadtdV
m=
Nh vy: Hm V s gim khi e khc 0. V vy c th kt lun l sai s e s tin
v 0. Tuy nhin cn ch l cc tham s t0 v s0 s hi t n gi tr cn bng nu
khng c cc iu kin khc tc ng vo. V vy lut ny tng t nh lut MIT nhng
nhy c thay i bi tn hiu khc. Lut hiu chnh cc thng s lm n nh cho h
thng m cc bin trng thi c th o lng c xy dng bng s tng qut ho trc tip
ca k thut dng trong v d sau . Lut hiu chnh theo phng trnh 2.17 t c
bng cch p dng l thuyt n nh tng t nh bng lut MIT ( so snh vi v d
2.2) trong c hai trng hp, lut hiu chnh c th vit nh sau :
dtd
= e
vi l vector cc tham s , = [-uc y]T khi s dng lut theo Lyapunov
v
= [-uc y]T/(p + am) nu s dng lut MIT vector c th c gii thch nh l
gi tr m ca gradient hm tn tht. Phng php Lyapunov by gi c p dng cho
vic hiu chnh h li nui tin. V d 2.7 y ch xt vn hiu chnh li nui tin.
Sai s c cho bi
e = G(p)( - 0 )uc gii thiu mt khng gian trng thi biu th cho hm
truyn G. Quan h gia tham s v sai s e c vit bi:
cuBAxdtdx )( 0 += (2.18)
e = Cx
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Chng 2 iu khin thch nghi
Trang 115
Nu h ng nht x = Ax l n nh tim cn v c tn ti 2 ma trn P v Q xc nh
dng sao cho: QPAPAT =+ (2.19) Chn hm Lyapunov nh sau :
V = 21 [xTPx + ( - 0)2]
o hm V v s dng phng trnh sai phn 2.18 c :
dtdV
=
2 (
dtd
dtdxPxPx
dtdx TT )() 0++
S dng phng trnh 2.18 ta c :
dtdV
=
2 {[Ax + B cu ( - 0)]TPx + xTP[Ax + B cu ( - 0)]}
+( - 0)dtd
= -
2
xTQx + ( - 0)(
dtd
+ cu BTPx)
Nu lut hiu chnh tham s c chn l :
PxBudtd T
c
= (2.20)
th o hm ca hm Lyapunov s m khi x 0. Vi lut hiu chnh theo phng
trnh 2.20 vector trng thi x v c sai s e = Cx v vy s tin n khng.Tuy
nhin ch l sai s tham s - 0 khng cn thit l phi tin n khng.
V d h bc hai MRAS
V d 2.8 Xt G(s) = )( assK+
v m hnh l Gm(s) = AmBm
= 22
2
2
++ ss
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Chng 2 iu khin thch nghi
Trang 116
a thc A0, R, S v T c chn bi : A0(s) = s + 0a R(s) = s + 1r
S(s) = 0s s + 1s T(s) = 0t s + 1t
Phng trnh Diophantine 2.7 cho li gii sau : 1r = 2 + 0a - a
0s = (2 0a + 2 - a 1r )/K 1s = a0
2/ K
0t = 2/ K
1t = Ka /2
0
n gin ha, ta chn : Q(s) = A0(s).Am(s) P1(s) = Am(s) P2(s) =
A0(s) Bi tp v nh (dng lm bi tp trong phn Cu hi n tp v bi tp cui
chng): M phng bng Matlab h bc hai MRAS trong v d 2.8 (V d 4.8) vi =
1, = 0.7, = 1, a0 = 2, a =1 v K = 2.Gi s rng 00 bb = . H thng MRAS
ri rc H MRAS c thc hin cho h lin tc khng c nhiu, nhng c th thc hin
c MRAS cho h ri rc. Thut gii trn c th c dng cho trng hp h ri rc. B
c lng c th da vo chun bnh phng ti thiu. Phn ny dnh trnh by trong b
iu khin t chnh nh trong phn 2.3
MRAS cho h thng ch bit c tng phn Trong phn trc ta gi s tt c m
hnh ca i tng l cha bit.Trong mt s trng hp c tnh ng hc ca h thng c
bit mt phn, cn li l khng bit. S bit trc ny c th c kt hp vo h MRAS.
iu ny c th thc hin tu thuc ch yu vo tham s v cu trc ca m hnh i tng.
Phng php ny c minh ha bng v d .
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Chng 2 iu khin thch nghi
Trang 117
iu khin thch nghi cho tay my Gi s cc bin trng thi c o lng y , c
th tm c mt bin sai s tuyn tnh i vi cc tham s, iu ny lm d dng trong
vic xy dng h thch nghi s dng m hnh chun n nh. iu ny c minh ha bng
vic iu khin tay my khi m c tnh ng hc l phi tuyn. Mt thao tc trc tip
c m t bi m hnh : H(q) q + C(q, q ) q + G(q) = T (2.21) vi q l
vector ta tng qut. H l ma trn qun tnh, C l ma trn tt, G l vector
trng trng. Bin iu khin l moment t vo c cu chp hnh.Phng trnh m t tay
my c tnh cht :
21
dtd ( q TH q ) = q TH( q ) q + q TC( q, q ) q = q T( T G )
(2.21a)
iu ny c gii thch l o hm ca ng nng q TH q bng vi cng sut c cung
cp bi c cu chp hnh v moment trng lc. V d:Tay my hai khp ni Xt tay
my hai khp ni vi ti cha bit trong hnh di y. Khp ni th hai vi ti cha
bit c xem nh l c thm mt khp ni vi 4 tham s cha bit: khi lng me,
moment qun tnh Ie, khong cch t trng tm n khp ni th hai cel , gc e
so vi khu lin kt th hai. H thng c m t bi phng trnh (2.21) vi
++
++++=
224232
2423224231
sincossincossin2cos2
qqqqqq
H
+
+++=
4433
121212413 cos)(YY
qeeYYG
vi:
12
2111111
2122124
2122123
222122
2222121
/
)sin()cos()cos()sin(
)cos()cos(2)sin()sin(2
lgelmllme
qqeqqY
qqeqqY
qqqqY
qqqqqY
cc
=
=
++=
++=
+=
=
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Chng 2 iu khin thch nghi
Trang 118
Tay my hai khp ni vi ti cha bit vi g l gia tc thng v bn tham s
cha bit 41 ,, l nhng hm c cc tham s vt l cha bit.
ecee
ecee
cee
eceeec
llmllm
lmIlmlmIlmI
sincos
14
13
212
21
221111
=
=
+=
++++=
Bn tham s cha bit ceee lIm ,, v e c xc nh duy nht bi 41 ,, . H
thng c th c vit li:
TqqqT = ),,( vi T c cho bi:
e
l1
me
l c1 q1
m 1
l ce
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Chng 2 iu khin thch nghi
Trang 119
( )
=
=
+++
+++++
2
1211
4321
41231221
222121221221
)cos(,,,
)sin()cos(0)sin()sin(2)cos()cos(2
qee
T
YqqYqqqqYqqqqYqqqqeqeq
T
vi )cos( 12 qee = v 21 , l cc moment tc dng vo. c tnh ng hc c th
c vit di dng tuyn tnh theo cc tham s vi gi s l tt c cc trng thi v
gia tc c th o lng c. V d c th c tng qut ho v phng trnh (2.21) c th
c vit thnh:
0),,()(),()( qqqqGqqqCqqHT T= vi GCH ,, v l bit trc hay c th o
lng c. D l m hnh khng tuyn tnh, n vn tuyn tnh theo cc tham s c th
thay i. Mt iu quan trng l kin thc bit trc c dng v h thng khng xem
nh l m hnh hp en vi tham s thay i theo thi gian. M hnh th vn cn cha
tho mn bi v gia tc phi c o cng vi v tr v vn tc.
t qu o tham kho cho v tr v vn tc l qm v mq . a ra phng trnh
Lyapunov nh sau:
( ) ~~~~~)(~21 ++= Tp
TT qKqqqHqV
Trong : 0~,~,~ === mm qqqqqq ; pK v l nhng ma trn xc nh dng. Ly
vi phn V , s dng phng trnh (2.21a) cho ta:
~~)~(~
~~)~~(~
~~~~~~
21
~~
++=
+++=
+++=
Tpmm
T
Tpm
T
Tp
TTT
qKqCqHGTq
qKqCqHqHq
qKqqHqqHqV
a ra lut iu khin:
qKqKGqCqHT dpmm ~~ ++= (2.21b)
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Chng 2 iu khin thch nghi
Trang 120
Lut iu khin bao gm thnh phn nui tin t thnh phn bit ca m hnh v
thnh phn t l v hi tip vn tc, ngha l:
~~)~~~~(~ ++= TdmmT qKGqCqHqV Trong :
)()()(~),(),(),(~
)()()(~
qGqGqGqqCqqCqqC
qHqHqH
=
=
=
t:
~~~~ Tmmm GqCqH =+ t c nh vy l do m hnh tuyn tnh i vi cc tham s.
Hn na
),,,( mmmm qqqq = c ngha l ch c gia tc ca m hnh phi c bit, khng
phi l gia tc thc. Dn n:
)~~(~~~ qqKqV TmTdT ++= ngh cp nht thng s:
)(~~ 11 mTmTm qqq === (2.21c) Hm V tho tnh cht ca hm Lyapunov l
xc nh dng v o hm:
qKqV d ~~=
l bn xc nh m. iu ny c ngha l h vng kn n nh v vn tc khi xc lp bng
khng. B iu khin cng c th c b sung m bo l sai s v tr bng 0. Lut iu
khin theo phng trnh (2.21b) v tham s c cp nht theo phng trnh
(2.21c) l cc hm ca bin mm qqqq ,,, v mq , nhng gia tc ca khp ni
khng cn thit phi o c. rng lut iu khin l trng hp c bit ca h MRA tng
qut vi mqqe = .
2.2.5 Kt lun
Cc tng c bn da trn MRAS c trnh by trong phn ny bao gm :
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Chng 2 iu khin thch nghi
Trang 121
1. Phng php gradient 2. Thit k theo Lyapunov v siu n nh 3. S gia
sai s Trong mi trng hp lut cp nht tham s cho di dng :
dtd
= (2.22)
hay di dng chun ho :
dtd
=
T+
(2.23)
Trong phng php gradient, vector l gi tr m ca gradient sai s theo
cc tham s. c lng thng s hay xp x c th c dng trong phng php
gradient. Trong nhng trng hp khc l vector li c c bng cch lc ng vo,
ra v tn hiu t. S hng l s gia sai s c gii thch l sai s d bo ca vn c
lng.Thng dng s gia sai s tuyn tnh theo cc thng s. Phng php gradient
linh hot v n gin p dng vo mi cu trc h thng. Cch tnh ton i hi phi xc
nh c hm nhy bi v lut hiu chnh da vo vic tnh gradient, c th khng nh
l phng php s hi t, c cho bi li thch nghi c chn l nh. Hn na, gi tr
ban u ca tham s phi chn h thng vng kn l n nh. Phng php ny s gy khng
n nh nu h s li thch nghi ln. Vn l kh tm c gii hn n nh trc. H MRAS
tng qut c a ra da vo vic thit k m hnh km theo. Thut gii ny bao gm
nhng trng hp c bit ca vic thit k MRAS c trnh by trong cc phn trn.
Vic c lng tham s c th c thc hin vi nhiu cch khc so vi phng trnh
2.22 v 2.23.
2.3 B t chnh nh (STR Self Tuning Regulator) 2.3.1 t vn S tng ng
chc chn Thng s c lng Phng php gradient Bnh phng cc tiu Cc phng php
thit k b iu khin
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Chng 2 iu khin thch nghi
Trang 122
PID V tr cc LQG (Linear Quadratic Gaussian) B t chnh nh (STR) da
trn quan im phn tch, nh gi cc thng s cha bit. tng c bn c minh ho
trong hnh 2.9 . Cc thng s cha bit c nh gi trc tuyn (on-line) bng
cch dng phng php c lng qui. Cc thng s c lng c xem nh l thng s thc,
khng tin cy ca cc c lng l b qua. y gi l qui tc tng ng nht nh
(certainty equivalence principle).
Hnh 2.9 M hnh t chnh nh
Nhiu phng php c lng khc nhau c th c vn dng nh xp x c on, bnh
phng ti thiu..... Khi design hnh 2.9 tng trng
Thit k b iu khin
S thch nghi
Qu trnh B iu khin
Ng vo
Tham chiu
Cc tham s b iu khin
Cc tham s qu trnh c tnh
Ng ra
B t chnh nh
-
Chng 2 iu khin thch nghi
Trang 123
cho bi gii trc tuyn cc bi ton thit k h thng vi cc thng s cha bit
trc. y l bi ton thit k c bn. in hnh cho phng php ny l phng php khc
bit cc tiu, bnh phng tuyn tnh, t cc, model following. Phng php thit
k c la chn ph thuc vo c tnh ca h thng vng kn. Mc tiu ca mc ny l a
ra quan im c bn v tnh cht ca cc b t chnh nh. B t chnh nh ban u ch p
dng cho cc h thng ly mu d liu, nhng cc thut ton lin tc v hn hp
(hybrid) cng c pht trin. Trong mc ny, gi s h thng l SISO : A(q)y(t)
= B(q)u(t) + C(q)e(t) (2.24)
y : u ra u : u vo {e(t)} : chui phn b Gausse A, B, C : cc a thc
theo q (ton t sai phn ti). Gi thit bcA = bcC = n v bcA - bcC = d0.
Qu trnh iu khin thng c m t dng ton t q-1. a thc c tnh c dng:
)()( 1* = zAzzA n n = bcA. Khi m hnh (2.24) c m t nh sau:
)()()()()()( 1*01*1* teqCdtuqBtyqA += B t chnh nh da trn quan im
c lng cc thng s ca qu trnh. Phng php d hiu l c lng cc thng s ca hm
truyn ca qu trnh v nhiu (thut ton thch nghi gin tip). Cc thng s ca
b chnh nh s khng c cp nht trc tip m l gin tip thng qua c lng m hnh
ca h thng. B iu khin thch nghi loi ny da trn phng php bnh phng ti
thiu v iu khin bm theo (Kalman 1958). Phng php ny khng da vo c tnh
vng kn ca h thng. Cc thng s ca b chnh nh cng c th c lng trc tip gi
l thut ton thch nghi trc tip. C 2 phng php trc tip v gin tip u gi l
iu khin t chnh nh.
2.3.2 B t chnh nh gin tip Trong phn ny, gi s m hnh ca h thng c
phng trnh 2.24. Cch d dng nht l to b t chnh nh theo nh phn 2.3.1 c
lng cc thng s ca a thc A, B, C.
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Chng 2 iu khin thch nghi
Trang 124
Xt trng hp xc nh (e(t) = 0). Nhiu phng php qui cp c th c s dng c
lng cc thng s ca A, B.
T = [b0 b1 ... bm a1 ... an ] T(t 1) = [u( t d0) ... u(t d0 m )
y(t 1) ... y(t n)]
trong 0dmn = . Khi b c lng bnh phng cc tiu c cho bi:
[ ][ ] )28.2(/)1()1()()(
)27.2()1()1()1()1()1()()26.2()1()1()()()25.2()()()1()(
1
=
+=
=
+=
tPttKItPttPtttPtK
tttyt
ttKtt
T
T
T
Trong trng hp nhiu l ngu nhin, phng php bnh phng ti thiu cho ra
cc c lng sai lch nu C(q) qn. Lc ny, chng ta phi dng cc phng php nh
cc i qui, bnh phng cc tiu tng qut.
Tnh hi t Nu tn hiu u vo c kch thch y v cu trc ca m hnh cn c lng
thch hp th cc c lng s hi t n mt gi tr thc nu h thng vng kn n nh. iu
kin hi t cho cc phng php khc nhau l khc nhau. Trong c 2 trng hp
nhiu xc nh (e(t) = 0) v nhiu ngu nhin (e(t) khng ) th iu kin hi t
ph thuc tn hiu u vo, qu trnh v nhiu ca h thng. Tn hiu iu khin u(t)
c pht i qua khu hi tip. iu ny lm phc tp vic phn tch nhng n cn thit
yu cu h thng vng kn phi n nh. Trong MRAS vic phn bit tnh hi t s c
cp r hn chng 6 (TLTK[1]).
Bi ton thit k nn tng cho nhng h thng bit trc Nhiu phng php thit
k c s dng trong cc b t chnh nh ph thuc vo c tnh ca h thng vng kn.
Phng php thit k thng s dng l t cc (pole placement). Phng php da
theo m hnh mu (mode following) v phng php t cc c cp phn 2.2 v ph lc
A (TLTK[1]).
-
Chng 2 iu khin thch nghi
Trang 125
Xt m hnh ca h thng c phng trnh 2.24 v p ng ca h thng vng kn mong
mun l : Am(q).y(t) = Bm(q).uc(t) (2.29) B iu khin l: )()()()()()(
tyqStuqTtuqR c = (2.30) R1 v S l gii php cho phng trnh Diophantine
mAASBAR 01 =+
(2.31) trong
)35.2()34.2()33.2()32.2(
1
0
RBR
BAT
BBBBBB
m
mm
+
+
=
=
=
=
Mt vi iu kin phi tho mn chc rng b iu khin l nhn qu (causal) (xem
ph lc A TLTK[1] ). Cc phng trnh trn l c bn cho nhiu bi ton thit k
khc nhau.
Mt kiu mu cho mt b t chnh nh gin tip B t chnh nh gin tip da trn
thit k t cc c th biu din trong thut ton sau:
Thut ton 2.1 - B t chnh nh gin tip D liu : Hm truyn p ng xung
vng kn mong mun Bm/Am v a thc quan st mong mun A0 c cho trc. Bc 1:
c lng cc h s ca a thc A, B, C trong phng trnh (2.24)
dng phng php bnh phng ti thiu t cc phng trnh (2.25) (2.28) Bc 2:
Thay A, B, C bng cc c lng t c bc 1 v gii phng trnh (2.31) tm R1, S.
Tnh R bng phng trnh (2.35) v T bng phng trnh (2.34). Bc 3 : Tnh tn
hiu iu khin t phng trnh (2.30) Lp li bc 1, 2, 3 mi chu k ly mu.
-
Chng 2 iu khin thch nghi
Trang 126
Mt s vn cn ch vi thut ton ny : + Bc ca cc a thc phng trnh 2.24
hoc gii hn bc cao nht phi bit trc. + Tha s chung ca cc c lng A, B c
kh nng gii c phng trnh 2.31 + Phi m bo h thng vng kn l n nh. + Cc
tn hiu nn kch thch lin tc m bo s hi t ca cc thng s.
.
V d 2.9 B t chnh nh gin tip vi nhiu xc nh Xt h thng c hm truyn
:
G(s) = )1(1+ss
Hm truyn ny c xem nh l m hnh c bn ca ng c. Hm truyn p ng xung vi
chu k ly mu h = 0.5 l :
H(q) = AB
=
61.061.1090.0107.0
2 +
+
qqq
= )61.0)(1()84.0(107.0
+
qqq
H thng c ly mu c 1 zero = -0.84 bn trong vng trn n v vi h s tt
nh. Gi s h thng vng kn mong mun l :
m
m
AB
=
50.032.118.0
2 + qq
iu ny tng ng vi h thng c tn s dao ng t nhin 1 rad/sec v h s tt =
0.7 Gi s a thc quan st l :
A0 = (q 0.5)2 Bi tp v nh (dng lm bi tp trong phn Cu hi n tp v bi
tp cui chng) ng dng Matlab m phng h thng trong v d 2.9 (V d 5.1
(TLTK[1]).Kt qu nhn c c m t hnh (5.2), (5.3) v (5.4) trong TLTK[1].
Hnh 5.2 biu din tn hiu u ra v tn hiu iu khin ca h thng thc khi mt b
t chnh nh gin tip c s dng vi phng php bnh phng cc tiu v zero z = -
0.84 ca h thng thc b kh.
-
Chng 2 iu khin thch nghi
Trang 127
Hnh 5.3 ch ra vic c lng cc thng s ca h thng hi t nhanh n cc thng
s ca m hnh thc.C s dao ng ln ca tn hiu iu khin do vic kh zero. Dao
ng ny l kt qu ca s chn la km trong bi ton thit k c bn ch khng phi
ph thuc vo b t chnh nh. Dao ng ny c th trnh c bng cch thay i thit k
m khng kh zero ca h thng thc ( chng hn Bm = B). Hnh 5.4 ch ra kt qu
khi thay i thit k khng c zero no b kh. p ng ca h thng vng kn by gi
c tho mn.
V d 2.10 B t chnh nh vi nhiu ngu nhin : Xt h thng c m t nh sau
:
y(t) + ay(t 1) = bu(t 1) + e(t) + c e(t 1) vi a = - 0.9, b = 3,
c = -0.3. Bi ton thit k c bn c s dng l iu khin sai lch cc tiu. B iu
khin sai lch cc tiu c cho nh sau :
u(t) = - b
ac y(t) = - 0.2y(t)
iu ny dn n h thng vng kn : y(t) = e(t) Phng php cc i qui c s dng
c lng cc thng s cha bit a, b v c. Cc c lng t c t phng trnh 2.25
2.28 vi :
)1()1()()()]1()1()1([)1(
][
=
=
=
tttyt
ttytut
cab
T
T
T
B iu khin l:
)()()()(
)()()(
0
0
tbtatc
ts
tytstu
=
=
Bi tp v nh (dng lm bi tp trong phn Cu hi n tp v bi tp cui chng):
ng dng Matlab m phng b t chnh nh trong v d 2.10 (V d 5.2 TLTK[1]).
Xem kt qu m phng trong hnh (5.5), (5.6) v (5.7) ca TLTK[1]. Hnh 5.5
ch ra kt qu ca m phng thut ton ny. Hnh 5.6 biu din hm chi ph :
-
Chng 2 iu khin thch nghi
Trang 128
V(t) = =
t
iiy
1
2 )(
Khi s dng b iu khin sai lch cc tiu ti u v b t chnh nh gin tip.
ng cong cho tn hao tch lu ca STR gn vi ng cong ti u. iu ny c ngha b
t chnh nh gn nh ti u ngoi tr khong t qu khi khi ng. Hnh 5.7 biu din
thng s ca b iu khin )(0 ts . Tm tt Thut ton t chnh nh gin tip l
nhng ng dng n gin ca tng t chnh nh. Chng c th c p dng ti nhiu phng
php thit k b iu khin v c lng thng s. C 3 kh khn chnh vi phng php
ny. Phn tch tnh n nh l phc tp bi v cc thng s chnh nh ph thuc vo cc
thng s c lng. Thng th cn phi gii cc phng trnh tuyn tnh trong cc
thng s b iu khin. L trnh t cc thng s qu trnh n cc thng s t chnh c
th c cc im k d. iu ny xy ra trong cc phng php thit k da vo phng php
t cc, chng hn, nu m hnh c lng c chung cc v zero. Cc cc v zero chung
cn phi loi b trc khi tin hnh phng php t cc. Do vic phn tch tnh n nh
ch thc hin trong mt s t trng hp. m bo cc thng s hi t n cc gi tr
chnh xc th cu trc ca m hnh phi chnh xc v tn hiu u vo phi kch thch
lin tc.
2.2.3 B t chnh nh trc tip Khi lng tnh ton cho cc thut ton phn
trc tn nhiu thi gian v tnh n nh rt kh phn tch. Nhiu thut ton khc c
xut vic tnh ton thit k n gin hn. tng l dng cc c tnh, cc cc v zero
mong mun vit li m hnh h thng sao cho cc bc thit k l khng ng k. iu
ny dn ti vic thng s ho li m hnh. Nhn phng trnh Diophantine (2.31)
vi y(t) v dng m hnh c phng trnh 2.24 th :
[ ] )()()()36.2()()()(
)()()(
1
11
10
tCeRtSytRuBtCeRtSyBtBuR
tSyBtAyRtyAA m
++=
++=
+=
Ch rng phng trnh 2.36 c th c xem nh l mt m hnh ca h thng c thng
s ho trong B-, R v S. Vic c lng cc thng s ny to ra cc a thc R v S
ca b chnh nh mt cch trc tip. Kt hp
-
Chng 2 iu khin thch nghi
Trang 129
phng trnh 2.34 , tn hiu iu khin c tnh t phng trnh 2.30 . Lu m
hnh phng trnh 2.36 l phi tuyn tr phi B- l hng s. Cch khc thng s ho
l vit m hnh phng trnh 2.36 nh:
)37.2(10 CeRySuRyAA m ++=
trong RBR = v SBS = Ch a thc R phng trnh (2.36) l monic (a thc c
h s bc cao nht bng 1) nhng R phng trnh (2.37) th khng phi monic. Cc
a thc R v S c mt tha s chung tng trng cho cc zero tt km. Tha s
chung ny nn kh b trc khi tnh ton lut iu khin.
Thut ton 2.2 - B t chnh nh trc tip : Bc 1: c lng cc h s ca a thc
R v S m hnh phng trnh (2.37). Bc 2: Kh cc tha s chung trong R v S t
c R v S. Bc 3: Tnh tn hiu iu khin t phng trnh 2.30 m R v S c c bc
2. Lp li bc 1, 2, 3 mi chu k ly mu. Thut ton ny trnh vic c lng phi
tuyn nhng cn phi c lng nhiu thng s hn khi dng phng trnh 2.36 v cc
thng s ca a thc B- c c lng 2 ln. Bc 2 do rt kh thc hin. V vic c lng
cc thng s phng trnh 2.36 tng i kh nn ta xt trng hp c bit B- l hng
s. Gi s tt c cc zero c th b kh ( 0bB = )
[ ] )38.2()()()()( 100 tCeRtSytRubtyAA m ++=p ng mong mun nh
sau:
)()( 0 tTubtyA cmm = Trong : bc(A) = n v 0A chia ht cho T. Sai s
(t) = y(t) - ym c cho bi:
-
Chng 2 iu khin thch nghi
Trang 130
[ ] )()()()()(0
1
0
0 teAACR
tTutSytRuAA
bt
m
c
m
++=
By gi ta xem xt cc trng hp khc nhau. u tin gi s e = 0. a thc
quan st c th c chn t do, khi dng m hnh lin tc theo thi gian th iu
cn thit phi gi s b0/(A0Am) l SPR (Strictly Positive Real = Thc dng
cht) t c mt MRAS n nh. Ta cng cn lu rng hm truyn c cc h s l s thc
dng tho iu kin cn n nh c gi l PR (Positive Real). Hm l SPR nu n n
nh vi d tr dng nh tu . Mt iu kin tng t cng l cn thit cho cc m hnh
ri rc theo thi gian. Vit li m hnh nh sau:
trong
)()()(1)(
)()()(1)(
)()()(1)(
1*1*0
1*1*0
1*1*0
tuqAqA
tu
tyqAqA
ty
tuqAqA
tu
c
m
cf
m
f
m
f
=
=
=
iu ny tng ng vi trng hp P = Q = A0Am phn 2.2 . Tnh hi t by gi s
ph thuc vo du ca b0. iu ny ch ra mi lin h gia MRAS v STR.
Thut ton 2.3 - B t chnh trc tip vi nhiu xc nh D liu : Cho trc
gii hn thp nht ca thi gian tr d0 v du ca b0, p ng xung hm truyn vng
kn mong mun b0/A*m v a thc quan st mong mun A0. Bc 1 : c lng cc h s
ca a thc R*, S*, v T* phng trnh 2.38 dng phng php c lng qui. Bc 2 :
Tnh tn hiu iu khin t :
R*u(t) = - S*y(t) + T*uc(t)
)]()()([
])()()([)(
0*
0*
0*
0
0000
dtuTdtySdtuRbAAtu
TAAtyS
AAtuRbt
cfff
m
c
mm
+=
+=
-
Chng 2 iu khin thch nghi
Trang 131
Lp li cc bc 1, 2 mi chu k ly mu. Thut ton ny tng ng vi b iu khin
thch nghi dng m hnh chun phn 2.2 . Ch thut ton yu cu b0 phi bit
trc. Nu khng bit trc b0 th cng c th c lng c bng cch thay phng trnh
2.38 bng :
A0Amy(t) = Ru(t) + Sy(t) +R1C.e(t) m R by gi khng phi l
monic.
Cc b iu khin thay i cc tiu v mc trung bnh di chuyn (Minimum
Variance and Moving average)
Cc thut ton iu khin trong trng hp nhiu ngu nhin cho h thng c m t
bi phng trnh 2.24 s c xem xt. u tin gi s m hnh bit trc, e l mt nhiu
ngu nhin v uc = 0. a thc ca b quan st ti u cho m hnh phng trnh 2.24
l A0 = C. Tiu chun thit k l thay i cc tiu hoc trung bnh di chuyn.
Nu qu trnh l cc tiu pha, b chnh nh thay i cc tiu c cho bi: R*(q
-1)u(t) = - S*(q -1)y(t) (2.39) Trong R* v S* l nghim c bc cc tiu
ca phng trnh Diophantine A* (q -1)R* (q -1) + q 0d B* (q -1)S*(q
-1) = B* (q-1)C* (q -1) (2.40) vi d0 = Bc (A) - Bc (B). B iu khin
thay i cc tiu tng ng vi m hnh mong mun vi mt khong tr d0 bc, A*m =
1. T phng trnh 2.40 th R* phi chia ht cho B* :
R* = R*1.B*
Trong : Bc(R1) = d0 1. Phng trnh 2.40 c vit li : A*R*1 + q 0d S*
= C*
C*y(t) = A*R*1y(t) + S*y(t d0) = B*R*1u(t d0) + S*y(t d0) +
R*1C*e(t)
= R*u(t d0) + S*y(t d0) + R*1C*e(t) phng trnh ny c th c vit li
:
y(t + d0) = *1
C[R*u(t) + S*y(t)] + R*1e(t + d0) (2.41)
-
Chng 2 iu khin thch nghi
Trang 132
vi b iu khin phng trnh 2.39 th u ra ca h thng vng kn tr thnh
:
y(t) = R*1(q-1).e(t) Ng ra v vy l mt trung bnh di chuyn vi bc
(d0 -1). Trong strm (1970) ch ra rng b chnh nh s cc tiu s thay i ng
ra. Mt c im quan trng l ng ra tr thnh mt trung bnh di chuyn bc (d0
1). Ch s t nhin d0 c din t nh l s mu tri qua u ra thay i khi u vo
thay i. B iu khin thay i cc tiu c hn ch l tt c cc zero ca qu trnh u
b kh. iu ny c ngha s l kh khn nu B c cc zero bn ngoi vng trn n v.
Cc kh khn ny s trnh c b iu khin trung bnh di chuyn. B iu khin ny lm
cho ng ra c bc ln hn (d0 1). B iu khin c xut nh sau: tha s B+ v B-
trong B vi B+ c cc zero tt nhanh ( zero well damped). Xc nh R* v S*
t :
A*R* + q- 0d B*S* = B+ *C*
Phng trnh 2.41 cho ta:
y(t + d) = *
1C
[R*u(t) + S*y(t)] + R*1e(t + d) (2.42)
Trong : +
= BRR *1*
V ng ra c iu khin l mt qu trnh trung bnh di chuyn vi bc (d 1) nn
chng ta gi l iu khin trung bnh di chuyn. Ch khng c zero no b kh nu
B+ * = 1, c ngha d = bc (A) = n. C 2 lut iu khin thay i cc tiu v
trung bnh di chuyn dn n m hnh tng ng ca phng trnh 2.41 v 2.42 . S
khc nhau duy nht l gi tr ca d m s iu khin s zero ca qu trnh b kh.
Vi d = d0 = Bc(A) - Bc(B) : tt c zero b kh. Vi d = Bc(A) : khng c
zero no b kh. Lc vi A*0 trong phng trnh 2.38 cng c th to ra m hnh
ca phng trnh 2.42 :
y(t + d) = *
*
0
CA [R*uf(t) + S*yf(t)] + R*1e(t + d) (2.43)
Nu B+ cha tt c cc zero n nh ca h thng th n s tng ng nh b iu khin
thay i cc tiu cn ti u trong strm (1970)
-
Chng 2 iu khin thch nghi
Trang 133
B t chnh nh thay i cc tiu v trung bnh di chuyn Thut ton 2.4 -
Thut ton t chnh nh trc tip c bn
D liu : Cho trc khong d bo d. Gi k v l tng ng l s thng s trong
R* v S*. Bc 1: c lng cc h s ca a thc R* v S*
y(t + d) = R*(q -1)uf(t) + S*(q -1)yf(t) + (t + d) (2.44) trong
: R*(q -1) = r0 + r1q -1 +. . . + rkq k S*(q -1) = s0 + s1q -1 + .
. . + slq l V
uf (t) = )(1
1*0
qA u(t)
yf (t) = )(1
1*0
qA y(t)
s dng cc phng trnh 2.25 2.28 vi
(t) = y(t) - R*uf (t d) S*yf (t d) = y(t) - T(t d) (t 1)
T = )(1
1*0
qA[u(t) . . . u(t k) y(t) . . . y(t l)]
T = [r0. . . rk s 0 . . .sl] Bc 2: Tnh lut iu khin )()()()( 1*1*
tyqStuqR = (2.45) Vi R* v S* c thay bng cc c lng tng ng trong bc 1.
Lp li cc bc 1 v 2 mi chu k ly mu. Ch : Thng s r
0 c th c lng hoc gi s bit trc. cc trng hp sau thun li ta vit R*
nh sau:
R*(q -1) = r 0 (1 + '1r '1 ... krq ++ kq )
V s dng
(t) = y(t) 0r uf(t d) - T(t d) (t 1)
-
Chng 2 iu khin thch nghi
Trang 134
)()()()1([)(1
001*0
ltytykturturqA
T=
]......[ 0''1 lkT ssrr= Tnh cht tim cn M hnh phng trnh 2.41 v
2.42 c din t nh l vic thng s ho li m hnh phng trnh 2.24 . Chng tng
ng vi m hnh phng trnh 2.44 trong thut ton 2.4 nu A0 c chn bng C.
Vector hi qui khng tng quan vi sai s v phng php c lng bnh phng ti
thiu s hi t ti thng s tht. Mt kt qu ng kinh ngc l cng t chnh nh
chnh xc khi A0 C. Kt qu sau ch ra cc thng s t chnh nh chnh xc c ga
tr tng ng vi thut ton 2.4 khi A0 C.
nh l 2.1 Tnh cht tim cn Xt thut ton 2.4 vi A*0 = 1 dng phng php
c lng bnh phng cc tiu. Thng s b0 = 0r c th c nh hoc c c lng. Gi s
vector hi qui c gii hn, v cc c lng l hi t. H thng vng kn t c trong
iu kin gii hn c c im
kdddtutyldddtyty
++==+
++==+
,...,1,0)()(,...,1,0)()(
(2.46)
trong du gch ch gi tr trung bnh theo thi gian; k, l l s cc thng
s c lng trong R* v S*.
Chng minh: M hnh ca phng trnh 2.44 c th c vit li: y(k + d) =
T(k) + (k + d)
v lut iu khin tr thnh:
0)()( =+ dkkT Ti mt trng thi cn bng, cc thng s c lng l nhng hng
s. Hn na, chng tho mn cc phng trnh chun, trong trng hp ny c vit li
nh sau:
==
+=+t
k
Tt
kdtkk
tdkyk
t 11)()()(1)()(1
S dng lut iu khin
-
Chng 2 iu khin thch nghi
Trang 135
==
++=+
t
k
Tt
k ttdkdtkk
tdkyk
t 11)]()()[()(1lim)()(1lim
Nu thng s c lng )( t hi t khi t , v cc vector hi qui b gii hn th
v phi s tin ti zero. Phng trnh 2.46 by gi ko theo
1*0 =A v xc nh v s hi qui vector trong thut ton 2.4
nh l 2.2 Tnh cht tim cn 2 Gi s thut ton 2.4 vi phng php c lng
bnh phng cc tiu c p dng cho phng trnh 2.24 v: min(k, l) n 1 (2.47)
C ngha tn hiu ra l qu trnh c mc trung bnh di chuyn bc (d -1). Nu cc
c lng tim cn ca R v S lin quan vi nhau, nghim trng thi cn bng
l:
)()( tyty + = 0 = d, d + 1,..... (2.48)
Chng minh: H thng vng kn c m t nh sau: )()(* tSytuR =
A* y(t) = B*u(t d0) + C*e(t) V vy
(A*R* + 0dq B*S*)y = R*C*e (A*R* + 0dq B*S*)u = eCS **
Tn hiu c nh ngha (A*R* + 0dq B*S*) = C*e (2.49) V vy:
*Ry = v *Su = iu kin ca phng trnh 2.46 a n
)()( tytR + = 0 = d, d + 1, ..., ld + )()( tytS + = 0 = d, d +
1, ..., kd +
t
-
Chng 2 iu khin thch nghi
Trang 136
)()()( tytC y += cc phng trnh trn c th c vit li:
l
l
l
k
k
k
ssss
ssss
ssss
rrrr
rrrr
rrrr
210
210
210
210
210
210
00
000
00
000
++
)(
)(
dC
lkdC
y
y
= 0
Cy( ) = 0 = d, d + 1, . . . , d + k + l Hm tng quan tho mn phng
trnh:
F*(q -1)Cy( ) = 0 0 H thng phng trnh 2.49 c bc
n + k = n + max(k,l) Nu
k + l + 1 n + max(k, l) hoc tng ng vi
min(k, l) n 1 dn n
Cy( ) = 0 = d, d + 1,... l iu cn chng minh.
2.3.4 Kt ni gia MRAS v STR Cc h thng thch nghi dng m hnh chun
trc tip c cp trong phn 2.2. Trong ph lc A (TLTK[1]) cng ch ra m hnh
km theo v t cc l lin quan vi nhau. By gi chng ta s chng t b chnh nh
trc tip dng phng php t cc thut ton 2.2 l tng ng vi mt MRAS. Trong
trng hp nhiu xc nh, khi B- l hng s, m hnh ca qu trnh c vit li nh
sau:
-
Chng 2 iu khin thch nghi
Trang 137
y(t) = )( 0dtTf Trong thut ton gin tip, cc thng s c c lng bng cc
thng s ca b chnh nh. Phng php bnh phng cc tiu c s dng cho vic c lng
v (t) c vit li: (t) = y(t) - )( ty = )()( 0dtty Tf (2.50) Thng s cp
nht c th c vit li:
)()()()1()( 0 tdttPtt Tf += (2.51) Ch rng theo phng trnh 2.50
th
)()( 0 tgraddtTf = Vector )( 0dtTf din t nh l o hm ca nhy. Vic
cp nht thng s phng trnh 2.51 l mt phin bn ri rc theo thi gian ca
lut MIT. S khc bit chnh l sai s m hnh e(t) = y(t) - ym(t) c thay
bng gi tr thng d (t) v li MRAS c thay bng ma trn P(t) cho phng trnh
2.28. P lm thay i hng ca gradient v to ra mt chiu di bc thch hp.
Ngc li, lut MIT cng c th xem nh l mt thut ton gradient cc tiu e2,
phng trnh 2.51 dc xem nh l mt phng php Newton cc tiu 2(t). Gi tr
thng d c xem nh s gia sai s. Ch rng trong cc k thut nhn dng nh cc b
t chnh nh chng ta thng c gng t c mt kiu mu tng t vi
Tfty =)( Vi phng php m hnh chun th thng xuyn ch c th t mt m hnh
kiu ))(()( TfpGty = Vi G(p) l SPR.
V d 2.11 - B t chnh nh trc tip vi thay i cc tiu M hnh ca qu trnh
phng trnh 2.44 l :
y(t + 1) = )1()()( 00 +++ ttystur Gi s 0r c nh ti gi tr 0r = 1.
Ch iu ny khc vi gi tr tht l bng 3. Thng s s0 c c lng dng phng php
bnh phng cc tiu. Lut iu khin tr thnh:
-
Chng 2 iu khin thch nghi
Trang 138
)(
)(0
0 tyr
stu =
Bi tp v nh (dng lm bi tp trong phn Cu hi n tp v bi tp cui chng):
Dng Matlab m phng cho v d 2.11 (V d 5.3 TLTK[1]). Xem kt qu m phng
hnh 5.8 v 5.9 trong TLTK[1]. Hnh 5.8 biu din t s 00 / rs , n nhanh
chng hi t n mt gi tr ca b iu khin thay i cc tiu ti u thm ch 0r khng
bng gi tr tht ca n. Hnh 5.9 biu din hm tn hao khi dng b t chnh nh v
b iu khin thay i cc tiu ti u.
2.3.5 iu khin d bo thch nghi Thut ton 2.4 l cch thc hin mt b iu
khin vi tm d bo thay i. Bi ton iu khin c bn l b iu khin trung bnh
di chuyn. B iu khin trung bnh di chuyn cng c th p dng c cho cc h
thng khng cc tiu pha nh c minh ha phn B chnh nh trc tip. Nhiu cch
khc c iu khin d bo s c cp trong ti liu, mt vi trong s ny s c tho
lun v phn tch. Cng nh i vi cc thut ton trc, xc nh bi ton iu khin c
bn l rt quan trng hiu r cc tnh cht tim cn ca thut ton. Trc tin ta s
phn tch trng hp cc thng s l bit trc. Thut ton iu khin d bo da trn
mt m hnh ca qu trnh gi thuyt v cc tn hiu iu khin tng lai. iu ny to
ra mt chui cc tn hiu iu khin. Ch c mt tn hiu u tin l c p dng cho qu
trnh v mt chui cc tn hiu iu khin mi c tnh ton khi thc hin php o c
mi. y gi l b iu khin li tm (receding horizon controller).
D bo ng ra tng c bn trong cc thut ton iu khin d bo l vit li m
hnh qu trnh c c mt biu thc r rng cho ng ra mt thi im tng lai. Xt m
hnh : A* (q -1) y(t) = B* (q -1) u(t d0) (2.52) 1 = A* (q -1)F*(q
-1) + q d G*d(q -1) (2.53) trong
bc( F*d ) = d 1 bc( G*d) = n 1
-
Chng 2 iu khin thch nghi
Trang 139
Ch s d l tm d bo vi d bc. Gi s d d0. Vic ng nht a thc phng trnh
2.52 c s dng d bo ng ra d bc pha trc. V vy : y(t + d) = A*F*d y(t +
d) + G*d y(t) = B*F*d u(t + d d0) + G*d y(t) B* (q -1)F*d (q -1) =
R*d (q -1) + q (d - d 0 + 1) R *d (q -1) Bc(R*d) = d d0 Bc( R *d) =
n 2 Cc h s ca R*d l nhng gii hn d d0 + 1 u tin ca p ng xung ca h
thng vng h. iu ny c th thy nh sau:
q - 0d B*/A* = q - 0d B*(F*d + **
AG
q dd )
= )( 1*0 qRq dd + q ( d + 1) *dR (q -1) + )()()(
1*
1*1*
qAqGqB d q ( d + 0d ) (2.54)
y( t + d) = )( 1* qRd u(t + d d0) + *dR (q -1) u(t 1) + G*d (q
1) y(t) = )( 1* qRd u(t + d d0) + dy (t) (2.55)
)( 1* qRd u(t + d d0) ph thuc vo u(t), . . . , u(t + d d0), dy
(t) l hm ca u(t 1), u(t 2),... v y(t), y(t -1)....Bin dy (t) c hiu
nh l iu kin d bo ca y(t + d) vi gi s u(t) v cc tn hiu iu khin tng
lai l zero. Ng ra thi im (t + d) v vy ph thuc vo cc tn hiu iu khin
tng lai ( nu d > d0), tn hiu iu khin, cc ng vo v ng ra thi im
trc. Cng c th gi s tn hiu iu khin duy tr hng s: u(t) = u(t + d) =
.... = u(t + d d0) (2.56) Cch khc xc nh lut iu khin l mang y(t + d)
n mt gi tr mong mun trong khi cc tiu mc tiu iu khin theo tm d
bo:
+
=
dt
tkku 2)( (2.57)
iu khin khng thay i theo thi gian: Chn ng ra c d bo bng vi ng ra
mong mun ym v gi s vn gi phng trnh 2.56 :
[R*d(1) + q -1 *dR (q -1)]u(t) + G*d (q 1) y(t) = ym(t + d)
-
Chng 2 iu khin thch nghi
Trang 140
Lut iu khin l:
u(t) = 11**1*
)()1()()()(
+
+
qqRRtyqGdty
dd
dm (2.58)
Tn hiu iu khin ny s c s dng cho qu trnh. ln ly mu k tip, mt php
o mi t c v lut iu khin 2.58 dc s dng tip. Ch gi tr ca tn hiu iu
khin thay i theo thi gian ch khng phi c nh. y ta s dng qui tc iu
khin li tm. Ch lut iu khin l khng i ngc vi b iu khin LQ c nh tm. By
gi chng ta s phn tch h thng vng kn khi s dng phng trnh 2.58 iu khin
qu trnh 2.52.Vic thc hin cc php tnh ton t sai phn ti l cn thit c th
quan st cc cc ban u. Phng trnh 2.53 dc vit li theo ton t sai phn ti
nh sau: q n + d - 1 = A(q)Fd(q) + Gd(q) (2.59) a thc c tnh ca h
thng vng kn l:
P(q) = A(q) [q n 1Rd(1) + dR (q) ] + Gd (q) B(q) Bc(P) = Bc(A) +
n - 1 = 2n 1 Phng trnh thit k 2.59 c th c s dng vit li hm P(q):
B(q)q n + d - 1 = A(q) B(q)Fd(q) + Gd (q) B(q)
= A(q)[q n -1 Rd (q) + dR (q)] + Gd (q) B(q) V vy :
A(q) dR (q) + Gd (q) B(q) = B(q) q n + d -1 - A(q)q n 1Rd(q) Cho
ta :
P(q) = q n 1A(q)Rd(1) + q n 1[qd B(q) - A(q)Rd(q)] Nu h thng n
nh th cc s hng pha sau ca 2.54 s bin mt khi d . Do :
dlim P(q) = q n -1 A(q)Rd(1) nu A(z) l mt a thc n nh.
V d 2.12 - iu khin d bo Xt qu trnh :
y(t + 1) = ay(t) + bu(t) Phng trnh 2.59 cho ta :
-
Chng 2 iu khin thch nghi
Trang 141
qd = (q a)(q d 1 + f1q d 2 + . . . + fd 1) + g0 V vy:
F(q) = q d 1 + aq d 2 + a2q d 3 +. . . + a d 1 G(q) = a d Rd(q)
= bF(q) dR (q) = 0 v khi ym = 0, lut iu khin tr thnh:
u(t) = - )...1( 1+++ dd
aaba y(t) = - )1(
)1(
d
d
abaa y(t)
Phng trnh c tnh ca h thng vng kn l:
P(q) = q a + 1
)1(
d
d
a
aa
c cc:
pd = 1
d
d
a
aa
V tr ca cc c cho bi: 0 pd < a a 1 (h thng n nh)
0 pd < 1 a > 1 ( h thng khng n nh)
Cc vng kn vi cc gi tr khc nhau ca a v b c ch hnh 5.16 (TLTK[1]).
V d cng cho thy vic quan st l y th tm d bo phi t 5 10 mu. Cng c th
tng qut ho kt qu v d 2.12 cho cc h thng bc cao hn. i vi cc h thng
thay i chm hoc khng n nh th p ng vng kn ca n s rt chm khi tng tm d
bo. V vy gii hn 2.56 khi s l khng hu ch.
N lc iu khin cc tiu Thut ton iu khin l s mang y(t + d) ti ym(t +
d) trong khi cc tiu phng trnh 2.57 . Phng trnh 2.55 c vit li:
-
Chng 2 iu khin thch nghi
Trang 142
y(t + d) = R*d(q -1)u(t + d d0) + )(tyd = rd 0 u(t + ) + . . .+
rdu(t) + )(tyd
= d d0. Gii thiu hm Lagrange: 2J = u(t)2 + . . .+ u(t + )2 +
2[ym(t + d) - )(tyd - R*d(q -1) u(t + )]
Cho o hm ring i vi cc bin u(t), . . . ,u(t + ) v bng 0 ta c:
u(t) = rd
.
.
.
u(t + ) = 0dr ym(t + d) - )(tyd = 0dr u(t + ) + . . . + rd
u(t)
Cc phng trnh ny cho ta:
u(t) =
)()( tydty dm +
trong :
=
d
idi
r
r=0
2
S dng nh ngha )(tyd cho ta: u(t) = ym(t + d) - *dR u(t 1) - *dG
y(t)
hoc
u(t) = *1
* )()(d
dm
RqtyGdty
+
+
= )(
)()()1(1 qRq
tyqGndtyd
n
dm
+
++
(2.60)
S dng phng trnh 2.60 v m hnh ca phng trnh 2.52 cho a thc c tnh
vng kn:
P(q) = A(q) [q n - 1 + dR (q)] + Gd(q)B(q) Phng trnh ny c dng nh
2.58 vi Rd(1) c thay bng . iu ny c ngha cc cc vng kn tin gn ti zero
ca q n 1A(q) khi A(q) l n nh
-
Chng 2 iu khin thch nghi
Trang 143
v khi d . iu g s xy ra khi h thng khng n nh Hy xt v d sau y:
V d 2.13 - iu khin n lc cc tiu Xt h thng tng t nh v d 2.12 . B
iu khin n lc cc tiu trong trng hp ny c cho bi:
= b 1)1(22
...1
+++d
d
a
aa = )1(
)1(21
2
aa
abd
d
cho ta ( khi ym = 0)
u(t) = -
da y(t) = - )1()1(
2
212
d
d
abaa
y(t)
Cc ca h thng vng kn l:
pd = a - 1)1(
2
212
d
d
a
aa =
1212
d
d
a
aa
cho ta:
dlim pd = a | a | 1 (h thng n nh)
dlim pd = 1/a | a | > 1 (h thng khng n nh)
v d ny, b iu khin n lc cc tiu s to ra mt h thng vng kn tt hn nu
iu khin tng lai c gi s l hng s.
iu khin d bo tng qut: Cc b iu khin d bo cp t trc ch xem xt gi tr
ng ra ch mt thi im tng lai. Nhiu tng qut ho khc nhau ca iu khin d
bo c xut m trong hm tn hao l cc tiu:
J(N1, N2, Nu) = E{ = =
++++2
1 1
22 )1()]()([N
Nk
N
km
u
ktuktykty } (2.61)
Trong = 1 q -1 l ton t vi phn. S la chn cc gi tr khc nhau ca N1,
N2, Nu s a ra cc phng php khc nhau. Phng php iu khin d bo tng qut c
minh ho bng cch dng hm tn hao 2.60 v m hnh qu trnh: A*(q)y(t) =
B*(q -1)u(t d0) + e(t) / (2.62)
-
Chng 2 iu khin thch nghi
Trang 144
M hnh ny c gi l CARIMA ( Controlled AutoRegressive Intergrating
Moving Average). N c thun li l b iu khin bn thn s cha mt khu tch
phn. Ging nh phng trnh 2.53 ta c ng nht: 1 = A*(q)F*d(q - 1)(1 q
-1) + q d G*d (q 1) (2.63) Cng thc ny c s dng xc nh ng ra d bc k
tip:
y(t + d) = F*dB* u(t + d d0) + G*dy(t) + F*de(t + d) F*d c bc d
-1. B d bo vi sai s qun phng ti u vi ng ra c o c n thi im t v chui
ng vo bt k l: )( dty + = F*dB* u(t + d d0) + G*dy(t) (2.64) Gi s u
ra mong mun ym(t + k), k = 1, 2, ...l c sn. Hm tn hao 2.61 s c cc
tiu cho ra mt chui cc tn hiu iu khin tng lai. Ch gi tr mong i 2.61
s c c tng ng vi d liu c c ti thi im t vi gi s cc o c tng lai khng c
sn. iu ny c ngha ch c tha s u tin ca chui iu khin l c s dng. Cc php
ton s lp li khi c c mt o c mi. B iu khin vi kt qu nh th gi l iu
khin hi tip ti u vng h. Nh tn ca n, gi s s dng hi tip nhng n ch c
tnh ton ch da vo thng tin c sn thi im hin ti. Dng phng trnh 2.55
:
y(t + 1) = R*1(q 1) u(t + 1 d0) + 1y (t) + F1*e(t + 1) y(t + 2)
= R*2(q 1) u(t + 2 d0) + 2y (t) + F2*e(t + 2)
.
.
.
y(t + N) = R*N(q 1) u(t + N d0) + Ny (t) + *NF e(t + N) Mi gi tr
ng ra bao gm cc tn hiu iu khin tng lai ( nu d > d0), ng vo o c v
tn hiu nhiu tng lai. Cc phng trnh trn c th c vit li:
y = Ru + y + e trong :
y = [y(t + 1) . . . y(t + N)]T u = [u(t + 1 d0) . . . u(t + N
d0)]T
y = [ y 1(t) . . . y N(t)]T
-
Chng 2 iu khin thch nghi
Trang 145
e = [F1*e(t + 1) . . . *NF e(t + N)]T
T phng trnh 2.54 ta thy cc h s ca R*d chnh l (d d0 + 1) s hng u
ca p ng xung q d 0 B*/ (A*) v cng ging nh (d d0 +1) s hng u ca p ng
bc q d 0 B*/ A*. Do ma trn R l ma trn tam gic di:
R =
021
01
0
000
rrr
rr
r
NN
Nu h thng c thi gian tr (d0 > 1) th (d0 1) hng u ca R s l
zero. Gi:
ym = [ym(t + 1) . . . ym(t + N)]T Gi tr mong i ca hm tn hao c
vit li: J(1, N, N) = E{( y ym)T(y ym) + uTu}
= (Ru + y - ym)T(Ru + y - ym) + uTu Cc tiu ho biu thc ny theo u
ta c: u = (RTR + I ) 1RT(ym - y ) (2.65) Thnh phn u trong u l u(t)
l tn hiu iu khin ng dng cho h thng. Ch b iu khin t ng c mt khu tch
phn. iu ny l cn thit b cho s hng nhiu sai lch phng trnh 2.62 Vic
tnh ton phng trnh 2.65 lin quan ti ma trn nghch o NxN, m N l tm d
bo ca hm tn hao. gim khi lng tnh ton th ta c th gii hn cc tn hiu iu
khin tng lai. Chng hn, ta gi s vic tng tn hiu iu khin l bng zero
sau Nu bc (Nu < N):
u (t + k 1) = 0 vi k > Nu iu ny c ngha tn hiu iu khin sau Nu
bc s l hng s. So snh vi iu kin khng ch phng trnh 2.57 . Lut iu khin
( phng trnh 2.65) s thay i: u = (R1TR1 + I ) 1R1T(ym - y ) (2.66)
R1 l ma trn NxNu
-
Chng 2 iu khin thch nghi
Trang 146
R1 =
uNNNNrrr
r
rr
r
21
0
01
0
000
Ma trn ly nghch o by gi c bc NuxNu. Ng ra v cc tm iu khin c chn
nh sau: N1: Nu thi gian tr bit trc th N1 = d0, ngc li chn N1 = 1.
N2: Tm ng ra cc i N2 c chn sao cho N2h c gi tr bng vi thi gian ln
ca h thng, trong h l thi gian ly mu ca b iu khin. Nu: Thng Nu = 1 s
c c kt qu tt i vi nhng h thng n gin. i vi cc h thng phc tp, Nu t
nht phi bng vi s cc khng n nh hoc s cc gy dao ng tt yu. b iu khin d
bo tng qut c kh nng thch nghi th iu cn thit l phi c lng A* v B* mi
bc thi gian. Cc gi tr d bo ng vi cc tm d bo khc nhau s c tnh ton v
tnh tn hiu iu khin phng trnh 2.66 . B iu khin d bo thch nghi v vy s
l mt thut ton iu khin gin tip. Phng trnh 2.64 c tnh bng cch qui n
gin khi lng tnh ton.Cui cng, Nu thng c gi tr nh ma trn nghch o c bc
thp. Tn hiu iu khin u(t) t phng trnh 2.66 l:
u = [ 1 0 . . . 0] [R1TR1 + I ] 1R1T[ym - y ] = [1 . . . N] [ym
- y ] Hn na, t phng trnh 2.62 , s dng phng trnh 2.54
y =
+
+
)()1(
)()1(
**
*
1*
1
tyGtuR
tyGtuR
NN
=
+
+
*1
*
**
*
11
*
**
1
0
0
NdN
d
GqBAR
GqBAR
y(t)
H thng vng kn c phng trnh c tnh:
-
Chng 2 iu khin thch nghi
Trang 147
A* + [1 . . . N]
+
+
**1**
*
1*1**
1
0
0
Nd
N
d
GBqAR
GBqAR
ng nht phng trnh 2.63 cho ta: B* = A* B*F*d + q dGd* B*
= A*[ *dR + q ( d - 0d + 1) *dR ] + dq ** BGd iu ny cho ta phng
trnh c tnh:
A* + [1 . . . N]
NN qRAB
qRAB
)(
)(
***
*
1**
= A* + =
N
ii
ii RABq
1
*** )( (2.67)
Phng trnh 2.67 cho ra mt biu thc ca phng trnh c tnh vng kn nhng
vn cn kh khn a ra mt kt lun tng qut v tnh cht ca h thng vng kn ngay
c khi qu trnh bit trc. Nu Nu = 1 th:
i =
=
+N
jj
i
r
r
1
2
Nu ln, h thng vng kn s khng n nh khi h thng vng h khng n nh. Tuy
nhin nu c 2 tm iu khin v tm d bo u tng th bi ton s tng t nh bi ton
iu khin LQ vi tm c nh v do n s c c tnh n nh tt hn.
2.3.6 Kt lun Trong phn ny chng ta xem xt nhiu b t chnh nh khc
nhau. tng c bn l c lng cc thng s cha bit ca h thng v thit k b iu
khin. Cc thng s c lng gi s bng vi thng s thc khi thit k b iu khin.
Thnh thong cng bao gm cc c lng cha chc chn vo trong thit k. Bng cch
kt hp cc phng php c lng khc
-
Chng 2 iu khin thch nghi
Trang 148
nhau v cc phng php thit k khc nhau ta s c c cc b t chnh vi cc
tnh cht khc nhau. Trong phn ny ta ch cp tng c bn v cc tnh cht tim
cn. Tnh hi t ca c lng v tnh n nh ca h thng s c tho lun trong chng 6
(TLTK[1]). Kha cnh quan trng nht ca cc b t chnh nh l a ra cc thng s
ho. Mt thng s ho li c th t c bng cch s dng m hnh h thng v p ng vng
kn mong mun. Mc tiu ca vic thng s ho li l thc hin c lng trc tip cc
thng s ca b iu khin sao cho m hnh mi tuyn tnh vi cc thng s. Ch c vi
thut ton t chnh nh c cp v gii quyt trong phn ny. Vic kt hp cc phng
php c lng khc nhau v vn thit k c bn s to ra cc thut ton vi cc tnh
cht khc nhau. Mc tiu ca phn ny l a ra mt cm nhn cch pht trin v phn
tch cc thut ton. Khi thc hin mt b t chnh th vic la chn bi ton thit
k c bn l rt quan trng. Mt phng php thit k m khng ph hp cho h thng
bit trc th cng s khng tt hn khi h thng cha bit trc. B t chnh nh cng
c kh nng p dng cho cc h thng MIMO. Trng hp MIMO l rt kh phn tch. Kh
khn chnh l xc nh c kin thc u tin cn thit trong h MIMO l g. Cng tng
i n gin khi a ra mt thut ton t chnh tng ng vi b t chnh nh trc tip
tng qut cc trng hp hn ch khi cc ma trn tng tc ca h thng bit trc.
2.4 Chnh nh t ng v lch trnh li
2.4.1 Gii thiu 1. Chnh nh v thch nghi 2. Kin thc u tin 3. Gi tr
ban u ca b iu khin thch nghi 4. iu khin PID 5. Cc vn vn hnh 6. Giao
tip iu khin Mt loi c bit ca thch nghi vng h hay s thay i cc tham s
b iu chnh c cp trong phn ny. Trong nhiu trng hp, c th bit c s thay
i ng hc ca qu trnh theo cc iu kin vn hnh. Ngun gc ca s thay i ng hc
c th l tnh phi tuyn. C th thay i tham s ca b iu khin bng cch gim st
cc iu kin vn hnh
-
Chng 2 iu khin thch nghi
Trang 149
ca qu trnh. Khi nim ny gi l lch trnh li, v m hnh u tin c s dng
ch iu chnh li ca qu trnh.
2.4.2 K thut chnh nh 1. Phng php Zeigler Nichols Lut iu khin
PID:
++=
t
di
c dtdeTdsse
TteKtu
0
)(1)()(
2. Phng php p ng qu M hnh 3 thng s:
sLesT
ksG
+=
1)(
Phng php p ng nc:
Phng php Zeigler Nichols:
Thi gian
L T
a
k 0.63k
-
Chng 2 iu khin thch nghi
Trang 150
B iu khin aKc Ti / L Td / L Tp / L P PI
PID
1 0.9 1.2
3 2
0.5
4 5.7 3.4
Nhng kh khn i vi phng php Zeigler Nichols: - Kh xc nh cc thng s
- Tt qu chm - Hai thng s th khng Phng php din tch:
keAT
kA
LT
1
0
=
=+
3. Phng php p ng tn s tng: Cho chy b iu khin t l, tng li cho n
khi h thng bt u dao ng. Quan st li Ku gii hn v Chu k gii hn Tu . Lp
li: Xc nh c tnh p ng tn s.
A0
A1
k
L + T
-
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Cc thng s b iu khin:
B iu khin Kc / Ku Ti / Tu Td / Tu Tp / Tu P PI
PID
0.5 0.4 0.6
0.8 0.5
0.12
1 1.4
0.85 Thc nghim:
G(j)
)(1N
PID
Relay
-1
A
T
y u Qu trnh
-
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Kt qu thc t - Thng tin bit trc? - Bt u thc nghim nh th no? - Hi
tip n bin gii hn ca dao ng. - Hiu chnh lut Zeigler Nichols: Thay i
cc gi tr trong bng. S dng 3 thng s: Ku, Tu v Kp. - Lm sao ng u vi
nhiu c Nhiu ti Nhiu o T tr S lp li trc tuyn tng: Tm cc nt c trng ca
p ng trc tuyn i vi im t hoc cc nhiu ti. Hiu chnh b iu khin da trn
cc c tnh quan st c.
c tnh: h s tt d v vt l o
21
23
ee
eed
=
1
2
e
eo =
B iu khin hiu chnh da trn lut th v sai. D dng i vi PI v kh khn
hn i vi PID.
Tp
e1
e2
e3
-
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Thng tin bit trc Tin chnh nh
2.4.3 Lch trnh li V d cc bin lch trnh Tc sn xut Tc my S t l v p
lc ng
Thnh thong c th tm thy nhng bin i ph c tng quan tt vi nhng thay
i ca qu trnh ng hc. V th c th lm gim nh hng ca tham s bin ng ch n
gin bng vic thay i tham s ca b iu chnh nh cc hm ca cc bin ph (xem
hnh 2.10)
Hnh 2.10 M hnh lch trnh li
Lch trnh li c th c xem nh h thng iu khin hi tip m li hi tip c
chnh bi b b c cung cp trc.
Qu trnh
Lch trnh li
B iu khin
Tn hiu iu khin
Ng ra
iu kin vn hnh
Cc thng s b iu khin
Tn hiu vo
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Chng 2 iu khin thch nghi
Trang 154
u, khuyt im ca lch trnh li Mt hn ch ca lch trnh li l b vng h.
Khng c hi tip b cho sai s lch trnh. Hn ch khc ca lch trnh li l vic
thit k tn nhiu thi gian. Tham s b iu chnh phi c chn cho nhiu iu kin
vn hnh v c tnh k thut phi c kim tra bng nhiu qu trnh m phng. Nhng
kh khn ny trnh c nu lch trnh da vo cc php chuyn i phi tuyn. Lch
trnh li c u im l cc tham s b iu chnh c th p ng rt nhanh vi s thay i
ca qu trnh. Khi khng c c lng tham s, nhn t gii hn ph thuc vo tc p
ng cc php o ph vi s thay i ca qu trnh.
2.4.4 Xy dng lch trnh La chn cc bin lch trnh Hon thin vic thit k
iu khin cho nhng iu kin vn hnh khc nhau. S dng vic chnh nh t ng. S
bin i. Tht kh tm lut chung cho vic thit k b iu chnh theo lch trnh
li. Vn chnh l vic quyt nh cc bin s dng lm bin lch trnh. R rng cc tn
hiu ph phi phn nh iu kin vn hnh ca i tng. S c nhng trnh by l tng n
gin cho cc tham s b iu chnh lin quan n cc bin lch trnh. V th cn c
kin thc tt v h ng hc ca qu trnh nu lch trnh li c s dng. Cc khi nim
tng qut sau c th phc v cho mc ch ny. - Tuyn tnh ho c cu dn ng phi
tuyn. - Lp trnh li da vo o c cc bin ph - Vn hnh da vo hiu sut - Cc
php bin i phi tuyn. Cc khi nim ny c minh ho trong cc v d sau.
V d 2.14 Xem h thng vi 1 valse phi tuyn.Tnh phi tuyn c gi s
l:
v = f(u) = u4 , u 0
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Chng 2 iu khin thch nghi
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Hnh (a)
Hnh (b) t 1 f l hm ngc xp x ca c tnh valse. b cho tnh phi tuyn,
ng ra ca b iu chnh c cung cp thng qua hm ny trc khi n c p vo valse
(xem hnh (a)). ta c quan h : v = f(u) = f [ 1 f (c)] Vi c l ng ra
ca b iu chnh PI. Hm f [ 1 f (c)] c li t thay i hn hm f.
Nu 1 f chnh xc l hm ngc ca f th : v = c. Gi s f(u) = u4 c xp x
bi 2 ng thng: mt ng ni t im (0 , 0) n im (1.3 , 3) v ng thng th hai
ni gia 2 im (1.3 , 3) v (2 , 16), c v trong hnh (b) . Khi :
-1
c u y v
Qu trnh
f PI 1 f
0 0.5 1 1.5 2
u
v
f
f 5 10
15
20
yr
-
Chng 2 iu khin thch nghi
Trang 156
1 f (c) =
+
163,139.10538.0
30,433.0
cc
cc
Hnh (c) Hnh (c) cho thy s thay i trong tn hiu chun ti 3 iu kin
vn hnh khc nhau khi s dng hm 1 f nh hnh (a) . So snh vi h thng
trong hnh 2.2 (TLTK[1]) . Ta thy c s ci thin trong c tuyn ca h thng
vng kn. Dng hm ngc 1 f trong h thng s cho p ng bng phng hn trong cc
bi ton iu khin valse phi tuyn. V d trn cho thy tnh n gin v tin dng
trong vic b cho h thng phi tuyn tnh bit trc. Trong thc t thng xp x
h phi tuyn
y
20 40 60 80 100 0
yr
0.2
0.3
20 40 60 80 100 0
yr
1.1
1
20 40 60 80 100 0
yr
y
5.1
5
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Chng 2 iu khin thch nghi
Trang 157
bng mt vi on thng (nhiu hn 2). C nhiu b iu khin vng n thng mi s
dng phng php b ny. Trong v d trn khng c s o c no ca iu kin vn hnh
ngoi tr vic iu chnh ng ra. Trong cc trng hp khc, tnh phi tuyn c xc
nh t s o c mt vi bin s.
2.4.5 ng dng Lch trnh li l phng php rt hu dng. N yu cu phi c kin
thc tt v qu trnh v cc bin ph c th c o c. Mt thun li ln ca phng php
ny l b iu chnh thch nghi (p ng) nhanh khi cc iu kin thay i. Mt s ng
dng nh: nh hng cho tu, kim sot nng pH, kim sot kh t, iu khin ng c v
iu khin bay. c im ca van
c tnh van ph thuc vo vic ci t.
tuyn tnh
Dng chy
V tr
M tnh theo %
M nhanh
A
B C
-
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Trang 158
Lch trnh cho ng ra b iu khin
Lch trnh cho bin qu trnh
FT
FIC
LT
LIC
-
Chng 2 iu khin thch nghi
Trang 159
Lch trnh cho bin ngoi
2.4.6 Kt lun Lch trnh li l cch tt b cho c tnh phi tuyn bit trc.
B iu chnh c th phn ng nhanh vi s thay i ca cc iu kin. Mt hn ch ca k
thut ny l thit k tn nhiu thi gian nu khng dng php chuyn i phi tuyn
v t ng chnh nh. Mt hn ch khc l cc tham s iu khin c thay i trong vng
h, khng c hi tip t c tnh lm vic ca h thng. Phng php ny khng th dng
c nu c tnh ng hc ca qu trnh hoc nhiu khng c bit trc y , chnh
xc.
2.5 Bi tp ng dng Matlab
1.M hnh: H thng ga t ng trn t
ng lc hc ca t trn ng: t vn hnh trn ng nh moment sinh ra t ng c,
thng qua h thng truyn ng, chuyn thnh lc ko tip tuyn ti cc bnh xe ch
ng y t dch chuyn ln pha trc. Lc ko tip tuyn ny lun cn bng vi cc lc
cn tc ng vo t theo nh lut I Newton:
Fko = Fcn ln + Fcn khi ng + Fcn leo dc + Fcn qun tnh
TIC
TT FT
-
Chng 2 iu khin thch nghi
Trang 160
Tng cc lc cn i vi t khng ph thuc tuyn tnh vo vn tc ca t v cc
thnh phn lc cn ny c nhng h s ph thuc vo iu kin lm vic ca t nh loi
ng, mp m, nghing ca mt ng, loi lp xe, nhit mi trng, gi, ti trng ca
xe, tnh trng ca ng c, ca h thng truyn ng, mn ca lpCc iu kin lm vic
ny khng c nh m thay i mi khi t vn hnh v trong lc t vn hnh.
2.Phng trnh trng thi: i tng vn hnh trn ng l mt i tng phi tuyn ch
bao gm mt tn hiu iu khin vo l m cnh bm ga ca ng c ( hay