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BO CO TH NGHIM IU KHIN TNG
Bi 4: M PHNG V NH GI CHT LNG H THNGIU KHIN TC NG C DC
NHM 7
1. Nguyn Vn Tin 090764812. Thi Thanh Tn 090794413. Phan Trung Tn 09179281I. MC CH. Gip sinh vin hiu c m hnh ton ca ng cDC. Kho st i tng ng c trn m hnh ton, nh gi cht lng ca cc phng php iu
khin c s dng.II.TH NGHIM.
Cho hm truyn ca m hnh ng c c dng nh sau:
.
(+ .)(+ .)
II.1 Kho st m hnh iu khin tc ng c dng phng php iu khin PID Mc ch:
Kho st nh hng ca hiu chnh PID ln tc ng c.
Th nhim:
Trong :- Tn hiu u vo l hm nc Uu(t) = 1200 (tc 1200 RPM)- Transfer Fcn v Transfer Fcn1 th hin m hnh ca ng c.
a, Kho st h thng sdng khu hiu chnh P (Ki = 0, Kd = 0), ta sc vt l, sai s xc lp,
thi gian xc lp trn h thng da vo bng di y.Ta c thxc nh cc gi tr vt, sai s xc lp, thi gian xc lp nh sau:u tin ta s nhp hm truyn ca ng c DC vo Matlab vo bng cng c sisotool v ly
cc thnh phn nh hnh trn.Sau ta s thit lp u vo l tn hiu hm nc vi Final Value = 1200, cc thng s cn li s
bng 0 nh hnh sau Tip theo, ta s thit lp cc thng s Ki = 0, Kd = 0, ri ln lt theo i cc gi tr ca Kp nh
bng. t ta sxc nh c cc gi trnh vt l, sai s xc lp, thi gian xc lp.
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Mun vc xem c p ng ca h thng sc vtng tnh cc bi trc.
Bn c th vp ng bn ca hm nc, Sau ta sangbn ca s chnh ca Matlab ri g lnh.
>> plot(ScopeData.time,ScopeData.signals.values)
>> grid on;
Tcc bc trn ta c s nhn c p ng ca h thng vi Kp = 2, Ki = 0, Kd = 0.
Da vo p ng ca h thng ta c thxc nh c vt l, sai s xc lp, thi gian xc lp.Nhn hnh trn ta c th d dng nhn thy, vt l ca h l khng c. Thi gian xc lp l khng c,Sai s xc lp s bng 15
- vt lc tnh nh sau: (1298-1200)/1200 *100 = 8.16%(~8.2%).- Sai s xc lp: 1200 1125 = 75- Thi gian xc lp c tnh l thi gian cn thit sai lch gia p ng ca h thng v gi tr
xc lp ca n khng vt qu 5%(0.05) hoc 2%(0.02). D ta nhn thy rng h thng nhhnh di khng c thi gian xc lp.
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Lm tng tnh trn cho cc trng hp cn li. ta c c bng tng kt di y.
Kp 1 2 3 10 20
POT(%) Cha c Cha c 8.2 36,25 50.58
exl 200 109 75 23 12
txl (s) Cha c Cha c Cha c 0.23 0.2
Nhn xt: H thng sthay i trng thi hot ng ca mnh khi thay Kp. Khi Kp nh th h thnghot ng n nh nh m khng t c gi trt ban u(sai s xc lp ln). Khi Kp t cc gi trln thi lc h thng c tht c gi trt ban u nhng b li h thng s mt n nh vi vt l ca h thng qu ln
Gii thch: Ta c th thy rng khi s dng khu hiu chnh P(khu t l) th s lm cho sai s xc lpt c kt qu tt nht l tin v0, nhng b li vt l ca h thng stng cao nh th s lm hthng mt n nh. Do khng th mun sai s xc lp ca h thng bng 0 khi tng Kp n v cng.
b,Kho st h thng sdng khu hiu chnh PI (Kp = 2, Kd = 0), ta sc vt l, sai s xclp, thi gian xc lp trn h thng da vo bng di y.
Nh ta bit, th h thng s c mi gi tr sai s xc lp ng vi mi gi tr ca Kp. By gita schn 1 gi tr Kp thch hp th nghim khu hiu chnh PI. Chn Kp = 2 v khi ta c th nhnthy rng p ng ca h thng l tt nht so vi cc trng hp cn li(da vo cc tiu ch nh vt l, sai s xc lp).
By gita s vp ng ca h thng vi Kp=2, Kd = 0, Ki = 2(Ki thay i)
Lm tng tnh phn trn ta sc hnh nh sau:
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By gita stnh vt l v sai s xc lp tng tnh phn a. tnh thi gian xc lp nh sau:
(1200 1177)/1200*100 = 1.92 (%) tha iu kin 2% ca thi gian xc lp. sau ta ly gi trthi gian ngay im bit c thi gian xc lp. y c th thy rng. Thi gian xc lp l
bng 1.1(s).
Lm tng tnh cc phn trn(thay i Ki ln lt cc gi tr theo bng) ta s c bng tnghp cc gi trnh sau
Ki 1 2 3 10 20
POT(%) 1.25 3.80 6.20 21.80 39.70
exl 0.001 0.001 0.001 0.001 0.001
txl (s) 2.8 1.10 0.38 0.30 0.40
Nhn xt: da vo p ng ca h thng v bng s liu trn ta c th thy rng h thng t cgi trt. Nhng cng ging nh khu hiu chnh P, gi tr Ki cng ln th h thng cng mt nnh( vt ltng cao) v cn lm chm p ng qu ca h thng(h thng i vo trng thi nnh)
Gii thch:Nh ta bit khu hiu chnh PID l 1 trng hp c bit ca khu tr pha v th ncng sc cc c tnh ca khu trpha nh lm chp p ng qu , tng vt l, gim sai s xc
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lp. Do khi tng Ki qu ln th c thp ng v sai s xc lp nhng n vt ltng cao(mt nnh) tt nht ta nn chn 1 gi tr Ki sao cho h thng ph hp nht.
c, Kho st h thng sdng khu hiu chnh PID (Kp = 2, Ki = 2), ta sc vt l, sai s xclp, thi gian xc lp trn h thng da vo bng di y.
Ta s chn gi trKi = 2 v khi thi gian xc lp cho php trong khong 2% v vt l ca hthng khng qu ln(trng hp ny h thng tt nht theo em nhn nh)
Vy by gita s kho st khu hiu chnh PID vi Kp = 2, Ki = 2, K = 0.005
Ta c hnh.
Khu hiu chnh PID vi Kp = 2, Ki = 2, K = 1
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Lm tng tcho cc trng hp cn li. thay i ln lt cc gi tr Kd ta c bng tng hp kt qudi y.
Kd 0.005 0.01 0.5 1 2
POT(%) 0.02 0.6 3 20.83 37.83
exl 0.001 0.001 0.001 0.001 0.001
txl (s) 1.1 1.2 1.4 2.1 2.7
Nhn xt: ta c th nhn thy rng khi thay i cc gi tr Kd th h thng cng sthay i trng thi.H thng s hot ng tt khi gi tr ca Kd khng c ln. cn cc gi tr cn li th h thng hotng n nh v i vo trng thi xc lp sm nht c th.
Gii thch: Khi c 1 gi tr Kd thch hp th khu PID s hot ng rt tn a h thng vo trngthi n nh. Ta c th xem khu hiu chnh PID c to thnh t cch mc ni tip khu PI vi khuPD do n cng c cc u im m 2 khu c th mang li nh ci thin p ng qu ( vt lv thi gian xc lp), gim sai s xc lp
d, Nhn xt nh hng ca cc khu P, I, D n cht lng ca h thng.
Khu P: thc cht khu p l khu t l nn n sc cc c tnh nh hng n cht lng cah thng nh sai s xc lp cng nhv vt ltng khi Kp tng v ngc li. Do , takhng th c 1 h thng m va c sai s xc lp nhv vt l thp vi khu ny.
Khu I: l 1 dng ca khu trpha, nn n cng sc cc c tnh ca khu tr pha nh hngn cht lng h thng nh tng vt l,lm chm p ng qu , gim sai s xc lp. H
thng khi s dng khu ny ch c thp ng c 1 s tiu chun cht lng ch khng thp ng cng 1 lc c ht.
Khu D: l 1 dng ca khu sm pha. Nn n cng c cc c tnh nh hng n cht lng hthng ging nh khu sm pha nh p ng ca h thng nhanh, gim thi gian qu . Hthng cng ch c thp ng c cc c tnh m mnh mang li
c 3 khu trn u c nhng u im m ngc im ring p ng c 1 h thng nnh tt nht. ta c th kt hp 3 khu trn to thnh khu hiu chinht PID vi cc thng snh gi cht lng h thng m c 3 khu mang li.
II.2 Kho st m hnh iu khin vtr phng php iu khin PID
Cho hm truyn nh sau:
.
(+ .)(+ .)
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Th nghim:
Trong :- Tn hiu u vo l hm nc Uu(t) = 10 (v tr 10)- Transfer Fcn, Transfer Fcn1 v Transfer Fcn2 th hin hm truyn ca hiu khin v tr.
a, Kho st h thng sdng khu hiu chnh P (Ki = 0, Kd = 0), ta sc vt l, sai s xc lp,thi gian xc lp trn h thng da vo bng di y.
Thit lp cc thng s ca phn mn Matlab nh phn a. ri ta s thc hin ln lt
Tcc bc trn ta c s nhn c p ng ca h thng vi Kp = 4, Ki = 0, Kd = 0.
Da vo p ng ca h thng ta c thxc nh c vt l, sai s xc lp, thi gian xc lp.
Lm tng tcho cc trng hp cn li, ta c bng sau.
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Kp 1 2 4 8 16
POT(%) Cha c Cha c 7.9 22.8 40.6
exl 189 109 0.001 0.001 0.001
txl (s) 11 4.2 5.2 3.8 4.2
Nhn xt: H thng sthay i trng thi hot ng ca mnh khi thay Kp. Khi Kp nh th h thnghot ng n nh nh m thi gian xc lp ln. Khi Kp t cc gi tr ln thi lc h thng c tht c thi gian xc lp nhanah hn nhng b li h thng s mt n nh vi vt l ca h thngtng nhanh
Gii thch: Ta c th thy rng khi s dng khu hiu chnh P(khu t l) th s lm cho thi gian xclp t c kt qu tt nht, nhng b li vt l ca h thng stng cao nh th s lm h thngmt n nh. Do khng th mun thi gian xc lp ca h thng nh th khng th ctng gi tr caKp.
b,Kho st h thng sdng khu hiu chnh PI (Kp = 4, Kd = 0), ta sc vt l, sai s xclp, thi gian xc lp trn h thng da vo bng di y.
Chn gi tr Kp = 4 v lc ny h thng t trng thi tt nht
By gita s vp ng vi Kp=4, Kd = 0, Ki = 0.5
Lm tng tcho cc trng hp khc ta s c bng sau:
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Ki 0.1 0.5 1 2 4
POT(%) 10.1 18.3 28.1 45.5 69.7
exl 1.15 0.001 0.001 0.001 0.001
txl (s) 4.26 12.39 9.1 6.59 14.9
Nhn xt: da vo p ng ca h thng v bng s liu trn ta c th thy rng h thng t cgi trt. Nhng cng ging nh khu hiu chnh P, gi tr Ki cng ln th h thng cng mt nnh( vt ltng cao) v cn lm chm p ng qu ca h thng(h thng i vo trng thi nnh thi gian xc lp)
Gii thch:Nh ta bit khu hiu chnh PID l 1 trng hp c bit ca khu tr pha v th ncng sc cc c tnh ca khu trpha nh lm chp p ng qu , tng vt l, gim sai s xclp. Do khi tng Ki qu ln th c thp ng v sai s xc lp nhng n vt ltng cao(mt nnh) tt nht ta nn chn 1 gi tr Ki sao cho h thng ph hp nht
c, Kho st h thng sdng khu hiu chnh PID (Kp = 4, Ki = 0.5), ta sc vt l, sai s xclp, thi gian xc lp trn h thng da vo bng di y.
Chn gi trKi = 0.5 v khi sai s xc lp gn bng 0, v c thcoi p ng trong trng hp ny ttnht cho h thng iu khin v tr.
By gita s vp ng vi Kp=4, Kd = 2, Ki = 0.5
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Kd 0.1 0.5 1 2 4
POT(%) 17.3 13.1 9.77 7.40 6.17
exl 0.001 0.001 0.001 0.001 0.001
txl (s) 12 12 13.8 14.22 15.5
Nhn xt: ta c th nhn thy rng khi thay i cc gi tr Kd th h thng cng sthay i trngthi( vt l, thi gian xc lp). H thng s hot ng tt khi gi tr ca Kd khng c qu ln. cncc gi tr cn li th h thng hot ng n nh v i vo trng thi xc lp sm nht khi Kd nh.
Gii thch: Khi c 1 gi tr Kd thch hp th khu PID s hot ng rt tn a h thng vo trngthi n nh. Ta c th xem khu hiu chnh PID c to thnh t cch mc ni tip khu PI vi khuPD do n cng c cc u im m 2 khu c th mang li nh ci thin p ng qu ( vt lv thi gian xc lp), gim sai s xc lp
d, Nhn xt nh hng ca cc khu P, I, D n cht lng ca h thng.
Khu P: thc cht khu p l khu t l nn n sc cc c tnh nh hng n cht lng cah thng nh sai s xc lp cng nhv vt ltng khi Kp tng v ngc li. Do , takhng th c 1 h thng m va c sai s xc lp nhv vt l thp vi khu ny.
Khu I: l 1 dng ca khu trpha, nn n cng sc cc c tnh ca khu tr pha nh hngn cht lng h thng nh tng vt l,lm chm p ng qu , gim sai s xc lp. Hthng khi s dng khu ny ch c thp ng c 1 s tiu chun cht lng ch khng thp ng cng 1 lc c ht.
Khu D: l 1 dng ca khu sm pha. Nn n cng c cc c tnh nh hng n cht lng hthng ging nh khu sm pha nh p ng ca h thng nhanh, gim thi gian qu . Hthng cng ch c thp ng c cc c tnh m mnh mang li
c3 khu trn u c nhng u im m ngc im ring p ng c 1 h thng nnh tt nht. ta c th kt hp 3 khu trn to thnh khu hiu chinht PID vi cc thng snh gi cht lng h thng m c 3 khu mang li.
e, thay i tn hiu u vo l hm nc thnh tn hiu song song vung c chu k 10s. ta s khost li cc khu hiu chnh trn.
Ta sc s khi trong similink l:
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iu ch thng s cho xung vung c chu k 10s. Ta Click vo khi Pulse Generator ri sau cit thng snh hnh:
Vy l ta c tn hiu u vo l xung vung vi ch k l 10s
+ Kho st h thng sdng khu hiu chnh P (Ki = 0, Kd = 0), ta sc vt l, sai s xc lp,thi gian xc lp trn h thng da vo bng di y.
Ta lm tng tnh phn trn. Ta s c bng sau y
V chu k ca xung l 10s nn sau 5s th xung si trng thi t 1 sang 0(hoc ngc li) nn ta s c2 p ng xung ng vi mi trng thi l 1 p ng xung, nh c th nhn thy 2 p ng ny ngcnhau nh hnh.
Vi Kp = 5, Ki = Kd = 0;
Ta sc hnh nh trang bn
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Vi N l s ln i trng thi ca h thng. N bt u t 1
Kp 1 5 10 15 20
POT(%) Cha c 12.2 23.8 38.9 46.7
exl Cha c 0.001 0.001 0.001 0.001
txl (s) Cha c 4.059+N5 4.2+N5 4.3+N5 4.6+N5
Nhn xt:
- c th thy rng p ng ca h thng sp ng vi tng trng thi ca xung vung vi chuk l 10(s). mi ln xung vung i trng thi th h thng cng s tng p ng ph hp vitrng thi mi ca n
- H thng sthay i n nh ca mnh khi thay Kp. Khi Kp nh th h thng hot ng nnh nh m thi gian xc lp ln. Khi Kp t cc gi tr ln thi lc h thng c tht c thigian xc lp nhanh hn nhng b li h thng s mt n nh vi vt l ca h thng tng nhanh.
Gii thch:
- V trong khong thi gian 5 s th h thng sthay i nn khu hiu chnh ca mnh cng phihiu chnh cho n ph hp vi trng thi mi ca xung vo( 0 or 1)
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- Ta c th thy rng khi s dng khu hiu chnh P(khu t l) th s lm cho thi gian xc lpt c kt qu tt nht, nhng b li vt l ca h thng stng cao nh th s lm h thng mtn nh. Do khng th mun thi gian xc lp ca h thng nh th khng th ctng gi tr ca Kp.
+ Kho st h thng sdng khu hiu chnh PI (Kp = 5, Kd = 0), ta sc vt l, sai s xclp, thi gian xc lp trn h thng da vo bng di y.
Ta s c Kp = 5, Kd = 0; Ki = 1
Ta s c bng sau:
Ki 0.1 0.5 1 2 5
POT(%) 13.8 20 27.1 40.1 70.3
exl 0.001 0.001 0.02 0.05 0.1
txl (s) 3.5 Kx Kx Kx Kx
Note: Kx= khng xc nh c
+ Kho st h thng sdng khu hiu chnh PID (Kp = 4, Ki = 0.5), ta sc vt l, sai s xclp, thi gian xc lp trn h thng da vo bng di y.
Vi Kp = 5, Ki = 0.1 Kd = 0.5
Ta sc hnh nh sau:
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Vi Kp = 5, Ki = 0.1 Kd = 10
Ta s c bng kt qunh sau:
Kd 0.1 0.5 1 5 10
POT(%) 12.6 2.36 4.51 Khng c Khng c
exl 0.001 0.001 0.001 0.001 0.037
txl (s) 3.44 3.56 3.67 3.55 Kx
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NHN XT:
Nhn chung cc nhn xt trn nu ht c tt ccc c im ca khu P, I, D, PI, PIDnhng l trng hp h thng hot ng 1 ch lm vic nht nh. Khi h thng chuyn sang haitrng thi lm vic(hc u vo l xung vuong vi chu k l 10s) th cc khu hiu chnh ca chngta thit kcng phi tng thay i thch nghi vi trng thi mi ca h thng, gip h thng nnh trong tt c cc trng thi ca h thng c th gp.
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