TRNG I HC BCH KHOATHNH PH H CH MINH

BO CO TH NGHIMC S T NG

GVHD: H V TN: MSSV: NHM : T :

BI TH NGHIM 1PHN A: NG DNG MATLAB PHN TCH CC H THNG IU KHIN T NG

I.TM HM TRUYN TNG NG H THNG

Tm hm truyn tng ng h thng>> G1 = tf([1 1],conv([1 3],[1 5])) >> G2 = tf([1 0],[1 2 8]) >> G3 = tf(1,[1 0]) >> H1 = tf([1 2],1) >> G13 = parallel(G1,G3)>> G2H1=feedback(G2,H1);>> G0=series(G2H1,G13);>> G=feedback(G0,1) Hm truyn h thng l:

II KHO ST H THNG DNG BIU D BODE* kho st h thng phn hi m n v c hm truyn vng h :

a. Vi K = 10 V biu Bode trong khong tn s (0.1, 100):

>> TS = 10 >> MS = conv([1 0.2],[1 8 20]) >> G = tf(TS,MS) >> bode(G,0.1:0.01:100) >> grid on

Tn s ct bin =0.49rad/sTn s ct pha =4.71rad/s tr bin =25dB tr pha =101 Da vo biu bode ta thy h thng vng kn n nh do c d tr bin GM > 0 v d tr pha M > 0

V p ng qu ca h thng vi u vo hm nc n v trong khong thi gian t = [0 10]s>> Gf= feedback(G,1)

>> step(Gf,10) >> grid on

b.Vi K = 400 V Bode trong khong tn s (0.1, 100)>> TS = 400 >> MS = conv([1 0.2],[1 8 20]) >> G = tf(TS,MS) >> bode(G,{0.1,100}) ) >> grid on

Tn s ct bin =6.93rad/sTn s ct pha =4.71rad/s tr bin =-7.04dB tr pha =-25

Da vo biu bode ta thy h thng vng kn khng n nh do c d tr bin GM < 0 v d tr pha M < 0V p ng qu ca h thng vi u vo hm nc n v trong khong thi gian t = [0 10]s>> Gf= feedback(G,1) 10---------------------------s^3 + 8.2 s^2 + 21.6 s + 14 >> step(Gf,10) >> grid on

III. KHO ST H THNG DNG BIU NYQUIST:1.Vi K=10,: dng biu Nyquist>> G = tf(10,conv([1 0.2],[1 8 20]))>> nyquist (G)>> grid on>> hold on>> t=(-1:0.001:1)*2*pi;>> x=cos(t);>> y=sin(t);>> plot(x,y);>> hold off

Tn s ct bin =0.49rad/sTn s ct pha =4.71rad/s tr bin =25dB tr pha =101

H thng trn n nh : ng cong Nyquist ca h h khng bao im (-1, j0) (theo chiu m cng chiu kim ng h)2.Vi K=400, dng biu Nyquist:

Tn s ct bin =6.93rad/sTn s ct pha =4.71rad/s tr bin =-7.04dB tr pha =-25

H thng trn khng n nh v theo tiu chun Nyquist: ng cong Nyquist ca h h khng bao im (-1, j0) (theo chiu m cng chiu kim ng h) th h thng kn n nh. Nhng biu trn ta thy ng cong Nyquist c bao im (-1,j0).

IV. KHO ST H THNG DNG PHNG PHP QNS: TS = 1>> MS = conv([1 3],[1 8 20]) >> G = tf(TS,MS) >> rlocus(G) >> grid on

1. K=4121. n = 4 >>K=51.11. = 0,7>>K=211. P=25%>>K=81.61. Txl=4>>K=186

V.NH GI CHT LNG H THNG:

1.Vi K=412

H thng n nh

2.Vi Kgh=81.6, u vo l hm nc n v, v p ng qu ca h thng trong t = 0 5sec

vt l =22.4%txl=2 s

KP = G(s)H(s) = =1.36

exl = = = 0.424

3.Kgh=186

Txl = 3.38 (sec).Kim chng li ta thy h thng c thi gian xc lp l 3.38 (sec) ch khng phi l 4 s4. V hai p ng qu cu b,c trn cng 1 hnh v:

PHN B:NG DNG SIMULINK M PHNG V NH GI CHT LNG H THNG

I.KHO ST M HNH H THNG IU KHIN NHIT :1.Kho st h h,nhn dng h thng theo m hnh Ziegler-Nichols:

L = 20 (sec)T = 173 (sec)

2.Kho st m hnh iu khin nhit ch ON-OFF:

a)Kho st qu trnh qu h thng vi cc gi tr ca khu Relay:

Kho st vng tr +5/-5

b) Sai s ng ra so vi tn hiu t v thi gian ng ngt ng vi cc trng hp khu RelayVng tre1-e2Chu k ng ngt (s)

+1/-152.363

+5/-5127.597

+10/-101912.5127

+20/-2030.522.5170

Nhn xt: Ta thy vng tr cng ln th sai s ng ra cng ln th chu k ng ngt cng ln (tc l tn s ng ngt cng nh) v ngc li. V vy sai s ng ra gim xung xp x bng 0, th vng tr phi gim xung xp x bng 0, lc chu k ng ngt xp x bng 0 v tn s ng ngt tng ln rt ln. Trong thc t ta khng th thc hin b iu khin ON OFF nh vy v b iu khin ch c kh nng p ng tn s gii hn v hot ng vi tn s ng ngt qu cao lm cho li in cung cp khng th p ng, lm h li in. Do cn phi la chn vng tr sao cho ph hp vi kh nng p ng tn s ca b iu khin v va phi nh ph hp vi yu cu thit k (gim sai s ng ra).3.Kho st m hnh iu khin nhit dng phng php Ziegler-Nichols (iu khin PID)

L = 20 (sec) T = 173 (sec) K = 300a) Cc gi tr Kp, KD ,KI

KP = = = 0.0346 KI = = = 2.883*10^-6KD = 0.5 Kp L = = 0,5.0,0395.18=0.355

b) Chy m phng h thng

- vt l:PID > ON-OFF: b iu khin PID c vt l rt ln m trong khi b ON-OFF c th thit k cho vt l b- Sai s ng ra: PID < ON-OFF: b iu khin PID sai s ng ra la 0 nh hn b iu khin ON-OFF- Thi gian xc lp: PID > ON-OFF.- p ng ng ra trng thi xc lp ca b PID khng dao ng,cn i vi b ON-OFF th dao ng quanh gi tr t.

II.KHO ST M HNH IU KHIN TC ,V TR NG C DC:1.Kho st m hnh iu khin tc ng c DC

a) Thc hin kho st h thng vi b iu khin P (KI = KD = 0)

Kp1102050100

POT %0.241.17.90.670.5

exl16.7210.40.2

txl0.410.50.510.510.517

Nhn xt: Khi KP tng ta thy: vt l gim Sai s xc lp gim Thi gian xc lp thay i ng k Khi KP tang lm gim sai s xc lp nn s ci thin c cht lng h thng,tuy nhin nu KP qu ln s lm h thng km n nh hn,nu KP ln hn Kgh th h thng s mt n nh

b) Thc hin kho st h thng vi b iu khin PI (KP = 2; KD = 0)KI 0.10.50.812

POT (%)000.52.612.7

exl00000

txl (s)23.11.40.520.522.2

Nhn xt: Khi tng khu tch phn lm KI tng dn n chm p ng qu , tng vt l, gim sai s xc lp ng ra..nu tng qu mc th h thng km n nh hn.

c) Thc hin kho st h thng vi b iu khin PID (KP = 2; KI = 2)KD0.10.20.512

POT (%)11.410.710.416.325.2

exl00000

txl (s)2.22.232.553.56.9

Nhn xt: B hiu chnh PID tt hn PI v PD Trong , c thm khu PD tng ng vi thm 1 zero c phn thc m vo h thng, ko QNS ri xa trc o nn lm gim vt l h thng. Khu PD l mt trng hp c bit ca b hiu chnh sm pha, nn n c tc dng ci thin p ng qu , gim thi gian xc lp. B iu khin PID, vi cc thng s KP, KI c chn trc (nh trong trng hp ny) th vic la chn KD phi ph hp tho mn yu cu v POT, txl nu KD qu ln li lm cho h thng c cht lng xu hn. Tm li PID c tc dng : Gim sai s xc lp, gim vt l, gim thi gian qu . Gim nhiu tn s cao. Gim c s thay i t ngt ng ra ca b PID nn h thng iu chnh c m hn, ko di tui th ca i tng m h iu khin.

2.Kho st m hnh iu khin v tr ng c DC

a) Thc hin kho st h thng vi b iu khin P (KI = KD = 0)b) Nhn xt: Khi KP tng t 1 ln 100 (KI = KD = 0) ta thy: vt l tng. Sai s xc lp gim Thi gian xc lp gim Khi KP tng -> gim sai s xc lp -> ci thin c cht lng h thng. Khi Kp cng ln -> p ng ca h thng cng dao ng, vt l cng cao. Nu KP qu ln s lm h thng km n nh hn, nu KP > Kgh th h thng s mt n nh.

c) Thc hin kho st h thng vi b iu khin PI (KP = 2,KD = 0)

KI0.10.50.812

POT (%)001.13.9

exl2.10.2200

txl (s)685.97.4

Nhn xt: Khu tch phn tham gia vo lm chm p ng qu , tng vt l, gim sai s xc lp ng ra. Do khu PI l 1 trng hp c bit ca b hiu chnh tr pha nn n c c im ca b hiu chnh tr pha. Mt khc thm vo h thng khu PI tng ng vi vic thm vo 1 cc ti gc to v 1 zero c phn thc m QNS b y v pha phi mt phng phc - h thng km n nh hn. So vi vi b iu khin P th b iu khin PI vi h s KI thch hp s cho cht lng tt hn.d) Thc hin kho st h thng vi b iu khin PID (KP = 2,KI = 1)KD0.10.20.50.81

POT (%)4.13.23.1

exl0.040.040

txl (sec)7.989.1

Nhn xt:Khi KD tng ln (KP = 2; KI = 1) ta thy: vt l gim xung. Sai s xc lp gim xung. Thi gian xc lp gn nh khng i Khu vi phn (khu gia tc) tham gia vo lm gim thi gian xc lp ng ra. Khi ta tng h s KD ln vi 1 gi tr thch hp th cht lng h thng c ci thin, vt l gim xung v thi gian xc lp cng gim xung. V khu vi phn lm tn hiu qua n tr nn phng hn do p ng cng phng hn (khng b thay i t ngt) nn cht lng h thng tt hn. B hiu chnh PID c cc u im ca PI v PD v loi tr c sai s , lm gim vt l lm cho vic iu khin chnh xc hn v gim c s thay i t ngt ng ra ca b PID nn h thng iu chnh c m hn, ko di tui th ca i tng lin tc m h iu khin. Trong , c thm khu PD tng ng vi thm 1 zero c phn thc m vo h thng, ko QNS ri xa trc o nn lm gim vt l h thng. B iu khin PID, nu c cc thng s KP, KI c chn trc (nh trong trng hp ny) th vic la chn KD phi ph hp tho mn yu cu v POT, txl nu tng KD qu ln li lm cho h thng c cht lng xu hn. Tm li, phi la chn ph hp c 3 thng s KP, KI, KD, Gim sai s xc lp, gim vt l, gim thi gian qu .Gim nhi u tn s cao.Gim c s thay i t ngt ng ra ca b PID nn h thng iu chnh c m hn, ko di tui th ca i tng m h iu khin.

Kt lun v vai tr ca cc khu P,I,D: Khu t l P: lm gim sai s xc lp v vt l, tuy nhin KP tng li lm h c dao ng. nu KP>Kgh th h s mt n nh. Khu tch phn I: lm gim mnh sai s xc lp, p ng chm li, tng vt lKhu vi phn D: gim vt l, gim thi gian xc lp.