bao cao thi nghiem cstd - bai 1

8/2/2019 Bao Cao Thi Nghiem CSTD - Bai 1

1/23

BO CO TH NGHIM C STNG

BI TH NGHIM 1

PHN A: NG DNG MATLAB PHN TCH CC H THNG IU KHIN

TNG

TH NGHIM:

1. Tm hm truyn tng ng ca h thng:Th nghim: S dng cc lnh c bn conv,tf,series,parallel,feedback, tm biu thc hm truyntng ng G(s) ca h thng sau:

Bi lm:

>>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)

Transfer function:

2 s^2 + 9 s + 15


2/23

------------------

s^3 + 8 s^2 + 15 s

>> Ght=feedback(G2,H1)

Transfer function:

s

---------------

2 s^2 + 4 s + 8

>> Gnt=series(G13,Ght)

Transfer function:

2 s^3 + 9 s^2 + 15 s

-----------------------------------------

2 s^5 + 20 s^4 + 70 s^3 + 124 s^2 + 120 s

>> G(s)=feedback(Gnt,1)

Transfer function:

2 s^3 + 9 s^2 + 15 s

-----------------------------------------

2 s^5 + 20 s^4 + 72 s^3 + 133 s^2 + 135 s

2. Kho st h thng dng biu bode:Th nghim: kho st h thng phn hi m n v c hm truyn vng h:

G(s) ()( )

a. Vi K=10, v biu bode bin v pha h thng trn trong khong tn s (0.1, 100).b. Da vo biu bode, tm tn s ctbin, d tr pha, tn s ct pha, d tr bin ca

g thng.


3/23

c. H thng trn c n nh khng, gii thch.d. Vp ng qu ca h thng trn vi u vo hm nc n v trong khong thi gian

t=0->10s.

e. Vi K=400, thc hin li t cu a->d.Bi lm:

a. Vi K=10.>> Gs=tf([10],conv([1 0.2],[1 8 20]));

>> bode(Gs,{0.1,100})

>> grid on

Kt qu:

b. Tn s ct bin: 0.455(rad/sec). d tr pha: 103Tn s ct pha: 4.65(rad/sec). d tr bin: 24.8

c. H thng trn l n nh v c d tr bin v pha u dng.d. p ng qu ca h thng:


4/23

Vi K=400:a. Biu bode:


5/23

b. Tn s ct bin: 6.73(rad/sec)

d tr pha: - 24.3Tn s ct pha: 4.65(rad/sec). d tr bin: - 7.27

c. H thng trn l khng n nh v c d tr bin vpha u m.d. p ng qu ca h thng:

3. Kho st h thng dng biu nyquist:


6/23

Th nghim: kho st h thng phn hi m n v c hm truyn vng h:

G(s) ()( )

a. Vi K=10, v biu Nyquist ca h thng.b. Da vo biu Nyquist tm d trbin, d tr pha ca h thng.c. H thng c n nh khng, gii thch.d. Vi K=400, thc hin t cu a->c.

Bi lm:

a. K=10:>> Gs=tf([10],conv([1 0.2],[1 8 20]));

>> nyquist(Gs)>> grid on

>> hold on

>> t=(-1:0.001:1)*2*pi;

>> x=sin(t);

>> y=cos(t);

>> plot(x,y) % vvng trn n v.

>> hold off


7/23

b. Tn s ct bin: 0.455(rad/sec). d tr pha: 103Tn s ct pha: 4.65(rad/sec). d tr bin: 24.8.

Kt qu ging ht nh i vi biu bode.

c. H thng trn l n nh v c d trbin v pha u dng.d. Vi K=400:

Biu Nyquist ca h thng.

>> Gss=tf([400],conv([1 0.2],[1 8 20]));

>> nyquist(Gss)

>> grid on

>> hold on

>> plot(x,y)


8/23

Tn s ct bin: 6.73(rad/sec)

d tr pha: - 24.3

Tn s ct pha: 4.65(rad/sec). d tr bin: - 7.27

H thng trn l khng n nh v c d trbin v pha u m.

4. Kho st h thng dng phng php QNS.Th nghim: H thng hi tip m n v c hm truyn vng h.

G(s) ()( )

, K0.

a. VQNS ca h thng. Da vo QNS tm Kgh ca h thng.b.

Tm K h thng c tn sdao ng tnhin n=4.c. Tm K h thng c h s tt =0.7.

d. Tm K h thng c vt l POT= 25%.e. Tm K h thng c thi gian xc lp (tiu chun 2%) txl=4s.

Bi lm:

a. VQNS:


9/23

>> G=tf([1],conv([1 3],[1 8 20]));

>> rlocus(G)

>> grid on

Nhn vo QNS ta thy Kgh=422.

b. h thng c tn sdao ng tnhin n=4 ta tm giao im ca vng trn n=4 viQNS. Theo thta c K=55.c. h thng c h s tt =0.7 ta tm giao im ca ng =0.7 vi QNS. Theo thta c K=19.8.

d. Theo thta xc nh t cng thc:POT=exp(

) => =0.4

Vy ta c K=76.9

e. Vi txl=4s => n=1. Vy ta tm giao im ca QDNS vi ng thng n=1. Tac K=174.

5. nh gi cht lng h thng.Th nghim: Vi h thng nh phn 4.

a. Vi K=Kghtrn, vp ng qu ca h thng vng kn vi u vo hm nc n v.Kim chng p ng ng ra c dao ng khng?


10/23

b. Vi gi trK tm c cu d, phn 4, vp ng qu ca vng kn vi u vo hmnc n v trong khong thi gian t 0->5s. T hnh vtm vt l v sai s xc lp cah thng. Kim chng li h thng c vt l bng 25% khng?

c. Vi gi trK tm c cu e, phn 4, vp ng qu ca vng kn vi u vo hmnc n v trong khong thi gian t 0->5s. T hnh vtm vt l v sai s xc lp cah thng. Kim chng li h thng c txl=4s khng?

d. V2 p ng qu ca cu b. v c. trn cng mt hnh v. Ch thch trn hnh vpng no l tng ng vi K .

Bi lm:

a. Vi K=422>> G=tf([422],conv([1 3],[1 8 20]));

>> Gk=feedback(G,1);

>> step(Gk,10)

>> grid on

Nhn xt: H thng n nh, ng ra c dao ng nhng khng ng k ta c thxem nh ln nh.


11/23

b. Vi K=76.9>> Gb=tf([76.9],conv([1 3],[1 8 20]));

>> Gbk=feedback(Gb,1);

>> step(Gbk,5)

>> grid on

vt l bng 20.9%, khng bng 25% na.

Sai s xc lp: 2.1s.

c. Vi K=174>> Gc=tf([174],conv([1 3],[1 8 20]));

>> Gck=feedback(Gc,1);

>> step(Gck,5)

>> grid on


12/23

vt l bng 45.4%.

Sai s xc lp: 3.46s.

H thng lc ny khng cn txl=4s na.

d. V2 p ng qu ca cu b. v c. trn cng mt hnh v>> Gb=tf([76.9],conv([1 3],[1 8 20]));

>> Gbk=feedback(Gb,1);

>> step(Gbk,5)

>> grid on

>> hold on

>> Gc=tf([174],conv([1 3],[1 8 20]));

>> Gck=feedback(Gc,1);

>> step(Gck,5)


13/23

---------------------------------------------------------------------------------

PHN 2: NG DNG SIMULINK M PHNG V NH GI CHT LNG

H THNG

TH NGHIM:

1. Kho st m hnh h thng iu khin nhit :1.a) Kho st h h, nhn dng h thng theo m hnh Ziegler-Nichols:

Th nghim: Dng SIMULINK xy dng m hnh h thng l nhit vng hnh sau:


14/23

Step l tn hiu hm nc.

a. Chnh cc gi tr ca hm nc l 1 cng sut cung cp cho l l 100% (Step time = 0,Initial time = 0, Final time = 1). Chnh thi gian m phng Stop time = 600s. M phng vvqu trnh qu ca h thng trn.

b. Trn hnh v ca cu trn, v tip tuyn ti im un tnh thong sL v T theo nhhng dn trong bi th nghim 5. Ch r cc gi tr ny trn hnh v. So snh gi tr L v Tva tm c vi gi tr ca m hnh l nhit tuyn tnh ha.

Bi lm:

a. M phng v vqu trnh qu ca h thng.

b. Theo hnh v ta c:T=165s, L=25s.

Theo m hnh l nhit tuyn tnh ha, ta c:

T=120s, L=30s.

1.b) Kho st m hnh iu khin nhit ON-OFF:


15/23

Th nghim: Xy dng m hnh h thng iu khin ON-OFF:

Trong :

Tn hiu t vo hm nc u(t)=100. Khi relay l biu khin ON-OFF. Gi tr li khi Gain l 150.

a. Chnh Stop time = 600s. Kho st qu trnh qu ca h thng vi cc gi tr ca khurelay nh sau:

b. Tnh sai s ng ra vi tn hiu t v thi gian ng ngt ng vi cc trng hp cakhu relay cu a theo bng sau:

Nhn xt snh hng ca vng trn sai s ng ra v chu kng ngt ca khu relay.


16/23

c. Lu qu trnh qu ca trng hp vng tr (+5 / -5). Trn hnh v ch ra sai s+e1 /-e2 quanh gi trt v chu kng ngt.

d. sai s ng ra st s bng khng th tat hay i gi tr vng tr bng bao nhiu? Chu kng ngt lc ny thay i nh th no? Trong thc t, ta thc hin biu khin ON-OFF nhvy c c khng? Ti sao? Vng tr la chn bng bao nhiu l hp l. Giithch.

Bi lm:

a. Kho st:Vng tr: +1/-1.

Vng tr: +5/-5.


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Vng tr: +10/-10

Vng tr: +20/-20


18/23

b. Tnh sai s ng ra:Vng tr e1 -e2 Chu kng ngt

+1/-1 5 1.5 54

+5/-5 11.7 7 97

+10/-10 19 12 125

+20/-20 30 22 170

Nhn xt: khi vng tr cng ln th sai sng ra cng tng v chu kng ngt cng

tng. Do vng tr ca khu Relay lm cho h thng p ng chm hn vi s thay i ca tnhiu sai s, vng tr cng ln th p ng ca khu Relay cng chm.

c. Vng tr: +5/-5


19/23

d. Thng qua m phng ta nhn thy cng gim rng vng tr th sai s cng tin v0. Chu kng ngt rt nh v tin v 0. Trong thc t ta khng th thc hin biukhin ON-OFF nh vy c v phn t chp hnh bao gicng c vng tr v thigian p ng ca phn t chp hnh. Ta c th s dng gi tr sao cho dung ho giasai s ng ra v chu kng ngt.

1.c) Kho st m hnh iu khin nhit dng phng php Ziegler-Nichols (iukhin PID):

Th nghim: xy dng m hnh iu khin nhit PID nh sau:


20/23

a. Tnh gi tr cc thng s KI, KP, KD ca khu PID theo phng php Ziegler-Nichols tthng sL v T tm c phn 1.a.

b. Chy m phng v lu p ng ca cc tn hiu Scope vit bo co. C th chn liStop time cho ph hp. Trong hnh v phi ch thch r tn cc tin hiu.

c. Nhn xt v cht lng ng ra 2 phng php iu khin PID v ON-OFF.Bi lm:

a. Thng s:KP =

=

=0.0193

KI=

=

=0.000345

KD=0.5* KP*L=0.5*0.0193*28=0.2702

b. Kt qu m phng:

c. Nhn xt:kt quhon ton nh ta mong mun, h thng nhanh chng t gi trt.Phng php iu khin ON-OFF th tn hiu ra lun dao ng quanh gi trt.

Phng php PID cho cht lng tt hn phng php ON-OFF.2. Kho st m hnh iu khin tc , vtr ng c DC:2.a Kho st m hnh iu khin tc ng c DC:

Th nghim: xy dng m hnh h thng iu khin ng c PID nh sau:


21/23

a. Thc hin kho st h thng vi biu khin P v tnh vt l , sai s xc lp, thigian xc lp ca ng ra theo bng sau:

Kp 1 10 20 50 100POT 0% 0.01% 0.01% 0 0

exl 16.7 2 1 0.4 0.2

txl 0.3s 0.45s 0.5s 0.55s 0.6s

Nhn xt: khi KP cang tng th sai s xc lp cng b, nhng thi gian xc lp li tng them, vt lcng c tng nhng rt t. Nh vy khi KP cng ln th lm cho cc ca h thng ra xa trcthc nn thi gian xc lp cng tng ln.

b. Thc hin kho st h thng vi biu khin PI( KP=2, KD=0) v tnh vt l, sai sxc lp v thi gian xc lp theo bng sau:

KI 0.1 0.5 0.8 1 2POT 0% 0% 0.5% 2.5% 12.5%

exl 0 0 0 0 0

Txl 90s 18s 3.5s 7s 5s

Nhn xt: khi KIthay i th sai s xc lp gn nh khng i v bng khng. Khu PI c cim ca khu hiu chnh tr pha lm chm p ng qu lm tng vt l. Khi Ki cng lnth vt lcng tng, v thi gian xc lp cng tng.Nhng vi nhng gi tr KI nh th khuPI s lm cho sai s xc lp tin v 0 chm.

So snh cht lng h thng dng khu hiu chnh P v h thng dng khu hiu chnh PI ta

thy rng khi dng khu hiu chnh P th thi gian xc lp nh, vt l nh tuy nhin sai s xclp khc 0 v h thng c th khng n nh nu Kp qu ln, ngc li khi s dng khu hiuchnh PI th h thng lun c sai s l 0 tuy nhin thi gian xc lp ca h thng l kh ln sovi lhu P v vt l kh cao.

c. Thc hin kho st h thng vi biu khin PID( KP=2, KI=2) v tnh vt l, sai sxc lp v thi gian xc lp theo bng sau:


22/23

KD 0.1 0.2 0.5 0.8 1

POT 11% 10.5% 10.4% 14% 16.3%

exl 0 0 0 0 0

txl 5s 4.7s 4.4s 4s 6.5s

Nhn xt:Sai s xc lp ca b hiu chnh PID l 0.Khi Kd tng th vt l v thi gian xc lpgim nhng khi Kd ln hn gii hn cho php th vt l v thi gian xc xc lp ca h thngtng.

Khu hiu chnh PID c thc xem nh hai khu PI v PD mc ni tip v vy n ccu im v khuyim ca khu PI v khu PD c khnng ci thin p ng qu v lmcho sai s xc lp bng 0 do hm truyn h thng khng c khu vi phn l tng, tuy nhin taphi la chn cc thng s Kp,Ki,Kd cho ph hp nu khng cht lng h thng s khng tht c nh mong mun.

2.b Kho st m hnh iu khin vtr ng c DC:

Th nghim: xy dng m hnh HT iu khin PID vtr ng c DC nh sau:

Thc hin cc phn nh phn 2.a.

Bi lm:

a. iu khin PKP 1 10 20 50 100

POT 0 0 0 12%exl 2 0.2 0.1 0.04

txl 4s 3s 3.5s 10s

Nhn xt :Khi Kp cng tng th sai s xc lp cng gim v tn hiu vo l hm dc n v cnhn truyn h thng c khu vi phn l tng v vy h s sai s xc lp s ch ph thuuc vo1/Kp, v vy khi Kp cng ln th sai s xc lp cng gim.Tuy nhin khi Kp cng tng th thi


23/23

gian xc lp cng ln, h thng cng dao ng v vt l ngy cng cao. Nu Kp tng qu lnth h thng sdao ng v khng cn n nh.

b. iu khin PI(KP=2, KD=0):KI 0.1 0.5 0.8 1 2POT 0 0 0 1% 6%

exl 0.6 0 0 0 0

txl 2s 7.5s 8s 8.5s 9s

Nhn xt : Khu hiu chnh PI c tc dng lm cho sai s xc lp ca h tin dn v0 Ki tngth sai s xc lp bng 0. Tuy nhin khu PI li lm cho h thng km n nh v vt l cngtng khi Ki tng, nguyn nhn l do khu PIc c tnh ca khu hiu chnh tr pha lm chmp ng qu , tng vt l v lm gim sai s h thng.

So sng vi khu P th khu PI lm cho cht lng h thng tt hn v sai s xc lpbng 0, tuy nhin hn ch ca khu PI l lm chm p ng qu v vy thi gian xc lp tngln.

c. iu khin PID( KP=2, KI=2):KD 0.1 0.2 0.5 0.8 1

POT 2.3% 2% 0.5% 0 0.5%

exl 0 0 0 0 0

txl 10s 8s 12s 12s 12s

Nhn xt : Khi KDthay i th sai s xc lp lun bng 0 cn vt lban u gim sau litng ln, tng t nh vy i vi thi gian xc lp. Khu hiu chnh PI l cnh ch c khuhiu chnh PD thay i lm thay i cht lng ca h thng. Ta bit rng khu hiu chnh PDc c im ca mt khu hiu chnh sm pha v vy n lm nhanh p ng qu ca hthng, lm gim thi gian qu v vt l.

Kt lun : c ba khu hiu chnh P, PI,v PID u c tc dng ci thin cht lng hthng. Khu P lm cho sai s xc lp ca h gim tuy nhin nu Kp qu ln th h thng s mtn nh do khu PI(tch phn l tng) c s dng lm gim sai s xc lp v trong khuPI c khu hiu chnh tr pha ci thin p ng xc lp.Nhng nu Ki qu ln cng s lm chothi gian xc lp ca htng v vt l ln v vy khu hiu chnh PID(vi tch phn) c sdng lm gim thi gian xc lp v khu PID c khu hiu chnh sm pha PD ci thin png qu .

bao cao thi nghiem cstd - bai 1

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