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Bi th nghim s 4

4.1 > mc ch :

4.2 > yu cu :4.2.1 > kho st m hnh iu khin nhit :

a.hm truyn gn ng ca l nhit :

=> Gs =180

250

s

*20=5000/(80s+1)

b.dng Simulink xy dng m hnh iu khin vng hl nhit :

M hnh kho st vng h

p ng qu ca h trn :

= > nhn xt :

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c.thit k biu khin PID cho l nhit dng phng php Ziegler Nichols :

Bng thng s PID ca phng php ZieglerNichols th nht

Thngs

B iu khin

KP TI TD

PKT

T

1

2 0

PIKT

T

1

29.0 T1/0.3 0

PIDKT

T

1

22.1 2T1 0.5T1

Tp ng h h ta c T1= 9(s) , T2=6(s) , k =5000

K p =KT

T

1

22.1 = 1.25000*9

6= 1.6*10^-4

Ki=Kp/(2*T1)=(1.6*10-4)/(2*9)=8.89*10-6

Kd=0.5*Kp*T1=0.5*1.6*10^-4*9=7.2*10^-4

d.xy dng m hnh iu khin nhit nh sau :

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d.1.thc hin kho st h thng :

tc ng ca vic tng mt thng sc lp

Thng s Thi giankhi ng

Qu Thi gianxc lp

Sai s nnh

n nh

kp Gim Tng Thay inh

Gim Gim cp

ki Gim Tng Tng Gim ngk

Gim cp

kd Gim t Gim t Gim t V l thuytkhng tc

ng

Ci thinnu kd nh

vi biu khin p (ki=0,kd=0)

- Kp =0.01

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- Kp=0.02

- Kp=0.03

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- Kp = 0.04

- Kp =0.05

kp 0.01 0.02 0.03 0.04 0.05

%max 0 31 81 240 300

exl 0.3 0.17 He khong

on dinh

He khong

on dinh

He khong

on dinh

txl 100 500 He ko on

dinh

He khong

on dinh

He khong

on dinh

=>khi thay i kp thi h thng gim sai s ,gim vt l,nhng thi gian xlthay i tng ln, Kp cng ln h d mt n nh.

V Khu t l c cho bi:

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trong

: tha s t l ca u ra: li t l, thng s iu chnh

: sai s: thi gian hay thi gian tc thi (hin ti)

Da vo cng thc trn ta c ng ra h thng ko i .ma kp tang th e(t) gim.

vi biu khin pi (kp = 0.024, kd = 0)

- Ki = 0.0001

- Ki =0.0003

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-ki = 0.0005

- Ki = 0.0007

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- Ki = 0.001

Ki 0.0001 0.0003 0.0005 0.0007 0.001

%max 62 83 115 139 170

Exl 0 0 H ko nnh

H ko nnh

H ko nnh

Txl 400 H ko nnh

H ko nnh

H ko nnh

H ko nnh

=>

V Tha s tch phn c cho bi:

trong

: tha s tch phn ca u ra: li tch phn, 1 thng s iu chnh

: sai s: thi gian hoc thi gian tc thi (hin ti)

: mt bin tch phn trung gian

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vi biu khin pi (kp = 0.024, ki = 0)

- Kd = 0.05

- Kd = 0.1

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- kd = 0.3

- kd = 0.5

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- kd = 0.9

Kd 0.05 0.1 0.3 0.5 0.9

%max 50 45 45 31 34

Exl H daong

H daong

H daong

H daong

H daong

Txl H daong

H daong

H daong

H daong

H daong

Tha s vi phn c cho bi:

trong

: tha s vi phn ca u ra: li vi phn, mt thng s iu chnh

: Sai s: thi gian hoc thi gian tc thi ( hin ti)4.2.2. kho st m hnh iu khin tc ng c:a> h pt trng thi :Ta c iu kin ban u

U(t) = i(t)R + Ldt

tdiu )(+ E(t)

dt

tdiu )( = - iR / L - ke n/L +

U / L (1)

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Phng trnh cn bng moment

M(t) = kft + j (d (t)/dt)

Dn / dt = km i/ 2j (2)

T (1) (2) ta c h pt trng thi :

dt

tdiu )( = - iR / L - ke n/L + U / L

Dn / dt = km i/ 2j

b> hm truyn h thng : = 0.5

ta c

T = L / R l hng s thi gian in t ca ng cTC = J / Kf l hng s thi gian in c ca ng c

T (1) => i = u/ [R (1 + Ts)]T (2) => (s) = km i/ Kf(1 + Tc s)

Vy hm truyn ca h thng

G(s) = 0.25/[(1 + 0.25s)(1+0.1s)]

c>thit k b iu khin pi theo tiu chun modul ti u :ta c

R(s) = (1 + 0.25s)/2*0.25*0.1s

= (1 + 0.25s)/0.05s

d>thit k m hnh iu khin tc ng c:tn hiu u vo hm nc n v r(t) = 200 l gi tr final value = 200s khi

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D1> khu p(ki=0 , kd=0)

kp 1 10 20 30 50

max % 0 42 90 118 145

exl 160 H daong

H daong

H daong

H daong

txl 27 H daong

H daong

H daong

H daong

- Vi kp = 1

- Vi kp = 10

- Kp =20

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-kp=30

- Kp=50

d.2>pi(kp=33.323,kd=0)

ki 1 50 100 200 300

max % 127 Mt n

nh

x x x

exl H daong

Mt nnh

x x x

txl H daong

Mt nnh

x x x

- Ki=1

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- Ki=50

- Ki=100

- Ki=200

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- Ki= 300

Ki cng ln th h cng mt n nh

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