Tomtat: con lắc ngược

8/12/2019 Tomtat: con lc ngc

1/26

B GIO DC V O TO

I HC NNG

MINH TIN

THIT K, CH TO M HNH IU KHIN

CN BNG CON LC NGC HAI BC T DO

Chuyn ngnh : Cng ngh Ch to myM s : 60.52.04

TM TT LUN VN THC S K THUT

Nng - Nm 2013


2/26

Cng trnh c hon thnh ti

I HC NNG

Ngi hng dn khoa hc: PGS.TS. PHM NG PHC

Phn bin 1: PGS.TS. NGUYN VN YN

Phn bin 2: PGS.TS. PHM PH L

Lun vn c bo v trc Hi ng chm Lun vn tt

nghip thc s K thut hp ti i hc Nng vo ngy 18

thng 04 nm 2013.

C th tm hiu lun vn ti:

- Trung tm Thng tin - Hc liu, i Hc Nng

- Trung tm Hc liu, i Hc Nng


3/26

1

M U

1. L do chn ti

Trong thc t, nhiu cng trnh c m hnh dng con lcngc nh: nh cao tng, cn bng trong ch to robot ngi, thp

v tuyn, gin khoan, tu thy, cng trnh bin. S gia tng v quy

m kt cu s dn n cc p ng ng lc phc tp ca kt cu v

s sinh ra cc dao ng lm gim bn ca cng trnh, v vy

nghin cu cc dao ng ny v lm cn bng h thng c m hnh

dng con lc ngc l vn ang c quan tm.Vi iu khin ti u pht trin mnh m trong nhng nm

gn y to ra c s xy dng cc h thng my mc phc tp,

nhng h c kh nng cung cp kinh nghim iu khin h thng

hay cn gi l cc h tr gip quyt nh.

T cc vn trn, ta thy cn thit phi nghin cu v con lc

ngc nhm nm bt v pht trin k thut iu khin phc v chonhu cu sn xut, phc v hc tp, nghin cu.

2. Mc ch ca ti

iu khin cn bng con lc ngc nc ta c nghin

cu nhm ch to m hnh ng dng cho cc lut iu khin hin

i t lm c s ng dng vo trong sn xut.

ng dng l thuyt iu khin ti u thit k b iu khin

gi cn bng con lc ngc.

Thit k, ch to m hnh thc nghim

3. Phm vi v ni dung nghin cu

3.1. Phm vi

Ngin cu con lc ngc hai bc t do.

iu khin cn bng con lc ngc hai bc t do bng b iu

khin s dng cc phng php iu khin ti u.


4/26

2

nh gi kt qu da trn m hnh thc nghim.

3.2. Ni dung nghin cu

Nghin cu l thuyt v phng php xy dng m hnh tonhc, lp phng trnh vi phn chuyn ng ca con lc ngc hai

bc t do trn c s phng php bin phn Lagrange-Euler

S dng phn mm Matlab lm cng c xy dng m hnh v

m phng h thng;

kim nghim kt qu nghin cu, ta ch to m hnh con

lc ngc hai bc t do. Thng qua qu trnh hot ng ca m hnh,ta nh gi kt qu nghin cu c.

4. Phng php nghin cu

ti nghin cu c thc hin theo phng php kt hp

gia l thuyt v thc nghim. C th nh sau:

Nghin cu cc ti liu lin quan, trn c s tnh ton

thit k b iu khin cn bng con lc ngc hai bc t do.Ch to m hnh kim chng cc kt qu.

5. ngha khoa hc v thc tin

Con lc ngc l c s to ra cc h thng t cn bng nh:

xe hai bnh t cn bng, cn bng robot ngi, thp v tuyn, gin

khoan, cng trnh bin

Khi l thuyt v cc b iu khin hin i ngy cng hon

thin hn th con lc ngc l mt trong nhng i tng c p

dng kim tra cc l thuyt .

To ra phng php hc tp nghin cu trc quan bng m

hnh c th. Bc u tip cn k thut iu khin chnh xc.

6. Cu trc ca lun vn

Cu trc ca lun vn gm c bn chng.

- Chng 1: M hnh ha con lc ngc hai bc t do


5/26

3

- Chng 2: L thuyt iu khin ti u

- Chng 3: Thit k b iu khin cn bng con lc ngc

hai bc t do- Chng 4: Thit k, ch to m hnh iu khin cn bng

con lc ngc hai bc t do

CHNG 1

M HNH HA CON LC NGC HAI BC T DO

1.1. CC NGHIN CU HIN NAY TRN TH GII

1.2. M HNH CON LC NGC

Xt h thng con lc ngc c gn vo xe v c ko bi

ng servo DC. Yu cu ca bi ton l iu khin v tr xe v gi

cho con lc ngc lun thng ng (con lc lun cn bng).

Hnh 1.7: M hnh con lc ngc hai bc t do


6/26

4

1.3. M HNH TON HC CA H CON LC NGC HAI

BC T DO

Con lc 11 1 1 1 1 1 1

1 1 1 1 1 1 1

z z l sin z z l cos

y l cos y l sin

= + = + = =

Con lc 2

2 1 1 2 2 2 1 1 1 2 2 2

2 1 1 2 2 2 1 1 1 2 2 2

z z L sin l sin z z L cos l cos

y L cos l cos y L sin l sin

= + + = + + = + = +

ng nng xc nh theo cng thc

21T mv

2= Trong 2 2 2v z y= +

ng nng ca xe 20 01

T m z2

=

Hm tiu tn ca xe 20 01

D c z2

=

ng nng ca con lc 12

2 2 2 21 1 1 1 1 1 1 1 1 1 1

1 1 1T m z l cos m l sin J

2 2 2 = + + +

Hm tiu tn ca con lc 1

2

1 1 1

1D c

2=

ng nng ca con lc 2

2 2

2 2 2 2 2

2

2 1 1 1 2 2 2

22

2 1 1 1 2 2 2 2 2

1 1T m v J2 2

1m z L cos l cos

2

1 1m L sin l sin J

2 2

= +

= + + +

+ +

Hm tiu tn ca con lc 2 22 2 21

D c2=


7/26

5

ng nng ca h con lc ngc hai bc t do2

2 2 2 2

0 1 2 0 1 1 1 1 1 1 1 1

22

1 1 2 1 1 1 2 2 2

2

2 1 1 1 2 2 2 2 2

1 1 1T T T T m z m z l cos m l sin

2 2 2

1 1J m z L cos l cos

2 2

1 1m L sin l sin J

2 2

= + + = + + + +

+ + + +

+ +

Hm tiu tn ca h con lc ngc hai bc t do

2 2 2

0 1 2 0 1 1 2 2

1 1 1

D D D D c z c c2 2 2= + + = +

Th nng ca xe 0V 0=Th nng ca con lc 1 1 1 1 1V m gl cos= Th nng ca con lc 2 ( )2 2 1 1 2 2V m g L cos l cos= + Th nng ca h con lc ngc hai bc t do

( )0 1 2 1 1 1 2 1 1 2 2V V V V m gl cos m g L cos l cos= + + = + +

Phng trnh Lagrange

( ) ( ) ( )

( ) ( )

( )

2 2 2 2 2 20 1 2 1 1 2 1 1 1 2 2 2 2

1 1 2 1 1 1 2 2 2 2 2 1 2 1 2 1 2

1 1 2 1 1 2 2 2

L T V

1 1 1L m m m z m l m L J m l J

2 2 2

m l m L z cos m l z cos m L l cos

m l m L gcos m l gcos

=

= + + + + + + + +

+ + +

+

Dng phng php Lagrange Euler tm phng trnh vi phnchuyn ng ca h khi xt n ma st gia xe-thanh trt v ma st

ti cc khp

( ) ( )

( )

0 1 2 1 1 2 1 1 1 2 2 2 2

2 20 1 1 2 1 1 1 2 2 2 2

L L Df

t z z z

m m m z m l m L cos m l cos

c z m l m L sin m l sin f

+ = + + + + + +

+ =


8/26

6

( ) ( )( ) ( )

( )

11 1

2 21 1 2 1 1 1 1 1 2 1 1 1 1

22 1 2 1 2 2 2 1 2 2 1 2

1 1 2 1 1

L L D0

t

m l m L J m l m L zcos c

m L l cos m L l sin

m l m L gsin 0

+ =

+ + + + + + + + =

( ) ( )( )

22 2

22 2 2 2 2 2 2 2 2 2 1 2 1 2 1

22 1 2 1 2 1 2 2 2

L L D0

t

m l z cos m l J c m L l cos

m L l sin m l gsin 0

+ =

+ + + + =

t cc s hng nh sau:

( )

1 0 1 2 2 1 1 2 1

2 23 2 2 4 1 1 2 1 1

25 2 1 2 6 2 2 2

7 1 1 2 1 8 2 2

h m m m ;h m l m L

h m l ;h m l m L J

h m L l ;h m l J

h m l m L g;h m l g

= + + = +

= = + +

= = +

= + =a h phng trnh v dng

( ) ( )

( ) ( )

2 21 2 1 1 3 2 2 0 2 1 1 3 2 2

22 1 4 1 5 1 2 2 1 1 5 1 2 2 7 1

23 2 5 1 2 1 6 2 2 2 5 1 2 1 8 2

h z h cos h cos c z h sin h sin f

h z cos h h cos c h sin h gsin 0

h z cos h cos h c h sin h sin 0

+ + + = + + + + =

+ + + =

Chuyn tip v dng cc ma trn:

M( ) N( , ) H( ) Rf + + =

Trong :

1 2 1 3 2

2 1 4 5 1 2

3 2 5 1 2 6

h h cos h cos

M h cos h h cos( )

h cos h cos( ) h

=


9/26

7

( )

( )

0 2 1 1 3 2 2

1 5 2 1 2

5 1 1 2 2

c h sin h sin

N 0 c h sin

z h sin c

=

7 1

8 2

0

H h sin

h sin

=

[ ]T

R 1 0 0=

Chng ta thy y l mt h phi tuyn. Do thit k b

iu khin vi mc tiu n nh cc thng s trong h thng trong

min gi tr cn bng, chng ta tuyn tnh ha h vi gi thit cc

gc1

,2

nh. Khi ta c c:

( )

( )

1 2 1 2

1 2

2 2

1 2

1

2

1 1

2 2

sin

cos 1

0cos 1

cos 1

sin

sin

=

=

= = = =

=

= H phng trnh tr thnh

1 2 1 3 2 0

2 1 4 1 5 2 1 1 7 1

3 5 1 6 2 2 2 8 2

h z h h c z f h z h h c h 0

h x h h c h 0

+ + + =+ + + =

+ + + =

Cc ma trn tr thnh:

1 2 3

2 4 5

3 5 6

h h h

M h h h

h h h

=

;0

1

2

c 0 0

N 0 c 0

0 0 c

=

;7 1

8 2

0

H h

h

=

Trong M l i xng v khng suy bin.


10/26

8

a phng trnh vi phn chuyn ng ca h v dng ma trn

1 2 3 0

2 4 5 1 1 1 7 1

3 5 6 2 2 2 8 2

z zh h h c 0 0 0 f

h h h 0 c 0 h 0h h h 0 0 c h 0

+ + =

1.4. KIM NGHIM KT QU M HNH HA BNG

MATLAB

>> Kt qu tnh bng phn mm Matlab hon ton trng khp

vi kt qu tnh bng tay. Vy kt qu ca phng trnh vi phn

chuyn ng ca h con lc ngc hai bc l ng.

CHNG 2 : L THUYT IU KHIN TI U2.1 CHT LNG TI U

2.1.1 c im ca bi ton ti u

2.1.2. iu kin thnh lp bi ton ti u

2.1.3. Ti u ho tnh v ng2.2 XY DNG BI TON TI U

2.2.1. Ti u ha khng c iu kin rng buc

2.2.2. Ti u ha vi cc iu kin rng buc

2.3 CC PHNG PHP IU KHIN TI U

2.3.1 Phng php bin phn c in Euler_Lagrange

2.3.2 Nhn xt

2.4 IU KHIN TI U CC H TUYN TNH VI

PHIM HM DNG TON PHNG

2.4.1 n nh Lyapunov i vi h thng tuyn tnh

2.4.2 iu khin ti u h tuyn tnh vi ch tiu cht lng

dng ton phng _ Phng trnh Riccati i vi h lin tc

2.4.3 Cc bc gii bi ton ton phng tuyn tnh

2.4.4 Nhn xt


11/26

9

CHNG 3 : THIT K B IU KHIN GI CN

BNG CON LC NGC HAI BC T DO

3.1. CC THNG S M HNH CON LC NGC

Tham s K hiu Gi tr n v

XeKhi lng ca xe 0m 1.037 kg

H s cn nht gia xe v thanh trt 0c 0.005 kgm2s

-

1

Conlc 1

Moment qun tnh 1J

0.0017 kgm

2

Khi lng 1m 0.088 kg

Chiu di 1L 0.2 m

Chiu di t tm quay n trng tm 0.102 m

H s cn nht ti khp quay 1 1c 3x10-3 kgm

2s

-

1

Conlc 2

Moment qun tnh 2J 0.059 kgm2

Khi lng 2m 0.110 kg

Chiu di 2L 0.4 m

H s cn nht ti khp quay 1 2c 5x10-3 kgm

2s

-

1

3.2. THIT K B IU KHIN LQR

3.2.1. B iu khin LQR

H phng trnh tuyn tnh m t h thng lc ny tr thnh:

x(t) Ax(t) Bu(t)

y(t) Cx(t) Du(t)

= + = +

Trong


12/26

10

0 0 0 1 0 0

0 0 0 0 1 0

0 0 0 0 0 1A1.2460 0.0641 0.0459 0.0031 0.0001

0 63.8739 16.6718 0.1948 0.1598 0.0232

0 24.7046 28.3039 0.0149 0.0618 0. 393

0

0

=

B0.9179

3.896

0.2971

0

0

0

=

;

1 0 0 0 0 0

0 1 0 0 0 0

0 0 1 0 0 0C0 1 0 0

0 0 0 0 1 0

0

0 0

0 0 0 0 1

=

;

0

D

0

0

00

0

=

Vi cc thng s ca mt h thng c cc cc nh sau:

1

2

3

4

5

6

8.6345

8.4630

4.3544

p 0

4.3874

0.

p

p

p

0405

p

p

=

= =

= =

=

H thng c 2 cc nm bn phi mt phng phc, do hthng khng n nh.

kim tra tnh iu khin v quan st c ca h thng, ta

tnh hng ca ma trn:2 3 4 5rank B AB A B A B A B A B 6 =

2 3 4 5 Trank[C CA CA CA CA CA ] 6=

Chng ta thy hng ca cc ma trn ny u bng 6, nh vy

h thng chng ta kho st iu khin c v quan st c.


13/26

11

Hnh 3.1-. M hnh n nh h thng s dng b iu khin LQR

Tt c trng thi ca h thng c hi tip v qua ma trn

li K. Xd l gi tr t vo b iu khin.

1 1d

12

23d d

4

15

6 2

zx x

x 0

x 0X ; X ; E X X

zx 0

x 0

0x

= = = =

Ma trn hi tip tm c:

K =[37.9; -739.1; 1330.7; 91.3; -12.5; 252.1]T

Vi gi tr cu K, h thng n nh vi cc cc:

1

2

3

4

5

6

16.1012 7.0897i

16.1012 7.0897i

15.5221

2.5665

2.1551 1.9498i

2.1551

p

p

p

p

1.949p i

p

8

+

= = =

=

==

+

Cc cc ca h thng khi c b iu khin nm bn tri mt

phng phc, do h thng l n nh.

3.2.2. Dng Matlab v Gii thut di truyn tm ma trn hi

tip ti u cho b iu khin LQR


14/26

12

3.3. THIT K B IU KHIN PD

Hnh 3.2- M hnh n nh h thng s dng b iu khin PD

Tn hiu iu khin u c xc nh qua biu thc sau:

1 D1 P1

2 D2 P2

3 D3 P3

D (S) k s k

D (S) k s k

D (S) k s k

= + = + = +

Hm truyn ca h con lc ngc hai bc t do4 3 2

1 6 5 4 3 2

4 3 2

2 6 5 4 3 2

4 3 2

3 6 5 4 3 2

0.9s 0.2s 79s 4s 1130G (S)

s 0.25s 92.2s 9.4s 1395.8s 56s

3.9s 0.2s 115.2sG (S)

s 0.25s 92.2s 9.4s 1395.8s 56s

0.3s 0.3s 115.2sG (S)

s 0.25s 92.2s 9.4s 1395.8s 56s

+=

+ + + + =

+ + + + =

+ + +

3.4. THIT K B IU KHIN PID

3.4.1. iu khin PID

3.4.2. B iu khin PID

Mt m hnh b iu khin PID cng c xy dng tng t

nh b iu khin PD c th hin trn hnh 3.7.


15/26

13

Hnh 3.7- M hnh n nh h thng sdng b iu khin PID

Tn hiu iu khin u c xc nh qua biu thc sau:2

D1 P1 I11

2D2 P2 I2

2

2D3 P3 I33

k s k s k C (S)

s

k s k s k C (S)

s

k s k s k C (S)s

+ +=

+ + =

+ +=

Hm truyn ca h con lc ngc hai bc t do

4 3 2

1 6 5 4 3 2

4 3 2

2 6 5 4 3 2

4 3 2

3 6 5 4 3 2

0.9s 0.2s 79s 4s 1130G (S)

s 0.25s 92.2s 9.4s 1395.8s 56s

3.9s 0.2s 115.2sG (S) s 0.25s 92.2s 9.4s 1395.8s 56s

0.3s 0.3s 115.2sG (S)

s 0.25s 92.2s 9.4s 1395.8s 56s

+=

+ + + +

= + + + + =

+ + +

Vi b iu khin PD, PID th vic la chn nhiu thng s

kh l kh khn, chng ti la chn cc thng s theo phng php

s dng gii thut di truyn.


16/26

14

3.5. KT QU M PHNG

3.5.1. B iu khin LQR

Kt qu m phng h con lc ngc hai bc t do trong thi

gian 5s

V tr xe Gc con lc 1

Gc con lc 2 Vn tc xe

Vn tc con lc 1 Vn tc con lc 2


17/26

15

Lc tc ng ln xe

3.5.2. B iu khin PD3.5.3. B iu khin PID

3.6. SO SNH CHT LNG CA CC B IU KHIN

v tr xe gc con lc 1

gc con lc 2 vn tc xe

(iu kin ban u: z = 0.02 [m], 1=0.087 [rad], 2=-0.087 [rad])


18/26

16

Cc kt qu m phng cho thy cc p ng ca h vi cc

thay i khc nhau ca v tr ca xe, chng ta thy cc b iu khin

vn cho cc p ng tt, thi gian xc lp ngnTrong kt qu ny, chng ti xem xt cc yu cu v thi gian

qu , thi gian xc lp v tng bnh phng sai s so snh. Cc

thng s ny c th hin trong cc bng B3.1, bng B3.2 v bng

B3.3.

B3.1.Bng so snh cc p ng ca v tr xe

Cc p ng ca h thng LQR PD PID

Thi gian qu 0.3 0.2 0.25

Thi gian xc lp (s) 3 2.2 2.8

*SSE [m2s] 5.9283 0.6753 0.7190

B3.2. Bng so snh cc p ng ca gc ca con lc 1


Thi gian qu 0.25 0.28 0.18

Thi gian xc lp (s) 1.5 1.3 1.2

SSE [m2s] 2.7485 4.8682 3.1875

B3.3. Bng so snh cc p ng ca gc ca con lc 2


Thi gian qu 0.35 0.15 0.16

Thi gian xc lp (s) 2.5 1.8 1.65

SSE [m2s] 0.8627 0.3768 0.3947

T bng trn, chng ta thy p ng ca cc b iu khin PD

v PID tt hn b LQG. Sai s cng nh thi gian xc lp ca b


19/26

17

iu khin PID tt hn c. Nhng y cng l b iu khin kh la

chn cc thng s nht.

CHNG 4THIT K, CH TO M HNH IU KHIN CN

BNG CON LC NGC HAI BC T DO

4.1. THIT K KT CU C KH

Do kt cu khng chu ti trng ln nn ta chn vt liu ch

to cc gi l nhm hp kim, cc chi tit ny c gia cng trn

my tin v my phay thng thng

Hnh 4.3- M phng lp ghp gia gi bn tri v thanh nh v

Hnh 4.4- Gi bn tri v thanh nh v


20/26

18

Hnh 4.9- M phng lp ghp gia gi bn phi v thanh nh v

Vi tng c th thay i c khong cch trc gia haipuli nhm thay i sc cng dy cp nn ta gn ng c Servo DC

trn mt c cu c th trt theo phng ngang vi gi bn phi

v trn trc ng c gn puli

Hnh 4.10- Gi bn phi v thanh nh vKhi thit k h con lc ngc hai bc vi tng dng

Encorder o gc lch, vic gn Encorder c xc nh nh sau :

Encorder th nht c gn c nh trn xe (chi tit 8), trc

ca Encorder gn cng vi trc quay (chi tit 9) bng mi ghp c

di. Encorder th nht o gc lch ca con lc 1


21/26

19

Encorder th hai c gn c nh trn con lc 1, dng trc

ca Encorder th hai lm khp quay th hai nn ta c th o c

gc lch con lc 2

Hnh 4.13- M phnglp ghp con lc hai bc t do ln xe vbn trt

Hnh 4.14- Con lc hai bc t do ln xe v bn trt


22/26

20

Hnh 4.15- M phng lp ghp h con lc ngc hai bc t do

Hnh 4.16- H con lc ngc hai bc t do


23/26

21

4.2. THIT K MCH IU KHIN

4.2.1 La chn h vi iu khin thit k

PIC 30F4012 ca hng Microchip cho mc ch iu khin bi

l n c mt s u im sau:

- Tc x l nhanh 16bit, dung lng Ram ln thch hp vi

cc ng dng iu khin m, LQR, PID

- Trnh bin dch l ngn ng C thng dng v gn gi vi

ngn ng matlab m phng nn thch hp khi chuyn i.

- y l chip chuyn dng iu khin ng c.- Mch np c th t lp rp d dng vi chi ph thp v h tr

kt ni USB

- Microchip cung cp y cc thng s k thut ca cc

dng PIC

- S lng ngt x l nhiu.

- Dung lng SRAM: 512 Bytes- Su knh chuyn i A/D 10-12 bit nhanh v chnh xc

- H tr Quandrature Encoder Interface

Hnh 4.17- S chn dsPIC 4012

- Ngoi ra, mt kh nng ng tin cy cao ca b nh Flash c

kh nng lu tr d liu trn 40 nm, vi kh nng ghi v xa ln

n 1 triu ln, kh nng chu ng nhit cao (85 C).


24/26

22

S dng trnh bin dch CCS np chng trnh cho

dsPIC4012

4.2.2 Thit k mch iu khinS nguyn l v mch in ca mch driver iu khin

ng c Servo DC 24V c thit k trn phn mm OrCAD

C4100V

C5100V

OUT+OUT-

G_POWER

F1FUSEC1225V/100uF

C1315V/100uF 24V124V

D14

BACK

D15 FOR

J5

P-CONTROL

1234

R35

10K

J6

CON4

1234

D114007

R36

10K

D124007

12VG_12

5V

PGD

INT2

PGC

INT1

VCC

INT0

INT3

HIN2

INDX

LIN2

QEA

HIN1

QEB

LIN1

INT2INT1INT0

TXRX

J10

CON6

123456

U5

dspic30f4012MCLR

1

AN0/VREF-/CN2/RB02

AN1/VREF+/CN3/RB13

AN2/SS1/CN4/RB24

AN3/INDX/CN5/RB35

AN4/QEA/IC7/CN5/ RB46

AN5/QEB/IC8/CN7/ RB57

VSS

8

OSC1/CLKI9

OSC2/CLKO/RC1510

T2/U1ATX/CN1/RC1311

T1CK/U1ARX/CN0/RC1412

VDD 13

IC2/INT2/RD114

EMUC2/OC1/IC1/INT1/RD015

FLTA/INT0/SCK1/OCFA/RE8

16

PGD/EMUD/U1TX/SDO1/SCL/C1TX/RF3 17

PGC/EMUC/U1RX/SDI1/SDA/C1RX/RF2 18

VSS

19

VDD 20

PWM3H/RE5 21PWM3L/RE4 22

PWM2H/RE3 23

PWM2L/RE2 24

PWM1H/RE1 25

PWM1L/RE0 26

AVSS

27

AVDD 28

PGDPGC

C17104

LIN1

U2

IR2110

LO 1HO 7

HIN10

SHDN11

LIN12

VSS13

COM 2

VB 6

VCC 3

VDD9

VS 5

U3

IR2110

LO 1HO 7

HIN10

SHDN11

LIN12

VSS13

COM 2

VB

6

VCC 3

VDD9

VS 5

12V

Q1IRF540N/TO

Q2IRF540N/TO

Q3IRF540N/TO

Q4IRF540N/TO

5V

-

+

U4A

LM393

3

21

8

4

INDX

HIN1

R10.22

12V

J2

ENC

12345

24V

C2

224

Q5

2N5551

C3

224

LIN2

VCC

Q6

2N5551

12V

Q7

2N5551

Q8

2N5551

12V

G_12D16

DZ4.7V

J3

PROG

12345

HIN2

C11104

R18

10

VCC

R19

10

R20

10

G_12

R21

10

R38

R

C10

105

R22

220

R23220

R24

220

R25

220

R26

R

C14104

R27

20K

R31

10K

C15104

VCC

PAD1

PINPAD

PAD2

PINPAD

VCCJ4

UART

1234

G_12C6

10uF

R28

2K

R37

1M

D1

LED

VCCY1

40Mhz

J11

CON2

12

R 29 0

QEB

C7

33

ISO5PC817

1

2

4

3

QEA

C8

33

VCC5V

5V

C1

51

D2

BYV28-1501

2

D3

BYV28-1501

2

R3910K

D4

BYV28-1501

2

R40

2K

D5

BYV28-1501

2

G1

C16104

OUT+ OUT-

R41

RESISTOR SIP 9

1 23456789

INDX

AN1

QEA

AN0

QEB

INT3INT2INT1INT0

AN1

G_12

AN0

RESET

RXTX

INT3

Hnh 4.19- S nguyn l mch driver iu khin ng c

Hnh 4.20- S mch in driver iu khin ng c


25/26

23

Hnh 4.21-Driver iu khin ng c

4.3. LP TRNH IU KHIN CON LC NGC HAI BC

T DO

KT LUN V KIN NGH

1. KT QU NGHIN CU CA TI

Mc tiu ca cc h thng iu khin l ngy cng nng cao

cht lng cc h thng iu khin t ng. Trn thc t c rt nhiu

i tng cn iu khin nhng khng c cc tham s cn thit.

V vy, vic thit k cc b iu khin da trn l thuyt kinh in

gp rt nhiu kh khn. Chnh v l do ny i hi chng ta phi ng

dng cc l thuyt iu khin hin i vo trong thc t. Lun vn

ny ch trng nghin cu xy dng h iu khin ti u cho h conlc ngc hai bc t do da trn nn tng phng php bin phn

Euler_Largrange.

ti thc hin vic la chn m hnh con lc ngc hai

bc t do, trn c s , thit lp thnh cng phng trnh vi phn

chuyn ng ca h v m phng h trn Matlab cho cht lng p

ng h thng tt.


26/26

24

ti cng thc hin thnh cng vic a ra m hnh ton

hc cho vic iu khin cn bng con lc ngc hai bc t do bng

cch s dng phng php iu khin ti u tuyn tnh dng tonphng (LQR) cho i tng iu khin l ng c Servo DC.

Vi kt qu t c, c th ng dng ti vo vic tnh ton

cho robot ngi di chuyn bng hai chn nh ngi hoc cc my s

dng ng c servo. Cc kt qu t c ca ti, c th ng

dng vo vic ging dy v l thuyt iu khin hin i v cc mn

hc v t ng ha, iu khin Bn cnh , c th p dngphng php iu khin ny p dng iu khin ng c theo v

tr mong mun, t c th ng dng vo trong thc t cho cc h

thng yu cu chnh xc cao.

2. HNG PHT TRIN CA TI

ti gii quyt c vn v iu khin cn bng con

lc ngc hai bc t do. Tuy nhin, do thi gian hn ch, vn chathc hin hon chnh. Tuy nhin, vi tin nghin cu ny, ch cn

ci thin v ng c v phn cng mch iu khin, c th l thit b

c tn hiu nhiu Encorder cng lc l c th p dng ch to m

hnh thc hnh cho cc trng i hc.

Vic thit k iu khin ch mi tnh ton thit k vi nguyn

l iu khin tuyn tnh, cha so snh kt qu vi cc phng php

iu khin hin i khc nh Fuzzy, iu khin bn vng,

AdaptiveV vy hng pht trin ca ti l iu khin h con

lc ngc hai bc hoc nhiu bc hn bng cc phng php iu

khin cho h phi tuyn nh: Fuzzy, iu khin bn vng, Adaptive

Documents

Tomtat: con lắc ngược