nghien cuu va ung dung card dieu khien so dsp de thiet ke bo dieu khien so trong dieu khien chuyen...

S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn

I HC THI NGUYN

TRNG HKT CNG NGHIP

CNG HO X HI CH NGHA VIT NAM

c lp - T do - Hnh phc

-----------***-----------

THUYT MINH

LUN VN THC S K THUT

TI

NGHIN CU V NG DNG CARD IU KHIN S DSP

THIT K B IU KHIN S TRONG IU KHIN

CHUYN NG

Hc vin: inh Vn Nghip

Lp: CHK10

Chuyn ngnh: T ng ho

Ngi HD Khoa hc:TS. Bi Chnh Minh

Ngy giao ti: 01/02/2009

Ngy hon thnh: 31/07/2009

KHOA T SAU I HC CB HNG DN

TS. Bi Chnh Minh

HC VIN

inh Vn Nghip


I HC THI NGUYN

TRNG I HC K THUT CNG NGHIP

----------------***----------------

LUN VN THC S K THUT

NGHIN CU V NG DNG CARD IU

KHIN S DSP THIT K B IU KHIN

S TRONG IU KHIN CHUYN NG

THI NGUYN 2009

Ngnh: T NG HA

M s:

Hc vin: INH VN NGHIP

Ngi HD Khoa hc: TS. BI CHNH MINH


I HC THI NGUYN

TRNG I HC K THUT CNG NGHIP

----------------***----------------

LUN VN THC S K THUT

NGNH: T NG HO

NGHIN CU V NG DNG CARD IU

KHIN S DSP THIT K B IU KHIN

S TRONG IU KHIN CHUYN NG

INH VN NGHIP

THI NGUYN 2009


1

LI CAM OAN

Tn ti l: inh Vn Nghip

Sinh ngy 25 thng 12 nm 1981

Hc vin lp cao hc kho 10 - T ng ho - Trng i hc K thut

Cng nghip Thi Nguyn.

Hin ang cng tc ti khoa in - Trng i hc K thut Cng nghip

Thi Nguyn.

Xin cam oan: ti Nghin cu v ng dng Card iu khin s DSP

(Digital signal Processor) thit k b iu khin s trong iu khin chuyn ng

do thy gio TS. Bi Chnh Minh hng dn l cng trnh nghin cu ca ring

ti. Tt c cc ti liu tham kho u c ngun gc, xut x r rng.

Tc gi xin cam oan tt c nhng ni dung trong lun vn ng nh ni

dung trong cng v yu cu ca thy gio hng dn. Nu sai ti hon ton

chu trch nhim trc Hi ng khoa hc v trc php lut.

Thi Nguyn, ngy 31 thng 7 nm 2009

Tc gi lun vn

inh Vn Nghip


2

LI CM N

Sau su thng nghin cu, lm vic khn trng, c s ng vin, gip

v hng dn tn tnh ca thy gio TS. Bi Chnh Minh, lun vn vi ti

Nghin cu v ng dng Card iu khin s DSP (Digital signal Processor) thit

k b iu khin s trong iu khin chuyn ng hon thnh.

Tc gi xin by t lng bit n su sc n:

Thy gio hng dn TS. Bi Chnh Minh tn tnh ch dn, gip tc

gi hon thnh lun vn ny.

Khoa o to Sau i hc, cc thy gio, c gio thuc b mn T ng ho

Khoa in - Trng i hc K thut Cng nghip Thi Nguyn gip tc

gi trong sut qu trnh hc tp cng nh qu trnh nghin cu thc hin lun vn.

Trung tm Th nghim Trng i hc k thut Cng Nghip, c bit l cc

cn b phng th nghim t ng ho tn tnh gip tc gi xy dng h thc

nghim.

Ton th cc ng nghip, bn b, gia nh v ngi thn quan tm, ng

vin, gip tc gi trong sut qu trnh hc tp v hon thnh bn lun vn.

Tc gi lun vn

inh Vn Nghip


3

MC LC

Ni dung Trang

Trang ph ba

Li cam oan 1

Li cm n 2

Mc lc 3

Danh mc cc hnh v, th 6

CHNG 1. TNG QUAN V H IU KHIN S 11

1.1. L thuyt v h iu khin s 11

1.1.1. Cu trc in hnh ca h iu khin s 11

1.1.2. C s ca iu khin s 21

1.1.2.1. Bin i Z 21

1.1.2.2 Tn hiu v ly mu tn hiu trong h iu khin s 24

1.2. Tng hp h iu khin s 27

1.2.1. L lun chung. 27

1.2.2. iu kin tng hp c b iu khin s trong h. 29

1.2.3. Chn tn s ly mu. 30

1.2.4. Thit k b iu khin s theo phng php lin tc. 32

1.2.4.1. Phng php vi phn 32

1.2.4.2. B iu khin s c xc nh theo hm truyn t 34

1.2.4.3. Phng php dng bin i z 36

1.2.4.4. Tng hp b iu khin c tnh phn t lu gi (ZOH) 37

1.2.5. Thit k b iu khin s theo phng php trc tip 38

1.2.5.1. Phng php qu o nghim s trn mt phng z.

1.2.5.2. B nh hng ca khu tr

38

1.2.5.3. H n nh v tn 40


4

1.2.6. Dng matlab tng hp h iu khin s 41

1.3. iu khin s trong iu khin chuyn ng 41

1.3.1. Mt s cu trc iu chnh c s dng 41

1.3.2. Thit k v m phng h thng bng my tnh 47

CHNG 2. GII THIU CARD DSP DS1104 49

2.1. Gii thiu chung 49

2.2. Cu trc phn cng ca DS1104 51

2.2.1. Cu trc tng quan 51

2.2.2. Ghp ni vi my ch (Host Interface) 53

2.2.3. Cc thnh phn ch yu ca DS1104 59

2.2.3.1. B x l tn hiu s DSP TMS320F240. 59

2.2.3.2. H con AD (Analog to Digital). 65

2.2.3.3. H con DA (Digital to Analog). 67

2.2.3.4. H con Vo/Ra s (Digital I/O) 70

2.2.3.5. H con b m ho so lch 73

2.2.3.6. Thanh ghi iu khin vo ra IOCTL 75

2.2.3.7. S chn I/O Connector ca DS1104 76

2.3. Phn mm dSPACE 78

2.3.1. Ci t dSPACE 79

2.3.2. Cc khi dSPACE trong Simulink 80

2.3.2.1. Cc iu khin vo/ra tng t 81

2.3.2.2. Cc iu khin vo/ra s 81

2.4. Mt s cc tnh nng c bn ca Card DS1104 cho iu khin

chuyn ng.

81

2.4.1. Cc iu khin v tr Encoder 81

2.4.2. iu khin PWM (Pulse Width Modulation) 82


5

2.5. To ng dng vi dSPACE v Simulink 88

2.5.1. To ng dng vi Control Desk 93

2.5.2. Hin th cc iu khin, quan st vi Instrumentation

Management Tools.

94

CHNG 3. XY DNG H IU KHIN CHUYN

NG S DNG CARD DS 1104

100

3.1. Tng hp h iu khin chuyn ng v tr DC servo(theo phng

php tng t)

100

3.1.1. M hnh ton hc ca h 100

3.1.2. Cu trc h iu khin v tr v phng php tng hp cc

mch vng 104

3.1.3. Tnh ton cc thng s h iu khin v tr v cu trc h iu

khin v tr

110

3.1.4. M phng h trn Matlab 114

3.2.H iu khin v tr ng c DC Servo dng b iu khin Fuzzy logic

ng dng Card DS1104

115

3.3. Xy dng h thng iu khin chuyn ng 121

3.3.1 Gii thiu cc thit b trong h thng thc 121

3.3.2. Lp trnh iu khin h 123

3.3.3. Cc c tnh thc nghim h iu khin chuyn ng 124

KT LUN VA KIN NGHI 129

TI LIU THAM KHO 129


6

DANH MC CC BNG, HNH V, TH

Bng 2.2. M t thanh ghi trng thi

Bng 2.3. M t thanh ghi ci t

Bng 2.4. Cc ngt cng ca DSP

Bng 2.5. Qun l cc ngt cng

Bng 2.6. Cc a ch thanh ghi ca h con AD

Bng 2.7. Cc a ch thanh ghi ca h con DA

Bng 2.8. M t thanh ghi ch DA

Bng 2.9. Thanh ghi cng vo/ra

Bng 2.10. Tn cc chn ca DS1104 trn P1A

Bng 2.11. Tn cc chn ca DS1104 trn P1B

Bng 2.12. Bng m t cc chn ca DS1104

Bng 2.13.Cc iu khin v tr encoder ca DS1104

Bng 2.14. Tn cc chn ca cc knh phht xung

Bng 2.15. Tn cc xung PWM 3 pha

Bng 2.16.Tn ca cc knh pht xung PWM 3 pha

Bng 2.17. Tn cc xung PWM 3 pha vector

Bng 2.18.Tn ca cc knh pht xung PWM 3vector

Bng 3.1. Cc thng s cho trc

Bng 3.2. Lut iu khin

Hnh 1.1. Cu trc h iu khin s

Hnh 1.2. S nguyn l b chuyn i s - tng t trong h iu khin s

Hnh 1.3. S nguyn l b DAC

Hnh 1.4. Tn hiu ra ca b DAC

Hnh 1.5. B bin i DAC vi mng in tr

Hnh 1.6. B bin i DAC dng mng in tr R v 2R

Hnh 1.7. S nguyn l chuyn i A/D

Hnh 1.8. S chuyn i A/D song song

Hnh 1.9. S chuyn i A/D theo phng php b


7

Hnh 1.10. B bin i A /D theo nguyn tc servo

Hnh 1.11 : Hm thi gian

Hnh 1.12. Tn hiu lin tc

Hnh 1.13.Tn hiu ri rc

Hnh1.14:B ct mu

Hnh 1.15: Mi quan h qu trnh gin on v lin tc

Hnh 1.16

Hnh 1.17

Hnh 1.18

Hnh 1.19

Hnh 1.20

Hnh 1.21

Hnh 1.22

Hnh 1.23

Hnh 1.24

Hnh 1.25

Hnh 1.26. Cu trc c bn ca iu chnh tc quay

Hnh 1.27. Cu trc ti gin phc v thit k xp x

Hnh 1.28. Cu trc c bn iu chnh gc

Hnh 1.29. Cu trc c bn iu chnh gc ti gin

Hnh 1.30. Cu trc iu chnh b sai s gi tr t

Hnh1.31. Cu trc iu chnh b nhiu

Hnh1.32. Cu trc iu chnh b ngc

Hnh 1.33. Cu trc iu chnh b xui bng phng php m hnh

Hnh1.34. Cc giai on ca mt qu trnh chuyn ng

Hnh 1.35. Cu trc iu khin tng qut ca mt nhnh truyn ng

Hnh 1.36.Cc lut thng dng nhm iu khin chnh xc chuyn ng

Hnh 1.37. Trnh t thit k v m phng h thng bng my tnh

Hnh 2.1- Card DS1104


8

Hnh 2.2. S khi ca DS1104

Hnh 2.3. Vi x l tn hiu s DSP TMS320F240

Hnh 2.4.Bn b nh ca DSP

Hnh 2.5.Bn b nh ngoi vi ca DSP TMS320F240

Hnh 2.6. S khi ca h con AD

Hnh 2.7. nh dng d liu ca ADC 16-bit


Hnh 2.9. Mch u vo ca ADC

Hnh 2.10. S khi ca h con DA

Hnh 2.11. nh dng d liu ca DAC 12-bit

Hnh 2.12. nh dng d liu ch DA

Hnh 2.13. Mch u ra ca DAC

Hnh 2.14. S cu trc ca giao din encoder so lch

Hnh 2.15. Mch u vo ca encoder

Hnh 2.18. To ngun 1,5V t ngun 5V

Hnh 2.16. nh dng ca thanh ghi IOCTL khi c

Hnh 2.17. nh dng ca thanh ghi IOCTL khi ghi

Hnh 2.18. Cc khi ca DS1104 Master PPC

Hnh 2.19. Cc khi trong th vin ca DS1104

Hnh 2.20. Tn hiu encoder v gii hn m

Hnh 2.21. Tn hiu PWM ca Card DS1104

Hnh 2.22. Tn hiu PWM ch i xng

Hnh 2.23. Tn hiu PWM ch khng i xng

Hnh 2.24. iu ch xung PWM ca Card DS1104

Hnh 2.25. iu ch vector khng gian

Hnh 2.26. Cc vector SPWM1, SPWM3, SPWM5 ca DS1104

Hnh 2.27. Lu thut ton thc hin mt ng dng vi Simulink v Control

Desk: (a)- Bc 1; (b)- Bc 2

Hnh 2.28. V d minh ho


9

Hnh 2.29. Thay i tham s khi Transfer Fcn

Hnh 2.30. Kt qu m phng

Hnh 2.31. Cu trc iu khin trn Matlab Simulink

Hnh 2.32. Downloading and Building

Hnh 2.33. Giao din Control Desk

Hnh 2.34. Ca s New Experiment

Hnh 2.35. Th Variable Manager v cc bin m phng

Hnh 2.36. Ca s New Layout

Hnh 2.37. Chn Slider v v hnh ch nht trong Layout1

Hnh 2.38. Thay i tham s ca Slider

Hnh 2.9. iu khin Slider sau khi gn bin cn iu khin

Hnh 2.40. V mt Plotter quan st tn hiu

Hnh 2.41.Thit lp c tnh cho th

Hnh 2.42. Thit lp thng s quan st

Hnh 2.43. iu khin s thc thi ca DSP (a) v iu khin Animation (b)

Hnh 3.1.S cu trc chung ca h iu chnh v tr

Hnh 3.2. S mch thay th ng c mt chiu

Hnh 3.3. S mch thay th mch in phn ng

Hnh 3.4. M hnh tuyn tnh ho ng c in mt

Hnh 3.5. M hnh tuyn tnh ho ng c in mt

Hnh 3.6. M hnh tuyn tnh ho m phng ng c mt chiu kch t c lp

Hnh 3.7. S khi mch chnh lu c iu khin

Hnh 3.8. S mch vng iu chnh dng in

Hnh 3.9

Hnh 3.10: S cu trc ca h iu chnh v tr.

Hnh 3.11

Hnh 3.12

Hnh 3.13

Hnh 3.14. Cu trc h iu khin v tr trong matlab Simulink


10

Hnh 3.15. c tnh m phng h iu khin chuyn ng

Hnh 3.16. Cu trc h iu khin v tr vi Card DS1104

Hnh 3.17 Cu trc iu khin m v tr vi Card DS1104

Hnh 3.18. H iu khin m v tr vi Card DS1104

Hnh 3.19. Hm lin thuc ca bin sai lch v tr

Hnh 3.20. Hm lin thuc ca bin thay i sai sv tr

Hnh 3.21. Hm lin thuc ca tn hiu iu khin

Hnh 3.22. Surface lut iu khin m

Hnh 3.23. Vi phn sai lch v tr

Hnh 3.24. Sai lch v tr

Hnh 3.25. Cu trc h iu khin v tr vi b iu khin m

Hnh 3.26. M phng lut iu khin m

Hnh 3.27.Cu trc h thng thc nghim

Hnh 3.28.Card DS1104 trong h thc nghim

Hnh 3.29. Driver DC servo motor

Hnh 3.30.DC servo motor

Hnh 3.31. Chn thi gian ly mu cho h

Hnh 3.32. Chn thi gian ly mu cho h

Hnh 3.33. Mn hnh ControlDesk vi h thc nghim

Hnh 3.34.Chng trnh iu khin h thng thc nghim

Hnh 3.35. Chng trnh iu khin h thng thc nghim dng b iu khin m

Bng 2.1. Dung lng cc b nh ca DS1104


11

CHNG 1. TNG QUAN V H IU KHIN S

1.1. L thuyt v h iu khin s.

1.1.1. Cu trc in hnh ca h iu khin s.

Ngy nay vi nhng thnh tu ni bt trong cng ngh my tnh, chng ta c

th thc hin cc b iu khin s bng my tnh thay th cc b iu khin

truyn thng. Do vy iu khin s lin quan ti thut ton iu khin trong thit b

iu khin s, c th l Card s v my tnh s. Chng ta c th tn dng s tin b

trong iu khin logic v s linh hot v mm do ca iu khin s thay v vic

thc hin cc b iu khin tng t truyn thng. Mt khc chng ta cng cn s

giao din kt ni gia i tng iu khin v my tnh. C th nh:

- cc php o c thc hin ti cc thi im ri rc

- cc d liu cng phi c ri rc ho cho php x l d liu s

Mt khc cc b iu khin s c th x l c d liu ri rc theo khng gian v

thi gian. Cch ri rc ho thng c thc hin bng cch ly mu v sau l

lng t ho. Vi hai c im ny khin h thng iu khin s khc hn vi cc

h thng thng iu khin tuyn tnh thng thng v h thng iu khin thi gian

bt bin.

a. B chuyn i s-tng t (D/A converter).

B chuyn i s-tng t bin i mt chui cc i lng u(kT) thnh tn hiu

lin tc u(t) iu khin h thng. B chuyn i D/A c m phng bi b lu

Hnh 1.1: Cu trc h iu khin s

My tnh s

Chng trnh iu khin

i tng


12

gi, nhn thi im kT xung c bin t l vi tr s u(kT) c rng rt b so

vi T (tn hiu ly mu) v duy tr hng s y sut c chu k T. Nh vy p ng

vi mt chui xung l mt chui bc thang c di T. Qu trnh bin i ny l

tc thi v khng c tr.

B lu gi bc khng y tng ng vi c cu nh hnh vi xung ch nht, h

s lp y =1. Nhng b lu gi bc cao to nn nhng dng sng phc tp hn

nhng chnh xc cao hn.

Nguyn tc lm vic ca DAC

Chuyn i s tng t l qu trnh tm li tn hiu t n s hng (n bits) bit

ca tn hiu s. B chuyn i s tng t (DAC) tip nhn mt m s n bits song

song u vo v bin i thnh tn hiu lin tc u ra.

Hnh 1.2: S nguyn l b chuyn i s - tng t trong h iu khin s

Hnh 1.3: S nguyn l b DAC

Hnh 1.4- Tn hiu ra ca b DAC


13

Tn hiu ny c a qua b lc thng thp. u ra ca b lc l tn hiu tng t

UA bin thin lin tc theo thi gian, l tn hiu ni suy ca Um. Vy b lc thng

thp ng vai tr l b ni suy.

Cc c tnh quan trng ca DAC

- phn gii: lin quan n s bit ca mt DAC. Nu s bit l m th s trng thi

tn hiu ca s nh phn a vo l 2n

v tn hiu ra s c 2n

mc khc nhau, do

phn gii l 1/ 2n

. phn gii cng b th tn hiu u ra c dng lin tc gn

vi thc t.

- tuyn tnh: Trong mt DAC l tng s tng tn hiu s u vo s t l vi

s tng tn hiu s u ra.

- chnh xc ca mt DAC cho bit s khc bit gia tr s thc t ca UA v tr

s l thuyt cho bi mt gi tr bt k ca tn hiu s u vo. S sai khc ny

cng nh th chnh xc cng cao.

- Thi gian thit lp: Khi tn hiu s u vo ca mt DAC thay i, tn hiu

u ra khng th thay i ngay lp tc m phi sau mt khong thi gian no gi

l thi gian thit lp. Thi gian thit lp phn nh tnh tc ng nhanh ca mt

DAC.

Mt s mch DAC in hnh

Bin i DAC vi mng in tr trng lng

Mch gm mt ngun

in p chun Uch, cc

b chuyn mch v in

tr c gi tr R, R/2,

R/4... v mt mch

khuch i thut ton.

Khi mt kho in no

c ni vi ngun

in th chun th s Hnh 1.5. B bin i DAC vi mng in tr


14

cung cp cho b khuch i thut ton dng in cng l:

chi i

UI =

R.2 (i=0n-1)

Cng dng in ny c lp vi cc kha cn li, c th thy ngay bng bin

in p Ura ph thuc vo ch kho no c ni vi Uch

tc l ph thuc vo gi

tr ca bit tng ng trong tn hiu s a vo mch chuyn i.

Mch c u im l n gin, nhng nhc im l chnh xc v tnh n nh

ca kt qu ph thuc nhiu vo tr s ca cc in tr v kh nng bin thin nh

nhau theo mi trng ca cc in tr ny. Ch to cc in tr theo ng t l

chnh xc nh vy thng kh khn v tn km. Ngoi ra Ura cn ph thuc vo c

chnh xc v tnh n nh ca ngun in p chun.

B bin i D /A dng mng in tr R v 2R

DAC vi thang in tr R - 2R khc phc c mt s nhc im ca DAC mng

in tr trng lng. Mch ch gm hai loi in tr R v 2R vi nhiu chuyn

mch (mi chuyn mch cho 1 bitm) v mt ngun in p chun Uch. i lng

cn tm l Ith vo mch khuch i khi c mt s chuyn mch ni vi Uch.

Lc ta c: Ura=-Ith.Rf

Hnh 1.6. B bin i DAC dng mng in tr R v 2R


15

Xt ti chuyn mch tng ng vi bit th i, nt tng ng trn mch l nt 2i. Khi

b chuyn i ng vo Uch th in th tng ng ti nt 2i

s l Uch/ 2 v ngun

tng ng c ni tr l R (theo nh l Thevenin). Nh vy ti nt 2i+1

ta c

ngun tng ng tr s l Uch/ 4 v ni tr l R.

T nhng kt qu trn ta suy ra rng khi di chuyn v pha mch khuch i

thut ton in th ti mi nt bng na tr s ca nt k cn bn tri n. Nh vy

nu t nt th 2i

n nt 2n-2

c k nt (k c nt th 2n-2) th in th ti nt 2n-2 do

chuyn mch 2i

gy ra l Uch/ 2k v dng in t-ng ng l Uch/(2k.2R). Ti nt 2n-1

do c tnh ca khuch i thut ton m in th ti y c coi l 0V.

Tm li, mt cch tng qut ta c cng thc tnh in p ra ca mt DAC n bit

(t B0 Bn-1) vi mng in tr R - 2R.

n-1 n-2 0fra ch n-i n-2 0nR

U =-U 2 B +2 B +...+2 B2 R

Trong B0 Bn-1

c gi tr 0 hoc 1.

Cc DAC theo phng php ny phi dng s in tr kh ln, v d nh

DAC n bit th phi dng 2 (n-1) in tr, trong khi theo phng php in tr trng

lng ch phi dng n in tr. Nhng b li n khng rc ri v ch cn dng c 2

loi in tr m thi. Nn chnh xc v tnh n nh ca tn hiu ra c m

bo.

b. B chuyn i tng t - s (A/D Converter)

Qu trnh chuyn i tng t - s khng th tc thi, cn c thi gian tr

bin i tn hiu tng t l mt i lng vt l (in p) u vo thnh tn hiu

s u ra.

Hnh 1.7. S nguyn l chuyn i A/D


16

B chuyn i A/D c ba chc nng: ly mu (lng t ho theo thi gian), lng

t ho theo mc v m ho (h nh phn).

Nguyn l lm vic ca ADC c minh ho trn s khi.

Tn hiu tng t UA c a n mch ly mu, mch ny c hai nhim v:

Ly mu nhng tn hiu tng t ti nhng thi im khc nhau v cch u.

Thc cht y l qu trnh ri rc ho tn hiu v mt thi gian.

Gi cho bin tn hiu ti cc thi im ly mu khng thay i trong qu trnh

chuyn i tip theo (qu trnh lng t ho v m hoq). Qu trnh lng t ho

thc cht l qu trnh lm trn s. Lng t ho c thc hin theo nguyn tc so

snh tn hiu cn chuyn vi cc tn hiu chun. Mch lng t ho lm nhim v

ri rc tn hiu tng t v mt bin . Trong mch m ho, kt qu lng t ho

c sp xp lai theo mt quy lut nht nh ph thuc loi m yu cu u ra b

chuyn i.

Nhiu loi ADC, qu trnh lng t ho v m ho xy ra ng thi, lc khng

th tch ri hai qu trnh, php lng t ho v m ho c gi chung l php bin

i AD.

Cc tham s c bn ca ADC

Cc tham s c bn ca b bin i ADC gm di bin i ca in p tng t

u vo, chnh xc ca b chuyn i, tc chuyn i.

- Di bin i ca in p tn hiu tng t u vo l khong in p m s t 0

n mt s dng hoc s m no , hoc cng c th l in p hai cc tnh:

-UAUA.

- chnh xc ca ADC: Tham s u tin c trng cho chnh xc ca ADC l

phn gii. Tn hiu u ra ca mt ADC l cc gi tr c sp xp theo mt

quy lut ca mt loi m no . S cc s hng ca m s u ra (s bits trong t

m nh phns) tng ng vi gii bin i ca in p vo cho bit mc chnh xc

ca php chuyn i. V d mt ADC c s bits u ra l n = 8 th s phn bit

c 28

mc trong di bin i in p vo ca n. Nh vy trong thc t dng s


17

bits nh gi chnh xc ca mt ADC khi gii bin i in p vo l khng

i.

Lin quan n chnh xc ca mt ADC cn c cc tham s khc nh: mo phi

tuyn, sai s khuch i, sai s lch khng, sai s lng t ho.

- Tc chuyn i cho bit s kt qu chuyn i trong mt giy, c gi l tn

s chuyn i fc. Cng c th dng tham s thi gian chuyn i Tc c trng

cho tc chuyn i. Vi mt ADC thng th fc < 1/Tc

v gia cc ln chuyn i

phi c mt thi gian cn thit ADC phc hi li trng thi ban u. Mt ADC

c tc chuyn i cao th chnh xc gim v ngc li.

Cc phng php chuyn i tng t - s : C nhiu cch phn loi ADC,

nhng hay dng hn c l phn loi theo qu trnh chuyn i v mt thi gian.

Trong n ny ch gii thiu mt s phng php in hnh.

Chuyn i A /D theo phng php song song

Nguyn tc hot ng. :Tn hiu tng t UA c ng thi a n cc b so

snh t S1 n Sm. in p chun Uch

c a n u vo th 2 ca cc b so

snh qua thang in tr R. Do cc in p chun t vo cc b so snh ln cn

khc nhau mt lng khng i v gim dn t S1 n Sm. u ra ca cc b so

snh c in p ln hn in p chun ly trn thang in tr c mc logic "1", cc

u ra cn li c mc logic "0". Cc u ra ca mch so snh c ni vi mch

AND, mt u mch AND c ni vi mch to xung nhp. Ch khi c xung nhp

a n u vo AND th cc xung trn u ra ca b so snh mi a vo mch

nh Flip_Flop (FF). Nh vy c sau mt khong thi gian bng chu k xung nhp

li c mt tn hiu c bin i v a n u ra. Xung nhp m bo qu trnh so

snh kt thc mi a xung nhp vo b nh. B m ho s bin i tn hiu v

di dng m m thnh m nh phn.

Mch bin i song song c tc chuyn i nhanh nn c gi l ADC nhanh

nhng kt cu ca mch rt phc tp v d nh ADC n bits cn phi dng 2n-1 b so

snh. V vy phng php ny ch yu dng trong cc ADC c tc chuyn i

cao nhng s bit nh.


18

Chuyn i A /D theo phng php b

Ti thi im ban u b m c t trng thi khng bi xung Cl, nh vy u

ra ca n cng c tn hiu khng. Mch so snh thit lp gi tr mt tn hiu nhp H

qua cng AND c a vo mch m. Mch m lm vic cho ra tn hiu s t

Q0Qm-1 ng thi qua b bin i D /A s c in p U0

cho n khi U0

= UA th

b so snh lt gi tr, u ra ca n c gi tr 0 cng AND s kho v b m s

dng. Trn u ra b m Q0Qm-1 dng s t l vi in p vo UA, s ny c

xp vo b ghi. Tip theo b m c xo v chun b cho chu k bin i tip

Hnh 1.8. S chuyn i A/D song song

Hnh 1.9. S chuyn i A/D theo phng php b


19

theo. Sau mi chu k b ghi s ghi s liu mi ca b m. Nu nh b m nh

phn c m bits th in p vo cc i UmaxA:m

maxAU =2 -1

in p UA c lng t theo gia s: maxA

A m

UU =

2 -1

in p UA c din t bng phng trnh: maxA

A m

UU = N

2 -1

Trong N l tng s bc ca b m v dung lng ca n y sau khi kt thc

qa trnh m.

Thi gian bin i: A

n

NT =

f ,Trong fn

l tn s xung nhp.

Thi gian bin i ph thuc ln in p. Tc thay i in p c th t gi

tr cc i.

Amax AmaxA A nnm m

max

U .N UdU U f= = . = f

dt T 2 -1 N 2 -1

Nu tc bin i in p UA ln hn tc cc i th pht sinh sai s ng ca

b bin i. Sai s tnh ca b bin i l sai s lng t U. gim thi gian

bin i, b m nh phn ta s dng mch iu khin chng trnh.

B bin i A /D theo nguyn tc servo

B bin i ny c ba phn t c bn: mch so snh, mch m hai chiu v b bin

i D /A.

Hnh 1.10. B bin i A /D theo nguyn tc servo


20

Tn hiu in p vo UA so snh vi in p ra D /A. Nu UA

> U0

th b bin m

m theo chiu tin. Nu UA < U0

th b m m theo chiu li cho n khi UA

=

U0 th b m dng, tng t nh c cu servo. Tuy vy tc bin i in p vo

UA lun lun phi nh hn tc ca b m v b bin i D /A. Nn thi gian

bin i ph thuc vo tn s xung nhp fH v phn ng ca b so snh.

c. My tnh s hoc b vi x l.

My tnh thc hin cc thut ton nh: dch chuyn, cng, nhn, lu gi: n

to nn tn hiu iu khin uk=u(kT) theo chu k, l hm ca cc i lng uk-1, uk-2,

uk-q cc thi im trc v cc i lng sai lch ek-1, ek-2, ek-q. Angorit

m t hm y c dng tuyn tnh nh:

n n

k k

k=0 k=1

u(mT)= b e (m-k)T - a u (m-k)T

k 1 k-1 2 k-2 q k-q 0 k 1 k-1 p k-pu =a u +a u +...+a u +b e +b e +...+b e

Yu cu l xc nh cc h s aj v bj sao cho p ng ca h s i vi i lng

t xd(kT) l thch hp mc d c nhiu tc ng n h thng hay n cm bin.

Trong angorit, sai lch e(kT) xut hin ng thi vi iu khin, i hi chu k

lng t ho T ln (t nht l 20 ln ) so vi thi gian tnh u(kT). Thi gian ly

mu v thi gian bin i tn hiu u cn tnh n ch T.

Chu k ly mu T nh hng rt ln i vi cht lng ca h kn. Nu T qu ln

h c th mt n nh. Nu T v mc lng t ho (m qu trnh phn tch khng

quan tm n) b th tn hiu s cng nh tn hiu ri rc c th xem nh lin

tc.

Ngy nay vi s pht trin vt bc ca cng ngh thng tin, in t cc nh

sn xut tch hp cc h vi x l tn hiu s thay th cc my tnh trong h

iu khin s. Cc h vi x cng vi cc b chuyn i A/D,D/A c tch hp trn

mt Board n (Card). C nhiu hng sn xut nhiu Card iu khin s DSP

ng dng trong cng nghip v nghin cu, in hnh l cc Card DS1102,DS1104,

DS1103, DS1105.


21

1.1.2. C s ca iu khin s. 1.1.2.1. Bin i Z

Khi phn tch h iu khin tuyn tnh lin tc th ta dng php bin i Laplace

lc hm truyn ca h thng l t s gia hai a thc theo bin t. Trong h iu

khin s th hm truyn ca h thng khng cn l mt a thc i s theo p m a

thc i s theo Tpe . n gin ta t zeTp lc hm truyn ca h thng

tr thnh a thc i s theo z. Ta c th s dng cc kt qu kho st h tuyn

tnh lin tc cho h iu khin s.

1.1.2.1.1. Php bin i z

Cho tn hiu ri rc x(nT) th bin i z ca tn hiu ny s l:

nznTxzX )()(

Cng thc trn c gi l cng thc bin i Z theo hai pha. Trong k thut

iu khin s ta thng dng bin i Z theo mt pha (0 +).

X(z)=

0

)( nznTX

Xt hm lin tc f(t) c hm ri rc l: f(nT)=

0

)()( nTttf

Trong )nTt( l xung irc

Bin i Laplace ta c: dte)nTt()t(fdte)nT(f)p(F0 0

tptp

0

*

0

*

)()( nTpenTfpF Vi Z = eTp p = T

1lnZ

*

F (p) =

ZT

p ln1

= F(z) =

0

)( nzntf

Bin i Z ca hm 1(t): f(t) = 1(t)

f(nT) = 1(n) vi T = 1

Z )(1 t =

0

2)(1 nn = 1 + 2

11

zz + =

1z

z


22

Bin i Z ca hm f(t) = e-aT vi a = const

f(nT) = f(n) vi T = 1 = e-na

Z ate =

0

nna ze = 1 + e-az-1 + e-2az-2 + ...

l cp s nhn li v hn vi q = e-nz-1 l cng bi

Vy Z ate = q1

1 =

ze a1

1

1

= 1ze

zea

a

= aez

z

1.1.2.1.2. Cc tnh cht ca bin i Z

a. Tnh dch gc

Nu hm f(n) c bin i Z l F(z) th hm f(n + 1) c nh l:

ZF(z) Zf(0) ( f(0) l iu kin u )

Tng qut: Z mnf ( = mZ F(z) -

1m

0j

)jm(z)j(f

b. Tnh cht tuyn tnh

Nu )z(F)n(fv)z(F)n(f2211

th:

)z(Fb)z(Fa)n(fb)n(faZ2121

c. Gi tr u ca hm gc ri rc

)z(FLim)0(f)0n(fz

Xut pht t bin i Z: V

n10

n

nz)n(f...z)1(f)0(fz)n(ffZ

)n(flim)0(f)z(FLim0nz

d. Gi tr cui ca hm gc ri rc

zF)z1(lim)n(flim 11zn

V

m

0n

n

mz)n(f)1n(flim)n(f)1n(fZ

m

0n

n

mz)n(f)1n(flim)z(F)z(FZ


23

)0(f)1m(flim)0(f)z(F)1z(limm1z

)z(F)z1(Lim)n(fLim 11zn

e. Bin i Z ca sai phn tin ( )n(f )

f(n)=f(n+1)-f(n)

Z f(n) =Z f(n+1)-f(n) =Z f(n+1) -Z f(n)

Z f(n) =zF(z)-zF(0)-F(z)=(z-1)F(z)-zf(0)

Tng t i vi sai phn cp hai:

)0(fz)0(f)1z(z)z(F)1z()0(fz)n(fZ)1z()n(fZ 22

f. Bin i Z ca sai phn li

)z1)(z(F)z(Fz)z(F)n(fZ)1n(f)n(f)n(f

11

1.1.2.1.3. Bin i Z ngc: Cho hm F(z) tm f(n). C ba cch thc hin:

a. Phn tch thnh nhng phn thc n gin

Phn tch thnh nhng phn thc n gin sau s dng bng nh gc v

cc tnh cht bin i Z s c kt qu.

b. Phn tch thnh chui lu tha

...z

f

z

)1(ffz)n(f)z(F

2

2

0

n

Suy ra f(n) thi im ly mu ta xc nh c gi tr thi gian.

c. Dng phng php tch phn ngc

e(nt) = j2

1

L

1n dzz)z(F

Trong ng cong L ly sao cho bao kn nghim (ng cong kn L l

ng trn n v). Phng php ny t dng.

d. S dng my tnh s

Chuyn F(Z) thnh phng trnh sai phn, sau gii phng trnh sai phn

bng my tnh.


24

1.1.2.1.4. Bin i Z pht trin

Bin i Z pht trin l mt cng c xc nh hm thi gian gia cc ln

ly mu khi m s ln ly mu khng phi l s nguyn ca tn s ly mu.

Trong trng hp ny ta thay php bin i Z thng thng bng cch thm

vo h thng d liu ly mu mt s tr hon thi gian tng tng. Khi php

bin i ny s m t cc chui xung c lm r bi cc hm thi gian, vi bi s

khng nguyn ca tn s ly mu.

Bng cch thay i thi gian tr ta c th tm c tn hiu lin tc gia cc

ln ly mu.

- Xt hm thi gian nh hnh v (Hnh 1.11). Hm c lm tr mt khong

thi gian giy. Nu l s nguyn th bin

i Z ca hm )Tt(e l :

)z(Ez)Tt(eZ

Nu chn n1n th sai s gia

nT v T l :

nTnTT

Trong l mt s dng v 10 .

Gi thit E(p) l bin i Laplace ca e(t) v E(p, ) l bin i laplace ca e(t-

T )

)e)p(E,p(E)Tt(eL Tp

Thay n ta c:

TpnTpee)p(E),p(E

Bin i Z pht trin:

Tpn e)p(EZz),z(E

0

nzTnE),z(E

1.1.2.2 Tn hiu v ly mu tn hiu trong h iu khin s

(n-1) t n (n+1)

e(t)

Hnh 1.11 : Hm thi gian e(t)


25

1.1.2.2.1. Ly mu tn hiu

Trong h iu khin s lun tn ti hai loi tn hiu l tn hiu lin tc v tn

hiu ri rc. Tn hiu a vo my tnh l tn hiu

ri rc, cn tn hiu a vo i tng iu khin

v i tng o lng l tn hiu lin tc.

tn hiu a vo my tnh s ta phi

bin i cc tn hiu o lng vn l lin tc

thnh tn hiu ri rc v n c gi l qu trnh

ct mu tn hiu.

Xt mt tn hiu lin tc nh hnh v (Hnh

1.12):

Ta gi thit ly mu tn hiu nhng im cch

u nhau. Vi cch ly mu nh th th hm x(t)

c m t bi chui cc con s ri rc x(0), x(T),

x(2T), x(3T), ., x(nT). N m t cc gi tr ca

hm x(t) ti cc thi im ri rc v thi gian.

Cc gi tr ca hm ti cc im khc nh

)T5

2(x . ch c th c c nh phng php

ni suy.

Trong thc t cc khu iu khin v i tng iu khin thng l tng

t, v vy tn hiu ri rc sau khi ly mu phi c xy dng thnh tn hiu lin

tc, trong sut khong thi gian gia hai ln ly mu. Qu trnh ny c gi l qu

trnh lu gi d liu (Hold), c hai cch lu gi d liu l: lu gi bc

khng v lu gi bc mt.

1.1.2.2.2. Cc c tnh ly mu

Mt b ly mu l tng c m t nh hnh v(H-21) sau:

t

x(t)

t

Hnh 1.12. Tn hiu lin tc

x(nT)

Hnh 1.13.Tn hiu ri rc

T 2T 3T nT

T 2T 3T nT


26

Vi b ct mu l tng trn s to ra mt chui xung n v ri rc t hm lin

tc. Gi thit thi gian p ng ca b ct mu nh hn nhiu thi gian gia hai ln

ly mu lin tip (chu k ly mu), khi gi tr ri rc x(nT) chnh l cc gi tr

ca hm khi b ct mu ng.

m t ton hc qu trnh ly mu ta c th coi b

ly mu nh mt cng c thc hin php nhn tn hiu

x(t) vi hm ly mu (t). Vic ny tng ng nh

vic iu ch tn hiu, trong sng mang l hm (t)

v ta c x(nT) =x(t).(t). Hm ly mu tt nht l chui xung n v, chui xung

ny c b rng v cng hp, bin v cng ln (chnh l o hm ca hm 1(t) )

n l cc hm (t), (t-T), (t-2T), (t-nT)

Trong thc t cc b ly mu vn c mt khong thi gian tc ng nht

nh, do hm ly mu thc t c mt din tch xc nh khc mt (din tch A).

Ta ch c th coi cc hm ly mu c din tch bng mt khi thi gian ly mu nh

hn nhiu hn so vi hng s thi gian ca h thng (thng gp trong thc t).

Gi thit hm ly mu c m t bi chui xung n v:

(t) =

n

nT)(t

Trong : (t-nT) =

nTtvi

nTtvi0

sao cho )nTt( dt =1 chnh l o hm ca dt

d1(t-nT)

Khi hm x(t) c iu ch nh sau:

n

*

)nTt()nT(x)t(x

t (t-nT)

0 T 2T 3T . nT

B ct mu x(t) x(nT) x(t) x(nT)

Hnh1.14:B ct mu


27

Trong : x(nT) l gi tr ca hm ti thi im ly mu. V hm (t-nT) ch

c gi tr xc nh ti thi im nT, do c th thay x(nT) = x(t). Mt khc, x(t)

xc nh t thi im t = 0.

T ta c: *

( ) ( ) ( )n

x t x nt t nT

1.2. Tng hp h iu khin s

1.2.1. L lun chung.

H iu khin s c tng hp theo hai hc ch yu: trong min tn s v

trong khng gian trng thi. Tng hp trong min tn s ch yu da vo m t

ng hc ca h tng bng cc bin i Laplace v Fourier (cn gi l phng php

tng hp dng k thut bin i). Hng th hai l tng hp h iu khin s trong

khng gian trng thi.

Phng php dng k thut bin i c cc phng php gin tip (phng

php tng t) v phng php trc tip.

phng php gin tip, mt b iu khin lin tc l tng Gc(s) c tng

hp sau mt t hp CAD - b iu khin gin on -DAC c chn sao cho

tng ng vi Gc(s) nh hnh 1.15. Phng php ny c nhng ngi quen

dng iu khin tng t a chung v ch cn bin i t k thut tng t sang

s. Tuy nhin vic gin on ho b phn iu khin s cho kt qu km chnh xc

v:

1. Tn hin lin tc dng bc thang t phn t lu gi khng th to nn tn

hiu l tng u*(t).

2. Tn hiu l tng y ph thuc lin tc vo y(t), cn b iu khin s ch o

c y(t) thi im ly mu.

Tuy nhin, nu so vi ph ca cc tn hiu u vo, u ra m chn tn s

lng t ho ln, c th chn c b phn iu khin gin on gn nh Gc(s).

Phng php chn gin n nht l theo:

Gc(z) = Gc(s)|s = (z-1))/T


28

Mt phng php chun xc hn:

Gc(z) = Gc(s)s = 2(z-1))/T(z+1).

Phng php th hai l phng php bin i n ng (bin i kp, bi i

Tustin) duy tr c iu kin n nh ca hm truyn: nu Gc(s) n nh th Gc(z)

cng n nh do php bin i chuyn min bn trong ng trn n v mt

phng z sang na mt phng tri ca s. Tuy nhin iu khng c ngha l nu

Gc(s) n nh c qu trnh th b iu khin gin on CAD - Gc(z) - DAC

cng s n nh c qu trnh. Do vy sau khi chn b phn iu khin s cn

nh gi li sai lch v tnh n nh ca h.

Ch rng phn t lu gi bc khng to tr trung bnh l 2

T (nh hnh

1.16) cho nn b iu khin Gc(s)e-

sT/2.

Lng t ho c tn s ln,

khong 10 n 20 ln tn s ring ca

i tng.

phng php trc tip qu

Gc(z)

Khi iu khin, s (theo thi gian gin on)

DAC

phn t lu gi

ADC

phn t ly mu

G1(s)

Qu trnh

lin tc

Qu trnh gin on (i vi b iu khin)

B iu chnh lin tc (i vi qu trnh)

Hnh 1.15: Mi quan h qu trnh gin on v lin tc

Hnh 1.16


29

trnh lin tc cng vi cc phn t lu gi v ly mu c xem nh mt qu trnh

gin on, tng hp trong min z, cho php khai thc tnh nng mm do ca my

tnh m phng php tng t b hn ch.

Tn hiu lin tc u vo u(t) c xc nh hon ton bi uk. Kt qu l y(t)

= G1(s)u(t) v tn hiu c ly mu yk c xc nh hon ton bi uk. Nh vy

vic dng b iu khin gin on iu khin mt qu trnh gin on c u

vo uk v u ra yk s khng cn n s xp x no. Phng php trc tip c s

trn p ng c xc nh trc (p t) i vi tn hiu vo hay nhiu nht nh,

nhm tho mn nhng yu cu t ra nh chnh xc, lng qu iu chnh, thi

gian qu hay nhng ch tiu c trng khc i vi h xung nh n nh v tn,

thi gian cc tiu

Tuy nhin cn ch rng vic gin on ho s lm mt kh nng quan st

c v iu khin c i tng. Mc d iu ny ch xy ra khi T=

r

n

(r l

tn s ring ca i tng) v ch h n bin. Do cn chn T< 2

iT

trnhtrng hp ngcng T=

r

iTn2

. Nh vy tn s lng t ho ln cn l loi

tr mt kh nng quan st c v iu khin c.

Nu n nh c qu trnh gin on (ngha l xk 0) th bo m c

s n nh ca qu trnh lin tc (ngha l x(t) 0).

1.2.2. iu kin tng hp c b iu khin s trong h.

hnh 1.17 c s khi ca h xung m my tnh s thc hin chc nng

ca h iu hnh Gc*(s).

H kn c hm truyn

Wk*(s) = )(*)(*1

)()(*

)(*

)(*

sGsG

sGsG

sX

sY

c

c

vi G*(s) l phn khng thay i ca h xung

Hnh 1.17


30

B iu khin c xc nh bi: G*(s) =)(*W1

)(*W

)(*

1

k

k

s

s

sG (*)

Gc*(s) c th thc hin c nu bc ca mu s ln hn hoc bng bc ca

t s; ni cch khc tn hiu ra khng vt trc tn hiu vo.

Nu Wk*(s) = wke-k thut

+ wk+1e-(k+1)T+ v sau khi chia t cho mu s ca

hm truyn G*(s) ta c: G*(s) = gnenT

+gne-(n+1)T+.

Biu thc (*) c dng : Gc*(s) = ...)W1...)((

...WW

k

)1(

1kk

kTnT

n

TkkT

eeg

ee

=Ck - ne-(k-n)T

+ Ck-n+1e-(k-n+1)+.

iu kin thc hin c l k n tn hiu ra ca b iu khin khng th c

c khi cha c tn hiu vo.

Nh vy, bc ca hm truyn h kn mong mun Wk*(s) khng thp hn bc

ca thnh phn khng bin i G*(s) ca h.

1.2.3. Chn tn s ly mu.

Vic chn tn s lng t ho 0 (hay thi gian ly mu T) rt quan trng.

Nu 0 qu b s c hin tng mo tn hiu, mt lng thng tin, gim cht lng,

thm ch cn c th mt n nh. Nu chn 0 qu ln (hay T qu b) mt mt h c

p ng gn vi h lin tc mong mun, tng hp theo phng php bin i, mt

khc i hi tc tnh phi nhanh, gi thnh s cao, tuy rng hn ch v phng

din ny ngy cng gim nh do cng ngh v my tn ngy cng pht trin. Vic

chn ng tn s lng t ho vn cn mang tnh cht ngh thut hn l tnh cht

khoa hc.

1. Vic chn tn s lng t ho hp l trc tin da vo bn cht ca qu

trnh.

- Cc phn ng ho hc l qu trnh chm c thi gian iu khin tnh bng

gi.

- Cc qu trnh nhit, thi gian iu khin tnh bng pht.

- Cc h iu khin tu thu chng hn tc ng vo cn li i hi nhiu giy

dn tu ng hng.


31

- Cc qu trnh c hc tc ng nhanh (nh ngi my chng hn) thi gian y

tnh n phn trm ca giy.

Trong trng hp th nht chu k lng t ho T khng phi b gii hn bi

cng sut v tc tnh ca my. Trong trng hp cui nhng hn ch v kinh t

(gi thnh ca my tnh cht lng cao) li t ra. Ngoi ra cn phi tnh n nhng

kh khn khc: Khi T 0 cc m hnh ca qu trnh tr nn th thin (F = I, G =

0) tt c cc nghim cc u bng 1. Do vic tnh ton cc b iu khin s c

kh khn. Cc phng trnh truy hi tr nn km chnh xc thi gian thc, cc

phng php xy dng h n nh v tn, thi gian cc tiu khng cn ngha.

Do vic xc nh T (hay 0) hp l l cn thit, tuy rng cc kh khn trn u

c bin php khc phc .

2. Tn s lng t ho 0 c chn phi tho mn nh l Kachenhicov .

Khi c tn hiu lin tc gin on ho cn c phc hi th tn s lng t

ho t nht phi gp i tn s ln nht ca tn hiu y 0. i vi h iu khin

kn, tn s lng t ho khng b hn hai ln di thng tn cn thit 0 m dung

lng ph tn hiu vo ph thuc vo 0 nn: 20 b

l gii hn thp nht c th. Trong thc t gii hn ny c th qu thp i

vi p ng thi gian chp nhn c. m bo chnh xc cn thit v ti ca

my tnh, thng c chn: 4 200 b

Hay T c chn khong 1/10 hng s thi gian b nht ca i tng.

qu trnh qu , khi lng t c tr s xc lp vi thi gian p ng tm

ca h cn c 2 n 4 chu k lng t ho T.

3. Trong nhiu trng hp, cn c trn cao i vi hm qu . Mc

trn tu thuc vo i tng c th; i vi ng c in, chu k gin on c th

ln hn i vi c cu tha hnh thy lc. i khi gia phn t lu gi (ZOH) v

c cu tha hnh thu lc c b lc h tng. Mc trn cn tu thuc vo phm

vi ng dng ca h. i vi con ngi, tc ng c nh hng trc tip, i hi


32

mc trn cao hn so vi cc thit b iu khin v tinh khng c ngi. V d

vi h c tm=1sec (gii thng tn l 0,5Hz) cn chn 0 t 3 n 20Hz p ng

trn v hn ch lng qu iu chnh. Do cn chn: 6 400 d

Nhiu tc ng vo i tng rt a dng, t nhiu bc thang n n trng

(Whitenoise). i vi tn s lng t ho th nhiu ngu nhin c tn s cao l c

nh hng nht. Mt h lin tc chng nhiu tt l h c sai s o nhiu to nn l

b. Nu dng iu khin s i vi h ny th cht lng y s gim. Nu t s

d

0 cng b th s suy gim cht lng do lng t ho ln khi c nhiu l n trng

tc ng. i vi h iu khin c b quan st th t s ti u d

0 20.

Nu chu k lng t ho ln hn thi gian p ng ca qu trnh th nhiu s tc

ng vo qu trnh trc khi b iu khin c tc ng hiu chnh. Do tn s

lng t ho c chn trn c s nh gi ng hc ca qu trnh v nhiu, ng

hc ca qu trnh v kh nng ca my tnh. Cc b iu khin trn thng trng

vi t mch vng iu khin c chu k lng t ho b v c nh.

1.2.4. Thit k b iu khin s theo phng php lin tc.

Phng php thng thng thit k h iu khin s l chn b iu khin

Gc(s) cho h lin tc tng ng, ri xp x ho b iu khin lin tc y vi b

lc s cn tm Gc(s) (hay Gc(z)). C nhiu phng php thc hin.

1.2.4.1. Phng php vi phn:

B iu khin s c m t bng phng trnh lp, rt gn vi phng trnh

vi phn ca b iu khin tng t. V d b iu khin PID c hm truyn v

phng trnh vi phn tng ng.

c p i d

t

p i d

0

1 U(s)G (s)=k +k +k s=

s E s

deu(t)=k e(t)+k e(t)dt+k

dt

(1-7)


33

Hnh 1.19

Hnh 1.20

C ba phng php xp x ho tn hiu lin tc e(t) thnh tn hiu gin on

e(kT).

1. Xp x sai phn hu hn bc mt i vi tch phn.

a) Lut ch nht theo tch phn tin

Din tch di ng cong e(t) c

xp x bng din tch ch nht nh hnh

1.18. Tch phn ca e(t) ti t=kT c xp x

bi:

u(kT) = u[(k-1)T] + Te(kT) (1-8)

Nu ly bin i z cho c hai v, hm

truyn ca khu tch phn gin on l:

Gi(z) ki1)(

)(

z

Tzk

zE

zUi

(1-9)

b) Lut ch nht theo tch phn li

Nh hnh 1.19 tch phn ca e(t) ti t = Kt

c xp x bi:

u(kT) = u[(k-1)T]+Te[(k-1)T (1-10)

v hm truyn ca khu tch phn gin on l:

Gi(z) ki 1

)(

z

Tk

zE

zUi

(1-11)

c) Lut hnh thang theo tch phn gia

Din tch di ng cong c xp x

bng hnh thang nh hnh 1.20.

u(kT) = u[(k-1)T] + 2

T{e(kT)+e[(k-1)T]}

(1-12)

Hm truyn ca khu tch phn gin on l:

Gi(z) ki 1

1

2

)(

z

zTk

zE

zUi

(1-13)

2. Xp x sai phn hu hn bc mt i vi o hm:


34

o hm ca e(t) ti t=kT c th c xp x theo sai phn li bng cch xc

nh e(t) thi im t=kT v (k-1)T:

))1()((1)( TkekTeTdt

tdekTt (1-14)

Ly bin i z cho c hai v ta c: u(z) = )(1

)()1(1 1 zE

Tz

zzEz

T

Phng php xp x ni trn tng ng vi:

Z = etS

1+Ts v s T

z 1

Tng hp thnh phn t l, tch phn v vi phn ta c b iu khin PID vi

hm truyn theo:

a) Lut ch nht tch phn tin:

Gc(z) = )1(

22

zz

T

kz

T

kkzTk

T

kdk ddpip

b) Lut ch nht tch phn li

Gc(z) = )1(

22

zz

T

kz

T

kkTkz

T

kdk ddpip

c) Lut tch phn hnh thang :

Gc(z) = )1(

2

22

2

zz

T

kz

T

kk

Tkz

T

kTkk ddp

dp

S khi thc hin b iu khin PID gin on nh hnh 1.21.

1.2.4.2. B iu khin s c xc nh theo hm truyn t

V z = eTs

nn hm truyn t ca b iu khin s v nguyn tc c th c

xc nh bng cch thay th s = T

1ln(z). Tuy nhin biu thc xc nh Gc(z) l siu

vit . tng hp b iu khin c th dng phng php khai trin ln(z) v ch gi

li thnh phn th nht hoc ch p dng biu thc z=eTs nghim khng ca Gc(s).

1. Bin i tuyn tnh kp:


35

Hnh 1.21

Khai trin ln(z) di dng

Ln(z) = 2

...

3

3vv |v| = 1

1

11

1

z

z

V biu thc s = T

1ln(z)

by gi c dng

s

-1

-1

2 1-z 2 z-1. = . =

T T z+11+zw(z)

(1.15)

w l i lng xp x

ca s.

Phng php xp x ny (phng php Tustin) tng ng vi phng php

tch phn gia - lut tch phn hnh thang.

u(kT) = (k-1)T kT

0 (k-1)T

e(t)dt+ e(t)dt (1.16)

t : u(kT) = u[(k-1]+T e (k-1)T +e(kT)

2 (1.17)

Bin i z ca phng trnh sai phn trn l:

1z

1z.

2

T

1

z1.

2

T

E(z)

u(z)

E(z) 1)+1-(z + 1u(z)-z = u(z)

1

1-

z

(1.18)

Biu thc (1-13) v (1.18) ch khc nhau h s ki m b iu khin tch phn

c cho trc.

2. Phng php tng nghch nghim cc v nghim khng

Nh bit, nghim cc v nghim khng sj ca Gc(s) nh x vo nghim cc

v nghim khng ca Gc(z) tng ng vi zj = esjT

, cn h s khuych i ca Gc(z)

th tho mn iu kin.


36

Hnh 1.22

G0(z)|z=1 = Gc(s)|a=0 (1.19)

Nu Gc(s) c nhiu

nghim khng kh

th Gc(s) 0. iu y

tng ng vi Gc*(s) = 0

di tn th nht

2,

2

00 v G0(z) = 0

vi z = 1. V vy, bc

ca t s v mu s nh

nhau, cn thm nhn t

(z+1) (z+1)q-p m q v p

l bc ca mu s v t s

ca Gc(s).

1.2.4.3. Phng

php dng bin i z

phng php ny,

Gc(z) c xc nh theo

bin i z i vi Gc(s)

sao cho hm trng lng

hay hm qu ca

chng nh nhau. y phn t lu gi

(ZOH) ch to nn dng bc thang ca hm trng lng hay hm qu , xp x vi

hm lin tc tng ng. H kn s cho cht lng xp x km. V d vi hm trng

lng Gc(z) = cz{Gc(s)} hng s c c xc nh theo iu kin (1.19).

p ng tn s ca b lc s v tng t khc nhau tn s cao nn phng

php ny ch dng cho cc b iu khin c p ng tt nhanh tn s cao vi thi

gian ly mu T b ph n chng ln nhau. Hnh 1.22 c p ng tn ca b lc

bc hai nhm so snh cc phng php xp x khc nhau ni trn. Chu k lng t


37

ho T=1s l nh so vi chu k ring ca b lc T0=2 . Phng php xp x theo

ch nht v bin v pha u khc xa vi p ng tng t lm chun. H s

khuch i tnh khng cn nh trc sau khi dng bin i z p ng hm trng

lng. Bin i z vi c phn t lu gi (p ng hm qu ) cho p ng tt v

bin nhng khng tt v pha do tr T/2. Bin i tuyn tnh kp bin dng

c p t cng bin tn s ring 1rad / secr cho kt qu chp nhn c

v bin cng nh pha, nhng tn s gii hn

T bin bng khng. Phng

php tng thch nghim khng v nghim cc c bin thp hn p ng tng

t. Kt qu so snh ny gii thch v sao bin i tuyn tnh kp thng c dng

cc b lc s.

1.2.4.4. Tng hp b iu khin c tnh phn t lu gi (ZOH)

cc phng php nu trn, phn t lu gi khng c tnh n khi xc

nh Gc(z).Phn t lu gi c th thay th bi Gca(s) = e-Ts/2

v tn hiu u ra ca

n chm sau mt thi gian T

2hoc bi Gob(s) =

T

Ts1+

2

suy ra t

Gob(s) = -Ts1-e

s v e

-Ts =

Ts1-

2Ts

1+2

.

Vic chn b iu khin tng t t trc G0a(s) G1(s) hay Gob(s), G1(s)

c thc hin nh h lin tc. Tuy nhin cn bit trc chu k lng t ho T.

Mt phng php khc c thc hin theo cc bc sau:

* Tnh phn khng bin i ca h :G(z) = (1-z-1) z 1G (s)

s

* Dng bin i tuyn tnh kp bng cch thay z bi wT-2

wT2 c G(w).

* V ng cong Bode L(*) v (*) .

* Chn khu hiu chnh dng Gc(w) = ka w

a w

chng hn, tho mn iu kin

n nh v chnh xc.


38

* Gin on ho khu hiu chnh Gc(w) c Gc(z)

1.2.5. Thit k b iu khin s theo phng php trc tip.

Nh nu, phng php gin tip khng khai thc ht kh nng linh hot ca

my tnh trong iu khin s. V d cc nghim khng v nghim cc ca thit b

b u nm trn phn m ca trc thc mt phng s. Cc nghim y tng ng

vi phn dng ca trc thc trn mt phng z. Th nhng cc b iu khin s cho

php c nghim cc v nghim khng c phn m v phn dng ca trc thc

trn mt phng z nn iu kin hn ch c m rng hn. iu khin s cn cho

php tng hp cc b iu khin c hm truyn h kn mong mun.

1.2.5.1. Phng php qu o nghim s trn mt phng z.

y ch nu nhng im chnh.

Hm truyn ca h gin on kn c xc nh bi:

Wk(z) = )()(1

)()(

0

0

zGzG

zGzG

vi: G(z) = (1-z-1)z

s

sG )(1

Phng trnh c trng : 1+Gc(z)G(z) = 0

Phng php qu o nghim s thng dng xc nh thng s K c cu

iu khin nn c th vit phng trnh c trng di dng:

1+K

r

i

i=1

n

j

j=1

(z-z )

(z-p )

= 0

M pj v zi l nghim cc v nghim khng ca h xung h. T :

- K =

n

j

j=1

r

i=1

(z-p )

(z-pi)

Qu o nghim s ca h gin on c xy dng theo nhng quy tc

tng t nh h lin tc.


39

Hnh 1.23

Hnh 1.24

Hnh 1.25

1. Qu o nghim s i xng vi trc thc v gm c n nhnh xut pht t n

nghim cc ca phng trnh c trng khi K = 0; trong s r nhnh kt thc r

nghim khng v n - r nhnh i v v tn khi K = .

2. Qu o tim cn khi K (n - r) tia i xng to vi trc honh mt gc

n-r.

n nh h thng, c th dng

c cu b dng:

Gc(z) = Kz-a

z-b ; 0 b < a < 1

vy K.Gc(z)G(z) =

K2

2

(z-a)(z+1) T;K=K'

(z-b)(z-1) 2

By gi qu o nghim s s c

ba nhnh v h c ba nghim cc p1 = p2

=1; p3 = b. Mt nhnh n nghim khng

z1 = -1, nhnh th hai n z2 = a v nhnh

th hai tin n - .

C th c hai trng hp:

1. C ba nghim u thc nm ng

thi trn hai on thng ca qu o [b, a] v

[-1, -], trong trng hp y hai nghim b

hn -1 trn on [-1, -] v h s khng n

nh.

2. Kh nng h n nh l ch mt

nghim thc duy nht nm gia a v b, hai

nghim khc l nghim phc c mun

nh hn 1, nm trong ng trn n v.


40

1.2.5.2. B nh hng ca khu tr .

Nu thnh phn khng bin i ca h c tr, trong trng hp thi gian tr l

bi s ca thi gian ly mu.

= n0T; (n0 = 1, 2, 3, _

G(z) = G1(z) z-n0

hnh 1.23 trng hp a) khu tr nm trong mch vng s lm nh hng

n tnh n nh ca h. hnh 1.24 trng hp b) khu tr nm ngi mch vng v

s khng nh hng n h. yu cu t ra l tm c cu iu khin Gc(z) sao cho

nh hng ca khu tr khng cn na, ngha l ta c th ng tr hai s khi hnh

1.24 v 1.25.

0

1

1

0

1

0

1

)(1

)(

)()(1

)()( nn

c

n

c zzG

zG

zzGzG

zzGzG

T )(1

1

)()(1

)(

1

0

1 zGzzGzG

zGn

c

c

Hay Gc(z) + Gc(z)G1(z) = 1 + Cc(z) G1(z) z-n0

Gc(z) [1 + G1(z) (1-z-n0

)] = 1

Cui cng ta xc nh c: Gc(z) = )1)((1

10n

c zzG

S thc hin c cu iu khin s nh hnh 1.24. Nh vy vic dng c

cu iu khin s nh trn tng ng vi vic a phn t tr ra ngoi mch hi

tip. Tht vy v:

WK(z) = 0

1

1

1

1

)(1

)(

)()(1

)()( n

c

c zzG

zG

zGzG

zGzG

Khi c tr, h s khuych i ca

h c th ln hn so vi h khng c phn t tr nn nhiu khi khng cn b ton

b thi gian tr m ch cn mt phn ca n.

1.2.5.3. H n nh v tn.

Hm truyn ca h xung kn c dng


41

Hnh 1.26. Cu trc c bn ca iu chnh tc quay

WK(z) = c

c

G (z)G(z)

1+G (z)G(z) vi E(z) = X(z) - Y(z) = X(z)[1-WK(z)]

T b iu khin c xc nh bi: Gc(z) = )(W1

)(W.

)(

1

K

K

z

z

zG

Vic chn Gc(z) t cht lng mong mun gp phi nhng iu kin hn

ch:

1. iu kin thc hin c i hi bc ca h kn ln hn hoc bng bc ca

phn lin tc quy i (kn):

2. Sai lch trng thi xc lp, theo (4-58b) v theo nh l ti hn

3. n nh v tn t c khi sai lch trng thi xc lp ca cc tr ri rc

bng khng, k c mt thi im hu hn.

1.2.6. Dng matlab tng hp h iu khin s

- Tng hp theo c tnh tn Bode

- Tng hp theo qu o nghim s

1.3. iu khin s trong iu khin chuyn ng.

1.3.1. Mt s cu trc iu chnh c s dng.

1-Khu C

2-iu khin mmen

3-ng c

4-Khu o


42

Hm truyn c trng ca vng iu chnh v tr:

Gi thit gi tr t c dng hm dc tuyn tnh:

Gc ra c dng:

d sai lch gc:

Hnh 1.27. Cu trc ti gin phc v thit k xp x lin tc

Hnh 1.28. Cu trc c bn iu chnh gc

Hnh 1.29. Cu trc c bn iu chnh gc ti gin


43

Hnh 1.30. Cu trc iu chnh b sai s gi tr t

Hnh1.31. Cu trc iu chnh b nhiu


44

Hnh1.32. Cu trc iu chnh b ngc

Hnh 1.33. Cu trc iu chnh b xui bng phng php m hnh chun


45

Hnh1.34. Cc giai on ca mt qu trnh chuyn ng

Hnh 1.35. Cu trc iu khin tng qut ca mt nhnh truyn ng


46

Hnh 1.36.Cc lut thng dng nhm iu khin chnh xc chuyn ng


47

1.3.2. Thit k v m phng h thng bng my tnh.

Hnh trn gii thiu v d khi s dng mi trng thit k trn nn MATLAB &

Hnh 1.37. Trnh t thit k v m phng h thng bng my tnh


48

Simulink vi phn cng c vi x l tn hiu (Digital Signal Processor: DSP) ca

tp on Texas Instruments. S ch ra r rng: kt hp vi MATLAB v cc

Toolbox, ta c th tin hnh cc bc:

Bc 1: Xc nh hm truyn ca i tng, thit k b iu chnh bng l thuyt.

Bc 2: M phng Offline bc u xc nh tham s ca thut ton C.

Bc 3: B xung thm cc khi xut/nhp d liu (v d:cc khi ADC hoc

DAC) vo s cu trc vng C.

Bc 4: S dng C-compiler to m C np xung card hardware, ci xen vi

h thng phn mm iu khin theo ngt.

Xu hng pht trin ca ngnh t ng ho l ngi ta tn dng trit

nhng thnh tu khoa hc k thut mi nht. Trong c k thut iu khin s,

do c nhiu u im hn hn k thut tng t v c kh nng linh hot cao nn

iu khin s c ng dng ngy cng nhiu, c bit l trong iu khin

chuyn ng.

- ng dng k thut iu khin s trong cc h iu khin chuyn ng mang

li nhiu tnh nng vt tri so vi k thut iu khin chuyn ng truyn

thng nh: linh hot trong vic thay i thng s b iu chnh khi yu cu

cng ngh thay i, thay i cc phng php iu khin tin tin; tng kh

nng chng nhiu. Tuy nhin thc hin mt b iu chnh s li mt nhiu

thi gian v gp nhiu kh khn.

- ng dng k thut iu khin s vo cc h iu khin chuyn ng, hin

nay ch yu ngi ta s dng cc h vi x l tn hiu s (DSP), cc my tnh

s.

- Trong cng nghip cc h iu khin chuyn ng s ng dng cc my tnh,

cc Card iu khin chuyn dng c tch hp h vi x l tn hiu s(DSP).

- Trong nghin cu, c bit trong cc trng i hc k thut vic nghin cu

cc h iu khin s thng c thc hin trn cc Card iu khin s a

nng nh: DS 1102, DS 1104, DS 1103, DS 1105


49

CHNG 2. GII THIU CARD DSP DS1104

2.1. Gii thiu chung

thc hin c cc thut ton phc tp trong cc h thng sn xut hin

i, linh hot, cc nh sn xut phi t ng ho qu trnh thit k, rt ngn thi

gian th nghim, nhanh chng a thit b vo sn xut v m bo ti u cht

lng sn phm. iu ny ch c c vi s tr gip ca my tnh, qua cc bc

sau:

- Trong giai on phn tch: M phng thng c s dng phn tch i

tng, phc v cho vic thit k h thng. Cho php gim chi ph trong qu trnh

nghin cu khi chun b cho mt sn phm mi.

c im ca m phng l my tnh cn c thi gian cn thit tnh ton

tin trnh ca h thng. Vi m hnh n gin, kt qu tnh ton nhanh v m hnh

m phng phn nh c c im ng hc ca i tng. Tuy nhin, vi m hnh

phc tp th vic tnh ton mt nhiu thi gian hn.

- Sau khi qua giai on phn tch: Kim tra b iu chnh thit k tm ra

thng s ti u trc khi em i sn xut mch cng. V vy, cn phi ni i

tng thc vi b iu chnh c m phng bng thi gian thc.

c im chnh ca m phng thi gian thc l qu trnh m phng phi din

ra nhanh nh h thng thc ang chy, do n cho php ta kt hp m phng v

i tng thc.

- Khi b iu khin c m phng: c th iu khin c i tng

thc, ta bt u sn xut b iu chnh thc. Bc th nghim cui cng, ta ni b

iu chnh thc vi m hnh ca i tng (c m phng bng thi gian thc)

m bo chc chn rng b iu chnh khng cn li c th dn n ph hng i

tng thc, k thut ny c gi l m phng c phn cng trong mch vng.

Trong c hai cng on trn th m phng thi gian thc l rt cn thit. Tc

tnh ton yu cu cho m phng thi gian thc ph thuc vo c im ca m


50

hnh c m phng. Vi nhng m hnh phc tp, s lng php tnh ln th thi

gian m phng l vn cn c quan tm.

DS1104 l Card iu khin s do hng dSPACE ca c sn xut da trn b

x l tn hiu s DSP (Digital Signal Processor) du phy ng (floating-point) th

h th ba, h TMS320Cxx ca hng Texas Instruments (M). DS1104 c thit k

c bit pht trin cc b iu khin s a bin tc cao v m phng thi gian

thc. N thng c dng trong cc lnh vc sau:

- Robot.

- Cc c cu chp hnh bng in v thu lc.

- iu khin servo cc truyn ng a (disk drive).

- iu khin truyn ng in.

- iu khin cc phng tin c gii.

- iu khin trn ng tch cc.

- Trong cc my CNC,

v n cng rt thch hp cho cc tc v c lin quan n x l tn hiu s ni

chung.

Ht nhn ca DS1104 l b x l tn hiu s du phy ng (floating-point)

th h th ba TMS320F240 ca hng Texas Instruments. B x l tn hiu s c

Hnh 2.1- Card DS1104


51

b sung thm mt lot thit b ngoi vi thng c s dng trong cc h thng

iu khin s. Cc b bin i tng t-s v s-tng t, mt b x l tn hiu s

da trn cc h con vo ra s v cc giao din cm bin so lch (incremental sensor)

lm cho DS1104 tr thnh mt gii php bo mch n l tng cho mt di rng

cc bi ton iu khin s.

DS1104 l Card c thit k theo chun PC/AT, do n c th cm vo

my tnh qua cng m rng ISA. N cng c th gn vo hp m rng dSPACE

giao tip vi my tnh. Hnh 2.1 l hnh dng bn ngoi ca DS1104.

2.2. Cu trc phn cng ca DS1104

2.2.1. Cu trc tng quan

DS1104 c xy dng trn c s vi x l tn hiu s TMS320F240 ca hng

Texas Instruments.

ON-CHIP MEMORY (WORDS)

Ngun nui

(V) Chu k (ns) S chn

RAM FLASH

EEPROM

DATA DATA/PROG PROG

288 256 16K 5 20 PQ 132P

Ngoi ra, n cn c h con ngoi vi khc phc v cho cc ng dng x l tn

hiu s, giao tip vi my tnh v bn ngoi,

B x l chnh:

MPC8240, PowerPC 603e core, 250 MHz

32 kByte internal cache

Timer:

Mt b Timer c lng ly mu, b m li 32 bit

Bn b Timer a mc ch, 32 bit

phn di 64 bit o thi gian

Bng 2.1. Dung lng cc b nh ca DS1104


52

B nh:

32 Mbyte RAM DRAM (SDRAM)

8 Mbyte b nh Flash cho cc ng dng

Cc ngt iu khin:

Cc ngt bi timer, giao tip ni tip, DSP t, incremental encoder, ADC,

PC ch, 4 u vo t bn ngoi.

Ngt ng b PWM

u vo tng t:

4 knh ADC, 16 bit, a thnh phn

Di in p u vo 10V

Thi gian ly mu 2us

H s tn hiu/ nhiu >80 dB

4 knh ADC , 12 bit

Di in p 10V

Thi gian ly mu 800ns

H s tn hiu/ nhiu >65 dB

u ra tng t:

8 knh DAC, 16 bit, thi gian n nh max 10us

Di in p ra 10V

Incremental Encoder:

2 u vo s, TTL hoc RS422

Knh encoder c phn di 24 bit

Tn s xung max u vo l 1.65MHz. gp 4 ln xung m ti 6.6MHz

Ngun sensor 5V/0.5A

Vo/ra s:

Vo/ra s 20 bit

Dng ra 5mA


53

Giao tip:

RS232, RS485 v RS422

H con DSP t:

Texas Instruments DSP TMS320F240

4 kWord of dual-port RAM

3 pha u ra PWM, 4 u ra n PWM

14 bit vo/ra s

c im vt l:

Ngun nui 5 V, 2.5 A / -12 V, 0.2 A /12 V, 0.3 A

Yu cu cn c khe PCI 32 bit

2.2.2. Ghp ni vi my ch (Host Interface): DS1104 ghp ni vi my ch qua mt khi gm 4 cng vo/ra (I/O port) 16-

bit v 3 cng vo/ra 8-bit. Giao din vo/ra c s dng thc hin vic ci t

Hnh 2.2. S khi ca DS1104


54

cho bo mch, ti chng trnh xung v truyn d liu thi gian thc. Vic ci t

b iu khin bus kim tra v truyn d liu cng c thc hin vi giao din

vo/ra.

ng b ho s thc thi ca DSP v cc chng trnh ca my ch DS1104

s dng mt cng ngt hai chiu cho php my ch c th ngt DSP v ngc

li.

Giao din vo/ra gia my ch v DS1104 bao gm mt khi vi 7 cng

vo/ra lin tip. chn cc a ch c s ca khi ny trong di a ch vo ra 64K

ca PC/AT (my ch), DS1104 s dng cc chuyn mch DIP (Dual In-line

Package v hai hng chn) gn trn bo mch.

Giao din vi my ch ca DS1104 cha nhng thanh ghi c di khc nhau

(8 hoc 16 bit). Khi truy cp vo mt thanh ghi c th th phi s dng lnh vo/ra

tng ng, chng hn nh mun truy cp vo thanh ghi 8-bit th phi s dng lnh

vo/ra 8-bit, cn mun truy cp vo thanh ghi 16-bit th phi dng lnh vo/ ra 16

bit. Nu s dng cc lnh vo/ra 8-bit cho mt thanh ghi rng 16-bit th kt qu s

b li. Nu s dng ngn ng cp cao lp trnh cho cc thanh ghi giao din vi

my ch th cn phi m bo rng chng trnh dch Compiler to ra cc dng lnh

chnh xc.

Mt s thanh ghi giao din vi my ch phi c truy cp theo mt th t

c bit. ghi hoc c b nh ca DSP th mt trnh t c bit l bt buc.

a. Thanh ghi d liu (Data Register): a ch Offset: 00H v 02H

Thanh ghi d liu l mt thanh ghi c/ghi rng 32 bit c s dng truy

cp vo cc b nh off-chip (bn ngoi chip) ca DSP. Cc hot ng ghi v c

trn thanh ghi d liu lun c thc hin ti v tr b nh hin ang c chn bi

cc thanh ghi a ch LAR (Lower Address Register) v UAR (Upper Address

Register). V my ch ti mt thi im ch c th truy cp 16 bit nn thanh ghi d

liu 32-bit c chia thnh hai thanh ghi 16-bit: thanh ghi d liu thp hn LDR

(Lower Data Register) v thanh ghi d liu cao hn UDR (Upper Data Register).

chuyn mt t d liu 32-bit gia b nh ca my ch v ca DSP cn c hai


55

php ghi hoc c lin tip. u tin, 16 bit thp hn c truy cp bng cch s

dng LDR. Sau , 16 bit cao hn c truy cp thng qua UDR. Mch chuyn i

rng bus trn bo mch (on-board) lu tr tm thi gi tr 32-bit v thc hin ch

mt truy cp 32-bit n vo b nh ca DSP. mch chuyn i rng bus hot

ng chnh xc th th t truy cp LDR-UDR nh c m t trn l bt buc.

Ni dung ca cc thanh ghi LAR v UAR phi khng i trong mt truy cp 32-bit.

V cc thit b ngoi vi trn bo mch ca DS1104 c sp xp trong b nh

ca DSP nn thanh ghi d liu cng c th dng truy cp vo cc thit b ny.

Thanh ghi d liu c th c truy cp thm ch c khi DSP dang chy cho php

chuyn d liu chy thc gia my ch v DSP.

b. Thanh ghi a ch (Address Register): a ch Offset: 04H v 06H

Thanh ghi a ch l mt thanh ghi ghi/c c rng 19-bit c s dng

chn v tr ca b nh chng trnh ca DSP. V tr b nh m thanh ghi a ch

ang tr ti c th c ghi v c thng qua thanh ghi d liu. Thanh ghi a ch

c xy dng bng hai thanh ghi, thanh ghi 16-bit cha 16 bit a ch thp

A0A15 (LAR) v mt thanh ghi 3-bit cha cc bit a ch cao A16A18 (UAR).

Thanh ghi a ch c mt ch t ng tng/gim cho php chuyn khi gia b

nh ca my ch v ca DSP. Mun cho php ch ny th bit AUTOEN trong

thanh ghi ci t (Setup Register) phi c t ln 1. Sau bit UPDOWN s

chn chiu m. Nu ch t ng tng/gim c cho php th ni dung ca

thanh ghi a ch s c t ng tng/gim sau khi hon tt mt php ghi hoc c

thanh ghi d liu 32-bit. iu ny cho php truy cp lin tip cc khi ca b nh

DSP m khng cn thay i thanh ghi a ch cho mi ln chuyn.

truy cp thanh ghi a ch thp LAR cn c mt ch lnh vo/ra my ch

16-bit, cn truy cp vo thanh ghi a ch cao UAR cn phi s dng mt ch

lnh vo/ra my ch 8-bit. truy cp ln sau vo cng mt v tr b nh thanh ghi

a ch ch cn c ghi mt ln. Ch t ng tng/gim phi c loi b

(disable) cho nhng ng dng kiu ny. Nm bit cao ca UAR khng xc nh khi

c v c gi tr 0 khi ghi.


56

c. Thanh ghi trng thi (Status Register): a ch Offset: 07H

Thanh ghi trng thi (STS) l mt thanh ghi ch c 8-bit cung cp thng tin

v trng thi ca DS1104. N cho php my ch c nhiu ng iu khin ca

DSP, b iu khin kim tra bus (TBC - Test Bus Controller) v mt phn ca

thanh ghi ci t.

STS:

Bit Tn Chc nng

0 RESET14 Trng thi ti lp (reset) Slave-DSP. RESET14=1 biu th Slave-

DSP b thit lp li, RESET14=0 biu th Slave-DSP ang chy.

1 TBCINT Trng thi ngt TBC. TBCINT=1, mt ngt TBC ti my ch hot

ng. TBCINT=0, my ch hon tt dch v ngt.

2 RSTDSP Trng thi ti lp TMS320C31. RSTDSP=1, DSP c reset.

RSTDSP=0, DSP ang chy.

3 TBCRST ng reset TBC. TBCRST=1, TBC c reset. TBCRST=0,

TBC ang chy.

4 TBCRDY ng sn sng ca TBC. TBCRST=0 khi TBC ang thc thi mt

lnh. TBCRST=1 khi TBC kt thc mt lnh.

5 DSPRDY14

C sn sng truyn thng Slave-DSP. DSPRDY14=1, Slave-DSP

kt thc vic thi hnh lnh. DSPRDY14=0, TMS320C31 ghi

mt lnh v Slave-DSP cha kt thc vic thi hnh lnh

6 AUTOEN AUTOEN =1, Cho php ch t ng tng/gim. AUTOEN =0,

loi b ch t ng tng/gim.

7 UPDOWN Chn ch tng/gim. Ch tng nu UPDOWN=1. Ch

gim nu UPDOWN=0.

Bng 2.2. M t thanh ghi trng thi

d. Thanh ghi ci t (Setup Register): a ch Offset: 07H

UPDOWN AUTOEN DSPRDY14 TBCRDY TBCRST RSTDSP TBCINT RESET14


57

Thanh ghi ci t (STP) l mt thanh ghi ch ghi 8-bit dng iu khin rt

nhiu ch hot ng v trng thi ca cc tn hiu iu khin ca DS1104, chng

hn trng thi khi ng li ca DSP, Slave-DSP, TBC, cc yu cu ngt t my

ch ti DSP v ch t ng tng/ gim ca thanh ghi a ch.

STP:

Bit Tn Chc nng

0 RESET14

Trng thi reset Slave-DSP. Ghi 1 reset Slave-DSP. Ghi 0 s

khi ng li (restart) Slave-DSP. RESET14 phi gi mc logic

cao t nht 2ms. Khi khi ng Slave-DSP b reset.

1 RSTDAC

Reset DAC. Ghi 1 s t DAC trong ch reset. in p ra c

t xung 0 v thanh ghi ch DA c t ch khuch i

ng nht mt cc. Ghi 0 s ngt ng reset DAC. Khi khi ng

RSTDAC=1.

Lu rng thanh ghi ch DA phi c t ch khuch i

ng nht v hai cc sau khi RSTDAC c p dng

2 RSTDSP

Reset TMS320C31. Ghi 1 reset DSP. Ghi 0 s ngt ng reset

v cho php DSP bt u thc thi chng trnh. Khi khi ng,

TMS320C31 b reset.

3 TBCRST Reset TBC. Ghi 1 s reset TBC. Ghi 0 s khi ng li TBC. Khi

khi ng TBCRST =1

4 IRQDSP Yu cu ngt ca my ch ti DSP. Ghi 1 s yu cu mt ngt DSP

trn ng DSPINT3. Ghi 0 s khng tc ng g.

5 IRQEOI Kt thc ca ngt my ch.

6 AUTOEN Cho php ch t ng tng/gim thanh ghi a ch.

AUTOEN=1, Cho php ch t ng tng/gim. AUTOEN =0,

loi b ch t ng tng/gim. Khi khi ng AUTOEN=1 7 UPDOWN Chn ch tng/gim. Ch tng nu UPDOWN=1. Ch

UPDOWN AUTOEN IRQEOI IRQDSP TBCRST RSTDSP RSTDAC RESET14


58

gim nu UPDOWN=0. Khi khi ng UPDOWN=1

Bng 2.3. M t thanh ghi ci t

e. Thanh ghi d liu TBC: a ch Offset: 08H

Thanh ghi d liu TBC (TBCDR) l mt thanh ghi ghi/c 16-bit dng truy

cp vo TBC trn bo mch. TBC c 24 thanh ghi a ch 16-bit, cc thanh ghi ny

c th c chn thng qua thanh ghi a ch TBC (TBCAR). truy cp vo mt

thanh ghi bt k, u tin a ch thanh ghi phi c t bng cch ghi vo thanh

ghi a ch TBC, sau php ghi hoc c c thc hin bng cch s dng thanh

ghi d liu TBC.

f. Thanh ghi a ch TBC: a ch Offset: 0AH

Thanh ghi a ch TBC (TBCAR) l mt thanh ghi ghi/c 8-bit gm 5 ng

a ch TBC A0A4. Trc khi c hoc ghi mt thanh ghi TBC, TBCAR phi

c t ti mt a ch thanh ghi tng ng. Sau khi thit lp TBCAR, d liu c

th c chuyn s dng TBCDR. Lu khi ghi th 3 bit cao ca TBCAR nn t

bng 0.

g. Cng ngt DSP ti my ch:

DS1104 c mt cng ngt hai chiu cho php DSP yu cu ngt my ch v

ngc li. Cng ngt DSP ti my ch bao gm hai bit iu khin (ATREQ v

IRQAT) trong thanh ghi IOCTL v bit IRQEOI trong thanh ghi STP.

yu cu mt ngt DSP ti my ch th DSP phi t bit ATREQ. iu ny

to ra mt yu cu ngt trn ng ngt my ch c chn bi chn chn ngt.

Sau khi hon thnh dch v ngt, my ch s t bit IRQEOI trong thanh ghi STP.

C IRQAT trong thanh ghi IOCTL biu din trng thi ca ng ngt my ch,

cho php DSP nhn ra khi my ch kt thc phc v ngt.

yu cu mt ngt my ch ti DSP, my ch phi t bit IRQDSP trong

thanh ghi ci t STP. iu ny s t c IRQDSP trong thanh ghi IOCTL v to ra

mt ngt INT3 ti DSP. Nu INT3 c cho php trong thanh ghi cho php ngt

(IE) ca TMS320F240 v bit cho php ngt ton cc (GIE) c t ln 1 trong


59

thanh ghi trng thi ca DSP, mt li gi ngt ti a ch 000004H c thc hin.

Sau khi ngt my ch ti DSP c phc v, DSP phi thng bo cho my ch bit

phc v ngt hon tt. iu ny c thc hin bng cch ghi vo bit DSPEOI3

trong thanh ghi IOCTL, c DSPINT3 v ng ngt INT3 s b reset. Sau c

ngt trong thanh ghi c ngt ca DSP phi c xo.

2.2.3. Cc thnh phn ch yu ca DS1104

2.2.3.1. B x l tn hiu s DSP TMS320F240:

B x l tn hiu s DSP TMS320F240

TMS320F240 (F240) l mt thnh vin ca h b iu khin DSP da trn nn tng

b x l tn hiu s 16 bit TMS320C2xx. H vi x l ny c ti u ho cho

cc ng dng iu khin s ng c v chuyn ng. Cc b iu khin s DSP

phi hp TMS320 c tng cng li CPU C2xLP khi thit k vi chi phi thp, c

nhiu kh nng x l hiu xut cao v mt s im ni tri trong ti u ho ngoi vi

cho cc ng dng iu khin ng c v chuyn ng. Ngoi vi bao gm module

qun l s kin to ra cc Timer a mc ch v b ghi so snh to ra 12 u

ra PWM, v cc b bin i tng t - s kp 10 Bit (ADC).

Mt s c trng quan trng ca TMS320F240:

- Hiu sut cao vi ng dng cng ngh CMOS

+Tng thch vi h TMS320C2xx

- Li l h CPU T320C2xLP

+ M ngun tng thch vi TMS320C25

+ C th nng cp tng thch vi TMS320C5x

+ Tch hp trong v Plastic 132 chn

+ Thi gian thc hin lnh 50ns

- Thch hp vi nhit trong cng nghip v cc phng tin chuyn ng

- B nh

+ 544 Words 16 Bits of On-Chip Data/Program Dual-Access RAM


60

+ 16K Words 16 Bits of On-Chip Program ROM (C240)/Flash EEPROM

(F240)

+ 224K Words 16 Bits of Total Memory Address Reach (64K Data, 64K Program

and 64K I/O, and 32K Global Memory Space)

- Module qun l cc s kin

+ 12 knh so snh v iu ch rng xung PWM

+ 3 b Timer a mc ch 16 bit vi 6 ch , bao gm c ch m tin li

+ 3 b so snh 16 bit vi vng cht

+ 3 b so snh n 16 bit

+ 4 b thu thp d liu

Hnh 2.3. Vi x l tn hiu s DSP TMS320F240


61

- Module kp bin i 10-Bit Analog-to-Digital

- 28 chn vo/ra c th lp trnh n v a nng

- Phase-Locked-Loop (PLL)-Based Clock Module

- Watchdog Timer Module (With Real-Time Interrupt)

- Module giao thc truyn thng ni tip (SCI)

- Module giao thc ngoi vi ni tip (SPI)

- 6 ngt m rng (Power Drive Protect, Reset, NMI, and Three Maskable Interrupts)

- 4 ch hot ng tt km nng lng

- c lng vng qut c bn

- Pht trin cc cng c sn c

+Texas Instruments (TI) ANSI C Compiler, Assembler/Linker, and C-Source

Debugger

+ Scan-Based Self-Emulation (XDS510)

+ h tr pht trin iu khin m, hng iu khin s ng c th 3

Trn y ch gii thiu mt s c trng cn thit c bn ca TMS320F240

gip hiu kin trc v hot ng ca DS1104. DS1104 s dng tnh nng qun l

bus ca TMS320F240 gip my ch c th truy cp vo tt c cc b nh off -chip,

cho php cc hot ng download nhanh m khng yu cu mt chng trnh gim

st chy trn DSP.

DS1104 cn bao gm mt giao din ni tip tc cao c th c s dng

cho truyn thng gia mt vi bo mch x l tn hiu s nhm hnh thnh h thng

nhiu b x l (multi-processor).

TMS320F240 h tr cc ngt mm c th lp trnh c mm do v cc ngt

ngoi thng c ng dng trong cc ng dng iu khin truyn ng thi gian

thc. TMS320F240 c 3 loi ngt chnh: Reset, ngt cng, ngt mm.

Ngoi ra cn cung cp mt ng tn hiu thng bo v trng thi sn sng

ca bus, c s dng lm cho TMS320F240 thch ng v thi gian vi cc thit

b ngoi vi khc nhau trn bo mch. Thanh ghi iu khin bus ca TMS320F240

c lp trnh s dng tn hiu sn sng t bn ngoi.


62

a. Cc ngt ngoi ca DSP:

Tn cc ngt Th t u tin

RESET 1

TI RESERVED 2

NMI 3

INT1 4

INT2 5

INT3 6

INT4 7

INT5 8

INT6 9

TI RESERVED 10

Bng 2.4. Cc ngt cng ca DSP

Cc ngt c iu khin bi module h thng v b qun l s kin

Ngoi vi ng ngt

System Module

INT1

INT5

INT6

NMI

Event Manager

INT2

INT3

INT4

b. Bn b nh ca DSP TMS320F240:

DSP TMS320F240 c b sung thm 3 khng gian a ch ring bit cho

vng nh chng trnh, vng nh d liu v cc cng vo/ra. Mi mt khng gian

Bng 2.5. Qun l cc ngt cng


63

c tng s 64K 16Bit Word. Trong khong 64K Word ca khng gian d liu, t

256 n 32K word phn nh ca vng nh c th c xc nh m rng b

nh chung, khi c ch nh bi vng nh a phng (GREG). Vic truy cp vo

vng nh chung c iu khin bi tn hiu o BR.

Cc ngt ngoi

Dng cho bn

ngoi

On-Chip DARAM B0

(CNF = 1)

hoc Bn ngoi (CNF = 0)

D tr

Cc ngt (On-Chip)

On-Chip ROM

(Flash EEPROM) (8 x 2K Segments)

Dng cho bn ngoi

On-Chip DARAM B0

(CNF = 1)

hoc Bn ngoi (CNF = 0)

D tr

Dng cho bn ngoi

D tr

Ghi ch iu khin

Flash

D tr iu khin b ghi khi ch i c to ra

B ghi bn b nh v d tr

On-Chip DARAM B2

D tr

On-Chip DARAM B0

(CNF = 0)

hoc D tr (CNF = 1)

On-Chip DARAM B1

D tr

Cm

Cm

Dng cho bn ngoi

D tr

B ghi bn b nh ngoi vi (h thng,

WD,

ADC, SPI, SCI,

cc ngt, I/O)

B ghi bn b nh ngoi vi

(Qun l s kin)

Hnh 2.4.Bn b nh ca DSP


64

c. B nh:

B x l tn hiu s TMS320F240 cung cp hai ch hot ng khc nhau:

ch vi x l v ch my vi tnh. Ch my vi tnh ti thiu ho cc yu cu

phn cng bn ngoi trong cc ng dng cui cng. Tuy nhin, trong khi pht trin

D tr

B ghi mt n v ngt

B ghi ton b vng nh a phng

B ghi cc c ngt

B ghi cc c ngt

Cm

B ghi cu trc h thng v iu khin

Watchdog Timer v b ghi iu khin PLL

ADC SPI

SCI

Cm

B ghi cc ngt ngoi

Cm

B ghi cc u iu khin vo/ra s

Cm

B ghi Timer a mc ch

D tr

B ghi so snh, PWM v vng cht

D tr

B ghi Capture & QEP

D tr

B ghi mt n che ngt, vector v c

D tr

B ghi bn b nh v d tr

On-Chip DARAM B2

D tr

On-Chip DARAM

B0 (CNF = 0)

D tr (CNF = 1)

On-Chip DARAM B1

D tr

Cm

Peripheral Frame 1

Peripheral Frame 2

D tr

Cm

M rng (Dng cho bn ngoi)

Hnh 2.5.Bn b nh ngoi vi ca DSP TMS320F240


65

chng trnh, ch vi x l li ph hp hn. Trong ch ny, tt c cc tm np

(fetch) lnh u c thc hin t b nh ngoi, thng l RAM trong cc h thng

pht trin DS1104 s dng ch vi x l ca TMS320F240 c c s iu

khin b nh y bi my ch. Cc chng trnh ca ngi s dng c th c

ti xung (download), gim st, hoc thay i ti bt k thi im no ngay c khi

DSP ang chy. B nh ca DS1104 nhanh cho php hot ng vi trng thi

i bng khng tc ng h gii hn l 60 MHz.

d. Giao din ni tip:

TMS320F240 bao gm mt cng ni tip cung cp truyn thng trc tip vi

cc thit b khc nhau nh cc b bin i s-tng t ADC ni tip hoc cc DSP

khc. Hot ng ca cng ni tip c iu khin bi nhiu bit ch , cc thanh

ghi ca TMS320F240 v c th c lp trnh cho chiu di d liu t 8-32 bit

trong rt nhiu ch hot ng ng b v khng ng b. Tc truyn v nhn

d liu c quyt nh bi mt my pht xung ng h c th lp trnh c bn

trong hoc mt ngun xung ng h bn ngoi.

2.2.3.2. H con AD (Analog to Digital):

DS1104 c hai loi ADC Analog to Digital Converter B chuyn i tng

t/s):

- Hai b chuyn i tng t - s ADC 16-bit c tn s ly mu l 256 KHz.

- Hai b chuyn i tng t - s ADC 12 bit c tn s ly mu l 800 KHz.

in p u vo l 10V, tt c cc ng tr v u phi ni t. trnh

cc vng lp t (ground loops) nn s dng cc ng tr v ring bit cho tt c

cc cm bin v im t ca cc cm bin nn c cch ly vi nhau.

S chuyn i bt u bng cch t cc bit t STROBE AD1 ti STROBE

AD4 trong thanh ghi IOCTL. Trng thi u ra ca cc ADC c th gim st c

bng cch c cc bit t BUSY AD1 ti IOCTL BUSY AD4 ca thanh ghi IOCTL.

iu ny cho php DSP c th theo di c s chuyn i dng v c d liu


66

ADC sau khi qu trnh chuyn i kt thc. Hnh 2.6 biu din s khi ca h

con AD.

Trong cc a ch ca cc thanh ghi d liu ADC c cho trong bng 2.2

a ch Thanh ghi

023000H Knh 1: Thanh ghi d liu ADC 16-bit

023001H Knh 2: Thanh ghi d liu ADC 16 bit



Bng 2.6. Cc a ch thanh ghi ca h con AD

a. ADC 16-bit

DS1104 gm hai ADC 16-bit c tch hp mch ly mu v gi chm. Mi

ADC c mt b bin i tng t /s AD kiu xp x lin tip SAR v mt mch

ly mu/gi chm. Thi gian bin i ca mi b bin i t 4s.

u ra ca cc ADC c cn l bn tri ca t DSP 32-bit (hnh 2.7) v c

th c c bng cch s dng cc thanh ghi d liu ADC tng ng.

Hnh 2.6. S khi ca h con AD


67

b. ADC 12-bit

DS1104 gm hai ADC 12-bit c tch hp mch ly mu/gi chm. Mi ADC

c mt b bin i tng t /s AD kiu xp x lin tip SAR vi mt mch ly

mu/gi chm v c iu khin s bi cc n v hiu chnh offset (lch khng).

Thi gian bin i ca mi b bin i t 1,25s.

u ra ca cc ADC c cn l bn tri ca t DSP 32-bit (hnh 2.8) v c

th c c bng cch s dng cc thanh ghi d liu ADC tng ng.

c. Hiu chnh lch khng (Offset Calibration)

Mch u vo ca ADC c c trng bi mt n v hiu chnh dng b

cc sai s lch khng. Mch ny c dng xo b cc sai s lch khng ca

mch tng t pha trc v cc mch ADC ch khng dng b cc lch khng

ca cm bin bn ngoi. n v hiu chnh bao gm mt thit b EEPROM

(Electrically Erasable Programmable Read-Only Memory) c ni vi mt DAC

(Digital to Analog Converter) nhm cung cp mt in p nh b vo in p vo

ca ADC. N c iu chnh khi ch to DS1104 v khng cn thay i trong iu

kin lm vic bnh thng.

d. Mch vo ca ADC

Hnh 2.9 biu din mch vo ca ADC:

2.2.3.3. H con DA

(Digital to Analog):



Hnh 2.9. Mch u vo ca ADC


68

DS1104 gm mt DAC 12-bit, 4 knh vi phm vi in p u ra c th lp

trnh c. H con DA bao gm 4 thanh ghi d liu, 4 thanh ghi u ra, mt thanh

ghi ch v mt bit STROBE (STROBE DA)trn thanh ghi IOCTL.

Trong cc a ch ca cc thanh ghi d liu DAC c cho trong bng 2.3

a ch Thanh ghi

022000H Knh 1: Thanh ghi d liu DAC




026000H Thanh ghi chn ch

026001H Thanh ghi chuyn ch

Bng 2.7. Cc a ch thanh ghi ca h con DA

Hnh 2.10. S khi ca h con DA


69

S cn l ca thanh ghi d liu DAC trong t DSP 32-bit nh c biu din

trn hnh 2.11.

Cc DAC c in p u ra mi knh l 10 V. Cc ng tr v ca cc u

ra c ni vi t h thng.

a. Hiu chnh khuch i v lch khng (Offset and gain Calibration)

Mch u vo ca ADC c trng bi mt n v hiu chnh khuch i v

lch khng. Mch ny c dng xo b cc sai s khuch i v lch khng ca

mch. n v hiu chnh bao gm mt thit b EEPROM (Electrically Erasable

Programmable Read-Only Memory) c ni vi mt DAC (Digital to Analog

Converter). N c iu chnh khi ch to DS1104 v khng cn thay i trong

iu kin lm vic bnh thng.

b. Thit lp ch DA

DS1104 bao gm mt cp thanh ghi ch iu khin di in p u ra

ca DAC. thay i ch ca DA, cn c hai hot ng ghi. Th nht l gi tr

ch ca DA phi c ghi vo hng u tin ca thanh ghi ch DA a ch

026000H, sau ghi vo a ch 026001H cp nht gi tr ch ca DA.

Hnh 2.12 biu din nh dng d liu ch DA.

Bit Name Chc nng

24

27

M4

M1

Ch DA. Ghi 1 cho hot ng hai chiu, ghi 0 cho

hot ng mt chiu

28 G4 Khuch i DA. Ghi 1 cho h s khuch i = 2, ghi 0

Hnh 2.11. nh dng d liu ca DAC 12-bit

Hnh 2.12. nh dng d liu ch DA

Ch


70

31

G1

cho h s khuch i =1

Bng 2.8. M t thanh ghi ch DA

c. Mch u ra ca DA

Mch in u ra ca DAC c biu din trn hnh 2.13

2.2.3.4. H con Vo/Ra s

(Digital I/O):

H con vo/ra s ca

DS1104 hot ng da trn vi

iu khin x l tn hiu s

25MHz TMS320P14 ca hng Texas Instruments. Bn cnh phn li DSP 16-bit c

nh, n cn bao gm mt cng vo/ra song song c th chn tng bit, 4 b nh

thi gian (timer), 6 mch iu ch rng xung PWM (Pulse Width Modulation) c

phn gii 40 ns, 4 u vo trc tip v mt mch vo/ra ni tip. P14 c sn

chng trnh phn mm c s trong PROM cho php TMS320F240 c th truy cp

vo tt c cc thit b ngoi vi trn chip. Sau khi khi ng DSP thc thi chng

trnh c s (firmware) phc v vo/ra thng tr trong PROM. Chng trnh

PROM ny s c m rng bi mt chng trnh bn ngoi c np vo

RAM cung cp tnh nng ti (download) chng trnh, cho php cc chng trnh

ng dng c th ca DSP c thc hin song song vi TMS320F240. Tnh nng

ny cho php tu chnh (customize) h con vo/ra s (Digital I/O) theo cc yu cu

ca

nghien cuu va ung dung card dieu khien so dsp de thiet ke bo dieu khien so trong dieu khien chuyen...

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