NH Đề Thi -KHMT-XDCHTN

8/2/2019 NH Thi -KHMT-XDCHTN

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HOC VI N CNG NGH BU CHINH VIN THNG

KHOA: Khoa hc my tnh

NGN HNG CU HI THI T LUN

Tn hc phn: XY DNG CC H THNG NHNG M hc phn:............

Ngnh o to : ........................................... Trnh o to: .....................

Ngn hng cu hi thi

Cu hi loi 1 im

CHNG 1 : Gii thiu chung v cc h thng nhng

Cu hi 1.1: Nu cc nh ngha tng i v h thng nhng (nh ngha tng

qut, nh ngha theo t chc IEEE).

Cu hi 1.2: Hy chn xem cc h thng sau y, h thng no thuc h thngnhng :

A. Cc thit b y t.

B. Cc h thng iu khin qui trnh cng nghip.

C. Cc h thng my tnh.

D. Cc thit b truyn thng k thut s.

E. Cc h thng c tin cy cao v rt cao

F. Tt c cc h thng trn.

G. Khng c h thng no trn .

Cu hi 1.3:

1

Mu 2


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A. Nu tn mt s lnh vc i sng v cng nghip, c s ngdng ca cc h thng nhng.

B. Cho bit t nht hai thit b nhng mi lnh vc.

Cu hi 1.4:

A. H thng nhng l mt my tnh a nng hay my tnh chuyn dng,c bit cho ng dng c bit?

B. Phn mm h thng nhng l phn mm kiu h iu hnh a dch v,rt phc tp hay ch l phn mm hng ng dng, hay c hai ?

C. Cho v d v phn mm h thng chy trn cc h thng nhng ?

Cu hi loi 2 im:


Cu hi 2.1:

A. Hy nu nhng c im ca cc mi trng m h thng nhnghot ng.

B. Cho v d v mt h thng nhng no v nu c im mi trngm h thng nhng ang hot ng.

Cu hi 2.2 : Ti sao ni hu ht cc h thng nhng hot ng vi s rng bucv thi gian ?

Cu hi 2.3 :

A. Nu cc kiu hot ng c bn ca h thng nhng.

B. Thng thng h thng nhng l h thng hot ng ch tchcc hay ch th ng? Cho v d v gii thch ti sao li l h tchcc hay ti sao l h th ng.

CHNG 2 : Cc thnh phn phn cng ca h thng nhng

Cu hi 2.4:

A. BUS ca CPU gm cc thnh phn no hp thnh ?

B. BUS h thng v BUS CPU c g khc bieejt ?2


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Cu hi 2.5: H thng nhng tng tc vi mi trng vt l nh th no, phngtin, cng c g ? Nu mt s v d ?

Cu hi 2.6: Cho mt m hnh qui trnh iu khin cng ngh c ng dng hthng nhng nh hnh sau :

Hy khoanh vng cho bit h thng nhng l phn no ? cc thnh phnhp thnh ca h thng nhng l g ? Chc nng ca cc thnh phn ?

CHNG 4 : Thit k v ci t H thng nhng

Cu hi 2.7:

A. xy dng mt kin trc cho mt h thng nhng, phi tun th 6bc c bn. Cc bc l cc bc no ?

B. Nu cc pha trong qu trnh thit k mt h thng nhng ?

Cu hi 2.8 :

A. Th no l phn hoch phn cng v phn mm khi thit k mt hthng nhng ?

B. Th no l qui trnh ng thit k phn cng v phn mm(hardware/software codesign) v ng kim nghim (co-verification) ?

Cu hi 2.9 :

A. Hy c t cc tc v khi thc hin khi ng h thng ngui (coldboot) v khi ng nng (warm boot) ?

B. Hy cho v d v cch khi ng vi mt loi CPU t chn ?3


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Cu hi 2.10 :

Khi thit k h thng nhng, cn xy dng mt m hnh chnh tc(formal model) vi cc yu cu t ra. Vy cc yu cu l nhng yucu g ?

Cu hi loi 3 im:


Cu hi 3.1: Nu cc thch thc phi i mt khi thit k mt h thng nhng ?

Cu hi 3.2 :

A. Hy nu cc tiu ch khi phn loi cc h thng nhng ?

B. Ti sao cc h thng nhng li c s khc nhau ?

Cu hi 3.3: Hy nu s khc bit khi thit k h thng nhng kiu trn cng mtbo mch :

A. H thng nhng xy dng t b vi x l (Microprocesssor-based

Embedded Systems) l : B. H thng nhng xy dng t cc vi iu khin (Microcontroller-

based Embedded Systems): l


Cu hi 3.4:

y l hai kiu kin trc my tnh :

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CPU 1

B nh lnh B nh d liu

Vo Ra

D liua chd liu

Lnhmy

a ch cholnh

Kiu 1

CPU 2

B nh

a ch ti b nh Lnh/D liu

Vo Ra

Kiu 2

A. Cho bit tn gi ca cc kiu kin trc l g ?

B. im khc bit c bn hai kin trc ny l ch no ?

C. Kiu kin trc no l thch hp hn xy dng cc h thng

nhng (kiu 1 hay kiu 2)? Ti sao ?

Cu hi 3.5:

A. ADC (Analog Digital Converter) l g ? Nu chc nng v lnhvc ng dng ?

B. DAC (Digital Anlog Converter) l g ? Nu chc nng v lnh vcng dng ?

C. chn mt ADC, cn cc thng c bn s no ? Da vo nhng

tiu ch no chn ADC cho mt h thng nhng ?

Cu hi 3.6 : Khi nghin cu CPU thit k mt h thng vi x l nh mt hthng nhng, c mt s khi nim sau y :

A. Th no l mt trng thi my ? Th no l mt chu k my ?

B. Th no l mt chu k lnh ?

C. Cc khi nim trn c tc ng g khi vit cc on m chngtrnh cho cc x l ti hn (critical code) khi cc x l mang tnhcnh tranh ti nguyn h thng ?

Cu hi 3.7: Cho m hnh kin trc h thng nh hnh v.

A. Hy gii thch chc nng ca khi GIAO DIN VI CPU, l do s c mt ca khi ?

B. Th no l BUS ng b ?

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C. Th no l BUS khng ng b ?

CHNG 3 : Cc hnh phn phn mm ca h thng nhng

Cu hi 3.8:

A. Trnh iu khin thit b l g ?

B. Chc nng ca trnh iu khin thit b l g ?

C. Nhn theo kin trc phn mm my tnh, trnh thit b c t u ?

Cu hi 3.9:

A. Th no l trnh iu khin xc nh theo kin trc.

B. Th no l trnh iu khin tng qut (hay trnh iu khin trn bomch).

C. Hy nu mt s loi trnh iu khin thit b in hnh ?

Cu hi 3.10:

A. Nu cc thao tc ca trnh iu khin thit b khi c kch hot ?

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B. Nu cc bc cn thc hin khi i vo thit k mt trnh iukhin thit b ?

C. Cho v d cc bc cn thc hin khi vit trnh iu khin ngt ?

Cu hi 3.11:

Hy chn cu tr li cho l ng khi ni v h thi gian thc:A. L mt h x l c tc cao ;

B. L mt h hot ng c tnh tin nh ;

C. L mt h c kh nng cho p ng kp thi v chnh cc.

Cu hi 3.12:

A. Cho biu thi gian lin quan ti cc thi im thc hin mttc v h thi gian thc sau y :

t

aisi fi di

ci

ri wi

Cc thng s ai , ri, si, Ci, fi, wi, di l biu din ca cc thiim no ?B. Th no l h thi gian thc cng (hard real time system) ?C. Th no l h thi gan thc mm (soft real time system) ?

Cu hi li 4 im


Cu hi 4.1:

A. Hy c t kin trc CPU kiu Von Neumman v kin trc CPU kiuHarvard.

B. Hai kin trc ny khc nhau im no ? Kin trc no thch hphn khi chn thit k mt h thng nhng ?

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Cu hi 4.2 :

A. Nu (v ) m hnh tng qut phn cng ca mt h thng nhng ?

B. Nu chc nng ca tng khi ca m hnh ?

Cu hi 4.3 : Ni H thng nhng l mt h thng ng tin cy (dependable), v ccc tnh sau y :

Tin cy (Reliability): Hy gii thch c th Kh nng duy tr (Maintainability):Hy gii thch c th Tnh sn sng (Availability): Hy gii thch c th Chc chn (Safety): Hy gii thch c th An ninh (Security): Hy gii thch c th


Cu hi 4.4:

A. T cch xc nh a ch cho dch v x l ngt (Interrupt ServiceRoutine-ISR) sau y :

1. a ch ca ISR c nh sn bn trong CPU ;

2. a ch ca ISR c ch nh mt ch no trong b nh, hocphi thc hin mt lnh nhy (JMP addr) ti a ch hin ti ca

ISR ;3. Thit b ngoi vi phi cung cp cho CPU a ch ca ISR thng

qua s hiu ngt.

Hy chn cc kh nng nh a ch ph hp vi kiu t chc ngtsau:

A. Ngt c nh l : 1 ? hay 2 ? hay 3?

B. Ngt vector l : 1 ? hay 2 ? hay 3 ?

B .Ngt cng l g ? Cho v d ng dng ?

C .Ngt mm l g ? Cho v d ng dng ?

Cu hi 4.5 :

A. Th no l ngt c che (maskable) ?

B. Th no l ngt khng che (non-maskable) ?

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C. Hy c t cc bc m CPU s thc hin khi chp nhn x l mtngt ?

D. Khi vit m cho dch v x l ngt (Interrupt Service Routine-ISR),cn c bit quan tm n g ? Ti sao ?

Cu hi 4.6:

Hy chn nhng tiu ch cho l ng vi c ch trao i d liu kiu truynhp trc tip b nh (DMA) :

A. Cn c mt vi mch (DMAC) iu khin qui trnh DMA.

B. CPU vn kim sot BUS h thng.

C. CPU trao cho vi mch DMAC quyn kim sot BUS h thng.

D. CPU vn thc hin chy chng trnh nu chng trnh khngc i hi truy nhp BUS h thng.

E. Trong khi xy ra ch DMA, loi CPU no vn c th tm lnhv thc hin lnh b nh lnh (code) nu khng truy nhp ti bnh d liu ? Ti sao ?

CHNG 3 : Cc hnh phn phn mm ca h thng nhng

Cu hi 4.7:

A. Th no l lp lch hng vo/ra (hng I/O) ?

B. Th no l lp lch hng CPU ?C. Th no l lp lch tnh (static/offline) ? Th no l lp lch ng

(online) ?

D. Th no l lp lch c th chen ngang (preemptive algorithms) vlp lch khng th chen ngang (non-preemptive algorithms) ?

Cu hi 4.8 :

A. nh thi (watch-dog) l g ? Vai tr ca nh thi trong h thi gian thc l g ?

B. Pht tho mt gii thut x l c tc nhn gim st ca nh thi ?

Cu hi 4.9 :

Di y l m hnh ca h iu hnh thi gian thc (RTOS) v h iuhnh chun, chung cho my tnh. S khc nhau l v tr ci t trnh iukhin thit b (device driver). Hy gii thch l do ca s khc bit ?

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Cu hi 4.10 :

A. C th ni trong h thi gian thc, hu ht cc hot ng x l u do b lpbiu kim sot (nh hnh di). Hy gii thch ti sao nh vy ?

T h c th i tv

I S R i u p h i

n g t

C c d ch v th i g i a n v

s k i n

C c d c h v (t o t i n t r n h,l u n g, c h u y n tr n g t h i,

n h n, p h t d l i u )

L p b i u v i u p h i t c

v

N h n R T O S

N g t n g o i

N g t n h t h i

G i h t h n g, b y

B. Th no l mt h thng nhng thi gian thc ?

Cu hi 4.11 :

A. Th no l phn mm trung gian trong mt h thng nhng ?

B. Trong cc hnh sau a, b, c, d, hnh no sai khi t phn mm trung gianvo cc lp kin trc phn mm ca h thng nhng ?

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Phn mm ng dng

Phn mmtrung gian

iu khinthit b

Phn cng

Phn mm h thng

Phn mm ng dng

Phn mm ng dng

Phn mm ng dng

Phn cng Phn cng

Phn cng

Phn mmtrung gian

Phn mmtrung gian

Phn mmtrung gian

iu khinthit b

iu khinthit b

iu khinthit b

Phn mm h thng

Phn mm h thng

Phn mm h thng

H iu hnh

H iu hnh

a b

c d

C. Cc thit b mng lp 3 nh chuyn mch lp 3 (SW L3), nh tuyn(Router), ADSL l cc thit b nhng. Phn mm mng lp 2 v lp 3l cc phn mm trung gian. Hy chn mt m hnh trong cc m hnhtrn c cho l ph hp xy dng cc h thng nhng cp ?

Cu hi loi 5


Cu hi 5.1:

A. Nu m hnh phn mm ca mt h thng nhng ?

B. c t cc lp ca m hnh phn mm h thng nhng ?

Cu hi 5.2: M hnh phn mm h thng nhng c th biu din bi 3 lp :-Trn cng l Lp phn mm ng dng, l ty chn (optional).

- gia l Lp phn mm h thng, l ty chn (optional).

- Di cng l Lp phn cng, l cn thit phi c(required).

Ni vy l ng hay cha ng, v sao ? Hy gii thch !

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Cu hi 5.3:

A. Pht tho m hnh hot ng vi h c s dng ngt kiu vector ?

B. c t cch hot ng ca m hnh ?

Cu hi 5.4 :

A. Pht tho m hnh hot ng vi vi DMA ?

B. c t cch hot ng ca m hnh ?

Cu hi 5.5:

A. Th no ghp ni do CPU ch ng ? Cho vi v d ?

B. C my k thut trin khai ghp ni do CPU ch ng ?C. Th no l ghp ni do ngoi vi ch ng ? Cho v d ?

D. C my k thut trin khai ghp ni do ngoi vi ch ng ?

E. Da vo tiu ch no la chn mt m hnh ghp ni cho phhp ? Cho v d ?

Cu hi 5.6:

A. Hy m t t chc b nh ca dng CPU kiu Havard ?

B. Trong kin trc Havard, m chng trnh cha RAM hayEPROM ?

C. Trong kin trc Havard, c th truy nhp ng thi vo RAM vEPROM ?

D. Trong kin trc Havard , ni rng s bit cho d liu v s bit cholnh c di khc nhau, v d d liu 8 bit, trong khi lnh c th

di ti 32 bit, ng hay sai ?E. S khc nhau c bn ca kin trc Von Neumann v kin trc

Havard th hin im no ?

Cu hi 5.7:

M hnh Module ghp ni c a ra nh sau :

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mdliu

NhnlnhtCPU

Chttrngthithitb

Cngghpint

Cngghpint

Giim,logiciukhin

Thit bi

Thit bi

BUS cua Module

BUSdliu

BUSiach

BUSiu

khin

Module ghp ni

A. Hy nu c t ca cc khi chc nng bn trong Module ?

B. Phng thc Module hot ng ?

C. Cho v d v mt hay vi vi mch tch hp loi ny ?

D. Th no l mt vi mch ghp ni kh trnh ?

E. Th no gi l t iu khin (Control Word), chc nng lm g ?

Cu hi 5.8: Hy thit k mt module ROM vi cc d kin sau y:- Dung lng ca module l 32 KB- S dng chip ROM loi 2732, dung lng 4KB/chip

- Gii a ch t FFFF-8000.- Hy v s thit k v gii thch nguyn l hot ng.

Cu hi 5.9:Cho trc mt bng cc gi tr thit k b nh RAM vi chip loi4KB/chip sau y:

A19 -A16

A15 A14A14 A12

A11 A10A9 A8

A7 A8 A5A4

A3 A2 A1A0

a chu/a ch

cui(HEX) ?

1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

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Chip 0 1 1 1 1 1 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1

Chip 11 1 1 11 1 1 1

1 0 0 11 0 0 1

0 0 0 01 1 1 1

0 0 0 01 1 1 1

0 0 0 01 1 1 1

Chip 21 1 1 11 1 1 1

1 0 1 01 0 1 0

0 0 0 01 1 1 1

0 0 0 01 1 1 1

0 0 0 01 1 1 1

Chip 31 1 1 11 1 1 1

1 0 1 11 0 1 1

0 0 0 01 1 1 1

0 0 0 01 1 1 1

0 0 0 01 1 1 1

Chip 41 1 1 11 1 1 1

1 1 0 01 1 0 0

0 0 0 01 1 1 1

0 0 0 01 1 1 1

0 0 0 01 1 1 1

Chip 51 1 1 11 1 1 1

1 1 0 11 1 0 1

0 0 0 01 1 1 1

0 0 0 01 1 1 1

0 0 0 01 1 1 1

Chip 61 1 1 11 1 1 1

1 1 1 01 1 1 0

0 0 0 01 1 1 1

0 0 0 01 1 1 1

0 0 0 01 1 1 1

Chip 71 1 1 11 1 1 1

1 1 1 11 1 1 1

0 0 0 01 1 1 1

0 0 0 01 1 1 1

0 0 0 01 1 1 1

a. Tnh a ch u v a ch cui ca mi chip (in vo bng)?b. Ln s thit k module nh ny.

Cu hi 5.10:

Cho mt thit b ngoi vi c kh nng ghp ni vi h vi x l vi cc thng ssau y:

Trao i d liu vo h vi x l mi ln 8 bit, C thng bo cho CPU bit qua tn hiu Strobe (STB) rng c sn d

liu CPU c vo :

D0.

D7

Strobe

Thit b

CPU cung cp hai cng vi a ch nh sau:300 (hex): cng thit b thng bo c d liu sn sng CPUc vo;301(hex): l a ch cng ghp ni thit b t d liu vo v CPU

s c d liu vo.

a. Hy chn gii php thit k, ln s thit k ghp ni.b. Pht tho lu trnh iu khin ghp ni .

Cu hi 5.11:

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c. Hy pht tho gii thut (logic ca chng trnh) cho phn mmny ?

Cu hi 5.13 :

Di y l gii thut vi nh thi (Watchdog), trong cn nh nghahai gi tr cho Twdv Ttask. Hy chn cu tr li ng ?

Main loop code :

main (){

Call (Set Twd );

Task 1(){...FLAG =(TRUE /FALSE );}

Task 2(){...FLAG =(TRUE /FALSE );}

Taskn (){..

FLAG =(TRUE /FALSE );

}}

If (AllFLAGs =OK){

Call (reset Twd );Jmp mainLoop

}else{

Log error ;reset System ; WD reset CPU

}

Ttask

A. Twd > TtaskB. Twd = Ttask

C. Twd < TtaskD. Th no l ti hn cng (hard deadline) ? S dng nh th no ?E. Th no l ti hn mm (soft deadline) ? S dng nh th no ?

Cu hi 5.14 :

Nu kch bn khi ng chy phn mm HTN t ROM, RAM cho dliu

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Cu hi 5.15 :

Nu kch bn khi ng chy phn mm RAM sau khi m copy tROM vo RAM

Cu hi 5.16 :

Nu kch bn khi ng chy phn mm RAM sau khi ti xung t hpht trin (ang pht trin h thng).

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