08～09数字逻辑期末考试试卷评讲

电子科大学二零技八至零二 零零九学第二学年期末考试期一T、O FLIL YOUR ANSWRES I THNE“[ 1. 2[]101= > [0010111 1]RAG Y Assume( the dumbnr esysetm s 8-ibit lng o )2. f I[X 10]= +91 ,hetn[ ] Xtw’o-scmolpemnte =[ 0011111 ]0,2[ -] twX’o-cospmlmeetn=[ 10 001101]2 ( sAumsd ethe unbmr esytes ism 8-bit ong l )3 .T oesdig na“ 101001”0 esria sequlecne geerntar bo syhftireg itsesr, te shift herigtesr hosud lned e 5 )(bit a letsta ]”. 2’(X1 )04.

I fu es te hsmipelt stats assiengmet nemtodhfo r1 3 s3taets,htenn ed et aeaslt[ 8 s]atet arviblesa。5. f Inubmre[ A ]tw os’-ocmlempent =11111001an d [] twB’oscomp-elemt=1n0110011, accllate [u--B A] wotsco’pmleent, m-[A+ ] Btw’soc-omplmene tnadin idctea whteehr or no tvoeflrw ooccrsu.[ A--B] t o’sw-cmolepmetn [= 00110010 , ]oevflrw:o [n ] [o-+AB ]to’s-wcompelent=[m11 11001 0],ov eflro:w no[] . 6nO estte atrasntiinoe qutiaoni s *QJ=Q’K+Q’I。fw eus e Dfilpflo-pto coplmet teehequ tainot,he iDnputt rmiean of Dl lfipf-lp should beo hva teh fenutcion D= JQ[+’K’ ]。Q

1A.3 in-pt uturh ttalb eissh on wblowe. Fni dhet minmail ums-f-prooucdste xpressin for theo utpuot F. [5]C 0 B A 00F 0:解 A B C卡诺图:0 0 01 1 11

1 0 1 0 101

101 01 0

0 101 0 111 00 0 1111 0 0100

01

11

0

1Theo tupt uufctinn o=FBAA++CC

2.Bo fTind het mniima slum-o-prodfucste pressxoniof F’ froF(X 1,2X,X,34)X (=,315,7,9,,1,12113,,14,5), 1(F is ’complmenet of F) [ 5] : 函数解F卡诺的图 00 0 0101 110 1 1 11 101 11 1 11 11 1 函数F’0的诺卡图 00 1011 010010 1 11

10 0 01

0 0 1100 0 0

1

00

1三 .Tr ty ofinihslo icagl fucntio n(FW,,YX,)Z =∑ wyz(3,x,671,011,1,)by4on e hip ofc74 1X38 nd onea -i8nut napdn atge.[1 ]0:解

(FW,,Y,ZX ) ∑ =WXY(Z,367,10,,1,114) W’=’XZ+ W’YYZX+ W’’YX+ ZX’YW’+ZWX YZ’+W XYZ =Y’(WX’’Z+ W’XZ’ +W’XZ+ W’XZ+’WX ’+Z WZX’

=) ·YW∑X(Z,21,34,,5,)6

四 ._bit a4dder 4xx7238s ’unctfino tbal eis hswnoas elbow D.eigns coabminaitnal coicriut itw h_b4i tinutp {YY(=yy2y31y0} )an 6db_it ouptu tZ{=z5,z(4…,1,z0)z }adn to ralize the fuencito nZ6=Y, u isngo en hcipof 74283 nda onn eloicg ates.g0≤Y≤9( )[01][EGNREL ADSECRIPTON FOR I74XX83:2T eh7 4HC82 add 3wo t4-it bbiary worns d(An lpusBn ) pus lte incohmnig acrr. yTh ebinrya sum paepra onst h sue mouputt sS(M1Ut o UMS4) nad ht eou-tgoing crary C(OT)Ucaorcidngt ot e equathoin: IC N+( A1 + B)1+ 2 A(2+ 2)B + 4A(3+ B3 +)8(A4 +B 4 ) =SM1 U +2SMU +2 SU4M3 +8UM4S+ 6C1UO WThree +)(= plu .]s

: 解=6 Z=(Y+42)Y=Y42+Y=( 3y2yy1y00)+(00y 32yyy001=) (5,zz,…4z,z10)

五 Cl.oced Sykcnronous Stathe aMhcie nesigD [n15 ]

. De1sgn i aclockdes nycrhonou stsat emcaine whthi th esatte/utoptu talb seowhn elbw, osiug n flDi-plfpos Us.e wtost taev raablie,Q1s Q2,wi t hth steaet ssigamennt hsonw a sfllowos. rWti traesitnion/otupt utbael ndaex ctaitoino/tuut tapleb.[ 6]st

at/eotuupt talb: Se 0X 1 stteaas isngenmt: ASB C D Q2 Q 1 0 001 1 0 11A C DBB0, ,C 0D,0 A,S*,Z

1,C A10 ,,0 B,0D

t

rnastiio/ontuut tpalebX Q2 1 Q0 001 1 011 001 /01 00 1/1/000/1 1 1 0/1 000 /11/ 00/10exicattonio/tuput tbal Xe 2 QQ10 0 1 10 10 0 0110 /0/10 1/01 00/1 1001/ 10/0 1/001 /0

Q12 Q*1/Z*

DD2/1Z

2. heT xceitatonio/upttut blaeo af clckeods ycnronhos suatt eamcinheusin gD lfipflo-s shpwon a sfollo,wW ire thteex ctation equitiansoand otuup tqeatiuo. n[]9 xeciattio/ontupt utbla X Qe Q01 0 00 1 0 11 010 /0111/0 0 01/10/ D10D/0Z1 0/11 0/001 /0 011/

0

-maK fop D0rQQ0 10 00 1 10 0 1 10 1 01 11 0 10

K-am fpro D1

XX

Q1Q 0000 10 1 011 011 01

10 01

-mapKf o r Z000 1 010 0 0111 00 10 1 0Tehe xitacton eqiatuionsa n duotptuequ taion:X

1D=Q0 X+ ’0QX’0=DQ1’X ’+ QX1 Z=Q1Q ’X0+ Q’1Q’’0X

六. Cloced Skychrnnoos utaSt Meachie nnaAlysi [s 5 1]1.An lyaez te hcriuct io facl ocek syncdrhnous stote machaie snhwn beloo, wriwt ehtee cxitatoine qations, uotuutpe uqtaon, trainstion eiqautino asdnconstr cu attr asitnoino/utpu ttbae. l[1]

解：o0tpuu teuatioqn: MX=EN· Q A1· Q0the xeciattione uqtiaon: Ds=E0N’·0+ENQ·Q0’ EN⊕Q0=D =1E ’·QN1+N·E 0Q’ Q+ 1EN ·Q0Q ’1 rantsiitn oqeutiona:sQ0 =EN’·Q*+0N· E0’=QEN⊕Q0Q 1*= N’·E1+QE·NQ0 Q’1+ EN·Q0 Q 1 ’

rtnasiiot/noutptut ble:aQ10Q00 1 10 11

0EN0 00 0 /0/01 1/00 11/ 1 01/0 10/0011/ 0001

/1*QQ*/MA0X

2.Th etanritsinoeq utions anad otpuu etqiutationo f

clockead yncsrohnou stast meahcie is shnwn oas ollfosw Co.mlpet eteh itinm dgigarm fao Qr,Q10 and ,Y sausmngi hattt eh amchin seattsri nsta t eQ1Q0=,0 0nd flaipfl-os paerpo istieve-gedritgegedr [5]. tanritsio nqeatious:nQ0=*QA0+Q’1A’ 1Q*Q=1+’1’QQ0AQ+Q0’1Ao uptt euqutitiona: YQ1’Q0=

解：

如 1图建.立换-转输出表

Q Q01

A01

00Y0 110 111011 00 0

1111 110 0

00 10

Q0*Q0*12.根转据-输出换画波形表图

。七.Cnstruotca inmialm tast/eutout taplbeO R sataetdiaragmfo r aM EAY sLeqeuntial amhcie, nhtta wil detletcthe olfowlingi nput sqeunecse :x=0011 r 1o10.0 I xf=0110or x= 1001 s ietdcteed, the Z=1.Then wto ipun teqseunecsma oveylrpa noe anthoe.r[10 F]r eoamxle:p 001 1 0 0 0 100 1 1 01 1 00 0

0:XZ:

0 0 1 001 0 0 0 0 01 00 001 0 0

解状：/态输出：表含义 初状始 最态新到0 最新收到1 收最新到收00最 新到收11最 收到0新01 SNII T0 SS 1S0 01S S0110 X0

10S0,0S,0 S0,00

1S0,1,0S 11,S00S,001S100,S ,0 0S00, 1*,SZ

S0100,11,0 S11S,1S1, 0最新收到110110S

八.74X63 1isa s nyhrcoonsu

Te 2h--241 B-DCcode tab leD B3B 21B 0B

4bi- bitarny cuotnrewit hsycnhornosu lao dad synnhrcnouso lcea inputs, rthe bsai functconit albe si howsna sf lolw. oDesgin a modulo-10 ountcr,e suin gno 7e4X613 ndas mo enecsseay grtea,s nd aht ceuntiongseq unceeis -422-B1DC.C omletpe ht edesig nan drdawa lgic oiagrdam .[0]1

01 23 4

0 0 0 0000 00 1 0

0 11 00

011

05 6 78 91

11 11

101 1 1

01 011 1

0 011

unFtcointa be folr a 7X146 I3nutps LC_R LLD_LEN T 01 1X 0 1 XX 0 E N P X X CurrenXtStat e QD QC QB Q AX X XX X XX X X X X XNxtestat e D* QC* QBQ QA* 0* 0 00D C B AQ DC QQB AQOu tut pRO C 00

1101 11

11 1 1

1X11 1 1

0 11 1 1X X X0X0 00 0 0 0 0 10 0 1 001 1QD QCQ BQ A00 0 1 00 1 000 110 01

00 00 0

101

11

11

11

………….1 1 11 ………..…00 0 0 0

1

解

1:根据题、目要建求计数立态转状换，确定状态表换转所控制输入需及据输数。入 D 0 Q 000 C Q00 0 0 Q B0 0 11QA 01 0 1DQ*0 0 00 QC* 0 0 1 0QB*0 1 0 Q1A* 1 0 1 LD1_ L 1 111 D ddd dCd d d d B d ddd Add d d

0 00 0

11 1100 1 1

010 11

00 1

011 0

111 000 100

11 11

d d

0dd dd 1d d

1dd dd

111 11 1

0100 0 1 11

00 1 10 0 1 0

1 100 1 0

1 1 111 11

00 1 01 11

1 1 0 001 1

0 1 011 1 1 1

11 1 1 11ddd dd dd

d d dd d dd

ddd d d d

ddd d d d dd2、根

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