EEL4767ComputerSystemDesign1,Spring2005

Exam1PartI: ClosedBookINotes(75pts) NAME

I. Whatregisterisusedtokeeptrackoftheaddressofthenextinstructiontobeexecutedis:

a.b..c)d.

IndexRegistersStackPointerProgramcounterConditioncoderegister.

2. WhichtypeofROM canbeerasedbyelectricalsignalsandreprogrammed

a. MROMb. EPROMc. PROMd. EEPROM

3. WhichinstructionisbestperformedbyanALU:

a. Storeb. Decc. Jmpd. Load

4. In 68HCII, whatarethesizeofthefollowingregistersintheorder:B, D, SP,CCR:

a. 8bits,16bits,8bits,16bitsb. 8bits,16bits,16bits,16bitsc. All are16bits.d. 8bits,16bits,16bits,8bits.

5. In atypicalmicro-processor,whichof thefollowingstatementsistrue:

a. Databusisbi-directionalandtheAddressbusisunidirectional.b. DatabusandAddressbusarebothunidirectional.c. Databusisunidirectionalandtheaddressbusbi-directional.d. DatabusandAddressbusarebothbi-directional.

6. In assemblylanguageprogramming,differentprefixesareusedtodifferentiatebetweendifferentkindsofdata.Whichstatementgivestherightprefix?

cd

Decimal##$

Nothin

Hexadecimal$

7. Whatkindofaddressingis SUBA$2000?

a. Immediateb. Registerc. Extendedd. Inherent

8. Whatkindofaddressingis ADDA #$32?

a. Immediateb. Registerc. Extendedd. Inherent

9. WhatkindofaddressingisINX?

a. Immediateb. Registerc. Extendedd. Inherent

10.RelativeModeisusedonlywith??????instructions?

a. Arithmeticb. Branchingc. Loadingd. Incrementing

11.Howmanydifferentmemorylocationscanthe68HCli address(hint:thinkofthesizeofaddressbus):

a. 256b. 262,144c. 65536d. 16384

12.Whenusingbranchinstructions,thebranchoffset(BRAoffset)hasarangeof:

,'i a. -64to63bytes.b. Dependsonthesizeoftheassemblyprogram.c. Dependsonthetypeofbranchinstructionused.d. -128to 127bytes.

I 13.WhatarethecontentsoftheD registerandthememorylocationsafterexecutingLDD $00001Beforetheinstruction:[D]=$1010;[$0000]=$20;[$0001]=$30.

14.ThestatementM[lOOO]f- Rl +[R2]

a. Addwhatis inRl andR2 andstoretheresultin memorylocation1000.b. UseRl asanaddresstoobtainthedatafomandaddwhatis inR2 and

storetheresultinmemorylocation1000.c. UseR1 asanaddresstoobtainthedatafomanduseR2 anaddressasan

addresstogetthedatafomandthenaddthemtogetherandstoretheresultinmemorylocation1000

d. Addwhatis inRl tothedatastoredinmemorygivenbytheaddressinR2andstorestheresultinmemorylocation1000

15.WhatdoesABX do:

a. AddaccumulatorB andtheindexregisterX andstoretheresultinX.b. AddaccumulatorB andtheindexregisterX andstoretheresultinB.c. AddaccumulatorB andthecontentsofthememoryaddresspointedbythe

indexregisterX andstoretheresultinX.d. AddaccumulatorB andthecontentsofthememoryaddresspointedbythe

indexregisterX andstorestheresultinB.

16.WhatisthedifferencebetweenLDX #$1000andLDX $10001

a. Theyarebasicallythesame.b. Thefirstinstructionplacesthecontentsofmemorylocations$1000and

$1001intoX andthesecondplacesthehexvalue$1000intoX register.c. Thefirstinstructionplacesthehexvalue$1000intoX registerandthe

secondplacesthecontentsofmemorylocations$1000and$1001intoX.d. Oneoftheinstructionsisnotvalid.

[D] r$OOO11 [$0000]a 1010 10 10b) 0000 30 20c 2030 30 20d) 3020 30 20

17.Whichcond~tionalbranchinstructioninthefollowinglistwill notbranchif theconditionflagsareN=C=1andZ=V=O:

a. BCS targetb. B~ targetc. :a,CCtargetd. Bve target

18.Findthevaluesof theconditionflagsN, Z, V ande afterexecutingSUBA #$50.Given:[A] =#$70,N=O,Z=I,v=oandC=1.(IDNT: N is setifMSB oftheresultis set;Z is setif resultis zero;V is setif tWoscomplementoverflowresults;e issetif acarryisgenerated.):

19.ComputethenewvaluesofaccumulatorA andthee flagafterexecutingtheinstructionASLA. TheoriginalvalueinA is $74(01110100)andtheC flagis 1.(Hint:0movesintoLSB andMSB movesintoC) :

a. 11101000andC flagisO.b. 10111010andC flagisO.c. 11101000andC flagis 1.d. 10111010andC flagis 1.

"~"I ~~~.; \.~ . " Q 0 J:-_. (;, .--l -~ 1

20.If thememorylocationat$00contains$F6andthee flagis 1,whatarethenewvaluesofthismemorylocationandthee flagaftertheexecutionofLSR $OO?(Hint:0 intoMSB andLSB intoC):

a. 11110110andC flagis 1.b. 11101101andC flagisO.c. 01111011andC flagisO.d. 01111011andC flagis 1.

1f b -::. \ \ " 0 I \0

0' / I.') ,~Q--"3--y LO

~~IC]---)? ,~!j21.ComputethenewvaluesofaccumulatorB andthee flagafterexecutingthe

instructionROLB. TheoriginalvalueinB is $BEandtheC flagis 1.

(Hint:einto LSB andMSB intoC): f PA~ -;:: " )~ ~ \0' \ \., 0

a. 01111101andC flagisO.b. 11011111andC flagisO.c. 01111101andC flagis 1.d. 11011111andC flagis 1.

'N Z V Ca 1 1 0 1b) 0 0 1 1c 1 0 0 0d) 0 0 0 0

22.A programrequiresadelayof 1msec.ThefrequencyofE clockcycleis2MHz.CalculatethenumberofE clockcyclesrequiredforthisdelay:

f ~ +~ 7. 1'1~l.-::---

,' ~ 2Mtil,..~1 -qJC) ~- ", - ~

- '11':r:l

I 28.WhathappensafterexecutingSTAA #$1001A ~ ~DI()9 ZZ2

a. Theaccumulatorremainsunchanged.b. Theccmtentsoftheaccumulatorarestoredattheaddress$0100.c. TheZ flagis setif theaccumulatorhas$00.d. InvalidInstruction

29. Whatarethecontentsoftheaccumulatorafterexecutingtheseinstructions:

LDX#VECLDAA I~XVEC FCB I, 2,3,4,5

y..-: )A -;. 1.

IIII

a. Thecodereturnsan"error.b. [A]=1c. [A]=2d. [A]=3

30.If thereg.SP =$F7C3,thentheinstructionPULA retrievesreg.A fromthestackandsetstheSP to:

a. Remainsunchanged.b. $F7C2c. $F7C4d. Dependsonthedatainthestack.

EEL4767ComputerSystemDesignI, Spring2005

Exam1PartI: ClosedBookINotes NAME

Ques. A B C D1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11 X12 X13 X14 X15 X16 X17 X18 X19 X20 X21 X22 X23 X24 X25 X26 X27 X28 X29 X30 X

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EEL4767ComputerSystemDesignI, Spring2005

E~l ~Part2:ClosedBook/2PagesofNotes(25pts) NAME LlA.tc{ trJt,

UsingtheINCHAR andOUTA subroutineswriteaprogramthatinputstwoASCnnumbersbetween0 and9. Yourprogramshouldcheckandmakesurethatthekeyboardinputisavalidnumberbetween(0-9).Onceyourprogramhasdeterminedthatthekeyenteredisavalidnumber,it shouldechoit tothescreen.Next,yourprogramshouldcompute A(.L:f1 0-4 :::.~J-?:'"

Result=INumber1- Numher2I whereI I is equaltotheabsolutevalue.

Finally,yourprogramshouldoutputtheresulttothescreenasanASCII number.

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ORG $0100LDS" #$OOFFJSR INCHARCMPA #$30BLO LPICMPA #$39BHI LPIJSR OUTATABJSR INCHARCMPA #$30BLO LP2CMPA #$39BHI LP2JSR OUTACBABHS SUBTRPSHAPSHBPULAPULB

SUBTR SBAADDA #$30JSR OUTA

EQ JMP STINCHAR PSHBRCHRI LDAB $102E

ANDB #$20BEQ RCHRILDAA $102FPULBRTS

OUTA PSHBWCHRl LDAB $102E

ANDB #$80BEQ WCHRlSTAA $102FPULBRTS

CRLF LDAA #$ODJSR OUTALDAA #$OAJSR OUTARTSEND

STLPI

LP2

GO GET THE INPUT CHARGO SEE IF IT'S A #INVALID INPUTGO SEE IF IT'S A #INVALID INPUTGO ECHO THE CHARACTERPUT 1ST NUMBER IN AGO GET THE INPUT CHARGO SEE IF IT'S A #INVALID INPUTGO SEE IF IT'S A #INVAL ID INPUTGO ECHO THE CHARACTER

SEE WHICH IS POSITIVE

SWAP ATHE #'S INTHE A AND B ACC

TAKE THE DIFFERENCECONVERT BACK TO ASCII

THIS ROUTINE INPUTS ONE CHAR IN AIS THE RECEIVE BUFFER IS EMPTY

SOURCE INPUT CHARACTER IN THE A

THIS ROUTINE OUTPUTS ONE CHAR IN AIS THE TRANSMIT BUFFER IS EMPTY

GO WRITE THE CHAR

GO WRITE THE LF AND CR