ECE 2030 11:00am Computer Engineering Fall 20035 problems, 6 pages Final Exam 12 December 2003 1 Instructions: This is a closed book, closed note exam. Calculators are not permitted. If you have a question, raise your hand and I will come to you. Please work the exam in pencil and do not separate the pages of the exam. For maximum credit, show your work. Good Luck! Your Name (please print) ________________________________________________ 1 2 3 4 5 total 44 28 26 30 22 150ECE 2030 11:00am Computer Engineering Fall 20035 problems, 6 pages Final Exam 12 December 2003 2 Problem 1 (6 parts, 44 points) Assembly Language FunPart A (7 points) Write an MIPS fragment that places 0x12345678 in $3. label instruction comment Part B (7 points) Write a MIPS fragment that exchanges $1 and $2 without modifying any other registers. label instruction comment Part C (7 points) Write a MIPS fragment that pops a value from the stack into $4. The SP is $30. label instruction comment Part D (7 points) Write a MIPS fragment that branches to SKIP if $5 is greater than -25. label instruction comment Part E (9 points) List three differences between a branch and a jump in the MIPS instruction set. 1: 2: 3: Part F (7 points) Explain the need for byte addressing in today’s microprocessors? Byte addressing is needed to support:ECE 2030 11:00am Computer Engineering Fall 20035 problems, 6 pages Final Exam 12 December 2003 3 instruction example meaning add add $1,$2,$3 $1 = $2 + $3subtract sub $1,$2,$3 $1 = $2 - $3add immediate addi $1,$2,100 $1 = $2 + 100multiply mul $1,$2,$3 $1 = $2 * $3divide div $1,$2,$3 $1 = $2 / $3and and $1,$2,$3 $1 = $2 & $3or or $1,$2,$3 $1 = $2 | $3xor xor $1,$2,$3 $1 = $2 xor $3and immediate andi $1,$2,100 $1 = $2 & 100or immediate ori $1,$2,100 $1 = $2 xor 1xor immediate xori $1,$2,100 $1 = $2 | 100shift left logical sll $1,$2,5 $1 = $2 << 5 (logical)shift right logical srl $1,$2,5 $1 = $2 >> 5 (logical)shift left arithmetic sla $1,$2,5 $1 = $2 << 5 (arithmetic) shift right arithmetic sra $1,$2,5 $1 = $2 >> 5 (arithmetic) load word lw $1, ($2)$1 = memory [$2]store word sw $1, ($2)memory [$2] = $1load upper immediate lui $1,100 $1 = 100 x 216 branch if equal beq $1,$2,100 if ($1 = $2), PC = PC + 4 + (100*4) branch if not equal bne $1,$2,100if ($1 ≠$2), PC = PC + 4 + (100*4) set if less than slt $1, $2, $3 if ($2 < $3), $1 = 1 else $1 = 0 set if less than immediate slti $1, $2, 100 if ($2 < 100), $1 = 1 else $1 = 0 jump j 10000 PC = 10000*4jump register jr $31 PC = $31jump and link jal 10000 $31 = PC + 4; PC = 10000*4 Problem 2 (4 parts, 28 points) Assembly LanguageConsider the following MIPS program fragment. The instruction set is listed below. address label instruction 1000 slt $3, $1, $0 1004 bne $3, $0, skip1 1008 jal foo 1012 addi $2, $0, 0x100 1014 sw $3, ($2) 1020 skip1: j Next Part A (7 points) What is the branch offset (in bytes) for the bne instruction? branch offset (in bytes): Part B (7 points) For what values of $1 will subroutine foo be called? values of $1 (in decimal): Part C (7 points) What does $31 contain when subroutine foo begins execution? contents of $31 (in decimal): Part D (7 points) If Foo produces a single result, which register is most likely to contain it when Foo returns? Explain your answer?ECE 2030 11:00am Computer Engineering Fall 20035 problems, 6 pages Final Exam 12 December 2003 4 Problem 3 (2 parts, 26 points) Karnaugh MapsPart A (13 points) For the follow expression, derive a simplified product of sums expression using a Karnaugh Map. Circle and list the prime implicants, indicating which are essential. CBADCBADCAOut ⋅⋅+⋅⋅⋅+⋅⋅= AABBCCCDD Dprime implicantsessential?yes no simplified POS expression Part B (13 points) For the follow expression, derive a simplified sum of products expression using a Karnaugh Map. Circle and list the prime implicants, indicating which are essential. )()()( DCBADCBDBAOut +++⋅++⋅++= AABBCCCDD Dprime implicantsessential?yes no simplified SOP expressionECE 2030 11:00am Computer Engineering Fall 20035 problems, 6 pages Final Exam 12 December 2003 5 Problem 4 (3 parts, 30 points) CountersPart A (10 points) Design a toggle cell using transparent latches and basic gates. Include a toggle enable TE, active low clear CLR, clocks PHI1 and PHI2. Label the output OUT. Part B (10 points) Use toggle cells to build a divide by six counter with active high inputs CE and CLR, and a max count output MAX. Do not draw the clock signals. Label all signals. CEOutClrToggleCEOutClrToggleCEOutClrToggle Part C (10 points) Use divide by six and ten counters plus needed gates to build a stop watch. Include an external count enable CE and clear CLR. Assume a one cycle/second (1 Hz) clock. CE CLRdivide by 6maxCE CLRdivide by 10maxCE CLRdivide by 6maxCE CLRdivide by 10maxCECLRECE 2030 11:00am Computer Engineering Fall 20035 problems, 6 pages Final Exam 12 December 2003 6 Problem 5 (3 parts, 22 points) Building BlocksPart A (8 points) Implement a 2 to 4 decoder using only AND gates. Assume inputs signals IN0, IN1, and En and their complements are available. Label all inputs and outputs. Part B (10 points) Complete the truth table below to describe the behavior of the following circuit. 2 to 4decoderIn0In1EnOut0Out1Out2Out3ABCDOutY1Z D C B A Z Y Out 1 1 1 0 0 0 0 0 1 1 0 1 0 1 0 0 0 1 1 0 0 0 1 0 1 1 0 0 1 0 0 1 0 1 1 1 Part C (4 points) Name the building block implemented in part B. It’s a
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