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Main points of this exam paper are: Block Input, Block Drawn, Memory System, Computer Engineering, Signals, ExpressAnswers in Decimal, Number of Columns, Column Decoder Required, Mux Required, Chips Needed
Typology: Exams
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5 problems, 7 pages Final Exam 6 May 2011
Instructions: This is a closed book, closed note exam. Calculators and other electronics 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
5 problems, 7 pages Final Exam 6 May 2011
Problem 1 (3 parts, 21 points) “A chip off the old block”
Part A ( 15 points) Consider the five definitions for the block drawn below. One block input is the
logical value A. The other input is the control value C. The output behavior for each of the five
definitions is given in the table. Complete the full truth table and state the logical (gate) names
for each definition. (hint: the first block one appears to mask A when its control input is low.)
A 0 0 A A A Zo
A 1 A A 0 1 A
Part B ( 6 points) The circuit below is built using these blocks. Describe its behavior. Also give
the circuits common name.
X
Y
IN Out
C O
I N C
I N O C
O
I N C
O (^) IN
C O
X Y Out
0 0
1 0
0 1
1 1
It's a
5 problems, 7 pages Final Exam 6 May 2011
Problem 3 (3 parts, 34 points) Memory and Maps
Part A (1 2 points) Consider a gigabit DRAM chip organized as 64 million addresses of 16 bit
words. Assume both the DRAM cell and the DRAM chip is square. The column number and
offset concatenate to form the memory address. Using the organization approach discussed in
class, answer the following questions about the chip. Express all answers in decimal.
number of columns
column decoder required ( n to m )
type of mux required ( n to m )
number of muxes required
number of address lines in column number
number of address lines in column offset
Part B (10 points) Consider a memory system with 512 million addresses of four byte words
using DRAM chips organized as 64 million addresses by 16 bit words
word address lines for memory system
chips needed in one bank
banks for memory system
memory decoder required ( n to m )
DRAM chips required
Part C (12 points) For the follow expression, derive a simplified sum of products expression
using a Karnaugh Map. Circle and list all prime implicants, indicating which are essential.
Out = A ⋅ B ⋅ D A ⋅ C ⋅ D A ⋅ B ⋅ C A ⋅ C ⋅ D
prime implicants
essential? yes no
simplified SOP expression
5 problems, 7 pages Final Exam 6 May 2011
Problem 4 (4 parts, 38 points) Number Systems
Part A ( 8 points) Convert the following notations:
binary notation decimal notation
1010 1011. 1010
41.8 75
binary notation hexadecimal notation
11 1100 0011.1100 0011 11
3F.
Part B (9 points) For the representations below, determine the most positive value and the step
size (difference between sequential values). All answers should be expressed in decimal notation.
Fractions (e.g., 3/16ths) may be used. Signed representations are two’s complement.
representation most positive value step size
signed integer ( 20 bits). (0 bits) unsigned fixed-point ( 15 bits). ( 5 bits) signed fixed-point ( 10 bits). ( 10 bits)
Part C ( 9 points) A 20 bit floating point representation has a 13 bit mantissa field, a 6 bit
exponent field, and one sign bit. Express all answers in decimal.
What is the largest value that can be represented (closest to infinity)?
What is the smallest value that can be represented (closest to zero)?
How many decimal significant figures are supported?
Part D (12 points) For each problem below, compute the operations using the rules of arithmetic,
and indicate whether an overflow occurs assuming all numbers are expressed using a five bit
unsigned fixed-point and five bit two’s complement fixed-point representations.
result
unsigned error? □ no □ yes □ no □ yes □ no □ yes □ no □ yes
signed error? □ no □ yes □ no □ yes □ no □ yes □ no □ yes
5 problems, 7 pages Final Exam 6 May 2011
MIPS Instruction Set
instruction example meaning arithmetic add add $1,$2,$3 $1 = $2 + $ subtract sub $1,$2,$3 $1 = $2 - $ add immediate addi $1,$2,100 $1 = $2 + 100 add unsigned addu $1,$2,$3 $1 = $2 + $ subtract unsigned subu $1,$2,$3 $1 = $2 - $ add immediate unsigned addiu $1,$2,100 $1 = $2 + 100 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 set if less than unsigned sltu $1, $2, $3 if ($2 < $3), $1 = 1 else $1 = 0 set if < immediate unsigned sltui $1, $2, 100 if ($2 < 100), $1 = 1 else $1 = 0 multiply mult $2,$3 Hi, Lo = $2 * $3, 64-bit signed product multiply unsigned multu $2,$3 Hi, Lo = $2 * $3, 64-bit unsigned product divide div $2,$3 Lo = $2 / $3, Hi = $2 mod $ divide unsigned divu $2,$3 Lo = $2 / $3, Hi = $2 mod $3, unsigned transfer move from Hi mfhi $1 $1 = Hi move from Lo mflo $1 $1 = Lo load upper immediate lui $1,100 $1 = 100 x 216 logic and and $1,$2,$3 $1 = $2 & $ or or $1,$2,$3 $1 = $2 | $ and immediate andi $1,$2,100 $1 = $2 & 100 or immediate ori $1,$2,100 $1 = $2 | 100 nor nor $1,$2,$3 $1 = not($2 | $3) xor xor $1, $2, $3 (^) $1 = $2 ⊕ $ xor immediate xori $1, $2, 255 (^) $1 = $2 ⊕ 255 shift shift left logical sll $1,$2,5 $1 = $2 << 5 (logical) shift left logical variable sllv $1,$2,$3 $1 = $2 << $3 (logical), variable shift amt shift right logical srl $1,$2,5 $1 = $2 >> 5 (logical) shift right logical variable srlv $1,$2,$3 $1 = $2 >> $3 (logical), variable shift amt shift right arithmetic sra $1,$2,5 $1 = $2 >> 5 (arithmetic) shift right arithmetic variable srav $1,$2,$3 $1 = $2 >> $3 (arithmetic), variable shift amt memory load word lw $1, 1000($2) $1 = memory [$2+1000] store word sw $1, 1000($2) memory [$2+1000] = $ load byte lb $1, 1002($2) $1 = memory[$2+1002] in least sig. byte load byte unsigned lbu $1, 1002($2) $1 = memory[$2+1002] in least sig. byte store byte sb $1, 1002($2) memory[$2+1002] = $1 (byte modified only) branch branch if equal beq $1,$2,100 if ($1 = $2), PC = PC + 4 + (1004) branch if not equal bne $1,$2,100 (^) if ($1 ≠ $2), PC = PC + 4 + (1004) jump jump j 10000 PC = 10000* jump register jr $31 PC = $ jump and link jal 10000 $31 = PC + 4; PC = 10000*
