Exam Solutions for ECE 2030 D Computer Engineering - Spring 2012 - Problem 1 to 4, Exams of Computer Science

The solutions to exam three for the computer engineering course ece 2030 d in spring 2012. It includes the solutions to four problems related to memory systems, datapath elements, microcode, and counters. The problems cover topics such as memory organization, barrel shifter, microcode writing, and toggle cell design.

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Uploaded on 04/08/2013

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ECE 2030 D Computer Engineering Spring 2012
4 problems, 5 pages Exam Three Solutions 12 April 2012
Problem 1 (3 parts, 30 points) Memory Systems
Part A (12 points) Consider a 1 Gbit DRAM chip organized as 32 million addresses of 32-bit words.
Assume both the DRAM cell and the DRAM chip are 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 (not powers of two).
total number of bits in address log(32M) = log(225) = 25
number of columns
230=215=32K
column decoder required (n to m)15 to 32K
number of words per column 215/25 = 210 = 1K
type of mux required (n to m)1K to 1
number of address lines in column offset 10
Part B (10 points) Consider an 8 Gbyte memory system with 1 billion addresses of 64-bit words using
a 32 million address by 32-bit word memory DRAM chip.
word address lines for memory system log(1B) = log(230) = 30
chips needed in one bank 2
banks for memory system 230/225 = 25 = 32
memory decoder required (n to m)5 to 32
DRAM chips required 64
Part C (8 points) Design a 8 million address by 4 bit memory system with 4M x 4 memory chips. Label
all busses and indicate bit width. Assume R/W is connected and not shown here. Use a bank decoder if
necessary. Be sure to include the address bus, data bus, and MSEL.
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4 problems, 5 pages Exam Three Solutions 12 April 2012

Problem 1 (3 parts, 30 points) Memory Systems

Part A Assume both the DRAM cell and the DRAM chip are square. The column number and offset concatenate (12 points) Consider a 1 Gbit DRAM chip organized as 32 million addresses of 32 -bit words.

to form the memory address. Using the organization approach discussed in class, answer the following questions about the chip. Express all answers in decimal (not powers of two).

total number of bits in address log(32M) = log(2^25 ) = 25

number of columns √ 230 = 215 =32K

column decoder required ( n to m ) 15 to 32K

number of words per column 215 /2^5 = 2^10 = 1K

type of mux required ( n to m ) 1K to 1

number of address lines in column offset 10

Part B a 32 million (10 points) Consider an address by 32-bit word 8 Gbyte memory DRAM chip. memory system with 1 billion addresses of 64 -bit words using

word address lines for memory system log(1B) = log(2^30 ) = 30

chips needed in one bank^2

banks for memory system^230 /2^25 = 2^5 = 32

memory decoder required ( n to m ) 5 to 32

DRAM chips required^64

Part C all busses and indicate bit width (8 points) Design a 8 million address by 4 bit. Assume R/W is connected and not shown here. Use a bank decoder if memory system with 4M x 4 memory chips. Label

necessary. Be sure to include the address bus, data bus, and MSEL.

4 problems, 5 pages Exam Three Solutions 12 April 2012

Problem 2 (3 parts, 26 points) Datapath Elements Part A (6 points) Suppose the following inputs (in hexadecimal) are applied to the 32-bit barrel shifter used in the datapath. Determine the output (in hexadecimal). Assume the shift amount is drawn from the 16-bit immediate value. Shift Type Shift Amount Input Value Output Value arithmetic 0x0008^ ABCD12EF^ 0xFFABCD rotate 0xFFF4^ DEAF7892^ 0xF7892DEA arithmetic 0xFFF0^ FACE2537^ 0x Part B (in hexadecimal) to correctly program the logical unit. (8 points) For each bitwise logical function specification below, determine the LF code

X Y Out logical function LF 0 0 LF 0 Y + X E 1 0 LF 1 X^ A 0 1 LF 2 X ⊕ Y 6 1 1 LF 3 XY 8 Part C (12 points) Given the following state table, draw the corresponding state diagram below. S 1 S 0 H/M NS 1 NS 0 Out S 1 S 0 H/M NS 1 NS 0 Out 0 0 0^0 1 0 1 0 0^0 0 0 0 1^0 0 0 1 0 1^0 0 0 1 0 1 0 0 1 1 0 x x x 0 1 1^0 0 0 1 1 1 x^ x^ x

S 1 S 0

S 1 S 0

S 1 S 0

H

H

H

M M

M/Out

Give the simplified Boolean expression for computing Out in terms of the current state and the input.

Out = S 0 M.

4 problems, 5 pages Exam Three Solutions 12 April 2012 Problem 4 (2 parts, 18 points) Counters Part A (8 points) Design a toggle cell using AND, OR, NAND, NOR, NOT). Use an icon for the transparent latches. Your toggle cell should only transparent latches and basic gates (XOR, have an active high toggle enable input TE , and an active low clear input CLR , clock inputs Φ 1 and Φ 2 , and an output Out. The CLR^ signal has precedence over TE. Label all signals. Also complete the behavior table for the toggle cell.

In Out En Latch In Out En

TECLR (^) Latch Out

Φ (^1) Φ 2

TE CLR CLK Out 0 0 ↑↓ (^) 0 1 0 ↑↓ (^) 0 0 1 ↑↓ (^) Q 0 1 1 ↑↓ (^) Q 0 Part B (10 points) Now combine these toggle cells to build a divide by thirteen counter. Your counter should have an external clear, external count enable, and four count outputs O O 3 , O 2 , O 1 , inputs to the toggle cells are already connected. Your design must support multi-digit systems.^0. Use any basic gates (AND, OR, NAND, NOR, XOR & NOT) you require. Assume clock