Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

Exam Two in Computer Engineering, Spring 2002, Exams of Computer Science

A computer engineering exam from spring 2002. It consists of 4 problems, each with multiple parts, totaling 100 points. The topics covered include numbers and arithmetic, storage, counters, and arithmetic units. The exam is a closed-book, closed-notes exam and is hand-written in pencil. It includes instructions for converting decimal and octal values to binary and hexadecimal, performing arithmetic operations, completing truth tables, designing latches and counters, and adding and subtracting four-bit numbers.

Typology: Exams

2012/2013

Uploaded on 04/08/2013

seetharaman_sehgal
seetharaman_sehgal 🇮🇳

4.4

(11)

127 documents

1 / 5

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
ECE 2030 1:00pm Computer Engineering Spring 2002
4 problems, 5 pages Exam Two 20 March 2002
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) ________________________________________________
1234 total
30 30 20 20 100
pf3
pf4
pf5

Partial preview of the text

Download Exam Two in Computer Engineering, Spring 2002 and more Exams Computer Science in PDF only on Docsity!

4 problems, 5 pages Exam Two 20 March 2002

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 total

4 problems, 5 pages Exam Two 20 March 2002

Problem 1 (3 parts, 30 points) Numbers and Arithmetic

Part A (9 points) Convert some decimal values (and powers of two) into binary notation:

decimal notation binary notation

942

512 billion 2 _________

Part B (9 points) Convert the following octal values into hexadecimal notation:

octal notation hexadecimal notation

555

1234

Part C (12 points) For each problem below, (a) compute the operations using the rules of arithmetic, (b) indicate whether an error occurs assuming all numbers are expressed using a four bit two’s complement representation, and (c) indicate whether an error occurs assuming all numbers are expressed using a four bit unsigned representation.

result

signed error? unsigned error?

4 problems, 5 pages Exam Two 20 March 2002

Problem 3 (2 parts, 20 points) Counters

Part A (10 points) Design a toggle cell using transparent latches and basic gates (AND, OR, NAND, NOR, XOR, XNOR, and NOT). Include an active high toggle enable TE, active low clear CLR, and a two-phase non-overlapping clock PHI1 and PHI2. Label the output OUT.

Part B (10 points) Now use copies your toggle cells (in icon form) to build a divide by five counter. This design should include an active high external count enable CE and an active high external clear CLR. You do not need to draw in the clock signals. Assume all toggle cells are connected to the two-phase clock. Label all of your outputs signals.

4 problems, 5 pages Exam Two 20 March 2002

Problem 4 (2 part, 20 points) Arithmetic Unit

Part A (10 points) The adder below adds two four bit numbers A and B and produces a four bit result S. Add extra digital logic to support subtraction as well as addition. Label inputs X 3 , X 2 ,

X 1 , X 0 , Y 3 , Y 2 , Y 1 , Y 0 , ADD / SUB and outputs Z 3 , Z 2 , Z 1 , Z 0.

Part B (10 points) Now define the behavior for a two’s compliment overflow detection unit.

Assume the inputs are X 3 , Y 3 , Z 3 , and ADD / SUB. The output Error is 1 for overflow.

ADD / SUB = 0 ADD / SUB = 1

X 3 Y 3 Z 3 Error X 3 Y 3 Z 3 Error

0 0 0 0 0 0

1 0 0 1 0 0

0 1 0 0 1 0

1 1 0 1 1 0

0 0 1 0 0 1

1 0 1 1 0 1

0 1 1 0 1 1

1 1 1 1 1 1