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ECE 2030 Exam Solution - Spring 2001: Binary to Hex, Signed Addition, Latches, Counters, Exams of Computer Science

The solutions to exam two of the computer engineering class ece 2030 held in spring 2001. The exam covers various topics including converting binary and octal values to decimal and hexadecimal notation, addition of signed numbers using two's complement and unsigned representations, implementing transparent latches, and designing toggle cells for counters.

Typology: Exams

2012/2013

Uploaded on 04/08/2013

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ECE 2030 2:00pm Computer Engineering Spring 2001
4 problems, 4 pages Exam Two Solution 21 March 2001
1
Problem 1 (3 parts, 30 points) Numbers and Arithmetic
Part A (9 points) Convert some binary values (and powers of two) into decimal notation:
binary notation decimal notation
110.011 4 + 2 + .25 + .125 = 6.375
11101100 128 + 64 + 32 + 8 + 4
225 32 Million
Part B (9 points) Convert the following octal values into hexadecimal notation:
octal notation hexadecimal notation
77 111111 = 3F
3546 11101100110 = 766
72.1 111010.001 = 3A.2
Part C (12 points) For each problem below, (a) compute the operations using the rules of
addition, (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.
1 0 1
+ 0 1 1
1 1 1 0
+ 1 0 0 1
1 1 1 0
+ 0 1 1 1
1 0 0 0
+ 1 0 1
addition
result 1 0 0 0 0 1 1 1 0 1 0 1 1 1 0 1
signed
error? yes yes no no
unsigned
error? no yes yes no
pf3
pf4

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Download ECE 2030 Exam Solution - Spring 2001: Binary to Hex, Signed Addition, Latches, Counters and more Exams Computer Science in PDF only on Docsity!

4 problems, 4 pages Exam Two Solution 21 March 2001

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

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

binary notation decimal notation

110.011 4 + 2 + .25 + .125 = 6.

11101100 128 + 64 + 32 + 8 + 4

225 32 Million

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

octal notation hexadecimal notation

77 111111 = 3F

3546 11101100110 = 766

72.1 111010.001 = 3A.

Part C (12 points) For each problem below, (a) compute the operations using the rules of addition, (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.

addition result

signed error?

yes yes no no

unsigned error?

no yes yes no

4 problems, 4 pages Exam Two Solution 21 March 2001

Problem 2 (3 parts, 28 points) Art of the State

Part A (10 points) Implement a transparent latch using only six two-input NOR gates. Label the inputs In and En , and the output Out. No other gates should be used.

Out

En

In

Part B (8 points) Implement register with write enable using transparent latches, NAND gates, and inverters. Use an icon for the transparent latches. Label the inputs In , WE, ΦΦΦΦ 1 , ΦΦΦΦ 2 and the output Out.

In Out

En

Latch

In Out

En

Latch

WE

In

Out

Part C (10 points) Assume the following signals are applied to your register. Draw the output signal Out. Draw a vertical line where In is sampled. Assume Out starts at zero.

WE

In

Out

4 problems, 4 pages Exam Two Solution 21 March 2001

Part C (10 points) Now use copies of your toggle cell (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. Your design should clear if (A) the external clear CLR is high, or (B) the maximum output count is reached and the count enable is high. You do not need to draw in the clock signals.

O 0

O 1

O 2

Ext Clr

Ext CE TE Out Clr

TE Out Clr

TE Out Clr

Problem 4 (4 parts, 12 points) Multiple Choice

Consider four different 32-bit representations: (A) an unsigned integer representation, (B) a twos complement integer representation, (C) a twos complement fixed point representation with sixteen bits on each side of the fixed point, and (D) a floating-point representation with a 23 bit mantissa and an 8 bit exponent. For each quantity below, circle which representation (A, B, C, or D) most accurately represents the quantity (i.e., which representation can represent the value with the smallest error).

π (pi)

A B C D

(dollars Bill Gates lost in stock market last week) A B C D

(mass of Pluto the planet) A B C D

(mass of Pluto the Disney dog) A B C D