Download Exam Solutions for ECE 2030 Computer Engineering Fall 2001 - Problem 1 to 4 and more Exams Computer Science in PDF only on Docsity!
4 problems, 4 pages Exam Two Solutions 19 October 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
101.101 4 + 1 + .5 + .125 = 5.
10111011 128 + 32 + 16 + 8 + 2 + 1 = 187
244 16 * 1K * 1K * 1K * 1K * = 16 Trillion
Part B (9 points) Convert the following hexadecimal values into octal notation:
hexadecimal notation octal notation
555 010101010101 = 2525
DEC 110111101100 = 6754
13.61 10011.01100001 = 23.
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 0 0 0 0 0 1 1 1 0 1 1 1 1 0 0 0
signed error?
no yes yes yes
unsigned error?
yes yes no yes
4 problems, 4 pages Exam Two Solutions 19 October 2001
Problem 2 (3 parts, 28 points) Art of the State
Part A (10 points) Implement a transparent latch using only two-input NAND gates. Label the inputs In and En , and the output Out. No other gate types should be used.
Out
En
In
Part B (8 points) Implement register with write enable using transparent latches, pass gates, and inverters. Use an icon for the transparent latches. Label the inputs In , WE, Φ 1 , Φ 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 Solutions 19 October 2001
Problem 4 (2 part, 22 points) Building Blocks
Part A (11 points) Consider the following circuit. For each combination of inputs listed below, describe the corresponding outputs.
4 to 2
priority
encoder
In 0
In 1
In 2
In^3 In 3 > In 2 > In 1 > In 0
Out 0
Out 1
V
A
B
C
D
2 to 4
decoder
In 0
In 1
En
Out 0
Out 1
Out 2
Out 3
W
X
Y
Z
A B C D W X Y Z
Part B (11 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. Do not consider error determination here.
A 3 A 2 A 1 A 0 B^3 B^2 B^1 B^0
S 3 S 2 S 1 S 0
Cin
X 3 X 2 X 1 X 0
Y^3 Y^2 Y^1 Y^0
Z 3 Z 2 Z 1 Z 0
ADD/SUB
