Download Boolean Function - Computer Engineering - Solved Exam and more Exams Computer Science in PDF only on Docsity!
STUDENT NAME and NUMBER:
(PLEASE PRINT CLEARLY)
REMARKS: 1. All questions should be answered (in the space provided).
- Books and notes may NOT be used, with the exception of one 8.5”x11” sheet of paper.
DATE: June 11, 2001 TIME: 12:00 p.m. - 1:10 p.m.
Marks:
Total:
Q
Q
Q
Q
Georgia Institute of Technology
Department of Electrical and Computer Engineering
Test
ECE2030: Introduction to Computer Engineering
Q
1. COMPUTER SYSTEMS
(MAX MARK: 15)
a) The diagram below shows output voltage ranges for typical TTL (transistor-to-transistor logic). Describe
the purpose for the three voltage ranges for computer design and give some justification for their size.
b) Convert the following radix-2 number (binary number) to hexadecimal and octal.
2
->
c) Expand the following hexadecimal number to decimal?
F5A.
16
->
V
Hmax
V
Hmin
V
Lmin
V
Lmax
5.5 V
4.0 V
1.0 V
-0.5 V
2
3
1
Region 1 is for the logic level 1 and region 3 for logic level 0.
Region 2 is a zone that separates the other two regions so that
noise or voltage variations will not make a logic level 1 a 0 or a
logic level 0 a 1. The size of the voltage ranges allow the
introduction of small signal noise or voltage variations without
affecting the measured logic level.
hexadecimal
octal
2DB.
16
8
/
/
/
/
/
/
2. STANDARD FORMS (MAX MARK: 10)
Use Boolean algebra to expand the following Boolean function into a product-of-maxterms standard form.
Once expanded, express the Boolean function in -notation.
3. GATE DESIGN (MAX MARK: 25)
a) Extract the Boolean expressions for the outputs in the following gate diagram.
b) For the previous question, mark each gate with the number of transistors required to implement this gate
diagram using CMOS technology (i.e. nMOS and pMOS transistors).
What is the total number of transistors required?
M
F a b c( , , ) a b
( ) c
a b
( )c ac bc
ac b
( ) ac c
= ( a +b) ( b +c) ( )c = ( a + b +cc) ( aa + b +c) ( aa + bb +c)
= ( a + b +c) ( a + b +c) ( a + b +c) ( a + b +c) ( aa + bb +c)
= ( a + b +c) ( a + b +c) ( a + b +c) ( a + bb +c) ( a + bb +c)
= ( a + b +c) ( a + b +c) ( a + b +c) ( a + b +c) ( a + b +c)
M 0 1 3 5 7( , , , , )
A
B
F
C
D
E
G = F + ( BC) ⊕( D ⊕E) = A + ( BC)+( BC) ⊕( D ⊕E)
G
F = A +( BC)
8
8
2
2
0/2/
4
4
6
22 + 24 + 6 + 2*8 = 34 transistors without buffer
4. SWITCH NETWORKS (MAX MARK: 20)
For the expression below, create a switch level implementation using CMOS technology with nMOS and
pMOS transistors. Assume the inputs and their complements are available. Your design should contain no shorts
or floats. Use as few transistors as possible, but do not simplify the expression.
out = ( AB +AB) ( C +D E( +F)) +GH
D
F E
C
out
A
B
G H
A
B
G
H
A B
A B
D
E
F
C
5. KARNAUGH MAPS AND SIMPLIFICATION (MAX MARK: 30)
Shown below are two Karnaugh maps. The ‘X’s indicate “don’t cares” and should be used to form a mini-
mum number of prime implicants. For the following questions, circle and list ALL of the essential and non-
essential prime implicants. Also, write out ALL simplified Boolean functions of the form requested if more than
one simplified Boolean function exists for the Karnaugh map.
a) Given the following Karnaugh map, derive the simplified sum-of-products (SOP) expression.
b) Given the following Karnaugh map, derive the simplified product-of-sums (POS) expression.
prime implicants
essential?
yes no
AB
CD
00
01
11
10
00 01
11
10
C
A
B
D D D
B
B
C
A
F A B C D( , , , ) = ABC + AD + BC +AC
(SOP)
X=1 1
0 X=
ABD
AD
BC
AC
CD
F A B C D( , , , ) = ABC + AD + BC +CD
or
prime implicants
essential?
yes no
AB
CD
00
01
11
10
00 01
11
10
C
A
B
D D D
B
B
C
A
F A B C D( , , , ) = ( B +C) ( A + B +D) ( B + C +D)
(POS)
X=1 0
X=0 0
X=1 X=1 0
X=
A + B +D
B +C
B C D
