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CS121 : Digital Design
Basic Information
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Title: Digital Design
Code: CS121
Lecture: 3
Tutorial: 1
Pre-Requisite: Computer Introduction
Overall Aims of Course
By the end of the course the students will be able to:
 Grasp basic principles of combinational and sequential
logic design.
 Determine the behavior of a digital logic circuit
(analysis) and translate description of logical problems
to efficient digital logic circuits (synthesis).
 Understanding of how to design a general-purpose
computer, starting with simple logic gates.
Contents
Topics
Contact
Hours
No. of
Weeks
-Introduction to the course content, text book(s),
reference(s) and course plane.
- Digital Systems and Binary numbers
9
3
- Boolean Algebra and Logic Gates
6
2
- Gate Level Minimization
6
2
- Combinational Logic
12
4
- Synchronous Sequential Logic
12
4
Total
45
15
Assessment schedule
Assessment Methods
Week
Weighting of Assessments
First Midterm Exam
6
10%
Second Midterm Exam
12
15%
5 & 10
5%
11
5%
2,4,6,8,10,12
5%
15
20%
After week15
40%
Quiz
Lab-project
Lab- Evaluation
Final Lab
Final Exam
Total
100%
List of References
Essential Books
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“DIGITAL DESIGN”, by Mano M. Morris, 4th edition, Prentice- Hall.
Recommended Books
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“FUNDAMENTALS OF LOGIC DESIGN”, by Charles H. Roth,
Brooks/Cole Thomson Learning.
“INTRODUCTION TO DIGITAL SYSTEMS”, by M.D. ERCEGOVAC, T.
Lang, and J.H. Moreno, Wiley and Sons. 1998.
“DIGITAL DESIGN, PRINCIPLES AND PRACTICES”, by John F.Wakely,
Latest Edition, Prentice Hall, Eaglewood Cliffs, NJ.
“FUNDMENTALS OF DIGITAL LOGIC WITH VHDL DESIGN”, by
Stephen Brown and Zvonko Vranesic, McGraw Hill.
“INTRODUCTION TO DIGITAL LOGIC DESIGN”, by John Hayes,
Addison Wesley, Reading, MA.
1. Digital Systems and Binary Numbers
1.1 Digital Systems
1.2 Binary Numbers
1.3 Number-Base Conversions
1.4 Octal and Hexadecimal Numbers
1.5 Complements
1.6 Signed Binary Numbers
1.7 Binary Codes
1.1 Digital Systems
1.2 Binary Numbers
In general, a number expressed in a base-r system has
coefficients multiplied by powers of r:
r is called base or radix.
1.3 Number-Base Conversions (Integer Part)
Example:
1.3 Number-Base Conversions (Fraction Part)
Example:
Binary-to-Decimal Conversion
Example:
Example:
1.4 Octal and Hexadecimal Numbers
Decimal-to-Octal Conversion
Example:
Decimal-to-Hexadecimal Conversion
Example:
Octal-to-Decimal Conversion
Example:
Example:
Hexadecimal-to-Decimal Conversion
Example:
Example:
Binary–Octal and Octal–Binary Conversions
Example:
Example:
Hex–Binary and Binary–Hex Conversions
Example:
Example:
Hex–Octal and Octal–Hex Conversions
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For Hexadecimal–Octal conversion, the given hex number is firstly
converted into its binary equivalent which is further converted into
its octal equivalent.
An alternative approach is firstly to convert the given hexadecimal
number into its decimal equivalent and then convert the decimal
number into an equivalent octal number. The former method is
definitely more convenient and straightforward.
For Octal–Hexadecimal conversion, the octal number may first be
converted into an equivalent binary number and then the binary
number transformed into its hex equivalent.
The other option is firstly to convert the given octal number into its
decimal equivalent and then convert the decimal number into its hex
equivalent. The former approach is definitely the preferred one.
Example
Arithmetic Operation
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Addition
augend
Added:
Sum:
101101
+ 100111
---------1010100
Subtraction
minuend:
subtrahend:
difference:
101101
- 100111
------------000110
Multiplication
1.5 Complements
Diminished Radix Complement ((r-1)‘s complement)
Given a number N in base r having n digits, the (r - 1)’s
Complement of N is defined as (rn- 1) -N.
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the 9’s complement of 546700 is 999999 – 46700=453299
the 1’s complement of 1011000 is 0100111
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Note:
The (r-1)’s complement of octal or hexadecimal numbers is
obtained by subtracting each digit from 7 or F (decimal 15),
respectively
Radix Complement
Given a number N in base r having n digit, the r’s
complement of Nis defined as (rn -N) for N ≠0 and as 0 for
N =0 .
The 10’s complement of 012398 is 987602
The 10’s complement of 246700 is 753300
The 2’s complement of 1011000 is 0101000
Subtraction with Complement
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The subtraction of two n-digit unsigned numbers M – N
in base r can be done as follows:
M + (rn - N), note that (rn - N) is r’s complement of N.
If M  N, the sum will produce an end carry x,
which can be discarded; what is left is the result M- N.
If M < N, the sum does not produce an end carry and
is (N - M). Take the r’x complement of the sum and
place a negative sign in front.
Example:
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Using 10’s complement subtract 72532 – 3250
M = 72532
10’s complement of N = 96750
sum = 169282
Discarded end carry 105 = -100000
answer: 69282
Example:
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Using 10’s complement subtract 3250 - 72532
M = 03250
10’s complement of N = 27468
sum = 30718
Discarded end carry 105 = -100000
answer: -(100000 - 30718) = -69282
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The answer is –(10’s complement of 30718) = -69282
Example
Using 2’s complement subtract (a) 1010100 – 1000011
M
= 1010100
N = 1000011,
2’s complement of N = 0111101
1010100

0111101
sum = 10010001
Discarded end carry 27=-10000000
answer: 0010001
Example
(b) 1000011 – 1010100
M
= 1000011
N = 1010100, 2’s complement of N = 0101100
1000011
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0101100
sum = 1101111
answer: - (10000000 - 1101111) = -0010001
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The answer is –(2’s complement of 1101111) = - 0010001
Example
Using 1’s complement, subtract 1010100 - 1000011
M = 1010100
N = 1000011, 1’s complement of
N = 0111100
1010100
 0111100
10010000
end-around carry = +
1
answer: 0010001
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Example
Using 1’s complement, subtract 1000011 - 1010100
M = 1000011
N = 1010100, 1’s complement of
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N = 0101011
1000011
0101011
1101110
Answer: -0010001
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1.6 Signed Binary
Numbers
1.6 Signed Binary Numbers
1 - Sign and Magnitude representation
2 - 1’s Complement Representation
3 - 2’s Complement Representation
Notes
1 - The previous representation are the same for positive numbers and
different for negative numbers
2 - For a signed binary number the most significant bit is used for
representing the sign of the number
We use 0 for positive numbers and 1 for negative numbers
Example : represent +76
 7610   010011002  Sign & Magnitude
 7610   010011002  1' s Complement
 7610   010011002  2' s Complement
Representing negative numbers in the previous three systems
1’s Complement of a negative number can be obtained by
flipping all bits of the positive binary number
2’s Complement of a negative number can be obtained by
adding 1 to the 1’s Complement or by flipping bits of the
positive binary number after the first one from the right
Example : represent -76
 76 10  11001100 2  Sign & Magnitude
 76 10  101100112  1' s Complement
 76 10  101101002  2' s Complement
Arithmetic Addition
Arithmetic Addition with Comparison
The addition of two numbers in the signed mgnitude system
follow the rules of ordinary arithmetic.
If the signed are the same, we add the two magnitudes and give
the sum the common sign.
If the signed are different, we subtract the smaller magnitude
from the larger and give the difference the sign of the larger
magnitude. EX. (+25) + (-38) = -(38 - 25) = -13
Arithmetic Addition without Comparison
The addition of two signed binary number
with negative numbers represented in
signed 2’s complement form is obtained
from the addition of the two numbers,
including their signed bits. A carry out of
the signed bit position is discarded (note
that the 4th case).
Arithmetic Addition without
Comparison
 06  00000110  06  11111010
 13  00001101  13  00001101
 19  00010011  07  00000111
 06  00000110  06  11111010
 13  11110011  13  11110011
 07  11111001  19  11101101
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Arithmetic Subtraction
(+/-) A – (+B)= (+/-) A + (-B)
(+/-) A – (-B)= (+/-) A + (+B)
Example
(-6) – (-13)= +7
In binary: (1111010 – 11110011)= (1111010 +
00001101)
=100000111
after removing the carry out the result will be : 00000111
1.7 Binary Codes
Binary Coded Decimal (BCD)
Binary Coded Decimal (BCD)
in this system each digit is represented in 4 bits
For example : to represent  94510 in BCD
9
1001
4
0100
5
0101
 94510  100101000101BCD
BCD Addition
Example : Evaluate the following operations in BCD System
1–3+4
2–4+8
3 - 148 + 576
0011

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BCD
 4 
  0100
Decimal
7  0111
3
BCD
BCD Addition
Example : Evaluate the following operations in BCD System
1–3+4
2–4+8
3 - 148 + 576
4

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0100
8

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 1000
BCD
BCD
1100
 0110
12
 00010010
Decimal
Error
We must add 6
(0110) to the
result
BCD Addition
Example : Evaluate the following operations in BCD System
1–3+4
2–4+8
3 - 184 + 576
184
 576
760

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BCD

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BCD
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Decimal

0001
1000
0100
0101
0111
0110
0111 10000
0110
1010
0110
0111 1 0110
1 0000
Notes
1 – In BCD Addition , we add (0110)=(6) if the result
value was greater than (1001)=(9) or if the result was
more than 4 digits
In previous Example we added 0110 when the result was
1 - greater than 9 (1001)
2 - more than 4 digits (10000)
Note : result more than 4 digit is greater than 9(1001) 
Decimal Arithmetic
Addition for signed numbers
Example:
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(+375) + (- 240) = + 135 in BCD
Apply 10‘s complement to the negative number only.
Addition is done by summing all digits,including the sign
digit,and discarding the end carry
0 375
+9 760
-----------0 135
Decimal Arithmetic
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Subtraction for signed and unsigned numbers
Apply 10‘s complement to the subtrahend and
apply addition (same as binary case)
Excess-3 (ex-3)
Excess-three (ex-3)is another system to represent a number
(ex-3) is like (BCD) in the way of representing number
i.e. each digit is represented in 4 bits
Except that : each digit is firstly incremented by three
For example : to represent  945 in ex-3
10
9
12
1100
4
7
0111
5
8
1000
 94510  110001111000ex 3
Gray Code
ASCII character code
ASCII : American Standard Code for Information Interchange
ASCII code is used to represent characters , Symbols , …
ASCII code consists of 7-bits (to represent 128 character)
Upper case Letters are
represented by ASCII (65 : 90)
Lower case Letters are
represented by ASCII (97 : 122)
#
65
66
ASCII
1000001
1000010
Ch
A
B
90
1011010
Z
97
98
1100001
1100010
a
b
122 1111001
z
Error Detecting Code
ASCII A 1000001
ASCII T 1010100
with even parity
01000001
11010100
with odd parity
11000001
01010100
For more information about Number
Systems and Conversations between them
Check these
1 – Our Logic Book
2 - Computer Organization's Lectures
3 – Any other References