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CS220 : Digital Design
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Basic Information
Title: Digital Design Code: CS220 Lecture: 3 Tutorial: 1 Pre-Requisite: Computer Introduction (CS201)
Ass. Prof. Sahar Abdul RahmanOffice: 1021 Building: 9Email: [email protected]: Wednesday 8:00-12:00
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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.
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Contents
TopicsContactHours
No. ofWeeks
-Introduction to the course content, text book(s), reference(s) and course plane.
- Digital Systems and Binary numbers12 4
- Boolean Algebra and Logic Gates 3 1
- Gate Level Minimization 6 2
- Combinational Logic 12 4
- Synchronous Sequential Logic 12 4
Total 45 15
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Assessment schedule
Assessment Methods Week Weighting of Assessments
First Midterm Exam 7 20%
Second Midterm Exam 12 20%
Performance (Quizzes) 5,10 5%
Home work Every week 5%
project 11 10%
Final Exam After week15 40%
Total 100%
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List of References
Essential Books “DIGITAL DESIGN”, by Mano M. Morris, 4th edition, Prentice- Hall.
Recommended Books “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.
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1. Digital Systems and Binary Numbers
1.1 Digital Systems1.2 Binary Numbers1.3 Number-Base Conversions
1.4 Octal and Hexadecimal Numbers
1.5 Complements
1.6 Signed Binary Numbers
1.7 Binary Codes
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1.1 Digital Systems
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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.
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1.3 Number-Base Conversions (Integer Part)
Example:
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1.3 Number-Base Conversions (Fraction Part)
Example:
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Binary-to-Decimal Conversion
Example:
Example:
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1.4 Octal and Hexadecimal Numbers
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Decimal-to-Octal Conversion
Example:
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Decimal-to-Hexadecimal Conversion
Example:
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Octal-to-Decimal Conversion
Example:
Example:
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Hexadecimal-to-Decimal Conversion
Example:
Example:
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Binary–Octal and Octal–Binary Conversions
Example:
Example:
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Hex–Binary and Binary–Hex Conversions
Example:
Example:
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Hex–Octal and Octal–Hex Conversions
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.
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Example
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Arithmetic Operation
augend 101101Added: + 100111 ----------
Sum: 1010100
Addition
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Subtraction
minuend: 101101subtrahend: - 100111 -------------difference: 000110
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Multiplication
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Diminished Radix Complement ((r-1)‘s complement)
Given a number N in base r having n digits, the (r - 1)’sComplement of N is defined as (rn- 1) -N.
the 9’s complement of 546700 is 999999 – 46700=453299the 1’s complement of 1011000 is 0100111
Note: The (r-1)’s complement of octal or hexadecimal numbers is obtained by subtracting each digit from 7 or F (decimal 15), respectively
1.5 Complements
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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 987602The 10’s complement of 246700 is 753300The 2’s complement of 1011000 is 0101000
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Subtraction with Complement
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.
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Example:
Using 10’s complement subtract 72532 – 3250
M = 72532
10’s complement of N = 96750
sum = 169282
Discarded end carry 105 = -100000 answer: 69282
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Example:
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
The answer is –(10’s complement of 30718) = -69282
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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=-10000000answer: 0010001
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Example
(b) 1000011 – 1010100M = 1000011
N = 1010100, 2’s complement of N = 0101100
1000011 0101100sum = 1101111
answer: - (10000000 - 1101111) = -0010001
The answer is –(2’s complement of 1101111) = - 0010001
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Using 1’s complement, subtract 1010100 - 1000011
M = 1010100
N = 1000011, 1’s complement of N = 0111100
answer: 0010001
1010100 0111100
10010000
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Example
end-around carry = + 1
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Using 1’s complement, subtract 1000011 - 1010100
M = 1000011
N = 1010100, 1’s complement of N = 0101011
1000011
0101011
1101110
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Example
Answer: -0010001