CS220 : Digital Design

Basic Information

Title: Digital Design Code: CS220 Lecture: 3 Tutorial: 1 Pre-Requisite: Computer Introduction (CS201)

Ass. Prof. Sahar Abdul RahmanOffice: 1021 Building: 9Email: sahr_ar@yahoo.comO.H: Wednesday 8:00-12:00

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

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

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%

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.

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

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

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

augend 101101Added: + 100111 ----------

Sum: 1010100

Addition

Subtraction

minuend: 101101subtrahend: - 100111 -------------difference: 000110

Multiplication

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

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

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.

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

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

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

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

Using 1’s complement, subtract 1010100 - 1000011

M = 1010100

N = 1000011, 1’s complement of N = 0111100

answer: 0010001

1010100 0111100

10010000

33

Example

end-around carry = + 1

Using 1’s complement, subtract 1000011 - 1010100

M = 1000011

N = 1010100, 1’s complement of N = 0101011

1000011

0101011

1101110

34

Example

Answer: -0010001