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Lecture 01: INTRODUCTION

Lecture 01: INTRODUCTION. PROF. INDRANIL SENGUPTA DEPARTMENT OF COMPUTER SCIENCE AND ENGINEERING. Main Objectives of the Course. Switching Circuits and Logic Design Learn about number systems, logic gates, and Boolean algebra.

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Lecture 01: INTRODUCTION

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  1. Lecture 01: INTRODUCTION PROF. INDRANIL SENGUPTA DEPARTMENT OF COMPUTER SCIENCE AND ENGINEERING

  2. Main Objectives of the Course Switching Circuits and Logic Design • Learn about number systems, logic gates, and Boolean algebra. • Learn about the representation, manipulation and minimization of Boolean functions. • Learn how to design combinational and sequential circuits. • Understand the concepts of finite state machines, state minimization, and algorithmic state machines. • Learn about analysis and synthesis of asynchronous circuits. • Learn about the basic concepts in testing and fault diagnosis of digital circuits..

  3. Number Systems • Systematic way to represent and manipulate numbers. • Some examples: • Decimal number system • Roman number system • Binary number system • Sexagesimal number system • Broad classification: • Weighted: decimal, binary, etc. • Non-weighted: Roman, Gray code, etc.

  4. Some Basic Concepts • We are accustomed to the so-called decimal number system. • Ten digits :: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 • Every digit position has a weight which is a power of 10. • Baseor radixis 10. • Examples: 234 = 2 x 102 + 3 x 101 + 4 x 100 250.67 = 2 x 102 + 5 x 101 + 0 x 100 + 6 x 10-1 + 7 x 10-2

  5. Binary Number System • Two digits: 0 and 1. • Every digit position has a weight that is a power of 2. • Baseor radixis 2. • Binary digits are called bits. • Examples: 110 = 1 x 22 + 1 x 21 + 0 x 20 101.01 = 1 x 22 + 0 x 21 + 1 x 20 + 0 x 2-1 + 1 x 2-2

  6. Generalization: Radix-based Number System • Radix (r): Number of distinct digits. • Assume that the digits are 0, 1, 2, …, (r-1) • Every digit position has a weight that is some power of r (say, rk). • k ≥ 0, for the integer part • k < 0, for the fractional part • A (n+m)-digit number representation: D = dn-1dn-2 …..d1 d0. d-1d-2 ….. d-m

  7. Binary to Decimal Conversion • Each digit position of a binary number has a weight. • Some power of 2. • A binary number: B = bn-1 bn-2 …..b1 b0 . b-1 b-2 ….. b-m where bi are the binary digits. Corresponding value in decimal: D = bi 2i n-1 i = -m

  8. Some Examples • 101011  1x25 + 0x24 + 1x23 + 0x22 + 1x21 + 1x20= 43 (101011)2 = (43)10 • .0101  0x2-1 + 1x2-2 + 0x2-3 + 1x2-4 = .3125 (.0101)2 = (.3125)10 • 101.11  1x22 + 0x21 + 1x20 + 1x2-1 + 1x2-2 = 5.75 (101.11)2 = (5.75)10

  9. Decimal to Binary Conversion • Consider the integer and fractional parts separately. • For the integer part: • Repeatedly divide the given number by 2, and go on accumulating the remainders, until the number becomes zero. • Arrange the remainders in reverse order. • For the fractional part: • Repeatedly multiply the given fraction by 2. • Accumulate the integer part (0 or 1). • If the integer part is 1, chop it off. • Arrange the integer parts in the order they are obtained.

  10. Examples • 239 • 2 119 --- 1 • 59 --- 1 • 2 29 --- 1 • 14 --- 1 • 2 7 --- 0 • 3 --- 1 • 2 1 --- 1 • 2 0 --- 1 • 64 • 2 32 --- 0 • 16 --- 0 • 2 8 --- 0 • 4 --- 0 • 2 2 --- 0 • 1 --- 0 • 2 0 --- 1 .634 x 2 = 1.268 .268 x 2 = 0.536 .536 x 2 = 1.072 .072 x 2 = 0.144 .144 x 2 = 0.288 : 37.0625 (37)10 = (100101)2 (.0625)10 = (.0001)2 (37.0625)10 = (100101 . 0001)2 (.634)10 = (.10100……)2 (64)10 = (1000000)2 (239)10 = (11101111)2

  11. END OF LECTURE 01

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