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UNIT I: DIGITAL ELECTRONICS LAST SIX YEARS' GATE ANALYSIS 6 Number of questions 5 4 Marks 1 Marks 2 Total number of questions 3 2 1 0 2015 2014 2013 2012 2011 2010 Concepts on which questions were asked in the previous six years Chapter 1.indd 1 Year Concepts 2015 K-maps, Half and full adder, JK flip-flop, Boolean operators, Prime implicates, Base values, Johnsen counter, Binary operator, D flip-flop 2014 Minterm expressions, Combinational functional blocks, Boolean expressions, flip-flops 2013 2’s complement, Truth tables, Logic gates, RAM chips 2012 K-maps, Truth tables, IEEE single precision 2011 Flip-flops, Boolean expressions, Logic gates 2010 K-maps, 2’s complement, Multiplexer, Boolean expression, Sequential circuits 4/9/2015 9:40:08 AM

Chapter 1.indd 2 4/9/2015 9:40:08 AM

CHAPTER 1 DIGITAL ELECTRONICS Syllabus: Digital electronics: logic functions, minimization, design and synthesis of combinational and sequential circuits, number representation and computer arithmetic (fixed and floating point). Number system 1.1 INTRODUCTION Digital electronics represents discrete signals instead of signals in a continuous range. It uses two binary levels 0’s (corresponding to false) and 1’s (corresponding to true). The main reason for advancement in digital electronics is integrated circuits (ICs). Radix/Base representation Numeric codes Codes representation Alphanumeric codes EBCDIC Weighted Non-weighted ASCII 1.2 NUMBER SYSTEM Number system is an age old method to represent numerals (Fig. 1.1). Decimal number system is the most common number system and binary number system is used by computers. Chapter 1.indd 3 Binary Gray code BCD XS3 code 2421 code Figure 1.1 | Number system. 4/9/2015 9:40:08 AM

4 Chapter 1: Digital Electronics Table 1.1 | Types of number systems Base/Radix Unique Numbers Terminology Example 10 0 to 9 Decimal number system 10.47 8 0 to 7 Octal number system 65.32 2 0 to 1 Binary number system 110.11 16 0 to 15 Hexadecimal number system BAD.1A 0 to 9 are numerals and 10 to 15 are represented with alphabets 10 = A, 11 = B, 12 = C, … 15 = F (111001.0100)2 = 1 × 25 + 1 × 24 + 1 × 23 + 0 × 22 + 0 × 21 + 1 × 20 + 0 × 2-1 Radix or base is the number of unique digits, so the different base or radix can be expressed as represented in Table 1.1. For example, (57)10 − here 10 is the base (or radix) and 57 is the number. Another example is of (101.11)2 = (5.75)10 = (5.6)8 = (5.C)16. Example 1.1 Let us understand the logic of representation by considering (57.1)10. We know 57 means 50 + 7 + (1/10). Now, representing the same by making use of radix (which is 10), we get 1.2.1 Conversions of Number System A number of a particular number system can be converted into another number system, as follows: 1. Decimal number system to any number system: Consider the example of (57.3)10. To find the binary equivalent, we have 2 57 2 28 1 2 14 27 23 21 0 0 1 1 2 × 0.3 = 0.6 2 × 0.6 = 1.2 2 × 0.2 = 0.4 2 × 0.4 = 0.8 1.2.2 Complement of a Number Any given number will have two complements: 1. Radix minus one complement (R-1’s complement): It is obtained by subtracting the given number from the highest possible number. For example, if there is a two-digit decimal number, it will be subtracted from 99; whereas a 3-bit number in a binary system will be subtracted from 111. 5 × (10)1 + 7 × (10)0 + 1 × (10)-1 (111001)2 + 1 × 2-2 + 0 × 2-3 + 0 × 2-4 = 32 + 16 + 8 + 0 + 0 + 1 + 0 + 0.25 + 0 + 0 = (57.25)10 0 1 0 0 Note: The 1’s complement of a binary number is obtained by interchanging the 1’s and 0’s. For example, 1’s complement of (1011)2 is (0100)2. 2. Radix complement (R’s complement): This is also known as true complement. It is obtained by adding one to R-1’s complement. For example, (57)10 − its 9’s complement is 42 and its 10’s complement is 43. 1.2.2.1 Subtraction Using R-1’s Complement For the following subtractions, A −B (0.0100)2 ---------- Therefore, (57.3)10 = (111001.0100)2 2. Any number system to decimal number system: Consider the example of (111001.0100)2. To find the decimal equivalent we have (111001.0100)2 = 1 × 25 + 1 × 24 + 1 × 23 + 0 × 22 + 0 × 21 + 1 × 20 + 0 × 2-1 ---------the steps involved are as follows: Step 1. Find R-1’s complement of B. Step 2. Add A and B. Step 3. If there is carry, then add this carry to the answer. + 1 × 2-2 + 0 × 2-3 + 0 × 2-4 = 32 + 16 + 8 + 0 + 0 + 1 + 0 + 0.25 + 0 + 0 = (57.25)10 Chapter 1.indd 4 4/9/2015 9:40:09 AM

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