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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
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Chapter 1.indd 2
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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.
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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
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