NOTES FOR SUBJECT

1. BEEE102L BASIC ELECTRICAL
AND ELECTRONICS
ENGINEERING
Dr.S.ALBERT ALEXANDER
SCHOOL OF ELECTRICAL ENGINEERING
albert.alexander@vit.ac.in
1
Dr.S.ALBERT ALEXANDER-
SELECT-VIT

2. Module 5
Dr.S.ALBERT ALEXANDER-SELECT-
VIT 2
 Number base conversion
 Binary arithmetic
 Boolean algebra
 Simplification of Boolean functions using K-maps
 Logic gates
 Design of basic combinational circuits:
 Adders (Half adder & Full adder)
 Multiplexers & De-multiplexers

3. 5.1 Number Base Conversion
Dr.S.ALBERT ALEXANDER-SELECT-
VIT 3
 The number systems are used quite frequently in the field
of digital electronics and computers
 However the type of number system used in computers
could be different at different stages of the usage
 For example, when a user key-in some data into the
computer, he/she will do it using decimal number system
 i.e. the system we all have used for several years for doing
arithmetic problems
 But when the information goes inside the computer, it
needs to be converted to a form suitable for processing
data by the digital circuitry
 When the data has to be displayed on the monitor for the
user, it has to be again in the decimal number system

4. Types
There are several number systems but the following are the
important ones in the field of digital electronics:
 Decimal number system
 Binary number system
 Octal number system
 Hexadecimal number system
Dr.S.ALBERT ALEXANDER-SELECT-
VIT 4

5. Binary to Decimal Conversion
Dr.S.ALBERT ALEXANDER-SELECT-
VIT 5
 Following is the procedure for converting an integer (or
whole) binary number to its equivalent decimal number
 Step 1. Write the binary number
 Step 2. Directly under the binary number, write the position
values or weights of each bit working from right to left
 Step 3. If a zero appears in a digit position, cross-out the
weight for that position
 Step 4. Add the remaining weights to obtain the decimal
equivalent

6. Exercise-1
Convert each of the following binary numbers to their decimal
equivalents : (a)101, (b) 10101, (c) 01010110.
SOLUTION:
a) 1 0 1
22 21 20
= 22+ 20= 4+1=5
b) 10101= 24 + 22 + 20 = 16 + 4 + 1 = 21
c) 01010110 = 26 + 24 + 22 + 21 = 64 + 16 + 4 + 2 = 86
Dr.S.ALBERT ALEXANDER-SELECT-
VIT 6

7. Exercise-2
Convert each of the following fractional binary numbers to
their decimal equivalents: (a) 0.1010, (b) 0.1100.
SOLUTION:
a) .1 0 1 0
2-1 2-2 2-3 2-4
= 2-1+ 2-3= 0.5+0.125=0.625
b) 0.1100= 2-1+ 2-2= 0.5+0.25=0.75
Dr.S.ALBERT ALEXANDER-SELECT-
VIT 7

8. Decimal to Binary Conversion
Dr.S.ALBERT ALEXANDER-SELECT-
VIT 8
 The conversion from decimal-to-binary is usually
performed by a digital computer for ease of interpretation
by the person reading the number
 On the other hand, when a person enters a
number into a digitalcomputer, that number
decimal
must be
converted to binary before it can be operated on
There are two methods of decimal-to-binary conversion:
 (1) Sum-of-weights method and
 (2) Repeated division by-2 method

9. Exercise-3
Convert each of the following decimal numbers to their binary
equivalents using sum-of-weights methods: (a) 17, (b) 24, (c)
61, (d) 93.
SOLUTION:
a) 17= 16+1 = 24+20
24 23 22 21 20
1 0 0 0 1 = (10001)2
b) 24= 16+8 = 24+23 = (11000)2
c) 61= 32+16+8+4+1 = 25+24 +23 +22 +20 = (111101)2
d) 93= 64+16+8+4+1 = 26+24 +23 +22 +20 = (1011101)2
Dr.S.ALBERT ALEXANDER-SELECT-
VIT 9

10. Exercise-4
Convert the decimal fraction a) 0.375 and b) 0.625 by using
sum-of-weights method to its equivalent binary fraction.
SOLUTION:
a) 0.375= 0.25+0.125 = 2-2+2-3
20 2-1 2-2 2-3
0 0 1 1 = (0.011)2
b) 0.625= 0.5+0.125 = 2-1+2-3 = (0.1011)2
Dr.S.ALBERT ALEXANDER-SELECT-
VIT 10

11. Exercise-5
Convert each of the following decimal numbers
repeated-division by-2 method. (a) 19 and (b) 45.
SOLUTION:
using
a) 19 ÷ 2 = 9 with a remainder 1 (LSB)
 9 ÷ 2 = 4 with a remainder 1
 4 ÷ 2 = 2 with a remainder 0
 2 ÷ 2 = 1 with a remainder 0
 1 ÷ 2 = 0 with a remainder 1 (MSB)
= (10011)2
b) 45=(101101)2
Dr.S.ALBERT ALEXANDER-SELECT-
VIT 11

12. Exercise-6
Convert the decimal fraction a) 0.9028 and b) 0.8125 to its
using
equivalent binary fraction (up to 4 binary places)
repeated multiplication-by-2 method.
SOLUTION:
a) 0.9028 × 2 = 1.8056 = 0.8056 with a carry of 1 (LSB)
 0.8056 × 2 = 1.6112 = 0.6112 with a carry of 1
 0.6112 × 2 = 1.2224 = 0.2224 with a carry of 1
 0.2224 × 2 = 0.4448 = 0.4448 with a carry of 0 (MSB)
= (0.1110)2
b) 0.8125 =(0.11010)2
Dr.S.ALBERT ALEXANDER-SELECT-
VIT 12

13. Octal Number System
Dr.S.ALBERT ALEXANDER-SELECT-
VIT 13

14. Exercise-7
Convert the octal number a) 2374 and b) 326 to its equivalent
decimal.
SOLUTION:
a) 2 3 7 4
83 82 81 80
2x 83 +3x 82 +7x 81+4x80 = 1276
b) 3 2 6
82 81 80
3x 82 +2x 81+6x80 = 214
Dr.S.ALBERT ALEXANDER-SELECT-
VIT 14

15. Exercise-8
Convert the decimal number a) 266 and b) 435
equivalent octal.
SOLUTION:
a) (266)10= (412)8
 266 ÷ 8 = 33.25 with a remainder 0.25 0.25 × 8 = 2
 33 ÷ 8 = 4.125 with a remainder 0.125  0.125 × 8 = 1
 4 ÷ 8 = 0.5 = 0 with remainder 0.5 0.5 ×8= 4
to its
LSD
MSD
b) (435)10= (663)8
 435÷8 = 54.375 = 54 with a remainder 0.3750.375 × 8 = 3 LSD
 54 ÷ 8 = 6.75 = 6 with a remainder 0.75 0.75 × 8 = 6
 6 ÷ 8 = 0.75 = 0 with a remainder 0.75 0.75 × 8 = 6 MSD
Dr.S.ALBERT ALEXANDER-SELECT-
VIT 15

16. Exercise-9
Convert the octal number a) 321, b) 4653, c) 13274 to its
equivalent binary.
SOLUTION:
a) 3011
2 010
1 001
= 011010001
b) 4653  100110101011
c) 13274  001011010111100
Dr.S.ALBERT ALEXANDER-SELECT-
VIT 16

17. Exercise-10
Convert the binary number a) 100111010 and b) 10111001 to
its equivalent octal.
SOLUTION:
a) 1004
111 7
010 2
= 472
b) 010 2
111 7
001 1
=271
Dr.S.ALBERT ALEXANDER-SELECT-
VIT 17

18. Hexadecimal Number System
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VIT 18

19. Exercise-11
Dr.S.ALBERT ALEXANDER-SELECT-
VIT 19
Convert the hexadecimal number a) E5, b) 0.12, c) 2A6 to its
equivalent decimal.
SOLUTION:
a) (Ex161)+(5x160) =229
b) .(1x16-1)+(2x16-2) =0.0703
c) (2x162) (Ax161)+(6x160) =678

20. Exercise-12
Convert the decimal number a) 650, b) 151, c) 498 to its
equivalent hexadecimal.
SOLUTION:
 a) 650÷16 = 40.625 = 40 with a remainder 0.625  0.625×16 = 10 (= A) LSD
 40 ÷ 16 = 2.5 = 2 with a remainder 0.5  0.5 × 16 = 8
 2 ÷ 16 = 0.125 = 0 with a remainder 0.125  0.125 × 16= 2 MSD
= 28A
b) 151 97
c) 498 1F2
Dr.S.ALBERT ALEXANDER-SELECT-
VIT 20

21. Exercise-13
Convert the hexadecimal number a) 2D6, b) 9F2, c) 2A6 to its
equivalent binary.
SOLUTION:
 a) 20010
D1101
60110
=001011010110
b) 9F2 100111110010
c) 2A6 001010100110
Dr.S.ALBERT ALEXANDER-SELECT-
VIT 21

22. Exercise-14
Convert the binary number a) 10111, b) 1111 110000, c)
1110.101 to its equivalent hexadecimal.
SOLUTION:
 a) 10111  00010111  17
 b) 1111 110000  001111 110000
 3F0
 c) 1110.101  1110.1010
EA
Dr.S.ALBERT ALEXANDER-SELECT-
VIT 22

23. Exercise-15
Dr.S.ALBERT ALEXANDER-SELECT-
VIT 23
Convert the hexadecimal number a) 5C2, b) 8AD9, c) A7.3B
to its equivalent octal.
SOLUTION:
 a) 5C2  0101 1100 0010  2702
 b) 8AD9 1000 1010 1101 1001  42331
 c) A7.3B  01010 0111 0011 10110  247.166

24. Exercise-16
Dr.S.ALBERT ALEXANDER-SELECT-
VIT 24
Convert the octal number a) 321 and b) 1024 to its equivalent
hexadecimal.
SOLUTION:
 a) 321 011 010 001  D1
 b) 1024001 000 010 100  214

25. Exercise-17 (Fill the ?)
Dr.S.ALBERT ALEXANDER-SELECT-
VIT 25
Decimal Binary Octal Hexa-decimal
19.6875 ? ? ?
? 11101.111100 ? ?
? ? 255.454 ?

26. Exercise-17 (Solution)
Dr.S.ALBERT ALEXANDER-SELECT-
VIT 26
Decimal Binary Octal Hexa-decimal
19.6875 10011.10110 23.54 13.B
29.9375 11101.111100 35.74 1D.F
173.5859 10101101.10010110 255.454 AD.96

27. 5.2 Binary Arithmetic (Addition)
2 6 4
1 7 3
4 3 7
 The addition of two binary numbers is performed in exactly
the same manner as the addition of decimal numbers
 Let us review the decimal addition: 264+173
1
carry
 Unlike decimal addition, there are only four cases that can
occur in binary addition
 0 + 0 = 0
 1 + 0 = 1
 1 + 1 = 10 = 0 + carry of 1 into next position
 1+ 1 +1 = 11 = 1+ carry of 1 into next position
Dr.S.ALBERT ALEXANDER-SELECT-
VIT 27

28. Exercise-18
a) Add 101 and 110.
1
1
0
1
1
0
10 1 1
b) Add 11 and 11.
1 1
1 1
11 0
c) Add 100 and 10
1 0 0
1 0
1 1 0
1
carry
Dr.S.ALBERT ALEXANDER-SELECT-
VIT 28

29. Exercise-19
Dr.S.ALBERT ALEXANDER-SELECT-
VIT 29
a) Add 11.01 and 101.11.
1 1. 0 1
1 0 1. 1 1
10 0 1. 0 0

30. Binary Arithmetic (Subtraction)
Dr.S.ALBERT ALEXANDER-SELECT-
VIT 30
 0 - 0 = 0
 1 - 0 = 1
 1 - 1 = 0
 0 - 1 = 1 (with borrow of 1)

31. Exercise-20
Dr.S.ALBERT ALEXANDER-SELECT-
VIT 31
a) Subtract 101 from 1001.
1 0 0 1
1 0 1
0 1 0 0
b) Subtract 11 from 10000.
1 0 0 0 0
1 1
0 1 1 0 1

32. Exercise-21
Dr.S.ALBERT ALEXANDER-SELECT-
VIT 32
a) Subtract 100.10 and 110.01.
1 1 0. 0 1
1 0 0. 1 0
0 0 1. 1 1

33. Binary Arithmetic (x and )
Dr.S.ALBERT ALEXANDER-SELECT-
VIT 33
 0 x 0 = 0
 0 x 1 = 0
 1 x 0 = 0
 1 x 1 = 1
 0  1 = 0
 1  1 = 1

34. Dr.S.ALBERT ALEXANDER-SELECT-
VIT 34