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1. COURSE NAME: BASIC ELECTRONICS
COURSE CODE : EC 1001
LECTURE SERIES NO : 01(ONE)
CREDITS : 3
MODE OF DELIVERY : ONLINE (POWER POINT PRESENTATION)
FACULTY :
EMAIL-ID :
PROPOSED DATE OF DELIVERY:
B.TECH FIRST YEAR
ACADEMIC YEAR: 2020-2021

2. SESSION OUTCOME
“MATHEMATICAL UNDERSTANDING
OF NUMBER SYSTEM”

3. ASSESSMENT CRITERIA’S
ASSIGNMENT
QUIZ
MID TERM EXAMINATION –II
END TERM EXAMINATION

4. PROGRAM
OUTCOMES
MAPPING WITH
CO4
[PO1]
DEMONSTRATE DIFFERENT NUMBER SYSTEMS,
BOOLEAN EXPRESSIONS AND DIFFERENT
ELEMENTS OF COMMUNICATION SYSTEMS AND
TO PROMOTE DIFFERENT SKILLS TOWARDS
ELECTRONICS INDUSTRIES.

6.  We use numbers
. to communicate
. to perform tasks
. to quantify
. to measure
 Numbers have become symbols of the present era
 Many consider what is not expressible in terms of numbers is
not worth knowing

7.  A number system can be used to represent
the number of students in a class or number
of viewers watching a certain TV program
etc.
 The digital computer represents all kinds of
data and information in binary numbers.
 It includes audio, graphics, video, text and
numbers.
 The total number of digits used in a number
system is called its base or radix.
AKHILESH
MAITHANI

8.  Decimal Number System
 Binary Number System
 Octal Number System
 Hexadecimal Number System
AKHILESH
MAITHANI

9. COMMON NUMBER SYSTEMS
Number Systems
9
System Base Symbols
Decimal 10 0, 1, … 9
Binary 2 0, 1
Octal 8 0, 1, … 7
Hexa-decimal 16 0, 1, … 9,
A, B, … F

10. DECIMAL NUMBER SYSTEM
10
 The decimal number system is also known as base 10.
The values of the positions are calculated by taking 10 to
some power.
 Why is the base 10 for decimal numbers?
 Because we use 10 digits, the digits 0 through 9.

11. BINARY NUMBER SYSTEM
11
• The binary number system is also known as base 2.
The values of the positions are calculated by taking 2
to some power.
• Why is the base 2 for binary numbers?
o Because we use 2 digits, the digits 0 and 1.

12. BINARY NUMBER SYSTEM
12
• Example of a binary number and the values of the
positions:
 1 0 0 1 1 0 1
 26
25
24
23
22
21
20

13. Characteristics of octal number system are as
follows:
 Uses eight digits, 0,1,2,3,4,5,6,7.
 Also called base 8 number system
 Each position in an octal number represents a 0
power of the base (8). Example 80
AKHILESH
MAITHANI

14. HEXADECIMAL NUMBER SYSTEM
14
• The hexadecimal number system is also known as base 16.
The values of the positions are calculated by taking 16 to
some power.
• Why is the base 16 for hexadecimal numbers ?
– Because we use 16 symbols, the digits 0 to 9 and the
letters A through F.

15. QUANTITIES/COUNTING
Decimal Binary Octal
Hexa-
decimal
0 0 0 0
1 1 1 1
2 10 2 2
3 11 3 3
4 100 4 4
5 101 5 5
6 110 6 6
7 111 7 7
15
Decimal Binary Octal
Hexa-
decimal
8 1000 10 8
9 1001 11 9
10 1010 12 A
11 1011 13 B
12 1100 14 C
13 1101 15 D
14 1110 16 E
15 1111 17 F

16. CONVERSION AMONG BASES
 The possibilities:
Hexadecimal
Decimal Octal
Binary
16

17. EXAMPLE – DIFFERENT FORMS
2510 = 110012 = 318 = 1916
Base
17

18. 18
Decimal to Decimal (for
understanding)
12510 => 5 x 100
= 5
2 x 101
= 20
1 x 102
= 100
125
Base
Weight

19. BINARY TO DECIMAL
 Technique
 Multiply each bit by 2n
, where n is the “weight” of the bit
 The weight is the position of the bit, starting from 0 on the right
 Add the results
19

20. EXAMPLE- BINARY TO DECIMAL
20
1010112 => 1 x 20
= 1
1 x 21
=
2
0 x 22
=
0
1 x 23
=
8
0 x 24
=
0
1 x 25
=
Bit “0”

21. BINARY NUMBER SYSTEM
(Continued from previous slide..)
Example
101012 = (1 x 24) + (0 x 23) + (1 x 22) + (0 x 21) x (1 x 20)
= 16 + 0 + 4 + 0 + 1
= 2110

22. 22
CONVERT THE GIVEN BINARY NUMBER INTO DECIMAL EQUIVALENT
NUMBER
 (100010)2 =0×20
+1×21
+0×22
+0×23
+0×24
+1×25
=0+2+0+0+0+32
=(34)10
 (0.1011)2 =1×2-1
+0×2-2
+1×2-3
+1×2-4
=0.5+0+0.125+0.0625
=(0.6875)10
 (10101.011)2
Integer part: (10101)2=1×20
+0×21
+1×22
+0×23
+1×24
=1+0+4+0+16 = (21)10
Decimal part: (0.011)2 =0×2-1
+1×2-2
+1×2-3
= (0.375)10
= (21.375)10

23. OCTAL TO DECIMAL
 Technique
 Multiply each bit by 8n
, where n is the “weight” of the bit
 The weight is the position of the bit, starting from 0 on the right
 Add the results
23

24. EXAMPLE- OCTAL TO DECIMAL
7248 => 4 x 80
= 4
2 x 81
= 16
7 x 82
= 448
46810
24

25. OCTAL NUMBER SYSTEM
8)
are
• Since there are only 8 digits, 3 bits (23
= sufficient to represent any octal number in
binary
Example
20578 = (2 x 83) + (0 x 82) + (5 x 81) + (7 x 80)
= 1024 + 0 + 40 + 7
= 107110

26. 26
CONVERT THE FOLLOWING OCTAL NUMBER INTO DECIMAL NUMBER
SYSTEM
 (2376)8 = (?)10
= 2×83
+3×82
+7×81
+6×80
= 1024+192+56+6
= (1278)10
 (1234.567)8
=1×83
+2×82
+3×81
+4×80
+5×8-1
+6×8-2
+7×8-
=512+128+24+4+0.625+0.09375+0.01367
= (668.7324219)10

27. HEXADECIMAL TO DECIMAL
 Technique
 Multiply each bit by 16n
, where n is the “weight” of the bit
 The weight is the position of the bit, starting from 0 on the right
 Add the results
27

28. EXAMPLE- HEXADECIMAL TO DECIMAL
ABC16 => C x 160
= 12 x 1 = 12
B x 161
= 11 x 16 = 176
A x 162
= 10 x 256 = 2560
274810
28

29. HEXADECIMAL NUMBER SYSTEM
• Each position of a digit represents a specific
power of the base (16)
• Since there are only 16 digits, 4 bits (24 = 16) are
sufficient to represent any hexadecimal number
in binary
Example
1AF16
= (1 x 162) + (A x 161) + (F x 160)
= 1 x 256 + 10 x 16 + 15 x 1
= 256 + 160 + 15
= 43110

30. 30
CONVERT THE FOLLOWING HEXADECIMAL NUMBER INTO DECIMAL
NUMBER SYSTEM
 (269)16
 =2×162
+6×161
+9×160
= 2×256+96+9
= (617)10
 (2B8D.E2)16
=2×163
+11×162
+8×161
+13×160
+14×16-1
+2×16-2
=8192+2816+128+13+0.875+0.0078125
= (11149.88281)10

31. DECIMAL TO BINARY
 Technique
 Divide by two, keep track of the remainder
31

32. EXAMPLE- DECIMAL TO BINARY
12510 = ?2
2 125
62 1
2
31 0
2
15 1
2
7 1
2
3 1
2
1 1
2
0 1
12510 = 11111012
32

33. 33
 (734)10=(X)2
Take the numbers from
bottom to top,
 (734)10 = ( 1011011110)2

34. 34
 (0.705)10
0.705×2=1.410 ---1
0.410×2=0.820 ---0
0.82×2= 1.64------1
0.64×2= 1.28------1
0.28×2= 0.56------0
0.56×2=1.12-------1
0.12×2=0.24-------0
0.24×2=0.48-------0
0.48×2=0.96-------0
0.96×2=1.92-------1
Take the number from top to bottom,
(0.705)10 = (0.1011010001)2

35. 35
 (41.915)10 =(?)2

36. OCTAL TO BINARY
 Technique
 Convert each octal digit to a 3-bit equivalent binary representation
36

37. EXAMPLE- OCTAL TO BINARY
7058 = ?2
7 0 5
111 000 101
7058 = 1110001012
37

38. 38
 (632)8
 So, (632)8=(110011010)2
6 3 2
110 011 010

39. 39
 (7423.245)8
 (7423.245)8= (111100010011.010100101)2
7 4 2 3 . 2 4 5
111 100 010 011 . 010 100 101

40. HEXADECIMAL TO BINARY
 Technique
 Convert each hexadecimal digit to a 4-bit equivalent binary representation
40

41. EXAMPLE- HEXADECIMAL TO
BINARY
10AF16 = ?2
1 0 A F
0001 0000 1010 1111
10AF16 = 00010000101011112
41

42. DECIMAL TO OCTAL
 Technique
 Divide by 8
 Keep track of the remainder
42

43. EXAMPLE- DECIMAL TO OCTAL
123410 = ?8
8 1234
154 2
8
19 2
8
2 3
8
0 2
123410 = 23228
43

44. 44
 (2003)10=(X)8
 Take the numbers from bottom to top, (2003)10= (3723)8

45. 45
 (0.12)10 = (X)8
0.12×8=0.96---0
0.96×8=7.68---7
0.68×8=5.44---5
0.44×8=3.52---3
0.52×8=4.16---4
(0.12)10=(0.07534)8

46. DECIMAL TO HEXADECIMAL
 Technique
 Divide by 16
 Keep track of the remainder
46

47. EXAMPLE- DECIMAL TO
HEXADECIMAL
123410 = ?16
123410 = 4D216
16 1234
77 2
16
4 13 = D
16
0 4
47

48. BINARY TO OCTAL
 To convert, Starting from the binary point, the binary digits are arranged in groups of three on both sides. Each in
group of binary digit is replaced by its octal equivalent.
 Note: 0’s can be added on either side, if needed to complete a group of three.
48

49. EXAMPLE- BINARY TO OCTAL
10110101112 = ?8
1 011 010 111
1 3 2 7
10110101112 = 13278
49

50. 50
011 101 . 110
3 5 . 6
(011101.110)2 = (?)8

51. BINARY TO HEXADECIMAL
 Technique
 Group bits in fours, starting on right
 Convert to hexadecimal digits
51

52. EXAMPLE- BINARY TO
HEXADECIMAL
10101110112 = ?16
10 1011 1011
2 B B
10101110112 = 2BB16
52

53. 53
CONVERSION FROM HEXADECIMAL TO BINARY
 When a hexadecimal number is to be converted its equivalent binary number, each of its digits is replaced by
equivalent group of 4 binary digits.

54. 54
HEXADECIMAL TO BINARY
 (347.28)16 = (?)2
3 4 7 . 2 8
0011 0100 0111 . 0010 1000
(347.28)16= (001101000111.00101000)2

55. 55
HEXADECIMAL TO BINARY
(8BE6.7A)16 =(?)2
8 B E 6 . 7 A
1000 1011 1110 0110 . 0111 1010
(8BE6.7A) 16 = (1000101111100110.01111010)2

56. OCTAL TO HEXADECIMAL
Write down the three bit binary equivalent of octal digit and then rearranging into group of four bits with ‘0’s
added on either side of decimal point, if needed to complete the group of four.
56

57. 57
OCTAL TO HEXADECIMAL
(46) 8 = (?)16
Octal equivalent 4 6
100 110
Hexadecimal
equivalent
0010 0110
2 6
(46) 8 =(26)16

58. 58
OCTAL TO HEXADECIMAL
(764.352) 8 = (?)16
Octal equivalent 7 6 4 . 3 5 2
111 110 100.011 101 010
Hexadecimal
equivalent
0001 1111 0100.0111 0101 0000
1 F 4 . 7 5 0
(764.352) 8 = (1F4.750)16

59. EXAMPLE- OCTAL TO HEXADECIMAL
10768 = ?16
1 0 7 6
001 000 111 110
2 3 E
10768 = 23E16
59

60. HEXADECIMAL TO OCTAL
 Technique
 Use binary as an intermediary
60

61. EXAMPLE- HEXADECIMAL TO OCTAL
1F0C16 = ?8
1 F 0 C
0001 1111 0000 1100
1 7 4 1 4
1F0C16 = 174148
61