1. MMNSS COLLEGE,KOTTIYAM DEPARTMENT OF PHYSICS
DIGITAL ELECTRONICS

2. Digital devices are based on electronic circuitry It represents two states 1 or high & 0 or low In this system, terms like bit, nibble & byte are frequently used

3. NUMBER SYSTEMS

4. Objectives After studying this chapter, You should be able to:
Understand the concept of number systems.
Distinguish between non-positional and positional number systems.
Describe the decimal, binary, hexadecimal and octal system.
Convert a number in binary, octal or hexadecimal to a number in the decimal system.
Convert a number in the decimal system to a number in binary, octal and hexadecimal.
Convert a number in binary to octal and vice versa.
Convert a number in binary to hexadecimal and vice versa.
Find the number of digits needed in each system to represent a particular value.

5. 2-1 INTRODUCTION
A number system defines how a number can be represented using distinct symbols. A number can be represented differently in different systems. For example, the two numbers (2A)16 and (52)8 both refer to the same quantity, (42)10, but their representations are different.
Several number systems have been used in the past and can be categorized into two groups: positional and non-positional systems. Our main goal is to discuss the positional number systems, but we also give examples of non-positional systems.

6. Common Number Systems System Base Symbols Used by humans?
Used in computers?
Decimal 10
0, 1, … 9
Yes No
Binary 2
0, 1
Octal 8
0, 1, … 7
Hexa- decimal 16
0, 1, … 9,
A, B, … F

7. Quantities/Counting (1 of 3)
Decimal
Binary
Octal
Hexa- decimal 1 2 10 3 11 4
100 5
101 6
110 7
111

8. Quantities/Counting (2 of 3)
Decimal
Binary
Octal
Hexa- decimal 8
1000 10 9
1001 11
1010 12 A
1011 13 B
1100 14 C
1101 15 D
1110 16 E
1111 17 F

9. Quantities/Counting (3 of 3)
Decimal
Binary
Octal
Hexa- decimal 16
10000 20 10 17
10001 21 11 18
10010 22 12 19
10011 23 13
10100 24 14
10101 25 15
10110 26
10111 27
Etc.

10. Conversion Among Bases
The possibilities:
Decimal
Octal
Binary
Hexadecimal

11. Quick Example
2510 = = 318 = 1916
Base

12. Decimal to Decimal (just for fun)
Octal
Binary
Hexadecimal

13. Weight
12510 => 5 x 100 = x 101 = x 102 =
Base

14. Binary to Decimal
Decimal
Octal
Binary
Hexadecimal

15. 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

16. Example
Bit “0”
=> 1 x 20 = x 21 = x 22 = x 23 = x 24 = x 25 = 32
4310

17. Octal to Decimal
Decimal
Octal
Binary
Hexadecimal

18. 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

19. Example
7248 => 4 x 80 = x 81 = x 82 =

20. Hexadecimal to Decimal
Octal
Binary
Hexadecimal

21. 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

22. Example
ABC16 => C x 160 = 12 x 1 = B x 161 = 11 x 16 = A x 162 = 10 x 256 = 2560
274810

23. Decimal to Binary
Decimal
Octal
Binary
Hexadecimal

24. Decimal to Binary Technique Divide by two, keep track of the remainder
First remainder is bit 0 (LSB, least-significant bit)
Second remainder is bit 1
Etc.

25. Example
12510 = ?2
12510 =

26. Octal to Binary
Decimal
Octal
Binary
Hexadecimal

27. Octal to Binary Technique
Convert each octal digit to a 3-bit equivalent binary representation

28. Example
7058 = ?2
7058 =

29. Hexadecimal to Binary
Decimal
Octal
Binary
Hexadecimal

30. Hexadecimal to Binary Technique
Convert each hexadecimal digit to a 4-bit equivalent binary representation

31. Example
10AF16 = ?2
A F
10AF16 =

32. Decimal to Octal
Decimal
Octal
Binary
Hexadecimal

33. Decimal to Octal
Technique
Divide by 8
Keep track of the remainder

34. Example
= ?8 8
19 2 8
2 3 8
0 2
= 23228

35. Decimal to Hexadecimal
Octal
Binary
Hexadecimal

36. Decimal to Hexadecimal
Technique
Divide by 16
Keep track of the remainder

37. Example
= ?16
77 2 16
= D
0 4
= 4D216

38. Binary to Octal
Decimal
Octal
Binary
Hexadecimal

39. Binary to Octal Technique Group bits in threes, starting on right
Convert to octal digits

40. Example
= ?8
= 13278

41. Binary to Hexadecimal
Decimal
Octal
Binary
Hexadecimal

42. Binary to Hexadecimal Technique Group bits in fours, starting on right
Convert to hexadecimal digits

43. Example
= ?16
B B
= 2BB16

44. Octal to Hexadecimal
Decimal
Octal
Binary
Hexadecimal

45. Octal to Hexadecimal
Technique
Use binary as an intermediary

46. Example
10768 = ?16 E
10768 = 23E16

47. Hexadecimal to Octal
Decimal
Octal
Binary
Hexadecimal

48. Hexadecimal to Octal
Technique
Use binary as an intermediary

49. Example
1F0C16 = ?8
1 F C
1F0C16 =

50. BINARY ADDITION The following table gives the rules for binary addition. When 1 is added to 1 the sum is 10(binary)=2(decimal) A B
A+B 1 10

51. 1
In the above example when 1 is added to 1,there is a carry=1.This carry is added to the sum of the adjacent bits

52. Example 2 1

53. Binary Subtraction When 1 is subtracted from 0,there is a borrow from the adjacent bit1

54. 1’sComplement. The 1’s complement of a binary number is obtained by subtracting each bit of the number from 1. It is simply obtained by changing each bit 1 to zero and zero to 1
Binary number
1’s complement
100110
111110
011001
000001

55. 2’s complement The 2’s complement of a binary number=its 1’s complement+1
10011
01100
+1 (00001)
01101
111
000
+1(001)
001
0000
1111
+1(0001)
010110
101001
+1(000001)
101010

56. Addition of a binary number to its 2’s complement
Addition of a binary number to its 2’s complement. (In this case, the last carry is lost for a 4 bit processor)
Binary number
1’s complement
2’s complement
Number+2’s complement
1001
0110
0110+1=0111
1001 +
0111
0000
(carry 1 is lost)

57. Representation of numbers in 1’s complement
Positive numbers are represented by including a zero at the extreme left.
The negative numbers are represented by the complemented value of the positive signed binary.
Example.
+(6)10 is represented by (0110)2
-(6)10 is represented by (1001)2,the 1’s complement of (0100)2

58. Addition &Subtraction Using 1’s Complement
There are four possibilities
1.Both the numbers are positive
2.One number is positive and the other number is negative, the negative number being greater
3.One number is positive and the is negative, the positive number being greater.
4.Both the numbers are negative.

59. Decimal representation
Case 1. Add and +0100
Normal form
Representation
Decimal representation
+ 0011
0 0011 +3
+ 0100
0 0100 +4
+0111
0 0111
+ 7

60. Case II:One number is positive & the other number is negative, the negative number being greater. When a positive number is added to a larger negative number, the sum is the difference of the two numbers with sign of the negative number. The negative number is represented as its 1’s complement and is added to the positive number. The answer will be the complement of the sum.
Normal form
Representation
Decimal representation
+ 0101
0 0101 +5
- 1100
1 0011
-12
-0111
1 1000 -7

61. The answer is therefore the complement of 1000 which is --0111
In the above example when the larger number is subtracted from the smaller number the result is in complemental form.
The answer is therefore the complement of which is
1100
-0101
0111

62. Case 3: When the positive number is greater, the end around carry generated should added to get the final sum. The result directly gives the sum. (no complementing needed).Example add 1110 and -1001
Normal representation
Binary representation
Decimal representation
+ 1110
0 1110 14
-1001
1 0110 -9
+0101
(carry 1)
0 0001
0 0101 +5

63. Case 4: When both numbers are negative
Case 4: When both numbers are negative. In this case there will be an end around carry since the numbers are added in 1’s complement form. The carry should be added to the sum.The result will be in 1’s complement form
Normal representation
Binary representation
Decimal representation
-0011
1 1000 -3
-1011
1 0100
-11
-1110
with carry 1 1
1 0001
The final sum is the complement of the above ie
-14

64. Addition and Subtraction using 2’s complement
The addition of two positive numbers using 2’s complement representation is the same as the one described in 1’s complement method .If the numbers are opposite in sign, the method is different.
If the larger number is positive, add the 2’s complement of the smaller negative number to the larger number and ignore the carry
Decimal
Binary
2’s complement form +9
1001
0 1001 -8
-1000
1 1000 +1
0001
1 0001 ↓
Ignore the carry
The answer is

65. If the larger number is negative take its 2’s complement and add to the smaller positive number. It is seen that there is no carry. The sum is in 2’s complement form. To get the answer, take the 2’s complement and change the sign
Decimal
Binary
2’s complement method
+17
10001
-18
-10010 -1
-00001
11111
The sum is negative and is the 2’s complement of the answer. Taking the
2’s complement of the above number

66. FRACTIONAL NUMBERS 1.To convert a fractional decimal number in to binary repeatedly multiply the number by two. 2.The integer part of the product of the first multiplication gives the MSB. 3.The fractional part is again multiplied by two and the integer part of the product gives the next bit. 4.The process is repeated till the product is zero. The last bit retained is the LSB

67. Convert 0.812510 in to binary. The operation is tabulated as follows
Fractional number
Product
Fractional part
Carry
0.8125x2
1.6250
0.6250
1 MSB
0.6250x2
1.2500
0.2500 1
0.2500x2
0.5000
0.5000x2
1.0000
0.0000
The binary equivalent of (0.8125)10 is

68. Convert in to decimal =1x(1/2)+1x(1/4)+0x(1/8)+1x(1/16) =(1/2)+(1/4)+0+(1/16) = =0.8215

69. Convert 0.312510 in to binary Fractional number Product
Fractional part
Carry
0.3125x2
0.6250
MSB
0.6250x2
1.2500
0.2500 1
0.2500x2
0.5000
.05000x2
1.0000
0.0000
The required binary number

70. Binary Coded Decimal
Representation of decimal numbers using binary digits is defined as the binary coded decimal or BCD
It is a weighted code
The successive digits from right to left have specified weights
The required number can be obtained by adding the products of the weights with the corresponding binary digit

71. Example of BCD: 8421 code This is a 4 bit BCD code
The binary weights of the four bits are represented as 2³,2²,2¹, and 2
In 8421 code 16 combinations can be represented.
Only 10 combinations are used to represent the decimal digits.
The six combinations 1010,1011,1100,1101,1110,1111 are illegal

72. A decimal number can be expressed in BCD by simply replacing the digits with the corresponding 4 bit code
Decimal
8421 code 5
0101 12
38.5