The document discusses binary number systems and conversions between binary, decimal, hexadecimal, and octal numbers. It explains binary operations like AND, OR, XOR, and provides truth tables for different logic gate combinations. The next topics will be Boolean logic and software basics like the differences between system and application software.

2.  During that lecture we learnt about the various types of
computers with respect to their size, capability,
applications (FIVE TYPES)
 And its five essential components and various
subsystem
 Bus interface unit
 Port
 Modem

3. 1. To become familiar with number system used by the
microprocessors - binary numbers
2. To become able to perform decimal-to-binary
conversions
3. To understand the NOT, AND, OR and XOR logic
operations – the fundamental operations that are
available in all microprocessors

4. BINARY
(BASE 2)
numbers

5. DECIMAL
(BASE 10)
numbers

6. 0 1 2 3 4
5 6 7 8 9

8.  Octal
 base = 8
 8 symbols (0,1,2,3,4,5,6,7)
 Hexadecimal
 base = 16
 16 symbols (0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F)

9. The right-most is the least significant digit
4202 = 2x100 + 0x101 + 2x102 + 4x103
The left-most is the most significant digit

10. 1
4202 = 2x100 + 0x101 + 2x102 + 4x103
1’s multiplier

11. 10
4202 = 2x100 + 0x101 + 2x102 + 4x103
10’s multiplier

12. 100
4202 = 2x100 + 0x101 + 2x102 + 4x103
100’s multiplier

13. 1000
4202 = 2x100 + 0x101 + 2x102 + 4x103
1000’s multiplier

14. The right-most is the least significant digit
10011= 1x20 + 1x21 + 0x22 + 0x23 + 1x24
The left-most is the most significant digit

15. 1
10011= 1x20 + 1x21 + 0x22 + 0x23 + 1x24
1’s multiplier

16. 2
10011= 1x20 + 1x21 + 0x22 + 0x23 + 1x24
2’s multiplier

17. 4
10011= 1x20 + 1x21 + 0x22 + 0x23 + 1x24
4’s multiplier

18. 8
10011= 1x20 + 1x21 + 0x22 + 0x23 + 1x24
8’s multiplier

19. 16
10011= 1x20 + 1x21 + 0x22 + 0x23 + 1x24
16’s multiplier

20. Counting
in Binary
0 10 20 30 0 1010 10100 11110
1 11 21 31 1 1011 10101 11111
2 12 22 32 10 1100 10110 100000
3 13 23 33 11 1101 10111 100001
4 14 24 34 100 1110 11000 100010
5 15 25 35 101 1111 11001 100011
6 16 26 36 110 10000 11010 100100
7
17 27 . 111 10001 11011 .
8
9 18 28 . 1000 10010 11100 .
19 29 . 1001 10011 11101 .

21. ?
Because this system is natural for digital computers
The fundamental building block of a digital computer –
the switch – possesses two natural states, ON & OFF.
It is natural to represent those states in a number
system that has only two symbols, 1 and 0, i.e. the
binary number system
In some ways, the decimal number system is natural
to us humans. Why?

22. binary digit

24.  1 Kilobyte = 1,024 bytes
 1 Megabyte = 1, 024 KB = 1, 048, 576 bytes
 1 Gigabyte = 1,024 MB = 1, 048, 576 KB
= 1, 073, 741, 824 bytes.

25. Decimal Binary
conversion

26. 2 75 remainder
2 37 1
2 18 1
2 9 0
2 4 1
2 2 0
2 1 0
0 1
1001011

27. 1001011 = 1x20 + 1x21 + 0x22 + 1x23 +
0x24 + 0x25 + 1x26
= 1 + 2 + 0 + 8 + 0 + 0 + 64
= 75

28. 2 100 remainder
2 50 0
2 25 0
2 12 1
2 6 0
2 3 0
2 1 1
0 1
1100100

29. Binary Decimal
conversion

30. 10002 = 0x20 + 0x21 + 0x22 + 1x23
=0 + 0 + 0 + 8
=810

31. 1110012
1x20 = 1
0x21 = 0
0x22 = 0
1x23 = 8
1x24 = 16
1x25 = 32
5710

32. Hexadecimal
Decimal
conversion

33. 100AH
10x160 = 10
0x161 =0
0x162 =0
1x163 = 4096
4106

34. Binary
Hexadecimal
conversion

35. 1001 10102
10 = A
9
9AH

36. Hexadecimal
Binary
conversion

37. 1 0 0 AH
1010
0000
0000 = 0001000000001010
0001 or
1000000001010

38. Decimal
Octal
conversion

39. 1 1st quotient
8 15
8
7 1st remainder
0 2nd quotient
8 1
0
1 2nd remainder
178

40. 33 1st quotient
8 264
264
0 1st remainder
4 2nd quotient
8 33
32
1 2nd remainder
Check if the last quotient
(MSD) is still divisible by 8.
If not consider the MSD as
the leftmost value then
append the remainder(start
with the last remainder).
4108

41. Octal
Decimal
conversion

42. 1248
4x80 = 4
2x81 = 16
1x82 = 64
84

43. That finishes our first topic - introduction to
binary numbers and their conversion to and
from decimal, hexadecimal and octal
numbers.
Our next topic is …

46. 0 1
Off On
Low High
False True

47. Name Example Symbolically
NOT y = NOT(x) x´
AND z = x AND y x·y
OR z = x OR y x+y
XOR z = x XOR y xy

48. ?

49. x y = x´
0
1

50. x y = x´
0 1
1 0

51. x y z=x·y
0 0
0 1
1 0
1 1

52. x y z=x·y
0 0 0
0 1 0
1 0 0
1 1 1

53. x y z=x+y
0 0
0 1
1 0
1 1

54. x y z=x+y
0 0 0
0 1 1
1 0 1
1 1 1

55. x y z=xy
0 0
0 1
1 0
1 1

56. x y z=xy
0 0 0
0 1 1
1 0 1
1 1 0

57. z = (x + y)´
z = y · (x + y)
z = (y · x)  w
STRATEGY: Divide & Conquer

58. x y x + y z = (x + y)´
0 0 0 1
0 1 1 0
1 0 1 0
1 1 1 0

59. x y x + y z = y · (x + y)
0 0 0 0
0 1 1 1
1 0 1 0
1 1 1 1

60. x y w y · x z = (y · x)  w
0 0 0 0 0
0 0 1 0 1
0 1 0 0 0
0 1 1 0 1
1 0 0 0 0
1 0 1 0 1
1 1 0 1 1
1 1 1 1 0

61. n
2
n = number of input variables

62. A. Convert the following into base 2, 8 and H:
i. The last five digits of your cellphone number
ii. 256
B. x, y & z are Boolean variables. Determine the truth
tables for the following combinations:
i. (x · y) + y
ii. (x  y)´ + w
Whole sheet of yellow paper. Deadline: July 12 (Monday 12nn).
Show your solution.

63. 1. About the binary number system, and how it differs
from the decimal system
2. Positional notation for representing binary and decimal
numbers
3. A process (or algorithm) which can be used to convert
decimal numbers to binary numbers
4. Basic logic operations for Boolean variables, i.e. NOT,
OR, AND, XOR, NOR, NAND, XNOR
5. Construction of truth tables (How many rows?)

64. Next lecture will be the ?
The focus of the one after that, the 10th lecture,
however, will be on software. During that lecture
we will try:
 To understand the role of software in computing
 To become able to differentiate between system and
application software