Number system, 1's & 2's Complement, Boolean Algebra

2. Common Number Systems

3. Conversion Among Bases
• The possibilities:
Decimal
Hexadecimal

4. 2510 = 110012 = 318 = 1916

5. Decimal to Decimal
Decimal Octal
Hexadecimal
Binary

6. N=푎푚푟푚−1 + 푎푚−1푟푚−2 + ….. +푎2푟1 + 푎1푟0.+
푏1푟−1+푏2푟−2+ ……+ 푏푛푟−푛
r=10 (decimal)

7. 12510 => 5 x 100 = 5
2 x 101 = 20
1 x 102 = 100
125
Base
Weight

8. Binary to Decimal
Decimal Octal
Hexadecimal
Binary

9. 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

10. Example
1010112 => (1 x 20)+(1 x 21)+(0 x 22)+(1 x 23)+(0 x 24)
+(1 x 25)
= 1+2+0+8+0+32
= 4310
Bit “0”

11. Decimal to Binary
Decimal Octal
Hexadecimal
Binary

12. 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.

13. Example
12510 = ?2
2 125
2 6 2 1
2 3 1 0
2 1 5 1
2 7 1
2 3 1
1 1
12510 = 11111012

14. Example:
(0.3125)10
0.3125 x 2=0.625 0 MSB
0.625 x 2=1.25 1
0.25 x 2=0.50 0
0.50 x 2=1.00 1 LSB
Ans.:- (0101)2

15. Binary Arithmetic:
Binary Addition:
• Two 1-bit values
A B A + B
0 0 0
0 1 1
1 0 1
1 1 10
“two”

16. Binary Addition
• Two n-bit values
• Add individual bits
• Propagate carries
• E.g.,
1 1
10101 21
+ 11001 + 25
101110 46

17. Binary Subtraction
Rules for binary subtraction are:
0 – 0 = 0
1 – 0 = 1
1 – 1 = 0
0 – 1 = 1 , with 1 borrowed from the next column
Ex 1: 1100101 – 100111
Verify the result in decimal system.
Solution:
(10011)2 = 1*24 + 1*21 + 1*20 = (19)10

18. Complements of Binary Numbers
• 1’s complements
• 2’s complements
18

19. 1’s and 2’s Complement:
 The 1’s complement of a binary number is simply obtained by
replacing every 1 by 0 , and every 0 by 1.
 The 2’s complement of a binary number can be obtained in
two ways:
 By adding 1 to the 1’s complement.
 Start the binary number from right. Leave the binary digits
unchanged until the first 1 appear, then replace every 1 by
0 , and every 0 by 1.

20. Complements of Binary Numbers
• 1’s complement
• Change all 1s to 0s and all 0s to 1s
1 0 1 0 1 0 1 0
0 1 0 1 0 1 0 1

21. Ex 1: Obtain the two’s complement of the binary number 1011010.110
First solution
Second solution

22. Ex 2: Calculate the following binary Subtraction: 1101.101 – 11011.11 ,
then verify the result in decimal System.
Solution

23. Basic Logic Gates

24. NOT Gate -- Inverter
X Y
0
1
1
0
NOT
X Y
Y
= ~X

25. NOT
X ~X ~~X = X
X ~X ~~X
0 1 0
1 0 1

26. AND Gate
AND
X
Y
Z
Z = X & Y
X Y Z
0 0 0
0 1 0
1 0 0
1 1 1

27. OR Gate
OR
X
Y
Z
Z = X | Y
X Y Z
0 0 0
0 1 1
1 0 1
1 1 1

28. NAND Gate
NAND
X
Y
Z
X Y Z
0 0 1
0 1 1
1 0 1
1 1 0
Z = ~(X & Y)
nand(Z,X,Y)

29. NAND Gate
NOT-AND
X
Y
Z
W = X & Y
Z = ~W = ~(X & Y)
X Y W Z
0 0 0 1
0 1 0 1
1 0 0 1
1 1 1 0
W

30. NOR Gate
NOR
X
Y
Z
X Y Z
0 0 1
0 1 0
1 0 0
1 1 0
Z = ~(X | Y)
nor(Z,X,Y)

31. NOR Gate
NOT-OR
X
Y
W = X | Y
Z = ~W = ~(X | Y)
X Y W Z
0 0 0 1
0 1 1 0
1 0 1 0
1 1 1 0
Z
W

32. NAND Gate
X
Y
X
Y
Z Z
=
Z = ~(X & Y) Z = ~X | ~Y
X Y W Z
0 0 0 1
0 1 0 1
1 0 0 1
1 1 1 0
X Y ~X ~Y Z
0 0 1 1 1
0 1 1 0 1
1 0 0 1 1
1 1 0 0 0

33. De Morgan’s Theorem-1
~(X & Y) = ~X | ~Y
• NOT all variables
• Change & to | and | to &
• NOT the result

34. NOR Gate
X
Y
Z
Z = ~(X | Y)
X Y Z
0 0 1
0 1 0
1 0 0
1 1 0
X
Y
Z
Z = ~X & ~Y
X Y ~X ~Y Z
0 0 1 1 1
0 1 1 0 0
1 0 0 1 0
1 1 0 0 0

35. De Morgan’s Theorem-2
~(X | Y) = ~X & ~Y
• NOT all variables
• Change & to | and | to &
• NOT the result

36. De Morgan’s Theorem
• NOT all variables
• Change & to | and | to &
• NOT the result
• --------------------------------------------
• ~X | ~Y = ~(~~X & ~~Y) = ~(X & Y)
• ~(X & Y) = ~~(~X | ~Y) = ~X | ~Y
• ~X & !Y = ~(~~X | ~~Y) = ~(X | Y)
• ~(X | Y) = ~~(~X & ~Y) = ~X & ~Y

37. Exclusive-OR Gate
X Y Z
XOR
X
Y
Z 0 0 0
0 1 1
1 0 1
1 1 0
Z = X ^ Y
xor(Z,X,Y)

38. Exclusive-NOR Gate
X Y Z
XNOR
X
Y
Z 0 0 1
0 1 0
1 0 0
1 1 1
Z = ~(X ^ Y)
Z = X ~^ Y
xnor(Z,X,Y)

39. Boolean Algebra
• Boolean Algebra is a set of rules and theorems by
which logical operations can be expressed
mathematically.

40. Boolean Logic Operations
• AND function
Y=A.B
• OR function
Y=A+B
• NOT function
Y=Ᾱ

41. Arithmetic of Boolean Algebra
• Boolean Addition | Boolean Multiplication
0+0=0 | 0.0=0
0+1=1 | 0.1=0
1+0=1 | 1.0=0
1+1=1 | 1.1=1
|

42. Basic Laws
1. Commutative Property
A+B=B+A / A.B=B.A
2. Associative Property
(A+B)+C=A+(B+C) / (A.B).C=A.(B.C)
3. Distributive Property
A+BC=(A+B)(A+C) / A.(B+C)=A.B+A.C
4. Absorption Property
A+AB=A / A.(A+B)=A