Boolean algebra and logic gates

1. Boolean algebra and logic gates Lecture 4 Lecturer : May Aldoayan

2. 2.4 : Basic Theorems And properties of Boolean Algebra Lecturer : May Aldoayan

3. Boolean algebraic properties • Commutative property of (OR) A + B = B + A Lecturer : May Aldoayan

4. Boolean algebraic properties • Commutative property of (AND) A . B = B . A Lecturer : May Aldoayan

5. Boolean algebraic properties • Associative property of OR A + (B + C) = (A + B) + C Lecturer : May Aldoayan

6. Boolean algebraic properties • Associative property of and A . (B . C) = (A . B) . C Lecturer : May Aldoayan

7. Boolean algebraic properties • Distributive property A(B + C) = AB + AC Lecturer : May Aldoayan

8. Boolean Functions • Boolean function described by an algebraic expression consist of binary variables the constant 0 and 1, and the logic operation symbols. • A Boolean function can represented by in a truth table. • F1 = X + y’ z Lecturer : May Aldoayan

9. Boolean Functions Gate complementation of F1 = X + y’ z F1 = X + y’ z Lecturer : May Aldoayan

10. Boolean Functions • By manipulating a Boolean expression according to the rules of Boolean algebra, it is sometimes possible to obtain a simpler expression for the same function and thus reduce the number of gates in the circuit and the number of gates in the circuit and the number of inputs to gate. • Designers are motivated to reduce the complexity and number of gates because their effort can significantly reduce the cost of a circuit. F2 = x’ y’ z + x’ y z + x y’ A schematic of an implementation of this function with logic gates is shown in Fig2.2 Lecturer : May Aldoayan

11. Boolean Functions • FIGURE 2.2 Implementation of Boolean function F2 with gates Lecturer : May Aldoayan

12. Boolean Functions • Now consider the possible Simplification of the function by applying some of the identities of Boolean algebra: F2 = x’ y’ z + x’ y z + x y’ = x’ z (y’ + y) + x y’ = x’ z + x y’ The function is reduced to only two terms and can be implemented with gates as shown in following figure: Lecturer : May Aldoayan

13. Algebraic Manipulation • When a Boolean expression is implemented with logic gates, each term requires a gate and each variable within the term designers an input to the gate. • We defined a literal to be a single variable within a term, in complemented or uncomplemented form. The function of Fig2.2(a) has three terms and eight literals, and the one in fig2.2 (b) has two terms and four literals. By reducing the number of terms, the number of literals, or both in a Boolean expression, it is often possible to obtain a simpler circuit. Lecturer : May Aldoayan

14. Algebraic Manipulation • Example 2.1 :Simplify the following Boolean functions to a minimum number of literals: • x(x’ + y) = xx’ + xy = 0 +xy = xy • x + x’y = (x + x’) (x + y) = 1 (x + y) = x + y • (x + y ) ( x + y’) = xx + xy + xy’ + yy’= x + xy + xy’ + 0 = x (1 +y +y’) = x • xy + x’z + yz =xy + x’z + yz ( x+ x’) = xy +x’z+xyz +x’yz = xy (1 +z) + x’z( 1+y) = x y + x’z • (x + y) (x’ + z)(y + z)= (x + y) (x’ + z), by duality from function 4. Lecturer : May Aldoayan

15. Complement of a Function • The complement of a function F is F’ and obtained from an interchange of 0’s for 1’s and 1’s for 0’s in the value of F. • The complement of a function may be derived algebraically through DeMorgan’s theorems, listed in Table 2.1 for two variables. • DeMorgan’s theorems can be extended to three or more variables. • The three-variables form of the first DeMrgan’s theorem as follows, from postulates and theorems listed in table2.1: Lecturer : May Aldoayan

16. Complement of a Function (A + B + C)’ = ( A + x )’ let B + C = x = A’ X’ by theorem 5(a) (DeMorgan) = A’ (B + C)’ substitute B + C = X = A’ (B’ C’) by theorem 5(a) (DeMorgan) = A’ B’ C’ by theorem 4(b) (associative) Generalized DeMorgan’s Theorem ( A + B+ C+ D + ..........+F)‘ = A‘ B‘ C‘ D‘ .....F‘ (A B C D ....F)‘ = A‘ + B‘ + C‘ + D‘ + ......+ F‘ Lecturer : May Aldoayan

17. Complement of a Function • Example 2.2: Find the complement of the Function F1 = x’yz’ + x’y’z and F2 = x(y’ z’ + y z). By applying DeMorgan’s theorems as many times as necessary, the complements are obtained as follows: F’1= (x’yz’ + x’y’z)’ = ( x + y’ + z) (x + y+ z’) F’2 = [ x(y’ z’ + y z )]’ = x’ + (y’ z’ )’ (y z)’ = x’ + (y + z) (y’ + z’) = x’ + y z’ + y’z \ Lecturer : May Aldoayan

18. Boolean rules for simplification • A + AB = A Prove Lecturer : May Aldoayan

19. Boolean rules for simplification • A + A’B = A + B Prove Lecturer : May Aldoayan

20. Boolean rules for simplification • (A + B)(A + C) = A + BC Prove Lecturer : May Aldoayan

21. DeMorgan’s Theorems (x+y)’=x’y’ Implications and alternative symbol for NOR function (xy)’=x’+y’ Implications and alternative symbol for NAND function Lecturer : May Aldoayan

22. DeMorgan’s Theorems • Expand and simplify the following : (A+(BC)’)’ Lecturer : May Aldoayan

23. DeMorgan’s Theorems • Expand and simplify the following : (AB+CD)’ Lecturer : May Aldoayan

24. Circuit simplification examples • Reduce this circuit The first step in simplification must be to write a Boolean expression for this circuit Lecturer : May Aldoayan

25. Circuit simplification examples Reduce this circuit Now that we have a Boolean expression to work with, we need to apply the rules of Boolean algebra to reduce the expression to its simplest form (simplest defined as requiring the fewest gates to implement): Lecturer : May Aldoayan

26. Circuit simplification examples Reduce this circuit Lecturer : May Aldoayan

27. Circuit simplification examples