### SlideShare for iOS

by Linkedin Corporation

FREE - On the App Store

- Total Views
- 517
- Views on SlideShare
- 517
- Embed Views

- Likes
- 0
- Downloads
- 15
- Comments
- 0

No embeds

Uploaded via SlideShare as Microsoft PowerPoint

© All Rights Reserved

- 1. Boolean Expressions and Boolean Functions Let B = {0, 1}. The variable x is called a Boolean variable if it assumes values only from B. A function Bn, the set {(x1, x2, ….xn)|xi ε B, 1 ≤ i ≤ n}, to B is called a Boolean function of degree n x y F(x, y) 1 1 0 The values of Boolean function. 1 0 1 0 1 0 0 0 0 The Boolean expressions in the variables x1, x2, …,xn are defined recursively as follows: 0, 1, x1, x2, ….., xn
- 2. Example: Find the values of the Boolean function represented by F ( x, y, z ) = xy + z x 1 1 1 1 0 0 0 0 y 1 1 0 0 1 1 0 0 z 1 0 1 0 1 0 1 0 xy 1 1 0 0 0 0 0 0 z’ 0 1 0 1 0 1 0 1 xy + z’ 1 1 0 1 0 1 0 1
- 3. The Boolean functions F and G of n variables are equal if and only if F(b1, b2, …, bn) = G(b1, b2, …,bn) whenever b1, b2, …, bn belong to B. Two different Boolean expressions that represent the same function are called equivalent. ex: xy, xy + 0, and (xy)1 The complement of the Boolean function F is the function F , where F ( x1 , , xn ) = F ( x1 , xn )
- 4. The Boolean sum F + G and the Boolean product FG are defined by (F + G)(x1,…., xn) = F(x1,…, xn) + G(x1,…, xn), (FG)(x1,…, xn) = F(x1,…, xn)G(x1,…, xn) The Boolean functions of degree 2 x y F1 F2 F3 F4 F5 F6 F7 F8 F9 F10 F11 F12 F13 F14 F15 F16 1 1 0 0 1 0 1 0 1 1 1 1 1 1 1 0 1 1 0 1 1 1 0 0 1 0 1 1 1 0 1 0 1 0 0 1 1 0 0 0 0 1 1 1 0 1 1 0 0 1 0 1 0 1 0 0 0 0 1 1 0 0 1 0 0 0 0 1 0 0 0 0 To determine the number of different Boolean functions of degree n, we use the formula: 22n
- 5. Identities of Boolean Algebra Identity Name x=x Law of double complement x+x = x x.x = x x +0 = x x.1 = x x +1 =1 x.0 = 0 x+ y = y+ x xy = yx Idempotent Laws Identity Laws Dominance Laws Commutative Laws
- 6. Identities of Boolean Algebra Identity x + ( y + z) = ( x + y) + z x( yz ) = ( xy ) z Name Associative Laws x + yz = ( x + y )( x + z ) x( y + z ) = xy + xz Distributive Laws ( xy ) = x + y ( x + y ) = xy De Morgan’s Laws
- 7. Example: Show that the distributive law x(y + z) = xy + xz is valid x y z y+z xy xz 1 1 1 1 0 0 0 0 1 1 0 0 1 1 0 0 1 0 1 0 1 0 1 0 1 1 1 0 1 1 1 0 1 1 0 0 0 0 0 0 1 0 1 0 0 0 0 0 x(y+z) xy+xz 1 1 1 0 0 0 0 0 1 1 1 0 0 0 0 0
- 8. Duality The dual of a Boolean expression is obtained by interchanging Boolean sums and Boolean products and interchanging 0s and 1s. Find the duals of Answer: x( y + 0 ) and x .1 + ( y + z ) . x + ( y.1) and ( x + 0)( yz )
- 9. Note: The dual of the Boolean function F, denoted by Fd, represented by a Boolean expression is the function represented by the dual of this expression. This dual function does not depend on the particular Boolean expression used to represent F. An identity between functions represented by Boolean expressions remains valid when the duals of both sides of the identity are taken. This result, called the duality principle, is useful for obtaining new identities.
- 10. A Boolean algebra is a set B with two binary operations v and ʌ, elements 0 and 1, and unary operation ¬ such that the following properties hold for all x, y, and z in B: x ∨ 0 = x x ∧1 = x Identity Laws x ∨ x =1 x ∧ x = 0 Domination Laws
- 11. x ∨ y = y ∨ x x ∧ y = y ∧ x Commutative Laws ( x ∨ y) ∨ z = x ∨ ( y ∨ z) ( x ∧ y) ∧ z = x ∧ ( y ∧ z) Associative Laws x ∨ ( y ∧ z) = ( x ∨ y) ∧ ( x ∨ z) x ∧ ( y ∨ z) = ( x ∧ y) ∨ ( x ∧ z) Distributive Laws
- 12. Representing Boolean Functions Find Boolean expressions that functions F(x, y, z) and G(x, y, z) x 1 1 1 1 0 0 0 0 y 1 1 0 0 1 1 0 0 z 1 0 1 0 1 0 1 0 F 0 0 1 0 0 0 0 0 G 0 1 0 0 0 1 0 0 represent the To represent F, the value is 1 when x = z = 1 and y = 0 or xy’z. To represent G, the value is 1 when x = y = 1 and z = 0 or when x = z = 0 and y =1. We can form an expression with these values by taking the Boolean sum of two different Boolean products.
- 13. x 1 1 1 1 0 0 0 0 y 1 1 0 0 1 1 0 0 z 1 0 1 0 1 0 1 0 F 0 0 1 0 0 0 0 0 G 0 1 0 0 0 1 0 0 xyz’ x’yz’ 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 xyz’ + x’yz’ 0 1 0 0 0 1 0 0 xyz’ has value 1, iff x = y = 1 and z = 0 x’yz’ has value 1, iff x = y = 1 and z = 0 xyz’ + x’yz’ has value 1, iff x = y = 1 and z = 0 or x = z = 0 and y = 1
- 14. A literal is a Boolean variable or its complement. A minterm of the Boolean variables x1, x2, …, xn is a Boolean product y1, y2, …, yn, where yi = xi or yi = x’i. Hence, a minterm is a product of n literals, with one literal for each variable. Example: Find a minterm that equals 1 if x 1 = x3 = 0 and x2 = x4 = x5 = 1, and equals 0 otherwise. x1 x2 x3 x4 x5 The minterm showed has the correct set of values.
- 15. Sum-of-products expansion or the disjunctive normal form of the Boolean function is the sum of minterms that represents a function. A Boolean sum of minterms has the value 1 when exactly one of the minterms in the sum has the value 1. It is also possible to find a Boolean expression that represents a Boolean function by taking a Boolean product of Boolean sums. The resulting expansion is called the conjunctive normal form or product-ofsums expansion of the function. These expansions can be found from sum-of-products expansion by taking their duals.
- 16. Find the sum-of-products expansion for the function F ( x, y , z ) = ( x + y ) z x 1 1 1 1 0 0 0 0 Answer: y 1 1 0 0 1 1 0 0 z 1 0 1 0 1 0 1 0 x+y 1 1 1 1 1 1 0 0 z’ 0 1 0 1 0 1 0 1 (x + y)z’ 0 1 0 1 0 1 0 0 F ( x, y, z ) = xyz + xyz + x yz
- 17. Logic Gates Gates are the basic elements of circuits. Combinatorial circuits or gating networks are circuits have no memory capabilities. These circuits give output that depends only on the input, and not on the current state of the circuit. Types of Elements Inverter which accepts the value of one Boolean variable as input and produces the complement of this value as output. OR Gate, the inputs are the values of two or more Boolean variables. The output is the Boolean sum of their values. AND Gate, the inputs are the values of two or more Boolean variables. The output is the Boolean product of their values.
- 18. Basic Types of Gates x x x y x+y x y xy inverter OR gate AND gate
- 19. Combinations of Gates Combinational circuits can be constructed using a combination of inverters, OR gates, and AND gates. Example: x y x y xy xy + x y x xy
- 20. Exercises: 1.Find the Boolean product of the Boolean variables x, y, and z, or their complements, that has the value 1 if and only if a. x =y = 0, z = 1 b. x = z = 0, y = 1 c. x = 0, y = z = 1 d. x = y = z = 0 2. Find the sum-of-products expansions of the following Boolean functions. a ) F ( x, y , z ) = x + y + z b ) F ( x, y , z ) = ( x + z ) y c) F ( x, y, z ) = x + yz d ) F ( x, y, z ) = xy + z
- 21. Exercises: Construct circuits that produce the following outputs: 1)( x + y ) x 2) x ( y + z ) 3)( x + y + z )( x yz )

Full NameComment goes here.