Chapter 2: Princess Sumaya Univ. Computer Engineering Dept.
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1 hapter 2: Princess Sumaya Univ. omputer Engineering Dept.
2 Basic Definitions Binary Operators AND z = x y = x y z=1 if x=1 AND y=1 OR z = x + y z=1 if x=1 OR y=1 NOT z = x = x z=1 if x=0 Boolean Algebra Binary Variables: only 0 and 1 values Algebraic Manipulation 1 / 28
3 Boolean Algebra Postulates ommutative Law x y = y x x + y = y + x Identity Element x 1 = x x + 0 = x omplement x x = 0 x + x = 1 2 / 28
4 Boolean Algebra Theorems Duality The dual of a Boolean algebraic expression is obtained by interchanging the AND and the OR operators and replacing the 1 s by 0 s and the 0 s by 1 s. x ( y + z ) = ( x y ) + ( x z ) x + ( y z ) = ( x + y ) ( x + z ) Theorem 1 x x = x x + x = x Applied to a valid equation produces a valid equation Theorem 2 x 0 = 0 x + 1 = 1 3 / 28
5 Boolean Algebra Theorems Theorem 3: Involution ( x ) = x ( x ) = x Theorem 4: Associative & Distributive ( x y ) z = x ( y z ) ( x + y ) + z = x + ( y + z ) x ( y + z ) = ( x y ) + ( x z ) x + ( y z ) = ( x + y ) ( x + z ) Theorem 5: DeMorgan ( x y ) = x + y ( x + y ) = x y ( x y ) = x + y ( x + y ) = x y Theorem 6: Absorption x ( x + y ) = x x + ( x y ) = x 4 / 28
6 Operator Precedence Parentheses NOT (... ) (...) x + y AND OR x + x y x [ y + z ( w + x)] ( w + x) ( w z ( w y + z ( w x [ y + z ( w + x) + x) + x) + x)] 5 / 28
7 DeMorgan s Theorem a [ b + c ( d + e )] a + [ b + c ( d + e )] a + b ( c ( d + e )) a + b ( c + ( d + e )) a + b ( c + ( d e)) a + b ( c + d e) 6 / 28
8 Boolean Functions Boolean Expression Example: Truth Table F = x + y z All possible combinations of input variables Logic ircuit x y z F x y z F / 28
9 Algebraic Manipulation Literal: A single variable within a term that may be complemented or not. Use Boolean Algebra to simplify Boolean functions to produce simpler circuits Example: Simplify to a minimum number of literals F = x + x y = x + ( x y ) = ( x + x ) ( x + y ) ( 3 Literals) Distributive law (+ over ) = ( 1 ) ( x + y ) = x + y ( 2 Literals) 8 / 28
10 9 / 28 omplement of a Function DeMorgan s Theorm Duality & Literal omplement B A F + + = B A F + + = B A F = B A F + + = B A F B A F =
11 anonical Forms Minterm Product (AND function) ontains all variables Evaluates to 1 for a specific combination Example A = 0 A B B = 0 (0) (0) (0) = = 1 A B m m m m m m m m 7 Minterm A B A B A B A B A B A B A B A B 10 / 28
12 anonical Forms Maxterm Sum (OR function) ontains all variables Evaluates to 0 for a specific combination Example A = 1 A B B = 1 (1) + (1) + (1) = = 0 A B M M M M M M M M 7 Maxterm A + B + A + B + A + B + A + B + A + B + A + B + A + B + A + B + 11 / 28
13 anonical Forms Truth Table to Boolean Function A B F F = AB + AB + AB + AB / 28
14 anonical Forms Sum of Minterms F = AB + AB + AB + AB F = m + m + m + F m7 = (1,4,5,7) Product of Maxterms F = AB + AB + AB + F = AB + AB + AB + AB AB A B F F = AB AB AB AB F = ( A + B + )( A + B + )( A + B + )( A + B + ) F = M 0 M 2 M 3 M 6 F = (0,2,3,6) F / 28
15 Standard Forms Sum of Products (SOP) AB( + ) = AB(1) F = AB + AB + AB + AB = AB A( B + B) = A B ( A + A) = B F = B ( A + A) + AB( + ) + A( B + B) F = B + AB + A 14 / 28
16 Standard Forms Product of Sums (POS) A B( + ) F = AB + AB + AB + AB B ( A + A) A ( B + B) F = A( B + B) + AB( + ) + B( A + A) F = A + AB + B F = ( A + )( A + B)( B + ) 15 / 28
17 Two-Level Implementations Sum of Products (SOP) F = B + AB + A Product of Sums (POS) F = ( A + )( A + B)( B + ) B A B A A A B B F F 16 / 28
18 Logic Operators AND x y x y x y AND NAND (Not AND) x y NAND x y x y / 28
19 Logic Operators OR x y x + y x y OR NOR (Not OR) x y x + y x y NOR / 28
20 Logic Operators XOR (Exclusive-OR) x y x y x y + x y x y XOR XNOR (Exclusive-NOR) (Equivalence) x y x y x y x y + x y x y XNOR / 28
21 Logic Operators NOT (Inverter) x NOT x x Buffer x Buffer x x / 28
22 Multiple Input Gates 21 / 28
23 DeMorgan s Theorem on Gates AND Gate F = x y F = (x y) F = x + y OR Gate F = x + y F = (x + y) F = x y hange the Shape and bubble all lines 22 / 28
24 Homework Mano hapter / 28
25 Homework Mano 2-4 Reduce the following Boolean expressions to the indicated number of literals: (a) A + AB + A (b) (x y + z) + z + xy + wz (c) A B (D + D) + B (A + A D) (d) (A + ) (A + ) (A + B + D) to three literals to three literals to one literal to four literals 2-5 Find the complement of F = x + yz; then show that FF = 0 and F + F = 1 24 / 28
26 Homework 2-6 Find the complement of the following expressions: (a) xy + x y (c) (x + y + z) (x + z ) (x + y) (b) (AB + )D + E 2-8 List the truth table of the function: F = xy + xy + y z 2-9 Logical operations can be performed on strings of bits by considering each pair of corresponding bits separately (this is called bitwise operation). Given two 8-bit strings A = and B = , evaluate the 8-bit result after the following logical operations: (a) AND, (b) OR, (c) XOR, (d) NOT A, (e) NOT B. 25 / 28
27 Homework 2-10 Draw the logic diagrams for the following Boolean expressions: (a) Y = A B + B (A + ) (b) Y = B + A (c) Y = A + D (d) Y = (A + B) ( + D) 2-12 Simplify the Boolean function T 1 and T 2 to a minimum number of literals. A B T 1 T / 28
28 Homework 2-15 Given the Boolean function F = xy z + x y z + w xy + wx y + wxy (a) Obtain the truth table of the function. (b) Draw the logic diagram using the original Boolean expression. (c) Simplify the function to a minimum number of literals using Boolean algebra. (d) Obtain the truth table of the function from the simplified expression and show that it is the same as the one in part (a) (e) Draw the logic diagram from the simplified expression and compare the total number of gates with the diagram of part (b). 27 / 28
29 Homework 2-18 onvert the following to the other canonical form: (a) F (x, y, z) = (1, 3, 7) (b) F (A, B,, D) = (0, 1, 2, 3, 4, 6, 12) 2-19 onvert the following expressions into sum of products and product of sums: (a) (AB + ) (B + D) (b) x + x (x + y ) (y + z ) 28 / 28
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