Digital Fundamentals Logic Design PDF

CS221: Logic Design Instructors: Dr. Ahmed Shalaby http://bu.edu.eg/staff/ahmedshalaby14# Dr. Fatma Sakr Floyd Copyright ©2006 by Pearson Education, Inc. Di...

CS221: Logic Design Instructors: Dr. Ahmed Shalaby http://bu.edu.eg/staff/ahmedshalaby14# Dr. Fatma Sakr Floyd Copyright ©2006 by Pearson Education, Inc. Digital Fundamentals, 9/e Upper Saddle River, New Jersey 07458 All rights reserved. Slide 1  Digital Fundamentals CHAPTER 4 Boolean Algebra and Logic Simplification Floyd Copyright ©2006 by Pearson Education, Inc. Digital Fundamentals, 9/e Upper Saddle River, New Jersey 07458 All rights reserved. Slide 2  Boolean Operations and Expressions In Boolean algebra, a variable is a symbol used to represent an action, a condition, or data. A single variable can only have a value of 1 or 0. The complement represents the inverse of a variable and is indicated with an overbar. Thus, the complement of A is A. A literal is a variable or its complement. Floyd Copyright ©2006 by Pearson Education, Inc. Digital Fundamentals, 9/e Upper Saddle River, New Jersey 07458 All rights reserved. Slide 3  Boolean Operations and Expressions Boolean Addition Addition is equivalent to the OR operation. The sum term is 1 if one or more if the literals are 1. The sum term is zero only if each literal is 0. Inputs Output A X A B X 0 0 0 B 0 1 1 1 0 1 1 1 1 Determine the values of A, B, and C that make the sum term of the expression A + B + C = 0? Each literal must = 0; therefore A = 1, B = 0 and C = 1. Floyd Copyright ©2006 by Pearson Education, Inc. Digital Fundamentals, 9/e Upper Saddle River, New Jersey 07458 All rights reserved. Slide 4  Boolean Operations and Expressions Boolean Multiplication In Boolean algebra, multiplication is equivalent to the AND operation. The product of literals forms a product term. The product term will be 1 only if all of the literals are 1. Inputs Output A B X A X 0 0 0 B 0 1 0 1 0 0 1 1 1 What are the values of the A, B and C if the product term of A.B.C = 1? Each literal must = 1; therefore A = 1, B = 0 and C = 0. Floyd Copyright ©2006 by Pearson Education, Inc. Digital Fundamentals, 9/e Upper Saddle River, New Jersey 07458 All rights reserved. Slide 5  Laws and Rules of Boolean Algebra Floyd Copyright ©2006 by Pearson Education, Inc. Digital Fundamentals, 9/e Upper Saddle River, New Jersey 07458 All rights reserved. Slide 6  Laws Boolean Algebra Commutative Laws Associative Laws Distributive Law Floyd Copyright ©2006 by Pearson Education, Inc. Digital Fundamentals, 9/e Upper Saddle River, New Jersey 07458 All rights reserved. Slide 7  Laws Boolean Algebra Commutative Laws The commutative laws are applied to addition and multiplication. For addition, the commutative law states In terms of the result, the order in which A+B=B+A variables are ORed makes no difference. For multiplication, the commutative law states In terms of the result, the order in which AB = BA variables are ANDed makes no difference. Floyd Copyright ©2006 by Pearson Education, Inc. Digital Fundamentals, 9/e Upper Saddle River, New Jersey 07458 All rights reserved. Slide 8  Laws Boolean Algebra Associative Laws The associative laws are also applied to addition and multiplication. For addition, the associative law states When ORing more than two variables, the result is the same regardless of the grouping of the variables. A + (B +C) = (A + B) + C Floyd Copyright ©2006 by Pearson Education, Inc. Digital Fundamentals, 9/e Upper Saddle River, New Jersey 07458 All rights reserved. Slide 9  Laws Boolean Algebra Associative Laws For multiplication, the associative law states When ANDing more than two variables, the result is the same regardless of the grouping of the variables. A(BC) = (AB)C Floyd Copyright ©2006 by Pearson Education, Inc. Digital Fundamentals, 9/e Upper Saddle River, New Jersey 07458 All rights reserved. Slide 10  Laws Boolean Algebra Distributive Law The distributive law is the factoring law. A common variable can be factored from an expression just as in ordinary algebra. That is A(B+ C) = AB + AC Floyd Copyright ©2006 by Pearson Education, Inc. Digital Fundamentals, 9/e Upper Saddle River, New Jersey 07458 All rights reserved. Slide 11  Rules of Boolean Algebra 1. A + 0 = A 7. A. A = A 2. A + 1 = 1 8. A. A = 0 = 3. A. 0 = 0 9. A = A 4. A. 1 = A 10. A + AB = A 5. A + A = A 11. A + AB = A + B 6. A + A = 1 12. (A + B)(A + C) = A + BC Floyd Copyright ©2006 by Pearson Education, Inc. Digital Fundamentals, 9/e Upper Saddle River, New Jersey 07458 All rights reserved. Slide 12  Rules of Boolean Algebra 1. A + 0 = A OR Truth Table 2. A + 1 = 1 Floyd Copyright ©2006 by Pearson Education, Inc. Digital Fundamentals, 9/e Upper Saddle River, New Jersey 07458 All rights reserved. Slide 13  Rules of Boolean Algebra 3. A. 0 = 0 AND Truth Table 4. A. 1 = A Floyd Copyright ©2006 by Pearson Education, Inc. Digital Fundamentals, 9/e Upper Saddle River, New Jersey 07458 All rights reserved. Slide 14  Rules of Boolean Algebra 5. A + A = A OR Truth Table 6. A + A = 1 Floyd Copyright ©2006 by Pearson Education, Inc. Digital Fundamentals, 9/e Upper Saddle River, New Jersey 07458 All rights reserved. Slide 15  Rules of Boolean Algebra 7. A. A = A AND Truth Table 8. A. A = 0 Floyd Copyright ©2006 by Pearson Education, Inc. Digital Fundamentals, 9/e Upper Saddle River, New Jersey 07458 All rights reserved. Slide 16  Rules of Boolean Algebra = 9. A = A Floyd Copyright ©2006 by Pearson Education, Inc. Digital Fundamentals, 9/e Upper Saddle River, New Jersey 07458 All rights reserved. Slide 17  Rules of Boolean Algebra 10. A + AB = A AND Truth Table OR Truth Table Floyd Copyright ©2006 by Pearson Education, Inc. Digital Fundamentals, 9/e Upper Saddle River, New Jersey 07458 All rights reserved. Slide 18  Rules of Boolean Algebra 11. A + AB = A + B AND Truth Table OR Truth Table Floyd Copyright ©2006 by Pearson Education, Inc. Digital Fundamentals, 9/e Upper Saddle River, New Jersey 07458 All rights reserved. Slide 19  Rules of Boolean Algebra 12. (A + B)(A + C) = A + BC AND Truth Table OR Truth Table Floyd Copyright ©2006 by Pearson Education, Inc. Digital Fundamentals, 9/e Upper Saddle River, New Jersey 07458 All rights reserved. Slide 20  DeMorgan’s Theorem DeMorgan’s Theorem DeMorgan’s 1st Theorem The complement of a product of variables is equal to the sum of the complemented variables. AB = A + B Applying DeMorgan’s first theorem to gates: A A AB A+B Inputs Output B B A B AB A + B NAND Negative-OR 0 0 1 1 0 1 1 1 1 0 1 1 1 1 0 0 Floyd Copyright ©2006 by Pearson Education, Inc. Digital Fundamentals, 9/e Upper Saddle River, New Jersey 07458 All rights reserved. Slide 21  DeMorgan’s Theorem DeMorgan’s Theorem DeMorgan’s 2nd Theorem The complement of a sum of variables is equal to the product of the complemented variables. A+B=A.B Applying DeMorgan’s second theorem to gates: A A Inputs Output A+B AB B B A B A + B AB NOR Negative-AND 0 0 1 1 0 1 0 0 1 0 0 0 1 1 0 0 Floyd Copyright ©2006 by Pearson Education, Inc. Digital Fundamentals, 9/e Upper Saddle River, New Jersey 07458 All rights reserved. Slide 22  DeMorgan’s Theorem DeMorgan’s Theorem Theorem 1 XY = X + Y Theorem 2 X + Y = XY Apply DeMorgan’s theorem to remove the overbar covering both terms from the expression X = C + D. To apply DeMorgan’s theorem to the expression, you can break the overbar covering both terms and change the sign between the terms. This results in X=C =. D. Deleting the double bar gives X = C. D. Floyd Copyright ©2006 by Pearson Education, Inc. Digital Fundamentals, 9/e Upper Saddle River, New Jersey 07458 All rights reserved. Slide 23  Boolean Analysis of Logic Circuits Floyd Copyright ©2006 by Pearson Education, Inc. Digital Fundamentals, 9/e Upper Saddle River, New Jersey 07458 All rights reserved. Slide 24  Boolean Analysis of Logic Circuits Boolean Analysis of Logic Circuits Combinational logic circuits can be analyzed by writing the expression for each gate and combining the expressions according to the rules for Boolean algebra. Apply Boolean algebra to derive the expression for X. Write the expression for each gate: A (A + B ) C (A + B ) B C X = C (A + B )+ D D Applying DeMorgan’s theorem and the distribution law: X = C (A + B) + D = A B C + D Floyd Copyright ©2006 by Pearson Education, Inc. Digital Fundamentals, 9/e Upper Saddle River, New Jersey 07458 All rights reserved. Slide 25  Boolean Analysis of Logic Circuits Boolean Analysis of Logic Circuits Use Multisim to generate the truth table for the circuit in the previous example. Set up the circuit using the Logic Converter as shown. (Note that the logic converter has no “real-world” counterpart.) Double-click the Logic Converter top open it. Then click on the conversion bar on the right side to see the truth table for the circuit (see next slide). Floyd Copyright ©2006 by Pearson Education, Inc. Digital Fundamentals, 9/e Upper Saddle River, New Jersey 07458 All rights reserved. Slide 26  Boolean Analysis of Logic Circuits Boolean Analysis of Logic Circuits The simplified logic expression can be viewed by clicking Simplified expression Floyd Copyright ©2006 by Pearson Education, Inc. Digital Fundamentals, 9/e Upper Saddle River, New Jersey 07458 All rights reserved. Slide 27  Boolean Analysis of Logic Circuits SOP and POS forms Boolean expressions can be written in the sum-of-products form (SOP) or in the product-of-sums form (POS). These forms can simplify the implementation of combinational logic, particularly with PLDs. In both forms, an overbar cannot extend over more than one variable. An expression is in SOP form when two or more product terms are summed as in the following examples: ABC+AB ABC+CD CD+E An expression is in POS form when two or more sum terms are multiplied as in the following examples: (A + B)(A + C) (A + B + C)(B + D) (A + B)C Floyd Copyright ©2006 by Pearson Education, Inc. Digital Fundamentals, 9/e Upper Saddle River, New Jersey 07458 All rights reserved. Slide 28  Boolean Analysis of Logic Circuits SOP Standard form In SOP standard form, every variable in the domain must appear in each term. This form is useful for constructing truth tables or for implementing logic in PLDs. You can expand a nonstandard term to standard form by multiplying the term by a term consisting of the sum of the missing variable and its complement. Convert X = A B + A B C to standard form. The first term does not include the variable C. Therefore, multiply it by the (C + C), which = 1: X = A B (C + C) + A B C =ABC+ABC+ABC Floyd Copyright ©2006 by Pearson Education, Inc. Digital Fundamentals, 9/e Upper Saddle River, New Jersey 07458 All rights reserved. Slide 29  Boolean Analysis of Logic Circuits POS Standard form In POS standard form, every variable in the domain must appear in each sum term of the expression. You can expand a nonstandard POS expression to standard form by adding the product of the missing variable and its complement and applying rule 12, which states that (A + B)(A + C) = A + BC. Convert X = (A + B)(A + B + C) to standard form. The first sum term does not include the variable C. Therefore, add C C and expand the result by rule 12. X = (A + B + C C)(A + B + C) = (A +B + C )(A + B + C)(A + B + C) Floyd Copyright ©2006 by Pearson Education, Inc. Digital Fundamentals, 9/e Upper Saddle River, New Jersey 07458 All rights reserved. Slide 30  Boolean Analysis of Logic Circuits Karnaugh maps The Karnaugh map (K-map) is a tool for simplifying combinational logic with 3 or 4 variables. For 3 variables, 8 cells are required (23). The map shown is for three variables labeled A, B, and C. Each cell represents one possible product term. Each cell differs from an adjacent cell by only one variable. 3-Variable Karnaugh Map Floyd Copyright ©2006 by Pearson Education, Inc. Digital Fundamentals, 9/e Upper Saddle River, New Jersey 07458 All rights reserved. Slide 31  Boolean Analysis of Logic Circuits Karnaugh maps Cells are usually labeled using 0’s and 1’s to represent the variable and its complement. The numbers are entered in gray code, to force adjacent cells to be different by only one variable. Ones are read as the true variable Gray and zeros are read as the code complemented variable. 3-Variable Karnaugh Map Floyd Copyright ©2006 by Pearson Education, Inc. Digital Fundamentals, 9/e Upper Saddle River, New Jersey 07458 All rights reserved. Slide 32  Boolean Analysis of Logic Circuits Karnaugh maps Floyd Copyright ©2006 by Pearson Education, Inc. Digital Fundamentals, 9/e Upper Saddle River, New Jersey 07458 All rights reserved. Slide 33  Boolean Analysis of Logic Circuits Karnaugh maps 4-Variable Karnaugh Map Floyd Copyright ©2006 by Pearson Education, Inc. Digital Fundamentals, 9/e Upper Saddle River, New Jersey 07458 All rights reserved. Slide 34  Boolean Analysis of Logic Circuits Karnaugh maps 4-Variable Example Floyd Copyright ©2006 by Pearson Education, Inc. Digital Fundamentals, 9/e Upper Saddle River, New Jersey 07458 All rights reserved. Slide 35  Boolean Analysis of Logic Circuits Karnaugh maps K-maps can simplify combinational logic by grouping cells and eliminating variables that change. Group the 1’s on the map and read the minimum logic. C 0 1 AB 1 1. Group the 1’s into two overlapping 00 B changes groups as indicated. across this 01 1 1 2. Read each group by eliminating any boundary variable that changes across a 11 boundary. 10 C changes 3. The vertical group is read AC. across this 4. The horizontal group is read AB. boundary X = AC +AB Floyd Copyright ©2006 by Pearson Education, Inc. Digital Fundamentals, 9/e Upper Saddle River, New Jersey 07458 All rights reserved. Slide 36  Boolean Analysis of Logic Circuits Karnaugh maps Group the 1’s on the map and read the minimum logic. C changes across outer boundary CD AB 00 01 11 10 1. Group the 1’s into two separate 00 1 1 groups as indicated. B changes 2. Read each group by eliminating 01 1 1 any variable that changes across a 11 boundary. 1 1 B changes 3. The upper (yellow) group is read as 10 1 1 AD. 4. The lower (green) group is read as C changes AD. X X = AD +AD Floyd Copyright ©2006 by Pearson Education, Inc. Digital Fundamentals, 9/e Upper Saddle River, New Jersey 07458 All rights reserved. Slide 37  QUIZ The minimum expression that can be read from the Karnaugh map shown is C C AB AB AB 1 1 AB 1 1 Floyd Copyright ©2006 by Pearson Education, Inc. Digital Fundamentals, 9/e Upper Saddle River, New Jersey 07458 All rights reserved. Slide 38