1. ECE 301 – Digital Electronics Boolean Algebra and Standard Forms of Boolean Expressions (Lecture #4) The slides included herein were taken from the materials accompanying Fundamentals of Logic Design, 6 th Edition, by Roth and Kinney, and were used with permission from Cengage Learning.

2. Spring 2011ECE 301 - Digital Electronics2 Basic Laws and Theorems Operations with 0 and 1: 1. X + 0 = X1D. X 1 = X 2. X + 1 = 12D. X 0 = 0 Idempotent laws: 3. X + X = X3D. X X = X Involution law: 4. (X')' = X Laws of complementarity: 5. X + X' = 15D. X X' = 0

3. Spring 2011ECE 301 - Digital Electronics3 Basic Laws and Theorems Commutative laws: 6. X + Y = Y + X 6D. XY = YX Associative laws: 7. (X + Y) + Z = X + (Y + Z) 7D. (XY)Z = X(YZ) = XYZ = X + Y + Z Distributive laws: 8. X(Y + Z) = XY + XZ 8D. X + YZ = (X + Y)(X + Z) Simplification theorems: 9. XY + XY' = X 9D. (X + Y)(X + Y') = X 10. X + XY = X 10D. X(X + Y) = X 11. (X + Y')Y = XY 11D. XY' + Y = X + Y

4. Spring 2011ECE 301 - Digital Electronics4 Basic Laws and Theorems DeMorgan's laws: 12. (X + Y + Z +...)' = X'Y'Z'...12D. (XYZ...)' = X' + Y' + Z' +... Duality: 13. (X + Y + Z +...) D = XYZ...13D. (XYZ...) D = X + Y + Z +... Theorem for multiplying out and factoring: 14. (X + Y)(X' + Z) = XZ + X'Y14D. XY + X'Z = (X + Z)(X' + Y) Consensus theorem: 15. XY + YZ + X'Z = XY + X'Z 15D. (X + Y)(Y + Z)(X' + Z) = (X + Y)(X' + Z)

5. Spring 2011ECE 301 - Digital Electronics5 Simplification Theorems: Example #1 Use the simplification theorems to simplify the following Boolean expression: F = ABC' + AB'C' + A'BC' Simplification Theorems (9 – 11): X.Y + X.Y' = X (X+Y).(X+Y') = X X + X.Y = XX.(X+Y) = X (X+Y').Y = X.YX.Y' + Y = X+Y

6. Spring 2011ECE 301 - Digital Electronics6 Simplification Theorems: Example #2 Use the simplification theorems to simplify the following Boolean expression: F = (A'+B'+C').(A+B'+C').(B'+C) Simplification Theorems (9 – 11): X.Y + X.Y' = X (X+Y).(X+Y') = X X + X.Y = XX.(X+Y) = X (X+Y').Y = X.YX.Y' + Y = X+Y

7. Spring 2011ECE 301 - Digital Electronics7 Simplification Theorems: Example #3 Use the simplification theorems to simplify the following Boolean expression: F = AB'CD'E + ACD + ACF'GH' +ABCD'E +ACDE' + E'H' Simplification Theorems (9 – 11): X.Y + X.Y' = X (X+Y).(X+Y') = X X + X.Y = XX.(X+Y) = X (X+Y').Y = X.YX.Y' + Y = X+Y (See Programmed Exercise 3.4 on page 75)

8. Spring 2011ECE 301 - Digital Electronics8 Consensus Theorem: Example #1 Use the consensus theorem to simplify the following Boolean expression: F = ABC + BCD + A'CD + B'C'D' Consensus Theorem: (15)X.Y + Y.Z + X'.Z = X.Y + X'.Z (15D)(X+Y).(Y+Z).(X'+Z) = (X+Y).(X'+Z)

9. Spring 2011ECE 301 - Digital Electronics9 Consensus Theorem: Example #2 Use the consensus theorem to simplify the following Boolean expression: F = (A+C+D')(A+B'+D)(B+C+D)(A+B'+C) Consensus Theorem: (15)X.Y + Y.Z + X'.Z = X.Y + X'.Z (15D)(X+Y).(Y+Z).(X'+Z) = (X+Y).(X'+Z)

10. Spring 2011ECE 301 - Digital Electronics10 Consensus Theorem: Example #3 Use the consensus theorem to simplify the following Boolean expression: F = AC' + AB'D + A'B'C + A'CD' + B'C'D' Consensus Theorem: (15)X.Y + Y.Z + X'.Z = X.Y + X'.Z (15D)(X+Y).(Y+Z).(X'+Z) = (X+Y).(X'+Z) (See Programmed Exercise 3.5 on page 77)

11. Spring 2011ECE 301 - Digital Electronics11 DeMorgan's Law: Example DeMorgan's Law: (12)(X + Y + Z + … )' = X'.Y'.Z'... (12D)(X.Y.Z… )' = X' +Y' + Z' … Find the complement of the following Boolean expression using DeMorgan's law: F = (A + BC').((A'C)' + (D' + E))

12. Spring 2011ECE 301 - Digital Electronics12 Simplifying Boolean Expressions Boolean algebra can be used in several ways to simplify a Boolean expression:  Combine terms  Eliminate redundant or consensus terms  Eliminate redundant literals  Add redundant terms to be combined with or allow the elimination of other terms

13. Spring 2011ECE 301 - Digital Electronics13 Importance of Boolean Algebra Boolean algebra is used to simplify Boolean expressions. Simpler expressions leads to simpler logic circuits.  Reduces cost  Reduces area requirements  Reduces power consumption The objective of the digital circuit designer is to design and realize optimal digital circuits.  Thus, Boolean algebra is an important tool to the digital circuit designer.

14. Spring 2011ECE 301 - Digital Electronics14 Problem with Boolean Algebra In general, there is no easy way to determine when a Boolean expression has been simplified to a minimum number of terms or a minimum number of literals. Karnaugh Maps provide a better mechanism for the simplification of Boolean expressions.

15. Spring 2011ECE 301 - Digital Electronics15 Circuit Design: Example For the following Boolean expression: F(A,B,C) = A.B.C + A'.B.C + A.B'.C + A.B.C' 1. Draw the circuit diagram 2. Simplify using Boolean algebra 3. Draw the simplified circuit diagram

16. Spring 2011ECE 301 - Digital Electronics16 Standard Forms of Boolean Expressions

17. Spring 2011ECE 301 - Digital Electronics17 Standard Forms There are two standard forms in which all Boolean expressions can be written: 1. Sum of Products (SOP) 2. Product of Sums (POS)

18. Spring 2011ECE 301 - Digital Electronics18 Sum of Products (SOP) Product Term  Logical product = AND operation  A product term is the ANDing of literals  Examples: A.B, A'.B.C, A.C', B.C'.D', A.B.C.D “Sum of”  Logical sum = OR operation  The sum of products is the ORing of product terms.

19. Spring 2011ECE 301 - Digital Electronics19 Sum of Products (SOP) The distributive laws are used to multiply out a general Boolean expression to obtain the sum of products (SOP) form. The distributive laws are also used to convert a Boolean expression in POS form to one in SOP form. A SOP expression is realized using a set of AND gates (one for each product term) driving a single OR gate (for the sum).

20. Spring 2011ECE 301 - Digital Electronics20 Product of Sums (POS) Sum Term  Logical sum = OR operation  A sum term is the ORing of literals  Examples: A+B, A'+B+C, A+C', B+C'+D' “Product of”  Logical product = AND operation  The product of sums is the ANDing of sum terms.

21. Spring 2011ECE 301 - Digital Electronics21 Product of Sums (POS) The distributive laws are used to factor a general Boolean expression to obtain the product of sums (POS) form. The distributive laws are also used to convert a Boolean expression in SOP form to one in POS form. A POS expression is realized using a set of OR gates (one for each sum term) driving a single AND gate (for the product).

22. Spring 2011ECE 301 - Digital Electronics22 SOP and POS: Examples For each of the following Boolean expressions, identify whether it is in SOP or POS form: 1. F(A,B,C) = (A+B).(A'+B'+C').(B+C') 2. F(A,B,C) = A.B.C + B'.C' + A.C' + A'.B.C' 3. F(A,B,C) = A + B.C + B'.C' + A'.B'.C 4. F(A,B,C) = (A'+B'+C).(B+C').(A+C').(B') 5. F(A,B,C) = A.B.C + A'.(B+C) + (A+C').B 6. F(A,B,C) = A + B + C

23. Spring 2011ECE 301 - Digital Electronics23 Questions?