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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.

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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

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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

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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)

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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

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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

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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)

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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)

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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)

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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)

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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))

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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

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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.

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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.

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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

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

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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)

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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.

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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).

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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.

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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).

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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

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Spring 2011ECE 301 - Digital Electronics23 Questions?

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