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**ECE 331 – Digital System Design**

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

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**Basic Laws and Theorems**

Operations with 0 and 1: 1. X + 0 = X 1D. X • 1 = X 2. X + 1 = 1 2D. X • 0 = 0 Idempotent laws: 3. X + X = X 3D. X • X = X Involution law: 4. (X')' = X Laws of complementarity: 5. X + X' = 1 5D. X • X' = 0 Spring 2011 ECE Digital System Design

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**Basic Laws and Theorems**

Commutative laws: 6. X + Y = Y + X D. 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 D. (X + Y)(X + Y') = X 10. X + XY = X D. X(X + Y) = X 11. (X + Y')Y = XY D. XY' + Y = X + Y Spring 2011 ECE Digital System Design

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**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 D. (XYZ...)D = X + Y + Z +... Theorem for multiplying out and factoring: 14. (X + Y)(X' + Z) = XZ + X'Y 14D. 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) Spring 2011 ECE Digital System Design

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**ECE 331 - Digital System Design**

Duality (13) The dual of a Boolean expression can be written by Replacing AND with OR, and OR with AND Replacing 0 with 1, and 1 with 0 Leaving literals unchanged See the Boolean laws and theorems, previously discussed, for examples of Boolean expressions and their duals. Spring 2011 ECE Digital System Design

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**Distributive Law: Example #1**

Distributive law (8): X.(Y + Z) = X.Y + X.Z Use the distributive law to multiply out the following Boolean expression: F = (A+B).(C+D).(E+F) Spring 2011 ECE Digital System Design

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**Distributive Law: Example #2**

Distributive law (8D): X + Y.Z = (X+Y).(X+Z) Use the distributive law to factor the following Boolean expression: F = A.B + C.D Spring 2011 ECE Digital System Design

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**Simplification Theorems: Example #1**

X.Y + X.Y' = X (X+Y).(X+Y') = X X + X.Y = X X.(X+Y) = X (X+Y').Y = X.Y X.Y' + Y = X+Y Use the simplification theorems to simplify the following Boolean expression: F = ABC' + AB'C' + A'BC' Spring 2011 ECE Digital System Design

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**Simplification Theorems: Example #2**

X.Y + X.Y' = X (X+Y).(X+Y') = X X + X.Y = X X.(X+Y) = X (X+Y').Y = X.Y X.Y' + Y = X+Y Use the simplification theorems to simplify the following Boolean expression: F = (A'+B'+C').(A+B'+C').(B'+C) Spring 2011 ECE Digital System Design

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**Simplification Theorems: Example #3**

X.Y + X.Y' = X (X+Y).(X+Y') = X X + X.Y = X X.(X+Y) = X (X+Y').Y = X.Y X.Y' + Y = X+Y Use the simplification theorems to simplify the following Boolean expression: F = AB'CD'E + ACD + ACF'GH' +ABCD'E +ACDE' + E'H' (See Programmed Exercise 3.4 on page 75) Spring 2011 ECE Digital System Design

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**Consensus Theorem: Example #1**

(15) X.Y + Y.Z + X'.Z = X.Y + X'.Z (15D) (X+Y).(Y+Z).(X'+Z) = (X+Y).(X'+Z) Use the consensus theorem to simplify the following Boolean expression: F = ABC + BCD + A'CD + B'C'D' Spring 2011 ECE Digital System Design

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**Consensus Theorem: Example #2**

(15) X.Y + Y.Z + X'.Z = X.Y + X'.Z (15D) (X+Y).(Y+Z).(X'+Z) = (X+Y).(X'+Z) Use the consensus theorem to simplify the following Boolean expression: F = (A+C+D')(A+B'+D)(B+C+D)(A+B'+C) Spring 2011 ECE Digital System Design

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**Consensus Theorem: Example #3**

(15) X.Y + Y.Z + X'.Z = X.Y + X'.Z (15D) (X+Y).(Y+Z).(X'+Z) = (X+Y).(X'+Z) Use the consensus theorem to simplify the following Boolean expression: F = AC' + AB'D + A'B'C + A'CD' + B'C'D' (See Programmed Exercise 3.5 on page 77) Spring 2011 ECE Digital System Design

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**ECE 331 - Digital System Design**

DeMorgan's Law DeMorgan's Law: (12) (X + Y + Z + … )' = X'.Y'.Z'... (12D) (X.Y.Z… )' = X' +Y' + Z' … Prove (using a truth table): (X+Y)' = X'.Y' X Y X + Y (X + Y)' X' Y' X'.Y' 1 Spring 2011 ECE Digital System Design

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**ECE 331 - Digital System Design**

DeMorgan's Law Graphical representation of DeMorgan's Law x y (X.Y)' X' + Y' (X+Y)' X'.Y' Spring 2011 ECE Digital System Design

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**DeMorgan's Law: Example**

(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)').((AD)' + C.(B' + D)) Spring 2011 ECE Digital System Design

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**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 Spring 2011 ECE Digital System Design

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**Equivalency of Boolean Expressions**

Two Boolean expressions are equivalent iff both expressions evaluate to the same value for all combinations of the variables in the expressions. The equivalency can be proven using A Truth table Boolean algebra theorems to manipulate one expression until it is identical to the other. Boolean algebra theorems to reduce both expressions independently to the same expression. Spring 2011 ECE Digital System Design

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**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. Spring 2011 ECE Digital System Design

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**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. Spring 2011 ECE Digital System Design

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**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 Spring 2011 ECE Digital System Design

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**Standard Forms of Boolean Expressions**

Spring 2011 ECE Digital System Design

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**ECE 331 - Digital System Design**

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) Spring 2011 ECE Digital System Design

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**ECE 331 - Digital System Design**

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. Spring 2011 ECE Digital System Design

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**ECE 331 - Digital System Design**

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). Spring 2011 ECE Digital System Design

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**ECE 331 - Digital System Design**

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. Spring 2011 ECE Digital System Design

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**ECE 331 - Digital System Design**

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). Spring 2011 ECE Digital System Design

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**ECE 331 - Digital System Design**

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 Spring 2011 ECE Digital System Design

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**ECE 331 - Digital System Design**

Questions? Spring 2011 ECE Digital System Design

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