ECE 331 – Digital System Design Boolean Algebra (Lecture #3) 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.

Fall 2010ECE Digital System Design2 Boolean Algebra George Boole developed an algebraic description for processes involving logical thought and reasoning.  Became known as Boolean Algebra Claude Shannon later demonstrated that Boolean Algebra could be used to describe switching circuits.  Switching circuits are circuits built from devices that switch between two states (e.g. 0 and 1).  Switching Algebra is a special case of Boolean Algebra in which all variables take on just two distinct values Boolean Algebra is a powerful tool for analyzing and designing logic circuits.

Fall 2010ECE Digital System Design3 Boolean Algebra and Expressions A logic function can be represented by a  Boolean expression (i.e. equation)  Truth table  Circuit diagram A Boolean expression is composed of logic operations and literals  where the logic operations are represented by logic gates in the circuit diagram  and the literals are represented by the inputs to the logic gates.

Fall 2010ECE Digital System Design4 Boolean Algebra and Expressions Boolean algebra can be used to manipulate or simplify Boolean expressions. Why is this useful?

Fall 2010ECE Digital System Design5 Boolean Algebra and Expressions Manipulating a Boolean expression (using Boolean algebra) results in an alternate expression that is functionally equivalent to the original. The circuit diagram corresponding to the new expression may be  Easier to build (due to the logic gates required) than the circuit diagram corresponding to the original expression.  More cost effective than the circuit diagram corresponding to the original expression.

Fall 2010ECE Digital System Design6 Boolean Algebra and Expressions Simplifying a Boolean expression (using Boolean algebra) results in an expression with fewer logic operations and/or fewer literals than the original.  The two expressions are functionally equivalent. The circuit diagram corresponding to the new expression will be  More cost effective than the circuit diagram corresponding to the original expression.

Fall 2010ECE Digital System Design7 Basic Laws and Theorems

Fall 2010ECE Digital System Design8 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

Fall 2010ECE Digital System Design9 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

Fall 2010ECE Digital System Design10 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)

Fall 2010ECE Digital System Design11 Distributive Law Using a truth table, it is easy to show that the distributive law is valid: In addition to the ordinary distributive law, a second distributive law is valid for Boolean algebra, but not for ordinary algebra: Proof of the second distributive law follows:

Fall 2010ECE Digital System Design12 Multiplying Out The two distributive laws are used to multiply out an expression to obtain a Sum-of-Products (SOP) form. An expression is said to be in SOP form when all products are the products of single variables. A SOP expression is realized using one or more AND gates feeding a single OR gate.

Fall 2010ECE Digital System Design13 Multiplying Out Examples of Boolean expression in SOP: F 1 (A,B,C) = A'BC + ABC' + A'B'C' + AC F 2 (A,B,C) = ABC + A'B'C' + B'C Examples of Boolean expressions not in SOP: F 3 (A,B,C) = A'.(B+C) + A.B' + A.B.C

Fall 2010ECE Digital System Design14 Factoring The two distributive laws can be used to factor an expression to obtain a Product-of-Sums (POS) form. An expression is said to be in POS form when all sums are the sums of single variables. A POS expression is realized using one or more OR gates feeding a single AND gate.

Fall 2010ECE Digital System Design15 Factoring Examples of Boolean expression in POS: F 1 (A,B,C) = (A'+B+C).(A+B'+C').(A+B+C).(A'+B') F 2 (A,B,C) = (A+B+C').(A'+B'+C).(B'+C') Examples of Boolean expressions not in POS: F 3 (A,B,C) = (A'.B+C).(A'+B').(A+B'+C')

Fall 2010ECE Digital System Design16 Simplification Theorems The following theorems are useful in simplifying Boolean expressions:

Fall 2010ECE Digital System Design17 Simplifying a Logic Circuit F = A(A’ + B) By Theorem (2-14), (X + Y’)Y = XY, the expression F simplifies to AB. Which circuit is cheaper?

Fall 2010ECE Digital System Design18 Boolean Algebra Example: Using Boolean Algebra, simplify the following Boolean Expression: F = A'.B.C + A'

Fall 2010ECE Digital System Design19 Boolean Algebra Example: Using Boolean Algebra, simplify the following Boolean Expression: F = (A + B'C + D + EF)(A + B'C + (D + EF)')

Fall 2010ECE Digital System Design20 Boolean Algebra Exercise: Using Boolean Algebra, simplify the following Boolean Expression: F = (AB + C)(B'D + C'E') + (AB + C)'

Fall 2010ECE Digital System Design21 Boolean Algebra Exercise: Using Boolean Algebra, simplify the following Boolean expression. F(A,B,C) = A'.B.C + A.B'.C + A.B.C

Fall 2010ECE Digital System Design22 DeMorgan's Laws Can be stated as follows:  The complement of the product (AND) is the sum (OR) of the complements. (X.Y)' = X' + Y'  The complement of the sum (OR) is the product (AND) of the complements. (X + Y)' = X'. Y' Easily generalized to n variables. Can be proven using a Truth table

Fall 2010ECE Digital System Design23 Proving DeMorgan's Law (X. Y)' = X' + Y'

Fall 2010ECE Digital System Design24 x 1 x 2 x 1 x 2 x 1 x 2 x 1 x 2 x 1 x 2 x 1 x 2 x 1 x 2 x 1 x 2 += (a) x 1 x 2 + x 1 x 2 = (b) DeMorgan's Laws

Fall 2010ECE Digital System Design25 DeMorgan's Laws Example: Using DeMorgan's Laws, find the complement of the following Boolean Expression: F = (A' + B).C'

Fall 2010ECE Digital System Design26 DeMorgan's Laws Exercise: Using DeMorgan's Laws, find the complement of the following Boolean Expression: F = (AB' + C).D' + E'

Fall 2010ECE Digital System Design27 Duality Given an expression, the dual is formed by replacing AND with OR, OR with AND, 0 with 1, and 1 with 0. Variables and complements are left unchanged. The dual of AND is OR and the dual of OR is AND: (XYZ...) D = X + Y + Z +... (X + Y + Z +...) D = XYZ...

Fall 2010ECE Digital System Design28 Duality The dual of an expression may be found by complementing the entire expression and then complementing each of the individual variables. For example: (AB' + C)' = (AB')'.C' = (A' + B).C' Therfore, (AB' + C) D = (A + B').C complement expression complement individual variables

Fall 2010ECE Digital System Design29 Boolean Algebra Exercise: For the following Boolean expression F(A,B,C) = ABC + A'BC + AB'C + ABC' 1. Using Boolean algebra, simplify the Boolean expression. 2. Derive the Truth table for the simplified expression. 3. Confirm that the Truth table is also valid for the original expression. 4. Draw the circuit diagram for the “cheaper” expression.

Fall 2010ECE Digital System Design30 Questions?