9/15/09 - L5 Boolean AlgebraCopyright Joanne DeGroat, ECE, OSU1 Boolean Algebra

9/15/09 - L5 Boolean Algebra Copyright Joanne DeGroat, ECE, OSU2 Class 5 outline  Boolean Algebra Basic Boolean Equations Multiple Level Logic Representation Basic Identities Algebraic Manipulation Complements and Duals  Material from section 2-2 of text

History  George Boole 2 November 1815 Lincoln Lincolnshire, England – 8 December 1864 Ballintemple, Ireland Professor at Queens College, Cork, Ireland  “spring of 1847 that he put his ideas into the pamphlet called Mathematical Analysis of Logic.” from wikipedia.com 9/15/09 - L5 Boolean Algebra Copyright Joanne DeGroat, ECE, OSU3

Basic Boolean Equations  For the basic gates/functions  AND Z = A B X = C D E3 input gate Y = F G H K4 input gate  OR Z = A + B Y = F + G + H + K 4 input gate  NOT Z = A Y = (F G H K)actually 2 level logic 9/15/09 - L5 Boolean Algebra Copyright Joanne DeGroat, ECE, OSU4

2 Level Logic  Consider the following logic equation Z(A,B,C,D) = A B + C D The Z(A,B,C,D) means that the output is a function of the four variables within the (). The AB and CD are terms of the expression. This form of representing the function is an algebraic expression. For this function to be True, either both A AND B are True OR both C AND D are True. 9/15/09 - L5 Boolean Algebra Copyright Joanne DeGroat, ECE, OSU5

Truth table expression  Just like we had the truth tables for the basic functions, we can also construct truth tables for any function. 9/15/09 - L5 Boolean Algebra Copyright Joanne DeGroat, ECE, OSU6

Examples of Boolean Equations  Some examples F = AB + CD + BD’ Y = CD + A’B’ SUM = AB + A Cin + B Cin P = A 0 A 1 A 2 A 3 A 4 B 0 B 1 B 2 B 3 B 4 + … Equations can be very complex Usually desire a minimal expression 9/15/09 - L5 Boolean Algebra Copyright Joanne DeGroat, ECE, OSU7

Basic Identities of Boolean Algebra  1. X + 0 = X  3. X + 1 = 1  5. X + X = X  2. X · 1 = X  4. X · 0 = 0  6. X · X = X 9/15/09 - L5 Boolean Algebra Copyright Joanne DeGroat, ECE, OSU8

Basic Identities (2)  7. X + X’ = 1  9. (X’)’ = X  8. X · X’ = 0 9/15/09 - L5 Boolean Algebra Copyright Joanne DeGroat, ECE, OSU9

Basic Properties (Laws)  Commutative 10. X + Y = Y + X  Associative 12. X+(Y+Z)=(X+Y)+Z  Distributive 14. X(Y+Z) =XY+XZ AND distributes over OR  Commutative 11. X · Y = Y · X  Associative 13. X(YZ) = (XY)Z  Distributive 15. X+YZ=(X+Y)(X+Z) OR distributes over AND 9/15/09 - L5 Boolean Algebra Copyright Joanne DeGroat, ECE, OSU10

Basic Properties (2)  DeMorgan’s Theorem  Very important in simplifying equations 16. (X + Y)’ = X’ · Y’ 17. (XY)’ = X’ + Y’ 9/15/09 - L5 Boolean Algebra Copyright Joanne DeGroat, ECE, OSU11

Simplify, simplify  These properties (Laws and Theorems) can be used to simplify equations to their simplest form. Simplify F=X’YZ+X’YZ’+XZ 9/15/09 - L5 Boolean Algebra Copyright Joanne DeGroat, ECE, OSU12

Affect on implementation  F = X’YZ + X’YZ’ + XZ  Reduces to F = X’Y + XZ 9/15/09 - L5 Boolean Algebra Copyright Joanne DeGroat, ECE, OSU13

Other examples  Examples from the text 1. X + XY = X·1 + XY = X(1+Y) = X·1 = X  Use XY+XY’ = X(Y + Y’) = X·1 = X  Use X+X’Y = (X+X’)(X+Y) = 1· (X+Y) = X+Y  Use /15/09 - L5 Boolean Algebra Copyright Joanne DeGroat, ECE, OSU14

Further Examples  Examples from the text 4. X· (X+Y)=X·X+X·Y= X+XY=X(1+Y)=X·1=X  Use (X+Y) ·(X+Y’)=XX+XY’+XY+YY’= X+XY’+XY+0=X(1+Y’+Y)=X·1=X by a slightly different reduction 6. X(X’+Y) = XX’+XY = 0 + XY = XY 9/15/09 - L5 Boolean Algebra Copyright Joanne DeGroat, ECE, OSU15

Consensus Theorem  The Theorem gives us the relationship XY + X’Z + YZ = XY + X’Z Proof is on page 47. Note that in doing the reduction the first step is to and in a 1 to the YZ term. That 1 is in the form, (X+X’). 9/15/09 - L5 Boolean Algebra Copyright Joanne DeGroat, ECE, OSU16

Application of Consensus Theorem  Consider (page 47 of text) (A+B)(A’+C) = AA’ + AC + A’B + BC = AC + A’B + BC = AC + A’B 9/15/09 - L5 Boolean Algebra Copyright Joanne DeGroat, ECE, OSU17

Complement of a function  In real implementation sometimes the complement of a function is needed. Have F=X’YZ’+X’Y’Z 9/15/09 - L5 Boolean Algebra Copyright Joanne DeGroat, ECE, OSU18

Duals  What is meant by the dual of a function? The dual of a function is obtained by interchanging OR and AND operations and replacing 1s and 0s with 0s and 1s.  Shortcut to getting function complement Starting with the equation on the previous slide Generate the dual F=(X’+Y+Z’)(X’+Y’+Z) Complement each literal to get: F’=(X+Y’+Z)(X+Y+Z’) 9/15/09 - L5 Boolean Algebra Copyright Joanne DeGroat, ECE, OSU19

Getting XILINX software  Go to  In the upper line you have Sign in Language Documentation Downloads  Choose Downloads  On the right side of the page you will see “Logic design tools” Choose ISE WebPack TM 9/15/09 - L5 Boolean Algebra Copyright Joanne DeGroat, ECE, OSU20

Class 5 assignment  Covered section 2-2  Problems for hand in 2-7  Problems for practice 2-2a,b,c 2-6b,c,d  Reading for next class: section 2-3 9/15/09 - L5 Boolean Algebra Copyright Joanne DeGroat, ECE, OSU21