1. 9/15/09 - L5 Boolean AlgebraCopyright 2009 - Joanne DeGroat, ECE, OSU1 Boolean Algebra

2. 9/15/09 - L5 Boolean Algebra Copyright 2009 - 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

3. 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 2009 - Joanne DeGroat, ECE, OSU3

4. 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 2009 - Joanne DeGroat, ECE, OSU4

5. 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 2009 - Joanne DeGroat, ECE, OSU5

6. 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 2009 - Joanne DeGroat, ECE, OSU6

7. 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 2009 - Joanne DeGroat, ECE, OSU7

8. 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 2009 - Joanne DeGroat, ECE, OSU8

9. Basic Identities (2)  7. X + X’ = 1  9. (X’)’ = X  8. X · X’ = 0 9/15/09 - L5 Boolean Algebra Copyright 2009 - Joanne DeGroat, ECE, OSU9

10. 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 2009 - Joanne DeGroat, ECE, OSU10

11. 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 2009 - Joanne DeGroat, ECE, OSU11

12. 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 2009 - Joanne DeGroat, ECE, OSU12

13. Affect on implementation  F = X’YZ + X’YZ’ + XZ  Reduces to F = X’Y + XZ 9/15/09 - L5 Boolean Algebra Copyright 2009 - Joanne DeGroat, ECE, OSU13

14. Other examples  Examples from the text 1. X + XY = X·1 + XY = X(1+Y) = X·1 = X  Use 2 14 3 2 2. XY+XY’ = X(Y + Y’) = X·1 = X  Use 14 7 2 3. X+X’Y = (X+X’)(X+Y) = 1· (X+Y) = X+Y  Use 15 7 2 9/15/09 - L5 Boolean Algebra Copyright 2009 - Joanne DeGroat, ECE, OSU14

15. Further Examples  Examples from the text 4. X· (X+Y)=X·X+X·Y= X+XY=X(1+Y)=X·1=X  Use 14 6 14 3 2 5. (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 2009 - Joanne DeGroat, ECE, OSU15

16. 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 2009 - Joanne DeGroat, ECE, OSU16

17. 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 2009 - Joanne DeGroat, ECE, OSU17

18. 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 2009 - Joanne DeGroat, ECE, OSU18

19. 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 2009 - Joanne DeGroat, ECE, OSU19

20. Getting XILINX software  Go to www.xilinx.comwww.xilinx.com  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 2009 - Joanne DeGroat, ECE, OSU20

21. 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 2009 - Joanne DeGroat, ECE, OSU21