ECE 238L Boolean Algebra - Part I

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Transcription:

ECE 238L Boolean Algebra - Part I August 29, 2008 Typeset by FoilTEX

Understand basic Boolean Algebra Boolean Algebra Objectives Relate Boolean Algebra to Logic Networks Prove Laws using Truth Tables Understand and Use First 11 Theorems Apply Boolean Algebra to: Simplifying Expressions Multiplying Out Expressions Factoring Expressions Typeset by FoilTEX 1

A New Kind of Algebra Regular Boolean Algebra Algebra Values Numbers Zero (0) Integers One (1) Real Numbers Operators Complex Numbers + / AND,,, Log, ln, etc. OR, +,, Complement Typeset by FoilTEX 2

Complement Operation Also known as invert or not. x x 0 1 1 0 This is a truth-table. It gives inputoutput mapping by simply enumerating all possible input combinations and the associated outputs Typeset by FoilTEX 3

Logical AND Operation denotes AND Output is true iff all inputs are true A B Q = A B 0 0 0 0 1 0 1 0 0 1 1 1 Typeset by FoilTEX 4

Logical OR Operation + denotes OR Output is true if any inputs are true A B Q = A + B 0 0 0 0 1 1 1 0 1 1 1 1 Typeset by FoilTEX 5

Truth Tables A truth table provides a complete enumeration of the nputs and the corresponding output for a function. A B F 0 0 1 0 1 1 1 0 0 1 1 1 If there n inputs, there will be 2 n rows in the table. Unlike with regular algebra, full enumeration is possible (and useful) in Boolean Algebra. Typeset by FoilTEX 6

Boolean Expressions Boolean expressions are made up of variables and constants combined by AND, OR and NOT. Examples: 1, A, A B, C + D, AB, A(B + C), AB + C A B is the same as AB ( is omitted when obvious) Parentheses are used like in regular algebra for grouping. A literal is each instance of a variable or constant. This expression has 4 variables and 10 literals: a bd + bcd + ac + a d Typeset by FoilTEX 7

Boolean Expressions Each Boolean expression can be specified by a truth table which lists all possible combinations of the values of all variables in the expression. F = A + BC A B C A BC F=A +BC 0 0 0 1 0 1 0 0 1 1 0 1 0 1 0 1 0 1 0 1 1 1 1 1 1 0 0 0 0 0 1 0 1 0 0 0 1 1 0 0 0 0 1 1 1 0 1 1 Typeset by FoilTEX 8

Boolean Expressions From Truth Tables Each 1 in the output of a truth table specifies one term in the corresponding boolean expression. The expression can be read off by inspection... A B C F 0 0 0 0 0 0 1 0 0 1 0 1 0 1 1 0 1 0 0 0 1 0 1 0 1 1 0 0 1 1 1 1 F is true when: A is false and B is true and C is false OR A is true and B is true and C is true F = A BC + ABC Typeset by FoilTEX 9

Another Example A B C F 0 0 0 0 0 0 1 1 0 1 0 1 0 1 1 0 1 0 0 1 1 0 1 0 1 1 0 0 1 1 1 1 F =? Typeset by FoilTEX 10

Another Example A B C F 0 0 0 0 0 0 1 1 0 1 0 1 0 1 1 0 1 0 0 1 1 0 1 0 1 1 0 0 1 1 1 1 F = A B C + A BC + AB C + ABC Typeset by FoilTEX 11

Yet Another Example A B F 0 0 1 0 1 1 1 0 1 1 1 1 F = A B + A B + AB + AB = 1 F is always true. May be multiple expressions for any given truth table. Typeset by FoilTEX 12

Converting Boolean Functions to Truth Tables F = AB + BC A B C F 0 0 0 0 0 0 1 0 0 1 0 0 BC { 0 1 1 1 1 0 0 0 1 0 1 0 1 1 0 1 BC { 1 1 1 1 } AB Typeset by FoilTEX 13

A B F= A B 0 0 0 0 1 0 1 0 0 1 1 1 Some Basic Boolean Theorems From these... A B F= A 0 = 0 0 0 0 0 0 0 1 0 0 1 0 0 A B F= A 1 = A 0 1 0 0 1 0 1 1 1 1 1 1 A B F= A + B 0 0 0 0 1 1 1 0 1 1 1 1 We can derive these... A B F= A + 0 = A 0 0 0 0 0 0 1 0 1 1 0 1 A B F= A + 1 = 1 0 1 1 0 1 1 1 1 1 1 1 1 Typeset by FoilTEX 14

Proof Using Truth Tables Truth Tables can be used to prove that two Boolean expressions are equal. If the two expressions have the same values for all possible combinations of variables, they are equal. XY + Y = X + Y X Y Y XY XY + Y X + Y 0 0 1 0 0 0 0 1 0 0 1 1 1 0 1 1 1 1 1 1 0 0 1 1 These two expessions are equal. Typeset by FoilTEX 15

Basic Boolean Algebra Theorems Here are the first five Boolean Algebra theorems we will study and use: X + 0 = X X + 1 = 1 X + X = X (X ) = X X + X = 1 X 1 = X X 0 = 0 X X = X X X = 0 Typeset by FoilTEX 16

Basic Boolean Algebra Theorems While these laws don t seem very exciting, they can be very useful in simplifying Boolean expressions: Simplify: (MN } + M N) {{ P + P } X }{{ + 1} 1 + 1 Typeset by FoilTEX 17

Boolean Algebra Theorems Commutative Laws X Y = Y X X + Y = Y + X Associative Laws (X Y) Z = X ( Y Z) = X Y Z ( X + Y ) + Z = X + ( Y + Z ) = X + Y + Z Just like regular algebra Typeset by FoilTEX 18

Prove with a truth table: Again, like algebra Distributive Law X(Y+Z) = XY + XZ X Y Z Y+Z X(Y+Z) XY XZ XY + XZ 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 1 0 1 0 0 0 0 0 1 1 1 0 0 0 0 1 0 0 0 0 0 0 0 1 0 1 1 1 0 1 1 1 1 0 1 1 0 0 1 1 1 1 1 1 1 1 1 Typeset by FoilTEX 19

Other Distributive Law Proof: X + Y Z = (X + Y )(X + Z) (X + Y )(X + Z) = X(X + Z) + Y (X + Z) = XX + XZ + Y X + Y Z = X + XZ + XY + Y Z F actor Out X = X 1 + XZ + XY + Y Z = X(1 + Z + Y ) + Y Z = X 1 + Y Z = X + Y Z NOT like regular algebra! Typeset by FoilTEX 20

Simplification Theorems 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 These are useful for simplifying Boolean Expressions. The trick is to find X and Y. (A + B + CD)(B + A + CD) (A + CD + B)(A + CD + B ) A + CD Using the rule at the top right. Typeset by FoilTEX 21

Multiplying Out All terms are products of single variables only (no parentheses) A B C + D E + F G H A B + C D + E A B + C ( D + E ) Yes Yes No Multiplied out = sum-of-products form (SOP) Typeset by FoilTEX 22

Multiplying Out - Example ( A + B )( A + C)( C + D ) ( A + B C )( C + D ) A C + A D + B C C + B C D A C + A D + B C + B C D A C + A D + BC Use ( X + Y )( X + Z ) = X + Y Z Multiply Out Use X X = X Use X + X Y = X Using the theorems may be simpler than brute force. But brute force does work... Typeset by FoilTEX 23

Factoring Final form is products only All sum terms are single variables only (A + B + C )(D + E) Yes (A + B)(C + D ) E F Yes (A + B +C )(D+E) + H No (A + BC)(D + E) No This is called product-of-sums form or POS. Typeset by FoilTEX 24

Factoring - Example AB + CD (AB + C)(AB + D) Use X + YZ = (X + Y)(X + Z) Use X + YZ = (X + Y)(X + Z) again (A + C)(B + C)(AB + D) (A + C)(B + C)(A + D)(B + D) And again Typeset by FoilTEX 25

POS vs SOP Any expression can be written either way Can convert from one to another using theorems Sometimes SOP looks simpler AB + CD = ( A + C )( B + C )( A + D )( B + D ) Sometimes POS looks simpler (A + B)(C + D) = BD + AD + BC + AC SOP will be most commonly used in this class but learn both. Typeset by FoilTEX 26

Duality in Boolean Algebra If an equality is true Its dual will be true as well To form a dual: AND OR Invert constant 0 s or 1s Do NOT invert variables This will help you to remember the rules. Because these are true X + 0 = X X + 1 = 1 X + X = X X + X = 1 These are also true X 1 = X X 0 = 0 X X = X X X = 0 Typeset by FoilTEX 27