Copyright © 2004 by Miguel A. Marin Revised 2008-011 McGILL UNIVERSITY DEPARTMENT OF ELECTRICAL AND COMPUTER ENGINEERING COURSE ECSE - 323 DIGITAL SYSTEMS.

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Presentation transcript:

Copyright © 2004 by Miguel A. Marin Revised McGILL UNIVERSITY DEPARTMENT OF ELECTRICAL AND COMPUTER ENGINEERING COURSE ECSE DIGITAL SYSTEMS DESIGN

Copyright © 2004 by Miguel A. Marin Revised BOOLEAN LOGIC THEORY BOOLEAN FUNCTIONS CANONICAL FORMS, DUALITY, COMPLEMENTATION IMPLICANTS, PRIME-IMPLICANTS SIMPLIFICATION OF BOOLEAN FUNCTIONS K-MAP METHOD TABULAR METHODS PRIME IMPLICANTS DETERMINATION QUINE-McCLUSKEY METHOD SVOBODA METHOD PRIME IMPLICANTS SELECTION COVERING TABLE PETRICK FUNCTION SHANNON EXPANSION

Copyright © 2004 by Miguel A. Marin Revised BOOLEAN FUNCTIONS A BOOLEAN FUNCTION F OF N VARIABLES IS A MAPPING OF 2 N POINTS ONTO 0,1 EXAMPLE: N = 3, F = A B + !C ID A B C F Maxterms minterms

Copyright © 2004 by Miguel A. Marin Revised CANONICAL FORMS SUM-OF-PRODUCTS OR  -FORM IS OBTAINED BY SUMMING (OR’ING) ALL MINTERMS EXAMPLE: F(A,B,C) = A B + !C ID A B C F minterms F =  m ( 0,2,4,6,7) = !A !B !C + !A B !C + A !B !C + A B !C + A B C

Copyright © 2004 by Miguel A. Marin Revised CANONICAL FORMS PRODUCT-OF-SUMS OR  -FORM IS OBTAINED BY AND’ING ALL MAXTERMS EXAMPLE: F(A,B,C) = A B + !C ID A B C F MAXTERMS F =  M (1,3,5) = ( A + B +!C) (A + !B + !C) (!A + B + !C)

Copyright © 2004 by Miguel A. Marin Revised DUALITY THE DUAL F D OF A FUNTION F IS OBTAINED BY CHANGING IN F BOOLEAN CONSTANTS 0, 1 BY 1, 0 RESPECTIVELY OPERATORS “ “BY“+“ AND VICE VERSA THE POLARITY OF THE VARIABLES ARE UNCHANGED EXAMPLE 1: F = A B + !C. F D = (A + B) !C. EXAMPLE 2: A + 0 = A; (A + 0) D = (A) D ; A 1 = A; A = A THE HANTINGTON AXIOMS OF BOOLEAN ALGEBRA ARE INVARIANT UNDER DUALITY. PROBLEM: FIND ALL SELF-DUAL FUNTIONS OF N = 3 VARIABLES. HOW MANY SUCH FUNTIONS ARE THERE?

Copyright © 2004 by Miguel A. Marin Revised COMPLEMENTATION De MORGANS THEOREM THE COMPLEMENT !F OF A FUNTION F IS OBTAINED BY CHANGING IN F TRUE VARIABLES BY COMPLEMENTED VARIABLES COMPLEMENTED VARIABLES BY TRUE VARIABLES BOOLEAN CONSTANTS 0, 1 BY 1, 0 RESPECTIVELY OPERATOR “ “ BY “+“ OPERATOR “+“ BY “ “ EXAMPLE 1: F = A B + !C. !F = (!A + !B) C. PROBLEM: SHOW THAT F D (A,B,C,…,Z) = !F(!A,!B,!C,…,!Z).

Copyright © 2004 by Miguel A. Marin Revised IMPLICANTS, PRIME-IMPLICANTS LITERAL: a variable either in true or complemented form TERM: the product (AND’ing) of literals Example: Let F = A B + !C, AB and !C are terms MINTERM: a term containing all literals Example: F(A,B) = A B + !A !B; AB and !A!B are minterms IMPLICANT of F: a term,T, such that (T=1)  (F=1) Example: Let F = A B + !C; T = AB is an implicant of F because for A=1, B=1, T = 1 and also F = 1.

Copyright © 2004 by Miguel A. Marin Revised IMPLICANTS, PRIME-IMPLICANTS PRIME-IMPLICANT of F: is an implicant containing a minimum number of literals Example: Let F = A B C + A B !C. ABC is an implicant of F but it is not a prime-implicant. However, AB is an implicant and also a prime-implicant. RULE: An implicant becomes a prime-implicant when no literals can be removed by the application of the theorems of Boolean algebra. Example: Let F = ABC + AB!C = AB(C+!C) = AB. AB is a prime- implicant of F.

Copyright © 2004 by Miguel A. Marin Revised IMPLICANTS, PRIME-IMPLICANTS PROPERTIES : ANY MINTERM OF THE ∑∏ - CANONICAL FORM OF !F CANNOT BE IN THE ∑∏ - CANONICAL FORM OF F. Example: F(A,B,C) = ∑ m (0,2,4,6,7), !F = ∑ m (1,3,5) Thus, F and !F have no minterms in common. ANY TERM OF A ∑∏ - FORM (NOT NECESSARILY CANONICAL) OF !F CANNOT BE IN A ∑∏ - FORM OF F Example: Let F = A B + !C and !F = !A C + !B C; !BC cannot belong to a ∑∏ - FORM of F because it implies !F. Effectively, for B=0,C=1, !BC = 1, !F = !A = 1. !BC  !F. In other words a term cannot be a term of both, F and !F. AN IMPLICANT OF F CANNOT BE AN IMPLICANT OF !F AND VICE VERSA.

Copyright © 2004 by Miguel A. Marin Revised SIMPLIFICATION OF BOOLEAN FUNCTIONS: K-MAPS A K-map is formed by assigning points of a logic space to squares of a map, in such a way that any two adjacent squares, including the edges of the map, represent points differing only in one bit. Example: K-map of n = 2 variables ID Logic space A B A B K-map for n = 2

Copyright © 2004 by Miguel A. Marin Revised SIMPLIFICATION OF BOOLEAN FUNCTIONS: K-MAPS Example of K-map for n = 3 variables ID Logic Space A B C AB C

Copyright © 2004 by Miguel A. Marin Revised SIMPLIFICATION OF BOOLEAN FUNCTIONS: K-MAPS Example of K-map for n = 4 variables AB CD

Copyright © 2004 by Miguel A. Marin Revised SIMPLIFICATION OF BOOLEAN FUNCTIONS: K-MAPS REPRESENTATION OF BOOLEAN FUNCTIONS EXAMPLE: N = 3, F(A,B,C) = A B + !C ID A B C F AB C

Copyright © 2004 by Miguel A. Marin Revised SIMPLIFICATION OF BOOLEAN FUNCTIONS: K-MAP METHOD SUM-OF-PRODUCTS MINIMIZATION: OBJECTIVE: FIND MINIMUM NUMBER OF PRODUCTS, EACH PRODUCT CONTAINING MINIMUM NUMBER OF LITERALS. RULE: GROUP MAXIMUM NUMBER OF 2 K (K  N) ADJACENT 1’S, AND SELECT MINIMUM NUMBER OF THESE GROUPS TO FORM THE GIVEN FUNCTION EXAMPLE: F(A,B,C) = ∑ m(0,2,4,6,7) = AB + !C AB and !C are prime-implicants. THE ABOVE RULE USE THE THEOREM: A B+A !B = A(B+!B) = A EXHOSTIVELY AB C !C AB

Copyright © 2004 by Miguel A. Marin Revised SIMPLIFICATION OF BOOLEAN FUNCTIONS: K-MAP METHOD PRODUCT-OF-SUMS MINIMIZATION: OBJECTIVE: FIND MINIMUM NUMBER OF SUMS, EACH SUM CONTAINING MINIMUM NUMBER OF LITERALS. RULE: GROUP MAXIMUM NUMBER OF 2 K (K  N) ADJACENT 0’S, AND SELECT MINIMUM NUMBER OF THESE GROUPS TO FORM THE GIVEN FUNCTION EXAMPLE: F(A,B,C) = ∑ m(0,2,4,6,7) = =(A + !C)( B + !C). AB C (A + !C) (B +!C)

Copyright © 2004 by Miguel A. Marin Revised SIMPLIFICATION OF BOOLEAN FUNCTIONS: K-MAP METHOD MORE EXAMPLES FIND MINIMAL SUM-OF-PRODUCTS OF F(A,B,C,D) =  m (0,1,2,5,7,8,9,10,11,13,14). F MIN (A,B,C,D) = A !B + !C D + !A B D + !B !D + A C !D AB CD A !B !B !D FOUR CORNERS !C D A C !D !A B D

Copyright © 2004 by Miguel A. Marin Revised SIMPLIFICATION OF BOOLEAN FUNCTIONS: K-MAP METHOD MORE EXAMPLES FIND MINIMAL SUM-OF-PRODUCTS OF F(A,B,C,D) =  m (1,5,7,8,9,10,11,13,14) +  d (6,15) where  d (6,15) are the unspecified or DON’T care points, which may be assigned either 0 or 1 to help forming larger groups of adjacent 1’s F MIN (A,B,C,D) = !C D + A !B + B C AB CD x x1 1 A !B !C D B C

Copyright © 2004 by Miguel A. Marin Revised SIMPLIFICATION OF BOOLEAN FUNCTIONS:TABULAR METHODS TABULAR METHODS: PRIME IMPLICANTS DETERMINATION QUINE-McCLUSKEY METHOD SVOBODA METHOD PRIME IMPLICANTS SELECTION COVERING TABLE PETRICK FUNCTION TABULAR METHODS ARE ITENDED TO BE USED IN THEIR COMPUTERISED VERSION. THEY ARE TEDIOUS WHEN USED MANUALLY.

Copyright © 2004 by Miguel A. Marin Revised SIMPLIFICATION OF BOOLEAN FUNCTIONS:TABULAR METHODS PRIME IMPLICANTS DETERMINATION: THE QUINE-McCLUSKEY METHOD CONSISTS OF APPLYING THE THEOREM X A + X !A = X, SYSTEMATICALLY AND EXHOSTIVELY, TO ELIMINATE AS MANY LITERALS AS POSSIBLE FROM EACH MINTERM. A PRODUCT TERM IS REPRESENTED IN BINARY FORM. A MISSING LITERAL IS REPLACED BY THE SYMBOL “-”. EXAMPLE: FOR N = 4, (A B !C D) = ( ), (A B!D) = (1 1 – 0)

Copyright © 2004 by Miguel A. Marin Revised SIMPLIFICATION OF BOOLEAN FUNCTIONS:TABULAR METHODS PRIME IMPLICANTS DETERMINATION: QUINE-McCLUSKEY METHOD EXAMPLE: FIND THE PRIME-IMPLICANTS OF F(A B C) =  m(0,2,4,6,7) 1.- ORGANIZE THE MINTERMS IN GROUPS CONTAINING EQUAL NUMBER OF 1’S, ID A B C F

Copyright © 2004 by Miguel A. Marin Revised SIMPLIFICATION OF BOOLEAN FUNCTIONS:TABULAR METHODS 2.- COMPARE, IN PAIRS, EACH MINTERM IN ONE GROUP WITH EVERY MINTERM IN THE ADJACENT GROUP. IF THEIR BINARY REPERENTATION DIFFER ONLY IN ONE BIT, THEY CAN BE GROUPED. THE VARIABLE IN QUESTION GETS REPLACED BY A THE SYMBOL “- “. THE GROUPED MINTERMS ARE CHECK MARKED. WE OBTAIN, THUS, THE FIRST REDUCTION TABLE. ID A B C FGROUPS A B C  (0,2)  (0,4)  (2,6)  (4,6)  (6,7) 1 1 -

Copyright © 2004 by Miguel A. Marin Revised SIMPLIFICATION OF BOOLEAN FUNCTIONS:TABULAR METHODS 3.- REPEATE THE GROUPING PROCEDURE WITH THE GROUPED MEMBERS OF THE FIRST REDUCTION TABLE. CANDIDATES FOR GROUPING WILL BE THOSE PAIRS HAVING THE “-” SYMBOL IN THE SAME LOCATION AND DIFFERING IN ONLY ONE BIT. CHECK MARK THOSE MEMBERS GROUPED. THUS, WE OBTAIN THE SECOND REDUCTION TABLE. ID A B C FGROUPS A B CGROUPS A B C  (0,2)  (0,2,4,6)  (0,4)   (2,6)   (4,6)   (6,7) 1 1 -

Copyright © 2004 by Miguel A. Marin Revised SIMPLIFICATION OF BOOLEAN FUNCTIONS:TABULAR METHODS 4.-THE PROCESS IS ITERATED UNTIL NO FURTHER GROUPINGS CAN BE FORMED WITH THE MEMBERS OF THE REDUCTION TABLES OBTAINED. 5.-THE MEMBERS WITH NO CHECK MARKS IN ALL TABLES, INCLUDING THE MINTERM TABLE, ARE THE PRIME-IMPLICANTS OF THE FUNCTION. ID A B C FGROUPS A B CGROUPS A B C  (0,2)  (0,2,4,6)  (0,4)   (2,6)   (4,6)   (6,7) THE PRIME-IMPLICANTS ARE: (1 1 -) = A B, (- - 0) = !C

Copyright © 2004 by Miguel A. Marin Revised SIMPLIFICATION OF BOOLEAN FUNCTIONS:TABULAR METHODS PRIME-IMPLICANT DETERMINATION: Quine-McCluskey Method Example: Find the prime-implicants of F(A,B,C,D) =  m(0,1,2,5,7,8,9,10,11,13,14) Id. A B C D       Prime-implicants  !A B D  A C !D  !B !C  !B !D  !C D A !B

Copyright © 2004 by Miguel A. Marin Revised SIMPLIFICATION OF BOOLEAN FUNCTIONS:TABULAR METHODS PRIME-IMPLICANT DETERMINATION: SVOBODA METHOD A PRODUCT TERM IS REPRESENTED IN TERNARY FORM. A MISSING LITERAL IS REPRESENTED BY “0”. A TRUE VARIABLE BY “1 “. A COMPLEMENTED VARIABLE BY “2” EXAMPLE: FOR N = 4, (A B !C D) = ( ), (A B!D) = ( ) EACH TERM HAS AN IDENTIFIER WHICH IS THE DECIMAL EQUIVALENT OF ITS TERNARY REPRESENTATION EXAMPLE: T 43 = (A B !C D) = ( ), T 38 = (A B!D) = ( ) THE METHOD CONSISTS OF CANCELLING, IN THE SPACE OF 3 N TERMS, ALL THE IMPLICANTS OF THE MINTERMS OF !F. THE NON-CANCELLED TERM WITH SMALLEST IDENTIFIER IS AN IMPLICANT OF F AND IS A PRIME-IMPLICANT, BECAUSE IT CONTAINS MINIMUM NUMBER OF LITERALS (SMALLEST IDENTIFIER) THE NEXT NON-CANCELLED TERM IS A PRIME-IMPLICANT IF IT IS NOT PART OF (CONTAINED IN, OR IS AN IMPLICANT OF) ANOTHER, PREVIOUSLY FOUND PRIME-IMPLICANT.

Copyright © 2004 by Miguel A. Marin Revised SIMPLIFICATION OF BOOLEAN FUNCTIONS:TABULAR METHODS PRIME-IMPLICANT DETERMINATION: SVOBODA METHOD EXAMPLE: FIND THE PRIME-IMPLICANTS OF F(A,B,C)=  m(0,2,4,6,7) 1.- FORM !F =  m(1,3,5) 2.- CONSIDER EACH MEMBER OF THE SPACE OF 3 N TERMS. STARTING WITH IDENTIFIER 0 AND PROGRESSING IN ASCENDING ORDER, FIND THE FIRST TERM WHICH IS NOT IMPLIED BY ANY MINTERM OF !F. IN OUR EXAMPLE THIS TERM IS, T 2 = (0 0 2) = !C,WHICH IS THE FIRST PRIME IMPLICANT OF THE FUNCTION. 3.- TERMS T 3 TO T 11 ARE ALL CANCELLED, BECAUSE THEY EITHER ARE IMPLICANTS OF A MINTERM OF !F OR IMPLY T 2, THE ALREADY FOUND PRIME IMPLICANT. 4.- THE NEXT AND LAST NON-CANCELLED TERM IS T 12 = (1 1 0) = A B BCABCA xxPIxxxxxx xxx (1) x (1 x x(0) x (1) x xxxx(0) x (1 x x(0) x (1) x

Copyright © 2004 by Miguel A. Marin Revised SIMPLIFICATION OF BOOLEAN FUNCTIONS:TABULAR METHODS PRIME-IMPLICANT SELECTION COVERING TABLE TO COMPLETE THE MINIMIZATION PROCESS, THE SELECTION OF THE MINIMAL NUMBER OF PRIME- IMPLICANTS COVERING THE FUNTION IS ACCOMPLISHED USING A TWO ENTRY TABLE, CALLED THE COVERING TABLE. THE PROCEDURE CONSISTS OF: ENTERING A CHECK MARK IN THOSE ENTRIES OF THE TABLE WHERE A PRIME-IMPLICANT (ROW) CONTAINS OR COVERS THE CORRESPONDING MINTERM (COLUMN)

Copyright © 2004 by Miguel A. Marin Revised SIMPLIFICATION OF BOOLEAN FUNCTIONS:TABULAR METHODS SELECTING THOSE PRIME-IMPLICANTS CORRESPONDING TO COLUMNS WITH ONLY ONE CHECK MARK. THESE PRIME-IMPLICANTS ARE CALLED ESSENTIAL AND BELONG TO EVERY MINIMAL FORM THE TABLE IS REDUCED BY CANCELLING ALL THE MINTERMS (COLUMNS) COVERED BY THE ESSENTIAL PRIME-IMPLICANTS. CONSIDERING NOW THE REDUCED TABLE, A TRIAL-AND-ERROR PROCESS GIVES THE MINIMUM NUMBER OF PRIME-IMPLICANTS COVERING ALL THE COLUMNS OF THE TABLE. THERE MAY BE MORE THAN ONE POSSIBILITY TO DO THIS. THE NUMBER OF LITERALS OF THE CHOSEN PRIME-IMPLICANTS MUST BE TAKEN INTO CONSIDERATION.

Copyright © 2004 by Miguel A. Marin Revised SIMPLIFICATION OF BOOLEAN FUNCTIONS:TABULAR METHODS PRIME-IMPLICANT SELECTION COVERING TABLE EXAMPLE: FIND THE MINIMAL SUM-OF-PRODUCTS EXPRESION FOR F(A,B,C) =  m(0,2,4,6,7) THE SET OF PRIME-IMPLICANTS FOUND BY EITHER THE QUINE McCLUSKEY OR SVOBODA METHOD IS, {AB, !C}. BOTH PRIME-IMPLICANTS ARE ESSENTIAL F MINIMAL = A B + !C Min PI 0 !A!B!C 2 !AB!C 4 A!B!C 6 AB!C 7 ABC P1 = AB  P2 = !C  P1 COVERS 6 AND 7 AND IS ESSENTIAL FOR 7 P2 COVERS 0,2,4 AND ALSO 6; IT IS ESSENTIAL FOR 0,2,4

Copyright © 2004 by Miguel A. Marin Revised SIMPLIFICATION OF BOOLEAN FUNCTIONS:TABULAR METHODS PRIME-IMPLICANT SELECTION PETRICK FUNTION THE SAME RESULT MAY BE OBTAINED BY WRITING AND SIMPLIFYING A FUNCTION P, CALLED THE PETRICK FUNCTION, IN SUCH A WAY THAT P = 1 WHEN THE STATEMENT {ALL MINTERMS HAVE BEEN COVERED} IS TRUE. THE P FUNCTION CORRESPONDING TO THE PREVIOUS EXAMPLE IS P = P2.P2.P2 (P1 + P2) P1 = P1.P2 (P1 + P2) = P1.P2 WHICH IS EQUIVALENT TO THE STATEMENT, {P1 AND P2 ARE ESSENTIAL TO COVER ALL MINTERMS OF THE FUNCTION} THUS, F MINIMAL = P1 + P2 = A B + !C IN GENERAL, THE SIMPLIFIED SUM-OF-PRODUCTS FORM OF THE PETRICK FUNCTION HAS MORE THAN ONE TERM. THE TERMS HAVING MINIMUM NUMBER OF PI’S (PRIME-IMPLICANTS) CORRESPOND TO MINIMAL SUM-OF- PRODUCTS EXPRESSIONS OF THE GIVEN FUNCTION F.

Copyright © 2004 by Miguel A. Marin Revised SIMPLIFICATION OF BOOLEAN FUNCTIONS:TABULAR METHODS ANOTHER EXAMPLE FIND ALL MINIMAL SUM-OF-PRODUCTS EXPRESSIONS OF THE FUNCTION F(A,B,C,D) =  (0, 2, 4, 7, 9, 10, 14, 15) 1.- PRIME-IMPLICANTS DETERMINATION: Quine-McCluskey Method List of MintermsFirst and only reduction  (0,2)  (0,4)  (2,10) (10,14)  (7,15)  (14,15)  The prime-implicants are shown in bold  characters

Copyright © 2004 by Miguel A. Marin Revised SIMPLIFICATION OF BOOLEAN FUNCTIONS:TABULAR METHODS 2.- PRIME-IMPLICANTS SELECTION: COVERING TABLE METHOD Minterms --  Prime implicants P1:(9) 1001 x P2(0,2)00-0 x x P3:(0,4) 0-00 x x P4:(2,10)-010 x x P5:(10,14) 1-10 x x P6:(7,15)-111 x x P7:(14,15) 111- x x P1, P3 AND P6 ARE ESSENTIAL PRIME-IMPLICANTS

Copyright © 2004 by Miguel A. Marin Revised SIMPLIFICATION OF BOOLEAN FUNCTIONS:TABULAR METHODS COVERING TABLE METHOD F MINIMAL = P1 + P3 + P6 + … THE REDUCED TABLE IS Minterms --  _ Prime implicants P2: (0,2) 00-0 x P4: (2,10) -010 x x P5: (10,14) 1-10 x x P7: (14,15) 111- x__ To cover the minterms 2, 10, 14, we have three choices, namely, P4 and P5, or P2 and P5 or P4 and P7, each one requiring only two terms of equal number of literals; any other choice will require three terms.

Copyright © 2004 by Miguel A. Marin Revised SIMPLIFICATION OF BOOLEAN FUNCTIONS:TABULAR METHODS COVERING TABLE METHOD THUS, THERE ARE THREE MINIMAL SUM-OF-PRODUCTS EXPRESSIONS: F MINIMAL = P1 + P3 + P6 + P4 + P5. F MINIMAL = P1 + P3 + P6 + P2 + P5. F MINIMAL = P1 + P3 + P6 + P4 + P7. ESSENTIAL SECONDARY PRIME-IMPLICANTS PRIME-IMPLICANTS

Copyright © 2004 by Miguel A. Marin Revised SIMPLIFICATION OF BOOLEAN FUNCTIONS:TABULAR METHODS PETRICK FUNCTION METHOD Minterms --  __ Prime implicants P1:(9) 1001 x P2:(0,2) 00-0 x x P3:(0,4) 0-00 x x P4:(2,10) -010 x x P5:(10,14) 1-10 x x P6:(7,15) -111 x x P7:(14,15) 111- x x_ _ P = (P2 + P3) (P2 + P4)(P3)(P6)(P1)(P4 + P5)(P5 + P7)(P6 + P7). P = P3 P6 P1 (P2 P5 + P4 P5 + P4 P7 ) = P = P3 P6 P1 P2 P5 + P3 P6 P1 P4 P5 + P3 P6 P1 P4 P7. THESE THREE TERMS CORRESPOND TO THE THREE MINIMAL FORMS OF F F MINIMAL = P1 + P3 + P6 + P4 + P5. F MINIMAL = P1 + P3 + P6 + P2 + P5. F MINIMAL = P1 + P3 + P6 + P4 + P7.

Copyright © 2004 by Miguel A. Marin Revised SHANNON EXPANSION SHANNON POVED THAT ANY BOOLEAN FUNCTION CAN BE EXPANDED, WITH RESPECT TO A PARTRICULAR VARIABLE, IN THIS WAY: F(A,B,C,…,Z) = !A F(0,B,C,…,Z) + A F(1,B,C,…,Z) WHERE F(0,B,C,…,Z) = F A=0 (A,B,C,…,Z), AND F(1,B,C,…,Z) = F A=1 (A,B,C,…,Z) ARE CALLED RESIDUE FUNCTIONS EXPANDING NOW THE RESIDUE FUNCTIONS WITH RESPECT TO B WE OBTAIN: F = !A!B F(0,0,C,…,Z) + !A B F(0,1,C,…,Z) + A !B F(1,0,C,…,Z) + A B F(1,1,C,…,Z). Repeating the process for all variables, we obtain the standard canonical sum-of-products of F.

Copyright © 2004 by Miguel A. Marin Revised SHANNON EXPANSION EXAMPLES: Give the Shannon expansion of F(A,B,C) = A B + !C with respect to A and C, F = !A !B [!C] + !A B [!C] + A !B [!C] + A B [ 1]. Give the Shannon expansion of F(A,B,C,D) = AB + AC + AD+ BC + BD + CD with respect to B and D. F = !B!D[AC]+!BD[AC + A + C]+B!D[A + AC + C]+ BD[1]. F = !B!D [AC] + !BD [A + C] + B !D [A + C] + BD [1]. THE SHANNON EXPANSION HAS APPLICANTIONS IN MULTIPLEXER IMPLEMENTATION OF BOOLEAN FUNCTIONS