1 Minterm and Maxterm Expressions Definition: a minterm of n variables is a product of the variables in which each appears exactly once in true or complemented form. e.g.: minterms of 3 variables: - Each minterm = 1 for only one combination of values of the variables, = 0 otherwise. Definition: a maxterm of n variables is a sum of the variables in which each appears exactly once in true or complemented form. e.g.: 3 variables - Each maxterm = 0 for only one combination of values of the variables, = 1 otherwise. ABC A’BC AB’C 2 n terms, n = no. variables 2 3 = 8 A+B+C A’+B+C A’+B’+C

2 All possible minterms and maxterms are obtained from the truth table: e.g. recall from our votetaker: F = ABC + ABC' + AB'C + A'BC which can be written in terms of minterms as F = m 3 + m 5 + m 6 + m 7 Row # ABCMintermsMaxterms AB'C'=A'+B+C= 5101AB'C=A'+B+C'= 6110ABC'=A'+B'+C= 7111ABC=A'+B'+C'= e.g. 3 variables How do we write minterm and maxterm expansions? value =1 value = 0 A’B’C’ m 0 A+B+C M 0 A’B’C m 1 A+B+C’ M 1 A’BC’ m 2 A+B’+C M 2 A’BC m 3 A+B’+C’ M 3 m 4 M 4 m 5 M 5 m 6 M 6 m 7 M

3 Minterms & Maxterms (continued) which is abbreviated as F(A,B,C) = - For each F = 1 row of truth table, only one m i = 1. Therefore the minterm expansion is unique, i.e. there is a 1 to 1 correspondence between each minterm and each 1 in the truth table. - Alternate form of F: F = ( A + B + C ) ( A + B + C' ) ( A + B' + C ) ( A' + B + C ) or in terms of maxterms: F = or F(A,B,C) = - For each F = 0, only one M i = 0. Therefore maxterm expansion is unique.  m (3,5,6,7) all others = 0 ( ) ( )( )( ) M0M1M2M4M0M1M2M4  M (0,1,2,4)  F =  m (3,5,6,7) =  M (0,1,2,4)

4 Minterms & Maxterms (continued) (Note that we have been given a "simplified" expression, and we want to find the minterm expansion. This is moving in the opposite direction to what we did before, i.e. writing F from the truth table, and then simplifying). -Note that if m i is present in minterm expansion, then M i is not present in maxterm expansion, and conversely. - Note also that: F' = - To convert from a general expression to a minterm or maxterm expansion, use: a) truth table orb) algebraic manipulation e.g. Find the minterm expansion of: F = AB' + A'C ABCF  m (0,1,2,4) =  M (3,5,6,7)

5 Minterms & Maxterms (continued) b) Algebraically: use X + X' = 1 to introduce the missing variables in each term. Therefore F = AB' + A'C = Solution of F = AB' + A'C a) Using truth table: Therefore F = = ABCF Evaluate F A’B’C+A’BC+AB’C’+AB’C m 1 + m 3 + m 4 + m 5  m (1,3,4,5) AB’(C+C’) + A’C(B+B’) = AB’C + AB’C’ + A’BC + A’B’C = m 5 + m 4 + m 3 + m 1 =  m (1,3,4,5)

6 Example Find the maxterm expansion of F = ( A + B' ) ( A' + C ) b) Use XX' = 0 to introduce missing variables in each term Therefore F = = - Minterm and maxterm expansions are unique, therefore can prove equation F = G is valid by finding minterm or maxterm expansions of both sides, and demonstrating the equality. a)Truth Table  F =  (2,3,4,6) X+YZ=(X+Y)(X+Z) (A+B’+C)(A+B’+C’)(A’+B+C)(A’+B’+C)  (2,3,4,6) (A + B’+ CC’)(A’ + C + BB’) x yz

7 Incompletely Specified Functions In some applications, certain combinations of inputs never occur, or the output from certain combinations of inputs may be irrelevant. e.g.: The binary number in BCD should never occur. In a truth table, the function F (at the output) is not important in such cases and is said to be incompletely specified. We don't care what value (0 or 1) is assigned to F. e.g.: A function of 3 variables; consider the truth table: ABCF X10111X211X10111X21 F is incompletely Specified X 1 X 2 could be 00, 01, 10, 11 What values do we assign?? Ans. Choose values such that F is in simplest form.

8 Incompletely Specified Functions(continued) -When we expand F in minterm or maxterm, we must specify each x as 0 or 1. We should choose the values of x to produce the simplest form for F. Easiest to do this using a Karnaugh map (next topic). - - In previous eg., simplest form for F is obtained by assigning 1 to 1st X 1 0 to 2nd X 2 yielding F = after simplification Formal minterm expansion would be written: F = A’B + BC 4 times with X 1, X 2 = 0, 1  m ( 0, 3, 7) +  d ( 1, 6) Required terms + don’t care minterms