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ELEC 261 Digital Logic Design Lecture 11. Minterms and Maxterms - Read Chapter 5.1, 5.2, and 5.3 for next class 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.: maxterm of 3 variables: - Each maxterm = 0 for only one combination of values of the variables, = 1 otherwise Minterm and Maxterm Expressions

Minterm and Maxterm

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ELEC 261Digital Logic Design

Lecture 11.Minterms and Maxterms

- Read Chapter 5.1, 5.2, and 5.3 for next classDefinition: 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.: maxterm of 3 variables:

- Each maxterm = 0 for only one combination of values of the variables,= 1 otherwise

Minterm and Maxterm Expressions

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recall from our vote taker:F = ABC + ABC' + AB'C + A'BC

which can be written in terms of minterms asF = m3 + m5 + m6 + m7

Row #

A B C Minterms Maxterms

0 0 0 0

1 0 0 1

2 0 1 0

3 0 1 1

4 1 0 0 AB'C‘ = A'+B+C =

5 1 0 1 AB'C = A'+B+C‘ =

6 1 1 0 ABC‘ = A'+B'+C =

7 1 1 1 ABC = A'+B'+C‘=

All possible minterms and maxterms are obtained from the truth table:

How do we write mintermand maxterm expansions?

e.g. 3 variables

Minterms & Maxterms (continued)

which is abbreviated as F(A,B,C) =- For each F = 1 row of truth table, only one mi = 1Therefore 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 Mi = 0.Therefore maxterm expansion is unique.

Minterms & Maxterms (continued)

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(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 mi is present in minterm expansion, then Mi 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 tableor b) algebraic manipulation

e.g. Find the minterm expansion of: F = AB' + A'C

A B C F

0 0 0 0

0 0 1 0

0 1 0 0

0 1 1 1

1 0 0 0

1 0 1 1

1 1 0 1

1 1 1 1

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 =

=

=

A B C F

0 0 0

0 0 1

0 1 0

0 1 1

1 0 0

1 0 1

1 1 0

1 1 1

Minterms & Maxterms (continued)

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Find the maxterm expansion of F = ( A + B' ) ( A' + C )

b) Use XX' = 0 to introduce missing variables in each termTherefore 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.

Example

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 1010 - 1111 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 . We don't care what value (0 or 1) is assigned to F.

e.g.: A function of 3 variables; consider the truth table:

10 - 15

A B C F

0 0 0

0 0 1

0 1 0

0 1 1

1 0 0

1 0 1

1 1 0

1 1 1

Incompletely Specified Functions

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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 example, simplest form for F is obtained by assigning

1 to X1

0 to X2

yielding F =after simplification

Formal minterm expansion would be written:F =

Incompletely Specified Functions

A B C F

0 0 0

0 0 1

0 1 0

0 1 1

1 0 0

1 0 1

1 1 0

1 1 1

1X1

0111

X2

1