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Min and Max Terms Notation in Boolean Algebra

In this tutorial, we are going to learn about the Min and Max Terms in Boolean Algebra in Digital Electronics.
Submitted by Saurabh Gupta, on November 28, 2019

In short or for convenience purposes, we represent canonical SOP/POS form in min/max terms.

Minterm

Each of the product terms in the canonical SOP form is called a minterm. Minterm are represented as binary numbers in terms of 0s and 1s. The binary words are formed by representing each non-complemented variable by 1 and each complemented variable by 0, and the decimal equivalent of this binary word is represented as a subscript of m as m0, m1, m2, etc. We generally use the ∑ (sigma) notation to represent minterms.

Maxterm

Each of the sum terms in the canonical POS form is called a maxterm. Maxterm can also be represented using binary numbers where each non-complemented variable is represented using 0 and complemented variable using 1, and the decimal equivalent of this binary word is represented as a subscript of M as M0, M2, M2, etc. We generally use ∏ (pi) notation to represent the max terms.

Note: For n-variable logic function there are 2n minterms and 2n maxterms.

Min and Max terms for two literal binary expressions

Input Variable (A) Input Variable (B) Minterm Minterm notation Maxterm Maxterm notation
00A.Bm0A+BM0
01A.Bm1A+BM1
10A.Bm2A+BM2
11A.Bm3A+BM3

Min and Max terms for three literal binary expressions

Input Variable (A) Input Variable (B) Input Variable (C) Minterm Minterm notation Maxterm Maxterm notation
000A.B.Cm0A+B+CM0
001A.B.Cm1A+B+CM1
010A.B.Cm2A+B+CM2
011A.B.Cm3A+B+CM3
100A.B.Cm4A+B+CM4
101A.B.Cm5A+B+CM5
110A.B Cm6A+B+CM6
111A.B.Cm7A+B+CM7

Example: Express the following in corresponding minterm and maxterm expression

  1. Y = ABC + A.B.C + A.B. C + A. B. C
  2. Y= (A+B+C) (A+ B+C) (A+ B+ C)

Solution (a):

Y = ABC + A. B.C + A.B. C + A. B. C , is an example of canonical SOP expression, so its each term can be represented in minterm notation. Therefore,

    Y = ABC + A. B.C + A.B. C + A. B. C
    = m7 + m3 + m5 + m4
    = ∑m (3, 4, 5, 7)		[ ∑ is used to denote CSOP]

Solution (b):

Y= (A+B+C) (A+ B+ C) (A+ B+ C), is an example of canonical POS expression, so its each term can be represented in maxterm notation.

    Y= (A+B+C) (A+ B+ C) (A+ B+ C)
    = M0 + M5 + M7
    = ∏M (0, 5, 7)		[ ∏ is used to denote CPOS]

Note: If a truth table is given, and if the output is 1 then it corresponds to minterm and in case the output is 0 then it corresponds to 0.

Input Variable (A) Input Variable (B) Input Variable (C) Output (Y)
0000
0010
0101
0110
1000
1011
1101
1111

In the above truth table, the minterms will be m2, m5, m6 and m7 i.e.,

    F = ∑m (2, 5,6, 7)

and maxterms will be M0, M1, M3 and M4 i.e.,

    F = ∏M (0, 1, 3, 4)

Hence, from this we can conclude there is a complementary relationship between CSOP and CPOS.







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