Minterms three variables

Question

1. Express the following as the sum of minterms of three variables:

       _

a)    BC

b)   A

please help me im stuck please show work on how this is done would be much appreciated thanks.

Answer 1

Good question, and this whole topic comes down to one idea: a minterm has to contain every variable exactly once, either plain or complemented. So when your term is missing a variable, you put that variable back in without changing the value of the expression, and the way you do that is by multiplying by (X + X'), since that always equals 1.

I will write A' for NOT A, because overbars do not always display properly in this box.

First, set up the numbering for three variables. Treat A as the leftmost bit, B as the middle bit, and C as the rightmost bit, then read each row as a binary number.

m₀ = A'B'C' (000)
m₁ = A'B'C (001)
m₂ = A'BC' (010)
m₃ = A'BC (011)
m₄ = AB'C' (100)
m₅ = AB'C (101)
m₆ = ABC' (110)
m₇ = ABC (111)

Part a) B'C

The variable that is missing is A, so multiply by (A + A').

B'C = B'C(A + A') = AB'C + A'B'C

AB'C is the row A=1, B=0, C=1, which reads 101, so that is m₅. A'B'C is the row 001, so that is m₁.

B'C = m₁ + m₅ = Σm(1, 5)

Check it against a truth table. B'C is true exactly when B = 0 and C = 1, and A is free to be either value, so the only two rows that fit are 001 and 101. Those are m₁ and m₅, so it matches.

Your overbar got a little lost in the typing, so in case the bar was meant to sit over the C instead of the B, the very same method gives BC' = BC'(A + A') = ABC' + A'BC' = m₆ + m₂ = Σm(2, 6).

Part b) A

This time two variables are missing, B and C, so you multiply by both factors.

A = A(B + B')(C + C')

A = (AB + AB')(C + C')

A = ABC + ABC' + AB'C + AB'C'

Those four rows read 111, 110, 101 and 100, so

A = m₄ + m₅ + m₆ + m₇ = Σm(4, 5, 6, 7)

Check it the same way. A all by itself is true in every row where A = 1, and those rows are exactly 100, 101, 110 and 111, so it matches.

A fast way to check your own work. If a term is written with k of the n variables, it has to expand into 2^n−k minterms. In part a you had 2 of the 3 variables, so you should get 2¹ = 2 minterms, and you did. In part b you had 1 of the 3 variables, so you should get 2² = 4 minterms, and you did. If your count does not come out to a power of 2 like that, you know you dropped a term somewhere and can go back and find it.