Question

2. This question has two related parts 2.a) and 2.b). 2.a) Use the Quine McCluskey minimization method to find the minimal SOP for the following function: f(a, b, c) = m(0,2,6) 2.b) Find the ALL NAND representation of the minimized expression of f resulted from part 2.a). B IV A - - I E3 X X DE

Answer 1

a)

f(a,b,c)=.This function is simplified using Quine-McCluskey tabular method.All the steps are shown below.

Ση(0, 2, 6)

a	b	c
0	0	0	m0
0	0	1
0	1	0	m2
0	1	1
1	0	0
1	0	1
1	1	0	m6
1	1	1

Group the minterms based on the number of 1's.Group 0 means.minterm with 0 number of 1,similarly group 1 have 1 number of 1 and group 2 have 2 number of 1's.So total 3 groups are made by this way.

Group	minterm	Binary representation
0	m0	000
1	m2	010
2	m6	110

The below table shows the possible merging of the minterms from the adjacent groups.

Group	Matched pair	Binary representation
0	mo-m2	0_0
1	m2-m6	_10

So we get two merged groups which contain diiference in 1 bit position.No more possible grouping is possible in these minterms because there is no 1 bit position difference groups in above table.We have 2 rows and each row will give 1 prime implicant.So A'C' and BC' are prime implicants.

Prime implicant table is shown below.

Prime implicant	Minterms involved	0	2	6
A'C'	m0,m2	*	*
BC'	m2,m6		*	*

minterms 0 is covered by 1 prime implicant A'C'.So it is an essential prime implicant,which means it will be part of simplified boolean function.Similarly,minterm 6 is covered by 1 prime implicant BC',So it eillbe also an essential prime implicant and be the part of simplified boolean function.

Therefore,we have 2 essential prime implicants and the simplified boolean function is given below,

            F=A'C'+BC'

b) Now we want to find all NAND representation of F=A'C'+BC'.

Take double complement on both side of equation,(F')'=((A'C'+BC')')'

                                                                               F=((A'C')'(BC')')'

                                                                               F =(A'C')' NAND (BC')'

F = Ã c + B Ē ĀĞ + B (double complement both side of equations. Applying demorgan's theorm A+B+C - Ā B C FTĀE(BC) Boolean function with NAND gate only, CAT). (BC)