Boolean algebra Canonical Forms and Solution, Karnaugh Map

Hi, this is the first one in this year and once again failed to post on time but not going to tender apology for it, among you, who knows me will understand that it is examination time and I am little more pre-occupied.

Today we will solve a few minterm and maxterm using Karnaugh map as well as other aspect of Canonical form

Solution for Sum of Product (SOP)

Examples 1: Using three literals.

Reduce this term using K-Map : f(x,y,z) = Σm(0,2,3,4,6,7)

The sigma sign and as well as 'm' written in lowercase indicate that it is SOP (sum of Product), so we have to find the minterm.

First mark the given cell with 1's and rest with 0's. We have two Quad here, first one using 0,2,4,6 (Map rolling) and second 2,3,6,7.

Quad(0,2,4,6) having m0+m2+m4+m6, we get : x'y'z' +x'y'z+xy'z'+xyz', common factor is : z'.
Quad(2,3,6,7) having m2+m3+m6+m7, we get : x'yz' + x'yz+xyz'+xyz, common factor is : y
Quad(0,2,4,6) = z'
Quad(2,3,6,7) = y

f(x,y,z)=Σm(0,2,3,4,6,7) = y + z'

Examples 2: Using four literals.
Reduce this term using K-Map : f(a,b,c,d)=Σm(0,1,3,5,7,10,11,12,13,14,15)

Pair(0,1) = m0+m1 = a'b'c'd' + a'b'c'd = a'b'c'
Quad(1,3,5,7)=m1+m3+m5+m7 = a'b'c'd + a'b'cd + a'bc'd + a'bcd = a'd
Quad(12,13,14,15) = m12+m13+m14+m15 =abc'd' + abc'd + abcd' + abcd = ab
Quad(10,11,14,15) = m10+m11+m14+m15 =ab'cd' + ab'cd' + abcd' + abcd = ac

f(a,b,c,d)=Σm(0,1,3,5,7,10,11,12,13,14,15) = a'b'c' + a'd + ab + ac

Example 3: Using don't care

f(a,b,c,d)=Σm(0,1,3,8,9,12,13) + Σd(2,6)

Quad(0,1,2,3) = m0 + m1 + m2 + m3 = a'b'c'd' + a'b'c'd + a'b'cd' + a'b'cd = a'b'
Quad(8,9,12,13) = m8 + m9 + m12 + m13 = ab'c'd' + ab'c'd + abc'd' + abc'd = ac'
[We are using cell 2 to create a quad as it is given as don't care but we cannot use only don't care so left the cell No. 6 as it is failed make any pair or quad with other given cell.]

f(a,b,c,d)=Σm(0,1,3,8,9,12,13)+Σd(2,6) =a'b' + ac'

Solution for Product of Sum (POS)

Example 1: Using four literals.

f(a,b,c,d) = p(0,1,6,7,8,9,12,13)

Quad(0,1,8,9) = M0 . M1 . M8 . M9 = (a+b+c+d) . (a+b+c+d') . (a'+b+c+d) . (a'+b+c+d') = b+c
Pair(6,7) = M6 . M7 = (a+b'+c'+d) . (a+b'+c'+d') = a+b'+c'
Quad(8,9,12,13) = M8 . M9 . M12 . M13 = (a'+b+c+d) . (a'+b+c+d') + (a'+b'+c+d) . (a'+b'+c+d') = a' +c
f(a,b,c,d) = ΠM(0,1,6,7,8,9,12,13) =Π(b+c)+(a+b'+c')+(a'+c)

Convert Boolean expression to Canonical form using Boolean Laws

Sum of Products

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

= A.1 + B'C.1

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

=AB + AB' + AB'C + A'B'C

=AB.1 + AB'.1+AB'C + A'B'C

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

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

=A'B'C+ AB'C' +AB'C+ABC' + ABC (REMOVE THE DUPLICATE ENTRIES)

f(A,B,C) = m(1,4,5,6,7)

Using Truth Table

A	B	C	B'	B'C	A+B'C
0	0	0	1	0	0
0	0	1	1	1	0	A'BC'
0	1	0	0	0	1
0	1	1	0	1	1	A'BC
1	0	0	1	0	0
1	0	1	1	0	1	AB'C
1	1	0	0	1	1	ABC'
1	1	1	0	1	1	ABC

Sum of Products

F(A,B,C)=(A+ B)( B + C)

= (A + B+0).(0+B+C) (X+0 = X)

= (A + B + C.C').(AA'+B+C) (0 = XX')

=(A + B+ C)( A+B+C')(A+B+C)(A'+B+C) (REMOVE THE DUPLICATE ENTRIES)

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

F(A,B,C) = m(0,1,4)

Using Truth Table