Question

1.The algebraic minterm of F(A,B,C) = ABC + B’C is:

2.The digital minterm of F is:

3. The digital maxterm for F is:

4.The digital minterm of F’ is:

5.The digital maxterm of F’ is:

6.If the digital minterm of F(A,B,C,D) = Σ m(0, 2, 4, 6), the digital maxterm is:

7.The minimum sum of products for F(A,B,C) = Σ m(0, 1, 4) is:

Answer 1

Solution:

(1)

Given,

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

Explanation:

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

Converting into cannonical form:

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

=>F(A, B, C) = ABC + AB'C + A'B'C

(2)

Explanation:

Finding digital minterms of F:

=>F(A, B, C) = ABC + AB'C + A'B'C

=>F(A, B, C) = m7 + m5 + m1

=>Hence digital minterms of F(A, B, C) = (1, 5, 7)

(3)

Explanation:

Finding digital maxterms of F:

=>Digital minterms of F(A, B, C) = (1, 5, 7)

=>All the remaining values which are not present in the minterms will be present in the maxterms.

=>Digital maxterms of F(A, B, C) = (0, 2, 3, 4, 6)

(4)

Explanation:

Finding complement of F':

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

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

=>F'(A, B, C) = (ABC)'.(B'C)' using Demorgan's law

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

Multiplying terms

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

We know that B'B = 0, C'C' = C'

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

Taking C' common from second, third, fourth and fiifth terms

=>F'(A, B, C) = A'B + (A' + B' + B + 1)C'

We know that (A' + B' + B + 1) = 1

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

Finding cannonical form:

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

=>F'(A, B, C) = A'BC + A'BC' + ABC' + AB'C' + A'BC' + A'B'C'

=>F'(A, B, C) = m3 + m2 + m6 + m4 + m2 + m0

=>Hence digital minterm of F'(A, B, C) = (0, 2, 3, 4, 6)

(5)

Explanation:

Finding maxterms of F':

=>Digital minterm of F'(A, B, C) = (0, 2, 3, 4, 6)

=>Digital maxterms of F'(A, B, C) = (1, 5, 7)

(6)

Given,

=>F(A, B, C, D) = (0, 2, 4, 6)

Explanation:

Finding digital maxterm:

=>Digital maxterm = (1, 3, 5, 7, 8, 9, 10, 11, 12, 13, 14, 15)

(7)

Given,

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

Explanation:

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

Taking B'C' common from first and last terms

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

We know that A' + A = 1

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

Taking B' common from first and second terms

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

We know that A + BC = (A + B)(A + C)

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

We know that C' + C = 1

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

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

I have explained each and every part with the help of statements attached to it.