JNTU B.Tech II SWITCHING THEORY AND LOGIC DESIGN Previous Question Paper (November/December - 2020 - R13)

The given expression is F(A, B, C, D) = ABC + ABD + ABCD. First, express each term in canonical form by including missing variables:

1. ABC = ABC(D + D') = ABCD + ABCD' = m(15) + m(14)
2. ABD = ABD(C + C') = ABCD + ABC'D = m(15) + m(11)
3. ABCD = m(15)

Combining these, the minterms are Σm(11, 14, 15). The canonical sum-of-products form is F = m(11) + m(14) + m(15).

The maxterms are the remaining indices from the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13}. Thus, the product-of-sums form is ΠM(0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13).

For a 4-bit data word (1011), we require 3 parity bits (p1, p2, p4) since 2ⁿ ≥ m + n + 1 (2³ ≥ 4 + 3 + 1). Assigning positions: D3=1, D5=0, D6=1, D7=1.

Parity bit calculations (even parity):

• p1 checks positions 1, 3, 5, 7: p1 ⊕ D3 ⊕ D5 ⊕ D7 = 0 ⇒ p1 ⊕ 1 ⊕ 0 ⊕ 1 = 0 ⇒ p1 = 0
• p2 checks positions 2, 3, 6, 7: p2 ⊕ D3 ⊕ D6 ⊕ D7 = 0 ⇒ p2 ⊕ 1 ⊕ 1 ⊕ 1 = 0 ⇒ p2 = 1
• p4 checks positions 4, 5, 6, 7: p4 ⊕ D5 ⊕ D6 ⊕ D7 = 0 ⇒ p4 ⊕ 0 ⊕ 1 ⊕ 1 = 0 ⇒ p4 = 0

The resulting Hamming code is 0110011.