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Answer the following from the logic gate diagram.
From the logic gate diagram given below: Derive Boolean expression for (1), (2) and R. Reduce the…
From the logic gate diagram given below:
(a)[4.0]
Derive Boolean expression for (1), (2) and R. Reduce the derived expression.
(b)[1.0]
Name the logic gate that represents the reduced expression.
Answer
Answer (a)
Checked answer.
(1) $= (A \oplus B)' = A \cdot B + A' \cdot B'$, (2) $= A + B$, $R = (A \cdot B)' = A' + B'$
- The first gate is an XNOR gate with inputs A and B: (1) $= (A \oplus B)' = A \cdot B + A' \cdot B'$
- The second gate is an OR gate with inputs A and B: (2) $= A + B$
- The last gate is a NAND gate with inputs (1) and (2): $R = [(A \cdot B + A' \cdot B') \cdot (A + B)]'$
- Expanding: $(A \cdot B + A' \cdot B') \cdot (A + B) = A \cdot B + A \cdot B \cdot B + A \cdot A' \cdot B' + A' \cdot B \cdot B'$
- $A \cdot A' = 0$ and $B \cdot B' = 0$ (Complement law), and $A \cdot B + A \cdot B = A \cdot B$ (Idempotent law), so the product is $A \cdot B$
- $R = (A \cdot B)' = A' + B'$ (De Morgan's law)
Answer (b)
Checked answer.
$R = (A \cdot B)'$, so the reduced expression represents a NAND gate.
From ISC 2025 Computer Science Paper 1, question 4(ii).
Check your working with the Boolean algebra solver: the steps law by law, the K-map and the logic circuit for any expression.