Design a circuit using only NOR gates to implement the OR function for two inputs A and B.
Simplify the expression F = (A + B)(A’ + C) using Boolean algebra and verify with a truth table.
Construct a truth table for the expression F = (A XOR B) AND C, and draw its circuit.
Explain how an AND gate can be fabricated using NOR gates, including the logic behind it.
Use a 3-variable Karnaugh map to simplify the expression F = Σ(0, 2, 4, 6).
Answers and Descriptions for Group 1
Answer: Connect A and B to a NOR gate to get (A NOR B) = (A + B)’. Feed this output to another NOR gate with both inputs tied together, yielding ((A + B)’ NOR (A + B)’) = A + B.
Description: NOR is a universal gate, so OR can be implemented by first creating NOT (using NOR) and then applying De Morgan’s Law. This reinforces universal gate applications.
[Image Placeholder: NOR-based OR circuit diagram]Answer: F = AA’ + AC + A’B + BC = AC + A’B + BC. Truth table confirms outputs.
Description: Using distributive and identity laws, the expression is simplified. The truth table verifies equivalence, enhancing Boolean algebra skills. This exercise connects simplification to verification.Answer: Truth table:
A
B
C
A XOR B
(A XOR B) AND C
0 0 0 0 0 0 0 1 0 0 0 1 0 1 0 0 1 1 1 1 1 0 0 1 0 1 0 1 1 1 1 1 0 0 0 1 1 1 0 0 Circuit: XOR gate for A, B; output to AND gate with C.
Description: This combines XOR and AND operations, requiring students to build and verify complex expressions, strengthening circuit design skills.
[Image Placeholder: Circuit for (A XOR B) AND C]
Answer: (A NOR A) = A’, (B NOR B) = B’. Feed A’ and B’ to a NOR gate: (A’ NOR B’) = (A’ + B’)’ = A AND B.
Description: Using NOR’s universality, NOT gates are created first, then De Morgan’s Law constructs AND. This deepens understanding of gate fabrication.Answer: F = A’C’.
Description: Grouping minterms 0, 2, 4, 6 in a K-map yields A’C’. This advanced simplification exercise enhances K-map proficiency for complex expressions.
[Image Placeholder: 3-variable K-map for F]