Question Design a circuit 3.49 A 3-bit "comparator" circuit receives two 3-bit numbers, \( P=P_{2} P_{1} P_{0} \) and \( Q= \) \( Q_{2} Q_{1} Q_{0} \). Design a minimal sum-of-products circuit that produces a 1 output if and only if \( \mathrm{P}<\mathrm{Q} \).
U5WTSC The Asker · Electrical Engineering
Design a circuit
Transcribed Image Text: 3.49 A 3-bit "comparator" circuit receives two 3-bit numbers, \( P=P_{2} P_{1} P_{0} \) and \( Q= \) \( Q_{2} Q_{1} Q_{0} \). Design a minimal sum-of-products circuit that produces a 1 output if and only if \( \mathrm{P}<\mathrm{Q} \).
More Transcribed Image Text: 3.49 A 3-bit "comparator" circuit receives two 3-bit numbers, \( P=P_{2} P_{1} P_{0} \) and \( Q= \) \( Q_{2} Q_{1} Q_{0} \). Design a minimal sum-of-products circuit that produces a 1 output if and only if \( \mathrm{P}<\mathrm{Q} \).
Community Answer
SFJXA4
【General guidance】The answer provided below has been developed in a clear step by step manner.Step1/1To compare two 3-bit numbers P and Q, we need to check if P is less than Q, i.e., P<Q.We can perform this comparison bit-by-bit starting from the most significant bit. If at any bit position, P has a 0 and Q has a 1, then we can conclude that P<Q. Otherwise, if P has a 1 and Q has a 0 or if P and Q both have the same bit value, then we need to move to the next bit position to perform the comparison.Using this approach, we can design the following sum-of-products circuit ... See the full answer