4-bit parallel adder — adding 1011 and 1111 with Cin = 1 Two 4-bit binary numbers (A = 1011 and B = 1111) are applied to a 4-bit parallel adder with carry-in Cin = 1. What are the resulting 4-bit sum (∑4∑3∑2∑1) and the carry-out (Cout)?
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A∑4∑3∑2∑1 = 0111, Cout = 0
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B∑4∑3∑2∑1 = 1111, Cout = 1
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C∑4∑3∑2∑1 = 1011, Cout = 1
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D∑4∑3∑2∑1 = 1100, Cout = 1
Answer
Correct Answer: ∑4∑3∑2∑1 = 1011, Cout = 1
Explanation
Introduction / Context:Parallel adders compute multi-bit sums in one stage by chaining full-adders. Understanding how carries propagate and how the final carry-out relates to overflow or extended precision is essential for designing arithmetic units.
Given Data / Assumptions:
- A = 1011 (decimal 11), B = 1111 (decimal 15).
- Cin = 1.
- Unsigned addition with a 4-bit adder.
Concept / Approach:Add the operands plus Cin arithmetically, then separate the lower 4 bits as the sum and the 5th bit as Cout. This mirrors the ripple-carry adder behavior where carries ripple from LSB to MSB, producing a final carry-out if the total exceeds 15.
Step-by-Step Solution:
1) Convert to decimal: 1011 = 11, 1111 = 15.2) Add with Cin: 11 + 15 + 1 = 27.3) Express 27 in binary: 27 = 11011 (5 bits).4) Lower four bits (from LSB) → 1011; the 5th bit → Cout = 1.Verification / Alternative check:Perform bitwise addition: start from LSB, track carries, and verify final sum 11011, which splits into Cout=1 and 4-bit sum 1011.
Why Other Options Are Wrong:
- 0111, 1111, 1100 with Cout shown: none match the correct 5-bit result of 11011.
Common Pitfalls:Losing the final carry-out or misaligning bit positions when reading the 5-bit total back into 4-bit sum plus Cout.
Final Answer:∑4∑3∑2∑1 = 1011, Cout = 1