Question

A = \( a_1a_0 \) and B = \( b_1b_0 \) are two 2-bit unsigned binary numbers. If \( F(a_1, a_0, b_1, b_0) \) is a Boolean function such that \( F = 1 \) only when \( A>B \), and \( F = 0 \) otherwise, then \( F \) can be minimized to the form _________

• A. ( a_1 \overline{b_1} + a_1 a_0 \overline{b_0} \)
• B. ( a_1 \overline{b_1} + a_1 a_0 \overline{b_0} + a_0 \overline{b_0} b_1 \)
• C. ( a_1 a_0 \overline{b_0} + a_0 \overline{b_0} b_1 \)
• D. ( a_1 \overline{b_1} + a_1 a_0 \overline{b_0} + a_0 b_0 b_1 \)

Hint:

To minimize Boolean functions, use Karnaugh maps or Boolean algebra simplifications to reduce the number of terms.

Answer 1

The Correct Option is B

Step 1: Understanding the problem.
The Boolean function \( F \) outputs 1 when \( A>B \), and 0 otherwise. To simplify the function, we examine the conditions for each possible pair of inputs for \( a_1, a_0, b_1, b_0 \) and express them as a sum of minterms.
Step 2: Analyzing the options.
- (A) Incorrect, this expression does not account for all the conditions where \( A>B \).
- (B) Correct, this expression simplifies the Boolean function such that it represents the conditions for \( A>B \).
- (C) Incorrect, this expression excludes some cases where \( A>B \).
- (D) Incorrect, this expression is too complex and does not properly represent the required conditions. Step 3: Conclusion.
The correct answer is (B) \( a_1 \overline{b_1} + a_1 a_0 \overline{b_0} + a_0 \overline{b_0} b_1 \).