Find the minimal sum-of-products (SOP) Boolean expression for the condition A B, where A...
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Find the minimal sum-of-products (SOP) Boolean expression for the condition $A > B$, where $A$ and $B$ are unsigned 2-bit numbers represented by $A = A_1 A_0$ and $B = B_1 B_0$.
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Answer: A. A1 B1' + (A1 ⊙ B1) A0 B0'
For a 2-bit magnitude comparator, A > B when the most significant bit (MSB) of A is greater than the MSB of B, or when the MSBs are equal and the least significant bit (LSB) of A is greater than the LSB of B. The term A1 B1' handles the first case, and (A1 ⊙ B1) A0 B0' handles the second case, where ⊙ denotes XNOR (equality).
Step-by-step Derivation:
Let A = A1 A0 and B = B1 B0. The condition A > B is true if:
- A1 > B1, which occurs when A1 = 1 and B1 = 0, giving the term A1 B1'.
- A1 = B1 and A0 > B0. The equality A1 = B1 is represented by the XNOR operation (A1 ⊙ B1). The condition A0 > B0 occurs when A0 = 1 and B0 = 0, giving A0 B0'. Combining these gives (A1 ⊙ B1) A0 B0'.
The complete expression is therefore A1 B1' + (A1 ⊙ B1) A0 B0'.