SOP Explained for Boolean Function | Generated by AI
Let’s clarify the concept of Sum of Products (SOP) and address your question about why the SOP expression for the function \( F(A, B, C) \) is \( F = A \cdot B’ \cdot C + A’ \cdot B \cdot C \) when the function outputs 1 for the input combinations \((A=1, B=0, C=1)\) and \((A=0, B=1, C=1)\), and why \( F \neq 2 \).
Understanding SOP (Sum of Products)
- SOP is a standard form of expressing a Boolean function where the output is 1. It consists of a sum (logical OR) of product terms (logical AND).
- Each product term corresponds to a minterm, which is a combination of variables (including their complements) that makes the function output 1.
- For a function \( F(A, B, C) \), a minterm is a product of all input variables (or their complements) that evaluates to 1 for a specific input combination.
Step-by-Step Explanation
The problem states that the function \( F(A, B, C) \) outputs 1 for two specific input combinations:
- \( (A=1, B=0, C=1) \)
- \( (A=0, B=1, C=1) \)
We need to derive the SOP expression for \( F \).
Step 1: Identify the Minterms
- A minterm is a product term where each variable appears exactly once (either uncomplemented or complemented) corresponding to an input combination where the function is 1.
- For \( (A=1, B=0, C=1) \):
- \( A = 1 \), so we use \( A \).
- \( B = 0 \), so we use \( B’ \) (the complement of \( B \)).
- \( C = 1 \), so we use \( C \).
- The minterm is: \( A \cdot B’ \cdot C \).
- For \( (A=0, B=1, C=1) \):
- \( A = 0 \), so we use \( A’ \).
- \( B = 1 \), so we use \( B \).
- \( C = 1 \), so we use \( C \).
- The minterm is: \( A’ \cdot B \cdot C \).
Step 2: Form the SOP Expression
- In SOP form, the function is the logical OR of all minterms where the function outputs 1.
- Since \( F = 1 \) for the two input combinations above, the SOP expression is the sum of the two minterms: \[ F = A \cdot B’ \cdot C + A’ \cdot B \cdot C \]
Step 3: Verify the Expression
Let’s check if the expression \( F = A \cdot B’ \cdot C + A’ \cdot B \cdot C \) correctly outputs 1 for the given input combinations:
- For \( (A=1, B=0, C=1) \):
- First term: \( A \cdot B’ \cdot C = 1 \cdot (0)’ \cdot 1 = 1 \cdot 1 \cdot 1 = 1 \).
- Second term: \( A’ \cdot B \cdot C = (1)’ \cdot 0 \cdot 1 = 0 \cdot 0 \cdot 1 = 0 \).
- \( F = 1 + 0 = 1 \). (Correct)
- For \( (A=0, B=1, C=1) \):
- First term: \( A \cdot B’ \cdot C = 0 \cdot (1)’ \cdot 1 = 0 \cdot 0 \cdot 1 = 0 \).
- Second term: \( A’ \cdot B \cdot C = (0)’ \cdot 1 \cdot 1 = 1 \cdot 1 \cdot 1 = 1 \).
- \( F = 0 + 1 = 1 \). (Correct)
- For other combinations (e.g., \( A=0, B=0, C=0 \)):
- First term: \( A \cdot B’ \cdot C = 0 \cdot (0)’ \cdot 0 = 0 \cdot 1 \cdot 0 = 0 \).
- Second term: \( A’ \cdot B \cdot C = (0)’ \cdot 0 \cdot 0 = 1 \cdot 0 \cdot 0 = 0 \).
- \( F = 0 + 0 = 0 \). (Correct, since the function is not specified to be 1 here)
The expression correctly outputs 1 only for the specified input combinations.
Step 4: Addressing the Confusion About \( F = 2 \)
Your question suggests that if the function is in SOP form, then \( F = 2 \). This seems to stem from a misunderstanding. Let’s clarify:
- In Boolean algebra, the output of a function is either 0 or 1. The value \( F = 2 \) is not possible in standard Boolean logic, as it operates in the binary domain ({0, 1}).
- You might be thinking of the number of minterms or the count of input combinations where \( F = 1 \). Here, there are two minterms (corresponding to the two input combinations where \( F = 1 \)), which could lead to the number 2. However, this does not mean \( F = 2 \).
- The SOP expression represents the function’s output, not a numerical count. Each minterm contributes a 1 to the output for its corresponding input combination, and the OR operation combines them to produce the final output (0 or 1).
For example:
- If \( A=1, B=0, C=1 \), the function evaluates to 1 (not 2).
- If both minterms were true for some input (which isn’t the case here), the OR operation would still yield 1, because \( 1 + 1 = 1 \) in Boolean algebra (where \( + \) represents logical OR).
Final Answer
The SOP expression for the function \( F(A, B, C) \) that outputs 1 for the input combinations \( (A=1, B=0, C=1) \) and \( (A=0, B=1, C=1) \) is correctly given as: \[ F = A \cdot B’ \cdot C + A’ \cdot B \cdot C \] The output of \( F \) is always 0 or 1, not 2. The number 2 might reflect the count of minterms (two input combinations where \( F = 1 \)), but this does not mean the function outputs 2. In Boolean logic, the function’s output is binary, and the SOP form simply lists the conditions under which \( F = 1 \).
If you meant something specific by “F = 2” (e.g., a different context like counting minterms or a non-Boolean function), please provide more details, and I can refine the explanation!