Given a standard SOP (Sum of Products) expression, we're tasked with finding the binary values that make the expression equal to 1. However, you didn't provide a specific SOP expression, so I'll create a sample problem and walk you through the solution.
Consider the following Boolean expression in SOP form:
F(A, B, C) = A'B'C + AB'C' + ABC
Here, we want to determine the binary values for A, B, and C such that the SOP expression equals 1.
1-A'B'C: This term is true when A=0, B=0, and C=1.
Binary input: 001, Output: 1
2-AB'C': This term is true when A=1, B=0, and C=0.
Binary input: 100, Output: 1
3-ABC: This term is true when A=1, B=1, and C=1.
Binary input: 111, Output: 1
For the binary values 001, 100, and 111, the SOP expression equals 1.
Now let's convert a given Boolean expression to SOP form:
G(A, B, C) = (A + B)(A + C)(B' + C')
Step 1: Apply the distributive law (FOIL):
G(A, B, C) = (AA' + AC' + BA + BC + AB'C' + AC'B')
Step 2: Simplify the expression by eliminating terms that simplify to 0 or are redundant:
G(A, B, C) = (AC' + AB + BC + AB'C' + AC'B')
This is the Boolean expression G(A, B, C) in SOP form.