Given f1, f3 and f in canonical sum of products form (in decimal) for the circuit, if f1 = Σm (4, 5, 6, 7, 8), f3 = Σm (1, 6, 15), and f = Σm (1, 6, 8, 15), then f2 is what?

Steps to Solve

1. Identify the Inputs and Outputs We start with the given functions:

• $f_1 = \Sigma m(4, 5, 6, 7, 8)$
• $f_3 = \Sigma m(1, 6, 15)$
• $f = \Sigma m(1, 6, 8, 15)$

To find $f_2$, we analyze the relationships in the circuit, with $f_2$ being the output of the AND gate fed by $f_1$ and $f_3$.

2. Use the Logical Relationship The relation can be written as: $$ f = f_1 \cdot f_2 + f_3 $$

3. Rearrange the Equation To isolate $f_2$, rearrange the equation: $$ f_2 = \frac{f - f_3}{f_1} $$

4. Compute the Minterms for each Function Now we calculate the values:

• $f_1 = {4, 5, 6, 7, 8}$
• $f_3 = {1, 6, 15}$
• $f = {1, 6, 8, 15}$

5. Find Common Minterms Subtract the minterms of $f_3$ from the minterms of $f$: $$ f - f_3 = { 1, 6, 8, 15 } - { 1, 6, 15 } = { 8 } $$ Now check which minterms from $f_1$ correspond to this: $$ f_1 \cap { 8 } = { 8 } $$

6. Identify the Resulting F2 Minterms Since $f_2$ must produce $f = {1, 6, 8, 15}$, and we have found that $8$ is the only remaining value, we derive:

• Therefore, $f_2$ should yield $4$ and $6$ as remaining possibilities (based on conditions).

7. Finish with the Set of Minterms The final possible minterms for $f_2$ can be concluded as: $$ f_2 = \Sigma m(4, 6) $$